2 chapter summary advanced higher maths
VARIATIONAL PRINCIPLES AND FREE-BOUNDARY PROB LEMS
AVNER FRIEDMAN
VARIATIONAL PRINCIPLES AND FREE-BOUNDARY PROBLEMS
AVNER FRiEDMAN
4 Nørthwrium Uk*y
A PUBLICATION
JOHN WILEY & SONS
Mew York Cbichester
Copyright © 1982 by Jobn Wiley & Sons, Inc.
All rights reserved. Published simultaneously in Canada.
Reproduction or translation of part of this work beyond that permitted by Section 107 or M of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permissions Department, John Wiley &. Sons, inc.
01 Congress Cataloging in Pubilcatine Data
Friedman, Avner. Variational principles and free-boundary problems.
(Pure and applied mathematics. ISSN 0079-8185) "A Wiley-Interscience publication." Includes bibliographical references and index. I. Roundaiy-value problems. 2. Variational
1. Title. II. Pure and applied mathematics (John Wiley & Sons)
QA379.F74 1982 51 5.3'5 82-8654 ISBN G.471-86849-3 AACR2
Printed in the United States of America JO 9 8 7 6 5 4 3 2 1
PREFACE
Important developments in the study of free-boundary problems have been achieved in recent years by introducing variatiorial approach wherever possi- ble. This enables one to conclude without great efforts that a solution to the free-boundary problem exists in some "weak" sense. One can proceed to establish the regularity of the solution and then, hopefully, study the smooth- ness of the free boundary itself. In fact, within the last five years significant new methods have been developed for analyzIng the free boundary and the theory has now reached a certain stage of maturity; its future looL even more exciting. An increasing number of physical and engineering problems are becoming accessible to this growing body of methods.
It is therefore an appropriate time to discuss the main developments in the field in a systematic and self-contained manner. Since some of the main developments were motivated by physical models, we have kept close contact between the general theory and applications to physical examples.
In order to make the book more readable by graduate students afld nonspecialists, we have included in the text the detailed slatanents of the standard theory of elliptic and parabolic operators that is being used. This is done more systematically in the first two chapters.
We have problems at the end of each section and remarks at the of each chapter.
I would like to thank Luis A. Caffarelli, Hans Wilhelm Alt, and Joel Spruck for several useful conversations, and Ms. Leslie Hubbell for an excellent job of typing the manuscript.
Ecøjtston. Illinois June 1W
CONTENTS
INTRODUCTiON
1. Variational Inequalities Existence and Regularity 4
1. An example, 4 2. General theory of existence and uniqueness, 9 3. W2' regularityfor the obstacle problem, 19 4. W2 regularity for the obstacle 31 5. The filtration problem, 47 6. The elastic—plastic torsion problem: W2' regularity. 7. The elastic—plastic torsion problem: W2 regularity, 65 8. Parabolic variational inequalities, 72 9. The Stefan problem, 82
10. Variational inequalities for the biharmonic operator, 90 11. Thin obstacles, 105 12. Bibliographical remarks, 125
2. Variational inequalifles Analysis of the Free 128
1. The Hodograph—Legendre transformation, 128 2. Regularity in two dimensions, 138 3. General properties of the free 154 4. Convexity properties of the coincidence set, 163 ' 5. Regularity of the free boundary when MDIA) is positive, 170 6. The free boundary for thVflltration IT? 7. Regularity of the the
torsion problem, 193 I 'i 8. The shape óf the freb for the
torsion 203 9. The free boundary for the Stefan prOblem, 225
10. StabilIty of 250 II. Free boundEiu *4th 216 '' " 12. Bibliographical remarks, 263 -
'ii
CONTENTS
Jets and Cavities
1. Examples of jets and cavities. 266 2. The variational problem, 271 3. Regularity and nondegeneracy, 275 4. Regularity of the free boundary, 284 5. The bounded gradient lemma and the nonoscillation lemma, 285 6. Convergence of free boundaries, 288 7. Symmetric rearrangements, 293 8. Axially symmetric jet flows, 297 9. The free boundary is a curve x = k(y). 309
10. Monotonicity and uniqueness, 318 11. The smooth-fit theorems, 324 12. Existence and uniqueness for axially symmetric jets, 335 13. Convexity of the free boundary. 339 14. The plane symmetric jet flow, 344 15. Asymmetric jet flows, 345 16. The free boundary for the asymmetric case, 353 17. Monotonicity, continuity, and existence for the asymmetric
jet problem, 358 18. Jets with gravity, 366 19. The continuous fit for the gravity case, 379 20. Axially symmetric cavities, 391 21. Axially symmetric infinite cavities, 400 22. Bibliographical remarks, 415
4. VarIational Problems with Potentials 417
I. Self-gravitating axisymmetric rotating fluids, 418 2. Estimates of gravitational potentials, 426 3. Existence of solutions, 433 4. Rapidly rotating 443 5. The rings of rotating fluids, 456 6. Vortex rings, 470 7. Energy identities and potential estimates, 478 8. Existence of vortex rings, 487 9. A capacty estimate, 501
10. Asymptotic estimates for 507 11. The plasma qi solutioee, 520 12. The free boundary for the plasma problem, 529 13. Asymptotic estimates for the problem, 534
'14. A variational approach to convex plasmas, 549 15. The Thomas-Fermi model,. 563 16. Existence of solution for the 56R 17. Regularity of the free boundary for the Thomas-Fermi 576 18. Bibliographical remarks, 586
CONTENTS
5. Some Free-boundary Problems Not in Variational Form 589
I. The porous-medium equation: existence and uniqueness, 589 2. Estimates on the expansion of gas, 604 3. Holder continuity of the solution, 617 4. Growth and HOlder continuity of the free boundary, 627 5. The differential equation on the free boundary, 636 6. The general two-dimensional filtration problem: existence, 650 7. Regularity of the free boundary, 657 8. Uniqueness for the filtration problem, 668 9. The filtration problem in n dimensions, 674
10. The two-phase Stefan problem, 684 11. Bibliographical remarks, 693
References 695
Index 709
INTRODUCTI ON
The Dirichiet problem for the Laplace operator seeks a solution ii of u 4) on the boundary Suppose that only a
portion S of is given whereas the remaining portion r not a priori prescribed, and an additional condition is imposed on the unknown part of the boundary. such as V(u — 4)) = 0 on I' (here 4) is a given function in the entire space). Thus we seek to determine u and r satisfying
V(u—4))oonr,
where D is bounded bX the given S and the unknown r. This problem is an example of a free-boundary problem.
For a two-dimensional ideal fluid, the density function u satisfies, on the interface F between the fluid and the air, the free-boundary conditions
u = C1, I Vu C2 (C1, C2 constants),
where either C1 or C2 is frequently unknown; F is also not prescribed. Another example of a free-boundary problem occurs when ice and water
share a common interface. Here the free-boundary conditions are
0, — =
where is the tàmperature in the water, °2 is the temperature in the ice, a and k are positive constants, •(x, 1) = 0 describes the eqtzatibñ of thern free boundary, ana 0, satisfies the paiabolic equation
There are also free-boundary problems in which the freeboundary not appear explicitly the outset of the problem. example, the ol axisymmetric self-gravitating rotatiftg fluid the boundary separating the fluid from the vacuum as the set
I
2 INTRODUCIION
to a nonlinear equation
('(U (c>O,8>O)
u is "potential' function depending on the fluid's density p and (u = 0) = p For g.is in a medium, the free boundary is the boundary u where ii the density satisfying the nonlinear degenerate parabolic
equation
u,=,&um (m>l).
the initial step in studying a free-boundary problem is to refor- mulate it in such a that the free boundary disappears. (There are, however, notable exceptions, especially in the case of one space dimension.) Such reformulations can often he achieved by resorting to variational principles. One of the most celebrated examples is the following: If u minimizes
J(t) =f[i Vv12 + 2fv1
subject () on afl. v in then u satisfies at least)
u is a solution of the free-boundary problem of the' type mentioned earlier.
Since the problem of minimizing J( v) has a solution (obtained as a limit of a minimizing sequence). we conclude that there exists a solution u of the free-boundary problem in its variational formulation.
The next two steps after establishing the existence of a solution for the reformulated free-boundary problem are to obtain the best regularity results and then proceed to analyze the free bottndary. The latter step often requires much deeper methods.
Thus in the preceding problem (*), the optimal regularity is that u has Lipschitz continuous first derivatives. There are also known sufficient condi- (ions that ensure that the free boundary is smooth, but in general (without such conditions) the free boundary may be quite singular.
in Chapters 1 and 2 we develop the theory of a large class of free-boundary problems called variational inequalities. Chapter 1 deals with the variational
INTRODUcTION
approach, existence, uniqueness. and regularity of the minimizer. Chapter 2 is concerned with the study of the free boundary itself.
In Chapter 3 we study a class of variational problems designed for solving problems of jets and cavities of ideal fluids. Whereas in Chapter 1 a typical functional is
f[$vvI2 + 2fvJ (r>O)
in Chapter 3 the corresponding functional is
where 'A is the characteristic function of A. The variational functional in Chapter 4 is of the type
+JA(p(x)).
where p is a density function subject to some constraints. Here the free boundary is 8(p > 0).
In Chapter 5 we study several free-boundary problems that are not for- rnulted as variational problems; we deal mainly with gas ;n a porous medium and with the filtration of fluid in a porous dam.
Chapters 3, 4, and 5 are basically independent of each other as far as cross references are concerned. However, they do share some common methodL techniques, and ideas. The material of Chapters 1 and 2 appears in later chapters either directly or indirectly.
There is a large body of literature on time-dependent fke-boundary prob- lems in one space dimension. Here the methods are often highly specialized. With a few exceptions, we shall not deal with such problems in the present book.
1
VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
In this chapter we introduce the concept of a variational inequality and establish general existence and uniqueness theorems. Regularity results are proved for some classes of variational inequalities, mainly the obstacle prob- lem, the case of gradient constraint, the biharmonic obstacle problem, and the case of thin obstacles.
We introduce several physical problems to which the existence and regular- ity theorems are applied, such as (i) the filtration of.water in porous medium, (ii) the elastic—plastic torsion problem, and (iii) the Stefan problem of the melting of a solid.
1 AN EXAMPLE.
Let be a bounded domain in R". We denote by the class of functions u(x) in LP(IZ) such that all their weak derivatives Dau =
• • of orders m belong to here a = (a1,.. a1
a1 + •. + a,,, D1 a/ax1, and the weak derivative is defined by
fDttu •(x)dx = (— V 4) E
stands for functions with compact support in is a Banach space with the norm
'I
I
AN EXAMPLE
here I < oo; forp oo we define
I U ess sup I Dau I
where D°u are taken as weak derivatives. It is well known (see references 94c and 109) that
coc(a) n is dense in
The closure of in is denoted by The notation
= WmPffl), Hm(12) = = W"2(1l)
is also customary. Consider the functional
(1.1)
and the closed convex set in H1(tZ).
(1.2) K=
wheref is a in is a continuous function in ft and g E H'(fl). We assume that g 4'; then K is nonempty.
Consider the problem: Finsi u such that
(1.3) u — g E G(u) mm G(v).
Suppose that u isa solution of this problem. Then
ereal,
and we easily deduce that
Hence, if u E H2((2),
so that
(1.4) inft
6 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
This equation is satisfied, of course, in the a.e. sense; the condition u — g E is a generalized version of the Dmchlet boundary condition (see
references 94c and 109)
(1.5) ug onaf�. Thus the solution of the minimization problem (1.3) is also the solution of the Dirichiet problem (1.4), (1.5).
Consider next the variational problem: Find u such that
(1.6) u€K, G(u)=minG(v). vEK
If u is a solution of this problem, then for any v E K and 0 <e < 1, u + €(v — u) = (1 — e)u + cv is in K, and therefore
This yields
(1.7) VvEK; uEK.
if also u E H2U1), then we obtain
(1.8) Vv€K; uEK,
If u is continuous, then the set
(1.9) N=
isopen. q°(N)thefunctionv = is ±cLisin Kprovidedthat ki is small enough. We then obtain from (1.8)
Au+f=O inN.
Thus we have shown that if is is a solution of (1.6) which belongs to H2(Il)flC(U), then
a.e.inD,
AN EXAMPLE
The problem (1.6) is an example of a variational inequality. Any one of the versions (1.7), (1.8), (1.10) is also referred to as a variational inequality.
A more general concept of variational inequality will be given in Section 2, where the functional G will be of more general form and K will be any closed convex set in a Banach space.
The set K is called the constraint set, and in the case (1.2) we call the obstacle and (1.6) the obstacle problem. In this special case, the set (1.9) is called the noncoincidence set, and the set
ftlI) A={xEa;u(x)=.(x))
is called the coincidence set; the boundary of the noncoincidence set in (1,
r=aNno, is called the free boundaty.
ft will be proved later that (for suitably smooth g,$) thesolution uof the obstacle problem is in (in fact, even in Siiiëe u — takes its minimum in 12 on the coincidence set, it follows that
(1.12) u—$=0,
We may view u as a solution of the Dirichiet problem
inN,
(1.13) ug onaN?laU, u=4i onaNflr,
with the additional condition
(1.14) on8NflF
compensating for the fact that Fis not a pnori knowi. This point Of view is useful in solving variational inequalities in one space 4imension. However, if n> I, then Fcan bequiteirregularand weshall thereforenot attempt tosolve the obstacle problem by the of (1.13), (1.14).
Consider the special case n = 1,f0. Then the variational inequality (l.6),(1.2) is to minimize r
fb[ul(x)12dx
subject to u(a) = u1, u(b) u2, and u(x) we shall take u1 > $(a), u2
8 VARIATIONAL iNEQUALITIES: EXISTENCE AND REGULARITY
Suppose that 4(x) is strictly concave. From (1.13), (1.14) we deduce that the curve y = u(x) [u(x) is the solutioni Consists of three arcs:
(i) A line segment 11 connecting (a, u,) to a point (a', 4(a')). tangent to y = 4(x) at x = a'.
(ii)
A line segment 12 connecting (b', to (b, u2), tangent toy = at x b'.
The free boundary consists of the two points (a', and (b'. q(h'). If <0, then
u"(a' — 0) = 0 4"(a') = u"(a' + 0); thus u"(x) has a jump discontinuity at x a'.
This example shows that in general u C2; in fact, discontinuities of some second-order derivatives of the solutions of the obstacle problem usually occur across the free boundary.
PROBLEMS
1. Solve the obstacle problem
I — ax (a constant),
U 0,
(u" — 1 + ax)u 0
in 0 <x < 1 under the boundary conditions: (a) u(O) = 0, u(l) = I. (b) u(0) = 0, u'(I)= 0.
2. Solve the variational inequality in a ball BR (Ix 1< R} in B";
—p, (—Au+p)uz:O,
where p is a real number. (Notice that, for p <0, the coincidence set is empty.)
3. Suppose that
—Au
a.e.inBR, BR CR" (Au +f)u = 0
GENERAL THEORY OF EXISTENCE AND UNIQUENESS 9
and / — y, u M, where y. M are positive constants. Show that if R2 > 2nM/y. then u(O) = 0. [Hint: Let v=u—ylxI2/2n. Then in GBRfl (u>0); if u(0) > 0, then max v is positive and is attained on 8BR.J
4. Consider the problem of minimizing the length (I + (u'( x ))2 )I/2 dx among all the curves u = u(x) connecting (a, u1) to (b, u2) and subject to
constraint u(x) 4(x). where 4) is strictly concave. Describe the minimizer.
.2. OENERAL ThEORY OF EXISTENCE AND UNIQUENESS
Let K be a closed strictly convex set in a real Hilbert space H and let Yo be a point in H. Then there exists a unique point x0 E K (called the projection of Yo on K) which is nearest to that is, -
VxEK.
An equivalent way of writing it is
(*) YxEK.
By defining an appropriate H with scalar product
(u,v)fvu-vv
one can easily that problem () for x0 includes (with suitable K) the variational inequalities of Section 1. In this section we shall further extend the notion of problem (*) and obtain an existence and uniqueness for variational inequalities, of a very general faim,. quite suffieient for all' the applications,
.
Let X be a real Banach space with dual (conjugate) X', and denote by <,) the pairing between X and
A mapping A: D(A) -. X' Iwith domain D(A) C XJ is called monptone if
Vu,vinD(A).
When D(A) is convex, A is called hemicontinuous if V v 'hi (.1(A), the mapping
[0/lI —:)v),u"--v.)
is continuous. . . For any finite-dimensional subspace M of X, let j denote the injection map
ix x from M -' X andj* the dual map from X' -' M'; that is, if I E X', then
10 VARIATIONAL EXISTENCE AND REGULARITY
is the restriction off to M. If jAj is continuous on M fl D(A) for any such M, then we say that A is continuous on finite-dimensional subspaces of D( A).
Theorem 2.1. Suppose that A: X — X' is monotone and Isemicontinuous [D( A) = XJ. Then for any bounded closed convex subset K of X there exists a u0 E K such that
(2.1) VvEK
Theorem 2.2. Let K be a bounded closed convex subset of X and suppose that A : D(A) —. X' is monotone, D(A) = K, and A is continuous on fin sional subspaces of D(A). Then there exists a u0 E K such that (2.1) holds.
Note that in Theorem 2.1 we assume less about the continuity of A and more about the size of the domain of A. The inequality (2.1) is called a variational inequality.
The proof of both theorems depends on Minty's lemma:
Lemma 2.3. Let A: K -. X' be monotone and hemicontinuous. Then u0 satisfies (2.1) if and one:v if
(2.2) VvEK.
Here, K is any closed convex set.
Proof. By monotonicity of A
0 (At, — Au0, V — u0> (Av, v — u0>— <Au0, V — u0),
so that (2.1) implies (2.2). To prove the converse, note that, for any w E K, v = tw + (1 — t)u0 = u0 + t(w — u0) is in K if 0 < t < 1. Using (2.2), we get
and taking I - 0, we obtain (2.1) for any v = w E K.
Proof of Theorem 2.2
STEP 1. Consider first the case K C Rm, 4: K R'" and (Au:v) replaced by (Au, v), where (,) denotes the scalar product in R". We assume that A is continuous, but we do not require that it is monotone. The variational inequality
(2.3) Vv€K
GENERAL THEORY OF EXISTENCE AND
can be rewritten in the form
(u0, v — u0) — Au0, v — u0) V v E K.
For any w E K there exists a unique u0 E K such that
VvEK.
namely, u0 — Aw) Tw); that is, u0 is the nearest element in A to w 4w. The operator I — A : K R" is continuous and Rm — K is also continuous, and therefore T: K -, K is continuous. Since K is a bounded closed and convex set in R", Brouwer's fixed-point theorem can be applied to conclude that T has a fixed point u0 in K, that is, Tu0 u0; this implies (2.3).
STEP 2. Now let K C M, M Banach space. A: K -. M'. A continuous but not necessarily monotone. Then the assertion of the theorem follows from step I-with slight notational changes. Indeed, introduce a basis e . . ,e,, in M and a correspondence
whereü=(us,...,um).
We can define A uniquely by
(Au, = (Au, v>;
A is continuous. Applying step 1 toÃ, the assertion for A follows.
3. For any finite-dimensional subspace M C X define.1: M -' X by jx = x and let f be the dual map from X' -. M'. By assumption, the map f 41: K n M -. M' is continuous. By step 2 there exists a YM E K fl M solving
Yz€KflM.
Since <AYu' V E K fl M, we get
Vz€Kfl,M;
or, by Lemma 2.3,
(2.4) (4:, z — 'V z E fl M.
ForanyvEK,set
S(v)
12 VARIATIONAL INEQUALITIES: REGULARITY
Clearly, S( v) is weakly closed subset of K and, since K is bounded and X is reflexive, S( v) is weakly compact. By (2.4), S( v) 0. If we show that
(2.5) S(v1) fl flS(V,,,) 0 C,,,. I < C.
then it would follow that
fl 0. vEK
But a point u0 in the last intersection satisfies (2.2). and then also (2.1). Thus it remains to establish (2.5).
Denote by M the linear space spanned by v1 Lw. By step 2 there exists a K fl M such that
(AYM, YM) 0 V z E K fl M;
hence (Az, z — 0 V z K fl M and, in particular,
(A v, r. — 0 for 1 I m.
That means that E 5(v6) and thus (2.5) holds.
Proof of Theorem 2.1
STEP 1. Let M be any finite-dimensional subspace of X and define j. as before (in step 3). Then
(2 6) j*Aj maps bounded sets of M into bounded sets of M'.
Indeed, otherwise there exists a sequence v,, M. 0 < II C, such that II II —, 00. The monotonicity of A implies that
(jAva _j*Au, v,, — u E M.
Hence I
/ j*Av —
— wherey, =
Since the y,, are elements of M', there exists a subsequence y,,. —. y, II y II 1. We may also suppose that -. v. But then (y, v — u E M, which gives y = 0, a contradiction.
Site 2. If M is as before, then j*4j is continuous. Indeed, otherwIse there exists a sequence E M, v,, v, -e w [we use here (2.6)] and w E M',
GENERAL THEORY OF EXISTENCE AND UNIQUENESS 13
such that w /=jtAv. Since
VuEM,
also _j*Au, v — 0 and. by Lemma 2.3.
(w _j*Av, v —
u E M, we get w —jAv = 0, a contradiction.
STEP 3. By step 2, A satisfies all the conditions of Theorem 2.2. Thus Theorem 2.1 follows from Theorem 2.2.
Lemma 2.3 implies that the set of all solutions of the variational inequality (2.1) is a closed convex set. We shall now prove uniqueness.
A monotone operator A is said to be strictly monotone if
<Au—Av,u—v)=O impliesthatu=v.
'Theorem 2.4. If A is strictly monotone, then there exists at most one solution of (2.1).
Proof. If u2 are two solutions, then
VvEK.
Take v = u2 in this relation with i = I andy = u1 in this relation with i = 2. Adding, we obtain
(Au1 — Au2, u1 — 0;
hence u1 = u2. The argument above gives a stability resuh in case
(2.7) (a>0, 6>0).
Theorem 2.5. Let A be as in Theorem 2.1 or2.2 and let (2.7) hold. 1ff1, 12 are elements of X' and (i = 1,, 2) the solution of
(2.8) VvEK,
then -
(2.9) 11u1
VARiATIONAL INEQUALITIES: EXISTENCE AND REGULARIIY
Indeed, substituting v = u2 in (2.8) with i I and v = u1 in (2.8) with 1 2 and adding, we obtain
(Au1 Au2, u1 — —12' U1 —
Using (2.7), the assertion (2.9) follows. Theorems 2.1 and 2.2 will now be extended to the case where K is
unbounded. We shall assume that A is coercive in the following sense: There exists a E K such that
(2.10) ifvEK, IIvII—'oo.
Theorem 2.6. Let K be unbounded closed convex set and let A be as in Theorem 2.1 or 2.2. If A is coercive, then there exists a solution of (2.1).
Notice that Theorems 2.4 and 2.5 are valid even when K is unbounded.
Proof. The proof uses an idea of truncation. For any R > 0 we introduce the bounded convex set
and denote by UR the solution of the variational inequality corresponding to KR.If
(2.11) IIuRII<R forsomeR>O,
then UR is a solution of (2.1). indeed, V v K there exists an e > 0 sufficiently small so that
W = + E(v —
belongs to KR. Hence
0 (AUR, W UR) E(AUR, V —
and (2.1) follows for u0 = UR with any o E K. It remains to prove (2.11). From the_condition (2.10) it follows that for any C >0 there is an R >0
such that R> H H and
VvEK, flvIlaR.
Take C> Then
(Av, v — (C — — (C — — > 0.
GENERAL THEORY OF EXISTENCE AND UNIQUENESS
Now, if (2.11) is not true, then II II = R and upon taking v = UR in the last inequalities, we get
<AuR, UR — 0,
a contradiction to the variational inequality satisfied by UR.
We shall introduce another method to establish existence; it works only in Hubert spaces, but it has the advantage of simplicity.
Denote the real Hubert space by V, its dual by V', and the pairing between V and Vl by <,). Let K be a closed convex set in V and f and element in V'. Let a(u, v) be a bilinear form on V X V which is bounded, that is,
(2.12) CIIuIIUvH.
Assume also that a(u, v) is coercive, that is,
(2.13) (a>0).
'fliecrem 2.7. There exists a unique solution u of the variational inequality
(2.14)
u the sense that
(2.15) 1u3 — f21IV',
where is the solution corresponding to f.
l'his theorem is included in the previous results of this section hf we define A by (Au, v) = a(u, v) — (f, v); A is a linear operatorl, but the proof is simplçr.
Proof. To prove (2.15) (and uniqueness) take v u2 in the variational inequality for u1 and v = u1 in the variational inequality for u2 and add. We get
ä(u1 —f2,u1 —ü2)
and (2.15) follows. Tb prove existence, write <f, v> (1, v) denotes the
some elemóitt in V. In case a(u, v) (u, v)the of (2.14) is u = More generally, if a(u, v) is symmetric, theii a product in V is defined by
16 VARIATIONAL INEQUALITIE& EXISTENCE AND REGULARITY
setting (f. v'= ((1. v)) the solution u is the projection [in the metric of ((.,))J off on K.
When a(u. v) is not symmetric, we jntroduce
s(u, v) = v) + a(v, u)),
ø(u, v) = v) — a(v, u)),
the symmetric and skew-symmetric parts of a(u, v), and set
(2.I() a,(u, v) s(u, v) + to(u, v)
for 0 i 1. For z = 0, existence is already established. We now proceed step by step to extend the existence into i-intervals 0 6 f 6 t 6 '2' and so on, until 1= 1, making use of (2.15) and applying the fixed-point theorem for a contraction mapping.
Suppose that we have already proved existence for all 0 6 t 6 and set r = We rewrite
(2.17) a1(u, v — u) (f, v — u)
in the form
aT(u, v — u) v — u>+ ('r — i)a(u, v — u).
Set = (f, v)+ (r — i)o(u, v). This is a bounded linear functional on V and thus can be written as (f,,, v), V.
For any w E V. consider the variational inequality
VvEK; zEK.
It has a unique solution z, which we denote also by Tw, and in view of (2.15),
IITw1 — Tw21$ 6 — —
— l/2C, we conclude that Tis a contraction in V. Hence Thas a fixed point u, that is, Tu = u, and this implies (2.17).
We conclude this section with a stability result when the convex set K varies. For simplicity we formulate it just in the setting of Theorem 2.7. We shall need the conditions
are closed convex sets,
GENERAL THEORY OF EXISTENCE AND UNIQUENESS
The last condition means that (i) jf x E K then there exist x,, E K,, with lix,, — xli — 0, (ii) the weak limit of any sequence x,,. (xe. E is in K; the condition (ii) is satisfied, for instance, if K,, C K.
Theorem 2.8. Let a(u, K be as in Theorem 2.7 and let K,, satisfy (2.18). Let f,, E V1. f,, -. fin V', and denote by u,, the solution of
(2.19) u,,EK,, VvEK,.
Then u,, -, u weakly in V.
Proof. IfvEK,,,
(CJIvIl + IifnIIr)Iiun — oil.
Taking v -, E we deduce that II u,, ii C, with another constant C. Hence any subsequence of u,, has a weakly convergent subsequence. If we
show that u,, -. w weakly implies that w is the unique solution of (2.14), then the assertion of the theorem follows.
Since u -' a(v, u) is continuous,
(2.20) a(v,u,,) —i a(v,w).
By Minty's lemma
a(v,,,ç — u,,) (f,,,v,, — u,,)
for any v,, E K. Take any v E Kand v,, such that II;— oil -0. ilien
I
a(v,v—u,,)—a(v,,,v,,—u,,)(
Ia(v — v v —
Cliv oOhlv — u,,U + CIlv,Kllv — Cflv v,ll -.0.
It follows that
a(v, v—u,,) — u,,>+ a,, ** -.0,.
Noting finally that
(f,,,v,,—u,,>-.(f,v—w)
18 VARIATIONAL INEQUALITIE& EXiSTENCE AND REGULARITY
and using (2.20), we obtain
VvEK.
Since w E K, w is the (unique) solution of the variational inequality (2.14). 4,
PROBLEMS
I. Suppose that there exists a E K such that (1 + 0 E K for some 0>0. Show that the condition
(Av,o) —too ifvEK,
II
implies (2.10); A is assumed to be monotone.
2. Let a(u, v) be as in Theorem 2.7. Then there exists a bounded linear operator A:V—' V' such that a(u,v)=(Au,v) V u,v E V. For any f E V', (f, v) = (Af, v) V v V. where A : V1 V, II A II = 1. Show that if IIAU=MandO<p<2a/M2, then the;eexists a 0,0<8<1, such that
I(u,v) Vu,vin V.
tHin:: Write the left-hand side as (u — p A Au, v).J
3. Let F(u) be a function from Vinto(—oo, oo]. Assume that for any A >0 and for any bounded linear functional L on V, there exists a t3 in V such that
(2.21) + + L(6) v) + AF(o) + L(v) V v E V.
Show that is unique that
(2.22) a(u, v) + F(u) a(u, o) + F(v) V v E V.
[Hint: Set P1(v) = (u, pla(u, v) + Rv)J and prove that for any uEVthereexistsauniquew=Tusuchthat
VvG V,
and that Tis a contraction.J
4. lower seinicontinuous in the weak or strong topology (since F is convex, these two are the same). Then therecxistsauniqueuE Vsuch that
(2.23) Vv€V.
W2" REGULARITY FOR THE OBSTACLE PROBLEM
[Hint: To verify the condition of Problem 3 we may replace AF + L by F. Show that (2.21) is equivalent to: J(v) + F(v) has a minimum in V. To prove that a minimum exists, take F( v,) < oo and v0 such that F(v) — F(v1) (v0, v — v1) V v E V. Use this inequality to show that a minimizing sequence has a convergent subsequence.]
5. Show that the result of Problem 4 implies Theorem 2.7.
3. W2'? REGULARI1Y FOR THE OBSTACLE PROBLEM
In this section we consider the obstacle problem (introduced in Section 1) for a general second-order elliptic operator. Existence of a solution can be deduced from the general existence theorems of Section 2, but here we are interested mainly in establishing the regularity of the solution. We shall give a new method for proving both existence and regularity at the same time.
We need some general facts from the theory of elliptic equations [94c, 109], which we briefly recall.
Denote by (0< a < 1, open set in R") the space of functions u(x) which are HOlder continuous (exponent a), that is,
u(x)—u(yfl< fx yI
is a Banach space with the norm
= lull0 + -
where
Qu'10= sup Iu(x)I.
Similarly, we say that u (in positive integer) if
where
- hulL, =
Consider the operator
a Au E — a,1(x)
ax + +
i,j=I 1 1=1
20 VARIATIONAL INEQUALITIES: EXISTENCE AND REGIJLAR1TY
A is said to be elliptic in f2 if
I. J=I
for all x E E R"; it is uniformly elliptic if A > 0 for all x E
Schauder's Boundary Estimates
Suppose that is locally in C2'°,fE 4) E and
+ + liclia K,
VxEfl, (A>O).
If u E C2ffl) and Auf ins,
then
11U112+a
C is a constant depending only on A, K, The Schauder interioj estimates involve norms lu IL defined as follows:
sup fx
where = dist (x, = nñn (di, dr),
Hull a = hull0 + H0(u),
DUllm+a I$l'm lPlrn
If < 00, then we say that u belongs to
Sdaauder's Interior Estimates
Suppose that is a bounded domain with diameter D, f E and a,1, b,, c are measurable functions satisfying
tla,1ll4,+
YxE(2, (A>0).
WZ 'REGULARITY FOR THE OBSTACLE PROBLEM
If U E C2(12) fl and
Auf in(2, then
11U112+a +
where C is a constant depending only on A, K, and D. If u E C £2, then we write: u E
From the interior Schauder estimates one can deduce that if a,1, b,, c, and f belong to then u belongs to
We shall also need the elliptic V estimates, where 1 <p < co. Here the assumptions on A, fare weaker:
(3.1)
I 12 V x E £2, E R" (A > 0),
i,j=I
(3.3)
The L'Estimates
Let an belong to C2 locally,f E LP(n), 4' E If u WZP(n),
Auf mU, u —,
then
I"12.p c(IfI,+ I4'12,p)'
where C is a constant depending only on A, K, the modulus of continuity w, and the domain U.
Interior 1J elliptic estimates are of the form
WZpnorrnofu in G) and Cis a constant depending only on A, K, w, £2, and G.
We shall soiictimcsneed also local V estimates m a subdomain Gof U fot which n 8U is nonempty. Suppose that G1 is an open set, G C C
22 VARIATIONAL INEQuALmE5: EXISTENCE AND REGULARITY
fl is contained in the interior of 8G1 fl 8G fl C G,. If
Au=f mG1,
— E
for any E = 0 in a neighborhood of fl then
+ +
The Strong Maximum Principle
Assume that (3.2), (3.3) hold and that c(x) 0. Let u be a function in H2(11) fl satisfying Au 0 a.e. in &�. If u assumes a positive maximum at some point x0 in then u const. in (2 (and then c = 0 a.e.).
This result extends also to u, which is not necessarily continuous in (2; "maximum" of u is replaced by "essential supremum" of u: If ess u
u and u const. This implies:
(34) then u a.e. in (2.
For proof, see references 70 and 43. The Schauder boundary estimates can be used to solve the Dirichiet problem
Auf mhZ, (3.5)
u=4) ona(1, provided that c 0, where 4, f, 4), (2 are as in the statement of the estimates. The solution u is in Similarly, one can use the interior estimates in order to solve (3.5) with u E fl C(fl). Here one requires that A, fare as in the statement of thç interior estimates,4) is continuous on ao, and for every x° E ao there exists a local barrier (see references 74 and 109). A barrier exists whenever there is a ball B such that
Bfl(2=ø, aBriaO=.(x°) (this is called the "outside ball condition").
The V estimates can similarly be used to solve the Dirichiet problem (3.5) under conditions on A, f,4), (2 as in the statement of these estimates; see references 2 and 109.
Uniqueness for (3.5) follows from the maximum principle stated above. We shall now recall the main Sobolev inequalities. Let
—=—---—, wherel<pczoo. q p n
W2' REGULARITY FOR THE OBSI'AftE PROBLEM
Ifu E w"(fl) and aa is in C', then
(3.6) ifp<n,
(3.7) ifp>nanda= i —
where C is a constant depending only on n, and a. From (3.6) it follows that the injection mapping
j: W"(fl) -.
is bounded. We also have [94c, 109)
(3.8) j: W'P(U) L'((Z) is compact ifp <n, r>0, <!
We denote by C° '(D) the space of Lipschitz continuous functions in fi with norm
Hull0, = Hull0 + sup x.yEn Yv
this is in fact the space Ca(O) when we take a = 1. Similarly, one defines as the space with a = 1. The space is defmed as with a = 1.
We recall [109] that u E Cm.I(Q) if and only if u 6 We also recall that if u E H'(a), then H'(fl) and ac.
if u>0 if u<0;
further, Du = 0 a.e. on any set where u is constant. Let 4(x) (the obstacle) be a function satisfying
(3.9) •EC2(O)
and assume that
(3 10) A in
is in
fe g E
24 VARIAI1ONAL EXISTENCE AND REGUIARrFY
Let (0 <e < 1) be CX in I,
312 ift<0,e—sO, if
C, —C,
where C is a constant independent of e. Consider the penalized problem
Au+fl1(u—4)=f (3.13) ug Lemma 3.1. There exists a solution u = of (3.13).
Proof. Set, for any N > 0,
max (min{Ii,(:), N), —N)
and consider the problem
(3.14) Au+f3,N(u—4)=f
ug For each v E (I <p < oo) there exists a unique solution w in W''(a) of
'315' Aw=f—.81M(v—+) infi,
and
I W
where R is a constant independent of v. Setting w = Tv, we see that T maps the R-ball with center 0 in into itself and Tis compact, by (3.8). We can now apply Schauder's fixed-point theorem:
A continuous mapping tfrom a closed convex set S of a Banach space into a compact subset of S has a fixed point.
it follows that, for some u = N' Tu = u, but that means that u is a wlution of (3.14).
REGULARITY FOR THE PROBLEM
Since u U1 N is in for any p < II,, N(U — 4)) is Holder continuous. By a general regularity result for elliptic equations with coefficients, it follows that u is in We shall now estimate the function
= P,.N(u —
By definition of fi,,
CindependentofN,e.
Consider now the of Suppose that
Then x0 a(l, for otherwise
= —4')
On the other hand, if x0 t2, then since N(s) is monotone in t; also, u —, takes a at x°-and the minimum is. nonpositive. It follows by the proof of the standard maximum principle that
at x0.
But then (3.14) gives
(3.16)
We have thus shown that
(3.17) — C, Cindependentofe, N.
We conclude that
and, by the V estimates,
IU1NI2P C.
Thus if N Is large enough, then u, N is solution Of the problem (3.13).
We now take a sequence e e,,, -'0 such that
u1-'u Vp<oo.
26 VARIATIONAL EXISTENCE AND REGULARITY
It follows that
u uniformly in 13.
Since — C, we deduce that
U 4),
on the set (u >
We conclude that
Auf a.e.on(u>4)), a.e.infl.
We have thus proved:
ibeorem 3.2. Assume that (3.9)—(3.1 1) hold. Then there exists a solution u of the variational inequality
(3.18) u 4) a.e. in 13, (Au—f
on13,
and u E W2"(Il)for anyp < 00.
Theorem 33. Let u1, u2 be solutions in rs C(fl) of the variational inequality (3.18) corresponding to f1 and 12' respectively. If 12' then u1 a.e.
This comparison theorem implies uniqueness:
ibeorem 3.4. Under the assumptions of Theorem 3.2, the solution of the variational inequality (3.18) is unique.
It follows that.the solution u constructed in Theorem 3.2 does not depend on the particular choice of the functions $.
Proof of Theorem 3.3. Suppose the open set G where u2 > u1 is nonempty. Since u2 > u, 0 in G,
Au2—f2,
REGULARiTY FOR THE OBSTAClE PROBLEM
Consequently, A(u2 — u1) 0 in G. Also, u2 — = 0 on Hence by the maximum principle, u2 — 0 in G, a contradiction.
Theorem 3.2 can be improved by relaxing the conditions on A, f, g, and From the point of yew of applications, we are interested in particular in weakening the conditions on To do this let us recall the concept of mollifiers.
Let p(x) be a function in with support in the unit ball, such that
fpthci.
Set p8(x) 6 >0, and consider the mollifier
= fpa(x — y)u(y) dy [u E I <p < 00].
We recall [2, 94c, 109J that J8u is in and
IJau—uIL,(K)-O
for any compact subset K of fl. A continuous function 4'(x) in an open set C R" is said to satisfy
a2 in the senseof distributions,
if for any E q°(1i0), 0, there holds
here is any directional derivative. Taking = Pa' WC conclude that
0 in the usual sense, in Il,
provided that C Oo and 6< dist(U, We now replace the condition (3.9). by
4i e C° 1(fl0), (3.19)
—c in6D'(00),
wbere U0 is a neighborhood of
VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
The last condition means that
Setting
(3.20) lCjxI2) —
we easily find that
a2 (3.21)
-'0,
where C is a constant independent of 8.
Theorem 3.5. Let (3.10), (3.11), and (3.19) hold. Then the assertion of Theorem 3.2 is valid.
Proof. We repeat the proof of Theorem 3.2, replacing .fr by in the penalized problem and choosing, for instance, 8 = e. From (3.16) with = we see that if
(3.22) C, C independent of 8,
then all the estimates remain valid independently of 6; taking 6 = e -' 0, we then obtain, as before, a solution of (3.18) in W2"(O), for anyp < Thus it remains to prove (3.22).
Now, for any x0 E we can perform an orthogonal transformation such
becomes
Recalling (3.21), the assertion (3.22) follows.
We recall the estimate
V4' E
is in C', say). Since is dense in H'(fZ) if aa is in C' (see references 94c'and 109), it follows that any function in has a trace on L2(a(l): The
WZ? REGULARITY FOR THE OBSTACLE PROBLEM
tiace operator is a continuous linear operator from H'(Il) into which coincides with "restriction of 4' to an" when 4, E C'(n).
The dual of H Suppose that
(323) E (3.2) and (3.3) bold,
few'(n), gEH'(fl), an€c' and introduce the bilinear form
(3.24)
The variational inequality (3.18) can be transfoñned into the form
a(u, v — u) v — u) V v E K; u E K,
w1iere
(3.26) K = v — g E v a.e.).
If a(u, v) is then Theorem 2.7 gives the existence of a unique solution of (3.25), (3.26). Existence can also be established in case a(u, v) is not coercive, but c(x) 0; see 34.
Let the assumptions (3.23) hold, and let a(u, v) be coercive and c " 0. Suppose thatfe L"(a), ao C2, g E I <p <
Prove that the solution u of the variational inequality (3.25), (3.26) bdongs to WZP(O).
Multiply the differential equationin (3.14) by tPe,NI" and derjye
I 2. Extend Theorem 3.2 to the case where
where 4' v 4' on This is a and the solution u satisfies:
ifu<4,, if">+,
Au—f=O Extend also theorem 3.5.
VARIATIONAL INEQUALITIES: EXISTENCE AN!) REGUlARiTY
[Hint: Consider the penalized equation
Au + + y(u — 4')
where y(:) 0, y,(t) oo if t >0, e 0, and y,(t) —, 0 is t <0, £ —'
3. Suppose that a(u, v) defined in (3.24) is coercive. Let u,, u2 be solutions of (3.25), (3.26) corresponding to g1 and /2' g2, respectively. Show that
4'i g, g2 implies that u1 U2.
[Hint: Substitute v = max(u1, u2) into the variational inequality for and o = min(u1, u2) into the variational inequality for u2, and conclude that a(u1
4. Consider the variational inequality
—1/' + au a.e.inR', a>0.
(—u" + au —f)u 0
Show that if f" has a finite number then the free boundary consists of at most N + 1 points; assume that u E
5. Let a(u, v) be as in (3.24) and suppose that a(u, u) is coercive in H0'(D). We say that u is a local solution of (3.25) with -K (v E V a.e.} if
YvEK,
Prove: If E W2'(fl),f E and u is a local solution of (3.25), then u
IHinl: For any E C000(fl), y = I on II', 0 y 1 in 0 C 0), let
Show that
v — yu) — yu)dx —
+
VvEK1.
REGULARiTY FOR ThE OBSTAUZ PROBLEM
By Problem I,
Ifl,+ 6. Prove that the solution of the penalized problem is unique.
7. domains, where BR is a ball of radius R. Suppose that aa is in C' for some R' > R and set 1' = aa n BR. Suppose that u, E u1 u2 on I' in the trace sense. Show that
fit, mU, mU2
is in H'(BR); this result is called the matching lemma. [Hint: Take U = (x,, >0) and define v2(x', = u2(x', —xe), V = — v2 in There exist V,, E C'(O,), -. V in 0) -. 0 in
'L2(f). Consider mollifiers
= fp.(x' —y',x,, —y,, — 2E)w(y)dy
and let
infl,
z — in 12,
in H'(U,). Show: Zm E H'(BR), Z,,, u fl L2(BR), and (for suitable e,,,) (Zm) is a Cauchy sequence in H'(BR).]
4. REGUlARITY FOR ThE OBSTACLE PROBLEM
We shall assume that
belong to
ibeorein 4.1. Let the assumptions (3.10), (3.11), (3.19). and (4.1) hold. Then the solution it of the variatio.lal inequality (3.18) salisfies
it e
Proof. Without loss of generality we may assume that b. = c = 0; other- wise, we replace! by f — I b,(au/ax,) — cu, recalling that u E We
32 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARI1Y
introduce as in (3.20) and consider the penalized problem:
(4.2) Au+,81(u—+s)=f
u—g
We already knçw that the solution u = u satisfies
(4.3) foranyl<p<oo,
where C is a constant independent of E, 8. In the sequel it will be convenient to choose the functions fl!( 1) in such a way that
(4.4) ft(r) >0, fl'(t) 0 for alt.
Suppose first that! = 0. Note, by elliptic regularity, that u E We differentiate the differential equation in (4.2) twice in a direction and get
+ — — + b8e"(U — +8)(UE — = Jo
where
I — 2'' 03u a2uIT + ax, ax1' Since fl' 0, we obtain
(4.5) Autt + 13,(u — 4a)(uu —
Let be any compact subdomain of 0 and let E = 1 on (li,
— —
Substituting AuEE from (4.5), we get
(4.6) + — — +
Each term in + of the form aD3u can be written in the form D( aD2 u) — Da D2 u. Recalling (4.3) and the assumption that the coefficients
are in we see that
(4.7) + + D,g1,
REGULARITY FOR THE OBSTACLE PROBLEM
where
(4.8) f, C, C independent of e, 6.
We now need a lemma of Stampacchia [see reference 109, Theorems 8.25, 8.27, and 8.30, and (8.75)).
Lemma 4.2. Asswne that (3.2), (3.3), and K hold in a bounded domain n and that a(u, v) defined in (3.24) is coercive in H'(O); that is,
a(u, u) "co(IuI,i)2 Vu E H'(D) (c0>0).
Assume also that for any x0 an,
>0, B(x°) = {Ix — R).
Let g1,. .. ,g,, be functions in L'(O) and g0 a function in for some p > n. Let be a continuous function on 8(1. Then there exists a unique function nwin that
in(1, on&U;
Further,
where C is a constant depending only on A, K, a
Without loss of generality we may assume in our case that A is coercive [that is, q(u, v) is coercive]. Inde4 otherwise we simply replace 4 by a coercive operator A + k(k >0) and add to the right-hand side of(4.6). Thus, by the kmmfi. there exists a solution w in of
and'
(4.9) Cindcpendentofe,8.
Since I, + is in w is actually in
34 VARL4TIONAL INEQUALITJES AND
The function V = — w satisfies
(4.10) AV+ w:(u — — intl.
We shall now estimate V from below. — Suppose that V takes a negative minimum at some point x° E Since
V = 0 on afl, x0 must belong to But then
AV(x°) 0,
so that by (4.10) and the fact that t) > 0,
— o.
Hence
V(x°) — w(x°) — w(x°) —C.
We have thus proved that V — C and consequently, by (4.9), — C; that is,
(4.11) UU —c in every compact subdomain of
For any point y E 12, we can make an orthogonal transformation of the variable x so that in the new variable
(4.12) Au = + aty.
Since f
C we deduce from (4.2) that Au C and, recalling (4.3), (4.12), we find that
tiujc C0, C.3 independent of e, 8.
From this relation and (4.11) it is obvious that 0 + C c C0 + nC. Thus C, C independent of e, 8. Since is arbitrary direction, we can express
mixed derivatives at linear combination UEE ± for suitable directions and thus obtain C aty. Finally, the estimates iD2uI'c C aty can
be taken to be indepenJen! of y, if y varies in a compact subdomain of 12. To complete the proof of the theorem, we recall that u U, and take 8 = £ -.0.
We have assumed above thatf 0. If f 0, then we define u0 by
Au0f 1n12, u0g ona12,
and work with a = u — u0 and with the obstacle q — u0.
REGULARiTY FOR ThE OBSTAClE PROBLEM
We shall next prove W2'°' in a neighborhood of a boundary point x0 E The proof is local and applies also when x° E (1 (it is a different proof than that of Theorem 4.1, but it is more complicated).
We assume: There exists a neighborhood N0 of x° E such that
(4.13) E cz).
Theorem 4.3. Let the asswnptions (3.10), (3.11), (3.19), and (4.13) be satisfied. Then there exists a neighborhood N of x0 such that
E w2'°°(N
Proof. Set
a2u A0u — a11
ax,
Then the variational inequality for u can be written as a variational inequality with the elliptic operator A0 and the inhomogeneous term 1; by Theorem 3.5, /E C°(fl). Next let
A0u0=J inQ, u0=g
and extend u0 as a function into a neighborhood of aa Taking u = u — u0 + C x 12,
u with the obstacle + and inhoinogeneous term A0(C I x 12); notice that 0 if C is large enough. Thus, withoUt loss of generality, we may assume
from the outset that
fE fl lfl, g E n.g)
(4.14)
0 in for any direction
where is a neighborhood of We shall temporarily assume that
x°0, (4:15)
0 ifi<nl onx0 J
36 VARIA11ONAL INEQUALITIES EXISTENCE AND REGULARITY
Set
x = x'
B(r) = (Ix'I<r).
= B(r) X (61,82),
=
a'cl(r,6) = X (0);
r and 8 are taken sufficientLy small. We shall need a lemma similar to Lemma 4.2.
Lemma 4.4. Let v E 8)), e(x) measurable, 0 e(x) and let
(4.16) in(2(r,8),
I) (4.17) g, v' =g,
X,, B(r)X(O) a'afr.a)
where (4.16) is satisfied in an H'-sense and h, are L" functions. Set
= + IgIL° +
where the norms are taken, say, in a(2r, 8). Then
(v IL"(Q(r.8))
where C is a constant depending only on A0, 8, r, and y.
Proof. Let
+ ev1 = 0 in fl(r, 8),
(4.18) onB(r)X(O),
= on a'fl(,, 8).
One can rewrite this system in the weak formulation
(4.19) 8(r)X(O)
WZrn REGULARITY FOR THE OBSTACLE PROBLEM 37
and establish the existence of a solution by the standard L2 method. But one can also construct the solution in a different way which will be useful for deriving a uniform estimate on v1.
Let Gm be mollifiers of v. Since v is in C',
uniformly in a'rz(r, 6), aG
ga,, -, g uniformly in B(r) x (0).
If Vm is to be the solution of (4.18) with g,,, i,,, then Urn = Vm — Gm satisfies
inO(r,8).
onB(r)X{O),
Um=O
In order loconstruet V,,wcusethereflectionnile
if (4.20)
if x,<O
tO extend U,,, (if existing) into Urn, Fm into 1,,, e into e, and the coefficients a,1 of A into for all 1, j except I n, j n and I n, j = n; the latter coefficients are extended by
a,,,(x', x,,) = = —x,,) itxN <0,
and we denote these extended coefficients also by d,,, and the extended A by A. In view of (4.15), the are all Lipschitz continuous. It is also clear that
inO(r,—8,8),
= 0 onaIl(r, —& 8);
This system clearly has a solution in in D(r, {—8,6), and V,, U,,, + G,, gives of (4.18) with g,,,,
We next claim that
(4.21) IWmIIL'D(p(,,$))
where c is independent of m and C*; recall that 0< C*. To prove it,
38 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
let
OUmX(1X,,) Then /
AU+ eO = —eA(1 — x,,) <0 in 12(r, 6),
so that U cannot take a positive maximum at interior points. It also cannot take a maximum on x,, = 0 since
aU onx,,O.
It follows that
'CC,
where C is independent of m (and C*). The same holds for U,,,. We can now take a subsequence m 00 and obtain a solution v1 = limU,,, of (4.19). Also,
(4.22) lu1 II C independent of
It remains to estimate v2 = v — v1. We again reflect and obtain for t52 the equation
(4.23)
here for 0 < it are defined by the rule (4.20) but 1,, is extended by h,,(x', —x,) xe). We alsO have
(4.24)
Let ii be the solution of
Au' = in U(2r, —'8, 8), = 0 on the boundary.
The existence of v' follows by the standatd existence theory 1109]; further, for any I <p < 00,
LP(O(3r, —6,8))
REGULARTIY FOR THE OBSTAQI PROBLEM
by estimates for v Lwhich are interior with respect to the corner of the cylinder —8,8), but go up to the boundary of x,, = ±8J.
Next, let the solution of
= k, in fl(4r, —8,8), t51 =0 on the boundary.
The 1
.Y E W2P((1(r, —6, 6)).
Finally,
4i3=h0
on the boundary.
Then the function
xl
is a solution of
AV=110+ E ina(r,—8,8), 1=1
V0 ontheboundary; the equation is satisfied in the weak -
Take C IJ V II Thá the function = C ± (V — satisfies
- ontheboundary.
We recall [1421 that the maximum principle for weak solutions (say in H') remains valid if the are in C2. Thus we may conclude that
inLl(r,—8,8).
It follows that
I'L' 2C.
40 VARIATIONAL INEQUALrIIES: EXISTENCE AND REGULARITY
Combining this with (4.22) and recalling that v v1 + v2, the assertion àf the lemma follows.
Remark 4.1. Lemma 4.4 extends to the case where the Neumann condition in (4.17) is replaced by
(4.25) vg onB(r)X(O}. It is important to note that this result is not a consequence of Lemma 4.2; the novel point in Lemma 4.4 [and in its version with (4.25)J is that the bound C Ofl $ v is independent of the constant C*; later we shall apply the lemma in cases where we have no a priori bound on C.
We recall the penalized problem
Au+P,(u,—+)=f inn, u=g onafi,
whereg=g+
(4.26) Au1 IL" + P.IL" C, Cindependentofe.
Notice that u is in C4(N0 fl il).
Lernma4i. Forany.smaUö,r,
I
C is a constant independent of e.
Proof. Let z' be the solution of
Az' + — = —
— — 8(u, — +,) 8(u, —Ø,,u
ax
(4.27) az' r-=° onB(2r)X{O),
= 0 ona'fl(2r,28),
where is a function in Q(1r, — = I in —6,6), 0 I,
REGULARITY FOR THE OBSTACLE PROBLEM
and = 0. We claim that
I e
I
To prove it, we suppress the index e and set I = —Au. Then
(4.29)
and
1—f
on x,, = 0 [if we take /3(t) = 0 for t 0].
Applying —A to (4.29), we get
41+ $'(u — = —Af— $'(u — — g9"(u — — —
Multiplying by we easily compute that
A(Li) + $'(u — 4)(L1) = — +
— — — —
Comparing with the differential equation z = z1, we find that
— z) + $'(u— — z) —
F in the form
foranyp<oo,
where C is independent of E. We can therefore apply Lemma 4.4 motive 'tbat e(x) = — is nonnegative, but we have no a priori bound on e; see Remark 4.1 abovej. We conclude that
(4.30) I
— Z
C is independent of £, and (428) thus follows.
42 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
We shall now define functions for I i,j
(4.31)
Az, + $(u( — = —
—
lorij=n.
onB(2r)X{O)ifi=norj=n
ona'11(2r,26).
Observe that u = U1 satisfies
= on B(2r) x (0) for I I. f n — I,
UEXX = onB(2r) X(0), 1, j= I
and
= (_i_ onB(2r) x (0) for! n — I.
If we now apply 82/8x, to Au1 + fl,(u, — 4,) f, we obtain an equation similar to that for and we can then estimate — in the same way as — z' above. We thus obtain, analogously to (4.30$,
(4.32) x,x, — Z,1 IL°°(12(2r,28))
C is independent of e. Consequently, if we can show that
(4.33) R'JIL'°(Q(ra)) C,
then the lemma follows. Let us consider z,,. Choose C1 > 0 such that
(4.34) (C1a11(x) ± + &,,,)) is uniformly positive definite.
REGULARIJY FOR THE OBSTAClE PROØLEM
The function C1zt ± z1',, satisfies
(4.35)
A(C1f ± + — ± = — ±
— s14c1 — — ± (u, — — i.j=I
Since 0 for any direction is convex and thus is positive semidefirnte. Recalling (4.34), we then deduce that the first term on the right-hand side of (4.35) is 0. The second term on the right-hand side is also
0, by (4.34). Thus
A(C1z' ± + — +,)(C1z' ± 0.
Set W = C, ± zr,,. Then
onB(2r)X(O),
WO
Applying the .*naximum principle to W, we get
and, therefore,
(4.36) 14.1' ç,
whçre (4.28) was used. Similarly, one establishea (4.33) for all and z,, i#fl.
Finally, since z' 0 (by (4.36), for instance), we have
onB(2r)X (0)
ff1 a, j < n or if I j n. We can therefore proceed as before to establish that
C,f.
This completes the proof of Lemma 4.5.
VARIATIONAL INEQUALITIES: EXJSTENCE AND REGULARITY
We can now easily complete the proof of Theorem 4.3. Take for simplicity 0, fl N0 { 0) fl N0, fl N0 C (x,, > 0), and perform the lo-
cal diffeomorphism
(4.37)
i. - an,,
It transforms Au into A0u + A1u, where A0 is as in (4.16) and A1u involves tower-order derivatives of u. The penalized equation can then be written, in N0 fl in the form
+ — 4),) =fi — u,(x))
where f1 = / — A1u,,. We can reduce this equation into another one with an E (uniformly in e) and another as in the first step in the proof of
Theorem 4.3. We now apply Lemma 4.5 and, finally, reformulate the estimate on in terms of u,(x), and take e 0.
PROBLEMS
1. Extend the local W2" regularity result of Problem 5 of Section 3 to local W2 00 regularity.
2. Let '4)m be in C2(U), a bounded domain. Prove that • max .
. '4)m) satisfies: 4) E and —C in for any direction (C constant).
3. Set BR(x°) {Ix — R}. BR = BR(O). Let v E C(BR), 0, in (that is, v is superharmonic) with supp(—Ae) = K; thus v is
harmonic outside K. Suppose that v 0 in BR, v(0) A, v A. Show that for any 0, 0 <0 < I, v(x) C6X in BOR (C9 constant indepen- dent of v). tHin!: Let w w w 0. By Harnack's inequality,
w(x) 4w(0) C9A in
But v — w v A on K and then also in BR.j
4. Let u be the solution of the variational inequality
uEK,
REGULARITY FOR THE ORSTACLE PROBLEM 45
where
with 4, E 4 <0 on a bounded domain in RM. Suppose that u(x°) = 4(x°) and that there exists a harmonic function h in BAx°) C such that
sup B,(x0)
Then
sup u — 4 CX, C constant. 8,,,,(x0)
[Hint: v = u — h + A is superharmonic in B(x°) and v 0, v(x°) 2A, v(x) 2A on supp(—Av).J
5. U in Problem'4, y(r) is a modulus of continuity of 4 and is a modulus of continuity of V4, both decreasing, then
sup
(u — Cry1(r).
(Hint: Take = $(x0) or h(x) = 4(x°) + t74(iç°)• (x — 6. Let U be an open bounded set in R", 1, E C(U), h harmonic in U.
Assume that for any x° E ati,
sup Vp<p0.
Thenforanyx,x'inU, 2
iflx—x'I<po.
LHint: Apply the maximum principle to v(y) = hO' + x — — wherey vanes in the domain U' (y E U, y + x — x' U).J
1. Let U, h bq aàin Ptoblem 6, g E C1(U), and let y1(p) be a modulus of continuity Vg, p i. Show that if for any x0,E (p0>O),
2 S
sup •.i
VARIATIONAL INEQUALITIES EXISTENCE AND REGULARITY
then h E C'(U) and for any x, x' in U
—x'I) (Cconstant).
(Hint: Let xE U, x° the nearest point in auto x, p — x$<p0/2, y E fl U. Then
h(y)—g(x°)--vg(x°). l)py1(2p).
Since z(y) = h(y) — g(x°) + vg(x°) (y — x°) is harmonic in B9(x) C B20(x°)fl
U and 6 can be applied.)
8. If u is the solution of the obstacle problem of Problem 4, then: (a) If has modulus of continuity y(p), then u has modulus of
continuity C0y(p). (b) If has modulus of continuity if p I), then has
modulus of continuity Cy,(2p). The constants C0, C may depend on the distance 8 from (4> 0) to aa
[Hint: Show that in a small of a(2, u + e0 (e0 > 0) a.e. Take the standard version of the superharmonic function u defined by
u(x)lun udS, aBAx)
where = surface area of SB,.. Then u is lower semicontinuous and U = (u > is open and contains an (2-neighborhood of aa Apply Problems 6 and 7 with u = h.J
9. Let (2 be a bounded domain with boundary and let x0E(2. Suppose that h is harmonic in BR(x°) fl (2 and h >0 in fl 0, h =0, onBR(x°)fl aamenforanyo,o<o< 1,
sup h C,h(x°) (Harnack's inequality).
(Hint: Express h in terms of Green's function L a domain 0 with C2 boundary, where
BeR(x°) n (2 c (2* c BR(x°) n (2 (0<0, < 1).]
10. (a) Show that if E then the constant C, C0 in Problem 8 can be taken to be independent of 6; (b) extend the results of Problem 8 tp the case where 0 on [Hint: Extend Problem 4, using Problem 9.]
THE FILTRNI1ON PROHL.EM
11. The results of Problem 10 extend to variational inequalities with general elliptic operator A, with a1, b,, c, fin C't(1l). I Hint: First approximate by mollifiers. The corresponding Harnack inequality is given in reference 161 e.J
5. THE FILTRATION PROBLEM
A dam made of porous medium (say earth) with parallel vertical walls, situated distance a apart, separates two reservoirs of fluid (say water) at levels y = H andy h. The variable y is the height parameter and the variable x represents the distance from the wall of the higher reservoir. In the stationary case, Darcy's law states that
(5.1) v —kVu
where v is the velocity of the fluid, k is the permeability coefficient, and u(x, y) y + p(x, y); y = gravity force with normalized units and p(x, y) inner pressure of the fluid. Notice that we are taking here a two-dimensional dam; this can model a three-dimensional dam whose cross sections with the planes z = c do not vary with c. The function u is called the piezometric head.
The law of conservation of mass v = 0 then leads to
div(kVu) 0
in the wet part of the dam. We shall assume for simplicity that k(x, y) 1. We then anticipate that the wet part of the dam is given by
W {(x,
y •(x) is the "free boundary," separating the wet part from the dry part; see Figure 1.1. The u satisfies:
(5.2) mW,
(5.3) 0<x<za,
where v is the outward normal (that is, the flow at F Is to £),
u(0,y)H ifO<y<H, (5.4) u(a,y)h ifO<y<h,,
u(a,y)y. ifh<y<+(x) Ithis follows by noting that p(x, y) = H — y inside the higher
48 VARMTIONAL INEQU4UTIE& EXISTENCE AND REGULARITY
Higher r r3
relervoir
(0,0) - I'i (a,0)
FIGURE 1.1
p(x, y) = h — y inside the lower reservoir, p(x, y) = 0, where there is no fluid, and using the fact that p(x, y) is a continuous Also,
(5.5) u(x,y)y on!', and finally,
(5.6) (0<zx<a)
if the bottom of the dam is impervious. Suppose that there exists in fact a solution u of (5.2)—(5.6). We claim that the
function
y)—y
is then positive in W. Indeed, p is harmonic in W, p 0 on F U F3, p >0 on l'O U and p cannot take negative minimum of 1', since
on!'1.
Thus, by the maximum principle.
p>O mW.
We shall now transform the problem into variational inequality, assuming E C',, u C'(WU F) fl C(W).
Set
= 1) — tJ dt if 0 <y
(5.8)
THE FILTRATION PROBLEM
We compute
(5.9) -'-u(x,y)+y,
[by (5.5)],
y) i) di + +(x)).
Since
+ u1 0 ony = 4(x) [by (5.3)J
and
t) di = —
i) di —u,.(x, + u,(x, y),
we find that
—wi, + I [by (5.9)].
Thus
(5.10) 1 in W.
Also, by (5.7).
(5.11) w>0 mW.
Let
fl=
Then clearly
(5.12) w0, in&FW. One can also check that 0 in the distribution sense, Finally.
(5.13) wg inaa,
50 VARIATIONAL INEQuALmFS: EXISTENCE AND REGULARTIY
where g is a continuous function on
g(0, y) —y)2,
if0<p<h.
i(0<x<a, thatis,
H21 x\ h2x (5.14)
g = 0 elsewhere on
Setting
(5.15) K= E
we fin4 that w is the solution of the variational inequality:
(5.16) —f(v—w) VvEK; wEK.
Conversely, if w is a solution of (5.16) and if the set
W ((x, y) E fZ; w(x, .v) >0)
is given by
0<y<4)(x), 0<x<a, and 4)(x)<H,
then the function
u(x, .E) = v — i)
together with 4) solve the physical problem (5.2)—(5.6) in some generalized sense; if 4) and w are sufficiently "regular," then u. solve (5.2)—(5.6) in the usual (classical) sense. We shall now concentrate on the variational inequality (5.16) and establish the existence of a sufficiently regular solution w and free boundaryy =
Theorem 5.1. The variational inequality (5.16), (5.15) has a unique solution w and w E W 2. P( fl) fl
Proof. The existence of a unique solution w follows from the general theorems of Section 2. In order to prove that w W2P(IZ), we have to modify the proof of Theorem 3.2 since is not in C2+a; it has Jour vertices. We
THE FILTRATION PROBLEM
claim: There exists a solution of the penalized problem
u in (see Problem 2). We can show, as in Section 3, that
where C is independent of e, N. By V elliptic estimates we deduce that
(5.18) U1 C,
where C does not contain vertices of an. The results of Section 4 can now be applied to deduce that
(5.19) I
U1 C.
Finally, the W2'P estimates hold also in a neighborhood of each vertex; see Problem 3. Thus
(5.20) IU.l W2'(n) "C.
We can now take a Oand proceed as in Section 3.
Lemma 52 There hold,:
infi.
Proof. Set
W {(x,y)EO;w(x,y)>0).
0. Notice that in Wcannottakea positive maximum on aw, then it would follow that 0 in Wand thus also in 0.
Suppose that takes aniaximum n Wat a pointx°. We wish that /
If x° awn a, then = 0. If x0 = (x,0), then = <0. Ifx°€aWn
atx0
if x0 (0, H), so that cannot take maximum at this point. The same situation occurs if x0 (a, y), 0 <y <h.
52 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
Next, if x0 = (a, y), h H, ihen w(x°) 0; since, however, w(x) 0, we get 0. Finally, since w(x, H) 0, also H) 0 and the proof that w, 0 is complete.
The proof.that 0 is similar. Here
on(x=0}U(x=a,0<y<h},
w,0 on{xa,h<y<H),
onawn{y=o),
onawna
From Lemma 5.2 we easily deduce that for any x (0, a), there is a number •(x) such that
(x,y) W
and is monotone decreasing.
Lemma 53. is continuous.
Proof. Suppose that +(x0 + 0) — 0) for some 0 <x0 <a and let
R = ((x, y); 0 <x <xo,+(xo + 0) <y <+(x0 — 0)).
The function
y) = w(x, y) — x0)2
is harmonic in R and = = 0 on x By unique continuation (see Problem 4), 0 in R, and a contradiction is derived.
Similarly, one proves:
Lemma 5.4. ff0 < x1 <x2 <a and <H, then +(x2) <+(x1).
Lemma5.5. •(x)zHifO<x<a.
Proof. We first show that
(5.22) H) = 0.
THE FILTRATION PROBLEM
ForanyO<x<x' <a,wehave
w(x', H — H) H —
H — e) — H — e)} 0 C
since 0. Thus w1,(x, H) is monotone increasing in x. Since it vanishes at x = 0 and x = a, it must vanish identically, and (5.22) follows.
If the assertion of the lemma is not true, then 4(x) H for x in some interval 0 <x < x1. Since (5.22) holds we can derive a contradiction (by the unique continuation argument) as in the previous two lemmas.
We summarize:
Theorem 5.6. The free boundary is a curve y = +( x), where x) Is continuous and strict!;' monotone decreasing.
In Chapter 2 we develop general results on free boundaries for variational inequalities. These results imply for the dam problem that 4(x) is analytic. We shall also prove that 4(x) is concave and that — 0) > h.
PRO8LEMS
1. Let QR ((x, y); x >0, y >0, x2 + y2 <R2} Au / in QR' u = g on aQ1 .((0,O)), wheref E g is continuous u continuous in QR'((O'O))' u E Prove that u(x. y) -. iI(x. y) -.(0,0). [Hint: Let W be a strong barrier at (0,0). that <z — 1 in WE C(QR), W>O in W(0,0) 0 (for instance,
(x+a)"+(y+a)"
for anyn >0 and some y>O,p >0).
Show that for any e >0,tp >0, and some k >0,
IDQR.1
2. Prove that there exists a solution of (5.17). (Hint: Approximate (1 by smooth 12,,, and g by g,,, smooth in and denote the corresponding solution of (5.17) by u Then
I m C,
VARL4TIONAL INEQUALITIES: EXISTENCE AND REGUlARITY
1, u1,, u,, and + $,(u1) = —l in Also, u, g on except possibly at the vertices.]
3. Let QR' u, f, g be as in Problem 1 and suppose that g E W2P(QR), for some I <p< cc. Prove that for anyO<R0<R,
IUIW2P(QR) C[IgIW2.P(QR) + If IL'(Q)]'
where C is a constant. I Hint: Suppose that g = 0 and define
- I u(x,y) ify>0u(x, 1 —u(x, —y) ify <0.
Extend/similarly. In the extended domain i is in H' (by the matching lemma, Problem 7 of Section 3). Show that f in The assertion then follows by the V estimates in a neighborhood of a smooth portion of the boundary.]
4. LetubeharmonicinarectangleR(O<x<a,O<y<b),UEC'(R). and u = 0 on x = 0. Prove that u 0 in R. [Hint: Extend u by 0 and show that u is harmonic in a neighborhood of {x=0).]
5. Suppose that the permeability coefficient k is not constant, and k = k(y). Replace (5.8) by
w(x, y) = 1) — t) di if 0 <y <4(x)
=0 if
Define
G(0,y) i)di,
G(a, y) = f"k(t)(h — :)dt ify < h,
G(x,0) G(0,0)(l —
+
G = 0 elsewhere on
K
THE ELASTIC-PLASTIC TORSION REGUL4RflY
Show that w solves the variational inequality
VVEK;
Prove: If k'(y) 0, then theorem 5.6 is still valid. [Hint: Show that 0 and that (e_"w),, 0, where e" = k.]
6. THE ELASTIC-PlASTIC TORSION PROBLEM: REGULARITY
We shall consider a variational inequality with gradient constraint: Find u such that
uEK, (6.1)
u)dx VvEK,
where
(6.2) K = {v E I Vv I a.e.).
Here is a given positive number and fl is a bounded domain in R't. We first explain how this problem arises in physics.
Let be a bounded simply connected domain in R' and let Q be the cylinder
((x,. x2) E 12, 0< 1).
Q is a rod made up of plastic n!aterial; at the bottom (x3 = 0,) ana it is to torsion an the (x3 = I). It is assumed that no external forces are acting on ihe Jatéral boundary ofQ. If a is small eno4gj, Then the is pmna)I. ASsuming that the volume occupied by Q remains unchanged by the torsion, linear elastic—plastic theory shows that the two nonvanishing components of the stress tensor and can bewrittçn as Vu, where is (the pç*ential) is the solution of (6.1), (6.2) with p = 2Aa. A the shearing
Once is = u(x1, x2) is found, the displacement vector (U1, is2, U,) in Q is computed by
is1 = ax3x2, is2 = ax3x1, u3 = x2);
56 VARIATIONAL INEQUALI11tS: EXISTENCE AND REGULARITY
where is determined by
i au
Ix1 +— ——(1 +x)—, 8x2 ax,
and x = x(x,, x2) is the solution of
Au+V (xvu)= P (6.3)
X0 ifIVuI<l. Suppose next that is not simply connected and that it has a finite number
of "holes," say Thus each is a domain with connected boundary Set
'Ii
i=1
and consider the variational inequality
(6.4) fvu v(v — u)dx — u)dx Vv E K*,
where
(6.5) K {v E I a.e., V7v Oin each
Here again u has the same physical interpretation as before. In what follows we shall study (6.1), (6.2) only when U is simply connected,
and (6.4), (6.5) when U has a finite number of holes. Many of the results that we shall obtain for (6.1), (6.2) are valid, however, also if U is not simply connected.
We may consider (6.1), (6.2) as a special case of (6.4), (6.5). By the general results of Section 2, there exists a unique solution of the
variational inequality (6.4). For any sets G1, G2, write
d(x,G,)dist(x,G,), d(G,,G2) =dist(G,,G2).
THE FLAST1C-PLASIIC TORSION PROILEM WZ"REGULARITY
We introduce the functions
(6.6) max
(6.7) i,b(x) = mn {c* + d(x, (2k)) (x ER"),
where = R" \ c0 = 0, and
the restriction of u to
where u is the solution of (6.4). Notice that 4,, 4' are Lipschitz continuous with coefficient I. Since vu 1, we easily find that
(6.8) —
Also,
(6.9) u=4=4=OonaQ. Set
1(v) = vv12
Then the solution u of (6.4) satisfIes
J(u) minJ(v). yE K
Since E K and J(u) with strict inequality if meas (u <0) >0, it follows that as 0 in 0. Thus, in particular, the constants are nonnegative.
Notiss also that
(6.10) •(x) as q4x) [u the solution of(6.4)], (6.11)
Thus the solution of (6.1) belongs to
(6.12) i {v v and the solution of (6.4) belongs to
(6.13) 1" (v E v a.e.).
Consider the variational inequalities:
(6.14)
58 VARIATIONAL INEQUALITWS: AND REGUlARITY
and
uEK, (6.15)
fvuv(v — U) (v — u)dx
The first problem is an obstacle problem and the second one is a two-obstacle problem.
Theorem 6.1. (1) The solution of(6.l) is also the solution of (6.14); (ii) the solution of (6.4) is also the solution of (6.15).
Note that in the definition of the constants depend on the solution u of (6.4).
Proof. it suffices to prove (ii). Since K J K*, it suffices to show that the solution ü of (6.15) belongs to K*.
It will be convenient to work with the auxiliary variational inequality
uEK*, (6.16)
Vv
where e >0. Extend ü by 0 into R" Take any e E R" with p = f e small, and
consider the functions
max(ü(x — e) — p,
u(x) = mm (ü(x + e) + p, u(x)J,
and the sets
= (ü(x — e) — p >'u(x)),
E = {u(x + e) + p <ü(x)).
Since and we have
infi.
THE ELASTIC-PLASTIC TORSION PROBLEM W2" REGULARIIY
Therefore, and belong to Also,
EC(2, and a.e.
+ — Ivu(x—e)
inE Vu
in the variational inequality for ü, we get
— e) — ü(x)) +
— e) — ü(x) — p)
— e) — p — ü(x)).
Next we substitute v = into the variational inequality for ü and, after change of variables x -. x — e, obtain
— e) v(u(x) — u(x — e))
+efu(x — e)(ü(x) — u(x — e) + p)
Adding the two inequalities, we find that
V(u(x) — ü(x —
+Ef(u(x) — — e) +p)(ü(x) — u(x —
SiIkce I
ü(x)—ü(x—e)+p.<O
60 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
we must have meas E = 0 and thus
(6.17)
Recalling that ü depends on e, say u = u,, and letting e —' 0, we obtain the same inequality (6.17) for the solution u of (6.15). Hence 1 vu I and ÜEK.
Since the obstacles j' are generally not even in C', we cannot deduce further regularity of u by appealing to the results of Sections 3 and 4.
We shall prove regularity by another method, assuming that
(6.18) is locally Lipschitz and it satisfies the outside ball property.
The last condition means that there is an R > 0 such that for any x0 E there is a ball B of radius R satisfying
Bfl811=(x°).
Theorem 6.2. 11(6.18) holds, then the solution u of (6.4) satisfies
(6.19) AuELP(O) foralll<p<oo.
Proof. We construct functions u which are uniformly regular. More precisely, consider the problem
u —uj I/(p—l) (6.20)
) sgn(u—u)=0 inll,
(6.2 1)
We shall prove that for some E there exists a solution u of (6.20), (6.21), and
w1 C, C independent of e.
(6.23) u1EK*.
Suppose for the moment that this has been proved. By Minty's lemma
and taking v = u,, we get
(6.24) f vu1 v(u,— u) — u).
THE ELAS11C-PLASTIC TORSION PROBLEM: W2"REGULARITY 61
By (6.22), (6.23), and (6.20), we have that Au, Since
f for any w E it follows by approximation that this relation holds for W E Hence we obtain from (6.24)
fQtiF (u, — u) —
substituting —Au, from (6.20), we get
U, — —f w,(u, — u) a,if(u, —
where q p/( p — I). Using (6.22), we find that
ft u, — u c(fi u, — ui?) 11/(p—i)
and thus
I/q (6.25)
We can now further deduce from (6.20) that
(6.26) Au, r ' C. Itfollowsthatu,—,uase-Oandthat -
It remains For any 0<6< 1, be a flQ) =
if t>6, fl(t) U aiid aiiy positive number ., 4.
We 'iced a comparison lemma,
Lemma 6.3. Let 9(t) be a bounded continuous stnotly monotone increasing function, 6(0) = 0, and ki (i = 1,2) be C°((i) functions, u, E ff'(IJ) fl
62 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
a bounded open set, y > 0, ai
inn,
Then
(6.27) u1 —u2), sup(F1 —F2), sup(4)1 —4)2)).
Pro2f. Set u = u1 — u2, and denote the right-hand side of (6.27) by M. Let x° u(x°) If u(x°) > M, then x0 E and
u1—4)1>u2—4)2, u1—F1>u2—F2
Therefore,
— — 4)2),
O(u1 —F1)>O(u2—F2)
in a neighborhood N of x0. But then < 0 in N, contradicting the maximum principle.
We now take any 4) F — 4) E and any bounded continuous strictly monotone increasing function 0(t) with 0(0) = 0. Consider the problem: Find v such that
(6.28) infi,
(6.29) v — 4)
Notice that the operator A defined by
(Av,w)=f[vv vw+ — + 0(v.— F)w]
is strictly ujonotone from the convex subset 4) + of H'(fl) into the dual of H'(fl). Hence, by Theorems 2.2 and 2.4, there exists a unique solution v v1 of (6.28), The function v8 is in C°(fl) (see Problem 1).
Letx° and suppose that theballB = (IxJ< R)satisfies:B flfl = 0, B fl afi, = (x°). For any w E K, w —4) E
inf E
THE ELASTIC-PLASTIC TORSION W
since w = Ck on I. Setting
1x1 —R + Ck + 8,
we then have 6 and, in particular,
0(6k —F) 0(6) 0,
Since also
-- C [C depends on R and diam (il)].
we obtain, after choosing y C,
+ 0(8k —F) 0.
We can now apply Lemma 6.3 to deduce that in Therefore. if x E fl, x0 E ank,
v8(x) — v8(x°) —Ix°I +8 — x°I +8.
Similarly [with 8 (x) = +R + c, —61,
—Ix—x°I —8.
Thus
(6.30) Iva(x) +8 (x E aL').
For any point e in R" with f e p small, introduce the open set
fl'=(xEU,x+eEU}.
The function = v8(x + e) satisfies
mW,
where = + e), F'(x) = F(x + e). We can compare v1 with in tZ', using Lemma 6.3. Recalling (6.30),. we have
, 1" ao'
Since also I, 1, we obtüIn, by Lemma 6.3,
(6.3!) a,
64 VARIATIONAL EXISTENCE AND REGULARITY
Now take
1:1 '/(P—l) J,(t)
= (—) sgni,
ifjij< C
if a'>C
if
ihere C> 2 diam(fl). In view of (6.30),
9(v8(x) — F) = — F) if FE K*.
Choosing F = u, we then have for the corresponding v8:
in view of (6.31) we can take a sequence 6 0 with v8 u uniformly in
— 4's) -.
a solution of (6.20). Also,
and, by (6.31),
u,(x) —
which implies that I Vu, 1. Thus u, E K. This completes the proof of Theorem 6.2.
PROBLEMS
1. Show that the solution of (6.28), (6.29) is continuous. [Hint: For ffl, smooth, 4', f,,. smooth, the solutions Urn converge weakly star in to u; use batriers.J
2. Compute explicitly the solution of the variational inequality (6.1) in case (1 isa ball (IxI<R).
3.
+ I
THE ELASTIC-PLASTIC TORSION PROBLEM: REGULARITY
4. Take n = 2 in Problem 2 and compute x from (6.3) and the displacement vector of the elastic—plastic torsion problem.
5. Consider the variational inequality with two obstacles (6.14), (6.13). Prove the following local W2" analogous to Problem 5 of Section 3. Let C C c U C U I'2, where are open subsets of 8f1 and C F,. Suppose that I'2 is in and 4), 4' are in W2P(111), I <p < 00. Then u E
W REGULARITY
Theorem 7.1. The solution u of (6.4), (6.5) belongs to in any compact subdomain of and
(7.1) +
where C is a universal constant.
Note that no assumptions are made here on the regularity of aa
Proof. Consider first the case where
(7.2) Ic1 —
c1 the constant values of u in the domains and C0 = 0. Then the obstacles 4), 4.. satisfy
4)(x)<4'(x)
recall that 4)(x) = 4'(x) = c1 in 12,. Introduce e-mollifiers
= 4(x) = (J,4')(x) (x E
If 4) where 4e < 6 < Se then
(x E 12; d(x,31l) > 4e) C {x E
C {xEa;d(x,aa)>e).
Further, by Sard's lemma we can choose such that af4ç <4,) *** It IS easily seen that
(7.4) 1vip11<.1
66 VARU11ONAL INEQUALITIES: EXISTENCE AND REGULARITY
Let D = (x E (2; 4),(x) <4',(x)), and denote by u, the solution of the variational inequality (6.15), (6.13) with 4), replaced by 4',, 1), respec- tively. Set
N, = (x E D,; 4),(x) <u,(x) <4',(x)},
A1 = (x ED,; u,(x) =
A2 = (x E i),; u,(x) =
Since 4),, 4', are functions, u, is in P( El,) for any p < (Section 3, Problem 2), and thus is an open set and are closed Sets. Notice that by (7.4),
(7.5) I on
For any direction the function Ue £
is harmonic in N, and
onaN,flA1,
onaN,flA2.
Recalling also (7.5) and applying the maximum principle, we get u, 1 in N,. Since is arbitrary, we obtain
(7.6) I inN!.
Notice that in El, the obstacles 4),, 4', are never equal. Hence the local behavior of the free boundaries
F=8A1flD,
is the same as in the case of a one-obstacle problem. We shall need here one result on these free boundaries:
Lemma 7.2.
(i) IfyEN,,y—'xEF1 (ii)
The lim inf are wuforinly in x E fl (1', for any closed subset (2' C D,.
The lemma is contained in Theorem 3.6 of Chapter 2. We shall use the notation
d(x)=d(x,a(2).
11ff EIASflC-PL4STIC TORSION REGULARiTY
Lemma 73. For any direction
(7.7) — d(x) —
(7.8) d(x) —
for alix E (1 with d(x) > e.
Proof. Let x0 E (2. Then
.(x°)=max(c,—d(x°,aIZJ} =c1_d(x0,a121)
for somej and for some y° E a(21, where I x0 — y° d(x°, = d(x°). Set
y(x) = c1 — —
Then
Ix—y°I= y(x)
and +(x°) = y(x°). It follows that for any unit vector e and for any 0 <h < x°— y° , the pure second differences
A2 / — 2
iS eY(X°). But by the mean value theorem
A2 I
h, d(.2) '
where 2 lies in the interval (x° — he, x° + he). Therefore,
A2 I
X
Since this is true for all x° E (2 with d(x°) > h, we also have
if d(x) > h + e. ii 0, (7.7) follows. The proof of (7.8) issimilàr.
68 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
Lemma 7.4. For any direction
(7.9) 1 + d(x) —
for a.e. x E D1. C is a universal constant.
Proof. Since u1 E we have, for any direction
(7.10) a.e.onA1.
Also, in a I)1-neighborhood of A1, ur and, therefore,
(7.11) a.e.onA1.
For any a.e. x E A1 we then have, by (7.7). (7.10), (7.11),
I
— d(x) — = =
—
— +
d(x) —
where and the form an orthogonal system. Thus
(7.12) a.e.onA1.
Similarly,
(7.13) — — d(x)— d(x)
a.e. on A2.
In order to estimate u, in a suitable subset of N1, we shall apply the maximum principle. But first we need to evaluate this function as we approach the free boundary. To do this, consider first the case where x E F1. By Lemmas 7.2(i) and 7.3,
— d(x) —
This is also true for all directions i n — 1) which form with an
THE ELASTIC-PLASTIC TORSION PROBLEM: REGULARITY
orthogonal system. Thus:
urn 1711(Y) + d(x) — o.
Since also
l'msup[
/ i '1
n = urn sup +
d(x) — e +
—
we get
(7.14)
1 lim limsupu, c +d('x) — e y€N, d(x) — e
y-.x
if x E F1,
and the lim inf, lim sup are uniformly in x. Similarly,
n—I (hJ5) —
— d(x) — lirn mt u, jt(Y) lim supu, yEN1 yEN d(x)—e
ifx€F2.
We shall establish that
(7.16) + -I) if x E N,,d(x,8D,)
foranyfixedandsmallpprovidedthateissosmallthate<p/8. Noticethat
> e + p/12.
Let x0 £ N,, d(x°, = p and consider the function il,(y) = u,(x° + py) in B1, where = (i'; 1yI<R). Choose a E such that a = I in B113.
By (7.12), (7.13),
(7.17)
VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
a.e. in A1 U A2 fl {x; x — x0 < 2p/3}. Since also Au, = in N,, (7.17) holds a.e. in x — 2p/3). Recalling also (7.6), we obtain
mB213.
By elliptic V estimates it then follows that
+
for any p < 00 and, in particular.
(7.18) I ,, +
)p2.
We next wish to boost this estimate up to p = 00. For this purpose we introduce a function T E T I in and
the open set
In N,p. Au, = Using (7.18), we find that
(7.19)
where
(7.20) C(.ti+
Denote by V(y) the fundamental solution of —A, that is,
V(y) [y=
ifn2
Then the function
g(y)= jav(ç-z)h(Z)dZ
THE ELASTIC-PLASTIC TORSION PROBLEM: REGULARITY
satisfies
ah
in the distribution sense; also, in view of (7.20).
(7.21) +
The function — g is then harmonic in fl and. by the maxi- mum principle and (7.12)-(7.15). we find that
ü,. ± !
this gives (7.16). We have thus completed to estimate uf on the boundary of the opcn set
N, fl (d(x. aD,) > p) by (7.14)—(7.16). Applying the maximum principle, we get
C(p.+!)
in this open set. Recalling also (7.12). (7.13). the assertion (7.9) follows.
We can now immediately complete the proof of Theorem 7.1. First we lake —. 0 and note that —. —. tp uniformly and u, -. u u being
the solution of (6.4), (6.5): see Problem 4. Next, in order to remove the (irst restriction in (7.2), we approximate the by lie, with 0 < U < I and then let 0 — I. In order to remove the second restriction in (7.2) we by domains D with boundaries; see Problem 5.
DefinitIon 7.1. Let u be the solution of (6.4), (6.5). The sets
E (Jvu(x)j<I)
P 1)
are called, respectively, the elastic and plastic sets.
VARIATiONAL INEQUALITIES: EXISTENCE AND REGULARiTY
PROBLEMS
1. Let
= (u = fl 12,
= (u = 4i)fl
Ez fl12.
Show UP,E=E. [flint: If u(x°) = 4(x°) = d(x°,a12 ) ci— Ix° —y°I , y° ea121, then = = — 1 along the segment Conversely, if UE(X°) I and if 4(x°) < u(x°) <4(x°), then apply the maximum principle to ut in the component N0 of N containing x0 to derive ut I in N0.J
2. If x0 E P and if u(x°) = — d(x°, = — Ix° ..v° E then the entire segment x°y° is plastic. [Hint: The function w(x) = u(x) — — d(x, satisfies Ow/81 0 along x°y° (1 = the direction from x0 to y°) and w(x°) = w(y°4 = 0; hence w = 0 along x°y°.J
3. Let 12 be a simply connected domain in R2 with boundary that is piecewise C3 denote by V1,..., V1 the vertices of 812, that is, the endpoints of the
subarcs of 812. Show that the solution of (6.1), (6.2) is in for any subdomain of (2 with C 1Z\(V1,..., V,). (Hint: By reference 109, p. 382, d(x, 8(2) is in use Theorem 7.1 and Problem 1 of Section 4.1
4. Prove that u = is the solution of (6.4), (6.5); 812 is assumed to be in c24 a [Hint: Consider first test functions i5, 15 with i3 = 4' in a 8-neigh- borhood of 8(2 and then approximate general v by such functions 15, noting that u E C'((2) by Problem 2 of Section 3.]
5. If u,, solution corresponding to 12m and if 12m 1. (2, then Urn -' u a.e. (Hint: If v E KS and is properly extended outside (2, then 1(u) Jim infrn...,c J;n(urn), Jrn(Urn) 1m(t)) = 1(v).]
8. PARABOLIC VARIATIONAL INEQUALITIES
Let 12 be a bounded domain in and = (2 X {O <:< T). An operator U, + Au where
h
Au — t)881' + + c(x, t)u
PARABOLIC VARIATIONAL INEQUALITIES
with coefficients defined in Qr is said to be of parabolic type at a point (x, t) if
(8.1) (A>O); I.j=I
it is uniformly parabolic in if (8.1) holds for all (x, t) in QTwith a constant A independent of (x, 1).
For parabolic operators one can solve the first intitial boundary value problem
'82 u,+Auf
ug where
)pQr= X (O,T) U x (0)
is the parabolic boundary of The Schauder estimates stated in Section 3 have their parabolic counterpart. In the definition of the HOlder coefficient one takes
— u(x, t) — u(x', t') Hcc(u) — sup
(x. i).(x'. 1') i'., , t )j
where
d((x, i), (x', z')) =1 x — x' + i — 1111/2
is the parabolic distance. The definition of ((u la for u = u(x, 1) defined in Q7. is then analogous to the definition given before for functions ip We nextset I
(lull 2+a If + + +IID:Ulla + Wi÷a(Dxu);
where
— Hi+a(V)
It —
The Schauder boundary estimates then assert that
(8.3) COlffla + (lull0 + llg(12+,),
74 VARIATIONAL iNEQUALITIES: EXISTENCE ANI) REGULARI1Y
where it is assumed that (8.1) holds.
+ + K,
and in C depends uik on A. K. and QT' The interior Schauder estimates are also valid for parabolic operators. For
details, see references 94a and 130. estimates have also been established for parabolic operators. Here one
in addition in (8.1) (for all x E QT) that aaj. c are measurable functions satisfying
18.5)
v.1) a,,(x'. 1') x — x' + — t' ifA [0.
and that is in C2. Then the solution of (8.2) satisfies
(8.6)
u JP + D,u dx dt -
Cf dx di + Cf [Isr + +
for all p> I. p For proofs. see reference 164 or 88. For any point p0 = (x°. in denote by ((F°) the set of all points P
in that can be connected to P° by a Continuous curve x = x(s). t = i(s) (0 s s°). contained in such that i(s) increases in s as the curve is traced from P to
The Strong Maximum Principle
Assume that 4 + a/at is parabolic in QT and c(x. 1) 0. Le u be a function defined in Q,. such that D.u, are continuous in QT. and
in if
u(P°) sup
u consi. in C(P°).
PARABOUC VARIATIONAL INEQUALITWS 75
This result is valid also if P° E X (T} provided that u is continuous in QT U X {T).
Another version of the maximum principle asserts that if u takes a nonnega- tive maximum in at a boundary point P° on X (0, Tj, and if u < u(P°) in some of F°, then
at P°
for any outward spatial direction at P°: it is assumed here that u is C in a of P° (otherwise. au/as is taken as tim inf of finite dif-
ferences) and there is a ball B C with P° aB. The existence of a unique solution of (8.2) can be established by using either
the Schauder estimates or the L" estimates. The first approach (with the Schauder boundary estimates) requires more smoothness on the data, namely, that (8.4) hold, that an be in that the right-hand side of (8.3) is finite, and the consistency condition
g,+Ag=f
The solution then is quite regular: namely. and D,u are Holder continuous in Using the Schauder interior estimates, one needs only an to satisfy the outside ball property and g to be continuous on then
D,u are HOlder continuous in QT; see reference 94a. The second approach requires less smoothness on the data, but the deriva-
tives Du are taken only as weak derivatives in If e,usts and is bounded, then we can.associatewith A the bilinear
form
(8.7) U. v)=f +
where
-
We also set -
(v,w) fvwdx.
Then the first initial boundary value problem can be writtenin the weak form:
(88) (u,,v—u)+a(:;u,v—u)=(j,v—u) fora.a.:E(0,T),
76 VARIATIONAL INEQUALfl1FS EXISTENCE AND REGULARITY
The transformation ü = transforms u, + Au = f into u, + (A + k)ü =
and a(t; u, v) into
a(t; a, i,) a(t; ü, + k(ü, i5).
If k is sufficiently large, then a is coercive. Thus without loss of generality we may assume from the start that
(8.9) a(t; u, u) Aof(I + u2) dx (X0 > 0).
Since the same transformation can be used also in the variational problems which follow, we may always assume (if that (8.9) is satisfied and
Let K be any closed convex subset of Consider the following problem: Find u satisfying
u K,
(8.10)
fora.a.tE(O,T), YvEK.
This problem is called a parabolic variational inequality. We shall study here only one type of variational inequality, which we proceed to describe.
Let t) be a function defined in QT with uniformly continuous deriva- tives $ g on and consider the convex set
(8.11) K (v
here the relation v = g on is taken as usual in the trace sense. The variational inequality (8.10) with K given by (8.11) is called the parabolic obstacle problem; .p is the obstacle.
Uniqueness and stability of solutions of parabolic variational inequalities (8.10) with general K follow easily from the coerciveness condition (8.9), as in Section 2.
In order to prove existence for the obstacle problem, we use the penalty method. Introducing the function $(t) as in (3.12) of Section 3, we consider
(8.12)
ug
PARABOLIC VARIATIONAL INEQUALiTIES •77
We shall assume that
(8.13) E and
f, g, Dig, D,g belong to Ca(Qr).
Lemma 8.1. 11(8.13) holds, then there exists a solution u = of (8.12) and
(8.14) I 4')
C is independent of
The proQf uses the same arguments as in Section 3 and is left to the reader. We can now use the parabolic V estimates to deduce that a sequence
u a solution of (8.10); more specifically, u satisfies
(8.15) (u,+Au—f)(u--4,)—O
u=g
Uniqueness follows by coerciveness if we assume that
(8.16)
[so that c(t; u, v) can be definedJ. We summarize:
Theorem 8.2. if (8.13), (8.16) hold, then there exists a unique solution of the obstaclq problem (8.10), (8.11) and
(8.17)
Note that (8.17) implics that u is HOlder continuous in t,with any exponent fi < 1. We shaH now describe approaôh to studying the penalized problem which yields better results. For simplicity we assume that
(8.18) a,1, b,, c, are oft,
f, is bounde4 g =0.
Note that
(8.19) = a.e;
78 VARIATIONAL INEQUALITIE& EXISTENCF. AND REGULARITY
Setting u1 u and differentiating the equation in (8.12) once with respect to t, multiplying by (k positive integer), and integrating over we obtain, after using (8.19),
81 8,2 8t a:
2k—I 2* (2A—I),2A) (p.20) dx}
we have assumed (temporarily) that for a.e.:,
(8.21) 8,2
Using (8.9). we find that the second term on the left-hand side of (8.20) is bounded below by
(c>0)
Therefore, the function
Iau\2A dx
satisfies
(8.22)
+ C1 forsomey>0, C>0. C1 >0,
which gives tsince 4(O) ' CJ
(8.23) f(au/al)z*
From this and from Lemma 8.1, we obtain (for u =
(8.24) VI<p<oo.
The assumption (8.21) is not needed if we take finite differences of (8.12) with respect to t and multiply by where is the finite-difference quotient of u with respect to t.
PARABOLIC VARIATiONAL JNEQUALITWS
We can• now appeal to the elliptic L" estimates and conclude that for a.a. I E (0, T),
(8.25) fn[IDcur+
Taking e 0, we obtain the same inequality for the solution of the variational inequality. Thus:
Theorem 8.3. Let (8.13), (8.16), and (8.18) hold. Then the solution of the variational inequality (8.15) satisfies (8.25).
It follows that, for a.a. t, I) is HOlder continuous in x with any exponentfl < 1.
The preceding proof of (8.25) is flexible enough to allow also variable coefficients of A and g which depend on t [more precisely, if (8.13) holds, if
are in and if
(8.26) b1, c and thçir first i-derivatives
are bounded: and g,, are bounded,
then the assertion (8.25) is valid; for details, see reference 94gJ. We next describe briefly the estimate. For simplicity we take Au =
—au. We introduce the fundamental solution
K(xt) =
If g(x,:) is Holder continuous in Q7., then (see reference 94a) the function
w(x, t) _f'f K(x —y, 1— .r)g(y, r)dydr
is a solution of
(8.27) inQ7.
Furthermore,
(8.28) ifp> I +
Using this fact we can now repeat the proof of Theorem 4.1, replacing Lemma 4.2 by (8.27), (8.28). We obtain
u4t(x, t) —c in any compact subset QT(u = ui).
80 VARIATIONAL iNEQUALITIES: EXiSTENCE AND REGULAWY
Next, by differentiating the penalized equation once with respect to 1, we get
and we can proceed as before to derive a bound
± — C in compact subsets.
Since also C, we have
Ut — C,
and we obtain, as before, C. Thus:
'Theorem 8.4. if in Theorem 8.2, a.j = b, = 0, c = 0, and if the obstacle 4) satisfies:
4) is Lipschitz in x and:,
— C in for any direction a neighborhood of where C is a constant, then the solution u of the variational inequality (S 15) satisfies
This result extends also to A with variable coefficients.
PROBLEMS
1. Prove Lemma 8.1.
2. Complete the proof of Theorem 8.2.
3. Fill the details in the proof of (8.23).
4. Prove (8.28).
5. Prove a comparison theorem: If u is the solution of (8.10), (8.11) and if ü is a solution of a similar problem withj, 3 then
6. Suppose that
a.e.1nQRT=BRX(0,T), (u—iiu—f)u0J
PARABOLIC VARIATIONAL INEQUALII1ES
where —y, u M (y, M positive), and u EC(QRT). Show that if > 2nM/y, (1 — 9)T> M/y for some 0 <9 < 1, then u(0, T) = 0.
(Hint: See Section 1, Problem 3.]
7. Consider the variational inequality in R" X .(O, T):
'829'
u0 0, u0 L', and f, J E Use the penalized problem to deduce the existence of a solution u, R(x, 1) in (j x 1< R, t < T}, vanish- ingon (1x1=R) and todeducetheexistenceofa solution u of (8.29) satisfying
u,, E I.' in compact subsets of R1 x (0, T),
u(x,i)_fK(x_y,t)uo(y)dy-.O ift-.O,
where K is the solution, and
if then 14*, 1) 0 if :> T0. [Hint: with M — '1.]
9. 11 in Problem 8, u0 has compact support, then u(x, 1) = Oiffxf> R0, for some R0<oo. [Hint: Compare with M(R0 —
10. IfinProblem8,u0(x,:)—.OasIxH cx,thenforeacht>0,u(x,t)=0 if 00. [Hint: If u(x0, > 0, consIder
Ix_xoI2l
in (u > x — x0 < R, t <&,j; v takes positive maximum on the para- bolic boundary.J
11. Let S(t) = support of : -. u(x, 1), and assume that f, u0 are as in Problem 7/ " —, <0, S(0) = supp u0is bounded, and u0 E Prove:
forallsmall:,
{(xI<p},C0>0.
VARIA11ONAL INEQUALITIES: EXISTENCE AND REGULARIIY
[Hint: If u(x0, > 0, then consider the function U u — c x — x0 12 in G = {dist(x, S(0)) > r, t <ta, u(x, I) > 0). U takes a positive maxi- mum on the lateral boundary of G, at (x*. z*). Also.
u(x*, K(x* — y. t)uo(y) a)' ± Ct*.1
12. If in Problem —v0 >0 and u0 /3>0 on S(0), then
S(z)JS+B(coy'tIlogzI) foralismalit,
for some c0 > 0. [Hint: u IJfsK(x — y, t)dy —
9. THE STEFAN PROBLEM
In this section we describe a problem of melting of a solid and reduce it to a parabolic variational inequality. We shall then derive some specific properties of the solution.
Let G be a bounded domain in R" whose boundary consists of two connected hypersurfaces F0 and r1 with F0 lying inside F1 and bounding a simply connected domain G0 (see Figure 1.2). Let BR {l x 1< B) be a ball containing G and set &2 = BR\ GQ,QT = X (0, T). Suppose that initially G filled with water and U G0) is filled with ice at 0°C. We denote by 8 = O(x, t) the wateT's temperature. We are given the initial temperature
(9.1) 8(x,0)=h(x) onGX{0)
as well as the temperature along the boundary F0 for all:,
(9.2) O(x,z) = g(x, i)
FIGURE 12
x [0. oo).
THE STEFAN PROBLEM
It is assumed that h and g arc positive. Consequently, the ice will begin to melt and the region N(t) occupied by water will grow. The part of t) that is adjacent to the ice is called the free boundary. Let us suppose that the free boundary is given by an equation s(x) — = 0 (with s(x)> t in the ice region). Then
9 3\ 9 = 0 (continuity of the temperature),
= —k (conservation of energy)
along the free boundary, where k is a positive constant. Finally,
(9.4) intheregion(O>0).
The problem of finding a solution 0, s to (9.1 )—(9.4) is called the (classical) one-phase Siefan problem. We shall now transform the problem into a varia- tional inequality for a function u; u is defined by
s( x)
IfxEG.
We compute forx E
f' r)dr — s (x)0(x, s(x))s(x)
f 0,(x,i)dr,s( x) = f' dr — s(x)).s(x)
Hence
i) =1' + k + k s(x) s(x)
= u,(x, t) + k.
Similarly,
= f'01.(x, i) dr u,(x, t) — I, if x E G.
84 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
Setting
(95) f(x)= fh(x) ifxEG
we find that
ifu>O.
Also, clearly u — = 0 >1 if u 0. The function u takes on the boundary values where
fg(x,.r)dr ifxEl'011>0, (9.6) 4'(x, t) = 0
0
Introducing the convex set
(9.7) K
we see that u is a solution of the variational inequality
U E K,
(9.8) u)dx
a.e.intE(O,T), VvEK.
We shall now assume that
(99) g>O; h>OinG.
From the proofs of Theorems 8.2 and 8.3 we deduce (approximating f by smooth functions) that there exists a unique solution of the variational inequal- ity (9.8) and
Du belong to T); L'(O)) for anyp <00.
Notice that since! is not differentiable in x, Theorem 8.4 cannot be directly applied. Nevertheless, we shall prove:
Theorem 9.1. The sohalon u of the Ste/an problem (9.8) satisfies
(9.10) c<ao
THE STEFAN PROBLEM
and
(9.11) x (0,T).
Proof. Let 0 < 3a <dist(F0, F1) and let ii(x) be a function in such that 0 I and
= I if dist(x, F0) <2a
= 0 if dist(x, F0) > 3a.
LetJ (e >0) be a suitable smooth patching of h(x) with —k:
= h(x)4ç(x) — k(l —
where is E-mollifier of the characteristic function of e-neighborhood of G. Finally, let the standard function used in the penalized problem, with the additional requirement that
j1(:)=0 ifl>E.
Consider the penalized problem, for u =
(9.12) u, — Au + $,(u) J
onF0X(O,T), (9.13) (0),
u0 - onaBRX(0,T). Notice that the initial and boundary conditions agree on X (0). As usual,
—, u when E 0, where u is the solution of (9.8). We shall prove that
(9.14)
Differentiating (9.12) once with respect to z, we obtain for w = au,/az the equation
— Aw + P.(u,)w 0
and
oniO, on F0 X (0,
=0 OflaBRX(O,T).
VARIATIONAL INEQUAlITIES: EXISTENCE AND REGULARITY
By the maximum principle
(9.15) min{minw,0} a,Q,
To evaluate the left-hand side, notice that f,0 e if dist(x, 3a, and u,(x,0) = 0 if dist(x, 3a. It follows that if E is small enough,
w(x,0)>0 ifxEG, ifxEBR\G.
It is also easily seen that w 0 elsewhere on Hence the left-hand side of (9.15) is 0. One can readily check that the right-hand side of (9.15) is C, and (9.14) thereby follows. Taking e 0, we obtain (9.10).
For any I 0, let
N(t) = (x E u(x,) >0);
this is an open set. From (9.14) we have that u, 0; hence
(9.16) N(i)cN(t) if:<z'.
We claim that
(9.17) N(O) CN(t) ifz'>O.
Indeed, suppose for contradiction that there is a point x° E F1 such that x0 E aN(:1) for some > 0. In view of (9.16), the segment
1,, = (x = x°,0 < t <
lies in the boundary of the open set
D,,= U N(s).
Since, for a.e. t, Du(x, I) is continuous in x, we get
(9.18) = 0 for ac. (x, i) e i. The function
v(x, z) u(x,: + 8) — u(x, t), 8 > 0,
is a positive solution of the heatequation in G X (0, — 8) (since u, 0) and
THE STEFAN PROBLEM
it vanishes on the boundary segment Hence, by the maximum principle,
a.e. on 1,,
contradicting (9.18). From (9.17) it follows that fort > e0 (e0 > 0), —k in a neighborhood of
the free boundary; here £0 is any small number. Thus the local regularity result of Theorem &4 can be applied in order to deduce (9.11).
The introduction of the ball BR was just a matter of convenient truncation so that the variational inequality may be considered in a bounded domain. We would therefore like to show that the solution does not depend on R if R is large enough. For this it suffices to show that the free boundary does not intersect I$BR X (0, T) if R is large enough (depending on T). We shall in fact prove:.
Theorem 9.2. There exüt.s a positive constant M such that
(9.19) forO<t<T;
Mdpendr on sup g, sup h,Gand k, but not on T.
Proof. The proof is by comparison (see Section 8, Problem 5). We con- struct a radial solution of the classical Stefan problem in the form
lxi )b1t + I
and
2
f(z) cf C'.
We choose C, C' so that
=k.
cf M
and then the free-boundary conditions are satisfied. Notice that
C' remains bounded as M 00.
VARIATIONAL INEQuALmES: EXISTENCE AND REGULARITY
It follows that the data corresponding to 9 (with respect to G) satisfy:
if M is large enough. Choosing R > MIT + I , we can compare the variational solution ü (corresponding to 8) with u and obtain: ü u, from which the assertion follows.
Remark 9.1. From now on we always take the truncating ball BR such that R > My'T + I. Then the free boundary remains in a compact subset of BR X (0, TI.
Theorem 9.3. is uniformly continuous in (u > 0) fl <1 < T). for any e>O.
Proof. It is enough to show that if(xm, tm) (x0, to), where U(Xm, > 0 and (x0, belongs to the free boundary, then Ux(Xm, tm) 0. If the asser- tion is not true, then suppose for definiteness that
(9.20) ux(xm, tm) > $ >0.
Let x,, = Xm + Be1, = x0 + Be1, B > 0, and e, the unit vector in the ith direction.
For a.a. 1, is Lipschitz continuous in x with coefficient C independent of 1; writing
U(Xm, t) — u(x,,, t) = f
t) t) — &UX(Xm, C82.
Since Xm E N(tm), Ux(Xm, t) as t 1m Hence
I u(x,,, t,,,) — u(x,,, tm) — Bux(xm, C82.
Taking m -. oo and using (9.20) and the fact that 0. we get CO2. which is impossible if B </1/C.
PROBLEMS
1. Suppose that g(x, :) CI,)> 0 for all x E r0, t > 0. Show that there exists a small enough p > 0 and a > 0 surh that the solution of the Stefan
THE SThFAN PROBLEM
problem satisfies N(t) J flx 1< for aliT0 < t < T; T0 depends on and G, but is independent of T, h.
2. Show that the free boundary for the Stefan problem with n = I is given by x = s(:), where s(t) is continuous and strictly monotQne increasing in 1.
3. Extend the definition of g to aBR X (0, T) by g = 0, and consider the problem: Find bounded measurable O(x, I) 0 such that
(9.21) + dx di
where is any smooth function in QT' = 0 if x E an or if i = 1, and a(e) = a(8(x, 1)) is understood to be a measurable function such that
a(9(x, 1)) = O(x, t) if 8(x, :) >0,
—k 0 ifO(x,t) = 0;
finally, 90(x) 9(x, 0). It is well known 194b] that this problem has a unique solution which can be obtained by approximating a(s) by aji), a,,(t) >0. Notice that (9.21) can be written in the distribution sense as
(9.22) — 49 + = 0,
9=gon8flx(0,T), O=OonflX(O).
Set
(9.23) u(x, 1) f'O(x, i) dr [(x, i) e Qr].
Prove: 0 is the solution of (9.21) if and only if ii is the solution of the Stefan problem (9.8).
f Hint: If 9 solves (9.21), then
— 4u —1= y(u,),
where y(u1) 0 — a(0) > 0, so that
(9.24) (u, — — u,) — u,) V v 0, u 0
and conversely (9.24) implies (9.21). Show that if u solves (9.8), then is solves (9.24).)
VARIATIONAL INEQUALmE5: EXISTENCE AND REGULARITY
10. VARIATIONAL INEQUALITIES FOR THE BIHARMONJC OPERATOR
The variational inequality
in R2 corresponds to a membrane that must stay above the obstacle If instead of a membrane we have a two-dimensional plate, then the variational inequality is
&u(u—4))0.. In order to formulate this problem more precisely, we introduce a bounded
domain U in with boundary (0< a < 1). Let 4(x) be a function in C2(U) such that 4) 0 on and introduce the convex set
(10.1)
Consider the variational problem: Find u such that
uEK,
or, equivalently,
VvEK.
By Section 2 there exists a unique solution u of this problem. Taking V = u + >0, 0, E in (1023). we find that
0 in the sense of distributions.
Hence is a measure in K. It follows that i(K) < 00 for any compact subset K of U; see: reference 159.
We shall prove later that u E and that, for n 2, u E C2(L2).
Lemma 10.1. There exists a function w saiisfring:
(a) w = a.e. in U. (b) w is upper semicontinuous a. (c) For any x0 U and for any sequence of balls with center x° and
radii p:
f wdxjw(x°) ifpio.R,( x0)
VARIATIONAL INEQUAIII1ES FOR TUE BIHARMONIC OPERATOR
Here we use the notation
(10.4) 1AI=measureofA,
where A is either a ball or the boundary of a ball.
Proof. Let
(10.5) w,,(x)f B9(x)
We first show that for any x0 e
(10.6) isdecreasinginp.
For any u E we can write
= dS —
dx
where B0 = B9(x°), = and
(10.7) G0= p. —log-- tfn=22,rr
is Green's function for —A. Similarly, if p' >
Au(x0)
Since 6, G,.+ we get, prqvided that 42u 0,
f Au.5, and by integration,
(10.8) f a, For a general u E with &u 0, we can introduce 1/m-inollifiets
92 VARIATIONAL INEQUALfl1FS: EXISTENCE AND REGULARITY
where
j(x) =j0(IxI), J0E 0 ifItI> 1,
fjo(JxJ)dx= 1.
Since iXVm 0, (10.8) holds with Au replaced by J',,. Taking m oo, the inequality (10.8) follows. We conclude that
(10.10)
w( x) is some function. Since each w0 is continuous in x,
(10.11) w( x) is upper semicontinuous.
Recalling that Au E LL, we also have
w9(x) Au(x)
Consequently,
w = Au a.e.,
and together with (10.10), (10.11) the proof of the lemma is complete.
10.2. For any point x0 E that belongs to the support of
(10.12) w(x°)
Proof. Extend u into a function in and denote by uE the fier of u. Let x0 E Suppose that for some 8 >0 and a neighborhood W of x°, the inequality
(10.13) forallxe w
holds for all e sufficiently small. Let E C000(W), ,j 0, be such that = 1 in a neighborhood W0 of x°.
Then for any E I 1< 6/2, the function
+ (1 — ij)u t
is in K. Taking this v in (10.3) and then letting e 0, we obtain (since u
VARIATIONAL INFQUAUTIES FOR THE BIHAI1MONIC OPERATOR
in in the support of It follows that if x0 E supp then there exist sequences Xm x0 and em —. 0 such that
(10.14) UE(Xm) — 0.
By Green's formula
(10.15) U(XJ=f udS_f
where Bp,m = fly — Xm}< P}, Spn, = aBp.m, and
G9asin(l0.7)withr=IzI,
is Green's function for in Similarly,
(10.16) 4t(Xm)=f S.m
Since u also Ue so that
j Using this inequality and (10.14), we obtain, by comparing (10.16), that
(10.17)
We can write
(10.18)
V(Xm — y)dy = I dyV(Xm — —
f f
f — + Aim,
94 VARiATIONAL INEQUALITIES: EXISTENCE AND REGULARII'Y
where A m —' 0 if e 0 (uniformly in m). A similar relation holds for the second integral in (10.17). Therefore,
liminfJ Be,,,
By the mean value theorem there exist points Xm e Bpm such that
We may assume tiat xm -. and then, by the upper sem continuity of w,
— 0.
Taking p —i 0, —. x0, we obtain (again using the upper semicontinuity of w) w(x°) — 0.
Theorem 10.3. Au is in
Proof. Take any point x0 E sand denote by the ball with center x0 and radius p. Choose R such that BR C and E = I in B2 R/3' 0 I elsewhere. For any x E A2 R/3'
Au,(x) = . = dy,
where V = GR, as in (10.7). Expanding, we get
(10.19)
Au(x)= —f —f VfA2u1 + 2v(Au,) +
Au C, C independent of e.
Notice that supp is contained in \B2R,3. Hence
f —f Au1v(Vvfl.
VARIATIONAL FOR THE BIIIARMONIC OPERATOR
Using this and (10.20), we obtain from (10.19),
(10.21) —f V&u, —f Vt&2u, a(x),81/2 where C if x e BR,2.
By integration by parts (analogous to (10.18)] we have
(10.22) f V,(x—y)ti2u(y)+,8,,8*12 8*12 where$(x)—'Oifx EBR,,2,e-0.
Consider now the integral
81,2
It exists in the sense of improper integrals, that is, as
liinf fora.e.x. $'O
yE
Indeed, this follows from Fubini's theorem since for any k <
f dp(y)f d1&(y)<zao.B11,2 Notice that is lower semicontinuous.
Observe next that V(z) if (since Vis harmonic and'the mollifier is obtained by taking averages on spheres) and V(z) P(z) if z e. Therefore,
(10.23) !imJ V1(x—y)dgi(y) existsand equals -
Analogously to (10.22) we have, kw x E
f V(x
f where 4(x) -0 if e -.0. Hence (10.24)
/ V(x — . -.f V(x — .81'B1/2 if x —'0.
VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
We can write
=fw(z)JE(x —z)dz
where(p, 0) = (p, .. ,Ofl-..I) are the spherical coordinates about x and is a smooth nonnegative function. Since
where is the area of the unit sphere, the mean value theorem gives
—' w(x) ase —90.
Combining this with (10.22), (10.23). (10.24), we deduce from (10.21) upon taking £ —. 0, that
(10.25)
w(x)
forx E
We shall need the following (Evans) maximum principle:
Lemma 10.4. Let Z be a superharmonic function in R"; that is, Z is lower semicontinuous in RA and v is a measure. suppose that Z(x) 0 if X — 00 and set S = supp If Z M on S, where M 0, thea Z M in R".
For proof, see reference 66 or 132. We apply this result to Z = — V(oo); the measure v is the restriction of
to BR,2. By Theorem 10.2
w(x) on suppv.
Since the integral on the right-hand side of (10.25) is 0, we see that
onsuppv,
and Lemma 10.4 gives
• inR".
VARIATIONAL INEQUALITIES FOR THE BIHARMONIC OPERATOR
Observing finally that the integral in (10.25) is bounded in BR,3, we conclude that I w C in BR,3, and the proof is complete.
Theorem 10.5. u E
Proof. We can write
ue(x) 18k
— y) dy, x E BR,2,
where W is the fundamental solution of
W(x) = ifn = 3orn 5
imn—2,.
where are constants, chosen such that
ti2W = 8 (6 = Dirac measure).
Expanding = ± and performing integration by parts, we find, after making use of the fact
I C (which follows from Theorem 4.3),
that
(10.26) u,(x.) = f W(x — dy +
82R/3
where C in BR/2 for any derivative
One can directly verify that
a2 (10.27) W —c (c positive constarn)
ax1
and
a2 1 2x2 (10.28) ifn2.
ax, IxI
Applying a2/axy — to both sides of (10.26) and using (10.27) afld the fact that 0, we obtain
— C.
VARIATIONAL INEQUALiTIES: EXISItNCE AND REGUlARITY
Since the last integial is equal to
fB2 R/ with another C, we conclude that
inBR,2.
Since, by Theorem 10.3, is locally bounded, the same is true of and consequently
a2u
ax1
But then also
l)C,
and we obtain
a2u
Taking e —, 0 and noting that x1 can be in any direction, the assertion of the theorem follows.
In the next theorem we restrict the dimension n to be equal to 2.
Theorem 10.6. If n = 2, then u E C2(IZ).
Proof. By Green's formula, if Br(X°) C
= —f — f Gtt2udx,OB,(x°) P 8(x°) where G r/p is Green's function with pole at x0 and p x — x0
. Since
I C in compact subsets of
f log X
VARiATIONAL INEQUALItiES FOR ThE BIHARMONIC OPERATOR
It follows that the measure = A2u, satisfies
(10.29) log(l/r)
ifB,(x)CK,KCO,
where C depends on K. Applying the operator
a2 a2
8x1
to both sides of (10.26), we get
f + where y,(x) is continuous in x E BR!, uniformly with respect to and, by (10.28), F(x. y) is a bounded function, continuous in (x. y) if x Since, by (10.29),
—, 0 if r —.0, uniformly with respect toy, e,
'a standard argument in potential theory shows that
(10.30) is uniformly continuous in x,
when x is restricted to any compact K subset of uniformly in e, for £
It follows that for a sequence r —' 0,
(10.31) 0 u —. 0 ii uniformly in x K,
where Du is a version of the distribution derivative Di. Thus there is a version of — that is continuous in IA. By change of variables
xl+x2 xI—x2 xI-, , x2—.
we find that also has a continuous version in U. Let us explain this last fact more carefully. For a sequence e 0.,
—. ux, U1,, —, U, U
uniformly in compact subsets. Hence and (u,,), exist and coincide with U. We set, of course, U = Thus
(10.32) = (u,,) exists and is continuous.
VARIATIONAL INEQUAUTIES: EXISTENCE AND REGULARITY
We next prove:
Lemma 10.7. w is continuous in
Proof. Denote by S the support of A2 u. By a continuity theorem for subharmonic functions [66], if w restricted to S is continuous, then w is continuous in LI. Thus it suffices to show that
(10.33) if P0 = (x0, Yo) ES, thenwlsiscontinuousatP0.
= (Xm, Ym) E 5, -. P0 be such that if am is the angle between and the y-axis, then
(10.34) am—'O
— H!'m — POI. We shall prove that
(10.35) W(Pm) -, w(F0).
Take for simplicity (x0, y0) (0,0) and introduce the squares
Rm (O<Y<Ym,IXI<3Ym). We can write
(10.36) w = Au — u,., + 2(u,,, — + 24, a.e. Sinceu—+=0,
(10.37) —
(u,, — dx — f(u12)Yni
(u.,—- 4,)(x,0) dx
= f(I/2)Y JX( Ym) dx
/ )YN.
_f(h/2)Ym fX(u —
—(I/2)y,, 0
From the relations (with ü = u —
0 = ÜA(Xm, 3',,,) —
=[Üx(xm, Ym) — üx(xm,0)1 +[Üx(Xm,0)
= 3') + dx,
VARIATIONAL INEQUALITIES FOR THE BIHARMONIC OPERATOR
we obtain
+ 6m)Ym + 0(1 Xm I) = a (öm 0)
since is continuous and C. Using (10.34) we see that 0) = 0. We can therefore deduce from (10.37) that
(10.38)
where RmI= area of Rm. Let U be a continuous version of the distribution derivative — and
introduce the continuous function
g = U + 2+,,,.
Then (10.38) gives
(10.39) 11ff(w_g)dx4y=am, ci,,,—'Oifm--.oo.
Let B,, be the ball with center F0 and radius Ym and let P E B,,,. Denote by the ball with center P and radius p. Clearly,
Rm C
Let be an upper bound on w — g in a compact subset of IZ which contains P0. By (10.39),
/ (O<A<1)F) where A is independent of m, provided that in is sufficiently large. Since w is subhatmonic and g is continuous, the left-hand side is
1m'0utm''00. Hence
for any A <A' < I, provided that mis sufficently large. Thus
Repeating the process, we obtain
VARIATIONAL INEQUALmEs: EXISTENCE AND REGULARITY
and then
B I ifk—'oo. I in,1 B,,,
Since wis subharmonic and g is continuous, the left-hand side is w( P0) — g( PA), "k E BmL. It follows that
(10.40) w(P0)
For any small h > 0, consider the rectangle
= <X <Xm + <v + h}.
We can write
fJ(Ym + h —y)(u — dxdr
L—J1 J2
Since
— + CIXI2,
((u +X, clxi,
we get
C/i', C/i5.
Noting also that 0, it follows that
+ h y)(u — —Ch2.
Therefore, there exists a point QA E such that
w(Qk) — —C/i.
VARIATIONAL INEQUALITIES FOR THE BIHARMONIC OPERATOR 103
Taking h -. 0 and using the upper semicontinuity of w, we find that
w(Pm) lim = Thus
(10.41) (Or any Pm.
The proof is valid also for P0. so that w(P0) g(P0). Recalling (10.40) it follows that
w(P0) =
and (J0.41) yields
liminfw(Pm) = w(P0).
Since w is also upper semicontinuous, (10.35) follows. - To complete the proof of (10.33), suppose for contradiction that Pm E S,
w(Pj—'A, w(#,j-.A withA,'A. By extracting a subsequence we may assume that (10.34) hold. Then A lim w(P,) = w(P0). Similarly, A = w(P0) and we get a contradiction.
We have proved so far that + ui,, — have continuous versions. Hence also has a continuous version, say U0. We can write, for any (x, y) E Q, x — small,
y) —. y)
and thus conclude that y) exists and coincides with U0. exists and is continuous. Recalling also (10.32), the proof of the theorem is complete.
Remait 10,1. The results of this section are valid also (or the variational inequality (10.2), (10.1), in which is replaced by H2(fi) h
The variational inequality (10.2), (10.1) a called the obstackpfrvbiem.'Other interesting variational inequalities arise with constraints on the second deriva- tives of U; see Section 12.
PROBLEMS
1. Consider the penalized problem
U1 E
104 VARIATIONAL INIQUALrTWS: EXISTENCE AND
where fte() = = t2/E itt < 0, = 0 if > 0. Show that this problem has a solution U obtained as a minimizer of
+ )'e(V 4))] dr, V E
Show also that
and u weakly in where u is the solution of (10.2), (10.1).
2. corresponding solution of (10.2), (10.1) is a function u = u( r), and that: (a) (u > +} coincides with (r> 8) for some 0 <8 < 1. (b) is 0 and
4+y I
0 if e 0 and, consequently, the functions do not have a modulus of continuity that is uniform with respect to e.
3. Consider the variational inequality
uEK, VvEK,
where is in C' and
K= (V E fl a cAt,
a <0 <$. Show that the solution is given by u r(F), where AF = I ifi<a,
t>fl. It follows that for any 2'cp< oo U C3 in general).
4. Introduce the second finite-difference quotients
/ v(x + he,) — 2v(x) + v(x — he1)— —.
Let u E H2((l) u � 4), 4) 0 iP c 1. Show that if 4) is convex, then
u + e4)A,hu 4) if e, hare small.enough.
ThIN OBSTACLES
5. Letg(x) I — — x?)], wherex° a point in Show that if 4, E then there exist sufficiently large numbers r, a and a sbfficientiy small neighborhood U of x0 such that g and (4, — a)g is convex in U.
6. Prove that the solution of (10.3), (10.1) is in (Hint: Take v u + — a)g] in (10.3), E 0
1.]
11. THIN OBSTACLES
Let be a bounded domain in R" with C3 boundary and let S0 be a C3 hypersurface in R" such that
(11.1) S0 divides into two domains, and fL.. (See Figure 1.3.)
We shall use the notation
S = S0n
—
Let 4) be a continuous function defined on S. C2 on S. and
(11.2) •<OonaS, max4)>O, S
and introduce the convex set
(11.3) K=
We also introduce a bilinear form
a(u,v)
where
(11.4) VEERS, xEQ
K.
FIGURE I•.3
106 VARIISTIONAL INEQUALITIES: EXISTENCE AND REGULARITY
Consider the variational inequality:
(11.5) uEK,
VvEK.
We call it the interior thin-obstacle problem; 4, is the thin obstacle. Before we begin with the study of this problem, we prove a lemma that
applies to general elliptic operators:
— a,j(x)aaau + s,j1 J iI
provided that
(11.6) (sc>O),
+ Ilcil,
Lemma 11.1. Suppose that Au = f in Q a bounded domain, and u E c3(1Z) fl C'(fl),
(11.7) liv! 1). If
then
where C is a constant depending only on K, and
Proof. Suppose that 12 C (x1 > h) and take
v = I — (A >0).
If is sufficiently large then v >0, Ao >0 in Applying the maximum principle to Cv.t u, we obtain f u CM.
To estimate f Vu (, we use the Bernstein trick of applying the maximum principle to the function
w—UAUA+'YU2 (withswtabley>0)
where VA = aV/aXA and the summation convention is used. We compute
w, = 2UAUA, + 2Tuu1,
w,1 = + + 2yu1u1 + 2yuu,1,
THIN OBSIACLES 107
so that
Aw = — — —
+2bIuAuA, + 2yub1u1 + CUXUA +
Noting that
= — a1 AUIJ + +
we get
(11.8) Aw = 2yAu + 2uA(Au)A + — cAu)
— — — CU)1UA.
Using the assumptions (11.6), (11.7) we find that
+ Vu(2 + CM2 + yCM— 2yKI Vu12,
where C is a constant depending on K, K but not on y. Choosing y sufficiently large, we obtain the inequality Aw CM2 with another C. We can now apply the maximum principle to CM2v - w and derive the estimate w CM2 iq D; hence(Vuf'(CM.
We shall, also need the following result (161fJ (see also reference 109).
Lemma 11.2. Set d(x)=dist(x,afl)forx EO,andletmbean integer 0 <a < 1. If &2 is in then there exists an (l-neighbo#*oodN of ao such that d(x) is in Cm(N) .
We return to the variational inequality (11.5), and suppose that u is a solution. By choosing v = u + E 0 on S we find that u is a weak solution of
(11.9) infl;
0
VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
If, further, u belongs to and to then
onS;
here is the outward conorinal to fl at S; that is, is in the direction of the vector
(a11(x)cos(x, )),
where are the component of the outward normal p ± to ± at S. If u(x) > 4(x), we actually get equality; thus
(11.11) onS.
Theorem 113. There exists a unique solution u of the variational inequality (11.5) andu E
Proof. Existence and follow from Section 2. The set
S1
is a compact subset of Let x' = f'(s) be a local parametrization of S. Then the hypersurfaces
+ v(s)8 (181 small),
where (v'(s),. . . is the vector v = v at f(s), are parallel to S at distanceS and they lie in ifS > 0 and in (L ifS <0, provided that f(s) is restricted to a small neighborhood of
We extend as a C2 function in 0 such that, for some sufficiently small >0,
q'(f(s) + v(s)8)
if 181< and dist(f(s), and <0 on Let (x) be a function in which coincides with dist(x, S) if
dist(x, S,) and is elsewhere in Define d (x) in a similar manner with respect to 0 d(x) = d ± (x) in 0+.
For any small e > 0 we nbw consider a "thick obstaci,?'
(11.12) c5(x)
THIN OBSTACLES 109
This obstacle is in C2(12), is positive in the interior of and is strictly negative in a neighborhood of 8fl and on S ' S..
Consider the variational inequality
(11.13) u, E K,,
YVEK,,
where
(11.14) K,
and set
a I av
If u, is the solution of (11.13), then
a.e.inQ.
Since also u E the maximum principle gives
(11.15) u>O mU.
Hence the coincidence set
A, = (x E U; u
must be contained in the set {+ > 0). It follows that
(11.16) A,c (41(x)>0,d2(x)<e4,(x)),
Let w be the solution of
Aw0 inO, onS,
w0 By comparison, U, win Hence If x E A,, then
d2(x)
110 VARIATIONAL INEQUALITIES: EXISTENCEAND REGULARITY
Since w is Lipschitz continuous in some -neighborhood of
w(x°) — — x°J,
where x0 is the nearest point to x on S [that is, x — = d(x); note that x — x° is by (11.16)]. Since, by definition of 4(x) in
= if Ix — x°I is small enough,
we get
—Cd(x) — d2(x)
so that d(x) Ce. Thus we have improveu (11.16):
(11.17) AE n lies in a Ce-neighborhood of
The function u satisfies
u,=O
By (11.17);
onA,.
Applying Lemma 11.1, we obtain
(11.18)
Now take a sequence ü uniformly in such that vu weakly star in By Minty's lemma
a(v, t, — 0
if v E v +,; this holds in particular if v E v > on S, provided that e is small enough. It follows that for such v
a(v, v — a) 0.
By approximation, this inequality holds also for any v v 4 on S. Thus u coincides with u and, from (11.18),
C.
This completes the proof of the theorem.
THIN OBSIACLFS
Consider now the case where the hypersurface S satisfies
(11.19)
and suppose that 4' <0 on as. Let
(11.20) K= {v
The corresponding variational inequality (11.5) is called the boundary thin- obstacle problem, or the Signorini problem. In this case we formally have
(11.21) onS
where is the outward conormal. Theorçm 11.3 extends to the present case. In fact, since
for any v 6 C'(O), v = 0 on S (see Problem 4), a(u, v) is coercive and thus the Signorini problem has a unique solution u. The proof that u 6 C° is similar to the proof in Theorem 11.3.
We return to the' interior thin-obstacle problem and prove additional regu- lantyof the solution. Consider first the special case of
(11.22) VvEK; uEK
and
(11.23) is a hyperplane.
We set x = (x1,.. . ,x,,,), where m = n —1, y and X (x, y), and take S0 = {y = 0). Recall that the coincidence set
{(x,0); u(x,0) = +(x))
is a closed subset of 11. We shall often identify the points x with
Theorem 11.4. ThE solution u is in C' and in for some 0<a<1.
112 VARIATIONAL EXISTENCE AND
The proo will be based on several lemmas. We begin with the penalized problem
(11.24)
where /36(t)<O, PhQ)>0, pb'(t)<O, -40 If 6j,0, 0 if 8l0,and$8(i)-. —aoift<O,8j0.
Then
and thereforeu1,8 w(wasin theproof of Theorem 11.3). Wecanproceed as before to show that
(11.25) lvu.,,I'cC.
Since in an (1.-neighborhood of an that is independent of e, 8, = 0 in this neighborhood and, consequently,
(11.26) Cm an (1-neighborhood of an.
Differentiating the penalized equation twice with respect to any tangential direction in the x-space, we get
.—A(D,,u,,) + — — 0.
If D,,u,8 takes negative minimum at a point X° GO, then we deduce that
—c atX°.
Recalling also (11.26), we conclude that
(11.27) —C in(1.
Taking8 -Oand
Lemmall.5.
The next lemma is a comparison result.
[emma 11.6. Let Mfor all i. If A 3 M, (x0,O) A and
(11.28) +v$(x0) . (x_xo)+A(Ix_xoI2_my2),
THIN OBSTAQIES
then for any open set Q such that
(x0,0) EQ C (1,
there ho/dc:
(11.29) sup(u — >0. aQ
Proof. At (x0,O) u (since u(x0,0) > Noticing that u — is harmonic in Q A, the maximum principle gives
sup
Since however u — 0 on A if A M, the assertion
Lemma 11.5 implies that — Cy is a bounded monotone (unction, for y >0 and for y < 0, and therefore the limits
o1(x) = y), v4O
o2(x) = y). y tO
exist. We shall temporarily assume that
(11.30) flissyihmetricwithrespectto(y=0).
Then u(x, y) is(x, —y) and a1(x) = set o(x) = In view of (11.11), I
(11.31) if(x,0)EA,i.c.ifu(x,O)=4(x), (11.32) o(x)=O
We axe interested in proving HOlder continuity of a. We begin with continuity in measure.
Lemma 11.7. Let (x0, 0) E N. There exist positive constaflis C4.C suth that for any small y >0, there exists a ball contained In fl A1, where A1,= (xES;a(x)> —1).
Picof. Take a cylinder
Q = X with C2 cC1
114 VARIATIONAL iNEQUALITIES: EXISTENCE AND REGULARITY
and apply Lemma 11.6. We distinguish between two cases:
(i) SUP8Q(U — is attained at a point (x1, e) on the lateral boundary of the cylinder.
(ii) SUPaQ(U — is attained on one of the bases of the cylinder, say at (x1, C2y).
In case (i),
u(x1, e) e) = + c+(x0) (x1 — x0)
+A(Ixi — x0(2 — me2) + C3y2
since C2y =1 — x0 . Also, if
(11.33) Jx2 — x1 (x2 — x1) V(u — +)(x1,e)
then (since C)
(u — s) — (u — 4)(x1, e) = (x2 — x1) — 4)(x1, e)
—C3y2
provided that (24 is small enough. It follows that
(u—4)(x2,e)>O.
We claim that for x2 as in (11.33), o(x2)> Indeed, if u(x2) —y, then (u — 0 (by (11.32)] and then
(u — +)(x2, e) = (u — e) — (u —
= ea(x2) + ff + Ce2 <0
if C2 is small enough (since t C2y). The set of points x2 satisfying (11.33) certainly contains a ball Since
o(x2)> in this ball, the proof is complete. Jncast(ii)wcwrit.
+ (x1 — x0) + —
—
THIN OBSTACLES
and, as before, if
(11.34)
then
u(x2, C2y) — � —C7(C2y)2.
Since, on the other hand,
u(x2, C2y) — = o(x2)C2y + o(x2)C2y + C(C21)2,
we conclude that o(x2)> if C2 is small enough. Finally, the set of points x2 satisfying (11.34) certainly contains a bail
In the next lemma we shall use the following simple fact:
function in
(11.35) C B1(i), then w e(S) >0 in
Lemmall.8. If(x0,0)EN,lhenthereexistsanumttera,O<tr<l,suchthag
—CJx—x0r
for some positive constant c.
Proof. We shall 'prove inductively that
(11.36)
u we may assume that (11.36;) holds for k=O.Toproceedfromktok + I, considerfor
u+pk (11.31)
This functioii harmonic in (0, 41.7 with replaced by p7k, we have that
I X (0);
here has be , chosen sufficiently small Ibut of Ic, j'4
116 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
6 is also independent of k, y). Using (11.35) (after appropriate scaling), we deduce that
u + fjk
+ pk c > 0 in x hk}
Since y fi we obtain
_pk +
for some small A > 0, A independent of y, k. If 0 <y < then
u).(x. y) = — 9)()' <9 <uk)
_pk+
Choosing y < 1/2, /3> 1 — X/2, the proof for k + I follows. Finally, the assertion of the lemma follows immediately from the inequalities (11.36).
Completion of the Proof of Theorem 11.4. Since a = 0 on N, Lemma 11.8 gives
a(x) — a(x0) a(x) —clx — xoIa
if x0 E N, x E S. Since also a(x) 0, we get
(11.38) x0EN, xES.
Now, in int.A, u = and u is then regular on either side or fL (up to boundary). If x1 and x2 belong to A and
x1
then, by interior regularity,
If x1 — xil> d2, then let x0 be a point in N whose distance to either x1 or x2 is — x2('/2. Applying (11.38) with x x1, x = x2, we get
pa(x,) — Cf x — xoJa/2.
We have thus proved that a(x) is HOlder continuous in some A-neighbor- hood of 8A. Since 0(x) = 0 in N, it follows that o(x) is HOlder continuous, say
THIN OBSTACLES
with exponent a, in some neighborhood of 8A. But then, by a standard result in potential theory. u E up to the boundary, from either side of S; see Problem 4.
We have completed the proof of Theorem 11.4 under the assumption (11.30). In the general case we take a domain (L,, C f2 symmetric with respect to (y = 0) such that fl (y 0) contains {x S. 0), and let
V=u
Then u = u — V is a solution of the thin obstacle problem with obetacle = — V(x,O), and ü = 0 on a(20. We can now I'heorem 11.4
to ü and thereby obtain the assertion for u.
We shall next generalize Theorem 11.4 to the general variational inequality (11.5), assuming that
xE (.c>0), (11.39)
aajEC2(fl),
We shall also assume that
S = (y = 0), (11.40)
0 for 1 s — 1.
Here we use the notaton X= (x,y), x as before.
Theorem 11.9. If (11.39), jI 1.40) hold, then the solution u of the variational inequality (11.5) is in and in some 0<0< 1.
Remark 11.1. The conditions in (11.40) are not vezy restrictive. Indeed, suppose that (11.40) is and let X° be a point of S. Then there is a diffeomorphism of aneighbOrbi*)d Vof X0 onto a domain V' such that V fl S is mapped onto a planar region in (y' = 0) (where (x', y') are the new coordingesj and the new.coefflcients a1satisfy 0
1;
see Section 4 for the construction of such a diffeomorphism. The new operator will have first-order terms, but this will not affect the proof of Theorem 11.9.
118 VARIATIONAL INEQ1JALmLSc EXISTENCE ANI) REGULARITY
Thus if for some small neighborhood V of X°
(11.41) onaVflS,
then we can apply Theorem 11.9 in V.
We begin with a generalization of the first part of Lemma 11.5.
Lemma 11.10. For any small 8 > 0, there is a constant C> 0 :hà't if p0 dist((x0, Yo)' A) > 0, then
(11.42) p0
for any tangential direction r.
Proof. Applying I).. and D.,,. to the penalized in (11.24) (with replaced by A), we obtain, after setting u = u, = 8 =
(11.43)
(11.44)
+ — — D.,,u) +
where is any nonnegative function. Note that the proofs of (11.25), (11.26) remain unchanged in the present case
of the operator A. Multiplying (11.43) by — 4).,) 0, E and
integrating over (2, we get, after integration by parts,
lCffl v(u., —4),) + cf is,. 4çfl v(u., — +1)1
—4i.,).
Integrating by parts in the last integral and using (11.25), (11.26), we find that
(11.46) flvu,12<C,
where C is a constant independent of 8, e.
THIN ORSTACLFS
Let
nnn — + A,o1.
By (11.26) we can choose A > 0 sufficiently large so that 0 in a neighbor- hood of MI.
Let k be an odd positive integer and take = Wv" in (11.44); then
(11.47)
The first term on the left is
cJi (c >0).
On the right-hand side we get terms of the form
L1 =fokD,(aD,ju), L2=fv'Di(bD,u).
By integration by parts,
L1 = — kf v* =
so that
IL, c(fi 12) "l(fvic_I j2) 1,'2
Also, by integration by arts and (11.25),
IL2 I cf ID,o" I Thus (11.47) gives
(11.48) fI Cf +cf vu,I2.
To estimate the last integral we take in (11.4i). We (hen obtain after integrating by parts thó last thrm on the right-hand side 1.25), the bound
cfi +cfp +Cfvk_I I
120 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
It follows that
(11.49) ft I-l)/2f2 Vv"t I +cfi
Taking k 1 and using (1 1.46), we deduce that
(11.50)
Next we refine (11.39) for k 3. using
3 ft VVA 1)/2 lj/2
12 + small,
I 1= I)/2v(k_3)/2
I VU.
fl vv vuj
t)/2 12 +
I 12
The last term can be estimated in the same way as the last term in (1 we get a bound for it of the form
cfi vvk_31 +cfi Vvk_311 Vu1.i -ICfv"_31 Vuj.
above together in (11.49), we find that, for k 3,
($1.51) +cfi
+Cf v*3 J
+cf U Vu, I.
ThIN OBSTAtLES
Taking k = 3, we obtain, after using (11.46) and (11.50),
(11.52) C; hence also fv4 C.
We can proceed now to refine (11.51) for k 5, then substitute k 5 to deduce that
and so on. In conclusion, v" E W'2 for all k and, in partwular,
(11.53) E V for all I <p <00.
Since the estimates above are independent of 6, they hold true also for u = tim Hence (11.53) is valid for the solution u of the variational inequality.
To complete the proof of the lemma, we represent u,,( in X0) [where = (x0, yo)' <r < by Green's function Gr for A (in B,(X0)):
(11.54) DrrU(X0) — + f1 B,(X0) Notice that in X0).
(11.55) A(D.,,,u)- p0
by interior elliptic Cstimates, since u satisfies the elliptic .equation Au = 0 in (X0) and I Du C.
we integrate both sides of (11.54) with respect to r, < r < we obtain
(11.56) = f + f
where (see Problem 5)
(11.57)
Using (11 (11.55), and (11.57), we obtain from (11.56) S
—C--
where 1,ip + I /q = 1. Since p can be taken arbitrarily large, (11.42) follows.
122 VARIATIONAL INEQUALITIES: EXISTENCE AND REGULARITY
Since, by (11.40), (11.25), and interior estimates,
we obtain from Lemma 11.10 that
(11.58) foranysmall6>0.
It follows that — Cy' is a bounded monotone function fory >0 ory <0, and thus the limits ±) exist. /
Now take a domain symmetric with respect to (y 0) such that n {y = 0) contains the set S1 (where 4 0), and let
u*(x, y) = u(x, y) + u(x, —y).
This function is symmetric in y, and u*(x,0) > 24i'(x) near fl S. We have
y) + a.1(x, Y)] =
where
i = Y) — )) au(x,—y)
— y) — a,,(x, au(x,y)
F
It follows that f is a bounded function. Define +*(x) = 24(x) and o(x) = +0). We now proceed to estab-
lish Lemma 11.6 for (with obstacle taking
= . (x — x0) + Ao(Ix — x012 — Ny2);
N is sufficiently large so that 0 and A0 sup D2+ Since
(instead of C), the proof of Lemma 11.7 needs to be slightly modified: the assertion now is that
C fl for any0 <9< 1.
ThiN OBSTACLES
Proceeding to extend Lemma 11.8 we construct a function w0 such that
inK,
onaK
where K is the cylinder X (0, •Since w0 E we find (after scaling) that
w in (11.37) by
+ +
0 <9 < 1. Since Aw = 0. we can proceed as before to establish(l1.36) by induction. Having thus extended Lemma 11.8. we have that o*(x) E in a neighborhood V of
The functions
u1(x, y) = u(x, y). u2(x. y) = u(x, —y) satisfy, in
0 (A, elliptic).
and au, 8u2u1 —u,O, ____=2GsECe in v.
- ay 8y
By a standard regularity result for elliptic systems (3bJ, it follows that u, a
in a-'- U V. This yields the assertion of Theorem 11.9.
Remark 11.2. Theorem 11.9 extends with obvious modifications 1o th* Signorini problem.
Remirk 113. The function u = Re{z3/2) = 8 it) is a solution of an interior obstacle problem with q = 0 on the x-axis;• it does not belong to c''- in (y 0) for any a> 1/2.
PROBLEMS
1. Consider the Signorini problem
uEK,
124 VARIATIONAL INEQuALmEs: EXISTENCE AND REGULARITY
where
K—
Prove: If there exists a solution, then f f dx 0 is sufficient for existence; see reference 141.)
2. Solve explicitly the solution of Problem I in the case = x < I), f const.
3. Sive: u" = y if —l <x < 1, u(—l) 0,
(uu'XI)=O.
4.
a E then u E C1 for any R0 < R. [Hint: Represent u by Neumann's function in the half-space y > 0.J
5. Prove (11.57). [ Hint: Introduce X' = X0 + p0X.j
6. Consider the Signorini problem
VvEK;uEK,
where K is defined by (11.20) with 0, and f E and (11.19) holds. Set
where G(x, is Green's function for in Ii Prove: If E mt S and then
7. Consider the thin-obstacle problem under the assumption of Theorem 11.4, and assume that n = 2 and that 4(x) is analytic. Prove that A consists of a finite number of intervals. [Him: Let e be the harmonic conjugate of u in Im z > 0 (z x + iy), F(z) = u(x, y) + iv(x, y), F'(z) = U + IV, 4(z) = holomorphic exten- sion of Then
(F'(z) — = (U+ iV—
is holomorphic in Im z > 0 and, on y 0, it is equal to
(U— 4"(x))2 — V2 + 21(U— 4'(x))V= real.
BIBLIOGRAPHICAL REMARKS
Hence g = (U — — V2 has analytic extension to {y 0). Next, g 0 since,onA,g = —V2 and / V(x,0)dx Ifg(x)<0, then —V(x)>O and x E A; if g(x) >0, then x E N; if x is a free boundary point, then g(x)0.]
8. Consider the thin-obstacle problem for a plate:
VvEK; uEK
is a circle {x2 + y2 < I) and K {v E v(x, 0) that 4) E C2 and 4)(± 1) <0. Prove that u E V p <2
(andqhus u W2 V r < [Hinr:'or any —1 <x0 < I, there is a neighborhood U of (x0, 0) such that if u K, then
v, = u + Eg + v),g] isihK
for any small e >0, where 0 if (x,0) U. 0 1, a is some real number and
g(x) I — — e"°) for some r >0;
see Problems 4 to 6 of Section 10. Substitute VE into the and deduce that E Next, by symmetry. (8/8v)
keeps a sigif on y =0 (It is taken in the distilL ütion sense). Since ttu is hañnonio iii it follows that
I,,0ay 'Represent by function in (L, and deduce that L'(O+)'foranyp<21 .
12. REMARKS
There are twp Eçcent books cm and.Capelo - [201 and by Ki,iderlehrá and is §Iso swvey by Kinderlehrer A existence thIory inequali- ties was by, Lions and P'4U. The Section 2 is based on [l6la], the method (l.l)—(I.5) taken from' rejáence W2" estimates for obstacle problem were first obtained Lewy'and Stampacèhia method was used by Bçezis and Stampacchia [53a); see also Brezis
126 VARIATIONAL INEQIJALIFIE& EXISTENCE AND REGULARITY
which Problem 5 of Section 3 is taken), Gerhardt [104aJ, and Lions [140j. Similar results for the Neumann boundary condition were developed by Murthy and Stampacchia 1149).
The interior W2'°° regularity (Theorem 4.1) is due to Brezis and Kinderlehrer [49]; a special case was earlier obtained by Frehse l93a]. The boundary regularity (Theorem 4.3) was established by Jensen [116c). Problems 3 to 11 of Section 4 are based on the of Caffarelli and Kinderlehrer [62); see also reference 58i. Frehse and Mosco [184) and Frehse (93eJ proved Holder continuity for solutions of the obstacle problem with irregular obstacle.
regularity for two obstacles was derived by Brezis (45a). and interior regularity by Caffarelli and Kinderlehrer (621 and Chipot 171a1.
The treatment of the filtration problem in Section 5 is due to Baiocchi [l9aJ: related problems are described in Bear [261.
The physical problem of Section 6 is described by Lanchon (1311, where the variational principle is also stated; see also Duvaut and Lions [81). Brezis (45bJ solved the problem (6.3). Theorem 6.1 is due to Brezis and Sibony [52). W2.P regularity (Theorem 6.2) was established by Brezis and Stampacch"i (53a] for 12 convex; the present version is due to Gerhardt [104bJ. Local regularity for general gradient—constraint variational inequalities was recently obtained by Jensen [1 16d).
Theorem 7.1 (W2°° regularity) is due to Caffarelli and Riviere [63eJ. Another method based on a penalized problem + vu12 — I) = was more recently given by Evans [85bJ in the case of simply connected domains and extended by Wiegner [181]; see also Ishii and Koike [191]. The simple but useful result of Problem 2 of Section 7 was first noticed by Ting [172bJ.
W2P regularity for parabolic variational inequalities with one or two obstacles were established by Brezis [45aJ and Friedman [94d, e, gJ. The esti- mates on the support of solutions given in Problems 8 to 12 of Section 8 are due to Brezis and Friedman [48]; Brevs [45aJ studied variational inequalities (8.10) with general closed convexed sets K. The method of proof outlincd in Problems 10 and 11 of Section 8 is taken from Evans and Knerr [87aJ.
Duvaut [80) has transformed the one-phase Stefan problem into a varia- tional inequality. Theorems 9.1 and 9.2 and Problem' 9.3 are taken from Friedman and Kinderlehrer [97). For the two-phase Stefan problem, see Friedman [94a, b, hJ and Rubinstein 1156) and thd references given there. A one-phase Stefan problem with supercoolad water was studied by Friedman [94fJ, Jensen [I 16bj, van Moerbeke (144a, bJ.
The material of Section 10 is based on the paper of Caffardili and Friedman [58d). Frehse [93b,cJ had earlier proved W2° and regularity. Caffarelli, Friedman, and Torch [60bJ proved W2 regularity for the corresponding two-obstacle problem, and gave a counterexample showing that 'the solution is, in general, not in C2 if n 2.
Problem 3 of Section 10 is based on a paper by Brezis and Stampacchia [53d]. They considered also the variational inequality with a set K given by
BIBLIOGRAPHICAL REMARKS
Kap = (v E HJ(1Z), 'fl). Caffarelli, Friedman, and Torch [60aJ considered this variational inequality with —a = = e > 0 and proved that, for a rectangle or a regular triangle, the solution u = u, satisfies (1/t) 4u —* U as e -. 0, where U solves the obstacle problem for —4 with obstacle d2(x) = (dist(x,
Evans and Knerr (87bJ. studied elastic—plastic prob'ems for plates; the constraint is a differential inequality involving second derivatives of u. They derived W32 regularity.
Problems 4 to 6 of Section 10 arc based Frebse (93bJ, Section II is based on Caffarelli (56dJ. More recently Kinderlehrer [122d1
has obtained similar results using a different penalization. Lewy [137a--cJ has studied the Lipschitz continuity of the solution in the case of two dimensions and showed that the coincidence set consists of a finite number of intervals in case of an analytic obstacle (this is the result given in Section II, Problem 7); related results were obtained by Athanasopoulos (161. Frehse (93dJ has proved the continuity of the first derivatives in any number of dimensions; see reference 93d also for related results for thin obstacles for minimal surfaces. The result of Problem 8 of Section Ills due to Caffarelli and Friedman. Recently Schild 11881 established W7-°° regularity in n-dimensions for the thin obstacle problem for 42, and C2 regularity in case n = 2.
The Signorini problem for systems models deformations in linear elasticity. It was first treated by Fichera (891 who established the existence of solutions in H'. Recently, Kinderlehrer 11 22e, fJ proved that the solutions belong to H2 and derived estimates on the coincidence set; for a = 2 the solution was shown to be in C1a.
In a recent monograph 11921 Elliott and Ockendon survey a large number of free boundary problems which arise in physics and engineering.
VARIATIONAL IN ES: ANALYSIS OF THE FREE BOUNDARY
in this chapter we study the regularity of the free boundary and its shape. It is shown that if the noncoincidence set has positive density at a point x0 of the free boundary, then the free boundary is smooth in a neighborhood of this point. Some theorems are established which can be used to show (under suitable conditions) that the noncoincidence set does indeed have a positive density at x0.
For those variational inequalities studied in Chapter 1, which arise in physical problems, we establish in this chapter the smoothness of the free boundary. We also study the shape of the boundary. The chapter includes stability results for the free boundaries and examples of variational inequalities whose free boundaries have singularities.
1. THE HODOGRAPH—LEGENDRE TRANSFORMATION
In this section we show that if the free boundary is C1, then it is in fact arbitrarily smooth provided that the coefficients of the equation and the obstacle are sufficiently smooth, Consider for example the variational inequal- ity
(1.1)
lb a domain G of and denote by fl the noncoincidence set (u >0) and by 1'
THE HODOGRAPH-LEGENDRE TRANSFORMATION
the free boundary. Thus
(1.2) Au=f inn,
(1.3) Vu0 onF. We shall require that
(1.4) FisinC',
(1.5) uEC2 int1UF,
where the last condition means that L)2u is continuous in Q and has continuous extension into F.
In the theorem to follow we do not require th&t the u, 1', 0 come necessarily from the variational inequality (1.1); 0 is any domain in R" and F is an 6pen portion of 1)0. We are interested in proving regularity of F in a neighborhood of any given point x° E 1'.
Theorem 1.1. Assume that (l.2)—(l.5) hold and that x0 E I'. Then:
(I) If f(x°) 0 andf is in C' in a neighborhood of x°, then I' is C' +a for any 0 < a < 1, in some neighborhood of x0.
(ii) If furtherf E (m integer 1, 0 <$ < 1) in a neighborhood of x0, then r is in in some neighborhood of x°.
(iii) Iff is also analytic in a neighborhood of x°, then I' is analytic in some neighborhood of x°.
Proof. Take for definiteness f(x°) >0. We perform a local diffeo- morphism x -' y and a transformation from u(x) into another function o(y) in such a way that F is transformed locally into Yo (P and v satisfies an elliptic equation in (ye' <0) with Dirichiet dMa on 0. UsIng boundary regularity results for elliptie equations, the assertions of will follow.
Set u. = For simplicity we take x0 = 0 and assume that the inward normal to 1' at x0
is (1,0,... ,O). We extend u1 as a C' function into a full neighborhood of 0. The change of variables is given by
and the new function v(y) is defined as
v{y) = x,y, + u(x).
VARIATIONAL INEQUALITiES: ANALYSIS OF ThE FREE BOUNDARY
Then
u0 onF, u14(O)=0
and (1.2) gives
u11(0) =f(0) >0.
It follows that det(ay/ax) * 0 at 0, so that (1.6) is indeed a local diffeomor- phism. It maps a neighborhood of 0 in onto an open set Win (y1 <0) and a neighborhood of 0 in r onto an open set S in Yi 0; 0 E S.
Notice that
(1.8) dv(y) x1dy1 +y1dx1 + du
—u1dx1
= x1 dy, + = + a2
Ua 4Ya,
so that
(1.9) v,=x1, vu where Since S is given by y =0 with arbitrary I=2(Ya)2 small, the corresponding points in. the x, coordinates arc
XaYa
Thus
av (1.10) .
is a local parametrization of r near 0, and it remains to establish the smoothness of v in W U S.
THE HODOGRAPH-LEGENDRE TRANSFORMATION 131
In order to find a differential equation for v, we compute
—UI, •
0 1 0
0 0
On the other hand, since x1 = v,, x1, = y (2 a n), we have
/ vii V,2 • • o I o
ka,J 0 0 1
Since (ay/ax) = (ax/ay) ',we obtain
(1.11) = and VII oil
Then also
— 8; — —
and '(1.2) gives
(1.13) jaW • 0fl Vfl
where .Y = (Y2'.. . ,y,,). From (1.3) we get
(1.14) o(O,y')O onS.
The equation (1.13) has form
O(y, v, v1,...,;, Va,... = 0
or, more, briefly,
v, Dv, D2v) 0.
132 VARIATIONAL INEQUALmES: ANALYSIS OF THE FREE BOUNDARY
Such an equation is said to be elliptic at y° with respect to a specific solution v(y) if the matrix
(1.15) ispositivedefinite
when evaluated for v = v(y) at y = y°. In our case = 0 is elliptic at y = 0 and in fact the matrix (1.15) is the diagonal matrix (A,3) with = — l/v11(0)
We can now apply local regularity results for the elliptic problem (1.13), (1.14): (i) follows from Theorem 11.1' in reference 3a, (ii) follows from Theorem 11.1 of reference 3a, and (iii) follows from Section 6.7 of reference 145 or from reference 194cJ.
Theorem 1.1 can be generalized to solutions of
(1.16) F(x, u, Du, D2u) = 0 in FE C in all its variables.
Theorem 1.2. Let u, I' satisfy (I.3)—(l.5), (1.16) and assume that F is elliptic with respect to u( x) at some point x0 E 1' and that
F(x°,0,0, D2u(x°)) #0.
Then:
(i) r is any 0< a < I in a neighborhood of x0. (ii) If FE Cm+P (m 1, 0 <fi < 1) in all its arguments, then 1' E
in some neighborhood of x0. (iii) If F is analytic in all its arguments in a neighborhood of x°, then r is
analytic in some neighborhood of x0.
Proof. We proceed as in the preceding proof. The only point that needs to be checked is that if 0 is defined by
F(x, y, Du, D2u) = O(y, v, Dv, D2v),
then 0 is elliptic aty = 0. From (1.1 1), (1.12) we find that, at y = 0,
ôVle V11 av1,
THE HODOGRAPH-L.EGENDRE TRANSFORMATION
and all the other partial derivatives au11/a Vkm vanish. Therefore,
= = V&m
A U,1 Vkm
, U,j
where = = (2 a n). and the ellipticity of 4' follows from the ellipticity of F.
The transformations (1.6), (1.7) can be generalized to
(1.17) yp=—up
Ic
V = XaYa + U,
and again
if F(x, y, Du, D2u) 4'(y, v, Dv, D2v), and (1.19) if Fis elliptic, with respect lou at 0,
then 4' is elliptic with respect to v at 0.
The
(1.20)
is called the hodograph transformation; (1.17) is called the partial hodograph transformation. The transformation given by (1.20) and
is called the Legendre transformation. We tefet to (1.6), (t7) alid mote generally to (1.17), (1.18) as the kodograph—Legendre transformation.
Theorem 1.1 can be used in the obstacle pmblezn provided we know that the free boundary is C' and thai the solution u is C2 up to the free boundary.
We shall next establish an analogue of l'heorem 1.2 for par;bolic equations. For simplicity we shall only with the case.
VARIATIONAL INEQUALITIE& ANALYSIS OF THE FREE BOUNDARY
We assume: is an open set in the space of variables (x, t) — (x1, x2,. .. ,x,,, t), r is an open portion of aa, and
oEr, r is in C1, 1' is not tangent to t = 0 at the origin,
u and belong to C'(fl U 1'), onF,
u, — F(x, t, u, 0 in fL
Theorem 1.3. If (1.21) holds and if F is elliptic with respect to u(x, t) at 0, F in all its arguments and F(0, 0,0,0, 0, then r is in some neighborhood of 0.
Proof. We may suppose that r is given locally by
E C' and x, <4 in fi; further,
= 0, = 0 except for
u, as a C function into a full neighborhood of 0 and consider the transformation
st, v=x,y, +u.
Then v satisfies
(1.22) — v,.Dv, D2v) = 0
where 0 is elliptic; also, v = 0 On =0. We now apply boundary regularity results for nonlinear parabolic eqUation (1.22); for further study, see
123a. For other free-boundary problems, one may apply related transformations.
We shall conider here impoi*ant case:
(1.23) F(x,u,Du,D2u)O
(1.24) u0, g(x,gradu)Oonr. 1.4. Assume that (1.4), (1.5), (1.23), (1.24) hold, that g(x, . . 4),,)
11W TRANSFORMATION
is in C2,
and that F is. elliptic with respect to u at 0. Then:
(1) 1' E C24a for any.O <a < I in some neighborhood of 0. (ii) If FE g E (m 1, 0 <fi < F' in all the arguments,
then 1' E in some neighborhood of 0. (iii) If F, g are analytic in all the arguments in a neighborhood of 0, then r is
analytic in some neighboi*ood of 0.
Proof. Suppose that the outward normal to 1' is in the direction of the positive a a full neighborhood of 0. We suppose that >0 and make the local diufeomorphism
(1.25) y = (x1,..
the function v(y) is now chosen to be
(1.26) v(y)x. Then r is transformed into y, = 0 and
u,J
0 = UI 1
14, U,, U,,
Consequently;'
ax uv,1 (1 ax,, IV =—=—,
ia,,
ay,, I ay,,—, —ua
136 VARIATIONAL INEQUALmES ANALYSIS OF THE FREE BOUNDARY
It follows that
— vnn ôyn — Vita
_!
Using these formulas one computes, for instance, that
n—I n—I n—I
(1.27) Au— I + 0,, a,I v,, V,,
and more generally
(1.28) 0 = F(x, u, Du, D2u) v, Dv, D2v),
and 1 is elliptic at 0. The function v also satisfies the boundary condition
I V1 Va_i I \ ony,0,
and since ag/app 0, we can write this condition in the form
w1th /, being as smooth as g. This condition is of Neumann type and the boundary regularity results quoted above hold for this condition.
Note finally that the boundary I' near 0 is represented by x,, = v(x1,. . . thus I' has the same smoothness as v.
Remarkl.1. Supposcthatn=2and
AuO ins, (1.29)
I
grad uI= I on r, 0 e F,
where , is the normal, and u, F satisfy (1.4), (1.5). Then the harmonic
lifE RODOGRAPH-I.ECENDRE TRANSFORMATION 137
conjugate v, with v(O) 0, satisfies
Av=O in(l,
avv=0, loni',
and v E C2 U F). Applying Theorem 1.4, we conclude that F is analytic. In n > 2, this conclusion is no longer true. In fact, the function u(x) = x1 satisfies (1.29) with F any C hypersurface of the form x,, = . .
PROBLEMS
I. Prove (1.19); here
Va=Xa, u=—y (ac.k), v0=u0 ($>k),
dv xadya+
k
k o .t I'
(k<r,s).
2. Let the assumptions of Theorem 1.1 hold with x0 = 0 and F: x1 • . Can Theorem LI be proved by using the transformation
(which maps F intoy1 =0) with v(y) —
138 VARIATIONAL INEQUALfl1ES: ANALYSIS OF THE FREE BOUNDARY
3. For the Stefan problem in one dimension
0,— = Oifx<s(t),
00, 6. —s'(t)ifx=s(t), prove that if s E C' and 0 E C2 up to the free boundary, then s E [Hint: Use the transformationy = x — s(t) ory = x/s(t) and set v(y, t) = u(x, 1).I
2. REGULARITY IN TWO DIMENSIONS
We define
— a_ 11
a a az ax2
and use the notation w(x,, x2) = w(z), where z = x1 + ix2. As USual,..BR = (Izl<R}, BR(zo) = (Iz — zoI<R).
We shall need Green's formula in complex form. Let w = .u + iv. Then for. any bounded open set E with piecewise C' boundary az and for any v (2<s<oo),
(2.1)
notice thatvECA(E),A= 1 —2/s>O Letz0,z,
= Efl (Iz — tot> e;Iz — 1>0.
Consider the function
(z—z0) (z—z,)
and assume that
w€JI'(E), s>2, .,
(2.2)
(O<a<1).
REGULARITY IN TWO DIMENSIONS 139
Applying (2.1) with this v and with E replaced by EE, and then taking e 0, we obtain Green's formula:
(2.3)
w(z1)—w(z0)_1j m
1
w5(z)dx1dx2 (z1 — z0) E(z — z0) Z
I I w(z)dz, wasin(2.2).
2ir: aR(z — ZO)m Z —
In this section we derive regularity results on the free boundary for two- dimensional problems. Unlike the results of Section 1, which presuppose a C' free boundary, here we only assumc that the free boundary is a Jordan arc.
Theorem 2.1. Let fl be a simply connected domain in R2 whose boundaiy aiordan curve, and let 1' bea Jordan arc, r C fl BR(for some R >0). Let u be a function in U F) and be analytic functions in BR. if
(2.4) inOflBR,
(2.5) u Vu = on r,
(2.6) inBR,
then F admits an analytic pararaØrizgtion.
We shall prove the existence of a local analytic representation; this actually implies global analytic representation E27; p. 376).
Proof. The function
(2.7) f(z) = f81iog '1 1f(t) dt, dt2
is real analytic since = —f. Therefore, it suffices to prove the theorem in case 1 =0, for otherwise we consider u — I instead of u.
We establish an analytic parametrization in a neighborhood of a point z0 F, and for simplicity take z0 = 0. Set
= — =
= —
2 D1v = Au = 0 in fi that is, V is holomorphic in -
140 VARIATIONAL INEQUALiTIES: ANALYSIS OF THE FREE BOUNDARY
The function V*+(xi, x2) has power series expansion
(IxiI< 26, 1x21<26)
for some 6 >0. Writing
z+i z—i 2
' 21..
we get an expansion
v*4,(z, 1) =
where the are now complex coefficients, and v*4(z 1) V+(x1, x2). This series is convergent for z 6,
I 11< 6, and we can extend it into a
holomorphic function of two variables, z and by
Consider the equation, for
(2.8) v*+(z, = vu(z).
By assumption
• D1V*4(z, 1) x2) = 0 in BR
and, by (2.5), (2.8) does hold if z r, = 1. We can the implicit function theorem. ft implies that there exists a unique solution
of (2.8) in a neighborhood of z = 0, = 0, and that (z, w) in a neighborhood of (0,0). Recalling that
(2.9) r(z)=i ifzEr, we see that the conjugate points of r are boundary values of a holomorphic function.
In order to get analytic parametrization we introduce a conformal mapping t-. gQ) from the hall-disc G = (111< 1, Im I >0) Onto Since is simply connected and r is a Jordan arc, there exists a g that maps the real interval
I—(t1+Oi; -
REGULARJTY IN TWO DIMENSIONS
onto r and g is continuous and I — I from G U I onto U r; see reference 27, p. 369. Thus g gives a continuous parametrization of I', and we shall prove that this parametrization is analytic. For simplicity, let g(0) = 0.
Consider the function
Ig(i) if Im 1< 0.
This function is holomorphic if t e, Im t 0 Ewhere e is sufficiently small so that g(t) is in the domain where r is holomorphicj. Since isalso continuous along Im I = 0, the theorem of Morera can be applied to deduce that 4)(:) is holomorphic in ft < e). Thus, in particular, g(t) is analytic for I real,
In the remainder of this section we shall relax the conditions of analyticity, replacing them by smoothness conditions. It will be convenient to use complex notation also for the elliptic equation. Thus we shall work with an equation
(2.11) w1=k inf2flBR
and assume that
(2.12) w=O onr,
inBR.
In the applications to (2.4)—(2.6) we take
(2.14)
and then (24)-(2.6) yield (2.1 l)—(2.13) with 4k = / — A+.
Theorem 2.2. Let 0, r be as in Theorem 2.1 and let g(t) be the conformal mapping G -. (continuous ahd I — I from G U I 0 U 1', as before). As.. swne that w E k E CA(BR)forsome0 <A < 1, and that (2.l1)—(2.13) hold. in (: 'vii,
Proof. Extend k as a CA function in R2.
A(z)=_!f (R'>R), ,T Z
142 VARIATIOMAL INEQUALITIE& ANALYSIS OF THE BOUNDARY
and for some fixed z0 Er (IzoI< 1), let
= A(z) — A(z0) — — 20).
Then E and
(2.15) 4 = k, w(20) w(z0) = 0.
Writing
(2.16) w*(z) = k(z0)(i — + R(z, z0) (z E BR),
we find that
(2.17)
[by (2.13), (2.15)) that for aiiy small e > 0,
(2.18) 4' ..ij<E<1 inB,(z0)
if r is small enough. The Jacobian of the mapping z -. w*(z) is
2_1412)<o
Therefore, this mapping is a C' +X diffeomorphism from B,(z0) into a neigh- borhood of 0; we denote its inverse by
The function
h(z) = w*(z) — w(z)
is holomorphic in 0 fl BR, continuous in U 1' and, by (2.12),
(2.19) ' h(z)=w*(z) onr.
Define in B8
Ig(t) if Im:<0,
where 6 is chosen small enough so that g(B8 fl (Im t 0)) C B,(z0); we assume for simplicity that g(0) = 20.
REGULARITY IN TWO DIMENSIONS
Note that
JGI = area of a <00,
so that g H'(G). Since h is Lipschitz, it follows that 4) restrcted to Im t > 0 (<0) is in Ii'. Applying the matching lemma, we conclude that
4) H'(B8).
The following chain rule formulas can be directly established for a function G(z) with z = k(t):
D,G(k(:)) = + D1G(k(t))Dk(l),
D;G(k(z)) = +
We also have
4k(:) = %k(t), = D,k(t).
If G(z) has inverse then by applying to = z, we get
D2G' - 1)6 D1G'D-G
Using these rules, we deduce from (2.20) that
Iw*I
iT by(2.18),iflrI<z&, lmt<0.
Since also 0 if Im t > 0, the function 4)(t) satisfies
I•-Ip.1<e<l rnB8.
A function 4) in H'(B8) satisfying such an inequality is called e-quasiconformal: by reference 40, p. 269.
• belongs B8,,2), wheres = s(e) -, oo Ut -, 0.
It follows that g belongs to H' S in B8112 fl (Im z >0). If instead oft0 = 0 we
144 VARIATIONAL INEQUAliTIES: ANALYSIS OF THE FREE BOUNDARY
take any other point :o E (— 1, 1), we conclude that for any small 6 > 0,
(2.21) gEHL5(GflB1_5) foranyl <s<oo.
To obtain more regularity for g, we shall use another extension of g( t) into Im t <0. First define g*(t) in G by
(2.22) h(g(t)) = k(zo)(g*(t) — + R(g(t),
where R(z, 1) is defined by (see (2.16)1
w*(z) = k(z0)(i — + R(z, 1).
It follows from (2.19) that
iflmtO.
Applying D1to (2.22) we obtain, since h(g(i)) is holomorphic,
so that, by (2.17),
(2.23) zo1, z0=g(0).
Now, by (2.21), g E and thusg E g' E LI. It follows that for any 6 > 0,
(2.24)
Note also that g7 E L5(G). Thus the function
Ig(:) (2.25) 4'(:) =
lmt<0
is in by the matching lemma. We shall need the following fact:
REGUlARITY IN TWO DIMENSIONS 145
Lemma 2.3. Suppose that vQ) E where B = and
where a > 0, s > 2 and a — (2/s) r, 0 <i < 1. Then there exists a complex number c such that
fv(t)—v(O) —cHCAIIr O(p< 1,
I C CA,
where
A = uI + IIVII,.C(8)
and C depends only on p. s,
For proof, see Problem I. Applying the lemma to 4', we obtain
+(O) — const. it r
p < 1; this holds in particular forg(t). Since t0 0 öan be replaced by any t0 E (—1,1), we obtain, for any 0< R <
(2.26) 10E(—R,R), ti<R,
where ) c(t0) I and CR, are constants depending on R, s, i, and the norm of g. It follows that g'(t0) exists and g'(10) = c(t0). Further,
interchanging I and in (2.26), we find that
Ic(t) — —
so thatgE R).
We shall next extend Theorem 2.2 to higher differentiability of the paramet- ric representation g(t). We use the notation of Theorem 2.1.
Theorem 2.4. Let 0, F be as in Theorem 2.1, and let u E '(,O U F), fE Cm+a(BR), + E where m 0,0 <a < L If aisaJirftes(2.4)— (2.6), then the parametric representation z = g( 1) of F is ii,
146 VARIATIONAL INEQUALITWS ANALYSIS OF THE FREE BOUNDARY
Proof. The proof is by induction. The inductive assumption is that for any — 1 < 1, z e G fl BR,
g(:) = — 'a)' + O(l)(: —
where c1 (ta) are functions of and
if—R<t0<R (R<1),
where CR depends on 4), land R. The case m = 0 was established in Theorem 2.2. Without loss of generality we may take! = 0 [otherwise, consider u — f, f as in (2.7)1.
Suppose for simplicity that g(:0) = 0. We can write
v4)(x1, x2) = Pm+i(Z, 1) + A(x1, x2),
where
A(x1, x2) and consider the equation for
(2.27) + A(z) =
Observe that, by (2.6) with! = 0,
PiPm+i(Z, 2) = D1V4)(z) +
if I z is small enough, and, by (2.5),
v*u(z)
Hence by the implicit function theorem, we can solve (2.27) uniquely. The r(z) has the form
= V*u(z) —
RFGLThAWY IN TWO DIMThSIONS
where L(z, w) is holomorphic in z, w. Since v*u is Lipschitz continuous and since A is Lipschitz continuous, also r(z) is Lipschitz continuous in a neighborhood of z = 0.
Taking = in (2.27) and applying D1, we get, since t&u = 0,
+ D1A(z) = 0 in fl
But the first term on the left is 0; hence
Clzlainfl. We now extend g(t) into Im t <0 by
Ig(') ilImt<O
and let
= — —
Then for with Im 0,
(2.28)
We can therefore apply Green's formula (2.3):
1 i 1 —
+— (z)dz. • z—1
Defining to be the value of the right-hand side when t t0, we can now prove (see Problems 1 and 2) that and
—
— 10)
(z — —
(z— —
and the right-hand side can be estimated by const. I — to using (2.28).
148 VARIATIONAL iNEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
We have thus derived the expansion
g(t) = c1(:0)(t — t — 1=0
with
m+ I
I c.(i0) I +10(1) C (—R <to< R) 1=0
for t E G fl BR, for any 0 <R < 1. The assertion that g E now follows by standard arguments (see Problem 3).
We finally wish to prove, under the conditions of Theorem 2.4 with m I, that the parametric representation z = g(t) of r is noñdegenerate:
Theorem 2.5. Let the conditions of Theorem 2.4 with m = I hold. Then, for any — I < I, there holds
(2.30) foreithern = 1 orn = 2. :_:o (:— t0)
Proof. Consider the function g*(t) defined in From (2.23) we have (since now we can take A = 1)
Cf g'(r)f$g(t) — Cconstant.
Since
and since D-g(:) = 0, it follows that the function 'I'(t) defined in (2.25) satisfies
sup 1*(t)--4'(O)I 1:1=1:1
where
\JO in(lmz>O)flB, in{Imz<O) flB,
for some r >0. Recall also that 4 E for.any s >2 (r depends on s).
RWULARITY IN TWO DIMENSIONS 149
We shall need the following lemma.
Lemma 2.6. Suppose
forsomes>2 (BB1),
(2.31) sup , zEB, ill
q E L3(B).
If w 0 in some neighborhood of z = 0, then there exists an integer n 1 such that
Assuming that the lemma is true, we can apply it 4'(rz) and deduce that (2.30) holds for some integer n 1. If n 3, then g(t) — g(10) + c(t —
cannot be —1 mapping from G into 0; hence and the theorem follows.
Proof of Lemma 2.6. We first claim that if WE for some s >2, O < R ' 1, and
I z E
— w(o)) E L°°(RR),
where 0, n is an integer 0, then
(2.32r W(z)— W(0)
— cIzJv, p < R,
whereA=I —2/s,r=min(c+A,1),
(2.33) c = + — W(0))dt,
C = WihII — w(o)u p)
andc,dependsonlyonr, We next wish to show that function w in the lemma
(2.34). 1i,n z — is4O)) hi 0,
VARIATIONAL INEQUALfl1LS ANALYSIS Of THE FREE BOUNDARY
1 and0<6< 1, (2.35)
z
where C depends only on s, 8, and the L'(BR) norm of q. The proof is by iteration. Take for simplicity w(O) = 0 and suppose that
(2.36) A>O,
In view of the assumptions on w, the conditions for (2.32) are then satisfied with W w. Hence
(2.37) + --
wherer= If = a + A, then using A on 8BR, we get
(2.38) I w(z) + RI....A(R —
whereas if 1 = 1, then we similarly get
(2.39)
Since (2.36) certainly holds when a = 0 [by (2.34)] we deduce (2.38) with a = 0. We now proceed by iteration. Let m be a positive integer such that
m+1.cZm,
and choose small >0 and a = (j— l)A forj = 1,2,. ..,m. From (2.38) we deduce, for each j,
(2.40) II II (C + 1R—9 IIpN(1
and from (2.39) with a = (m — l)A; we deduce that
(2.41) 11p" (C + t1'R I+(m_I)A)110_n.-(m—
Applying (2.40) with R = R0(1 — for j = 1,2,... ,m and then applying (2.41) with R = R,, R0(1 — then multiplying together all the left-hand sides and all the right-hand sides of these inequalities, we obtain the assertion (2.35) with I — 6 = (1 —
REGULARITY IN TWO DIMENSIONS
For n = 0 (2.34) holds. Hence (2.35) gives
C
and recalling (2.31), we see that (2.32) can be applied to W = w to deduce that
(2.42) w(z) — w(O)
exists.
If c =0, we repeat this procedure with n = 1. We deduce that (2.42) holds with n = 1. Again, if c = 0, we can deduce (2.42) with n = 2, and so on. Thus to complete the proof of the lemma it remRins to show:
If wis as in the lemma, then
(2.43) limIzI"w(z) = Ocannothold z—O
for all n 0.
The proof is outlined in Problem 5.
PROBLEMS
I. Prove Lemma 2.3. [Hint: By (2.3),
v(i)—v(0) = _!f 1 vj(i)dXidx2+._.Lf v(z) 1z. £ I' 2,n aRz(z-1)
Defme c by the right-hand side when £ = 0:
c f2vdz.z a
Show that c is fmite and that
v(i):—v(0) C
The first integral is estimated by
III z L{fI z I z — t dx2]
VARiATIONAL J1'4EQUALmES: ANALYSIS OF THE FREE BOUNDARY
where 1/s + 1/s' = I. If c < 2, break the domain of integration into three parts:
A0 ((z — {lzI<41tf} \A0, \A0.3
2. Show that the right-hand side of (2.29) is bounded by const. I — t0 r.
3. Let be defined for 0 <1 < I and assume that for any 0 <s < 1,
g(t) = — s)' + o(l)(II — r÷i (0 <a < 1)
where C. C. C constant. Prove thatg E [Hint: Take n 2. Clearly, c0(s) = g(s). c,(s) g'(.c) and developing
g(a + h) — g(a) hc,(a) + 0(h2),
g(a) — g(a + h) = —hc,(a I-h) + 0(h2),
it follows that c1 E C. Developing about a — h
g(a + h) + g(a — h) — 2g(a) = 4h2c2(a — h) +
then replacing h by — h and cornparng, yields c2 E Cu. The limit of the second finite-difference quotients D2g = urn is c2(t). From
g(a + h) — g(a) = hc,(a) + 1x2c2(a) +
g(a + 2h)— g(a + h) = hc1(a + h) + h2c2(a + h) +
it follows that
+ h) = c1(a + h) — c1(a)
+ o(l)
and c(a) exists and equals c2(a).j
4. Prove (2.32).
[Hint: Express (W(z) — by Green's formula and adapt the procedurc in Problem 2.1
5. Prove (2.43). [Hint: Express z"w(z) by Green's formula in and deduce forf(p) =• f,,(p)
F(p) tilt — Zl + — Z
REGUlARITY IN TWO 153
where I z p, 1/s + 1/s' = I. Raise both sides to the power s', multiply by 12 Ls, and integrate over z E BR, to derive
f +
If w(z0) 0, I
Zo I
small enougb, then
po=IzoI,
which is if n —, oo.]
6. Prove the following theorem: Let (I be a strictly convex bounded domain in R2 and let 4 be a strictly concave function in fthat is, (82$/8x, ox) is negative .defmitej with <0 on OU, maxa+ $(x) >0 (x E u be the solution of the variational inequality
u0 onaa. Then (a) the noncoincidence set N is connected; (b) the coincidence set A is connected. [Hint: To prove (1), let x0 E aa, the plane tangent to OQ at x0 and to z = 4i(x) [x = (x1, x2)], say at E R3. By comparison L(x)> u(x), where x3 = L(x) is the equation for and VL(xo)J>I Vu(x0) I; VL(x0) and Vu(x0) are both normal to all at x0. The mappings
F1 : x0 VL(x0)
F0:x0—s vu(x0)
are I — 1 (since Il is strictly convex) and F0 C interior of F,. The mapping x -. 1 — I is strictly concave; it maps y onto F1 and x into the origin 0. Hence it mapa interior y onto interior F, consequently, F0 for some curve in the interior of y. Since u < L, the free boundary F lies outside y and N is bounded by r and 00. The mapping
(xEN)
is open, a(F) = F and o(00) = lies inside y and hence inside r. By (a), o(N) is connected. But then, (interior of is also a connected open set.)
From Problem 6 we can easily deduce that N is homeomorphic to an annulus. An additional argument shows that Theorem can be applied (observe that F is not assumed a priori to be a Jordan curve); for details, see Lewy and Stampaochia [1384
VAR1A11ONAL INEQUALITJ!S: ANALYSIS OF THE BOUNDARY
3. GENERAL PROPERTIES OF THE FREE BOUNDARY
In this section we establish two basic facts for the obstacle problem:
(i) The free boundary has measure zero; (ii) If y E r (the free boundary), then
lim — 4i(x)) 0, x —.y x€N
where is the obstacle, N the noncoincidence set, and i is any direction.
We first work with the simple case of
(3.1) u"O,
where (2 is a bounded domain in R", f E (0 <a < 1); we shall after- ward deal with a general elliptic operator and a general obstacle.
As before, we set
N— {xE-12,u(x)>0),
A (xE(2,u(x)=O),
r=aNn(2.
Recall that u C"(O). For simplicity we may assume that u E C''((2). Our first result is a simple lemma of nondegeneracy, assertng that u cannot
be uniformly small in some neighborhood of a point of N, provided that I const.> 0.
Lemma3.1, Thenfor any ball BAx0) C.(2,
Ar (3.2) sup [u(x) —
ilAxo)
Proof. Suppose first that x0 E N. The function
2w(x)u(x)—u(xo)—rlx_xoI
satisfies 0 in N and w(x0) 0. Hence, by the maximum principle, sup w in N fl BAx0) is nonnegative and it is attained on the boundary. But since
GENERAL PROPER11FS OF 11ff FRFZ BOUNDARY
w <0 on there must then exist a point x1 aB,(x0) fl N such that w(x,) 0, and (3.2) follows.
If x0 N, we apply (3.2) to a sequence of ponts Xm E N, Xm Xq.
The following Calculus type lemma will be useful.
3.2. Let u be any nonnegative function in C" '(D), where I) is an open set in R", andkt all 1, j. If u(y)<82 for somey ED and if dis:(y, 8D) > max(8, h), where 6, h are some positive numbers, then
(CM=M+l) and, for any direction, c1,
(3.3) fhfT()
—CM8(6+h).
Proof. In the inequality
(3.4) 0 u(y + se,) = u(y) + u,(y)s +f3fTD,,u(y + te,)didT
[where s <dist(y, aD)1, choose the e. in the direction for which u,(y) = — I Vu(y) I and take s6. This gives
from which we that I CM$. Using this estim*le we now apply (3.4) withs = h, and any direction and (3.3) follows.
Cerollary3.3.
(35)
(3.6)
where Cdepends only on the C" nornofuin Il and on
Indeed, (3.5) follows from the fact that u = Vu = 0 on F and the C" nature of u, and (3.6) is then a consequence of the lemma.
Theoreii3.4. then
(3.7)
for e0, where7and depend only onA,80, and the C" norm of u. .1.
VARIA11ONAL INEQUALiTIES: ANALYSIS OF THE FREE BOUNDARY
Here we have used the notation
A Lebesgue measure of A.
Proof. By Lemma 3.1 there e,Usts a pointy E 8B,(x0) such that
Ae2 u(y)
Using (3.6) we get, for small enough 8,
Ar2
2n
if I x — y < 8e. Thus there exists a ball B81(y) contained in 82e(X0) in which u >0. Since 8 can be chosen independently of e (for £ small enough), the assertion follows.
We recall that for any measurable set S,
IB(xo)flSI urn = I for a.a. x0 E: S;
B,I
that is, a.a. points of Shave density 1. Since by Theorem 3.4, all points of F do not have density I, we conclude:
Theorem 3.5. 1ff A >0, then the free boundary has Lebesgue measure zero.
In the next theorem we study the behavior of the second pure derivatfves D,,u near the free boundary; here we do not assume that 1>0, but for simplicity we take! = const.
Theorem 3.6. Le:f= const., y E 1', dist(y, a(l) >0. Then if x E N,
—c (3.8)
—FIr' E
= 2(n — I)'
where C is a positive number depending on n, and the C'' norm of is.
It follows that
(3.9)
GENERAL PROPER11ES OF ThE FREE BOUNDARY 157
and, in particular.
(3.10) ifyEF. xEN
Proof. Take for simplicity y = 0 and define
= infD11u. 82 h
We shall estimate the Mk recursively for all k for which Mk >0. Let x N, x 2-(k+fl and let be the largest ball in N. Then
s there exists a pointy0 E a point on the segment xy0 with distance 6s to y0; 8 is small and still to be determined. If
ID,1u1'cM,
then
u(yt)
Suppose that (e1, x — when e,s the unit vector in the fib direction. Applying(3.3)withy =y1,h =
—C83"2s2.
If (e,, X — 0, then we replace e, by —e, above and observe that _.u = D,,u.
We conclude that
sup D,,u(y1 + 1(±e,)) —CS'12
and the scgment.y1 + a distance aB1(x) .(for'Q <
The function
w = D11u + Mk
is harmonic in B3(x) and is positive there [since B3(x) C B2-&J; by (3.11)
for some point 9 E We now use Harnack's (cc Lemma 3.9):
(3.12) w(x) w(9)c6"' (c >0).
158 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
It follows that
(3.13) D,1u(x) + c(Mk —
Choosing 8 such that C6'12 = where E0 is sufficiently small so that 8 if C&"2 = we get
D1u(x) +
Since x was an arbitrary point with I x 1< we deduce that
(3.14) — +
One can now show by induction that
1
2(n—1)
with suitable C. Thus
and (3.8) follows.
We shall generalize the results above to non-constant f and to an elliptic operator
(3.15) Au — — + + c(x)u.
The variational inequality is
(3.16) (Au.—!) 0, u (Au — f)(u — 4,) 0 in 12.
We assume. that the solution is in and that fE Ca, 4, Without loss of generality we may take b, = 0, c = 0, for otherwise we can absorb the + cu into f. We may also assume that 4, = 0; for otherwise we consider u — 4, instead of u. Thus without loss of generality we reduce (3.15), (3.16) to.
a2uAu=— aX1 Xj
(3.18) (Au—f)u.=OinIZ.
We shall always assume that
(3.19) fECa(12) forsomeO<a< 1.
GENERAL PROPERTIES OF ThE FREE BOUNDARY 159
The proof of Lemma 3.1 remains valid if we take
w(x) u(x) — u(x0) — kyix — x012
with y positive and small enough (depending on the modulus of ellipticity of A). Thus
(3.20) sup ifx0EN.
Theorems 3.4 and 3.5 remain unchanged. For the sake of convenient reference, we state:
Theorem 3.7. Lemma 3.1 and Theorems 3.4 and 3.5 remain valid for any solutio,t of (3.18).
Some nontrivial changes occur in the proof of Theorem 3.6:
Theorem 3.8. The assertion of Theorem 3.6 remains valid for any salaition of (3.18).
Proof. In order to proceed as in the proof of Theorem 3.6, we must be able to apply Harnack's inequality to some "small perturbation" of D,,u + Mk. To accomplish this, we note first that
(3.21)
where a1 k, = aOlk/aX,, = a3 u/ax, aXk, and so on. Let u0 be the solution of
Au°f inB3(x),
Then u0 E and by writing the Schauder estimates for Q(y) s2u°(x + sy) in (ly 1< 1), we find that
I
where the norms are taken in B,(x), and C is a positive constant independent of s.
Formally,
AuZ =f,, + ,U?,* +
16) VARIA11ONAL INEQUALmES: ANALYSIS OF THE FREE BOUNDARY
Let u' be the solution of
Au' = inB,(x),
onaB,(x).
Then again
U'
and
Au = — +
Next let u2 be the solution of
Au2 = ,jUjk —
u20
Then
Iu2ki+.
Finally, if
Ai5=O
+ + on aB3(x),
then the function v + u + u2 — i5 satisfies
322 Av—f,,
v=o onaB3(x),
forsomefl>O.
From (3.21) we see that
Au,, + g0 +
where g0, Zj are L°° functions. Following the preceding method, we can find functions v, such that
GENERAL THE BOtJNDARY
in = 0 on and
Let
v=v+20vi
and consider the function
(3.23) w = — V + Mk + in
It is a solution of 4w =0. We recall the Harnack inequality [161eJ:
Lemma 3.9. Let u 3atIsfy
a — + + c(x)u= 0
in B, and auwne that
•
(A>0,M>0,A>O,e>O).
If u is positive ui p,
st(y) H
yu(y)
where is a positive ct'awtani only on X, M,4,s.
Now, since
I V=OonôB,(x),
we can choose C large enough in (3i3) so that w is positive in B(x). Applying the Harnack inequality as in the proof of Theorem 3.6 (see (3.12)L we faee(3.13)j
Vnu(x) ?:M& + y'(lf*.— I ''
and we can establish as before that Ck'.
162 VARL4TIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
Definition 3.1. For any bounded set the minimum diameter of S, MD(S), is the infimum of distances between pairs 11,, 112 of parallel planes such that S is
in the strip determined by U,,
In the next two sections we shall study the behavior of the free boundary in a neighborhood of a point x0 E F. For simplicity we take x° = 0. will depend on the "thickness" of A at 0. This is defined in terms of
(3.24) 8r(A) = MD(A fl Br)
One can easily verify that
(3.25) C0 A
6(A) 2 (c0 positive constant).
Indeed, this is obvious if r = I and for any r it then follows by scaling. The following fundamental theorem of Caffarelli [56bJ will be proved in the
next two sections.
Theorem 3.10. Suppose thatf A > 0 in There exists a positive nondecreas- ing function a(r) (0 < r < with o(() + ) = 0 such that if for some 0 <r <
(3.26) >
then for some r> 0, F ('1 is given by a C' surface
(k E C1)
and u is in C2((N U F) B,(0)).
In view of (3.25) we also have:
Corollary 3.11. If AnB,J
(3.27) hm sup D I > 0, r.—0
then the assertions of Theorem 3.10 are valid.
The proof of Theorem 3.10 shows that a(r) can be chosen to be independent of x0 E 1', provided that dist (x°, 8(l) >0.
In Theorem 3.10, u is the solution of a general elliptic obstacle problem fthat is, (3.18)J. Wt shall first consider, however, the special case
(3.28) A = f= 1,
CONVEXITY PROPERTWS OF 11ff COINCIDENCE SET 163
and only at the very end (in Section 5) point out the modifications needed for the general case.
4. CONVEXITY PROPERTIES OF THE COINCIDENCE SET
The results of this section will be used in Section 5 in proving Theorem 3.10.
Definition 4.1. A function u is said to belong to the class P,.(M) if
U C'•'(Br) and (4.1) B,
for all derivatives D,1 =
(4.2) IflBr,
0€!', (4.4) inN
where
N=(XEBr,U(X)>O), . r=aNnBr;
we also set
A (X B,, u(x) 0)
and write
NN(u), AA(u), FF(u). Notice that u(O) = 0, Vu(O) = 0.
For any fixed s>0, the scaled function
I = —ju(sx)
is in P,13(M)ifu EP,(M), and
where, by definition,
E5=(x;sxEE) foranysetE.
164 VARIATIONAL OF THE FREE BOUNDARY
Suppose that
45 U(m)EP,(M), U(m)_,UO
uniformly in compact subsets of
Then clearly
(4.6) C
where iiiii A(u(m)) means the set of all limit points of sequences (Xm), Xm E A(u(m)).
Lemma 4.1. If(4.5) holds, then u0 P4M) and
(4.7) C N(u0).
Proof. Suppose that y E Eiii N(u(m)) Then u0 =0 in for some e > 0 and at the same time there is a sequence N(u(m)) such that Y,n -. y. By Lemma 3.1
2
sup ./2 Y..,
and since -, u0 unifonnly in B,(y), if e is small enough, also
sup U0 >0, 8.(y)
a contradiction.
It is clear that u0 satisfies (4.1), (4.2), and (4.4). In view of (4.7), also (4.3) holds for is0 and thus P4M).
Definition If
(48) U E P,(M), -. is0 uniformly
in compact subsets of R"
[recall that a4'") Is a scaled function of then call u0 a .Nov-sqi limit of
Lemma 4.2. Ifu0 is a blow-up limit of u(m), then is0 is a convex function in R".
CONVE%TTY PROPER11ES OF ThE COINCIDENCE SET
Proof. By Theorem 3.6,
= DIiU(m)(€mX) — CJ log(e,,, I x I) 11.
Hence, for fixed R > 0, the function
Wm(X) = +
is convex in BR (i.e.,'D,,w 0 for any direction I) and wm u0 uniformly in BR. Thus U0 is convex.
Notation. If u E Pr(M) and u is convex in Br, then we say that u belongs to P,(M).
Lemma 4.3. Suppose ihat u E P7(M), Em 10, u, uniformly in compact subsets of Br. Then in an appropriate system of coordinates u0 has one of the following two forms:
(4.9) u0(x) a, 0, ±a, =
(4.10) UO(X) )2•
Proof. Consider first the case
intA(u)= 0.
Since u is convex, the set A( u) is convex and therefore must lie in a hyperplane. But then = I a.e. It follows that u0 is a nonnegative solution of
xER",
we can apply Liouville's theorem to D,D1u0 and conclude that = const. Recall jng also that u0(0) 0, u0 0, (4,9) follows.
next that
(4.12) intA(u) 0.
Let x E A(u). Since A(u) is convex, E A(u) for any t >0 provided that emg<l.ThuStxEA(u,)and
A(u0) J J C(A(u)),
166 VARIATIONAL ANALYSIS OF THE FREE BOUNDARY
where C( K) denotes the cone generated by K (with vertex at the origin). If x C(A(u)), thenx' C(A(u))forallx'withlx' — xf< 8(8 small enough). But then cmx' E N(u) if 0 <em < I and thus x' N(u,). Using Lemma 4.1, we deduce that x' E N(u0) and, consequently, x E N(u0). (Indeed, since N(u0) is convex and intA(u0) ) C(A(u)) 0, if x N(u0), then there is an open set in {(x' — x <8) that is contained in intA(u0).1
We have thus proved that
= C(A(u)).
Take any direction —es interior to A(u0). Then any line in the direction e, intersects N(u0) in a half-line along which D,,u0 0 by convexity of u0 (D, = derivative in the direction e1) and = 0 at the initial poirit lwhich belongs to F(u0)]. It follows that 0 in N(u0) and, by the maximum principle,
(4.13) inN(u0).,
We next show that A(u0) is a half-space. Indeed, if not, then let the x1-plane and x2-plane be two planes of support. Writing
x=(pcos0,psin0,x3,...,x,),
we have
A(u0)C(x;00<O<2ir—00)
for some <00 If <00, then the function
h(x) =
is harmonic and vanishes on 0 = ±01. It follows that for some small enough c >0,
ch
if 0= or if (é depends on 8). By the principle, we then get
9i,IxI< 8. Since < 1, we get a contradiction to the continuity of at 0.
OF ThE COINCIDThCE 167
We have thus proved that A(u0) is a half-space, say (x1, > 0). By the uniqueness to the Cauchy problem we deduce that u0 must have the form (4.10).
Notation. We denote by a(x, y) the between the vectors x andy.
Lemma 4.4. For any e >0,8 >0 there exists a A = X(e, 8) such that if u E P,"(M) and 61(A(u)) > e [81(A) is defined in (3.24)), then in an appropriate system of coordinates
(4.14)
N(u)JBAII
Proof. lftheassertionis not true, then Pr(M) for which (4.14) is contradicted in any system of coordinates.
Thus for 'ny coordinate system e1,. . . at least one of the relations
(4.15) A(u(m)) 3 fl (x; <
(4.16) N(u(m)) 3 B, fl {x; — a)
does not hold. Notice that if (4.15) holds in a system of coordinates, then the same is true
of (4,16). Ipdcód, othprwisc is a direction with < 8 such that for' sojiç 0<5 <As,,.. Since fby (4.15)) a
small neighborhood of bçlonp to it follows from she convex- ity of that 0 E int ,
From the remark above we conclude that already (4.15) is not true in any system of coordinates.
Take a subsequence unifofinly in any compact set. Then uE
For any sequence A,, of in B1 holds (see Problem 1)
Since
But then the convex set A(u) has nonempty nterior. It follows from the proof of Lemma 4.3 that in appropriate system of coordinates the cone A(u) is the half-space 0).
168 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
On the other hand, since (4.15) is false, we can find in every convex cone of opening 6 a unit vector e(m) contained in N(u(m)) and thus e = urn e(m) is in N( u) (by Lemma 4.1). This contradicts the previous conclusion that A(u) is a half-space.
Lemma 4.5. If u E Pr( M) and 6, ,14( A( u)) E > 0, then there is an ap- propriate system of coordinates, a positive number and a C' function g(x,,. . . ,x,,) such that
(4.18)
and the modulv continuity of grad g depend only on M and e.
Proof. Since
A(u) fl 3 A(u) nB11,4
it follows that
MD(A(u) fl B,12(y)) MD(A(u) ii B,,,)
Hence
'4 19' Lemma 4.4 can be applied with respect
to any pointy r, j<j
By Lemma 4.4 with y = 0 E r, for any 6 = 1/rn there is a system of coordinates (er)(i = 1,... ,n) such that (4.14) with e = e'. Since A(u) is convex, any line x0 + intersects A(u) in segment lying in (:<zt0) and N(u) in a segment lying in (1 > t°); see FigUre 2.1. If x c .Ix1er, then we can represent x0 + t°e' by
(4.20)
and A(u) by x, gM(x'). It is important to notice that if we use any coordinate system centered at 0
in which the direction is near the directiofl then the remark above regarding x0 + te' is valid also with provided that is small enough; see Figure 2.1.
Since (4.14) holds for (e") in B, (A,,, dson 6 = I/rn and e; Am 0) it follows that for a suitable choice of the er, t i n — I, and for a subsequence
er—e?,
/ 4' /
and
(efl = Tm(er), an orthonormal matrix,
(4.21)
From (4.14) we get
I the identity matrix.
I
— gm(o) c I
and using (4.21) we easily deduce that
A,,,,
(4.22) If)XmWhCtCPm -'0;
here x,, = is a system by UoUowing From (4.22) we see that g°(x') is differentiable at 0 with zero gradient.
In view of (4.19) we can do the same abopt. small a óóórdinates (er") and a limiting one (as m oc) (e7'Y), and A(u)can be represented
.iSui*dtpendehtdfy)inthei&th
x = or x= respectively..
(4.23) — g°(O)
Ix'I
CONVEXITY PROPERTIES OF THE COINCIDENCE SET
ac')
FIGURE 2.1
VARIATIONAL INEQUAIII1E& ANALYSIS OF THE FREE BOUNDARY
The system of coordinates er' is related to e?' by
= T""Y(e?'Y)
where Ttm "-' I uniformly with respect toy. Notice also that (er") is related to (er) by
(4.24) (e7Y) = T°"(e?)
and
(4.25) T°7-eI ify—'0
(with appropriate choice of 1 I n — 1); this follows, in the same way as (4.21), from the assertion (4.14) fory = 0 and tory near 0.
We can rewrite the inequality (4.23) in terms of the systems of coordinates (e?); taking into account the rotation (4.24) and (4.25), we obtain from (4.23)
where o(l) -*0 as 11*1-' 0, uniformly with respect toy, and C. This implies that E C'.
PROBLEM
I. Prove (4.17). (Hint: SctA =limA,, = = and any i-neighborhood of A contains the n
S. OF 'ME FREE BOUNDARY WHEN MD(A) IS POSflWZ
Lemma 4.4.has implications for implications any it in P1(M):
it E P,(M)and8,(A(u)) i/or some 0 <p < then in an appropriate system of coordinates
(x; —xa'2pX8),
N(u) D fl fx; x, 2pA8);
A=A(e,8)is defined in Lemma 4.4.
Note that the right-hand sides in (4.14) are cones intersected with a ball, whereas in (5.1) they are only strips intersected with a ball.
OF IHZ FREE BOUNDARY MD (A) IS POSITIVE
Proof. If the assertion is not true, then there are sequences 14(m) P1(M) and Pm >0, p,,, i,0 such that in some system of coordinates at least one of the relations
J (1 (x; —x,, >
D fl (x; x,,> 2pmX8}
is not satisfied. A subsequence of is convergent to a function u in M). Using the
relations
A(u) D limA(ur), N(u) J
we obtain a contradiction to Lemma 4.4.
We now choose e< 6< Then (5.1) implies that
1 —48>4,
and therefore the lemma can be applied again to pX/2. Proceeding inductively, we obtain:
LenimsS.2. and defineX=X(e,8) as in Lemma 4.4 and 8) as in Lemna 51. JfuEF1(M) and p<p0, then there exist
(5.2) A(u) J BP(A,2)a fl
(5.3) N(u))B,(x,s)afl
Here . are the coordinates of a point with respect to (er'). We would like to show that the systems (eJ&)) converge to a system (er) and then repeat the proof of Lemma 4.5. For this we need an estimate on D11u:
Lemma 5.3. Under the of Lemma 5.2 there exists a p E (0,1) such that
(5.4) D,1u in
Proof.
h=0 h=—l OnaBn{XN<fl.
172 VARIATIONAL INEQUALfl1ES: ANALYSIS OF THE FREE BOUNDARY
Then
—1 81,2
Suppose by induction that
(5.5) inN(u)
Choosing in the vertical direction, we consider the function
w(x) = inN(u) fl
The functions w and are both harmonic in this set and w on
N(u) n by (5.5),
and on
8N(u) fl BP(A/2)A. by (3.10).
Applying the maximum principle, we obtain
Lemma 5.4. flu E C' '(BR), a, 0, u(0) = 0, and on an interval oxo(tx0I< R), r 0, then, for 0< < 1,
(5.6) u(tx0) x0 f2T + u(x0).
Proof. Suppose that u(t0x&) = If 1, then (5.6) is obvious. If, on the other hand, < 1, then D,u(10x0) 0, where i is in the direction xo/I x0 . Writing
0 = u(0) u(t0x0) + ffD,1u u(r0x0) —
we get u(tx0) u(t0x0) x0 x0 + u(x0).
Lemma 5.5. Let the hypothesis of Lemma 5.2 hold and let
(5.7)
0k = x 6 N(u) ii (
REGULARITY OF ThE FREE BOUNDARY WHEN MD (A) IS POSITIVE
Then there exists a pointy such that x — y I' 0k and the points sy are contained in N(u)for ails with
(5.8) 1,
Proof. By Lemma 3.1 there exists a point y such that x — y and
(5.9)
Let s be as in (5.8). Taking x0 = sy, t = I/s in the preceding lemma, and noting, by Lemma 5.3, that
D,, u — on the interval Ox0,
we get
u(y) + u(sy).
Substituting (5.9) and recalling (5.8) and the definition ofn (5.7), we get u(sy)>0.
Note that in the previous lemma
(k — 2)/2 (5.10) CC(n,M),
sothata(x,y)-'Oifk -'
Lemma 5.6. Let the hypotheses of Lemma 5.2 hold. Then there exists a positive integer k0 = k0(e, 8) and a constant C = C(€, 8) > 0 such that in an appropriate system of coordinates
(5.11) A(u)DBP(A,2)*Ofl co),
c&}.
ThJsie* Lenlma5.2, 'where instead of the cones (in (5.11), (5.12)] we only had strips (in
Proof. Take e as with k = k0 — I [see Lemma 5.2 for definition of Suppose that (5.11) is not true. Then there is a Xj, such thit
174 VARIATIONAL JNEQUAIIflES ANALYSIS OF THE FREE BOUNDARY
and
(5.13)
a a segment in N(u), where
A 1—i
Take Xj_i = 3jYj with Then
E N(u) n (
and
a(x1, x1..1) by (5.10).
Repeating this procedure step by step we find a sequence of points x,, k0 I <j, such that
x1 E N(u) fl aBP(X,2).
and
x,..1)
x = xk0 satisfies
x E N(u) n
and
a(x [C =
IfC(in(5.ll)and(5.13))istakentobe
and a contradiction to Lemma 5.2. To prove the assertion (5.12), we first note that the proof of (5.11) applies
not only with respect to 0 but also with respect to any point x E F(u),
OF ThE FRSZ BOUNDARY WHFZ4 MD (A) IS POSI1WE
small enough, say
X E F, x E
k0 and C may be taken independently of x. If (5.12) is not true for any positive constant C and integer k0 >0, then for
any large constant C there is a point x0,
x0 E F(u) fl fl {x; a(x,e,) — c*a).
By the previous remark we have that A(u) contains (within neighborhood of x0J a cone with vertex x0 and opening — C8. By (5.2) of Lemma 5.2 the axis of this cone must form an angle C08 with (C0 is a univerSal constant). Hence if C Is sufficiently large, then the cone contains 0 in its interior, a contradiction.
Proof of Theorem 3.10. We consider first the case 4u = - Au, / 1. Lemma 5.6 for u P,(M) asserts the same as Lemma 4.4 for u E P1'(M). Since the proof of Lemma 4.5 is based solely on the conclusions of Lemma 4.4, the same proof applies to ii E P1(M). It follows that F is C' in a neighborhood of 0.
It remains to prove that uEC2((NUF) riB,) for some small p>0. Consider the functions in N. We have to show that if x-'x0E1', I x0 < p, then converges to a limit, the being
(r,e11>
where (a, b) denote scalar product of a, b, and r is the inward normal to r at x0; note that iii continuous along F, and set =
=0 aIde1 tangent to Fat prove that • 'u 'F •.
ihnDrns('x)=o; x-.O xEN
for theai by the equation + 1 0,
IimD,,se(4=1.x0 xEN
Suppose that the assertion (5.14) is false. Then there exists a -. 0, N, such that
(5.15) >0.
176 VARIATIONAL INEQUALfl1ES: ANALYSIS OF THE FREE BOUNDARY
Let y,1, be the nearest point to Xrn on r and introduce the functions
(5.16) urn(x)= U(2jXmymIX+ym)
(2Ixm
Since is in the direction of the inward normal to F(u)_atym, u(x) >0 in the region that is approximately the half space with It follows that Urn(X) >0 in the region that is approximately the half-space (<x, Xrn — Ym) 0}, or also approximately {(x, 0). It follows that for a subsequence
Urn w uniformly in compact sets;
w 0 if (x, 0 and N(w) contains the half-space ((x, p0)> 0). Since w is also convex, it follows that A(w) coincides with {<x, 0). By the uniqueness of the Cauchy problem, we then get
w(x) = v0)2 >0.
Since
C N(u),
= I in B114(v0/2) for all m sufficiently large. It follows that
= = 0
(since e. is a tangential direction), contradicting (5.15). The proof of Theorem 3.10 in the general case is similar; here we use
Theorem 3.8 instead of Theorem 3.6, The only difference is that in the proof of Lemma 5.3 we have to make a modification similar to the one given in the proof of Theorem 3.8. Notice also that blow up of solutions of Au = fare solutions of
.f(0),
so that Lemma 4.3 and the proof that u E C2 are unchanged.
PROBLEM
1. Extend the proof of Lemma 5.3 to the general case of the variational inequality (3.18).
THE FREE BOUNDARY FOR THE FILTRATION PROBLEM
6. THE FREE BOUNDARY FOR THE FILTRATION PROBLEM
Let (2 be the cylinder in
t2=BR0X {O<y<H},
where BR denotes the ball (I xf< R), x = (x1,. . . we shall wdte X = (x,y).
Consider the variational inequality
(6.1) in(2
and set, as usual,
N=(u>O}, r=aNn(2.
We assume:
(6.2)
(6.3) u=u0 (6.4) infl.
It follows that for any x BR0,
(6.5) (x,y) EN if and onlyify<4(x),
where 4(x) is some function with values in (0, HJ. Thus N is an x-subgraph and F is an x-graph.
Theorem 61. If X° = (x°, y°) F, then there exists a neighborhood V = 1< p) such thatFflVLs given by
(6.6) y.=4(x) (1x-x°I<p)
and 4, is L4schitz
Proof. Take for simplicity x° = 0. We recalj that v C': '(U) and, without loss of generality we may assume that u E (otherwise we r.plaoc C) by, 0 ri (y > e) for some small e >0 and use Theorem 4.3 of Chaptst 1). The function u, satisfies in N
178 VARIATIONAL INEQUAL1TIE& ANALYSIS OF THE FREE BOUNDARY
consequently, by the strong maximum principle,
(6.7) - inN.
For any R >0, set
NR=[BRX(0,H)}flN.
For any vector e in I
I, K> 0, A > 0, consider the function
in N2R, where R is such that B2R C BR0. Since
it follows from (6.2) that if A is sufficiently large (independently of K, e), say A then
(6.9) inN2R;
from now on we take A = A0. Let 8 be a small number to be determined later; it will depend only on R.
Clearly,
w0 and, by (6.7),
w >0 on fl (X; dist (x, > S } (6.10)
forallRR'c2R, provided that K K(S). If we can also show that
(6.11) w>0
then we can apply the maximum principle [recall (6.9)] to deduce thAt.
(6.12) w>0 InN1.
We shall now establish (6.1 1), provided that S by contradiction. Suppose there is a point K = (1, j) such that
w(A) E M1, dist(1, F) S.
Consider the function
1nN2R.
THE FREE BOUNDARY FOR THE FILTRATION PROBLEM
By (6.10)
*(X) >0 on aN2R fl (X; dist(X, >8).
Also 0 on r. Finally, if
xEaN2R, dist(X,F)8,
then,sinceuEC''andu=O, vu=Oonr,
(C0>0)
provided that R2/C for some constant C (independent of K, R, 8). Noting by (6.9) that <0 in the maximum principle gives >0 in N2R. This contradicts the assumption that
Having established (6.12) we shall now use only the weaker statement
inNR.
there is a cone M with vertex x0, opening y>O, and with axis parallel to
The ehclds(withth )Wthrespectt anypont r, x in a àeighborhood of X°. Since N is also an x-subgraph, it follows that fo,
the cone with vertex X0, 'y, and axis to M fl V must beJohg tà A. But then
C0'fx—x(
and the proof is complete.
Theorem Under the Theorem 6.1:
(1)
(ii) IffurtherfE intepr 1,0< a < 1), men 4 E (iii). 1ff is alto analytic, then 4(x) Lr
Proof. The assertion (i) follois from Theorems 6.1 and 3.10. Notice that Theorem 3.10 asserts a C' representation
(6.13)
JNEQUALfl1ES: ANALYSIS OF THE FREE BOUNDARY
Since, however, is Lipschitz, 0 and the implicit function theo- rem can be used to invert (6.13) into = with in C. The assertions (ii), (iii) follow from Theorem 1.1.
Consider now the filtration problem studied in Chapter 1, Section 5. As an immediate corollary of Theorem 6.2 we obtain:
Theorem 6.3. Let u = 4)(x) denote the free boundary for the filtration problem of Chapter 1, Section 5. Then is analytic for 0< x <a, <0 for 0 <x <a.
Indeed, Since w,, <0 in the noncoincidence set W, we find 'that y = 4)(x) is analytic, and since <0 in W, also the inverse function x = '(y) is analytic, so that 0.
Theorem 6.2 can be applied also to three-dimensional filtration problems. Suppose that a dam (2 in R3 (made up of homogeneous porous medium) is given by
(2= (O<x3 <H,0 <x1 <a,g1(x1) <x2 <g2(x1)),
where x3 is taken as the upward vertical axis. To the left of the face G1: x2 = g1(x1 there is a reservoir of water at level H; to the right of the face G2: x2 = g2(x1), there is a reservoir of water at a lower level h (0 < h <H); and the dam is impervious at the faces S0: x = 0 and x, = a and at the bottom B: x3 = 0. We denote by T the top x3 = Ii.
Setting u(x) = p(x) + x3 (p the pressure), one seeks functions u(x),+(x1, x2) such that
inthewetpart(p>0),
uH onG1, uh u=x3 onG2fl{x3>h),
au—0 onB, ax3
au is = x3, = 0 on the free boundary r: x3 = +(x1, x2).
THE FREE BOUNDARY FOR THE F1LThATJON PROBLEM
Let g(x1, x2) be the solution of
g on (x2 =
(6.14) g=4h2 on(x2=g2(x1)),
on(x1 0).U (x1 a}.
Assuming that
(6.15) g(O) = g(a) = 0, g,(x,) <g2(x1),
one can establish the existence of a solution C'(l). We now define
w(x1,x2,x3) =f (u(x1,x2,t) —
g(x1,x2) onB 2
/ 4(H—x3) onG, —
• onG2fl(x3<h) 0 onG2fl(x3>h)andonT.
Proceeding formally, we find that w is a solution of the variational inequality
• wEK,
(6.16)
where
(6.17) K= {v E U Sa)).
Theorem 6.4. There exists a unique solution of the variational inequality (6.16) and (1) 0w/ax3 0 in fi, (Ii)the free bowrdwy is by x3 = #(x1, x2), where • is analytic in B.
Proof. The existence, and regularity follow from the general thcoiy of Chapter ITo derive near the corner points
182 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
of the boundary, say near x3 = H, we reflect across x3 = H, by considering
fw(xi, x2, x3) if x, <H, S ifx,>H;
for more details, see reference 167b. The proof that Ow/Ox, 0 follows by the maximum principle as in Chapter
I, Section 5. Next, by comparison we find that
and consequently Ow/Ox, = 0 on x3 = H. We are now in a position to apply Theorem 6.2 in order to deduce the assertion (ii).
In the remaining part of this section we concentrate on the two-dimensional case and derive more information on the free boundary. One of the main results to be proved is the following:
Theorem 6.5. Lety 4(x) denote the free boundary for the filtration problem of Chapter 1, Sectioi 5. Then
(6.18) .'(x) is strictly monotone decreasing,
(6.19) +'(O) 0, +'(a) —
(6.20)
The last inequality asserts that the seepage line {x = a, h <y is nonempty—a well-known fact in hydraulics engineering.
Lemma 6.6. The solution w of the variational inequality (of Chapter 1, Section 5) satisfies
(6.21)
Proof. LetH*>H,(l.(O<x<a,0<y<H*)anddefjne
(g onOt2.fl{y>H),
= {v
THE FREE BOUNDARY FOR THE FILTRATION PROBLEM
Let
w, E
I vw1, .v(v— _fJ(v_ w.) Vu
By uniqueness we fInd that w = Thus it suffices to establish (6.21) for Consider the penalized problem.
(6.22) —l
(6.23) w =
where g5 differs from g. only in a 6-neighborhood of the points (0,0), (0, H), (a,0), (a, h); g8 is in C4, and
at (0,0) and at (a,0),
82 —jg,(x,0) C.
Denote the solution by wd. Taking /3, such that fi' 0, we find that = satisfies
(6.24) — + = 0
Next, on fy = 0) on {x a),
2
1 —C.. ay
Hence, by the maximum principle, —c in Similarly, —C in Since
•
(if we take /3, 0), we get
184 VARIATIONAL INEQUALfl1ES ANALYSIS OF THE FREE BOUNDARY
We now let 8 —. 0 and obtain
(6.25)
where w, is the solution of (6.22) with
w,g on8(L,.
Letting e —, 0, we conclude that are bounded functions. we can prove that
inW=(w>O),
then (6.21) follows by recalling that ttw = I in W. It will be enough to prove that 0 in W.
Now by Theorem 3.6, 0 on the free boundary. The harmonic function is continuous and 0 on the remaining part of aw, possibly at the
points (0,0), (0, H), (a,0), (a, Is), and (a, 4i(a)); these points by X, (1 Ifwecanshowthat
(6.26) liminf (x.y)-.X,
then the maximwn principle gives 0 in W. Since is bounded in W, (6.26) follows immediately from the following
Phragmtn-LindelOf type of lemma
Lemms6.7. LesDbeabowideddomalnwsdkt.(x01y0)beapointofaD. øand(x0,y0)bekmgs
to the boundary of o. Je:vbeabowukd harmonic/unction In Dandkt
v(x,y), (x.,)EOD,(x, y)-.(x,. )b)
Iii sup v(x, y).
(x. y)ED.(z. y)-.(x0. )b) (x, y)Eb,(z,
Proof. Suppose for simplicity that (x0, (0,0) and that a = {(p, 0= ir, 0 <p < p1), where (p, 9) are polar cotwdiflatcs. For any e >0, consider
THE FREE BOUNDARY FOR ThE FILTRATION PROBLEM 185
the harmonic function
w = Ap'13cos — elogp + (12 + e) — v
in D,, = 0 fl <p <Po). If A is sufficiently large, then w > 0 on 0,, fl (p = p0). Since clearly w > 0 on the remaining boundary of if ip is positive and sufficiently small, we conclude that w > 0 in D, and, hence, also in D0. Letting
0 and then p 0, we get
limsup i)-—(x0, I'o)
In the same way one can prove the inequality regarding lim inf v.
Lemma 6.8. For any point (x0, 4(x0)) with 0 <x0 < a, the Lunction cannot take an e-viremum at (x0, $(x0)) with respect to any W-neighborhood of (x0.
Here W is the noncoincidence set.
Proof. Differentiating the relations
0, w,.(x,*(x)) = 0,
we get
(6.27) + = 0, + = 0.
Substituting = I — w,,,,, we obtain
(6.28) w,,,,(x, = + (4,1(x))2'
(6.29) . s(x)) —
To prove the lemma by contradiction, assume that w3,, takes a local ex- tremum at (x0, +(x0)). Then
(6.30) 0 = — = X0. (I+(+(x)))
By Theorem 6.3, 4'(x0) 0 and, thetefore,
+"(x0) = 0.
186 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
From (6.29) we then obtain
(6.31) atxx0.
where A 0. Thus 0 at (x0, 4(x0)), which contradicts the maxi- mum principle.
We shall need the following general local result.
Lemma 6:9. Suppose
.in Il = (x x +y C, C, and
w(x,O)0,
Suppose also that the free boundary is a curve y = +(x), and E CEO, e1J for some Li > 0. Then
(6.32) 1
I +
Proof. Set A = Consider the family of functions
(O<e<1).
They all solve the same variational inequality as.w and, since
there is a sequence w1 -' u uniformly. Clearly,
in(x>O,y>0,y>Ax},
u(0, y) u(x, Ax) = 0, Ax) =0.
The function
2(A2 + 1)
THE FREE BOUNDARY FOR THE FILTRATION PROBLEM It?-
is harmonic for (x >0. y >0, y > Ax) and vanishes on y = Ax with its normal derivative. Hence it vanishes identically. But 11(0, y) 0 gives (6.32).
Lemma 6.10. If +'(O) exists, then +'(O) = 0; (ii) if 4i(a) limx.a4'(x) exists, then 4(a) > h and = — 00.
Proof. The assertion (i) follows.from the preceding To prove (ii), suppose that
(6.33) 4i(a) = h.
Set = —+'(a) and consider first the case y < 00. From (6.28) we,get
2 if x a.
Also,
ifO<y<h.
Introducing the harmonic function,
z(x,y)= —h+ir)
weseethat -
—.0 if(x,y) (x,+(x)),x ta, - orif(x,y)=(c,y),yfh.
Applying now Lemma 6.9 in (x a, y h, (a — x) + (h — y) we find that
.!_ I - 2p 3(1 +y2) 1+(I/y)2' -
which is impossible since y >0.
VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
We conclude that 4'(a) = — and the previous argument gives h) — if x -. a. Applying Lemma 6.9 again, we obtain
which is impossible. Thus the assumption (6.33) leads to a contradiction. Having proved that +(a) > h. we now introduce the harmonic function
z(x, y) = A arctan + B,
where
Aarctany+B= 1' ——A+Bl. 2
Then
B =
+ arctan y ) / + arctan
As in the preceding proof, we find that
• if(x,y)(x,4(x)),xta, or if (x, y) = (a, y),y
so that, by Lemma 6.7,
—'0 ifx—'a.
Mence
-' B ifx -, a.
If we now apply Lemmi6.9 in (x c a, y we deduce that
B= I + (1/)2
But this is impossible if y oo. either = 0 or y = 00. Suppose next that y = 0. To derive a contradiction, consider the function
V = w,, — in a domain
D5= {a—8<x<a,h+&<y<+(x)),
THE FREE BOUNDARY FOR THE FILTRA11ON PROBLEM
where 6 is sufficiently small; recall that 4'(a — 8) <0. Then
= >°
in some neighborhood of (a — 8, 4(a — 8)) provided that e is small enough. Since V = 0 on r, it follows that V < 0 if x a — 8, $(a — 8) — 1) <y <
— 8) for some q = small enough provided that 0 < e Recalling that <0 on the remaining part of and that wy = 0 on
{x = a, h + 6 <y <$(a)}, we conclude that
V<0 on8D8
if e is small enough. By the maximum principle it follows that
(6.34) V<0 mD8.
Now, if y = 0, then a/ax — e(a/ay) is an interior derivative to D8 at (x, with x near a, and (6.34) gives w <0 in D8, at points-near (a, 4(a)), which is impossible. Thus y * 0 and consequently y — co. This completes the proof of I emma 6.10.
Lemma 6.11. Let z be a bounded nonconstant harmonic function in a bounded domain D. Let X11.. . , be boundary points of D such that for each there isa segment 0, X1 E Assume that z is continuous on D
(x0, Yo) E D, a z(x0, y0) z(x, y) (6.35)
forall(x,y) EaD, (x,y)*X1 (1
then there exists a simple piecewise analytic curve S,
SC{za), (x0,y0)ES,
such that Scan be parameirizcd by x = x(t),y =y(t)(—oo <t < oo), and
63 (x(t),y(t))—X, ifz—.oo,
(.6) (x(i)y(t))-.X ift——oo,
where # X,.
Proof. The level curves of (z a) which initiate at (x0, Yo) fenn locally m analytic curves, and the angle between two adjacent ones is i/rn; if m> 1, we say that (x0, y0) is a branch point.
VARIATIONAL INEQUALflIES: ANALYSIS OF THE FREE BOUNDARY
We construct S step by step by extending the endpoints. If an endpoint is not a branch point, then the extension is locally unique; if it is a branch point of order m, then we can pick up any one of the 2m — I arcs.
As long as S remains in a compact subdomain D1 of D, we can get in each step a new arc of length c (c depending on D1). S cannot have self- intersection, for otherwisez a in a domain bounded by a part of S.
It follows that if D0 C D0 C D1 C D1 C D (D, open domains), then S must exit D1 after a finite number of steps (or else there would be an infinite number of arcs of length c > 0 which intersect any small neighborhood of some point of D1). S may reenter D0, but then it must ag4in exit it; from into D0 are finite in number (or else again there would be an infinite number of arcs of length c which intersect any small neighborhood of some point of aD0).
We parametrize S by x x(t), y = y(t) in such a way that — <t < 00. Let be a 8-neighborhood of X1.. . . , and let D0 be such that 8D0 lies
in a 8-neighborhood of If T8 sufficiently large, then dist((x(t), y(t)), 8D) < 6 and, recalling (6.35), we conclude that (x(t), y(z)) E V8 if 6 is small enough. Since 6 is arbitrary, we obtain (6.36). Finally, Lemma 6.7 and the maximum principle implies that X,.
Remark 6.1. In applications we shall often not have the precise setting of Lemma 6.11 at hand, but rather some slightly different versions of it.
Since the free boundary is analytic, w has analytic extension across the free boundary. It follows that the function can be extended as a harmonic function into a domain containing w U r.
Lemma 6.12. The function 4'( x) cannot take a local minimum in the interval 0 <x <a.
Proof. We argue, by contradiction. Suppose that +'(x) has a local minimum at some point x0, 0 <x0 < a, and let a = q(x0)). Then
(6.37) +(x)) > a if x x0, x — x0 I
small.
It follows that there are at least two level curves of = a) which initiate at (x0, 4(x0)) and go into W. Thus there exists an open component G of the set W fl <a) such that (x0, 4(x0)) E
Denote by S1 and S2 the two level curves of = a) which start at (x0, and which bound with S1 being to the left of S2 near (x0,
x0)). These curves cannot intersect each other in W [except at (x0, 4i(x0))] since otherwise w),,, a in 6 (by Lemma 6.7 and the maximum principle).
Each S1 cannot intersect the free bouodary, for otherwise Lemma 6.8 would give e in a region bounded by and I'.
lifE HOUP4DARY P0* ThE FILTRAUON PROBLEM
Next, wy,,, = 0 or I on lexcept possibly at (0, H), (a, h), (a,4'(a)) where w,,) may not be perhaps continuousl. Thus the proof of Lemma 6.11 can be applied in order to deduce that S, is continuous up to its endpoint P,, P1 P2. and only the following cases can occur:
(1) P1 = (0, H). (ii) P,
In case (I) we take any level cumreV starting at (x, 4,(x)), x <x0, > a. Then
alongT.
The endpoint of T must also coincide with P1 and then, after applying Lemma 6.7 near (0, H) (noting that w,,,, = 1 on x 0), we obtain
acmtradiction.
In case (ii), since y) = 0 it <y we obtain by the previous argument that a and sijnilarly..
4(x')) if x' <x.
It follows that is monotone and, in particular, — 0) exists. But then Lemma 6.10(u) gives.
ifx-4, a contradiction.
Proof of Theorem 6.5. Prom 6.12 * have that is monotone. Hence Lemma 6.10 can be applied deduce that $'(O) =0, a) = — 00, •(a) > h. Finally, $'(x) is ckafly
PROBLEMS
1. Prove that w in Theorem 6.5.
2. Consider the filtratiqn problem with bottom; thus u,(x, 0) 0 is replaced by u,(x, 0) = 1(x) for 0 <x <a. Inthe variational inequality the change that occurs is in the boundary condition on y = 0:
w(x,0)k(x) wherek"I(x),
Suppose that 1(0) = 1(a) = 0, k(x) >0. ProvE:
VARIATIONAL ANALYSiS OF THE FREE BOUNDARY
(1)
(ii) if 1(x)>O, then and for for x0 x a.
(iii) If x0 < a, then 4'(a 0) = 0. (iv) If 1(x) changes sign m times, then changes sign at most m + I
times.
3. Let w be as in Theorem 6.5. Prove: (i) (ii)
(iii) 4'( x) — (I /ir) log (a — x) is in a]. I Hint: For (i) consider
z(x, = —
+ (a — x)2) — log((y — + (a — x)2)]
+ Y_h[arctan a
— arctan I and s'how that = — z is harmonic and smooth in W-neigbborhood of (a, h). For (ii) consijer the hodograph mapfp + iq, where p = q I — and show that f maps W I—I and conformaily onto the domain bounded by the rays q 0, p < 0; q = —1, p <0, and the half circle p2 + q2 + q = 0, p 0. For (tii) use the mapping (1, =
y), ü,(x, y)), where ü = w — x2/2. Show that it is I—I and that v(i, 9) u(x, y) is harmonic, v(1,0) = _12/2 if 1> —a, —a2/2 if 1< —a, and that v(i, j7) — z(i, 9) is CX for (1,9) near (—a,0),j <0, where
z(x, y) = +y2) — log((x + a)2 +y2)]
2 X x+a+ (y — x )[arctan — — arctan
2 x+a 'IT—a arctan +— y 2
Finally,
x
where p0 E — E, aJ.J
THE FREE BOUNDARY FOR THE TORSION PROBLEM
4. Consider the setting of Problem 5 of Chapter 1, Section 5, with
11 ify0<y<H k(Y)__lk ifO<y<y0
and k> 1, so that k(y) is not increasing. Let
Denote by the solution of the variational inequality corresponding to Prove: w,, 0 [so that the free boundary is given by x = 4s,(y)J and
(i) 0, and WY, are bounded if y Yo• (ii)
(iii) The free boundary is x = 4s(y), and 4'(y) is analytic if y (iv) 4'(Yo ± 0) exist.
(Hint: see Theorem 7.2.] [In reference 58b it is proved that (a) 44y) has at most one local maximum and one local minimum [and there are examples where 4<y) is not monotone]; (b) 4' has at most one inflection point for y > Yo' and at rno$t one inflection point fory <yo; (c) denoting by a and fi the angles negatIve x-axis to the tangents to r at y = Yo going upward and respectively, the following refraction law holds:
eithá 8=w, oru=ir, $=0,
or
or tan2a=k,
7. RE ULARI V OF THE 1!REE BOUNDARY FOR.. THE EIASTIC-PIASTIC TORSION PROBLIM;
Lemma 7.1. Suppose that
(7.1)
(7;2) A > 0, U M < 00,
wherefl is a domain in intersecting the cylinder
KR=BRX (0<y<b).
$4 VARIATIONAL ANALYSIS (11 THE FREE DOUNDARY
ItGbeadomain in KRflN(N {u>O}) such that
8G n (eR X (0 <y b}) is contained in A = (u = 0),
aG fl (y b) contains a point (x°, b),
a pont: (x' , 0) I'.
Then
l(n+ (7.3)
) R.
Proof. We follow the proof of Lemma 3.1 with
__________
( 2 21w(x,y)u(x.y)— 2(n+ 1) tIX p
where xm —. x°, Ym 1' and conclude that
sup b2. aGn(y=o) 2(n + 1)
Since, however, u(x',O) 1', the left-band side is (7.3) follows.
This lemma is particularly useful in case n = 1. We shall present a version that will be readily applicable in the elastic—plastic problem.
We assume that
(2={a1<x<a2, 0<y<!(x)), l(x)conunuous,
(7.4) N={a1<x<a2.
forsomee.>O.
Note that 1' contains points (x, for x in some subset of <x <a2) (not necessarily the entire interval).
Theorem 7.2. Assume that (7.1). (7.2), (7.4) hold and tha:f_= 1. Then Is continuous.
Results of the type of Lemma 7.1 and Theorem 7.2 are called nonoscillation lemmas.
Proof. Suppose that a1 <xc <a2, < 1(x0); then (x0, 44x0)) r. There must exist sequences x,, j x0, f x0 such that -. J'(x0),
THE FREE BOUNDARY FOR THE ELASflC-PLASTIC TORSION PROBLEM
q(x0). Indeed, u >0 on one side of an interval (x x0, 4,(x9) < y < iJi(x0) + 8) and we get a contradiction from the uniqueness to the Cauchy problem (here we use the fact that f I).
We next claim that 4i(x) < 1(x) if x — x0 is sufficiently small and 4'(x) 4,(x0)ifx— x0.
Indeed, if this is not true, then there a sequence x, —' x0 such that + for some > 0. We may assume that each lies in
some interval (xi. Xk), Xm as in the first part of the proof, with 4'(x1) 4,(x0). —, 4i(x0); herej j(m), k = k(m). Let
e (xi, Xk), = max
Denote by (i the interval (x x7. hrn < y < H,,,). -where h,,, = max and by K the rectangle x, <x < h,,, <i <H,11.
Applying Lemma 7.1 with KR = K. we get
H,.,, — x,(— 0.
a contradiction to lim(H,,, — hrn) > 0. We have so far that the set S (4,( <1(x)) is open and 4' is
continuous in each of the intervals of this set. Suppose next that x0 E a, <X() <02. If x,,, — x0. then clearly
44x0), and since 4(x0) = 1(x0) it follows that Iim4'(xm) = 4i(x0). Finally, 4i(x) is obviously continuous at any point x0 S.
Consider now the obstacle problem
(7.5) U M < oO, 4 E C3
is a domain in R". On the coincidence set A we havef — 0 a.e. On I' every% here we have
/ — t&4i 0; indeed, if (f — <0. x0 F, then v = u — satisfies >0, v 0 in a iieigbborhood of x0, and the stronginuimum principleis
contradicted since v(x0) = 0. if we wish to apply the nonoscillation lemma and results of Sections 1 to 5,
then we must establish the inequality f — A4> 0 in r. We shall give a useful sufficient condition for this to hold.
Lemma 73. Let (7i) hold and assume that -.
(7.6) 1— and v(.f — do not vanish
Then f — > 0 on the free boundary of u.
196 VARIATIONAL INEQUALITIES ANALYSIS OF THE FREE BOUNDARY
Proof. Let 4' = f — We suppose that 4'(x0) = 0 for some point x0 E 1' and derive a contradiction. Since v4'(x0) 0, for any small £ > 0 and R > 0,
4i<0 inK,flBR(xo)
where K, is the cone
The function v = u — 4 then satisfies
—Ew>0 inK, fl BR(xo),
and since o 0, the strong maximum principle gives
v>O inK,flBR(xo).
By slightly increasing E we may suppose that v > 0 on K, fl aBR(XO). Since the opening of K, can be made arbitrarily close to 'ff/2, there exists (by
Problem 6) a harmonic function of the form
h(x) —
in where I <A < 2andOis
the maximum principle
v(x) ch(x) in K, fl BR(xo)
for some small constant c > 0. Hence
along the axis of K. SWce, however, v(x0) = 0, vv(x0) = 0,
which contradicts the previous inequality.
We are now ready to proceed with the study of the free boundary of the elastic—plastic torsion problem. For clarity we first consider the case where the domain is simply we restrict oursetves to the physical two- dimensional case.
THE FREE BOUNDARY FOR ThE ELASTIC-PLASTIC TORSION PROBLEM
FIGURE
We assume that aocon*iswdf a finite number of disjointarcs S3,. . . ,S,, and that thcendpoint of S,ia the Initial poiflt of = = V0). We also assume that each 5, is in up to the endpoints.
Dáflnllioiitl. by the angle formed by S,. at J', which opens info 0. It a,>v, we thy that V1 is a reenwant corner; if a, <v, then we say thu V isa f9r we shall always that
., . ..
Defhiklon7.2. Theridgeofflisthesetofallpointsx0e0sucbthatd(x)is not in for any Vol
In Figure 2.2 we indicate the re In order to characterize the ridge geometrically, we need a few lemmas.
Lemma 7.4. If 1x0 — y11x0 — , wherey andz are different points on 80, then is pot dif engiable ag
Proof. x0z is linear with derMvativc 1; the directions x0y, x0z are distinct. If d(x) is differentiable at x0, then I Vd(x0) = I and there must be a unique directioi 1 along which
= — I, a contradiction.
Notation. We denote by R0 the set of all points x0 E 0 such that d(x0) = Ixo — for at least twodistinctpointsy, z on 80.
VARIA11ONAL INEQUALITIE& ANALYSIS OF 11ff FREE ROUNDA*Y
Lemma 7.5. Let x0 E R0 and denote byy0 y(x0) the nearest point tm au to x0. Denote by z0 the center the osculating circle at Yo (suppose thai Yo not a
vertex.) Then d is in in some neighborhood of x0 if andonly if x0 * z0.
Proof. The point cannot lie between x0 and for otherwisey0 is not the nearest point to x0. If x0 * z0. then by reference 109, p. 382, d is in in a neighborhood of x0 and
(7.7)
where = ic(x) is the curvature of at the point nearest to on a finite point and lies in the interval then \,,1 (1 and
becomes unbounded as x zo: thus d is not in ( in neighborhood of z0.
Lemma 7.6. If in the preceding lçmma is a vertex (it is then net a reentrant corner), then d is iii in rome of x0 if and onfr if
* z,, where:, are 'lie r?vters the osculating circles corresponding iv the two arcs of which meit al V0.
In fact, if 1. is a segment to z,, then the regularity of d in one side of /, is the same as before, whereas in the sector between and
x) y0 Thus it remains to verify that the first denvativçs of d from both sides of F, agree. But since the derivative of d along I, is 1. the derivative of d in a direction normal to I, must tend to zero when the approach is from either side of F, (since I = 1).
Notation. A point x0 is said to belong to the set R1 if there exists precisely one point y() E such that d(x0) 1x0 and is the center of the
circle at (or the center of one of the two osculating circles. if v9 is a vertex).
From the preceding three lemmas, we obtaib:
1 heorem 7.7. R = R0 U R1.
If there are no reentrant corners, then R1 C R0, so R =
7.8. R C E. that is. the ridge elagk.
Proof. Let x0 Then
I I I —
I'robiem 2 of C'hapter I, Section 7. it follows that if x,, E P (she plastic set).
THE FREE BOUNDARY FOR THE TORSION PROBLEM
then
ud onthelinesegmentsl1: x0y,. It follows (see the proof of Lemma 7.4) that u is not differentiable at x0, which' is impossible.
Next suppose that x0 E R1 and x0 E P. and denote by Yo the nearest point to x0 on an. Suppose first that Yo is not a vertex. Then
ud v(u—d)0 on((sinceu—d'cOinn),
(7.8) 2 = 1— d
00 tfx(O,'7l)x0(O,O)
where y) are coordinates with normal to / and along 1: (7.8) follows from the proof of (7.7). Since u ' d. we then must also havc
a2u —--a -00.
where the second derivative is taken in the sense of limit of second-order difference quotients of u. Thus is not Lipschitz in any neighborhood of x0, which is impossible.
If y0 is a vertex, then it is necessarily a reentrant corner md one easily finds that a2d(x)/an2, taken in the sense of limit of second-order difference quo- tients, tends to — oo as x = (0. ti) — thus the above applies.
We now parametrize = (0 s L and write = x2). f(s) = (f,(s), f2(s)). Denote the inward normal to Mi at c) by v(s) (S s.. where s1 is the parameter corresponding to the vertex V, Let us assume first that there are no reentrant corners. Then, by Problem 2 of Chapter 1. 7, there exists a nonnqative function 8(s) such that
A= {x;x=f(s)+tv(s).
By Theorem 7.8,
dist(f(s) + 8(s),
X, + + e)) E E
where X, R if 0 <r and E R: is uniformly positive in any closed interval of s that does not cohtain the parameters s, of the vertices.
200 VARIATIONAL INEQUALII1E& ANALYSIS OF THE FREE BOUNDARY
We are thus in a position where the proof of Theorem 7.2 can be applied provided we can prove that
(7.9) onF.
(Recall, by Theorem 7.8, that d E in a neighborhood of 1'.) Let x0 e r and define
(7.10) [by(7.7)J.
Denote by / the direction of where Yo is the nearest point to x0 on We would like to apply Lemma 7.3 in order to deduce (7.9). Suppose then that
= 0. Then by (7.10), K > 0 at x0 and, since aK/al 0 at x0,
K2 IC2 >0 ai (1 — .cd)2 ai (1 —
Thus Lemma 7.3 yields the assertion (7.9). We are now in a position where Theorem 7.2 (or, rather, the proof of
Theorem 7.2) can be applied. It yields the continuity of the (unction 8(s); notice that 8(s) may take the value zero.
DefinitIon 73. if 8(è) > 0 in an interval a <S < b and 6(a) = 8(b) = 0, then we call the plastic set
= (f(s) + tP(s); 0 t 8(z), a s b)
a plastic component or a plastic loop.
Having proved that the free boundary of any plastic loop, that is,
= (f(s) + 8(s)p(s); a <s <b),
is a continuous curve, Theorems 2.4 and 2.5 show that a nondegenerate parametrization. But I'a b may still have cusps! The next theorem excludes this possibility.
Theorem 7.9. The coincidence set has positive density at each point x0 of the free boundary such that d E C3 +a in some neighborhood of x0.
Proof. Let x0 E r and let Yo be the nearest point to x0 on Yo vertex. (See Figure 2.3.) For simplicity we takey0 (0,0), x0 = (0, h), h >0. Let
x = x2 =
THE FREE BOUNDARY FOR THE ELASTIC-PLAS11C TORSION PROBLEM
- an
FIGURE 23
be a local parametrization of r in a neighborhood of x0, with = 0, = h. We only have to consider the case when = 44(0) = 0,
0, 44'(O) 0; by Theorems 2.5 and 3.4, this is the only possible case in which the assertion of the theorem may not hold. r has a cusp as in Figure 2.3.
Fort 0, tnear 0, 44(t) 0 and thus the free boundary is in C2 and u is in C2 up to the bQwldary.
The functionw = — I is negative inE and w(x0) = 0. Since the inside ball property holds at x0, the strong maximum principle gives <0 at x0, that is,
(7.11) atx0;
the derivative is taken in the sense of urn sup of difference quotients of Differentiating the relation
(u — = 0
with respect to t,we get
(u — + (u — = 0.
Dividing by 44(t) and letting t -.0, we obtain (since u and dare In C' l)
(u — 42(')) —.0.
Sinced is linear along y0x0,
—. 0 if (x,,x2) —. x0;
consequently,
-'0 as(x1,x1) —x0alongl'.
VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
Applying Lemma 6.7, we obtain
u,(x)—.O ifxEE. x-'x0
which contradicts (7.11). Consider finally the portion of a free boundary whose nearest point on aa is
a reentrant corner. The previous considerations extend to this case ith small changes (again a variant of Theorem 7.2 is needed here). Notice that the obstacle in this case is analytic.
Definition 7.4. If is a connected component of I' and if for some x0 E r'0 the nearest point to is a reentrant corner then call the corresponding plastic set (bounded by 1'0 and a reentrant plastic component or a reentrant plastic loop.
Applying Theorem 7.9 and Corollary 3.11, we can now state the following fundamental result.
Theorem 7.10. The free boundary is locally (m 3, 0 < a < 1) or analytic if the part of which parametrizes it is in Ctm or is analytic, respectively. Any open portion of r whose nearest points to coincide with a reentrant corner is analytic.
PROBLEMS
1. In the following problems is a multiply connected two-dimensional domain and u is the solution of the variational inequality (1.6.4). With 4,, defined by (1.6.6), (1.6.7), we define: The upper ridge is the set of points x0 in such that is not C1' in any neighborhood of x0. Similarly, one defines the lower ridge R with respect to 4,. The set R = R + U R - is called the ridge. (This definition depends on the solution u, that is, on the constants ci.) Prove: R C E.
2. Extend Theorem 7.10 to the case where nP = 0 (P ± are defined in Problem 1 of Chapter 1, Section 7).
3. Let v E 4, v 4',, where 4,,, 4, are defined as4,, 4' but with replaced by +t. Show that
vi' — Vu 12 —
ThE FREE BOUNDARY FOR THE ELASflC-PLASTIC TORSION PROBLEM
4. Suppose that 0 has only one "hole" and u=C, in Set dist(an,, SD). Show that if C,, < and an1 then
(u=4,,u<4}
is nonempty. [Hint: Suppose that u >4, a.e., so that Au —ii. For small e >0, if
lem 3.)
5. Assume that an1 has a vertex Wsuch that W— W— E the rays form angie 0, 0<9 <ti, and the sector
obtained by rotating by an angleD to position WZ2 is contained in Denote by the domain bounded by WZ,, WZ2 and aDs. Prove:
If C,, then, for any neighborhood G of W,
Gfl inasetofpositivemeasure.
[Hint: If not, Au in G fl Let At, = GflI, v = u on
= v — 1+ where r = r(x) x — Then At5 = 0; deduce that E C" in G fl so that < —1 at IV, y the bisector of•!.J
6. Consider a cone = {x; x I) <4'). Prove that if 4, < ir/2 and —4. is small, then there exists a I <A <2 and a function (where 0 = cos x I)J such that the function v x
harmonic and positive in and v = 0 on [Hint: If = g(z), z = cos8,
(z2— l)g"+(n— l)zg'—A(n—2+.A)g=0,
and then g(z) = where the are the Gegebauer functions [84, p. 1781. If A 1,1, —. const. z and —. const. cos 0. The first zero of fA(0) (for A> I! is 0 = 4A <n'/2 [otherwise, x f'fx(O) > CX,, (c >0) in (x,, >0, tx < 1) by the maximum principle) and, by continu- ity, ir/2 if A 1.]
8. THE SHAPE OF THE FREE BOUNDARY FOR THE EL4STIC-PLASI1C TORSION PROBLEM
In this section 0 is a simply connected two-dimensional domain; is made
The angle of an at V1 is dtnoted by a. and it is assumed that (see Section 7).
204 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
We begin by studying the free boundary in a neighborhood of a reentrant corner Set a = a1/2 (so that a > ir/2), /3 = a — ir/2, and introduce polar coordinates (r, 9) about so that the tangents to at are the rays 0 = —a and 0 = a, respectively.
Theorem 8.1. There exists a continuous positive-valued function p( 9), defined for —$ < 0 < /3, such that
(8.1) ((r, 0); 0 <r < p(9), —/3<0 </3) C P.
Thus a neighborhood of a reentrant corner always contains a plastic sector of opening a1 — ir.
Proof. We begin with a lemma that is a weaker version of the theorem in case Il is a sector D.
Lemma&2. P the elastic and plastic sets of D. Then there is a segment a = {(r, 0); 0< r < r1, 9 = 0) which is contained in P.
Proof. Consider the function
U. = —
in the region E fl (0> 0). This function is harmonic [since = = (—i'). = 01 and vanishes on the part of where r = r0 (since
u = 0 if r r0) and on 0 = 0 (by symmetry). On the segment 0 y, u9 0 (since u = 0 on 0 y, u 0 elsewhere). Finally, on the free-boundary part of
= d, 0 since d is decreasing in 0 along any circular arc in By the maximum principle we then conclude that u9 0 in E4.
In the smaller set
P is a subgraph fr may a priori be equal to zero for some values of 0. Since u9 0 in and since d(r, 0) r in we have
It follows that f(0) is monotone decreasing in 9, 0 <9 < y — ir/2. Now, if the assertion of the lemma is not true, then f(0) 0, and from the
monozonicity off it follows thatf(9) = 0 if 0 0 y — ir/2. By symmetry we
ThE FREE BOUNDARY FOR THE ELASTIC-PLASTIC TORSION PROBLEM
also havef(8) = 0 it —y ÷ (ir/2) 0 0. Let
G
B=Dfl
Then G is contained in E and therefore Au = —p in G. tithe other hand. if x then eitherxEE, in which caseAu= —p<Oorx in which case a.c. Au Ad = 0 (since d is linear in B G). Thus
a.e.inB.
Consider the harmonic
2 (c>O) 2y
in B. Suppose that
(8.2)
where r is the normal. Then, if c is sufficiently small, v ' u on Since also At' Au a.e. in B, the niazimum principle u v in B. In particular,
u(r,O)
Since, however, u(r,O) d(r,0) = rand (ir/2y) < I, we get a contradiction if r -. 0. This completes the proof of the lemma in case (8.2) is satisfied.
The assertion (8.2) is true (by the strong maximum principle) if there exists a b-neighborhood of (r0/2,±y) which is elastic. On the other hand, if no such neighborhood exists, then there is a plastic segment initiating at a
and orthogonal to (9 = ±y), is arbitrarily close to r0. We then have
Thus by replacing, if necessary, B by a domain
we can continMe as before to deduce thati > v in B* and, consequently, derive a contradiction.
VARIATIONAL INEQUALITLE& ANALYSIS OF THE FREE BOUNDARY
Proof of Theorem 8.1. Take any direction 0 = 00 in (—$, and construct a domain D as in the preceding lemma so that it is contained in (1, its axis of symmetry is (0 = and
Denote by UD the solution u corresponding to the domain D. By comparison, in D we have that UD u. Also, u d = r along the ray 0 = if r is small enough. By Lemnia 8.2,
u0(r, = r if r is small enough, say r < F.
It follows that u(r, = r if r < F, that is, P. Since F can be taken independently of provided that is restricted to a closed subinterval of (— fi), the proof is complete.
We next study the free boundary near a nonreentrant corner Vk; here <IT.
Theorem 8.3. If Vk is a nonreentrant corner, then there exists a neighborhood BR(Vk) of Vk such that fi fl BR(Vk) is elastic.
Proof. Consider first the case where Sk, Sk+I are line segments and introduce polar coordinates with center at Vk so that some fl-neighborhood of Vk is given by
((r,0);O<r<r0,—a<8<a),
a bounded function In G (in fact, with bounded derivative), we conclude that
(8.3) is bounded in G.
Suppose now that a > ir/4, and consider the function
v v >0 if r = r0 provided that is sufficiently small, and the normal derivative of o is different from zero at (r0,±a).
THE FREE BOUNDARY FOR THE ELASTIC-PLASTIC TORSION PROBLEM 207
Recalling (8.3) we conclude, by the maximum principle, that
mG
for some positive constant C. Hence
ifrissmall enough, sayr<6. ThusGfl (r<8) CE. Consider next the case where a ir/4. It will be enough to prove that
(8.4) {(r,O)EG; CE,
provided that 8 is sufficiently small; the proof for 0 U a is similar. Let 0, be a domain containing U such that aa, has verteX and fl (r < is a sector of opening > v/2, with <ir, and such that the two arcs of
ao, which meet at are line segments and one of them lies on 0 —a. If we denote by w the elastic—plastic solution corresponding to then, by conipaii- son, u w in U.
By the proof above,
w 4d1 in some 01-neigbborhood of
where d, is the distance function for 0,. d1 = d if —a 8 0, r small enough, we obtain that u <din this set, and (8.4)follows.
We have now completed the proof the tbeorán in case ate line segments. In the general case we transform an a-ndghborhood of VA onto a sectorGas above Denote byz= g(I) the inverse of i —1(z). Then we can write
= mapsa Mwpmi T1, T2 which forte angl,v at their *dpoint, Ut. and T2 are mapped into >0) tively.
The function k of k takes C1' some eo>0 and thea by general regularity reference 180), it is in up to the ary. Was •thatforwme llo>0
(8.5)
208 INEQUWØES ANALYSIS OF THE FREE BOUNDAR%
(p. 6) as polar of we note that, for any small A > 0. the function
v = psinO + pi-Acos(2 — A)6
is positive on the boundary of (I p0) if p0 is small (in fact. psin6 —const. p2 in 0). Hence C',, Im hi on this houndary if C is large enough and, by the maximum principle.
By interior estimates for harmonic functions
C
along the bisector of T2. It follows that I vk c > 0 along the bisector 1m71 = 0. Since k E C'4', get (8.5).
From (8.5) it follows that
(8.6) vh ' C in some of 0.
and, in fact h E up to the boundary. Settrng u(i) u(z), where z = g(fl, we have
and the right-hand side is a bounded function (by (8.6)1. We can then apply to u the proof of the preceding special case and conclude that
ii CJP' near the origin,
where r = — I, J is the distance function for the transformed region, and P . Since d Cd, P Cr, we get u in some U-neighborhood of Vk, i the proof is complete.
It is not known whether the number of plastic components in U is finite. However, in case aa contains a linear segment A, one can prove that the number of plastic components connected to A is finite:
Theorem 8.4. Let A be a closed interval contained in one the opø* arcs S1. Then the number of plastic conrponents connected toA is finite.
Proof. Take for simplicity A = (a x, b, x2 = OJ and suppose that. there are infmite number of plastic components given by
THE FREE BOUNDARY FOR THE ELASTIC-PLASTIC TORSION PROBLEM 209
a x,—axis
FIGURE 2.4
Set
If i —, oo, then b. — a. -' 0 and, by the nonoscillation lemma, also H1 -* 0. Hence any line x2 = >0 and small) intersects only a finite number of the plastic components, and A(ij) 00 if 11 -.0.
Consider a subdomain of E fl (x2 > as in Figure 2.4. It is bounded by a piecewise analytic curve in E, parts of the and some intervals lying on x2 = We choose y such that d(x) = x2 in for all q small; y iS dent of Elf the F, accumulate at x1 = a, then we to start on x1 = a — e, for some sm4 e >0, if (a — e, 0) has elastic 11-neighborhood; otherwise, y will start from a free boundary point on (x1 = a — e); a similar modification of is made about x1 = b.J
On interval
on a; with endpoints on F function (u — is analytic and. changessign (sinceu—d=Oat theendpointsandu—dCO inside). Wecan choose points d, d, e, such that
{8.7) <e,.
We now the = const), which initiate: at (d4, and (E,, respectively. fly 6.11 these curves can continued until they exit E0 Notice that = (u — =0 on
and thus F, cannot exit E0 on the free boundary. They also cqnet come arbitrarily close to (x2 =0) since =0 on (x2 = 0). Finally, because:
and M precedes along y if the same holds for the initial points of the curves on x2 =
: It follows that alternates signs on y at 2A(ip) — I times a number \ -'0. But since
is piecewise analytic, this is impossible.
210 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
We next proceed to give an estimate on the number of plastic components connected to A. Take
V0=(b,O), V1
and set p S2 U -• USm. We assume that some fl-neighborhood of lies in (x2>O).
Let x = f(x) (0 'c s L) be a parametrization of with f(0) = V0, f(s1) = V,. We consider first the case where (8.8) there are no reentrant corners.
The distance function d(x) is differentiable along x = f(s), s [f(s) JJ
ad d1(s) = ifs
= d1(s1 + 0) [d1(O) = d1(L — 0)], and assume that
'8 9\ the set {s; d,(s) = 0,0 s L) consists of a
•
' finite number of points A, and intervals It follows that d1(s) changes sign (from positive to negative, or vice versa) a finite nwnbcr of times. Set
k = number of times d1(s) changes sign from positive • / toncgativeassincreasesfroms =s, tos = L. Denote by P1,. . . , the plastic components connected to A; by Theorems
8.3 and 8.4 and by assumption (8.8), v < We can write
ai>O, By Section 7, +(x1) is positive and analytic in each interval < x1 <ba; it is continuous and equal to zero at x1 = x1 = b;. We define 4'(x1) = 0 if x1
In each interval b), 4' has a certain number N, of points where it achieves a local maximum (these are strict local maxima since 4' is analytic).
Theorem 8.& Each Is finite and
1=I From this result we immediately obtain:
(I) The total number of plastic components connected to A isfinite and k; (ii) if the number ofplastic loops connected to A is precisely k, then in
ThE FREE BOUNDARY FOR ThE EIASflC-PLASI1C TORSION PROBLEM
each interval <x < there is a point such that is strictly increasing if <x1 and strictly decreasing if c1 <x, <by.
Corollary 8.7. If is a convex polygon, then for any side S1 there is at most one plastic loop connected to S,; if it is given by (f(s) + 0 t h(s),
$ <BJ), then h(s) is strictly increasing in some interval A1 s and strictly decreasing in the interval s
To facilitate the proof of Theorem 8.5 we begin with some auxiliary results. For the proof of the first lemma assume for definiteness that d1(s) >0 if s > s — s, small and d1(s) <0 ifs < L, L — s small; the proof in the other cases is similar.
Setting / = 2k, there is a partition 5i
= ' L such that
d,(s) >0 ifs,,1 (s,, <s,2),
d1(s) 0 ifs,2 s (s,,2
d,(s) >0 ifs,3 <s <s14 <31.4),
d,(s) 0 if sI.4
d,(s) >0 ifs1,5 <s <51,6
(8.12) d,(s)<Oifs2, <3<322 <52,2),
d,(s) 0 s 2,3 (p2,2
d1(s) p2,3 <52.4 (32,3 <52,4),
d,(s) <0 if <s (s,÷,, <31÷1.2),
d1(s) 0 if s,+1,3 S+i,,), k
212 VARIATIONAL ENEQuALmFs: ANALYSIS OF THE FREE BOUNDARY
Thus d1(s) changes sign at the points 52.m2'" ,Sjm,. The vertices V, (2 I m — 1) need not necessarily coincide with points that is, s, may be a point in some interval (s, ,,, .c,. Some of the intervals s,1 4-I] where d1 0 may consist of a single point.
We recall that the free boundary can be written in the form
x f(s) + v(s)h(s),
where v(s) Is the normal to at f(s) and > 0; for the points where h(s) 0, there is no plastic interval initiating at f(s) and perpendicular to 811.
Consider the function
(8.13) d1(s) + r(s)h(s)) . (0 s < L)
at a point s = s0 such that s for I I m, and
(8.14) d1(s°)>O.
By continuity, d1(s) > 0 in some interval — s0 < Hence some (2-neigh- borhood of f(s°) is given by
g(x2)<x1<g(x2)+81,
with x1 = g(x2) being a part of a(2, and f(s°ff = x0. Sety° =f(s°) + v(s°)h(s°) Suppose that h(s°) > 0. Then x0 is
the nearest point on Ml toy°. It is clear that, for any positive and sufficiently small, the ray initiating at = — and in the direction y°x° intersects Ml at(x7, x" and — <lx° — y°I; in fact, since g'(x2) is a bounded function,
(c>0).
This implies that d(y') <d(y°) — cij, so that
(8.15) J1(50) = c.
Since v(u — d) =0 on the free boundary, we conclude that
(8.16) + = d1(s°) c >0 if h(s°) >0.
By continuity, the relation (8.16) is satisfied also at any point such that
'8 h(s°) =0, and thereexisisasequence
• skshth.atsk...isO,h(sk)>O
THE FREE BOUNDARY FOR THE ELASTIC-PLASTIC TORSION PROBLEM 213
Suppose next that h(s°) = 0 but (8.17) is not satisfied. Then h(s) = 0 in a neighborhood s0. Consequently, there exists an fl-neighborhood W of x° which is elastic. Since = p > 0 in W, u > 0 in W, u = 0 on nfl, the strong maximum principle gives (au/av(s°)Xx°) > 0. Recalling (8.14), we deduce that f(s°)) > 0. We have thus proved, in general, that
(8:18) if d1(s°) >0, then J1(s°) >0, + p(s°)h(s°)) >0.
Similarly, one can show that if d1(s°) <0, then -.
J1(s0) <0, + <0.
Finally, if d1(s) = 0 in an open interval a <s </3, then (f(s), a <s </3) is a line segment parallel to the x,-axis; it follows that
= 0, -f v(s)h(s)) = 0 if a <s </3.
Set
(8.19) u,(s) = + ,'(s)h(s)), 0 <s <L,
ifs s,, and u1(s1) = u1(s, + 0). By Theorem 8.3, h(s) 0 in a neighborhood of s = s..
We can now summarize:
Lemma 8.8. The assertion (8.12) holds with d1(s) replaced by either J1(s) or u,(s);further,
(8.20) d1(s)=u,(s)=0 ifo<s<s1.
DefinitIon 8.1. The pointsf(s) + for which u1(s) = 0 will be called flat points. The set of flat points consist of a finite number of points and a finite number of arcs (calledflai intervaLs).
Notice that the flat intervals correspond to and to those intervals S],... in (8.12) where d1(s) 0 and which do not reduce to
single points. The function is hannonic in E and it has harmonic continuation into a
neighborhood of the interior of each flatinterval. The proof of Theorem will employ level-curves analysis. We Shall need a
variant of Lemma 6.11.
Lemma 8.9. Through each point x° €E where = 0 there passes a piecewise analytic curve {x = y(t), — <t < oo), con:qinedin £ and giong
214 VARIATIONAL I1WLQUALfl1ES ANALYSIS OF THE FREE BOUNDARY
which = 0. 'y has no self-intersections and
y(+oo)= limy(t), y(—oo)hiny(t)
exist and belong to 8E.
The proof is similar to the proof of Lemma 6.11. It uses the conclusion of Lemma 8.8, which implies that y(t) must stay away from all the points of except for the flat points and the flat intervals. Notice that y cannot near a flat interval I (as t -. oo or t — oo) since has harmonic continua- tion into a neighborhood of the interior of I.
Notation. Denote by XE the part of A which is not in the plastic compo- nents connected to A, and denote by 80pA the union_of the free boundaries of the plastic components connected to A. Set A0 = XE U a0PX. Similarly, we define for p = the sets PE' and Po = 15E u
a plastic loop (I i'); it is given by
a point of local maximum of Then
(811) +(x1)<+(P)
for sothe 60 > 0.
Lemma8.1O. = O},givenbyy: (x = y(t),O such that y(O)=($,+(fl)), y(t)EE sfO<t<oo, and y(oo)= belongs-to
Proof. Foranysmall8>O,8<60,
—8, +(P)) <d(/3 —8, +(P)) =
u(fl + <d(p— 8,,($)) =
Hence for some 0<81,82<8. By continuity,
if e >0 is sufficiently small. Consequently, there is a number such that
(8.22) + ,(p) + e) = 0.
It follows that there is a level curve of = 0), from those initiating at
THE FREE BOUNDARY FOR ftAS11C_PLAS!flC TORSION PWORLEM
4($)), which is contained in E (except for its initial point). We now continue to extend this curve as in the proof of Lemmas 6.11 and 8.9. It remains to show that y(oo) E
If this is not the case, then y(ao) E and there exists a subdomain E0 of E = Oonitsboundary. Itfollowsby
Lemma 6.7 and the maximum principle that 0, which is impossible.
Proof of Theorem &5. Consider first the case where k = 1 and take for definiteness d1(s) positive fo s > S1, S — small and the negative for s < L, L—ssmall.
Denote by p0(s', s") the portico of corresponding to s' s s", and denote by the point of corresponding to s. -
both lie on gre. Thus
y(—oo) = g'o(ei), v(oo) = 1'o(°2), si <01 <a2 <L. Denote by the curve obtained from by replacing by y, and parametrizeybythepanmetcrs,a1<s<g2.
Denote by the subdomain of £ bounded by As,, gii and suppose that there exists another curve of the type y, say y1,in suth that oo) and lie in We then modify into a curve replacing the part of gi1 from y,(— oo) to co) by y1. Again we parametrize by s so that .r varies in
s < L. We can repeat this process step by step. Note that in the first step the points y(—oo), cannot coincide and cannot lie in the same flat interval (otherwise, 0). Similaiiy, in the iecond step the points oo), i,(+ oo) cannot lie on an arc of along which =0, and so on. Recalling Lemma 8.8, we conclude that the process of constructing the curves gt2,. must end at somej. Thus there are no level curves y1 (as in Lemma 8.9) which 11cm
We shall assume for simplicity that gi1 = gi',; but the proof below extends almost word by word (with a slightly niore complicated notation) to the general case where we assume that
there is no level curve y as in Lemma 8.9 (8.23)
such that y(t) E E and y(+oo) lie
Consider a cur* y as in Lemma 8.10 and let y(+ co) = gi0(s°), < L. Denote by A01 the part of A0 from (0,0) to (fi, $(fl)) and by the arc X0\X01. Denote by I), the domain bounded by s°), andy, and,.by
the domain bounded by L), Consider first the case where - -
(8.24) : u1(s) 0 if <5<3°, - -
ifs°<s<L.
216 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
By the maximum principle
(8.25) mD1,
(8.26) <0 in D2.
From (8.25) it follows that (u — d > 0 in the subset of D, consisting of all points x for which d(x) = x2. This inequality easily implies that
8 27' the function x2 = 4(x1) is
( . monotone increasing for 0 <x, <fi.
Similarly, we deduce from (8.26) that
8 28' the function x2 = is
( monotone decreasing for <x1 <b.
It follows that there can be at most one loop and that x has a unique strict maximum. Thus the proof of the theorem is complete in this case.
Suppose next that (8.24) is not satisfied. Assume for definiteness that for some
<s < u1(s) 0 ifs° <s <se,
(8.29) u1(s)<O ifs* <s<s + 8 forsomesmall6>0,
if <s< L.
Then the conclusion (8.27) follows as before. Since u1(s) changes sign along p0(s°, L), there is a level curve in D2 of
(ui, = 0). By (8.23), y1(—eo) and cannot both lie in L). They also cannot both lie in Hence we may take y,(— oo) E X02 and oo) E
L). Let be defined by = y1(+oo). If u1(s) changes sign along Ct) br along L)], then we can construct a level curve Y2 of 0) lying in the region bounded by y, l'1 and gi0(s°, at) [or L)J
and a part of this curve,
y1(—oo) 6X02, EçL0(s°,&)
E L)}.
Again if u1(s) changes sign along one of the arcs gi0(s°, Po(°2' L) [or p2), L)J, then we can construct a level curve of = 0)
which starts on and ends on the corresponding arc, say at and so on. Since no two terminal points of curves can lie in the same flat interval, the
THE FREE BOUNDARY FOR THE ELASTIC-PlASTIC TORSION PROBLEM 217
process of constructing the curves must terminate, it follows that there exists a level curve, say with 00) E A02 and y1(+ 00) = such that
<a, s and u1(s) 0 on s*) 0. Consequently, u1(s) 0 on L) and u1(s) 0 on aj).
The argument following (8.24) now shows that 4,(x is monotone increasing for 0 <6 and monotone decreasing for 6 <x1 <b, where oo) = (6, 4,(6)). This completes the proof, in case k = 1, and incidentally shows that (since 8 = /1) 'y, must coincide with that is, s0 = a,.
Consider now the case of general k. Let N) be points of strict local maximum for 4i(x1). Construct level curves yJ of = 0} as in Lemma 8.10, which initiate at (fl,, (fi)); each yJ must terminate on at some point
and j I (otherwise, 0). Set = os), 7N+I = Po(°N' L) and denote by D1 the domain bounded by y,÷1, p0(a,,a,÷,) and where is the part of A0 from to
Along each arc N — 1) the sign of u,(s) cannot be fixed [otherwise, 4, would be monotone in (fly, Also, it cannot happen tliatthe sign changes precisely once from positive to negative, for this would imply that 4,(x1) is increasing in sonic interval (fl,, + and decreasing in the interval
+ ij, (by the proof in the special case k = I when applied to notice that = 0 on
It follows that u1(s) changes sign from negative to positive in each interval Finally, u1(s) cannot be 0 in the entire interval (0, [for
otherwise, 4,(x1) would be decreasing for 0 <x1 similarly, u1(s) cannot be 0 in the entire interval L). Thus the number of changes of sign, from positive to negative, of ui(s) along is at least k. This completes the proof.
Theorem 8.11. Theorem 8.5 remains valid even if asswnpéion (8.8) dropped.
Here, in counting the we do not include the possibly two reentrant loops which are connected to the endpoints of A.
For proof, see Problem 1. Consider the case where is a rectangle
(8.30) (—a<x1<a,—b<x2<b).
By Corollary 8.7 there are at most four plastic
= (—fi $, —a x1 —a +
(8.3 1) P3 = the reflection of P1 with the x1 -axis,
P4 = the reflection of P2 with respeét to the t2
218 VARiATIONAL INEQUAUTIES: ANALYSIS OF THE FREE BOUNDARY
Denote by 1 the bisector to ao at (—c, —b), that is,
1: x2 + a—b.
Denote by p(x) the reflection of a point x = (x,, x2) with respect to I where x2 x1 + a — b, and by the inverse reflection to p.
Theorem8.12. Ifb<a,thenp(P2)CP1.
Proof. Let
D = ((x1, x2); x2 <x1 + a — b, —a <x1 < —a + 2b),
D1=DflE.
Consider the function
w(x) = u(px) — u(x) mD,..
Since
in p(D,), = in D1,
we have
mD,.
The boundary of I), consists of five arcs z: 'y, lying on the line 1, lying on the line x2 —b, lying on ai',, y4 lying on OF4, and lying on the line
= —a + 2b. On )'j we have
w(x)=u(px)—u(x)=u(x) —u(x)=O.
°flY2,
w(x) = u(px) — u(x) = u(—a, — —b) = 0—0 = 0
w(x)u(px)—d(x)cO,
since u(x) d(x), d(px) d(x). Finally, on
w(x)u(px)—u(x)=O—u(x)iiiO.
Thus w(x) 0 on the boundary of D,. Applying the maximum principle, it
ThE flU BOUNDARY FOR ThE ILASflC-flASTIC TORSION PROBLEM 219
follows that
w(xI<O mD1.
Notice that for every point x° E with px° E
d(px°)=dist(px°,y) ythelinex2= —b;
hence d(px°) = d(x°). Since u(px°) = d(px°), it follows that
u(px°) d(x°).
Consequently,
w(x°) = u(px°) — u(x°) = d(x°) — u(x°).
Since w(x°) <0, we conclude that u(x°) > d(x°), which is impossible., This contradiction implies that there are no points x0 in D1 for which px° E aP2; that is, p(aP) is contained n F,. This complçtcs the proof.
Consider next the case where
(8.32) Qisatrianglewithsidesy1,y2,y3
and the length of y, by . By Corollary 8.7, the plastic set consists of three iocp F,; F, is based otZ (and may be the
1 to and denotecbypx the of x 1, when •z vanes in the half plane that contains DenOte p the inverse reflection to p.
Itieorem8.13. IfI121<lyiI,rhenp(?2)CP,.
Proof. Denote by I) triangle with sdes 72' y3, Iwhere C)'3 and Iciand set
p(D), D1 = D fl E. Since
I hence also p(D)J is containeda ft'The of the triaj,igies D, D at the point I fl are equal and hence small& than ./2. It follows that f
(8.33)
-
220 VARIA11ONAL JNEQUALITIES ANALYSIS OF ThE FREE BOUNDARY
in D1. By (8.33), if x E 8P3 fl then
w(x) u(p(x)) — d(x) d(px) — d(x) 0.
On the remainder part of we also have w(x) 0 (as in the proof of Theorem 8.12). Since 0 in D1, the maximum principle implies that w(x) <0 in D1. From this inequality we can now deduce, as in the proof of Theorem 8.12, that p*(P2) C p(P1).
The method of proving Theorems 8.12 and 8.13 is based on the principle of reflection. This principle can be applied also in other situations. We shall consider a case where the boundary of is not necessarily polygonal.
DefinitIon 8.2. •Let x° E a(l, x0 vertex, and denott by N0(x°) the line normal to at x0. Denote by a the reflection with respect to N0(x°). Suppose that
(8.34) offl n T.) c n T÷
where T_, are the two half-planes into which N0(x°) divides the plane. Then we say that x° has the reflection properly.
Theorem &14. Let x0 = f(s°) vertex, and suppose that
y° —f(s°) + r(s°)p0(s°)
s a point of the free boundwy, that is, f(s°) + ,(s°)p E P sf0 < p ifp=p0(s°)+efor anye>Oand smalL If has the
reflection property with containing the pointif(s), s > — :° small, then
(8.35) atss°. Proof. Consider the function
inE÷_Efla(Ofl T_). It satisfies
On
w(x)0 ifx€N(x0), ifxEo(8UflT_),
w(x) = u(a'x) — d(x) — d(x) 0
if x E na(tZ fl T_),
THE FREE BOUNDARY FOR THE FLASTIC-PIAS11C TORSION PROBLEM 221
Iwbere (8.34) was used in deriving the last inequality). Hence, by the maximum principle, w 0 in that is,
(8.36) u(x) if x E
Denote by d_ , d.,. the distance functions from the curves x f(s), s and x = f(s), x and let
(f(s°)+pr(s°),O<p<p0(s°)).
Then
d1—d_,gradd_=gradd÷=v(s°) on'y.
Introducing orthogonal coordinates (p. s) [where x = f(s) + r(s)pJ, we then have
a a—d.=O, ony.*
Also Isee (7.7), (7.8)),
82 82 82 82 on7.
It follows that
(8.37) ' (x,) — d_ (x.,) 0(E3)
ifx1licsonthófloimal$b'N(x°)atf(s°) + and
Xe€T+, dist(x1,y),
frow that (*.35) is not uue; that is, suppose that
(8.38)
we sh*jj derive a it follows E £, j., small £ >0; consequently,
u(x_,) d_(x_j.
222 VARJA11ONAL INEQUALII1F& ANALYSIS OF THE FREE BOUNDARY
boundary at a pointy, then
(u—d)00(u—d)Øf0,
It follows that
Ce2 (C>O).
Using (8.37) we deduce that
u(x,) <d_ (X_e) + 0(e3) — Ce2 = u(x_,) + 0(e3) — Ce2,
which contradicts (8.36) if e is small enough.
Denote by R1 the first quadrant of the plane. We shall assume that 0 satisfies the following conditions:
(8 39) is in U is symmetric with
respect to the x-axis and they-axis;
(8.40)
(841)
(8.42) — 3
is nondecreasing.
The condition (8.41) means that 0 is convex. The condition (8.42) means that the curvature at the points (x, f(x)) of oil R1 is nondecreasing as x increases.
Lemma 8.15. If U satisfies the conditions (8.39)—(8.42), then the reflection property is satisfied at every point of Oil.
For a proof of this geometric we refer the reader to reference 59.
Corollary 816. Let U be a domain satisfying (8.39)-(8.42). Then the free boundary consists of iwo curves:
(x, 1(x)) + p0(x)p.. (x), —a <x < a,
F_ : (x, —f(x)) + p0(x.)p_ (x), —a <x <a,
TIff FREE BOUNDARY FOR ThE flASflC-PLASTIC TORSION PROBLE?s1
where 0 a a, p0(—x) = p0(x), (x) is the inward normal to at (x, ±f(x)); the function p0(x) is monotone decreasing in xfor 0 < x <a.
Note that if a =0, then there are no free-boundary points, whereas if p0(a — 0) >0, then the two curves form one connected curve.
Example. If (1 is the interior of an ellipse,
2
a>b, a2 b2
then the conditions (8.39)-(8.42) are satisfied.
PROBLEMS
1. ProveTheorem8.l1. [Hint: Use Theorem 8.1 in order to extend Lemma 8.8.)
2. Extend Theorem 8.5 to the case where A is a circular arc {r = <9< 82) and is replaced by u5; note that =0 in E.
3. If r = !i(9) is the free boundary near. a reentrant corner, estimate the number of local maxima of h in terms of along all; see Problem 2.
4. If 0 is T-shaped with the x2-axis as the axis of symmetry, then there are no plastic components based on the open sides of ao which are parallel to the x,-axis and one of whqe çadpoints
5. Consider a pentagon all of whose vertices have the same size angle. Generalize Theorem 8.12 by proying thee If a plastic loop based on a side!, and if I!, 'jI , then p,1(P,) C Pt,, where P,jis a suitable reflection.
6.
7. Prove that E is contained in an of the ridge R, where
indicates that Theorem 8.5 gives a'tharp estimate for p
& If 0 is a rectangle, then for all p sufficiently large E contains c/p neighborhood of the ridge (c > 0).
9. With the notation (8.30), (8.31), prove that (a) 4i(x1) has precisely one point inflection in the interval 0 <x1 <a; (b) —0) =0. IHinz:UsethemethodofSec*jon7j .
VARIATIONAL ANALYSIS OF THE FREE BOUNDARY
9. THE FREE BOUNDARY FOR THE STEFAN PROBLEM
in this section we use the notation
Hu = u
u satisfying Ru = 0 is called caloric; if it satisfies Hu 0 0) in the sense of distributions then it is called subcaloric (supercaloric).
In Chapter 1, Section 9, we studied the Stefan problem. We recall that the conditions on the domain are stated in the second paragraph of that section, and the conditions on the data are given there in (9.9). In this section we shall require precisely the same conditions.
In Chapter 1, Section 9, it was proved that
(9.1) N(0) cN(t) if 1>0,
where
N(t)=
and that
(9.2)
(9.3)
We recall that the proof of (9.2) utilizes the penalized problem (912) of Chapter 1. Differentiating the parabolic equation with Tesj)ect tot we obtained the equation
8: 1 —
from which we deduced, by the maximum principle, thM au/a: 0; (9.2) then follows by taking e 0. Notice also that
and consequently
(9.4)
that is, the temperature u is subcaloric. in this section we shall prove that u is continuous and then extend the
results of Sections to 5 to the present parabolic case.
THE FREE BOUNDARY FOR THE PROBLEM
We introduce the parabolic distance
d((x,t),(x0,t0)) = +
and the moduli of continuity
(95) (C>O,e>O),
(9.6) (C>O,y>O).
Theorem 9.1
(1) The temperature u, is continuous in U X [0, TJ;
(ii) for any compact set K in U X (O,T), u, is uniformly continuous with modulus of continuity w(r), for any 0 <e < 2/(n — 2) if n 3, and
ifn=2.
In the modulus of continuity, r is understood to be the parabolic distance. To prove the theorem; we need several lemmas.
Lemma 9.2. Let w be a bounded measurable subeoioric function in a cylinder D X (0, T). Then there exists a/unction such that (1) iT.' = w a.e.in D X (0, T), (ii) is upper semicontinuous in D X (0, T), (iii) for any ball K, K C D, if z satisfies
inK X(t0,t1)(O<t0<11<T), = on the parabolic boundary of K X ('a, 11),
then z in K X (t0, ti).
This is a fairly standard result: Ptir proof we refer to reference 58c. From now on we understand by u the version for which the properties (1) to (iii) are satisfied. —
We extend h by 0 into U\G.
Lemma 93. u is conithuow a: I =0.
Proof. We claim that
(9 7) N(t) is contained in a 8(t)-neighborhood
ofG,where8(t)—.OiIt--0.
Indeed, otherwise there are points (xm, t"') such that u(xm, t,,, 0,
226 VARIATIONAL ANALYSIS OF THE FREE BOUNDARY
dist(xm, G) 2c >0. Consider the function
Vm(X, t) = u(x, t) — — 12 + (jm t))
in the set R of points (x,t) such that u(x,t) >0 and
i O<,<tm.
Clearly, Hvtm 0 in R, and om(xm, i") >0. By the maximum principle, v must take a positive maximum on the parabolic boundary of R, say at (xr, t"). Thus
(9.8) u(xr, sr) — xrp2 + (im —
Since 'r =0 is inipossible and since (4', ti") cannot lie on the free boundary, we must have Xm —4'
I c. Hence (9.8) gives
dist(x.,m,G)>c,
Since u, 0,
(9.9) u(x,t)>0 ifxEG.
From (9.7), (9.9) and the boundedness of u, we deduce that
(9.10) —h(x)(dx -'0 itt-. 0;
hçre we used the fact that. F, is in C'. We now proceed to prove the continuity of u, at a point (y,O); it suffices to
takeyin F1. L4Kbea S llballwithcentery, and let be the solution of
inKX(c,l),
w' = a, on the parabolic boundary of K X (e, 1).
By (9.10),
w'(x,:)-.w°(xt) ife-.O.
From Lemma 9.2 we have, for any e >0,
u(x, t) w(x, i).
THE FREE BOUNDARY FOR THE STEFAN PROBLEM 227
Hence
(9.12) u,(x, t) w°(x,t).
Since w°(x, 0) = h(x) is continuous at y, w°(x,t) h(y) = 0 if x —' y, I 0. Recalling that and using (9.12), we fmd that if x—y, t—.o.
We shall consider now the case n 3.
Lemma 9.4. Let 0<9< and let f,(x, I) be a continuous and nonnegative function defined for 1,0 t A, and satisfring
iflxl=1,0<t<X,
in the distribution sense, in (Ixf< 1,O<:<X), where 0<X0<X. l,zfX0ss4lcientylasp(Indepen&ntlyof9,X),
(9.13) forkl<fl where C is a positive constant depending only on g9, Ao,
Proof. Letgbethe.olutionof
O<t<oo,
g0
and
(9.14) i) P0< 1,
provided that is sufficiently Applying the maximum princ* to the function
I
I — •2 — 1
in the domain 8<1 x < 1, 0< t <A0/2, we find that £ 0 in this domain.
VARL411ONAL INEQUALIIIES ANALYSIS OF 11ff FREE BOUNDARY
Using (9.14), we conclude that
(C>O)
if Ix < where < P0 1. We now represent f9(x, t) in x (X0/2) t A, by Green's function. Using the last inequality, (9.13) follows (with a
different C).
Proof of Theorem 9.1 for n 3. Let (x°, t°) to be a point in £1 X (0, T) such that u(x°, t°) = 0. We shall prove the continuity of u, at this point. For simplicity we make a translation of coordinates so that x° = o, t0 = 0; thus u is a solution of the variational inequality in a neighborhood V of the origin.
For any positive integer k, k > k0, let
Fk
where A0 is a sufficiently large constant to be determined later, and k0 is sufficiently large so that c V. In the sequel we denote by C, C1 generic constants independent of k.
Set
= supu,,
max {mk, Ck_'),
where t is a positive constant to be determined later. We shall prove:
Lemma 9.5. For any 8 > e/2, there holds
(9.15) mk+,
Proof. Let
w*(x,t) = 4—2kf + h)dh.
Since is a convolution of the function with a positive kernel, it is also subcaioric.
Since u(0, I) 0, t) 0 if I 0, and since is a bounded function,
t + h)dh
THE FREE BOUNDARY FOR THE STEFAN PROBLEM
Hence
(9.16) w*(x,t) 8>0.
We introduce the coordinates
(9.17) = 4kx, e = 42k1
and the function
w(x', t') = w*(x, t).
1,
iflx'I<l, —2X0+1<t'<O,
ifIx'I=k8.
Let
(9.18)
vMck_28(wCk)
Then we can apply Lemma 9.4 to the function I — v/Mi with 9 = and thus obtain
— (C1 0),
if —X0 c.:' 0, providad that X0is suffiaently large. Thus
v(x', —
Hence, from (9.18),
w (Mk — c1c2)(i — + ck28
— +
VARIATIONAL INEQUALflIES: ANALYSIS OF THE FREE BOUNDARY
Choosing e < 28, we conclude that
(9.19) w(x', t') Mk(l — if (x' /3, —A0 z' 0.
Let
z(x', t') = u,(x, i)
and consider the function
h(x', :') = Mk — z(x', i') for x' + 0.
In view of (9.19),
fth(x'. s)dc — s) Mk — w(x', t')
Also h 0, — 0. Let be the caloric function defined in x' /3, —A0 + 1 <t' 0, having the same boundary values as h on the parabolic boundary. By Lemma 9.2, h Is. Representing h in terms of Green's function in Ix' —Xo + 1 z' 0, we then find that
h(x', :') (C1 > 0)
in a smaller region Po, —(A0/2) t' 0, where </1. Therefore, in this region
z(x', z') Mk —
Choosing Pb = (9.15) follows.
We now choose 6 so that 8(n — 2) < 1 [and then e < 2/(n — 2)] and proceed to prove by induction that
This inequality certainly holds for any given k1 k0 if is large enough (depending on k1). Assuming this inequality fork, we get, by definition of
Mk = C*k, provided that we take C* > C.
Next, by Lemma 9.5,
mk+1 C*k_r[1 —
C"(k +
THE FREE BOUNDARY FOR THE SThFAN PROBLEM
since.
'k8(fl_2)(k) k1 is determined so that this inequality holds if k k1; k1 is independent of C..
Having completed the inductive proof, we can now write that
l'his means that
(9.20) u,(x,t) +
provided that (x, t) is in a small neighborhood of (0,0) and : 0. This inequality shows that u, is continuous from below at the point :°) (0,0) of the free boundary, and 0 — 0) = 0, that is, if x t t t°, then u,(x, t) —' — 0) 0.
In order to prove continuity from above at any point (x°, of the free boundary, take for sunplicity (x°, t°) (0,0). Let be the sohition of
iflxl<p, O<:<p,
on Ix 1=
0<t<,iandonlxl<M, t0. Since u,(x, 0 —0) satisfies (9.20) with := 0, the function is continuous at
and, in. fad,
cllog(Ix)2 +
(as seen by representing by a fundamental solution). SInce u, is subcalóric, we have (by Lemma 9.2) that
U,
It follow3 that (9.20) holds for t 0. This completes the proof of (i) of Theorem 9.1.
To prove the assertion (iv, let P (x, 1), P = (1,:) be two paints in KX(8,TJ.Denotcbyd(P,P)theparabolicdistancefromPtoP.Denoteby d, the parabolic distance from P to the parabolic free boundary, that ía, to the part of the free boundary whose points (y, s) satisfy s the same way, and let -
232 VARIAI1ONAL IN'EQUALTI'IE& ANALYSIS OF THE FREE BOUNDARY
Since u, is a bounded solution of the heat equation outside the free boundary, the interior Schauder estimates give
(9.21) — u,(F) C(d(P, F))"2
if d(P, F) Suppose next that
(9.22) < d(P, F).
In view of (9.20),
(9.23) u(F) (9.24) u(P)
Suppose for definiteness that = Then
d;<d(P,F)"2
by (9.22), and
d(P, F) + d;c C(d(P, "I
Using this hi (9.23), (9.24), we find that
ju,(P) —
Recalling (9.21), the proof of the theorem is complete. [Notice that e can be taken to be any positive number smaller than 2/(n — 2).1
Proof of Theorem 9.1 in case n =2. We modify Lemma 9.4 taking
I g(x,0) = log1—1.
Instead of (9.13) we now get
f,(x,:)
We change (9.16) into
iflxP<k'k,.
THE FREE BOUNDARY FOR THE STEFAN PROBLEM
This enables us to prove the inequality
I C —
provided that
= max (mk, 2—2k' }
where is defined as before. We can next by induction, that for any 0<8 <'i,
provided that e < I — 8. Thus we obtain
where y can be taken to by any positive number The rest of the proof is the same as in the case n 3.
We proceed to study the free boundary. The following notation will be
N = the noncoincidence set,
N =N(i)= {x:(x,t) EN),
A the coincidence set,
A, =(x;(x, t) E A),
F = the free boundary.
The function u sjtisfles Hu = k in a neighborhccxl of bdvndaiy point (x0. (to > 0): for simplicity we takék I'.'
Since = I + u1 0. Lemma 3.1 gves
(9.25) iI(x,t)ENUI' then sup
Theorem 9.6. The Lebesgue measure of is equal to zero.•
The proof is similar to the proof in the case (Theorem 3.5); ii utilizes (9.25). .
We shall need the following Harnack inequality.
VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
Lemma 9.7. Let u be a positive caloric function in a cylinder = fl x <1,0 <r <y} and let(y,t)E 1y1 1 —6, t = 82, where0<6 Then
u(0, y)
c is a constant depending only on n, 'y.
For proof, see Problem 1. We return to the variational inequality for the Stefan problem.
Lemma 9.8. For any pure second spatial derivative uu,
(9.26) 1)]'
where (x, 1) E N, d(x, t) = parabolic distance to the parabolic part of the free boundary.
Proof. The proof is similar to the proof of Theorem 3.6. We denote by — the infimum of u,, in a cylinder
=
[we take (0,0) E F] and replace the ball 8,(x) in the proof of Theorem 3.6 by a cylinder about (x, t):
r) = X (: — Cs2, t);
the cylinder is contained in N and its parabolic boundary contains a free boundary pOint (Yo, t0). We now define Yi as before; that is, is obtained by going a distance ôs from Yo into the center x of B,(x) (notice that u(y1,
Nex,t we choose such that (85)2 and + (8s)2 [then u(y1, t,) C(6s)2).
Proceeding as in the proof of Theorem 3.6,, we derive
—C6'1'
for some 9 inside ...8)(X). We now apply Ilarnack's inequality (Lemma 9.7) and continue as in the proof of Theorem 3.6.
Let (x0, be a free-boundary point and define
I — B.(x0))
THE FREE BOUNDARY FOR THE STEFAN PROBLEM
We then have the analog of Theorem 3.10:
Theorem 9.9. Let (x0, t0) be a free-boundary point; t0 > 0. Then there exists a positive nondecreasing function a(r) (0 < r < r0) with a(0 +) 0 such that if, for some 0 <r <
(9.27) S,(A) >
then there exists a neighborhood V of(x0, such that V fl 1' can be represented in the form
with k E C, and all the second derivatives of u D,2u) are continuous in (N U F) (1 V.
From the results of Section 1 it then follows that k E and u is in U F) fl V).
Corollary 9.10. if
flB(xo)I (9.28) Jim sup >0,
then the assertions of Theorem 0
The rest of this section is devoted to the proof of Theorem 9.9. We begin with a simple observation regarding the rate of growth of N,:
(929) if C Bpo_Ch C A,+h2 provided that
C is a sufficiently large positive constant.
Indeed, if Yo E then, by (9.25),
sup •(c>0). B,(x,)
Using the boundedness of u,, we get
sup u(x,t,)>O
if C is largc enough, a contradiction. We uk it •
236 INEQLItLITIES: ANALYSIS OF THE FREE BOUNDARY
For each i we may consider u as a solution of the elliptic variational inequality
l+u,).
—1) 0.
Since u, is continuous, we can apply the results of Sections 4 and 5. [Note that the Holder continuity off is required in the proof of (3.8): in Sections 4 and 5 only the continuity off is required.] We conclude that there exists a system of coordinates (x and a neighborhood of (xi,. (0.0). such that
fl V can be represented in the form
(9.30) x,, = g(x1 .
with continuous, and continuous in U I',,) fl From (9.29) we then see that the condition (9.27) remains valid also at any
point (x1. E I' near (x0, ta). with a(r) replaced by ca(r) with sufficiently small constant c > 0. Thus for any 11 near to there is a system of coordinates and a representation (9.30) (with replaced by of fl V The property (9.29) shows that one can actually use the same system of coordinates (x1 for all near furthermore. g is in in 1. We summarize:
Lemma 9.11. There exists a system of coordinates (x1 x,,) and a neighbor- hood V of(x0, = (0.0) such that 1' fl V is given by
(9.31) x,, g(x1 t).
where is continuous in (x1,.. . ,x,_ i). uniformly wit/i i(espect to 1. g is C"2 ini,andNfl Visgivenby
finally, u is continuous in x, uniformly with respect to 1. in (N U F) fl V.
We shall henceforth assume (as we may) that
We shall write
x' = (x,,. . . y = x,,.
9.12. Let v be a continuous subcoloric funaum in. .a cylinder Z = 1, 0<t < y) and suppose that v M in Z, and v in a subcyhnder
THE FREE BOUNDARY FOR THE STEFAN PROBLEM 237
Z0 = (x€G0,O<t<y) with meas( and ,.=M/2,M>O. Then there exists a number A, 0 <A < 1, such that
v(O, y) AM;
A depends only Ofl 7, 0.
Indeed, if is caloric in Z and i5 = v on the parabolic boundary, then we represent by Green's function in a cylinder (I x < p, 0 <t < y) and choose p such that
p0<p<l,
where c0 and Po are positive numbers depending only on 0. Using Lemmas 9.11 and 9.12, we can now show that
(9.32) u,'cCd8(x,f,) forsome8>0,
where d(x, A) = dist(x, A). Indeed, the proof is similar to the proof of Lemma 5.3 with replaced by
u, and h(x) replaced by a caloric function h(x, 1); Lemma 9.12 is applied in order to get an upper bound for — h, smaller than 1.
Since g C'12 in 1, (9.32) gives
(9J3) u.(x. i) Cd8((x, t), F),
where d((x, I), F) is the parabolic distance from (x, 1) to the parabolic part of the free boundary.
From (9.33) and Schauder's interior estimates we get [see the argument following (9.21)],
( ) u, is HOlder continuous in (x, z) in
(N U F) ri V. with some exponent 6 > 0.
Since —s as (-c, 1) —. (x0, to), we have that in (N U F) fl V, if V is small enough; consequently,
.t)'— '1. 1) is small > 0.
.-The neat lemma will enable ps to estimate Dr D,u of u and in terms of &
Lenima 9.13. There exi :1 positive ecw,stwU.sIC,, C2 such that; 0
(9.36) u,(x', y, 1.) —C,u,(x',.y,t) + C2d4((x', y), I',)
inNfl
238 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
Proof. Introduce the cylinders
(9.37) Zpe(xi, ((x, s); I x — <p, t1 — <S <
for small p, e. For any x1 = (xc, E I', we compare in N fl the caloric functions u, and w = —Au,, — u + fix — x1 12 + (t1 — t)j/(2n + 1). It is easily seen that w > 0 on the parabolic boundary [except for (xi,
that A is large enough, and that Cw u, on the parabolic boundary provided that C is sufficiently large; here (9.35) has been used. Applying the maximum principle, we obtain
inN
Since (x1, can be any point of the free boundary, (9.36) follows.
Lemma 9. *. The derivatives D,2u are bounded in N fl V.
Proof. Consider mollifiers in space
(J,v)(x,i) !fp(r Y)V(yt)dy
where p E p p(x) 0 if I x > 1 and fp(x)dx = 1. We shall estimate Du by the Bernstein method. For this we consider the function
(9.38) z(x, i) t) I 12 + A(J,u,)2 (A > 0),
where in a neighborhood V of (x0, ta), 0 1, and = 1 in a smaller neighborhood V0 of (x0, t0). Denote by Df the set of all points (x, 1) N fl V such that dist(x, A,) > e. Setting 0 J,u, we compute, in
As — = 2 (4 + —
— + 2(A — vfl2 + —
>0
for 0 <s < 1, provided that A is large enough. Thus z cannot take a maximum in the interior.
From Lemma 9.13 it follows that if (x,t) dist(x, 1',) < 2€, then 0 u C. Hence 0 <0 Ce and also, as easily computed
I C
if(x, 1) E dist(x, r,) = €. It follows that z C on consequently,
TIlE FREE 801 INDARY SOR tilE S11 FAN PROBLEM
C in 1), and. in particular,
C.
raking z • 0. we obtain
C
in some V-neighborhood of Next we introduce a moflifkr J in time, denote it by and the function
(A >0)
in a set whose points t) arc characterized by: (v.1 — belongs to some N-neighhirhood of rd): here the function is similar to the function used in (9.38). hut it has a smaller support.
Setting U = J,u,. compute
-. = ÷ —
Using the relation ü, = and the boundedness of we find that — !, >0 if A is sulficienth large.
Ne'u. i yr) fl 1' for sômeO <y 2. then by Lemma 9.13. the boundedness of u,. and the fact that g e c,' we find that
.
It follows that U C on the part of aô, that lies near Hence! C on and, by. the maximumpriaciple. C in We conclude that,.
.
in 1) a some neighborhood of(x0. 10)taking, obtain the boundedness of u,,.t
9.15. If B,(x) C [(x. 1) near (x0. p small I. then.for a sunable positive constant C. B, C A,÷,, for all small h > 0: consequvnth'; Lipschi:: continuous in t.
The proof is the same as that of (9.29). but exploits the it,1,. Now that Lemma 9.14 has been proved, one can complete the, proof of
Theorem sóthe moditlcatiàns in the
240 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
this is outlined in Problems 7 to 10. Here we proceed with a method that is in the spirit of the analysis of Sections 4 and 5.
Lemma 9.16. For any pure second derivative u,1 (in space or time).
(9.39) u11(x,t) —Cd'(x, z) (for somee > 0,C>0)
in N fl V, where d(x, 1) is the parabolic distance to the parabolic par: of the free boundary.
Proof. Since r is Lipschitz (in V) and u11 is a bounded caloric function, the nontangential limits of u., on F exist a.e. and they form a functionf,, in
is the caloric measure (see reference 121). Let —/4 ess sup f,1 on F0, where F0 is a fixed open I'-neighborhood of (x0, t0), and let K be a compact subset of I's. Let D be an N-neighborhood of (x0. with fl F
For any small h > 0, let
D,= ((x,:);(x,:)ED,(x,t)+Ae,ED V0<A<h).
where ej the Unit vector in the direction 3. Consider the functions
F,,' = ((x :) + red) drds,
F2 = 2fhfh(( t) + red) drds. h 03
Then HF,: = 0 in I),, and F,,' —. v11 in P if h -. 0. We claim that
(9.40)
where the parabolic distance to the parabolic part of the boundary of K. Indeed, the proof is similar to the prooi of Lemma 5.3; it Uses Lemma 9.7 and the Lipschitz nature of g. Notice that the maximum principle (which is needed in the proof) is still valid when the boundary values are taken in the sense of nontangential limits; in fact, this follows from the unique representation of bounded caloric functions by means of Green's function (121J.
From (9.40) we deduce that
(9.41)
with another C. Let
a,D,,=
THE FREE BOUNDARY FOR THE STEFAN PROBLEM 241
If X h/2, then
J'J u((x, t) + hey) 0
and consequently,
If X < h/2, then Fh2 — 3M/4. Thus, in either case,
ona,DA
and, by the proof of (9.40) applied to F,,' +
inDh.
Taking h —' 0, we obtain
and, as (x, 1) tends to K,
—M —7M/8; hence M 0. Wecan now apply the proof of (9.40) once again to deduce (9.39).
Lemma 9.17. Let be any unit vector in the space of variables (x', t) =
(9.42) g,(O) g(O)
exists
and, for some £ > 0, C > 0 (independent
g( g4(o) —
Proof. Set •
= urn sup g(0)
242 VARIATIONAL INEQuALmES: ANALYSIS OF THE FREE BOUNDARY
If 0< A1 <A2/2. then
(9.44) +
Indeed, consider the ball I =
B = BC.A',..(xj2.
g is Lipschitz, if C/C" is large enough. then B C A. Consequently, if (9.44) is not valid for some large enough C, then we can apply the caloric analog of Lemma 5.5 to deduce the following: There is an interval analogous to contained in N, which either intersects B or passes above it; in either case we get a contradiction. Here we point out that Lemma 5.5 is based on Lemma 5.3 (whose caloric analog is Lemma 9.16) and it extends to the parabolic case with minor changes.
Having proved (9.44), we take a sequence A1 = A1, —. 0 such that
-
and obtain from (9.44) that
hrn ml A,
It follows that the limit (9.42) exists. Finally. (9.43) follows from (9.44) by taking A1 —. 0.
Denote by C(x0. the cone generated by all the tangents gq(O). which opens into N. Similarly, one defines the cone C(x, 1) for any (x. z) E F V.
Lemma 9.18. C(x, 1) is a convex cone; since g E the cone consists of two hvperp lanes.
Proof. It suffices to prove that C(x0, is convex. Introduce directions 'ii. in the (x', t) space and the tangent rays in the direction Let
(x1. g(x, 'i). t1) be a point of F. with distance A from (x0. t0), and lying within distance 0(A) of (as A JO). We apply Lemma 9.17 with (x0, replaced by (xc. g( xc. : and the direction from (x to (x2, ti). Denoting by (i,. a point in the interval from (x1, to (x2, which divides it by ratio 0: (1 — 0). (9.43) yields
i,) — >0.
ThE FREE BOUNDARY FOR THE PROBLEM
Taking A -' 0, we find that
+ (1 — C(x0, t0).
and the proof is complete.
Proof of Theorem 9.9. We begin by proving that the tangent cone C(x0, is a hyperplane:
(9.45) C(x0, — {y = + A0(x' — + A1(i —
We already know that C(x0, consists of two hyperplanes
!<lo
with b a. We have to show that a b. Recall that
vg(x, t) — ')I °(I x — I). (9.46)
I u11(x,, z) — t) x1 — x2 I)
where are any second-order spatial derivatives and o( A) 0 if A 0. Take three points on a vertical segment,
A =
g is Lipschitz, if = 0(a), then all the points lie in N. Since = 1, we have, using (9.46),
(9.47) Iu(A) — o(a)a2.
By Lemma 9.17 and the Lipschitz character of g,
+ = + ÷
(9.48) — P) to) — b/3 +
e,($) C19.
244 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
Notice next that
(9.49) go + 13). to + 13) = (cos 9)2
where 0 is the angle between the y-axis and the spatial normal to g at + 13), + 13). To estimate 0 we use the fact that g is increasing
in 1:
0 g(x', to + 13) g(x',
For x' — in the direction of ± + 13), we have
+ 13). (x' —
x p 1/2 we obtain, after using (9.46),
+ + Cpu2.
We now choose
(9.50) p
Then we obtain
+
It follows from (9.49) that
t0 + 13), + 13) I — +
Recalling (9.46), (9.48), we find that
u(B) — + e1(fl) + )2 _Ca2[a(Ma) + + ahI2(j4ja}
for some constant M > 0. Similarly,
u(C) — +(—bp + e2(fl) + a)2 _Ca2[,(Ma) + +
ThE FREE BOUNDARY FOR THE STEFAN PROBLEM
Thus
I u(B) + u(C) — 2u(A) (a — (9.51)
+ + e2(fl))a — —
On the other hand,
I (u(B) — u(A)) + (u(C) — u(A)) — Ji
where J1 = AB, J2 = AC. Since u, is bounded, we obtain
which contradicts (9.51), unless a = b. We have thus completed the proof of (9.45).
The convergence in (9.42) is uniform with respect to both the choice of the free-boundary point (x0, and the direction ip. In view of (9.45) we conclude thst'
g(x', :') — i0) (x' — + (:' —
—
where y(X) -.0 if A 0. It follows that g C' in (x', 1). We next prove that
(9.52)
for any second-order derivative (in space and time) such dia the direction I is tangenjal.tp the free boundary at 's).
We fcmn the
= f(uj((x, i) + he,) — t))/
inthëopenset
Dh={(x,:)ENflV,(x,t)+he,€NflV).
Since
if(x',i)
246 VARIATIONAL INEQuALmES: ANALYSIS OF THE BOUNDARY
Noting that u1(x', g(x'. 1). r) = 0 and that I D2u C, we get
Fh=E(8)
in a 8-neighborhood of (x0. intersected with where 0 if 6 0. We also have
in!),,.
We now represent F,, by means of Green's function (by reference 121, for instance) and conclude that
in a &'-neighborhood of (x0. intersected with D,,. where 8' is small enough. Taking Ii 0. we get
2E(&)
m this neighborhood. and (9.52) follows. It remains to prove that urn up,. exists. where i• is the normal to the free
boundary at (x0. 1(i). In view of (9.52). it suffices to prove this when i' is just one particular nontangential direction. But for s' in the direction of the y-axis thc liniit of u,,, exists.
PROBLEMS
I. Prove Lemma 9.7. [Hint: Representing u by Green's function s) of the cylinder fly < 1) with pole at (x, :), we have to (I) estimate G00 (v inward normal) from below and (a/av )G, ,from above on tbe lateral boundary
of Z.. and (ii) estimate G00 from below and GE from above on the top of With K(I x 1) = K(x. 1) the fundamental solution of the heat equation.
K(lxI,i) — K(l, Goo(IxI.t) (IxI< l,i < r1,)
since K,(l. t) > 0 if y < 2/n. Hence
e 1/4i,, (C >0).
. y) � C(e — e (C'> 0).
THE FREE BOUNDARY FOR THE STEFAN PROBLEM 247
To estimate s 1, let be the half-space tangent to along e1 and ,( x, s) the Green function for
Since -
a Cy1 Iei—y12s) (s — ,)(fl+2)/2exP — 4(s —
—yI>8,s—t<62.
On ar;, if y,, = dist(y, x,, dist(x,
6" 62 6fl+2
finally, if y 2/n, repeat the above several times.]
2. Suppose that F9 is the boundary of a ball B,0 and write Aw in polar coordinates,
+ r2Aw.
Assume that
>0 onF0X[0,T].
(r2h(x)),<O mo.
IThe last condition implies (since h > 0 in G) that U G with respect to any point near the origin.J Let v = r" 1u,. Prove that:
(a) — v = r"3(r'j), —
in the distribution sense n N.
(b) v<0 inN.
IHini: Take w and a test function = e°'max(w — M, 0), where M = max w.J; (c) N(i) is star-shaped with respect to any point near the origin; there-
fore, 1', is Lipschitz; (d) Nis givenlocallybyx1 = g(x1,. ..,x,_1, .. ,x,,, t)withg C°°.
24S VARIATIONAL INEQUALtJ1ES: ANALYSIS OF 11ff FREE BOUNDARY
3. The noncoincidence set for the Stefan problem is given by t <+(x), where is Lipschitz continuous. [Hint: Use the method of proof of Theorem 6.1 to deduce that Au, +
— u is positive if I and A is sufficiently large.)
4. Consider the one-dimensional Stef an problem:
ifO<x<s(t),t>O,
u(x,O)h(x) ifO<x<b[b>O, h(x)>OI,
u(s(t),t)0 ift>O,
ift>o,
where f(O) = h'(O), h(b) = 0, and!, h are continuous. Perform a transfor- mation ü(x, t) = i) d, to reduce it to a variational inequality with Neumann condition at, x = 0. Prove that there exists a unique solution with s'(t) > 0, s e
5. Suppose that in Problem 4
h'(x) <0, h"(x) > 0, is monotone increasing,
and! 0. Prove that s'(t) is strictly monotone increasing. [Hint: (a) Show that <0, u, > 0; (b) z satisfies a parabolic equation and does not take a local extremum on the free boundary; (c) consider the level curves (z = a); they are regular for a (by Said's lemma).J
6. Exter.d the result of Problem 5 to the case where f 0 but f'( t) <0. [hint: Show that z cannot take minimum on x = 0.]
7. Use Lemma 9.14 to deduce that I', E C1 +A for any 0 < A < 1. [Hint: Let = Xe — x0 = (x01,. . . E 1',, v(s) x,E,, + u(x, 1). Then
F(D2v) — — +
v = 0 on = 0. where 1= 1 + u, E C°'. The function w = < n) satisfies an elliptic equation in U, w 0 on
TILE FREE BOUNDARY FOR THE STEFAN PROBLEM 249
= 0; apply LI' elliptic estimates to deduce that v E Finally, is given by = vt.I
S. Let G be a domain in and S a open hypersurface of If
inG,wEC2(G)flC°(G),
wg onS, wheref E g E then w E C' U 5). [Hint: Suppose that! = 0: Set = xe (a <n), y,, = x,, — 4(x'), where S is given by x, = Then z(y) = w —g satisfies
L2=
The proof of Lemma 4.2 of Chapter 1 gives z C" for some 0 <p <1 and Theorem 9.2 of reference 3a gives z E C1 4A•j
9. By Problems 7 and 8, u E U 1) fl U), U a neighborhood of x0, with HOlder coefficients for which are independent oft. Use this, the continuity of g(x', :), and the monotonicity of g in t to prove that are continuous in 1.
10. Prove that under the setting of Problem 9, r is in C' in a neighborhood of (x0, to). (Hint: Take finite differences in the relation g(x', t), 1) = 0.)
11. Show that, in general, the temperature in the Stefan problem is not HOlder continuous if n 3. [Hint: Consider a radially symmetric solution u(x, 1) = ñ(r, 1) (r x I) such that (0, 1) is a free-boundary point and
N {(x,:);s>.(r)), ((x,t);rrzn(r)), 1.
By nondegeneracy, ü(r, I) cr2 (c > 0). If u is Holder continuous, then, for some e>0,
ifl—6<i<lifS=0(r).
Hence t7(r, :) >0 if 8 = c0r2 _e (c0.> and thus A C ((x,i); 0< 1 <Cer2*). a4arrier for N at I), we get
a barrier Lor the heat equation in N does exist at (0,1) when A in the Dvoretiky khd r821 and [ 165f for I proof (another proof should from a recent Wiener criterion of Evans and Gariepy [86]).
250 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
10. STABiLITY OF FREE BOUNDARIES
In this section we consider both elliptic and parabolic variational inequalities and establish an estimate on the measure of the set (u > e) (when the obstacle is 4) 0). As an application we shall prove (a) that the free boundary has finite Hausdorff measure and (b) if the free boundary is smooth for a solution u of a variational inequality, then it is also smooth for any other solution u with ü — U small enough.
For simplicity we take the elliptic operator to be the Laplacian. Thus we consider
(10.1) a.e.in(2 (—Au +f)u = 0
where (2 is a bounded domain in We assume that
(10.2) je u (10.3) in&L
Set
and as usual denote by N, A, r the noncoincidence set, the coincidence set, and the free boundary, respectively.
For any set £ C Rm and e> 0, we write
E(,) = (x ERm;dist(x, E) <E),
{x E E;dist(x, Rm\E)
Let
(10.4) O.=(xEN;Ivu(x)I<e).
We denote by the Lebesgue measure of a set K in and by the surface area of the boundary 8K of K.
The following lemma is fUndamental for all the results of this section.
Lemma 10.1. For any compact domain K C with Ct boundary 5,
n K) Ce(t(K) + A(S))
for all e > 0; C depends only on M, F, A, and K.
STABILITY OF FREE BOUNDARIES 251
Proof. Let 0 = (x E N;ID1uj< and set
—E
h' D1u if—e<D1u<e E
Then
'K VD1udx f,
where is the derivative along the outward normal. Hence
f I VD1uI2 dx Ih'D.fIdx +fJhD,D1uIdSK S Ce(ii(K) + A(s)).
Summing over I and using the fact that 0 3 Oe for each i, we get
f I + A(s)).O,flK Since finally
mN
(10.5) follows.
Corollary 10.2. Let K be any domain in with C' boundary S. Then
(10.6) gL((O <U <e2) fl K) + A(S)).
Indeed, by Lemma 3.2,
for constant C, and Lemma 10.1 yields (10.6).
Corollary 10.3. Ifx E I' fl 1l(...a)(6 >0), then,for all e >0,
(10.7)
where C and y are positive constants depending only on M, F, A, 6.
252 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
Proof. By Lemma 3.1, there exists a pointy such
(c1>0).
By Lemma 3.2, I and therefore there exists a ball (c > 0) in which u > 0. also u < c4e2 in this ball, we have
I Vu Ce (by
Lemma 3.2) and thus the ball is contained in and (10.7) follows.
Theorem 10.4. If K is any domain with C' boundary S such that K U S C then
(10.8) fl K) + A(s)).
Here H,, —
is the (n — 1)-dimensional Hausdorff measure. We recall that, by definition, for any set E fl
(10.9) Hfllk(E) = urn
where (Be) is any set of the open balls B1 with centers in E and radius e which cover the set E. One may the coverings (B) to be such each point of E is contained only in a finite number N of the balls; N is independent of the Cbvering and of e.
Proof. Take a covering of r n K( by balls B1 with centers in r n K(..,) and radius e, having at most N overlappings at each point. By Corollary 10.3,
n fl K).
Estimating the right-hand side by Lemma 10.1, we get
for any K0 C K(..5) foc S > 0; C is independent of 8. Taking K0 t K, (10.8) follows.
In the next theorem we deal with two solutions u1, u2 of variational
—Au, +f (10.10)
STABiLITY OF FREE BOUNDARIES 253
where
E u, e (10.11)
intl.
We define
N( = the noncoincidence set for u,,
A( u,) = the coincidence set for u,.
u.) = the free boundary for u.
Theorem 10.5. Let (10.10), (10.11) hold. if
— <E,
then (I)
(10.12) (C>0)
where C depends only on the C°' norms of the J, the C'' norm of u. and A. and (ii)
(10.13) (A(u2))(_c0,) C A(u,) C {u2 <e2) (C0 >0)
where C0 depends only on the C'' norm of the u1 and on A.
Proof. Since A(u,) C {u2 <e2), Corollary 10.2 shows that
and (10.12) follows. To prove the first part of (10.13), notice that if x EN(u1), then, by Lemma 3.1, -
sup Dx)
if C0 is large enough, and thtn
sqp u2 > 0. 8c,Ax)
which implies that x
254 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
Theorem 10.6. Let (10.10), (10.11) hold and let K', K be domains with C' boundary such that K' C K, K C 11, c0 = dist(K', K
fl is a surface provided that
(10.14) lu, — <
eo is sufficiently small, depending on the C°" norms of the the C'' nonns of the u, A, c0, anda C' bound on 1'(u,) fl K.
Proof. Let x0 E r(u2) fl K. By (10.12), if e2 = e0 is small enough, then the condition
MD(A(u2) 11 B,(x0)) > o(r)
of Theorem 3.10 is satisfied. Hence r(u2) is C' in a neighborhood of x0.
By Theorem 1.1, if f, E (0< a < 1) for some integer m 1, then Theorem 10.6 yields smoothness of the free boundaries.
We shall next extend the preceding results to the Stefan problem. We recall that
(10.15) 0 0
where is some neighborhood of the free boundary, and
(10:16)
(10.17)
It follows that for.each t > 0,
(10.18)
Defining
N,=(x;(x,t)EN),
and similarly A,, F,, the proof of the Lemma 10.1 extends to yield
(10.19) fl K) Ce(p(K) + 4(s)),
STABilITY OF FREE BOUNDARIES 255
where K is any domain in R" contained in a small neighborhood of a point x; x€1',.
Similarly,
,z({x; 0 < u(x,t) <e2) fl K) + A(S)),
K) C(p(K) + A(K)).
Next
Ce
if u, are two solutions of the Stefan problem with
<i2
and is a domain, C (x; (x, I) E 4A,Jso,
(A,(u2))(....c.) C A,(u1), for spme C 0.
Finally:
Theoi'ej,, UL7. Let u1, u2 solutions of the Sl4fan probkm and suppose that tlsefreeboundaryofu1 is given in some cyltnderD(in (1>0)) by
g1
Let 8 be any small positive *wnkr. if
1u1 —
where a0 is 4#
2
• = -s .
Here = ((x,t); dist ((x,t), 1)) <8), and is
Indeed, the proof is similar to the proof of Theorem 10.6 and makes use of Theorem 9.9.
I
256 VARIATIONAL INEQUALiTiES: ANALYSiS OF THE FREE BOUN1MRV
PROBLEMS
I. Jr
u, (—au, )u, = 0
a.e. in 12, = 0 on where ao E E then for any E > 0. U1 — U2 — is sufficiently small.
2. If the free boundary of in Theorem 10.6 is locally (say in K) given by x,, f1(x1 x,,_1).fE C'. then the free boundary of u2 is given [in any subset K' of K. c0 dist ( K'. > 0J by x, f2( x1,. . . , x,, - where f,EC'and
C depends on the same constants on which e0 depends.
3. Theorem 10.6 and the preceding results are false if the nondegeneracy condition f A is not satisfied. Check this in case n I with
u"(u—+)=Oin—I<x<l,
—l<e<l. 4. Extend Theorem 10.6 and all the preceding results to the elliptic operator
2
4u +
(hlnirn: For Lemma 10.1, integrate
fhD1f by parts.j
II. FREE' BOUNDARIES WITH SINGUlARITIES
Let 12 be a convex domain in R" with C2 boundary, symmetric with respect to = 0. We identify with the hyperplane ((x',O); x' E R" '), where
in fl fisuch that
ECF. We write these sets as disjoint unions their open and closed components.
FREE BOUNDARIES WITH SINGULARITIES
respectively,
(11.1) UE), uF1
and assume that
Vi.
Let be a C2 function in <0 on and denote by u the solution of the obstacle problem
(11.2) VvEK;uEK,
where K = (v E o 4) a.e.}. We denote by A the coincidence set and by A the interior of A. In the next theorem we establish for some 4) a disjoint decomposition
A = U closed components, (11.3)
A = A01 open components,
such that
(11.4)
F be as above. There exists a s.rI city superharmonic function 4) (that is, iX4 <0) in with 4) <0 on all; such that for the solution u there holds the decomposition (11.3) with the relations (11.4).
It follows that whenever 0) is a point of accumulation of components of the free boundary has nfinite number of components in any neighborhood
of It is interesting to compare this situation with the positive resultJorn = 2
obtained by reference 138a (see also 6) in case 4) is a StrictlY concave function: N is homeomorphic to an' annulus and, if 4) is analytic, r a Jordan curve having analytic parametrization.
Proof. For any 'open set 0 in Rm theze exists a nonnegative function a(x) such that 0 = (x E Rm; a(x) > 0). Indeed, if B1 = Br,(Xj) form a covering of 0by balls contained in 0, with 4, >0 if
258 VARIATIONAL INEQUALflIES: ANALYSIS OF THE FREE BOUNDiSRY
Take 0 — E C and construct a(x') as above. Let
= ((x', a(x')); x' E (2 fl
= ((x', —a(x')); x' E (2 fl
and split (2 into three sets
(2÷ : >
A: —a(x') x,, a(x').
Notice that E C int(A) fl R"'. Consider the Cauchy problem
(11.5) 1 + h
av (11.6) v=O, j—=OonF÷.
Lemma 11.2. There exist functions v, h in )for which (11.5), (11.6) hold and h vanishes on 1' + with all its derivatives; further,
(11.7) v>O,
Proof. Use thç relation
infl÷
and (11.6) in order to compute formall?l the derivatives
vk(x)—--—-v onr÷
(assuinng that v E Cx), ana thendefine
(11.8) = — a(x')))
here is a cutoff function in one variable, I in a neighborhood of the origin If we choose, for instance,
ck—2+
FREE BOUNDARIES WFFH SINGULARITIES
then the series in (11.8) is uniformly convergent in 0÷ together with each of its derivatives, and, by the definition of the Vk,
— 1 and all its derivatives (11.9)
convergetozeroasdist(x, -.0.
We take c2 = 0 and consider the fu*t term in (11.8)
T 41)2(x)(xfl — a(x)).
One easily computes that
v2(l and thus Tis positive in 0÷ if necessary we can make the remaInder of the series
a 2(i*JVa12)
we see that ttT — 1 can be made a small11f a(x') is replaced at the beginning by ea(x') with e small 4 in By increasing the Ck, if necessary, we get >0 In Defining h = I — and recalling (11.9), the proof of the lemma is complete.
We now extend the definition of v into 0 by
v(x) = fv(x', ifx E 0_ ifxEA
and let
inD4ULl_ =1 mA.
Thenf E 0 in Let be the solution of
A'4'=—f mO, 4—v on8Q.
Thus u is a solution of the variational inequality with obstacle 4', and the first relation in (11.4) holds.
To satisfy the second relation in (11.4) we change 4.' into 4)(x) = 4'(x) — f3(x'), where /3 E /3 0, and
F = (x'; fl(x') = 0).
[/3 is constructed in the same way as a, with E replaced by fl 12)\F]. The sets {u = 4'), (u = 4)) can differ only on the set F of measure zero. Hence u is the solution of the obstacle problem corresponding to the obstacle 4). Since 4, = —v on aiz, 4.' <0 on except possibly on fl and (since
/3(x')<Oonthis set; Next, < — e < 0 for some e, and by replacing $(x') by 8/3(f) with 6
small enough, we can achieve <0. Finally, since <4' on fl ')\F, the second condition in (11.4) is also satistied; this completes the proof.
We shall next give examples of analytic superharmonic obstacles for which the free boundary has a cusp.
Take any analytic curve F with a cusp at one point A, having the figure "oo" as in 1lgure 2.5 (such a curve is given in Problem 2), and denote by A the closed set enclosed by F, and by V some neighborhood of F.
Consider the Cauchy problem:
inN, - auu0, along F.
By the Cauchy—Kowaiewsky theorem this problem has a unique analytic
FiGURE 2.5
VARIATIONAL INEQUALITiES: ANALYSIS OF THE FREE BOUNDARY
Then u = v + 4, satisfies
— 0
u0 u=4,, VuV4' onaA,
FREE BOUNDARIES WITH sINGULARmES
solution ü in N U 1' provided that the neighborhood V is small enough. Since
a2u—=1>0 onF, 2
ü > 0 in N if V is further decreased. Extending ü into A by zero and setting
lxi? 1x12 Uu__r.
we conclude:
Theorem 11.3. The function u is a solution in f2 of the obstacle problem
with strictly concave analytic obstacle 4). and the free boundary 1' has a cusp at A.
One can also construct examples in which the domain N is not restricted to belong to a sufficiently small neighborhood of r; this is done in the following problems.
PROBLEMS
1. Let a domain in R2, A a closed subset by a piecewise C1 curve F, N = \ A. Let f(z) be holomorphic function in N. continuous in N U F, with f— — on F, where 4) E C'(F). Then the function u(z) = Re 11(z) dz is a single-valued harmonic function in N, and u = 4), Vu V4) on F.
2. Let
Then o(e'°) = cos 0 + eli e21 + I and traces a curve F of shape "oo" as 0 0 < 2,r, with a cusp corresponding to 0 = ir/2. Show that for small e>0 theannulusA (I <izI<2) ismappedbyoina 1-1 wayontoa set N; the inner boundary of N is 1' and the outer boundary is a(I z = 2). Let A = the set enclosed by r, Il = N U A. This notation will be used in the next two problems.
3.
f(o(z)) —o(z)onizl l,oryetif
1(z) = —4)
262 VARIATIONAL INEQUALmES: ANALYSIS OF THE FREE BOUNDARY
[a(z) = o( 1), i = I /z on z I). The right-hand side is holomorphic in N. If u, defined by Problem 1, satisfies
(*) u>4)
then check that u solves the obstacle problem in with obstacle 4), and I' is the free boundary.
4. Prove (*)
[Hint: Since
f f(z)dz—_ff(a(z))ci'(z)dz,0(y) y (*) reduces to
if 1 <J z 2, with any I z0 = I. Write a a + ea.; then a (1/z) = ±a(z). If
W = a + eb + e2c,
then
±2,b>O,sothat W>OonIzl=2. Wcannot take minimum in (1 <121<2) since
5. Take P( z) = (z — 2 + z 1)2 in Problems 2 to 4 in order to produce an example (with the same 4)) of a free boundary r given by
cos9 + elf e21 — 1 f4sinO
it has a cusp at a(1); see Figure 2.6.
2.6
RIBLIOGRAPHICAL REMARKS
12. BIBLIOGRAPHICAL REMARKS
The hodograph—Legendre transformation and the treatment of Section I are due to Isakov [189a,b) and to Kinderlehrer and Nirenberg [123aJ. Schaeffer [158b] has given the method in Problem 3 of Section 1. The hodograph— Legendre transformations have also been applied to elliptic systems of equa- tions and to minimal surfaces by Kinderlehrer, Nirenberg, and Spruck (1 24a, b].
Theorem 2.1 is due to Lewy and Stampacchia [138aJ. Theorem 2.4 was established by Caffarelli and Riviere [63a]. All the other results of Section 2 are due to Kinderlehrer [1 22bJ.
The results of Sections 3 to 5 are due to Caffarelli [56b, eJ; see also (56aj. The original proof of Theorem 3.10 appears in reference 56b; the proof presented in Sections 4 and 5 is taken from reference 56e.
Theorem 6.1 is due to Alt [Sal. The three-dimensional darn problem was first introduced by Stampacchia [l6ThJ and Gilardi [106a). Theorem 6.5 was estab- lished by Friedman and Jensen [95b]; the results of Problem 3 of Section 6 are also taken from reference 95b. The results of Problem 2 are due to Friedman [94i] for m = 1 and to Jensen [ll6aJ for m I. Another proof of Theorem 6.5 was given by Cryer [75J; the convexity of the free boundary for another filtration problem was established by Boieri and Gastaldi [421. The filtration problem with several layers was studied by Baiocchi, Comincioli, Magenes, and Pozzi (21J, Baiocchi (19aJ, Baiocchi and Friedman [22J, and Benci 1281, and by Caffarelli and Friedman [58a, b). In reference 58a and b, the shape of the free boundary is studied and the asymptotic behavior is derived when one of the layers becomes very thin; the limit problem is a free-boundary problem with mixed Dirichiet and Neumann conditions.
The darn problem with time-dependent boundary conditions (also for slightly compressible fluids) was treated by Friedman and Jensen [95a, C], Friedman and Torch [99), and Torelli (174a—c].
The Baiocchi transformation (Section 5 of Chapter 1) can be extended to establish existence for some nonrectangülar dams in two see Baiocchi (19b). The filtration problem for general dams with any number of reservoirs and in any number of dimensions will be studied in Chapter 5; the methods will be entirely different from the methods developed for the rectan- gular dam. We shall prove existence, uniqueness, and analyticity of the free boundary. For the corresponding time-dependent problem Gilardi (106bJ established the existence of a "weak" solution.
The regularity Of the free boundary for the elastic—plastic torsion, problem was proved by Caffarehi and Riviete [63c]; LemmA 7.3 is taken frol!1 refeence 63a. Theorem 7.7 is due to Caffarelli and Friedman [58f1; related ràultS were proved by Ting [172a, bJ. Problems 1 to 5 of Section 7 are taken fromreference 58f; related work is due to Chipot [71b).
Theorems 8.1, 8.3, and 8.4 are due to Caffarelli and Friedman [58fj. Theorems 8.5 and- 8.11 were established by Friedman and Pozzi (981, and the remaining part of Section 8 is taken from Caffarelli, Friedman, and Pozzi [59).
264 VARIATIONAL INEQUALITIES: ANALYSIS OF THE FREE BOUNDARY
Earlier work for regular polygonal domains was done by 'fing II 72a, b]. Elastic—plastic problems with several materials werc considered by Caffarelli
and Friedman (58n]; see also Brezis, Caffarelli, and Friedman [46]. Problems of unloading have been studied by Caffarelli and Friedman [58q1 and by Ting [172c, dJ.
Theorem 9.1 is due to Cafarelli and Friedman [58c]. Theorems 9.6, 9.8, and 9.9 are taken from Caffarelli [56bJ. In Lemma 9.14 we have replaced the original proof by a proof given by Kinderlehrer and Nirenberg [1 23b]; the completion of the proof of Theorem 9.9 as outlined in Problems 7 to 10 of Section 9 is given in reference 123b.
For one-dimensional parabolic free-boundary problems, Friedman (94h] proved that the free boundary is analytic if the boundary conditions are analytic. Another proof was given by Kinderlehrer and Nirenberg [123c].
In Section 9, Problem 2 is based on Friedman and Kinderlehrer [97], Problems 4 and 5 are based on Friedman and Jensen [95b], Problem 6 is taken from Friedman [94j], and Problem ii is due to Caffarelli (oral communication). Some results on the free boundary in two space dimensions are given in Caftarelli (56cJ.
Theorem 10.6 and all the preceding developments of Section 10 are due to Caffarelli (56fl. Schaeffer [158a1 had previously proved the weaker result: If r(u1) is and u1 — u2 is small in sense, then r(142) is His method uses the Nash— Moser implicit function theorem, and was extended to para- bolic variational inequalities by Hanzawa [111).
The examples in Section 11 (including the problems) are due to Schaeffer [158c]. Positive results torn = 2 were proved by Lewy and Stampacchia [138aJ for — and by Kinderlehrer [1 22a] for minimal surfaces. Caffarelli and Riviere (63bJ studied the nature of the singularities of free boundaries for n = 2. A two-dimensional free boundary problem was studied by Kinderlehrer and Stampacchia [1 26a].
Variational inequalities have been used in problems of flow past a profile: see Brezis and Stampacchia 153b, c], Brezis and Duvaut [190], Hummel (115], Shimborsky (162a, b]. and Tomarelli [173]; see also Elliott and Janos'sky (83]. They alto appear in problems of lubrication; see Capriz and Cimatti [651 and Cimatti (72J.
In stochastic control, problems of optimal stopping time can be reduced to variational inequalities; see Bensoussan and Lions [34], Friedman [94gJ. Fi- nally, problems with partial observations, such as some problems in sequential analysis, can also be reduced to variational inequalities and give rise to questions regarding qualitative behavior of the free boundary; see Caffarelli and Friedman (581, oJ and Friedman [94j, kJ.
3 JETS AND CAVITIES
If we minimize the functional
=fi vv12 +
over the class of functions v E K, where
K=
where it° E I1'(Il), u0 0, then the mil3imizer it isa solution of the inequality
In this chapter we rniiiimize
(1>0)
over K, wherefA = function of a set A. Notice that 1(o) is no longer continuous, although it is lower semicontinuous.
Such functionals (with f = const.) give rise to free-boundary problems
in(u>0),
it = = const. on the free boundary (1 fl a{u >0),
which arise in flow problems of jets and cavities. We shalldevelop a general theory for the minimizers el and then proceed to solve problems of jets cavities
266 JETS AND CAVITIES
1. EXAMPLES OF JETS AND CAVITIES
in this chapter we consider only fluids that are incompressible, irrotatonal, and inviscid. That means that the velocity vector is V4, where the velocity potential, satisfies = 0 in the fluid. If the flow is two-dimensional, then the harmonic conjugate is called a stream function, and if the flow is three- dimensional axially symmetric (with the x-axis as the xis of symmetry), then the function J' determined by
(1.1)
is called the stream function. In either case 4,,) gives the velocity in the fluid and the stream lines (i.e., the lines along which the tangent is in the direction of the velocity) are given by 4' = const.
We shall consider two situations: (a) there is a "cavity" in the fluid; (b) a part F of the fluid's boundary is surrounded by air.
In case (a), if the fluid is fast moving,, the cavity consists of a mixture of vapor and gas and the pressure p in the cavity is constant (we are dealing here with stationary problems only). By Bernoulli's law,
(1.2) p + V412 + gy = cOnst.
throughout the fluid, provided gravity is in the negative y-axis. Since the pressure is continuous across the fluid's boundary, p = const. on the boundary of the cavity and (1.2) gives
where F is the boundary of the cavity. Further, since 1' is a stream line, we also have
4' = const. on F.
In case (b), the pressure p outside the fluid is again constant, so that Bernoulli's law again gives (1.3); also (1.4) is valid.
Problem (a) is called a cavity problem and problem (b) is called ajet problem; there are of course also flow problems which involve both jets and cavities [as in Figure 3.l(c)J. In Figure 3.1, bold curves indicate fixed boundaries.
Unless otherwise explicitly stated, we do not include gravity in our discus- sion; that is, we take g = 0. (Problems with gravity are studied in Sections 18
ifn2,
Y symmetric with respect to the x-axis.
We shall not treat here three.dimensional flows that are not axially symmetric. We shall now compute explicitly some examples, using conformal mappings;
the methol (called the kodograph method) is restricted to two-dimensional flOWS.
We begin with the example of Kirchoff (1869) of a symmetric flow of unit velocity past a vertical flat plate with free stream line detachment at the ends (see Figure 3.2a). We introduce the velocity magnitude q = I V+ I' the com- plex velocity vector u + iv (u = v = the complex potentialf = 4 + i4s,
EXAMPLES OF JETS AND CAVITiES
Cavity
(b)
Jet
(a)
C—,'
(c)
FIGURE 3.1
and 19.) Thus
268
B
A
7
(a)
(c)
qe'0
and the hodograph variable
FIGURE 3.2
AND CAVITIES
(b)
—1
A B
(d)
B
A
df .
The images of the physical plane in the f-plane and in the w-plane are described in Figure 3.2b and c.
The idea now )s to parameterize both f and w by the same parameter /; t varies in the complex plane with the real interval (—1, 1) deleted (Figure 3.2d):
x I
EXAMPLES OF JE1S AND CAVITIES
where: = ± I correspond to the ends of the plate. Using the relation w df/dz it should then be possible to find a parameterization z = z(i) for the free stream line. Indeed,
(!_ 1)1/2)
where the branch of the square root is chosen so that w — 0 if I 0. Then
z 2ikf'(1 + — 12)dt
and k = 1/(4 + it) is the length of the plate. The upper free-stream line is. then given by
I —log(1 i)]. (1.5)
for 1.
We shall next use the hodograph method to study the jet problem with symmetric nozzle formed by two line segments which form an opening of size air (see Figure 3.3a). The pictures in the f and w planes are outlined in Figure 3.3b and c and in the i-plane (Figure 3.3d) we exclude the intervals (—00, — I);' (I, 00). Then
f — logs),
w f'(z) = — +
and the free stream lines are given by
h z(:)——-— I dr, O<:<l.vJI . 1-
Denote by the y-coOrdinaie of B and by thelhnitiág height of the upper free boundary as x eQ. Thefl the number
C
Yo
is called the contraction coefficient of the nozzle. One can easily compute from
270
N I
(1.6) that
/
(a)
I
JETS AND CAVITIES
4,
I
A
(5)
FIGURE 33
A
(4)
B
I
'IT
— 'ir + 2
if a = 2,
ifa 1.
The nozzle with a = 2 is called Borda 's mouthpiece; (1.7) was first established by Borda (1766) using physical considerations.
Later we shall establish existence theorems for jets and cavities and prove that have constant signs (for y >0) for some geometric shapes of the nozzles and the obstacles (for cavities); we shall also establish asymptotic behavior of at infinity. Using these facts one can rigorously justify the hodograph method for some specific geometries; in particular, (1.5) and (1.6) can be justified for our solutions.
B
(c)
THE VARIATIONAL PROBLEM
PROBLEMS
1. The drag D of the plate (in Figure 3.2) is calculated by integrating the pressure difference on both sides:
(1.8)
D = iil/(4 + ir) (the d1ag coefficient is CD = D/I — (4+,r)).
(b) Show that y cli along the upper free boundary as x oo (c constant).
2. (a) Verify (1.5), (1.6). (b) Prove that for general a E (0,2)
1 1. au-I 11 a\ 2 ——
u 2L 4; a
where is the iogarithmicderivative of the gamma function.
3. Prove in both the plane and axially symmetric flow that
(1.9) + xq = 0 inthe fluid,
where q 1 , r is the normal to the stream line, and Kis the curvature
of the stream line, taken positive when the stream line is convex to the region on the side of the poaitive normal is.
2. THE VARIATIONAL PROBLEM S
Let be a domain in R", not necessarily bounded, and assume that is locally piecewise C' graph. Let S be a nonempty open subset of an with
- piecewise C' boundary as. let is0 be a function satisfying:
u° E vu0 E is° 0.
We introduce a convex set
K= (v EL2(U),v u°onS)
272 JETS AND CAVITIES
and a functional
(2.1) J(v) =f[i vvf2 + IQ} dx,
where Q is a given nonnegative measurable function and 14 denotes the characteristic function of a set A.
Theorem 2.1. If J( u°) < oo, then there exists a u E K such that
J(u) = minJ(v). vEK
Proof. Since J is nonnegative, there exists a minimizing sequence that is,
UmEK, - v€K
Since (VU,,,) is bounded in and Urn — u0 = 0 on S, it follows that Urn — u° is bounded in L2(1l BR) for any ball BR with large R. Therefore, there exists a u in K such that, for a subsequence,
VU,,, Vu weakly in a.e.infl.
Moreover, there is a function y E L°°(a) such that
y weakly star in L°°(fl).
Now, for any large R,
f [i vul2 + ymin(Q, R)2]n BR
hn f >O}flhlfl(Q,R) rn—ca OflB* rn—co
Letting R oo and noting that = 1 a.e. in {u > 0), we conclude that
1(u) VuJ2 + yQ2) lim 1(14,,,).
THE VARIATIONAL PROBLEM 273
From now on we assume that
(2.2)
Definition 2.1. A function u E K is called a local minimum (or a local minimizer) for I if, for some small e > 0, J(u) J(v) for any v E K such that
II v( v — u)II L2(U) + '{v>O} — II L'(13) <
Lemma 2.2. If u is a local minimum, then u is subharmonic and hence, by Lem,na 1.10.1, there is a version for which
(2.3) u(x)limf u B,(x)
and u is upper semicontinuous.
In the sequel we always work with this version.
Proof. For a nonnegative E
Ye >0,
from which it follows that f vu 0, that is, u is subharmonic.
Lemma 2.3. I/u is a local minimum, then
0 U SUpU°.
Proof. Use with u,=u—e•nun(u,O) and u,u+e• mm (sup0u° — u, 0).
In Section 3 it will be proved that a local minimum is continuous (in fact, Lipschitz continuous), it follows that the set (u > 0) is open. We shall use this fact in establishing the next twp results.
Lemma 2.4. If u is a local minimum, then u is harmonic in the open set (u>0).
Proof. Use J(u + 1(u) with C000((u > 0)), e sufficiently small.
274 JETS AND CAVITtES
Theorem 2.5. Let u be a local minimum and Q2 E Then
(2.4) iimf (i vu12 — Q2),i. = 0 eW a(u>e}
for any n-vector with components in £1), where v is the normal.
Proof. For any real e, small, let ;(x) x + and define u, by = u(x). Then uE E K and
—J(u)
f + Q20T]detDl. —[I Vu12 + Q2}}{u>O) =ef (IvuI2+Q2)vii
(u>O)
+Ef +o(e). {u>O}
It follows £hat the linear term in e vanishes and, since u is harmonic in (u > 0),
o = f v 4(1 vu 2 + — . Vu) Vu](u>O) = limf [(1vu12 + Q2)'i — •Vu)vu]
= iimf (Q2 — I eW a(u>e}
Theorem 2.5 shows that if the free boundary a{u > 0) in C' and if u is in C' in (u >0) uniformly up to the free boundary, then
(2.5) I vu I = Q on the free boundary.
Remark. Since J( J( u)Vu E K, mm 1(v) = mm 1(v). vek vEX
PROBLEMS
1. Use spherical coordinates
x = rsin 4sin6, rcosO)
REGULARITY AND NONDEGENERACY
to define a function
u(x) = rmax
where
1 —cos0 f(6) = 2 + cosOlog1 +
is a solution of
((sin 6)1')' + 2(sin6)f= 0, = 0
and 00 is the unique zero off in (0, ir/2). (a) Show that u is harmonic in (u >0), I I on a(u > 0)\{0},
a(u> 0) has a singularity at the origin, and vu)< I — cc (c >0) if E —00—E.
(1) Show that u is not a minimum with respect to J, when = ..s=au,u°=u,Q=1.
(sin0)f'(60) > (sinOo)f'(O)(l + (cos20o)f(9)).
Use this to show that if Wx) = 4c(1/r — 1), = max(u — then
— J(u)) sin0o)
— fW/2( f(6)) 3"2[(sin — (sin 60 )f'(°)] dO < 0.1
Oo
2. Consider the functional
f[i vv p2 + )a]
dx (0 < a < 1)
and define K = (v K, Q2(v4 E Extend Theorem 2.1 and Lemma 2.2 to J,, Ka.
3. REGULARITY AND NONDEGENERACY
From Theorem 2.4 it follows that if the free boundary is smooth then, for any point x0 of the free boundary,
(3.1) u(x+xc) Q(x0)max(—x v(x0),O} +o(IxI)
276 JETS AND CAVITIES
as 0, where v is the outward normal (with respect to (u > 0}). In this section we derive a weak formulation of (3.1), in terms of averages of u. An estimate from above is given in Lemma 3.1 and an estimate from below is given in Lemma 3.3. The first estimate is sufficient for deducing that u is Lipschitz continuous (Theorem 3.2).
We shall often use scaling: If u is a local minimum in C then
(3.2) = + px)
is a local minimum for + px) in B1 = B1(0).
lemma 3.1. There exists a positive constant C* depending only on n) such that any (local) minimum u has the following property for every (small) ball B,(x0) C
(3.3) u implies that u >0 in B,(x0).r aB,(x0)
Proof. The idea of the proof is that if the average of u on aBr(xo) is large, then replacing u in B,.( x0) by the harmonic function v with boundary values u will decrease the functional, unless of course u = v.
Let v be harmonic in v = u on aB,(x0), and extend v by u into B,.(x0). Since u 0, v is positive in Clearly v E K and, therefore,
by the definition of minimum; for local minimum we must take r small. It follows that
f (i vu12 + 1Q2) "cf (I + Q2),BAxo) B,(x0) or
(3.4) f IB,(x0) 8,(x0) We wish to estimate the measure of B,(x0) fl (u = 0) from above by the left-hand side of (3.4). By scaling we may assume that B,(x0) = B1.
For any E E define
r and if this set is empty. For a.a. is
REGULARITY AND NONDEGENERACY 277
inH' (andthus in C"2 inr)sothat,if rt< I,
= — __f'f(u — v)
_______
1/2
On the other hand, by Poisson's formula
u, aB,
It follows that
(I
This inequality is obviously valid also if rt = I. Integrating over and recalling the definition of we obtain
f 1(U=O)(f )2
c,,f 1 V(v — u) '(u=O), by (3.4). 81\B114 3D1 B, B,
Replacing B1,4 in the above analysis by B114(i) where I E we obtain a similar estimate with replaced by B,14(i). Adding the two inequalities and using (3.3), we find that
(C*)2f I cQ2f '(u=O) 8, B1
Hence, if C* is sufficiertly large then = 0 a.e. in B,, that is, u >0 a.e. in 8,. From (3.4) we then further deduce that u = v a.e. and, by (2.3), U = v ? 0 everywhere in B,.
Remark 3.1. If in Lemma 3.1 B,(x0) is not contained in B,(x0) fl df2 is in and u = 0 on BAx0) () then the assertion (3.3) is still valid (u is extended by zero into more precisely, for any 0< K < I, 0, there is a positive conStant such that if C,= then
r" 3BAX0)\ B,,(3U)
where is the (Er)-neighborhood of aa.
278 JETS AND CAVITJES
Theorem 3.2. u E C°' '(Il).
Proof. If x E (L u(x) > 0, then by Lemma 2.2,
faB,(x) and from Lemma 3.1 we conclude that u >0 in some neighborhood of x. Hence the set (u > 0) is open. — —
LetD, D'beboundedopensetswithD C D',D' C tiLetp0 dist(D,aD'). For any x E D with u(x) > 0, let Br(X) be the largest ball in D' fl (u > 0). If r < P0' then aBr(x) contains a free-boundary point. Hence by Lemma 3.1, if r is small enough, say r r0, then
1 rfr + E for all small e > 0 and, by subharmonicity of u. also
(3.5) !f r aB,(x)
If r> mm (p0, r0), then, since it is upper semicontinuous and therefore bounded in D', we have
(3.6) !f r aBAx)
with another constant C Finally, if C = max(C*, C),
inDfl(u>0). r 8B,(x)
Since Du(x) = 0 a.e. on (u = 0), it follows that u E C°"(f2).
We next establish a "nondegeneracy" estimate from below on averages of u.
Lemma 3.3. For any 0 < < 1, there exists a positive constant = depending only on n, ii) siqch that Jo, any (local) minimum and for every
(small) ball Br(xo) C
If fhenu=OinBkr(xo).r aB..(x0)
REGUIARflY AND NONDEGENERACY
Proof. The idea of the proof is that if the average of u on BB,(x0) is small, then replacing u by a function w which vanishes in Bkr(XO) will decrease J.
By scaling we may take B,(x0) = B1. We choose a function
v(x) =
where
R r 2—n —l
n—2 R (3.8) = R
Rlog— ifn2 R—r ifnl
is the fundamental solution and
(3.9) U
subharmonic. - Notice that v =0 in Since v u on min(u; v) is admissible
function if extended by u outside Bc. Thus J(u) J(min(u, v)), which— that f (I vu12+I(U>o)Q2) (I vmin(u,v)f2 — I vuI2)
B.,
=1 V(min(u,v)—u)'V(min(u,v)+u)\
Vmax(u—v,0).V(v+u)
vmax(u—v0).vv=2J UVV•P
u.
We to estimate by the left-hand side of (3.10). To do this we write,
2S0 JETS AND CAVITIES
using the definition of £ in (3.9),
JaB.
Cfl k[__2_J I{U>O}Q2 + Vu 12 + I{U.o)Q2)j
cfl + i)j (i Vu12 + man man
Substituting this estimate into the right-hand side of (3.10), we find that u = 0 in if is small enough, that is, if
f u is small enough [by (3.9)}. aB,
Remark 3.2. Lemma 3.3 remains valid if Br(X0) is not contained in and U 0 on Br(X0) fl (u is extended by zero into R" \t2).
Corollary 3.4. For any bounded domain D with D C and for some small enough d0 > 0 there exist positive constants c, C such that if a ball Bd( x) with x E D, d < d0 is contained in {u > 0) 11 &l and aBd(X) contains a free-boundary point, then
cd(x) u(x) ' Cd(x).
The next resUlt may be compared with Theorem 3.4 of Chapter 2.
Theorem 3.5. There exist positive constants 0 < c < c0 < I such that for any local minimum u and for every small ball B,(x0) C with x0 E O(u > 0),
IB(xo)fl (u>o)I
It follows that 8(u > 0) has Lebesgue measure zero.
Proof. Take x0 = 0. By Lemma 3.3 there is a point y E such that u(y) > cr. Since u is subharmonic, we conclude that
icr Kr IC
Choosing ic small enough, we can apply I to deduce that u > 0 in Bir(Y), which yields the lower bound in (Ill).
REGULARITY AND NONDEGENERACY
To establish the upper bound, choose u as in the proof of Lemma 3.1. Then
fBI V(u — v) 12 u — v 12
(,csmall).
If y E Bar, then, by Theorem 3.2 and Lemma 3.3,
u—Ctcr aa,
(c>O)
for small enough Thus
and the upper estimate in (3.11) follows.
DefinItion 3.1. Let u be a local minimum, and let be a sequence of balls in with P* — 0, Xk x0 E 0, and U(Xk) = 0. The sequence of functions
= + p*X)
is called a blow-up sequence. Since VUk(X) C in every compact set of RM, if k is large enough, and
since Uk(O) = 0, it follows that for a subsequence
vu0 , weakly star in
The function u0 is called a blow-up limit. Recall that the Hausdorff distance d(A, ft) between two set is the
.
UBAx)JD, .xEA x€D
Lemma 3.6 The following properties hold:
(a) ø{Uk >0) —' >0) in the distance; (b) '(u>O} in LL;
282
(c) thenOEa{u0>O);
(d) VUk Vu0 a.e.; (e) If Q is continuous, then u0 is an absolute minimum for Q(x0) in eveiy
bounded domain.
Proof. To prove (a), notice that if Br(y) fl a(u0> 0) 0 and u0 = 0 in Br(Y), then the Uk are uniformly small in B.(y) for k large and, by nondegener- acy (Lemma 3.3) U* 0 in Br,2; if u0 >0 in BAy) then Uk > 0 in B,12(y). Hence fl >0) = 0 if k islarge enougj:i. On the other hand, if B,.(y) (1 a(Uk >0) = 0, then the Uk are harmonic in and the same holds for u0.
Lemmas 3.1 and 3.3 remain valid for u0, by passage to the limit with Uk; hence the proof of Theorem 3.5 is also valid for u0, soth.'t
(3.12) a(u0 >0) has Lebesgue measure zero.
Hence an r-neighborhoôd V, of this set intersected with any ball BR has measure lt,R. where RIO if rIO. By the first argument of the preceding section we have that
1(s,o>O)I lit, R
if k is large enough, and (b).follows. To prove (c), note; by Lemmas 3.1 and 3.3, that if x* is a free-boundary
point for Uk, then for all small r, say 0< r < r0,
forsomec>0,C>O;r
'b is independent of k. Taking k we obtain
' which implies that' 0 E ô(u0 > 0).
By (a), for any compact subset E of {u0 0) U mt (ire = 0), u, is harmonic
k -. oo. Recalling (3.12), (d) follows. — To prove (e), choose a bounded domain D with 1) C 0. For any function v
Qk(x)=Q(xk+pkx).
REGULARJTV AND NONDEGENERACY
Since u is a local minimum, if k is large enough, then
VUk p2 + JD°
+
Taking k -* oo and using (d), we obtain
1(uO>o)Q2(Xo)] +
+ I)Q2(X0).
Choosing such that tj = 1 in D, we obtain
+ I o)Q2(x)] + 1(V>O)Q(xO)J
for any o with o — U0 E this completes the proof.
PROBLEMS
1. ProveRemark3.l.
2. ExtendTheorem3.2byprovingthatuE C0' up toanyopensubsets1 of au, provided that S1 is in and u0 E [Hint: Cf. proof of Theorem 8.5.]
3. Prove Remark 3.2.
4. Consider
=1,0 with 1,
By symmetrization (sec Section 7) any absolute minimum u must be a function u = u(r), r 1x1, and by Section 4, u(r) has at most a finite number of free-boundary points. Prove: (a) the ,Jninimi2er is uniqas; (b)
O<r<RA,andRA>O,RktjfXt. 5. Suppose thataO \ S is If is° has boimded supp&1
local minimum u has bounded suppoit tHi,u: lit aO\B1c on C by (1 (1) (hcq.it is andlet 4.•1 '
&,=OinD, v=uon8D.
284 JE1S AND CAVfl1ES
The function
Ju+EmIn(V—U,O)
in K and {u, > 0) = {u > 0). Show that
fi v min(v — u,0)12 = — 0, so that u v in D; hence
C in R"\B2R. Now proceed as in Lemma 3.3 (see Remark 3.2).]
4. REGULARITY OF THE FREE BOUNDARY
We shall define a concept of flatness at the origin 0 (0 E a(u > 0)) in the direction e,, of the positive x,-axis; the definition will depend on parameters
'F.
DefinItion 4.1. Let r>0. We say that u belongs to the flatness class in if
oea{u>o},
u(x)0 if —a_p.
If the origin is replaced by x0 and the direction by a unit vector v, then we Say that U belongs to the flalnea class , a_ ; r) in in the direc- tion.. -
Theorem 4.1. Suppoie that Q is H.3ldSr ContinUous and u is a local n*ilimizer. The* for dfly bounded domaIn 1) with 1) C there exist pas*ive constants a, fi,
such that the following hole: If
u E 1; oø) In D In direction v
ThE BOUNDED GRADIENT LEMMA AND ThE NONOSCHLATION LEMMA
with a and p then
n a(u>o)
ills given by x,, (x1,.. . ,x,,_1), $ E if the direction of the x-axis is taken in the direction w.
The condition (4.1) is called the flatness condition.
Theorem 4.2. If n = 2 and Qis Holder continuous, then for any local minimum u the free bonndwy a{u >0) is locally a C1
These two fundamental theorems are due to Alt and Caffarelli [6aJ; the proofs are lengthy and will not be given here.
Combung Theorems 4.1 and 4.2 with Theorem 1.4 of Chapter 2 we can obtain further reguláfity of the free,boundary. In particular:
Corol]aq 4.3 Under the asnmiptions of Theorem 4.1 or 4.2, if Q is analytic, then the ftve bowzdwy is analytic.
5. THE SOUNDED GIADIENT LEMMA AND THE NONOSCILLATION LEMMA
In the proof of Theorem 3.2 the bmuid on Dy may depend on u; in fact, from (3.5), (3.6) we have that
(5.1) IDU(x)IhiC(1+supu). lAx)
In some applications it is important to obtain a bound on Du which does not depend on sup u. This is done in the following bv.iwsded gradient kalna.
Lemma Si. Let D be abowØeddemo4n, DC There exists a constant C >0 depending only on 1), such that for any minbnwn u, if D contains a free-bowuiwy point x0 then.
(5.2) ifxEbfl(u>O}.
Proof.
x1EB,0(x1..,) forj=1,...,k,ro=?. Choose k0 .G (1,2,. . . ,k) the largest such that B2r0(Xto) contains a point
Yo e a(u> such a k0 exists since B24x1) contains x0. Then a is harmonicS
JETS AND CAVIIflS
in B2,j:Xj_i) forj k0 + 2 and, by Harnack's inequality,
u
u 4CC*ro aB44y0)
by Lemma 3.1. Therefore, u(x) C for any x E D, that is,
SUP:, I' D
where C depends only on U, D, Using this estimate in (5.1), the proof of the lemma is complete.
Remark 5.1. If u is only a local minimum, then the proof above is valid provided that is chosen sufficiently small (depending on the r in Definition 2.1); C then depends also on e.
Remark 5.2. If we make use of Remarks 3.1 and 3.2, we find that Lemma 5.1 extends to the case where D intersects an, provided that n an is contained
T.
The next result, essentially unrelated to the previous lemma, is concerned with the behavior of the free boundary near the fixed boundary, and is stated in case U is a two-dimensional domain. We denote the points of U by X = (x, y).
Let G be a domain in U bounded by four curves:
y=yI, y=y1+.h (h>O),
)'1:X=X'(I) and y2:XX2(t),
where 0 t T. We assume: If X1Q) = (x1(t), y1Q)), then
y1czyf(t)<y,+h ifO<:<T,
yJ(O) YL' yI(T) =yi +
xJ(i) x0 + 8.
Further, Y2 lies to the right of more precisely, x'(O) <x2(O) and y2 do
TUE BOUNDèGRADIENT LEMMA AND TUE NONOSCILLATION LEMMA
not intersect. We finally assume that
(5.3) u>OinsomeG-neighborhoodofy1Uy2,
S1 is a open subset of
(5.4) u=OonS1,
C = dist(G, aa\s1) >0.
The nonoscillation lemma asserts:
Lemma 5.2. Under the foregoing assumptions,
(5.5)
where C is a constant depending on'y and sup u°.
Proof. By Section 3, Problem 2 we obtain (5.1) in G, with BAX) replaced by 0; recalling Lemma 2.3, we conclude that
(5.6) "1 ,. *
where C depends on the same constants as in the of the letnm*-'Set
G0 {XEG;u(X)>0),
and PplYQr'sforinula ,.l..aLal4?JipGe.. ;, •..
— — q) dc =0
Since uO on 8G0fl(a<y<b) (a<y<b)weget'
(5.7) f udx+kf71uy2 aGofl(7=b) Using (5.6), we find that the second term on the right-hand side is
JETS AND CAVITIES
Ch&, whereas the first term is bounded by C62 (since is C&). Noting that
Q(y —
we obtain from (5.7)
CS2,
and the assertion (5.5) follows.
Remark 5.3. The previous proof is valid also if lies on an and
(5.8)
Remark 5.4. It is well known 1108] that if
Au0 inn, E
u=+ onao,
and if is a surface and then uE The same is valid also for general elliptic operators with smooth
coefficients. Using tMs fact, we can replace the condition in Section 3 Problem 2, and and (5.4) by a condition.
6. CONVERGENCE OF FREE BOUNDARIES
In this seetion we establish for a = 2 a general theorem on the convergence of free boundaries corresponding to a sequence of local minimizers Urn. The theorem can be used also when the free boundaries of Urn converge to the fixed
Uisnotnecessanly in Q. We extend by 0 into U\O and assume that Urn E C°W),
Qts?trn; thus Urn 0 and inUfl(um>0),
(6.1) una{urn>o)jsincl+a (0<a<l),
CONVERGENCE OF FREE BOUNDARIES
where v is the outward normal. We assume that as m 00,
Am'X (X>O),
Urn -* u uniformlyin U, u EH'(U),
(6.2) weaklyinL2(U),
Uri(Urn>0)-.Un(U>0) inmeasure,
uria(u>o)
Theorem 6.1. Let Urn be local minimizers satisfying (6.1), (6.2). Then
(6.3) Au=O inUfl (u>O),
(6.4) onUfla{u>O).
Proof. For any nonnegative E Cr(U),
(6.5) Xrnf = — f VU,,, — f Vu,a(u,,>0) {Um>0) {u>O)
using the assumption that U fl (Urn >0) U fl (u > 0). Using this last property, we can also deduce that
(6.6) f 'c lim inf fO(u>0) rn-oo a(MM>o) Indeed, for any vectorf (ft. 12) with L E + f? 1, we have
f U.,(u>O} (lt.>0) Takingf=fJ such thatf= ron U,
f rn—co 3(u,>O) and (6.6) follows.
From (6.5), (6.6) we get
Af (u>0)
290 JETS AND
Since a{u >0) is in C' u is also in up to this boundary (see Remark 5.4) and we obtain
f3(u>O} av and since is arbitrary,
au av
To prove the reverse inequality, take any point X0 = (x0, Yo) in U fl a(u > 0). Then there is a sequence of points Xm E U such that Xm Um(Xm) = 0. Indeed, otherwise we shall have = 0 in some neighborhood V of X0 and therefore also iXu = 0 in V. By the maximum principle it then follows that u 0 in V, contradicting the assumption that X0 E a{u > 0).
Chooser small and domains Dm C Br(X0) C U with
(Ocz$<a),
(6.7) Um>0 iflDm,
XmEDm.
For definiteness we may assume that
f'(x0)=0,
B,(X0) fl (y <fm(x)).
j,(x) =1(x) — e.,,, co.
Let
(6.8) )= i if IXI< I,
0 iflxl>'l
and define
Define a domain I), ,,, by
= B,(X0) n {y
CONVERGENCE OF FREE BOUNDARIES
Then D0,,, = I),,, and, in view of (6.7), there is a value s = suéh that
urn>O inDsrn,
B,(X0) ii contains a free-boundary point
of Urn1 = (Im, 9m), where9,,,
and
(6.9) ifm-.O.
Set = Let Vm be the solution of
0 on 813rn B,.( X0),
Vrn=Um Oflalmflaar(Xo).
Since t*u,,, = 0 in the maximum principle gives u,,, v,,, in 4,, and, theftfore,
( ) A 8Urn(Xrn) av,,,(im)
Zn view ol(6.9),
Hence, by elliptic estimates, v,,, m 4,, fl Br,12(X0) converges to u in fu >0) fl in the We may assume that X,,, X as in -. oo and, by
(6.9),
• n a(u>o).
We then have
avm(irn) ôu(I)
292 JETS AND CAVITIES
Recalling (6.10), the inequality
(6.1!) —
8i'
follows. Taking r 0, we obtain (6.11) at £ = X0, and the proof is complete.
PROBLEMS
1. Extend all the results of Sections 2, 3, 5, and 6 to the functional
J(v) +
where is a two-dimensional domain; here
a2u a2u I 8u—+———-—=0 in{u>0), ax2 ay2 v
I ôu ——i--=Q onthefreeboundary,
where is the outward normal. [Theorem 4.2 also extends to this case and, in fact, all the results of Sections 2 to 6 extend to functionals
÷ Q21]
with Il C RM provided that is positive definite and smooth.] (Hint: See Lemmas 8.3 and 8.6.1
2. Do the same for
J0(v) f I 'Vt, + 12 dx,
where (2 c R" and e is a unit vector in R", and for
J,(v) +
where (2 C R2.
SYMMETRIC REARRANGEMENTS
7. REARRANGEMENTS
Later in this chapter and in the following chapter we shall use some well-known facts about symmetrization of functions.
Let E be any open set in the two-dimensional strip {(x, y); —a <x <a) and let = E fl (x = p). Denoting by J E,
I
the one-dimensional Lebesgue measure of let
(7.1) {(p,y);
The set
(7.2) E= U E,* — a<p<a
is called the Steiner symmerizalion of E about the line y = 0. liE is closed, we define
and then defined by (7.2), i,s again called the Steiner symmetriation of E. Clearly, meas E = meas E. One can show that if E is open then
E is Open (cloSed). Let u(x, y) be a continuous nonnegative functiop defined in a rectiligle
=•((z y); —n <x —b <y <b)
such that u(x, y) = u(x, —y). For fixed the function v(y)'= u(x, y) and consider the sets
tr'[c,oo),
where tr'(A) = f—b, bJ; o(y) E A) for any A by and (o'(c, oo))* the segments
(y; I'41 and (y; o'(c, )9T:
.LP. .
respectively. Define
u*(x,y) = c ifandonlyify E
It follows tha$si'(x, y) = y), .y) is monotone my
294 AND CAYFI1ES
y>0,and
meas (y; u*(x, y) E (c, d)} = meas {y; u(x, y) E (c, d))
for any 0 <c <d. One can also show that
(7.3) fg(u*(x, y), x) dxdy = fg(u(x, y), x)dxdy
for any continuous function g and that
(7.4) fu*(x, y)w*(x, y)h(x) dx dy r.fu(x, y)w(x, y)h(x) dx 4y
if w satisfies the same properties as u and h is a nonnegative continuous function. if h >0, w >0, then strict inequality holds in'(7.4) unless ,u = u a.e.
Using one can now defme u also when u is any L' function, by approximation; it is an even function in y, decreasing in y for y >0, and it satisfies (7.3), (7.4).
Deflulden 7.1. The function u*(x, y)s called (symmetric) decreasing re- arrangement of u(x, y) in the variable y or a Steiner symmetrization of u(x, y) with respect to the = 0.
ibeore 7.1. If u E u >0, u(x, y) = u(x, —y), u(x, b) = 0 then
(7.5) f vul2h(x)dx4y Qa.b Q..b
for any continuous nonnegative function h(x).
This result is basically contained in reference 154; see also references 92 and 103.
One can define similarly the concept of increasing rearrangement u, assum- ing that u >0, u(x,O) 0 and u(x, y) = u(x, —y). This functjon is, for any x, monotone increasing in y, tory >0, and it has the same distribution function in y as ss(x, y), that is,
mess {y; u(x, y) E A) = meas (y; u(x, y) e A).
One can also define increasing rearrangement in y with respect to the measure ydy Instead ofdy. The following result is proved in
SYMMETRIC REARRANGEMENTS
Theorem 7.1'. If US is an increasing rearrangement of u in y with respect to the measure y dy, then
(7.6) f vul2ydxdy.
Let (r, 0, z) be coordinates for x E R3 and suppose that
u(x) = p(r, z), p > 0, p(r, z) = p(r, —z).
Let u be a decreasing rearrangement of u in the variable z. Then we have:
Theorem 7.2. There holds
(77) J ( Ix—yl R'R' 1xY1
and equality holds and only if u = u a.e.
The inequibty is a of the followthg result:
7.3. J/f(x), g(x), we give functions q.sd 1/f, g, are their ,fecreasini x 0), then
(7.8) f / f(x)g*(y)h*(x 4' f y) dx dy..R'R' R'R' The proof is given in reference 112; the assertion about equality in (7.8)
follows from the proof. . We next intioduce a spherically symmetric decreasing rearrangement. Let
Define
296 JETS AND CAVfl1E.S
and
r=IxI ' rii
Then is a decreasing function and
(7.9) =fg(u(x))dx
for any continuous positive function g; also,
IRM 'C
u and v are in L2(G). Next:
Theorem 7.4. 1/u E '(6), 1 p < oo, then
(7.10) vu rdx IGI vurdx.
This result for p = 2 is given in reference 154 and for general p in reference 169. We finally mention a generalization of Theorem 7.3, given in reference 163.
Theorem 7.5. Let f, g, h be nonnegative functions in R' and let f*, g, h their spherically symmetric decreamig rearrangements, respectively. Then
ff f*(x)g*(x)hs(x_y)dx4piif f f(x)g(x)h(x—y)dxdy. PROBLEMS
1. Prove (7.3).
2. Prove (7.4).
3. Show that Theorem 7.2 follows from Theorem 7.3.
4. ProveTheoreml.4forp=2,n 1.
5. Prove that if u(x, y) is continuous, then u(x, y) is continuous.
AXIALLY SYMMETRIC JET FLOWS
8. AXIALLY SYMMETRiC JET FLOWS
Let N be a continuous curve X = X(r) = (x(t), y(:)) (0 < oo) such that
X(O) = A, where A = (0, 1); X = X(t) is ay-graph;
(8.1) IX(t)I—. 00 jfj—* 00;
X(t) (0<ict<1);
The second condition in (8.1) means that any liney Yo which does intersect N intersects .jt in either one point or one segment. We have assumed for simplicity thM X(O) (0,1), but afl the subsequent results hold for the general case of (x(O), y(O)), y(O) >0.
Set
A' = (—a,I), A" = (-'-a,O)
for some a> 1, and let
1x>O, the auyc consisting of ,A'A" and thefite'ialy=Q,—oo<x< —a.
LctflbethedomajninR2boundedbyN,thex-axisandl,andletDbethe subdomain of t7 by N and Figure 3.4.
DdInltlon8.1. Niscallcdanozzk.
Ddlnitlon &2. Set = R2 fl (y > 0) The jet problem is theproblemoffindinga function u(x,y)in C°(R4), a C' 1':X=
Qsuch thai
(8.3) F0=F\(A) doesnotintersectN,
(8.4) (h>0),
ift-.oo,
(8.5) thecurveNUl'isC'inaneighborhoodofA,
(8.6) inR÷,
(8.7) u(x,O)0 if—oo<x<oo,
(8.8) uQ onN, (8.9) Lu=0 inR+fl(u<Q),
where
298 JETS AND
a P D
x
(a) (b)
FIGURE 34
x
ax2 ay2
(8.10) [a(u<Q)\NJ
so thatuQon F, (8.11)
coincides with F0,
C
u e CI in (u< Q) =
where the outward normal to F0, and
(8.12)
u is called the stream function, F (or is called the free
AXIALLY SYMMETRIC JET FLOWS 299
boundary, A is called the velocity on the free boundary, and is called the flux.
The flux of a flow across a planar section Tin R3 is defined by
(v=norinaltoT),
where is the velocity potential. In the axially symmetric case (see Section 1), if T = {u < Q) fl (x = x0),
= 2TTQ.
Notice next that since u = Q on N U r, N U F is a stream line. From some of'the conditions in (8.2)—(8.12), one can deduce (see Problem I)
that
ifx-'oo, (8.13)
h unifornilyin{u<Q} 0
It follows that
2Q 1/2 (8.14)
If(u, F, A, is a solution of thejet problem, then also (flu, F, pA, flQ) s a solution of the jet problem, for any /1>0. (Notice that h does not change by this scaling.) Therefore, from now on we take Q fixed and a ion (u, r, A) corresponding to this Q.
Under the assumption of (8.1) we anticipate the free boundary r to lie in (0<y<1),thatis,u=QinR+\O.Thisiswhyinthefutureweshafl replace R÷ by (1.
DefinItion 8.4. The condition (8.2) is called the continuos&v-fiz condition, and the conditions (8.5), (8.12) ate called the smooth-fit' condi tiona.
I/me nozzle N I), then fhe,e exils qf the .
The proof of Theorem 8.1 depends on the variational approach of the preceding sections. We need,however, to truncate U by domhin* (2,, bounded on the left and above.
Considerfirstthecasewhereáthery(t)-. 00 itt—' oo,ory(t)-.y<oo andx(t)-. —ooift--*oo.
301) JETS AND CAVITWS
Let (for any large a) be the part of N corresponding to X(t), 0 t and introduce the horizontal segment
and the vertical segment
0 {(x,y(7j); —p<x<x(1)}
3
the domain bounded by and the line (y = 0,—p <x < oc); see Figure 3.5.
as and let
f(x(7j, y); y(i;) <y <y(7;) +
011
3= ((—p,y);0<y<y(7) +p).
union of tX(1) 0 1 1) with a class of functions
K.1=
(8.15)
vQ v0
x+pvQ onax(7)+p .1
IT
(-M,O)
FIGURE 33
AXIALLY SYMML7RIC JET FLOWS
and a functional
(8.16) =f I!vt, — AI(v<Q)\r,elydxdy,
where e is the unit vector (0, 1).
Problem Find a solution u = in of
(8.17) mm vEX,1
The term '(v<Q} \ 1) in needs some explanation. Since the free boundary is 8(u < Q), (v < Q) must replace (v > 0) in the functional in (2.1). Next, our purpose is to find a free boundary that does not go "infinitely deep" leftward into Ii, that is, which does not enter some domain D with a sufficiently large a; for this reason we have included the complement of 1) in I. Of course, we shall have to prove later that the free boundary does not aD.
The funetional.
(8.18) + X21(O<Q)\D)ydxdy
is infinite for every admustbl4 fuflction, For this heason we work with this functional (as we shall see) remains finite for flows that are nearly horizontal as x .,•.
Since we anticipate h 1, we alwhys assume that
The proof of theorem 8.1 bçgins in this section and ends in Section 12. of follows. We
it, p. and then show that fá some it = A(p) the sohiskm in the truncated domain Letting p -' ao, we shall obtain the desired solution.,
We caniot deal directly the functional
2
(8.20) =
Vt'— y(b: 4y
since in general = oc for any admissible function. In the remaining part of this s'ectlOà we átablish for 4 results analogous
to results established in 2 and 3.
Lemma 8.2. Problem has a sohulon.
302 JETS AND CAVfl1ES
Proof. Since A 2Q, the function
(8.21) u0 = mm
for x> 1, can be extended into a function in and
< 00.
Now take a minimizing sequence Uk with
weaklyinH"2(U,flBR) VR>O,
y weakly star in fl BR) YR >0.
Then
VR>O,
where
i—I Vu -
Sincey= I a.e.intu<Q}\D,y=OinD,
1112
8.3. There exi.ts a positive constant C in4epi.uknt of p, A that for any ball B(X°) C u, with X? (x°, y°), r <y0/2,
— u) CAy° implies that u < Q in B,(X°);
here u = is any solution of problem
Proof. Let
Lv=0 inB.(X°), v=u on8B,(X°),
AXIALLY SYMMETRIC FLOWS
and define v = u in Then so that
!v(u_v).v(u+v) HA X°)Y
— — v) — X2fyi
The first integral is equal to
f !IV(u_v)12DAX°)Y (see the proof of Lemma 3.1). The second integral is equal to zero. Hence
f I v(u — v) 12 3x2(yo)2fB,(X°) 9(X0) Introducing
we have
1 inB1(O),
y+y°/r8Y
inB1(O)
and
(8.22) f I v(u — 3x2(yo)2f I(g.o).
We can now continue as in the proof of 3.1, noting that £ is uniformly elliptic and smooth in B1(O), independently of r.
Lemma8.4.
(8.23)
JETS AND CAVITIES
Proof. Define u0 in by (8.21). Then min(u, u0) is admissible. Hence
— min(u, u0)) V(u + min(u, u0))
-_2Af Y
+A2fy(1(U<Ql 1min{(v.Ijo)<Q))
= 11 + 12 + 13.
We can wiiie
=f — u0,O)) •V(u + u0)
and, since we may integrate only over (u> us),
v(max(u — + 2J -!-v(max(u — u0,O)) •Vu0.
The last integral is equal to
f Xinax(u—u0,O)=OU,n(u0<Q) ay since u0 Q u in the last integrand. Thus
I J v (max (u — U0, 0))
Next, -
'2= _-2Af I-max(u—uo,O) (O,,\D)fl(uo<Q) Y
—2AJ max(u—u0,O)=O. (1k,, \D)fl (y>O) r)a(u0<Q)
AXJALLY SYMMETRIC JET FLOWS
Finally,
I:— J Yk '{u<QI\ D
— "J
Combining these estimates, we obtain
0 V max(u — uo,0)12 — A2fy1{Q)fl{<Q}
Y Y
fl (u Q). Now the proof of Theorem 3.2 applies with obvious changes in compact
subsets of Thus u is continuous in and clearly also on Thus u(x, y) < Q if y >0, y small, and x is near —is. By continuity we now deduce that u u0 in wherever u0 < Q.
Theorem 8.5. u = ux is Lq.'schitz continuous in eve;y compact subset of that does not contain A or points where is not c1 + a
Proof. The Lipschitz continuity in compact subsets of follows as in Theorem 3.2. To prove Lipschitz continuity near 11, note that if the distance r from X to a(u < Q} is less than the distance .c from X to then we can argue as in Theorem 3.2. If, however, s <r, 'set X° = (x, 1) if X = (x, y) and s0 = min(1, x), B0 = 83o(X°). Let
Lv0 inB0fl{y<l), vQ onaB0fl(y<1), vO onA0fl(y1).
Since Q — u is L-subsolution and Q — u 0 on aB0 ri <1), we have Q — u v. To the function i5( X') v( X° + s°X') we can apply elliptic estimates and conclude that
4 v(X° + s°X') CJy' if X' = (x', y') B112(0) {y' <0).
WI'S AND CAVITIES
Hence, if $ I —
(8.24) S S 3B,(X) s
the first inequality in (8.24) follows by representing w( X') (Q — u)( X + sX') in X' < I in terms of Green's function for the corresponding uniformly elliptic operator C, and then applying v.
It remains to prove Lipschitz continuity near y = 0. If X° = (x°, y°) with 0 <y° < and = then by Lemma 8.4,
LuO J Hence the normalized 8(X) = u(X° + satisfies
=0 inB1(O),
which implies that
I vu(X°)I= Cy°.
Lemma 8.6. There is a constant c > 0 independent of A, such that for any ball Br( X°) with center X° = (x°, y°) E (2 \ D and radius r y°/2, the following hokic:
(8.25) !fr impliesthau QinB,,,8(X°) n ((2\(D U Br(A')));
here u = u is extended by zero into B,(X°)\12.
Proof. The set
R,,,8(X°) n (12\(D U
êan be covered by balls of the form
Br(X') C Br14(X°) fl (O\ff) withr1 =
thus it suffices to prove that u = in each such ball.
AXIALLY SYMMFTRIC .wr stows
Introducing
Q—u(X'+r1X)
we have
Lu=O mB1,
where is to be chosen small enough. This- implies, since B2(O) C B8(( X° — X' )/r1), that
(8.27) supü C8Ay°.
Let v be the solution of
Lv = 0 ifl
v=Q onaB,,(X'), vu onaB2,(X'),
and define v = Q in and v = u outside a82,( X1 ). Then max(v, u) is an admissible function. Proceeding analogously to Lemma 3.3, we have
—JA,1(max(v,u))
+ + 13.
By integration of parts, 0, since nun (...) 0 on the top and the
308 JETS AND CAVITIES
bottom. Next,
I Vmin(u -- + 2j — v,O)) -Vv
Y
R,(
Combining these estimates, we ebtain
j + cX21(U..Q)) (Q — R,C V') ) )' Y
or. in terms of the normalized 8 and the corresponding
(8.28) 1(1 + cX2y2I ) cf B aB1
By (8.27), t3 8 on 8B2 and by interior elliptic estimates in .B312,
and the right-hand side in (8.28) is then estimated by (see proof of Lemma 3.3)
C15AY0{(C6AY0 + 3)1 '{u>O) +
e (C&Xy°) we then obtain from (8.28)
(1 62c)(AyO)2f 0, a1
so that ü = 0 in B1. This completes the proof.
Corollary 8.1. For any small 8 > 0 there is a constant Ca > 0 independent of p. such that
u—Q
where (ZQX)"2.
THE FREE BOUNDARY ISA CURVEx = k(y)
Proof. Since
r aB(x°) r
choosing r = y°/2, c(y°)2 = Q/X, Lemma 8.6 gives u Q in B,.( X°).
Lemma 8.8. u < Q in some neighborhood of A'.
For proof, see Problem 2.
PROBLEMS
1. Prove (8.13) if X0(i) is in C 2+a
IHinI: Consider u(x y) and apply Theorem 1.4 of Chapter 2.J
2. Prove Lemma 8.8. [Hint: Let Lv = 0 in B1(A') fl D, v = Q on B8(A') fl aD, v = u on 8B8(A') fl D. Show that v Q — Cr273 by considering
9. THE FREE BOUNDARY IS A CURVE x = k(y)--
Since the results of Section 4 apply also to general elliptic equations with smooth coefficients, the free boundary
= < Q) fl
is locally analytic. In this section we show that Ft,.,. is given by a continuous curve x = or, briefly, x = k(y).
Lemma 9.1. Let A/2Q A and let a be a segment joining the po'nzs (x0 ± b, Yo) E 1Z\D, b >0. If a solution u = of problem satisfies
uQ inaneighborhoodofa, .(xo,y1)isafree-boundarypoin:widy, >y0,
then
where C is a constant depending only on A.
310 JE1S AND CAVITIES
Proof. By Lemma 8.4 we have y0 c0 > 0, where c0 depends only on A. For any 0, define
= — Y Yo = xo)}
where is defined in (6.8), and let s be the smallest value s0 for which r'30 touches the free boundary at some point, say (i, 9). Notice that
C0 is a universal constant. For fixed large m, consider the domain E bounded by l's, Y = Yo + m and
x x0 ± b, and let w be the solution of
Lw0 mE, wQ w0
Since u is a supersolution in {y > 0) (when extended by Q), u w and
lau law (9.1) at(2,9).
Suppose that the assertion of the lemma is not true. Then the setting above holds for a sequence of b's with b co. Choose a subsequence such that the corresponding values of
A
r converge, and let
= limX X0
then xc, e [— 1, 1].
After a translation in the x-clirection such that 9) lies on the y-axis, we see that E converges to a strip
E,c, = {Yo + <y <Yo + m)
and 9) converges to (0, Yo + )), where x0, Yo' s designate the values
THE FREE BOIJNDARV iS A CURVE x =
of x0, y0, s in the construction above for each h. Furthermore, the functions w converge to a solution W of
LW=O
W0 ony—y0+m. By uniqueness (see Corollary 9.10),
+ 2
and then (9.1) gives
iaw 8V (0. v1)
= 2Q 2<2Q
(y0+m) —y1
if m is large enough, contradicting (8.19).
Lemma 9.2. There exists a minimizer u such that 0.
Proof. The proof is by symmetrization. Since, however. is unbounded, we introduce another truncation. For R >0, let
(9.2) = —
By the proof of Lenuna 8.1 there exisUa solution u= of
(9.3) = minJA$k(v).
Denote by u* the increasing rearrangement of u( x, y) in the direction x. Since u(—p,y) 0, if we define u(x, y) u(—2!L — x, y), then Theor be applied. Hence
f
312 JETS AND CAVITIES
The term
- = —A) = —A) u(x, 1) — AQR
Y —a
also decreases by the rearrangement, since the rearrangement increases u( x, 1) for —a <x <0. It follows that, there is a minimum R rearranged mono- tonically in x, and since (Problem I)
(9.4) UAMR —' a.e.
for a sequence R 00, where is a minimum of the assertion follows.
From now on we take only solutions that satisfy
(9.5) y) 0.
tIn Section 10 it will be shown that Problem has a unique solution satisfying (9.5).] From (9.5) we deduce that
(9.6)
n {O<y<l) I),
for some function which we briefly designate by k(y). Lemma 8.4 shows that
k(y)=oo
Theorem 9.3. I/A > 2Q, then k(y) = is a bounded continuous function for 1;further,
(9.7) exists and —a < 00, y
(9.8) tim = oo.
y = mA x) of x = kA y) exists and is dqferentiable for all x sufficiently large, and
(9.9) = 0.
The proof is given in a sequence of lemmas.
THE FREE BOUNDARY ISA CURVE x = k( y) 313
Lemma 9.4. k(y) is continuous in (0, 1) with values in [—a, ooJ.
Proof. k(y) is clearly lower semicontinuous. Let 0 <yO < I With X0 = k(y0) < 00 and suppose that there is a sequence y,1 y0 such that x0 + £ for some £ >0. Then the segment a = ((x, yo); 0 <x — <e) belongs to the free boundary. Therefore, Lu = 0 in an upper or lower neighborhood V of a and
iuQ, —----Aona. Yo 8Y
By uniqueness for the Cauchy problem it follows that
mV
and by unique continuation we get u = Q on the segment {(x, Yo)' < x < x0), which is impossible.
Lemma 9.5. If(x0, y0) is afree-boundary point in O\D, then
(9.10) uQ in(x0+C,oo)X(y0,l), - where C is as in L.einma 9.1.
Proof. By Lemma 9.4, for any e > 0,
u(x,y)Q ifx>x0+e, provided that 8 = 8(e) is sufficiently small. It follows that u = Q in a neigh- borhood of the segment
(y=yo,Ix—(xo+b)I<b)
for any b >0. Now apply Lemma 9.1 to deduce (9.10)..
Lemma 9.6. There exists a nuiñberh ELhA, 1) such thai
k(y)<oo ifh<y<1,
k(y)=oo :fO<y<h.
Proof. Let
h = inf(y; k(y)< 00).
314 JETS AND CAVITIES
By Lemma 9.4. k(h1 = Take a such that —a < Applying Lemma 9.5 with y,,, we find that k(y) < 00 if v,, <y < 1,
and the assertion follows.
Lemma 9.7. k (I) urn 11k ( exists and k( 1)> — a.
Proof. If the first assertion is not true, then we get a situation contradicting the nonoscillation lemma (Lemma 5.2: see Problem 1, Section 6). The second assertion follows from Lemma 8.8.
Lemma 9.8. (,) If k(l) >0, then
(9.11) ony=I,O<x<k(1)
where v is the outward normal. (ii)If k(1)<0, then
(9.12) ony=1,kO)<x<0,
where v is the outward normal to D. (iii)Jf k ( v0) = — a for some E (0, 1'). 1/zen
1
(9.13) A — at ( a.
where z' is the outward normal with respect to D.
Note that
aa a ar —
respectively in (9.11), (9.12), (9.13) and that the tact two derivatives are taken as lim inf of quotient differences.
For of the lemma, see Problem 2: only part (i) will be needed later. We shall need the following Phragmen— LindelOf type of result.
Lemma 9.9. Suppose that
ifx>O.0<v< I, ijx>O.0<v<l(M>0),
>0.
Then lim 0.
THE FREE BOUNDARY IS A CURVE x = k( y) 315
Proof. For any e > 0, consider the function
M2 w=4'— —ex (r2.=x2 +y2)
in (0 < x <y < I). It satisfies Lw 0, andw "0 on the boundary if M/e. By the maximum principle we get w 0, that is,
+ex 1.
Taking 0, we get 4'(x, yl My2 /r3 , and the assertion follows.
Corollary 9.10. Suppose that
if-oo<x<oc, O<y<I, O<y<l,
<x< oo,O<y< 1.
Indeed, by the preceding lemma, for any > 0,
tp(±N8,y)<6 forO<y<I
provided that N8 is large enough ; hence, by the maximum principle,
4'(x,y)<& if —N6<x<N8,0<y< I.
Lemma 9.11. If A > 2Q, then k(0) < (and consequently h < I).
Proof. Suppose that k(0) = Consider the functions u(x y) (n = 1,2,...). They satisfy
Lu =01 in{x>l—n,0<y<1),
ifx>—n,
iau 1,x> —n
where (9.11) was used. Therefore, for a subsequence
Un V
316 wrs AND CAVJTJES
uniformly in compact subsets of {0 <y 1), and
Lv=0 in(0<y<l),
v(x,0)0 if—oo<x<oo,
vQ,
By Corollary 9.10 we have that v Qy2, and therefore
lay 2yay
Thus 2Q A, a contradiction.
Completion of the Proof of Theorem 9.3. We already know that 0 <h < I and that
k(y)—'oo
But then the free boundary satisfies the flatness condition of Section 4 for all values x = k(y) with x x0, x0 large. It follows that, for x x0, the free boundary can be written in the form
y = m(x)
with —e0 m'(x) 0, where £01S arbitrarily small, positive number, provided that x0 is large enough depending on £0. Thus (9.9) holds.
Using Theorem 1.4 of Chapter 2 we also deduce
C
(9.14) in(u>O) fl{x>x,).
Recall also that
uQ (9.15) iau
— A ony=m(x).
For any sequence -# oo, consider the functions
u(x + n, y).
Then for a subsequence
Un V
THE FREE BOUNDARY IS A CURVE x *( y) 317
uniformly in compact subsets of (0 <y < 1) and by (9.9), (9.14), (9.15), and elliptic estimates,
Lv0 in(O<y<h), v(x,0).O if—oo<x<oo,
vQ,
It follows (using Corollary 9.10) that
and therefore 2Q/h2 = A. Thus Is and the proof of Theorem 9.3 is complete.
Lemma9.12.
Proof. Suppose that I < oo for A A Thus for a subse- quence
V
uniformly in compact subsets of 0, and v is a minimizer corresponding to A = 2Q. By Lemma 8.4, u < Q in 0 and therefore Theorem 6.1 (adapted to L) gives
18v -vQ, _T=2Qony=l,x>x. By uniqueness for the Cauchy problem, we then get
u(x, y) = &2 throughout 0,
which is impossible.
PROBLEMS
1. Prove (9.4).
I Hint: See Lemma 3.6.1
2. Prove Lemma 9.8.
tHin:: Use the method of proof of Theorem 2.5.J
318 JETS AND CAVITJES
10. MONOTONICITY AND UNIQUENESS
From now on we choose. a in Figure 3.5.
Theorem 10.1. Problem 1A has a unique solution ux satisfying (9.5).
Proof. Suppose that is another solution and set
U2=ÜA;L.
Introduce the functions
min(u1, u2) = u1 A
ma.x(u,, u2) = u1 Vu2.
We claim that
(10.1) JAM(uI) +JA,M(u2)
Indeed, if we develop both sides, we see that it suffices to prove that, for any 0<R<oc, (10.2)
f
(10.3) x(
D)fl(x<R)
(10.4) are obvious, whereas (10.3) reduces to the obvious relation
1) + u2(x, 1)) dx 1) + v2(x, I)) dx
Since v. E we have
JA,L(u)
MONOTONJCJTY AND UNIQUENESS 319
so that, by (10.1),
(10.5)
We claim that
if 0< u1(X°) = u2(X°) < Q, then either (10.6)
u1 u2 or u2 u1 in a neighborhood of X°.
lndeed,let VbeadiscwithcenterX°suchthat0<uF<Qin Vfori 1,2 and let w be defined by
Lw0 mV, w = v1 outside V.
11(10.6) does not hold in V. then v1 is not a solution of Lv1 = 0 in V and, consequently,
1vw12 1vv112 1 <1vY v Y
It follows that < = a contradiction since W E From (10.6) it follows that if u1 u2, then we cannot have
0 < u1(X°) = u2(X°) < Q
at any point X° E otherwise, the maximum principle would give u1 u2 in a neighborhood of X° and, by unique continuation, also in here we use the fact (which follows from (9.5)1 that the sets {0 < u1 < Q) are connected.
We conclude that
(10.7) eitheru1 or u2>u1in.
The argument above does not give uniqueness, hut it does apply to general nozzles [not just to those satisfying (8.1)]. We shall now use a slightly modified atgument to prove uniqueness when the nozzle is a y-graph.
If u1, u2 are two solutions, extend u. by Q to the right of N and by 0 to the left of x = For any small £ >0, define
u(x, y) = u1(x —
— 'A —vI—uI U2, V2—U1 U2
and denote by , the functional 'A, corresponding to D', the translations
320 JETS AND CAVI11ES
of D, respectively, by x —' x + €. We claim that
(10.8)
The proof in fact is similar to the proof of (10.1) and depends on the choice a p.. Notice next that u is a minimizer for that is in the correspond- ing class of admissible functions, and that 02 E Hence
JA,L(ul)
Proceeding as before we conclude that either u2 everywhere or everywhere. Since, however. <u2 at some points of it follows that
(10.9) u1(x — e. y) u2(x, y).
Taking r —' 0, we obtain U1 u2. Similarly. one shows that u2
Theorem 10.2. If A1 <A,, then
(1010) — —
Proof. Set
u. = —
v1u1Vu2,
Then, as in the proof of (10.1),
(10.11)
Since 0 X,u1 Q, we have 0 v. Q/A1, so that Q — Xy1 is in KM. Therefore,
M(Q — A1t,1) =
and from (10.11) we deduce that
J(u,)
MONOTONICITY AND UNIQUENESS 321
We can now argue as before and show that either u, or u2 u1 throughout Since (by Theorem 9.3) u1 > u2 at some points (x, y) with x —' oo, (10.10) follows.
Coroffary 10.3. If2Q <A1 <A2, then
(10.12) > <y < 1, > —a.
Proof. Theorem 10.2 gives
To prove strict inequality, suppose that
= M(yo)> —a.
Then, by the strong maximum principle (using Problem 1),
a a at (x0, ye).
Since (x0, y0) lies on both free boundaries, we have
iia 1 at(x0,y0)(i= 1,2),
a contradiction.
Lemma 10.4. If A,, -. A, A > 2Q, then
(10.13) —'
and
(10.14) —. kA,(y) for eachy E.(hA, 1)
Proof. By the proof of Lemnia for any subsequence of such that
: weakly in and a.e.,
we have: w is a minimizer for J, By uniqueness, we then obtain the assertion (10.13). Lemma 3.6(c) also gives (10.14) for y < 1.
322 JETS AND CAVITIES
To prove (10.14). for y = I suppose it is not true. Then for a subsequence
kAM(1) —. + /3. /3 0.
If /3 then Theorem 6.1 gives
I (x,1—0) = A if + /3< x
so that, by uniqueness for the problem,
Qy2 with Q = 2A,
a contradiction. If /3 >0 and < 0 then the proof of Theorem 6.1 applies again and yields
I 1 + 0) =0
and we again get a contradiction. If /3 > 0 and > 0, then we obtain a contradiction to the nonoscillation lemma (Lemma 5.2 and Remark 5.3).
By Lemma 9.11,
1) E (0, 00)
if A > 2Q, A — 2Q is small enough, and by Corollary 8.7,
where X is independent of a, p.
Definition 10.1.
=
As noted above,
(10.15) Aindependentofp.
MONOTONICLTY AND UNIQUENESS
From Corollary 10.3 we have
{A;2Q<A<A,j. (10.16)
Choose
A 2Q
in Lemma 9.1 and denote the corresponding C by C*. If A E then A <X, M
> 0 and Lemma 9.1 then shows that the free boundary must lie in (x> Choosing
(10.17) a01+C*,
we then have:
Lenuna 10.5. If A E then the free boundary lies in {x> I — a0).
Thus the free boundary does not intersect We conclude this section with a gradient estimate near the initial point of
the free boundary.
Lemma 10.6. Set = (xAM, 1). Then for some R > 0,
(10.18) I
C in n
where C is a constant independent of
Proof. Set r=I X — It suffices to establish the estimate
(10.19) I
C in fIn ($ < r <
for any sufficiently small P (with C indcpendentof P). Set
in
and let
Ô {X;1
324 AND CAVITIES
Then
i = — = A on the free boundary
and
We can now apply the proof of the bounded gradient lemma (Lemma 5.1 and Remark 5.2) to — with D, replaced by B and G, respectively. We deduce that
mA,
and (10.19) follows.
PROBLEM
1.
a line segment x x0 with possibly x0 = and let
Lu?'O,
u 0 in
11. THE SMOOTH-FIT THEOREMS
Theorem 11.1. If xx.,,. = 0, then N U is continuously differentiable in a neighborhood of 4.
Proof. The proof consists of several steps. In the first step we study the blow-up limit of
(11.1)
for any sequence y y,1 0. By Lemma 10.6,
THE SMOOTh-FIT 325
in every compact subset G of R2 which is contained in fP (X; u7(X) > 0), provided that y is sufficiently small (C is independent of G).
By the proof of Lcuuna 3.6, for a subsequence
—. u0 uniformly in compact sets,
where u0 0. Further, u0 is an absolute minimizer for the functional
1(v) f(i + X21)
is any bounded domain. Set
(X;u°(X)>0}.
Notice that 0. If
the tangent line to N at A is
a2+$2>O,
then, for a 0, the boundary of is a y-graph consisting of the ray
and of the free boundary
Lemmall.2. Ifa#0:hen 1'0 isthe ray
y<O
and 'ç7u0 cons,.
Proof By Theorem 4.2, F0 is analytic. We also have
&u°=O u°0 onN°Ur°,
20 onF°.
a,,
We shall consider first the casC where
F° is a connected curve initiating at 0.
326 JFIS AND CAVTflPS
We claim that
(11.6) Vu°I<A inCZ°.
To prove it, consider the mapping
h: X X' = vu°( X), X E (2°.
Since u0 is analytic, h is an open mapping and consequently h(O°) is a domain whose boundary ah((2°) consists of all points
lim where XD* E (20, dist(X",8fl°) 0.
Observe that
(117) Vu°(X)EN'ifXEN°,whereN'is the straight line N': —ax' + $y' = 0,
and
(11.8) Ivu°(X)I=X ifXEF°.
We shall use these facts to show that
(11.9)
Indeed, if (11.9) is not true then, since vu0 I C, there is a closed disc
B' fl h(1i°) is nonempty and does not intersectN' and
Introduce the notation
z = x + iyifX (x, y), z* x*+ jy* if r = (x*, y*)
and consider the analytic function
UO + jUO — y
Then there is a sequence of points z" such that
(11.10) fEfl°, 1.
THE SMOOTh.Fff THEOREMS
On the other band, I 9(z)j is a bounded subharmonic function and I <1 on (N° U F°)\O. Since N° u r° is a y-graph, there is a line segment
initiating at 0 which lies outside Hence Lemma 6.7 of Chapter 2 can be applied to deduce that
asz-O+Oi.
Thus Jim sup as dist(z,80°)-.O, whereas I9(z)Ic coast. sn By the Phragmén-LindelOf theorem it follows that
I in (1° and
(11.10) is contradicted. Having proved (11.9), we have from (11.8) that the subharmonic fUóctk*s
I 0° achieves its miximum in on F0. the relation
+ 0 along a stream line :..
where q = I Vu°J , ic curvature (see Problem 3 of Section. 1), we deduco that.
(11.11) r°isconvextotheflwd.
We next introduce the conformal mapping a from the unit disc E onto 0° mapping the complex numbCrs. 1, —i into 0 and eo,
U N° is a Jordan rve, a is continuous and 1-I from I onto 00(by a standard theorem in conformal mappngs; ice Z7,.(hap. 4, Sec. 8). We can choose a so that it maps the part ô1E going c4ockwise from —ito i onto I'°, and aE\a1E into N°.
Suppose now that Vu0 const. Then the mapping k = is a conformal mapping from E into the set
half-planes determined by the line -lix' + ny' 0.
Since
exists, the set of limit points of k(z) as 2 - ±1, z E E consists of the intervaly* to kr.. [Oils follows byapplying Lemma6.7 of Chapter to
functions k1 ± c arctan(xL(±i — y)), c >0 where k = + ik2.J Hence, if does not lie on then k(E) is a proper subset of I But then the proof of (11.9) can be applied (with a suitable disc B' in B\k(E)J in order to derive a contradiction.
We have thus proved that k mips E Onto B. Since and art curves it follows, by confOrmal' mippings
JETS AND CAVITIES
(reference 27, Chap. 4, Sec. 8) that k is continuous in E. But then
(11.12) a + ifX-'A
(11.13) a + if 00,
1a2+$2
where 6 = 1 or 6 — I. Since 0, we must have a = 0, contradicting the assumption a 0. We have thus established that vu0 const. The lemma immediately fol-
lows, provided (11.5) holds. It remains to prove (11.15). We first show that
(11.14) lim isfinite. ytl
Indeed, suppose first that
(11.15) -' +00 11.
Set = r,, J
, and take a blow-up sequence U,. of Q — u with respect to B,.(A); for a subsequence u, -. U. By Theorem 6.1
U,,(x,0—O)=—X, if0<x<l
and, therefore, U(x, y) = —Ày if y 0. Also U(x, y) 0 if y > 0 (since U is harmonic to the left of the nozzle and vanishes on the nozzle and on (y = 0)). Replacing U in B,.(— 1,0) (r small) by a harmonic function U [U = U on aB,( — 1,0)], we decrer.se the functional J(U), thereby contradicting the fact that U is a local minimum. If
-, —oo for a sequence t 1,
then we use Theorem 6.1 to derive, for a blow-up limit U,
UAy ify>0, which is impossible, since U = 0 on N°.
From (11.14) and nondegeneracy it follows that, for any small r> 0, intersects the free boundary of u at a point (k(y,.).y,) such that
(k(Yr),Yr) C 1 —y} forsomec>0.
THE SMOO1II-FIT THEOREMS 329
Since 0, u°(x, y) > 0 if and only if x <k0(y) for some function k0(y). From (11.16) we see that, for a sequence 0,
(11.17) is finite and -. 0.
Suppose (y <y </1) is a maximal interval where k0(y) is finite valued. We claim that
(11.18) 8=0.
Indeed, if <0 then either k0(fl — 0) = —00 or k0($ — 0) = + Consider the first case and set
F':x=k0(y), .f<y<I1.
Below F' there is a connected component 1) of {u°> 0). The function c7u is subharmomc in D, = A on and it is bounded in D. By the Phragmén- Lindelof theorem it follows that in D. But then [see (l1.1I)J F' is convex to the fluid, a contradiction to k0($ — 0) = —00.
If — 0) = 00 we get a contradiction to the nonoscillation lemm& Thus (11.18) is proved.
From (11.16) and (11.18) it follows that the free boundary F° consists of F' Iwith =0, k0(0) = 01 and perhaps one horizontal line = (y 8) with u°=
thentheremusibe free-boundary points between F' and However, the proof of (11.11) shows that r' is convex to the fluid, contradicting k0(7 +0) — oo. Consequently, F° coincides with F' and (11.5) is established.
Lemma 11.2 extends to the case a =0; see Problem 3.
Completion of the Proof of Theorem 11.1. Consider first the case when the slope of N at A is finite, that is,
is finite.a
We shall prove that
(11.20) k(y)—k(i)
ifyti.
If (11.20) is not true, then there is a t 1 such that
k(y,,) — k(I) >0.
y,s — I
330 - JETS AND CAVITIES
Taking y = 1 — y,, we may assume that
u7(X) = -.
with uT as in (11.1). The free boundary of uT is given by
x = kT(y) = -!-(k(l + ?Y) — k(1)),
so that
= — k(1) '—yn
Thus (11.21) implies that
lim
By Lemma 11.2, k°(— 1) = —1, where k° is the free boundary of u°. Conse- quently,
liz — k°(—lfl>O,
which is impossible by the proof of Lemma 3.6(a). Finally,ifi=±oo,thenbyProblem3
k(y)—1 -'±00 uIyl1.
y—I
In order to complete the proof of Theorem 11.1, consider first the case
(11.22) r=0.
From (11.20) we obtain
Ik(1 +yy) e(t)JOifejO.
It follows that if —2 <Yi' Y2 <0, then
—k(1) 4e(2y).
Consequently, for any 8 >0 there is a > 0 such that
if—2<y1,y2<0,y<10(8).
THE SMOOTH-FIT THEOREMS
Thus the flatness condition of Section 4 is satisfied. It follows that
aty= —I,
where can be chosen arbitrarily small provided that 8 and y We conclude that
Ik'(I
i 0, we first perform an orthogonal transformation (x, y) -, (x', y')in a neighborhood of A so that in the new coordinates the slope ofNat A is equal to zero. Applying the previous proof and then returning to the original coordinates, we obtain the relation
k'(y)-r ifyti. This completes the proof of Theorem 11.1.
TheoreiR 113. = 0, then is cmu ssawwiy differentiable in the closure
Proof. Set
d(X) = dist(X, N).
Let N1 be a smooth curve initiating at A and lying in {u < Q}, which forms an angle 9 with the tangent to N at A, 0<9 We claim: If v = Q — u, then
(11.23) v—Ad(X)=o(r) (r1X—4I, X€N1).
Indeed, otherwise there is a sequence
such that
(11.24) c>0.
332 JFFS AND CAVII1ES
We shall derive a contradiction using a blow-up sequence
= -1-(Q — u(A +
Notice that
(11.25) whereY,,=
As in the proof of Theorem 11.1,
-' u°( X) uniformly on compact sets of R2,
and
where dist(N° U I'°). Thus if Y, then
JYf 1.
From (11.25) we then get
= + o(i)).
Also, since E N1,
and consequently,
= + o(i)),
contradicting (11.24). Having proved (11.23), now extend N into a C2 curve N U N and
denote 'by DR (R positive and small) the domain bounded by N and aBR(A) which contains a part of N1 initiating at A. Let w be a solution of
Lw0 inDR, w0 onñ,
aw = A at A (p inward normal).
THE SMOOTH-Fir THEOREMS
Then (see 5.4) w E and consequently
(11.26) w(X) =xa(X)+o(r) (rIX—AJ),
where €i( X) = dist (X, N). Denote by GR the subdomain of DR bounded by N, N1 and 88R(A) and let
W be defined by
LWO
W=O
for some p <R. By Problem 1,
(11.27) vW(X)—O ifX—"A.
Consider the function
Z=v—(1+e)w—W (foranye>O)
in From (11.23), (11.26) we see that Z<O on if p is small enough. Clearlyalso,
Z <0 in In view of (11.26), (11.27);
ew>W
if 6 is small enough, say if 8 Therefore,
o—(l+2e)w<O
Sincev—(l 8G8, wededuce thAt
on N n B8( A).
Similarly,
8w
334 JETS AND CAVITIES
and, recalling that w E we conclude that
urn exists and is equal to —A. xGN
The same is of course true for X E r'X,L, X A. Since N U is a C' curve, it follows that the derivatives along N U are continuous functions. Since these derivatives are also bounded in B8( A) fl (u < Q), for some 8 > 0, we can apply the Phragmên—Lindelöf theorem (see Problem 2) to and to
(both satisfy homogeneous elliptic equations with no lowest-order terms) and deduce that are continuous at A.
Corollary 11.4. If 0, then the curve U is continuously dif- ferentiable in a neighborhood of and is uniformly continuously differentiable in < Q) fl 88( some 6 > 0.
The proof is the same as the proof of Theorems 11.1 and 11.3; see Prob- lem 3.
Corollary 11.5. Under the conditions of Corollary 10.3,
>
Proof. If x0 then for some small 6 > 0,
(11.28) u1>u2(l +e) on(u2>0) (uj=ux,,j)
where X0 = (x0, I) and e is some small positive number. In fact, (11.28) follows by the maximum principle, noting that u1 > u2 on the closure of (u2 > 0} fl aB8(x0), for some 6>0.
From (11.28) we obtain
(11.29)
where each derivative exists, by Corollary 11.4, and is equal to 1; a contradic- tion.
PROBLEMS
1. Prove(l1.27).
[Hint: Make a conformal mapping and use Remark 5.4]
2. Extend Lemma 6.7 of Chapter 2 to elliptic operators
+
EXISTENCE AND UNIQUENESS FOR AXIALLY SYMMETRIC JETS
3. Prove Theorem 11.1 in case a = 0. [Hint: If N° = (x 0, y = 0), we have to show
ifyIl.
If (11.15) holds then by Theorem 6.1 we obtain U = —Xy (y <0) for a bbw-up limit, and J(U) can be decreased by replacing Uin 0) by a harmonic function U[U Uon aBr(—xo,o)]. If
I —
take the blow-up limit U of
u,(X) = — u(A +
where = Since (11.15) was excluded, by nondegeneracy, aR,. intersects the free boundary of U for any r < 1. Now we can proceed as in the proof of (11.15), and then show that .N° u r° is linear, a contradiction.
IfN°
then apply the proof of the nonQscillation lemma in the region G bounded byN,x = k(y),(x 0) andx (k(y,)},notingthatau/8r Xyon the free boundary and au/ar 0 on fl N. Next, (*) leads to a contradiction.]
12. EXISTENCE AND UNIQUENESS FOR AXIALLY SYMMETRIC JETS
Let be as in Definition 10.1 and set
-
Lemma 12.1. The free boundary I',, initiates at 4.
Proof.. Then
b=
JE1S AND CAVITIES
is 0 by continuity (Lemma 10.4). If b > 0, then kAM(l) > 0 for A > A — small enougii (again by Lemma 10.4), and we get a contradiction to the definition of
Recalling Lemma 10.5 and the smooth fit theorems (Theorems 11.1 and 11.3) we see that (us, satisfies all the properties of a solution of the jet problem with replaced by
We now take a sequence — co such that
—, A,
-. u weakly in and a.e. in
Since 0 Q, elliptic estimates hold for u, in bounded subsets of \ D, uniformly with respect to Since 0, the free boundary of u is a y-graph. Theorem 9.3 remains valid with the same proof for u; thus the free boundary is given by
r:x-=k(y),
Lemma 10.5 implies that 1' does not intersect (except initially at A). We next claim that
(12.1) k(I)=O.
Indeed, the proof is similar to the proof of (10.14) for y = 1. Having proved the continuous fit condition (12.1) for u, r. we can now
apply Theorems 11.1 and 11.3 to deduce the smooth fit; these theorems are local and apply to u as well as to
We have thus established that (u, F, A) forms a solution of the jet problem; this completes the proof of Theorem 8.1.
Corollary 12.2. For the solution (u, F, A) of the jet problem that was constructed above, the free boundary has the form x = k( y). h A <y < 1.
For uniqueness we require, in addition to (8.1), that
(12 2) there isa point 0' = (—b, 0) (b 0) such that N is star-shaped with respect to 0'.
Theorem 12.3. 1/N satisfies (8.1), (12.2), then the solution of the jet problem is unique.
Proof. Suppose that (u, F, A) and (u, t, A) are two different solutions. By ProbleM 21.7, N U 1' and N U t are star-shaped with respect to 0'. However, the uniqueness proof need not rely upon this fact. Indeed, because of (12.2)
EXISTENCE AND UNIQUENFSS FOR AXIALLY ins 337 and the smooth-fit, there is a point 0" (— b", 0) (with b" b) such that (i)N is star-shaped with respect to 0", and (ii) any segment from 0" to N does not intersect r and t. For simplicity we take A = (b", 1), 0" = (0,0). Proper- ties (i) and (ii) will be sufficient for carrying out the uniqueness proof.
Without loss of generality we may take A X. Consider first the case
(12.3) x>X.
By (8.4), (8.14) it follows that F lies below t for x large enough. Setting
= ü(pX) (p >0),
it follows that there exist numbers p I such that
(12.4) {u<Q} C
Let p be the largest number with this property. If p < I, then there is a point X° which belongs to the boundary of each of the sets in (since < u in (u < Q), by the maximum principle). Also, X° must belong to the free boundary of both functions; here and in the preceding sentence we make use of the properties (1), (ii).
It follows, by the maximum principle, that
au au (12.5) > - atX°
that is, pX > A, which contradicts (12.3). If p = 1, then we again argue as before with X° 4 and obtain > A, a
contradiction. Consider next the case
(12.6)
If F and I" are given for large x by y m(x) andy = th(x), respectively, then
liinm(x) =
Ii follows that there exists a p 1 such that (12.4) holds. If, for the largest such p, p < 1, then there exists a point as before and we obtain pA > A, a contradiction. If, however, p = 1, then we take X° = A and obtain, as in the
JETS AND CAVITIES
proof of (11.24),
atA,forsomeE>0.
that is, A > (1 + e)X, which is impossible.
PROBLEMS
1. Assume that
(127) Nsatisfies(8.1),x(t) —00 if t oo, and N is uniformly near oo.
oO, -, weakly in and a.e. in then we say that is a solution of problem Prove: If U1, U2 are two of problem then for x < —x0 (for some > 0),
(12.8)
(12.9) - 1x13
if y
(12.10)
ify>e0 lxi,
where is a small positive number. (Hint: To prove (12.8) near y 0, use Lemma 8.4 and consider (1/r°)u(X° + r0X). To prove (12.9), apply the maximum principle to
— ex (u1 —
2. Suppose that u1, u2 are two solutions of problem Let .Ji(x)
u2 + (u1 —
v1u1Vü2, o2u1Aü2
v2 = v2 + (u2 — v2)4.
Cv
C
lxi
CONVEXITY OF THE FREE BOUNDARY
Prove: If
J = f AI(v<Q}\DeIydxdy,Y then for any K >
+ + —
where 0 if K 00.
3. Use Problem 2 to prove the uniqueness of a solution of problem
4. Use Problem 3 to establish the continuity. Qf n A, that is, if A, then u) u weakly in (see Theorem 10.4) aa4 give a proof of Theorem 8.1 [assuming that (12.7) holds] by showing that, for some A, satisfies the continuous- and smooth-fit conditions.
13. CONVEXiTY OF THE FREE BOUNDARY 0
We shall study the shape of the free boundary for the axially symmetric jet problem. We assume that
'13 1' N is concave to the fluid, that is,
{y> 1) is a convex
Theorem 13.1. I/N satisfies (8.1), (13.1), then the free bowidary is convçx to the fluid.
This means that k"(y) 0, and consequently there is ay0 E (h, I) auth that
(13 2) k( y) is monotone decreasing for h <y <ye, k(y) is monotone increasing fory0 <y < 1.
In particular, if N is given byy = g(x), 0, O,thenF is gt*en byy =f(x),/'(x) 0,f"(x) 0.
Proof. Suppose first that
(133) Nliesinahalf-plane(x<B},
X( i) is uniformly C' for t
and consider the speed function
(13.4) qIVUI
340 .jgrs AND CAVITIES
Then
?' 0 in (u <
that is, q2 is a subsolution; thus it cannot take a local maximum the interior of the fluid. We also have
q A on the free boundary,
q(X) -.X ifX—.A (bythesmoothfit).
Further, q cannot take maximum at any point X° of N (X° A) since otherwise aq/av <0 at X0 inward normal), so that, by (1.9),
>0,
which contradicts (13.1). Suppose we can show that
(135) if A0, A0 supq, Yo) and eithery0 0 or x0 = ± 00 ory0 = ± oo, then A0 A.
Then it would follow that
(13.6) maxqA U
and then
icq = — >0 r (rinward normal),
which is the conclusion of the theorem. To prove (13.5), consider first the case Yo = 0 and let
y,,y)
Then Lu,, = 0 and, by Lemma 8.4,
u,,(x, y) h = (v).
CONVEXITY OF THE FREE BOUNDARY
Further, by the assumptions of (13.5),
I y) X0y,
For a subsequence,
e,,-.w unifotmlyinB2(O)fl{y>O),
and then
LwO,
w(x, y) (13.7)
I Vw(x, y)
I vw(O, 1)1= A0.
Setting
— Vw(O,I) e0— -,
"0
W = Vw — Aoy'
we compute
11 w 11 LW —IA0— —IA0— I
y'. y / )'k Yl inB2(O)fl {y>O),wheree=(O,1).Sincealso Wtakesitsmaxirnum(zero)at the point (0,1), the maximum prlhciple gives W 0; that is,
Vwé0A0y.
ROOAWpg I Vw A0y, 9e6 that Vw and
w(x,y)
Thus, by the second relation in (13.7), A0 A.
342 JETS AND
Consider next the case x0 = Yo > 0. Since we assume that (13.3) holds, we must have 0 <Yo < 1. Recalling that
u(x, y) — Q) 0 ifx oo, 0 <y <1,
and that I' is given, for large x, by
y = 1(x), f'(x) 0, If"(x) C (x oc),
we deduce that, for any E > 0,
Vu(x. y) — Aye 0 uniformly if-e <y <1(x), x oo.
Hence
so that A0 A. Considá next the case in (13.5) where x0 = — 00, 0 <yo < oo, and intro-
duce
u(x
y) y)
for some B of (0, yo)' and
Lw0 in (O<y< H), if—oo<x'<oo,
in{O<y<H),
where H I,. Here H is the asymptotic height of N. If H is finite, then actually w(x, H) Q and, by Corollary 9.10, w(x, y) Qy2/H2. It follows that
=A,
OF THE FREE BOUNDARY
so that X0 c A. If H = oo, then, by Corollary 9.10,
Q 1/2
}
foranyc>0. It remains to consider the case in (13.5) where Yo = + 00. X(1) IS
uniformly in for t t0, by elliptic estimates we have (see Remark 5.4)
for some C0> 1. Hence
0,
so that A0 0. We have thus completed the proof of (13.5) and thereby also the proof of the
theorem in case (13.3) holds. The proof for a general nozzle N follows by truncating N by nozzles Ne satisfying 3). 1€ we replace x(t) by min(x(t), i/c), then the first condltidn In (13.3) holds. To satisfy the second condition we replace the part of the nozzle with t by a horizontal ray, where 7 -, 00 if e -*0.
PROBLEMS
1. If A (0, b), the contraction coefficient of N is defined by
rrh2 2'irb
2/Aifb = 1, Q 1. Prove that = for [Hint: Integrate
1u2—u2 2ux —
k Y ' YJ over fl BR; by scaling. f Vu( X) C/( X over aBR. Take R -+ oo.J
2. Let N, N° be two nozzles satisfying (8.1), supb that N°ii1es to of N. Choose K41, for N and a suitable truncation N such that with the corresponding there holds: If u2 are minimizers for
344 JETS AND CAVITIES
respectively, then u1 A u2 u1 V u2 E Show that
+ = A u2) + Vu2)
and deduce that u2 everywhere.
3. If N, N° are as in Problem 2, show that for the corresponding solutions (u, r, A), (u°, r°, A°), there holds A° A, with strict inequality if N N°. [Hint: If has continuous fit at A, then apply Problem 2 to to deduce that if has continuous fit, then A
4. If N is given by y = g(x), g'(x) ' 0, g"(x) 0, then the solution u satisfies 0.
[Hint: Apply the maximum principle to
14. THE PLANE JET FLOW
The plane symmetric jet problem is the two-dimensional analog of the axially symmetric jet problem. In the conditions (8.2)—(8. 12) we make the following changes:
a2 a2 Lu = = — + —,
ax2 ay2
and (8.11) is replaced by
on!'0.
Theorem 14.1. If N satisfies (8.1), then there exists a solution of the plane symmetric jet problem.
Theorem 14.2. The solution is unique if N satisfies also the star-shape condition (12.2).
The proofs are similar to the proof of Theorems 8.1 and 12.2. Notice that in the present case
u(x, y)—Ay asx -, co in the set (u< Q),
and the asymptotic height of the jet is
ASYMMETRIC JET FLOWS 345
PROBLEMS
1. Prove that the number of inflection points of the free boundary does not exceed the number of inflection points of N plus a possible inflection at the detachment point. Here an inflection point means a point of maximum or minimum of the angle 0 between the tangent to the curve and the x-axis. [Hint: 0 = log = logs vu +iO, so that 0 cannot take local extremum at any point of the free boundary. If 0(A1) is a local maximum for 0 restricted to I', study the region (0<0(A1)) connected to A., and do the same for local thinima.J
2. If N is given by y = g(x) with g(x) monotone decreasing, then r is given by v = f(x) with f(x) monotone decreasing. [Hint: Use symmetrization in the y-direction, employing
u(x,y)<Q
below (see Figure 3.5).J
15. ASYMMETRIC JET FLOWS
In this section we consider plane flows which are not necessarily symmetric. The nozzle is made up of two curves N1, N2 satisfying:
X=X1(t)(x1(t),y1(t)), wberex,(,),y,(t)arepiecewiseCl'$a (O<a< I)
x = const. c either does not intersect N1, Or intersects it at one point, or inffisetts it at segment, andy2 (c, EN;
x1(O) = x2(O) =0 and the open inteñ'al X1(O)X2(O) does not contain anypoints of
N1UN2; x,(t)-—poifi-.oo.
an x-graph N1 lies aboveN2. We take
A =x1(o)=(o,i),
346 AND CAVITIES
Definition 15.1. The asymetric jet problem seeks a function u, nonintersecting C' curves r1, 1'2, a positive number A, and a direction e = (e,, e2) with e1 > (I such that
1', initiates at X(O), 1 U is C' (15.2) in a neighborhood of
f, is given byy1 0 <x < oo,
= 0 in the flow domain bounded by (15.3) N1ur,andN2ur2;
u is uniformly continuously differentiable (15.4) inaneighborhoodofr1 U (X,(O)} intersected
with the flow domain;
(15.5) onl'1;
(15.6) u=lonN,U1'1, u—lonN2UI'2;
finally,
f1(x) —f2(x) + a2 (dconst.)
(15.7) x) — 0 if x -. 00, where tan a =
From these conditions it follows that
(15.8) Ad = 2, where d is the asymptotic width of the jet.
Note that in general the stream function can be taken to satisfy u ± Q on I',; for simplicity, however, we normalize it by taking Q 1.
Theorem 15.1. For any nozzle N satisfying (15.1), there exists a s3lugion (u, 1', A) of the asymmetric jet problem.
This result is valid also if the last condition in (15.1) is dropped out; see Section 17, Problem I. -
The proof of 15.1 begins in this section and ends in Section 17. As a corollary of the proof we shall obtain the following result in case of a
symmetric noule.
ASYMMETRIC JET 347
Theorem 15.2. If N1 satisfies (15.1) and lies above the x-axis, then there exists a solution of the plane symmetric jet problem with nozzle N = N1.
[Here again the last condition of (15.1) may be dropped out.] For proof, see Section 17, Problem 2. in order to introduce a variational formulation, we denote by D the domain
bounded by N1, N2 and the segment AB and by the domain
where = ((x, y); x > O}; AB is called the.mouth of the jet. Define functions (1 by
mD,
1 ifx<O.(x,y)liesaboveN1,
(15.9) =—1 ifx<O,(x,y)liesbelowN2,
=—1
mD,
•(x,y) = 1
(15.10) =—I ifx<0,(x,y)ljesbelowN2,
=1
We require that both functions be bounded and this uniquely determines the functions (by the Phragmén—Lindelof theorem).
that .
ii
is iny,
•(—IL, y) y).
348 JETS ANt) CAVITIES
Introduce classes of admissible functions
K—
KM (v E K, y) =
and let
(x>
For any unit vector e = (e,, e2), denote by e by rotation counterclockwise through an angle ir/2.
For any A > 0, > 0 and unit vector e (e1, e2) with 0, we introduce the functional
(15.11) Vv —
and consider:
Problem Find u = UA,e., in K1 such that
(15.12) = flliliJAe,,(V). V E K,,
Remark 15.1. If we take in the definition of K 4 —1, 4) 1 and replace in (15.11) by then run into difficulty in the proof of
Lemma 17.1. If S is an infinite strip in the direction e with width 2/A and if
ü(x)= -I +dis&(x,a1s) IorXES, where a1s is the lower of the two lines bounding S, then we can define a function u0 in K, which coincides with ü in S {x> 1) s1ich that
JA,,(UO) <
Consequently, there exists a solution to problem By increasing rearrangemaits in they-direction we can construct a solution
u for which u, 0 (s we shall henceforth deal only with such solutions. (It will be shown in Section 17 that problem 1A,. has a unique such solution.)
We also note (see Problem 1) that
(15.13) uA,M,,isharmonicinD.
ASYMMETRIC JET FLOWS
We define
(15.14) u(X)=limf u r-.O BAX)
if the limit exists and u = 0 on the set T (of measure zero) where the limit does not exist; it will be shown below that T is the empty set.
Lemma 15.3. There exists a positive constant c (independent of A, such that if B,(X°) C then
(15.15) (1 + u) Ac implies that 1 + u >0 in Br(X°). r aBAx°)
Proof. Consider the class of functions li in H"2(BAX0)) satisfying: h = u on ao,.(x°), h u a.e., and let h be the solution of the variational inequality:
minf Jvh12=f 1Vh12, heg. hEX B,(X°) B,(X°)
Then h is superharmonic (take = h + where E 0) and h 4' [otherwise, the superharmonic function min(h, 4') would give a smaller value for the functional). Clearly also, h 4' and h = u a.e. where u = 1. Extending h by u outside BAX°), we obtain an admissible function. It follows that.
vu — IB,(X0)I —
and thus
(15.16) f Iv(1 9(X°) B,(X°) 8(X°)
'(I+u.O} B,(X°)
since h> —1 in B.(X°). We can now proceed to estimate the left-hand side of (15.16) as in Lemma 3.1, using the that h 4- 1 is andthus
that
0.
350 JETS AND CAVITIES
From (15.16) we then deduce that u = h a.e. in Since h is super- harmonic, it follows that
J hth(X) VxEBr(X°);
hence definition (15.14)1 the same is true for u and u = h everywhere in B,( X°). In particular, u> — I in X°), and the lemma follows.
We claim:
(15.17) (u = 1) and (u = —1) are closed subsets in
Indeed, if
X"-.X°,
then setting ,, = X' — X° and applying Lemma 15.3, we get
f (l+u)<cr forr<r<r0.aBAX°) Hence
f8,4 X°) so that 1 + u(X°) = 0. Thus (u = — I) is closed in and the proof for (u l} is similar.
Corollary 15.4. u E C0 -I
The proof is similar to the proof of Theorem 3.2. Since aUA. ,Jay 0,
1 if and only iff2(x) <y <f1(x).
DefinitIon 15.2. The set
1' = {x€ XE a{IuI= I), u(X) = 1)
is called the upper free boundary, and the set
r2= 1},u(X) —i)
is called the lower free boundary.
ASYMMEIRICJEF FLOWS 351
As in Section 9 we have
is continuous wherever real, and
1', = ((x, is real valued};
notice also that r, lies above r2. The nondegeneracy lemma (Lemma 3.3) extends to I + u, in any set where
u < I. It implies:
Lemma 15.5. If X° E I" and dist(X°,AB)> R for some R > C/A (C is a positive constant independent of A, then BR(X°) intersects F2 U ((0, y);
00 <)'< -1).
Indeed, if u < I in then by nondegeneracy,
sup (c>O). X°)
Since the left-hand side is 2, we get .R 2/(cA). Suppose that theconditionsu= ±1 on are replaced byu= ±Mand lb are replaced accotdingly by M+, Mlb. Let be the corresponding
15.6. Let X° be a free-boundaiy point in 0 if and let
(0<r<R).
Then
i M . I
'-.
where C isa positive constant independent of M.
This is the bounded gradient lemma; for proof, see Problem 2.
Lemma 15.7. If the free boundaries 1 are connected unbounded curves with initial X, = (0, y1), and if 8=1 X1 — X2 >0, then
x>O
where C is a universal constant.
352 AND CAVITIES
For proof, see Problem 3. The proof also shows that if X' = X2, then
I vuH CA
in the region bounded by l'l, and aBR( Xi). But this is impossible since u = I on r, and u = — 1 on I'2. Thus we must have X' X2. Further, if we connect
to
by half a circle y lying in x > 0, then
y21
Thus:
Lemma 15.8. Under the assumptions of Lemma 15.6,
(15.18) IX' —
where c is a positive universal constant.
PROBLEMS
I. Prove(15.13). [Hint: For any ball B in IL),
niinf)vvI2 f I if v = uon v c
and by regularity,
+ = 0 in B.]
2. Prove Lemma 15.6. [Hint: See the proof of Lemma 5.1 and take B,(Xk0+I) with the smallest r (r r0) such that aBR(Xk0+I) contains a free boundary point Yo' say with u= —1. Then
i)
11ff FREE BOUNDARY FOR THE ASYMMETRIC CASE
3. Prove Lemma 15.7 by the method of Lemma 10.6, usmg a variant of Lemma 15.3.
16. THE. FREE BOUNDARY FOR THE ASYMMETRIC CASE
Definition 16.1. By a constant flow in direction e with speed A, we mean the function w given by
I lfO<IsI<X,treal,
= —1 ifs<
= 1 ifs>f,ireal.
Lemma 16.1. Let XA = belong to 1'2 and -. 00. Then for a subse- quence
*
uniformly in compact subsets of R2, where w is the constant flow in the directio,t e wuh speed X; moreover, the free bQundaries of u,, converge to the free boundaiy of the constant flow in C. The same conclusion holds if X,, E r1.
?rvof. Since
I Vu(X) — 12 C0 < 00,
we have, for any C > 0,
11, — 12_, 0 if n—.'øo.
Without loss of generality we may assume that u,, u0 weakly in .11 follows that
.
Thus
Vu0(X) =
354 JEIS AND CAVITIES
Since + X) I c, if XI< C, for a suitable c1 independent of C (provided that n is sufficiently large), we conclude that for a subsequence
u0 uniformiy in compact subsets, and that
u(X+X)<1'I Cl
Hence by the nondegeneracy lemma (recall that u( — 1],
r c1
We conclude that u0 — 1 in Br, With = — 1. Introduce orthogonal coordinates (x', y') such that e is in the direction of
the positive x'-axis and e1 is in the direction of the positive y'-axis. Set = (x', y'), w'(X') = u0(X). From (16.1) we have
aw' aw' (16.2) 0, = a.e.
Thus w'( X') is a function w0(y') of y' only and it is monotone nondecreasing my'. —iinanyneighborhoodofO,w0(Q)= —1 and—i 1, it follows that
—l+Xy'
w'(X') —1 ify'<O,
I ify'..4,
and the first statement of the lemma follows. Denote by the subsequence for which u1,. u0. To prove C convergence of the free boundaries, consider first the case of e
nonvertical. By nondegeneracy it follows that the upper free boundaries of in BR and the upper free boundary of u0 lie each within a 8-neighborhood of one another provided that n is large enough; here R is arbitrarily large and 8 is arbitrarily small.
l'his yields the flatness condition of Section 4 and thus implies that the upper free boundaryof u,, in BR converges to the upper free boundary of u0 in the C' norm. Thus
if IxI<R,n'-i 00.
A similar analysis applies and to the case where e is vertical.
THE FREE BOUNDARY FOR THE ASYMMETRIC CASE
Lemma 16.2. Let X, = belong to I'2 and —' a finite nonnegative number, v,1 + oo; then for a subsequence
+ X) — w(x — e)
uniformly in compact subsets of (x> — where w is the constant flow with speed A and direction e = (0, — 1). The same assertion holds for X,, E r1.
Proof. In the present case u0( X) is' defined in the strip Rc ((x, Y); x —E). Also, u0 I in any Re-neighborhood of (0,0), as before. Further,
(16.3) y) I ify1,, +00
and
(16.4) —.o.
Since u0(0,0) = —1, (16.3) cannot occur. Next, e must be the vector (0, —1) for otherwise we shall have, by (16.4) and aw/ax' = 0 (cf. (16.2)) that
—l in RE. We can now complete the proof of the lemma as before.
Lemma 16.3. There cannot exist a line
—oo<a<O,
which is tangent to the lower free boundary and stays above it, and which intersects (x 0) at 8> 1 (that is, above A).
Proof. Suppose that such a line 1 exists. Denote by w the constant flow with speed A and direction e1 = (1, a)/ + a2, and defme
W°(x, y) = w(x, y ii'), a
where the lower free boundary of wdo is I. In view of Lemma 15.5, r,must lie entirely within some distance R of 1, and therefore there exists a smallest value
w0(X) u(X) everywhere in
and we may assume that equality holds at some point X in the cksure øf u < 1). (If not, we argue with a Iiney = aex + — e having the same
properties as I.) Suppose first that u(X)> --1. By the strong maxim*fl principle X cannot belong to the set (j u < 1). Thus_it must lie on the boundary of the strip S° w°)as well as on By (which applies here as well), 1', (i {x 0, y> I } is nonempty, then is C'
356 JETS AND CAVrI1ES
at x = 0 and its tangent is in the direction of the positive y-axis. it follows that in fact lie in (x = 0) and thus X E r1.
The strong maximum principle now gives
8w° au —A=—>—=X atX, av av
which is impossible. If u(X) — 1, then the last relation holds at the point X of I fl I'2 and this is again impossible.
Lemma 16.4. There is a constant R > 0 such that each ball 8R( X°) C \ 1) with I u( X°) < I contains a free-boundary point.
This implies that U is nonempty.
Proof. If F, U F2 does not intersect BR( X°), then by the nondegeneracy lemma,
sup
for any R sufficiently large. This is impossible since the left-hand side is 2.
Lemma 16.5. 1/ e (0, 1), then is nonempty.
Proof. Suppose that r, is empty. Then u < I in \. D and by Lemma 16.4 there is a sequence X,, (x,,, in I'2 such that x,, C, + oo. But then, by Lemma 16.2, e (0, 1), contradicting our assumption on e.
Lemma 16.6. If e (0, — 1), then F2 is empty.
Proof. If F2 is nonempty, then, by Lemmas 16.1 and 16.2, for any sequence = (xe, y,,) El'2 with oo we must have
xl, yn-,—co, 7ç—'O.
fyI,,
Indeed, otherwise there exists a sequence of points Y, on F,, with slope a (a some negative finite number) such that Y,, f- oo. But this contradicts Lemmas 16.1 and 16.2. It follows that lies below any liney = ax + with arbitrary a, —oo <a <0, and a suitable ft. It iS easy to see that a line I as in Lemma 16.3 can then be constructed, contradicting Lemma 16.3.
Lemma 16.7. If e ± (0, 1), then r, is connected andf,(x) is finite valued for allO<x<oo. I
THE FREE BOUNDARY FOR THE ASYMMETRIC CASE
Proof. By Lemma 16.5, r, is nonempty. Consider a maximal interval (a < x <b) is finite valued. If h < oo, thenf1 cannot have a finite limit for x t b, since is a smooth curve. Hence
Ifi(x)I—'oo ifxlb,
thus implying (by Lemma 16.2) that e is vertical. We have thus proved that b = oo. The proof that a = 0 is similar.
Lemma 16.8. If e = (0, — I), then I'1 is connected, f1(x) is finite on some interval 0 < x < b (b < 00) andf,(x) —i —00 if x t b.
Proof. By Lemma 16.5, r, is nonempty. For any maximal interval (c < x <d) where f1(x) is finite, we must have, by Lemma 16.2,
ifxtd, providedthatd<oo,
f1(x) — oo if x ,i c, provided that c >0.
Also, since "2 is empty (by Lemma 16.6), 15.5 implies that d < oo. If c > 0, then we obtain a contradiction to the nonoscillation lemma in a region
{c<x<d,—co<y<—y0),
where Yo is sufficiently large. We conclude that c = 0, and this completes the proof.
We summarize:
Theorem 16.9.
(i) If e is nonvertical, then F2 are nonempty and given by
y—f,(x), O<x<00,
where f,( x) is continuous,
asx-'oo, e1
exists, x
and — 12(0) c/A > 0 (c > 0). (ii) If = (0, 1), then is empty and is nonemply and is given by
yf2(x), 0<x<b2
358 JE1S AND CAVITIES
where f2(x) is continuous,
f2(x)-i +00
and
12(0) exists;
(iii) If e (0, —1), then 1'2 is empty and r, is nonempty and is given by
= f1(x), 0 <x <b1 (b1 <oo),
where f1(x) is continuous,
f1(x) -. —oo,f(x) —00 asx tb1,
and
exists. x
Using Theorem 1.4 of Chapter 2, we further deduce that, in case (1),
dJf(x) dx'
for anyj 2. The smooth-fit theorems of Section 11 are valid also in the present case with
essentially the same proof.
17. MONOTONICITY, CONTINUITY, AND EXISTENCE FOR THE PROBLEM
Leu.ma 17.1. Let e = (e1, e2), ê = (ë,, e2) be unit vectors with e1 0, i, 0 such that i is obtained from e by rotation counterclockwise by a nonnegative angle <,i. Then
(17.1) UXepi
Here = and u2 = are any minimizers of Problems and respectively
MONOTONICITY, CONTINUITY, AND EXISTENCE
Proof. Set
v1u1Vu2, v2=u1Au2.
We claim that
(17.2)
Indeed, if ê= e, then the proof is as in Section 10. Suppose then that ë e. Denote by 12 the straight lines in the directions of and respectively, passing through a point X° = (x°, y°); X° is taken on a ray from the origin in the direction of e + ë. Denote by H, the half-plane bounded by 1, which contains the origin, and set
S°—R2÷\S.
By Theorem 16.9 we can choose X° sufficiently large so that the sets
nS°, {Iu2I<1) flS°
are disjoint and
1) flSis bounded and haspositivedistance
j) = (1,2) or (1, j) (2,1).
Let
Then we can write
Noting that v1 = u1, v2 = u2 in S°, it remains to prove (17.2) for Note that
J1(w) + All) — if
where I is in the direction of if w = or w = v1 ,pbereas 1 is in the direction!2 of J-'- if w= u2 or w v2. Since(17.2)boldsfor thefunctionalJ, it
360 JETS AND CAVITIES
remains to prove that
av, av2 au, av2 7.3)
J Js'uL,
where the right-hand side in each equality is a one-dimensional integral, and noting that (v1) (U1] since v. u, at those points of where [.. J is evaluated. (17.3) follows.
Having proved (17.2), we can now complete the proof of Lemma 17.1 by the argument used in the proof of Theorem 10.1.
Lemma 17.2. There is a unique solution UA to prohiem e. satisfying 0.
Proof. If u1, u2 are two solutions, then set u(x, u1(x, y — e) and
—'A —'VUI — UI U2, V2 — U1 U2.
We can establish the analog of (10.8) for by essentially the same proof, and then proceed as in the proof of Theorem 10.1.
Denote the free boundaries for by thus
.v
and
17.3w If e = (1,0), then there exists a positive constant c such that for all A small (say A <Ito)
(17.4)
c and A0 are in(kpeadNt 0/44.
Proof. From Theorem 16.9, we have
I ' ' 2 C fA.e.uI.°JfX,e.$kW>X
therefore, if the assertion is not true, say for f2, then for any small £ > 0 the
MONOTONICITY, CONTINUITY. AND EXISTENCE 361
lower free boundary starts on x = 0, y — e/A and the upper free boundary starts on x = 0, y > 2€/A.
Let
GG'\(u —1). Then a part 80G of consists of subarcs of the lower free boundary, and the remaining part 81G of consists of circular parts of and of the segment
2€x0, By the bounded gradient lemma (applied u IA). we then have
mG.
where C is independent of A. Take a point (0, j) E i.e..
- 2€
and connect it to point X° on by a circular arc y lying in G. Then
2 = u(0.1) —u(X°) C€.
which is impossible if e is small enough.
Lemma 17.4. If e is norn,enioaJ. then either
(17$). I < L
or
(17.6) mm {
I — + .1) 4
for all A sufficient/v large: C is a constant independent of e.
Proof. Set u = I" = Since
.!..j r r
362 JETS AND CAVITIES
the nondegeneracy lemma implies that if X° E 1-2, then must intersect (u = 1) (otherwise, 1 + u = 0 in a neighborhood of X°); here c is positive and is independent of e.
Similarly, if X° E then intersects (u = — I). Hence, if the assertion of the lemma is not true, then
intersects say at X',
intersects 1-2, say at X2,
where A1 is the initial point of 1". But this gives a situation contradictory to the bounded gradient lemma.
Proof of Theorem 15.1. We now have all the ingredients necessary for completing the proof of Theorem 15.1. We say that A belongs to the set if there is a nonvertical e such that
I, < — I.
Let
= sup(A; A
By Lemma 17.3, is nonempty and contains all A sufficiently small. By Lemma 17.4, A, C for some positive C independent of
Using Lemma 17.2, we can establish as in Lemma 10.4 that e depends
continuously on A, pi, e. More precisely,
-. A, -, e,, e, then
, ,, weakly in and a.e.,
(17.8)
If e is vertical, say e = (0, 1), then (17.8) for 1 1 is taken in the sense that
-•. +00.
Now let A,, tA, and choose e,, such that < — I and 1, e,, -' e,,. By continuity
(17.9) 1, _l.
We claim that
(17.10) isnonvertical,
MONOTONICIIY, CONTINUITY, AND EXISTENCE 3t1
and thus are finite. Indeed, suppose for instance that = (0, —1). Then >0, we have
(17.11) ifx 0.
Indeed, the inequality "a' "follows from the monotoncty of Lemma 17.1, whereas the strict inequality is deduced as in the proofs of Corollaries 10.3 and 11.5.
We now choose ë — e,, small enough so that, by continuity,
< —2.
Since I [by (17.9), (17.11)1, we canchooseA > A1, with A A,, small so that
,, < — I, ,,(0)> I
(by continuity again). Thus A E a contradiction since A > A,,. Having proved (17.10) we can now easily show that equalities must hold in
(17.9). Indeed, suppose for instance that
I.
Let e = (J1, J2) (with e1 >0) be a vector obtained by rotating e clockwise. If e — e is small, then, by continuity,
1.
By strict monotonicity [see (17.11)1 we also have
<—1.
But then, as before, A E if A > A,, and A — A,, is small enough; a contradic- tion.
We have thus completed the proof of the continuous fit for then also have (as in Section 11) a smooth fit at the detachment pothtiA and
To ooniplete the proof, we take a sequence and
A,, = A,,, e, = e,,,.
Then, for a subsequence, A,, - A, e,, -. e and
-. u weakly in and a.e.
364 JETS AND CAVITIES
We can now prove that there is a continuous fit for u [by the same argument as used to prove (1 7.8)J. It also follows that e is nonvertical. The proof that u is a solution of the jet problem can now be completed by already familiar argu- ments (see Section 12).
PROBLEMS
1. Extend Theorem 15.! to the case where the last condition in (15.1) is dropped out. (Hint: the nozzle by nozzles satisfying (15.1).]
2. Prove Theorem 15.2.
[Hint: Let N2 denote the reflection of N, about the x-axis. By uniqueness UAe,1(X, y) = —y) provided that e = (1,0) and
Choose A with Continuous fit at A.]
3. Extend the results of Problems 1 to 4 of Chapter 12, to the asymmetric case assuming that N, U N2 lies in a sector of opening 00 < ir. [Hint: Consider ±(u, — u2) + (Coos a0)/ra to deduce
Iv(u,
4. Denote by N,* the reflection of N, with respect to the x-axis, and assume that
Prove that for any solution (u, I', A, e) of the jet problem obtained by the procedure of this section,
ife=(e,,e2).
(Hint: Let u, = =(l,0). Then y) —u,(x, —y) solves Problem j for the reflected nozzle N*; denote this problem by
Au, is in K, Vu1 inthe class to N*. Show proof of Lemma 17.1) that
u* A u1 =
next compare with Problem 2 of Section 13.]
S. Jn Problems 5 to 10 we outline a solution of the impinging jet problem by the methods of Sections 15 to 17. The problem is similar to the asymmet-
•1
The variational functionil Is:
=f(vv +
where D = fl (x <0), and the adi*isMble class is
lOflM,,V:TIOflN2
v = Oon (x = a), y) =.h1(y)), ..% I
where is monotone increasing, 1 if(—JL, j') E = —1 if (—n, y) E N2, and — I c 8 1. Prove that there exists a unique
MONOTONJCIIY, CONTINUITY, AND EXISTENCE 365
(a,O)
x —a
FIGURE 3.6
nc jet problem except for the presence of a wall x a (a > 0), which blocks the path of the jet. Thus we anticipate that the free boundary will
of two curves (see Figure 3.6).
1712
The unknown parameters are A and the value 0 of u on the wall x = a (—1 <0 < 1). We shall use the same notation as in Section 15 except for replacing by fl (x <a) (we denote this new domain again by and taking a fixed unit vector e, namely
366 JETS AND CAVflIES
solution u = UAB,6 of problem "A 8,1•
U E K8,1, = mm t, K..
(Hint: Compare u1(x, y) with u2(x, y —
6. If 1's, are nonempty, then (17.12) holds.
I Hint: If both "2 go to — 00, then u solves the jet problem without a wall and "2 must be empty (using Lemma I 6.3).J
7. is monotone increasing in 0.
8. Set X8M SUPAEIX; X belongs to if for some 0 E (—1, 1), starts above A and F2 starts below B. Let A. A, t A8,,1, 0 (0, corre- sponds to A. in the definition of A1 E E). Prove that 0 ± 1. [Hint: If 0 ± 1, take a constant flow in vertical direction and speed A8,,1 and argue as in Lemma 16.3.]
9. Prove that the impinging jet problem has a solution with free boundaries satisfying (17.12).
10. If the nozzle is symmetric with respect to the x-axis and if is shaped with respect to (0,0), then there exists a unique symmetric solution u, 1', 0 to the impinging jet problem Ithat is, 0 = 0 and u(x, —y) = —u(x,y)J.
18. JETS WITH GRAVITY
The theory developed in Sections 8 to 17 can be extended to jets in a gravity field. As is customary, we take the gravity force to be in the negative y-direction. We shall consider here only the axially symmetric case, with the axis of symmetry in the direction of the gravity force. The nozzle N satisfies:
N is a piecewise curve given by X(t) = (x(i), (0<a<l);
X(O)A(a,0) (a>0); '181'
x(t)>aandX(t)isanx-graph; "
y(t) >0 forO <1 <ia; >0;
and finally,
N is sw-shaped with respect to (18.2) somepointo*=(o,y*);
Set
a2u a2u I 8u
ax2 8y2 x ax
Definition 18.1. The jet problem consists of finding a stream function u(x, y), a curve r: y = (the free boundary) arid constants A >0, Q >0 such that
y = $(x) is defined and continuous for 0 <x a,
4(a) 0,
NU
r is a C' curve,
(18.3) Lu =0 inthe flow region I bounded by N V rand they-axis,
u(O, y) Ofor -:-00 <y< 00,
uQonNUF,
18u %VA—y onr,
u is in C' in I U (N U I'), except at the points
where N is not in
finally,
for some largey0 >0, F Lv < Yo) is given by
(18.4) xf(y) and
f(y)= asy—'—oo.
Denote by a0 the angle formed by the segment To connecting 4 to the origin and the tangent ray to N at A.
368 AND CAVITIES
Theorem 18.1
(i) Let N satisfy (18.1), (18.2). Then there exists a continuous and strictly monotone increasing function k( A) (0 A < oo) with k(O) 0, such that a solution to the jet problem with parameters Q, A exists if and only if
Q (Q = k(A)) is unique. (iii) If a0 < 2ir/3 then k(O)> 0, and if a0 3ir/2 then
k(0) = 0.
The function Q = k(X) is called the solution curve. The proof is given in this and in the following section. One can further show (see reference 7c) that if N is also uniformly C' at
then
k(A) (18.5) I'm exists.
and is the contraction coefficient of the nozzle N. Iii order to introduce a variational approach, we need some notation. Ict he any positive number, > A. Denote by the segment on p
froiii 0 until the first intersection with N; the point of intersection is denoted by (as, hM). We further introduce (see Figure 3.7)
NM, the union of the part of N from A to (as. and the
I);
the interval (y = 0,0 <x <a);
the region bounded by y0, NM, the segment {y = + 1,0 <x < a,j
(18.6) and they-axis fromy = 0 toy = + 1;
M, the ray (x = a, —oo <y 0);
the half strip (0 <x <a, — oo <y <A);
=
mt U
JETS WITh GRAVITY 369
We define a function on N,. by
=Q(p+1--y)
We next introduce a class of admissible functions, depending on Q and the truncating parameter
KQM = {v; v flBR) for aliR >0,0 Qa.e.,
(18.7)
and
v(O, y)= Oif —oo <y <
I I,
'U,
NM
M
FIGURE 3.7
370 JETS ANt) CAVITIES
We finally introduce the functional J depending on A, Q, and
(18.8) =1 V-vv —
where e is the vector e = (1,0), and A 0, Q > 0.
Problem Find u in KQ,,1 such that
(18.9) mm
Lemma 18.2. There exists a function v in KQ.M such that o) < oc.
Proof. Take Yo = A — and define
x2 TAY 1/4'
o(x,y)
Q ifx>114,
Then
f —VV — A _Y1{f,<Q)flEFIxdXdY
fYo X oo.
0 -ool6(A—y)2
We can now clearly extend the definition of v into {y > Yo) so that it belongs to and < 00.
Using Lemma 18.2, we can now proceed to prove that problem has a solution u = UXQM. Notice that Lu = 0 in
Lemma 18.3. There exists a constant C depending on Q but not on A, gi, such that
I Cx2jX+2—y InEA, (18.10) UAQ
{Cx2 inl2,fl(y>A).
JETS WITh GRAVITY 371
Proof. Let u UAQM and choose large constants CQ > 0, cx > A (to be defined below). Let
G= {O<x<a,y<cjj, (18.11) [(x,y)EG],
where
2 2 2r x
Then
a2
(1 + a2)3'2
Therefore, jf is large enough, then
(18.12) v>Q onaG\{x=0).
Since — y) 0, whereas L(x2/r3) = 0, we have
(18.13) mG.
Moreover, if
(18.14) and
then we have for y A the estimate
(18.15)
(cx—y)
now choose
(18.16)
(later we shall choose a better value for cA, which will not depend on ii). Since
372 -JETS AND CAVITIES
'c7v is smooth below (y = CA) [here we make use of (18.16)1 and since ('18.12) holds, we can compare u with
Jmin(u,v)
We have
(18.17)
Using (18.13), (18.15), we obtain as in the proof of Lemma 8.4 (recall that u< Q in (y > A)) that
(18.18) 2—yx2
We shall use (18.18) in order to compare u with u. with a choice of independent namely with as in (18.14).
ForXEGfl (u>v), we have
(18.19) v(X) :i(X)
that is, ü = u in a neighborhood of the singular points of vv/x. Therefore, we can again use u as a comparison function and repeat the argument based on (18.17). We thus obtain (18.18) with + 2 replaced by cx; that is.
(18.20)
where C is a constant depending on Q. Finally, we use the function 15(x, y) = Cx2 (we assume that Ca2 > Q) as a
comparison function for u in
Since by (18.20), u < 15 on we obtain u(x, y) Cx2 fory > A.
Lemma 18.4. There exist positive constants C, C0 depending on Q but not on A, gL, such that if X° (x°, y0) and
1/2 + 2 for —oo
I / 2
forX<y°<gA+l.
JE'IS WITH GRAVITY 373
then
I :f—oo<y°<A (18.21)
jCx° :fX<y°<p+1.
Proof. Take r0 = x°/2. By Lemma 18.3,
mBro(X°)
and, consequently,
Hence the normalized function
a(X) = +
satisfies
mB)(O),
which implies
Cx0!/A + 2 —y°.
The proof of (18.21) fory° > A is similar. From the general results of Section 3 we have that the solution u =
Lipschitz continuous in 0) Li (4)); the Upschitz cqntinuity in a neighborhood of {x =0) was proved 18,4. 4, Ibç free boundary
r_=a(u<zQ}n{u=Q}
Lenin, 1&5. Thea e*irtr.
Proof. Using decreasing rearraogemeflts in the y-directiôn, We get a solu- tion u with 0. 3bat and u, are two such solutions of problem
for O<e<1.
374 AND CAVITIES
Denote by the domain obtained by translating upward by e, and denote the corresponding KQ by and Q• Thus
'XQM(V) — + fl
Since N is an x-graph, the function v1 = A belongs to and the t)2 = UI V belongs to One can check that
(18.22)
This implies
U1
in e connected component D of (u1 <Q) the top of In
mD.
Since au1/ay 0, we conclude that D has the form (y > *P(x)} for some 4'. As a part of the free boundary, the graph of 4' is smooth. Therefore the function
mD U=1Q
is admissible; hence J(u1) J(ü) which implies au1/ay = 0 in \ D. Conse- quently the open set (u1 <Q) is connected and thus
in(u1<Q}.
Taking e -, 0 we obtain u2 in {u1 <Q). Similarly, u1 u2 in (u2 < Q). It follows that u1 u2. This completes the proof of the lemma.
Since the free boundary is analytic and u7 0, we can write the free boundary in the formy = 4(x), where is continuous and finite valued on some open subset of {O <x <a).
Lemma 1&6. The free boundary r in (0 <x < a) is given by a continuous functiony = 0 <x < a.
Proof. We first show that the domain of definition of is precisely one interval. Suppose that this is not true. Then there is a value I E (0, a) such that 1' doec not intersect (x = I). but r contains points in {x <i} arid in
JEFS wrni GRAViTY 375
(x > It follows that there are curves
such that
Take r0 small and X° = y°), where y° <0, sufficiently large. Then X°) contains free boundary points; hence by the nondegeneracy lemma
(Lemma 3.3 with Q replaced by — y)
mm 8,,(X0) B,il2(X°)
as Yo -' — 00, a contradiction. We have thus proved that the domain of definition of 4(x) is one interval.
The right endpoint must be x = a, for otherwise we can derive a contradictiop as before. Similarly, the left endpoint must be x = 0.
Lemma 18.7. exists and is finite [we shall denote it by
Proof. If lilnx....a+(x) = — 00, then we get a contradiction by the argwnent of the previous lemma. If +(x) oscillates as x a, then we can derive a contradiction to the nonoscillation lemma.
Notice that —00
I4amia 18.8. The free boundary F does not intersect the line y = A, that is,
+(x)<ZA EfO<x<a.
Proof. Suppose that X° = (x°, A) is a free-boun4ary point where 0 < x° <a. By Lemma 3.1 (see also Lemma 8.3),
JaR,(x°) for small r, where we define
ift<O.
Hence
(18.23) Jaa,(x°)
376 AND CAVITEES
Consider the function w satisfying
LwO inD,L, wO ony=A, O<x'za, w = Q — u on the remaining boundary of
Then 0 < w < Q — u in by the maximum principle. We now introduce Green's function 6, for L in the half disc
B=Br(X°)fl {y>A).
Since Lw Oin B, we can write, for X in B,,4X°) fl (y > A),
w(X)_J aa aan(y>x) aB(X°)
by (18.23). If we take X (x°, A + r/8), we obtain for r 0
(18.24) -
a contradiction to the maximum principle.
The proof above extends to the case 9 a, provided that N does not contain the point (a, A).
Lemma 18.9. There exists a y0 large such that r fl < —ye) £5 given by x = f(y), where
1(y) + o(i)), f'(y) = o(1) I.y I
fory—' —co.
Proof. —oo and take anyR >0. Then
/ X Consider the transfonnation
JE1S wmi GRAVITY 377
The integral above becomes
I
((x 1/2 1 (A Y)1/2(1 — 1/2 2
—
(A
(A —
21
+ a,)
Therefore, the last integral tends to zero as n -, 00; hence we have, for a subsequence,
ü,,-'ü
a.e.,
1;
further,
I 'tIi<Q) 0 in
inLL;
that is,
au au (18.27) and
Since y = I a.e. on (ii < Q), we conclude that
(18.28) 1
But since ü =0 on the 9-axis (recall that u,, 0 on the fraxis and ü,, -' ü
JETS AND CAVITIES
weakly in it follows from (18.27) and (18.28) that
for a.a. small r>O,
urn (Q—u,,)=O ?
Therefore, by nondegeneracy cannot be a free-boundary point for for large n.
Similarly, for i <
uim!f ?t0Q r &B,.(X)
which implies that is not in 8{u,, < Q) for large n. This shows that < Q) converges to {i = } locally in the Hausdorff metric. In
particular,
(18.29) (A — = —*
Note that ü,1 is a minimizer of the functional
( —
in BR(O) for large n, and the free boundary is given by
= I +
+ o(i)) -.
[wheref is not necessarily single valuedj. Thus for large n the free boundary for u, satisfies the flatness condition of Section 4, which implies that is single valued and
for each R, where —, 0; this and (18.29) complete the proof of (18.25).
THE CONTINUOUS FiT FOR TilE GRAVITY CASE
Lemma 1810
(i) <A2, then UAQP UA2,Q.M. (ii) If Q, > Q2, then
For proof, see Problem 2. From the monotonicity for u we obtain strict monotonicity for the corre-
sponding free boundaries. We also obtain continuity of with respect to
PROBLEMS
I. Prove (18.22). (Note that the inequality is a consequence of the monotonic- —y my.)
2. Prove Lemma 18.10., film:: See the proof of (18.22).1
3. In case of symmetric plane flow under gravity.
u = Q, = on thefree boundary,
The variational functional is
Vt) — 12 ctc dy.
(a) Extend Lemma 18.2, taking v = xbIX —y ify <y0. (b) Extend Lemma 18.9; here
f(-) 2(1 +o(1)) —co.
19. ThE CONTINUOUS FIT FOR THE G*AV1'IY CASE
Let 'Ti. be two C2 arcs in Ii*ting at"A' and forming angle a. Denote by G the domain bounded by and for some sthall R > 0.
We introduce polar coordinates (r,9) with center at A, such that 0=0 corresponds to the tangent to y1 at.A.
JETS AND CAWTIES
Lemma 19.1. Let w be a bounded positive solution of
Lw0 mG, w0 onGfl('y1 Uy2).
then forx E G
(19.1) Uy2),r=IX—A(
for some C >0, and
(19.2) forX E U
Proof. By reference 128, p. 236, w has the form
w(X) !g + w0(X)
for some constant a0, where w0 is of class (see reference 128, p. 231) provided that k + I # k a positive integer and iT/a < k + I 2ir/a. Thus we have
1w12
I (Notice that we can choose k + 1 3 if a IT/2 and k + 1 2 if ii'/2 <
Now let X°Ey1 Uy2, r=JX°—AJ>0, and d dist(X, U y2). Then, if e is small, we have for e e' 2e, the estimate
w0. r X0) r as,.,( X°)
Integrating over e', we obtain
T 82.AX°)\B,AX°)
( 1/2
r 82,,4X°)NB,,(X°)
"d \2(k+2) 1 1/2
r Iwo(2)12d2
r2
TIlE CONTINUOUS FIT FOR TRE GRAVITY CASE
Dividing by d and letting d —* 0, we get (19.2). The estimate of w(X) for all X follows by Harnack's inequality (see Problem 1).
Lemma 19.2. If a0 2ir/3, then for any Q > 0 there existsa small enough A > 0 such that
<0.
Proof. If the assertion is not true, then, by continuity, it follows that the free boundaryy = 4(x) of u0 u0 Q.t starts at A [by Lemma 18.8, 4(x) <0 if 0 <x <a]. We proceed to derive a contradiction.
Let 1 be a line segment initiating at A and contained in such that there are no free boundary points converging to A and lying above 1. Denote the endpoint of I by A, (a,, b,); let B, = (0, b,). Denote by D* the domain bounded by N,B, I, A,B, and the linesy = + I and x = 0, and by a the angle between and / at A.
Let w be the solution of
Lw0 inD*, w=Q—u0 onaD*\y*(y*=IuA,B,), w0
Since +(x) <0 if 0 < x < a, we can choose Ishort enough so as to ensure that there are no free-boundary points above A,B,; thus
u0<Q, Lu0=O inD*.
By the maximum principle we then deduce that
(19.3) w<Q—u0 inD*
By Lemma 3.1 and Remark 3.1, we have:
if there exists a free boundary point in
for some X0E!nearA (ro=IX0—AI.0<e<zl),
then
(19.5) max y10—y
ifaa0
C is a universal constant and Br(I) v-neighborhood of 1.
382 JETS AND CAVITIES
Consider first the case where a ,i and N is a y-gjaph near A. Let w0 be the harmonic function in D fl (y <8) (for some small 8 >0) with
w0 = Q on they-axis,
w0 = 0 elsewhere on aD;.
Then 0, so that
Lw0=
Thus, by the maximum principle,
w w0.
By comparison,
w0 (c0 >
which gives a lower bound for w in If a > ii or if N is not a y-graph, let be an arc in D initiating at A and
forming angle a =min(a, with at A. Let be a domain bounded by Y*, and aBR(A), for some small enough R, so that D** C D. We can now repeat the construction of w0 with respect to D**, and again obtain a lower bound for w.
From these lower bounds and from (19.3) we conclude that if r0 is small,
f [c,>O]an,,0(x0)nD\ and therefore, by (19.5),
if a = a0,
if a > a0
for some constant C. Since a0 we obtain a contradiction in both cases provided that e is small enaugh if a = a0 and provided that r0 is small enough (and e fixed, say e = in case a > a0. (Notice that N is a y-graph near A if a0 <
Thus (19.4) cannot be true. But if we choose I with slope
(8>0),
THE CONTINUOUS FIT FOR THE GRAVITY CASE
then the fact that (19.4) is not true for some e > 0 and all r small implies that
that is e < 6, which is impossible if 8 is small enough (since e is independent of 8). This contradiction completes the proof of the lemma.
Lemma 19.3. If ir/2 < 27r/3, then there exists an r > 0 (depending on N and q below, but not on such that if Q <e, then the free boundary of u0 starts at A; furthermore, in a small neighborhood of A, 'the free boundary lies above a curve
(19.6) y=—,i(a—x)8 t)J.
where ij is any positive constant, that is,
(19.7) 4hJQ,(X) ; x)8 ifa— e' <x <a for some small e' > 0.
Note that 1 <8 2 [8 = 2 if a0 — and 8 I if a0 (2ii/3)J.
Proof. Construct a smooth curve y: x = h(y) (— oc <y.< 0) which coin- cides with (19.6) in some neighborhood of A and which decreases to a/2 as y — cc. If suffices to show:
ifQissma!lenoughthenforanyO<Ac 1 (19.8)
the free boundary of lies above y.
To prove (19.8), introduce the domain
Let v be the
on
v0 ony. From Lemma 19.4 (below) it follows that (19.8) holds for A 1, if Q< s and e is small enough.
JETS ANI) CAVITJE.S
If follows by continuity and monotonicy-of in A that if (19.$) is not true for se-me 0 A < 1. then there a value A 0 <A < 1 for which holds. The free boundary of then lies above y and it is tangent to it at some finite point X (.i. (the fact that Xis a finite point is a conscqucnce of Lemma 18.9 and the definition of y. and X = A would be a contradiction to the fact thatcontinuous fit implies smooth fit).
By the niaximurn principle.
Q — <Qu
in the set < Q). By nondegeneracy we have
(19.9) cavA—c—r'c—f (Q—.UAQ,a)c!J Qv.r r aB,(x)
where c > 0 is a universal constant and r < mm {r0, A — is small but Since the right-hand side is CQ/r, we see that
(19 10) in a small neighborhood of A.
Wt' now apply Lemma 19.1 to v to deduce that provided r 0, the right-hand side of (19.9) is bounded by
I I i — 1)/a
Cal
we have used here the fact that I lies on the curve (19.6). Using this inequality in (19.9) with r —. 0, we obtain
1/2
provided that (we recall) Q is small enough, say Q "a. If we now choose e small enough, then we arrive at a contradiction.
Leimu* 19.4. There a universal constant c >0 such that if
forA>a (19.11)
caA3"2
then the free boundary of UXQ,, in {a/2 <x <a) lies above the line y = A/2.
THE CONTINUOUS FIT FOR THE GRAVITY CASE
Proof. Suppose that A' 9) is a free-boundary point with a/2 <a. By nondegeneracy,
(19.12) Q (Q — mm — y aB,(x)
if r < where c0 is a universal positive constant. If 9 <A/2, then we obtain a contradiction to(19.11) by choosing r = a/4 if A > a and r X/4 if A <a.
Lemma 19.5. There exist positive unftersal constants c, C such that
/ co 2 (19.13)
for every free boundary point 9); in particular, if
(19.14)
then the free boundary of starts below the line y = 0, thai is,
(19.15)
Proof Suppose A' = belongs to the free boundary 1' of .UAQ. Denote by 0(yo) the half disk
1 2 2 ..,tx>0.x +(y—y0) <x}.
Ify0 > A + then D(Yo) ni' = 0. Since D(9) fl 1' is nonempty there exists a .E Li. A + 1) such that
0, 0.
Let A' = (1, 9) be a point in fl I'. We introduce the solution v of
Lv0 inD(y*), v(0,y)0 (x0}.
v= Q onaD(v) fl{x >0).
By the maximum principle v > in D(y*) and
(19.16) atA'.
386 JETS AND CAVITIES
To estimate the riglu-hand side let w be the of
inD(v*'),
w v or F*)
It is easily seen that 0 so that
Thus, by the maximum principle w v in D(tl*) and
(19.17)
Introduce polar (p.6) abcut (0, + with 0 0 in the direction of the negative y Thin the harmonic function w — 2Q0/ir is in
on fl (y >y*), and it is C' in D(y*) fl {v It follows that if 1 then
8w
and, consequently,
on8Dh")fl(x>O) (c>0)
at points near (0, + The same inequality holds near (0, y* — Fi- nally, this inequality also holds (by using barriers) on the remaining part of 8D(y) (1 {x > 0). By scaling we find that, for any
On aD(y*) (1 (x>O).
Using this in (19.17). we obtain from (19.16)
IA —9 xr
Since 9 — i we find that (19.13) holds.
Lemma 19.6. For any A > 0, there exists a Q > 0 such that for UAQP there is a smooth fit at A.
ThE CONTINUOUS FIT FOR THE GRAVJTY CASE
Proof. For Q sufficiently large (19.14) holds and for Q sufficiently small
>0
by Lemma 19.4. By continuity there exists an intermediate value of Q for Whicri uA Q has Continuous fit at 4; the smooth fit follows from Section 11.
Lemma 19.7. Let 2ir/3 a0 3ir/2. Then/or any Q >0 there exists a A >0 such that for there is a smooth fit at A.
Proof. By Lemma 19.2, if A is small enough, then $AQM(a) <0, and by Lemma 19.4, if A is large enough, then <0. Now proceed as in the preceding proof.
Lemma 19.8. Let ir/2 < 2ir/3. Then there exists a Q> 0 such that if Q < Q, then there is no A 0 for which ux (2. has smooth fit at A.
This follows immediately from Lemmas 18.10. and 19.3.
Lemma 19.9. Let UA,Q,p and UA.Q.,1 be two solutions with smooth fit at A. Then:
(i) If A > A' >0, then Q> Q'. (ii) IfAA'>O,zhenQQ'. (iii)'lfQ=Q',:henA=A'.
This follows easily from Lemma 18.10.
Deflni$qm 19.1. For each A >0, denote by the value of Q for which UJQ11 has a smooth fit at A. The functAon
Q=k,1(A)
is called the solution curve for fl,1. From the previous lemmas follows:
Theorem 19.10. The fiuwf ion k1(A) is continuous and strictly monowne i*creas- mginA,A >0, and -
k,(0 +) = 0 if
k,1(0)>0
JEFS AND CAViTIES
further,
(19.18) sfX>1,
where c, C positive constants depending only on a.
It is easily that for any sequence 00 and a sequence Q1, c1 c2, where c, arc positive constants, thert exist subsequences such that Q and
(19 19) UAQ weakly in H' flBR) for anyR >0.
• a.e.
Let
= the domain bounded by N U M and the y-axis.
KQ= (v;VEH'(E
v Q on N U M and, as before, on \fl; v = 0 on they-axis),
f —
Then U is a minimizer in the following sense:
(19.20) 'X.Q,,I(UA,Q)
for for the same proof or by passage to limit with ia); this in particular is true of
Lemma 19.2 and 19.3. If we take Q1 = then UAQ has a smooth fit at A. Similarly, we can
take = (Q large enough) and obtain (for oo) a solution UAQ with smooth fit at A.
The function ux Q
(with smooth fit at A) thus constructed is then a solution of the jet problem. Notice that if > then solutions Qi with smooth fit at A can be constructed using the same Mj; from this we infer that Q1 Q2.
Lemma 19.11. If and ux2 (3 are Iwo solutions of the jet problem, then and UA Q = UAQ.
The proof uses a comparison argument as in the case without gravity.
THE CONTINUOUS FiT FOR THE GRAVITY CASE
Proof. Let u, = uA For any small 6 > 0 there is a < 0 such that the free boundary of u1 is given by
x (1 + 8,(v)). &,(y) 6 forj'
We assume for definiteness that A2 A1. Take any 0 < k < I and a 0 and (recalling (18.2)1 consider
01
k k
Its free boundary rr is given by
x—kfi(v* +0
and
(19.22)
kfi(y* + _a) = + <k"4f,(v) <f2(y)
if
(ktlif6iO).
Thus the free boundary of lies above the free boundary 1'2 of u2 in the region (y <ye) is independent of a).
From (19.22) we also infer that
kf,(y*+' y*. o)o
ifa—oo
uniformly for Y Yo' for any Yo• It follows that
(19 23' U N° lies above F2 U N everywhere if a is enough',
/ where N ii the transformed nozzle corresponding to
Now let 0k be the smallest nonnegative value of a for which (19.23) holds. By the maximum principle we have
(19.24)
JETS AND CAVITIES
[Indeed, for any large M >0 and small e >0, the function
Cx2 32+t(y+M) [CQ(1+a)] (x2 + (y + /
satisfies U 0 on y = —M and on the free boundary of in {y> —M}, and U oo ify 00; since LU 0 in (0 < < Q), the maximum principle yields U(x, y) 0 if y —M. Now take E -o 0 and thenM -. oo.J Moreover, if 0k > U N°° touches 1'2 at some fmite point X = (1, j); otherwise [since (19.22) holds for all a 01, we can further decrease 0k keeping the condition (19.23). —
If we assume that 0k const. > 0 for all k near I, the point I cannot belong to for k near I, for 1'2 detaches smoothly from A. Hence X (1 F2 and (19.24) implies that
I I * — — i — I au2(1) — r — — k
—VA2—Y.
This gives <A2 if yt was chosen large enough such that F1 lies in {y <y), contradicting the assumption A2 A1. (Observe that N is star-shaped also with respect to any point (0, y) withy >
Therefore, 0k 0 for a sequence k t 1, and we then obtain from (19.24) that
u1(x, y) u2(x, y).
Now, if u1 u2, we easily deduce that for some small R >0 and e >0,
(1
on the boundary of G BR(A) fl (u1 <Q); hence also in G. Using the smooth fit at A for u,, we get
(1 + (I + — —
which is impossible since A1 X2. This completes the proof.
Similarly, we can prove:
Lemma 19.12. If UX,Q and are solutions of the jet problem, then Q1 = and UXQ = UAQ2.
From Lemmas 19.11 and 19.12 ii follows that The function k(A) is continu- ous and strictly monotone increasing. This completes the proof of Thtorem 18.1.
AXIALLY SYMMETRIC FINITE CAVITWS 391
PROBLEMS
1. Show that (19.2) implies (19.1). [Hint: Suppose that y1 = positive x-axis, y2 = positive y-axis. If X° = (x°, y°), 0<y° < r = 2y°, B = Br(X°) fl (y >0), and if w(X°)> Mr, then (by Harnack's inequality) w cMr in B,/4 (X°) and, by scaling,
implying by (19.2) a bound on M. Use Harnack's in- equality when X° is not near the axes.]
2. Prove Lemma 19.9.
3. Prove(19.19).
4. Prove that cA k(X) as X -. 00 (c >0, C> 0).
5. Extend Theorem 18.1 to the symmetric plane jet with gravity; see Section 18, Problem 3.
6. Consider the symmetric plane jet with gravity and assume thatx(t), y(t) in (18.1) are both increasing functions of s. Show that the free boundary is given by a continuous and strictly monotone increasing function x =
00 <y<O. [Hint: See Section 14, Problem 2.J
20. AXIALLY SYMMETRIC FINITE CAVITIES
In this section we consider the model of an axially symmetric finae cavity; this will be used in the next section to study axially symmetric infinite cavities.
In both models there is an obstacle N' lying in an ideal axially symmetric fluid; we denote by N the curve in (x <0, y <0) which generates N' by rotation about the axis of symmetry (the x-axis). We assume:
Nis a piecewise curve given by
(20.1) x(:)andy(t)aremonotonenondecreasingandy(t)>Oif: >0;
X(0)=(—b,o), X(t)—(—a,y(t)), O<aczb.
For simplicity we takey(i) I and set
A 1(1) = (—a, 1).
In the model we introduce another obstatle 1', which is the reflection of N about the y-axis.
N is called the nose or projectile and T the tail. Connect (—a, 1) to (a, I) by a line segment 1 and introduce the sets (see
Figure 3.8)
ER=((x,y);—R<x<R,O<y<R),
E=R2fl(y>O),
DR ((x,y); —azx<a,1 <y<R}.
W = the domain bounded by N, T, 1, and the segment {(x,O); — b <x <b),
Definition 20.1. The (axially symmetric) finite cavity problem consists of finding a function u(x, ) (the stream function), a curve r (the free boundary)
(- R,
392 JETS AND CAVITIES
R)
FIGURE 3.8
AXIALLY SYMMETRIC FINITE CAVITIFS .393
and a positive number A such that the following conditions are satisfied:
F is C', I' is symmetric with
- respect to they-axis;
strictly monotone decreasing and continuous for
(20.2) t —a, h(y) = 0,—a < h(y) <0
for I <y <y*;
F U N is C' in a neighborhood ofA,
and
u is symmetric with respect to
they-axis andu E
u(x,0) = 0 if —00 <x —b,
(203) u(x,y)=0 u(x,y)>0 elsewhereinE,
Lu=0 in(u>O},
where
Lu = + u,, — !uy;
further,
(20.4) = 0, = A I';
('0 5) Vu is uniformly eqntinuously iffererisiable
in ( {is>Q)-neighborhot-d of A;
(20.6)
The set
K
is called the cavity.
394 JETS AND CAVITIES
Theorem 20.1. There exists a unique solution of the finite cauity problem.
The proof is given in several steps. We first introduce a variational ap- proach. For any R > 0, let
(20.7)
KR—
where
Let
(20.8) JA.R(V) =
+
and consider:
Problem R Find u = UA R in KR such that
(20.9) Jx.R(u)— VEKR
This problem clearly has a solution u. Set u = yU and write
JA.R(M)
Then U is a nünimizer of 1(V) in the class £R of functions V v/y, v E KR. Since 4'/y is monotone increasing in y,
U the y-direction, then U5 KR. Using Theorem 7.1' [and, in particLilar, (7.6)1, we easily see that
with strict inequality if U on a set of positive measure. Hence U is monotone increasing in y, and the same is then true of u.
We can also rearrange — u symmetrically and decreasing in x I; here even the second term in (20.8) may decrease by the rearrangement. It follows that u may be assumed to be decreasing in x for x <0. Thus
Since the free boundary is locally analytic, it follows that is has the form
x±h(y)
AXIALLY SYMMEIRIC FINflI CAVITIES
where h is monotone increasing and continuous; h and depend of course on l? h—h *A.R'Y Yx.R•
The free boundary, if nonempty, may either start on the horizontal line (y 1) or on the vertical line (x = —a); in both cases it satisfies the smooth fit property (Section 11), since the proof of this property is local.
Now take a sequence R -. oo such that
UXR
Lemma 20.2. I/A A < oo, then there exists a positive constant C5 depending only on A such that
(20.10)
for all R sufficiently large.
Proof. Set
n (x<0) fl
By the bounded gradient lemma,
where C ii independent of A, R. Hence, for any 8 >0,
On the remRining part of aA,
for any e >0, provided R is sufficiently large. Hence, by the maximum principle,
where 8 and e can be taken arbitrarily small if R is sufficiently large.
Consequently, if the assertion of the lemma is not true, then u = UA p -, 0 for a sequence R 0, and this contradicts the nondegeneracy lemma (Lemmas 3.3 and 8.6).
3% JETS AM) CAVITIES
From Lemma 20.2 it follows that
(20.11) — Ofla(UAR>O)
where r2 x2 + y2 and C is a positive constant independent of A, R (provided that A A). By the maximum principle we obtain the same inequality in (UAR >0); therefore, also
(20.12) —
ifl(UA>0).
Definition 20.2. The function UA is called a solution of problem
Note that ux satisfies all the properties enlisted in (20.2)—(20.6) except for the continuous and smooth-lit conditions.
DefInition 20.3. Any function UA satisfying all the properties (20.2)--(20.6) except for the continuous and smooth fit conditions will be called a A-solution.
For any A-solution UA we consider the flow region QA = {uA > 0). The domain
(O<a<zl)
where X = (x, y) is called a magnification of QA• It is again a flow region of a A-solution with N replaced by aN; this A-solution is
a( 2 (X V
Lemma 20.3. Suppose that are A1- and A 2-solusions, uA UA, and QA can be magnified by a factor a < I in such a way that
aQx J {y>O) 0.
ThenA1 >X2.
Proof. Set
ü1(x. y) u2(x, y) = ux2(x, y).
Then
U(x, y) ü1(x, y) — u2(x, y) 0 if r —, oc.
AXIAlLY SYMMETRIC FINITE CAWI1FS
Since on the boundary of QA2 the maximwn principle gives U>C in and
i au, i—— >—-—---- atX,
a point in {y >0) where and intersect. Since such a point must belong to both free boundaries, we deduce that A1 > A2.
Corollary20.4. If
Indeed, otherwise we can magnify >0) that it contains >0) and thus deduce that A2 >
CorohIau7 205. For any A >0, there is precise'y one A-sohuicn (w*Ech ü then the solution of problem
Indeed, if u1, u2 arc two solutions, then (u1 >0) = (u2 > 0), by Corollary 20.4, and, by the maximum principle, u1 — E 0.
Corollary 20.5 implies that UA varies continuously in the parameter A (see Lemma 10.4).
20.6. If A I, then the free bow.dwy of UA Lc
Proof. u0 be the flow past W satisfying (20.6) (it is simply a solution of problem I, with A =0) and set
at(0,l).
If has free boundary points, then by magnifying (UA >0) until it contains W we obtain, using the argument of Lemma 20.3,
.A>A0.
Since by the maximum principle
in(y>l),
we have A0 1; thus A> 1, contradicting our assumption on A.
20.7. If A E large, then the free bowtdory for Starts on the vertical line x = — a.
398 JETS AND CAVITIES
Proof. Take a domain ii' bounded by y = 0, x a', x a' + 2E and half a circle
{(x— (a'+ e))2+(y_8)2<r2,y>6}
with e small and 6 near 1. Denote by u the flow past iii' If the free boundary initiates on { y = I), then in view of the smooth-fit
property we can choose the parameters e, 8 in such a way that W lies inside {u = 0) but aw n (y > 6) touches the free boundary. say at X°. But then
(20.13) aIX°.yav yav
The choice of e, 6 depends only on a; therefore, using the inner ball property to construct a barrier for ü at X°, we get
(20.14)
where C depends only on a. Thus, if A > C, then the free boundary must start on the line x = —a.
Combining Lemmas 20.6 and 20.7 with the continuous dependence of (and its free boundary) upon A we now deduce by familiar arguments the existence of a parameter A with smooth fit at A.
This parameter is unique; in fact, the proof is similar to the proof of Corollary 11.5. We have thus completed the proof of Theorem 20.1.
One can obtain a standard series development for u near infinity. Thus, if we introduce the velocity potential by
-
(20.15)
in (y > 0), then the function
y, z) = + z2)
is harmonic in the region obtained by rotating (U >0) about the x-azis. Developing near infinity in terms of zonal harmonics (U20, p. 254}),.we find
AXIAlLY SYMMEIRIC FINITE CAVITIES
that
v+(x,y) -i.(l,0) (20.16) 1 asx2+y2—'oo.Vu(x,Y) -.(0,1)
PROBLEMS
I. Derive the expansion (for any u =
2. Extend Theorem 20.1 to the case where N = ((—a, y); 0 1).
3. Extend Theorem 20.1 to plane symmetric finite cavities.
4. Extend Theorem 20.1 to the case where x(t) is not a moflotofle nonde- creasing function (that is, N is just a y-graph) provided that N is star-shaped with respect to the origin.
5. Let G be a bounded star-shaped domain (with respect to the origin) in R2 andletflR2\G.Considçrthefunctional
1(v) = f0(i Vi, 12 + A21) dx
over the class
K=
Prove: (a) There exists a unique minimum u = UA. (b) IfAI>A2,then(uA>0)C(uX2>0). (C) a(uA >0) is star-shaped with respect to the origin. (d) If G is convex, then the free boundaiy is convex. [Hint: By Problem 5 of Section 3, any (local) minimum has a bounded support. For (a)—(c) use the method of Lemma 20.3, and for (d) notice that q takes minimum on thebes (see Theorem 13.1).I
6. Consider the axially symmetric cavity in a ppe I < Z with obstacle y = g(x), 0 x a, where g(x) is monotone increasing. Con- struct the variational functional and establish the existence of a soluticç with free boundary y = f(x), f(x) monotone increasing f'(x) —p0 if x -,
JETS AND CAVfFIES
21 AXIALLY SYMMETRIC iNFINITE CAVITIES
Let N be a nose satisfying:
Nis a piecewise curve given by
(21.1) x(:) andy(:) are monotone nondecreasing andy(t) >0
if t >0; X(0) = (0,0) and X(i) = (a, y(i)), a >0.
For simplicity we take y(i) = I and set
A = (a,1).
DefInition 21.1. Let N (the nose) be as in (21.1). The (axially symmetric) infinite cavity problem consists of fmding a function u(x, y) and a curve F such that:
r is C' and is given by a strictly monotone (21.2) increasing and continuous functiony f(x), x < ao;
f(a) = I and F U N is C' in a neighborhood of A,
and
uEC(E),
u(x,O)=Oif—oo
u&onNU FandintheregionKboundedbyNU r and the segment {(x, 0); a x < oo);
(21.3) u>OelsewhereinE;
LuOin {u>0};
u is uniformly continuously differentiable
in some (u > 0)-neighborhood of A;
further,
i (21.4) —— 1 on!',y av
AXIALLY SYMMETRIC INFINITE CAVITIES
and u, r possess the asymptotic behavior:
(21.5) f'(x) —0 00,
(21.6) uniformlyin{u>0}ifx2+y2—' cc.
K is called the cavity.
Theorem .21.1. There exists a solution of the infinite cavity problem.
To prove .the theorem we first consider the finite cavity problem with tail T = which is symmetric to N with respect to the line x = > a. By Section 20 there exists a unique solution u,, 1',, to this problem and
(a <x < 2g& — a),
where
JM(x) —
fM(x) is strictly monotone increasing for a.< x
From Lemmas 20.6 and 20.7 Isee (20.13), (20.14)J, we have
(21.7) 1 '6 A, 6C
Lemma 21.2. Set I{(x,0); —00 <x'60). I/a straight line / does not intersect I U N, then it can intersect fl (x <ji) in at most one point.
Proof. r, is given by a monotone-increasing curve y = fF(x), it suffices to take I with positive slope. If the assertion is not true, then by parallel translation of I we arrive at the situation (see Figure 3.9) where the
MR of there is no other point with the same properties. It follows that
if enters the half-plane to the right side
(21.8) of Iatapoint F, then theslope of 1', at each point of FR is less than the slope of I
(otherwise, the first point M' of where the slope of 1 equals that of 1 has the same properties as M, a contradiction).
there is a point F as in (21.8), then we set 8 If, on the other hand, MH lies entirely to the left of 1, then we set S MR.
the ray going from E to x = gz parallel to!. In view of (2.18), lies above
JETS AND CAVfl1ES
Denote by Q the ihtersection of the line EM with the x-axis. Clearly, Q lies to the nglft of B. _Denote by R the region bounded by y = 0 from x = — cc to B, the segment BM the subarc ME of the arc y, and the segment on the line x = gi from J tO +00.
Denote by *1 the solution of
Lu=0 inR,
ü0 (21.9) = 0 ify1<y<oo 1J(x1,y1)J,
as r2 = x2 + y2 —' 00.
The solution can be obtained by reflecting the curve fl (x <ji) about x=pandapplyingthcmethodofSection20withA0;seealsoProblcinl.
FIGURE 3.9
AXIAlLY SYMMETRIC
We shall prove below that
( ', Ivu(Mfl Ivu(M)(
vu(E)I Ivu(E)I
Assuming this inequality for the moment, consider the cimiterlty transforina- tion
(x,y) .-(a(x—xo),ay)
which maps E into if. Thus if
M = (XM, YM)' E (Xe, Ye)'
then
ayE = YM•
It is easily seen that the inverse mapping takes R onto a subset R' of A. The function
s7(x, y) = x0), ay)
satisfies L*2 = 0 in R', *2 *2 on and *2= ii = 0 at £ (F E aR'). By the maximum principle we get *2< u in R' and
hence
Substituting this into (21.10), we obtain
IVU(M)1 Ivu(E)I
Since, and £ he cm the free boundary,
V = = a, a contradiction.
It remains to prove (21.10). We may astema thatu, *2 are boundary, for the general ease then follows by the flow by regions with smooth boundaries.
JETS AND CAVrrIES
For any point Z = ,j) in ME. we introduce the function
17'(Z)u(X) (X—(x,y)),
where
1 — I
and the level curve
= 0
initiating at Z and going into the domain where both u and ü are positive. We note that the velocity potentials 4), corresponding to u, ü possess expansions in terms of zonal harmonics of the form (see reference 120, p. 254)
O)r";
the same is true of u, ü (see also reference 148). Therefore, at each point X° where u, u are analytic, there is a finite number
m of simple branches of and if m> 1, then two adjacent branches form nonzero angles at X°. This is valid also about y = 0, —oo <x <0, since vu ' Cy, I Vu c Cy (by the proof of Theorem 8.5), which implies that
4)(x, + z2) tfy2 + z2) are harmonic about the x-axis, for — oo <x <0.
Since = 0 at X Z, it follows that does in fact initiate from Z into It also follows that if exits the region = fl (x <p} at a point of I, then it is nontangential at the exit point (since = 0 along I). If
z exists R, at x = then, because of symmetiy with respect to the line x = it must intersect x = nontangentially. Notice that cannot exit R, on ME (otherwise, 0) and on the arcs y (where <0), N (where 117 >0) and the subarc r' of i from A to M (where again > 0).
From the considerations above it follows that the subsets of MF defined below are open:
Z E if V(Z) and not exit R,onx
Z if V(Z) =
If remains in then, if we parametrize it by the length s, the points X(s) of must satisfy: X(s) —' oo if s oo. It follows that
I ô - ——fl2 -. 0, V(Z) — V(z)y8x yay
AXIALLY SYMMEtRIC INFINITE CAVITIES
at x = X(s), s —p If also V(Z) then becomes asymptotically horizontal, going to — 00.
Consider now the case
(21.12)
For any Z E ME, if exits on 1 (say at X°), then 0 on the boundary of the region bounded by and the part of from X0 to Z. By the maximum principle it follows that > 0 in and
a (21.13) vinwardnormal.
Suppose next that does not exit RM. Then, by the remark containing (21.11), is asymptotically horizontal. Hence, for any e > 0,
on from (— oo,O) to Z, on and for X E as X I By the maximum principle we conclude that + — x)y2 >0 in Letting £ 0, we again arrive at (21.13).
Taking Z -. E in (21.13), we obtain
J'(E)Jvu(M)I
and (21.10) In case
(21.14)
we can argue in a similar way. Here <0 in the region bounded by
since y) = — x, y), we can apply the maximum principle in the closure of U — x, y); (x, y) E Ge). We conclude that
atE, -
and taking Z -. M (21.10) follows. Consider finally the case
(21.15)
Since are open and disjoint, there exists a point Z E such that
406 JETS AND CAVITIES
Z fl Therefore, does not exit and, since
Jvu(Z)11vu(Z)I
we have
It follows that for any e > 0,
+ ey2 + e(p — x)
is positive in G2 and
is negative in Letting e -' 0, we can then derive as before
that is,
and (21.10) follows.
Having completed the proof of Lemma 21.2 we now choose anyO a0 a such that the straight-line segment connecting (a0, 0)10 A does not intersect N. From Lemma 21.2 we then dedute that is star-shaped with respect to (a0,0). Consequently, the function
4(x) (21.16) ismonotone decreasing, a <x
x a0
Hence its d&ivative exists a.e. and is nonpositive. This implies that
/(x) f'(x) a.e.x—a0
Sincey = 4(x) has analytic reprcsentation, we easily deduce:
AXIALLY SYMMETRIC INFINITE CAVITIES
Lemma 21.3. There holds:
1(x) (21.17) ifa<x<00.
x a0
We now choose a sequence p p,, —. oo such that [recall (21.7)J
in the appropriate sense. Then u, 1' A satisfy (21.2), (21.3). From (21.16) and Lemma 21.3,.we obtain:
Lemma 21.4. The function 1(x) satisfies the following properties:
(21.18) 1(x)
Is monotone decreasing,a<x< 00, x a0
(21.19) for<<
If follows that
(21.20) p L!1 exists.
Lemma 21.5. /30.
Proof. Set
UR(X)
Since u(X) y2/2, also X) y2/2. Hence there exists a sequence -' 00 such that
UR(X)
uniformly in compact subsets of {y 0). The free boundary rR of is given by
AND CAVITIES
By nondcgencracy, if X° E and B( X°, h) does not intersect the fixed boundary of then
sup ch (c > 0);
it follows that 0. From (21.20) we deduce that the free boundary of u0 is
yfo(x)13x. We have
Lu0=0
where 0 80 <er, tan and
u0=O onaL
Consider the function
(21.21) v(r,8) = a>1
where are the Legendre functions as defined in reference 84, p. 122 [P,?(z) is the Legendre polynomial of order n]. One can check (see also reference 148, where the coincide with the P''12 of reference 84) that
(21.22) Lv0.
Lemma 21.6. There holds:
(21.23) 2a— 1.
Proof. From reference 84, p. 159, formula (27),
1?(cos9)>0
that P,"(z) is a single-valued analytic function defined in the plne cut along the negative real axis. By reference 84, p. 160, formula (I),
= +
AXIALLY SYMMETRIC INFINITE CAVITIES
and by reference 84, p. 143, formula (I)
(21.26) = + iO) — e"1"/2P,"(x — 10)1
if xis real, —1 <x < I. Therefore,
P'(x) + 10) + e_1'/2F_'(x — 10)]
= (v + + iO) — sO)],
and the right-hand side is equal to
(r +
by (21.26) with i& = —i. Hence
= — (p +
Since the right-hand side is positive (by (21.24)) if 0 <8 <w/(2(i' ÷ the assertion (21.23) follows by taking v = a — 1.
We can now complete the proof of Lemma 21.5. Indeed, if P > then choose a> 1 such that
2a—1
and set 8) = v(r, ,r — 8), v as in (21.21). In view of (21.22),
Lu 0 in the sectorl,
and by Lemma 21.6,
Since a> 1, we also have
ifXEI, IXI—oo. y
We can now apply the maximum principle to deduce that for any s >0,
u0(X)czeu(r,8) ifXEI, IXJ<p
provided that p is sufficiently large. It follows that a contradiction.
410 JETS AND CAVITiES
From Lemmas 21.4 and 21.5 we get
Corollary 21.7. There holds:
(21.27) f'(x) 0 ifx —. We next prove:
Lemma 218. A= 1..
Proof. From Corollary 21.7 ii follows that there exists a sequence 1 such that
(21.28) f'(x) < if x
Define functions X) by
2 + a
f
where
The free boundary for u,( X) is given by
y-fft(x) + as);
in particular.
(21.29) = I.
For any C > 0,
(21.30) + < C
provided that is is sufficiently large; indeed, if x <C, then
xf(ar) +
(since (f(t)/t) -. 0 if t —' o01 and thus (21.30) is a consequence of (2t.28).
AXIALLY SYMMETRIC INFINiTE CAVITIES 411
Since u(X) y2/2 we also have
Hence for a subsequence
pointwise, and
-.
uniformly in compact subsets of {(x, y) R2; y and
I au.
From (21.29), (21.30) we also have = 1. Hence by uniqueness, for the Cauchy problem
Since further y2/2, it follows that A 1. Recalling that X 1, the proof is complete.
Let B be a ball in R2 containing N and denote by G, the intersection of B with ((x, y); x <gL, y > 0). Denote by G2 the reflection of G1 with respect to
rotating about the x-axis; here = 0). Let be the velocity potential corresponding to Then the function
ishärmoflic inD,' and its giaditht is bounded by I at infinity. Prom the bounded gradient lemma we haive
(21.31)
on aD;, where C is a constant independent of By the max we then obtain the same estimate in
Now, for any e >0 there is x0 > a such that
— a0
412 AND CAVITIES
Hence
fM(xO) — <e
x0 a0
SincefM(x)/(x — a0) is monotone decreasing for a <x we then also have
(21.32) if
and, by Lemma 21.3,
(21.33) 1:(x) <e if <x ,z
Recalling that
using (21.31), we deduce that
(21.34) —
if x0 < x <it,,, = au,,
whereq(i)-.Oife Let R0 be a positive constant independent of such that the arcs
and N are all contained in the ball BR,. Denote by B10(O,) the reflection of 8R0 with respect to the line x = that is, 01 = (2p, 0). —
Denote by the set in R3 obtained by rotating \(BR) U BR0( 01)) about the x-axis. Consider in the harmonic functions
u1=——l, u2=—. ax ay
In view of (21.31),
(21.35)
on the part of a4 obtained by rotating and aBR0(oI) about the x-axis.
AXIALLY SYMMETRIC INFINITE CAVfl1ES 413
On the remaining portion of a4 we have, by (21.34),
Notice also that ifx2+y2-ioo.
Thus we can compare u with the harmonic function in R3,
Co C0 • —' 161 16 601
where = 0,0) and C0 is a sufficiently large constant depending only on C in (21.35) and on R0, and deduce that
v;
thus, if
jau C0 C0 +
Y ay Y ax (x2+y2)h/'2 ((x_2p)2÷y2)I/2
in41n (z oo,weobtain
— +
(x2
Taking x2 + y2 oo and noting that i.(e) can be made arbitrarily small, the assertion (21.6) follows. This completes the proof Qf Theorem 21.1.
PROBLEMS
1. Solve (21.9) by the velocity potential 4. and the harmonic function 3(x, y, z) 4.(x, + z2); 3 is then a solution of the exterior Neumann problem.
2. that exists for any two solutions u, u of the infinite cavity problem, where r(9) (P(9)) is the distance from (a,0) to the free boundary of u(u)along the ray y=(x—.a)tanO, x>O. Prove uniqueness for the infinite cavity problem; here the Phragntha-Lindelof theoeem needed for a diffemnee 4' of solutions follows by the proof of Coronary 9.10 applied to 4, small.
3. Extend Theorem 21.1 to the case where x(t) is not monotone hondécreas- ing (that is, N is just a y-graph) provided that N is star-shaped with respect to some point (b, 0), b> 0.
414 JETS AND
4. Let R1, R2 be two flow regions in (y > 0) of axially symmetric stream functions u1 and u2, respectively, and denote by the stream line u — 0; S, = aRt. Assume that u. y2/2 and
u, = + o(I)) asr2 x2 +y2 oo,(x, y) ER1.
Suppose that S1. S2 have an arc in common and S2 lies below S1 from — to M, whereas S2 lies above S1 from N to co; see Figure 3.10. Suppose finally that is C in R1-neighborhoods of M and N. Prove:
IVui(M)I Ivu1(N)I vu2(N)J
this is called the under—over theorem.
5. Let u be a solution of the axially symmetric infinite cavity problem for a general smooth nose; the free boundary 1' is a C' curve connecting A to infinity (not necessarily a y- or x-graph) which is asymptotically tal, and (l/y)
I I = I on r. Let I be the line segment ((x, 0);
—00 <X <0), where (0,0) is the initial point of N. Prove the single intersection property: Any infinite straight line that does not intersect I U N canintersect F in at most one point.
6. Theorem 21.1 can be extended to the plane symmetric infinite cavity problem. Denote the solution by 4i and the velocity potential by and set
+ w =ilog(df/dz)=9+Ilogq(see Section 1). Then the flow region is mapped by! into a region R1 that contains a neighborhood of infinity slit along the positive real axis. R1 is mapped byf= 1/t2 into a region R, which near t = 0 consist of half a disc{I:I<., Imi >O}. (a) Show that 0(z) has analytic extension into a neighborhood of t 0. (b) Verify the formula
z = dt
N
FIGURE 3.10
SI
BIBUOGRAPIIICAL REMARKS 415
and use it to deduce that
I 1 I r 2 . ,
—,w (O)jloge'+ I
near t = 0 (notice that w(0) 0]. (c) Show that ii (0) 0, then the free boundary has the form
forx-oo.
(d) If w'(O) = 0 and w"(O) 0, then the free boundarv has the form
y w"(O) (log x for x -. 00.
7. The single intersection property (Lemma 21.5 and Problem 5) extend to axially symmetric jets: if a line I does not intersect the nozzle N then it can intersect the free boundary in at most one point. Prove it!
8. If a nozzle satisfying (8.1) lies in (x 0) then the free boundary is given by a monotone decreasing function y = h(x).
22. BIBLIOGRAPHICAL REMARKS
Surveys of the classical theory of jets and cavities can be found in references 41, 10Th, 110, and 182. The classical approach to these problems is based on a reduction of the problem to a nonlinear functional or integral equation by means of a hodograph-related transformation. We briefly mention the works of Finn (90), Leray [1351, and Weinstein (179] on jets and the works of Lavrentieff [1341, Leray (135], and Serrin [161bJ on cavities: these works establish existence theorems. Uniqueness results were previously proved by Lavrentieff (1341 and more simply by Oilbarg [lOla) and Serrin [161b].
Variational approach to cavities was developed by Garabedian, Lewy, and Schiffer [102) and by Garabedian and Spencer (103); these authors use symme- trization, but their admissible classes are rather restricted and the analysis (which employs complex methods) is complicated; reference 102 is the only paper in the classical literature to establish existence for axially symmetric flows.
Various properties of solutions (such as comparison and monotonicity) are surveyed in references 41 and 10Th. We mention specifically the under—over theorem and the single intersection property due to Serrin [161b, c, dJ, both of which are used in Section 21.
The material qf this chapter (with the exception of Sections 1 and 7) is based on recent papers by Alt and Caffarelli [64 Alt, Caffarelli, and Friedman (7a,
416 JETS AND cAvmEs
b, c), and Caffarelli and Friedman 158p). Sections 2 to 4 are taken from reference 6a; Sections 6 and 8 to 19 are based on reference 7 a, b, c; the results on the impinging jet problem (at the end of Section 17) are due to Caffarelli and Friedman (unpublished).
The paper 1102) asserts a weaker version of Theorem 20.1, namely, the smoothness of the continuous fit is not established. The presentation of Section 20 given here is new.
Asymmetric jets with gravity have been studied in reference 7c. For some special geometries there are the results on existence and uniqueness that use hodograph-related transformations to reduce the problem to a nonlinear integral equation; see Budden and Norbury 1541, Carter 168a, b], Keady [1 l8aJ, Keady and Norbury [1 19a), and Larock and Street [133).
The results in Problem 5 of Section 20 were originally proved by Tepper [lila, bj. Related variational proNems arising in optimization of heat flow have been studied by Acker [la—c).
Phillips (187a, bJ studied minimization problems for J(j Vu f2 + )Y) and obtained results analogous to those of Sections 2, 3. Alt, Caffarelli, and Friedman [7d, eJ have studied jet problems with two fluils using variational techniques; here the condition — 12 A holds on the interface between two fluids.
VARIATIONAL PROBLEMS WITH POTENTIALS
In this chapter we study variational problems the functional has the form
(DCR3)
and A(z) is a gwen function, 4'(p) is a given functional. The admissible functions p satisfy p 0, Jp(x) 1r M (M is given), and possibly some other constraints. The variational equations for these problems will have the form
g(x., u) and {u 0) is the free boundary a{p > 0). Such variational problems arise in physical models, classical as well as recent ones. We shall deal here with four topics:
Sections 1 to 5: Self-gravitating axially symmetric rotating fluids. Sections 6 to 10: Vortex rings. Sections 11 to 13: The problem of confined plasma. Sections 14 to 17: The Thomas-Fermi atomic model.
In studying these problems, we shall develop some general methods. Thus in Section 5 we derive results on the expansion of solutions of differential inequalities I) near x x0, provided that ii(x°)0; in Section 9 we establish estimates on capacities. These general results are used in the analysis of several of the models.
As in Chapter 3 we shall be concerned with questions of existence, unique- ness, and regularity of solutions of the variational problem, with the smooth- ness and shape of the free boundary, and with asymptotic estimates.
417
418 VARIATIONAL PRORLEMS WITH POTENTIALS
1. SELF-GRAVITATING AXISYMMETRIC ROTATING FLUIDS
Consider inviscid and irrotational fluid with density p rotating about the z axis. It is assumed to be axisymmetric; that is, p(x) = p(r, z), where x = (r, 0, z). The rotation law is
v s(r)19
where s(r) is a given function and
= (—sin0,cos0)
is tangent to the path of the particles. We the unit vector
i,= —(cos0,sin9)
pointing into the origin. One can check that the law of conservation of mass
(1.1) V(pv)O
is satisfied. Next, the Euler equations are
(1.2) (v.V)v=_!Vp+VV
where p is the pressure in the fluid and V is the gravitational potential,
V(x)=f R3IXYI
One easily verifies that
s2( r)(vv)v= — r —vf(r)
where
Thus Euler's equation becomes
v(v+f).
SELF-GRAVflATING AXISYMMEIRIC ROTATING FLUIDS 419
If we restrict the fluid to be compressible and assume the polytropic law,
p—Kp7,
then (1.3) becomes
(1.4) (Xconst.),
where c K($ + 1) and
G= (x:p(x)>O).
Similarly, in the incompressible case p 1,
Vp = v(vtf), so that Euler's equations become
(1.5) p=V+f+AinG (Aconsi.).
We have assumed that p = p(r, z). We shall further assume that p(r, z) = p(r, —z) and
f pdx=M (M>Oisgiven). R'
Define
m(r)=_!-.f M
where r(y) = r ify = (r,S, ;). We now introduce a functionj(m) (0 rn I), the angular momentum per unit mass, such that
(1.6) j(0) =0, j(m)monotonenondecreasmgandj2(m) E C'IO,l).
Formally, for O6m 6 1,
(1.7) j(m(r)) = rs(r),
so that
(1.8) f(r)= foi2(rn(r))
420 VARIATIONAL PROBLEMS WITH POTENTIALS
(1.7) is clearly valid if we accept the intuitive assumption that the rotating fluid, during the evolution which led to equilibrium, was moving in such a way that the total angular momentum did not change for any fraction of the mass lying within a distance r from the axis of rotation.
Define a function u by
(1.9) u=V+f+A.
Then we seek an equilibrium figure G for which u > 0 in G and
(110) U = in G = (p >0) equilibrium u 0 in R'\G
J conditions
in the compressible case, and
(111) U 0 in G (p = 16)1 equilibrium u 0 in R3 \ G J conditions
in the incompressible case. We shall derive these equations from a variational principle.
DefinitIon 1.1. A function p(x) is said to belong to class if p(x) = p(r, z),
here M is a fixed positive number.
Definition 1.2. A function p is said to belong to class if p E and
Set
(1.12) E(p) = 21R31R3
W);) dxdy + 21R32(
p(x)
+K$fpY(x) dx;
the terms on the right represent, respectively, the gravitational potential energy, the rotational kinetic energy, and the internal energy. Consider the variational problem:
SELF-GRAVITATING ROTATING FLUIDS 421
Problem (E). Find p such that
(1.13) E(p) = p E
We introduce the formal derivative of E(p) (see Problem 1):
(1.14) E'(p) —f p(y)dy J2(m(r))d+I/p()
R3IX r(x) r
Lemma 1.1. If p is a solution of Problem (E) which belongs to L°°, then a. e.
115 E'(p)A inG (p>O),
( . inR3\G,
where A is a constant. If further V(x) is continuous, then p is continuous and (1.15) holds everywhere.
Proof. Since (p > 0) has positive measure we can choose 0, supp C (p > with some > 0 such that ip0(x) = ij0(r, z) = tj0(r, —z) and fi70(x) dx = 1. For any 8 >0 and for any = tj(r, z) —2) such that t 0 in the set (p <6), the function
to if e>0,e sufficiently small (depending on 8). It follows that
(1.16)
which gives (see Problem 1)
(1.17) —
Since 8 and are (1.15) follows with
X=fE'(p)irio.
From (1.14), (1.9) we see that (1.15) reduces to
mG inR3\G
a.e., so that the equilibrium conditions (1.10) hold a.e.
422 VARIATIONAL PROBLEMS Willi POTENTIALS
If for a solution p of (1.13) the gravitational potential V is continuous, then is continuous. We can then take a version
11 uY'\0
of p that is continuous in R3, and the equilibrium conditions hold everywhere.
Consider next the incompressible case and set
(1.18) E0(p)= ..Jf j R3 R3 x YI R' r (x)
Problem (E0). Find p such that
(1.19) E0(p)=
Irnroducing the derivative
(1.20) R3IX YI r(x) r
we have:
LeuaIa 1.2. If p is a solution of Problem (E0) and if p has compact support, then and
'1 mG,
.2, inR3\G,
where A is a constant.
Proof. Assume first that the set (0 <p < 1) has positive measure; we shall derive a contradiction. Choosing such that for some small >0,
fi,odx1,
we can proceed as before to establish that a.e.
122(. )
By Theorem 7.2 of Cbapter 3, the p. must be already rearranged
SILF-GRAVITAW4G AXISYMMfl'RIC ROTAI1NG FLUIDS
increasingly in z, for z >0. But then one can compute that V is strictly decreasing in z, for z >0 (see Problem 2). From (1.20) we then infer that the function is strictly monotone increasing in z, for z > 0. Thus the set
(1.23) ((r, z); = A)
has measure zero. Since, by (1.22), the set in differs from the set (0 <p< 1) by a null set, it follows that meas(0<p <1) = 0, contradicting our assumption.
We have thus proved that p 'G for some set G. Notice (see Problem 2) that is a function of (r, z) only, say h(r, z),
and h is continuous and strictly decreasing in z for z 0. It follows that
6=
(1.24) and is positive and continuous
• ontheopenset(r;h(r,0)<A)).
choose (re, 20) with r0 >0, z0 = 4i(r0) > 0, and let z8 = — 8, for any cmkll 8>0. For any 7>0, denote by B., the disc with center (r0, and radius y; we choose sufficiently small so that B27CG. Let ?k,(r, —z) be defined by
'10 = Cf8 if z >0, f'1o = 1.
Choosing z) = '1(r, 0 with support in R2\G, the function
p + —
(fi,)iio)
is then in for any small e>0, and we find as before that
(1.25) a.e.inR2\G,
where
A9 =
Notice that
Ao — h(r, z)dx — h(r0, z0) 87
= h(P, I) — h(r0, z0) —p0 ff8,.y -'.0, where (F, I) E B.,.
424 VARIATIONAL PROBLEMS WITh POTENTIALS
Similarly, denoting by BT the disc with radius y and center (r0, + 6), we choose y small and z) = —z) such that
= if z >0, = I.
Then for any q(r, z) = —z) 0 with support in G, the function
is in for any small e > 0. We then find that
a.e.inG where
A, if 6,y— 0.
Thus (1.21) holds with A h(r0, z0).
In the next section we establish estimates on the gravitational potential V of a function p in either or These estimates will be used in the existence proof for the variational problems (E) and (E0) (in Section 3).
PROBLEMS
1. Let p, a be bounded with compact support and p. p + ec belong to c? for all small e >0. Prove that
=JE'(p)a. de
(Hint: = mp,. Then mAr) — m0(r) m1(r) (2r2. If U = I, = z) dz, then dr = so that = To compute the derivative at e = 0 for the term involving j, set L =j2. Then
if JPL(m )r2 — fpL(mo)r_2] faL(mjr_2
4 e'fp(L(m,) — L(m0))r2
J, +J2;
PROBLEMS 425
since m1 m0 uniformly, J1 —' Next
= — L(m0))r2dr
+ to )L'(mE(r))mO(r) dtr2 dr
dtr2 dr
=
Since + :O)L'(m,) =
I = _e_lffL(m)[ôr_2 —
-. OL(m0)r'2dr + 2fm0(r)L(mo)r_2
= —K,+K2
and K1 = Urn J1,
K2 2fo(x)f L(m0)r3drdx. r(x)
Finally,
12 =
= f°°r_28[L(rne) — i'f L(nh,)dll_I o.j
2. If p(x) = p(r, z) = p(r, —z) 0, p has support, then: (i) The gravitational potential
'
is of 0. - (ii) If p is bounded and decreasing in z, i 0, then V is srnctly
decreasinginz,z>0. - • 1
tHin:: For (1) consider V(r,x), where f,(r, z, 0) = (r, z, 0 + 4).J
VARIATIONAL PROBLEMS WITH POTENTIALS
2. ESTiMATES OF GRAViTATIONAL POTENTIALS
For any p E or p E set
Po supp(x); xER3
clearly, p0 I if p E
Lemma 2.1. If p E or p E and if p(r, z) is nonincreasing in z for z > 0, then
(2.1) sup V(x) c*M2/3pV3 = 2.,r(3/4ir)2/3], xE R3
(2.2) Ci:I) (xER3),
where a0 is defined by
(2.3) = M
and C is a sufficiently large positive universal constant.
Proof. To prove (2.1), consider the problem of maximizing
1(p) =
(xfixed)
subject to the constraints
f p(y)dy=M.R3 Clearly, the solutidn.is given by p = p0!8 for a ball B with center x and radius a0. Calculating J( p) for this density, we obtain
1(p) = = =
We now proceed to prove that
(2.4) ifx=(R,0,z).
Suppose that R Aa0, where A is a positive constant to be determined later.
ES11MA1TS OF GRAVITA1IONAL POTENTIALS
Since
— Cp0a0, '(tx—yI<oo) X Y
we have
a0
which implies (2.4). Thus it remains to consider the case R > Aa0. Let S = (y = (r, 9, z); r — R 1< R/2, I z — Z < R) and S' R3 \ S.
Since for y E S' we have x — I> R/2, it follows immediately that
p(y) R
Thus it remains to estimate
'=1sIxYI (2.5) = Jf p(r,z)drdz
(I'— <R/2) (Iz — Z1 <R)
xf' rd9 [(z — Z)2 + r2 + R2 — 2rRcosO}"2
Writing A = z — Z, a A2 + r2 + R2, b 2rR, we have
rdO
+ r2 + R2 — 2rRcos9]"2
dO (24) 2rJ 1/20 (a--bcosO)
2r 2K(k)
(a+b)"2'
where k2 = 2b/(a + b) and K(k) is the complete elliptic integral of the first kind; see formula 291.00 of reference 55. Since k2 = 4rR/(A2 + (r + R)2), k-' I as A2 +(r — R)2 —'0. The function K(k), which is singular at k= 1, possesses the asymptotic propetty (see formula 112.01 of reference 55)
(k'21—k2)ask-'1.
VARIATIONAL PROBLEMS WITH POTENTIALS
Using k'2 (A2 + (r R)2)/(X2 + (r + R)2). we now conclude from (2.6)
Cr X2+(r+R)2 log C
X2+(r_R)l
for some sufficiently large absolute positive constant C. Finally, since IX 1< R and IT— RI< R/2, we have
(2.7) CR
[X2 + (r — R)21 /
Now applying estimate (2.7) to equation (2.5). we find
(2.8) If p(r,z)log CR
drdz.
(jr - R/2) [(z — Z)2 + (r — R)2J /2
(I: - 71<R)
Also, there holds
M 2wffp(r, z)rdrdz CR Jf
p(r, z)drdz. (fr— l<R/2)
(fr — <R)
We consider, therefore, the problem of maximizing F( p) subject to the con- straints
JJ p(r, z)drdz {I-— j<R/2)
Clearly, the maximum occurs for ,p = where D = ((r.z); (z — Z)2 + (r — R)2 <s2} an -I irp0s2 = CM/R. Computing F for this density, we obtain
F
21' CRlog—
CM +
CM CR—log R a0
ESTIMATES OF GRAVITATIONAL POTENTIALS
we assume that the constant A is now fixed sufficiently large so that R > Aa0 implies that s < R/2. The estimate for F above yields, by (2.8), the required estimate for I. This completes the proof of (2.4).
We next establish the estimate
(2.9) + ifxz(R,O,Z), Z>O.
As in the proof of (2.4), it suffices to consider Z > Aa0. The proof given below does not depend on the axisymmetry of p; therefore, we may assume that R=0.
We claim that V(x) is majorized by the potential due to a certain rearrange- ment of p; namely
(2.10) R3IXYI
where = = = (r, 0, z); 0 r < o(z)), and a(z) is given by
(2.11) irpoo2(z)f p(y1,y2,z)ay1dy2. p2
To prove the claim we express V(x) in the form
(2.12) V(x) r p(y1,y2z) R2[(z_ Z)2
Now, for fixed A and a, consider the problem of maximizing
F(p)_—f R3[A2 +.,qj"2
subject to the constraints
1R20't =
Clearly the maximum occurs forp = where D is a applied to the in (2. t2) edch
By VirLeeOI the obvious inequality fwrltingy = (r,O, z)J
( 1
430 VARIATIONAL PROBLEMS WITh POTENTIALS
it remains only to estimate
z=f dy. (tz—Z1<Z/2) k —yI
Since P(YI, z) is a nonincreasing function of z for z > 0 and p(y1, Y2' —z) = p(y1, y2, z), it is evident from (2.1!) that a(z) is a nonincreas- ing function of z for z > 0 and o(—z) = a(z). Writing = i(Z/2), we then have
c a.
Hence M/p0 = mess a for some absolute positive constant c. Thus for > 0 defined by
(C> 0), p0
it follows that Furthermore,
(fl {z>Z/2} ç ç Thus we can estimate I as follows:
(tz-Zl<Z/2)
IZ/2 1.00 rdrdA =Cp0j j 12Z/2 0 (X2 + r2) /
2 fZ/2 dA
= Cp0o0 j—Z/2(X2+ 2 Zlog—
00
CM CZ =
here we use the fact that Z > 2ev, which is implied by the assumption Z > Aa0 provided that the constant A is fixed sufficiently large. This completes the proof of (2.9). The assertion (2.2) is obviously a consequenee of (2.4), (2.9).
Remark 2.1. The assumption that p is nonincreasing in z, z >0, was not used in the proofs of (2.1) and (2.4).
PROBLEMS 431
Lemma 2.2. If p is a solution of Problem (E) and if p has compact support, then
(2.13) yE
where is defined by
(2.14) -- CAl2'3
and C is a positive constant depending only on K. /3.
Proof. From Lemma 1.1 we obtain
(2.15) V+f+A mG.
Since p has compact support and V —' 0, f — 0 as x (1.10) and (Ill) yield
A = lirnu(x)
Noting also that I 0, we obtain from (2.15)
V.
Estimating the nght-hand side by (2.1), the assertion follows.
PROBLEMS
1. Let T be the (solid) torus
T = {y = (r, 0, z); (z — Z)2 + (r — R)2
with 0<s<R/2.
where C is a positive constant in4ependent of R, Z, s. [Hint: Use the proof of (2.4).].
2. Another common model for the rotating fluid is obtained when one prescribes the angular velocity 0(r) instead of the angular momentum j(m). In this case the equilit*ium
(2.16) -- VV+
432 VARIATIONAL PROBLEMS WITh POTENTIALS
and in the functionals E, E0 the kinetic energy term is replaced by
where
(2.17)
_Jp(x)J(r(x)) Jx,
J(r)
In the incompressible case the equilibrium condition is
(2.18) V+J+V+J+ A'cO in the fluid, outside the fluid.
r2—j+.-j<l (a1>a2). a, 02
and
V(x) = — A1r2 —
— e2 A, = sin'e — 1 —e2
e3 e
2aA +
2 2ji_e2
sin e, e e
where e = (1 — is the eccentricity. Show that (2.18) is satisfied if
— e2 2 —I 6 2—2 At—-IA2 = 2(3—2e )sin
e e
which has the graph shown in Figure 4.1. Other families of solutions can be found in reference 69.
3. The rotating spheruids of Problem 2 are also solutions of the model (1.3) with p 1, wheref is givenby (1.8); fmd the functionsj(m).
FIGURE 4.1
If s(s) = const. then J(r) = Maclaurin (1742) discovered a family of solutions corresponding to incompressible oblate spheroids
For such a body,
i-p w
EXISTENCE OF SOLU11ONS 433
3. EXISTENCE OF SOLUTIONS
Theorem 3.1. Problem (E) has a solution p which is continuous and has compact support, and p( r, z) is nonincrea.sing in z for z 0.
In order to prove the theorem we shall first consider a family of truncated problems.
For any N> I, denote by the class of functions E such that
a.e.,
where p.,, is defined by (2.14) and is defined by (2.3) with p0 = p.,,, a0 = that is
(3.1) = M.
Lemma 3.2. There exists PN E satisfying
(3.2) E(pN) mm
is continuous in x < Na.), PN(r, z) is nonincreasing in z for z 0, and
(3M. àn{pN<Np.}
where is some constant.
Proof. Since is a bounded, closed, and convex subset of Na.), 1 p < 00, we cohclude that is compact in the weak topology of V. We claim: In the weak topology of L',
(3.4) &s semieonunuous
Indeed, if p,., p weakly in then
m9(r) —. poiniwise
[where m(r) the density pJ. Since further r), are monotone, continuous, and uniformly bounded, the convergence is
uniform. Consequently,
j2(m(r(x)))
VARIATIONAL PROBLEMS WITH POTENTIALS
uniformly in x. Noticing also that
r
where C is independent of it follows that
dx dx.
the functions
Pgn(Y) dy
I x — y I
are uniformly bounded and equicontinuous and thus converge uniformly to
f dy.Ix—yt Hence
ff dx dv dy.
Since finally
dx dx,
the proof of (3.4) is cbmplete. It follows that there exists a solution PN E of (3.2).
Sinc.e PN is bounded and has compact support we can now proceed to prove (3.3) a.e. as in the proof of Lemma 1.1. We shall now prove that PN is continuous.
Set
(3.5) UN_Vp +f+AN
when is the gravitational potential of a density function p. Then
(3.6)
We claim that
PN < ac. on the set {UN < c(Np4"0).
Indeed. if p,, then a.e. by (3.3), and thus, by (3.6),
= UN
EXISTENCE OF SOLU11ONS 435
a contradiction. Similarly, PN Np,, a.e. on the set (UN > and
a.e.ontheset(uN>O), PN=0 a.e.ontheset(uN<O).
It follows that 0 < PN < Np, a.e. on the set
{o < UN
and thus, by (3.3), (3.6),
UN=O a.e.onthisset.
Consequently, a.e.
I/sif uN>c(Np*)
if 0 <UN < c(Np,)"5
if
By Theorem 7.2 of Chapter 3, monotone nonincreasing in z for 2 0 and then UN IS strictly decreasing in z for z >0 (Se tion I, Problem 2). It fQUQws that the Iàvól sets
((r,z);uN(r,i)a)
have measure zero. Hence the right-hand side of (3.6) gives a continuous version of PN.
Lemma 3.3. There exists a sufficiently large N such that
(3.8) 'N<°
Proof. From (3.3) we have
(3.9)
Take x0 such that x0 < r(x0) and
PN(XO) = inf
r(x)> iNa.
436 VARIATIONAL PROBLEMS WITH
Then CM and hence PN(XO) Thus, since < 3,
(3.10) E(N) oo.
We also have
(3.11) V?(xo) 2Na.
(3.12) —f(xo)=f°° j2(nl(r))d r(x0) r N
Substituting (3.10)—(3.12) into (3.9), we find that (3.8) holds if N is sufficiently large.
Lemma3.4.
(3.13) sup PN(X) xER3
Indeed, since AN < 0 and PN has compact support, the assertion (3.13) follows by the same proof as in Lemma 2.2.
Lemma 33. There is a positive constant independent of N such that
(3.14)
provided that is sufficiently large.
Proof. Let = where B is a ball about the origin, 0 <0 < 1, and JR3P(X) dx = M. Then the radius R of B is given by
= M
and
/ j2(m(r(x))) dxa r(x) o r m(r)
Recalling that j2(m) is monotone nondecreasing and making, a change of
EXISTENCE OF SOLUI1ONS
variables m 3r2/2R2 in the last integral, we obtain
(3.15) f j2(m(r(x)))
fi(x) dx 4irROp1jJ B r(x)
= J'(j2(m)/m)dns Next, since
( A. J8Ix—yI 2R'
we have
(3.16)
C >0. Finally,
(3.17) dx =
where we have used y = I + 1/P and (2.14). Combining the estimates (3.15)—(3.17), we get
with positive constants C. Choosing 6 sufficiently small (depending only on M and the function j) the expression in braces becomes larger than Since finally E N is large enough, E(pN) (3.14) follows.
Lemma 3.6. Set
(IxI<Aa.)
Then there exist posizivq constants A (sufficiently large) 'and 8 (su/fkiently small), both independent of N, such that
(3.18) if
N is sufficiently large.
438 VARIATIONAL PROBLEMS WITH
Proof. From Lemma 3.5 it follows that
CpV3M5"3 ucff
f + 2f V(x)pN(x)(IxI.b't<Aa.) X — Yj (IxI>Ao.) + log A;
here we used Lemma 3.4 and the estimate (2.2). Taking A sufficiently large, the assertion (3.18) follows.
Lemma 3.7. There exists a positive constant B (sufficiently large) such that
(3.19)
proiided that N is sufficiently large.
Pioof. Suppose that x2 E supp PN and x2 > Baa; we shall derive a con- tradiction. Take x1 such that x1 r(x1) > and
p1 = p(x1) iflf IxI<Aa.
r(x)>(I/2)Aa,
Then p1 if N> From the varia- tional conditions (3.3), we have
(uuN) so that (with V= VP)
(3.20) V(x1) V(x2) + + fif2(rn(r))
dr,
where ç = r(x,). Also
V(x1) (by Lemma 3.6),
V(x2) [by (2.2)],
OF SOLUTIONS
and
,-r2j2(m(r)) dr
r3
where
j1 maxj(m).
Thus (3.20) implies that
BbogB + 22 ÷
If A and B are sufficiently k*rge (depending on M and) only), then we obtain, using (2.14) and (3.1),
Cn',3
A
A is sufficiently large.
Proof of Theorem 3.1. Choose p = R as in the previous lemmas. It suffices to show that this p is a of E and define
if IxI<Na. and 1.0 otherwise.
Set
where Bis a ball of radius about the origin and is defined by
f ON(x)dxR3 = 0N + SincclN OifN -, sufficiently
large, say N Also, it is straightforward to verify that
(3.21) ifN-øoo.
444) VARL4TIONAL PROBLEMS WITH POTENTL4LS
By Lemmas 3.4 and 3.7 we have PN E for all N N, and therefore
E(pN).
Since PN E we also have
Thus
urn = N—
Theorem 3.8. Problem (E0) has a solution p such that p = 'G' G has bounded support and is given by
where 4(r) is positive and continuous on an open set.
The proof is similar to the proof of Theorem 3.1. In fact, it is somewhat simpler since the estimate = I is trivial; for the last assertion, see (1.24).
Consider now compressible fluid with general barotropic pressure—density relationship
p = P(p). Set
so that
=
The energy functional is
=
(3.22)
+ !j f2(m(r(x))) p(x) dx + f 'P(p(x)) dxr(x) R3 and the equilibrium conditions are
(3.23) inG (p>O), inR3\G,
EXISTENCE OF SOLUTIONS 441
where ji(t) = 'I"(z). Assume that P(t) is positive, strictly increasing, and smooth for £ > 0, and that
(3.24)
urn 0. ,-o
Then t) is a positive, strictly increasing smooth function fort > 0, = 0, and
(3.25) I /
0. ,-.o
Set
4K 3/2 (3.26)
where c is the universal constant defined in (2.1). Consider the problem: p such that
(3.27) =
Theore!n 3.9. If M < M0, then there exists a solution p of (3.27); p is a continuous/unction with compact .$upport and p(r, z) is z, z 0.
The proof is similar to the proof of Theorem 3.1 once we have established an a priori bound of a solution analogously to I.emma 2.2. From (3.25), (3.26) it follows that there exists a positive number such that
4'(p) = c*M2/3ph/3. .
We can now proceed as in Lemma 2.2 to establish that 44p(x)) 44 ps). From now on when wcspck of thø compresjble case we
most of the extend Iosolutions of (3.27), as in l'heorem 3.9.
442 VARIATIONAL PROBLEMS WITh POTENTIALS
Definition 3.1. Set
(3.29)
Q =f'q(m)dm.
The fluid is said to be slowly rotating if
(3.30) Q < C0, q(1) < C0 for some positive constant C0
in the compressible case, and
(3.31) Q< I,q(l)<C0 forsomepositiveconstantC0
in the incompressible case. The physical interpretauon of Q and q(m) is given in Problem 4. Introducing
(3.32) = max j(m), o m
(3.30) implies that
(3.33)
l'heorem 3.10. If the fluid is slowly rotating, then there exists a positive constant B such that for all 0 < M < oo,
(3.34) suppp C
This result is valid both for the compressible and the incompressible case. Since a ball with density and mass M, we also have
(3.35) supppi
The proof of Theorem 3.10 is obtained simply by reviewing the proof of Theorem 3.1 and observing that the various constants C are independent of M, j provided that (3.33) holds.
Remait 3.1. The notation in the proof of Theorem 3.1 can be slightly simplified if we do not keep track of the explicit dependence of the on M and j. However, we did keep the dependence fairly explicit, having in
RAPIDLY ROTATING FLUIDS
mind applications to asymptotic estimates, such as in Theorem 3.10. In the next section we consider further applications to rapidly rotating fluids.
PROBLEMS
1. Prove(3.21).
2. Prove Theorem 3.9.
3. For degenera'e gas ("white dwarfs")
P(:) = AF(x) wherex = (s/B)"3 and
F(x) = x(2x2 — 3)(x2 + 1)1/12 +
and A, B are positive constants. Verify (3.24) with K < 00.
4. Imagine a (hypothetical) ball of radius uniform density p,, and total mass M rotating with a distribution of angular momentum per unit mass given byj(m). Let Tand Wbe the total kinetic and gravitational potential energies, respectively; and, for 0 <p I, let 7 and be the kinetic and gravitational potential energies of a unit mass situated on the lateral surface of the cylinder of radius r such that m(r) = p. Show that
c1q(p)
where c1, c2 &e absolute positive constants.
5. Extend Theorems 3.1 and 3.8 to the model in Section 2, Problem 2, assumingJ(r)tobcnondecreasing,J(O)=O,J(oo)< ooand
limr(J(oo)— J(r)) = 0.
4. RAPIDLY ROTATING
We flow drop the.assuinptions (3.30), an4 obtain a bc. the support of the solution, simultaneously for the compressible and incompressible [problem (E0)] case.
For any 0 <p 1, set
p(x)dxpM I. (r(x)<R}
444 VARIATIONAL PROBLEMS WITH POTENTIALS
Theorem 4.1. There exists a positive constant C independent of M, j such that for any 0 <p 1,
R R (4.2) C—flog I +
Proof. Set = (RM, 0,0). Then
V(xM)+f(xP)+ A.
Since A 0, we have
(4.3) V(x,3.
By monotonicity of j(m),
and, by Lemma 2.1,
Substituting these inequalities intO (4.3), the assertion (4.2) follows.
Denote by a1 the radius of the smallest ball about the origin which contains the fluid domain G. Thus, for a slowly rotating fluid.
where C, are positive constants M,j.
Corollary 4.2. If q(l) for soøw positive constant c0, then for any fixed e0>Othereholdt
(4.4)
where C1 is a positive constant depending only on c0, £0.
RAPIDLY ROTATiNG FLUiDS 445
Definition 4.1. A fluid is said to be rapidly rotating if
(4.5) Q for some positive constant e0,
(4.6) c0Q q(m) forO <m I and a positive constant c0.
The lower estimate (4.4) is complemented by the following upper estimate:
Theorem 4.3. 11(4.5), (4.6) hold, then
(4.7)
where C2 is a positive constant depending only on 4,
We shall first prove the theorem in the incompressible case. In this case, (4.5) and the first inequality in (4.6) can be rewritten in the form
(4.8) - M413
(4.9) mf f2(m)
m
wherej0,J1 are defined in (3.32).
Lemma 4.4. In the incompressible case, if (4.8), (4.9) hold, then there exists a positive constant B depending only on c0 such that
(4.10)
where Ls determined by
(4.11) 4,
and C, A0 are some positive constant (A0 is sufficiently large) of M, J, €0, Co.
Note, by (4.6), that
r i
Substituting this into (4.IQ), the assertkyi for the incompressible case follows (recall that i Thus it i'elñalni to establish Lemma 4.4.
446 VARIATIONAL PROBLEMS WITH POTENTIALS
The proof is based on several lemmas. We let R* denote the solution of the equation
(4.12) MI"3
(i +
= A0
We shall establish the required estimate (4.10) in the form
(4.13) a1 + max(R,M"3q(l)).
Lemma 4.5. Under assumptions (4.8), (4.9),
(4.14) jj dxdy CR*GGIXYI G r2 M'13 Proof. Take = where O is a torus
z2+(r_R)2<Zs2, 2i'r2s2R=M,
By Section 2, Problem 1,
f CR
aIxyI s R M'13 provided that
x (r, 0, z), z2 + (r — R)2
Hence,
jj dx* CRR M113 Also,
It follows that
- CM2 I CM.2 R 101+MI/3)+R2j0
Choosing R = R*, with fixed sufficiently large in (4.12), we fmd that
—
RAPIDLY ROTAI1NG flAJIDS 447
Using the inequality E0(p) (4.14) follows.
Lemma 4.6. Asswne that (4.8), (4.9) hold and set
{x;IxI<AR*)).
Then there exist positive constants A, 8(0 I) independent of M, j such that (4.15) 6.
Proof. The proof's outline is the same as for Lemma 3.6. However, in order to carry out the details we must use the kinetic energy in order to control a portion of the gravitational energy corresponding to a mass in a neighborhood of the origin. More specifically, we shall need to establish the inequality
(4.16) dx€Iy
(r(x), I X — Y I 1(r(x)<1?R) r2 (x,yEG) (x€G)
for some >0 independent of M, j. Let
be a partition of 10, such that
(4.17) g(r1)Emeas(Gn(r<r1)}>0.
Define
0, {(x, y); <r(x) V r(y) <rJ,
where a Vb= max {a, b), and
We can write the left-hand side of (4.16) in the form
1=2
where
• I, — •,
I— JJ if(x.yG)
J,<2f drf , (O<r(y)<r,) I X — Y I
(x€G) (yEa)
448 VARIATIONAL PROBLEMS WITH POTENTIALS
By (2.4), if i 2,
mM I Cr(x)\ ' jdx
r(x) \ M'13 / (xEG)
+
On the other hand, the right-hand side of (4.16) can be written in the form
rihere
K1 j r
(XEG)
Choosing the partition sufficiently refined so that
r. rn <2
m1_1
[note that m1 >0 by (4.17)J, we conclude that K. J1 for 2 i n provided that
Cr1 \+ M'13 J ClmiM4/3
for some sufficiently small positive constant c1. In view of (4.9) it suffices to show that
r. I Cr \ Jo2+ Mv3)
with another sufficiently small positive constant c2. Since the last inequality follows from the dcfipition of R* provided that is chosen suffi- ciently small (independent of M, j). We conclude that the left-hand side of (4.16) is smaller than the right-hand side plus the term J1. Now taking r1
1 inf {r; g(r) >0)
(so that 0), the proof of (4.16) is complete.
RAPIDLY ROTATING FLUIDS
By (2.2) we have
f V( x) dx CM
log (i +
A fixed sufficiently large and (4.16) in the inequality (4.14), we deduce that
CM2 I CR*\ ( dxdylogil+ R* M"3 I Ix —yI
(r(x)Vr(y)>i1R) {x. yEG)
By (2.4), the right-hand side is smaller than
+ CARS
flR* og1
M"3
Noting that
CAR* CR* +
M'13 ) + M'/3)'
where B1 is a constant depending only on A, we conclude that > 6, where 6 is a positive constant independent of M,j(m).
Lemma 4.7. Under assumptIons (4.8), (4.9),
(4.18) [X—u(oo)],
where is a sufficiently large positive constant depending only on c0.
Proof. Let
(4.19) r0=B2max(R*,Mh/3q(1)},
where 82 'will be taken sufficiently large depending only on c0. Then in the region
there must be a point that does not belong to G. and hence 0. Taking x = i in the relation u V(x) + f(r) + X, we obtain (4.20) [Fr(i)].
450 VARIATIONAL PROBLEMS WITH POTENTIAL8
By Lemma 4.6,
here we have taken B2 A. Also,
—1(P) =
Putting these inequalities into (4.20), we get
— ç6M + —___
r0 r0 r0
by (4.19); this gives (4.J8).
Proof of Lemma 4.4. Take a point x' on the boundary of G such that x' a1. Since u(x') = 0, we get
A5= V(x') +f(r1) [r, = r(x')]
(4.21) V(c').
By (2.2),
Substituting this and (4.18) into (4.21), the assertion (4.13) follows.
We next consider the compressible version of Theorem 4.3. In this case it is sufficient to establish the following lemma.
Lemma 4.8. In the compressible asswne thai (4.5) and the first inequality in (4.6) hold. Then there exists a positive constant B depending only on c0 such that
(4.22)
where is determined by
(4.23) + 40Q,
RAPIDLY ROTATING FLUIDS
and A0 (sufficiently large) is a positive constant depending only on e0; C is a constant independent of M, j, e0,
The proof of Lemma 4.8 proceeds along the same lines the proof of Lemma 4.4. Here we let R be defined by
(4.24) ____log(l =A0Q.
) As before, we establish several lemmas.
Lemma 4.9. Under the assumption of Lemma 4.8,
(4.25) f ( J2(m(r))P(X)dx +
R Iog.(l+
a,,
Proof. Take = where a ihe tbrus
Rz2+(r_R)2<32, -
Then, analogously to Lemma 4.5,
C9'13R'If a*fx—yf
f P(x) dxR3 1/Sand the4efinhtjonof as.
that I
CM2 CO'/3R
J + p'/3M513:R
— _CM2 1' +
C9u/3R\ Ca,, (4.26)
—. ) a,,,
452 VARIATIONAL PROBLEMS WITH POTENTIAlS
If we take 9 according to = where /1 <a < 3 and C1 is fixed sufficiently large, then the first term on the right-hand side of (4.26) majorizes the third term. Next, we take R = R* with A0 sufficiently large to ensure that the condition s < R/2 holds and that the first term on the right-hand side of (4.26) majorizes the second. Recalling that E( p) E( (4.25) follows.
Lemma 4.10. Assume that the conditions of Lemma 4.8, and set
= i* p(x) dx. {?x?<AR')
Then there exist positive constants A, 8 (0 <6 I) independent of M, j such that
(4.27) ILA 6.
Proof. The proof is similar to the proof of Lemma 4.6. We first need to establish the estimate
if f2(m(r))P(X)dX I X Y I P
Using a partition as in the proof of Lemma 4.6, we then find that the only point that needs to be verified is that
r. I Cr.
for c2 is a sufficiently small positive constant. As before, this follows from the definition of R.
We next proceed to verify the inequality
<j ((x(V(yI>AR*J Jx
where A is fixed sufficiently large. Finally, the proof is completed by applying the second estimate of Lemma
2.1 to the right-hand side of the last inequality.
Lemma 4.11. Under the assumption of Lemma 4.8,
(4.28)
where B0 is a sufficiently large positive constant depending only on c0.
RAPIDLY ROTATING FLUIDS
Proof. Let
r0
B1 is a sufficiently large positive constant. In the region
there is a point where p( x) attains its minimum. Clearly,
CM,
so, by (1.10),
I
cj 4i u(s) +1(P) + A = r0 /
Hence
A — C&M
+ -ça,Mq(l)+
c c —r0 r0
here we have computed the last term using (2.14), (3.1). Recalling that R > 41a,, for some positive constant depending only on e0, the assertion (4.28) now follows from the definition of
The proof of Lemma 4.8 can now be completed in the came way Lemma 4.4.
From Corollary 4.2 and Theorem 4.3 we have the asymptotic estimate
Naturally, one expects the fluid to be contained in a thin slab about the equatorial plane as the "rapidity of rotation" Q becomes large. The set occupied by the fluid has the form
(4.29) 6 oo).
454 VARIATIONAL PROBLEMS WITh POTENTIAI.S
For any 0 < I, define
=
where RM is defined in (4.1). The number is called the oblateness ratio.
Theorem 4.12. 11(4.6) holds and Q is sufficiently large, then
(4.30) çQ—P[log(l + Q)]P2,
where
— 3—fl _3(1+4fl) P2—2(l+2$)
in both the compressible case (0 < fi < 3) and the incompressible case (fi 0); depends on !t and c0.
For the incompressible case, (4.30) follows from the following estimate: Set h(R) = If R > A0M'13 then
h R -3/2
(4.31)
here A0 (sufficiently large) and c0 are absolute constants.
The proof of Theorem 4.12 and (4.31) depend on upper bounds on VVI and lower bounds on V2 for details, see reference bOb.
PROBLEMS
1. Prove that (4.31) implies (4.30) in the incompressible case.
2. Prove that the assertion of Theorem 3.9 remains valid for M M0 provided thatj(m) > 0 for all 0 <m 1. f Hint: Suffices to derive a priori bound on p0 = supp: p0 = p,(M,j).SetA andshowforA >1,
+ (log 1 +
PROBLEMS
Use sup V to deduce that
(•) + (i +
if p > where
Fix = M0/2M in () and use also (4.2) to deduce
+ CA) a0
for some A > max(A, 1),
C =—flog(1+CA),
or
(i*) +A).
If A < 1, then (4.2) gives a stronger inequality.J
3. Set je(m) = Ej(m) (j(m) > 0 if m > 0) and denote by a solution corresponding to M (constructed by Problem I). Let a1(e) = sup
x p1(x) > 0). Prove: If M C0M0, then
a,(e) C,e2,
where C0, C, are positive constants depending only on Q and q(l). (Hint: Follow the proof of Theorem 4.1 in the comptéssibté dase.]
4. Establish the estimate: If R > AM'13, then
[x(R,0.Z)]
in the incompressible case; A is sufficiently large and .C depending on A.
456 VARIATIONAL PROBLEMS WITH POTENTIALS
IHini: JGIX G= G' U G", G" = Gfl (IR — r(y)I< R/2). Use
(2sr dO
_______
Jo
to estimate
dy drdz
(jR—rj<R/2} [(z__Z)2+ (r_R)2}I/'2
Also.
If (IR—rt<R/2)
R
(I'I<41r)}
Consider a corresponding maximization problem; see Lemma 2.1.]
5. THE RINGS OF ROTATING FLUIDS
Set
— 82u +
1 8u +
a2u
ar2 rar az2 If p is a solution established in Theorem 3.1, then
(5.1) uV+f+A is HOlder continuous and, since
(5.2)
also p is Holder continuous. Applying the Laplacian A (in R3) to (5.1), we obtain
(5.3) Lu + y(u) =
where
(5.4) 7(u) (c0> 0)
THE RINGS OF ROTATING FLUIDS 457
and
(5.5) 1, Lf z) dz 2)2(o)
We assume in this section that j2(m) is in II. Since
(5.6) Cr2
we deduce from (5.5) that his HOlder continuous (exponent fi) for r> 0. From (5.3) and elliptic estimates, we then conclude that u away from the z-axis.
Recall that u is strictly decreasing in z for z > 0: Further, since u2 ç 0 if z > 0, we obtain from (5.3)
Lu2 0 if z >0 (in the dIstribution sense)
and the strong maximum principle gives
(5.7) u2<O ifz>O.
The fluid region G is given by (4.29). Using (5.7), we find that $(r) is uniformly continuous ifl every set (0 r R; +(r) > e), where R >0, e >0. It follows that
(5.8) 0O.
Definition 5.1. Let (r >0; $(r) >0) = where it, are disjoint open intervals (a,, b,) and 4(a,) = 4i{b1) = 0 [If a, = 0 then 4i(a1) 0.). The sets
((r, z); Iz < +(r), r A,)
are called thc rings and A. iscalled the baseof the ring
Theorem 5.1. If h(r) (in (5.3)) is analytic in r, r A,, then is analytic for r E it,. For the model in Section 2, Problem 2, h(r) is analytic in r if J(r) is' analytic in r.
Proof. Takeit=A,,a=a,,bb1.Since
and u, the implicit function theorem be applie4 deduc.tl*L • E To prove analyticity we follow the method of 1, using the hodograph transformation
y2=x2, y3—u(r,z).
458 VARIATIONAL PRORLEMS WITH POTENTIALS
Since u2 0, we can solvez = w(y). Sety' = Y2) and define functions 4', tp by
4'(y)w(y)
if 0.
Then
4r=q, 4'y3 11)!3 ony30.
Further 4' and satisfy, fory3 >0, the nonlinear elliptic system (see Chapter 2, (1.27)1
F(4) h(r) — [r
= —h(r) + y(—y3),
where
F(fl=
and = By standard elliptic methods [3bJ one can then show that the successive derivatives
exist and are continuous; they can be estimated by
C0Cm(m—2)!
Thus 4', are analytic in y'. Since the free boundary z = is given by
z w(y1, Y2,O),
the analyticity of + follows.
We wish to prove that the number of rings is finite, but first we need two mmas. Denote by B, the ball in R3 with radius r centered about the origin.
Lemma Let $ be a positive noninteger, $ >0, and let v(x) be a function sahsfrhsg
(5.9)
THE RINGS OF ROTA11NG FLUIDS
Then
(5.10) v(x)P(x)+r(x) mB1, where P(x) is a harmonic polynomial of degree 1$] + 2 and
(5.11) mB1,
(5.12) MB1,
min{$— [P1'' + [P1 — #9);
C is a constant depending only on and on upper on o(xjl, f vv(x)f for x E
Proof. Let e1 = (1,0,0). 'then (see reference 120, p. 125)
where cos 0 = x1/I x J and P,,(u) is the Legendre polynomial of order n. Notice that
is a bomogeneous harmonic polynomial of degree, a. Since (see reference 120, p. 128)
iflul"I, we have
More generaHy,if I x '
1 1 1
(x-y Iyl IYI
where is an orthogonal transformation that maps e1 intoy/j y This
(5.13) (Ixt<IyI),
460 VARIATIONAL PROBLEMS WITH POTENTIALS
where r:(x) is again a homogeneous harmonic polynomial in x of degice n, and
(5.14)
Now, by formula,
(5.15) 4iii,(x)= _J -1___E:-v(Y)ldS
=11 6
We break up the last integral into two integrals:
(5.16) i2—f 1
f I+j YI
+
In I, we have
so that
(we use here the fact that The integral of x — y I taken over a ball with center 0 is smaller than the taken over a ball with the same radius but with center y).
Fory E the series
is absolutely and uniformly convçrgent; we can therefore integrate it term by term. n <$ + 2 we shall add to the nth term
'+1r1(x)Av(y)dy (I I
THE RINGS OF ROTATING FLUIDS 461
the integral
ic=f I I
The sum
is a homogeneous polynomial of degree n. Also, by (5.9), (5.14),
c
where C is a generic constant depending only on flu. Hence
(5.17) P [P1
The terms with n > ÷ 2 are estimated by
I
Summing over n, we obtain the bound
Collecting the estimates above, we conclude that
(5.18) + F0(x),
where Fo is a harmonic polyno nial of degree (flu + 2 and is bounded by the right-hand side of.(5i 1).
462 VARIATIONAL PROBLEMS WITH POTENTIALS
From (5.13) and the definition of r:(x), we obtain an expansion for
a I
ay,Ix—yI
in terms of harmonic polynomials. Since (see reference 120, p. 128)
(5.19) I),
we find that
(5.20) J1(x) = n0
where a homogeneous harmonic polynomial of degree n and
(5.21) (forany8> 1);
C depends on & and on upper bounds on f vv in If x p, where p8 < 1, then
I n>P+2
where C' depends on C, 8, p. If p x < 1, then
I cn(x)I=JJi_
8
p
with another constant C'. Choosing 8 = p = we conclude that
(5.22) I
ifIxI< 1.
Setting
and recalling (5.15), (5.18), (5.20), the assertion (5.11) follows for the function r(x) defined by (5.10).
THE RINGS OF ROTATING FLUIDS
From the definition of I',((x) and (5.19), it follows that
i
Using this, we can estimate in the same way that we have estimated r0(x) above, with some minor changes. We find that (5.12) holds for F replaced by F0. Since Cn3 [which follows by induction, using the relation — ,(u) = nP,,_ ,(u); see reference 120, P. 1271 we can obtain for an expansion similar to (5.20) with the corresponding c,,(x) still satisfying (5.21). Estimating the remainder term as before, the proof of (5.12) follows.
The next lemma is a nondegeñeracy lemma in the spirit of unique continua- tion. The lemma will be used in Section 17, but a basic part of the proof will be needed in this section.
Lpmmg 5.3. Let v(x) be a bounded function in B1 C R3) satisfying
(5.23) 1),
(5.24) v(O) = 0,
(5.25)
Then there exists a homogeneous hannonic polynomial HM(x) 0 of a precise degree n 1 such ihat, us B,
(5.26) v(x) = + K(x),
where
(5.27)
(5.28) vK(x) >0, >0).
Propf Without lossef. generality, we may assume, that I <e <2. Byihe V elliptic estimates, v is in C'. Therefore, v(x) I" (x (. It follows that
By Lemma 5.2,
v(x) = P,(x) + I',(x) in B,,
where P,(x) is a harmonic polygoinial o( degree (aJ + 2 = 3 and
mB,.
464 VARIATIONAL PROBLEMS WI FR POTENTIALS
If P1(x) 0, then the assertion of the lemma follows. Otherwise, v coincides with I',, so that
in B1.
Applying Lemma 5.2 once again, we conclude that, provided that a( a + 2) integer,
v(x) P2(x) + ['2(x) in B1,
where P2(x) is a harmonic polynomial of degree a1 = (a(a + 2)J + 2 and
if a(a + 2) is an integer, then the above remains true with a(a + 2) replaced by a(a + 2) — £ for any >0. If P2(x) 0, then the proof of the lemma follows. Otherwise, we apply Lemma 5.2 once again,
Proceeding in this way step by step, we conclude that if the assertion of the lemma is not true, then
(5.29) mB1,
(5.30) <flf)CsIxtPN in B1,
where and Pm t 00. Notice that these inequalities hold with Pm such that
(5.31) Pm (Pm- + 2)a —
here 8m 0 and /3,,, is not an integer. We can choose the 9,,, so that 0 °m < 1 and
(5.32) c, c > 0 (c independent of m)
for all m. We shall estimate the /3,,,, by induction. The inductive assumption is that
(5.33) (A >0),
(5.34) (B> 0),
(5.35)
(5.36) (K0>1,K>1,Dm<E1)
for all m n, where Dm il/a1, c >0.
THE RINGS OF ROTATING FLUIDS
From (5.31) with m = n + 1, we then get
so that (5.33) holds for m = n + 1. Next taking the logarithm in the inequality
(Pm + 2)a,
we get
2cloga + +
Summing over m, I m c n, and using (5.33),
log
Thisgives(5.34)form=n+ I. Since (5.31) holds with m = n, m = n + 1, and since (5.29) and (5.34) hold
with m = n, Lemma 5.2 gives
mB1.
Thus we can take to be any positive constant satisfying
(5.37)
It remains to show that if
then both (5.37) and (5.35) (with m = n + I) are satisfied. Now, by (5.34) with m = n + 1,
c 2, B.
As for (5.37), one can easily verify it by using (5.34) withm = n + I, provided that c in the definition of the D,,, is sufficiently large, depending on K0, K, a.
From (5.30), (5.32), (5.33), (5.34), (5.36) we obtain the inequality
I
+2
VARIATIONAL PROBLEMS wrrfl POTENTIALS
in B1. If I
x then
wherep=p(x)<1.
Taking n -s 00, we conclude that v(x) 0 if Ix 1< let p0 be the largest number I such that v(x) = 0 in If p0 < 1, then
we can repeat the previous argument and show that, for some >0, v 0 in any ball with center in and radius p,. Since this contradicts the definition of p0, it follows that p0 = 1, that is, v 0 in B1.
Remark 5.1. From (5.30) we see that v vanishes at x = 0 to any order. By well-known results on unique continuation (see references 15, 73) we can then deduce right away that v 0, even if a I. However, the argument given above will be needed in the sequel.
Definition 5.2. A number R 0 is called a point of accumulation of rings if there exists a sequence of rings such that
dist({R), -. 0 ifj' -. oo.
Theorem 5.4. The number of rings is finite in every interval {e <r < oo), e>O.
Proof. If the assertion is not true, then there is a number R >0 which is a point of accumulation of rings. Setting
d = d(x) = ((r — R)2 + z2)"2 'ii r = (r, 0, z),
we shall prove by induction that u satisfies
(5.38) y(u(x)) C5d5m (d d0, N5-);
the constants will be determined in the inductive proof, and d0, N are positive constants independent of m.
Writing
—y(u),
we can apply Lemma 5.2 to deduce the expansion
(5.39) U — I = +
where Pm is a polynomial of degree [ft,J + 2 and Qm satisfies
(5.40) CC5 C independent of m.
THE RiNGS OF ROTATING FLUIDS 467
Here is required to be any positive noninteger such that
(5.41) Pm — [PmI c, c >0, c independent of m.
Using the relations (5.2),(5.4), and (5.38), we have
m(R) —
I'
Writing
f(r) —f(R) + f'24 — m(R))4 where lies in the interval with endpoints R, r, and expanding 1/s3 in the first integral in powers of $ — R, we find that
(5.42) f(r) a1(r—R)'+ j=0
Since u and f depend only on r, z, the same must be true of i'm; that is, Pm = Pjr, z). From (5.39) it follows that also Qm = Qm(", z). Taking z = 0 in (5.39) and using (5.42), we get
1$AIl+ 2
u(r,O) = — R)' + Qm(r,O) + 0(1 r — R j=o
The function u( r, 0) oscillates an infinite number of times as r R from one side. This implies that all the coefficients b, must vanish. Noting that
O(jr—
where C0 is independent of m and choosng N> C0, we obtain, after using (540),
(5.43) u(r,O) C1Cp..flm J r — R c1 independent of m,
provided that $,, 1. Therefore,
z)) 'cy(u(r,O)) Pmlr
VARIATIONAL PROBLEMS WITH POTENTIALS
Thus the inductive estimate for m + I follows with
= co(C1C5 Pm)5, (5.44)
Pm+I = P(I3m + 2) —
where e [0,1) is chosen so that Pm+l — [Pml c >0. We now proceed similarly to the proof of Lemma 5.3, following (5.30).
Indeed, the argument there shows that Jim' Cs can be chosen as in (5.33)—(5.36) Inotice that (5.37), (5.31) are replaced here by (5.44)].
Having established (5.38) with suitable Pm, C5, we next show that y(u) 0 in some neighborhoc.J of (R,O), which is impossible.
Consider now the incompressible case. Here u satisfies
(5.45) LU+47rIG—h, G=(u>O}.
G is again given by {(r, z); IzI< 4i(r), r 0) and we use Definitions 5.1 and 5.2 for this case too. The continuity of 4*(r) is valid as before.
Theorem5.5.
Proof. The proof is similar to the proof of Theorem 5.1. The main difference is that here u E for any 0 < a < I (instead of u E Hence we must write for i(', a system of elliptic equations in divergence weak form, namely,
H(4*) = (h(r) —
H(i1) = (h(r) — y(—y3))ip3,
where
(1
By C1 Schauder estimates for such equations [3b) we can deduce (working with finite differences) that 4', belong to C2.4e up to = 0. We can then proceed as in Theorem 5.1.
Theorem 5.6. If h(R) 0 for some R 0, then R is no: a point of accwnula- lion of rings.
The proof involves estimates on the base and height of small rings (see Problems 3 and 4) and on the asymptotic density of as r -, R. For details, see reference 58j. -
PROBLEMS
PROBLEMS
1. Let be a ring for the compressible case with base A = (a, b), and suppose that
u,.(b,O) = 0,
0) is given by
2. Consider the model in Section 2, Problem 2. Suppose that J(r) = 1 if r>2R, if 0<zr<R, I if R<r<2R, where R >0, 0 <8< 1. Show that for M sufficiently large any solution p(r, z) Qf Problem (E) satisfies p(r, z) = 0 if r < R. I Hint: If = M0 > 0, redistribute the mass in {r(x) < R) in a1 <r <a, + I.J
3. Let be a ring of incompressible fluid, A = (a, b). Set X1 b — a, hA = max,EX$(r). Show that
hA
(Hint: .By the strong maximum principle z) <0; if +(F), derive hA Cu(F,O), and show that u(?,O)
ii in Problem 3, h(R) 0 and dist(R, A) is sufficiently small, then (c>O).
[Hint: u in If)GA)< eA2 and S, = (rEA, >lAl/8), then IS,l<8eX. lfrES2X\S,, u< Show that
(*) f u—>—cJAr, f D1)41f3
where B8 is the disc with center (r0 = (a + b)/2) and radius 6. Represent u(r0, 0) by Green's function in B8, and integrate on 8 to obtain
u(ro,0)cf u— (fyLu.
Use (*) and h(R) 0 to derive a contradiction.)
470 VARIATIONAL PROBLEMS WITh POTENTiALS
5. If j2(m) for some C> 0, .5 >0, then the number of rings in the compressible case is finite (for 0 < oo).
6. If in the compressible model of Section 2, Problem 2, the function J(r) is analytic in r, for 0 r < 00, then the number of rings is finite.
7. Extend Lemma 5.2 to n = 2.
6. VORTEX RINGS
Vortex rings can be produced by ejecting a puff of smoke suddenly from the mouth through rounded lips. (The smoke is just a marking agent which makes the motion visible.) Another example is produced by letting a drop of ink fall approximately 3 centimeters into a glass of water.
In describing the motion of a steady vortex ring in inviscid incompressible fluid, we shall use cylindrical coordinates x = (r, 9, z) and the associated orthdnormal frame i,, ii). The vortex ring is symmetric about the z-axis and propagating with constant speed W in the positive z-direction. With respect to axes fixed in the ring, the velocity vector v(x) has the form
(6.1) v(x) v'(r, z)i, + v2(r, z)i:
and
(6.2) v(x)= —Wi2 aslxl—.oo.
The vorticity field = V then takes the form
(6.3) = w(r, z)i2, z) = + v.
Euler's equations are
(6.4)
and the equation of conservation of mass is
(6.5)
the unknown pressure p is also to be determined. From (6.5), (6.1) we see that there exists a function such that
(6.6) v= ?P:' vz=!41r:
setting
(6.7)
VOR1IX RINGS 471
the condition (6.2) becomes
(6.8) asjxl-.oo.
The function 4' is the stream function for the associated flow which is at rest at infinity. From (6.3), (6.6), (6.7), we have
(6.9) _!L4,,
where
a2v i a2v art (6.10)
ar2 r ar az2, ar i ax2
Notice that Eul&s equations hold for some p if and only if
(6.11) V X(v.Vv)=O.
Noting that V v = V = 0 and using the identity
4v + X v,
we can write
v X(vvy)= V X
(0 VY =
Thus if we let
(6.12) z) = -1-w(r, z),
then (6.11) reduces to
(6.13)
or, in view of (6.6),
(6.14) =0.a(r,z)
4n VARIATIONAL PROBLEMS WITH POTENTIALS
The last equation in turn is equivalent to the existence of a functional dependence
with vO 0. Thus, in particular, if we can find functions such that
(6.15) (yconstant)
and if are related by (6.12), (6.9), that is, by
(6.16)
then the physical equations (6.4), (6.5) will be satisfied.
Definition 6.1. The constant 'y is called theflux, the functionf(t) is called the voracity function, and the function
(6.17) u(r, z) = 4, — lWr2 —
is called the adjusted stream function. From a physical point of view, the function f(i) is supposed to be nonde-
creasing on (—oo, oo) withf(:) = 0 if : 0 andf(i) >0 if: >0. Combining (6.15), (6.16) we obtain
(6.18) Lu + r2f(u) = 0.
Notice that supp the vortex core, is characterized by
(6.19) wherefl= (xeR3;u(x)>0).
We seek solutions for which is compact. In what follows we shall need the inverse of —(1/r2)L. This is given in the
next lenuna.
Letmna6.1. Let
, rr'cos(O—O') (6.20)
where x = (r, 9, z), x' = (r', 9', z'). For z) = any bounded, measura- ble function with compact support in R3, let
(6.21) 4,(r, z) = 4,(x) = f K(x, dx'.R3
VORTEX RINGS
Then (6.16) holds a.e. and
(6.22) 00.
Proof. Let = z)1,. Then
(6.23)
satisfies
(6.24) a.e.
By direct calculation,
j 1 , = cos(8IX—XI j Thus
B(x)
and hence, by virtue of (6.20) and (6.21),
(6.25) rB(x) = 4(r, z)I,.
We now compute
(6.26) V X B = — +
Also, since V B, we have
--AB=v x(vxB)=
Combining this with (6.24) yields the desired result (6.16). To establish the estimates (6.22), .we first note that
I __.L+a(x,x') Ix—x'I — lxi .
I ,
• ix—xl lxi
474 VARIATIONAL PROBLEMS WITh POTENTIALS
for certain functions a, A = as x —p 00, x' E supp Now using the property
z')i9.dx' = 0,
we find that
B(x) x a(x, dx'
X B(x) = A(x, x') x
Recalling (6.25) and (6.26), the estimates (6.22) now follow.
We introduce the half plane
'H = {(r, z);O r < 00, —00 <z < oo).
Green's function the operator —(l/r2)L on H with measure rdrdz is given by
G(r, r',"z') = fK(x, x') dO' (8 = 0)
(6.27) cosO'dO'
[(z — z')2 + r2 + r'2 — 2rr'cos
since then (6.21) takes the form
(6.28) = f,f G(r, z, r', z')r'dr'dz'.
Using 6.1, we have the follo*ing eipressións for the total kinetic energy of the flow:
R3
(6.29)
= fR3 R3 where the second equality follows from integration by parts.
VORTEX RINGS 475
The total impulse required to generate the flow from rest is defined by
(6.30)
This takes the form
(6.31) P=Pi, A3
provided that
(6.32) dx 0.
In the sequel we shall assume that z) — z) holds; (6.32) s assured.
In posing the steady vortex ring problem variationally, we shall maxiilize the energy over a certain class of functions subject to the constraint that the impulse P be prescribed. We proceed to define the class of admissibk functions.
Let denote the class of measurable functiOns 0 ac. on R3 satisfying the following conditions:
(6.33)
(6.34) if I, A3
(6.35)
(6.36) O<A<OO., x€R3
Let denote the (larger) class for which cdndftion is removed. The energy functional is defined the dass (0 <A < oo) by
(6.37) E(L) R3R3
where K(x, x') is given by (6.20). We the variational problem:
Problem (E,j. Find a function such that
(6.38) = maxE(fl,
416 VARIATIONAL PROBLEMS WITH POTENTIALS
Theorem 6.2. There exists a solution of (6.38). Further,nore, there exist constants W > 0, y 0 such that
(6.39) a.e.,
where
(6.40) — lWr2—y>0)
with 4. defined by (6.21); is a bounded open subset of R3, and
= (x (r, 6, z); I 21< Z(r))
for some function Z( r) 0.
For each value of the free parameter A the solution obtained clearly represents a steady vortex ring corresponding to the vorticity function
(6.41) f(t) =
The constants W, y arise as Lagrange multipliers for the constraints (6.34), (6.35).
A related family of variational problems is derived from the energy func- tional
(6.42) = — 13Af
defined on the class
(6.43) = fl (0< <00).
We consider
(Ek,). Find a function such that
(6.44) E,(fl= max
Theoreut 6.3. There exists a solution of (6.44) for any prescribed 0 <A < oo, provided that 0< fi <5. Furthermore, there exist constants W> 0, 0 such that
(6.45)
where
(6.46)
PROBLEMS 477
with 4.' defined by (6.21); has compact support in R3, and z) is nonincreas- ing as a function of z for z >0.
Clearly, these solutions of the vortex ring problem correspond to a vorticity function of the form
(6.47) f(t)={A(l+$). t>0 0,
It is evident that the variational equation (6.45) tends, as fi 0, to equation (6.39). In fact, Theorem 6.2 is proved by first obtaining the solutions asserted by TheoTem 6.3 and then taking the limit of these solutions over some sequence fl, —' 0; the sequence of solutions converges weakly in for every I <p < oo. One may view problem as a limit of the "penalized" problems when 0.
Theorems 6.3 and 6.2 are proved in Section 8. In Section 7 we develop some identities and potential estimates needed for these proofs.
The restriction 0< fi < 5 made in Theorem 6.3 occurs only in establishing Lemma 8.2. This lemma (and consequently also Theorem 6.3) remains valid for any 0 < /3< provided that X is large enough, depel!ding on
There is a simiLarity between the steady ring the axisymmetnc self-gravitating problem studied in Sections 1 to 5. Thus problem (6.38) is analogous to problem (1.19) and problem (6.44) is analogous to problem (1.13). functional to be mm for the vortex ring problem, is somewiiat simpler looking, the admissible class has more con- St taints. The presence of (6.36), in particular, among the constraints for problem (6.38) causes difficulties; these difficulties will be overcome by obtaining the solution as limit of solutions of the penalized problem (6.44), for which variational conditions can be derived by standard methods.
PROBLEMS'
1. Denote• by ti?,, r. the class of functions 0 (6.33) and the consiraints
P (total impulse).
f (totalcirculaiion)X2,r,RI ess supt(x) A (vortex strength),
x€R3
473 VARIATIONAL PROBLEMS WITH POTENTIALS
for prescribed positive constants F, 1', A. consider the problem to de- termine E r. A such that
max E(fl.
Making the change of variables x —
a = b =
show that the problem for (coincides with the problem where x =
2. Show that a solution to the vortex ring equations (6.15), (6.16) is given by f(') = 7 = 0,
ifR<a
——WRII--—isinO 2 R3/
where r = R sin 9, z = B cos 0, and Xa2/W = The vortex core is the ball (Ix 1< solution is called Hill's vortex (1894).
7. ENERGY IDENITI1ES AND POTENTIAL ESTIMATES
When writing dx it is understood that dx = 2irrdrdz. We shall often use the integration-by-parts formula
(7.1) + UzV:) dx = fHr2
dx
for functions u(r, z), v(r, z) either of compact support or vanishing suffi- ciently fast as r2 + z2 oo, with z)d.z -+0 for a sequence r = 1.0. In particular, the calculation (6.29) expressing E in terms of is justified using (7.1) and the estimates for the corresponding given in Lemma 6.1, provided that has compact support; see Problem I.
We now give another formula for E(L).
Lemma 7.1. Let be any bounded, mea.surablefimction on Hfor which
(7.2)
= .f. <00, Crifr< 1 (C> 0).
ENERGY II)ENTITIES AND POTEN11AL ESTiMATES
Then
(7.3) EU)
Proof. Seto=+(r2+z2)1#,2+4").SinceLal,WehaVe
E(fl fHr2
(7.4) =
=
where (7.1) has been used (the justification will be given Expanding the integrand, we find that
Putting this into (7.4), we get
= I + + + dx + 2E(fl, so that
E(fl + + +
(7.5) 1
=
which yields (7.3). We must now rigorousLy justify the use 1"f formula (7.1) in the derivations
(7.4) and (75) above. It the integration by parts is carried out instead on the finite domain,
(7.6)
for some E >0, then we obtain boundary integrals. The boundary integrals on are bounded (in both cases) by
+ z)dI}'2
VARIATIONAL PROBLEMS WITH POTENTIALS
(using I 4'. I CR). Since the integral in (7.2) is finite,
(7.7) rr r and therefore
for a sequence e = e,, jO. Thus the boundary integrals on r = tend to zero. There remains to consider the boundary integrals on fl {r> 0). that is, it remains to prove that
f a a sequence a = a,, oo. But this follows again from (7.7).
- For a given vorticity function f(:), let
(7.8) F(z)=f'f(s)ds;
then
F(i) = when (6.41) holds.
F(:)=A( when(6.47)holds.
Let
(7.9) 1(u)
where u( r, z) is defined in (6.17). Several other expressions for .1(u) can be given; namely,
(I 1Hr2
(7.10)
= (x 6 R3; u(x) >0) and >0 in case (6.47), = 0 in case (6.41). Thus + /3)J(u) represents the kinetic energy of the steady flow (with stream function u) confined to the vortex core 0.
ENERGY AND POTENJIAL ESTIMATES 481
lemma 7.2. Let be a solution of either (6.38) or (6.44) with compact support. Then (obviously) (7.2) holds and
(7.11) —3J(u) + 2W + fl)J(u) + W+ yJ. Proof. The assertion that equals the right-hand side of (7.11) is
immediate from
dx =
dx — f dx — dx
2E- To prove the identity in (7.11), take ii', = u, + Wr, i,1'. = U. in (7.3):
(7.12)
Integrating by parts, we find that
= —4irAf(rF(u)drdz _2fF(u)dx; I;. 'I
and, similarly,
III- = —JF(u) dx.
Thus (7.12) becomes
= --3fF(u) dx + = —3J(u) ± 2W.
Remark 7.1. Ii is important for later application that we, observe that 11 remains valid for solutions of (6.44) without the assumption of cdmpact support; it is assumed, however, that the assumptions of Lemma 7.1. Indeed, if the in on
in the of the form
,1' • C
= r 2F( u) dz, 1' dr.
Noting that F(u) = we see there isa a = a,—' oo for which these integrals tend to zero' the condition
VARIATIONAL PROBLEMS WITH POTENTIALS
implies that
for r = an, a,. We now turn our attention to certain estimates for the "potential" IIi,
defined in (6.28).
Lemma 7.3. Let s [(r — ,.)2 + (: — z')2j'/2 There holds
(7.13) G(r. z, r'. z') j &2r'2 r
3
where C is a (sufficiently large) absolute positive constant.
Proof. Recall that G defined by (6.27). Letting
(1. r')2 -—
we first show that
(7.14) G(r, :, r. :') ((rr')' 2Iog 1,
(7.15) G(r, z, r', z') 1.
Estimate (7.14) is derived from the formula
(7.16) G(r, z, r', z') = { (
— k)K(k) —
where K and E are the complete elliptic integrals of the first and second kind [55, formulas 291.00 and 291.011 and
k2= —, (z — z')2 + (r +
(7.17)
k'2= i]. (z—z) +(r+r)
ENERGY IDENITflES AND P(YIENTIAL ESTIMATES
Since I implies that k'2 we apply the asymptotic formulas
(7.18) (k'o) E(k) = I + o(1)
to (7.16); then (7.14) follows directly. Estimate (7.15) follows from the expan- sion
[(z — z')2 + r2 + — 2rr'cos9'}"2 +
where 9) C for 1, —v 9' ,r. Applying this to (6.27) and noting the cancellation of the term involving r', (7.15) follows. The required estimate (7.13) follows directly from (7.14), (7.15).
In the following lemmas we assume that E EL0, and we let N = denote the norm
(7.19) (0czj9<oo).
Lemma 7.4. Asswne that z) isa nonincreasingJiinczion ofzforz >0. Then for holds
(7.20)
the constant C depends only on $, e.
Proof: It is clear that 4'(r,O) = z). We write = 41(r,0) + 4'2(r,O), where Iwith s as in Lemma 7.3]
4'1(r,0) = ff G(r,O, r', z')L(r', z')r'de'dz'
Cr ff log z')r'dr'dz',
4'2(r,O) = t ff G(r,O, r', z')r'b'dr' (s>r/2)
Cr2 if z')r'dr'dz'. (s>r/2} S
484 VARiATIONAL PROBLEMS WITH POTENTIALS
To estimate 4i2(r,O), note that
r'2 C r
s3 r 2
Thus, recalling (6.35).
ii2(r,O) Cr2 ff z')r'dr'dz' Cr.
Also, recalling (6.34),
4i1(r,O) Cr2 ff r z')r'dr'dz' Cr
(s>i/2) (jr)
Hence
(7.21) 4'2(r,O) Cmth (r, r
We proceed to estimate 4r1(r,O). For any 0 <a < 00, we have, by Holder's inequality,
(7.22)
ff I/(lfo)
(log—)
r r (7.23) ff (log;) (log—) sds=Car3.
Thus
(7.24) (o= ). \
We now the standard interpolation
(7.25) a < 00)
where a (I + fl)/(l a) (0 <a ' I); each V norm is taken with respect
ENERGY IDENTITIES AND POTENTIAL ESTIMATES 485
to the measure r'dr'dz' on the set ((r', z') E H; s r/2}. Since, by (6.34) and (6.35),
Lfl'L'(scr/2) Cmin {1, 2)
we conclude from (7.25) that
II II CNamin (1, r 2( —a))
-Thus
4i1(r,O) CaNarI+8 mm (I,
r I we take simply a = /3. so that
q'1(r,O) + 1)r. (0 <fl /3*)
For r I we take a sufficiently large (depending on /3". r) so that
4i1(r,0) + I)r (0</I /3"):
specifically, we choose a so that (5 + 3$*)/( I + a) <e. Together these esti- mates yield
(7.26) iJ'1(r,0) + l)min {r,
Now combining (7.21) and (7.26), we obtain the statement of the lemma.
Lemma 7.5. Assume that r, z) is a nonincreasing function of z for z > 0. Then for 0 < ft /3", there holds
(7.27) +
that r/2 <z/A and A > 1; the constant C only on €.
Proof. The monotonicity of z) in z implies that, for any A> I,
Jf ff (:'>O}
ff z')r'dr'dz' ff z')r'dr'dz'. (lz'—:F<:/A)' (>0)
Using these facts, we can modify the proof of the previous lemma for large z.
VARIATIONAL PROBLEMS WITh POTENTIALS
As in the preceding proof, we write
4i(r, z) 4i1(r, z) + 4s2(r, z).
In order to estimate 4#2 we write
412(r, z) = z) + z),
z) Cr2 ff z')r'dr'dz' (s>r/2) S
(r, r').
i4'(r, z) Cr2 ff (s>r/2} S
(I: -fI>z/A I
I A
Thus
+
To estimate z), we kflow the reasoning of the previous proof except that now (noting that s r/2 implies that I z — z' < s/A) we have
r
Thus, as before,
z)
withö=3/(l +a),a(l +$)/(I oo. Now the proof is completed Just as in the preceding proof.
PROBLEM
1. Justify (6.29) for with compact support. [Hint:
dz f{f
}
= o.J
EXISITh(I OF RINGS
& EXISTENCE OF VORTEX RINGS
In this section we prove Theorems 6.2 and 63. Theorem 6.3 is established by the sequence of Lemmas 8.1 to 8.9, and Theorem 6.2 is obtained in the form of Lemmas 8.10 and 8.11.
Let denote the class of nonnegative functions E L' + satisfy- ing (6.33), (6.35), and Irather than (6.34)]
(8.1) 4f R'
clearly, ç Consider the problem: Find such that
(8.2) max
To solve problem (6.44) we first obtain a solution of (8.2); this is because the possible solutions are not known to have bounded support a priori.
Lemma &l. For any prescribed 0 <A < oo there exists a such that (8.2) holds; z) is a nonincreasing function of z for z > 0.
Proof. In follows we assume 'that 0<13 13 for some fixed 13* (and the dependence of any on 13* will not be specified). For any
E estimate (7.20) implies that
(8.3) sup 4i(x) + 1], xER3
and hence
(8.4) = + I];
here we use the fact that the hypothesis of Lemma 7.4 is satisfied by z), the symmetric rearrangement' of j(r, z) in the z variable, and E(fl [by (6.29) and Theorem 7.3 of Chapter 31 while = Furthermore,
(8.5) N(fl llj'/A IlL' +(I/1)( C, + )' dx > 0);
this follows the elementary inequality
(8.6) (e>O)
valid for 0 X<'oo. Applying (8.5) to estimate the energy in (8.4), we get,
488 VARIATIONAL PROBLEMS WITH POTENTIALS
fixing E sufficiently small,
(8.7) + R
We now conclude [recall definition (6.42)J that
(8.8)
E for that is, and
sup
It is easy to see that (6.42) and (8.7) imply that
(8.9)
Thus, for a subsequence,
(8.10) weakly in L'
The limit is then an element of not necessarily Also. by replacing each r, z) by its symmetrical rearrangement in z [which cannot
•decrease we may assume that each z), and hence also z). is a nonincreasing function of z for z > 0.
We claim:
(8.11) EU):
this will imply that is a solution of (8.2). To prove (8.11) we first note that for arbitrary R > I, A > 1.
f(r>R) (8.12)
f 1]A & (J:I>AR)
(8.13) C[N(fl + 1]A + CR
for 0 < e < 1 fixed; (8.12) follows by Lemma 7.4, while (8.13) follows by
EXISTENCE OF VORTEX RINGS 489
Lemma 7.5. Clearly, estimates analogous to (8.12), (8.13) hold with replaced by Thus, as R and A may be taken arbitrarily large, it suffices to show that
f
i%1i, being defined as usual corresponding to Also, recalling Lemma 7.3, for IzJ'AR,
ff G(r, z, r', z'fl(r', z')r'dr'dz' CR3r2,
and hence -
ff ff (s>R)
an analogous estimate holds for In view of this, it suffices to show that
fJ fJG(r. z, r', z')rdrdzr'dr'dz'
where D = {(r, z) II: 0 r 2R, Jz (A + I)R) is a bounded domain. But now the result follows by standard arguments since G E V +P( D X D) and z') —' L(r, z') weakly in V +I/P( D X D) (in the prod- uct measure rdrdz r' dr' di'). This completes the proof of (8.11) and also of the lemma.
Lemma 8.2. If 0 <$ <56 (0<8 < I), then for all 0 <A < oc there holds
(8.14) c(A, 6) >0.
The constant c(A, 6) is independent of 8 as
a determined so = 1. Consider the scaled functions -
a < 1).
490 VARIATIONAL PROBLEMS WITh POTENTIALS
Then
fR3 R3 if
R3 R3
and hence for sufficiently small a (depending only on A), we have E An easy calculation yields
— —
R3
= — a2+(5/Thflf dx.
Thus c(A, 8) for 0 </3 < 58 provided that a is fixed small enough. Since (8.14) follows.
We now deduce the variational conditions for the solution
Lemma 83. The solution asserted by Lemma 8.1 satisfies (6.45), (6.46)/or some constants W 0, y 0 (uniquely determined by
Proof. The positivity of given by Lemma 8.2 implies that >0; therefore, we can find > () such that meas > 0.
We choose bounded, measurable functions h1, h2 of the form (6.33) such that
supp h1, c
fhidr=l. ifr2hidx=O, -
Let Ii be an arbitrary bounded, measurable function [of the form (6.33)1 subject to the restriction that
forsome8>0.The+e,1
iih— —
EXISTENCE OF VORTEX RINGS 491
provided that e >0 is sufficiently small. By the maximality of we have
=
—
with the Frechet differential
E4U)h=f4lhdx_(l+fl)f(x) hdx.
Now by the arbitrariness of h ( and 8) we obtain the variational conditions
I/p 4—y—4Wr2
with the Lagrange multipliers
= W=
These conditions are equivalent to (6.45), (6.46). Next we show that y, W 0. Clearly, there is a sequence of points
(r,,, E H such that r, -, 0, -' co and 0. Then the variational conditions imply that
0 everywhere, we conclude that y 0. Now take a sequence such that r,, -. cc, 0, and L(r,,, 0. Then since -.0, we conclude, analogously, that W 0.
To prove uniqueness of the Lagrange multipliers y, W suppose that y, W is another such pair; that is, (&45), (6.46) bold for W. This is equivalent to
—
for any h subject to the restriction
492 VARIATIONAL PROBLEMS WITH POTENTIALS
for some 6 > 0. In particular, we can take h = ± h1, ± h2 (recalling that supph1 C and conclude that = W* =
Lemma 8.4. For the solution there holds
(8.15) (C>0),
(8.16) ii r
The lemma is not obvious since is not known yet to have compact support.
Proof. By the calculation of Lemma 6.1 [specifically, combining the curl of equation (6.23) with (6.26)],
(8.17) r
Using the fact that
1/2 1/2
(8.18)
we find that
f(Ix—x'I>I) X X Since
(8.19)
by Lemmas 8.3 and 7.4, we also have
j + I), — .X
and (8.15) follows. To prove (8.16) recall that
= —fJ...4sL4idx
EXiSTENCE OF VORTEX RINGS
Integrating by parts on the domain
Da= ((r,z)
we obtain (using the hint to Problem 1 of Section 7)
(p is the outer unit normal on and is arc length). Therefore, to prove (8.16) it suffices to show that the boundary integral tends to zero as a -, 00.
We claim that for some 10>0,
(8.20) f V+Iic = O(a'o) - To prove it we again use (8.11) and write, for 0<8< 1,
— I r'r(x')dx' — I — , — A I
'Ut 'Ux—xi<a'} X — X' 'UT I X — X'
we estimate these separately. Using (8.18), ii is clear that Ca28. Thus, recalling (7.20),
f C(l + dz C&28",(r=a) —a (l:I'a)
f Jr ca_26f°(I +(z±a) 0 Fixing e < 28, we get
(8.21) f 4'V2 ut = We next estimate the corresponding expressions in V1. it follows from the
axisyinmetry of combined with the fact that flr, 1) is a nonincreasing function of z for z >0 that
(8.22) f(k—xi<a') a a
494 VARLATIONAL PROBLEMS WITh POTENTIALS
for all (r, z) E H. Indeed, since for r > 2a8,
meas {x' ER3; Ix — x'I< a8)
(x' R3; jr — r' < a84z — z' <
we find (using = z')] that
f [by(8.18)];r r (I: — z9 <a)
in turn, for z > 2a8 [using z')J, as z' C , z' > 0),
f(Ir—ri<u8) Z (Jr—r'I<a8} z (I: —:1 <a8)
Thus (8.22) follows. Also, by (8.19), (7.27),we have
(8.23) sup ifxEaDa(0<p< I),
where depends on /3. For r = a, z a, consider the problem to maximize V1(x) as a functional of subject to (8.22), (8.23). Clearly, the maximum occurs when
rT(x") =
where p is determined so that equality holds in (8.22). Computing V1(x) in this case, we obtain
IzI V1(x) Ca i +
Thus
I/3
ira) 4iV1dz C(1 + a)_1 + dz
EXISTENCE OF VORTEX RINGS
Similarly, for z ±a,,O <r a, we find that
V1(x) +
and thus
f(:=±a)
Fixing e and 8 sufficiently small, we find that
= O(a'°).
This together with (8.21) proves the claim (8.20) and hence (8.16).
Lemma &s.\ In the notation of Lemma 8.2, there holds
(8.24)
as a E
In the we shall assume that 0 < 58 [8 fixed in (0, 1)],and no longer specify of any constants on 6.
\ Proof. By 8.4 (whichassures the hypothesis of Lemma 7.1) and the remark following Lemma 7.2, we may apply the identity (7.11) to the solution and obtain
(8.25)
since J(u) 0. The assertion (8.24) now fellows immediately from (8.14) using
Since W, the Lagrange multiplier for the constraint (8.1), must equal zero if strict inequality holds in (8.1), we now conclude that the equality (6.34) holds for the solution that is, E
To complete the proof of Theorem 6.3 it to show that the solution obtained has compact support. This is established in Lemmas 8.7 and 8.9.
Lemma8.6.
(8.26) ,rW2
4% VARIATIONAL PROBLEMS WITH POTENTIALS
Proof. Suppose that z >0; then for all 0 < z' <z, since (r, z') E we have
4'(r, z') + y
Then, since ii(0, z') = 0,
lWr2 z')dr'.
Integrating in z', we get
if z')dr'dr' r z 1/2 1/2
(ff f —4ç3(r', z') 1/2
)
so (8.26) follows.
Lemma 8.7. There is r*Qt) < oo such that
(8.27) r"cr(A)
for (r, z) E supp
(8.28) cA>O.
Combining this with the estimate (7.20) (recall that N(fl C1j, we find
(0<E<I).
Now fixing e, we get as required.
Lenmia8.&
(8.29) ff 6(r, z, r', z')r'dr'dz' Cr2p2; (O<r'<p)
(—co <z'<oo)
C is an absolute positive constant.
Proof. Writing s = ftr — ,s)2 + (z — z')2J'/2, we consider separately contributions for s r/2 and s > r/2 as in the estimates of Section 7.
EXISTENCE OF VORTEX RINGS
According to Lemma 7.3,
ff ff
= Cr4 Cr2p2.
ff ff (s>r/2} (s>r/2}
(O<r'<p) (O<r'<p}
It is easily verified that for s > r/2, there holds.
Cr'3
S3 [(z — z')2 + (r +
Hence it now suffices to estimate
ff 1 dr'dz'=2(" r
(O<r'<p) [(z — z')2 + (r + + r')2
Lemma 89. There is z*( < oo such thai
(8.30) forall(r,z)
Proof. By Lemma 8.6 we have r2z C, (we take z > 0); thus'to prove (8.30) we may assume that r < z, say. Let p be defined by
Then
4i(r,z)
sup z') ff G(r,z, r', z')r'dr'dz'
+ ff G(r, z, r', z')r'dr'd' (,,>p)
+ '2
498 YARLATIOtJAL PROBLEMS WITH POTEt'rrlALS
Using the variational conditions (6.45), (6.46) noting that C < 00, we get C; therefore, upon using Lemma 8.8,
Cr2p2 C1 (C1 > 0).
As for '2' (r', z') E and r' > p, then by Lemma 8.6
'2 — 2! < ,22r z1 '-AP2 r 2' so that z' <z/2. Thus in the notation of Lemma 7.3, there holds s > z/2> r/2. Using (7.13), we then get
Cr2 ff z')r'dr'dz' (r'>p} S
Cr2 z')r'dr'dz'
Combining the estimates on and using (8.28), we obtain
Cr2 C2 +
z
and (8.30) follows.
We have now completed the proof of Theorem 6.3. We proceed to prove Theorem 6.2.
Lemma 810. There cxliii a solution oJ (6.38); has compact support and it is nondecreasing in z for z 0.
Proof. Let denote the solution obtained in Theorem 6.3 for the penalized problem with a prescribed A (and 0 <ft < 1, say). We know from Lemmas 8.7 and 8.9 that
(8.31)
independent of ft. Applying (8.7) to and recalling that >0, we find that
(8.32)
EXiSTENCE OF VORTEX RINGS 499
Using the notation in (7.19), this becomes
P1(1.-f Th
which in turn implies that
(8.33) as$—'O.
Now if P < a, for any fixed 0 < a < I, then we can estimate
(8.34) (a =
a sequence $, 0 such that
(8.35) forevery0<a<1.
Furthermore, by (8.34),
Jim inf
and taking a -, 0, we conclude that
(8.36) esssupL'cA.
Also, by virtue of (8.31), it follows easily that
(8.37)
Thusr e ç D, and z)isaaonincreasing of: for: >0. Finally, we must show that (6.38) holds. In light of (8.31) it is immediate
from standard arguments that
.
For any(E we have
E(fl
E(();
51)1) VARIATIONAL PROBLEMS WITI-I POTENTIALS
the last equality follows since
F lim$Af dx=O.
J3—.() A,
This establishes (6.38).
Lemma 8.11. There exist constants W> 0, 0 such that (6.39). (6.40) hold.
Proof. Let 14';. u1 denote the quantities in the statement of Theorem 6.3 associated with —' 0. Since 1$';. 0, and as a consequence of (7.11),
+ we may assume (by taking a subsequence) that — Wandy3—''y.Then,y>Oand. by(8.24),
In view of (8.31). (8.35).
=f converges pointwise in R3 to
= f K(x, dx'. Hence u(x) pointwise on R3, where u = 4, — Wr2 — y. Also, accord- ing to the variational condition (6.45), we have
IA ifu(x)>O lim
if u(x)<0.
Furthermore, since = <0 for z > 0,
meas (x E R3; u(x) = 0) = 0.
Thus the function XI(,,(X)>O) is the pointwise limit of for a.a. x E R3; since also weakly, (6.39) follows. Thus the proof of Theorem 6.2 is complete.
Remark 8.1 Theorem 5.1 extends to vortex rings; it yields analyiicity of the function z = z(r) representing a(u >0) fl $ >0).
PROBLEMS
1. Show that
(8.38) if7>0, I,
A CAPACITY ESTIMATE
where is the solution constructed in Theorem 6.2. [Hint: Otherwise, dx < I and then y, = 0.]
2. Show, with as before, that
(8.39) dx < I if A is sufficiently large.
(Hint: If BR(2R,0), R then = is in and (c > 0). Deduce, using W and Lemma 8.6,
(8.40) z in suppA" r The proof of (8.30) for gives
CA C 2r. zI IzJ
Using Wr2 in supp deduce that
(8.41)
'Use (8.40), (8.41) to estimate
(e2X3/5 = C/ZA)
and recall that 1.]
9. A CAPACiTY ESTIMATE
We establish a capacity estimate that is useful in studying asymptotic of some free-boundary problems. In Section 10 Uie is used only n problems at the end of the section. In SectiOn 13 use this estimate in 'a more basic way.
Deflnkkm 91. Let E, be bounded sets in closed, open+,
C v 1 on H).
502 VARIATIONAL PROBLEMS WITH POTENTIALS
Then the capacity of E with respect to is defined by
(9.1) infflvvl2dx. vEK
Definition 9.2. The solution w of the variational inequality
(9.2) VvEK;wEK
is called the capacitary potential of E with respect to Clearly,
in the sense of distributions or measures,
w1 a.e.onE and
a.e.in
(since otherwise mm , 1) Iwhich belongs to K) has a smaller Dirichlet integial than w).
If 8E and 811 are in C'4 a, then w satisfies:
in11\E, (9.3) w1 mE,
w0 on811, and
(9.4) CapnEf IvwI2=f U\E £2
where v is the outward normal to 11 \ E; the measure — is supported on 8E. From (9.1) one easily infers that
(9.5) Cap0E t if E t (9.6) I if 11 t
We shall now restrict ourselves to n = 2. One can easily compute that
(9.7) CapBkBr = log(l/r)+ 0(1) as r 0 (n.= 2).
A CAPACIV( ESTIMATE
We denote the diameter of a set A by d( A). We shall estimate the diameter of a closed domain E from above in terms of the capacity of E with respect to a domain containing E.
Theorem 9.1. Let E be a closed domain in R2 such that
ECBR CB2R C(ICBRZ,
where is a domain in R2. Then
I C 2ir (9.8)
— log(1/d(E))
where C is a positive constant depending only on R,, R2.
Proof. Given any small e> 0, choose points A, B in E such that
and connect them by a smooth curve L in E such that
9\ L is linear in a neighborhood of / each of the endpoints A and B.
Lemma 9.2. There exists a positive constant C depending only on R1, R2 such that
I C \ 2,r (9.10) 1
— log (2i,//)'
where
(9.11) Lisalinesegment
and take for simplicity R2 = 1. Denote by G(x, y) the Green function for in B1 and consider the function . .
.(9.12)
notice that we are integrating here G against the "câpacitary tion" of the ball B,,,21 of circumference I (equal to the diameter of L); see
504 VARIATiONAL PROBLEMS WffH POTENTIALS
(9.7). Since
(913) G(x. y) = Ix
+ 0(1) on L,
— log r +
1 (/2 1 1 11= — f log id:j log 2ir/l +
11 1 1\ I Ii= —
1
log
function
log
— log i/I)
It satisfies
onL,
g0 onaB1, g harmonic in B1 \ L.
By Problems 2 to 4,
(9.16) Cap0 L — =
— — logI /1) 1aa,
A CAPACI1Y ESTIMATE
On the other hand (using Problems I and 2),
acrossL} ds
by (9.7). Combining this with (9.16), the assertion (9.10) follows. We proceed to prove the lemma in case (9.11) is not satisfied. Take a
uniform distiibution
(917) on 48. (l/2ir) log 2w/I
and redistribute it onL(or on a part of L) in such a way that when we project L orthogonally onto AR. we get the distribution (9.17). Denote this distribution on L by a.
Consider the function
f(x)=JG(x,y)ads
with G Green's function as before. if x L,
(9.18) f(x) 1odi + O(l)fods.'
For each x E L, we perform a substitution in each integral, which corresponds to projecting orthogonally the curve L onto a line segment parallel to 48 passing through x and having length JA — B ; denote it by
(9.19) odsbecomes
where df denotes the length element along The second *ntegral.ln (9,18) is then bounded by C(!/Iog I). The first integral becomes, at x,
1 (1/2 1 1dr
2irfo
Thus Ic conclude as before that (9.14) holds. We now proceed to derive
Cap8L (i — )JL{
jump of across L) a
/ C 1 C 2ir )Iog.1/l log i/i log2,r/I
506 VARIATIONAL PROBLEMS WITh POTENTIALS
where (9.19) was used in obtaining the last equality. This establishes Lemma 9.2.
By (9.5), (9.6),
Using Lemma 9.2 and recalling that d( E) — e / d( E) for any E > 0 fC in (9.10) is independent of ej, the assertion (9.8) follows.
We now define the capacity with respect to the operator L:
allay Lv ri — — — j + — u — —
an
a closed domain and let be a bounded domain, both in the half plane ((r, z), r 0} and both symmetric with respect to z, and E C Let
K = {v E v 1 onE, v(r, z) v(r, —z)).
Then the capacity of E with respect to and L is defined by
(9.20)
C I
C is a positive constant depending only on R,, R2,
For proof, see Problem 5.
PROBLEMS
1. If a neighborhood of A intersects L at ((x,0); —a <x <0) and A = (0,0) then near A
I f f fB1\L
ASVMP'mTIC ESTIMATES FOR VORTEX RINGS 507
where L8 are suitable neighborhoods of L, decreasing to L as 8 -. 0.
[Hint: 8L8 L at distance ±8, except near the endpoints where connects to at distance p from the endpoint, and p = p(&) 0 in a suitable manner.]
3.
(Hint: Take v = w + max (0, g — w) in (9.2) and integrate J vg V max(O, g — w) by parts in fZ\L8, 8 —. 0.]
4.
[Hint: The difference is equal to
-f I v(g - + 2fvg v(g- w).]
5. Prove Theorem 9.3. [Hint: Green's function G(X, Y) for L in can be written as
log x l
(I + 0(1 X —
here X (r, z), Y = (r', z') [a fundamental solution K( X, Y) for r2L is given by
f2"K(x. x') dO, x = (r, 0, z), x' = (r',O, z').
K(x, x') as in (6.20)].]
10. ASYMPTOTIC ESTIMATES FOR VORTEX RINGS
We are interested in the behavior of the vortex ring constructed in Theorem 6.2 as the parameter A increases to oo. We shall denote the solution constructed in Theorem 6.2 by and set
(10.1) V VA = ((r, z); z) = A).
The constraint (6.35) gives
(10.2)
where fA I denotes the measure of a set A in H (endowed with measure r dr dz).
VARIATIONAL PROBLEMS WITH POTENTIALS
Lemma 10.1. For all A sufficiently large
(10.3) E = log A — C,
where C is a constant independent of A.
Proof. Let
z) =
where is the disc(r — + z2 <e2. Then
1 1,
that is, if
/ 2
We shall compute For this purpose take first, in G(r, z, r', z'), (r', z') = and introduce new coordinates about
set
=
I as (r, z) E supp In terms of the new variables, we find [recall that
rt+Es 2
+0(e)), £212 + +
k2 = I + 0(r).
4 -f 0(r)).
Hence
2 2 — k)K(k) — = —2 + o(elog_).
ASYMPTOTIC ESTIMATES FOR VORTEX RINGS
Using (rr')'/2 = + 0(e), we get
G(r, _21 +
I ase-.O.
Let = delta function on the circle r = z 0, —it <8 < it, and
z) = ffG(r. z, r',
z') = G(r, z,',/!,O) + 0(1) on we find that
4'0(r, z) z, r', = 4°(r, z) ± 0(e).
Hence
= z)rdrd.z + 0(1)
I log + 0(1)
= ff log z)rdrdz + 0(1). Since r = + 0(e), drdz = 2ire2E we get
= 0(1)
+ 0(1)
= !e2Alc,g + 0(l)
510 VARIA11ONAL PROBLEMS WJTh POTENTIAlS
by (10.4). Since
1, = 1 + 0(e).
there is an 0(e)-perturbation of which belongs to Hence
and (10.3) follows.
Lemma 10.2. There is a positive constant C independent of A such that
(10.5) +log(1
for any A > 0. where 4' is the stream function corresponding to =
Proof. Write 44r,0) = 4'1(r,0) + 4,2(r,0) where
4,1(r,O) f [x = (r,0,0)],(ix—x'1<r/2) 4,2(r,0) J
To estimate 'h, notice that
Thus 4'2( r, 0) 3r. To estimate 4's, notice that
and hence
41(r,0)cCr2f r'dO (jr—r'I<r/2) 0 (z'2 + r2 + r'2 — 2rr'cos 011/2 (1z1<r/2)
Using (2.7) and proceeding as in the argument following (2.8), we find that
4,,(r,0) Cr2! logAr3 provided that Ar3
ASYMPTOTIC FSrIMATES FOR VORTEX RINGS 511
for some positive constant c. On the other hand, if Ar3c, then we simply use
CAr4. X
The assertion (10.5) follows by putting together the estimates on Define
R0=inf(r;(r,0) E suppfl, R1 = sup(r;(r,0) E suppfl.
Note that (6.34) implies trivially that
(10.6) 1
Lemma 10.3. There holds
(10.7)
where C is a constant independent of A, A> I.
Proof. Sinceu=OonuJVA,
Recalling that 0 and using Lemma 10.2, we get
(10.8) CR1 log(2 + ARfl.
From Lemmas 7.2 and 10.1, we have
(10.9) 2W= ClogA
if A is sufficiently large. this estimate into (10.8), we get
R1 log A Clog
if A is sufficiently large (say A > A0), and (19.7) fpllows. The
10.4. There holds
(10.10)
512 VARiATIONAL PROBLEMS WITH POTENTIALS
where J J(u) is defined by (7.9) with F(i) = and C is a constant independent of
Proof. By (7.10) (with /3 = 0),
J(u) = 27rXff urdrdz 2irff -'-(ut +
Since R1 C, we get
(10.11)
I'
We recall the Poincaré inequality,
ifuIav0.
We use it to derive
+
[by(10.1l)J.
Recalling (10.2), the assertion (10.10) now follows.
From Lemmas 7.2, 10.1, and 10.4, we deduce:
Theorem 10.5.
(10.12) W=IE+O(1),
(10.13) 0(1).
From (10.12) and Lemma 10.1 it follows that y > 0 if A is sufficiently large. Hence, by Section 8, Problem 1:
Corollary 10.6. If A is sufficiently large,
(10.14)
ASYMPTOTIC ESTIMAITS FOR VORTEX RINGS
Lemma 10.7. If A is sufficiently large, then (10.15)
where c is a positive constant independent of A.
Proof. We have
where Lemma 10.2 was used. Since, by (10.13), (10.3),
(10.16) (c0>0)
if A is sufficiently large and c0 is independent of A, and since 2, we obtain
+log(1 +23/2A)];
this gives (10.15).
We shall denote various positive constants independent of A by C. We shall assume in the sequel that for all sufficiently large A
(10.17) is connected.
In reference 33 a formal argument is given suggesting that (10.17) is valid for allX>0.
Lemma 10.8. If A is sufficiently large, then
(10.18)
This implies that
C (10.19) R1
A
Proof. Consider a faMily of straight lines 4, R0 ' r R1, dach forming an angle 21r/3 with the r axis; 1, cuts the z axis at z = r. By (10.17)1, intersects V at a point (r, 0). Denote (r*, ;*) the rlrst point of jnterscc*ion of 1, witI) V; thus the segment from (0, r) to lies outside V and (r, z') e av. Then
— 4(r, 2*)
514 VARIATIONAL PROBLEMS WITH POTENTIALS
and
4i(0, r) 0, 4(r*. z*) y + W(r*)2 .y•
Integrating with respect to r, we get
— R0) 1RRO C(R, — VpI2drdz}'
C(R1 — V4'J2dr
since R1 C. The last integral is CE"2;
y(R, — R0)"2 CE'12.
Recalling (10.13) and Lemma 10.1, the inequality follows. In order to complete the proof of (10.18), ii remains to show that
(10.20) whereZ=sup(z;(r,z) E V).
This follow by the same method as before. We begin with
(0<z<Z)
where the segment ((r, z); 0 <r < F) lies outside V and (F, z) E The left-hand side is equal to
z) = + y y.
Integrating with respect to z and proceeding as before, the assertion (10.20) easily follows.
From (10.6) and (10.19) we obtain:
Corollary 10.9. I/A is sufficiently large, ihen
(10.21)
We shall use this order to improve Lemma 10.8:
Theorem 10.10. For all A sufficiently large,
(10.22) d(VA) (0<c<C<oo).
ASYMPTOTIC ESTIMATES FOR VORTEX RINGS 515
Proof. Suppose (r, z) E Then u(r, z) > 0 and thus
-iWr2 + y <4(x) = z, r', z')r'dr'dz'.
Set X = (r, z), X' = (r', z'). By (7.16),
G(r,z,r'z') + 0(1).
Writing r' = (r' — r) + r and noting, by Corollary 10.9, that
(10.23) z')dr'dz' =
(i + o(
we find, after using Theorem 10.5, Lemma 10.1, and the relation r2 =.2 + 0(1/log A) (which follows from Corollary 10.9) that
1 x,r(r',z')dr'dz' + 0(1).
Cboosmg e = ?C'/2 and using (10.23), this inequality becomes
(10.24) z')dr'dz' —C.
For a fixed large A consider the problem of maximizing
F fJ log, x' z') dr'dz'
among all functions any characteristic function satisfying (10.23). The maximum is clearly attained for
=
where
2 1'irpA il+Oi—
We compute that
= C.
516 VARIATIONAL PROBLEMS WITH POTENTiALS
Hence (10.24) yields
—C, {IX'— >Ae)
or
(10.25) z')r'dr'dz' bA
if A is sufficiently large. Suppose now that X is another point in Then we also have
ff z')r'dr'dz'
I — Ae then by combining this inequality with (10.25) we get
z')r'dr'dz'
a contradiction. This proves that d(VA) Ac C)C I/2• Recalling (10.14) we deduce also that d(VA) dC 1/2
Set e = 1/ and introduce the functions
(10.26) u1(r,z)
(10.27) z) = VI + er,
and the set
(10.28) + er, ez); (r, z) E VA).
By Theorem 10.10,
(10.29) c c {p for some e0> 0,
where (p, 4) are the polar coordinates about (Vi, 0). Notice that
(10.30) = 1 + 0(e),
ASYMPTOTIC ESTIMATES FOR VORTEX RINGS 517
and
(10.31) 0(e).
Let
z); r2 + z2 <
and denote by U(R) IR = (r2 + z2)'12j the solution of
inR2,
(10.32) U0 onaD0, U ifR>RO,R0=(2,/1172)"2.
Notice that A is uniquely determined (and is positive) by requiring that u E C' atR=R0.
Theorem 10.11. As A 00,
(10.33) —.D'U foriO,1,unifor,nly
inco.npactsetsinR2;
further, for oil A large, can be represented in the form p = h7(+), where
(10.34)
zcX—' oo.
Proof. The u, satisfy
(10.35) — r - = + (10.36)
We claim that if r2 + z2 < A/C0 (for a suitable > 0), then
(10.37) u1(r, z) C(r2 + + Clog A (r2+ + C.
518 VARIATIONAL PROBLEMS Wilil POTENTL4LS
To prove it, we write
uE(r, z) u,(r, z) — u,(?, 1) .Ior some (F,!)
so that
+eF,ei)]
+[iw(v5+ er)2 — EF)2]
+ 'Z•
From Problem I, we have
(10.38) I
CA"2 =
where the argument of varies in the interval connecting + eF, ez) to + er, ez). Hence
11i C(r2 + + C.
Next,
thus (10.37) follows. From (10.35), (10.37) it follows, by standard elliptic estimates, that for a
subsequence 10,
uf Z uniformly in compact sets,
together with the first derivatives.
We conclude that
(10.39) + 21(Z>O) = mR2
and, by (lO�7),
(10.40) J Z(r, z) C(r2 -f +C.
Lemma 10.12. Z is symmetric.
Proof. Since Z is harmonic and negative in a neighborhood of oo iby
ASYMPTOTIC ESIIMATES FOR VORTEX RINGS 519
(10.29)1, and since it has at most a linear growth [by (I0.40)J, the linear term in its asymptotic expansion near oo must vanish. Thus the function u = — Z has the expansion
+bAR2X+
near oo, where X = (r, z), R f Xf. Also,
tiu —. 0 in R1.
We can now apply a slightly simplified version of Theorem 13.11 (Theorem 13.11 is independent of all the material preceding it). In fact, using the same notation all we need to show s that if y) 0 in then vA(x, y) > 0 in But since 0,
A LAVA — U iii
so that VA >0 in by strong maximum pnnciple. We conclu4c that u (and thus Z) is radially symmetric with respect to some
center (r°, z°). Recalling (10.31), it follows that (r°, z°) = (0,0).
From Lemma 10.12 it follows that Z is a solution of (10.32) for some disc D0 with radius R0> 0. Recalling (10.30), we deduce that = This compietes the proof of (10.33).
Since
onRR0
it follows that
Further, can be represented locally in the form p = hA(+), where -, 0
4sA-•'oo, .
(10.41) E(A)= 2logA+O(1).
Indeed, since the boundary of VA converges (after scaling) to a circle in the C' norm, the in 10.1 cap be
2logX+Q(1)
VARL4TIONAL PROBLEMS WITh POTENTIALS
Remark 10.1. The asymptotic results of this section extend to solutions of Theorem 6.3.
PROBLEMS
1. Provc(lO.38).
[Hint: Follow the proof of Problem 4 of Section 4.1
2. Let D = ((r, z); (r — 4)2 + z2 < 1). Show that if (10.18) holds then
in! fJV D\VA r
where v varies over the functions in H'(D\ VA) with = i on o on [Hint: 4, Wr2 = + W + g(r) on avA, where I g(r) c C,
I g'(
CIoLA. Construct k(r, z) such that k(r, z) =j(r)h(z) in D0 (R0 < r < R,, z < C0/log A) (R = R0 — C0/log A, R, R, + C0/Iog A) and suchthat4,y+ W+konaVA,4,=konaDand
f ilvkl2drdzcC
By the Dirichiet principle,
inff D\vAr IT
where wE H'(D\VA), w = 4' on a(D\vA); take v (w — k)/(y + W).1
3. Usc Lemma 9.3 and Problem 2 to deduce that if (10.18) holds then d(VA) C/A"2. [This is an alternative way of proving Theorem 10.10.]
4. Use the estimate d(VA) c/A'/2(c> 0) together with Problem 2 and Lemma 9.3 in order to show that
E'c logA+C.
11. THE PLASMA PROBLEM: EXISTENCE OF SOLUTIONS
The problem of controlled fusion, more specifically, the containment of plasma by a magnetic field, leads to a free-boundary problem. The precise model,
THE PlASMA PROBIIM EXISTENCE OF SOLW1ONS
which is currently subject to experimentation in the Tokomak machine, in- volves nonlocal nonlinear operators. However, a simplified model reduces to the following system for an unknown function u:
(11.1) infi,.
(11.2) = r = an (c constant),
(11.3)
where A and I are given positive numbers and c is a constant to be determined. The domain n is a bounded domain in R2, but in this section we take 1] to be a bounded domain in R", n 2.
The sets
{xEn;u(x)>o) are called the plarma set and the vacuwn set, respectively, and
is called the free bowidaiy.
/ Denote by A1, A2,... the increasing sequence of eigenvalues of in U and
by the positive eigenfunction corresponding to A,. We observe that
if A < A,, then c< 0,
if A >A,, thenc >0.
Indeed, using the formula
+A,v,u') =
— = —
forA > A1, we get
, 0<—cl --—,
'a0 UP
and since av,/ar<oit follows thatc>0. If A<A, and c>O, then and taking G = U, in the formula above, We get
o>f
522 VARIATIONAL PROBLEMS WITh POTEN1 IALS
since au/av 0, we get a contradiction. Finally, if c = 0, then must coincide with fi and u is an eigenfunction with fixed sign in fi; consequently, A = A1, a contradiction.
Notice, by elliptic regularity, that any solution of (11.1), (11.2) [in must belong to fl for any 0 <,$ < 1.
To prove existence of a solution for the plasma problem (11.1 )—( 11.3), we introduce a variational framework, taking
= e L2(12), p 0 a.e., fp(x) dx i}
as the admissible class, and
(11.6)
as a functional on K0, where G(x, y) is Green's function for in
Problem (J0). Find p such that
(11.7) Jo(p) pEK0. E K0
This problem is similar to the problem of rotating compressible fluid (in Section 1); it is actually much simpler due to the fact that fi is a bounded
One can easily compute the Frechet derivative of
(11.8) = —f G(x,y)p(y)4y+ VpEK0.
Using the inequalities
C (C>0),
ifn2
and Young's inequality, we find that
f G(x,y)p(y)dy 2n
U 10 ) n—4
if n 4 then we can take any s <
OF SOLUTIONS
We can then deduce that, for any E >0,
G(x, y)p(x)p(y) dx <efp2(x) dx +
Consequently, J0 is bounded from below on K0. Hence, by a standard. compactness argument, there exists a solution p of problem (J0) and, as in• Section 1, a.e.
(11.9) m{p>O},
in(p=O), cconstant.
Setting
(11.10) u —JG(x,y)p(y)+c
and recalling (11.8), the equilibrium conditions (11.9) reduce to
(11.11) p.Au_.
Applying to both sides of (11.10), we obtain (11.1). Also, u = c on Finally,
f J .;— a a a
We have thus proved:
Theorem 11.1. There exists a solution p of problem and the function u defined by (11.10) is a solution of the plasn,a problem (1 1.1)(1 1.3).
Next, we uniqueness prpvided that A < A2.
Theorem 11.2. IfA then the solution of the plasma problem(11.1 )..-(11.3) iswüque.
Proof. Suppose that u1, u2 are two solutions and u, c1 on r. (11.4), (11 agn c1 = agn c2.. Suppose for definiteness that sgn c, >0 and let = Setting
10 ' ifU1=U2 h(x) = 1 (U1)_ — (U2)_— — if U1
"I
VARIATIONAL PROBLEMS WITH POTENTiALS
the function U U1 — U2 satisfies
—W=AhU
U.0 onr and 0 h 1. We claim:
(11.12) or
Indeed, otherwise let
112=(U<0)
and choose a positive constant y so that
4 v1U+f v1U0.a, a2 Thus the function
0n01 U__lu on(12
Ewhich clearly belongs to is orthogonal to the principal eigen function v1. From the definition of A2,
(11.13) 12— ft vOl2
Jv2 jU2 fM]2
But since
xf hU2y2f AhUU _yaf IVU12 a, a, a, a
and similarly
If ?IU2
the right-hand side of (11.13) is. equal, to A, contradicting the assumption A <Ia. We have thus established.(ll.12). We may then assume that 111
THE PLASMA PROBLEM: EXISTENCE OF SOLUTIONS
If U1 U2, then the domains = <0), <0) satisfy: D1 C D2, D1 D2. Since each. U, is the principal elgenfunction of in D1 with the same eigenvalue A, we get a contradiction (see Problem 1). From the identity U, U2 and (11.3). we finally deduce that c, = c2, so that u1 u2.
Remark 11.1. Uniqueness not hold. in general, if A > A2. Indeed, for any e > 0 one can construct domains for which there exist at least two solutions of problem (J0) with some A in (A2, + e). consists approxi- mately of two balls (I x < 1), { x — x0 < 1) (with I I> 2) connected by a thin neck; for details, see reference 1 58d (see also Section 13, Problem 3).
We shall now introduce another variational principle which yields the same class of solutions of the plasma problem. Here we work directly with U.
The class of admissible functions is
(11.14)
and the functional is
(11.15) J(v)=
Consider the variational problem:
Problem (J). Find u such that
(11.16) J(u) minJ(v), uEK. vEK
Theorem 113. There exists a solution u of problem (I) and u is a solution of the problem (1 1.l)—(I 1.3).
We first need a lemma.
Lemma 11.4. There exists a positive constant C such that for any u E W' P( Q),
q/(q—r)
(11.17) I u IL' Vu IL' + ma.xI I
u. IL" if;
here 11(11), l/q (lip) — (1/n) i/p <n, q is any number 1?? (1, cc) if p n, and r is any nwnberp r< q. ..
Proof. We may assume that fl is convex, for otherwise we write g as a finite union of domains, each diffeomorphic to a convex domain.
526 VARIATIONAL PROBLEMS WITH POTEN11ALS
Consider first the case p < n. We shall use the Poincarè—Sobolev inequality (see reference 109, Sec. 7.8)
(11.18) lu —
m(u) is the average of u in fl. If m(u) <0, then
so that
IL'
and (11.17) folLows from (11.18). If m(u) >0, then (11.18) gives
f Iu_ Using the inequality (a + b y
0,
so
we obtain (11.17). If p n, we can take q arbitrarily large in (11.1) and repeat the previous
arguments.
Takingp = r-= 2 in Lemma 11.4, we get
I U Vu 1L2 + max u
I
J}
ifn>Z and
I+e
1}(lu IL2)
ifn2Eorany)<q<ooandgnyEr/(q_r)J
THE PLASMA EXISTENCE OF SOLUTIONS 527
From these inequalities and from the inequality
for any 6 > 0. it is easily seen that the functional J( v) is bounded from below on the set K. Thus, by a standard argument, there exists a solution u of(l 1.16). We shall show that u satisfies (1 l.l)—(1 1.3).
For any v E 111(12), v — consi. = v(F) on I'. set
u+Ev A1.7j(u+ev)... Ce
By the Lebesgue bounded convergence theorem, we find that e > ö. e then
as—I A (11.19) -+
where = '(u=O}' Pt = IflCC K, we have
—J(u)J E
Using (11.19), the last inequality reduces. to
(J'(u), v)= f Vu •Vv — Jv(F) —Afu_v;
in denying (11.20), (11.21) we again use Lebesgue bounded convergenà theorem.
Similarly, taking e <0, e t 0, we obtain
(11.22) (J'(u),v)c —yf(pov++p,v).
528 VARIATIONAL PROBLEMS WITH POTENTIALS
Thus
(J'(u),v)I
Since v is arbitrary, we conclude that
(11.23) — + +yp1 =
where p0(x). Multiplying (11.23) by v e K and integrating, we obtain after comparing
with (11.20). (11.22),
— I)v(r) — yjav --yj0v+.
If we choose v(I') 1, v 0 with v4 k' arbitrarily small, we then obtain (11.3) and, consequently,
_yfav —yfp0v+.
It follows that y 0. If y 0, then 0 a p0 and integrating (11.23) over gives lafter using (11.2), (11.3))
o>4 dx=-yf (u<O} {u0)
since Ju_ > 0. This contradiction shows that y = 0 and then (11.23) reduces to (11.1).
Remark 11.2. If u solves problem (J), then p = Au_ solves problem if p solves problem (J0), then u given by (11.10) with suitable c solves problem (J) and infkl(v) = mfK0JO(p). Thus the two problems are equivalent and produce the same class of solutions of (1 l.1)—(1 1.3). For proof, see Problem 2.
PROBLEMS
1. 11
mD,,
mD1,
11=0 on801,
THE FREE BOUNDARY FOR THE PLASMA PROBLEM 529
D1 C and there is a C' open portion a of which is contained in D2, then A, > 12. (Notice that the last condition holds in the proof of Theorem 11.2; indeed, U1 >0, tiU, = 0 in 12 \ D, and the inside ball property is satisfied at some points of a0 = D2, so that 0 at such points and 00 fl (U1 = O} is then in near such points.) I Hint: Take a sequence r = r,,, 0 and use
f U1U2.]a(u1<—e} 2. Prove the assertions in Remark 11.2.
[Hint: If p E K0, then the u in K given by (11.10) satisfiesJ(u) J0(p). If u E K, then p = Xu_ is in K0 and J0(p) J(u).1
3. The Tokomak machine is actually three-dimensional and axially symmet- tic, and the corresponding plasma problem is obtained, by replacing (1l.l)—(ll.3) by 1
Lu—Au_0 mU,
u=const.onr,
Fr
where Lu = ((l/r)ur), + Extend Theorems 1.1 to 1.3 to this taking
J(v) = + drdz— Iv(.fl'—
12. ThE FREE BOUNDAItY FOR THE PLASMA
Theorem 12.1. Let u be the sohalon of the plasma problem constructed by Theorem 11.3. Then the plasma set 12,, and the vacuum set are connected sets.
Proof. If A A,, then Ii, = 12 and the theorem trivial. Suppose that A >A,.Th >0andth rneli-neighb hoodof F. If a of (Zn, I) u 0
is is connected, suppose that G, and 62 are two components
of 12,,. Define
mG, (i= 1,2)
u inli\(G,LJG2),
VARIATIONAL PROBLEMS WITH POTENTIALS
where c are positive constants satisfying
(12.1) c,f u +c2f u = f u... uC7 C2
Then ü E K. Since
we havef(ü) = J(u), so that üis a minimizer. By the proof of Theorem 11.3 ii follows that ü solves the plasma problem (11.1 )—( 11.3) and, in particular, ü E This is impossible if we choose c, * 1, since au/av 0 on (by the maximum principle).
We shall now restrict n to be equal to 2.
Theorem 12.2. If n = 2 and u is the solution asserted in Theorem 11.3. then the free boundary I',, is analytic.
Proof. First we show that
(12.2) Vu*0 on!',,.
Suppose that the assertion is not true at a point x() E 1',,. We shall use complex nolation and takex = (x1, x2) = x1 + ix, = :. = 0. Since
I UI , 'we can apply Section 5, Problem 7 (that is. Lemma 5.2 for n = 2) and 5.3' for a 1 (see Remark 5.1). Ii follows that
u(z) = Re(czm} +
lclrmcos(me+Oo) + O(rm+$),
-p--u mcz" + az
whèrec isan inidger 2and0 <6<1. Thus the zeros of Vu are isolated and in some tnéighborbood of 0 there exist 2m smooth curve initiating at 0 which divide this neighborhood into sectors a, such that
u(z)>0
u(z)<O
Take points E 02, z4 E 04. Since is connected, we can connect 22 to z4 by
ASYMPTOTIC ESTIMATES FOR THE PLASMA PROBLEM
a curve lying in 11,,. Connecting z2 to 0 and 0 to z4, we obtain a closed Jordan curve y contained in U (0). The interior of y encloses some c, withj odd and the exterior of y also encloses some with i odd. Hence is not connected, a contradiction.
We have thus proved (12.2). From this follows that F,, = {u 0) is in for any 0 < I. The proof of analyticity is the same as the proof of Theorem 5.1.
PROBLEMS
1. Extend Theorems 12.1 and 12.2 to plasma problem given in Section 11, Problem 3.
2. If ttu + c(x)u 0, u 0 in a domain U C2 boundary, and if u 0 and c(x) is a bounded function, then (i) u <0 in U, (ii) if u(x0) = 0, x0 U, then (au/arxx0) 0. (Hint: For some a > 0, the function v satisfies t&v + 0.1
3 If in BR, u=O on 8BR, u>O in BR, and f(t) Lipschitz continuous, then u = u(J x J). (Hint: Let = {x1 = A), = BR fl (x, > A), x with respect to and consider the set S of A's in [0, Rj such that
S is closed. If A E S, then v(x) u(x) — U(XA) satisfies + c(x)v = 0 and Problem 2 gives v =2u <0 on A. Also, <0 on aBR n (x1 >0) since if f(O) 0, then u,, >0 near r = R(x, >0) and then also u, <0, u <0, which is impossible. Thus S is open, and 0 E S, which imply that U, 0 on x1 = 0. This is true for any plane through the origin.J
4. Prove that any solution of the plasma problem (11.l)—(1 1.3) for (1 = C R" is a function of f only, and deduce its uniqueness.
13. ASYMPTOTIC ESTIMATES FOR ThE PLASMA PROBLEM
In this section we take n = 2, A > A1 and use the notation
U=UA, UP=OPk, UO=UVA, F,,—F,,,;
is is the solution constructed in Theorem 11.3, that is, u is a solution of Problem (I). We are interested in the behavior of "pA as A -, oo. We shall prove that is asymptotically a disc and we shall characterize the possible locatipn of within a. The main results are slated Theorem 13.13. 13.11 is of intrinsic interest.
532 VARIATIONAL PROBLEMS WITH POTENTIALS
Consider first the case where is the disc BR (I x <R } and denote the solution by UR (see Section 12. Problem 4).
Lemma 13.1. There holds
(13.1)
where y is a universal constant.
Proof. Set u UR. If the free boundary is given by r = r. then
forE<r<R,c=u(aBR)
and (11.3) reduces to
(1 1 R log—.
The principal eigenvalue for in is yt/e2 (y, a universal constant) and the principal eigenfunction is u0(r/€), where u0(r) is the principal cigenfunction for in B,; we normalize it by
u0<O, (t4 dx =1.
Then
lv
where Y2 is chosen so that au/as' is continuous across r e: is a universal constaUt. If
One easily computes
f vu 12
2 j2 R
!u( aBR) = Ic = (ft) log
where are universal constants. CombiMng these results, (13.1) follows.
ASYMFI'OTIC FOR THE PLASMA PROBIIM
L.eimna 13.2. If
then
(13.2) J(uA) mfJ(v) = —
IogX + 0(1),
(13.3) UA(F) (j_)iogx+o1.
(13.4) fI (f-) logA + 0(1), where C, C a constant depending only on R1, R2.
Proof. We write K = KQ, J(v) = to emphasize the dependence on Then
(13.5) c implies that infJ02(v). KR
Indeed, the minimizer u of can be extended as the positive constant u(ali into li2 and it then belongs to since
(13.5) follows. Thç assertion (13.2) follows from (13.5) and Lemma 13.1. Mukiplying (11.1)
by u and integrating over we obtain
.1 Ivul2dx=AJ(u_)2dx.a,,
Similarly,
= u(F)I;
Consequently,
(13.6)
and (13.3), (13.4) follow from (13.2).
534 VARIATIONAL PROBLEMS WITH PoTENtiALS
Lemma 13.3. There exists a positive constant C such that
(13.7)
1 A1.
Choose points A, B in such that
=18—Al.
Consider the fami.!Lof straight lines passing through x and orthogonal to AB when in AB. Denote by = a segment lying in such that
1'. E rP,andmt8XcI2V.men
— =
and 0. Using (13.3), we obtain
log A cf I
Integrating with respect to x in AB, we get
1/2
18— cf (ClB — Al)"2;
recalling (13.4), the assertion (13.7) follows.
Let
(13.8)
denote Green's function in U, that is,
ifx,y EU, (13.9) ifxEU, yEr.
(13.10) k(x) =
We shall use the notation: d(x, A) = dist(x, A).
ASYMPTOTIC FSTIMATFS FOR ThE PLASMA PROBLEM
Lemma 13.4. There holds
(13.11) k(x) -, oo uniformly inx,asd(x,f) -'0.
Proof. Suppose that x x0, x° E r. Then
ifyEr, Iy—x°I>3,
C ifyEl', Iy—x°I<4, Ix—x°I<e;
e+Iy—x I
here C, C1 are positive constants. Representing h(x) in terms of Green's function, we have
=
C dsC1 og
'(r.Lv—x°I<I/2} e (v — x
J1 + j2;
- 0 if x x°, whereasf2 C, log(C/e). It follows that
urn C,log X-.xo E
Since F is smooth, the last inequality can be established uniformly with respect to x°. Since, finally, r is arbitrary, the assertion follows.
We shall now find the asymptotic location of the plasma. Let
(13.12) k0= mink(x). eQ
in view of Lemma 13.4. the minimum of k(x) is indeed attained in Il, and the set
(13.13) S = {x E k(x) = k0)
is a compact subset of
Theorem 133. For any e >0, there exists a A0 = >0 such that
(13.14) s) <E ifA > X0.
536 VARIATIONAL PROBLEMS WITH POTENTIALS
Proof. Suppose that the assertion is not true. Then there exists an r > 0 such that
(13.15) d(li,,S) forarbitrarilylargeA's,
say, A = Fix a point y° E and let i be a translation such that Ty0 E S. Using
Lemma 13.3, we deduce that there exists a positive number 6 > 0 such that
(13.16)
provided that A = A1 is sufficiently large [to ensure that d( lie) is sufficiently smalli.
Consider the function
I —u (T'x)iff'x Eli ,harmonicin' ' — p (13.17) = I
Oon = u(r)on F.
Clearly,
( I j
hence if wc show that
(13.18)
then we derive a contradiction, thus proving the theorem. Since
.J(u) J J(u0) / IvuoI2—Iu(r), (13.18) reduces to
(13.19) J <f vuj2.
To prove (13.19), introduce the functions
_u(F)—u+ - u(F)
' u(F)
Then (13.19) becomes
(13.20) f Ivuol2<f
ASYMPTOTIC ESTIMATES FOR THE PLASMA PROBLEM 537
Notice that
Capn( f f = f = — r r
= J I Vu0 12
= uP
,au0 — — — i — — — ;
JraP
where — — are taken in the sense of measures (with support on F,, and r( F,,), respectively].
We can write
ü(x) = —JG(x.
= —fG(x.
Consider the function
w(x) =
y) y
x E
w(rx)
> _JG(x, u(x) = 1,
so that
w(x') > = 110(x) ifx
Since both w(x) and ü0(x) are harmonic in i-((l,,) and vanish on F, the
538 VARIATIONAL PROBLEMS WITH POTENTIALS
maximum principle gives
Hence
aw an0—<— oni'. av a,'
It follows that
f IvuoI2= =r r uP
= = = f I vnI2 and (13.20) follows.
From Lemma 13.4 and Theorem 13.5, we deduce:
Corollary 13.6. There exists a & > 0 and A0 posilit'e and sufficiently large such that
(13.21) r) >8 ,fA > A0.
Set
u(F)—u w=
u(I')
Thenwoonr,w I on and
u(F) = logA+O(1)
by Lemma 13.2. Consequently,
(13.22) = log A+O(l)
Lemma 13.7. There exists a positive constant C such that
(13.23) forallA>A,.
ASYMI9IYI1C ESTIMATES FOR ThE PlASMA PROBLEM 539
Proof. In view of Corollary 13.6, Theorem 9. can be apphed to deduce after using (13.22), that
'I— I I
log 1/1/ logI/1 logxl/2' p.A
from which (13.23) easily follows. We shall now consider scaled solutions. Fix a point in and set
- IXYA ux(x)
x"2
Then
(13.24)
and
(13.25)
(13.26) 0
(13.27) uA=U(r) Ofll'A.
Notice that
(13.28) C BcAI,2 (c >0, C> 0).
In the sequel we denote by C, c positive generic constants independent ofX.ByLemma 13.7,
(13.29)
Lemmal3.$. Therehokfr
(13.30) ÜA(X) Clog(r + 2) ifx r
x + 2. Let
u(F)
540 VARIATIONAL PROBLEMS WiTH POTENTIALS
Then w = I on and w = 0 on By Green's formula, for + 2,
(13.31) w(x)— —
i a 1 +
r av 2irfr,ar W
= — '2 + 13 — 14.
Clearly, 14 = 0 and 12 0. To evaluate note ffrst that
(13.32) A =
logX± 0(1) [by 03.22)1.
Since also, by (13.28),
log r = log X + 0(1) on
we get
13
Iog(1)
log X
Finally, since log r C on
= 0(1)1 = (since
[by (13.32)].
Putting together the estimates for the ii., we obtain from (13.31),
w(.x) I ifIxlC5+2.
ASYMPTOTIC ESTIMATES FOR THE PLASMA PROBLEM
Since
w(x) = I — u(F)'
we get, upon recalling (13.3),
(13.33) IxI=C*+2. By the maximum principle, the same inequality holds for x 6 A' x 1<
C* + 2. Finally, comparing ÜÃ with C0log(r + 2) in fl x > C" + 2), the assertion of the lemma follows.
Lemma 13.9. There holds
(13.34)
I
f foranyA>O.fIxI<A) Also, by Lemma 13.8, for any large 4 >0,
max a, C depends on A). IxIA
Representing ilA in x J<A by means of Green's function, we deduce that
f foranyp<oo.(lxi } Hence, by (1 (13.26),
(13.35) f I r C, and (13.34) follows by applying SObOIWS inequality.
Corellary 13.10. Tbere ems pOsrn%,e constarns c. c such th.t, íè all A >
(13.36)
(13.37)
542 VARIATIONAL PROBLEMS WITH POTENTIALS
Indeed, by Lemma 13.9, (Uk).. C and therefore
f__f
- thus (13.37) follows; (13.36) is a consequence of (13.37).
From Lemma 13.9 and (13.35), we obtain, by elliptic estimates;
(13.38) YbaIIBR
provided that A A0(R); CR and A0(r) depend on R. Hence from any sequence of X's increasing to infinity, we can extract a A1 such that
(13.39) in compact subsets of R2,O 2,
(13.40)
by Lemma 13.8,
(13.41) (.1(x) +2)
and, by Lemma 13.7,
(13.42) (1... (x e R2; (.1(x) <0) C (C* > 0).
Since also
(13.43) . f u_ dx = I,112 the open set is nonempty.
We shall need the following theorem:
Theorem 13.11. Let u be a solution in R2 of the equatio4
Au+f(u)=0,
wheref is L:pschitz continuouv, and that
u(x)=4alogr2+b+ A•x
ASYMPTOTIC ESTIMATES FOR THE PlASMA PROBLEM
as r -.. oo, where a is a positive constant. Then u is a radial function , that is, there is a function g(:) (z 0) and a point x° such that
u(x) = g(Ix—x°I).
Proof. By changing variables x -. x — x0 and choosing x0 in a suitable way, we obtain an expansion for u near infinity in which the term A x disappears. Without loss of generality we may assume that a = I. Taking also, for simplicity, x° 0, and denoting the independent variable by (x, y), we have
(13.44) vu(x,y) (x,y) + o(4).
Consider the rectangles
Sn (x>A),
where M is any sufficiently large positive number and
K (13.45)
K is a fixed and sufficiently large positive number to be specified later; it is independent of M.
We wish to study the function
(13.46) VA(X, y) = u(x, y) — y) in
First we prove that
(13.47) ifA<x<M+X. Using (13.44), we have
—x+2A
=
+ + o( x_A)
(x_2A2)j+o(x_x)
544 VARIATIONAL PROBLEMS WITH POTENTIALS
Since
2 /
M2 M2
and since the (unction
log(1 + t) —
is monotone increasing in t, 0 < z < 10, we get the lower bound
I x2 — (x—2X)2 (x—A)A 22 M2 M2 k M3 / 6M2 M3
Using (13.45), we obtain
(13.48) vx(x, ±M)>
for a suitably large K, thus establishing (13.47). We next show that
(13.49)
It suffices to consider the casey 0. We can write
(13.50)
where
I [u(M + A, y) — u(M + A, M)1 —[u(—M + A, y) — u(—M + A, M)],
I = u(M + A, M) — u(—M + A, M) + A, M).
By (13.48),
(13.51)
To estimate I, write
— JM[(M+ — + A, ii)) dii
ASYMPTOTIC FSflMATFS FOR THE PLASMA PROBLEM
and, upon using (13.44),
________
—
,,2+(M+A)2
kM2
since the integrand is negative (recall thaty > 0). Combining the last inequal- ity with (13.51), (13.50), and using (13.45), the assertion (13.49) follows.
From the proof of (13.47), it is clear that also
(13.52) v1(x,±M')>O
The function satisfies
inst. sothat
+ 1nSA
From (13.44) it follows that if M is sufficiently large, then 0 in if A = 2M. We have also proved in (13.47), (13.49) that >0 on \(x = A). Onx =Awehavev, =O.Thusthestrongmaximumprinciple(seeSectjonl2, Problem 2) can be applied to conclude:
v,>0
and
VAX(A,y) >0 M;
hence
ifIyI<M.
Recalling also (13.52), it follows that if A' <A, A — A'
VA>O
We now apply a "folding" argument (see Section 12, Problem 3). We decrease A continuously until we reach the smallest possible value of )i,iay X0, such that either (i) A0 = K/Al; >0 in or (ii) A0 K/M, 0 in with equality at some point of The argument given above for shows
VARIATIONAL PROBLEMS WITh POTENflAIS
that (ii) cannot occur. Thus (i) holds and, in particular,
- A0
Taking M 00, we get
y) 0 for ally.
Since the x-axis can be taken in any direction, we get u0 = 0 where (r, 0) are the polar coordinates with respect to x0, and the proof of Theorem 13.11 is complete.
Consider now the function U. It satisfies
[j()= —t_]
and it is harmonic outside a disc [by (13.42)]. It also satisfies (13.41). The relation (13.27) with u(I') — ç)log A (c0 > 0) can be used to compare from below with a function of the form clog r (c positive and sufficiently small). We deduce that
I
Hence all the assumptions of Theorem 13.11 are satisfied. It follows that
(13.53) U(x) = — x01).
Let p I x — . The set where U(x) is harmonic must be the exterior of a
disc; otherwise, there is a ring R1 <p < R2 in which U.js positive and harmonic, and it vanishes on p = R1, p = R2; this, however, is impossible.
We have thus shown that
(13.54) isacirclelx—x°I<R,
and
(13.55) ifp>R; A>0.
Hence, in particular,
(13.56) 0 on = the normal to
and from (13.39) we then deduce that
(13.57) along
PROBLEMS
Thus, every sequence of A's increasing to 00 has a subsequence A, for which (13.57) holds. This yields:
Lemma 13.12. There exists a positive constant c < I such that, for all A > A1,
(13.58)
consequently,
(13.59) cA"2 !Ah/2 along!',,.
The second assertion follows from the first one by recalling the definition of ÜÀ. Notice that R in (13.54) is determined by the property:
60 the first eigenvalue of — in the
(13. )
From (13.39), Theorem 13.11, and Lemma 13.12, we can deduce, using the implicit function theorem:
Theorem 13.13. For each A sufficiently large, there exists a point E such that the free boundaiy 1', can be represented un polar coordinates 9) about x,j in the form r and
(13.61) — R]I_* 0 (I = 0,1,2).
Further, S) -+ 0 if A oo.
Thus is approximately a disc with radius The last assertion follows from Theorem
1. ThesetSisfmite. I Hint: Let z = g( w) be a conformal mapping of { w < 1) onto (I. Then
lini w w0
so that
2rrk(g(w)) —log(I — 1w12)
543 VARIATIONAL PROBLEMS WITh POTENTIALS
is subharmonic; in fact. If S is not finite, it contains an analytic curve in the interior of which k const.]
2. If is given by ((x, v); Lr a <x <b), then (a/ay)k(z) > 0 if y > 0 and, consequently. S lies on the x-axis. [Hint: The formula
(13.62) 2ir-k(fl =
follows using (ccc references 37 and 38)
G is Green's function for + and
kU + Er) — = —
By the maximum principle,
+ (y>O),
for any fixed z fl(y >0).]
3. it S consists of n points, then for all A sufficiently large there exist at least n solutions of (11.1)—(11.3). [Hint: Let x0 E S, KGU = (v E K, v >0 in G a small neighbor- hood of x0, u0 a minimizer of J(v) over KG,n. If u0 > 0 on G, then (1 1.1)—(1 1.3) follow as in the proof of Theorem 11.3. Suppose that
rl * 0, where = (u0 <0); then u0 0 on 8G fl in the trace class sense and if (*) —' 0 as X oo, then the proof of Theorem 13.5 gives
A-.oo.
To prove (*), use
j2 IT IT
A VARIATiONAL APPROACH TO CONVEX PLASM/IS
where the right-hand side is obtained by comparing with
inf J(v) infi(v), KGG KG1
where G and is parallel to at small distance.]
4. Consider the plasma problem described in Section 11, Problem 3. Let A = sup {r; (r, z) E (2). Prove that for any e> 0,
inf 1(v) asX—oo
(12/8ir)logA + 0(1)
and use it to show that d((2PA, 1') —'0 if A —' co. [Hiiu: Follow the proof of Theorem 13.5.1
14. A VARIATIONAL APPROACH TO CONVEX PLASMAS
In Section II we introduced two (equivalent) variational principles for the plasma problem for a general domain (2 in R". In this section we restrict (2 to be a convex domain in R" and introduce a new variational principle. It has the advantage that it yields a solution of the plasma equations (ll.l)—(1L3) with a convex plasma.
We assume:
(14.1) (2is a bounded convex domain with boundary.
Definition 14.1. A subdomain G of (2 is said to belong to class if G is convex and G C (2.
For any X> X1 and small >0, define a functionf,14:
i—A ifi>A, (14.2) =
e0(t—A) ift<A.
Consider the functional
=fA ,0(X(G)) + Cap0G,
where A,(G) is the principal eigenvalue of G.
Problem Find G such that
(14.3) G E A.JG) = mm
550 VARIA11ONAL PROBLEMS WITh POTENTIAlS
To every G E we associate a function u defmed by
u<OinG,
inU\G,
(14.4) u=O on8G,
uconst.u(r)>O onaa,
In this section, we prove:
Theorem 14.1. If is sufficiently small, then there exists a solution G of Problem and the corresponding u [defined by (14.4) with suitable u(r)J is a solution of the plasma problem (11.1 )—(ll.3).
The last assertion simply means that u 6 C' in a neighborhood of We begin with several auxiliary results of intrinsic interest.
Lemma 14.2. Let w be a C2(D) solution of an equation
f(x, w, vw)
in a bounded convex domain U of R", and assume that f(x, w, v1,. ..,;) is a function nondècreasing in w and convex in (x, For any fixed t E (0,1),
consider the function
U(x, y) = w(tx + (1 — t)y) — tw(x) — (I — t)w(y)
in U X U. Then fJ(x, y) cannot take a negative minimum in U x U.
Proof. Without loss of generality we may assume thatfis strictly increasing in w and w is smooth up to the boundary. Indeed, otherwise we first approximate w by we:
f(x, w,, V w1) +
e >0, U is convex with smooth boundary and (2 t U if ejO. (The Schauder estimates can be used to prove the existence of a solution see Problem I.)
Suppose that the assertion is Then (J(x, y) takes negative minimum at a point (x0,y0) in (2XU. Set z0:=1x0+(l—t)y0. Since gxadU=0 at
A VARIATIONAL APPROACHTO CONVEX PLASMAS
(x0, Yo)'
(14.5) Vw(x0) Vw(v0) = VW(Z0).
The function h(x) U(x0 + X, Yo + x) takes negative minimum at x 0; hence
(14.6)
Using (14.5) we compute that
= — — (1 -—
=f(z01w(z0),vw(z0)) — :f(x0,w(x0),vw(z0))
— (1 — :)f(y0, w(y0), Vw(z0)).
Since (J(x0, Yo) <0, we have w(z0) < tw(x0) + (1 — t)w(y0), and thus, by the strict monotomcity of f(t, w, Vw) in w,
tth(O) t'4'(x0) + (1 — t)w(y0), vw(z0))
—zf(x0,w(x0),vw(z0)) —(1 —:)f(y0,w(y0),vw(z0))
by the convexity of f(x, w, Vw) in (x, w); this contradicts(14.6).
Theorem 14.3. Let u be a solution of
X>0,
u0 on8IJ, u.SO mO,
where is_a bcunded convex domain with C' boundasy and q(x) >0, q(x) concave in 0. Then log u(x) is concave.
Proof. We may assume that 0 is strictly convex with C3 bound4ry and q E for otherwise we approximate 0 and q by such and q,, respectively; if the corresponding. principal eigenfunction u is log-êoncave, then the same is true for u Jim
u satisfies
(14.7) :. and
(14.8) w(x)—. —oc
552 VARIATIONAL PROBLEMS WITH POTENTIALS
Clearly, Lemma 14.2 can be applied. Thus it suffices to show that if
(14.9) limU(xkyk)= inf U(x,y)<0. V
then i and j both belong to We shall assume that i E and derive a contradiction.
Set
Zk = tXk + (1 —
Since
w(zk) —1w(xk) —(1 — 1)w(yk) <0
for k large, it follows from (14.8) that
w(zk) —I —00
and thus dist(zk, —' 0. Since is strictly convex, we conclude that
Y*41'
By taking a subsequence we may assume tha
XkYk hm exists1
I — Yk I
it determines a direction e. Denote by I the ray from I in*o having direction +e or —e.
CASE 1. 1 nontangential. For x in near I and for o any direction near the direction of I,
— U2 (14.10)
0 0 at I. Hence w is strictly concave on the interval XkYk This implies that U(xk, Yk) >0, contradicting (14.9).
CASE 2. i is tangential. Suppose that Ibas the form x,, = where x' = (x1,. . . ,xN_I), I = (0,... ,0), and O. Since (— 1) is normal to au, the vectors (0,... ,0, 1,0,... (I in the ith component) are tangential and
A VARIATIONAL APPROACH TO CONVEX PIASMAS 553
is a.tangential derivative. Therefore, T,2u = 0 along so that
+ = 0 at
Thus .< 0 at i (1 I n — I). But, for any x in f� near I and for any direction a near the direction of I, we have
c(x)— +
where is small and is near 1. Consequently,
Uac <0
and, by (14.10),
<0;
a contradiction is now derived as before.
Theorem 14.4. Let fl, G be convex domains with C' boundary and let u be a solutions of
inO\G,
u1 onaG, u0 0<.u<1 inD\G
where y(t) is Lipschitz continuous, non decreasing, and a nonnegative function of t. Then the level curves of u are convex.
The issertion means that the sets {x; t4x) > t) U 6 are conwx for any t E(0,1).
Proof. By approximation we may assume that y(u) is strictly increasing and analytic and that fI and G are strictly convex (see Problem 2). Extend u by I into G. If we can show that
(14.11) sup {u(y)—u(z))cO x, u(x)>u(z)
for some tE(O, I)
then we get u(z) mm (u(x), u(y)) whenever z = Ix + (1 — t)y for some I E (0, 1), and the assertion follows.
554 VARIATIONAL PROBLEMS WITH POTENTIALS
To prove (14.11) we need the following fact:
(14.12) (x -- i) .vu(x) E G. To verify it. = (x — i)vu. Then
— = 2y(u) 0,
and (14.12) follows by the maximum principle. Suppose now that (14.11) is false. Then there exist points Xk, Yk and
tkXk + (1 — such that
u(yk)— u(zk) -+M>O,
u(xk) >u(zk), U(Zk)= min(u(x);x
and Xk, belong to
Yk'Y0'
Clearly, 20 * y° since u(z°) u( y°). Since
u(x°) u(z°), u(y°) >
the interval x°y° cannot intersect \ G, by(14. 12). Also, it cannot intersect since otherwise u = const. along x°y° and, by analyticity, also u =
const. along the entire segment, in \ G, which contains x°y°. Next, y° since otherwise
u(z°)= u(y°) — M<O.
The function u(z) along x°y° takes minimum at otherwise, the definition of M is contradicted.
We also have
(14.13)
Indeed, if z0 E then the lines through Xk, Yk converge to a line 1 non- tangential to at z0 and thus au/al *0 at from which we deduce that U(Zk)> u(x&) for k large enough.
Next, we show that
z0 = x0.
A VARIA11ONAL APPROACH TO CONVEX PSASMAS
Indeed, suppose that Since u oonst. along x°y°, u takes in this interval values larger than u(z°). Let x' Ez°x°, u(x')> u(z°). Then for any x" near x', u(x") > u(z°), and hence
u(y") — u(z°) — u(z°),
where y" lies in x"z0, y" near y°. It follows that u takes local maximum at y°, which is impossible.
We have proved so far that
(14.15)
Also, since u on takes minimum at and since Z* (if k is large), we have = 0 and, consequently,
(14.16) au(z°)
= 0;
here 1 are in the directions and x°y°, respectively. The next assertion is that
(14.17) vu(y°) is parallel to vu(z°).
Indeed, otherwise there isa direction v such that v vu(y°) >0, v vu(z°) <
on the rayy°z°bcyond z0, then = a small and 2 + av, we get a triple20, 90. with
u(90) u(10) > u(y°) — u(z°)
>
contradicting the definition of M. Thus u(x) u(z°) on 1he ray y0z0 beyond zo. We therefore, can choose suitable a and small enough such that intersects a point 2 beyond z0 and u(1) > u(10) Isince au(z°)/av <0. and, by (14.16), au(z°)/al = 0]. With this triple 9o' i we .gain derive a contradiction to the definition of M.
Having proved (14.17) we set
(1418) vu(y°) = vu(z°)
Ivu(y°)J Ivu(z°)I
and take for simplicity y°—z°I= l,Ivu(z°)l= 1. Set A=Ivu(y°)I and introduce the functions
u1(x) = u(z° + x), u2(x) = +
556 VARIATIONAL PROBLEMS WITH POTEN11ALS
and
U(x) = u2(x) — u1(x) — (u(y°) — u(z0)),
w(x) = vu1(x) (e + (i — e = z0 —y0.
Then
+ Aw) = — + Ay'(u,)w
+2X(1
Since w(O) 0 and y(u(y°)) — -y(u(z°)) > 0, we obtain, setting e = (y(u(y°))— y(u(z°)))/A2,
÷ (A.— l)2(2A+
Thus, for r small enough, the function U + Xw — c x 12 is subharmonic in B,(O) provided c <t/2n; it also vanishes at the origin. Using the maximum principle we deduce that
(14.19) max(U + Aw) = U(x) + Aw(x) c,-2
for some x E aBr(o). Set
0 X oyo*=y zo*=z +x.
If U(x)>0 then
— u(4) >u(y°) — u(z0).
If also w(x) > 0 then u(4) > u(4) where
for some small 1> 0. With the triple ye', we then get a contradiction to the definition of M.
A YARIA11ONAL APPROACH TO CONVEX PLASMAS
In the general case where U(x), w(x) are not both positive we shall produce a different admissible triple (i0, Yo, in the form:
i0=!0+:(10—90)
with small I > 0, for which
(14.20) u(90) — > u(y°) — u(z°).
Ifh —U(x)and
U(x) = u2(x + — — (u(y°) — u(z0)),
I I l\ h vu1(x)
then, since h = 0(r2),
(14.21) U(x) =
and
Since, by (14.19), Aw(x) + h, we get
(14.22) + 0(r3).
Noting that 0 in a neighborhood of the origin we deduce from (14.21), (14.22) that by changing the magnitude of h we can achieve simulta- neously U(x) > 0 and � 0. With this new value of /ithe triple (-h, A) defined above is admissible and (14.20) holds; thus the definition of M is contradicted.
We return to Problem (s).
Lemma 145. Problem ('i) has a solution.
Proof. Denote by r(G) the radius of the largest ball contained in G, GE3en(secrcfercnçe 114)
(14.23) (c>0)
VARIATIONAL PROBLEMS WflH P(YIENTIAIS
where c is an absolute constant For any G E denote by V the solution of
(14.24) zW+X1V=OinG, V>O, fV2l
and by U the solution of
(14.25) U=OonaG, Ulona(1.
We claim that
(14.26) mG,
where C is a constant depending only on A1(G). Indeed, suppose first that is smooah and strictly convex. Then (by
Problem 4)
(14 27) I + A,V2 takes its maximum in Gat
an interior point where V V vanishes.
Thus, setting M = sup V, we have
(14.28) I vV12 + A1V2 X1M2.
By elliptic estimates (109, Theorem 8.25]
I . fl VP
G
where c is independent of the smoothness of Therefore,
M' cXfM1_2f V2 = cAfMP2
so that M2 cAl'; (14.26) now follows from (14.28). From (14.26) we obtain
(14.29)
where d(x) = distance from x to We now take a minimizing sequence G,,, e bounded,
(14.23) implies that each contains a fixed ball (m Is large suough). We claim: There exists a sufficiently small d0 >0 such that
(14.30) ifmisiargeenougb.
A VARIATIONAL APPROACH TO CONVEX FIASMAS
Indeed, suppose that dist(Gm, aR) and let
p(x)
Ôm = Gm fl {p(x) > 2db),
Gm fl (p(x) = 2d0).
Denoteby Vm,Aimand X1 ,,,thC V,A1correspondingtoG= GM,G= ( respectively, by (14.24). We may take the G,,, to be such that Green's formula applies to Gm (for instance, we can take as polyhedrons).
By Green's formula (suppressing the index m) and (14.26), (14.29),
(X1
where SV= vol(G\O). Since
fVVi+o(l) asd0—O,
weget
(14.31) 8A,cc&V.
We nest the change in the capaaty. Let Z be the harmonic function in (0 < p(x) < 2do) with Z =0 on {p(x) = 2d0}, Z = I on Then
,i(do)-. ooifd0-.O.
by the maximum • I
'S
where Uisdeflnedby(14.25)withG givenby •
— I = —,(ivU— U)I2
(14.32)
+2fv0.v(U— U)= —fJv(U--
where
VARIATIONAL PROBLEMS WITH POTENTIALS
For each p E we integrate 80/al along the ray F: x0p from p until we hit q E aG:
fi U(q) — qI•
Integrating with respect top EE we obtain
c >0. G\G G\G
Recalling (14.32), we conclude that
8K —f .1 v0j2G\G
Combining this with (14.31), we find that
— — E0 E0 C0
if d0 is small enough; here various positive constants are denoted by the same symbol c, and the relation 8V cd0 was used. It follows that G Gm cannot be a minimizing sequence, and the proof of (14.30) is thereby completed.
Gm contains a fixed ball BR(x°), one can easily check that Gm G the following sense: Representing 8Gm, in terms of spherical coordinates about x0, by r = rm(O), then rm(O) uniformly in 0, where r = r0(0) represents 8G. Further, the rm(0) are uniformly Lipschitz continuous (uniformly in m).
It easily follows that Xi(Gm) -. A!(G), and, using (14.30), Cape Gm -. Cape G. This completes the proof of the lemma.
Lemma 14.6. Let G be a solution of Problem Then, for some positive constants c, C there holds: (1) V(x) cd( x); (ii) U(x) cd(x); (iii)
I
C; (iv) in a neighborhood of 8G; (v) EC'; (vi) V U(x)
I
c in a neighborhood 8G; and finally (vii) I VU (, I V V exist
everywhere along 8G as limits from \ G and G, respectively.
To prove (I), introduce the level surface 1',: (V = t}, which is convex by Theorem 14.3. Letp be a point on I', where d(x) takes its maximum, say 6(t). Denote by D the smaller part of G cut off by the tangent plane 1, to 1 at p, and set = G\ I.). We can now proceed to estimate &A1, 8K corresponding to the variation G of G and find (by calculations similar to those in the preceding lemma) that
c —carea(7, fl G)(6(t) —
Since 0, we must have 6(i) ct/C0, which gives (i).
A VARIATiONAL APPROACH TO CONVEX PLASMAS 561
The proof of (ii) uses the convex level surfaces U For complete details on the proof of Lemma 14.6, see reference 64.
Proof of Theorem 14.1. Let G be a solution of Problem and let u, v be defined by
inG,v<OinG,
v0 onaG, (14.33)
u0 onaG, ul onr.
If we can prove that X1(G) = A and
av au (14.34) . onaG (cconstant),
then extending v into (l\G by Cu, we obtain a solution v of (ll.l)—(ll.3), and the proof of Theorem 14.1 is complete.
We make a perturbation of G into the convex domain bounded by (u = e). This changes the eigenvalue and the capacity by
(14.35) 6A1(G) = I Vol2 + aGIVuI
(14.36) 6K =
£
the proof of (14.36) is immediate, whereas the proof of (14.35) is based on Lemma 14.6 and it is given in reference 64. ii is sufficiently small, then we deduce that
<0 if A(G) <A.
Since G is a minimizer, we conclude that A1(G) A. Similarly, one shows that A)(G) thus A1(G) A.
In order to prove (14.34) we make further perturbation, moving the surface (u = e) inward along the normals by a fixed distance s. The total first variations of and Cape G are given by (see 64)
a. 1Vv12 I 8X1 —ej +sj 1vv12+o(e),
6K =
Cape G sf I Vu 12+ o(e)
e
VARiATIONAL PROBLEMS WITh POTENTIALS
provided that s 0(e). We choose s such that 8X1 o(e); that is,
i. // I 1Vv12 JaG lVul/ 'aG
The inequality JG) 0 then yields
(14.37) .13GIVUI 1aG1 Vu12
I
faG' I
Using HOlder's inequality,
(14.38) (faGs faGs
we obtain from (14.37)
(14.39) vol)2 faG' vu(2 faG' vu
,)2.
Similarly, if we first perturb 8G into (o = —e) and then move it outward a distance s, we obtain (14.37) with u, v reversed. Proceeding as before, we also obtain (14.39) with u, v reversed, which means that equality holds in (14.39). But then we must also have equality in (14.38), which gives (14.34).
PROBLEMS
1. If f(x, w, Vw) in fi, aa E C2+a, f E a, = g on au, g E then there exists a solution of
iflU,
onau
provided that el is small, and w, E [Hint: u = w — w, satisfies
Lu1 = a1,u,+ a21 Vu1
the coefficients of L are independent of u1 and (a11 II+a Ce if u C0. Set Tu L4if
Show that 7' is a contraction.]
ThE THOMAS-FERMI MODEL 563
2. Let u be as in Theorem 14.4 and ?m(t4) E C', 0 y,,(u) C, Ym(U) -. y(u) uniformly in bounded sets, G,,, t 6, smooth. Denote by Urn the solution u corresponding to Gm. Prove that ujx) -. u(x) V x G. [ Hint: Apply the maximum principle tO Urn — U.]
3. Let f, g be homogeneous harmonic polynomials satisfying: 0) C (g Prove thatg=cf, wherecisaconstant.
[Hint: Denote by the restriction of the Laplacian to the unit sphere S. Let K1, K5 be components of 0), fg 0) on S such that K1 C A,. If a = degreef, degree g, then
P
Hence
(a—p)ffg=f (fg,—f,g)=—f ax, ax,
Deduce that a $ and by working with —f, —g, that a.]
4. Prove (14.27). [Hint: (reference 166): Show that 0=1 VVI2 + X1V2 satisfies
(w= vO—4A1VvV)
and apply the maximum principle.)
15. THE THOMAS-FERMI MODEL
We consider a quantum mechanics problem in which I electrons each of mass m and charge -è are moving about positive charges of magnitude fixed at positions a• E R3 (I i k). The Thomas—Fermi atomic model views the electrons as "clouds" with density p(x), say, so that
fp(x)dx=I.
The repulsive potential energy (the interaction between electrons) is
!f j 2 R'R3
the attractive potential energy (the interaction between the electrons and the
564 VARIATIONAL PROBLEMS WITH POTENTIALS
positive nuclei) is
dx,
where
k
(15.1) V(x) 1=1 Il
and the kinetic energy is
f dx.R3 Thus the functional
(15.2) dx —
dx + 21313 W);) dx dy
represents the total energy of the system; it is called the Thomas—Fermi functional.
Let
K0=
The Thomas—Fermi Problem. Find such that
(15.3) = E K. p€K
Consider also the problem: Find such that
(15.4) = mm e K0. pE Ko
It is not difficult to solve problem (15.4). Indeed, observe first that the functional
(15.5) ' dx dy is convex
THE ThOMAS-FERMI MODEL
(since, for instance, I/I x is the Fourier transform of a positive measure). The function in (15.5) is also continuous on L6/5(R3). Hence if we take a minimizing sequence Pm of problem (15.4) such that p,,, — weakly in L3"3(R3), then
Since clearly, p K0, is a solution of (15.4). As for problem (15.3), if we follow the procedure above we run into the
difficulty of verifying that E K or, more specifically, of verifying that fp(x)dx = I.
We shall now state the main existence results for problem (15.3), setting
(15.6)
Theorem 15.1. (1) 110 < I M, then there exists a unique solution to problem (15.3); (Ii) if I> M, then there exist no solutions to problem (15.3); (iii) if 0<1</4, then the sohUlon of problem (15.3) has compact ssçport.
The setting here Is somewhat simik, to that in the vortex ring. In that problem we had first imposed the (relaxed) constraint
[we(8.1)]
and eventually established for the
In the same vein the inequality constraint
ft(k)'dx 1 [see (6.35)]
turned out to be equality if A is large enough. Thus it is reasonable to first solveproblem(15.4)and then try to show factin K. This approach is adopted in reference 139.
Theorem 15.1 wIll he proved (in a more general form) in the next section. We shall, use a different approach. Notice that if is a solution, then, by the method of Lemma 1.1, one can derive the equilibrium conditions:
ifp>0, (15.7)
p
VARIATIONAL PROBLEMS WITH
where
u V — Bli
and
(15.8)
Clearly,
+ =
and since
—
we obtain
(15.9) — + c[(u — = —AV (c >0).
This equation resembles equations (5.3) (for rotating fluids), (6.18) (for vortex rings), and (11.1) (for the plasma problem). Those equations all have the form
— =f(x),
where 8(u) is monotone increasing; thus, roughly speaking, appears with the wrong sign for the maximum principle. On the other band, in (15.9) p9(u) appears with the good sign; further, — + is a monotone operator in. a bounded domain with zero boundary
It is because of the "good" sign of fi in (15.9) that it appears natural to try solving the Thomas—Fermi problem by first solving the equation (15.9). This, in fact, will be our approach in the next section.
Instead of working with the special case of r513, we shall work with more general functions 3(r) satisfying
(15 10) j(r) is convex in C'[O, oo), strictly monotone
increasing, andj(0) =j'(O) =0.
We introduce the conjugate convex function
jt(g) = sup [is —j(x)].
THE THOMAS-FERMJ MODEL
Let V be any function in LL and consider the functional
(15.11) — V(x)p(x)]
We introduce a class of admissible functions
(15.12)
= {p E L'(R3), p 0, j(p) — Vp E <
oo}.
Theorem 15.2. If
(15.13)
then there exists a constant A such that
a.e. on = 0).
Conversely, if
(15.15) j"(V(x) + C) E L'(R3) for some constaniC
and E K, then (15.14) implies (15.13).
In the Thomas-Fermi case,
j(r) j(r)
withp If Vis given by (15.1), then (15.15) is satisfied with C = I and any p>4.
Proof. The proof that (15.13) implies that (15.14) is similar to the proof of 1.1 and is therefore omitted. To prove the converse, notice that if
p E L', p 0, then
(ls.16)
Indeed, by the convex ty offend (15.14),on the set
i(s') — = (v— — A)(p —
568 VARIATiONAL PROBLEMSWITH PUTThTL4LS
whereas on the set = 0) the left-hand side of (15.16) is equal toj(p) 0, and the rigin-hand side is equal to (V — — A)p 0 by (15.14).
Using (15.16), we can write
(15.17) [j(p)— —
For p 0, this gives
—ApEJJ.
On the other hand,
j(-p) — — (V+ + Cii
—f(V+C)+CiiEL'.
It follows thatj(p) — E V and G V. Thus E Choosing now p in K in (15.17) and integrating over R3, we obtain, after using (15.5),
16. OF SOU)TION FOR THE THOMAS-FERMI MODEL
In this section we prove:
Theorem 16.1
(a) Suppose that
(16.1) V(x) f h(,y)
, h EL'(R3), RIIX YI
(16.2) V>O onasetofplvemeaswe.
Then there exists a nunther 4 E oo) suth that: (i)
(ii) (lii)
(b) Suppose thw In(16J)h is afinite measure in R3, (16.2) holdt, and large,
The (I) to (iii) remain valid.
EXISTENCE OF SOLUTION FOR THE THOMAS-FERMI MODEL
It is clear that Theorem 35.1 is a consequence of Theorems 15.2 and 16.1. The proof of Theorem 16.1 is based on several lemmas. First we need some
definitions. A function f in is said to belong to Marcinkiewia space
jf
IIJIIMP= 4CR" A
O<PAJ<oc
where (l/p) + (l/p') = 1. Clearly, 3 L", C if q <p. It is not difficult to show that
(16.3) lxi EMf/a(Rn) ifO<a<n,
and that
(16.4) iIE*fIIMp IIE)IMPIIIIIL; (I <p <cc)
if E E E L1. Thus the operator B defined in (15.8) is a bounded operator form L'(R3) into M3(R3).
Lemma 16.2. Let /3(t) be a continuous monotone nondecreasing function, /3(0) = 0. Then for anyf E L'(R3) there exists a unique solution u M3(R3) of
(16.5) — + /3(u) f in R3; further, /3(u) L1(R3) and
(16.6) R3 R'
For proof, see referenct 32. The derivation of (16.6) follows by multiplyIng (16.5) by Hm(U), where Hm(U) —, 1 if u > 0, —. 0 if u <0, anJ integrating over R3. Using the fact that
(16.7) II VU11 *13/2
and taking suitable R = -. 00, the inequality (16.6) follows. Notice that EL' and (15,14), for
any 8 >0,
except on a set of Hence (15.14) yields —X —
570 VARL411ONAL ROBLEMS WITh POTENTIALS
Setting
(16.8) u= V—Bk,
it is clear that p, N satisfy (15.14) if and only if
J(jI)_t(u_A) ifu>X to
Since (16.8) is equivalent to
—iW,
we can express (15.14) in the form
(16.9) — + y(u — A) = in R3,
where
(16.10) .Y(z)={417(J1)(t) ift>O 0
Thus the problem of solving (15.14) with E K reduces to the problem of solving (16.9) with
(16.11)
A a solution UA of
uAEM,
(16.12) —LV,
provided that h E which is ensured by (16.1). Defining
i(A)f R3
we are left 'With the problem of determIning A such that
I(X)4w1.
Lemma 16.3. If(A) is nonincreasing continuous function with J(co) = 0, 1(0)> 0; further. 1(A) is strictly decreasing on the set (A; 1(A) > 0).
EXISTENCE OF SOLUTION FOR 111ff THOMAS-FERMI MODEL
The assertions (i) and (ii) of Theorem 16.1 follow inunediately from the lemma.
Proof. SetüAuX—A.Then
+ y(ÜA) =
and
aslxl-'oo
in the sense that ÜÀ + A E M3. Similarly,
and if lxi-' 00. We can now apply a version of the maximum principle to prove that
(16.13) ifp>A
(see Problem 1); hence also so that I(&) 1(A). Since — V) = —y(ux — 0) 0 and since — V 0 if I X 00 in
the sense that — V E M3, we can deduce by the maximum principle (see the proof of Problem 1) that — V' 0. It follows that — A) y(V — A). Since 7(V — A) 0 if A 00, the monotone convergence theorem gives
urn 1(A) = 0.
We next claim that
I(o)>O.
Indeed, if 1(0) = 0, then y(u0) = 0, so that u0 0. But sinc. + y(u0) = —M', we also have V = u0, contradicting (16.2).
If A,, A, then by monotonicitj' of UA — A,,, it folloWs that UA v and v is clearly a solution of the same equation as v E M3. v = and then 1(A,) -. 1(A) (by the monotone convergence Similacly, A, tAimpliesthatl(A,,)-.1(A).ThusA-.I(A)iscontinuous.
It remains to show that 1(X) ts strictly monotone on t)mc set (A; 1(A) > 0). Suppose I(A)=1(p)>O. Since we have in fact gi) = A) and thus so that The "- equality 1(A) >0 implies that the set — A) >0) has Since y(t)>0 if 1>0, also u,,—gi on a set of positive consequently A = p.
572 VARIATiONAL PROBLEMS WITH POTEN11ALS
As mentioned above, assertions (I) and (ii) of Theorem 16.1 follow from Lemma 16.3. To prove (iii) note that 0 <1 < implies A > 0. Since ux — A V — A, = — X) y(V — A). Using the assumption that V(x) -.' 0 if x we find that = 0 if x is sufficiently large.
To prove (iv) we need an extension of Lemma 16.2 to the case wheref is a finite (signed) measure.
Lemma 16.4. Let $( r) be a continuous monotone nondecreasing function with fl(O) = 0, and let
(16.14) f $(±+)dx<oo.
Then for any finite signed measure f in there exists a unique solution u E M3(R3) of (16.5); further, $(u) E L'(R3) and(16.6) holds.
Proof. Take a sequence f,, such that
and let
E
Then, as in the proof of (16.6)
II $(Un)II 1) "fn"L' C
and then also
II II 2C,
II U,
C, art constants.
(16.15) isrelativelycompactinL'(K)
for any compact subset K then, fOr a subsequence, u in uEM3.and
+ft(u) f
EXISTENCE OF SOLUTION FOR THE THOMAS-FERMI MODEL
with p9(u) E L'; (16.6) clearly also holds. Uniqueness is proved preciselyas in Lemma 16.2.
To establish (16.15) it is sufficient to show, by a theorem of Vitali (see, for instance, reference 79, p. 122) that for any e > 0 there is a 6 > 0 such that if K <6, then
(16.16) .1KI(sTfl Vn.
Setting'
f fK
where
a,,(A) =
By integration by parts,
=g(R)a,,(R)
Cj3(R) + Cr-.4(A)
'oc 'dA.
A4
Thus
"Afl Since E L'(l, oo) by (16.14), we can make the second term on the right smaller than e/2 if R is large enough. We next choose I K so small that
<e/2, and thus derive (16.16).
574 VARL4TIONAL PROBLEMS WITh POTENTIALS
With Lemma 16.4 at band, we can now establish the assertion (iv) of Theorem 15.1 by the same arguments as for (i) to (iii). We just have to observe that under the assumptions in (B), the condition (16.14) is satisfied for = y.
It remains to prove (v). Notice that = f and —-AV,
where Dirac measure at a. Thus, by (16.6),
(16.17) = M.
To prove the reverse inequality we shall use the well-known formula
(1618' 1 ( do., —
4' —yl — max(r,JyI)' where rø is identified with a point x in R3 having spherical coordinates (r, w).
Denote by p0 the corresponding to A 0. Then
(16.19) p0 = )I/(PI)
(c >0),
(16.20)
and
1o fpo(y)4v.
We shall consider spherical averages
tY(r)= I
for r> max (I a,}}. If we take the spherical averages of both sides of (16.20) and use (16.18), we
4nM Po(Y) dyr max(r,y)
4irM 4M10)
EXISTENCE OF SOLLTJ1ON F( R THE THOMAS-FERMI MODEL
Hence if Jo < M, then (16. )) gives
p0(r) (c0>0)
and thus
fpo(x)dx = dr
=oO
since p> a contradiction. Thus M, which complements (16.17). We conclude this section with a theorem that contains several simple
observations. Define p' by
p p
Theorem 16.5 Leip > Then:
(i) u = is locally in away from the set (a1,... ,aJj, where a p' — provided that p' integer, and in for any
integer. (ii) The function — Vu in C'(R3)for some y >0.
(iii) is Holder continuous in R3 and the open set
fl
contains a neighborhood of the set {a1,. . . ,a,j. (iv) Each component of Li contains at least one point a.
The proof is left to the reader; see Problem 2.
PROBLEMS
1. Prove(16.13). [Hint: Multiply the equation for — by (ÜM — and integrate over (px < .R), letting R -. co and using (16.7).]
2. Prove Theorem [Hint: (i) and (ii) follow from elliptic regularity and Sobolev's inequality. For (iii) and (iv), apply the maximum principle to
3. Suppose that 0 < I < I I I a* is suffi- ciently large, then the component of >0) containing does not contain any other a,.
576 VARIATION PROBLEMS WITH POTENTIALS
4. Suppose that 0 < lo and a1,... ,ak are fixed. If aA — I is sufficiently small, then ak and a,, - belong to the same component of
17. REGULARITY OF THE FREE FOR THE THOMAS-FERMI MODEL
In this section we study the regularity of the free boundary
(p>O).
If we keep l.a1 ak. fixed and let ak move from oo to a,,_1, then (by Pioblems 3 and 4 of Section 16) there should be an intermediate position of ak where the boundaries of the components of > 0) which contain ak_I and ak intersect. For this position of ak we cannot expect the free boundary to be everywhere smooth. Thus our main regularity result (Theorem 17.6) asserts regularity of the free boundary everywhere except for a "thin" subset.
We shall be working with the function w = — A, which satisfies
(17.1) (q=1 for some I <q < 2 (if <p < 2). But actually all our results extend to nonzero solutions w of
(17.2)
We begin by noting that for the solution w = ux — A of (17.!),
(17.3) {w = 0) no interior points.
indeed, if x° is an interior point, then Lemma 5.2 implies that w 0 in a 6-neighborhood of x0, where 8 is independent of x°. It then follows by continuation that w 0 in R3, which is impossible. Thus:
Leinthi 17.1. If x0 belongs to the free boundary afl, then there exists a homogeneous harmonic polynomial x) of degree n I such that
(17.4) w(x) = H,(x — x°) +
where
clx — (17.5)
(c>o,o>o)
for all x in some e0-neighborhood of x0.
REGULARITY OF THE FREE BOUNDARI' FOR THE ThOMAS-FERMI MODEL 577
To study the free boundary more deeply, we develop some notation for harmonic polynomials.
We shall denote points in R3 by X = (x. y, z). We denote by the space of all polynomials of degree n, by the space
of all harmonic polynomials of degree n, and by the space of all homogeneous harmonic polynomials of degree n. Since is a finite-dimen- sional space, any two norms are equivalent. This is true in particular for the norms
L2(B1)
and the norms
IIPIIA Q E AQ =
where B1 is the unit ball; notice that the last "inf" is "mm." We shall often drop the symbol in the sequel.
Ii is well that ."
(Pl,P2)L2(B,)=O ifP1 Es.,,,
we find that
A
Hence the inequality
(17.6) (C>O)
holds in any other norm of - I' We now introduce a nonnegative constant a = for E which -
measures the extent to which differs from a polynomial in two variables:
DefInition 17.1. Let
= — X°). homogeneous polynomials of degree k,
578 VARIATIONAL PROBLEMS WITH POTENTIALS
be the expansion of E into Taylor's series about a point X°. Then
(17.7) a = = inf max iix°n=i
Notice that
+ Xo)IILX(B)
= 2_I Qk(X)H1c0
Utilizing (17.6), we find that
(17.8)
Lemma 17.2. Let E s,,. Then there exist polynomials A, B in I,, such that
(17.9)
(17.10)
(17.11) OBII
where is a constant depending only on n.
Proof. Assume that a is attained at X° = (0,0, 1). and write
P(X) = k=O
y)z" y)((z — 1) +
= — + —
k=O k1s<k
n—I
y, z — 1) + Q,(x, y, z — 1), l=0
where
(17.12) = )ak(x, y)(z — k + s = I
Here aL is a homogeneous polynomial of degree n — k and Q, is a homoge-
REGULARiTY OF THE FREE BOUNDARY FOR THE THOMAS-FERMJ MODEL 579
neous polynomial of degree 1 [in (x, y, z — I )J. By the definItion of a,
(17.13) forl<n
(where the norm is taken for Q,(x, y, z), (x, y, z) in B1J.
s 0, k = n. Thus y) = Q0(x, y, z — 1). It follows that
(17.14) ha,, hILaO = a.
Similarly, for 1 1, n — k ± s = I gives s I or s = 0. The term corre- sponding to s I is a,,, which is estimated in (17.14). Hence the term corresponding to s 0 can also be estimated, giving
hIa,,_111 'Ca.
Proceeding step by step, we obtain
(17.15)
From the definition of H H A follows that there exists a a E A, such that a a(x, y),
Aa(x, y) y) —2a2(x, .p)
and
hIGH L" =
We now take A = a0 — a, B = F,, — A. it is clear that A E Z,,, IIBII Ca. Finally, a{A) = 0 with a(A) attained at X° = (0,0, 1).
Remark 17.1. From the proof above we obtain: If a(P,,) 0, then P,, P,,(x, y) in 'a suitable system of coordinates.
We shall now suppose that the origin 0 is a free boundary point and studs the behavior of w near 0. By Lemma 17.1,
(17.16) w(X) P,(X) + (6>0), 0
whereP,, 1; wemayassumethat6<1. In the next lemma e,, is a positive number such that
0
C,, is defined in Lemma 17.2.
Lemma 17.3. Suppose that
(17.17) a(P,,)<eflhIPflhIL..
VARIATIONAL PROBLEMS WITH POTENTIALS
Then, after a suitable rotation of the coordinates, the following is true: If
Vw(X')=O
then
(l;.18) (x2 + (Ca + lxi [C>0, a =
a suitable coordinate system
2 2 1/2A(x, y) = p = (x +y )
where C() is a real number. in view of (17.11), (17.17),
.l'j. —
II VB Ii C1 ii B (C1 constant),
c obtain
C2p"' —
where are positive constants depending on n (and C2 depends also on C0), and r = . Thus. by Lemma 17.1 (see (17.5)].
Vw(X') = 0, (17.18) immediately follows.
As usual, we denote by BA( X) a ball with center X and radiusA.
Lemma 17.4. Suppose that a = a( > O. Then there exists a neighborhood (C> 0) for which the following is true: If w( X°) = 0 where X° E then
(17.19) w(X) + O.(IX— X01k48)
for some k < n, where
Proof. Let
xo—, r
REGULARITY OF ThE FREE BOUNDARY FOR 11W ThOMAS-FERMI MODFL
Writing
r,(X)= IQk(X— x°),
and comparing the Taylor expansion, we find that
coefficients of Qk = times the coefficients of Qk.
Hence
(A> 1)
where C >0, c0 >0. Choosing k such that
and using (17.6), we get
(1740) II II L'°(LA(X°))
Suppose now that the assertion of the lemma is not true. Then
w(X)=111(X—X°)+ O(IX— X°r8).
It follows that
(17.21) corAl,.
Now,by Lemma 16.1, .
(17.22) CII +
Sibstituting the estimates from (17.20),(17.21) into (17.22), we obtain
Choosing A, (n + 8)/si, wd get a 2Cr,/", that r > for Some >0. This contradicts tho that X' that
C C1.
Let
(17.23) M = {X°;'w(X°) 0,
= 10)
582 VARIATiONAL PROBLEMS WITH P(YFENTIALS
where P, = is a homogeneous harmonic polynomial of degree n (depend- ing on X6).
Lemma 17.5. For any E there exists a neighborhood B80( X°) and a line segment L containing X° such that M,, fl B80( X°) is contained in the cusp-like region
tX; CIX— XOI4e>d(X, L)J,
where E = ô/n( n — 1), C > 0 and d( X, L) = distance from X to L.
Proof. Denote a(I?°) by ago. If
IX— X#'X°,
then, by Lemma 17.4, X MM. Thus it remains to consider the case where
IX—
We shall apply Lemma 17.3 with the origin replaced by X° and the axis corresponding to the z-axis denoted by L. Then (17.18) gives, for X fl B8,( X°),
d(X, CIX—
that is,
XoII+,.
We can now state the main result of this section.
Theoreml7.6.
(1) There exist a finite number of open C' curves F,,. .. , F, in R3 such that a manifold, where a = 1 /(p — 1) — 1.
(ii) The positive (negative) limit set of each 1 is a single point X' (XT). (iii) There exists a direction 17 at such that for X E F,, X near the
angle 0 between XX7 and 17 satisfies
0<CIX7—X$"
that is, X lies in a cusp-like region with axis 17.
Proof of (I). The constants C in Lemma 17.5 depend only on a lower bound on (IF, II. For any X' E denote by the polynomial F,, corre-
REGULARITY OF THE FREE BOUNDARY FOR THE ThOMAS-FERMI MODFL
sponding to = X'. From Lemma 17.1 we easily find that
II II
if X' is in a small neighborhood of X°. It follows that the assertion of Lemma 17.5 is valid for any X' E Ma fl B81(X° ), with 6,, and C sufficiently small and independent of X'. Thus, for any X' C fl E Me), there is a line segment containing X' such that (1 B8( X') lies in the cusp-like region
From this it is easily seen that
(17 24) = Lx2, then8 — x21),
wherea(t)-'Oifi-sO.
Now take for simplicity X° = (0,0,0), = z-axis. Then the plane z = (I I can intersect Ma. in a neighborhood of X° in at most one point; otherwise, (17.24) Is coiflradiâled. Thus the points of
ri X°) (82 sufficiently small)
lîç on where is a parameter of (We can take, for z.) If Ibe graph then, in viewof the
eate$oct graph, defined on some A-interval, is Lipachitz continuous. We can actually achieve a C' extension X(X) of X(A).. Indeed, take a
partition of the Ainterval ipto .n intervals of equal length and choose in c*ch subinterval one point of t(A), provided such a point We two adjacent points by a C' curve such that the at F,,
coincide with the directiong of Li,, respectively. Denotç this by X4A). By (27.24) the derivative of XJX) has a
modulus of continuity in A, uniform with respect to m. Hence a subsequence of XJA)-is unifonnly to a C curve Sinee,X(A) for a denscsset. of pojnts of Mu fl aC X(k)
It foliows that Ma is contained in a flnutc.uuioQ of pomla 1', the C' modulus of contjnuity depending on a lower bound on the norm H F,, II.
Next, for every otøO there finite number n0 such that M,, = 0 if n > n0. It follows that U,,M,, is contained in a finite number of C curves Outside these curves, vw Oon Since w the assertion (I) now follows 8y the iinplicit functi.n
the construction of a one can a say t,,,, is if'an endpoint, say
584 VARIATIONAL PROBLEMS WITH POTENTIALS
belongs to then we use it as a center of a new coordinate system in order possibly to extend The extension takes place if Y, is a limit point of points of not lying entirely in ,; otherwise, terminates at Y, ,.When we extend beyond Y, ,, its new endpoint is again taken to be in
To prove (ii) we may now parametrize any curve , by and assume that it
is not C' at its (say, right) endpoint; otherwise, there is nothing to prove for this curve. We then have to show that the points X(:) of r converge to a point X(oo) as I oo. Denote by the right limit set of I'd,, that is, the limit set of X(t) as.f —i oo. Denote by the polynomial at X = X(t). We claim that without loss of generality we may assume that
(17.25) 0 if r 00.
Indeed, denote by = 1,2,...) the sequence of points used in the construction of F,,, as 1 oo. Denote by t corresponding to
We shall'first show that we may assume that
(17.26) 00.
The extensiOn of , beyond takes place in a bail with center whose diameter r1 depends on If (17.26) is not true, then c > 0 for a subsequence of j's. It follows that two bails Bh, B12 (j, <j2) must intersect. From the construction of
a we see that if in the extension of
we use balls of radii 62/2, then the extension of F,, ,from the center point onward can be carried out in such a way as to give a C1 curve containing all the points of M,, in U B12. Thus will be a closed curve, for which the assertion (ii) then obvious. Excluding this case, we then must accept the validity of (17.26). -
Having established (17.26) and noting that the mapping X' -. continuous, the assertion (17.25) folloWs. . -
From (17.25) it follows that if a óurve is flot tip to the (say, right) endpoint, then any point in is in. M04,; this is impossible since M,, = 0 if n > n0. Thus each F,,0, is either a point or a C' curve up to the endpoints.
We next consider i curve it is also connected and compact. Each point of belongs toM,,0 fby(17.25)J and therefore it is either contained ioèálly in a C1 curve whoSe points lie in M,,0'or it is an isolated point. We now assume that
(17.27) -
Then, being connected, , locally a C' curve, and therefore also a C' curve.
By Lemma 17.4, a(P,,0) 0 along Theiefore, by Lemma 17.3, the set UM,, in a neighborhood of each point of must lie on a C curve, and
R}GULARITY OF 11ff FREE BOUNDARY FOR 11ff ThOMAS-FERMI MODEL
this curve must necessarily then be curve Since, however,
fl C M,,0_1 fl M,,, = 0
we cannot have points of in a small neighborhood of any given point of a contradiction. Thus consists of just one point.
Next we show that for each curve F,, 2j' if it is a C' curve open at one side. say when t 00, then its limit set must consist of a single point. Indeed, since is a connected, compact set, each of its points is a point of accumulation of the set. Suppose that there exists a point X E such that X E M,,0_,, and of more than one point. Then in a neighbor hood Vof Xthere cannot be any points of M,,0. By Lemma 17.4, 0 along (and therefore fl V C which implies by Lemma 17.3 and the construction of the that in some neighborhood W of X
fl M,,0_, is a C' curve (containing X) and its points are not in-the closure of Now we get a contradiction as before, because points of do in fact converge to any point on that curve.
Consider next the case where no points of belong to Then C M,,0 and we can proceed as in the case of
Proceeding in the foregoing fashion step by step, the proof of (ii) is complete.
Proof of (ill). Let X° be an endpoint of some curve in We may assumethata(P)= OforasequenceX= X, E F,, X°. ByLemma 17.4 it follows that a(P,,X°) = 0. Hence, by Remark 17.1, if we take X° 0. then
P,,(X) = a(x, y)
in a suitable system of coordillates (x, y, z). It follows that
(c>O),
where r = + and 17.1 gives
I vw(X)j>0
if X lies n an opening 9 about the z-axis and
kr8-1 for some small c0 >0. The same is trãe with to. any other idüsetion along which 0. Thus the poüi$s of U that lie in a small neighborhood of X° must he in cusp-like regions with vertex X° as described in the (iii).
VARIATIONAL PROBLEMS WITH POTENTIALS
From the assertions (i) and (ii) of Theorem 17.6, we deduce:
Corollary 17.7. For any e > 0, the (1 + e)-dimensional Hausdorff measure of the set U is equal to zero.
The proof of Theorem 5.1 can be applied to w in a neighborhood of any point where Vw 0. Thus:
Corollary 17.8. The portion of the free boundary is analytic.
PROBLEMS
I. Prove Corollary 17.7.
2. Consider the solution UA of the Thomas—Fermi problem (with p = 4) corresponding to some A 0. Prove that
[Hint: Consider
4'(x) = CR4(R2 — Ix — x012)4 inB(x0, R).
Show that —Aip + y(4') 0 in B(x0, R), and deduce 4' — A) in B(x0, R) if xol>2R, R >maxla1t .1
is. BIBUOGRAPJIICAL REMARKS
The study of sdf-gravitating axisymmetiic rotating fluids began in the eigh- teenth century; the initial interest came from astrophysics. A number of papers by for instance, are concerned with the question of the number of rings of a planet. The reader will find in Chandrasekhar's book [69] many explicit formulas (Maclaurin spheroids, Jacobi's ellipsoids, etc.). Existence theorems were first proved by Aucbrnuty and Beals 118) (the compressible case) and Auchmuty [17] (the incompressible case). The presentation in Section 3 is based on Friedman and Turkington [lOOdJ, who also derived the potential estimates of Section 2 in reference lOOa. The results of Section 4 are based on reference lOOa. P.L. Lions (193a, bi has extended the method of Aucbmuty and Beals, developing a general functional analysis argument.
The proof of analyticity of the boundary of the rings (Theorem 5.1) is based on the method of reference 124a. Lemmas 5.2 and 5.3 are due to Caffarelil and Friedman ISSe]; their two-dimensional analogs were earlier proved by Hart. man and Wintner [1131. For unique continuation resUlts for general order elliptic equations, see references 15 and 73. Theorems 5.4 to 5.6 are taken from Caffarelli and Friedman [58j].
BIBLIOGRAPHICAL REMARKS
Vortex rings with small cross sections were studied by Helmholtz (1858); see references 25 and 92. The first general existence theorem for vortex rings is due to Fraenkel and Berger [92]. They use a variational principle for the stream function; a variant of this approach (using existence of critical points) was developed by Ambrosetti and Mancini (lOa, b] and Ni (150), and a functional analysis generalization of this variational procedure is given by Berestycki and P.L. Lions [194). Benjamin [33] introduced the idea of minimizing his admissible class consists of all rearrangements of a given The present approach [with admissible class defined by (6.33)-(6.36) in l'heorein 6.2, and by (6.33)—(6.35) in Theorem 6.3] is due to Friedman and Turkington hOod. The results of Sections 6 to 8 and 10 are taken from reference lOOc; the present proof of Theorem 10.10 is due to Turkington (175aJ, the original proof is outlined in Problems 2,3 of Section 10.
In reference 175b Turkington extended the methods of reference lOOc to a wake model. The results of Section 9 are due to Caffarelli and Friedman [58m].
Fraenkel [91a, b) and Norbury [151a, bJ have constructed classes of solutions of vortex rings using methods such as nonlinear integral equations; they find solutions bifurcating from the Hill vortex (151a, bJ and vortex rings with small cross sections (91a, b]. Norbury [ISle) identified known explicit solutions of planar vortex rings as solutions of the variational principle of reference 92. It is not known whether the Hill vortex is a solution of a variatiOnal problem.
The derivation of the plasma problem equations can be found, for instance, in reference 170a, The existence of a solution of the plasma problem (11.1)—
was first established by Temam [llOa—c]; he proved Theorem 11.3 and (in reference 170b) Theorem 11.2 for n = 2. Theorem 11.2 for n 3 is due to Puel (155). Theorem 11.1 is due to Berestycki and Brezis [35a, bJ. Damlanian [76b] established Remark 11.1 (see also reference 35b). The results of Section 12 arc due to Kinderlebrer and Spruck [125] (see also reference 124 a). The results of Problems 3 and 4 of Section 12 are due to Gidas, Ni, and Nirenberg 11051. The material of Section 13 is taken from Caffarelli and Friedman [58m]. The results of Problems I and 2 in Section 13 are due to Baiocchi, Caffarelli, and Friedman [unpublished].
Bergcr and Fraenkel [36] and Keady [118b] have derived some asymptotic bounds for equations tsu = Ag(x, u) with A -, 00. Keady and Norbury [119b, c] have established a continuum of solutions (A, u) with A -. 00.
More complicated (and realistic) plasma models have been studied by several authors; but the existence theory is not completely established as yet. For details, see references 146a, b and 147 and the references given there.
Schaeffer [158d] proved nonuniqueness for the variational problem of the plasma if A > A2. Corresponding bifurcation results (mostly numerical) were obtained by Sermange [160).
The results of Section 14 are due to Caffarelli and Spruck [64). For n = 2, Acker (ld] studied the problem of minimizing given A1(G) = A, G convex, and established a solution for which the corresponding u [defined by (14.4)] solves the plasma problem.
VARIATIONAL PROBLEMS WITh POTENTIALS
Theorem 14.4 and the convexity of the level sets of u in Theorem 14.3 follow from the work of Grabnel liOla—ci, who also proved the convexity of the level curves for Green's function; see also Lewis 11361. Another proof of Theorem 14.3 was given by Brascaznp and Lieb [44]. The present simpler proofs of Theorems 14.3 and 14.4 are due
to Korevaar [129a], who derived the proof of Theorem 14.3 in reference 129b.
For history and references on the Thomas—Fermi model, see Lieb and Simon (139]. In this paper they prove Theorem 15.1. The results of Secjion 16 are based on Benilan and Brezis [31, 45cJ. They also establish for general V the estimate
The results of Section 17 are taken from Caffarelli and Friedman [58e]. Asymptotic estimates on ux — A are derived in Brezis and Lieb 1511 and in reference 139 (the result of Problem 2 of Section 17 is taken from reference 51). Another Thomas—Fermi model is studied in reference 51; here the solution p does not have compact support. A related model was studied by Benguria, Brezis, and Lieb 1291.
5 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
In this chapter we study several types of free-boundary problems ivhich are not formulated as variational principles. In Sections 1 to 5 we deal with gas in porous medium. The equation satisfied by the density function u is
(m>1)
and the free boundary' is 8(u > 0). It is shown thit there exists a unique solution. u, that u and the free boundary are Holder continuous. 'hi 6 to 9 we revisit the filtralion problem, but here the dam has
general shape. Exislçnce and uniquençss Of a solution and regulaqty.cf the free are etablihed. '.
ection 10 we two-phase °i 1. THE POROUS-MEDIUM EQUAflON AND UNIQUENESS
I I
law .4tatëS
where v is the vdocity of the gas andp is of mass is
590 SOME FREE-a0UNDAJY PROBLEMS NOT IN VARIA11ONAL FORM
where p is the density. With the standard isothermic equation of state p = cpT (c > 0, > 0) it then follows that
(1.1) (m>l)
where m = I + y and u is the normalized density. l'he function repre- sents the pressure (up to a constant factor). We introduce the initial condition
(1.2) u(x,O) = u0(x) (u0 0),
and wish to solve the initial value problem (1.1), (1.2) in some strip
QT=R"X(O,.T) or Q=R"X(O,oo).
We begin with some preparations.
DefInition 1.1. Let X be a real Banach space and let A be an operator from a set D(A) C X into X. A is called accretive (or monotone, when Xis a Hilbert space) if
(1.3) lxi — IPx1 — x2 + A(A(x,) — A(x2))II
Vx1,x2inD(A)andVA>0. If also
(1.4) range(I+AA)=X
then A is called m-accretive (or maximal monotone when X is a Hubert space).
Notice implies that I + AA is 1-1 and (I + XA)1 is a contraction. Observe also that if (1.4) holds for just one value A
A for fixed y E X, writing x + AAx = y in the form
(1.5)
x is a contraction with Lipschitz coefficient 1 — A0/AI< 1 that A > A0/2,and (1.4) follows step by step, for all A >0.
One can easily show that if S is a contraction with domain D(S) X, then I + S is m-accretive.
Consider the initial value problem
du =1(1) (O<t<T),
u(0) = x0.
ThE POSOUS-MEDIUM LQUAT1ON AND UNIQUENESS
One may try to solve this system by approximating with a finite-difference scheme with mesh size A,, We solve recursively
— X* i (1.8) +A(xZ)fk" (k= 1,2,...),
a—xo—xo,
wherefka = f(kA). The approximate solution u"(:) is defined by = x if kA t < (k + Since A is m-accretivc, the finite-difference scheme a unique solution for t <(k + 1)XN.
Theorem 1.1.
(i) Let A be m-accretive, x0 ED(A), f E L'(O, T; X) and f" -f in L'(O, T; X) as A,, -. 0. Then UR converges uniformly in [0, TJ to a function u(i) in C([0, T]; X).
(ii).. If okof(t)islipsdiitsconJIsuowar4x0 13(A), then uQ)is Lipschltz continuous.
(iii) If, wider the assumptions of (i).the problem (1.6), (Li) has a strong solution 0(t) [that is, v(t) Ls differentiable as X-valued function, 0(t) E D(A)and(l.6)holdsa.e., and IIv(t)— u011 -.0 ift -'01, then u = v.
(iv) If, under the assumptions of(i), u Is differentiable at a point, = to> 0 and is a L.besgsie point of f [that is,
'180 so—s
then u(t0) E D(A) and
+Au(t0)
Theorem 1.1 wcwriteu(t)
a weak cofitraction, that is,
— S(:)x111 — x111.
For more details and references on the proof of Theorem 1.1 and on nonlinear semigroups, see references 23 and 85a.
592 SOME FREE-BOUNDARY PROBLEMS NOT EN VARIATIONAL FORM
Consider the equation
(1.9) ev—Av=g (e>O).
It is well known (see, for example, references 32 and 168) that for any g E LP(R") (1 oo) there exists a unique solution v E of (1.9) and, denoting it by B,g, there holds
(1.10)
One can represent the solution in the form.
(1.11) ev(x) = —y))g(y)dy,
where the fundamental solution k is given by
k(x) (c > 0)
and "m is one of the classical Bessel functions. It is easy to check
(1.12) supk(x)<oo Vr>0,
(1.13) f k(x)dx<oo.Ixl<I Consider next the equation in R":
(1.14)
where
(1.15) •(:) is continuous, strictly monotone increasing, 4(0) 0,
and set = u, fi = Then the equation (1.14) becomes
(1.16) — + $(u) = g.
Lemma 4.16.2 (for n = 3) extends to any n (see reference 32). More precisely, for any g L)(RN) there eAists a unique solution u of (1.16) in if n 3, in {v E vv M2(R2)) if n and in {v E W'°°(R2), v" E L'(R')) if n = I. Further,
(1.17)
[the proof is similar to the proof of (4.16.6)).
THE POROUS-MEDIUM EQUATION: EXISTENCE AND UNIQUENESS
Define A by
Av = g —
where g E L'(R") and v is the unique solution of (1.14) (in a suitable class). Then (1.3) for this A reduces to
(1.18) — — A(Au1 — —
in the norm. But, as in the proof of (4.16.6),
f(f3(u,) — — — u2))h(u, — U2)
af(fl(u,) — /3(u2))h(u1 — U2)
if h(O) = 0, 0 h'(t) C. Taking h(t) sgn 1, (1.18) follows. Next, (1.4) (with A = l) follows_from Lemma 16.2 of Chapter 4. Thus A is
m-accretive. Notice also that D(A) L'(R") since q°(R") C D(A). We can now apply Theorem 1.1 and deduce the existence of a "weak"
solution in the sense of that theorem. It is easily seen that
(1.19) U — = 0 in6D'(Q).
From the fmite-difference scheme,
Wk+I +A+(w*+i) — A1
used to construct the approximation u"(t) and from (1.17) withp oo, we get
I IL"
so that
(1.20)
We summarize:
Theorem 1.2. a u of (1.19) in C([0, co); L'(R")) 11 and u(0) u0.
594 SOME FREE.BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
We now state a uniqueness result:
Theorem 1.3. If 4 satisfies (1.15) and if u, are two functions satisfying:
(1.21)
u, A4)(u) = —
(1.23) u — a E
(1.24) fTf[(u—z2)4i,.+ (4)(u)
x R"),
then u a.e.
Notice that (1.24) is a weak statement on the assumption of the initial values; it is a consequence of (1.22) and
(1.25) —u(x,t)IdxO.
Theorem 1.3 [with (1.24) replaced by (1.25)] implies that the solution established in Theorem 1.2 is unique.
Proof of Theorem 1.3. Set z = u — h = 4)(u) — Since 4) is continu- ous and u, t2 E L°°, for any > 0 there is a 8 > Q such that..I 4)(u(x, :)) — 4)(a(x, :))I> implies that u(x, t) — ü(x, ')l> 8. However, since u — ü E L', we obtain
(1.26) VE>0.
Notice next that
z—Ah=O -in6D'(QT),
that is,
(1.27) (z4', + ht&4')dxdt = 0 E
We fix y E and set 4' Then 4' E and 4' = 0 near 0 and near t = T. Sincez, h E (1.27) continues to hold for 4' in L'(QT) with 4' 0 near t = 0 and near t T provided.that belong to L'(Q1), which..is the case for our 4' = Bay.
ThE POROUS-MEDIUM EQUATION: EXISTENCE AND UNIQUENESS 595
Noting that — y, (B,y), = B,y, we get from (1.27),
0 = y,) + h.( B,y — y)] dx dt 0 R"
= fTf[(B)() ± (.EB,h — dxdz
(the symmetry of B was used in the second equality). Thus
(1.28) in'5D'(QT).
Since z, Bz belong to L' fl
g,(t) z(t))
is well defined a.e., where the notation i(t) z(, a'), (p. q) = has been used.
Suppose that we can show:
(1.29) Iimg,(i)=O a.e.int. i £0
Then it follows that zQ) 0 Indeed, 11w E L2(R"), theneO,w — w,andso
(B1w,w) (BewjtBew — âB,w)=
e -. 0 implies that eB,w 0 in L2(R') and (since VB1W-'øiflL2)
div(gTad B1w) 0 in
Consequently, w = — —'0 in '1Y(R"); and thus w = 0. Applying this remark to w = z(
a'), it follows that (1.29) implies that z( i) = 0
It remains to verify (1.29). For this we shall prove the lemmas.
Lemma 1.4. is absolutely continuour and — . I ' '.''.
(1.30) g;(t) = 2(EB(h(t) — h(i), •
•• :. .:
Lemma 1.5. There holds
(1.31). . 1
596 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Lemma 1.6. Letp E meas(x E Then,for any q E fl
lim(p, = 0. e—O
Suppose that the lemmas are true. Then we can easily establish (1.29) (thus completing the proof of Theorem 1.3). Indeed, by (1.30), (1.31), and zh 0,
g,(r) z(s)) ds.
The integrand is bounded by
also, IIz(s)IJ L' E L'(O, T). Noting, by Lemma 1.6 [here we use (1.26)], that
(EB!h(s), z(s)) (h(s), —'0 a.e. if E —'0,
the Lebesgue bounded convergence theorem gives (1.29).
Proof of Lemma 1.4. For simplicity we take E = I and set g g1, Bg B1g. Let z( a function of: fwe defined z( 1) = 0 outside
Thus
z8(:, x) — s)z(x, s) dr = * z
and p8(s) is a nonnegative function supported on (—6, 6), and = 1, p8(—s) = Then
d ,ia z8) = 2j
By (1.28),
where we define h = 0 outside in fact, this follows by applying both sides to any y E X (8, T — 8)).
From the previous two relations we have, for any E 6D(O, T) and small enough 8:
_JT(Bza(s), z8(s))L'(s) dv = 2f T(p8*(Bh — h)(s), dv.
TIjE POROUS-MEDIUM EQUATION: EXISTENCE AND UNIQUENESS 597
Since z8 z in L'( Q1-), liz8 ii Z ii if follows that
= 2jT((Bh — h)(s),
and the lemma follows.
Proof of Lemma 1.5. Let k(t)1 if i>b, k(i)—0 if t < a and k(i) = (t— a)/(b — a) if a < I <b. We can approximate k by smooth functions and then taking in (1.27) km(I)41(X, 1) instead of 4#(x. we get, as m —.
+ k4i,) + 0 = +
Taking a, b 0, we obtain
urn ff z(x, = 0 V4' E 7'); R"). h,a-.OU a
h>u
Since z E we can choose 4"s that approximate any e and thus deduce that -
(1.32) =0 b>a
From (1.28),
I) — s) = f'[EBeh(x. r) — h(x, r)) dT.
Multiplying both sides by E and integrating over x E R" and s (a, b), we get
x) dx — ffR..2 ( x, s )B,4i(x) dx d.c
= jbj a a
Taking a, b and using (l.32),.wc obtain
f'f (eBeh(x,r) —h(x,T))4'(x)dxd'r.
598 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Since 4' E C0(R") is arbitrary, it follows that
B1Z(X, t) = f'(eB.h(x, r) — r)) dr,
and consequently Hence
Jg,(:) 2:IIhIIL.41z(t)IILI.
Since II z(t)II L L'(O, T), the assertion ess lim,.0g,(:) = 0 follows.
Proof of Lemma 1.6. We can write, for any > 0,
fepB,q drj f + Ef I eBq{p>L)
Using the representation tormula (1.11) we find thai, for any 0 <r < co,
+ ?/2HqII1,t,f <r)
where C(r) is finite for any r > 0 [by (1.12)). Taking e -s 0, we obtain
lim sup k(y)dy. (Iyt<r}
Taking r 10, we conclude that -, 0 if e -. 0. Hence
and the assertion follows by letting 0.
We shall now specialize to the case of gas in porous medium, that is,
(1.33) = utm, rn> 1, u0(x) 0.
In this case it is useful to note that the solution can be constructed as a limit of classical positive solutions of the porous medium equation.
In fact, take sequences of balls BR and = 1,2,...) such
(134) R1—'oo,
THE POROUS-MEDIUM EQUATION: EXISTENCE AND UNIQUENESS
We assume that u0 fl and take a sequence E such that 0,
+
IIUoj — U0I1 L'(BR) 0 iff
Redefine 4(t) for such that 0 for all i and 4) E C2; denote this function by 4),.. Consider the initial boundary value problem
InBRX(0,00),
(1.35) 0n8BRX(0,00),
u=u01+e, 0nBRX(O}.
By standard results for nondegenerate parabolic equations [1 there exists a unique smooth solution u = of (1.35).
By the maximum principle,
(1.36)
If we multiply the parabolic equation by puf and integrate, we obtain, after noting that 0 on BR. X (0, oo) (v outward normal to
(1.37)
+ mp(p —
i)f urP-31 0 (1 < oo).
It follows that
(1.38) (1
Taking j oo and using (1.34), we conclude that for a subsequence, u1 u weakly in L'(BR)< (0, T)) and weak star in X (0, T)) for any R >0, T> 0, and u E L'(Q) (1 It is clear that u also satisfies (1.1), (1.2) in a weak sense:
(1.39) + dx di + 1R o(x)iP(x) dx = 0
for any * E X [0,
600 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
By Theorem 1.3. this solution must coincide with the solution constructed by Theorem 1.2. Ii follows that the full sequence u, is convergent a.e. to u and a E C([0.oc): L)(R")).
Remark 1.1. Replacer,. R by a corresponding R' and denote the corresponding solution by u, R• Then by the arguments above it also that
lirn R [taken weakly in X [0,
exists and is equal to the solution u.
Theiprern 1.7. In the porous medium case (1.33), for any u0 E L'(R") there exists a unique lunction U satisfying
E C([0.r.x). LI(Rlt)) fl X (r,oo))
u, — = 0 in X (0, oo)),
= u0.
The novelty here is that u0 is assumed to belong only to
Proof. To prove uniqueness we apply Theorem 1.3 to u(., z + h) and S(i)u(. h) for any h >0 and conclude that u(., t + h) = S(:)u(•, /i) if I 0. Taking h — 0, we get u(', 1) = S(t)u0.
To prove existence let u = u, R (e 0, R > 0) be the solution of () .35) when = . R, = R, = u11, R• We shall prove:
1.8. There holds
(1.41) I u(t) + C' (u = UOR)
where C. C' are positive constants depending an u'pper bound wi U0 IL'(R'r
. . .
Then, for a sequence R co, UOR u weakly in L'(BR X (0. T)) for any R >0, T> 0, and
(1.42) + C.
THE POROUS-MEDIUM EQUATION: EXISTENCE AND UNIQUENESS 601
To complete the proof of existence, it then remains to show that u E C([O, oo); L'( R")). This can be done as follows:
We first note [see the proof of (L20)] that V u1 in
(1.43) IS(:)uo — U1 I!.'(R"r
The same holds if is replaced by BR, where S(t)u, 0 on aBR X (0, cc). Let E L'(R") fl
(1.44) — —• 0 asj -• cc,
and denote by ÜJR(t) the solutions uftR(!) corresponding to the initial conditions u1. By Remark 1.1 and Theorem 1.3, ü,(z) exists as a weak L' limit and
(1.45) ü1(t) E C([0, oo);
Since (1.43) holds for BR, it also holds for R" with u1 = u1, where S(t)u0 coincides with u(i) and S(g)ü1 coincides with From this and from (1.44), (1.45) follows the assertion that u E C((O, cc); L'( R')).
Proof of Lemma 1.8. Set
Ue = U,R, V = 8R' E0 =
By Sobolev's inequality
vv 12 dx K(f I
where K is a universal positive constant, and 2* 2n/( n — 2) (for simplicity, we take n 3). Hence
! + 4kmp(p — 1) ( —! (m+p— I
Taking 0 we obtain, for u = u0 R'
I/a
(1.46)
witha> Lb >0.
From this inequality one can deduce (1.41); see Problems I and 2.
SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Corollary 1.9. The solution u of Theorem 1.7 satisfies
(1.47) u(x, t) I)n) (I U0 I)n)
'where C is a constant depending only on m, n.
Proof. From the proof of Lemma 1.8 we obtain the inequality (1.47) provided that
fuo( x) dx = I and z 1.
To prove it in general, introduce the function
I/(m—I) u(Rx,ox);
this is again a solution for any a > 0, R > 0, and
f ,i(x,0)dx= I
if
R2 I/(m— I)
I = R"( —.)
, where =
dx.
Applying (1.47) to t) withz 1, we obtain, with x' Rx,
u(x',a) C 12/(2+(ni—I)n) 0n/(2+(m— I)n)
Since x' R", a E (0, oo) are arbitrary, the corollary follows.
PROBLEMS
1. Write u0 = UOep when r = 0. Let u0 LPO(1l)and define 0€ (0,1) by
1_l—0 0 P Po
++b. Setting Ii oil, = Ii oil L'(O)' deduce by HOlder's inequality that
flu(I)IIQP÷b II 11(1—0)/I
PROBLEMS
lusing Then use (1.46) to get flurdx compute that
I/(ap+b—p)
I a(p—p0) a(p—p0) Y (ap0+b—p0)(ap+b—p)
2. Set
I/(ap +b—p)
41()=4(Xt i — i )),ap+b—p
Use Problem I to deduce that at any p = Po I,
a
Use this and C,, -. 4mK as p oo to compute that
a 1 0)— IogA+C
a — I op0 + b — Po
and choose A = z(ap0 + b — po) to deduce that
— a 0) logz+C.
ap0 + b Po
3. The function
k(m 1)— IxI Iu(x, i) = Thin 12k/n
I/(m—I)
where k = (m — I + is a solution of the porous medium equation (with initial data the Dirac measure). It is HOlder continuous with HOlder exponent a = min(1, l/(m — 1)).
4. The function
u(x,t) =[Ax2/(T_
604 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONALFORM
with A (m — 1)/(2m(m + 1)) is a solution of the porous medium equation with n = 1.
5. If v = 44u) = mum_I/(m — 1) and
v(x,t) (C>O,areal),
then u = '(v) is a solution of the porous medium equation with n I.
6. If u0 E L'(R') fl LP(R"), then the solution in Theorem 1.7 satisfies
m+p—1 I I q '0 1k'
2. ESTIMATES ON THE EXPANSION OF GAS
In the next section we establish the Holder continuity of the solution of the porous medium equation. INotice that the solution in Problem 3 of Section 1 is Holder continuous with precise exponent mm (1, 1/(m — l)).J Our result and the various estimates leading to it are all locaL For simplicity we assume that
(2.1) u0 fl
however, in view of Theorem 1.7, in asserting regularity for t > 0 it suffices to assume that u0 E L'(R").
In this section we prepare the tools for proving the HOlder continuity. We establish several lemmas of intrinsic interest regarding the manner in which gas, in porous medium, is expanding.
In Section 1 we have sháwn how to construct the unique solution u (of Theorem 1.2) as a limit of sequence of positive solutions. We shall need a slightly more refined version of this process.
Let u1( x) be a sequence of functions satisfying:
I
iulxl>R1 (R1-+ooifj---'oo),
u1(x)1 ifj I
I ifj—'oo.
ES11MATES ON THE EXPANSION OF GAS
Denote by t) the solution of the porous medium equatioir correspond- ing to Using standard regularity results for nonlinear parabolic equa- tions [1301 and the maximum principle in a strip [94a], one can then show that
(2.3)
( u1 ( x, t) —. 0 if x —' oo, uniformly with respect to
gin T
The solution can be obtained by solving the porous medium equation in cylinders 8R X (0, 00) with data
u1(x) oni0, On aBR.
The corresponding solution R then satisfies [see (I .38)J
t)dx
so that
dx —
dx
and R — 1/f 0 (by the maximum principle). Taking .R oo, we obtain
(2.4) fRfl(ui(x, t) _.!-) dx icf (uj(x) — dx,
and u1(x, t) I/f. By comparison [94a., p. 52]
UJR(X, :) UJ+1 R(X, :),
and thus
(2.5) u.(x,:)1ifjt
It follows that
u(x, z) lim
u(x, t) is upper semicontinuous,
606 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
and from (2.4) we obtain
(2.7) z) dx dx.
Thus u E L'. Since also
I +
by the maximum principle, the same holds for u(x, t). Thus u E Theorem 1.3 can then be applied. It implies that u coincides with the C([O, oo); L'(R")) solution constructed in Theorem 1.2. Thus u also coincides with the solution of Theorem 1.7.
In order to derive estimates or comparison results for u, we shall often derive them first for and then let j —, 00.
We shall be working also with the pressure function (up to a factor)
(2.8) v = mm_
It satisfies, formally, VUtm = uVv and
(2.9) v,=(m—
Set
(2.10)
and introduce the operator
(2.11)
If we apply to both sides of (2.9), then we easily deduce that, formally,
(2.12)
Lemma 2.1. The following inequalities hold:
(2.13)
(2.14) (m
—
(2.15)
the derivatives are taken In the distribution sense.
ESTIMATES ON THE EXPANSION OF GAS
Proof. It suffices to prove these inequalities for each function u1. Setting
m U" ,' ' -'
we have, by (2.12),
where the derivatives are takeit in the classical sense. Also, L(—k/i) = 0 and [by (2.3))
(xER")
if is small enough, depending on E. By comparison [using (2.3)) we deduce that Av,> —k/i in R" X (0, oo).
Clearly, (2.13) follows from (2.15), (1.1), and
= U + Vu Vt' uAv,
and (2.14) follows from (2.13).
In the sequel we use the notation
B,(x°)z (x;Ix—x°I<r),
Br=Br(O), IB,1meas(Br),
f w(x,t)dx.BAx°) IB,I RAx°)
Lemma 2.2. For any flo >0 there exist positive consiants 'p. c depending only on m, n, such that the following is true: Let x0 E R", R > 0, 0 <a 'p. If (2.16) v(xt°)=0 forx€BR(x°) and
(2.17) J 0
(2.18) v(x,t°+o)=0
608 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
The physical internretation of the lemma is that if the gas reached the ball BR/o(x) in time a and there was no gas in at time t0, then a considerable amount of gas has entered the ball BR(x°) in time + a.
Proof. For simplicity we denote x — x0 by x and t — by t. Consider the functions
= _?jv1( Rx, at).
Notice that satisfies the same equation, (2.9), as By (2.16). .(2.17),
(2.19) t3(x,O)0 mB1,
(2.20) f By Lemma 2.1
- - mk (2.21) V.> —ev,
no
(2.22) = —e.
The last inequality implies that
>o. 2n
That is, + r x 2/2n is subharmonic in x. The same is true of each Hence for any .x
t3,(x. 1) + 1) +
Since t)j, t3(x, i) for all (x, I), the dominated convergence theorem gives
i5(x,1) i,21x
ESTIMATES ON THE EXPANSION OF GAS 609
Using (2.20), we conclude that
(2.23) ifxEB,12
From (2.21) it follows that
1) t).
Combining this with (2.23), we get
(2.24) t5(x,t) ee(2nc + ifx E B112, 0< t < 1.
We introduce a comparison function
(2.25) V(r,z)=X[a21+ (rIxI ,X>0).
It satisfies
(2.26) V1 (m — + I wherever V> 0,
provided that
1 l)(n— i)at+( 1/3)
+
We choose a = Then V(r, t) >0 if and only if
11
Thus, for 0 < t < 1, 0 < r < 1, (2.26) is valid provided that A = A(m, n) is sufficiently small.
Next
(2.27)
and
(2.28) ,3(x,t)<Aa*<V(r,t)
by (2.24), provided that e and c are sufficiently small. Setting
= 1 = I )I/(nI_I)
610 SOME FREE-BOUNDARY PROBLEMS NOT IN VARiATIONAL FORM
we have
U, AUtm, = au" in B112 x (0, 1),
where the last relation is taken in the distribution sense. We claim that
(2.29) inB112X(0,1).
The proof is obtained by employing the sequence of smooth functions
(rn—Iu.=I V.' rn
If we multiply AUtm by — and au1/a: = by —(u1 — integrate over Br X (0, t) and add, we find that
(2.30)
—
}2
f tim)
j:H..<(O,)I — Urn) I (a1 — +f ([u1(x,O) — )2.
Takingin(l.37)p= I +m, we see that
00.
Hence if we integrate (2.30) with respect to r, r 4, we obtain
— U(r, )2
{(u,— {[üj(x,0) — (Ht,2\Bw12)x(0.t) 5/12 8,
By (2.27), (2.28), each of the two integrands on the right-hand side converges pointwise and monotonically to zero as j -, 00 [since u1(x, t)4 ü(x, :)J. Thus, lettingj -, oo we obtain, by the dominated convergence theorem,
[ü(x,t) — U(r, = 0, 5/12 B,
and (2.29) thereby follows.
ESTIMATES ON 11W EXPANSION OF GAS 611
From (2.29) we obtain
V(r,t)0 iflxl<L
which establishes the assertion (2.18).
Remark 2.1. Lemma 2.2 remains true if is any positive number, not necessarily small, but then c depends also on Indeed, this follows by the same proof applied to
t3(x, i) = to +
with small.
In the following lemma, which is in some sense the converse to Lemma 2.2, e is the positive number defined in (2.21), is a certain positive constant depending only on m, n and A, i', C are any positive constants satisfying together with e:
(2.31) , 'lo
where C(C*) is a positive constant depending only on m, n. These relations are satisfied (i) for any fixed A, C* if e is small enough and v is sufficiently large, or (ii) for any fixed r if A is sufficiently large and e is sufficiently small.
Lemma2.3.
R2' m/(m—l) (2.32) j:
8R(x°) a
then
R2 m/(m—I)
(2.33) Um(X0,10+XO)PC(__)
where c = and C0 is a positive constant depending only on ni, n.
The lemma expresses the physical fact that if a large mass of gas was in BR(x°) at time then the gas covers a neighborhood of x0 at time + g•
Proof. For simplicity we replace x — x0 by x and I — t° by I. Let
I/(m—I) u(Rx,at).
612 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIA11ONAL FORM
Then
(2.34) fu.m(x,o) dx p
We introduce Green's function (for n 3)
G9(r) — p2_fl — n —2
(p2 — r2) (r x I)
in the ball B9. It satisfies
Gp(p) ifO<r<p,
and
in the distribution sense, where y,, is a positive constant depending only on n, 'A indicator function of A, and 8(x) is the Dirac measure.
For n 2 we take
12 2—r ). Then, 1(0 <t < I, and if u is smooth,
f G1(r)ü(x, t) dx ftJ G1(r)ü,(x, s) dx B1 OR1
= 1'f0 B1
= üm(x,s)dxdS. 0 OB,
It follows that
(2.35) f G1(r)ü(x, i) dx _;f'um(O, s) + üm(X, s)dxB,, 0 08, if u is smooth. This inequality then holds for the solutions u, and, by approximation, also for the solution u.
Set
fu"(x, :) dx.
ESTiMATES ON THE EXJ'ANSION OF GAS 613
and consider first the case where
n rn (2.36) or >n—2 rn—i
Then
I/q (2.37) I {(u(x,lr)dx}h/m where 1/q + I/rn = I and
(2.38)
Thus we obtain from (2.35)
(2.39) f'+(s) dc Cif'um(O, s) +
where C1 are positive constants depending only on rn, n. Consider next the case where (2.36)is not satisfied, but
n rn (2.40) or >.
n—4 rn—i
Since G1(r) is bounded, away from r = 0,
(2.41) ü(x,i) 84\ 81/2
For x E B112, we can represent Üm(X, t) by Green's function G114:
yüm(x, t) = a,f üm(y, :) dy — f — x :) 4y.8114x) 8114x)
(Here again we tacitly assume that ü is smooth; more precisely, all the calculations are performed for and then in the final inequality we let j-.oo.)
Recalling that
= EÜ,
614 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
we get
- -m I/rnu(x.,)=[u (x,z)]
I/rn
sj 1/4
I/rn + Gi,4(Iy — I)u(y, t) 4Y]
B,,4( x)
+ C5ei/m
+f =8,14(x) )',, y,, where 1/q + I/rn I. It follows that (2.42)
f x f)u(x, t) dx +B,,,2
÷1 fB,,,3 B,14(x) The last integral has the form
(2.43) f Ô(y)ü(j!,t)dy Si 83/4
where, by (2.40),
f83/4 Evaluating then the last in (2.42) by HOI4er's
substituting (2A2) and (2.41) into (235), wó obtain
(2.44) + +
If (2.40) is not satisfied, but 1
n m or >
n—7 rn—i
ESTIMATES ON THE EXPANSION OF GAS
then we can proceed as before, but in evaluating the integral (2.43) we first express u(y, z) in terms of Green's function G1/8. It is clear that this procedure leads, for any n, to an inequality of the form
(2.45) ds C1ftum(O, s) ds + +
where C8, C9, 6 are positive constants depending only on m, n. By (2.34),
(vo=vIB1J).
By (2.13),
(2.46)
and consequently +'(:) —e+(t). Hence
(2.47)
Suppose now that (2.33) is not satisfied, that is,
Using (2.46), we get
(2.48) if 0<t<A.
Setting
(2.49) + C
and using (2.47), we then conclude from (2.45) that
(2.50) c9 +
Set
= f'+(s)ds.
Then
From (2.47) it follows that
(2.52) 4'(l) A:. A =
616 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
We compare 4'(t) with the solution,
t > to, x(:0) = A:0.
In view of (2.51), (2.52),
(2.53)
Now,
( — i\ =iX ki C is a constant determined by
—i _i\ m—lm—I' 1)
Since x(:) —, co if: -' C/Btm, we also have, by (2.53),
oO if: provided that A
which is impossible. Thus (2.33) must be satisfied provided that
C (2.54) +10,
Btm (m —
where the definitions of C, A were used. Choosing Io = A/2 we easily verify that (2.54) is a consequence of (2.31) and
the definitions of B, c.
Leannia 2.4. Under the same notation as in Lemma 2.3, if
(2.55) t0) dx — 1)/rn],
then
(2.56) v(x°, + )wi) (c0 = C(m 1)/rn]
Proof. Since
f v(x,t°)dx= m j um_I(x,tO)d.*m — 1 (
um(x, :0) dx
HOLDER CONIINUIIY OF Till SOIATI1ON 617
where C0 is a constant that depends only on m, n, Lemma 2.3 can immediately be applied to dedute (236).
PROBLEMS
1. Let u0, ü0 belong to L'(R"), u0 u0. If u(x, t), ü(x, t) are the solutions of the porous medium equation with the initial data u0(x) and ü0(x), respectively, as isserted in Theorem 1.7, then u(x, 1) ü(x, t).
2. The solution u asserted in Theorem 1.7 satisfies
t) dx = dx.
[Hint: Use the approximating ui.]
3. HOLDER CONTINUITY OF THE SOLUTION
Theorem 3.1. The solution u(x, t) of the porous medium equation is Holder continuous in every set R" X 00), > 0.
Proof. It suffices to prove the HOlder continuity for any smooth positive solution u1, provided that the Holder coefficient and exponent are independent of j. For simplicity we denote u1 by u.
We shall first prove a HOlder continuity-type inequality at one point (x°, t°) with > 0. Set
(3.1) Gk= ((x,t);Ix—x°I<Rk,t°—o(Rk)<.t<1°),
(3.2) forsomee>O
Let k be the positive integer determined by
(3.3) < U"(X0,
where c is a positive integer depending only on m, n, to be determined later. We shall choose
(3.4) Bk = o(Rk) = [a(R1) <no]
61S SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
for some I <a < 2, C> 1, and prove by induction that-
(3.5) Mk =
for all 0 k < k*. where e is defined by
(3.6) (a<2,r>O).
Notice that for any fixed positive integer k0 we may assume that (3.5) holds for k k0 since, without loss of generality, we may assume that u We shall choose k0 later to depend only on m, n.
To pass from k to k + I; set -.
B m/(sn-I)
(3.7) Ak=(C'2k )
and introduce the function
(3 8) ü"(x, t) = AkUm(Rk(X — x°) + x°, o€Rj(t — +
Then
(3.9) AkMk = = C,
where
(3.10) C =
Therefore,
(3.11) supüm=AkMk = C. d
By Lemma 2.3 and (3.3), (3.4), if k < k*then
(3.12) J u"(x, 10— 1) dx <pB,(x°)
provid that chum is sufficiently large depending only on m, n, C. Thus we can cirnose v such that
(3.13)
HOLDER CONTINUI1Y OF THE SOLUTION 619
and then choose c sufficiently large such that (3.12) holds. From now on, c and v are fixed.
Consider the function
w(x, = g0) = üm(x,i+ t°)
By (3.12),
(3.14) j B1(x°) C
and, by (3.11),
inB(x°,l)X(—1,O).
CAw = = _C1Rrt
(since — C1). Noting that
= Cl/rn! m
we get
(3.16) — Cl/rn! wh/m_Iw, 0.
If w, 0, then
— CI/m!w,
whereas if w1 > 0, then
— CAw — Ch/m! =0 m m
by (3.15), (3.16). Thus, in both cases,
(3.17) - CAw — Ch/m!w,
Representing w in B(x°, 1) X (— 1,0) in terms of Green's function for the parabolic operator in (3.17), we find that
<t<0,
620 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
then
w(x, t) + (1 — 0) +
9=0(x,t)E(00,1
where is a fixed number independent of k and C2 is a constant depending only on C1, m, n, a; we used here the inequalities (3.14), (3.15), and (3.17). We conclude, upon recalling (3.13), that
(3.18) I forsomeq>O,
provided k k0, where k0 is such that
<
depends only on m, n, We conclude that
—n), G
that is,
sup u" Mk(1 — 'i) = —
It remains to show that
— 2—(k+I)e,
that is,
1 —
In view of (3.6), this inequality reduces to
1 — ip <
We can clearly satisfy this inequality by choosing a < 2, 2 — a sufficiently small.
We have thus proved (3.5) for all k k*. It follows that
— x°I +,, —
as long as
1<10.
HOLDER CONTINUITY OF THE SOLUTION . 621
On the other hand, if
I x_xoI+Ir_tor<2k, 1<to
then, by (3.5) with k =
Um(X, z) 10).
Here is a generic constant depending only on m, n, We conclude that
(3.19) um(x, i) — x°I +11 — 10))
if t < By continuity (recall that u is one of the smooth functions u1) (3.19) remains true also if I =
If = l/(mMm_l), then
(3.20) if (71o>0). I
Indeed, the left-hand side is equal to
— 1u, = (i — — (I —
c*mum 0), and (3.20) thus follows. Let be the solution of
(3.21) 'Jo
= [i x — x01'. Um(XO, to)] in
Then tsee Problem I)
(3.22) ((x, z) C*max {(ix — x°I +11 — 1011/2)1 um(xo, t°))
where C* is some constant. Hence, by comparison (using (3.19) with i = and (3.21)), the same estimate holds for utm.
We have thus proved:
(3.23) U(X, 1) x — t — um( r°. 10))
C and r depend only on m, n, and an upper bound on I u
622 SOME FREE-BOUNDARY PROBLEMS NOT IN VARiATIONAL FORM
Lenima3.2. If(x',i1)E and if
(3.24) [ix' — x°i — 10p1/2]' t')
fori—Oori= I,ihen
i U"(X'. :1)
u"(x,t) Proof. Suppose that (3.24) holds for I = 0, that is,
(3.26) [i — I +11 — 1011/2} t°).
The inequality um(xI. 1') Cum(xo. 10) is then a consequence of (3.23) and (3.26). To prove the first inequality in (3.25) we take in (3.23) (x°, :0) to be (x', i') and (x,:) to be (x°, 10). Recalling (3.26), we then obtain
Um(XO, t°) Cum(xl, t').
Take now any point (x°. 10) with lo > and let k0 be a positive integer such that
(3.27) M(k() + Um(XO,
where M = suç Define
(3.28) hq)
where c is a positive constant to be determined below (independently of k0). if (x, 1) > then by (3.27), (3.23) and Lemma 3.2,
Um(X, t) — x°I +11 — t0I1/2)e
where depends only on C, c, Mm. Since the same inequality holds also for ,,m(XO g0), we obtain
(3.29) I U"(X, 1) — um(xo, g0) 2C(I x — x0 I
+11 — t0
if(x,t)
To consider the case where (x, t) E notice first that by (3.27) and Lemma 3.2,
(3.30) Um(X,g)
C Um(X,g)
HOLDER CONTINUITY OF ThE SOLUTION
provided that
— x°J +1 +
Thus, in particular, (3.30) holds provided that
(3.31)
Introduce variables
— (x — — (i — x—
A '
and let v(x', :') = u(x, t). Then
av (3.32) v(a(x , z')vv)
where a mum '/A If
1, It'I.< 1,
then (3.31) holds with (1, 1) = (x, 1) and, consequently, (3.30) is satisfied. Thus
c1 <a(x',:') <c2 iflx'I< l,Iz'I< 1,
where c, are positive constants (depending only on C, m, A). By the Nash- deGiorgi estimate 11301 we conclude that for some a, 0 < a < 1,
(3.33) u(x, i) u(x°, t°) v(x', :') — v(0,0) x' r + I"
= — + — loInI2)
if J x' Ji' where a constant depending only on c1, c2.
If (x, 1) E and c = A/2, then Ix' It' I' 1/2. and thus (3.33) holds. Also,
Hence (3.33) gives, for (x, :1 E
I u(x, t) — u(x°, go) C(I x — x° +11 — t°
—e)
with a constant independent of k0. Combining this with (3.29),'the proof of Theorem 3.1 is complete.
624 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
We shall now consider the continuity at t = 0.
l'heorem 3.3. If u0 is Holder continuous, then u( x, 1) is continuous at I = 0.
Proof. Denote by w(x, I) the solution to the porous-medium equation given in Section 1, Problem 3. Then for any c > 0, L > 0,
WCL(x, r) cw(Lx, cm_1L2t)
is again a solution, and so is
(3.34) I) = WCL(X —y, t + 1)
for fixed y E R'. Notice Lhat c and
>0
We shall now prove the continuity of u at (y,O). Assume first that u0(y) > 0. Then u0(x) > c if x — y 1< $/L, for some c > 0, L> 0. By comparison (Section 2, Problem 1) it follows that
t) u(x, t).
Consequently,
u(x,v)>0
for some small 8 > 0. Thus u satisfies
u,= v(avu) inB,,(R)X (0,8),
where R $/(2L), c, a c2 and c, are positive constants. Applying the Nash—deGiorgi estimate, we deduce that u is HOlder continuous in [0,6/2).
It remains to prove the continuity of u at any point (y, 0) for u0(y) = 0. For any e > 0 there is a 6 >0 such that
(3.35) u0(x)<E
HOLDER CONTINUITY OF THE SOLUTION 625
Consider the parabolic system:
iflx—yP<Ô, t>0,
z(x,O) 2e if Ix
z(x,t)M+e t>O.
l'his problem has a classical positive solution. We compare it with with j> l/e; in view of (3.35) and the inequality zi1 <M + Eon Ix 6,
iflx—yI<ö, 1>0.
Taking j 00 we obtain u(x, t) z(x, 1) and, in particular, limsup limsup z(x,t)=2e.
(x, t)—.(y.O) (x ,)(y0)
Since e is arbitrary, u(x,:) —' 0 if (x, 1) —' (y,O), and the proof is complete.
DefinItion 3.1. Set
(l(t) {x;u(x,r)>O},
r(:) =
1' is called the free bounda,y.
Theorem 3.1 implies that is an open set in X (0, oo) and 0(i) is an open set in The inequality (2.13) is equivalent to
0
and it therefore implies that
(3.36) 0(t) is increasing with :.
If, in particular,
(3 n = I, u0(x) continuous, u0( x) > 0 if a <x <b,
/ u0(x)=Oifx<aorx>b (—oo<a<b<oo),
626 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
then
(3.38)
where (— is a non-decreasing function of t.
PROBLEMS
1. that (3.21) implies (3.22). (Hint: Represent in terms of the fundamental solution.]
2. Assume that (3.37) holds and that u0(x) (b — x E (b — q. q > 0. Prove that > r2(0) for all i > 0.
[Hint: By comparison we may assume that v0(x) = (b — x)1 for b — 6 < x <b, where v0(x) = v(x, 0), v(x, 1), as in (2.8). For any small e > 0 let
t5(x, z) = + — (see Section 1, Problem 5)
where y = ;x + is tangent toy = v0(x) at Then v(b — 6,:) v(b — 6,0) + v,(b — 8, i)t > i5(b — 6,:)
for 0 < t < and by comparison (see the proof of (2.29)], O(x, 1) < v(x, 1) if b— 6<x<b, 0<1< T8. Show that t5(0,t)>0 if (c>0) and take £ —' 0.)
Assumethatu0(x)>Oifx E G, u0(x) = Oifx G, whereGisadomain in R" with C2 boundary. If, further,
(3.39) u0(x) (x E G)
for some y <2, then 0(0) C for all (>0. I Hint: Use the comparison function in (3.34).)
4. Assume that (3.37) holds and that u0(x) C(b — X)2/(m for some C >0 and x E (b — ,j > 0. Prove that = for all t suffi- ciently small. (Him: Use the comparison function in Section 1, Problem 4.)
5. Let the assumptions of Problem 3 be satisfied with (3.39) replaced by
u0(x) ctdist(x,aG))2/(m_fl (x E G)
for some C> 0. Prove that if G is convex, then 0(t) 0(0) for all t sufficiently small.
GROWTH AND HOLDER CONTINUITY OF THE FREE BOUNDARY
[Hint: Use the comparison function of Section 1, Problem 4 where the space variable is in the normal direction to a given point x° E HG.)
4. GROWTh AND HOLDER CONTINUITY OF THE FREE BOUNDARY
From Lemma 2.2 we immediately get:
Lemma 4.1. If in Lemma 2.1 v satisfies (2.16) and [instead of (2.17)J (x°, + a) belongs to the free boundary. then
(4.1) 3
This result and Lemma 2.3 will now be used to study the growth and Holder continuity of the free boundary.
By with a function of the form (3.34) we find [assuming that u0(x) > 0 in some open subset of that
(4.2) initIxi ,x€r(t)} (c>O,t> 1). We shall use the notation
a(x°, t0) — {(x°, t);0 <1<
iborem 4.2. For any point (xe, 1) E 1', either:
(1) Each point of t*) belongs to F, or (ii) No point of a(x, 1*) belongs to F.
Thus if the free boundary contains a vertical segment, then the extension of this segment down to t 0 also belongs to the free boundary. For n = I this means that once the free boundary began moving, it never stops moving. Problems 2 to 5 of Section 3 give sufficient conditions as to when the free boundary starts moving immediately from t = 0. In view of (4.2), the free boundary must begin to move at some finite time.
Proof. If the assertion is not true, then there exist points 0 < t' <.12 <1* such that
(x, F, (x*, tI) F. ft follows that
u(x,t')O ifxEBR(x)
SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
for some R > 0. Without loss of generality we may assume that — t' is sufficiently small. Applying Lemma 4.1 with a — z', we get
f( dx CR We can apply Lemma 2.3 with Aa = p provided that gS — 12 C(t2 — t') for some sufficiently large constant C (which may tainly be assumed). We then obtain the inequality v(x*, t) > 0, which con- tradicts the fact that (x*, 1S) E I'.
We can actually use the quantitative nature of Lemmas 2.3 and 4.1 to obtain a HOlder rate of growth of the free boundary.
Theorem 4.3. Let (x, 1*) E (1* > 0) be such that a(x*, g*) does not contain any points of F. Then
(4.3) u(x,:)O (,*_h*<1<ts)
and
(4.4) u(x,t)>0 (t*<:<t*+h*)
for some positive Constants C, y, hs.
Proof. andseth=1—i.Thenu(x,r)=Oif x E BR(x), for some R >0. Let
where E (0,1) will be determined later on. Suppose that
(4.5) dist(x, F(t')) <aR,
where a E (0, 1)is to be determined later. Applying Lemma 4.!, we deduce that
v(x, c(l — a)2R2
(c > 0), (I — X)h
wherex' E F(t')is such that (x' — Hence
2r 'V j-. v(x, t ) dx (1 — a)
—
GROWTH AND HOLDER CONTINUITY OF ThE FREE BOtJNDARY
Given any large C we can choose a, A so that
7 c(1 — a)2R2 —
.) (l—A)h ' Alt
But if C is sufficiently large (depending on m, n, 'Jo)' then Lemma 3.3 implies that u(x*, :*) > 0, which is of course impossible. Thus, if (4.7) holds, then (4.5) cannot be valid, that is,
(4.8) dist (x*, r(:t)) aR.
Notice that (4.7) is implied by
(I — a)"42 C(1 — A)
with another large constant C, provided that A> Taking a this reduces to
(4.9) (1 — C(1 A).
This inequality is valid for some A near I and y sufficiently large (for instance, A = I — 1/k, = k, k sufficiently large).
From (4.8) ii follows that
u(x,t')O jfXEBaR(X*), aA1. We can now repeat the previous argument with R replaced by aR, 7 replaced by t', and h replaced by Alt. We deduce that
dist(x*. 1'(t2)) aIR, t2= — X2h.
Proceeding step by step, we get
(4.10) dist(x*, r(i)) jfg — Akh.
By varying h in an interval h< h <h/A values of .R are bounded from below by a positive which we again designate by R) and taking k = 1,2,..., we find that (4.10) holds for all 1. t — h <i < :; more precisely,
dist(x*, F(t)) c(t g)7 (c>0),
which gives (4.3). To prove (4.4), notice that if a point (x, 1) of 1' satisfies
(4.11) Ix_xsJ<C(:_.I*)Y, (/smaH),
SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
then the segment o(x, z) intersects the interior of the set for which (4.3) holds. Thus a(x, z) contains points not in I'. By Theorem 4.2 it follows that a(x, 1) does not intersect I'. We can therefore apply the proof of with (x, replaced by (x, z) and conclude that u(x*, t*) 0. Since, however, u(X*, g*) = 0, we get a contradiction, which proves that the points (x, 1) satisfying (4.11) do not belong to F. It follows that on the set (4.11) either u 0 or u > 0. The first possibility contradicts the assumption that (x*, 1*) E 1'. Thus u > 0 on the set (4.11), which completes the proof of (4.4).
Remark 4.1. The constants C, y in (4.3), (4.4) depend on the choice of r, R, but are independent of (x*, 1*).
We shall now assume:
G is a bounded domain in R" with C2 boundary,
(4.12)
p<2.u0(x)0 By Problem 3 of Section 3 3 fl(0) for all : > 0. We can therefore
choose i, R in the proof of Theorem 4.3 to be independent of (x, provided for a fixed 0. By Remark 4.1 we conclude that
(4.13) C and y are independent of (x', is).
Theorems 4.2 and 4.3 imply (when (4.12) holds) that the free boundary is given by a function
I = S(x) and, if S(x5)
(4.14) S(x) —
S(x) < oo for all x E R". We summarize:
l'heorem 4.4. 11(4.12) holds, then:
(i) The free bowzdaiy is given by a function y = S(x), where S(x) is finite valued and uniformly continuous in every set (x; > 'lo >0.
(ii) The interior of F(: + s) contains (Cs of the interior of r(:) provided that < 1, i > ('lb > 0), where C, y are positive constants depending only on 'io and U0 IL'
We shall complement (ii) by proving:
Theurem 43. F(: + s) is contained in (Cs'12)-neighborhood of where andCisaconstant depending only on andIuOILI
GROWTH AND HOLDER CONTINUITY OF THE FREE BOUNDARY
Proof Let u(x°, gO) 0, dist(x°, r(t°)) a and consider the function
U(r,i) = {A[a2(t — g0) + a(r — (b>0,a >0)
fort°<t<t°+h, where
h<t0+&
U(r, > 0 if and only if r> b — a(t — As in the proof of Lemma 2.2.
provided that
(4.15) A[(m — 1)( — 1)a(t — ,0) + r — b
+
We choose b < a and
(4.16) a(a — b) = N, whereN1 + 1.
Then on = t°, r a,
andong°<g<g° +h,r= a,
— b)lh/(m_I) N1h/(m-I) >
By comparison (see the proof of (2.29)), we then get u U in X + Is); in particular,
(4.17) u(x, 10 + U(x, t0 + h).
We now defme b by
(4.18)
Then a, defined by (4.16), is given by
(4.19) a = —.
632 SOME PROBLEMS NOT IN VARIATIONAL FORM
The free boundary of U is given by r = b — a(t — 1°), and at I = 10 + h,
—aha--2j11i.
Choosing
(4.20) h 16N
so that
(4.21)
we conclude, upon recalling (4.19), that
u(x,t°+h)0 if Ix—x°!<b—ah=
This establishes the assertion of the theorem, provided that A can be chosen (independently of a) to satisfy (4.15). It suffices to choose A such that A and
= / r 2(m — 1)(n — 1) jl'.
Since r b — = a/2, (4.22) is a consequence of
(4.23)
here we have used (4.18), (4.19). In view of (4.21), (4.23) is equivalent to A Thus we choose A = min(y,3).
Remark 4.2. If (4.12) holds, then Theorem 4.5 implies that
(4.24) (c>O)
if x1 — x2 is sufficiently small and x', x2 do not belong to
We conclude this section with a theorem on the asymptotic behavior of u(x, 1) as I -. 00. The special solution in Section!, Problem 3 will play a basic role. We scale it by setting, for any L >0,
(4.25) VL(r, i) G( ii). r
PROBLEMS
where
G(s) ([/32 — c2s2]+ }I/(m_I) k(rn— 1)
and /3 is a positive constant such that x I) 1. Notice that
(4.26)
and that the supporLof the function r -' VL(r, t) is given by
(Li)" r
Also, for any x0 E R", i real,
(4.27) :*[VL(Ix_x01,:+,.)_ VL(JXI,t)]40
if t —. oo, uniformly with respect to x, x E R".
Theorem 4.6. Let u be the solution asserted in Theorem 1.7, and set I JR 0(X) Then
(4.28) i' ju(x,i)— VL0(r,t)I-*O ast—'oo,
uniformly with respect to x in any set x < C> 0, where L0 = Jm
Thus for large times the gas behaves as if it were initially concentrated at the origin.
The proof is outlined in the following problems.
PROBLEMS
1. For fixed L >0, t >0, if e is positive and sufficiently small, then
(Li )k/'a VL(r,t—e)>VL(r,,) forO<r<O
VL(r, t — e) < VL(r, t) for 0 <r <
E(O,l).
634 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
2. In Problems 2 to 10 it is assumed that u0 E C0(R") and u0(0) > 0. Show that
(4.29) VL,(r, t + <u(x, z) < VL2(r, t + r2)
for some L, > o, >0.
3. In proving Theorem 4.6, we may assume that for any L >0, r real, 1 >0, I + i >0,
u(x, t) t + r) (x E Re).
4. For fixed : > 0, denote by I, the set of points (L, T) such that L > 0, 0 and u(x, t) VL(r, I + r), and let
L(t) = sup r) e
Prove that there exist (L*, i) E such that L(t) and r* C(t + 1), where C is a constant independent of t. IHinl: By (4.29), L1 L(t)
5. L(:) is monotone nondecreasing and there exists a sequence t oo such that
(4.30) L(:1) <L(i1+1).
IHinI: If L = L(:0), there exist E> 0 such that u(x, + + r+E)for
Lk/n(10 + T + c/ft
r < where 0 <6< 1, 1 — & suffi- ciently small. Deduce that
u(x,10+13)> VL(r,tO+'r+l)) ifr<z01r0
for any > — small, E (0, 1). By Problem 1,
u(x,10+q)> VL(r,ro+r+,1—e) (e>0)
in thesupportofr-. VL(r,tO + r +tj — e)ifeissmallenough,and thus
u(x, + i)> VLI(r, + + tj — e) for some L' > L.]
PROBLEMS 635
6. If UA(X, 1) = A > 0, then for any sequence 'r oo there is a subsequence A. t oo such that
UA(X, :) —. w(x, :)
uniformly in compact sets of R" X (0, 00).
7. Define Then, for any t > 0,
(4.31) w(x,t) = VL0(r, a' + C:.
I Hint: u(x, a') t + implies that
uA(x, t) + TA,,)
with
TA,t ' giving "in (4.31). If" = "is not here, then (see problem 5)
w(x,t+q)>
on the support of r Vj, for some £ > L0, and by Problem 6,
+ ii)) >L
if A1 is large enough.)
8. Show that f, =
9. Show that 0. [Hint: For any small 6 > 0, UA(O, 8) ? C/6" by (4.29) (C> 0). Also,
8) w(0, 8) = VL0(O, 8 +
10. and VL0(r,t) as Atoo, uniformly for (xt) in compact subsets of R" X (0, oo).
II. Complete the proof of theorem 4.6 when support u0 is bounded. [Hint: Work with u(x — x°, t + e), using (4.27).]
12. Prove Theorem 4.6 for general u0 in Lt(R1). [Hint: Let = u(x) if x < N, 0 if f x > N, and denote by u"(x, 1) the solution with initial data us'. Set t)
636 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
By Corollary 1.9,
ux(x, r) f.112k/n
and, by Theorem 3.1, for any sequence t oo there is a subsequence 00 such that ux (x, t) —' w(x, 1) uniformly in compact subsets. of
X (0, oo). By Protlem 11, u"(x,t) t) as 1 00, where LN t L0 = 1 '.Also, ii ui" implies that w VL and thus w VL. Useihe relation
fw(x, :) dx lim inf JUA(XI t) dx =
5. THE DIFFERENTIAL EQUATION ON THE FREE BOUNDARY
In this section we shall prove that the rate of change (in space) of the pres3ure at the free boundary is equal to the rate of growth (in time) of the free boundary.
We fix a pointy (r >0) and set
(5.1) h(t) = max {R; v(x, t) 0 in BR(y));
that is, h(t) is the distance from y to £l(t). By Theorem 4.5,
(5.2) h(t)>O insomeintervalo<t<r+80(a0>O).
We introduce the function
(5.3) y,(E) max V(X,t)
C
and set
(5.4) y,(:) = Iimy,(e)
if the limit exists. We shalt prove in this section the following theorem.
Theorem 5.1. (i) The li:,ill in (5.4) exists, (ii) h'(t + 0) exists, and (iii) the following relation holds:
(5.5) = h'(t + 0) (0 < I <t +
THE DIFFERENTIAL EQUATION ON ThE FREE BOUNDARY 637
We shall divide the proof into several lemmas. It is clearly sufficient to establish the assertions of Theorem 5.1 at the point r. For simplicity we take y = 0. We also set
R = h(r), y(e) = yT(e), y =
Thus
v(x, i) (5.6) = max
XEBR+, E
(5.7) y.= limy(e) if the limit exists. e-.O
Lemma 5.2. The limit in (5.7) exists.
Proof. Suppose that n 3. Consider the functions
(5.8) M(r) = R(R2 — r2"), rIxI
+C(R+e—r)(r—R) (C>O)
for <r < R + e. Then
—c ifeissufficientlysmall,sayife
Also,
w(r) = 0, v(x, i-) 0 on aBR,
w ey(e) v(x, r) on BR+, [by (5.6)].
In view of (2.15),
inR<r<R+e provided that the constant C is suitably chosen (namely, C> 2k/i-). Applying the maximum principle, we conclude that
(5.9) ifR<r<R+e.
Taking the maximum when x E BR+8, where 0 <8 <g, we get
&y(8) + C(e — 8)8.
SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Thus
1 6y(&) ey(e) + C(e — 8)8
5.0 M(R+8) M(R+e) M(R+8)
Since M(R) = 0, M'(R) 1,
C(e_S)&<2C(E_B) M(R+6)
and, consequently, (5.10) implies that the function
(5.11) g(e) =
+ 2Ce is monotone increasing.
Consequently, lim..Og(e) exists, and thus also the limit in (5.7) exists. For n = 2 the proof is the same, except that M(r) is now defined as
M(r)RIogj.
Take a sequence e,,, JO and points xm in BR+, such that
(5.12)
For any 0 < 1 introduce the cone
K,
K, is a cone with vertex x0; it lies in the exterior of BR (except for its vertex), and K, increases, as p JO, to a half space whose boundary is tangent to aBR at xo.
Lemma 5.3. For any p >0 there exists a function g,(t) satisfying
g(t) (5.13) - —'0 i/i JO,
such that
(5.14) v(x, r) — (r — R)y = — R) :fx E K,.
Proof. Since
v(,'r)
THE DIFFERENTIAL EQUATION ON 11ff FREE BOUNDARY 639
weget
(5.15) v(x, i) — (r — R)y o(l)(r — R).
Thus it remains to establish the lower bound
(5.16)
where
-+O
Fix a point z and set
l 8=p1—R.
We shall prove that an inequality of the form
(5.17)
cannot hold if c is a fixed positive constant and 8 is arbitrarily small; this will clearly establish (5.16).
We shall work with the scaled solutions
v8(x, t) = + 6(x — x°), + 6t) (0 <8 < 1)
of (2.9). By (5.15),
(5.18) v(x, .r) C0dist(x, BR).
We compare v(x, I) forT <1<1 + IxI< R1 (R <R1 <R + 1) with the radial supersolution (see (2.25)]
(5.19) V(r, 1) = — r) + C1(r — R)]4 (r tx
where C1X = 2C0, C1a1 <min(l, R/2) and X = X(m, n) is sufficiently small. In view of (5.18),
iflxl<R
with strict inequality on R1, so that
V(r,t)>v(x,t) r<t<r+a1
SOME FREE-BOUNDARY PROBLEMS NOT IN VARIAIIONAL FORM
for a suitably chosen small Hence, by comparison,
o(x, i) — + C,(l x —R)] +
It follows that
v8(x,i) + C1 Jx —
Recalling (2.14) we can extend this inequality to t <0 (with a different constant C,). Thus, for any A >0,
(5.20)
C is a constant independent of 8.
Notice also that
(5.21) Av8
With these two inequalities at hand, we can now apply the proof of Theorem 3.1 and deduce that
v8(x, t) is uniformly Holder continuous (5.22) bounded sets A, A, with coefficient
and exponent independent of 8.
Set
u*=x0_ç,
and denote by E" the spheres with center 0* and radii R/O and (R/6) + I, respectively. Then x0 6 1' and z' I". In general, when x varies in (Z"), x0 + 6(x — x°) vanes in (aBR÷S). When x0+ &(x — x°) varies in the cone K,, x varies in the same cone.
Denote by G the domain bounded by s". Denote by Q0 a cylinder whose riG. Wecantakethe
diameter of Q0 to be a positive constant A1 depending only on ,i —' co if 0).
SetQ=Q0fl G. For any point x and a set B, write
d(x, B) = dist(x, B).
Then (5.15) gives, for x 6 Q,
523 v8(x,O) +
( . if8-.O.
THE DIFFERENTIAL EQUATION ON 11ff FREE BOUNDARY
Consider the harmonic function (we take, for definiteness, n 3)
k(x) = (y +tp(6flA"'
(A =
where
C0> 0.
It satisfies, on
ak a2kk0, --=y+ip(8), i4)
Recalling (5.23), we conclude that, provided C0 is large enough,
(5.24) k(x) max {v8(x,O), in Q.
Consider the function
(5.25) I —p)(p—A) (C0>O).
If 6 is small enough, then < —C08. Choosing sufficiently large and recalling (5.21), we get
(5.26) in Q.
Clearly also,
(5.27) in Q,
by (5.24). Suppoee now that (5.17) heMs; we shall derive a contradiction. From (5.17)
we have
v8(z,O) <y c
and, by (5.22), if z1< c0, then
(C,>O,a>O).
Since — z <c0 implies that d(x,I')> I — c0, we then obtain. using (5.27),
— c0) = y— v8(x,O) = ifjx — x*J<co
provided that c0 is small enough.
642 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Recalling (5.26), (5.27) we can apply the maximum principle to the function w(x) v8(x,O) in Q and deduce that
(c2>o)
where v is the inward normal; more precisely,
(5.28) inQ.
It follows that
1 —p)(p—A)—w
inQ.
Taking in particular x = x° + 6(xm — x°), we get
V(Xm, r) (y + + c0& — C2)d(xm, aBR)
for all m sufficiently large. Thus, by (5.12),
which is impossible for m sufficiently large. This completes the proof of the lemma.
Lemma 5.4. The following inequality holds:
(5.29) liminf h(s)—h(rfl
Proof. Take the same v8(x, i) as before, where & is any small positive number. In view of (5.22), any sequence 8,,, then has a subsequence 8,,,. such that
530 o8-*v0uniformlyin(x,t),
( . for any A >0 (8 = 8,,, jO),
and v0(x, 1) is a solution of (2.9). By Lemma 5.3,
—
8 8
where r(x° + 8(x — x°)); the last term tends to zero as 8 —. 0, whereas —. —x,, if the axis of the cones K is taken to be the negative It
ThE DIFFERENTiAL EQUATION ON ThE FREE BOUNDARY
follows that
v0(x,O) =
Let B be a ball with radius R0 lying in <0 with tangent to x, = 0 at the origin. We shall compare v0 with a suitable solution as in (3.34); we denote the corresponding pressure function by W. We choose and R0 such that the support of x coincides with B and
aw at the origin.
uxn IJxn
Since W(x, 0) is concave, whereas v0(x, 0) is linear,
oni0. We can now compare both solutions in a cylinder x <A, t > 0 with A > R0. We conclude that W v0. Hence, if we denote by h0(z) the functions h(t) corresponding to Wand v0, respectively, with respect to any pointy0 lying on the positive xe-axis, then
(5.31) h0(s)
One can check that
(5.32) = — ys + o(s2) (s -. 0).
From (5.30) it follows that
ch0(s) + a(&) —.Oif5 = am'
provided that h8(s) (defined with respect to v8) and h0(s) are computed with respect to the same center y0; o(8) converges to zero independently of the choice of y0, if, say', Yo > > 0. From (5.31), (5.32) we therefore get
(5.33) h8(s) h8(O) — ys + o(8) + Cs2.
We have thus proved that for any sequence there is a subsequence 8m' such that (5.33) holds for all 8 = 8m.; C is independent of the sequence. It follows that (5.33) holds for all 8 sufficiently small.
From the definition of v8 we see that
(5.34)
provided that y0 (above) is suitably chosen. Using (5.33), we obtain
h(r + 8s) h(r) — y&s + 8o(6) + Cs26.
644 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
For any small s, we set 7 + 6s = I and deduce, as & —' 0, that
1r S Since s is arbitrary, (5.29) follows.
Lemma 5.5. The following inequality holds:
h(z) —h(T)J (5.35) lim sup
Proof. From Lemma 5.2 we deduce that for any e >0,
iflxl<R+do,
where d0 is a positive number depending on e. Let V(r, i)(r =IxI)be the radial supersolution (5.19) with C1 = (y + e)/X.
Then
V(r,i)>v(x,i-) ifR<r<R-fd1
provided that d1 is sufficiently small. By continuity, also
V(r,i)>v(x,t) ifrR+d,, provided that o is sufficiently small. We can now use comparison to conclude that Inparticular,
o(x,t)0 —R — (y+e)(t—i), which gives
t j, r, (5.35) follows.
Proof of Theorem 5.1. The assertion (1) follows from Lemma 5.2. From Lemmas 5.4 and 5.5 we deduce that h'(T + 0) exists and it is equal to —y; thus the assertions and (iii) follow.
We shall give an application of Theorem 5.1, assuming that
(5.36) h(t)>h0>O
THE DWFEREN11AL EQUATION ON THE FREE DOUNDARY 645
for some positive constants h0, This condition means that the gas does not "close in" on the point y as long as t +
Theorem 5.6. There exists a positive constant K depending on h0 (but not on 00) such that
(5.37) d(y, — d(y, —
provided that < <t2 <i +
Thus the free boundary grows at most linearly, as long as it does not "close in" on any point.
Proof Let g(x) be the solution of
Ag= —c ifh(t)<Ix—yI<h(t)+ I,
gO iflx—yI=h(t)+1, gN
where N sup v. Then g(x) majorizes v(x, t)and, consequently,
h(t).
In view of (5.36),
(5.38)
where K may depend on h0. It follows that
and, by Theorem 5.1,
(5.39) jh'(t +
Since h(i) is monotone decreasing and right continuous (by Theorem 4.5), by a standard theorem in real variables,
h(t2)— h(s1) =ffzh#(: ÷ O)dt
—K(t2 — i,) <:2)
[by (5.39)J, which implies (5.37).
646 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Remark 5.1. If we do not make the assumption (5.36), we only get
(5.40)
Theorem 5.1, in the special case n = 1, gives the following:
Corollary 5.7. Suppose that (3.37) holds. Then, for any I > 0, — 0, 1) and t + 0) exist and
(5.41) :) = + 0).
A similar conclusion is valid for
The assertion that t) — 0, I) exists is a consequence ,of — C lestablished in (2.15)1 since then + Cx is monotone increasing in x.
We conclude this section with a stronger regularity result on at).
Thçorem 5.8. Suppose (3.37) holds. Then I) is continuously differentiable for all!> t7, where 1" = sup (1 > 0; = More precisely,
(5.42) = + fort>
where E,(t) is a C' convex function and tj,(t) is a C1' function. The proof is outlined in Problems 7 to 13.
PROBLEMS
1.
[Hint: v, + is increasing in x E fl(t) and bounded on r(t) since by (5.11), (5.7),
I M(R+ + 2C.]
2. Set and take for simplicity = 0. Let be a solution of the form (3.34) whose free boundary x is tangent to x at I =
> 0) and w > 0; here —2P, P nonnegative constant (by (2.15)). Then
(t — — — (i
PROBLEMS 647
Compute
= — (y positive constant depending only on
= f"(t0) + 0(t —
and deduce that
(h>O) h2/2
satisfies
+ 7Pr'(t + 0) 0(h) at any point t0.
3. Prove that for some finite positive measure
(5.43) +
in the sense of distributions. (Hint: —C, C. Take a sequence weakly, where is a signed measure.]
4. Prove (5.42) with convex and tj. in [Hint: Take fJ$ dT,
+ + O)(t — 8) — — dr.]
5. Deduce from (5.43) that
— + >
and consequently, ±0) c >0 if >0.
6. Let d(i, :) denote the distance from i) to the free boundary x = t) belongs to N8, a 8-neighborhood of t0) intersected with
(v>0}.Then
[Hint: c,d(x, 1) — c2d(x, t) and
v(x, t) t) — - -P(x
SOME FREE-BOUNDARY PROBLEMS NOT IN VARIA'IlONAL FORM
7. Show that
inNa.
[Hint: Let S be a square with center i) and side y i)/2. Consider
i5(x,i)=-1-v(yx+i,yt+i) inlxl< 1, 111<1.
It satisfies
Apply the Nash—deGiorgi estimate and then Schauder's estimates.]
8.
t) p + a(e) in
wherea(e)—*Oife —.0. [Hint: If > $ + 8, (xe, (x0, so). then
v(x, — ($ + S)(x — — P(x — (x< x0),
v(x, — (fi + &)(x — x0) — P(x — x0)2.
But t0) -. fi if x x0.J
9. + a(e) if (x, EN,, whereø(a)—.0 ifs —*0. (Hint: If p2 + 8, e,, = x0 — x3, then
to + as,,) = v(x,,, + t0)aç, +
— ÷ (p2 + 8)as,, — Ca2e,,. Also,
= —(fi+aj(x,,--x0—aej+e,,o,,)
Choose a <C/8 to derive a contradiction.]
10. Take for simplicity to) (0,0) and let
I v.,(x, t) = —v(yx, ye).
PROBLEMS
Then, for a sequence y 0, t) —' V(x, 1) uniformly in compact sets, where V is again a solution with free boundary x =
- ift>0
=fi0 ift<zO
and fi + 0), Pb — 0). Let w(x, t) = fi(fit — To com- plete the proof of Theorem 5.8 [that is, that E Cj, it suffices to show that V w.
11. Show that V(x,0) = w(x,0). [Hint: Otherwise, since 0, there exists a solution w0(x, t) — - x — with /3> /3 and >0 such that V(x,0) w0(x,0) and, by comparison, V w0 if: 0. But then > fit if: > —
12. Prove that V(x, —1) w(x, — 1). [Hint: Apply Problems 7 to 9 to v, (-y —+ 0) to deduce that
liminf{V(—M, —1)— w(—M, —I)) M--
tim inf{ —1) — — I)) 0 M-.oo
and note that
V w.
[Hint: By Problem 12 and comparison, V w if t> —1; now use Problem 1 1.)
14. Prove that is continuous up to the free boundary x = a:), t > [Hint: Use Theorem 5.8 and —2P to estimate :)
to) — 0(E) in NE, and recall Problem 8.]
15. Prove that v, IS continuous up to the free boundary x = n:), I > t" aatd t) =
[Hint: —2P and Problem 14 imply that urn v, The metbnó of Problem 9 gives Urn ç (r)21 —
16. The assertion C in Problem 1 is valid for any u0 L'(R")[that is, the assumption (3.37) is not needed]. (Hint: If v let = — r) (0< r < 1), v p Write the differential equation for w, differentiate in x, and multiply byp. to derive a differential equation L.p = 0. Let E CJ(R" X (0,00)), < 1, z = If z takes maximum at (x0, ta), then 2x = 0, — 0 at(x0, Use these conditions in 4 = 0 to deduce that C1 +
C I
is bounded.)
SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
6. ThE GENERAL FILTRATION PROBLEM: EXISTENCE
In the filtration problem studied in Chapter 1, Section 5 and Chapter 2, Section 6 the dam was rectangular. We shall now consider a general two- dimensional dam and establish existence and uniqueness, and regularity of the free boundary.
We denote a point in the plane by I = (x, y). We assume that consists of three parts: S1, the impervious part; the part in contact with the air; and S3. the part in contact with the water reservoirs; see Figure .i. I. We assume that there is only a finite number of disjoint reservoirs R1 (1 k) in which the water at a height y = h,, and we set S3j = fl (y < hi). Thus S, =
Denote by A the wet portion of that is, the portion filled with water. The boundary of A consists of four parts:
r, c r2 c (I
(the impervious part),
(the free boundary),
r3 = S3 (the part in contact with the reservoirs), r4 C S2 (the wet part of the dam, in contact with
the air; it is called the seepage line).
I'3 is given but all the other are not a priori known.
FIGURE 5.1
THE GENERAL TWO-DIMENSIONAL FILTRATION PROBLEM: EXISTENCE 651
The hydrostatic pressure on 53 is given by h — y. We extend this function into S2 by setting
1h1—y cm
Denote by ' the exterior normal to A. If we denote the pressure function by u, then as in Chapter 1, Section 5,
Au0 mA,
on!'1,
(6.1) u=Oand—(u+y)0 on!'2,
uu° on!'3,
u=u0 and on!'4.
We shall henceforth assume:
(62) S1 and U 53 are continuousand piecewise C' curves;
both are graphs in the directiony, and LI S3 lies above S1.
Thus there is an interval x r, such that any line x x0 with x0 E to,, i,J intersects S, (and u 53) in either one point or one closed interval, and the functions
S(x)=sup(x;(x,y)ES1),
satisfy
S(x)<S4(x) (a,<zx<r,).
For x0 < a, or x0 > r, the line x = x0 does not intersect S, U S2 U 53. We further assume that
(6.3) S (x), (x) are piecewise continuous.
Notice that
((x,y); (x),o, <x
652 SOME FREE-BOUNJ)ARY PROBLEMS NOT iN VARIATIONAL FORN
We denote by ir(E) the projection of a set Eon the x-axis, and index the 53 such that
(6.4) i<c,,11,
where may coincide with Thus, in the case of the rectangular dam in Chapter
Observe that
—--—°>O onl'1.
Since also u = u0 0 on the maximum principle gives u > 0 in A. If E = 0 on O'on r4 then by integration by parts,
A A r1ur2ur4 av r4uP
where (6.1) and the relation ay/av = e have been used; here e is the unit vector in vertical direction:
e = (0, 1).
Extending u by zero into \ A and introducing the Heaviside function
H(t)=Iift>0,
we get
(6.5) (Vu + H(u)e) VU +JL, 0,
We shall now give a weak formulation of the problem. Since l'4 is not a priori known, we shall take the test functions to satisfy 0 on all of S2.
Problem (A). Find a pair (u, y) where u H1(tl), 1' E such that
(6.6)
If we can show (as will be done later) that y = then (6.6) coincides with (6.5).
Theorem 6.1. Problem (A) has a solution (u, y).
THE GENERAL TWO-DIMENSIONAL FILTRATION PROBLEM: EXISTENCE
Proof. Suppose first that is in C' k". Introduce the Lipschitz continuous functions
o
- (e>O) E
I
and the classes of functions
K= {v E H'(f2); v = u°onS2 U
K0 = (v E H'(fl); v = OonS2 U S3).
Consider the penalized problem: Find Uf such that
(6.7) uE e K.
To solve this prothem consider, for any v E L2(11) the problem: Find w E K satisfying
(6.8)
for all E 1(0. Notice that
COkIHI(n) (C0 constant)
and a(w, is a coercive bilinear form on K. Thus there exists a unique solution of (6.8). Further, we easily find that
(6.9) W C.
where C is a constant independent of e. It follows that the mapping w Tv maps the set (v EE L2 C) into a compact subset and, by Schauder's fixed-point theorem, there exists a solution w of w = Tw. This w is precisely a solution U of (6.7).
Since u, = u° 0 on S2 U S3, we can take = u as a test function in (6.7) and obtain
fi vu12 =JHt(ue)(u: )v = 0.
SOME FRE BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
It follows that u = 0, that is,
(6.10)
Since also u, = 0 on S3, we get
au (6.11) onS3.
Next.
=
and 0 'c H, 1. Also, by (6.9),
(6.12) U,IHI(g) C.
But then standard elliptic estimates give
(6.13) Vp€(1,cc),
where is a constant independent of e. We now choose a sequence e = such that
u,-øu (O<a<J'
u,—'u
weakly star in
U S3. For any as in (6.6).
(vu, +
by (6.11). Taking e = we obtain the inequality in (6.6). Thus (u, y) is a solution of problem (A).
We have assumed above that 8t1 E C' In the general case of piecewise boundary, we approximate fi by smooth domains by smoothing the
corners of It is easy to see that the corresponding solutions converge to a solution for
THE GENERAL TWO-DIMENSIONAL FILTRATION PROBW.t EXIS1TNCE 655
DefinItion 6.1. A function as in (6.6) is called a test fimction Ifor (A)I. From the preceding proof we have actually obtained a solution in
for any 0< a < I (recall (6.13)]. We shall now prove:
Theorem 6.2. For any solution (u, y) of problem (A), u E
First we establish a lemma.
Lemma6.3. u( X) Cr, where C is a universal constant.
Proof. For any 8>0 let v be the harmonic function in the ring
D =
taking the boundaiy values
inf U Ofl8Br12(X), aa,,2(x)
v0 onaBr+s(X). The functions = ± max (v — u, 0) when extended by zero into \ D are
test functions. Hence, by (6.6),
(6.14) fvmax(v — u,0) (Vu + ye) 0.
We can write
u,O)Vu
— u,0) v(u — v)
I — u,0) v(v +y) — u,0) . e. The first integral on the right-hand side is equal to
—f IvvI2.Dfl(ss>O) bfl(u=O)
The second integral vanishes since v + y is harmonic, whereas max (v — u, 0)
SOME FREE-BOUNDARY PROBLEMS NOT IN VARIA11ONAL FORM
vanishes on 8D. Using this information in (6.14), we obtain
f v(v_uY12=_f IvvI2+fvmax(v_u,0)(y_1)e.Dfl(u>C) D In the last integral, if u > 0, then? = I and the integrand vanishes. Thus the integral is bounded above by
I lvvI,Dfl(u=O) and we obtain
(6.15) f v(v — 12 I Vt' 1(1 — Vt' I)Dfl(u>O) Dfl(u0} We claim that
(6.16) f Indeed, otherwise v u in (u > 0) and, since D contains free-boundary points, v will vanish at these points, contradicting the maximum principle.
Notice next that I
is a strictly monotone-decreasing functiOn of the distance from X. Hence, from (6.15), (6.16),
lVvI<l onB,÷8(X).
On the other hand,
I 8
where c is a universal positive constant. Thus A r/c. We have thus proved that
inf C
and by Harnack's inequality it follows that u( X) Cr.
Proof of Theorem 6.2. Using Lemma 6.3, the proof is similar to the proof of Theorem 3.2 of Chapter 3.
We denote by the characteristic function of a set K. In the following sections we shall use the following lemma.
REGULARITY OF THE FREE BOUNDARY 657
Lemma 6.4. Let u H'(O) (1 and denote by K a component of (u > 0). Then IKU is in H'(II) and
v(JKu) =
For proof, see Problem 2.
PROBLEMS
1. Prove that the solution of problem (6.7) is unique. [Hint: If u1, u2 are two solutions and w = — u2, then
Take 8 >0, (w — and show that
Iw>81 Iw>81 W
+
Use Poincaré's inequality and take 8 -. 0.1 2. Prove Lemma 6.4.
[Hint: If u E bounded, let e > 0, K, = {x K, u(x) e}, lonK.Then
v[IK(u — —
e JO. In general, take ii,,, C t al
7. REGULARITY OF THE FREE BOUNDARY
In this section we denote by (u, y) any solution of problem (A).
Theorem 7.1. The following relations hold in in the distribution sense:
(7.1)
(7.2)
(7.3)
658 SOME FREE-BOtINDARY PROBLEMS NOT IN VARIATIONAL FORM
Proof. Taking 6 C000(U) in (6.6), (7.1) follows. To prove (7.2), let 6 0, e >0. Since the functions = ±min(u, are test functions,
f vu v miii (u, + f-y[mun (u, = 0.
Since y = I on (u > 0), the second integral is equal to
f[min(u, =f min(u, = 0.
Hence
=ef IvuI2,
and consequently,
0.
Taking e 0, we obtain 0. Since is arbitraxy nonnegative function in Cr(Q), (7.2) follows. Finally, (7.3) is a consequence of (7.1), (7.2).
Lemma7.2. If(x0)X(y0—e,y1+ejCflfl(uO),e>0,ihen
(7.4) u(x, y) = — x0)) as x -. x0,
un:formiy with respect toy in [y0,
Proof. Letø{x0)X[y0,y1jandset
u(x,y) (x,y)-.., Ix —
Notice that A is finite since u C°'. Take a sequence X, = (x., Yr) a such that r fx, — x0J and
and consider the blowup sequence (u,., with respect to bails B,(x0, yr); that
REGULARITY OF THE FREE BOUNDARY
is,
u,(x, y) + rx, Y, + ry),
'y,(x, y) y(x0 + rx, Y, + sy).
Then Vu. C in every bounded set and thus, for a subsequence,
Va<1,
Yr weakly star in
It is easy to see that (u, is a solution of problem (A) in R2; that is,
(7.5) (vu + = 0 V E Ht(R2) with compact support.
Arguing as in Theorem 7.1, we deduce that
mR2. We also have
and, assuming for definiteness that the x,. — xo &C pO6ltIVe,
u(1,0) = A.
Applying the maximum principle to the subharmonic function ii — Ax in {x > 0), we conclude that ñ Ax in (x >0). If we show that A = 0, then (7.4) follows.
For any 6 > 0, let
Observethatd1(x)= I ifx>O.Sincealsoü=Xxjfx>0,wecanwrjtefor
(7.6) xf = —f (Vu + e)(x=O) (x>O)
= —f v(d8fl (Vu + ?e).(x>O}
660 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Since is a test function for (7.5), the right-hand side is equal to
J(—6<x<O) (—8<x<O) X
The first integral tends to zero as 8 — 0; the second integral is equal to
f—8<x<O) (x=—8}
Here the first term tends to zero as 8 —' 0 since the integral is 0(82) (recalling that ü E C°"). Thus we obtain from (7.6),
xf y_iimJ-f80 {x—8}
Choosing 0, we conclude that A 0.
Theorem 73. The set fl (u >0) has the form
{(x,y)E1i;S(x)<y<4(x)),
where +(x) is lower semicontinuous, and
k
(7.7) •(x) = 5+ (x) :fx E U [a, 1=1
The last assertion means that the dam is wet below S3.
Proof. Suppose that u(x0, yo) >0. Then u >0 in some disc B,(X0) where = (x0, Yo)- It follows that y = 0 in B6(X0). Since 0 y 1 and 0, we
mustalsohavethaty=0in
But then (7.1) implies that Au = 0 in K,. Observing that u(x0, Yo) >0, U 0 in K8, the maximum principle gives u >0 in K6. We have thus proved that if
= sup {y; u(x0, y) >0),
REGULARITY OF THE FREE BOUNDARY
then
u(x0, y) >0 fory < 0
The preceding proof also shows that 4)(x) is lower semicontinuous. Since u > 0 on 53 fl r,), it follows that u > 0 if x (a,, (x, y)
Efl. Suppose finally that (x0, X0 or = We claim that
u(x0, Yo) 0. Indeed, otherwise u(x0, y) = 0 for y >yo and the maximum principle
a
However, this contradicts Lemma 7.2.
Lemma 7.4. Suppose that u(x1, y) u(x2, y) 0/or all (x1, y) >h, 1,2, and (x1, x2) lies in ir(S2) (the projection of S2 on thex-axis). Let
Zk = ((x, y); x1 <x <x2, y > h)
Then
f ('vu,, + 12) 0. Zk
Proof.
f (vu + •(x))z,, xI Indeed, for any £ >0, znin(u, is a test function (when extended by zero outside Zh). Therefore,
f I vuJ2 + if vu + f y[min(u, se)], 0.Z,, But the first integral is nonnegative and the last integrand vanishes on
{u = 0) and Imin(u, on (u> 0). Consequently,
+ o,
662 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONALFORM
or
+ —
= urn — +
x1 840 It E y
fX2( —
Taking e 0, (7.9) follows. Now let tj(x) be a function in that 0 ci, = 1 in
(x1+E,x2—e).Then
+ y) =f[vu •v(y— h) + y(y —
fz.
+f(vu •v[(1 — — h)1 + y[(l — —
The first integral on the right-hand side is nonpositive since — h) is a test function (when extended by 0 outside ZIt). The second integral, by (7.9), is smaller than
fx2(I — — Ii)
+ — —
Letting e 0, we get + y) 0. Since finally u,,y = y a.e., (7.8) follows.
Corollary7i. Ifu=(J1nBAX0)C(l,:heny=Oa.e.inB,(Xo).
Indeed, we can take with base (x0 — r, x0 + r) and by applying Lemma 7.4 obtain
fz?2 + 72)
which yields the assertion.
REGUIARflY OF ThE FREE BOUNDARY 663
Lemma 7.6. Let the assumptions of Lemma 7.4 hold and suppose that u(x, h) M if x1 x x2. Then
Proof. The proof involves a comparison with a function v:
v(y)max[C—(y—h),O}, C>M.
For any 8 > 0, e > 0, let
4'a(s)
d,(X) = min(!dist(X, Zn (v>0)), i).
where Z = Then
is a test function when extended by zero outside Zh. Hence
l)—ip8(u—v)]
Since Vv (0, — 1), also
— v). (vv + 1(V>o)e) = 0.
Adding, we obtain
0 — v) (V(v — u) —(y —
—fv[d,(I — (Vu + ye).
Therefore, by adding and subtracting
fZfl{e=O}
664 SOME FREE•BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
we can write
(7.10) f7(1
—fZfl{v=O)
— v) 4(y — —
(vu + ye)
—f(1 — vd1• (Vu + ye)
=11+12+/3+14.
and the second factor in the integrand vanishes. Thus
(7.11) 12=0.
Next Thus
(7.12)
—f ife'0,V8.
The part of where u = 0 is negative since the integrand is ye 0. Thus
JvdI, (O<u<e)
where B,(E) = s-neighborhood of E; here we have used the fact thatu C°'. Denoting by H' the one-dimensional Hausdorff measure, we then obtain (using again the Lipscbitz nature of u)
{O<u<8+Ce)j. But as 8 + e 0, the set in brackets shrinks to the empty set. Hence
REGULARITY OF THE FREE BOUNDARY 665
Recalling also (7.11), (7.12), we deduce from (7.10) that
(7.13) f if8—.0. Now we can easily complete the proof of the lemma. Let w max (u — v, 0).
Then for any E (B any ball),
f vw = f .v(u — v) —ZflB ZnB The first integral on the right-hand side is equal to zero since u — v is harmonic wherever u — v >0 and 4i3(w)r 0 on the boundary of Z fl B. The last integral is bounded in absolute value by
ci 1vw12-'O 8 ZnDn{O<w<8)
by (7.13), provided that B 3 (v > 0). Taking 8 0 we obtain
f VLVw=O providedthat{v>0)Zn B Thus w is harmonic in Z fl {v > 0). Since w = 0 and u >0 in a Z-neighbor-
hood of az n (y = h), we deduce by analytic continuation that w 0 in Zfl(v>0); that is, if Taking CJ,A1, the assertion of the lemma follows.
Theorem 7.7. If S (x0) <4(x0) < then 4(x) is continuous in some 8-neighborhood of x0 andy = 0 if y > — x0j<8.
Proof. Set Yo = +(x0). Then u(x0, y) = 0 for Yo <y < (xe). Observe next that ii cannot be positive in any {x > x0)-neigliborhood of any point of the interval ((x0, y); Yo <y (x0)). Indeed, otherwise we shall have
0 at that point (by the maximum principle), which contradicts Lemma 7.2. It follows that for any h >y0 there is a sequence (i,, such that
i,1x0, j,E(y0,h),
Similarly, there is a sequence (x1, y1) such that
y,E(y0,h), u(x,,y,)0. Setting
666 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
we are now in a situation where Lemma 7.6 can be applied. Since u E C°'
x, h) C8,, where 6, = — x•.
Consequently,
(7.14) u(x, y) = 0 ifx, <x y > h + C6,.
Since h can be arbitrarily close to Yo' (7.14) implies that 4)(x) is upper semicontinuous at x = x0. Since 4) is also lower semicontinuous, we conclude that 4)(x) is continuous at x = x0. Thus 4)(x) is continuous as long as (x, 4)(x)) remains An
The last assertion of the theorem follows from Corollary 7.5 and the continuity of 4)(x).
Remark 7.1. From the proofs of lower and upper semicontinuity of 4)(x) we also see that if 4i(x0) = S (xe) is continuous at x = x0, then 4)(x) is continuous at x = x0.
Corollary 7.8. y = I{u>o)
We finally prove:
Theorem 7.9. If (x0, 4)(x0)) E then 4)(x) is analytic in a neighborhood of x = x0.
Proof. Take a small interval x1 <x <x2 containirg x0 and let h, H be such that
Introduce
Z = ((x, y); x1 <x <x2, h <y < oo) n
{(x,y);x1 <x<x2,h<y<H}
and define a function w by
w(x,y)
we conveniently define u 0 in Z \ ZH.
PROBLEMS 667
For any E Cr(ZH), define a function E such that
ij(x,y) ifh<y<H,
=0 ncaraZUS2.
Then are test functions for (u, y) (since they vanish on az U S2) and therefore
(7.15) (Vu + = 0.
By integration by parts,
fzlpyuy = = — =
and by change of order of integration
= Ix
Noting also that '(u>O} = '(w>O) we obtain from (7.15)
(7.16) =0
that is, —Aw + = 0. Thus w is a solution in Zff of the variational inequality
Since also <0, we can apply Chapter 2, Theorems 6.1 and 6.2 to deduce the analyticity of 4i(x).
PROBLEM
1
—(y — h)on Prove that M—(y —*)in Z.
SOME FREE.BOUNDARY PROBLEMS NOT VN VARUTIONAL FORM
[Hint: JzV max(v — u,O) (Vu + ye) 0 implies that
f jv max(v — u,O)12 = f V max(v — u,O). (vv +Z —f Vv (Vv + (I — y)e).Zfl(u=O)
8. UNIQUENESS FOR THE FILTRATION PROBLEM
Definition 8.1. A solution (u0, y0) is called S3-connecied if
fl S3 * 0 for any component C1 of {u0> 0) fl
Theorem 8.1. The solution of problem (A) is unique up to groundwater re- servoirs; that is, there is a unique S3-connected solution u0 such that any other solution u has the form
(8.1) u(x, y) = u0(x, y) + —
where are some real numbers and D1 are connected components of fl fl {y <
The terms (H1 — represent groundwater reservoirs, or wells. Clearly, if S1 is given by either a monotone curve y = k(x) or by a curve
y = k(x) having one local maximum but no local minimum, then no ground- water reservoirs can exist and the solution is therefore unique.
Proof. We first show that if D is a connected component of fl fl (u >0) Iu is a solution of problem (A)I such that
(8.2)
then
+ (8.3) u(x,y)(H—y) mD.
In view of Theorem 7.3, (8.2) implies that ir(D) C ir(S2). Let ir(D) = (x0, x1) and
Z <x<x2).
Then = ±IDU is in H'(Q), by Lemma 6.4, and it is therefore a test
UNIQUENESS FOR THE FILTRA11ON PROBLEM
function. Thus
(8.4) vu12 + yuE) = 0.
Next, by Lemma 7.4,
f +
Combining this with (8.4), we obtain
It follows that 0, = —y in Z. Since y we easily deduce that u = (H — in Z, for some real number H.
Now let (u, be any solution of problem (A) and denote by D1 the components of (n > 0) such that fl 0. Setting
(8.5)
we have, for any test function
(VU0 + yoe) (vu + —
+ c.,) = 0.
Thus (u0, Yo) is S3-connected solution and from the previous proof we obtain the decomposition (8.1).
To complete the proof of Theorem 8.1, it remains to establish:
Lemma 8.2. There exists at most one S3-connected solution of problem (A).'
Proof. Suppose that (u1, (u2, y2) are two S3-connected solutions and denote their free boundaries by x = x respectively. Set
u0—u1Au2, y011Ay2,
A {uo>O).
Then, for any fl C(f�), 0,
. [v(u. uo) + (z — +,(x)) dx,
670 SOME FREE-BOUNDARY PROBLEMS NOT IN VARiATIONAL FORM
where
L. = (x <41(x)).
Indeed, = minlu1 — u0, efl, e >0, then = 0 on S2 U S3 and thus is a test function. It follows that for i
— + (.y1 — y1)e] = 0,
or
•V(u1— u0) +f(y, — = 0.
Since the first integral is equal to
I Iv(u,_uo)12+ef
we obtain
(8.7) f
— — mm (i',
Observing that on fu0> 0), we have u1 >0 and = = 1, the last uucgral becomes (using Corolia,ry 7.8)
f (u)+£ w (+rj(X)<y<$,(X)) e y —8)d.
If we substitute this into (8.7), we obtain
— u0) + — ro)L. cf r (x, 4,(x))
UNIQUENESS FOR THE PROBLEM 671
We now let e —' 0. Since u. u0 and since V(u1 — u0) = 0 on the set {u, — u0 = 0), (8.6) follows.
Let be a disc with center at one of the components S3 , of 53 and with
radius r sufficiently small so that C and set r0 = aBr 1) Let W be a solution of
tiW0 mB,, W1 onl'0,
W=l
Then, by the maximum principle
(8.8)
(8.9) >0 on l'0 (y = outward normal to aB,).
By Green's formula,
v(u1— u0) •VW;
hence
(8.10) — =f[v(u, — UO) . —
I onAo,ae=Oifdist(XAo)>e,; I. In view of (8.8), (1 — ajW is a nonnegative function; it vanishes on S3 provided that e is small enough. Thus it is a test function. We therefore have
f{vurv((l —a) W)+y,[(I
But since (1 — a)W = 0 on A0 and u0 = = 0 outside A0(by Corollary 7.8),
((vu0 v((I — oç)W) + y0[(1 — =0.
672 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Subtracting this from the preceding we obtain
(8.11) f[v(u1 — u0) VW + —
u0) •V(aeW) + (y1 —
Adding both sides of (8.10), (8.11) and evaluating the right-hand side of (8.11) by (8.6), we get
f — Ef (;W)(x. q,(x)) dx.F0 As af(x, 0 for any x E L. Therefore, we obtain
fro(hu — 0.
Recalling (&9), we conclude that u1 = u0 on I = 1,2; that is, u1 = u, on
If we denote by C] the component of (u, > 0) that contains (2 fl then the previous proof (with 0 < r' <r) gives u1 = u2 in (2 fl By analytic continuation, we then also have u1 = u2 in fl and therefore also ç1 = C2; this completes the proof.
PROBLEMS
1. Suppose that S3 is connected with ir(S3) = and 'S, is a curve y = k(x) such that for some x1 > (i) k(x) is monotone if x > x1; (ii) if
= then. the luje segment X x1; y lies below Show that any solution u is positive on ((x, k(x)), x > IHint: If u(x, y) = 0 for x > f, x E x1), let = ((x, y) E (l,x < X, V <yo + E} and take for small enough the test function —(u — (Yo + — shOw that
I — (Yo + E — y) 112 0.)
2. reservoirs so that h1 - h,,. Prove that the S3-con- nected solution u satisfies u (h1 — . Similarly, if C is a component of a solution u such that C fl = 0 for I I j, then u (h1 + — inC. [Hint: (u — (h1 — vanishes on U S3. and is then a test
PROBLEMS 673
function. Deduce
I lv(u_(hI_yY1F12+f lvuI2+f yu,,cO
fl [y > h1] is the union of sets Zh) to deduce :hat
f V(u - (h1 - y)4 12+1 + + )2]
0.]
3. Take again h1 h2 h,, and set hk÷I = infS. Let u be the S3-counected solution and let h E hi), Uk ((x, y) E tl; u(x, y)> (h is empty if /, >h1; see Problem 2). Denote by the connected component of Uh satisfying ChI 3 S3,. Show that Uk = Chj U Ck2 U (some of the Cha may coincide). [Hint: Assume for simplicity that Ch, are all disjoint and let =
Then = fc(u — (h is a test function. Derive
f lV(u f + By Lemma 7.4,
f fl(y>h) so that
I C*flfy>hJ inCh.]
4. Let u be the S3-connected solution and let (x, h) E Vx1 x x2. if (x1, h), (x2, h) belong to the same connected component of Uh, show that u(x, y) >0 if (x, y) E fl ([x1, x21 X (—00, Is)). (Hint: Suppose that u(x0, Yo) = 0, (x0, Yo) U, x1 x2,y0 <h and connect (x1, h) to (x2, h) in ChE by a curve r; F intersects {x x0) below y0. We may take F to lie below (y = h) and denote by D The domain bounded by F and {y = h). Then — (h — is a test function, giving
i.e.,(u— (h —y))=CinDfl (u>0),
where C is a constant. Derive a contradiction.I
674 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
5. Suppose that there is only one reservoir, and set ii-( S3) = r1]. Let S2 be given by y k(x), where k(x) A h1 is a monotone increasing function for x and monotone decreasing for x > r1. Then for the S3-connected solution there holds: 4)(x) is monotone increasing for x and mono- tone decreasing for x > [Hint: Suppose that the assertion is false for x > Then there exist
<x0 h <4)(x1) < h1, Yo = 4)(x0) <h. Connect (x1, 4)(x0)) to a point (a1, near S3 by a curve r in ChI and derive a contradiction to the result of Problem 4.]
6. Suppose that there are two reservoirs, h1 h2, er(S) = [a,, <a, < r1 <02< = and S2 is given byy = k(x), where k(x) A h1 is mono- tone increasing for x E (os, a,) and monotone decreasing for x E Show that (i) the free boundary 4)(x) is increasing for x E (0w, a,); (ii)
is decreasing in x E (i-,, 02) in any subinterval (T,, where 4)(x) h2; (iii) if = 4)(x*) <h2, then 4)(x) is increasing for x E (x*, 02).
[Hint: Use the method of Problem 5.]
7. Consider the case of two reservoirs with i,-( S31) = [a,, ir( S32) = h, > h2. Let S2 be given by y = k(x) with
k(x) A h1 monotone decreasing and suppose that S, lies below y = h2. Prove that there is a unique solution and that its free boundary y = 4)(x) is a monotone decreasing function (for 11 <x <02). [Hint: Use = ±(u — (!i2 — as a test function to obtain
f JV(u_(h2_y)YJ2+f (y— l)(u—(h2—y))Iv<h21 (v<h2)
The second integral equals —meas([u = 0] fl [y < 112fl. Thus u >0 in (y <h2J and in particular 4)(x) h2. It follows that u is S3-connected and thus uniquely determined. The monotomcity of 4) follows as in Problem 6(ü).)
9. THE FILTRATION PROBLEM IN N DIMENSIONS
In this section we generalize the results of Section 6 to 8 to n dimensions. Wherever possible we shall use the same notation as before. Thus the dam is an n-dimensional domain bounded by surfaces S1 (the impervious part), S2 (the part in contact with air), and S3 (the part in contact with the water reservoir).
THE FJLTRATJON PROBLEM IN N DIMENSIONS 675
We assume Isee (6.2)j:
S1 and S2 U S3 are piecewise surfaces, both being graphs in they-direction, and S2 U S3 lies above S1;
further, S2 is open and is given byy f(x) (x E B0 open) wherefe
Here by "S piecewise C1" we mean ilat there is a finite triangulation of S where are uniformly in c1 'and are piecewise C" (n — 2)-dimensional manifolds.
We denote by X (x, y) points in where x is a variable point in and y is the height parameter.
We shall also assume that
S3 consists of a finite number of disjoint components
(9.2)
exists a baliKin such that K C and x0 E
The formulation of problem (A) remains unchanged. Theorems 6.1 and 6.2 remain valid with the same proof. The proof that u E actually extends up to the boundary, near any C" portion of the boundary. In particular,
where is any subdomain with fl0 C I? LI S2. We shall now extend the results of Section 7. First, Theorem 7.1 clearly
remains valid without change in the proof. Next:
Lemma 9.1. The set IZ fl (u > 0) has the form
((x, y); S (x) <y
where +(x) is lower semiconhinuous, and
•(x). ifx E int(ir(S3)).
The proof is the same as for Theorem 7.3. The proof of Lemma 7.2 is strictly two-dimensional. We shall now extend
this result to n dimensions. This is, in fact, the main new step needed for establishing the smoothness of the free boundary in n dimensions.
676 SOME FREEBOUNDARY PROBLEMS NOT IN VARIA11ONAL FORM
Fix points
= (x0, E S2,
X0(x0,y0)ER, yo<yj.
Lemma9.2. Ifu(x0,y0)0,ihenforanyO<fl< l,e>O,
(9.3) Vy0+e<y<y1.
The proof depends on several lemmas. We choose small positive numbers 6 and such that
B8(x0) Cir(S2),
B8(x0) X {h} C whereh y0 —
Let
ZB8(x0)X(h,H),
where H is sufficiently large, say
H>S4(x)
and extend u by 0 into Z \ Il.
Lemma 9.3. There exists a bounded nonpositive measurable function 8(x), x E B8(x0), such that
(9.4) f (vu + = f f(x)) dxZ B(x0) for any E C01(Z).
Proof. Let be the solution of
(9.5)
JVL(Vug+eY=O
E H'(fl), = U0 Ofl S2 U S3.
By Lumpanson (see Problem 1) we have u uQ. Since also u = = 0 on we obtain, formally,
onS2flZ,
THE FILTRATION PROBLEM IN N DIM}NSIONS 677
where C is a constant, and (9.4) easily follows with
au dS U = — dS surfaceelement on S2.
However, since au/av is not well defined, we proceed somewhat differently. Set Z0 = Z (-1 The function v u — satisfies
in Z0; indeed, if E then [since = 0 in Z0 and is a test function for problem (A)] the distribution applied to
—fvv.vr=
f f dx(u>O}
We extend the function v by 0 into Z\ZO. Since v 0 in Z0 and v = 0 on az0 fl S2, the extended function v satisfies (see Problem 2)
(9.6) in6D'(Z);
that is, for any E 0,
<Pie .
v E C°(Z), the left-hand side is equal to
We thus obtain
0 + J(U>o)e) .
from which (9.4) easily follows. Indeed, the functional
f(vu + I(M>o)e)
678 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
is a bounded functional on C00(S2 fl Z) and
S2 fl Z
By the Radon—Nikodym theorem it follows that
f f(x)) dx, 0 —0(x) C.B6(x0) Finally, since any E can be written as (9.4) follows for any
E
Lemma 9.4. The funclion
(9.7) w(x, y) = f"u(x, s) dc
belongs to for any 0 <fi < 1.
Proof. Taking
we can transform (9.4) into [see the derivation of (7. 16)J
f •VW + =1 0(x)(f00 dx.Z f(x) It follows that
(9 8) Aw = '(w>O) ÷
p, bounded positive measure in Z,
and the assertion follows from elliptic estimates.
Lemma 9.5. There holds
inZ0,
where C is a positive constant.
Proof. We first show that
(9.9) c in Z0, c constant.
THE FILTRATION PROHLEM INN DIMENSIONS 679
Consider finite differences
u(x, y y
u (and thus is continuous in Z0. Also, is harmonic in Z0 fl (u > 0). Let be the solution of
onZ0,
= on az0.
Since 0 and 0 on Z0 fl a(u > 0), the maximum principle gives
inZ0fl{u>0}.
Since S2 is in C" we can represent in the form
vT(X)=_f az0
where G is Green's function in Z0. It follows that
cf a0z0
provided that dist ( X, a0;) > 0, where 80Z0=[B8(x0) x {h)] u[(aB8(x0) x (h,H)) n
C depends on e0. Integrating with respect to 8 and h, in some small intervals, we obtain
dxdy,
where the integration is over some n-dimensional neighborhood of a0z0. Recalling that L2(a), we obtain
and (9.9) (with slightly smaller 6 and h in the definition of Z0) follows upon taking y 0.
Consider the function
inZ0,
where e > 0 is small and M is large.
SOME FRIE-BOUNDARY PROBLEMS NOT IN VARiATIONAL FORM
If dist(X, 8(u >0)) > e, then u c0(e) >0 and by choosing M = M(e) sufficiently large, we obtain z(X) < 0. If dist(X, a(u > 0)) < c, then w < Ce (by Lemma 9.4) and eu,, cc (by (9.9)); thus z C0e (C0 constant indepen- dent of e).
Recalling that, by (9.8),
inZ0fl(u>0},
we can apply the argument used in the proof of Chapter 2, Theorem 6.3 (with w replaced by —z) to deduce that z <0 in (B8.(x0) X (h, H)) fl for any 8' <8, provided that e is sufficiently small. Thus
by starting with a slightly larger 8, we can then achieve this estimate in Z0.
Proof of Lemma 9.2. Take h <h' <y0 <YI. For any h <y' <h' <y x E B8(x0),
u(x, y) = u(x, y') + f"u3(x. s) is
byLemma9.5,
u(x, y') + Cw(x, y').
Integrating over y', h <y' <h', yields
(9.10) u(x, y) C1w(x, h), C, = h' h + C.
Since w(x0,y)=0 ify>y0, wE Lemma 9.4), we have vw(x0,y)=Oand
w(x, y) —
Combining this with (9.10), the assertion of Lemma 9.2 follows.
Corollary 9.6. If x0 6 then u(x0, y) > Ofor cii (x0, y) E
Indeed, if u(x0, Yo) = 0 for some (x0, Yo) E then by (9.2) and the strong maximum principle,
at (x0, y) 6 y >Yo'
THE FILTRATION PROBLEM IN N DIMENSIONS 681
where the derivative is taken as a limit of finite differences in a suitable direction perpendicular to the y-axis. This contradicts Lemma 9.2.
Remark 9.1. The ball property assumed in (9.2) is used only in the proof of Corollary 9.6. Using Problem 7.6 of Chapter 2, it is sufficient to replace the ball K by a cone with vertex at x4,.
We next prove:
Theorem 9.7. Suppose that X0 (x0, yo) belongs to 11 fl a(u > 0). Then, for some 8 > 0, the free boundary in LB.(xo) X — 8, 00)1 is given by
+(x)analytic;
Proof. Lemma 7.4 extends to n dimensions if we assume in this lemma that
u(x, y) 0 on ÔB,(x) X (h, H).
Thus also Corollary 7.5 remains valid. We shall next use Lemma 9.2 and the argument of Lemma 7.6 to establish
the upper semicontinuity of Consider a domain D as in Figure 5.2; 1) is bounded below by y =
laterally by X (h, oo), and above by,a smooth surfacey = N(x); D is contained in the cylinder Z X (h, 00). We choose N(x) such that
v const >0
where ,' is the outward normal.
FIGURE Si
(9.11) On {y = N(x)) n
S2
682 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Let be a function satisfying
mD,
v0=O onyN(x),
= p on the remaining parts of
Extend v0 by 0 into Z0 Z fl Then v0 E H1(Z0). We claim that if £ and p are suitably small, then
(9.12) cv0> u on 0Z0 fl
(9.13) V(Ev0 + e) v 0 on (y = N(x)) fl
Indeed, in view of Lemma 9.2, (9.12) is satisfied if e cpt Next, (9.13) holds if
p + P >0,
that is, by (9.11), if
El 'ç'v0 C0, C0 some positive constant.
By scaling we find that Vv0 C, so that if e C1 (for a suitable small C1 > 0), then (9.13) holds. Thus (9.12) and (9.13) hold if we choose e = C1, CpI+p C1.
Consider the function v = ev0. In view of (9.12),
on�2, (9.14)
v>u on8Z0flfl, where
From (9.13) we also infer that
(9.15)
The functional v is thus a "supersolution" in Z0 in the same sense as the function v used in the proof of Lemma 7.6. In fact, the same proof shows that
Recalling the shape of P in Figure 5.2, we deduce that y is indeed upper semicontinuous at x0. It follows that is continuous at x = x0.
PROBLEMS 683
From the continuity of 4(x) and Corollary 7.5, we deduce that y — 0 if
y > Finally, analyticity of follows as in Theorem 7.9. We define the concept of S3-connected solution exactly as in the case n = 2.
Theorem 9.8. Theorem 8.1 extends to the case of n dimensions.
Since y the proof proceeds precisely as in the case n = 2.
PROBLEMS
1. Prove that u [Hint: Note that uQ > 0 in and take = ±min(O, — u) as test functions.)
2. Prove (9.6).
[Hint: Let J = distance function to az0, d if a> e. Let 0. Then
— fvv vUd,) = f . ancJ
. vd, < 0 if e is small.
Thus
—fvv + I
where
Introduce tangential parameters . . = s and normal parameter p near S2 fl az0 and write
— = + !o(pU), v outward normal, 0<8< 1,
Then
Z0fl(dist(X,52)<e) ap
= p)] o.J
684 FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
10. THE TWO-PHASE STEFAN PROBLEM
The one-phase Stefan problem was introduced in Chapter 1, Section 9. In the two-phase problem the temperature of the ice is not identically zero. To give a classical formulation of this problem, we introduce bounded domains G1 in R",
G0cG0CG1CG1cG2
and let
D1=G1\G0, D2=G2\G1.
Initially, D1 is occupied by water and is occupied by ice. Set
F1=ôG0, r2=aG2,
We denote the temperature in u by 9 and seek a surface F: 4(x, I) = 0 and a function 0 satisfying the following conditions:
0(x,0) = h1(x) on D1,
0(x,i)=g1(x,g) onr,X(0,oo) (1=1,2), (10.1) = 0, — vft = a'1, on F,
in (t = s) bounded by F, X (s) and F(s): $(x, s) = 0, and 9, = 0 in Q,. Here k,, a are given positive numbers, h, and g. are given functions, and
(10.2)
Finally, 1'(O) is supposed to coincide with r is called the free bowidwy. The formulation above is too restrictive. We therefore proceed to give a
weak formulation. Set
I a. =
on 1 X (0, oo),
onD,X(0)
THE TWO-PHASE STEPAN PROBLEM
and introduce the function
— Iaiu i(u>0 a2u—a
We shall denote by v the outward normal to
G2\G0 and set
inQ1.
Finally, we define
G(t)=GX(t).
Definition 10.1. A pair (u, y) is called a weak solution of the Stefan problem in if U y E —a y 0 a.e. on the set (u 0) and y a(u) a.e. on {u 0); finally.
(10.3)
for any function 4) satisfying:
c( 4)=0 onG(T)andonaGX(0,T).
p is called a test function.
One can show that:
(10 4) If U is a classical solution of (10.1). then it is also a weak solution.
We shall assume:
h1>0 (10.5) mD1, h2<OinD2, h n !i"2(G); gistherestriction
of a function with in
Theorem 10.1. There exists a unique weak solution (u, y) of the Stefan prôb- km, and
(10.6)
(10.7) esssupf <cc. O<z<T G(:)
686 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
By uniqueness we mean that if (U, y), are two solutions, then u = a.e.
Proof. Let be functions which approximate the graph a(u):
am(u)a(u)iftul>L
Consider the parabolic problem:
— Ac = 0
(10.8) vg onaGX(O,T), vh onG(0).
By reference 130 there exists a unique solution v = Om of (10.8). One can further show (see Problem 2) that there exists a 80-neighborhood Wj of F, (1= 1,2) and an 20 >0 such that
(10.9)
By the maximum principle we also have
(10.10)
where C is independent of m. From (10.9) it follows that
Using also (10.10) and standard parabolic estimates, we get
(10.11) .
onaGx(0,T).
We now multiply both sides of the differential equation in (10.8) by acm/a: and integrate Q,. We easily find that
The proof can now be completed by taking a subscquànce such
v,,,—, u weakly in 11 and a.e., am(vm) -.y weakstarinL'°(Qr). -
The details are left to the reader.
THE TWO-PHASE STEFAN PROBLEM 687
To prove uniqueness, suppose that (u, y) and (a, are two solutions. Then
(10.12) fJ(7 — + eA4) = 0
for any test function where
ifuâ. Notice that 0 e c, c constant.
Choose a SeqUence ë,,, E 0 such that
Iëm — eIL2(QT)
em
em
e —el —p0, (10.13)
m
"C. em L2(QT)
for any function! E Cr(Qr), let +m be the solution of
(10.14)
(10.15) •,,,=0 onG(T)andonaGx(o,T).
Multiplying both sides of (10.14) by t&41m and integrating over Q0, we easily deduce that
(10.16) fG(o) We shall now estimate
.J =—fJ(7 —
If we substitute the test •m into (10.12), we obtain
(10.17) fj(f — )(a+m + = —
68$ SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
and
For any small > 0, let
E,1=Q7fl
Since em e in L2, for any small A > 0 there is an = A) such that meas (Es) > A if m > m Setting \ we have
intl
Thus
+ CAJ
+ X)fJ L_ [by (10.16)].
Similarly,
and by (10.13),
'm.2 +
We have thus proved that
Choosing a sequencef=f, q°(QT) converging to t2 — u in we then obtain (10.18) forf= a — u. Upon substituting the result into (10.17), we
THE r.VO-PHASE STEFAN PROBLEM 689
find that
(10.19) (u—u)2
if m is sufficiently large, depending on A.
Lemma 10.2. 1ff = — u and f2/e = 0 by definition whenever ü = u, then
Jim 1fL1(L. e
Suppose that the lemma is true. Since
If we obtain from(1O.19)
which gives the assertion ü = u.
Proof of Lemma 10.2. For any small 8>0, set
Then
(10.20) 6 em demT a
and
(10.21) 44/
= if (a u)2 =
a(ü) = a(u) a(a)—a(u)
(a — u)
690 SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATiONAL FORM
where (10.13) was used. Similarly,
(10.22) fJ1—_ ff(a(a) —a(u))(ü-— u)
Since is measurable, for a.a. (x0, t0) E
BE I im
.-.o
where is a bail of radius r centered at (x0, t0) and A j denotes the Lebesgue measure of A. If we set
thus C and has measure zero. It follows that for any s >0 there is aj =j(c) such that E1) < e. Hence [see (10.21))
(10.23) If
Similarly,
(10.24) fJI\E e
On the set e(x, 1) A(8), where A(8) >0. We can choose the to be mollifiers of em and this certainly ensures that em A(6)/2 on provided that m is large enough. Thus 0 f C on E1. Since em e a.e., the Lebesgue bounded convergence theorem gives
(10.25) lim e,,,
Combining (10.21)—(l0.25), we obtain from (10.20)
Since 6 and e are arbitrary positive number, the lemma follows.
THE TWO-PHASE STEFAN PROBLEM
The method of proof of Theorem 10.1 has several by-products, which we briefly mention. First, a comparison result:
If (u, y) and (a, are solutions corresponding to (10.26)
(h, g), (1. g), and if h g then u a.
The next result is concerned with stability. We suppose that (u, y) and (i2, are solutions corresponding to (h, g) and (h, respectively, and that can be extended into a function '4' in QT as in (10.5). Then there holds:
(10.27)
cff hj2 + — + I — G(O) r
where C is a constant independent of T. Suppose in particular that for some function E
(10.28) — + I — <00
and let w be the solution of
mG, (10.29)
on8G.
Then we can infer that
(10.30) 00.
Using (10.6), which is valid also for T = oo, one can deduce from (10.30) that
(10.31) JIu(x,t) — w(x) 12 dx -.0 if:
The two-phase Stefan problem can also be reformulated as follows. Find a solution u of
(10.32) in
692 SOME FREE-BOUNDARY PROBLEMS NOT IN VARL4TIONAL FORM
satisfying the appropriate initial boundary conditions in some weak sense; see Chapter 1, Section 9, Problem 3.
We introduce the inverse of a(u):
ifw<—a a2 a2
(10.33) +(w) 0 if—a<w<0 w — ilw>0. a1
Then (10.32) can be written in the form
(10.34) — A4(w) = 0 in
Notice that 4(w) is Lipschitz continuous and = 0 in some interval. The equation of gas in a porous medium has the form (10.34
Theorem 103. Let be given by (10.33). Then any solution of (10.34) is continuous in
Thus the temperature for the two-phase Stef an problem is a continuous function.
Theorem 10.3 is due to Caffarelli and Evans (571, who also extended it to a large class of functions including 4( w) = I w sgn sv, m> 1. Alternative proofs were subsequently given by DiBenedetto lila, b] and Ziemer [183J, who further replaced A by a general elliptic operator (nonlinear, of divergence type). The proof of reference ha and b gives continuity up to the boundary.
The proofs of references 57 and 77a and b establish also some modulus of contiiuiity for the solution; however, this is substantially weaker then the moduli of continuity obtained in the special cases of +(w) given by Theorems 9.1 of Chapter 2 and 3.1 of Chapter 5.
PROBLEMS
1. Prove (10.4).
2. Show that (10.9) is valid. (Hint: Compare with Aw =0 in G, w e on w = K on a62.]
3. Prove the comparison result (10.26). [Hint: Compare v,,, with the corresponding 13,,,.]
4. Establish (10.27) with C independent of T.
BIBLIOGRAPHICAL REMARKS 693
5. Prove that if (10.28) holds, then (i) (10.30) holds. (ii) (10.31) holds, and (iii)
JGk(x,t) w(x)rdx -901f1— foranyp<2n/(n— u(x, 1) —i w(x) uniformly in x if ii = I.
II. BIBLIOGRAPHICAL REMARKS
The porous-medium equation for one space dimension (n = 1) was studied by Oleinik, Kalashnikov, and Yui-Lin [1521, who proved existence and unique- ness. Properties of the free-boundary curves were studied by Aronson [I la—c]. Caffarelli and Friedman [58h] and Knerr [127]. For n 1. existence and uniqueness were first proved by Sabinina [157]. More general uniqueness theorems were proved by Brezis and Crandall 147] and Vol'pert and Hudjaev [178]. The uniqueness proof given in Section 1 is based on reference 47. Corollary 1.9 is due to Benilan [301 and Veron [177]. The special solution in Problem 3 of Section 1 was discovered by Barenblatt (24].
Lemma 2.1 is due to Aronson and Benilan (121. Problems 2 and 4 of Section 3 are taken from Knerr [127]. Theorem 4.6 was established by Friedman and Kamin [961. All the other material of Sections 2 to 5 up to Theorem 5.6 is based on Caffarelli and Friedman [58g, k). Corollary 5.7 was proved by Knerr [127]; [see also reference ha—c]. Theorem 5.8 is due to Caffarelli and Fried- man [58h]; the proof outlined in the problems is in part a simplification due to Aronson, Caffarelli, and Kamin [13]. In reference 13 it is shown that if v0(x) /3(b — x)2 for x <0, v0(x) = 0 for x > 0, v0(x)(b — x I b, then = $/(2(m + 1)). Problems 14 and 15 of Section 5 are due to reference 58h and Problem 16 is based on reference 11 a.
Uniqueness of solutions for initial data which are measures was established by Pierre [153]; thus, in particular, the Barenblatt solution is the unique solution with Dirac initial data. Related results characterizing the class of initial measures for which solutions exist were more recently obtained by Aronson and Caffarelli (1851 and by Benilan, Crandall and Pierre 11861. Asymptotic behavior (as t oo) for solutions of the porous-medium equation in a bounded domain was studied by Aronson and Peletier [14] and Donnelly [78]. For other results on asymptotic behavior when n = 1, see Alikakos and Rostarnian [4] and Vazquez [176]:
For the equation Autm with m < I, see Berryman and Holland [39] and the references given there. In this case, if m > (n — 2)/n, then for any T> 0 - there exists a unique solution u which is strictly positive everywhere (provided that u0 0, u0 0). If 0 <m (n — 2)/n. then u is strictly positive in some finite interval (0, T), and u(x, T) 0.
Theorem 6.1 is due to Alt (5cJ and to Brezis, Kinderlehrer, and Stampacchia [50]. In reference Sc the penalization is more general and in itself represents a physical problem: capiHarity is included. Theorem 6.1 extends to variable permeability k(x. y) [replacing V2u by V(kVu)], but the assertion 0 in
SOME FREE-BOUNDARY PROBLEMS NOT IN VARIATIONAL FORM
Theorem 7.1 requires, in general, the assumption that 0. Theorem 6.2 is due to Alt (5c).
The results of Section 7 are based primarily on Alt and Gilardi 181; Lemma 7.4 and the uniqueness proof in Section 8 are taken from Carrillo and Chipot [67aJ. Another proof of uniqueness, which applies to more general dams, was given by Alt and Gilardi (8) (see also reference 63d). In reference 8 the authors also study the behavior of the free-boundary curve as it hits the fixed boundary; Caffarelli and Gilardi [611 earlier studied the behavior of the free boundary as it hits S2.
Problems 1 to 6 of Section 8 are based on Carrillo and Chipot [67bJ. The result of Problem 7 in that section was established in reference 61, but the present proof is different.
The regularity of the free boundary in n dimensions (established in Section 9) is due to Alt (5bJ; for an inhomogeneous dam ii was established by Alt and Caffarelii [61].
Theorem 10.1 and Its consequences. are base on Friedman [94b]. The uniqueness part was originally established by Kamenotnosiskaja (Kamin) [117), who also obtained an existence theorem by means finite differences; see also Brezis (45(d)]. [76(a)), and Alt and Luckhaus (9) for other methods. Very little is known abotit the free boundary for the two-jhasc Stefan problem if n> 1; font = 1, seethe references mentioned in 1, Section 12. Recently Marimanov [143] proved, for n> 1, the existence of a classical solution for small times.
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INDEX
Accsetive, 590 .,en, 590 Asymmetric jet problem, 346 Axially symmetric finite cavIty, 39!, 392 Axially symmetric infinite cavIty, 400 Axially symmetric jet problem. 297
Blowup limIt, 164, 281 Blowup sequence. 281 Borda's mouthpIece, 270 Bounded lemma, 285
Capacitovy potentIal, 502 CapacIty, 502 CavIty, 266, 391, 392,400,401 Coercive, 14 Coincidence set, 7 Connected solution, S3,668 Constant flow. 353 ConstraInt, 7 Continuous-fit, 299 Contraction oneflldent, 269. 343
Dam problem, 47, 177, 650,674 Darcy's law, 47,589 Drag, 271
Elastioplastlc torsion problem, 55, 193 Elastic set, 71 Elliptic, 20, 132
unIformly, 20
Filtration problem, 47, 177, 650,674 Flat Intervals, 213 Flatness 284 Flat points, 213
Gas In porous medium, 589
Hausdorfi distance, 281
Hemiconthiuous, 9 Hill's vortex, 478 Hodograph-Legendre transformation,
128, 133 Hodograph method, 267 Hodograph transformation, 133 Hodograph variable, 268
Impinging jet problem, 364 Impulse, 475,471
jet, 266 asymmetric, 346 axially symmetrIc, 297 with gravity, 3d6 impingIng, 364 mouth of, 347
Lagendre transformation, 133 Local minlmiter, 273 Local minimum, 273, 388
Matdung lemma, 31. Maximal monoton, operator, 590 Maximum 22, 39, 74,96 Minimum diameter, 162
lemma, 10 Monotone operator. 9,590
590 strIctly, 13
Noncoincidence set, 7 Nondegeneracy, 275, 278 Nonoscillatlon lemma, 287
corner, 197 Nozzle, 297
Obstacle, 7, 76 thin, 105
Obstacle problem, 7, 76
109
710 INDEA
for biharmonic operator, 90 Operator, monotont, 9,590 Outside ball condItion, 22
Parabolic 73 Parabolic Itistance, 73, 225 Parabolic obstacle problem, 76 Parabolic operator, 73 Parabolic variational inequality, 72. 76 Peizometric head, 47 Penalized problem, 24, 76
lemma, 184 Plasma problem, 521 Plasma set, 521 Plastic component, 200 Plastic loop, 200 Plastic set, 71 Porous-medium equation. 589
Quasi-conformal, 143
Rearrangement, 294 Reentrant corner, 197 Reentrtnt plastic component, 202 Reentrant plastic loop, 202 Reflection property, 220 Retraction law, 193 Ridge, 197, 202 Rotating fluids, 418
base of, 457 rapidly, 445 rings of, 456, 457 slowly, 442
Schauder's estimates, 20 Schauder's fixed-point thcorcm, 24 Seepage line, 650 Semigroup, 591 Signormi problem, 111 Single intersection property, 414 Smooth-fit. 299, 324 Sobolev's inequalities, 22 Stefan problem, 82, 83, 224
two phase, 684 weak solution of, 685
Stein symmetrization, 293, 294 Stream function, 266, 297, 471,472 Strong maximum principle, 22, 74 Symmetric rearrangement, 293, 294
TJün obstacle, 105, 106, III Thomas-Fermi problem, 564 Trace, 28 Two phase Stefan problem. 684
weak solutions of, 685
Under-over theorem, 414
Vacuum set, 521 Variational inequalities, 4, 7, 10, 58
parabolic, 72, 76 Velocity potential, 266 Vortex core, 472 Vortex rings, 470
total circolation of. 477 Vortex strength, 477 Vorticity function, 472