numerical method PDE
Lecture13p.pdf.pdf
Thedeepness of freedom are threevalues at thenude functional Notconforming
patrtaf.us vi sci x
I beease ittouch
41 u VCsci inhalfedgeL
U VCI't x Since u CPz are sci sc 7 that it have 3 Zeusunless e o
E is P anisolvent it forgiven I lop C P s t 4 p di same degree
y i l N Yi C E Sabi n ofsystem YCp g
This is equivalent to say theonlypolynomial C PthetinterpolateZero
data Yifp o is the Zeno poly vcpi.POTFF.gg In Edem
e e I CRIvalue VCR Ca Ya
metfunctor p E3
pjJ Chip J Shun E is p unisolvent
Y Cul VCR7 0
Xz V UCR o rf VI UCB 0
Then over the edge PP we hone C P havingtworootsPR
D This implies we 0
If e consider the other to edges G e b thesame argument we can see Eo tht means W Lv o hersonly trivial Solution Then Yi CUI Ri for any Xi
E is P unisowent
y csiy
Ya f P Y cnn.PT
III Ldj Pg I Pre 2 ily a PyO ein a
451214 7 f p i y g d CP f ftp b f CRI I B so fickle Cps O y Cp 7 L
Escaple5 in lectureto
Lecture_03_S08.pdf
LECTURE # 3: ABSTRACT RITZ-GALERKIN METHOD
MATH610: NUMERICAL METHODS FOR PDES: RAYTCHO LAZAROV
1. Variational Formulation
In the previous lecture we have introduced the following space of functions defined on (0, 1): (1)
V =
v :
v(x) is continuous function on (0, 1); v′(x) exists in generalized sense and in L2(0, 1); v(0) = v(1) = 0
:= H10 (0, 1)
and equipped it with the L2 and H1 norms
‖v‖ = (v,v)1/2 and ‖v‖V = (v,v) 1/2 V =
(∫ 1 0
(u′2 + u2)dx )1
2
.
We also introduced the following variational and minimization problems:
(V ) find u ∈ V such that a(u,v) = L(v), ∀ v ∈ V,
(M) find u ∈ V such that F(u) ≤ F(v), ∀ v ∈ V,
where a(u,v) is a bilinear form that is symmetric, coercive and contin- uous on V and L(v) is continuous on V and F(v) = 1
2 a(u,u) −L(v).
As an example we can take
a(u,v) ≡ ∫ 1
0 (k(x)u′v′ + q(x)uv) dx and L(v) ≡
∫ 1 0 f(x)v dx.
Here we have assumed that there are positive constants k0, k1, M such that
(2) k1 ≥ k(x) ≥ k0 > 0, M ≥ q(x) ≥ 0, f ∈ L2(0, 1).
These are sufficient for the symmetry, coercivity and continuity of the bilinear form a(., .) and the continuity of the linear form L(v).
The proof of these properties follows from the following theorem:
Theorem 1. Let u ∈ V ≡ H10 (0, 1). Then the following inequalities are valid:
(3) |u(x)|2 ≤ C1
∫ 1 0
(u′(x))2dx for any x ∈ (0, 1),∫ 1 0 u2(x)dx ≤ C0
∫ 1 0
(u′(x))2dx.
with constants C0 and C1 that are independent of u. 1
2 MATH610: NUMERICAL METHODS FOR PDES: RAYTCHO LAZAROV
Proof: We give two proofs. The simple one proves the above inequali- ties with C0 = 1/2 and C1 = 1. The better proof establishes the above inequalities with C0 = 1/6 and C1 = 1/4.
Indeed, for any x ∈ (0, 1) we have:
u(x) = u(0) + ∫ x
0 u′(s)ds.
Since u ∈ H10 (0, 1) then u(0) = 0. We square this equality and apply Cauchy-Swartz inequality:
(4) |u(x)|2 = ∣∣∣∫ x
0 u′(s)ds
∣∣∣2 ≤ ∫ x 0
1ds ∫ x
0 (u′(s))2ds ≤ x
∫ x 0
(u′(s))2ds.
Taking the maximal value of x on the right hand side of this inequality we get the first inequality (3) with C1 = 1. Further, increasing the r.h.s by taking the integral in the whole interval and then integrating the obtained inequality for x ∈ (0, 1) we get the second inequality (3) with C0 = 1/2.
This simple proof uses only one boundary condition u(0) = 0. We can improve the constants if we use also the second boundary condition u(1) = 0. Namely, we derive in the same manner the inequality
(5) |u(x)|2 = | ∫ 1 x u′(s)ds |2 ≤
∫ 1 x
1ds ∫ 1 x
(u′(s))2 ≤ (1 −x) ∫ 1 x
(u′(s))2ds.
Now one multiplies (4) by 1 −x and (5) by x and adds the two inequalities to get the estimate:
|u(x)|2 ≤ x(1 −x) ∫ 1
0 (u′(s))2ds.
This will allows us to show (3) with C0 = 1/6 and C1 = 1/4, correspondingly. The appropriate inequalities for an arbitrary l are obtained by change of
the variable. Now the coercivity of the bilinear form follows easily from these inequal-
ities and the assumptions (2). Indeed,
a(u,u) ≥ k0 ∫ 1
0 u′2dx ≥
k0 2
∫ 1 0
(u′2 + u2)dx ≥ k0 2 ‖u‖2V .
2. Abstract form of Ritz-Galerkin method
Instead of (V ), we shall consider its approximation. Namely, let Vh be a n-dimensional subspace of V . We consider the following simpler problem:
(Vh) find uh ∈ Vh such that a(uh,v) = L(v), ∀ v ∈ Vh. One can show that this problem is equivalent to the following minimization problem in Vh:
(Mh) find uh ∈ Vh such that F(uh) ≤ F(v), ∀ v ∈ Vh. There some simple but important for the applications properties that
are easily obtaind from the equivalence of the problems (Vh) and (Mh). Using the fact that a(u,u) = L(u) and a(uh,uh) = L(uh) we get from the inequality F(u) ≤ F(uh) that
1 2 a(u,u) −L(u) ≤
1 2 a(uh,uh) −L(uh) =⇒ −
1 2 a(u,u) ≤−
1 2 a(uh,uh)
LECTURE # 3: ABSTRACT RITZ-GALERKIN METHOD 3
which gives a(u,u) ≥ a(uh,uh). This inequality has a clear physical inter- pretation.
Since Vh is n-dimensional, we can assume that it is spanned by n linearly independent functions φj(x) ∈ V, j = 1, ...,n; i.e.
Vh =
v : v(x) =
n∑ j=1
cjφj(x), cj are arbitrary constants
.
We may relate the parameter h to n by h = 1/n.
Examples of Spaces Vh:
Example 1:
φj(x) = sin(jπx), j = 1, ...,n. This will produce the so-called spectral method.
Example 2:
φj(x) = xj(1 −x), j = 1, ...,p. The so-called p-version of FEM.
Example 3:
The construction of the space is done in the following manner: split the interval [0, 1] into n + 1 subintervals by introducing the points xj = jh, j = 0, ...,n + 1, where h = 1
n+1 . The space Vh consists of all continuous
functions on [0, 1] that are linear on each subinterval (element) [xj−1,xj] and vanish at x = 0 and x = 1. Obviously, the functions in the space Vh are determined by their values at the nodes xj, j = 1, ...,n. Solving (Vh) with such space Vh will lead to the finite element method.
The following set of functions can serve as a basis for Vh:
(6) φj(x) =
x−xj−1 h
, x ∈ [xj−1,xj]; xj+1 −x
h , x ∈ [xj,xj+1];
0, elsewhere ;
j = 1, ...,n. The functions are constructed in such way that φj(x) is 1 at the node xj, 0 at all remaining nodes, and linear over the finite elements. This basis is called a nodal basis. Note, that this is just one possible basis. Another example is the so-called hierarchical basis.
Obviously, the solution uh of the problem (Vh) is in the form
uh(x) = n∑ j=1
ξjφj(x), where ξj are unknown constants.
4 MATH610: NUMERICAL METHODS FOR PDES: RAYTCHO LAZAROV
0 1x
j
j
φ
Figure 1. A nodal basis function for linear finite elements
Then the method (Vh) can be written in the form
a(uh,v) = l(v), ∀v ∈ Vh =⇒ a
n∑ j=1
ξjφj(x),φk
= L(φk), k = 1, ...,n.
(7)
This produces a linear system called also (Ritz or Galerkin system) for the unknown ξ ∈ Rn:
n∑ j=1
ξja(φj,φk) = L(φk), k = 1, ...,n, in matrix form Aξ = b,
where A ≡{ajk}nj,k=1 = {a(φj,φk)} n j,k=1, is a square n×n matrix
and b = {L(φj)}nj=1, and ξ = {ξj} n j=1 are vector-columns in R
n.
The matrix A is often called “stiffness” matrix while b is the “load” vector which is computed from the data. Since the bilinear form a(., .) is coercive the matrix A is nonsingular (show this) and therefore the system Aξ = b has unique solution for any b. However, the condition number of A play an importnat role in the numerical methods for solving the system and there is a necessity to discuss this in details.
3. Mixed boundary conditions
For the boundary value problem
(D) −(k(x)u′)′ + q(x)u = f(x), in (0, 1)
u(0) = 0, u(1) + k(1)u′(1) = β1
we introduced the space V :
(8) V =
v :
v(x) is continuous function on (0, 1); v′(x) exists in a generalized sense and is in L2(0, 1); v(0) = 0
and the variational formulation (V ) with
a(u,v) ≡ ∫ 1
0 (k(x)u′v′ + q(x)uv) dx + u(1)v(1)
LECTURE # 3: ABSTRACT RITZ-GALERKIN METHOD 5
and
L(v) ≡ ∫ 1
0 f(x)v dx + β1v(1).
Examples of Spaces Vh for the problem (D):
Example 1:
φj(x) = sin((j − 0.5)πx), j = 1, ...,n. This will produce the so-called spectral method.
Example 2:
φj(x) = xj, j = 1, ...,p. This will produce the so-called p-version of Galerkin method.
Example 3: (the finite element method)
The construction of the space is done in the following manner: split the interval [0, 1] into n subintervals by introducing the points xj = jh, j = 0, ...,n, where h = 1
n . The space Vh consists of all continuous functions on
[0, 1] that are linear on each subinterval (element) [xj−1,xj] and vanish at x = 0. Obviously, the space Vh is determined by the values of a function at the nodes xj, j = 1, ...,n. Solving (Vh) with such space Vh will lead to the finite element method.
In this case, a nodal basis will consist of all functions from the previous example corresponding to internal nodes as well as one more function that will involve the value at end xn = 1: The following set of functions can serve as a basis for Vh:
(9) φn(x) =
x−xn−1 h
, x ∈ [xn−1,xn];
0, elsewhere.
Another possible basis in Vh is the so-called hierarchical basis which utilizes hierarchy of grids.
4. Neumann boundary conditions
Now we shall consider the following simple model problem for the un- known function u(x):
(D) −u′′ + u = f(x), in (0, 1)
u′(0) = 0. u′(1) = 0.
In the previous lecture we have introduced the set of functions defined on (0, 1) that are is continuous function, have piece-wise continuous deriva- tive.This set has been equipped with the norms
||v||2 = (v,v) and ||v||2V = (v,v)V = (v,v) + (v ′,v′).
6 MATH610: NUMERICAL METHODS FOR PDES: RAYTCHO LAZAROV
After completing the set V in the norm || · ||V we get the Sobolev space H1(0, 1) of functions having generalized first derivatives in L2(0, 1). Note, that the functions in V do not satisfy any boundary conditions. Therefore, V ≡ H1(0, 1).
We also introduced the following variational problem:
(V ) find u ∈ V such that a(u,v) = L(v), ∀ v ∈ V, where
a(u,v) ≡ ∫ 1
0 (u′v′ + uv) dx and L(v) ≡
∫ 1 0 f(x)v dx.
We shall study the Ritz system for this particular BVP. It is obvious, that a(u,v) = (u,v)V so this form is trivially coercive.
We have introduced the following finite dimensional space: split the in- terval [0, 1] into n− 1 subintervals by introducing the points xj = (j − 1)h, j = 1, ...,n, where h = 1
n−1 ; the space Vh consists of all continuous on [0, 1] functions that are linear on each subinterval (element) [xj−1,xj]. Obviously, the functions in the space Vh can be determined by their values at the nodes xj, j = 1, ...,n. The approximate problem (Vh) for such space Vh will lead to the finite element method with linear elements.
The following set of functions can serve as a basis for Vh:
(10) φj(x) =
x−xj−1 h
, x ∈ [xj−1,xj]; xj+1 −x
h , x ∈ [xj,xj+1];
0, elsewhere ;
j = 2, ...,n− 1 and two additional functions defined at the end-points: (11)
φ1(x) =
x2 −x h
, x ∈ [x1,x2];
0, elsewhere ; φn(x) =
x−xn−1 h
, x ∈ [xn−1,xn];
0, elsewhere .
The functions are constructed in such way that φj(x) is 1 at the node xj, 0 at all remaining nodes and linear over the finite elements. This basis is called nodal basis.
The solution uh of the problem (Vh) is in the form
uh(x) = n∑ j=1
ξjφj(x), where ξj are unknown constants.
Then the method (Vh) can be written in the form (7). This basis of the space Vh will produce a tridiagonal matrix A. Indeed,
a(φj,φk) = 0, for |j −k| > 1. Also, for j = k we get
a(φj,φj) = ∫ xj+1 xj−1
( 1 h2
+ φ2j (x) ) dx =
2 h
+ 2h 3 , for 1 < j < n,
a(φ1,φ1) = ∫ x2 x1
( 1 h2
+ φ21(x) ) dx =
1 h
+ h
3 , for j = 1,
a(φn,φn) = ∫ xn xn−1
( 1 h2
+ φ2n(x) ) dx =
1 h
+ h
3 , for j = n.
LECTURE # 3: ABSTRACT RITZ-GALERKIN METHOD 7
Similarly, for k = j + 1 we get
a(φj,φj+1) = ∫ xj+1 xj
( −1 h2
+ φj(x)φj+1(x) ) dx =
−1 h
+ h
6 .
The coefficients below the main diagonal are recover from the symmetry of the matrix A.
Thus, the matrix A of the Ritz-system has the form A = A0 + A1, where:
(12) A1 = 1 h
1 −1 0 . . . 0 −1 2 −1 . . . 0
0 −1 2 . . . 0 . . . . . . . 0 0 0 . . . 1
, A0 = h6
2 1 0 . . . 0 1 4 1 . . . 0 0 1 4 . . . 0 . . . . . . . 0 0 0 . . . 2
.
Matrix A1 is called “stiffness” matrix, while the matrix A0 is called “mass” matrix. Both matrices are symmetric and A0 is positive definite while A1 is semi-definite.
5. Issues to be addressed
In the genral case we are facing the following issues: • to assemble the matrix A and to solve the system Aξ = b; • alternatively, if an iterative method is used that requires only the
matrix-vector multiplication Aξ, then one should prepare a pro- cedure of matrix vector multiplication (possibly without explicitly forming the matrix A); • estimate the condition number of the matrix A for a prticular chice
of the basis of the space Vh; • estimate the error e = u−uh; • to develope an algorithm that adaptively choses the mesh so that
the error is uniformly distributed in the domain and is dreven below a desired level.
We need to develop the mathematical tools for studying these problems. This includes: estimate for the condition number of A, deriving/finding fast methods for solving the system, proving various integral inequalities, deriv- ing the approximation error with piece-wise polynomial functions, estimates in various Sobolev norms, etc.
6. An estimate of the condition number of the global matrix A for Neumann BC
Further, we shall use the following definition of a condition number of a symmetric and positive definite matrix:
cond(A) = max λ(A) min λ(A)
,
where λ(A) is an eigenvalue of A, i.e. Aξ = λξ, for some ξ a nonzero vector in Rn.
Often it is not possible to compute the eigenvalues and the condition number, but for practical purposes it is enough to have an upper bound for cond(A). For this we need upper and lower bounds for the eigenvalues of A.
8 MATH610: NUMERICAL METHODS FOR PDES: RAYTCHO LAZAROV
Simple calculations show that h
6 ≤ λ(A0) ≤ h and 0 ≤ λ(A1) ≤
4 h .
So we produce the following bound from above for the condition number of the matrix A
(13) cond(A) ≤ max λ(A0) + max λ(A1) min λ(A0) + min λ(A1)
≤ 4/h + h h/6
= 24 h2
+ 6 = O(h−2).
Remark 1. Note, that A0 and A1 are square matrices of size n and one finds that cond(A0) ≤ 6 i.e. the condition number of A0 does not depend on the size of the matrix. Such matrices are called well-conditioned. In contrast, A1 has condition number O(h−2) which increases quadratically, when h → 0. Such matrices are called ill-conditioned.
7. Exercises
The following matrices play essential role in the finite element, finite vol- ume and finite difference methods for two-point boundary value problems and the solution of the corresponding linear systems. The spectral properties of these matrices are used very often in the computational practice.
(1) Find the exact eigenvalues of the matrices B1,B0 ∈ Rn×n given by
(14) B1 =
2 −1 0 . . . 0 0 −1 2 −1 . . . 0 0
0 −1 2 . . . 0 0 . . . . . . . . 0 0 0 . . . −1 2
,
and
(15) B0 =
4 1 0 . . . 0 0 1 4 1 . . . 0 0 0 1 4 . . . 0 0 . . . . . . . . 0 0 0 . . . 1 4
.
Hint: Show that λj(B1) = 4 sin2 πj
2(n+1) , j = 1, . . . ,n and then use
the fact that B1 +B0 = I, where I is the identity matrix in Rn. From these calculations follow that both B1 and B0 are positive definite.
(2) Estimate that eigenvalies of the scaled “stiffness” matrix B1 ∈ Rn×n
(16) B1 =
1 −1 0 . . . 0 0 −1 2 −1 . . . 0 0
0 −1 2 . . . 0 0 . . . . . . . . 0 0 0 . . . −1 1
,
and the scaled “mass” matrix B0 ∈ Rn×n
(17) B0 =
2 1 0 . . . 0 0 1 4 1 . . . 0 0 0 1 4 . . . 0 0 . . . . . . . . 0 0 0 . . . −1 2
.
LECTURE # 3: ABSTRACT RITZ-GALERKIN METHOD 9
Remark 2. Using the technique applied above we can show that in this case the eigenvalues are λj(B1) = 4 sin2
πj 2(n−1) , j = 0, . . . ,n− 1.
Remark 3. The eigenvalues and eigenvectors of these algebraic problems and problems obtained by approximation of the same differential operator with third type boundary conditions could be found in the monograph of Samarskii [6, pp. 104–109].
References
[1] L. C. Evans, Partial Differential Equations, Graduate Studies in Mathematics, vol. 19, AMS, 1998.
[2] Ch. Grossmann, H.-O. Ross, and M. Stynes, Numerical Treatment of Partial Differ- ential Equations, Springer, Berlin, 2005.
[3] M. Renardy and R. Rogers, An Introduction to Partial Differential Equations, Texts in Applied Mathematics, Springer-Verlag, 1993.
[4] P. Knabner and L. Angermann, Numerical Methods for Elliptic and Parabolic PDEs, Springer-Verlag, New Yrok Inc, 2003.
[5] S. Larsen and V. Thomee, Partial Differential Equations with Numerical Methods, Springer, 2003.
[6] A.A. Samarskii, The Theory of Difference Schemes, Monographs and Textbooks in Pure and Appled Mathematics, Marcel Dekker, Inc, New York, 2001.
Lecture_08_S08.pdf
LECTURE # 8: MULTIDIMENSIONAL SECOND ORDER ELLIPTIC
PROBLEMS
MATH610: NUMERICAL METHODS FOR PDES – R. LAZAROV
1. Introduction and preliminaries
First, we introduce some notations that will be used further. Here Ω will denote a polygonal bounded domain in Rd, d = 2, 3 with boundary ∂Ω. Further, for the vector q = (q1, . . . , qd) and for a scalar function v we define the divergence ∇ · q and the gradient ∇v, correspondingly, by
∇ · q = ∂q1 ∂x1
+ · · · + ∂qd ∂xd
and ∇v = (
∂v
∂x1 , . . . ,
∂
∂xd
) .
The Stokes theorem will be used in the following form: ∫
∂Ω q · n ds =
∫
Ω ∇ · q dx.
Here, n is the outward unit vector to ∂Ω and q·n denotes the inner product of two vectors on Rd.
We shall use the Hilbert space H1(Ω) of functions defined on Ω and having their generalized derivatives in L2(Ω). The subspace of those functions in H1(Ω) that vanish on the boundary ∂Ω will be denoted by H10 (Ω). The L2 and H1-inner products of these spaces and the corresponding norms are defined as follows:
(u, v) = ∫
Ω uv dx, (u, v)1 = (u, v) + (∇u, ∇v),(1)
‖u‖ = (u, u)1/2, ‖u‖1 = (u, u)1/21 .(2) For the elements in the space H1(Ω) we shall use the following Poincare
inequality:
(3) ∫
Ω u2 dx ≤ M0
∫
Ω |∇u|2 dx
where the constant M0 > 0 does not depend on u. We shall give proof of this inequality for d = 2.Without loss of generality,
we can assume that Ω is contained in the unit square Π, i.e. Ω ⊂ Π := (0, 1) × (0, 1). Then we can extend a function u ∈ H1(Ω) to Π by zero
1
2 MATH610: NUMERICAL METHODS FOR PDES – R. LAZAROV
outside Ω. The extended function is denoted by ū. It belongs to H10 (Π) and obviously,
∫
Π ū2 dx =
∫
Ω u2 dx.
Next, we write the equality
2 ∫
Π ū2 dx =
∫
Π
{(∫ x1 0
∂
∂x1 ū(ξ, x2) dξ
)2 dx(4)
+ (∫ x2
0
∂
∂x2 ū(x1, ξ) dξ
)2 } dx
and apply Cauchy-Schwarz inequality to each of the line integrals:
2 ∫
Π ū2 dx ≤
∫
Π
{ x1
∫ 1 0
( ∂
∂x1 ū(ξ, x2) dξ
)2 (5)
+x2 ∫ 1
0
( ∂
∂x2 ū(x1, ξ) dξ
)2 } dx.
Using Fubini theorem,we get finally:
2 ∫
Ω u2 dx = 2
∫
Π ū2 dx ≤ 1
2
∫
Π |∇ū|2 dx(6)
= 1 2
∫
Ω |∇u|2 dx,
which is the required inequality with M0 = 1/4. If the domain Ω is contained in a rectangle (0, l1)×(0, l2) the required inequality follows by change of the variables.
Further, we shall need the following two inequalities valid for functions in H1(Ω):
(7) ∫
∂Ω u2 ds ≤ C‖u‖21,
and
(8) ∫
Ω u2 dx ≤ C
{∫
Ω |∇u|2 ds +
∫
∂Ω u2 ds
} .
Here the constant C does not depend on u but depend on the domain Ω. One can prove these inequalities for rectangular domains simply by using the corresponding estimates from the one-dimensional case. The proofs are left as an exercise for this part of the class (see, e.g. [3, 7]).
MULTIDIMENSIONAL ELLIPTIC PROBLEMS 3
2. Problem formulation
In this lecture we shall consider the following Dirichlet boundary-value problem: find u(x) such that:
(D) Lu := ∇ ·
( −K(x)∇u + b(x)u
) + q(x)u = f (x), x ∈ Ω
u(x) = 0, x ∈ ∂Ω.
where the coefficients K(x), b, q and f are given functions on Ω. We assume that Ω is a bounded domain with Lipschitz boundary ∂Ω, K(x) is a symmetric and uniformly in Ω positive definite matrix and the coefficients K(x), b(x), q(x) are measurable and bounded function in Ω.
This is the divergent form of the problem. Quite often second order problems are given in the following non-divergent form:
(9) Lu := ∇ · (−K(x)∇u) + b̃(x)∇u + q(x)u = f (x), x ∈ Ω
u(x) = 0, x ∈ ∂Ω.
If the vector field b̃(x) is differentiable then these two forms are equivalent. In case when b ≡ b̃ and ∇ · b = 0, then these two form coinside.
In some applications this equation describes: (1) deflection of an elastic membrane under transverse load f (then K = I, b ≡ 0, q ≡ 0); (2) the pressure distribution in a porous media (K is the permeability tensor, b ≡ 0, q ≡ 0); (3) concentration distribution of a chemical in a flow with velocity b and absorption coefficient q. The quantity
q(x) = −K(x)∇u + b(x)u is often called total flux (mass, thermal, etc) with −K(x)∇u the diffusive part and b(x)u convective part of the flux.
For deriving the variational formulation of this problem we follow the standard approach used in the 1-dimensional problems. We multiply the differential equation (D) by a test function v ∈ H10 (Ω) and integrate over Ω:
∫
Ω
( ∇ · (−K(x)∇u + b(x)u) + q(x)u
) v dx =
∫
Ω f (x)v dx.
We use the identity ( ∇ · (−K(x)∇u + b(x)u)} v = ∇ · {(−K(x)∇u + b(x)u) v
) (10)
− ( − K(x)∇u + b(x)u
) · ∇v,
4 MATH610: NUMERICAL METHODS FOR PDES – R. LAZAROV
so that after applying the Stokes theorem we transform the right hand side of the above identity to the form:
∫
∂Ω
( K(x)∇u − b(x)u
) · n v ds +
∫
Ω
( K(x)∇u − b(x)u
) · ∇v dx.
Now we use the fact that v vanishes on ∂Ω to get ∫
Ω
( (K(x)∇u − b(x)u) · ∇v + q(x)uv
) dx =
∫
Ω f (x)v dx.
We rewrite this integral identity in the abstract form
a(u, v) = L(v) ∀ v ∈ H10 (Ω), where
a(u, v) = ∫
Ω
( K(x)∇u · ∇v − ub(x) · ∇v + q(x)uv
) dx
and
L(v) = ∫
Ω f (x)v dx.
Thus, we have shown that the solution of the problem (D) satisfies the following variational problem:
(V ) find u ∈ H10 (Ω) such that a(u, v) = L(v), ∀ v ∈ H10 (Ω) .
So we have reformulated the differential problem (D) in terms of integral identity involving the bilinear form a(·, ·) and the linear form L(·). Again, we can use the general theoretical framework and Lax-Milgram theorem to show the existence and the uniqueness of the solution u ∈ H10 (Ω). We shall prove that under reasonable conditions on the coefficients the bilinear form a(·, ·) is coercive and continuous in V = H10 (Ω) so we can apply the general theoretical framework for such problems.
Now we give conditions on the coefficients of the differential equation (D) that are sufficient for the coercivity and the continuity of the bilinear form a(·, ·):
(C) ξT K(x)ξ ≥ k0ξT ξ, ∀ ξ ∈ Rd, k0 = const > 0,
q(x) + 1 2 ∇ · b(x) ≥ 0, ∀ x ∈ Ω.
Theorem 1. Assume that the conditions (C) are satisfied. Then the bilinear form a(·, ·) is coercive and continuous in V , i.e. there are positive constants α and C such that
(11) a(u, u) ≥ α‖u‖21, (coercivity) a(u, v) ≤ C0‖u‖1 ‖v‖1. (continuity)
MULTIDIMENSIONAL ELLIPTIC PROBLEMS 5
Proof: First, we note that
(12) −u b · ∇v = −1 2 ∇ · (bu2) + 1
2 u2 ∇ · b.
Then applying the Stokes theorem and the condition (C) and the fact that v vanishes on ∂Ω we get the following for for a(u, u)
a(u, u) = ∫
Ω (K∇u · ∇u + (q + 0.5∇ · b)u2) dx ≥ k0||∇u||2.
Using Poincare inequality (3) we get the desired result regarding the coer- civity. The continuity of the bilinear from is a simple consequence of the boundness of the coefficients.
Let Vh be a finite dimensional subspace H10 (Ω). The Ritz-Galerkin method can be formulated in the already discussed abstract form:
(Vh) find uh ∈ Vh ⊂ H10 (Ω) such that a(uh, v) = L(v), ∀ v ∈ Vh.
Our goal now is to construct the space Vh and to show how the Ritz-system derived from (Vh) is computed and solved.
3. Other types of boundary conditions
Instead of Dirichlet boundary conditions one can put various other types of boundary conditions on ∂Ω. Below we give two natural boundary condi- tions that are widely used in the applications.
Case b ≡ 0; Then we have diffusion-reaction equation and the following Robin condition can be prescribed on the whole boundary ∂Ω or on part of it:
(13) K(x)∇u · n + σ(x)u = g(x) ∀ x ∈ ∂Ω. Here σ(x) ≥ 0 and g(x) are given functions on ∂Ω. If σ(x) ≡ 0 then this is the classical Neumann boundary condition. The meaning of this boundary condition is that we prescribe the diffusive flux on ∂Ω. If σ(x) ≡ g(x) ≡ 0 then no flux is allowed through ∂Ω. This is typical insulated boundary (in thermal problem) or no-flow boundary in porous media applications.
The weak formulation of this boundary-value problem is obtained in the same way as in the case of Dirichlet boundary conditions. Then after inte- gration by parts and using the boundary conditions we get that u ∈ H1(Ω) satisfies the following integral identity:
a(u, v) = L(v) ∀ v ∈ H1(Ω), where
a(u, v) = ∫
Ω {K(x)∇u · ∇v + q(x)uv} dx +
∫
∂Ω σu v ds
and L(v) =
∫
Ω f (x)v dx +
∫
∂Ω g(x) v ds.
6 MATH610: NUMERICAL METHODS FOR PDES – R. LAZAROV
Note, that the functions in the solution space doe not satisfy any boundary conditions. The bilinear form is coercive in H1(Ω) under the condition that σ(x) ≥ σ0 = const > 0 ∀ x ∈ ∂Ω. In fact, it is enough that σ(x) ≥ σ0 > 0 on a part of the boundary with a positive measure. Indeed, we have
a(u, u) ≥ k0 ∫
Ω |∇u|2 dx + σ0
∫
∂Ω u2 ds.
Next, we use the embedding inequality (8) to get the missing ‖u‖2-term in the coercivity.
Similarly,
|L(v)| ≤ ∫
Ω |f v| dx +
∫
∂Ω |g v| ds(14)
≤‖f‖‖v‖ + (∫
∂Ω |g|2 ds
)1/2 (∫
∂Ω |v|2 ds
)1/2 .(15)
Finally, we use the estimate (7) to get the required continuity of the linear form L(v).
Case b 6≡ 0; This diffusion-convection-reaction equation and the fol- lowing boundary conditions are quite natural (together with Dirichlet BC). First, split the boundary ∂Ω into two parts: ∂Ω = Γin ∪ Γout, where
Γin = {x ∈ ∂Ω : b(x) · n(x) < 0}, Γout = {x ∈ ∂Ω : b(x) · n(x) ≥ 0}.
Then the following boundary conditions are natural:
(16) −K(x)∇u · n = 0, x ∈ Γout,
−K(x)∇u · n + u b · n = g(x), x ∈ Γin. The physical meaning of these boundary conditions is the following: on the part of the boundary where the flow enter the domain, i.e. b(x) · n(x) < 0 we can prescribe either the function or the total flux q.
Then one gets the following form
(Lu, v) = ∫
Ω {K(x)∇u · ∇v − b · ∇vu + q(x)uv} dx(17)
+ ∫
Γout b · nu v ds +
∫
Γin g v ds
Now we define the bilinear from a(·, ·) and the linear form L(·) as
a(u, v) = ∫
Ω {K(x)∇u · ∇v − b · ∇vu + q(x)uv} dx +
∫
Γout b · nu v ds
and
L(v) = ∫
Ω f vdx +
∫
Γin g v ds.
MULTIDIMENSIONAL ELLIPTIC PROBLEMS 7
Ωb
Γ out
Γ out
Γ out
Γ out
Γ in
Γ in
Γ in
Figure 1. Domain with inflow Γin and outflow Γout boundaries
One can easily prove that if the coefficients of the differential equation satisfy one of the conditions
(A) q(x) + 1 2 ∇ · b(x) ≥ c0 = const > 0 ∀ x ∈ Ω,
(B) q(x) + 1 2 ∇ · b(x) ≥ 0 and the measure of the set Γin is nonzero,
(C) q(x) + 1 2 ∇ · b(x) ≥ 0 and the measure of the set Γout is nonzero,
then the corresponding bilinear form will be coercive in H1(Ω)-norm. Indeed, using (12) by Stokes’ theorem
a(u, u) = ∫
Ω (K∇u · ∇u + (q + 0.5∇ · b)u2) dx(18)
− 1 2
∫
Γin b · nu2ds + 1
2
∫
Γout b · nu2ds.
which is the same as
a(u, u) = ∫
Ω (K∇u · ∇u + (q + 0.5∇ · b)u2) dx + 1
2
∫
∂Ω |b · n|u2ds.
Obviously one of the conditions (A) – (C) guarantee the coercivity of the bilinear form in H1(Ω)-norm. Then, by Lax-Milgram Theorem, we get the desired result about existence and uniqueness of the solution of the equation (D) with boundary conditions (16).
For this problem the following maximum principle could be shown
Theorem 2. ([6, Theorem 31., p. 26]) Consider the differential operator L of problem (D) and assume that u ∈ C2(Ω̄) and
Lu ≤ 0 (Lu ≥ 0) in Ω.
8 MATH610: NUMERICAL METHODS FOR PDES – R. LAZAROV
(i) If q = 0, then
max x∈Ω̄
u(x) ≤ max x∈∂Ω
u(x) (
min x∈Ω̄
u(x) ≤ min x∈∂Ω
u(x) )
(ii) If q ≥ 0, then
max x∈Ω̄
u(x) ≤ max(max x∈∂Ω
u(x), 0) (
min x∈Ω̄
u(x) ≤ min( min x∈∂Ω
u(x), 0) )
This theorem allows us to study the uniqueness and the stability of the solution of the problem (D) in maximum-norm. This is quite useful in many applications. However, the natural way to numerically address this problem is to use the variational form of the problem (D) and derive approximation schemes generated by the finite element method.
4. Abstract Galerkin method
We take Vh is a finite dimensional subspace of V (denote the dimension of Vh by n) and consider the variational problem on Vh:
(Vh) find uh ∈ Vh such that a(uh, v) = L(v), ∀ v ∈ Vh. Let {φj (x)}ni=1 be a basis for Vh, so that for uh, v ∈ Vh we have:
(19) uh(x) = n∑
i=1
Uiφi(x) and v(x) = n∑
i=1
Viφi(x).
The parameters in the vector-column U T = (U1, . . . , Un1)T are called de- grees of freedom for the finite element method and are obtained from the Galerkin system AU = b. Here the entries of the matrix A are aij = a(φi, φj ) and the vector-load b has components bj = L(φj ). Since we have assumed coercivity of the bilinear from a(·, ·), the matrix A is nonsingular and there- fore the Galerkin system has unique solution.
Note that the matrix A is non-symmetric (as long as b 6= 0). Moreover, for convection dominated problems that is when b is much larger than K (in some norm) and this causes a number of serious problems for any numerical method.
Our main goal in this class will construction of appropriate (and practi- cally feasible) finite dimensional spaces Vh, the spaces of piece-wise polyno- mial functions over a partition of Ω into finite elements. The basic texts we shall use in this class are the textbooks of Larsen and Thomée [6] or [5] and the monographs of Ciarlet [1], and Ern and Guermond [2].
MULTIDIMENSIONAL ELLIPTIC PROBLEMS 9
References
[1] P.G. Ciarlet, The Finite Element Method for Elliptic Problems, Classics of Applied Mathematics, v. 40, SIAM, 2002.
[2] A. Ern and J.-L. Guermond, Theory and Practice of Finite Elements, Series of Applied Mathematical Sciences v. 159, Springer-Verlag, 2004.
[3] L. C. Evans, Partial Differential Equations, Graduate Studies in Mathematics, volume 19, American Mathematical Society, 1991.
[4] D. Kinkaid and W. Cheney, Numerical Analysis. Mathematics of Scientific Comput- ing, Third Edition, Brooks/Cole, 2002.
[5] P. Knabner and L. Angermann, Numerical Methods for Elliptic and Parabolic PDEs, Springer-Verlag, New Yrok Inc, 2003.
[6] S. Larsen and V. Thomée, Partial Differential Equations with Numerical Methods, Springer-Verlag, Texts in Applied Mathematics 45, 2003.
[7] M. Renardy and R. Rogers, An Introduction to Partial Differential Equations, Texts in Applied Mathematics 13, Springer-Verlag, 1993.
5. Appendix: Abstract variational problem:
We recall for the previous lectures the following general framework for elliptic equations.
Let V be a Hilbert space with an inner product (·, ·)V and corresponding norm || · ||V . Let the bilinear form a(u, v) defined on V D × V and the linear form L(v) defined on V are such that:
(1) a(u, v) is coercive in V , i.e., there is a constant α > 0 such that a(v, v) ≥ α||v||2V , ∀ v ∈ V ;
(2) a(u, v) is continuous, i.e., there is a constant C > 0 such that a(u, v) ≤ C0||u||V ||v||V , ∀ u, v ∈ V ;
(3) L(v) is continuous in V , i.e., there is a constant Λ > 0 such that L(v) ≤ Λ||v||V , ∀ v ∈ V .
The following theorem is a particular case of the well-know Lax-Milgram theorem for NON-SYMMETRIC bilinear forms (see e.g. [7] for the proof of the symmetric and your lecture notes, Lecture # 7, for non-symmetric forms a(·, ·)): Theorem 3. (Lax-Milgram) Let V be the Hilbert space with an inner product (u, v)V and let the conditions (1) - (3) holds true. Then the problem find u ∈ V s.t. (20) a(u, v) = L(v), ∀ v ∈ V has unique solution u ∈ V . Furthermore, the solution satisfies the stability estimate:
‖u‖V ≤ Λ α
.
Lecture_09_S08.pdf
LECTURE # 9:
INTRODUCTION TO FEM FOR SECOND ORDER ELLIPTIC EQUATIONS
MATH610: NUMERICAL METHODS FOR PDES – R. LAZAROV
As introduced before, Ω will be a polygonal bounded domain in Rd, d = 2, 3 with boundary ∂Ω. Further, for the vector q = (q1, . . . ,qd) and for a scalar function v we define the divergence ∇ · q and the gradient ∇v, correspondingly, by
∇· q = ∂q1 ∂x1
+ · · · + ∂qd ∂xd
and ∇v = ( ∂v
∂x1 , . . . ,
∂
∂xd
) .
No we consider the following model boundary-value problem:
(D)
find u(x) such that:
Lu := −∇·∇u + u := −∆u + u = f(x), x ∈ Ω
∇u · n := ∂u
∂n (x) = 0, x ∈ ∂Ω.
where a(u,v) = ∫
Ω {∇u ·∇v + uv} dx and L(v) =
∫ Ω f(x)v dx
The variational formulation of this problem is:
(V ) find u ∈ V := H1(Ω) such that a(u,v) = L(v), ∀ v ∈ H1(Ω),
where a(u,v) = ∫
Ω {∇u ·∇v + uv} dx and L(v) =
∫ Ω f(x)v dx.
So we have reformulated the differential problem (D) in terms of integral identity involving the bilinear form a(·, ·) and the linear form L(·). Again, we can use the general theoretical framework and Lax-Milgram theorem to show the existence and the uniqueness of the solution u ∈ H1(Ω). Obviously, the bilinear form a(·, ·) is coercive and continuous in V = H1(Ω) so we can apply the general theoretical framework for such problems.
Date: September 20, 2010.
1
2 MATH610: NUMERICAL METHODS FOR PDES – R. LAZAROV
Ω
τ 1
τ 2
τ 3
Figure 1. Left: Triangulation of a polygonal domain; Right: non-conforming triangulation, which is not allowed in our current considerations
Let Vh be a finite dimensional subspace H1(Ω). The Ritz-Galerkin method can be formulated in the already discussed abstract form:
(Vh) find uh ∈ Vh ⊂ H1(Ω) such that a(uh,v) = L(v), ∀ v ∈ Vh.
Our goal now is to construct the space Vh and to show how the system of linear equations derived from (Vh) is computed and solved.
We point out that the boundary conditions of the differential problem are not imposed on the functions from the space H1(Ω). This condition is weakly contained in the variational formulation itself. It is more natural to begin with such a problem, since we do not need to impose any boundary conditions on the finite dimensional subspace of H1(Ω).
1. FE partition of the domain and FE spaces
We partition the domain Ω into triangular (tetrahedral) finite elements τ. The finite elements τ are considered open sets and we denote their closure by τ, i.e. τ = τ ∪ ∂τ. This triangulation is denoted by Th. We shall consider conforming types of triangulation, i.e. triangulations that satisfy the following conditions:
(1)
(a) τ are disjoint, i.e. τi ∩ τj = ∅, i 6= j;
(b) τi ∩ τj is either: (i) a vertex of τi & τj; (ii) an entire edge of τi & τj; (iii) empty .
An example of a triangulation of the domain is shown on Figure 1. The right Figure 1 shows a non-conforming triangulation that is NOT considered in our current setting.
FEM FOR ELLIPTIC PROBLEMS 3
P i
Pj
P k (Pk)=Uk
τ
u (P )=Uh u (P )=Uh
uh
i jji
Figure 2. Liner triangular finite element
Together with set of all triangles Th we shall use the sets of all edges Eh and the sets of all vertices Vh. Further, we define the set Pm of polynomials of degree m with real coefficients:
Pm =
∑
0≤i+j≤m cijx
i 1 x
j 2, cij are real numbers
.
Now we consider the simpleast case of linear finite elements and define the finite-dimensional subspace Vh ⊂ H1(Ω) in the following way:
Vh = { v : v ∈ C0(Ω), v ∈P1(τ), for τ ∈Th
} .
The functions in Vh can be uniquely determined by its values at the vertices of the triangulation Th. We shall use the nodal basis in Vh. If the number of the vertices in Vh is N, then we define N linearly independent functions φj(x), j = 1, . . . ,N by:
φj(x) =
1 if x = Pj, Pj ∈Vh; 0 if x = Pk, Pk ∈Vh, Pk 6= Pj; linear over each τ ∈Th.
2. Finite Element Computations
Each function in Vh can be presented in the form:
uh(x) = N∑ j=1
Ujφj(x), where Uj = uh(Pj), Pj ∈Vh.
Then the finite element method for the problem (V ) reduces to solving the Ritz-Galerkin system of linear equations for the unknown values UT = (U1,U2, . . . ,UN ):
(2) N∑ i=1
Uia(φi,φj) = L(φj), j = 1, . . . ,N, or AU = b.
4 MATH610: NUMERICAL METHODS FOR PDES – R. LAZAROV
The entries of the matrix A are a(φi,φj) and the entries of the load-vector b are L(φj), i,j = 1, . . . ,N. Since the bilinear form a(·, ·) is symmetric and coercive the matrix A is symmetric and positive definite.
The matrix A of the system (2) is computed element-wise. Namely, the contributions of a particular finite element τ to the global “stiffness” and “mass” matrices are done by element-wise computations.
In each element we introduce local notations: let the triangle τ has vertices Pi, Pj, Pk and let the restrictions of the nodal basis functions to τ be denoted again by φi, φj, φk. We denote
Uτ =
UiUj Uk
, Vτ =
ViVj Vk
, Φτ (x) := Φτ =
φi(x)φj(x) φk(x)
,
and similarly for the element functions
∇Φτ (x) := ∇Φτ =
∇φi∇φj ∇φk
:=
∂φi ∂x1
∂φi ∂x2
∂φj ∂x1
∂φj ∂x2
∂φk ∂x1
∂φk ∂x2
.
This allows us to write the following presentations:
uh(x)|τ = ΦTτ Uτ, v(x)|τ = Φ T τ Vτ, ∇uh(x)|τ = ∇Φ
T τ Uτ, ∇v(x)|τ = ∇Φ
T τ Vτ,
so that ∫ τ ∇uh(x) ·∇v(x) dx =
∫ τ Vτ
T∇Φτ∇ΦTτ Uτ dx := V T τ A
1 τUτ,∫
τ uh(x)v(x) dx :=
∫ τ V Tτ Φτ Φ
T τ Uτ dx := V
T τ A
0 τUτ,
and similarly for the r.h.s.∫ τ f(x)v(x) dx :=
∫ τ V Tτ Φτf(x) dx := Vτ
Tbτ.
Here A1e and A 0 e are 3 × 3 matrices, called element “stiffness” and “mass”
matrices, correspondingly, and bτ is a vector of dimension 3, called the element load vector.
3. Element Stiffness and mass matrices for linear FE
One gets a very simple formula for these matrices, namely:
A1τ = ∫ τ
∇φi ·∇φi ∇φi ·∇φj ∇φi ·∇φk∇φj ·∇φi ∇φj ·∇φj ∇φj ·∇φk ∇φk ·∇φi ∇φk ·∇φj ∇φk ·∇φk
dx,
FEM FOR ELLIPTIC PROBLEMS 5
8
9
i−1
y j−1
y j
y j+1
τ1
τ 2
τ 3
τ 4
τ 5
τ 6
0 1
23
4
5 6
h h
h
h
x x i
x i+1
Figure 3. Uniform rectangular grid
and
A0τ = ∫ τ
φiφi φiφj φiφkφjφi φjφj φjφk φkφi φkφj φkφk
dx =
∫ τ φiφi
∫ τ φiφj
∫ τ φiφk∫
τ φjφi
∫ τ φjφj
∫ τ φjφk∫
τ φkφi
∫ τ φkφj
∫ τ φkφk
.
Further, we shall show that in fact the element “mass” matrix A0τ has very simple form, namely:
A0τ = |τ| 12
2 1 11 2 1
1 1 2
,
where |τ| denotes the area of the triangle τ. Obviously, the elements of the mass matrix are of order h2, where h is the diameter of the element τ.
Below we give the nodal basis function associated with the node (xi,yj) (note we are using the notations (x,y) instead of (x1,x2)).
For the element τ1, which is a right triangle and the vertices are ordered in the following way (P0, P1, P2) (see, Figure 3) we can easily compute the element “stiffness” matrix A1τ :
A1τ = 1 2
2 −1 −1−1 1 0 −1 0 1
.
Assembling the local matrices will give the global matrix of the Ritz system. For example, the equation for the internal point point P0 shown on Figure
6 MATH610: NUMERICAL METHODS FOR PDES – R. LAZAROV
finite element hφ0(x,y) h ∂φ0(x,y)
∂x h ∂φ0(x,y)
∂y τ1 h− (x−xi + y −yj) −1 −1 τ2 h− (y −yj) 0 −1 τ3 h + (x−xi) 1 0 τ4 h + (x−xi + y −yj) 1 1 τ5 h + (y −yj) 0 1 τ6 h− (x−xi) −1 0
Table 1. Analytic presentation of the nodal function at the node 0 and its derivatives
3 will be
4U0 −U1 −U2 −U4 −U5 + h2
12 (6U0 + U1 + U2 + U3 + U4 + U5 + U6)(3)
= ∫
Ω f(x)φ0 dx.
Similarly, assembling the equation for the node P4 that is on the Neumann boundary we get the equation:
2U4 −U0 − 1 2 U3 −
1 2 U8+
h2
24 (6U4 + 2U0 + U3 + 2U5 + U8)(4)
= ∫
Ω f(x)φ4 dx.
References
[1] L. C. Evans, Partial Differential Equations, Graduate Studies in Mathematics, v. 19, American Mathematical Society, 1991.
[2] Ch. Grossmann, H.-O. Ross, and M. Stynes, Numerical Treatment of Partial Differ- ential Equations, Springer, Berlin, 2005.
[3] D. Kinkaid and W. Cheney, Numerical Analysis. Mathematics of Scientific Comput- ing, Third Edition, Brooks/Cole, 2002.
[4] S. Larsen and V. Thomee, Partial Differential Equations with Numerical Methods, Springer-Verlag, Texts in Applied Mathematics 45, 2003.
[5] M. Renardy and R. Rogers, An Introduction to Partial Differential Equations, Texts in Applied Mathematics 13, Springer-Verlag, 1993.
Lecture11p.pdf.pdf
Stokes th takes u to
be grade'auf
g nds fagn Fdr e curve
Ingn 509 09 ff arlf.n S ds
T a i SScure Is S
q Cq i9d g Poincare Inqually i I
proof O I n
w L o G Lef r c t Co.in CqD 11 IT
anddefine thefunction b
a Iy U int u CHo Cri even if not
in 4 IF u HoC n bears If r toner seasci D whateverI DX Ludoc ycausesstill zero
Using F T o C but if ue Tr ut CE y2 Wy I Ceeog dg IL takeo x then Ineed
compactsupport otherwisethen
5,69 Gci y sci CSg day will be apushinuco y u o7 uh Io uc 1
TG ya try o gig Y dy is's fess yD Es y of x 92dy
Sum
a can g cf yids of candy
Squareboth sides
u Coca Is 2 g side t III I 2 I'Caine x Go y def y I candy
s y S Integrateboth sides withrespectto S and y how oto
I g
2 f nut ishere2 I Chigidity g d G
2 extend 2 it cany.idcx.DE f so I d ta Y Eydba sa II d f else ay y.AEdy ds de
I f ff's de f db dy.tofbdy.jojfdydc e tf f Eads dy Eff dy da E d e s diet s E f o off desert
I
I may DX E tu 11TH dksue.in So Mo tu f cure
Assure I n is square or f Uf Uf UG
G is i
f y o X I G 251
Problem Fourrmulation
Comparing to 1 di
Ceceei box u I gcs u tangy ED Ucsc ci blow go u f
prince here is in our work
Derivalm weak formula
T C ucsoouxbcxiulvde fqcnuvdx ffg.net
J Kcs put 3414 v T.C uGDouabGDuT
xfkCDJTUxebcgguJ.TV
a y Ckeen Tu bcm.ci VdX J0 CC kc boauB
Kcsi outbox a OV
butusing stocked theorem
I y f new outboard V doc kc.ci Tu b can v nets
so we pick vch sounds
0
Substitute in 1 i
www.ovdx fiiigtuuovxfqcwaud
Lecture12p.pdf.pdf
Substituteincl Kes Tu or else fu bag vdsi fgc.ci ur DX
D e C kcsdtu bc.ca qce1U f go seer 467 0 on Jr
d Find UE VEH cr sit amine w Tve V
where alum ice yu.yv ubcsi.yv qoduvt.de
cu Fcs doc
T X o acu ul Z X114112
a Cail J cues I ul Ubu 4 96144doc n
Since F bcs u
U2 T.b.CH 24 b c TU
D Ubc S2 U Yz U'T bCsc
Yz T Cbc us
D u bCH TU de u F beside
Yz T Cbcse w di
T b esDu d g
f wisps FILII
g r
Substitue in a cu od we obtain i d
a cu ul C kcsgpuTt fsci IzT.bctfu fwm g
7 J k ca I Tum d g n
7 Ku f 10412 d e r
7 ko B foul else I D lourdse
using Poincare inquably
x ko p tout de Ce B 4 ur doe
we want p Li B y
5 Ps U B Ys 7 01 tour aide
I 211411,2
Now the continuity acqul EC Hull Hull
ACUVH1uke ou v ubc TV 19 uv bcs 1330 end ofCS Q 70
sniff tours wifi G f f g
i
B w H7IQ u7 v t
f mas zu Q B Stanford Isaiah de
f C hull hull
unique Sul of this form
Other Erm
LU T C ucsouxbc.ci a of asf i b o
ie T C Kase u g caus f Tn SC GR
Boundwy k ca TU n Gcse a yes conditions you IN
flux 0 No flude going out
weak formulation
y C u cool ou v1194mV f cards
since Tfeu y v T L new INV 1 l ke TU TV
J T C ucsc out v Asa Su e Tu Vds r
g O Ck purr else
Since Rex 7 Uan
we use stock's theorom egcse Gu
L icon on vote u g vols
substitute in
ad v Kc au Tv Asc qcocsuvels
Guards L f Cade girds
Acqui Kcsi 104121gcse u2 g a dao m 7 Ko I tout't G
ul de
7cm in C ko al lairds vide
Trace inquality
U2 Is E C Kull
u E a foul diet guide or or
D a cu u min Ko El 421 loud x
Yz I I Tal de U2 Is
7 Min CKo 6 lout d s
y th f J lo d s S u or dir
xcoflouidx.su 7 A 114112
A
continently I c o
a cu v