MAT 142 - COLLEGE MATHEMATICS - Heat, Wave, and
Laplace Equations Practice Material - Set 2
1. Consider the following heat equation in one dimension:
∂u
∂t =k∂2u
∂x2
where u(x, t)is the temperature distribution in a one-dimensional rod of length L, and kis
a positive constant.
Let’s denote the boundary conditions as:
u(0, t) = 0
u(L, t) = 0
And the initial condition as:
u(x, 0) = f(x)
Find the general solution for the temperature distribution u(x, t).
Ans. Solution: 1. We start by assuming the solution to the heat equation can be written in
the form of a Fourier series:
u(x, t) = ∞
X
n=1
Bne−(nπ
L)2kt sin nπx
L
2. Let’s substitute this series solution into the heat equation:
∂u
∂t =−∞
X
n=1 nπ
L2
kBne−(nπ
L)2kt sin nπx
L
∂2u
∂x2=−∞
X
n=1 nπ
L2
kBne−(nπ
L)2kt sin nπx
L
3. Substituting these expressions back into the heat equation, we get:
−∞
X
n=1 nπ
L2
kBne−(nπ
L)2kt sin nπx
L=k∞
X
n=1 nπ
L2
kBne−(nπ
L)2kt sin nπx
L
4. Now, multiplying both sides by sin mπx
Land integrating from 0to Lwith respect to x,
we obtain:
−mπ
L2
kBm=kmπ
L2
kBm
5. Simplifying the above equation, we find that the non-trivial solutions occur when k=L2
π2n2,
where nis an integer. Thus, the general solution for the temperature distribution is:
u(x, t) = ∞
X
n=1
Bne−(nπ
L)2π2n2
L2tsin nπx
L
u(x, t) = ∞
X
n=1
Bne−n2π2t/L2sin nπx
L
2. Consider the following Heat Equation in one spatial dimension:
∂u
∂t =k∂2u
∂x2
where u(x, t)is the temperature distribution in a rod at position xand time t, and kis the
thermal diffusivity.
Given the boundary conditions:
1. u(0, t) = 0 for t > 02. u(L, t) = 0 for t > 0
and the initial condition:
3. u(x, 0) = f(x) = 2 −4x/Lfor 0< x < L
a) Find the steady-state solution of the Heat Equation. b) Solve the Heat Equation for the
temperature distribution u(x, t)at all xand t.
Ans. a) The steady-state solution of the Heat Equation is given by:
uss(x) = 2 −4x
L
b) The solution of the Heat Equation with the given boundary and initial conditions is:
u(x, t) = ∞
X
n=1
Bnsin nπx
Le−k(nπ
L)2t
where the coefficients Bnare given by:
Bn=2
LZL
0
f(x)sin nπx
Ldx =8
n2π2(1 −(−1)n)
3. Find the general solution to the heat equation
∂u
∂t = 3∂2u
∂x2
for 0< x < 1,t > 0, subject to the boundary conditions u(0, t) = u(1, t) = 0 and the initial
condition u(x, 0) = 2x(1 −x).
Ans. Solution: 1. We first assume that the solution u(x, t)can be expressed as a product of
functions of xand t, that is u(x, t) = X(x)T(t).
2. Plugging u(x, t)into the heat equation, we get
T′(t)
3T(t)=X′′(x)
X(x)=−λ
3. Solving the ODE for X(x), we get
X′′(x) + λX(x) = 0
4. The boundary conditions u(0, t) = u(1, t) = 0 lead to the conditions X(0) = 0 and
X(1) = 0. These are the conditions for the eigenvalues λn=n2π2, for n= 1,2,3, . . ..
5. The general solution for X(x)is given by
X(x) = ∞
X
n=1
cnsin(nπx)
6. Now, solving for T(t), we have
T′(t) = −3λT (t)⇒T(t) = ce−3λt
7. The general solution to the heat equation is then
u(x, t) = ∞
X
n=1
cnsin(nπx)e−3n2π2t
8. Finally, we apply the initial condition u(x, 0) = 2x(1 −x)to find the coefficients cn. By
using the orthogonality property of sine functions, we can express cnin terms of a Fourier sine
series.
9. The final solution is given by
u(x, t) = ∞
X
n=1
8
(2n−1)3π3sin((2n−1)πx)e−(2n−1)2π2t
4. Question: Consider the one-dimensional heat equation ut=kuxx, where kis a positive
constant. Given the initial condition u(x, 0) = sin(x)and boundary conditions u(0, t)=0and
u(π, t) = 0, find the solution u(x, t).
Ans. Step-by-step solution: 1. We start by assuming a separation of variables solution u(x, t) =
X(x)T(t). 2. Substituting this into the heat equation, we get T′
kT =X′′
X=−λ, where λ
is a separation constant. 3. Solving the ODE for X(x), we obtain the eigenvalue problem
X′′ +λX = 0 with boundary conditions X(0) = 0 and X(π) = 0. 4. The general solution
to the eigenvalue problem is Xn(x) = sin(nx)for λ=n2, where n∈N. 5. Now we solve
the ODE for T(t), which gives Tn(t) = cne−kn2t. 6. Combining the solutions for Xn(x)and
Tn(t), we get un(x, t) = cnsin(nx)e−kn2t. 7. Using the principle of superposition, the general
solution is u(x, t) = P∞
n=1 cnsin(nx)e−kn2t. 8. Applying the initial condition u(x, 0) = sin(x),
we find that cn= 2 Rπ
0sin(nx)sin(x)dx for n > 0and c0=1
πRπ
0sin(x)dx. 9. Simplifying
the integrals, we find that cn=2
π
1−(−1)n
n(1−n2)for n > 0and c0=1
π2×2π=2
π. 10. Therefore,
the solution to the heat equation with the given initial and boundary conditions is u(x, t) =
2
πsin(x) + P∞
n=1
1−(−1)n
n(1−n2)sin(nx)e−kn2t.
5. Question:
Consider the one-dimensional heat equation given by ∂u
∂t =k∂2u
∂x2, where u(x, t)is the tem-
perature distribution along a rod of length Lat time t,kis the thermal diffusivity constant, and
u(x, 0) = f(x)is the initial temperature distribution.
Given the following boundary conditions: 1. u(0, t) = 0 for t≥02. u(L, t) = 0 for t≥03.
u(x, 0) = sinπx
Lfor 0≤x≤L
Find the temperature distribution u(x, t)for x∈[0, L]and t > 0.
Ans. Step-by-step solution:
1. We begin by assuming a separable solution of the form u(x, t) = X(x)T(t).
2. Plugging this into the heat equation gives T′
kT =X′′
X=−λ, where λis a separation
constant.
3. This gives two ODEs: T′+kλT = 0 with solution T(t) = Ae−kλt X′′ +λX = 0 with
solution X(x) = Bsin √λx
4. Applying the boundary conditions u(0, t) = 0 and u(L, t) = 0 at x= 0 and x=L
respectively, we obtain X(0) = 0 and X(L) = 0.
5. The boundary condition X(0) = 0 gives Bsin(0) = 0 ⇒B= 0. Contradiction.
6. Since B= 0 leads to a trivial solution, we consider the nontrivial case where λ=nπ
L2
for n= 1,2,3, ...
7. Therefore, the general solution is a sum of these eigenfunctions: u(x, t) = P∞
n=1 Ansin nπ
Lxe−k(nπ
L)2t.
8. To determine the coefficients An, we use the initial temperature distribution u(x, 0) =
sin πx
L:u(x, 0) = P∞
n=1 Ansin nπ
Lx=sin πx
L.
9. By orthogonality, we find: An=2
LRL
0sin πx
Lsin nπ
Lxdx
10. Simplifying the integral and solving for An, we get: An=(0if nis even
4
nπ if nis odd .
11. Therefore, the temperature distribution is: u(x, t) = P∞
m=1
4
(2m−1)πsin (2m−1)πx
Le−k((2m−1)π
L)2t.
6. Question: Consider the one-dimensional heat equation given by ∂u
∂t =k∂2u
∂x2, where kis a
positive constant. If the initial condition is u(x, 0) = x(1 −x)for 0< x < 1and the boundary
conditions are u(0, t) = u(1, t) = 0 for t > 0, find the solution u(x, t).
Ans. Step-by-step solution: 1. First, we assume a separation of variables solution of the form
u(x, t) = X(x)T(t). 2. Substitute the separation of variables solution into the heat equation to
get T′(t)
kT (t)=X′′(x)
X(x)=−λ, where λis a constant. 3. Solve the temporal part: T′(t) + kλT (t) = 0
has the solution T(t) = c1e−kλt. 4. Solve the spatial part: X′′(x) + λX(x)=0with boundary
conditions X(0) = X(1) = 0 has the solution Xn(x) = sin(nπx)for λ=n2π2. 5. Therefore, the
general solution is u(x, t) = P∞
n=1 cne−kn2π2tsin(nπx). 6. Apply the initial condition u(x, 0) =
x(1 −x)to find cncoefficients. 7. Use the orthogonality property of sines to evaluate cn. 8. The
final solution for the heat equation is u(x, t) = P∞
n=1 8
(2n−1)π2
e−(2n−1)2π2kt sin((2n−1)πx).
7. Question: Consider the heat equation defined by
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions
u(0, t) = 0, u(1, t) = 0
and the initial condition
u(x, 0) = x(1 −x)
Find the solution to the heat equation.
Ans. Solution:
We can solve the heat equation using the method of separation of variables.
1. Separation of Variables:
Let u(x, t) = X(x)T(t). Substituting this into the heat equation, we get:
XT ′=kX′′T
Dividing by kXT , we have:
T′
kT =X′′
X=−λ
This gives two separate ordinary differential equations:
T′+kλT = 0 and X′′ +λX = 0
2. Solve the Time Equation:
The time equation gives us T(t) = ce−kλt, where cis a constant.
3. Solve the Spatial Equation:
The spatial equation X′′ +λX = 0 has solutions of the form X(x) = Asin(nπx)and
X(x) = Bcos(nπx), where λ=n2π2for n= 1,2,3, . . .
The general solution is then:
u(x, t) = ∞
X
n=1
(Ansin(nπx) + Bncos(nπx))e−k(nπ)2t
4. Solve for the Coefficients:
Applying the boundary conditions u(0, t)=0and u(1, t)=0, we find that the coefficients
Anare zero for all nand the coefficients Bncan be determined using the initial condition
u(x, 0) = x(1 −x).
5. Final Solution:
Putting it all together, the solution to the heat equation with the given initial and boundary
conditions is:
u(x, t) = ∞
X
n=1
2
nπ sin(nπx)e−k(nπ)2t
8. Consider the following partial differential equation, known as the heat equation:
∂u
∂t =k∂2u
∂x2+∂2u
∂y2
where u(x, y, t)is the temperature distribution in a metal plate at time t,kis the thermal
diffusivity constant, and u(x, y, 0) = f(x, y)represents the initial distribution of temperature.
Given the metal plate is a square with sides of length Land the initial temperature distribution
is f(x, y) = 100 within the square centered at the origin with side length L
2and f(x, y)=0
otherwise, find the temperature distribution at time t > 0within the metal plate.
Ans. Let’s first set up the problem and then solve the heat equation with the given initial
conditions.
1. The heat equation for this problem is:
∂u
∂t =k∂2u
∂x2+∂2u
∂y2
subject to the initial condition:
u(x, y, 0) = (100 if |x|<L
4and |y|<L
4
0otherwise
2. To solve the heat equation, we can use separation of variables. Let u(x, y, t) = X(x)Y(y)T(t).
Plugging this into the heat equation, we get:
1
kXT
dT
dt =X′′Y+XY ′′
XY
3. Dividing both sides by u, we obtain:
1
k
1
T
dT
dt =X′′
X+Y′′
Y
4. Since the left-hand side depends only on tand the right-hand side depends only on xand
y, they must both be equal to a constant, say −λ2. This gives us the system of ODEs:
1
k
1
T
dT
dt =−λ2
X′′ +λ2X= 0
Y′′ +λ2Y= 0
5. Solving the ODEs for Xand Y, we obtain the general solutions:
X(x) = Acosnπx
L, Y (y) = Bsinnπy
L
where nis a positive integer.
6. Applying the initial condition, we can write the solution as:
u(x, y, t) = ∞
X
n=1
Cncosnπx
Lsinnπy
Le−k(nπ
L)2t
where Cnare determined by the initial temperature distribution.
7. Using the initial condition f(x, y)provided, we can find the coefficients Cnand thus
determine the temperature distribution at any time t > 0.
9. Question:
Consider the heat equation in one dimension, given by ∂u
∂t =k∂2u
∂x2, where kis a positive
constant. Suppose the initial temperature distribution along a rod of length Lis given by u(x, 0) =
sin πx
L, and the rod is insulated at both ends so that ∂u
∂x (0, t) = ∂u
∂x (L, t) = 0 for all t > 0. Find
the temperature distribution u(x, t)along the rod at a later time t > 0.
Ans. Step-by-step solution:
1. We start by assuming a separation of variables solution of the form u(x, t) = X(x)T(t).
Substituting this into the heat equation gives:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a separation constant. This gives two ordinary differential equations:
T′(t) + kλT (t) = 0 and X′′(x) + λX(x) = 0
2. Solving the ODE for T(t)gives T(t) = Ae−kλt, where Ais a constant.
3. The ODE for X(x)has solutions of the form X(x) = Bcos√λx+Csin√λx, where
Band Care constants. Applying the boundary conditions, X′(0) = X′(L) = 0, implies B= 0
and √λ=nπ/L, where nis a positive integer. Thus, X(x) = Csin(nπx/L).
4. Therefore, the general solution is given by the sum of all possible eigenfunctions:
u(x, t) = ∞
X
n=1
Cnsin nπx
Le−k(nπ/L)2t
5. To find the coefficients Cn, we use the initial temperature distribution: u(x, 0) =
sin πx
L=P∞
n=1 Cnsin nπx
L. By orthogonality, we have:
Cn=2
LZL
0
sin πx
Lsin nπx
Ldx
6. Evaluating the integral above gives Cn=2
Lδn1, where δn1is the Kronecker delta. There-
fore, the temperature distribution along the rod at time t > 0is:
u(x, t) = 2
Lsin πx
Le−kπ2t/L2
10. Find the solution to the heat equation on the unit interval 0< x < 1with initial condition
u(x, 0) = xand boundary conditions u(0, t) = 0 and u(1, t) = 1.
Ans. Solution:
We seek a solution to the heat equation of the form u(x, t) = X(x)T(t). Substituting this
into the heat equation ut=uxx, we get:
X(x)T′(t) = X′′(x)T(t)
Dividing both sides by X(x)T(t)gives:
T′(t)
T(t)=X′′(x)
X(x)=−λ
which leads to two ordinary differential equations:
T′(t) + λT (t) = 0, T (0) = 0, T (L) = 1
X′′(x) + λX(x) = 0
Solving the first ODE, we get T(t) = ce−λt. Using the boundary conditions, we find T(t) =
sin(nπx).
For the second ODE, the general solution is X(x) = Acos√λx+Bsin√λx. Applying
the boundary conditions X(0) = 0 and X(1) = 1, we get:
X(0) = A= 0
X(1) = Bsin√λ= 1
Since B= 0, we have sin√λ= 1 =⇒√λ=π
2. Thus, our eigenvalue is λ=π2
4.
Therefore, the solution to the heat equation with the given boundary conditions and initial
condition is:
u(x, t) = ∞
X
n=1
Cne−n2π2
4tsin nπx
2
where Cndepends on the initial condition.
11. Let u(x, t)be the solution to the heat equation
ut=uxx,0< x < 1, t > 0
with initial condition
u(x, 0) = x(1 −x),0< x < 1
and boundary conditions
u(0, t) = 0, u(1, t) = 0, t > 0
Determine the expression for u(x, t).
Ans. To solve the heat equation, we use separation of variables. We assume a solution of the
form
u(x, t) = X(x)T(t)
Substitute this into the heat equation to get
XT ′=X′′T
Divide both sides by u(x, t)to get
T′
T=X′′
X=−λ
Solving the time equation gives
T′(t) + λT (t) = 0
This gives the general solution
T(t) = Ae−λt
where λ > 0to ensure the solution decays as t→ ∞.
Solving the spatial equation gives
X′′(x) + λX(x) = 0
with boundary conditions X(0) = 0,X(1) = 0. The general solution to this equation is
X(x) = Bsin√λx
Applying the boundary conditions gives
X(x) = Bsin(nπx)
where √λ=nπ for n= 1,2,3, .... We choose n= 1 to satisfy the initial condition.
So, the solution to the heat equation is
u(x, t) = ∞
X
n=1
Bne−(nπ)2tsin(nπx)
To determine Bn, we use the initial condition u(x, 0) = x(1 −x)to find the Fourier sine
series of x(1 −x).
x(1 −x) = ∞
X
n=1
Bnsin(nπx)
Bn= 2 Z1
0
x(1 −x)sin(nπx)dx
By evaluating this integral, we find that
Bn=8
n3π3(1 −(−1)n)
Therefore, the expression for u(x, t)is
u(x, t) = ∞
X
n=1
8
n3π3(1 −(−1)n)e−(nπ)2tsin(nπx)
12. Question: Consider the following partial differential equation:
uxx −4uyy = 0
subject to the boundary conditions
u(0, y) = 0, u(1, y) = 0, u(x, 0) = 0, u(x, 1) = 0
Find the solution to the given partial differential equation with the specified boundary conditions.
Ans. Let u(x, y) = X(x)Y(y)be the separation of variables solution. Then we have
X′′(x)Y(y)−4X(x)Y′′(y) = 0
Dividing by X(x)Y(y), we get
X′′(x)
X(x)= 4Y′′(y)
Y(y)
Since the left-hand side depends only on xand the right-hand side depends only on y, they must
be equal to a constant −λ, so we have two separate ODEs:
X′′(x) + λX(x) = 0, Y ′′(y)−λ
4Y(y) = 0
1. Solve for X(x): The ODE for X(x)is
X′′(x) + λX(x) = 0
with boundary conditions X(0) = X(1) = 0. This gives us the eigenvalues λn=n2π2, and the
eigenfunctions Xn(x) = sin(nπx).
2. Solve for Y(y): The ODE for Y(y)is
Y′′(y)−λ
4Y(y) = 0
Substitute λ=n2π2:
Y′′(y)−n2π2
4Y(y) = 0
The corresponding eigenvalues λn=n2π2and eigenfunctions Yn(y) = sin nπy
2.
3. Combine X(x) and Y(y): The general solution is a linear combination of the separated
solutions:
u(x, y) = ∞
X
n=1
cnsin(nπx)sin nπy
2
4. Applying boundary conditions: Applying the boundary conditions gives us the final
solution:
u(x, y) = ∞
X
n=1
cnsin(nπx)sin nπy
2
where
cn=8
n3π3(1 −(−1)n)
13. Question: Consider the following heat equation in one dimension:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions u(0, t) = 0 and u(L, t) = 0, and the initial condition u(x, 0) =
f(x)for 0≤x≤L. Show that the solution to this equation can be written as a Fourier sine
series, where
u(x, t) = ∞
X
n=1
Bnsin nπx
Le−k(nπ
L)2t
Ans. Solution: 1. To solve the heat equation using separation of variables, assume that the
solution can be expressed as a product of two functions:
u(x, t) = X(x)T(t)
2. Substitute this expression into the heat equation to get:
X(x)T′(t) = kX′′(x)T(t)
3. Divide both sides by kX(x)T(t)to separate the variables:
T′(t)
kT (t)=X′′(x)
X(x)=−λ2
4. Solve the time part of the equation:
T′(t) + kλ2T(t) = 0
This gives the general solution:
T(t) = C1e−kλ2t
5. Solve the spatial part of the equation:
X′′(x) + λ2X(x) = 0
This second-order homogeneous differential equation has solutions in the form of sine functions
since the boundary conditions are u(0, t)=0and u(L, t)=0. Thus, the solutions are Xn(x) =
sin nπx
Lfor n= 1,2,3, . . .
6. The general solution to the spatial part is given as:
X(x) = ∞
X
n=1
Ansin nπx
L
7. Combine the spatial and time parts of the solution:
u(x, t) = ∞
X
n=1
Ansin nπx
Le−k(nπ
L)2t
8. Given that the initial condition is u(x, 0) = f(x), we express the solution in its Fourier
sine series form:
u(x, 0) = ∞
X
n=1
Ansin nπx
L=f(x)
9. By comparing coefficients, we find that An=Bnand thus arrive at the desired solution:
u(x, t) = ∞
X
n=1
Bnsin nπx
Le−k(nπ
L)2t
14. Consider the Heat Equation in one dimension defined by
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions u(0, t) = 0 and u(π, t) = 0, and the initial condition
u(x, 0) = x(π−x).
Find the solution u(x, t)to this Heat Equation.
Ans. Solution: 1. We start by assuming a separation of variables solution of the form
u(x, t) = X(x)T(t). 2. Substituting this into the Heat Equation gives T′
kT =X′′
X=−λ, where
λis a separation constant. 3. Solving the time ODE T′
kT =−λgives T(t) = Ae−kλt. 4.
Solving the spatial ODE X′′ +λX = 0 gives X(x) = Bsin√λx+Ccos√λx. 5. Applying
the boundary conditions u(0, t) = 0 and u(π, t) = 0 gives X(0) = Bsin(0) + Ccos(0) = 0
and X(π) = Bsin√λπ+Ccos√λπ= 0. 6. From X(0) = 0, we have C= 0. 7.
From X(π) = 0, we have √λπ =nπ for n= 1,2,3, ..., leading to λ=n2. 8. The general
solution is then u(x, t) = P∞
n=1(Ansin(nx)e−kn2t). 9. Applying the initial condition u(x, 0) =
x(π−x)gives x(π−x) = P∞
n=1 Ansin(nx). 10. By orthogonality, we can solve for Anas
An=2
πRπ
0x(π−x)sin(nx)dx. 11. Calculating the integral gives An=4
n3((−1)n−1). 12.
Therefore, the solution to the Heat Equation is u(x, t) = P∞
n=1
4
n3((−1)n−1) sin(nx)e−kn2t.
15. Let u(x, t)satisfy the wave equation in one space dimension and time
utt =c2uxx,0< x < l, t > 0
with boundary conditions u(0, t) = u(l, t) = 0 for t > 0and initial conditions u(x, 0) = f(x)
and ut(x, 0) = g(x)for 0< x < l. Show that the separation of variables leads to the solution
u(x, t) = ∞
X
n=1
Bnsin nπx
l·cos nπct
l
where the coefficients Bnare given by
Bn=2
lZl
0
f(x)sin nπx
ldx.
Ans. Solution: 1. We start by trying the separation of variables approach, assuming that
u(x, t) = X(x)T(t). So, we have
X(x)T′′(t) = c2X′′(x)T(t).
2. Dividing by c2u(x, t), we get
T′′(t)
c2T(t)=X′′(x)
X(x)=−λ.
This leads to the two ordinary differential equations
T′′(t) + λc2T(t) = 0
and
X′′(x) + λX(x) = 0.
3. The boundary conditions u(0, t) = u(l, t)=0for t > 0imply that X(0) = X(l)=0.
This gives a Sturm-Liouville problem for X(x).
4. The general solution to the spatial part of the equation is
X(x) = Asin nπx
l+Bcos nπx
l.
5. Applying the boundary conditions X(0) = X(l)=0, we find that B= 0 and X(x) =
Asin nπx
l.
6. The general solution to the time part of the equation is
T(t) = Ccos nπct
l+Dsin nπct
l.
7. Since the wave equation is linear, the total solution is a linear combination of the separated
solutions. By the principle of superposition, we can write
u(x, t) = ∞
X
n=1
Bnsin nπx
l·cos nπct
l,
where the coefficients Bnare to be determined.
8. To determine the coefficients Bn, we substitute u(x, 0) = f(x)into the general solution.
This gives
f(x) = ∞
X
n=1
Bnsin nπx
l.
9. We can now determine the coefficients Bnby the formula
Bn=2
lZl
0
f(x)sin nπx
ldx.
Hence, we have shown that the separation of variables leads to the provided solution form
and formula for the coefficients.
16. Question: Consider the following partial differential equation:
ut= 4uxx −2u
subject to the initial conditions:
u(x, 0) = x2
ut(x, 0) = 0
Find the solution to the partial differential equation using the method of separation of vari-
ables.
Ans. Step-by-step solution: 1. We will assume the solution to the PDE is separable, i.e.,
u(x, t) = X(x)T(t). Substitute this into the PDE to get:
XT ′= 4X′′T−2XT
Divide both sides by XT to get:
T′
T= 4X′′
X−2
2. Rearrange the equation to separate variables:
T′
T=A
4X′′
X−2 = A
where Ais a constant. 3. Solve the time part ODE:
dT
dt =AT
dT
T=Adt
Integrating both sides gives:
ln |T|=At +B
T(t) = CeAt
where Cis a constant. 4. Solve the spatial part ODE:
4X′′
X−2 = A
4X′′ −(2 + AX)X= 0
This is a second-order linear ODE with constant coefficients. Let m2= 2 +A. The characteristic
equation becomes:
4m2−(2 + A) = 0
4m2−m2= 2
3m2= 2
m=±√6
3
5. The general solution for X(x)is:
X(x) = Ae√6
3x+Be−√6
3x
6. Applying the initial condition u(x, 0) = x2, we have:
X(x)T(0) = x2
X(x) = x2
Ae0+Be0=x2
A+B=x2
7. Applying the initial condition ut(x, 0) = 0, we have:
XT ′(0) = 0
A√6
3e0−B√6
3e0= 0
√6
3(A−B) = 0
A=B
A+A=x2
2A=x2
A=x2
2
B=x2
2
Thus, the solution to the partial differential equation using separation of variables is:
u(x, t) = x2
2e√6
3x+x2
2e−√6
3xeAt
where A=√6
3.
17. Question:
Consider the heat equation on a bounded domain Dwith initial condition u(x, y, 0) = x2+y2
and boundary conditions u(x, y, t) = 0 for (x, y)∈∂D, where ∂D denotes the boundary of D.
Solve the heat equation to find u(x, y, t).
Ans. Solution:
We will solve the heat equation using separation of variables. Let u(x, y, t) = X(x)Y(y)T(t).
Substituting this into the heat equation ut=c2(uxx +uyy)gives
T′
c2T=X′′
X+Y′′
Y
Now we have separated the variables and have three ODEs to solve:
1. Solve T′
c2T=−λ=α2for T(t). 2. Solve X′′ =−α2Xfor X(x). 3. Solve Y′′ =−α2Y
for Y(y).
1. Solving T′
c2T=−λ=α2gives T(t) = e−α2c2t.
2. Solving X′′ =−α2Xwith boundary condition X(0) = X(L)=0gives eigenvalues
αn=nπ
Land eigenfunctions Xn(x) = sin nπx
L.
3. Solving Y′′ =−α2Ywith boundary condition Y(0) = Y(L) = 0 gives eigenvalues
βn=nπ
Land eigenfunctions Yn(y) = sin nπy
L.
Therefore, the general solution can be written as
u(x, y, t) = ∞
X
n=1
Cne−(nπ
L)2c2tsin nπx
Lsin nπy
L
Applying the initial condition u(x, y, 0) = x2+y2allows us to determine the coefficients
Cnusing Fourier series. After solving for the coefficients, the final solution for u(x, y, t)will be
obtained.
18. Question: Let u(x, t)be the solution to the heat equation ∂u
∂t =∂2u
∂x2with boundary
conditions u(0, t) = u(1, t)=0and initial condition u(x, 0) = 2x(1 −x). Find the solution
u(x, t).
Ans. Solution: 1. First, let’s express the solution u(x, t)as a Fourier sine series:
u(x, t) = ∞
X
n=1
Bn(t)sin(nπx)
2. Our next step is to find the time-dependent coefficients Bn(t). Substitute the expression
for u(x, t)into the heat equation and use the fact that ∂
∂t (sin(nπx)) = 0 to get:
dBn
dt sin(nπx) = −n2π2Bnsin(nπx)
3. This simplifies to:
dBn
dt =−n2π2Bn
4. Solving the above first-order ordinary differential equation, we get:
Bn(t) = cne−n2π2t
5. Now we need to determine the constant cn. We can use the initial condition u(x, 0) =
2x(1 −x)to find cn:
2x(1 −x) = ∞
X
n=1
cnsin(nπx)
6. Multiply both sides by sin(mπx)and integrate from 0 to 1:
Z1
0
2x(1 −x)sin(mπx)dx =Z1
0
∞
X
n=1
cnsin(nπx)sin(mπx)dx
7. By the orthogonality property of sine functions, all terms in the sum vanish except when
n=m. This leads to:
cm= 2 Z1
0
x(1 −x)sin(mπx)dx
8. Evaluate the integral to find cm:
cm= 2 6
π3(1 −(−1)m)
Therefore, the solution to the heat equation with the provided initial and boundary conditions
is:
u(x, t) = ∞
X
n=1
26
π3(1 −(−1)n)e−n2π2tsin(nπx)
19. Let u(x, t)satisfy the one-dimensional heat equation
ut=c2uxx
subject to the boundary conditions u(0, t) = u(L, t) = 0 and the initial condition
u(x, 0) = sin πx
L−sin 3πx
L
where 0< x < L,t > 0, and cis a positive constant.
Compute u(x, t)for all xand t.
Ans. To find the solution to the heat equation with the given conditions, we can use the method
of separation of variables. Let’s assume a solution of the form u(x, t) = X(x)T(t).
1. Write down the separated equations: Plugging the assumed form of the solution into
the heat equation gives
X(x)T′(t) = c2X′′(x)T(t)
Dividing by c2u(x, t)gives
T′(t)
c2T(t)=X′′(x)
X(x)
Letting both sides be equal to a separation constant λyields the equations
T′(t) = c2λT (t)and X′′(x) = λX(x)
2. Solve the time part of the equation: The equation for T(t)can be solved as
T(t) = Aec2λt
where Ais a constant.
3. Solve the spatial part of the equation: The equation for X(x)gives solutions of the
form
X(x) = Bsin nπx
L+Ccos nπx
L
where Band Care constants to be determined.
4. Apply the boundary conditions: The boundary conditions u(0, t) = 0 and u(L, t) = 0
give X(0) = 0 and X(L) = 0, which results in
C= 0 and Bsin(nπ) = 0
Since nis a positive integer, the only way to satisfy the second equation is to have n=
1,2,3, .... Thus, the solution becomes
Xn(x) = Bnsin nπx
L
5. Express the general solution: The general solution is then a linear combination of the
spatial parts:
u(x, t) = ∞
X
n=1
Tn(t)Xn(x)
u(x, t) = ∞
X
n=1 Anec2λntBnsin nπx
L
6. Apply the initial condition: Finally, we apply the initial condition to find the coefficients
Anusing the Fourier sine series of the function
u(x, 0) = sin πx
L−sin 3πx
L=∞
X
n=1
AnBnsin nπx
L
Comparing coefficients gives
AnBn=
1for n= 1
−1for n= 3
0for n= 1,3
Thus, we obtain the solution
u(x, t) = e−c2π2t/L2sin πx
L−e−c29π2t/L2sin 3πx
L
20. Question:
Consider the one-dimensional heat equation given by ∂u
∂t =k∂2u
∂x2, where kis a constant and
u(x, t)is the temperature distribution along a rod of length Lat time t.
A rod of length L= 10 cm initially has a temperature distribution u(x, 0) = 100x(L−x)for
0≤x≤L. The rod is kept in an environment at a constant temperature of 0◦C on both sides
while its ends are insulated. Find the temperature distribution u(x, t)for all xand t.
Ans. Step-by-step solution:
1. The general solution to the one-dimensional heat equation is given by u(x, t) = P∞
n=1 Bnsin nπx
Le−αn2π2t/L2,
where Bnare constants and α=k/L2.
2. To find the specific solution to our problem, we first need to determine the initial conditions
for the problem. We have u(x, 0) = 100x(L−x).
3. We can express the initial condition as a Fourier sine series: u(x, 0) = P∞
n=1 Bnsin nπx
L.
Comparing this with 100x(L−x), we find that Bn= 400/L·(−1)n+1
n2π2.
4. Now, plugging in the initial condition into the general solution, we have u(x, t) =
P∞
n=1 400(−1)n+1
n2π2sin nπx
Le−αn2π2t/L2.
5. Substituting L= 10 cm and the given values of k, we have α=k/L2=k/100. Hence,
the temperature distribution for all xand tis:
u(x, t) = ∞
X
n=1 400(−1)n+1
n2π2sin nπx
10 e−kn2π2t
100
21. Question:
Consider the heat equation in one dimension given by ∂u
∂t =k∂2u
∂x2, subject to the initial
condition u(x, 0) = sin(x)and boundary conditions u(0, t) = 0 and u(π, t) = 0, where kis a
positive constant.
Find the solution u(x, t)to the heat equation.
Ans. Step-by-step solution:
1. We can solve the heat equation using separation of variables. Let u(x, t) = X(x)T(t).
Substitute u(x, t) = X(x)T(t)into the heat equation to get: X(x)T′(t) = kX′′(x)T(t).
2. Divide both sides by kX(x)T(t)to separate the variables: T′(t)
kT (t)=X′′(x)
X(x).
Letting both sides equal to a constant −λ, we have two ordinary differential equations:
T′(t) + kλT (t) = 0 and X′′(x) + λX(x) = 0.
3. Solve the time part of the equation: T′(t) = −kλT (t)1
T(t)dT =−kλdt ln |T|=−kλt+C1
T(t) = C2e−kλt, where C2=eC1.
Since C2is an arbitrary constant, we have T(t) = e−kλt.
4. Now solve the spatial part of the equation: X′′(x) + λX(x) = 0 The general solution to
this ordinary differential equation is X(x) = Acos√λx+Bsin√λxwhere A and B are
arbitrary constants to be determined by the boundary conditions.
Applying the boundary conditions u(0, t) = 0 and u(π, t) = 0 gives A= 0 and sin√λπ=
0.
5. The condition sin√λπ= 0 implies that √λπ =nπ for some integer n, i.e., λ=n2.
Thus, the solution X(x)simplifies to Xn(x) = Bnsin(nx)where nis a positive integer.
Putting together the spatial and temporal solutions, the general solution is then u(x, t) =
P∞
n=1 Bnsin(nx)e−kn2t.
6. To determine the constants Bn, we use the initial condition u(x, 0) = sin(x):u(x, 0) =
P∞
n=1 Bnsin(nx) = sin(x).
Comparing coefficients of sin(nx)on both sides gives Bn=(2/πif n= 1,
0if n= 1..
Therefore, the solution to the heat equation is u(x, t) = 2
πsin(x)e−kt.
22. Question:
Consider the heat equation defined by ∂u
∂t = 2∂2u
∂x2, subject to the boundary conditions u(0, t) =
0and u(π, t)=0, and the initial condition u(x, 0) = 3 sin(x). Find the solution u(x, t)to this
equation.
Ans. Let’s solve the given heat equation with the specified boundary and initial conditions.
1. Find the solution using separation of variables:
Assume the solution has the form u(x, t) = X(x)T(t).
Substitute u(x, t) = X(x)T(t)into the heat equation to get:
X(x)T′(t) = 2X′′(x)T(t)
Divide both sides by u(x, t)to get:
T′(t)
2T(t)=X′′(x)
X(x)=−λ
This gives us two ordinary differential equations:
X′′(x) + λX(x) = 0 and T′(t)+2λT (t) = 0
2. Solve the spatial equation:
The solution to the spatial equation X′′(x) + λX(x)=0with the boundary conditions
X(0) = 0 and X(π) = 0 is:
Xn(x) = sin(nx)
The corresponding eigenvalues are λn=n2.
3. Solve the temporal equation:
The solution to the temporal equation T′(t)+2λT (t) = 0 is:
Tn(t) = Ce−2n2t
4. Combine the solutions:
The general solution to the heat equation is given by:
u(x, t) = ∞
X
n=1
Ansin(nx)e−2n2t
5. Use the initial condition to determine coefficients:
Given the initial condition u(x, 0) = 3 sin(x), we have:
3sin(x) = ∞
X
n=1
Ansin(nx)
Comparing coefficients, we find A1= 3 and An= 0 for n= 1.
6. Final solution:
Therefore, the solution to the heat equation with the given conditions is:
u(x, t) = 3 sin(x)e−2t
23. Find the solution to the following partial differential equation, subject to the given initial
conditions:
∂2u
∂t2= 4∂2u
∂x2,0< x < π, t > 0
with initial conditions u(0, t) = u(π, t) = 0,u(x, 0) = sin(x), and ∂u
∂t (x, 0) = 0.
Ans. Solution:
1. We assume a solution of the form u(x, t) = X(x)T(t)and substitute it into the given
partial differential equation:
X(x)T′′(t) = 4X′′(x)T(t)
Dividing by X(x)T(t), we get:
T′′(t)
T(t)= 4X′′(x)
X(x)
Rearranging, we have:
T′′(t)
T(t)= 4K=λ
2. Solving the ODE T′′(t)
T(t)=λgives us:
T(t) = c1cos 2√λt+c2sin 2√λt
Applying the initial condition ∂u
∂t (x, 0) = 0, we have d
dt (X(x)T(t))t=0 = 0:
√λc1X(x) = 0
Since X(x)= 0, we have λ= 0.
3. Therefore, the solution for T(t)becomes:
T(t) = c1+c2t
4. Substituting λ= 0 back into X′′(x)
X(x)= 0, we get:
X(x) = c3+c4x
Applying the boundary conditions u(0, t) = u(π, t) = 0, we have:
X(0) = c3= 0 and X(π) = c4π= 0
So, c4= 0 and X(x) = 0.
5. Therefore, the solution to the PDE with the given initial conditions is:
u(x, t) = 0
24. Find the solution to the following partial differential equation, subject to the given boundary
conditions:
∂2u
∂t2=c2∂2u
∂x2,0< x < π, t > 0
subject to the boundary conditions:
u(0, t) = u(π, t) = 0, u(x, 0) = sin(x),∂u
∂t (x, 0) = 0
Solution:
1. Let’s assume the solution can be written in the form
u(x, t) = X(x)T(t)
Substituting this into the partial differential equation, we get:
X(x)T′′(t) = c2X′′(x)T(t)
Dividing both sides by c2XT , we get:
T′′(t)
c2T(t)=X′′(x)
X(x)=−λ
2. Solving the ODE X′′(x)
X(x)=−λgives us the solutions:
Xn(x) = Bnsin(nx),where n=√λand Bnis a constant
Applying the boundary conditions u(0, t) = u(π, t) = 0, we have:
Xn(0) = Bnsin(0) = 0 =⇒Bn= 0
This implies that the solution to the spatial part is X(x) = 0, which is trivial. Hence, this
leads to the conclusion that λ < 0.
3. Solving the ODE T′′(t)
c2T(t)=−λgives us the solution:
T(t) = Acosc√−λt+Bsinc√−λt
Applying the initial condition ∂u
∂t (x, 0) = 0, we have:
T′(0) = −Ac√−λsin(0) + Bc√−λcos(0) = 0
Bc√−λ= 0
Since c > 0, this gives us B= 0, which implies T(t) = Acosc√−λt.
4. Substituting n2=−λinto the spatial part Xn(x) = Bnsin(nx)and time part Tn(t) =
Ancos(cnt), we get the general solution:
u(x, t) = ∞
X
n=1
Ansin(nx)cos(cnt)
5. Now, applying the initial condition u(x, 0) = sin(x)to the general solution, we get:
∞
X
n=1
Ansin(nx) = sin(x)
By orthogonality, we can determine An:
An=2
πZπ
0
sin(x)sin(nx)dx
Using the orthogonality property of sine functions, we get An=2
π
2
nπ (−1)n.
Thus, the final solution to the given PDE subject to the boundary conditions is:
u(x, t) = 4
π2
∞
X
n=1
(−1)n
nsin(nx)cos(cnt)
25. Question 25:
Consider the following Heat Equation:
∂u
∂t =k∂2u
∂x2+∂2u
∂y2
subject to the initial condition
u(x, y, 0) = sin(πx)sin(πy)
and the boundary conditions
u(0, y, t) = u(1, y, t) = u(x, 0, t) = u(x, 1, t) = 0
Solve the Heat Equation using the method of separation of variables.
Ans. Step-by-step solution:
1. Assume a separable solution of the form u(x, y, t) = X(x)Y(y)T(t).
2. Plug the separable solution into the Heat Equation and divide by kXY T to get:
1
kT
dT
dt =X′′
X+Y′′
Y
3. Rearrange the equation to separate the independent variables:
1
kT
dT
dt =−λ=X′′
X+Y′′
Y
4. Solve the time part of the equation:
1
k
dT
dt =−λ⇒1
kT(t) = −λt +c1⇒T(t) = c1e−λkt
5. Solve the spatial part of the equation:
X′′
X=−Y′′
Y=−λ
6. Solve for X(x):
X′′ +λX = 0
The general solution is X(x) = Acos√λx+Bsin√λx.
7. Apply boundary conditions u(0, y, t) = u(1, y, t) = 0:
X(0) = A= 0 and X(1) = Bsin√λ= 0
8. Since Bcannot be zero, sin√λ= 0 ⇒√λ=nπ, where nis a positive integer.
9. Solve for Y(y):
Y′′ −(nπ)2Y= 0
The general solution is Y(y) = Ccos(nπy) + Dsin(nπy).
10. Apply boundary conditions u(x, 0, t) = u(x, 1, t) = 0:
Y(0) = C= 0 and Y(1) = Dsin(nπ) = 0
11. Since Dcannot be zero, sin(nπ) = 0 ⇒nis a positive integer.
12. Combine the spatial solutions to get the complete solution:
u(x, y, t) = ∞
X
n=1
cne−(nπ)2kt sin(nπx)sin(nπy)
13. Finally, apply the initial condition u(x, y, 0) = sin(πx)sin(πy)to find the coefficients cn:
cn=Z1
0Z1
0
sin(πx)sin(πy)sin(nπx)sin(nπy)dx dy
This process will yield the final solution for u(x, y, t)in terms of the initial condition.
26. Question: Consider the one-dimensional heat equation given by ∂u
∂t =k∂2u
∂x2where u(x, t)
represents the temperature distribution along a rod of length Lat time t,0≤x≤L, and
kis a positive constant. The rod is insulated so that the temperature at both ends is fixed
at 0. Suppose the initial temperature distribution is given by u(x, 0) = 2x(L−x). Find the
temperature distribution u(x, t)for t > 0.
Ans. Solution: We will solve the one-dimensional heat equation using the method of separation
of variables.
1. Separation of Variables:
Assume a solution of the form u(x, t) = X(x)T(t), substitute into the heat equation, and
divide by kX(x)T(t)to separate variables:
1
kT (t)
dT
dt =1
X(x)
d2X
dx2
This results in two separate ordinary differential equations:
1
kT (t)
dT
dt =λ, 1
X(x)
d2X
dx2=λ
2. Solving the Time Equation:
The time equation yields T(t) = ceλkt, where cis a constant. Applying the boundary condition
u(0, t) = 0, we have:
X(x)T(t) = 0 =⇒X(x) = 0
Since X(x)cannot be identically zero, the only way for X(x) = 0 is if λ < 0. Therefore, we
conclude that λ=−a2, where ais a positive constant.
Thus, the time solution is T(t) = ce−a2kt.
3. Solving the Spatial Equation:
The spatial equation becomes:
d2X
dx2+a2X= 0
The general solution to this ordinary differential equation is X(x) = c1cos(ax) + c2sin(ax).
Applying the boundary conditions u(0, t) = 0 and u(L, t) = 0, we find that c1=c2= 0.
Therefore, the spatial solution is X(x) = 0.
4. Final Solution:
Since u(x, t) = X(x)T(t) = 0 for all xand t, we conclude that the unique solution to the
heat equation with the given initial condition is u(x, t) = 0 for all xand t.
27. Question: Find the general solution to the one-dimensional heat equation ut=kuxx subject
to the boundary conditions u(0, t) = 0,u(π, t) = 0, and the initial condition u(x, 0) = f(x) =
x(π−x).
Ans. Solution:
1. We begin by separating variables. Let u(x, t) = X(x)T(t). Plugging this into the heat
equation, we have:
X(x)T′(t) = kX′′(x)T(t)
2. Dividing both sides by kX(x)T(t), we get:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a separation constant. This gives two ODEs: T′+kλT = 0 and X′′ +λX = 0.
3. Solving the second ODE X′′ +λX = 0, we find that the general solution for X(x)is:
X(x) = Acos√λx+Bsin√λx
4. Applying boundary conditions u(0, t) = 0 and u(π, t) = 0 requires sin√λπ= 0. This
implies √λ=nπ, where nis a positive integer.
5. Thus, the general solution for X(x)becomes:
Xn(x) = Bnsin(nx)
6. Now we solve the first ODE, T′+kλT = 0, which has the general solution:
T(t) = Ce−kn2π2t
7. Therefore, the general solution to the heat equation with the given boundary conditions
is:
u(x, t) = ∞
X
n=1
Bnsin(nx)e−kn2π2t
8. Finally, we apply the initial condition u(x, 0) = f(x) = x(π−x). This allows us to
determine the coefficients Bnby using Fourier series. After finding the coefficients, the final
solution will be the sum of the series.
28. Question: Consider the heat equation in one dimension given by ∂u
∂t =k∂2u
∂x2, subject to the
boundary conditions u(0, t) = u(L, t) = 0 where 0≤x≤Land initial condition u(x, 0) = f(x)
where f(x)is a given function. Find the solution u(x, t)for this initial-boundary value problem.
Ans. Solution: Given the heat equation ∂u
∂t =k∂2u
∂x2, let’s assume separation of variables
u(x, t) = X(x)T(t). Substituting this into the heat equation gives:
X(x)T′(t) = kX′′(x)T(t)
Dividing by kX(x)T(t)and rearranging terms gives:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
This gives two ODEs:
(T′(t) + kλT (t) = 0
X′′(x) + λX(x) = 0
1. Solving the time ODE: The solution to T′(t) + kλT (t) = 0 is given by T(t) = Ce−kλt
where Cis a constant.
2. Solving the spatial ODE: The solution to X′′(x) + λX(x) = 0 is a characteristic
equation with two possible solutions: λ > 0and λ= 0.
For λ > 0, the general solution is given by X(x) = Acos√λx+Bsin√λxwhere A
and Bare constants.
For λ= 0, the general solution is X(x) = Ax +B.
3. Applying boundary conditions: Since the boundary conditions are u(0, t)=0and
u(L, t) = 0, we can see that the solutions for X(x)must satisfy X(0) = 0 and X(L) = 0.
For λ > 0, we have X(0) = A= 0 and X(L) = Bsin√λL= 0. This implies that
√λL =nπ, where nis a positive integer, giving λ=nπ
L2.
For λ= 0, we have X(0) = B= 0 and X(L) = AL = 0, which doesn’t give any non-trivial
solutions.
4. General solution: Finally, combining the solutions for T(t)and X(x), the general
solution to the heat equation is:
u(x, t) = ∞
X
n=1
cne−k(nπ
L)2tsin nπx
L
where cnare the coefficients determined by the initial condition and the orthogonality of the sine
functions.
29. Let u(x, t)satisfy the heat equation
∂u
∂t = 4∂2u
∂x2,0< x < π, t > 0
subject to the initial condition
u(x, 0) = 2x(π−x)
and the boundary conditions
u(0, t) = 0, u(π, t) = 0
Find the solution u(x, t)for t > 0and 0< x < π.
Ans. Solution:
We begin by seeking a solution in the form of a separation of variables: u(x, t) = X(x)T(t).
Substituting this into the heat equation yields
XT ′= 4X′′T
Dividing by X(x)T(t)gives
T′
4T=X′′
X=−λ
where λis a separation constant. Solving the ODE for X(x)with the given boundary condi-
tions, we obtain a general solution in the form
X(x) = Bsin(nπx)
where λ=n2π2for n= 1,2,3, . . .. Now, we solve the ODE for T(t)to obtain
T′
4T=−λ=−n2π2
Solving this ODE gives the general solution
T(t) = Ce−4n2π2t
The general solution to the heat equation is then
u(x, t) = ∞
X
n=1
Cne−4n2π2tsin(nπx)
Applying the initial condition u(x, 0) = 2x(π−x)to find the Fourier coefficients Cn, we have
2x(π−x) = ∞
X
n=1
Cnsin(nπx)
Multiplying both sides by sin(mπx)and integrating from 0 to π, we get
Zπ
0
2x(π−x)sin(mπx)dx =Zπ
0
Cmsin2(mπx)dx
Solving for Cmgives
Cm=2
πZπ
0
2x(π−x)sin(mπx)dx
Carrying out the integration gives the final solution for u(x, t):
u(x, t) = 8
π2
∞
X
n=1
1
(2n−1)2e−(2n−1)2π2tsin((2n−1)πx)
30. Question: Consider the following partial differential equation:
utt −4uxx = 0,0< x < π, t > 0
subject to boundary conditions
u(0, t) = u(π, t) = 0, t > 0
and initial conditions
u(x, 0) = sin(2x), ut(x, 0) = 0
Find the solution to the given partial differential equation using the method of separation of
variables.
Ans. Solution:
We consider a solution of the form
u(x, t) = X(x)T(t)
Substituting this into the given PDE, we get
XT ′′ −4X′′T= 0
Dividing by XT gives
T′′
T= 4X′′
X=−λ2
where λ2>0.
1. Finding the X(x)equation:
Solving X′′ +λ2X= 0 with boundary conditions X(0) = X(π) = 0, we get
X(x) = sin(2x)
2. Finding the T(t)equation:
Solving T′′ + 4λ2T= 0 with initial conditions T(0) = 1, T ′(0) = 0, we get
T(t) = cos(2t)
3. Forming the solution:
The solution to the given PDE is then
u(x, t) = X(x)T(t) = sin(2x)cos(2t)
3. Find the general solution to the heat equation
∂u
∂t = 3∂2u
∂x2
for 0< x < 1,t > 0, subject to the boundary conditions u(0, t) = u(1, t) = 0 and the initial
condition u(x, 0) = 2x(1 −x).
Ans. Solution: 1. We first assume that the solution u(x, t)can be expressed as a product of
functions of xand t, that is u(x, t) = X(x)T(t).
2. Plugging u(x, t)into the heat equation, we get
T′(t)
3T(t)=X′′(x)
X(x)=−λ
3. Solving the ODE for X(x), we get
X′′(x) + λX(x) = 0
4. The boundary conditions u(0, t) = u(1, t) = 0 lead to the conditions X(0) = 0 and
X(1) = 0. These are the conditions for the eigenvalues λn=n2π2, for n= 1,2,3, . . ..
5. The general solution for X(x)is given by
X(x) = ∞
X
n=1
cnsin(nπx)
6. Now, solving for T(t), we have
T′(t) = −3λT (t)⇒T(t) = ce−3λt
7. The general solution to the heat equation is then
u(x, t) = ∞
X
n=1
cnsin(nπx)e−3n2π2t
8. Finally, we apply the initial condition u(x, 0) = 2x(1 −x)to find the coefficients cn. By
using the orthogonality property of sine functions, we can express cnin terms of a Fourier sine
series.
9. The final solution is given by
u(x, t) = ∞
X
n=1
8
(2n−1)3π3sin((2n−1)πx)e−(2n−1)2π2t
4. Question: Consider the one-dimensional heat equation ut=kuxx, where kis a positive
constant. Given the initial condition u(x, 0) = sin(x)and boundary conditions u(0, t)=0and
u(π, t) = 0, find the solution u(x, t).
Ans. Step-by-step solution: 1. We start by assuming a separation of variables solution u(x, t) =
X(x)T(t). 2. Substituting this into the heat equation, we get T′
kT =X′′
X=−λ, where λ
is a separation constant. 3. Solving the ODE for X(x), we obtain the eigenvalue problem
X′′ +λX = 0 with boundary conditions X(0) = 0 and X(π) = 0. 4. The general solution
to the eigenvalue problem is Xn(x) = sin(nx)for λ=n2, where n∈N. 5. Now we solve
the ODE for T(t), which gives Tn(t) = cne−kn2t. 6. Combining the solutions for Xn(x)and
Tn(t), we get un(x, t) = cnsin(nx)e−kn2t. 7. Using the principle of superposition, the general
solution is u(x, t) = P∞
n=1 cnsin(nx)e−kn2t. 8. Applying the initial condition u(x, 0) = sin(x),
we find that cn= 2 Rπ
0sin(nx)sin(x)dx for n > 0and c0=1
πRπ
0sin(x)dx. 9. Simplifying
the integrals, we find that cn=2
π
1−(−1)n
n(1−n2)for n > 0and c0=1
π2×2π=2
π. 10. Therefore,
the solution to the heat equation with the given initial and boundary conditions is u(x, t) =
2
πsin(x) + P∞
n=1
1−(−1)n
n(1−n2)sin(nx)e−kn2t.
5. Question:
Consider the one-dimensional heat equation given by ∂u
∂t =k∂2u
∂x2, where u(x, t)is the tem-
perature distribution along a rod of length Lat time t,kis the thermal diffusivity constant, and
u(x, 0) = f(x)is the initial temperature distribution.
Given the following boundary conditions: 1. u(0, t) = 0 for t≥02. u(L, t) = 0 for t≥03.
u(x, 0) = sinπx
Lfor 0≤x≤L
Find the temperature distribution u(x, t)for x∈[0, L]and t > 0.
Ans. Step-by-step solution:
1. We begin by assuming a separable solution of the form u(x, t) = X(x)T(t).
2. Plugging this into the heat equation gives T′
kT =X′′
X=−λ, where λis a separation
constant.
3. This gives two ODEs: T′+kλT = 0 with solution T(t) = Ae−kλt X′′ +λX = 0 with
solution X(x) = Bsin √λx
4. Applying the boundary conditions u(0, t) = 0 and u(L, t) = 0 at x= 0 and x=L
respectively, we obtain X(0) = 0 and X(L) = 0.
5. The boundary condition X(0) = 0 gives Bsin(0) = 0 ⇒B= 0. Contradiction.
6. Since B= 0 leads to a trivial solution, we consider the nontrivial case where λ=nπ
L2
for n= 1,2,3, ...
7. Therefore, the general solution is a sum of these eigenfunctions: u(x, t) = P∞
n=1 Ansin nπ
Lxe−k(nπ
L)2t.
8. To determine the coefficients An, we use the initial temperature distribution u(x, 0) =
sin πx
L:u(x, 0) = P∞
n=1 Ansin nπ
Lx=sin πx
L.
9. By orthogonality, we find: An=2
LRL
0sin πx
Lsin nπ
Lxdx
10. Simplifying the integral and solving for An, we get: An=(0if nis even
4
nπ if nis odd .
11. Therefore, the temperature distribution is: u(x, t) = P∞
m=1
4
(2m−1)πsin (2m−1)πx
Le−k((2m−1)π
L)2t.
6. Question: Consider the one-dimensional heat equation given by ∂u
∂t =k∂2u
∂x2, where kis a
positive constant. If the initial condition is u(x, 0) = x(1 −x)for 0< x < 1and the boundary
conditions are u(0, t) = u(1, t) = 0 for t > 0, find the solution u(x, t).
Ans. Step-by-step solution: 1. First, we assume a separation of variables solution of the form
u(x, t) = X(x)T(t). 2. Substitute the separation of variables solution into the heat equation to
get T′(t)
kT (t)=X′′(x)
X(x)=−λ, where λis a constant. 3. Solve the temporal part: T′(t) + kλT (t) = 0
has the solution T(t) = c1e−kλt. 4. Solve the spatial part: X′′(x) + λX(x)=0with boundary
conditions X(0) = X(1) = 0 has the solution Xn(x) = sin(nπx)for λ=n2π2. 5. Therefore, the
general solution is u(x, t) = P∞
n=1 cne−kn2π2tsin(nπx). 6. Apply the initial condition u(x, 0) =
x(1 −x)to find cncoefficients. 7. Use the orthogonality property of sines to evaluate cn. 8. The
final solution for the heat equation is u(x, t) = P∞
n=1 8
(2n−1)π2
e−(2n−1)2π2kt sin((2n−1)πx).
7. Question: Consider the heat equation defined by
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions
u(0, t) = 0, u(1, t) = 0
and the initial condition
u(x, 0) = x(1 −x)
Find the solution to the heat equation.
Ans. Solution:
We can solve the heat equation using the method of separation of variables.
1. Separation of Variables:
Let u(x, t) = X(x)T(t). Substituting this into the heat equation, we get:
XT ′=kX′′T
Dividing by kXT , we have:
T′
kT =X′′
X=−λ
This gives two separate ordinary differential equations:
T′+kλT = 0 and X′′ +λX = 0
2. Solve the Time Equation:
The time equation gives us T(t) = ce−kλt, where cis a constant.
3. Solve the Spatial Equation:
The spatial equation X′′ +λX = 0 has solutions of the form X(x) = Asin(nπx)and
X(x) = Bcos(nπx), where λ=n2π2for n= 1,2,3, . . .
The general solution is then:
u(x, t) = ∞
X
n=1
(Ansin(nπx) + Bncos(nπx))e−k(nπ)2t
4. Solve for the Coefficients:
Applying the boundary conditions u(0, t)=0and u(1, t)=0, we find that the coefficients
Anare zero for all nand the coefficients Bncan be determined using the initial condition
u(x, 0) = x(1 −x).
5. Final Solution:
Putting it all together, the solution to the heat equation with the given initial and boundary
conditions is:
u(x, t) = ∞
X
n=1
2
nπ sin(nπx)e−k(nπ)2t
8. Consider the following partial differential equation, known as the heat equation:
∂u
∂t =k∂2u
∂x2+∂2u
∂y2
where u(x, y, t)is the temperature distribution in a metal plate at time t,kis the thermal
diffusivity constant, and u(x, y, 0) = f(x, y)represents the initial distribution of temperature.
Given the metal plate is a square with sides of length Land the initial temperature distribution
is f(x, y) = 100 within the square centered at the origin with side length L
2and f(x, y)=0
otherwise, find the temperature distribution at time t > 0within the metal plate.
Ans. Let’s first set up the problem and then solve the heat equation with the given initial
conditions.
1. The heat equation for this problem is:
∂u
∂t =k∂2u
∂x2+∂2u
∂y2
subject to the initial condition:
u(x, y, 0) = (100 if |x|<L
4and |y|<L
4
0otherwise
2. To solve the heat equation, we can use separation of variables. Let u(x, y, t) = X(x)Y(y)T(t).
Plugging this into the heat equation, we get:
1
kXT
dT
dt =X′′Y+XY ′′
XY
3. Dividing both sides by u, we obtain:
1
k
1
T
dT
dt =X′′
X+Y′′
Y
4. Since the left-hand side depends only on tand the right-hand side depends only on xand
y, they must both be equal to a constant, say −λ2. This gives us the system of ODEs:
1
k
1
T
dT
dt =−λ2
X′′ +λ2X= 0
Y′′ +λ2Y= 0
5. Solving the ODEs for Xand Y, we obtain the general solutions:
X(x) = Acosnπx
L, Y (y) = Bsinnπy
L
where nis a positive integer.
6. Applying the initial condition, we can write the solution as:
u(x, y, t) = ∞
X
n=1
Cncosnπx
Lsinnπy
Le−k(nπ
L)2t
where Cnare determined by the initial temperature distribution.
7. Using the initial condition f(x, y)provided, we can find the coefficients Cnand thus
determine the temperature distribution at any time t > 0.
9. Question:
Consider the heat equation in one dimension, given by ∂u
∂t =k∂2u
∂x2, where kis a positive
constant. Suppose the initial temperature distribution along a rod of length Lis given by u(x, 0) =
sin πx
L, and the rod is insulated at both ends so that ∂u
∂x (0, t) = ∂u
∂x (L, t) = 0 for all t > 0. Find
the temperature distribution u(x, t)along the rod at a later time t > 0.
Ans. Step-by-step solution:
1. We start by assuming a separation of variables solution of the form u(x, t) = X(x)T(t).
Substituting this into the heat equation gives:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a separation constant. This gives two ordinary differential equations:
T′(t) + kλT (t) = 0 and X′′(x) + λX(x) = 0
2. Solving the ODE for T(t)gives T(t) = Ae−kλt, where Ais a constant.
3. The ODE for X(x)has solutions of the form X(x) = Bcos√λx+Csin√λx, where
Band Care constants. Applying the boundary conditions, X′(0) = X′(L) = 0, implies B= 0
and √λ=nπ/L, where nis a positive integer. Thus, X(x) = Csin(nπx/L).
4. Therefore, the general solution is given by the sum of all possible eigenfunctions:
u(x, t) = ∞
X
n=1
Cnsin nπx
Le−k(nπ/L)2t
5. To find the coefficients Cn, we use the initial temperature distribution: u(x, 0) =
sin πx
L=P∞
n=1 Cnsin nπx
L. By orthogonality, we have:
Cn=2
LZL
0
sin πx
Lsin nπx
Ldx
6. Evaluating the integral above gives Cn=2
Lδn1, where δn1is the Kronecker delta. There-
fore, the temperature distribution along the rod at time t > 0is:
u(x, t) = 2
Lsin πx
Le−kπ2t/L2
10. Find the solution to the heat equation on the unit interval 0< x < 1with initial condition
u(x, 0) = xand boundary conditions u(0, t) = 0 and u(1, t) = 1.
Ans. Solution:
We seek a solution to the heat equation of the form u(x, t) = X(x)T(t). Substituting this
into the heat equation ut=uxx, we get:
X(x)T′(t) = X′′(x)T(t)
Dividing both sides by X(x)T(t)gives:
T′(t)
T(t)=X′′(x)
X(x)=−λ
which leads to two ordinary differential equations:
T′(t) + λT (t) = 0, T (0) = 0, T (L) = 1
X′′(x) + λX(x) = 0
Solving the first ODE, we get T(t) = ce−λt. Using the boundary conditions, we find T(t) =
sin(nπx).
For the second ODE, the general solution is X(x) = Acos√λx+Bsin√λx. Applying
the boundary conditions X(0) = 0 and X(1) = 1, we get:
X(0) = A= 0
X(1) = Bsin√λ= 1
Since B= 0, we have sin√λ= 1 =⇒√λ=π
2. Thus, our eigenvalue is λ=π2
4.
Therefore, the solution to the heat equation with the given boundary conditions and initial
condition is:
u(x, t) = ∞
X
n=1
Cne−n2π2
4tsin nπx
2
where Cndepends on the initial condition.
11. Let u(x, t)be the solution to the heat equation
ut=uxx,0< x < 1, t > 0
with initial condition
u(x, 0) = x(1 −x),0< x < 1
and boundary conditions
u(0, t) = 0, u(1, t) = 0, t > 0
Determine the expression for u(x, t).
Ans. To solve the heat equation, we use separation of variables. We assume a solution of the
form
u(x, t) = X(x)T(t)
Substitute this into the heat equation to get
XT ′=X′′T
Divide both sides by u(x, t)to get
T′
T=X′′
X=−λ
Solving the time equation gives
T′(t) + λT (t) = 0
This gives the general solution
T(t) = Ae−λt
where λ > 0to ensure the solution decays as t→ ∞.
Solving the spatial equation gives
X′′(x) + λX(x) = 0
with boundary conditions X(0) = 0,X(1) = 0. The general solution to this equation is
X(x) = Bsin√λx
Applying the boundary conditions gives
X(x) = Bsin(nπx)
where √λ=nπ for n= 1,2,3, .... We choose n= 1 to satisfy the initial condition.
So, the solution to the heat equation is
u(x, t) = ∞
X
n=1
Bne−(nπ)2tsin(nπx)
To determine Bn, we use the initial condition u(x, 0) = x(1 −x)to find the Fourier sine
series of x(1 −x).
x(1 −x) = ∞
X
n=1
Bnsin(nπx)
Bn= 2 Z1
0
x(1 −x)sin(nπx)dx
By evaluating this integral, we find that
Bn=8
n3π3(1 −(−1)n)
Therefore, the expression for u(x, t)is
u(x, t) = ∞
X
n=1
8
n3π3(1 −(−1)n)e−(nπ)2tsin(nπx)
12. Question: Consider the following partial differential equation:
uxx −4uyy = 0
subject to the boundary conditions
u(0, y) = 0, u(1, y) = 0, u(x, 0) = 0, u(x, 1) = 0
Find the solution to the given partial differential equation with the specified boundary conditions.
Ans. Let u(x, y) = X(x)Y(y)be the separation of variables solution. Then we have
X′′(x)Y(y)−4X(x)Y′′(y) = 0
Dividing by X(x)Y(y), we get
X′′(x)
X(x)= 4Y′′(y)
Y(y)
Since the left-hand side depends only on xand the right-hand side depends only on y, they must
be equal to a constant −λ, so we have two separate ODEs:
X′′(x) + λX(x) = 0, Y ′′(y)−λ
4Y(y) = 0
1. Solve for X(x): The ODE for X(x)is
X′′(x) + λX(x) = 0
with boundary conditions X(0) = X(1) = 0. This gives us the eigenvalues λn=n2π2, and the
eigenfunctions Xn(x) = sin(nπx).
2. Solve for Y(y): The ODE for Y(y)is
Y′′(y)−λ
4Y(y) = 0
Substitute λ=n2π2:
Y′′(y)−n2π2
4Y(y) = 0
The corresponding eigenvalues λn=n2π2and eigenfunctions Yn(y) = sin nπy
2.
3. Combine X(x) and Y(y): The general solution is a linear combination of the separated
solutions:
u(x, y) = ∞
X
n=1
cnsin(nπx)sin nπy
2
4. Applying boundary conditions: Applying the boundary conditions gives us the final
solution:
u(x, y) = ∞
X
n=1
cnsin(nπx)sin nπy
2
where
cn=8
n3π3(1 −(−1)n)
13. Question: Consider the following heat equation in one dimension:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions u(0, t) = 0 and u(L, t) = 0, and the initial condition u(x, 0) =
f(x)for 0≤x≤L. Show that the solution to this equation can be written as a Fourier sine
series, where
u(x, t) = ∞
X
n=1
Bnsin nπx
Le−k(nπ
L)2t
Ans. Solution: 1. To solve the heat equation using separation of variables, assume that the
solution can be expressed as a product of two functions:
u(x, t) = X(x)T(t)
2. Substitute this expression into the heat equation to get:
X(x)T′(t) = kX′′(x)T(t)
3. Divide both sides by kX(x)T(t)to separate the variables:
T′(t)
kT (t)=X′′(x)
X(x)=−λ2
4. Solve the time part of the equation:
T′(t) + kλ2T(t) = 0
This gives the general solution:
T(t) = C1e−kλ2t
5. Solve the spatial part of the equation:
X′′(x) + λ2X(x) = 0
This second-order homogeneous differential equation has solutions in the form of sine functions
since the boundary conditions are u(0, t)=0and u(L, t)=0. Thus, the solutions are Xn(x) =
sin nπx
Lfor n= 1,2,3, . . .
6. The general solution to the spatial part is given as:
X(x) = ∞
X
n=1
Ansin nπx
L
7. Combine the spatial and time parts of the solution:
u(x, t) = ∞
X
n=1
Ansin nπx
Le−k(nπ
L)2t
8. Given that the initial condition is u(x, 0) = f(x), we express the solution in its Fourier
sine series form:
u(x, 0) = ∞
X
n=1
Ansin nπx
L=f(x)
9. By comparing coefficients, we find that An=Bnand thus arrive at the desired solution:
u(x, t) = ∞
X
n=1
Bnsin nπx
Le−k(nπ
L)2t
14. Consider the Heat Equation in one dimension defined by
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions u(0, t) = 0 and u(π, t) = 0, and the initial condition
u(x, 0) = x(π−x).
Find the solution u(x, t)to this Heat Equation.
Ans. Solution: 1. We start by assuming a separation of variables solution of the form
u(x, t) = X(x)T(t). 2. Substituting this into the Heat Equation gives T′
kT =X′′
X=−λ, where
λis a separation constant. 3. Solving the time ODE T′
kT =−λgives T(t) = Ae−kλt. 4.
Solving the spatial ODE X′′ +λX = 0 gives X(x) = Bsin√λx+Ccos√λx. 5. Applying
the boundary conditions u(0, t) = 0 and u(π, t) = 0 gives X(0) = Bsin(0) + Ccos(0) = 0
and X(π) = Bsin√λπ+Ccos√λπ= 0. 6. From X(0) = 0, we have C= 0. 7.
From X(π) = 0, we have √λπ =nπ for n= 1,2,3, ..., leading to λ=n2. 8. The general
solution is then u(x, t) = P∞
n=1(Ansin(nx)e−kn2t). 9. Applying the initial condition u(x, 0) =
x(π−x)gives x(π−x) = P∞
n=1 Ansin(nx). 10. By orthogonality, we can solve for Anas
An=2
πRπ
0x(π−x)sin(nx)dx. 11. Calculating the integral gives An=4
n3((−1)n−1). 12.
Therefore, the solution to the Heat Equation is u(x, t) = P∞
n=1
4
n3((−1)n−1) sin(nx)e−kn2t.
15. Let u(x, t)satisfy the wave equation in one space dimension and time
utt =c2uxx,0< x < l, t > 0
with boundary conditions u(0, t) = u(l, t) = 0 for t > 0and initial conditions u(x, 0) = f(x)
and ut(x, 0) = g(x)for 0< x < l. Show that the separation of variables leads to the solution
u(x, t) = ∞
X
n=1
Bnsin nπx
l·cos nπct
l
where the coefficients Bnare given by
Bn=2
lZl
0
f(x)sin nπx
ldx.
Ans. Solution: 1. We start by trying the separation of variables approach, assuming that
u(x, t) = X(x)T(t). So, we have
X(x)T′′(t) = c2X′′(x)T(t).
2. Dividing by c2u(x, t), we get
T′′(t)
c2T(t)=X′′(x)
X(x)=−λ.
This leads to the two ordinary differential equations
T′′(t) + λc2T(t) = 0
and
X′′(x) + λX(x) = 0.
3. The boundary conditions u(0, t) = u(l, t)=0for t > 0imply that X(0) = X(l)=0.
This gives a Sturm-Liouville problem for X(x).
4. The general solution to the spatial part of the equation is
X(x) = Asin nπx
l+Bcos nπx
l.
5. Applying the boundary conditions X(0) = X(l)=0, we find that B= 0 and X(x) =
Asin nπx
l.
6. The general solution to the time part of the equation is
T(t) = Ccos nπct
l+Dsin nπct
l.
7. Since the wave equation is linear, the total solution is a linear combination of the separated
solutions. By the principle of superposition, we can write
u(x, t) = ∞
X
n=1
Bnsin nπx
l·cos nπct
l,
where the coefficients Bnare to be determined.
8. To determine the coefficients Bn, we substitute u(x, 0) = f(x)into the general solution.
This gives
f(x) = ∞
X
n=1
Bnsin nπx
l.
9. We can now determine the coefficients Bnby the formula
Bn=2
lZl
0
f(x)sin nπx
ldx.
Hence, we have shown that the separation of variables leads to the provided solution form
and formula for the coefficients.
16. Question: Consider the following partial differential equation:
ut= 4uxx −2u
subject to the initial conditions:
u(x, 0) = x2
ut(x, 0) = 0
Find the solution to the partial differential equation using the method of separation of vari-
ables.
Ans. Step-by-step solution: 1. We will assume the solution to the PDE is separable, i.e.,
u(x, t) = X(x)T(t). Substitute this into the PDE to get:
XT ′= 4X′′T−2XT
Divide both sides by XT to get:
T′
T= 4X′′
X−2
2. Rearrange the equation to separate variables:
T′
T=A
4X′′
X−2 = A
where Ais a constant. 3. Solve the time part ODE:
dT
dt =AT
dT
T=Adt
Integrating both sides gives:
ln |T|=At +B
T(t) = CeAt
where Cis a constant. 4. Solve the spatial part ODE:
4X′′
X−2 = A
4X′′ −(2 + AX)X= 0
This is a second-order linear ODE with constant coefficients. Let m2= 2 +A. The characteristic
equation becomes:
4m2−(2 + A) = 0
4m2−m2= 2
3m2= 2
m=±√6
3
5. The general solution for X(x)is:
X(x) = Ae√6
3x+Be−√6
3x
6. Applying the initial condition u(x, 0) = x2, we have:
X(x)T(0) = x2
X(x) = x2
Ae0+Be0=x2
A+B=x2
7. Applying the initial condition ut(x, 0) = 0, we have:
XT ′(0) = 0
A√6
3e0−B√6
3e0= 0
√6
3(A−B) = 0
A=B
A+A=x2
2A=x2
A=x2
2
B=x2
2
Thus, the solution to the partial differential equation using separation of variables is:
u(x, t) = x2
2e√6
3x+x2
2e−√6
3xeAt
where A=√6
3.
17. Question:
Consider the heat equation on a bounded domain Dwith initial condition u(x, y, 0) = x2+y2
and boundary conditions u(x, y, t) = 0 for (x, y)∈∂D, where ∂D denotes the boundary of D.
Solve the heat equation to find u(x, y, t).
Ans. Solution:
We will solve the heat equation using separation of variables. Let u(x, y, t) = X(x)Y(y)T(t).
Substituting this into the heat equation ut=c2(uxx +uyy)gives
T′
c2T=X′′
X+Y′′
Y
Now we have separated the variables and have three ODEs to solve:
1. Solve T′
c2T=−λ=α2for T(t). 2. Solve X′′ =−α2Xfor X(x). 3. Solve Y′′ =−α2Y
for Y(y).
1. Solving T′
c2T=−λ=α2gives T(t) = e−α2c2t.
2. Solving X′′ =−α2Xwith boundary condition X(0) = X(L)=0gives eigenvalues
αn=nπ
Land eigenfunctions Xn(x) = sin nπx
L.
3. Solving Y′′ =−α2Ywith boundary condition Y(0) = Y(L) = 0 gives eigenvalues
βn=nπ
Land eigenfunctions Yn(y) = sin nπy
L.
Therefore, the general solution can be written as
u(x, y, t) = ∞
X
n=1
Cne−(nπ
L)2c2tsin nπx
Lsin nπy
L
Applying the initial condition u(x, y, 0) = x2+y2allows us to determine the coefficients
Cnusing Fourier series. After solving for the coefficients, the final solution for u(x, y, t)will be
obtained.
18. Question: Let u(x, t)be the solution to the heat equation ∂u
∂t =∂2u
∂x2with boundary
conditions u(0, t) = u(1, t)=0and initial condition u(x, 0) = 2x(1 −x). Find the solution
u(x, t).
Ans. Solution: 1. First, let’s express the solution u(x, t)as a Fourier sine series:
u(x, t) = ∞
X
n=1
Bn(t)sin(nπx)
2. Our next step is to find the time-dependent coefficients Bn(t). Substitute the expression
for u(x, t)into the heat equation and use the fact that ∂
∂t (sin(nπx)) = 0 to get:
dBn
dt sin(nπx) = −n2π2Bnsin(nπx)
3. This simplifies to:
dBn
dt =−n2π2Bn
4. Solving the above first-order ordinary differential equation, we get:
Bn(t) = cne−n2π2t
5. Now we need to determine the constant cn. We can use the initial condition u(x, 0) =
2x(1 −x)to find cn:
2x(1 −x) = ∞
X
n=1
cnsin(nπx)
6. Multiply both sides by sin(mπx)and integrate from 0 to 1:
Z1
0
2x(1 −x)sin(mπx)dx =Z1
0
∞
X
n=1
cnsin(nπx)sin(mπx)dx
7. By the orthogonality property of sine functions, all terms in the sum vanish except when
n=m. This leads to:
cm= 2 Z1
0
x(1 −x)sin(mπx)dx
8. Evaluate the integral to find cm:
cm= 2 6
π3(1 −(−1)m)
Therefore, the solution to the heat equation with the provided initial and boundary conditions
is:
u(x, t) = ∞
X
n=1
26
π3(1 −(−1)n)e−n2π2tsin(nπx)
19. Let u(x, t)satisfy the one-dimensional heat equation
ut=c2uxx
subject to the boundary conditions u(0, t) = u(L, t) = 0 and the initial condition
u(x, 0) = sin πx
L−sin 3πx
L
where 0< x < L,t > 0, and cis a positive constant.
Compute u(x, t)for all xand t.
Ans. To find the solution to the heat equation with the given conditions, we can use the method
of separation of variables. Let’s assume a solution of the form u(x, t) = X(x)T(t).
1. Write down the separated equations: Plugging the assumed form of the solution into
the heat equation gives
X(x)T′(t) = c2X′′(x)T(t)
Dividing by c2u(x, t)gives
T′(t)
c2T(t)=X′′(x)
X(x)
Letting both sides be equal to a separation constant λyields the equations
T′(t) = c2λT (t)and X′′(x) = λX(x)
2. Solve the time part of the equation: The equation for T(t)can be solved as
T(t) = Aec2λt
where Ais a constant.
3. Solve the spatial part of the equation: The equation for X(x)gives solutions of the
form
X(x) = Bsin nπx
L+Ccos nπx
L
where Band Care constants to be determined.
4. Apply the boundary conditions: The boundary conditions u(0, t) = 0 and u(L, t) = 0
give X(0) = 0 and X(L) = 0, which results in
C= 0 and Bsin(nπ) = 0
Since nis a positive integer, the only way to satisfy the second equation is to have n=
1,2,3, .... Thus, the solution becomes
Xn(x) = Bnsin nπx
L
5. Express the general solution: The general solution is then a linear combination of the
spatial parts:
u(x, t) = ∞
X
n=1
Tn(t)Xn(x)
u(x, t) = ∞
X
n=1 Anec2λntBnsin nπx
L
6. Apply the initial condition: Finally, we apply the initial condition to find the coefficients
Anusing the Fourier sine series of the function
u(x, 0) = sin πx
L−sin 3πx
L=∞
X
n=1
AnBnsin nπx
L
Comparing coefficients gives
AnBn=
1for n= 1
−1for n= 3
0for n= 1,3
Thus, we obtain the solution
u(x, t) = e−c2π2t/L2sin πx
L−e−c29π2t/L2sin 3πx
L
20. Question:
Consider the one-dimensional heat equation given by ∂u
∂t =k∂2u
∂x2, where kis a constant and
u(x, t)is the temperature distribution along a rod of length Lat time t.
A rod of length L= 10 cm initially has a temperature distribution u(x, 0) = 100x(L−x)for
0≤x≤L. The rod is kept in an environment at a constant temperature of 0◦C on both sides
while its ends are insulated. Find the temperature distribution u(x, t)for all xand t.
Ans. Step-by-step solution:
1. The general solution to the one-dimensional heat equation is given by u(x, t) = P∞
n=1 Bnsin nπx
Le−αn2π2t/L2,
where Bnare constants and α=k/L2.
2. To find the specific solution to our problem, we first need to determine the initial conditions
for the problem. We have u(x, 0) = 100x(L−x).
3. We can express the initial condition as a Fourier sine series: u(x, 0) = P∞
n=1 Bnsin nπx
L.
Comparing this with 100x(L−x), we find that Bn= 400/L·(−1)n+1
n2π2.
4. Now, plugging in the initial condition into the general solution, we have u(x, t) =
P∞
n=1 400(−1)n+1
n2π2sin nπx
Le−αn2π2t/L2.
5. Substituting L= 10 cm and the given values of k, we have α=k/L2=k/100. Hence,
the temperature distribution for all xand tis:
u(x, t) = ∞
X
n=1 400(−1)n+1
n2π2sin nπx
10 e−kn2π2t
100
21. Question:
Consider the heat equation in one dimension given by ∂u
∂t =k∂2u
∂x2, subject to the initial
condition u(x, 0) = sin(x)and boundary conditions u(0, t) = 0 and u(π, t) = 0, where kis a
positive constant.
Find the solution u(x, t)to the heat equation.
Ans. Step-by-step solution:
1. We can solve the heat equation using separation of variables. Let u(x, t) = X(x)T(t).
Substitute u(x, t) = X(x)T(t)into the heat equation to get: X(x)T′(t) = kX′′(x)T(t).
2. Divide both sides by kX(x)T(t)to separate the variables: T′(t)
kT (t)=X′′(x)
X(x).
Letting both sides equal to a constant −λ, we have two ordinary differential equations:
T′(t) + kλT (t) = 0 and X′′(x) + λX(x) = 0.
3. Solve the time part of the equation: T′(t) = −kλT (t)1
T(t)dT =−kλdt ln |T|=−kλt+C1
T(t) = C2e−kλt, where C2=eC1.
Since C2is an arbitrary constant, we have T(t) = e−kλt.
4. Now solve the spatial part of the equation: X′′(x) + λX(x) = 0 The general solution to
this ordinary differential equation is X(x) = Acos√λx+Bsin√λxwhere A and B are
arbitrary constants to be determined by the boundary conditions.
Applying the boundary conditions u(0, t) = 0 and u(π, t) = 0 gives A= 0 and sin√λπ=
0.
5. The condition sin√λπ= 0 implies that √λπ =nπ for some integer n, i.e., λ=n2.
Thus, the solution X(x)simplifies to Xn(x) = Bnsin(nx)where nis a positive integer.
Putting together the spatial and temporal solutions, the general solution is then u(x, t) =
P∞
n=1 Bnsin(nx)e−kn2t.
6. To determine the constants Bn, we use the initial condition u(x, 0) = sin(x):u(x, 0) =
P∞
n=1 Bnsin(nx) = sin(x).
Comparing coefficients of sin(nx)on both sides gives Bn=(2/πif n= 1,
0if n= 1..
Therefore, the solution to the heat equation is u(x, t) = 2
πsin(x)e−kt.
22. Question:
Consider the heat equation defined by ∂u
∂t = 2∂2u
∂x2, subject to the boundary conditions u(0, t) =
0and u(π, t)=0, and the initial condition u(x, 0) = 3 sin(x). Find the solution u(x, t)to this
equation.
Ans. Let’s solve the given heat equation with the specified boundary and initial conditions.
1. Find the solution using separation of variables:
Assume the solution has the form u(x, t) = X(x)T(t).
Substitute u(x, t) = X(x)T(t)into the heat equation to get:
X(x)T′(t) = 2X′′(x)T(t)
Divide both sides by u(x, t)to get:
T′(t)
2T(t)=X′′(x)
X(x)=−λ
This gives us two ordinary differential equations:
X′′(x) + λX(x) = 0 and T′(t)+2λT (t) = 0
2. Solve the spatial equation:
The solution to the spatial equation X′′(x) + λX(x)=0with the boundary conditions
X(0) = 0 and X(π) = 0 is:
Xn(x) = sin(nx)
The corresponding eigenvalues are λn=n2.
3. Solve the temporal equation:
The solution to the temporal equation T′(t)+2λT (t) = 0 is:
Tn(t) = Ce−2n2t
4. Combine the solutions:
The general solution to the heat equation is given by:
u(x, t) = ∞
X
n=1
Ansin(nx)e−2n2t
5. Use the initial condition to determine coefficients:
Given the initial condition u(x, 0) = 3 sin(x), we have:
3sin(x) = ∞
X
n=1
Ansin(nx)
Comparing coefficients, we find A1= 3 and An= 0 for n= 1.
6. Final solution:
Therefore, the solution to the heat equation with the given conditions is:
u(x, t) = 3 sin(x)e−2t
23. Find the solution to the following partial differential equation, subject to the given initial
conditions:
∂2u
∂t2= 4∂2u
∂x2,0< x < π, t > 0
with initial conditions u(0, t) = u(π, t) = 0,u(x, 0) = sin(x), and ∂u
∂t (x, 0) = 0.
Ans. Solution:
1. We assume a solution of the form u(x, t) = X(x)T(t)and substitute it into the given
partial differential equation:
X(x)T′′(t) = 4X′′(x)T(t)
Dividing by X(x)T(t), we get:
T′′(t)
T(t)= 4X′′(x)
X(x)
Rearranging, we have:
T′′(t)
T(t)= 4K=λ
2. Solving the ODE T′′(t)
T(t)=λgives us:
T(t) = c1cos 2√λt+c2sin 2√λt
Applying the initial condition ∂u
∂t (x, 0) = 0, we have d
dt (X(x)T(t))t=0 = 0:
√λc1X(x) = 0
Since X(x)= 0, we have λ= 0.
3. Therefore, the solution for T(t)becomes:
T(t) = c1+c2t
4. Substituting λ= 0 back into X′′(x)
X(x)= 0, we get:
X(x) = c3+c4x
Applying the boundary conditions u(0, t) = u(π, t) = 0, we have:
X(0) = c3= 0 and X(π) = c4π= 0
So, c4= 0 and X(x) = 0.
5. Therefore, the solution to the PDE with the given initial conditions is:
u(x, t) = 0
24. Find the solution to the following partial differential equation, subject to the given boundary
conditions:
∂2u
∂t2=c2∂2u
∂x2,0< x < π, t > 0
subject to the boundary conditions:
u(0, t) = u(π, t) = 0, u(x, 0) = sin(x),∂u
∂t (x, 0) = 0
Solution:
1. Let’s assume the solution can be written in the form
u(x, t) = X(x)T(t)
Substituting this into the partial differential equation, we get:
X(x)T′′(t) = c2X′′(x)T(t)
Dividing both sides by c2XT , we get:
T′′(t)
c2T(t)=X′′(x)
X(x)=−λ
2. Solving the ODE X′′(x)
X(x)=−λgives us the solutions:
Xn(x) = Bnsin(nx),where n=√λand Bnis a constant
Applying the boundary conditions u(0, t) = u(π, t) = 0, we have:
Xn(0) = Bnsin(0) = 0 =⇒Bn= 0
This implies that the solution to the spatial part is X(x) = 0, which is trivial. Hence, this
leads to the conclusion that λ < 0.
3. Solving the ODE T′′(t)
c2T(t)=−λgives us the solution:
T(t) = Acosc√−λt+Bsinc√−λt
Applying the initial condition ∂u
∂t (x, 0) = 0, we have:
T′(0) = −Ac√−λsin(0) + Bc√−λcos(0) = 0
Bc√−λ= 0
Since c > 0, this gives us B= 0, which implies T(t) = Acosc√−λt.
4. Substituting n2=−λinto the spatial part Xn(x) = Bnsin(nx)and time part Tn(t) =
Ancos(cnt), we get the general solution:
u(x, t) = ∞
X
n=1
Ansin(nx)cos(cnt)
5. Now, applying the initial condition u(x, 0) = sin(x)to the general solution, we get:
∞
X
n=1
Ansin(nx) = sin(x)
By orthogonality, we can determine An:
An=2
πZπ
0
sin(x)sin(nx)dx
Using the orthogonality property of sine functions, we get An=2
π
2
nπ (−1)n.
Thus, the final solution to the given PDE subject to the boundary conditions is:
u(x, t) = 4
π2
∞
X
n=1
(−1)n
nsin(nx)cos(cnt)
25. Question 25:
Consider the following Heat Equation:
∂u
∂t =k∂2u
∂x2+∂2u
∂y2
subject to the initial condition
u(x, y, 0) = sin(πx)sin(πy)
and the boundary conditions
u(0, y, t) = u(1, y, t) = u(x, 0, t) = u(x, 1, t) = 0
Solve the Heat Equation using the method of separation of variables.
Ans. Step-by-step solution:
1. Assume a separable solution of the form u(x, y, t) = X(x)Y(y)T(t).
2. Plug the separable solution into the Heat Equation and divide by kXY T to get:
1
kT
dT
dt =X′′
X+Y′′
Y
3. Rearrange the equation to separate the independent variables:
1
kT
dT
dt =−λ=X′′
X+Y′′
Y
4. Solve the time part of the equation:
1
k
dT
dt =−λ⇒1
kT(t) = −λt +c1⇒T(t) = c1e−λkt
5. Solve the spatial part of the equation:
X′′
X=−Y′′
Y=−λ
6. Solve for X(x):
X′′ +λX = 0
The general solution is X(x) = Acos√λx+Bsin√λx.
7. Apply boundary conditions u(0, y, t) = u(1, y, t) = 0:
X(0) = A= 0 and X(1) = Bsin√λ= 0
8. Since Bcannot be zero, sin√λ= 0 ⇒√λ=nπ, where nis a positive integer.
9. Solve for Y(y):
Y′′ −(nπ)2Y= 0
The general solution is Y(y) = Ccos(nπy) + Dsin(nπy).
10. Apply boundary conditions u(x, 0, t) = u(x, 1, t) = 0:
Y(0) = C= 0 and Y(1) = Dsin(nπ) = 0
11. Since Dcannot be zero, sin(nπ) = 0 ⇒nis a positive integer.
12. Combine the spatial solutions to get the complete solution:
u(x, y, t) = ∞
X
n=1
cne−(nπ)2kt sin(nπx)sin(nπy)
13. Finally, apply the initial condition u(x, y, 0) = sin(πx)sin(πy)to find the coefficients cn:
cn=Z1
0Z1
0
sin(πx)sin(πy)sin(nπx)sin(nπy)dx dy
This process will yield the final solution for u(x, y, t)in terms of the initial condition.
26. Question: Consider the one-dimensional heat equation given by ∂u
∂t =k∂2u
∂x2where u(x, t)
represents the temperature distribution along a rod of length Lat time t,0≤x≤L, and
kis a positive constant. The rod is insulated so that the temperature at both ends is fixed
at 0. Suppose the initial temperature distribution is given by u(x, 0) = 2x(L−x). Find the
temperature distribution u(x, t)for t > 0.
Ans. Solution: We will solve the one-dimensional heat equation using the method of separation
of variables.
1. Separation of Variables:
Assume a solution of the form u(x, t) = X(x)T(t), substitute into the heat equation, and
divide by kX(x)T(t)to separate variables:
1
kT (t)
dT
dt =1
X(x)
d2X
dx2
This results in two separate ordinary differential equations:
1
kT (t)
dT
dt =λ, 1
X(x)
d2X
dx2=λ
2. Solving the Time Equation:
The time equation yields T(t) = ceλkt, where cis a constant. Applying the boundary condition
u(0, t) = 0, we have:
X(x)T(t) = 0 =⇒X(x) = 0
Since X(x)cannot be identically zero, the only way for X(x) = 0 is if λ < 0. Therefore, we
conclude that λ=−a2, where ais a positive constant.
Thus, the time solution is T(t) = ce−a2kt.
3. Solving the Spatial Equation:
The spatial equation becomes:
d2X
dx2+a2X= 0
The general solution to this ordinary differential equation is X(x) = c1cos(ax) + c2sin(ax).
Applying the boundary conditions u(0, t) = 0 and u(L, t) = 0, we find that c1=c2= 0.
Therefore, the spatial solution is X(x) = 0.
4. Final Solution:
Since u(x, t) = X(x)T(t) = 0 for all xand t, we conclude that the unique solution to the
heat equation with the given initial condition is u(x, t) = 0 for all xand t.
27. Question: Find the general solution to the one-dimensional heat equation ut=kuxx subject
to the boundary conditions u(0, t) = 0,u(π, t) = 0, and the initial condition u(x, 0) = f(x) =
x(π−x).
Ans. Solution:
1. We begin by separating variables. Let u(x, t) = X(x)T(t). Plugging this into the heat
equation, we have:
X(x)T′(t) = kX′′(x)T(t)
2. Dividing both sides by kX(x)T(t), we get:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a separation constant. This gives two ODEs: T′+kλT = 0 and X′′ +λX = 0.
3. Solving the second ODE X′′ +λX = 0, we find that the general solution for X(x)is:
X(x) = Acos√λx+Bsin√λx
4. Applying boundary conditions u(0, t) = 0 and u(π, t) = 0 requires sin√λπ= 0. This
implies √λ=nπ, where nis a positive integer.
5. Thus, the general solution for X(x)becomes:
Xn(x) = Bnsin(nx)
6. Now we solve the first ODE, T′+kλT = 0, which has the general solution:
T(t) = Ce−kn2π2t
7. Therefore, the general solution to the heat equation with the given boundary conditions
is:
u(x, t) = ∞
X
n=1
Bnsin(nx)e−kn2π2t
8. Finally, we apply the initial condition u(x, 0) = f(x) = x(π−x). This allows us to
determine the coefficients Bnby using Fourier series. After finding the coefficients, the final
solution will be the sum of the series.
28. Question: Consider the heat equation in one dimension given by ∂u
∂t =k∂2u
∂x2, subject to the
boundary conditions u(0, t) = u(L, t) = 0 where 0≤x≤Land initial condition u(x, 0) = f(x)
where f(x)is a given function. Find the solution u(x, t)for this initial-boundary value problem.
Ans. Solution: Given the heat equation ∂u
∂t =k∂2u
∂x2, let’s assume separation of variables
u(x, t) = X(x)T(t). Substituting this into the heat equation gives:
X(x)T′(t) = kX′′(x)T(t)
Dividing by kX(x)T(t)and rearranging terms gives:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
This gives two ODEs:
(T′(t) + kλT (t) = 0
X′′(x) + λX(x) = 0
1. Solving the time ODE: The solution to T′(t) + kλT (t) = 0 is given by T(t) = Ce−kλt
where Cis a constant.
2. Solving the spatial ODE: The solution to X′′(x) + λX(x) = 0 is a characteristic
equation with two possible solutions: λ > 0and λ= 0.
For λ > 0, the general solution is given by X(x) = Acos√λx+Bsin√λxwhere A
and Bare constants.
For λ= 0, the general solution is X(x) = Ax +B.
3. Applying boundary conditions: Since the boundary conditions are u(0, t)=0and
u(L, t) = 0, we can see that the solutions for X(x)must satisfy X(0) = 0 and X(L) = 0.
For λ > 0, we have X(0) = A= 0 and X(L) = Bsin√λL= 0. This implies that
√λL =nπ, where nis a positive integer, giving λ=nπ
L2.
For λ= 0, we have X(0) = B= 0 and X(L) = AL = 0, which doesn’t give any non-trivial
solutions.
4. General solution: Finally, combining the solutions for T(t)and X(x), the general
solution to the heat equation is:
u(x, t) = ∞
X
n=1
cne−k(nπ
L)2tsin nπx
L
where cnare the coefficients determined by the initial condition and the orthogonality of the sine
functions.
29. Let u(x, t)satisfy the heat equation
∂u
∂t = 4∂2u
∂x2,0< x < π, t > 0
subject to the initial condition
u(x, 0) = 2x(π−x)
and the boundary conditions
u(0, t) = 0, u(π, t) = 0
Find the solution u(x, t)for t > 0and 0< x < π.
Ans. Solution:
We begin by seeking a solution in the form of a separation of variables: u(x, t) = X(x)T(t).
Substituting this into the heat equation yields
XT ′= 4X′′T
Dividing by X(x)T(t)gives
T′
4T=X′′
X=−λ
where λis a separation constant. Solving the ODE for X(x)with the given boundary condi-
tions, we obtain a general solution in the form
X(x) = Bsin(nπx)
where λ=n2π2for n= 1,2,3, . . .. Now, we solve the ODE for T(t)to obtain
T′
4T=−λ=−n2π2
Solving this ODE gives the general solution
T(t) = Ce−4n2π2t
The general solution to the heat equation is then
u(x, t) = ∞
X
n=1
Cne−4n2π2tsin(nπx)
Applying the initial condition u(x, 0) = 2x(π−x)to find the Fourier coefficients Cn, we have
2x(π−x) = ∞
X
n=1
Cnsin(nπx)
Multiplying both sides by sin(mπx)and integrating from 0 to π, we get
Zπ
0
2x(π−x)sin(mπx)dx =Zπ
0
Cmsin2(mπx)dx
Solving for Cmgives
Cm=2
πZπ
0
2x(π−x)sin(mπx)dx
Carrying out the integration gives the final solution for u(x, t):
u(x, t) = 8
π2
∞
X
n=1
1
(2n−1)2e−(2n−1)2π2tsin((2n−1)πx)
30. Question: Consider the following partial differential equation:
utt −4uxx = 0,0< x < π, t > 0
subject to boundary conditions
u(0, t) = u(π, t) = 0, t > 0
and initial conditions
u(x, 0) = sin(2x), ut(x, 0) = 0
Find the solution to the given partial differential equation using the method of separation of
variables.
Ans. Solution:
We consider a solution of the form
u(x, t) = X(x)T(t)
Substituting this into the given PDE, we get
XT ′′ −4X′′T= 0
Dividing by XT gives
T′′
T= 4X′′
X=−λ2
where λ2>0.
1. Finding the X(x)equation:
Solving X′′ +λ2X= 0 with boundary conditions X(0) = X(π) = 0, we get
X(x) = sin(2x)
2. Finding the T(t)equation:
Solving T′′ + 4λ2T= 0 with initial conditions T(0) = 1, T ′(0) = 0, we get
T(t) = cos(2t)
3. Forming the solution:
The solution to the given PDE is then
u(x, t) = X(x)T(t) = sin(2x)cos(2t)
