MAT 142 - COLLEGE MATHEMATICS - Heat, Wave, and
Laplace Equations Practice Material - Set 1
1. Find the solution to the following heat equation initial value problem:
∂u
∂t =k∂2u
∂x2,
subject to the initial condition u(x, 0) = 2x(1 −x)and the boundary conditions u(0, t) =
u(1, t) = 0, where kis a positive constant.
Ans. Solution:
1. We start by assuming a solution of the form u(x, t) = X(x)T(t).
2. Substituting this into the heat equation, we get:
X(x)T′(t) = kX′′(x)T(t).
3. Dividing by kX(x)T(t)and rearranging gives:
T′(t)
kT (t)=X′′(x)
X(x)=−λ,
where λis some constant. This gives two separate ordinary differential equations.
4. Solving the time component, we get T′(t) = −kλT (t), with solution T(t) = Ae−kλt.
5. Now solving the spatial component X′′ (x) + λX(x) = 0 gives X(x) = Bsin(√λx)+
Ccos(√λx).
6. Applying the boundary conditions u(0, t) = u(1, t) = 0, we find C= 0 and sin(√λ)= 0.
7. This implies that λ=n2π2for n= 1,2,3, . . ., and thus Xn(x) = Bnsin(nπx).
8. The solution to the heat equation can then be written as:
u(x, t) =
∞
∑
n=1
Ane−kn2π2tsin(nπx),
where Anare coefficients to be determined from the initial condition u(x, 0) = 2x(1 −x).
9. Applying the initial condition gives:
2x(1 −x) =
∞
∑
n=1
Ansin(nπx).
10. Multiplying by sin(mπx)and integrating from 0 to 1 for m= 1,2,3, . . . gives:
Am= 2 ∫1
0
2x(1 −x)sin(mπx)dx.
11. Finally, we can substitute the coefficients Anback into the solution to get the final solution
to the heat equation.
2. Question: Consider the wave equation on a finite domain 0< x < L with initial conditions
given by u(x, 0) = f(x)and ut(x, 0) = g(x). Solve the wave equation subject to these initial
conditions.
Ans. Let u(x, t)satisfy the wave equation utt =c2uxx on the domain 0<x<Land t > 0,
subject to initial conditions u(x, 0) = f(x)and ut(x, 0) = g(x).
Solution: 1. Write the general solution to the wave equation: We seek a solution of the
form u(x, t) = F(x−ct) + G(x+ct). Plug this into the wave equation:
F′′(x−ct)−c2F′′ (x−ct) + G′′(x+ct)−c2G′′ (x+ct) = 0
2. Solve for F(x)and G(x): The general solutions are given by
F(x) = 1
2(f(x) + 1
cg′(x))
G(x) = 1
2(f(x)−1
cg′(x))
3. Combine F(x)and G(x)to obtain the final solution u(x, t): Therefore, the solution to
the wave equation with the given initial conditions is
u(x, t) = 1
2(f(x−ct) + f(x+ct)) + 1
2c∫x+ct
x−ct
g′(s)ds
3. Find the general solution to the heat equation ∂u
∂t =k∂2u
∂x2subject to the initial condition
u(x, 0) = sin(2x)and boundary conditions u(0, t) = 0 and u(π, t) = 0 for t > 0.
Ans. To solve the heat equation with the given initial and boundary conditions, we first find
the eigenvalues and eigenfunctions of the problem using separation of variables, then use the
method of Fourier series. 1. Separation of Variables: Assume a solution of the form u(x, t) =
X(x)T(t). Substitute u(x, t) = X(x)T(t)into the heat equation and separate variables to get:
T′(t)
kT (t)=X′′ (x)
X(x)=−λSolve the ODE for X(x):X′′(x) + λX(x) = 0 with boundary conditions
X(0) = X(π)=0. The eigenvalues are λn=n2, and the corresponding eigenfunctions are
Xn(x) = sin(nx)for n= 1,2,3, ...
2. Finding the Temporal Equation: Now we solve the temporal equation T′(t)
kT (t)=−λ.
The solution is Tn(t) = Ce−n2kt for n= 1,2,3, ...
3. General Solution: The general solution is given by u(x, t) = ∑∞
n=1 Xn(x)Tn(t) =
∑∞
n=1 sin(nx)Ce−n2kt. Applying the initial condition u(x, 0) = sin(2x), we find that u(x, 0) =
∑∞
n=1 Cnsin(nx) = sin(2x). By comparing coefficients, we find that C2= 1 and Cn= 0 for
n= 2. Thus, the solution to the heat equation with the given initial and boundary conditions is
u(x, t) = e−4kt sin(2x).
4. Find the general solution to the partial differential equation:
∂2u
∂x2−4∂2u
∂t2= 0
where u(x, t)is a function of xand t.
Ans. To solve the partial differential equation ∂2u
∂x2−4∂2u
∂t2= 0, we assume a solution of the
form u(x, t) = X(x)T(t). Substituting this into the equation gives us:
X′′(x)T(t)−4X(x)T′′ (t) = 0
Dividing by X(x)T(t)gives:
X′′(x)
X(x)= 4T′′ (t)
T(t)
Since the left side of the equation depends only on xand the right side depends only on t,
they must be equal to a constant. Letting this constant be −λ, we get two ordinary differential
equations:
X′′(x) + λX(x) = 0, T ′′(t)−4λT (t) = 0
Solving the first ODE gives us X(x) = Acos(√λx)+Bsin(√λx), while the second ODE
gives us T(t) = Ccos(2√λt)+Dsin(2√λt).
The general solution is then:
u(x, t) =
∞
∑
n=1 [(Ancos(√λnx)+Bnsin(√λnx))(Cncos(2√λnt)+Dnsin(2√λnt))]
where λnare the eigenvalues that satisfy the boundary conditions, and An,Bn,Cn,Dnare
constants determined by the initial conditions.
5. Question:
Consider the wave equation ∂2u
∂t2=c2∂2u
∂x2, where u(x, t)represents the displacement of a string
at position xand time t, and cis a positive constant representing the speed of propagation of
disturbances in the string.
A string of length Lfixed at both ends is initially at rest, i.e., u(x, 0) = 0 and ∂u
∂t (x, 0) = 0.
At time t= 0, the string is given an initial velocity v0sin (πx
L), where v0is a positive constant.
Find the solution u(x, t)for 0≤x≤Land t > 0.
Ans. Step-by-step solution:
1. We will use the method of separation of variables to solve the wave equation. Let’s assume
u(x, t) = X(x)T(t).
2. Plugging u(x, t) = X(x)T(t)into the wave equation gives us T′′
c2T=X′′
X.
3. Rearranging, we have T′′
c2T=−λ2=X′′
X, where λis a separation constant.
4. Solving for X(x), we have X′′ +λ2X= 0, which gives us the general solution X(x) =
Asin(λx) + Bcos(λx).
5. The boundary conditions u(x, 0) = 0 and ∂u
∂t (x, 0) = 0 give us X(x) = 0 and T′(0) = 0.
6. Since X(x) = 0, we get Asin(λx) + Bcos(λx) = 0. This implies A= 0 and Bcos(λx) =
0, which gives B= 0.
7. Therefore, the solution for the spatial part is X(x) = 0.
8. Now, we solve for the temporal part using T′′ +c2λ2T= 0, which has the general solution
T(t) = Ccos(cλt) + Dsin(cλt).
9. The initial condition ∂u
∂t (x, 0) = 0 gives dT
dt (0) = 0, which implies D= 0.
10. Hence, the temporal part solution is T(t) = Ccos(cλt).
11. Finally, combining the spatial and temporal parts gives us the solution u(x, t) =
∑∞
n=1 Bnsin (nπx
L)cos (cnπt
L), where Bnare the coefficients determined from the initial condition.
12. Given the initial condition u(x, 0) = v0sin (πx
L), we find Bn=2v0
nπ .
13. Therefore, the final solution to the wave equation is u(x, t) = ∑∞
n=1
2v0
nπ sin (nπx
L)cos (cnπt
L).
6. Consider the following Heat Equation:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions:
u(0, t) = 0, u(L, t) = 0
and the initial condition:
u(x, 0) = sin (πx
L)
Find the solution to the Heat Equation using the method of separation of variables.
Ans. Solution: 1. We assume a solution of the form u(x, t) = X(x)T(t)and substitute it
into the Heat Equation:
X(x)T′(t) = kX′′(x)T(t)
2. Divide both sides by kX(x)T(t)to separate variables:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis the separation constant.
3. Solve the time part of the equation:
T′(t) + kλT (t) = 0
The general solution is T(t) = Aexp (−kλt).
4. Solve the spatial part of the equation:
X′′(x) + λX(x) = 0
The general solution is X(x) = Bsin (√λx)+Ccos (√λx).
5. Apply the boundary conditions u(0, t) = 0 and u(L, t) = 0 to the spatial part:
X(0) = Bsin(0) + Ccos(0) = 0 =⇒C= 0
X(L) = Bsin(√λL)= 0 =⇒√λL =nπ where n= 1,2,3, ...
6. Therefore, the eigenvalues are λ=(nπ
L)2and the eigenfunctions are Xn(x) = Bnsin (nπx
L).
7. The solution to the Heat Equation is then:
u(x, t) =
∞
∑
n=1
Anexp (−k(nπ
L)2
t)sin (nπx
L)
8. Apply the initial condition u(x, 0) = sin (πx
L)to find the coefficients An:
sin (πx
L)=
∞
∑
n=1
Ansin (nπx
L)
By orthogonality of sine functions, we get An= 2 ∫L
0sin (πx
L)sin (nπx
L)dx.
9. Therefore, the final solution to the Heat Equation is:
u(x, t) = sin (πx
L)−2
∞
∑
n=1
sin (πx
L)sin (nπx
L)exp (−k(nπ
L)2
t)
7. Question:
Consider the one-dimensional heat equation:
∂u
∂t =c2∂2u
∂x2
subject to the boundary conditions u(0, t) = 0 and u(L, t) = 0, as well as the initial condition
u(x, 0) = f(x)for 0≤x≤L, where cis a positive constant and Lis the length of the rod.
Show that the solution to this equation can be written as:
u(x, t) =
∞
∑
n=1
Bnsin (nπx
L)e−(nπct
L)2
where the coefficients Bnare given by:
Bn=2
L∫L
0
f(x)sin (nπx
L)dx
Ans. Solution:
1. First, we assume that the solution u(x, t)can be expressed as a product of functions of x
and t:
u(x, t) = X(x)T(t)
2. Substituting this into the heat equation gives:
X(x)T′(t) = c2X′′(x)T(t)
3. Dividing both sides by c2X(x)T(t)gives:
T′(t)
c2T(t)=X′′(x)
X(x)
4. Since the left side depends only on tand the right side depends only on x, both sides must
be equal to a constant −λ. This gives two ODEs to solve:
T′(t) + c2λT (t) = 0 and X′′(x) + λX(x) = 0
5. The solutions to T′(t) + c2λT (t) = 0 are:
Tn(t) = Ane−(nπct
L)2
6. To solve the second ODE, we find that the solutions X(x)are proportional to sin (nπx
L).
This gives:
Xn(x) = Bnsin (nπx
L)
7. Combining the solutions for X(x)and T(t)with the separation constant −λ=−(nπc/L)2
gives:
un(x, t) = Xn(x)Tn(t) = Bnsin (nπx
L)e−(nπct
L)2
8. Finally, by linearity of the heat equation, the solution to the initial-boundary value problem
is the sum of the individual solutions:
u(x, t) =
∞
∑
n=1
Bnsin (nπx
L)e−(nπct
L)2
where the coefficients Bnare given by:
Bn=2
L∫L
0
f(x)sin (nπx
L)dx
This completes the derivation of the solution to the heat equation with the given initial and
boundary conditions.
8. Question:
Consider the following wave equation in one dimension:
utt =c2uxx,
with boundary conditions u(0, t) = 0,u(L, t) = 0 for t > 0, and initial conditions u(x, 0) =
f(x),ut(x, 0) = g(x)for 0<x<L. Find the solution to this wave equation using the method
of separation of variables.
Ans. Solution:
We assume the solution to the wave equation can be written as a product of two functions,
u(x, t) = X(x)T(t). Substituting this into the wave equation gives:
X(x)T′′(t) = c2X′′ (x)T(t).
Dividing both sides by c2X(x)T(t)gives:
T′′(t)
c2T(t)=X′′(x)
X(x).
This can only hold if both sides are equal to a constant, say −λ. Then we have two ODEs
to solve:
1. The time part: T′′ (t) + λc2T= 0 with solutions T(t) = Acos(√λct)+Bsin(√λct).
2. The spatial part: X′′ (x) + λX = 0 with solutions X(x) = Ccos(√λx)+Dsin(√λx).
Applying the boundary conditions u(0, t)=0and u(L, t)=0gives us X(0)T(t)=0and
X(L)T(t)=0. This implies that C= 0 and sin(√λL)= 0, so √λ=nπ
Lfor n= 1,2,3, . . ..
Therefore, the spatial part becomes Xn(x) = Dnsin (nπx
L).
The general solution to the wave equation is then:
u(x, t) =
∞
∑
n=1 (Ancos (nπct
L)+Bnsin (nπct
L))sin (nπx
L).
Finally, we can determine the coefficients Anand Bnusing the initial conditions u(x, 0) =
f(x)and ut(x, 0) = g(x).
9. Question:
Consider the following partial differential equation:
∂2u
∂t2= 9∂2u
∂x2
with initial conditions:
u(x, 0) = 2 sin(2x)
∂u
∂t (x, 0) = 3 sin(2x)
Determine the solution to the partial differential equation.
Ans. Step-by-step solution:
1. To solve the given partial differential equation, we first assume a solution of the form:
u(x, t) = X(x)T(t)
2. Substituting this into the partial differential equation, we get:
T′′(t)X(x) = 9T(t)X′′ (x)
3. Dividing both sides by T(t)X(x), we obtain:
T′′(t)
T(t)= 9X′′ (x)
X(x)
4. Since the left side is a function of tonly and the right side is a function of xonly, they
must both be equal to a constant −λ. This gives us two ordinary differential equations:
T′′(t)
T(t)=−λand 9X′′(x)
X(x)=−λ
5. Solving the first ODE we get:
T′′(t) + λT (t) = 0
which has solutions of the form T(t) = Acos(3t) + Bsin(3t)if λ= 9.
6. Solving the second ODE we get:
X′′(x) + λ
9X(x) = 0
which has solutions of the form X(x) = Ccos(3x) + Dsin(3x)if λ= 9.
7. The general solution to the partial differential equation is then:
u(x, t) = (Acos(3t) + Bsin(3t))(Ccos(3x) + Dsin(3x))
8. Applying the initial conditions u(x, 0) = 2 sin(2x)and ∂u
∂t (x, 0) = 3 sin(2x), we find the
specific solution to be:
u(x, t) = 2e−9tsin(3x)+3e−9tsin(3x)
which is the solution to the partial differential equation.
10. Question:
Consider the heat equation in one dimension, given by ∂u
∂t =k∂2u
∂x2, subject to the boundary
conditions u(0, t)=0and u(π, t)=0, and the initial condition u(x, 0) = sin(x). Find the
solution u(x, t)for t > 0.
Ans. Solution:
We will solve the given heat equation using separation of variables.
1. Guess the form of the solution: We assume that the solution can be expressed as a
product of functions of xand t:
u(x, t) = X(x)T(t)
2. Substitute the solution form into the heat equation: Substitute u(x, t) = X(x)T(t)
into the heat equation ∂u
∂t =k∂2u
∂x2:
XT ′=kX′′T
3. Divide by the product of X(x)T(t):Divide both sides by XT :
T′
kT =X′′
X=−λ
This results in two ordinary differential equations:
T′+kλT = 0
X′′ +λX = 0
4. Solve the ODE for T(t):The solution to T′+kλT = 0 is:
T(t) = Ae−kλt
5. Solve the ODE for X(x):The solution to X′′ +λX = 0 is:
X(x) = Bsin(√λx)+Ccos(√λx)
6. Apply the boundary conditions: From the boundary conditions u(0, t)=0and
u(π, t) = 0, we can determine the values of λ,B, and C.
For u(0, t) = 0:
X(0)T(t) = 0
C= 0
For u(π, t) = 0:
X(π)T(t) = 0
Bsin(√λπ)= 0
This gives us the condition sin(√λπ)= 0, which implies √λπ =nπ ⇒√λ=n, where nis a
positive integer.
7. Find the complete solution: Therefore, the solutions are of the form:
un(x, t) = e−kn2tBnsin(nx)
where Bnare constants that can be determined from the initial condition.
The complete solution is then the sum of all the individual solutions:
u(x, t) =
∞
∑
n=1
e−kn2tBnsin(nx)
Applying the initial condition u(x, 0) = sin(x)gives Bn= 2 ∫π
0sin(x)sin(nx)dx.
Therefore, the solution to the heat equation with the given conditions is:
u(x, t) =
∞
∑
n=1
e−kn2t2∫π
0
sin(x)sin(nx)dx sin(nx)
11. Question: Solve the heat equation in one dimension, given by ∂u
∂t =k∂2u
∂x2, subject to the
initial condition u(x, 0) = sin(πx)and boundary conditions u(0, t) = u(L, t) = 0, where u(x, t)
is the temperature distribution along a rod of length Land kis a positive constant.
Ans. Solution: We will solve the heat equation using the method of separation of variables.
1. Formulating the solution Assume a solution of the form u(x, t) = X(x)T(t). Substi-
tuting this into the heat equation gives 1
kT
dT
dt =1
X
d2X
dx2.
2. Solving the time ODE Solving the time ODE, we have 1
kT
dT
dt =−λ, where λis a
separation constant. This gives us T(t) = Ae−kλt.
3. Solving the spatial ODE Solving the spatial ODE, we have 1
X
d2X
dx2=−λ. The general
solutions are X(x) = Bsin(√λx)+Ccos(√λx).
4. Applying the boundary conditions Applying the boundary conditions u(0, t) = u(L, t) =
0, we get X(0) = X(L) = 0. This implies C= 0, and sin(√λL)= 0, which gives √λL =nπ
for n∈Z+. Thus, λ=(nπ
L)2.
5. Obtaining the solution Therefore, the general solution is u(x, t) = ∑∞
n=1(ansin(nπx
L)e−k(nπ
L)2t).
6. Calculating the coefficients Applying the initial condition u(x, 0) = sin(πx), we can
find the coefficients an. This involves solving for an= 2/L∫L
0sin(πx)sin(nπx
L)dx.
7. Final solution Finally, the solution to the heat equation with given initial and boundary
conditions is u(x, t) = ∑∞
n=1 (2
(2n−1)πsin((2n−1)πx
L)e−k((2n−1)π
L)2t).
12. Question:
Consider the one-dimensional heat equation on the interval [0, π], given 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) = sin2(x).
Determine the solution u(x, t)to the heat equation.
Ans. Solution:
We will solve the one-dimensional heat equation using separation of variables.
1. Separation of variables: We assume that the solution can be written as a product of
functions of xand t, that is, u(x, t) = X(x)T(t).
2. Substitute into the heat equation: Substitute u(x, t) = X(x)T(t)into the heat
equation to obtain: T′(t)X(x) = kX′′(x)T(t).
Divide by kX(x)T(t)to get: T′(t)
kT (t)=X′′ (x)
X(x).
3. Solving the time-dependent ODE: Since the left-hand side is purely a function of t
and the right-hand side is purely a function of x, they must be equal to a constant −λ, which
gives two separate ODEs: T′(t) + kλT (t) = 0,X′′ (x) + λX(x) = 0.
4. Solve the ODE for X(x):The ODE for X(x)is a simple second-order linear ODE with
solutions of the form X(x) = Ancos(√λx)+Bnsin(√λx).
Applying the boundary conditions u(0, t) = 0 and u(π, t) = 0, we obtain Bn= 0 and
cos(√λπ)= 0, which implies √λ=n, where nis a positive integer.
Therefore, the eigenfunctions are Xn(x) = Ancos(nx).
5. Solve the ODE for T(t):The ODE for T(t)is a simple first-order linear ODE with
solution T(t) = e−kλt =e−kn2t.
6. General solution: The general solution to the heat equation can be written as a sum of
the separated solutions: u(x, t) = ∑∞
n=1 Ancos(nx)e−kn2t.
7. Determine the coefficients: Applying the initial condition u(x, 0) = sin2(x), we can
determine the coefficients Anby taking the inner product of u(x, 0) and Xn(x).
The solution to the heat equation is: u(x, t) = 1
2+∑∞
n=1
(−1)n+1
n(2n−1) cos(nx)e−kn2t.
13. Question: Consider the one-dimensional heat equation defined for 0< x < 1and t > 0:
∂u
∂t =k∂2u
∂x2,
subject to the boundary conditions u(0, t) = 0 and u(1, t) = 0, where u(x, t)is the temperature
distribution in a one-dimensional rod, kis the thermal diffusivity constant, and u(x, 0) = sin(πx).
Find the solution to the heat equation using separation of variables.
Ans. Step-by-step solution: 1. We assume that the solution to the heat equation can be written
as a product of two functions, X(x)and T(t), i.e., u(x, t) = X(x)T(t). 2. By substituting this
assumption into the heat equation, we get:
1
k
T′(t)
T(t)=X′′(x)
X(x)=−λ
where λis a constant. This leads to two ordinary differential equations: T′(t)
kT (t)=−λand X′′ (x)+
λX(x)=0. 3. Solving the ODE for X(x), we find that X(x) = Acos(√λx)+Bsin(√λx),
where Aand Bare constants. 4. Applying the boundary conditions u(0, t) = 0 and u(1, t) = 0,
we have X(0) = A= 0 and X(1) = Bsin(√λ)= 0. This implies that λ=n2π2, where
n= 1,2,3, .... 5. Therefore, the solution for X(x)is given by Xn(x) = Bnsin(nπx), where Bn
is a normalization constant. 6. Next, solving the ODE for T(t), we have T′(t)
kT (t)=−λ=−n2π2,
which gives Tn(t) = Cne−kn2π2t, where Cnis a constant. 7. Assemble the solution by taking the
sum of all the modes:
u(x, t) =
∞
∑
n=1
Xn(x)Tn(t) =
∞
∑
n=1
Bnsin(nπx)Cne−kn2π2t
8. Finally, we can determine the coefficients Bnand Cnby using the initial condition u(x, 0) =
sin(πx). By expressing the initial condition in terms of the Fourier sine series, we find that Bn=
2∫1
0sin(nπx)sin(πx)dx. 9. After evaluating the integral in step 8, we obtain the coefficients
Bn=2
(n2−1)π. 10. Hence, the solution to the heat equation with the given boundary and initial
conditions is:
u(x, t) =
∞
∑
n=1
2
(n2−1)πsin(nπx)e−kn2π2t
14. Question:
Consider the wave equation on the interval 0< x < π with the boundary conditions u(0, t) =
u(π, t) = 0 and the initial conditions u(x, 0) = sin xand ∂u
∂t (x, 0) = sin x. Solve for u(x, t).
Ans. Step-by-step solution:
1. We will first assume a solution of the form u(x, t) = X(x)T(t)and substitute it into the
wave equation:
X(x)T′′(t) = c2X′′ (x)T(t)
Dividing both sides by c2X(x)T(t)gives:
T′′(t)
c2T(t)=X′′(x)
X(x)=−λ
where λis a separation constant.
2. Solving the ODE X′′ (x)
X(x)=−λwith boundary conditions X(0) = X(π)=0gives eigen-
functions Xn(x) = sin(nx)for n∈Z+and eigenvalues λ=n2.
3. For the temporal equation, we have T′′ (t)
c2T(t)=−λ=−n2, which has solutions Tn(t) =
Ancos (nc2t) + Bnsin (nc2t).
4. Now, the general solution is given by:
u(x, t) =
∞
∑
n=1
(Ancos (nc2t) + Bnsin (nc2t)) sin(nx)
Applying the initial conditions u(x, 0) = sin xand ∂u
∂t (x, 0) = sin x, we find:
An= 0 and Bn=2
π∫π
0
sin xsin nx dx
5. Evaluating the integral Bn=2
π∫π
0sin xsin nx dx gives Bn={0if nis even
4
π
(−1)
n−1
2
nif nis odd
Therefore, the solution to the wave equation with the given boundary and initial conditions
is:
u(x, t) = 4
π
∞
∑
n=1
(−1)n−1
2
nsin (2n−1)xcos ((2n−1)c2t)
15. Question: Consider the heat equation in one dimension, given by ∂u
∂t =k∂2u
∂x2, where u(x, t)
represents the temperature distribution in a rod of length Lwith initial condition u(x, 0) = f(x)
and boundary conditions u(0, t)=0and u(L, t)=0. Find the solution to this boundary value
problem.
Ans. Solution: 1. We begin 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′(t)
kT (t)=X′′ (x)
X(x)=−λ, where λis a separation
constant.
3. Solving the ODE for X(x)with boundary conditions X(0) = 0 and X(L)=0gives
X(x) = Bsin (nπx
L)for n= 1,2,3, ....
4. The time part T(t)satisfies T′(t)
kT (t)=−λ, which implies T(t) = Ce−kλt.
5. Using the separation constant λ=(nπ
L)2, the general form of the solution becomes
u(x, t) = ∑∞
n=1 Bnsin (nπx
L)e−k(nπ
L)2t.
6. Applying the initial condition u(x, 0) = f(x)yields the Fourier sine series of f(x):f(x)∼
∑∞
n=1 Bnsin (nπx
L).
7. Comparing coefficients in step 5 and 6 gives Bn=2
L∫L
0f(x)sin (nπx
L)dx.
8. Therefore, the solution to the given heat equation with the specified boundary and initial
conditions is u(x, t) = ∑∞
n=1 (2
L∫L
0f(s)sin (nπs
L)ds)sin (nπx
L)e−k(nπ
L)2t.
16. Question: Consider the following heat equation with mixed boundary conditions:
∂u
∂t =k(∂2u
∂x2+∂2u
∂y2)
subject to the conditions:
u(0, y, t) = u(1, y, t) = 0, u(x, 0, t) = u(x, 1, t) = 0,
and the initial temperature distribution:
u(x, y, 0) = 4xy(1 −x)(1 −y).
Find the steady-state solution of the heat equation.
Ans. Solution: 1. To find the steady-state solution of the heat equation, we need to find
the solution when t→ ∞. This means that the time derivative will be zero, so the equation
simplifies to Laplace’s equation:
0 = k(∂2u
∂x2+∂2u
∂y2)
2. The general solution to Laplace’s equation in 2D is given by:
u(x, y) =
∞
∑
n=1
∞
∑
m=1 (Anm sin (nπx
L)sin (mπy
L))
3. Apply the boundary conditions to determine the coefficients Anm:
u(0, y) =
∞
∑
n=1
∞
∑
m=1
Anm sin(0) sin (mπy
L)= 0
u(1, y) =
∞
∑
n=1
∞
∑
m=1
Anm sin(nπ)sin (mπy
L)= 0
u(x, 0) =
∞
∑
n=1
∞
∑
m=1
Anm sin (nπx
L)sin(0) = 0
u(x, 1) =
∞
∑
n=1
∞
∑
m=1
Anm sin (nπx
L)sin(mπ) = 0
4. Applying the boundary conditions simplifies the general solution to:
u(x, y) =
∞
∑
n=1
∞
∑
m=1
Anm sin (nπx
L)sin (mπy
L)
5. Next, find the coefficients Anm by substituting the initial temperature distribution into the
general solution:
u(x, y, 0) = 4xy(1 −x)(1 −y) =
∞
∑
n=1
∞
∑
m=1
Anm sin (nπx
L)sin (mπy
L)
6. Comparing coefficients on both sides, we can determine the values of Anm. This will give
us the steady-state solution of the heat equation.
17. Question:
Consider the heat equation in one dimension given by ∂u
∂t =c2∂2u
∂x2, where cis a positive
constant. Solve the initial-boundary value problem for 0<x<1and t > 0with u(0, t)=0,
u(1, t) = 0, and u(x, 0) = x(1 −x).
Ans. Step-by-step solution:
1. We begin by assuming a solution of the form u(x, t) = X(x)T(t). Substituting this into
the heat equation gives T′
c2T=X′′
X=−λ, where λis a separation constant.
2. Solving the time part of the equation gives T(t) = Ae−c2λt. If λ < 0,T(t)grows
unbounded as t→ ∞, which is not suitable for a physically realistic temperature distribution.
Therefore, we must have λ≥0.
3. Solving the spatial part of the equation gives an eigenvalue problem X′′ +λX = 0. The
solutions to this equation are Xn(x) = sin(nπx)for n= 1,2,3, . . . corresponding to λn= (nπ)2.
4. The general solution to the PDE is then u(x, t) =
∞
∑
n=1
(Ansin(nπx))e−(nπc)2t.
5. To find the specific solution satisfying the initial condition, apply the initial condition
u(x, 0) = x(1 −x) =
∞
∑
n=1
Ansin(nπx).
6. Multiplying both sides by sin(mπx)and integrating from 0 to 1 gives Am= 2 ∫1
0x(1 −
x)sin(mπx)dx.
7. Evaluating the integral gives Am=4(−1)m+1
m3π3for modd and Am= 0 for meven.
8. Therefore, the solution to the initial-boundary value problem is u(x, t) =
∞
∑
n=1,3,5,...
4(−1)n+1
n3π3sin(nπx)e−(nπc)2t.
18. Question: Consider the following heat equation on the interval 0≤x≤πwith initial
condition and boundary conditions:
ut=uxx + 2 sin(x),0< x < π, t > 0
u(0, t) = u(π, t) = 0, t ≥0
u(x, 0) = x(π−x),0≤x≤π
Find the solution u(x, t)of the given initial/boundary value problem.
Ans. Step-by-step solution:
Step 1. Find the separated solution of the heat equation: Let u(x, t) = X(x)T(t). Then we
have: T′
T=X′′ + 2 sin(x)
X
This yields the pair of ODEs:
T′=cT, X′′ + 2 sin(x)X=cX
Step 2. Solve the ODE for T(t): The first ODE T′=cT has a general solution of the form
T(t) = Aect. Since we are interested in a bounded solution, we must have c≤0. Let’s set
c=−λ2for some λ > 0. Thus,
T(t) = Ae−λ2t
Step 3. Solve the ODE for X(x): The second ODE X′′ + 2 sin(x)X=−λ2Xis a Sturm-
Liouville eigenvalue problem. The corresponding eigenfunctions of X(x)are the solutions to
X′′ + (2 −λ2)sin(x)X= 0. To find such solutions, we solve for the eigenvalues λnand
corresponding eigenfunctions Xn(x).
Step 4. Solve for eigenvalues and eigenfunctions: Set 2−λ2
n=m2
n, then we get λn=
√2−m2
nfor mn= 1,2,3, .... The corresponding eigenfunctions Xn(x)are Xn(x) = sin(nπx)
for n= 1,2,3, ....
Step 5. General solution: The general solution to the heat equation is given by:
u(x, t) =
∞
∑
n=1
Bne−λ2
ntsin(nπx)
Step 6. Calculate the coefficients: Applying the initial condition u(x, 0) = x(π−x)gives:
x(π−x) =
∞
∑
n=1
Bnsin(nπx)
Multiplying both sides by sin(mπx)and integrating from 0to πgives:
Bm=2
π∫π
0
x(π−x)sin(mπx)dx
Step 7. Final solution: By solving for Bnand substituting back into the general solution,
the final solution of the given initial/boundary value problem is:
u(x, t) =
∞
∑
n=1
2
π((−1)n+1
n3)e−n2tsin(nπx)
19. Question: Consider the wave equation in one dimension: ∂2u
∂t2=c2∂2u
∂x2, where cis a constant.
Solve the wave equation subject to the initial conditions u(x, 0) = f(x)and ∂u
∂t (x, 0) = g(x),
where f(x)and g(x)are given functions defined on the interval 0< x < L.
Ans. Solution:
We first need to find a general solution to the wave equation. Let’s assume the solution can
be written as a product of two functions X(x)and T(t):u(x, t) = X(x)T(t).
1. Substituting this into the wave equation, we get:
T′′
c2T=X′′
X=−λ2
This gives us two ordinary differential equations:
2. Equation for T(t):
T′′ +c2λ2T= 0
This has the general solution:
T(t) = Acos(λct) + Bsin(λct)
3. Equation for X(x):
X′′ +λ2X= 0
This has the general solution:
X(x) = Ccos(λx) + Dsin(λx)
4. Combining the solutions for X(x)and T(t), we get:
u(x, t) = (Ccos(λx) + Dsin(λx))(Acos(λct) + Bsin(λct))
5. Applying the initial conditions:
u(x, 0) = f(x)⇒C=f(x)
∂u
∂t (x, 0) = g(x)⇒B=1
λcg(x)
6. Therefore, the solution to the wave equation with given initial conditions is:
u(x, t) =
∞
∑
n=1 (fncos (nπx
L)+1
nπc [∂fn
∂x cos (nπx
L)+gnsin (nπx
L)])cos (nπct
L)
where fnand gnare the Fourier coefficients of f(x)and g(x), respectively.
20. Let u(x, t)satisfy the wave equation on the domain 0< x < π and t > 0with initial
conditions given by:
u(x, 0) = x(π−x)
and boundary conditions given by:
u(0, t) = 0
u(π, t) = 0
Find the solution u(x, t)to this problem.
Ans. Solution: 1. We can start by separating the variables u(x, t) = X(x)T(t). Substituting
this into the wave equation, we have:
X(x)T′′(t) = c2X′′ (x)T(t)
2. Dividing by c2X(x)T(t), we get:
T′′(t)
c2T(t)=X′′(x)
X(x)=−λ
3. Solving the ODE X′′ (x) + λX(x) = 0, subject to the boundary conditions X(0) = X(π) = 0,
we obtain λ=n2and Xn(x) = sin(nx). 4. Solving the ODE T′′ (t) + n2c2T(t) = 0, we obtain
Tn(t) = Ancos(nct) + Bnsin(nct). 5. Combining the spatial and temporal parts, the general
solution will be:
u(x, t) =
∞
∑
n=1
(Ancos(nct) + Bnsin(nct)) sin(nx)
6. Applying the initial condition to find the Ancoefficients, we have:
u(x, 0) =
∞
∑
n=1
Ansin(nx) = x(π−x)
7. Multiply both sides by sin(mx)and integrate from 0to π, we find:
Am=2
π∫π
0
x(π−x)sin(mx)dx =8
π
(−1)m+1
m3
8. Therefore, the final solution to the problem will be:
u(x, t) =
∞
∑
n=1 (8
π
(−1)n+1
n3sin(nx)cos(nct))
21. Question: Consider the heat equation given by ∂u
∂t =k∂2u
∂x2where u(x, t)is the temperature
distribution in a rod of length Lwith initial condition u(x, 0) = x(L−x)and boundary conditions
u(0, t) = u(L, t) = 0. Find the solution to the heat equation for 0< x < L and t > 0.
Ans. Solution: 1. We will start by assuming a solution of the form u(x, t) = X(x)T(t). 2.
Substitute u(x, t) = X(x)T(t)into the heat equation and divide by kX(x)T(t)to get 1
kT (t)
dT
dt =
1
X(x)
d2X
dx2. 3. Rearranging the equation gives 1
kT (t)
dT
dt =λ=1
X(x)
d2X
dx2, where λis some constant.
4. Solve the ODE for T(t):dT
dt =kλT (t)gives T(t) = Aekλt, where Ais a constant. 5. Solve
the ODE for X(x):d2X
dx2+λX(x) = 0. 6. The general solution to the ODE for X(x)is X(x) =
C1cos(√λx)+C2sin(√λx), where C1and C2are constants. 7. Apply the boundary conditions
u(0, t) = u(L, t) = 0 to find that C1= 0 and √λ=nπ/Lfor n= 1,2,3, . . .. 8. Using the found
eigenvalues λn= (nπ/L)2and general solution for T(t), the solution u(x, t)becomes a sum over
the orthogonal eigenfunctions: u(x, t) = ∑∞
n=1 Bne−k(nπ/L)2tsin(nπx/L). 9. Apply the initial
condition u(x, 0) = x(L−x) = ∑∞
n=1 Bnsin(nπx/L)to find Bn=2
L∫L
0x(L−x)sin(nπx/L)dx.
10. Solve the integral to find the coefficients Bn=4L
n3π3((−1)n−1). Therefore, the solution to
the heat equation is
u(x, t) =
∞
∑
n=1
4L
n3π3((−1)n−1)e−k(nπ/L)2tsin(nπx/L).
22. Question: Consider the heat equation ∂u
∂t =k∂2u
∂x2subject to the boundary conditions
u(0, t) = 0 and u(π, t) = 0, with the initial condition u(x, 0) = sin(x). Find the solution u(x, t)
for t > 0.
Ans. Solution:
1. We first note that the heat equation is a separable partial differential equation, so we
assume a solution of the form u(x, t) = X(x)T(t).
2. Plugging this into the heat equation, we get T′(t)
kT (t)=X′′ (x)
X(x)=−λ, where λis a separation
constant.
3. Solving the two ordinary differential equations T′(t)+kλT (t) = 0 and X′′ (x)+λX(x) = 0,
we find T(t) = Ae−kλt and X(x) = Bsin(√λx).
4. The boundary conditions u(0, t) = 0 and u(π, t) = 0 lead to X(0)T(t) = 0 and
X(π)T(t) = 0, respectively.
5. From X(0) = 0, we have Bsin(0) = 0 =⇒B= 0, so we have X(x) = 0.
6. From X(π) = 0, we have sin(√λπ)= 0, which gives √λ=nwhere nis a positive
integer.
7. Therefore, the solution u(x, t)can be written as a series of sine functions: u(x, t) =
∑∞
n=1 Cnsin(nx)e−kn2t.
8. To find the coefficients Cn, we use the initial condition u(x, 0) = sin(x):u(x, 0) =
∑∞
n=1 Cnsin(nx) = sin(x). This implies Cn=2
π∫π
0sin(x)sin(nx)dx.
9. Evaluating the integral, we find Cn=2
π∫π
0
1
2i(eix−e−ix)1
2i(einx−e−inx)dx =2
π[π(−1)n−(−1)n+1
1−n2].
10. Finally, the solution to the heat equation with the given boundary and initial conditions
is: u(x, t) = ∑∞
n=1
2(−1)n
(1−n2)sin(nx)e−kn2t.
23. Find the solution to the Heat Equation, subject to the initial condition u(x, 0) = x(1 −x)
on the interval 0< x < 1, where the boundary conditions are u(0, t) = u(1, t) = 0.
Ans. We start by expressing the solution u(x, t)as a Fourier series:
u(x, t) =
∞
∑
n=1
Bne−(nπ
l)2kt sin (nπx
l)
1. To find the coefficients Bn, we apply the initial condition:
x(1 −x) =
∞
∑
n=1
Bnsin (nπx
l)
2. We multiply both sides by sin (mπx
l), where mis an integer from 1 to ∞, then integrate
from 0 to 1:
∫1
0
x(1 −x)sin (mπx
l)dx =
∞
∑
n=1
Bn∫1
0
sin (nπx
l)sin (mπx
l)dx
3. Simplifying the integral on the left side:
∫1
0
x(1 −x)sin (mπx
l)dx =8
π3((−1)m−1)
4. Calculating the right side integral:
∫1
0
sin (nπx
l)sin (mπx
l)dx ={0if n=m
l
2if n=m
5. Substituting the results from steps 3 and 4 into the integral equation, we get:
8
π3((−1)m−1) = Bm
l
2
6. Solving for Bm, we find:
Bm=16
m3π3(1 −(−1)m)
Hence, the solution to the Heat Equation with the given initial and boundary conditions is:
u(x, t) =
∞
∑
n=1
16
n3π3(1 −(−1)n)e−(nπ
l)2kt sin (nπx
l)
24. Find the general solution to the partial differential equation:
uxx −4uxy + 4uyy = 0
where u(x, y)is a function of xand y.
Ans. Solution: 1. We first compute the characteristics of the PDE by solving the system of
ODEs given by:
dx
dt = 1,dy
dt =−2,du
dt = 0
This system gives x=t+C1and y=−2t+C2for constants C1and C2. Therefore, the
characteristics are given by y=−2x+Cfor some constant C.
2. Let v(x, y) = u(x, y)along the characteristic curves, which gives us d2v
dx2−4d2v
dxdy +4d2v
dy2= 0.
3. Using the change of variables p=x+yand q=x−y, we rewrite the PDE in terms of
v(x, y)as:
∂2v
∂p∂q = 0
4. The general solution to this PDE is v(p, q) = f(p) + g(q)for arbitrary functions fand g.
5. Substituting p=x+yand q=x−yback, we have u(x, y) = f(x+y) + g(x−y)as
the general solution to the original PDE.
25. Question: Consider the one-dimensional heat equation
∂u
∂t =c2∂2u
∂x2
with the initial condition
u(x, 0) = {0if 0≤x < L/2
sin (π(x−L/2)
L)if L/2 ≤x≤L
and boundary conditions u(0, t)=0and u(L, t)=0. Find the solution u(x, t)for 0≤x≤L
and t > 0.
Ans. Solution: We will solve this problem 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 this into the heat equation to get
X(x)T′(t) = c2X′′(x)T(t)
Divide by c2X(x)T(t)to obtain
T′(t)
c2T(t)=X′′(x)
X(x)=−λ
where λis a separation constant.
This gives two ODEs: 1) T′(t) + c2λT (t) = 0 with general solution T(t) = Ae−c2λt. 2)
X′′(x) + λX(x) = 0 with general solution X(x) = Bcos(√λx)+Csin(√λx).
2. Applying Boundary Conditions: From the boundary conditions u(0, t) = 0 and
u(L, t) = 0, we have X(0)T(t) = 0 and X(L)T(t) = 0.
This implies X(0) = 0 and X(L)=0. From X(0) = 0, we have B= 0. From X(L) = 0,
we have Csin(√λL)= 0.
For non-trivial solutions, we need sin(√λL)= 0, which gives √λL =nπ where nis a
positive integer.
Hence, λ=(nπ
L)2with corresponding eigenfunctions Xn(x) = Cnsin (nπx
L).
3. Superposition: The general solution is given by
u(x, t) =
∞
∑
n=1
Tn(t)Xn(x) =
∞
∑
n=1 (Ane−c2(nπ
L)2t)Cnsin (nπx
L)
4. Finding Coefficients: To find the coefficients An, we use the initial condition u(x, 0).
Given u(x, 0), we find the Fourier sine series representation of u(x, 0) and compare coefficients.
By matching coefficients, we can find An.
Therefore, the solution to the heat equation with the specified initial and boundary conditions
is
u(x, t) =
∞
∑
n=1
Ane−c2(nπ
L)2tsin (nπx
L)
26. Question 26:
Consider the following heat equation in one dimension:
∂u
∂t =k∂2u
∂x2,
subject to the initial condition:
u(x, 0) = 2x(1 −x),0< x < 1,
and the boundary conditions:
u(0, t) = 0, u(1, t) = 0.
Find the solution to the initial-boundary value problem described above.
Ans. Solution:
1. Using separation of variables:
Let’s assume a solution of the form u(x, t) = X(x)T(t). Substituting this into the heat
equation gives:
X(x)T′(t) = kX′′(x)T(t).
Dividing both sides by kX(x)T(t)gives:
T′(t)
kT (t)=X′′(x)
X(x).
Since the left side only depends on tand the right side only depends on x, they must both
be equal to a constant −λ. This gives two equations:
T′(t) + kλT (t) = 0 and X′′ (x) + λX(x) = 0.
2. Solving the time equation:
The solution to the time equation is:
T(t) = Aexp (−kλt).
3. Solving the spatial equation:
The solution to the spatial equation will depend on whether λ > 0, λ = 0,or λ < 0.
If λ > 0, the general solution is of the form:
X(x) = Bsin(√λx)+Ccos(√λx).
If λ= 0, the general solution is:
X(x) = Ax +D.
If λ < 0, the general solution is:
X(x) = Fsinh(√−λx)+Gcosh(√−λx).
4. Applying the boundary conditions:
Given the boundary conditions u(0, t)=0and u(1, t)=0, we can use them to find the
common values of λfor which these conditions hold.
Since u(0, t) = 0, the spatial solution must satisfy X(0) = 0.
Since u(1, t) = 0, the spatial solution must satisfy X(1) = 0.
5. Solving for λ:
By applying the boundary conditions X(0) = 0 and X(1) = 0, we find that the eigenvalues
λnare given by:
λn=(nπ
1)2=n2π2,
where n= 1,2,3, . . ..
6. Finding the complete solution:
The complete solution is given by:
u(x, t) =
∞
∑
n=1
(Bnsin(nπx) + Cncos(nπx)) exp (−k(nπ)2t).
7. Determining the coefficients:
Finally, we determine the coefficients Bnand Cnusing the initial condition u(x, 0) = 2x(1−x).
Since the initial condition u(x, 0) = 2x(1 −x)can be expressed as a Fourier sine series, we
can equate coefficients to find the values of Bnthat satisfy the initial condition.
Therefore, the solution to the initial-boundary value problem described by the heat equation
is:
u(x, t) =
∞
∑
n=1 (8
n3π3sin(nπx))exp (−k(nπ)2t).
27. Find the general solution to the one-dimensional heat equation
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions:
u(0, t) = u(1, t) = 0
and the initial condition:
u(x, 0) = f(x) = 1 −x2
where 0< x < 1,t > 0, and kis a constant.
Ans. Step 1. First, we need to find the general solution to the heat equation.
The general solution to the one-dimensional heat equation can be found using separation of
variables. Let u(x, t) = X(x)T(t). Substituting this into the heat equation gives:
X(x)T′(t) = kX′′(x)T(t)
Dividing both sides by kX(x)T(t)gives:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
Where we have introduced the separation constant λ(which could be positive, negative, or
zero).
Step 2. Now we solve the spatial part of the equation.
We have the ODE for X(x):
X′′(x) + λX(x) = 0
Applying the boundary conditions X(0) = X(1) = 0 gives the solution:
Xn(x) = sin(nπx)
where n= 1,2,3, ....
Step 3. Next, we solve the temporal part of the equation.
For the temporal part, we have the ODE for T(t):
T′(t) + kλT (t) = 0
which has the solution:
Tn(t) = e−kλt
Step 4. Finally, we combine the spatial and temporal solutions.
The general solution is then given by the infinite series:
u(x, t) =
∞
∑
n=1
Ansin(nπx)e−k(nπ)2t
where the coefficients Ancan be found using the initial condition. Substituting the initial
condition into the general solution gives:
1−x2=
∞
∑
n=1
Ansin(nπx)
To find the coefficients, we can use the orthogonality properties of sine functions.
The final solution is then:
u(x, t) =
∞
∑
n=1 (2
nπ )sin(nπx)e−k(nπ)2t
28. Question 28:
Consider the one-dimensional heat equation with periodic boundary conditions:
∂u
∂t =k∂2u
∂x2,0< x < L, t > 0
Subject to the initial condition:
u(x, 0) = x(L−x)
and the periodic boundary conditions:
u(0, t) = u(L, t)and ∂u
∂x(0, t) = ∂u
∂x(L, t)
where kand Lare positive constants.
Find the solution to the heat equation with the given initial and boundary conditions.
Ans. Solution:
1. Separation of Variables: Assume a solution of the form u(x, t) = X(x)T(t).
Substitute into the heat equation to get:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a constant.
2. Time Component: Solve the ODE for T(t):
T′(t) + kλT (t) = 0
The solution to this is T(t) = Ae−kλt.
3. Spatial Component: Solve the ODE for X(x):
X′′(x) + λX(x) = 0
The general solution to this ODE is:
X(x) = c1cos (√λx)+c2sin (√λx)
4. Applying Boundary Conditions:
Given the periodic boundary conditions, we have:
X(0) = c1=X(L) = c1cos(√λL)=c1
This implies that √λL must be an integer multiple of 2π, i.e., √λL =nπ for n∈Z+.
5. Finding Eigenvalues and Eigenfunctions:
Solve λ=(nπ
L)2for n= 1,2,3, ... to find the eigenvalues. The corresponding eigenfunctions
are:
Xn(x) = cos (nπx
L)for n= 1,2,3, ...
6. Forming the General Solution:
The general solution to the heat equation is:
u(x, t) =
∞
∑
n=1
Cne−k(nπ
L)2tcos (nπx
L)
7. Applying Initial Condition:
Substitute u(x, 0) = x(L−x)into the general solution and use orthogonality properties to
find the coefficients Cn.
The final solution is:
u(x, t) =
∞
∑
n=1
4L2(−1)n+1
n3π3e−k(nπ
L)2tsin (nπx
L)
29. Consider the heat equation in one dimension given by:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions:
u(0, t) = 0 and u(π, t) = 0
with the initial condition:
u(x, 0) = f(x) = sin(2x)
Determine the solution to this heat equation.
Ans. Solution: 1. We first need to find the general solution to the heat equation. Let
u(x, t) = X(x)·T(t).
2. Plugging u(x, t)into the heat equation gives:
X(x)·dT
dt =k(d2X
dx2)·T(t)
3. Dividing both sides by kX(x)T(t)gives:
1
k·1
T(t)·dT
dt =1
X(x)·d2X
dx2
4. Since the left side depends only on tand the right side depends only on x, they must be
equal to a constant −λ. This gives two equations:
1
k·1
T(t)·dT
dt =−λand 1
X(x)·d2X
dx2=−λ
5. Solving the first equation gives T(t) = e−λkt.
6. Solving the second equation gives X(x) = Acos(√λx)+Bsin(√λx).
7. Applying the boundary conditions:
u(0, t) = 0 =⇒A= 0
u(π, t) = 0 =⇒Bsin(√λπ)= 0 =⇒√λ=n, where nis a positive integer
8. Therefore, X(x) = Bsin(nx).
9. The general solution to the heat equation is:
u(x, t) =
∞
∑
n=1
Cne−kn2tsin(nx)
where Cnare constants to be determined from the initial condition.
10. Applying the initial condition u(x, 0) = sin(2x)gives:
sin(2x) =
∞
∑
n=1
Cnsin(nx)=⇒C2= 1,and Cn= 0 for n= 2
11. Therefore, the solution to the heat equation is:
u(x, t) = e−4kt sin(2x)
30. Let u(x, t)be a solution to the one-dimensional heat equation
∂u
∂t =k∂2u
∂x2
subject to the initial condition u(x, 0) = x(1 −x)for 0< x < 1, where kis a positive constant.
Determine the solution u(x, t)for t > 0.
Ans. We will solve the heat equation using separation of variables and Fourier series.
1. Let u(x, t) = X(x)T(t). Substituting this into the heat equation gives
T′(t)X(x) = kX′′(x)T(t)
Dividing by kX(x)T(t)yields
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a constant (which we will determine later).
2. Solving the ODE for X(x)gives
X′′ +λX = 0
The general solution is
X(x) = c1cos(√λx)+c2sin(√λx)
Applying the boundary conditions X(0) = X(1) = 0, we obtain
c1= 0 and √λ=nπ
for positive integer n. Thus, Xn(x) = cnsin(nπx)for n= 1,2,3, . . ..
3. Solving the ODE for T(t)gives
T′(t) + kλT (t) = 0
The solution is
T(t) = ce−kλt =ce−n2π2kt
for n= 1,2,3, . . ..
4. By the principle of superposition, the general solution is
u(x, t) =
∞
∑
n=1
cne−n2π2kt sin(nπx)
Finally, applying the initial condition u(x, 0) = x(1 −x)gives
u(x, 0) =
∞
∑
n=1
cnsin(nπx) = x(1 −x)
Multiplying both sides by sin(mπx)and integrating over [0,1] yields
cm=∫1
0
x(1 −x)sin(mπx)dx
Thus, the solution is
u(x, t) =
∞
∑
n=1 (∫1
0
x(1 −x)sin(nπx)dx)e−n2π2kt sin(nπx)
3. Find the general solution to the heat equation ∂u
∂t =k∂2u
∂x2subject to the initial condition
u(x, 0) = sin(2x)and boundary conditions u(0, t) = 0 and u(π, t) = 0 for t > 0.
Ans. To solve the heat equation with the given initial and boundary conditions, we first find
the eigenvalues and eigenfunctions of the problem using separation of variables, then use the
method of Fourier series. 1. Separation of Variables: Assume a solution of the form u(x, t) =
X(x)T(t). Substitute u(x, t) = X(x)T(t)into the heat equation and separate variables to get:
T′(t)
kT (t)=X′′ (x)
X(x)=−λSolve the ODE for X(x):X′′(x) + λX(x) = 0 with boundary conditions
X(0) = X(π)=0. The eigenvalues are λn=n2, and the corresponding eigenfunctions are
Xn(x) = sin(nx)for n= 1,2,3, ...
2. Finding the Temporal Equation: Now we solve the temporal equation T′(t)
kT (t)=−λ.
The solution is Tn(t) = Ce−n2kt for n= 1,2,3, ...
3. General Solution: The general solution is given by u(x, t) = ∑∞
n=1 Xn(x)Tn(t) =
∑∞
n=1 sin(nx)Ce−n2kt. Applying the initial condition u(x, 0) = sin(2x), we find that u(x, 0) =
∑∞
n=1 Cnsin(nx) = sin(2x). By comparing coefficients, we find that C2= 1 and Cn= 0 for
n= 2. Thus, the solution to the heat equation with the given initial and boundary conditions is
u(x, t) = e−4kt sin(2x).
4. Find the general solution to the partial differential equation:
∂2u
∂x2−4∂2u
∂t2= 0
where u(x, t)is a function of xand t.
Ans. To solve the partial differential equation ∂2u
∂x2−4∂2u
∂t2= 0, we assume a solution of the
form u(x, t) = X(x)T(t). Substituting this into the equation gives us:
X′′(x)T(t)−4X(x)T′′ (t) = 0
Dividing by X(x)T(t)gives:
X′′(x)
X(x)= 4T′′ (t)
T(t)
Since the left side of the equation depends only on xand the right side depends only on t,
they must be equal to a constant. Letting this constant be −λ, we get two ordinary differential
equations:
X′′(x) + λX(x) = 0, T ′′(t)−4λT (t) = 0
Solving the first ODE gives us X(x) = Acos(√λx)+Bsin(√λx), while the second ODE
gives us T(t) = Ccos(2√λt)+Dsin(2√λt).
The general solution is then:
u(x, t) =
∞
∑
n=1 [(Ancos(√λnx)+Bnsin(√λnx))(Cncos(2√λnt)+Dnsin(2√λnt))]
where λnare the eigenvalues that satisfy the boundary conditions, and An,Bn,Cn,Dnare
constants determined by the initial conditions.
5. Question:
Consider the wave equation ∂2u
∂t2=c2∂2u
∂x2, where u(x, t)represents the displacement of a string
at position xand time t, and cis a positive constant representing the speed of propagation of
disturbances in the string.
A string of length Lfixed at both ends is initially at rest, i.e., u(x, 0) = 0 and ∂u
∂t (x, 0) = 0.
At time t= 0, the string is given an initial velocity v0sin (πx
L), where v0is a positive constant.
Find the solution u(x, t)for 0≤x≤Land t > 0.
Ans. Step-by-step solution:
1. We will use the method of separation of variables to solve the wave equation. Let’s assume
u(x, t) = X(x)T(t).
2. Plugging u(x, t) = X(x)T(t)into the wave equation gives us T′′
c2T=X′′
X.
3. Rearranging, we have T′′
c2T=−λ2=X′′
X, where λis a separation constant.
4. Solving for X(x), we have X′′ +λ2X= 0, which gives us the general solution X(x) =
Asin(λx) + Bcos(λx).
5. The boundary conditions u(x, 0) = 0 and ∂u
∂t (x, 0) = 0 give us X(x) = 0 and T′(0) = 0.
6. Since X(x) = 0, we get Asin(λx) + Bcos(λx) = 0. This implies A= 0 and Bcos(λx) =
0, which gives B= 0.
7. Therefore, the solution for the spatial part is X(x) = 0.
8. Now, we solve for the temporal part using T′′ +c2λ2T= 0, which has the general solution
T(t) = Ccos(cλt) + Dsin(cλt).
9. The initial condition ∂u
∂t (x, 0) = 0 gives dT
dt (0) = 0, which implies D= 0.
10. Hence, the temporal part solution is T(t) = Ccos(cλt).
11. Finally, combining the spatial and temporal parts gives us the solution u(x, t) =
∑∞
n=1 Bnsin (nπx
L)cos (cnπt
L), where Bnare the coefficients determined from the initial condition.
12. Given the initial condition u(x, 0) = v0sin (πx
L), we find Bn=2v0
nπ .
13. Therefore, the final solution to the wave equation is u(x, t) = ∑∞
n=1
2v0
nπ sin (nπx
L)cos (cnπt
L).
6. Consider the following Heat Equation:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions:
u(0, t) = 0, u(L, t) = 0
and the initial condition:
u(x, 0) = sin (πx
L)
Find the solution to the Heat Equation using the method of separation of variables.
Ans. Solution: 1. We assume a solution of the form u(x, t) = X(x)T(t)and substitute it
into the Heat Equation:
X(x)T′(t) = kX′′(x)T(t)
2. Divide both sides by kX(x)T(t)to separate variables:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis the separation constant.
3. Solve the time part of the equation:
T′(t) + kλT (t) = 0
The general solution is T(t) = Aexp (−kλt).
4. Solve the spatial part of the equation:
X′′(x) + λX(x) = 0
The general solution is X(x) = Bsin (√λx)+Ccos (√λx).
5. Apply the boundary conditions u(0, t) = 0 and u(L, t) = 0 to the spatial part:
X(0) = Bsin(0) + Ccos(0) = 0 =⇒C= 0
X(L) = Bsin(√λL)= 0 =⇒√λL =nπ where n= 1,2,3, ...
6. Therefore, the eigenvalues are λ=(nπ
L)2and the eigenfunctions are Xn(x) = Bnsin (nπx
L).
7. The solution to the Heat Equation is then:
u(x, t) =
∞
∑
n=1
Anexp (−k(nπ
L)2
t)sin (nπx
L)
8. Apply the initial condition u(x, 0) = sin (πx
L)to find the coefficients An:
sin (πx
L)=
∞
∑
n=1
Ansin (nπx
L)
By orthogonality of sine functions, we get An= 2 ∫L
0sin (πx
L)sin (nπx
L)dx.
9. Therefore, the final solution to the Heat Equation is:
u(x, t) = sin (πx
L)−2
∞
∑
n=1
sin (πx
L)sin (nπx
L)exp (−k(nπ
L)2
t)
7. Question:
Consider the one-dimensional heat equation:
∂u
∂t =c2∂2u
∂x2
subject to the boundary conditions u(0, t) = 0 and u(L, t) = 0, as well as the initial condition
u(x, 0) = f(x)for 0≤x≤L, where cis a positive constant and Lis the length of the rod.
Show that the solution to this equation can be written as:
u(x, t) =
∞
∑
n=1
Bnsin (nπx
L)e−(nπct
L)2
where the coefficients Bnare given by:
Bn=2
L∫L
0
f(x)sin (nπx
L)dx
Ans. Solution:
1. First, we assume that the solution u(x, t)can be expressed as a product of functions of x
and t:
u(x, t) = X(x)T(t)
2. Substituting this into the heat equation gives:
X(x)T′(t) = c2X′′(x)T(t)
3. Dividing both sides by c2X(x)T(t)gives:
T′(t)
c2T(t)=X′′(x)
X(x)
4. Since the left side depends only on tand the right side depends only on x, both sides must
be equal to a constant −λ. This gives two ODEs to solve:
T′(t) + c2λT (t) = 0 and X′′(x) + λX(x) = 0
5. The solutions to T′(t) + c2λT (t) = 0 are:
Tn(t) = Ane−(nπct
L)2
6. To solve the second ODE, we find that the solutions X(x)are proportional to sin (nπx
L).
This gives:
Xn(x) = Bnsin (nπx
L)
7. Combining the solutions for X(x)and T(t)with the separation constant −λ=−(nπc/L)2
gives:
un(x, t) = Xn(x)Tn(t) = Bnsin (nπx
L)e−(nπct
L)2
8. Finally, by linearity of the heat equation, the solution to the initial-boundary value problem
is the sum of the individual solutions:
u(x, t) =
∞
∑
n=1
Bnsin (nπx
L)e−(nπct
L)2
where the coefficients Bnare given by:
Bn=2
L∫L
0
f(x)sin (nπx
L)dx
This completes the derivation of the solution to the heat equation with the given initial and
boundary conditions.
8. Question:
Consider the following wave equation in one dimension:
utt =c2uxx,
with boundary conditions u(0, t) = 0,u(L, t) = 0 for t > 0, and initial conditions u(x, 0) =
f(x),ut(x, 0) = g(x)for 0<x<L. Find the solution to this wave equation using the method
of separation of variables.
Ans. Solution:
We assume the solution to the wave equation can be written as a product of two functions,
u(x, t) = X(x)T(t). Substituting this into the wave equation gives:
X(x)T′′(t) = c2X′′ (x)T(t).
Dividing both sides by c2X(x)T(t)gives:
T′′(t)
c2T(t)=X′′(x)
X(x).
This can only hold if both sides are equal to a constant, say −λ. Then we have two ODEs
to solve:
1. The time part: T′′(t) + λc2T= 0 with solutions T(t) = Acos(√λct)+Bsin(√λct).
2. The spatial part: X′′ (x) + λX = 0 with solutions X(x) = Ccos(√λx)+Dsin(√λx).
Applying the boundary conditions u(0, t)=0and u(L, t)=0gives us X(0)T(t)=0and
X(L)T(t)=0. This implies that C= 0 and sin(√λL)= 0, so √λ=nπ
Lfor n= 1,2,3, . . ..
Therefore, the spatial part becomes Xn(x) = Dnsin (nπx
L).
The general solution to the wave equation is then:
u(x, t) =
∞
∑
n=1 (Ancos (nπct
L)+Bnsin (nπct
L))sin (nπx
L).
Finally, we can determine the coefficients Anand Bnusing the initial conditions u(x, 0) =
f(x)and ut(x, 0) = g(x).
9. Question:
Consider the following partial differential equation:
∂2u
∂t2= 9∂2u
∂x2
with initial conditions:
u(x, 0) = 2 sin(2x)
∂u
∂t (x, 0) = 3 sin(2x)
Determine the solution to the partial differential equation.
Ans. Step-by-step solution:
1. To solve the given partial differential equation, we first assume a solution of the form:
u(x, t) = X(x)T(t)
2. Substituting this into the partial differential equation, we get:
T′′(t)X(x) = 9T(t)X′′ (x)
3. Dividing both sides by T(t)X(x), we obtain:
T′′(t)
T(t)= 9X′′ (x)
X(x)
4. Since the left side is a function of tonly and the right side is a function of xonly, they
must both be equal to a constant −λ. This gives us two ordinary differential equations:
T′′(t)
T(t)=−λand 9X′′(x)
X(x)=−λ
5. Solving the first ODE we get:
T′′(t) + λT (t) = 0
which has solutions of the form T(t) = Acos(3t) + Bsin(3t)if λ= 9.
6. Solving the second ODE we get:
X′′(x) + λ
9X(x) = 0
which has solutions of the form X(x) = Ccos(3x) + Dsin(3x)if λ= 9.
7. The general solution to the partial differential equation is then:
u(x, t) = (Acos(3t) + Bsin(3t))(Ccos(3x) + Dsin(3x))
8. Applying the initial conditions u(x, 0) = 2 sin(2x)and ∂u
∂t (x, 0) = 3 sin(2x), we find the
specific solution to be:
u(x, t) = 2e−9tsin(3x)+3e−9tsin(3x)
which is the solution to the partial differential equation.
10. Question:
Consider the heat equation in one dimension, given by ∂u
∂t =k∂2u
∂x2, subject to the boundary
conditions u(0, t)=0and u(π, t)=0, and the initial condition u(x, 0) = sin(x). Find the
solution u(x, t)for t > 0.
Ans. Solution:
We will solve the given heat equation using separation of variables.
1. Guess the form of the solution: We assume that the solution can be expressed as a
product of functions of xand t:
u(x, t) = X(x)T(t)
2. Substitute the solution form into the heat equation: Substitute u(x, t) = X(x)T(t)
into the heat equation ∂u
∂t =k∂2u
∂x2:
XT ′=kX′′T
3. Divide by the product of X(x)T(t):Divide both sides by XT :
T′
kT =X′′
X=−λ
This results in two ordinary differential equations:
T′+kλT = 0
X′′ +λX = 0
4. Solve the ODE for T(t):The solution to T′+kλT = 0 is:
T(t) = Ae−kλt
5. Solve the ODE for X(x):The solution to X′′ +λX = 0 is:
X(x) = Bsin(√λx)+Ccos(√λx)
6. Apply the boundary conditions: From the boundary conditions u(0, t)=0and
u(π, t) = 0, we can determine the values of λ,B, and C.
For u(0, t) = 0:
X(0)T(t) = 0
C= 0
For u(π, t) = 0:
X(π)T(t) = 0
Bsin(√λπ)= 0
This gives us the condition sin(√λπ)= 0, which implies √λπ =nπ ⇒√λ=n, where nis a
positive integer.
7. Find the complete solution: Therefore, the solutions are of the form:
un(x, t) = e−kn2tBnsin(nx)
where Bnare constants that can be determined from the initial condition.
The complete solution is then the sum of all the individual solutions:
u(x, t) =
∞
∑
n=1
e−kn2tBnsin(nx)
Applying the initial condition u(x, 0) = sin(x)gives Bn= 2 ∫π
0sin(x)sin(nx)dx.
Therefore, the solution to the heat equation with the given conditions is:
u(x, t) =
∞
∑
n=1
e−kn2t2∫π
0
sin(x)sin(nx)dx sin(nx)
11. Question: Solve the heat equation in one dimension, given by ∂u
∂t =k∂2u
∂x2, subject to the
initial condition u(x, 0) = sin(πx)and boundary conditions u(0, t) = u(L, t) = 0, where u(x, t)
is the temperature distribution along a rod of length Land kis a positive constant.
Ans. Solution: We will solve the heat equation using the method of separation of variables.
1. Formulating the solution Assume a solution of the form u(x, t) = X(x)T(t). Substi-
tuting this into the heat equation gives 1
kT
dT
dt =1
X
d2X
dx2.
2. Solving the time ODE Solving the time ODE, we have 1
kT
dT
dt =−λ, where λis a
separation constant. This gives us T(t) = Ae−kλt.
3. Solving the spatial ODE Solving the spatial ODE, we have 1
X
d2X
dx2=−λ. The general
solutions are X(x) = Bsin(√λx)+Ccos(√λx).
4. Applying the boundary conditions Applying the boundary conditions u(0, t) = u(L, t) =
0, we get X(0) = X(L) = 0. This implies C= 0, and sin(√λL)= 0, which gives √λL =nπ
for n∈Z+. Thus, λ=(nπ
L)2.
5. Obtaining the solution Therefore, the general solution is u(x, t) = ∑∞
n=1(ansin(nπx
L)e−k(nπ
L)2t).
6. Calculating the coefficients Applying the initial condition u(x, 0) = sin(πx), we can
find the coefficients an. This involves solving for an= 2/L∫L
0sin(πx)sin(nπx
L)dx.
7. Final solution Finally, the solution to the heat equation with given initial and boundary
conditions is u(x, t) = ∑∞
n=1 (2
(2n−1)πsin((2n−1)πx
L)e−k((2n−1)π
L)2t).
12. Question:
Consider the one-dimensional heat equation on the interval [0, π], given 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) = sin2(x).
Determine the solution u(x, t)to the heat equation.
Ans. Solution:
We will solve the one-dimensional heat equation using separation of variables.
1. Separation of variables: We assume that the solution can be written as a product of
functions of xand t, that is, u(x, t) = X(x)T(t).
2. Substitute into the heat equation: Substitute u(x, t) = X(x)T(t)into the heat
equation to obtain: T′(t)X(x) = kX′′(x)T(t).
Divide by kX(x)T(t)to get: T′(t)
kT (t)=X′′ (x)
X(x).
3. Solving the time-dependent ODE: Since the left-hand side is purely a function of t
and the right-hand side is purely a function of x, they must be equal to a constant −λ, which
gives two separate ODEs: T′(t) + kλT (t) = 0,X′′ (x) + λX(x) = 0.
4. Solve the ODE for X(x):The ODE for X(x)is a simple second-order linear ODE with
solutions of the form X(x) = Ancos(√λx)+Bnsin(√λx).
Applying the boundary conditions u(0, t) = 0 and u(π, t) = 0, we obtain Bn= 0 and
cos(√λπ)= 0, which implies √λ=n, where nis a positive integer.
Therefore, the eigenfunctions are Xn(x) = Ancos(nx).
5. Solve the ODE for T(t):The ODE for T(t)is a simple first-order linear ODE with
solution T(t) = e−kλt =e−kn2t.
6. General solution: The general solution to the heat equation can be written as a sum of
the separated solutions: u(x, t) = ∑∞
n=1 Ancos(nx)e−kn2t.
7. Determine the coefficients: Applying the initial condition u(x, 0) = sin2(x), we can
determine the coefficients Anby taking the inner product of u(x, 0) and Xn(x).
The solution to the heat equation is: u(x, t) = 1
2+∑∞
n=1
(−1)n+1
n(2n−1) cos(nx)e−kn2t.
13. Question: Consider the one-dimensional heat equation defined for 0< x < 1and t > 0:
∂u
∂t =k∂2u
∂x2,
subject to the boundary conditions u(0, t) = 0 and u(1, t) = 0, where u(x, t)is the temperature
distribution in a one-dimensional rod, kis the thermal diffusivity constant, and u(x, 0) = sin(πx).
Find the solution to the heat equation using separation of variables.
Ans. Step-by-step solution: 1. We assume that the solution to the heat equation can be written
as a product of two functions, X(x)and T(t), i.e., u(x, t) = X(x)T(t). 2. By substituting this
assumption into the heat equation, we get:
1
k
T′(t)
T(t)=X′′(x)
X(x)=−λ
where λis a constant. This leads to two ordinary differential equations: T′(t)
kT (t)=−λand X′′ (x)+
λX(x)=0. 3. Solving the ODE for X(x), we find that X(x) = Acos(√λx)+Bsin(√λx),
where Aand Bare constants. 4. Applying the boundary conditions u(0, t) = 0 and u(1, t) = 0,
we have X(0) = A= 0 and X(1) = Bsin(√λ)= 0. This implies that λ=n2π2, where
n= 1,2,3, .... 5. Therefore, the solution for X(x)is given by Xn(x) = Bnsin(nπx), where Bn
is a normalization constant. 6. Next, solving the ODE for T(t), we have T′(t)
kT (t)=−λ=−n2π2,
which gives Tn(t) = Cne−kn2π2t, where Cnis a constant. 7. Assemble the solution by taking the
sum of all the modes:
u(x, t) =
∞
∑
n=1
Xn(x)Tn(t) =
∞
∑
n=1
Bnsin(nπx)Cne−kn2π2t
8. Finally, we can determine the coefficients Bnand Cnby using the initial condition u(x, 0) =
sin(πx). By expressing the initial condition in terms of the Fourier sine series, we find that Bn=
2∫1
0sin(nπx)sin(πx)dx. 9. After evaluating the integral in step 8, we obtain the coefficients
Bn=2
(n2−1)π. 10. Hence, the solution to the heat equation with the given boundary and initial
conditions is:
u(x, t) =
∞
∑
n=1
2
(n2−1)πsin(nπx)e−kn2π2t
14. Question:
Consider the wave equation on the interval 0< x < π with the boundary conditions u(0, t) =
u(π, t) = 0 and the initial conditions u(x, 0) = sin xand ∂u
∂t (x, 0) = sin x. Solve for u(x, t).
Ans. Step-by-step solution:
1. We will first assume a solution of the form u(x, t) = X(x)T(t)and substitute it into the
wave equation:
X(x)T′′(t) = c2X′′ (x)T(t)
Dividing both sides by c2X(x)T(t)gives:
T′′(t)
c2T(t)=X′′(x)
X(x)=−λ
where λis a separation constant.
2. Solving the ODE X′′ (x)
X(x)=−λwith boundary conditions X(0) = X(π)=0gives eigen-
functions Xn(x) = sin(nx)for n∈Z+and eigenvalues λ=n2.
3. For the temporal equation, we have T′′ (t)
c2T(t)=−λ=−n2, which has solutions Tn(t) =
Ancos (nc2t) + Bnsin (nc2t).
4. Now, the general solution is given by:
u(x, t) =
∞
∑
n=1
(Ancos (nc2t) + Bnsin (nc2t)) sin(nx)
Applying the initial conditions u(x, 0) = sin xand ∂u
∂t (x, 0) = sin x, we find:
An= 0 and Bn=2
π∫π
0
sin xsin nx dx
5. Evaluating the integral Bn=2
π∫π
0sin xsin nx dx gives Bn={0if nis even
4
π
(−1)
n−1
2
nif nis odd
Therefore, the solution to the wave equation with the given boundary and initial conditions
is:
u(x, t) = 4
π
∞
∑
n=1
(−1)n−1
2
nsin (2n−1)xcos ((2n−1)c2t)
15. Question: Consider the heat equation in one dimension, given by ∂u
∂t =k∂2u
∂x2, where u(x, t)
represents the temperature distribution in a rod of length Lwith initial condition u(x, 0) = f(x)
and boundary conditions u(0, t)=0and u(L, t)=0. Find the solution to this boundary value
problem.
Ans. Solution: 1. We begin 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′(t)
kT (t)=X′′ (x)
X(x)=−λ, where λis a separation
constant.
3. Solving the ODE for X(x)with boundary conditions X(0) = 0 and X(L)=0gives
X(x) = Bsin (nπx
L)for n= 1,2,3, ....
4. The time part T(t)satisfies T′(t)
kT (t)=−λ, which implies T(t) = Ce−kλt.
5. Using the separation constant λ=(nπ
L)2, the general form of the solution becomes
u(x, t) = ∑∞
n=1 Bnsin (nπx
L)e−k(nπ
L)2t.
6. Applying the initial condition u(x, 0) = f(x)yields the Fourier sine series of f(x):f(x)∼
∑∞
n=1 Bnsin (nπx
L).
7. Comparing coefficients in step 5 and 6 gives Bn=2
L∫L
0f(x)sin (nπx
L)dx.
8. Therefore, the solution to the given heat equation with the specified boundary and initial
conditions is u(x, t) = ∑∞
n=1 (2
L∫L
0f(s)sin (nπs
L)ds)sin (nπx
L)e−k(nπ
L)2t.
16. Question: Consider the following heat equation with mixed boundary conditions:
∂u
∂t =k(∂2u
∂x2+∂2u
∂y2)
subject to the conditions:
u(0, y, t) = u(1, y, t) = 0, u(x, 0, t) = u(x, 1, t) = 0,
and the initial temperature distribution:
u(x, y, 0) = 4xy(1 −x)(1 −y).
Find the steady-state solution of the heat equation.
Ans. Solution: 1. To find the steady-state solution of the heat equation, we need to find
the solution when t→ ∞. This means that the time derivative will be zero, so the equation
simplifies to Laplace’s equation:
0 = k(∂2u
∂x2+∂2u
∂y2)
2. The general solution to Laplace’s equation in 2D is given by:
u(x, y) =
∞
∑
n=1
∞
∑
m=1 (Anm sin (nπx
L)sin (mπy
L))
3. Apply the boundary conditions to determine the coefficients Anm:
u(0, y) =
∞
∑
n=1
∞
∑
m=1
Anm sin(0) sin (mπy
L)= 0
u(1, y) =
∞
∑
n=1
∞
∑
m=1
Anm sin(nπ)sin (mπy
L)= 0
u(x, 0) =
∞
∑
n=1
∞
∑
m=1
Anm sin (nπx
L)sin(0) = 0
u(x, 1) =
∞
∑
n=1
∞
∑
m=1
Anm sin (nπx
L)sin(mπ) = 0
4. Applying the boundary conditions simplifies the general solution to:
u(x, y) =
∞
∑
n=1
∞
∑
m=1
Anm sin (nπx
L)sin (mπy
L)
5. Next, find the coefficients Anm by substituting the initial temperature distribution into the
general solution:
u(x, y, 0) = 4xy(1 −x)(1 −y) =
∞
∑
n=1
∞
∑
m=1
Anm sin (nπx
L)sin (mπy
L)
6. Comparing coefficients on both sides, we can determine the values of Anm. This will give
us the steady-state solution of the heat equation.
17. Question:
Consider the heat equation in one dimension given by ∂u
∂t =c2∂2u
∂x2, where cis a positive
constant. Solve the initial-boundary value problem for 0<x<1and t > 0with u(0, t)=0,
u(1, t) = 0, and u(x, 0) = x(1 −x).
Ans. Step-by-step solution:
1. We begin by assuming a solution of the form u(x, t) = X(x)T(t). Substituting this into
the heat equation gives T′
c2T=X′′
X=−λ, where λis a separation constant.
2. Solving the time part of the equation gives T(t) = Ae−c2λt. If λ < 0,T(t)grows
unbounded as t→ ∞, which is not suitable for a physically realistic temperature distribution.
Therefore, we must have λ≥0.
3. Solving the spatial part of the equation gives an eigenvalue problem X′′ +λX = 0. The
solutions to this equation are Xn(x) = sin(nπx)for n= 1,2,3, . . . corresponding to λn= (nπ)2.
4. The general solution to the PDE is then u(x, t) =
∞
∑
n=1
(Ansin(nπx))e−(nπc)2t.
5. To find the specific solution satisfying the initial condition, apply the initial condition
u(x, 0) = x(1 −x) =
∞
∑
n=1
Ansin(nπx).
6. Multiplying both sides by sin(mπx)and integrating from 0 to 1 gives Am= 2 ∫1
0x(1 −
x)sin(mπx)dx.
7. Evaluating the integral gives Am=4(−1)m+1
m3π3for modd and Am= 0 for meven.
8. Therefore, the solution to the initial-boundary value problem is u(x, t) =
∞
∑
n=1,3,5,...
4(−1)n+1
n3π3sin(nπx)e−(nπc)2t.
18. Question: Consider the following heat equation on the interval 0≤x≤πwith initial
condition and boundary conditions:
ut=uxx + 2 sin(x),0< x < π, t > 0
u(0, t) = u(π, t) = 0, t ≥0
u(x, 0) = x(π−x),0≤x≤π
Find the solution u(x, t)of the given initial/boundary value problem.
Ans. Step-by-step solution:
Step 1. Find the separated solution of the heat equation: Let u(x, t) = X(x)T(t). Then we
have: T′
T=X′′ + 2 sin(x)
X
This yields the pair of ODEs:
T′=cT, X′′ + 2 sin(x)X=cX
Step 2. Solve the ODE for T(t): The first ODE T′=cT has a general solution of the form
T(t) = Aect. Since we are interested in a bounded solution, we must have c≤0. Let’s set
c=−λ2for some λ > 0. Thus,
T(t) = Ae−λ2t
Step 3. Solve the ODE for X(x): The second ODE X′′ + 2 sin(x)X=−λ2Xis a Sturm-
Liouville eigenvalue problem. The corresponding eigenfunctions of X(x)are the solutions to
X′′ + (2 −λ2)sin(x)X= 0. To find such solutions, we solve for the eigenvalues λnand
corresponding eigenfunctions Xn(x).
Step 4. Solve for eigenvalues and eigenfunctions: Set 2−λ2
n=m2
n, then we get λn=
√2−m2
nfor mn= 1,2,3, .... The corresponding eigenfunctions Xn(x)are Xn(x) = sin(nπx)
for n= 1,2,3, ....
Step 5. General solution: The general solution to the heat equation is given by:
u(x, t) =
∞
∑
n=1
Bne−λ2
ntsin(nπx)
Step 6. Calculate the coefficients: Applying the initial condition u(x, 0) = x(π−x)gives:
x(π−x) =
∞
∑
n=1
Bnsin(nπx)
Multiplying both sides by sin(mπx)and integrating from 0to πgives:
Bm=2
π∫π
0
x(π−x)sin(mπx)dx
Step 7. Final solution: By solving for Bnand substituting back into the general solution,
the final solution of the given initial/boundary value problem is:
u(x, t) =
∞
∑
n=1
2
π((−1)n+1
n3)e−n2tsin(nπx)
19. Question: Consider the wave equation in one dimension: ∂2u
∂t2=c2∂2u
∂x2, where cis a constant.
Solve the wave equation subject to the initial conditions u(x, 0) = f(x)and ∂u
∂t (x, 0) = g(x),
where f(x)and g(x)are given functions defined on the interval 0< x < L.
Ans. Solution:
We first need to find a general solution to the wave equation. Let’s assume the solution can
be written as a product of two functions X(x)and T(t):u(x, t) = X(x)T(t).
1. Substituting this into the wave equation, we get:
T′′
c2T=X′′
X=−λ2
This gives us two ordinary differential equations:
2. Equation for T(t):
T′′ +c2λ2T= 0
This has the general solution:
T(t) = Acos(λct) + Bsin(λct)
3. Equation for X(x):
X′′ +λ2X= 0
This has the general solution:
X(x) = Ccos(λx) + Dsin(λx)
4. Combining the solutions for X(x)and T(t), we get:
u(x, t) = (Ccos(λx) + Dsin(λx))(Acos(λct) + Bsin(λct))
5. Applying the initial conditions:
u(x, 0) = f(x)⇒C=f(x)
∂u
∂t (x, 0) = g(x)⇒B=1
λcg(x)
6. Therefore, the solution to the wave equation with given initial conditions is:
u(x, t) =
∞
∑
n=1 (fncos (nπx
L)+1
nπc [∂fn
∂x cos (nπx
L)+gnsin (nπx
L)])cos (nπct
L)
where fnand gnare the Fourier coefficients of f(x)and g(x), respectively.
20. Let u(x, t)satisfy the wave equation on the domain 0< x < π and t > 0with initial
conditions given by:
u(x, 0) = x(π−x)
and boundary conditions given by:
u(0, t) = 0
u(π, t) = 0
Find the solution u(x, t)to this problem.
Ans. Solution: 1. We can start by separating the variables u(x, t) = X(x)T(t). Substituting
this into the wave equation, we have:
X(x)T′′(t) = c2X′′ (x)T(t)
2. Dividing by c2X(x)T(t), we get:
T′′(t)
c2T(t)=X′′(x)
X(x)=−λ
3. Solving the ODE X′′ (x) + λX(x) = 0, subject to the boundary conditions X(0) = X(π) = 0,
we obtain λ=n2and Xn(x) = sin(nx). 4. Solving the ODE T′′ (t) + n2c2T(t) = 0, we obtain
Tn(t) = Ancos(nct) + Bnsin(nct). 5. Combining the spatial and temporal parts, the general
solution will be:
u(x, t) =
∞
∑
n=1
(Ancos(nct) + Bnsin(nct)) sin(nx)
6. Applying the initial condition to find the Ancoefficients, we have:
u(x, 0) =
∞
∑
n=1
Ansin(nx) = x(π−x)
7. Multiply both sides by sin(mx)and integrate from 0to π, we find:
Am=2
π∫π
0
x(π−x)sin(mx)dx =8
π
(−1)m+1
m3
8. Therefore, the final solution to the problem will be:
u(x, t) =
∞
∑
n=1 (8
π
(−1)n+1
n3sin(nx)cos(nct))
21. Question: Consider the heat equation given by ∂u
∂t =k∂2u
∂x2where u(x, t)is the temperature
distribution in a rod of length Lwith initial condition u(x, 0) = x(L−x)and boundary conditions
u(0, t) = u(L, t) = 0. Find the solution to the heat equation for 0< x < L and t > 0.
Ans. Solution: 1. We will start by assuming a solution of the form u(x, t) = X(x)T(t). 2.
Substitute u(x, t) = X(x)T(t)into the heat equation and divide by kX(x)T(t)to get 1
kT (t)
dT
dt =
1
X(x)
d2X
dx2. 3. Rearranging the equation gives 1
kT (t)
dT
dt =λ=1
X(x)
d2X
dx2, where λis some constant.
4. Solve the ODE for T(t):dT
dt =kλT (t)gives T(t) = Aekλt, where Ais a constant. 5. Solve
the ODE for X(x):d2X
dx2+λX(x) = 0. 6. The general solution to the ODE for X(x)is X(x) =
C1cos(√λx)+C2sin(√λx), where C1and C2are constants. 7. Apply the boundary conditions
u(0, t) = u(L, t) = 0 to find that C1= 0 and √λ=nπ/Lfor n= 1,2,3, . . .. 8. Using the found
eigenvalues λn= (nπ/L)2and general solution for T(t), the solution u(x, t)becomes a sum over
the orthogonal eigenfunctions: u(x, t) = ∑∞
n=1 Bne−k(nπ/L)2tsin(nπx/L). 9. Apply the initial
condition u(x, 0) = x(L−x) = ∑∞
n=1 Bnsin(nπx/L)to find Bn=2
L∫L
0x(L−x)sin(nπx/L)dx.
10. Solve the integral to find the coefficients Bn=4L
n3π3((−1)n−1). Therefore, the solution to
the heat equation is
u(x, t) =
∞
∑
n=1
4L
n3π3((−1)n−1)e−k(nπ/L)2tsin(nπx/L).
22. Question: Consider the heat equation ∂u
∂t =k∂2u
∂x2subject to the boundary conditions
u(0, t) = 0 and u(π, t) = 0, with the initial condition u(x, 0) = sin(x). Find the solution u(x, t)
for t > 0.
Ans. Solution:
1. We first note that the heat equation is a separable partial differential equation, so we
assume a solution of the form u(x, t) = X(x)T(t).
2. Plugging this into the heat equation, we get T′(t)
kT (t)=X′′ (x)
X(x)=−λ, where λis a separation
constant.
3. Solving the two ordinary differential equations T′(t)+kλT (t) = 0 and X′′ (x)+λX(x) = 0,
we find T(t) = Ae−kλt and X(x) = Bsin(√λx).
4. The boundary conditions u(0, t) = 0 and u(π, t) = 0 lead to X(0)T(t) = 0 and
X(π)T(t) = 0, respectively.
5. From X(0) = 0, we have Bsin(0) = 0 =⇒B= 0, so we have X(x) = 0.
6. From X(π) = 0, we have sin(√λπ)= 0, which gives √λ=nwhere nis a positive
integer.
7. Therefore, the solution u(x, t)can be written as a series of sine functions: u(x, t) =
∑∞
n=1 Cnsin(nx)e−kn2t.
8. To find the coefficients Cn, we use the initial condition u(x, 0) = sin(x):u(x, 0) =
∑∞
n=1 Cnsin(nx) = sin(x). This implies Cn=2
π∫π
0sin(x)sin(nx)dx.
9. Evaluating the integral, we find Cn=2
π∫π
0
1
2i(eix−e−ix)1
2i(einx−e−inx)dx =2
π[π(−1)n−(−1)n+1
1−n2].
10. Finally, the solution to the heat equation with the given boundary and initial conditions
is: u(x, t) = ∑∞
n=1
2(−1)n
(1−n2)sin(nx)e−kn2t.
23. Find the solution to the Heat Equation, subject to the initial condition u(x, 0) = x(1 −x)
on the interval 0< x < 1, where the boundary conditions are u(0, t) = u(1, t) = 0.
Ans. We start by expressing the solution u(x, t)as a Fourier series:
u(x, t) =
∞
∑
n=1
Bne−(nπ
l)2kt sin (nπx
l)
1. To find the coefficients Bn, we apply the initial condition:
x(1 −x) =
∞
∑
n=1
Bnsin (nπx
l)
2. We multiply both sides by sin (mπx
l), where mis an integer from 1 to ∞, then integrate
from 0 to 1:
∫1
0
x(1 −x)sin (mπx
l)dx =
∞
∑
n=1
Bn∫1
0
sin (nπx
l)sin (mπx
l)dx
3. Simplifying the integral on the left side:
∫1
0
x(1 −x)sin (mπx
l)dx =8
π3((−1)m−1)
4. Calculating the right side integral:
∫1
0
sin (nπx
l)sin (mπx
l)dx ={0if n=m
l
2if n=m
5. Substituting the results from steps 3 and 4 into the integral equation, we get:
8
π3((−1)m−1) = Bm
l
2
6. Solving for Bm, we find:
Bm=16
m3π3(1 −(−1)m)
Hence, the solution to the Heat Equation with the given initial and boundary conditions is:
u(x, t) =
∞
∑
n=1
16
n3π3(1 −(−1)n)e−(nπ
l)2kt sin (nπx
l)
24. Find the general solution to the partial differential equation:
uxx −4uxy + 4uyy = 0
where u(x, y)is a function of xand y.
Ans. Solution: 1. We first compute the characteristics of the PDE by solving the system of
ODEs given by:
dx
dt = 1,dy
dt =−2,du
dt = 0
This system gives x=t+C1and y=−2t+C2for constants C1and C2. Therefore, the
characteristics are given by y=−2x+Cfor some constant C.
2. Let v(x, y) = u(x, y)along the characteristic curves, which gives us d2v
dx2−4d2v
dxdy +4d2v
dy2= 0.
3. Using the change of variables p=x+yand q=x−y, we rewrite the PDE in terms of
v(x, y)as:
∂2v
∂p∂q = 0
4. The general solution to this PDE is v(p, q) = f(p) + g(q)for arbitrary functions fand g.
5. Substituting p=x+yand q=x−yback, we have u(x, y) = f(x+y) + g(x−y)as
the general solution to the original PDE.
25. Question: Consider the one-dimensional heat equation
∂u
∂t =c2∂2u
∂x2
with the initial condition
u(x, 0) = {0if 0≤x < L/2
sin (π(x−L/2)
L)if L/2 ≤x≤L
and boundary conditions u(0, t)=0and u(L, t)=0. Find the solution u(x, t)for 0≤x≤L
and t > 0.
Ans. Solution: We will solve this problem 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 this into the heat equation to get
X(x)T′(t) = c2X′′(x)T(t)
Divide by c2X(x)T(t)to obtain
T′(t)
c2T(t)=X′′(x)
X(x)=−λ
where λis a separation constant.
This gives two ODEs: 1) T′(t) + c2λT (t) = 0 with general solution T(t) = Ae−c2λt. 2)
X′′(x) + λX(x) = 0 with general solution X(x) = Bcos(√λx)+Csin(√λx).
2. Applying Boundary Conditions: From the boundary conditions u(0, t) = 0 and
u(L, t) = 0, we have X(0)T(t) = 0 and X(L)T(t) = 0.
This implies X(0) = 0 and X(L)=0. From X(0) = 0, we have B= 0. From X(L) = 0,
we have Csin(√λL)= 0.
For non-trivial solutions, we need sin(√λL)= 0, which gives √λL =nπ where nis a
positive integer.
Hence, λ=(nπ
L)2with corresponding eigenfunctions Xn(x) = Cnsin (nπx
L).
3. Superposition: The general solution is given by
u(x, t) =
∞
∑
n=1
Tn(t)Xn(x) =
∞
∑
n=1 (Ane−c2(nπ
L)2t)Cnsin (nπx
L)
4. Finding Coefficients: To find the coefficients An, we use the initial condition u(x, 0).
Given u(x, 0), we find the Fourier sine series representation of u(x, 0) and compare coefficients.
By matching coefficients, we can find An.
Therefore, the solution to the heat equation with the specified initial and boundary conditions
is
u(x, t) =
∞
∑
n=1
Ane−c2(nπ
L)2tsin (nπx
L)
26. Question 26:
Consider the following heat equation in one dimension:
∂u
∂t =k∂2u
∂x2,
subject to the initial condition:
u(x, 0) = 2x(1 −x),0< x < 1,
and the boundary conditions:
u(0, t) = 0, u(1, t) = 0.
Find the solution to the initial-boundary value problem described above.
Ans. Solution:
1. Using separation of variables:
Let’s assume a solution of the form u(x, t) = X(x)T(t). Substituting this into the heat
equation gives:
X(x)T′(t) = kX′′(x)T(t).
Dividing both sides by kX(x)T(t)gives:
T′(t)
kT (t)=X′′(x)
X(x).
Since the left side only depends on tand the right side only depends on x, they must both
be equal to a constant −λ. This gives two equations:
T′(t) + kλT (t) = 0 and X′′ (x) + λX(x) = 0.
2. Solving the time equation:
The solution to the time equation is:
T(t) = Aexp (−kλt).
3. Solving the spatial equation:
The solution to the spatial equation will depend on whether λ > 0, λ = 0,or λ < 0.
If λ > 0, the general solution is of the form:
X(x) = Bsin(√λx)+Ccos(√λx).
If λ= 0, the general solution is:
X(x) = Ax +D.
If λ < 0, the general solution is:
X(x) = Fsinh(√−λx)+Gcosh(√−λx).
4. Applying the boundary conditions:
Given the boundary conditions u(0, t)=0and u(1, t)=0, we can use them to find the
common values of λfor which these conditions hold.
Since u(0, t) = 0, the spatial solution must satisfy X(0) = 0.
Since u(1, t) = 0, the spatial solution must satisfy X(1) = 0.
5. Solving for λ:
By applying the boundary conditions X(0) = 0 and X(1) = 0, we find that the eigenvalues
λnare given by:
λn=(nπ
1)2=n2π2,
where n= 1,2,3, . . ..
6. Finding the complete solution:
The complete solution is given by:
u(x, t) =
∞
∑
n=1
(Bnsin(nπx) + Cncos(nπx)) exp (−k(nπ)2t).
7. Determining the coefficients:
Finally, we determine the coefficients Bnand Cnusing the initial condition u(x, 0) = 2x(1−x).
Since the initial condition u(x, 0) = 2x(1 −x)can be expressed as a Fourier sine series, we
can equate coefficients to find the values of Bnthat satisfy the initial condition.
Therefore, the solution to the initial-boundary value problem described by the heat equation
is:
u(x, t) =
∞
∑
n=1 (8
n3π3sin(nπx))exp (−k(nπ)2t).
27. Find the general solution to the one-dimensional heat equation
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions:
u(0, t) = u(1, t) = 0
and the initial condition:
u(x, 0) = f(x) = 1 −x2
where 0< x < 1,t > 0, and kis a constant.
Ans. Step 1. First, we need to find the general solution to the heat equation.
The general solution to the one-dimensional heat equation can be found using separation of
variables. Let u(x, t) = X(x)T(t). Substituting this into the heat equation gives:
X(x)T′(t) = kX′′(x)T(t)
Dividing both sides by kX(x)T(t)gives:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
Where we have introduced the separation constant λ(which could be positive, negative, or
zero).
Step 2. Now we solve the spatial part of the equation.
We have the ODE for X(x):
X′′(x) + λX(x) = 0
Applying the boundary conditions X(0) = X(1) = 0 gives the solution:
Xn(x) = sin(nπx)
where n= 1,2,3, ....
Step 3. Next, we solve the temporal part of the equation.
For the temporal part, we have the ODE for T(t):
T′(t) + kλT (t) = 0
which has the solution:
Tn(t) = e−kλt
Step 4. Finally, we combine the spatial and temporal solutions.
The general solution is then given by the infinite series:
u(x, t) =
∞
∑
n=1
Ansin(nπx)e−k(nπ)2t
where the coefficients Ancan be found using the initial condition. Substituting the initial
condition into the general solution gives:
1−x2=
∞
∑
n=1
Ansin(nπx)
To find the coefficients, we can use the orthogonality properties of sine functions.
The final solution is then:
u(x, t) =
∞
∑
n=1 (2
nπ )sin(nπx)e−k(nπ)2t
28. Question 28:
Consider the one-dimensional heat equation with periodic boundary conditions:
∂u
∂t =k∂2u
∂x2,0< x < L, t > 0
Subject to the initial condition:
u(x, 0) = x(L−x)
and the periodic boundary conditions:
u(0, t) = u(L, t)and ∂u
∂x(0, t) = ∂u
∂x(L, t)
where kand Lare positive constants.
Find the solution to the heat equation with the given initial and boundary conditions.
Ans. Solution:
1. Separation of Variables: Assume a solution of the form u(x, t) = X(x)T(t).
Substitute into the heat equation to get:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a constant.
2. Time Component: Solve the ODE for T(t):
T′(t) + kλT (t) = 0
The solution to this is T(t) = Ae−kλt.
3. Spatial Component: Solve the ODE for X(x):
X′′(x) + λX(x) = 0
The general solution to this ODE is:
X(x) = c1cos (√λx)+c2sin (√λx)
4. Applying Boundary Conditions:
Given the periodic boundary conditions, we have:
X(0) = c1=X(L) = c1cos(√λL)=c1
This implies that √λL must be an integer multiple of 2π, i.e., √λL =nπ for n∈Z+.
5. Finding Eigenvalues and Eigenfunctions:
Solve λ=(nπ
L)2for n= 1,2,3, ... to find the eigenvalues. The corresponding eigenfunctions
are:
Xn(x) = cos (nπx
L)for n= 1,2,3, ...
6. Forming the General Solution:
The general solution to the heat equation is:
u(x, t) =
∞
∑
n=1
Cne−k(nπ
L)2tcos (nπx
L)
7. Applying Initial Condition:
Substitute u(x, 0) = x(L−x)into the general solution and use orthogonality properties to
find the coefficients Cn.
The final solution is:
u(x, t) =
∞
∑
n=1
4L2(−1)n+1
n3π3e−k(nπ
L)2tsin (nπx
L)
29. Consider the heat equation in one dimension given by:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions:
u(0, t) = 0 and u(π, t) = 0
with the initial condition:
u(x, 0) = f(x) = sin(2x)
Determine the solution to this heat equation.
Ans. Solution: 1. We first need to find the general solution to the heat equation. Let
u(x, t) = X(x)·T(t).
2. Plugging u(x, t)into the heat equation gives:
X(x)·dT
dt =k(d2X
dx2)·T(t)
3. Dividing both sides by kX(x)T(t)gives:
1
k·1
T(t)·dT
dt =1
X(x)·d2X
dx2
4. Since the left side depends only on tand the right side depends only on x, they must be
equal to a constant −λ. This gives two equations:
1
k·1
T(t)·dT
dt =−λand 1
X(x)·d2X
dx2=−λ
5. Solving the first equation gives T(t) = e−λkt.
6. Solving the second equation gives X(x) = Acos(√λx)+Bsin(√λx).
7. Applying the boundary conditions:
u(0, t) = 0 =⇒A= 0
u(π, t) = 0 =⇒Bsin(√λπ)= 0 =⇒√λ=n, where nis a positive integer
8. Therefore, X(x) = Bsin(nx).
9. The general solution to the heat equation is:
u(x, t) =
∞
∑
n=1
Cne−kn2tsin(nx)
where Cnare constants to be determined from the initial condition.
10. Applying the initial condition u(x, 0) = sin(2x)gives:
sin(2x) =
∞
∑
n=1
Cnsin(nx)=⇒C2= 1,and Cn= 0 for n= 2
11. Therefore, the solution to the heat equation is:
u(x, t) = e−4kt sin(2x)
30. Let u(x, t)be a solution to the one-dimensional heat equation
∂u
∂t =k∂2u
∂x2
subject to the initial condition u(x, 0) = x(1 −x)for 0< x < 1, where kis a positive constant.
Determine the solution u(x, t)for t > 0.
Ans. We will solve the heat equation using separation of variables and Fourier series.
1. Let u(x, t) = X(x)T(t). Substituting this into the heat equation gives
T′(t)X(x) = kX′′(x)T(t)
Dividing by kX(x)T(t)yields
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a constant (which we will determine later).
2. Solving the ODE for X(x)gives
X′′ +λX = 0
The general solution is
X(x) = c1cos(√λx)+c2sin(√λx)
Applying the boundary conditions X(0) = X(1) = 0, we obtain
c1= 0 and √λ=nπ
for positive integer n. Thus, Xn(x) = cnsin(nπx)for n= 1,2,3, . . ..
3. Solving the ODE for T(t)gives
T′(t) + kλT (t) = 0
The solution is
T(t) = ce−kλt =ce−n2π2kt
for n= 1,2,3, . . ..
4. By the principle of superposition, the general solution is
u(x, t) =
∞
∑
n=1
cne−n2π2kt sin(nπx)
Finally, applying the initial condition u(x, 0) = x(1 −x)gives
u(x, 0) =
∞
∑
n=1
cnsin(nπx) = x(1 −x)
Multiplying both sides by sin(mπx)and integrating over [0,1] yields
cm=∫1
0
x(1 −x)sin(mπx)dx
Thus, the solution is
u(x, t) =
∞
∑
n=1 (∫1
0
x(1 −x)sin(nπx)dx)e−n2π2kt sin(nπx)
