Homework 7 Solutions
#1 (Section 10.4): The following functions are defined on an interval of length L. Sketch the even and
odd extensions of each function over the interval [−L, L].
1
(a)
f (x)
f (x) = 1 − x2 ,
0≤x≤1
0
0
1
x
Even extension of f(x)
Odd extension of f(x)
1
1
f (x)
o
f (x)
e
0
0
-1
0
x
1
-1
0
x
1
2
(b)
f (x)
f (x) =

 x, 0 ≤ x < 2

1,
1
2≤x≤3
0
0
1
2
3
x
Even extension of f(x)
Odd extension of f(x)
2
2
f (x)
e
1
f (x)
o
1
0
-1
0
-2
-3
-2
-1
0
x
1
2
3
-3
-2
-1
0
x
1
2
3
#2 (Section 10.4): The following functions are defined on an interval of length L. Find the Fourier Sine
series and the Fourier Cosine series for each and sketch the functions to which the
series converge over the interval [−3L, 3L].
1
(a)
f (x)
f (x) = 1,
0≤x≤1
0
0
1
x
Fourier Cosine series
Even extension of f(x)
1
L=1
fe (x)
bn = 0, n = 1, 2, . . .
0
-1
0
x
1
The Fourier coefficients are calculated as follows.
2
=
L
a0
Z
L
0
2
f (x) dx =
1
Z
1
1 dx = 2
0
a0 = 2
an
=
=
2
L
Z
0
L
f (x) cos
nπx L
dx =
2
1
Z
1
1 · cos(nπx) dx
0
1
2
2
sin(nπx) =
[sin(nπ) − sin 0] = 0
nπ
nπ
0
an = 0
n = 1, 2, . . .
Then,
∞
f (x) = FC (x)
=
nπx a0 X
+
an cos
2
L
n=1
=
∞
X
2
+
0 · cos(nπx)
2
n=1
f (x) = FC (x) = 1
for 0 < x < L
0<x<1
Fourier Cosine series representation of f(x)
1
FC (x)
0
-3
-2
-1
0
x
1
2
3
Fourier Sine series
Odd extension of f(x)
1
f (x)
o
L=1
0
an = 0, n = 0, 1, 2, . . .
-1
-1
0
x
1
The Fourier coefficients are calculated as follows.
Z
Z
nπx 2 L
2 1
bn =
f (x) sin
dx =
1 · sin(nπx) dx
L 0
L
1 0
1
2
2
2
= −
cos(nπx) = −
[cos(nπ) − cos 0] =
[1 − (−1)n ]
nπ
nπ
nπ
0
bn =
2 1 − (−1)n
·
π
n
n = 1, 2, . . .
Then,
f (x) = FS (x) =
∞
X
bn sin
n=1
f (x) = FS (x) =
nπx L
for 0 < x < L
∞
2X
1 − (−1)n
sin(nπx)
π n=1
n
0<x<1
Fourier Sine series representation of f(x)
1
FS (x)
0
-1
-3
-2
-1
0
x
1
2
3
(b)
1
f (x) =

 1,
0≤x<π
f (x)
0, π ≤ x ≤ 2π

0
0
pi
x
2*pi
Fourier Cosine series
Even extension of f(x)
1
L = 2π
fe (x)
bn = 0, n = 1, 2, . . .
0
-2*pi
-pi
0
x
pi
2*pi
The Fourier coefficients are calculated as follows.
a0
=
2
L
Z
L
f (x) dx =
0
2
2π
Z
2π
f (x) dx =
0
1
π
Z
π
0
1 dx +
Z
2π
π
1
(π + 0) = 1
0 dx =
π
a0 = 1
an
=
2
L
=
1
π
=
1
π
Z 2π
nx 2
f (x) cos
dx =
f (x) cos
dx
L
2π 0
2
0
Z π
Z 2π
nx nx dx +
0 · cos
dx
1 · cos
2
2
π
0
i
nx π
nπ 2
2
2 h nπ sin
−
sin
0
=
sin
sin
+
0
=
n
2 0
nπ
2
nπ
2
Z
L
nπx an
2
nπ
=
sin
nπ
2
n = 1, 2, . . .
Then,
∞
f (x) = FC (x)
=
nπx a0 X
+
an cos
2
L
n=1
for 0 < x < L
∞
2X
1
1
nπ
nx
+
sin
f (x) = FC (x) =
cos
2
π n=1 n
2
2
0 < x < 2π
Fourier Cosine series representation of f(x)
1
FC (x)
0
-6*pi
-4*pi
-2*pi
0
x
2*pi
4*pi
6*pi
Fourier Sine series
Odd extension of f(x)
1
f (x)
o
L = 2π
0
an = 0, n = 0, 1, 2, . . .
-1
-2*pi
-pi
0
x
pi
2*pi
The Fourier coefficients are calculated as follows.
Z
Z 2π
nπx nx 2 L
2
bn =
f (x) sin
dx =
f (x) sin
dx
L 0
L
2π 0
2
Z π
Z 2π
nx nx 1
0 · sin
1 · sin
dx +
dx
=
π 0
2
2
π
i
nπ i
nx π
2
2 h
1
2 h nπ =
− cos
cos
−
cos
0
=
1
−
cos
+
0
=
−
π
n
2 0
nπ
2
nπ
2
bn =
nπ
2
1 − cos
nπ
2
n = 1, 2, . . .
Then,
f (x) = FS (x) =
∞
X
bn sin
n=1
f (x) = FS (x) =
nπx L
for 0 < x < L
∞
nπ
2X
1
1 − cos
π n=1 n
2
sin
nx
2
0<x<1
Fourier Sine series representation of f(x)
1
F
S
(x)
0
-1
-6*pi
-4*pi
-2*pi
0
x
2*pi
4*pi
6*pi
(7th edition: pages 579–580 # 7,9)
#3 Section 10.5: (page 610) # 7, 9
Problems 7 and 9 both involve solving a heat conduction problem of the form
ut = α2 uxx ,
u(0, t) = 0,
0 < x < L,
u(L, t) = 0,
u(x, 0) = f (x),
t>0
(1)
t>0
(2)
0 ≤ x ≤ L,
(3)
where α, L, and f (x) are given. It is efficient to first solve the general case, and then substitute the
specific quantities given in problems 7 and 9.
Separation of Variable Method:
Let u(x, t) = X(x) T (t).
Plugging this expression for u into the PDE (1) gives
X(x) T 0 (t) = α2 X 00 (x)T (t).
Divide by α2 X(x) T (t) to get
X 00 (x)
1 T 0 (t)
·
=
.
α2 T (t)
X(x)
Since the left hand side depends on t alone and the right hand side depends on x alone, they can be
equal only when both are equal to the same constant, call it −λ.
X 00 (x)
1 T 0 (t)
·
=
= −λ.
α2 T (t)
X(x)
This expression “separates” into two equations: one for T (t) and one for X(x),
T 0 + α2 λ T = 0,
X 00 + λ X = 0.
(4)
We can use (2) to determine boundary conditions for X(x). At x = 0,
u(0, t) = X(0) T (t) = 0,
Since this equality holds for all t, it must be that
X(0) = 0.
(5)
At x = L,
u(L, t) = X(L) T (t) = 0,
and by the same argument we have
X(L) = 0.
(6)
Equations (4)–(6) suggest the following eigenvalue problem for X(x).
X 00 + λ X = 0
X(0) = 0,
X(L) = 0.
We solved this eigenvalue problem in Section 10.1 (see lecture notes) and found the following eigenvalues
and eigenfunctions.
2
λn = nπ
L
n = 1, 2, . . .
Xn (x) = sin
nπx
L
For each λn , equation (4) can be used to determine the corresponding solution Tn (t),
Tn0 + α2 λn Tn = 0.
The is the familiar first order exponential ODE with a decay rate of α2 λn . It is both separable and
linear so it can be solved by either the separable variable technique or the integrating factor method.
In either case, we get
2
αnπ 2
Tn (t) = Cn e−α λn t = Cn e−( L ) t .
We have found infinitely many solutions of the PDE which are given by
un (x, t) = Xn (x) Tn (t) =
e−(
αnπ
L
)2 t sin
nπx ,
L
n = 1, 2. . . . .
It is easy to check that each solution satisfies the PDE (1) and the boundary conditions (2). The last
step is to satisfy initial condition (3). Note that any linear combination of the un ’s also satisfies the
PDE and the boundary conditions (principle of superposition). We can attempt to satisfy the initial
conditions by considering the linear combination of all the un ’s given by
∞
X
u(x, t) =
Bn e−(
αnπ
L
n=1
2
) t sin nπx
L
.
(7)
Applying the initial condition (3) yields
u(x, 0) =
∞
X
n=1
Bn sin
nπx
L
= f (x).
By comparing this equation to the Fourier Sine series of f (x) on the interval (0, L), it can be seen that
the Bn ’s are the Fourier Sine series coefficient, and therefore
Bn =
2ZL
dx .
f (x) sin nπx
L
L 0
(8)
Then the solution given by equations (7) and (8) satisfies the PDE, the boundary conditions, and the
initial conditions.
Now equations (7) and (8) can be used to solve problems 7 and 9.
#7
In this problem, equations (1)–(3) are given with
L = 1,
α = 10,
f (x) = sin(2πx) − sin(5πx).
Then by equation (7), the solution is
u(x, t) =
∞
X
2
Bn e−(10nπ) t sin(nπx),
n=1
where the Bn ’s are the coefficients of the Fourier Sine series of f(x) on the interval (0,1). Though
the Bn ’s can be calculated using equation (8), the initial condition has a special form and it is
more convenient to apply the initial condition directly to get
u(x, 0) =
∞
X
Bn sin(nπx) = sin(2πx) − sin(5πx)
n=1
The Bn ’s can be determined by equating the coefficients of like terms (sin(nπx), n = 1, 2, ...),
which yields
B2 = 1,
B5 = −1,
and Bn = 0 otherwise.
Therefore all terms of the series are zero except for the second (n = 2) and fifth (n = 5) terms,
so the solution is
2
2
u(x, t) = e−(20π) t sin(2πx) − e−(50π) t sin(5πx) .
#9
In this problem, equations (1)–(3) are given with
L = 40,
α = 1,
f (x) = 50.
Then by equation (7), the solution is
u(x, t) =
∞
X
n2 π 2
Bn e− 1600 t sin
n=1
nπx
40 ,
(9)
where the Bn ’s are the coefficients of the Fourier Sine series of f(x) on the interval (0,40). The
Bn ’s are determined from equation (8).
Bn
=
=
2
L
Z
L
f (x) sin
0
nπx
L
dx =
2
40
Z
0
40
50 sin
nπx
40
dx
40
5 −40
100
100
nπx ·
cos 40 = −
[cos(nπ) − cos 0] =
[1 − (−1)n ]
2 nπ
nπ
nπ
0
100
[1 − (−1)n ] , n = 1, 2, . . .
nπ
The solution is given by equations (9) and (10) and can be written as
Bn =
u(x, t) =
∞
100 X
1 − (−1)n − n2 π2 t
1600
sin nπx
.
e
40
π n=1
n
(10)
#4 Section 10.6: Consider the heat flow problem
ut = uxx ,
u(0, t) = 4,
0 < x < 2,
u(2, t) = 0,
u(x, 0) = −2x,
t>0
(11)
t>0
(12)
0 ≤ x ≤ 2.
(13)
(a) Find the steady state (time independent) solution, call it v(x), by setting ut = 0 in the PDE and
solving the resulting boundary value problem.
With ut = 0 and u(x, t) = v(x), equations (11) and (12) reduce to
vxx = 0,
v(0) = 4,
0 < x < 2,
v(2) = 0.
(14)
(15)
The general solution of equation (14) is v(x) = C1 +C2 x. The boundary conditions (15) determine
the constants to be C1 = 4 and C2 = −2, and therefore
v(x) = 2 (2 − x) .
(b) Let w(x, t) = u(x, t) − v(x), where v(x) is the steady state solution found in part (a). Find the
resulting PDE, boundary conditions, and initial condition for w(x, t).
The temperature u(x, t) can be written as u(x, t) = w(x, t) + v(x) and substituted into equation
(11) to give
wt + vt = wxx + vxx ,
and since vt = vxx = 0, we get
wt = wxx .
Equation (12) determines the boundary conditions for w.
w(0, t) = u(0, t) − v(0) = 4 − 4 = 0
w(0, t) = 0
w(2, t) = u(2, t) − v(2) = 2 − 2 = 0
w(2, t) = 0
Equation (13) determines the initial condition for w.
w(x, 0) = u(x, 0) − v(x) = −2x − 2 (2 − x) = −4
w(x, 0) = −4
(c) Let w(x, t) = X(x) T (t) and find the separated equations for X(x) and T (t). Solve the eigenvalue
problem for X(x), and then determine T (t).
Note that w satisfies equations (1)–(3) with
L = 2,
α = 1,
f (x) = −4.
The separated equations were already determined and solved above. By equation (4), the separated equations are
T 0 + λ T = 0,
X 00 + λ X = 0,
and the boundary conditions for X(x) are
X(0) = 0,
X(L) = 0.
The eigenvalues and eigenfunctions are given by
λn =
nπ
2
Xn (x) = sin
2
n = 1, 2, . . . ,
nπx
2
and the corresponding solutions for T (t) were found to be
Tn (t) = Bn e−
n2 π 2
4
t
n = 1, , 2 . . . .
,
(d) Complete the solution by solving the intial-boundary value problem for w(x, t) and writing the
solution for u(x, t).
It follows from equations (7) and (8) that
w(x, t) =
∞
X
Bn e−
n2 π 2
t
4
n=1
sin nπx
2
and
Bn
=
=
2
L
Z
0
L
f (x) sin
nπx
L
dx =
2
2
Z
0
2
−4 sin
nπx
2
dx
2
8
8
−2
nπx cos 2 =
[cos(nπ) − cos 0] = −
[1 − (−1)n ]
−4 ·
nπ
nπ
nπ
0
Bn = −
8
[1 − (−1)n ] ,
nπ
n = 1, 2, . . . .
Therefore,
∞
1 − (−1)n − n2 π2 t nπx 8X
w(x, t) = −
e 4 sin 2 ,
π n=1
n
and since u(x, t) = w(x, t) + v(x), the solution is
u(x, t) = 2 (2 − x) −
∞
1 − (−1)n − n2 π2 t nπx 8X
e 4 sin 2 .
π n=1
n
#5 Section 10.6: Suppose that a metal rod of unit length is completely insulated and that the temperature u(x, t) of the rod satisfies
ut = 3 uxx,
0 < x < 1,
ux (0, t) = ux (1, t) = 0,
t>0
u(x, 0) = f (x)
where f (x) is the initial temperature of the rod.
(a) Let u(x, t) = X(x) T (t). Find the separated equations, solve the eigenvalue problem for X(x)
(don’t forget to check λ = 0), and determine T (t).
The equations are separated as before.
T 0 + 3 λT = 0
X 00 + λ X = 0
The boundary condition for X(x) are found as follows.
ux (0, t) = X 0 (0) T (t) = 0
X 0 (0) = 0
ux (1, t) = X 0 (1) T (t) = 0
X 0 (1) = 0
The eigenvalue problem for X(x) was solved on homework 6 (problem #2(c) with L = 1). The
eigenvalues and eigenfunctions were found to be
λn = (nπ)2
n = 0, 1, 2, . . . ,
Xn (x) = cos(nπx)
The corresponding solutions for T (t) are found by solving
Tn0 + 3λn Tn = 0,
which gives
Tn (t) = Cn e−3λn t = Cn e−3n
2 π2 t
, n = 0, 1, 2, . . ..
(b) Find the general solution and show that u(x, t) approaches a constant as t → ∞. How is the
constant related to the initial temperature f (x)?
We have infinitely many solutions given by
un (x, t) = Xn (x) Tn (t) =
2
2
e−3 n π t cos(nπx),
n = 0, 1, 2. . . . .
The initial conditions can be satisfied by considering the following linear combination of the un ’s.
u(x, t) =
∞
X
A0
2 2
An e−3n π t cos(nπx)
+
2
n=1
Applying the initial condition yields
∞
X
A0
An cos(nπx) = f (x)
+
2
n=1
u(x, t) =
By comparing this equation to the Fourier Cosine series of f (x) on the interval (0, 1), it can be
seen that the An ’s are the Fourier Cosine series coefficient, and therefore
An = 2
Z
0
1
f (x) cos(nπx) dx , n = 0, 1, 2, . . .
As t → ∞, all terms of the series approach zero except for the constant term
A0
2 ,
so that
A0
as t → ∞ .
2
u(x, t) →
Recall from calculus that the integral
1
A0
=
2
L
Z
L
f (x) dx
0
gives the average value of f (x) over the interval (0,L). It makes sense that the temperature in
the rod approaches this value as t → ∞. Since the rod is completely insulated on all sides, it
neither gains nor loses heat, so the total heat in the rod is always equal to the initial amount. As
t → ∞, the heat diffuses through the rod until it is uniformly distributed and thermal equilibrium
(constant temperature) is attained. Since the rod is insulated, this constant temperature must be
the average of the initial temperature distribution f (x).