University of Alberta
Math 300 Q1, Advanced Boundary Value Problems I, Winter 2012
Solution of Assignment # 5,
Due time: 5:00 pm, Wednesday April 4, 2012
1. Exercises 8.2: 8.2.1 (a)+(b):
Consider the heat equation with time-independent sources and boundary conditions:
∂u
∂2u
= k 2 + Q(x)
∂t
∂x
u(x, 0) = f (x)
if an equilibrium exists. Analyze the limits as t −→ ∞. If no equilibrium exists, explain why and
reduce the problem to one with homogeneous boundary conditions (but do not solve). Assume
∂u
∂x (L, t)
(a) Q(x) = 0, u(0, t) = A,
(b) Q(x) = 0,
∂u
∂x (0, t)
= 0,
= B,
∂u
∂x (L, t)
= B 6= 0.
Solution. (a) Since Q(x) = 0, u(0, t) = A and
∂u
∂x (L, t)
= B, the equilibrium solution satisfies
d2
uE (x) = 0
dx2
with boundary conditions uE (0) = A and u0E (L) = B. So, UE (x) = A + Bx. Let v(x, t) = u(x, t) −
uE (x). Then we have
∂v
∂2v
=k 2
∂t
∂x
with v(0, t) = 0,
∂v
∂x (L, t)
= 0 and v(x, 0) = f (x) − uE (x). Let v(x, t) = φ(x)G(t). We have
with φ(0) = 0 and φ0 (L) = 0. Thus, λn =
So, Gn (t) = e−λn kt = e−(
(n−1/2)φ
L
)
2
kt
v(x, t) =
d2 φ
= −λφ
dx2
2
(n−1/2)π
L
and φn = sin (n−1/2)πx
for n = 1, 2, 3, · · · .
L
. Thus,
∞
X
n=1
an sin
2
(n − 1/2)πx −( (n−1/2)φ
) kt
L
e
L
if initially
v(x, 0) = f (x) − uE (x) =
∞
X
n=1
1
an sin
(n − 1/2)πx
.
L
Then we have
2
an =
L
Z
L
(f (x) − uE (x)) sin
0
(n − 1/2)πx
dx.
L
Since v(x, t) −→ 0 as t −→ ∞,
u(x, t) = v(x, t) + uE (x) −→ uE (x)
as t −→ ∞.
(b) Assume that uE (x) is an equilibrium solution of
∂u
∂2u
=k 2
∂t
∂x
with
∂u
∂x (0, t)
= 0 and
∂u
∂x (L, t)
= B 6= 0. Then the general solution can be written as uE (x) = ax + b
and u0E (x) = a. The condition
∂u
∂x (0, t)
= 0 implies a = 0 but
∂u
∂x (L, t)
= B 6= 0 requires a 6= 0.
This is a contradiction. Thus, no equilibrium solution exists for this problem. We can understant
this situation in the following way. Q = 0 means that there is no heat nenergy nenerated inside,
∂u
∂x (0, t)
= 0 tells us that there is no energy lost at the left endpoint and
∂u
∂x (L, t)
= B 6= 0 means
there is energy lost at the right endpoint. Thus the temperature can not reach a stady state.
We need to find a function r(x, t) which satisfies the boundary conditions
∂r
∂x (L, t)
= B 6= 0. We any choose r(x, t) =
Bx2
2L .
∂r
∂x (0, t)
Define v(x, t) = u(x, t) − r(x, t) = u(x, t) −
Then
∂v
∂u
∂2u
∂2
=
=k 2 =k 2
∂t
∂t
∂x
∂x
Bx2
u−
2L
∂2
+k 2
∂x
Bx2
2L
=k
kB
∂2v
+
.
∂x2
L
Thus
∂v
∂2v
kB
=k 2 +
∂t
∂x
L
with
∂v
∂x (0, t)
= 0 and
∂v
∂x (L, t)
= 0 and
= 0.
2. Exercises 8.2: 8.2.2 (a)+(b):
Consider the heat equation with time-dependent sources and boundary conditions:
∂u
∂2u
= k 2 + Q(x, t)
∂t
∂x
u(x, 0) = f (x).
2
Bx2
2L .
Reduce the problem to one with homogeneous boundary conditions if
(a)
∂u
∂x (0, t)
= A(t), and
(b) u(0, t) = A(t), and
∂u
∂x (L, t)
∂u
∂x (L, t)
= B(t),
= B(t).
Solution. (a) Let r(x, t) = A(t)x + B(t)−A(t)
x2 . Then
2L
∂r
∂x (0, t)
= A(t) and
∂r
∂x (L, t)
= B(t). Define
v(x, t) = u(x, t) − r(x, t). Then we have
∂v
∂t
∂u ∂r
−
∂t
∂t
∂2u
∂2r
∂2r
∂r
k 2 − k 2 + Q(x, t) + k 2 −
∂x
∂x
∂x
∂t
∂2v
B(t) − A(t)
B 0 (t) − A0 (t) 2
k 2 + Q(x, t) + k
− [A0 (t)x +
x ].
∂x
L
2L
=
=
=
Thus
∂2v
B(t) − A(t)
B 0 (t) − A0 (t) 2
∂v
= k 2 + Q(x, t) + k
− [A0 (t)x +
x ]
∂t
∂x
L
2L
with
∂v
∂x (0, t)
= 0 and
∂v
∂x (L, t)
= 0.
(b) Let r(x, t) = A(t) + B(t)x. Then r(0, t) = A(t) and
∂r
∂x (L, t)
= B(t). Let v(x, t) = u(x, t) −
r(x, t). Then we have
∂2v
∂2v
∂v
∂2r
∂r
= k 2 + Q(x, t) + k 2 −
= k 2 + Q(x, t) − (A0 (t) + B 0 (t)x)
∂t
∂x
∂x
∂t
∂x
with v(0, t) = 0 and
∂v
∂x (L, t)
= 0.
3. Exercises 8.3: 8.3.2:
Consider the heat equation with a steady source
∂u
∂2u
= k 2 + Q(x)
∂t
∂x
subject to the initial and boundary conditions described in this section:
u(0, t) = 0, u(L, t) = 0,
and u(x, 0) = f (x).
Obtain the solution by the method of eigenfunction expansion. Show that the solution approaches
a steady-state solution.
Solution. The corresponding homogeneous problem
ut = kuxx
3
nπ 2
L
with u(0, t) = u(L, t) = 0 has eigenvalues λn =
and eigenfunctions φn = sin nπx
L .
Assume the solution of the nonhomogeneous problem u(x, t) has the form
∞
X
u(x, t) =
an (t) sin
n=1
nπx
.
L
We will need the same series expansion of Q(x) and f (x) :
∞
X
Q(x) =
qn sin
nπx
L
bn sin
nπx
.
L
n=1
∞
X
f (x) =
n=1
So,
2
L
Z
2
L
Z
qn =
bn =
L
Q(x) sin
nπx
dx
L
f (x) sin
nπx
dx.
L
0
L
0
Substituting the series form of u(x, t) into the PDE, we have
a0n (t) = −k
nπ 2
L
an (t) + qn
with an (0) = bn . So,
Z
2
an (t) = bn e
−k( nπ
L ) t
t
+ qn
e−k(
nπ
L
2
)
(t−s)
qn
ds =
nπ 2
L
k
0
+ e−k(
nπ
L
qn
2
) t (b −
n
k
)
nπ 2
L
as t −→ ∞.
Thus
∞
X
u(x, t) =
n=1
qn
k
−k
+e (
nπ 2
nπ
L
qn
2
) t (b −
n
L
k
!
) sin
nπ 2
L
converges to
r(x) :=
∞
X
n=1
qn
k
sin
nπ 2
L
nπx L
as t −→ ∞. Here r(x) is the steady state.
4. Exercises 10.3: 10.3.1:
Show that the Fourier transform is linear operator; that is, show that
(a) F [c1 f (x) + c2 g(x)] = c1 F (ω) + c2 G(ω),
(b) F [f (x)g(x)] 6= F (ω)G(ω).
4
nπx L
qn
−→
k
nπ 2
L
Solution. (a)
F (c1 f (x) + c2 g(x))(ω)
=
=
=
Z ∞
1
(c1 f (x) + c2 g(x))e−ωx dx
2π −∞
Z ∞
Z ∞
1
1
c1 f (x)e−ωx dx +
c2 g(x)e−ωx dx
2π −∞
2π −∞
c1 F (f (x))(ω) + c2 F (g)(ω).
(b) It follows from F −1 (F (ω)G(ω))(x) =
1
2π f
∗ g(x) 6= f (x)g(x) that F [f (x)g(x)] 6= F (ω)G(ω).
5. Exercises 10.3: 10.3.5
If F (ω) is the Fourier transform of f (x), show that the inverse Fourier transform of eiωβ F (ω) is
f (x − β). This result is known as the shift theorem for Fourier transforms.
Proof.
F −1 (eiωβ F (ω))
Z
∞
F (ω)eiωβ e−iωx dω
=
−∞
Z ∞
=
F (ω)e−iω(x−β) dω
−∞
=
f (x − β).
6. Exercises 10.4: 10.4.4 (a)+(b):
(a) Solve
∂2u
∂u
= k 2 − γu,
∂t
∂x
−∞ < x < ∞
u(x, 0) = f (x).
(b) Does your solution suggest a simplifying transformation?
Solution. (a) Let U (ω, t) = F (u(x, t))(ω). Then the PDE implies
U 0 (ω, t) = −kω 2 U (ω.t) − γU (ω, t) = −(kω 2 + γ)U (ω, t)
which has solution
U (ω, t) = F (f (x))(ω)e−(kω
2
+γ)t
.
So,
u(x, t)
= F −1 (F (f (x))(ω)e−(kω
2
+γ)t
2
)
= e−γt F −1 (F (f (x))(ω)e−kω t )
Z ∞
(x−y)2
1
−γt
√
= e
f (y)e− 4kt dy.
4πkt −∞
5
(b) It follows from the previous formula of u(x, t) that
Z ∞
(x−y)2
1
eγt u(x, t) = √
f (y)e− 4kt dy
4πkt −∞
which is a solution of the heat equation without the term γu. This leads us to introduce the transform
v(x, t) = eγt u(x, t). Then we have,
∂v
∂t
= γeγt u + eγt ut
= γeγt u + eγt (kuxx − γu)
= keγt uxx
= kvxx
with initial condition v(x, 0) = u(x, 0) = f (x).
7. Exercises 10.5: 10.5.12:
Solve
∂2u
∂u
= k 2 (x > 0)
∂t
∂x
∂u
(0, t) = 0,
∂x
u(x, 0) = f (x).
Solution. Since we have a Neumann boundary conditions, we can apply the Fourier cosine
transform. So, let U (ω, t) = C(u(x, t))(ω). Then C(ut (x, t))(ω) = Ut (ω, t) and C(uxx (x, t)) =
− π2 ux (0, t) − ω 2 U (ω, t) = −ω 2 U (ω, t) since ux (0, t) = 0. So,
ut (ω, t) = −kω 2 U (ω, t).
Thus,
U (ω, t) = c(ω)e−kω
2
π
R∞
2
t
f (x) cos ωxdx which is even in ω. Then we have
Z ∞
2
u(x, t) =
c(ω)e−kω t cos(ωx)dω
0
Z
2
1 ∞
=
c(ω)e−kω t (cos(ωx) − i sin(ωx)dω
2 −∞
Z ∞
2
1
=
c(ω)e−kω t e−iωx dω
2
−∞
2
1
= F −1 ( c(ω)e−kω t ).
2
with c(ω) = C(f (x))(ω) =
0
6
Note that if we use feven to denote the even extension of f , then we have
1
c(ω)
2
=
=
=
=
Z
1 ∞
f (x) cos ωxdx
π 0
Z ∞
1
feven (x)(cos(ωx) + i sin(ωx))dx
2π −∞
Z ∞
1
feven (x)eiωx dx
2π −∞
F (feven (x))(ω).
Thus u(x, t) is a solution of the heat equation ut = kuxx for −∞ < x < ∞ with initial data
u(x, 0) = feven (x) the even extension of f. So,
u(x, t)
= feven ∗ G(x)
Z ∞
(x−y)2
1
= √
feven (y)e− 4kt dy
4πkt −∞
Z 0
Z ∞
(x−y)2
(x−y)2
1
= √
feven (y)e− 4kt dy +
f (y)e− 4kt dy
4πkt
−∞
0
Z ∞
Z ∞
(x−y)2
(x+y)2
1
− 4kt
dy +
= √
f (y)e− 4kt dy
f (y)e
4πkt
0
0
Z ∞
(x+y)2
(x−y)2
1
− 4kt
− 4kt
= √
f (y) e
+e
dy.
4πkt 0
7