ECE 205  Winter 2017
Assignment 8 solutions
Due 02042017
Note:
Please submit the full solutions to all questions.
In Questions 1 and 2, we consider the heat equation in one spatial dimension, for the
temperature,
U
, of a rod of length
L
:
U
t
=
α
2
U
xx
,
where
α
2
is the thermal diffusivity.
1: Heat equation
(
I
) Consider the initial condition
U
(
x,
0) =
100
x
L
.
a) (
i
) Solve the PDE with the boundary conditions:
U
(0
, t
) =
U
(
L, t
) = 0
,
(
t >
0)
.
(
ii
) What is the equilibrium state for this boundary value problem, i.e. solve the PDE
U
t
=
α
2
U
xx
,
U
t
= 0,
U
(0
, t
) =
U
(
L, t
) = 0,
t >
0.
(
iii
) Is your solution from a (
i
) asymptotic to your solution to a (
ii
) as
t
→ ∞
?
b) (
i
) Solve the PDE with the boundary conditions:
U
x
(0
, t
) =
U
x
(
L, t
) = 0
,
(
t >
0)
.
(
ii
) What is the equilibrium state for this Boundary value problem, i.e. solve the PDE:
U
t
=
α
2
U
xx
,
U
t
= 0,
U
x
(0
, t
) =
U
x
(
L, t
) = 0
.
(
t >
0).
(
iii
) Is your solution from b (
i
) asymptotic to your solution to b (
ii
) as
t
→ ∞
?
(
II
) a) Consider now the general Boundary value problem:
U
t
=
α
2
U
xx
,
U
(0
, t
) =
U
(
L, t
) = 0
,
t >
0
.
Let us consider the mean square temperature:
M
(
t
) =
1
L
L
Z
0
U
2
(
x, t
)
dx.
Use the Leibniz rule for differentiating under the integral and integration by parts to show that
this function is not increasing.
b) Consider now the general boundary value problem:
U
t
=
α
2
U
xx
,
U
x
(0
, t
) =
U
x
(
L, t
) = 0
,
t >
0
.
Let us consider the total energy:
E
(
t
) =
k
L
Z
0
U
(
x, t
)
dx,
for some positive constant
k
. Use the Leibniz rule for differentiating under the integral and
integration by parts to show that the function is constant, and explain your answer.
1
ECE 205  Winter 2017
Assignment 8 solutions
Due 02042017
Solution
(
I
) We try
U
(
x, t
) =
X
(
x
)
T
(
t
) and we get:
˙
T
α
2
T
=
X
00
X
=
k,
where
K
is a constant.
X
(0) =
X
(
L
) = 0
→
k
=

λ
2
for a non trivial solution. Therefore
X
(
x
) =
A
sin(
λx
) and
X
(
L
) = 0 implies
λ
=
nπ
L
for
n
∈
Z
. A solution is
X
n
(
x
) =
A
n
sin
(
nπx
L
)
.
Then
˙
T
α
2
T
=

nπ
L
2
→
T
n
=
c
n
e

nπα
L
2
t
.
We absorb
c
n
into
A
n
and we get:
U
n
(
x, t
)
A
n
e

β
2
n
t
sin
nπx
L
,
where
β
n
=
nπα
L
. The general solution is a sum of such terms due to the linearity so,
U
(
x, t
) =
n
=
∞
X
n
=1
A
n
e

β
2
n
t
sin(
λ
n
x
)
,
where
λ
n
=
nπ
L
and
β
n
=
αλ
n
. Note that we can combine the solutions with negative integer
values with those of positive integer values.
We want
U
(
x,
0) =
100
x
L
, then
n
=
∞
X
n
=1
A
n
sin(
λ
n
x
) =
100
x
L
⇐⇒
A
n
=
2
L
L
Z
0
100
x
L
sin
nπx
L
dx
=

200
nπL
[
L
cos(
nπ
)] =
200
nπ
(

1)
n
+1
.
The above integration is performed by parts. Thus,
U
(
x, t
) =
n
=
∞
X
n
=1
200
nπ
(

1)
n
+1
e

β
2
n
t
sin(
λ
n
x
)
.
(
ii
) If we solve the PDE with
U
t
= 0 we get,
U
xx
= 0. Therefore,
U
(
x, t
) =
c
1
x
+
c
2
.
With
U
(0
, t
) =
U
(
L, t
) = 0, we get
c
1
=
c
2
= 0. Therefore, the equilibrium solution is
U
e
(
x, t
) = 0
.
(
iii
) We have
lim
t
→∞
e

β
2
n
t
= 0
.
Thus for our general solution from (
I
), all terms die away and
lim
t
→∞
U
(
x, t
) = 0 =
U
e
(
x, t
)
.
The general solution tends to the equilibrium solution.
2
ECE 205  Winter 2017
Assignment 8 solutions
Due 02042017
b) As before, we have
U
(
x, t
) =
X
(
x
)
T
(
t
) with
˙
T
α
2
T
=
X
00
X
=
k,
where
k
is a constant. Now we also have
X
0
(0) =
X
0
(
L
) = 0. Therefore, if
k
= 0 then
X
=
c
1
is a
solution. If
k
=
λ
2
>
0, then
X
(
x
) =
c
2
e
λx
+
c
3
e

λx
is a solution. But
X
0
(0) =
X
0
(
L
) = 0 implies
c
2
=
c
3
= 0. If
k
=
λ
2
<
0, then we have
X
(
x
) =
c
4
sin(
λx
) +
c
5
cos(
λx
). The condition
X
0
(0) = 0
gives
c
4
= 0. The condition
X
0
(
L
) = 0 implies
λ
=
λ
n
=
nπ
L
where
n
∈
Z
. Thus
X
n
(
x
) =
d
n