So, the ODE becomes
(
L
2

L
2
cos
2
θ
)

cos
θ
L
3
sin
3
θ
dy
d
θ
+
1
L
2
sin
2
θ
d
2
y
d
θ
2


1
L
sin
θ
dy
d
θ
L
cos
θ
+
n
2
y
=
0, 0
< θ < π

cos
θ
sin
θ
dy
θ
+
d
2
y
dx
2
+
cos
θ
sin
θ
dy
θ
+
n
2
y
=
0.
Thus
d
2
y
dx
2
+
n
2
y
= 0, 0
< θ < π
.
(10 Marks)
b.
The equation
d
2
y
dx
2
+
n
2
y
= 0, 0
< θ < π
, has eigenfunctions,
y
1
(
x
) =
y
1
(
L
cos
θ
) = cos
n
θ
and
y
2
(
x
) =
y
2
(
L
sin
θ
) = sin
n
θ
(explain).
(5 Marks)
c.
dy
2
dx
(
x
) =
d
θ
dx
dy
d
θ
=

1
L
sin
θ
cos
θ
. When
x
=
L
=
⇒
θ
= 0. Thus
dy
2
dx
(
x
) =
1
0
which is undefined.
Hence,
y
1
(
x
) = cos
n
θ
is the only solution satisfying
y
(
L
) =
L
and
y
0
(
L
) =
L
.
(5 Marks)
[20 Marks]
QUESTION 4
Consider the temperature
u
(
x
,
t
) at time
t
at each point
x
of a rod length 5, if one of its end is in
contact with ice at 0
◦
and the other end is exposed to warm air at 35
◦
. Assume that initially the rod
is submerged in boiling water at 100
o
.
a.
Derive the initialboundary value problem of the problem described above. Explain all the steps,
do not repeat the derivations done in the study Guide and Prescribed book.
(8 MARKS).
6
APM3701/201
Show that the solution of the initialboundary value problem obtained in part (a) has a unique
solution.
[Hint: apply the maximum principle to the heat equation.]
(12 MARKS)
.
[20 MARKS]
SOLUTION
a.
The par partial differential equation describes the heat distribution in a rod, with temperature
u
(
x
,
t
) is given by
∂
u
∂
t
=
k
∂
2
u
∂
x
2
0
≤
x
≤
5,
t
>
0.
The initial condition is
u
(
x
, 0) = 100, 0
≤
x
≤
5.
The boundary conditions are given by
u
(0,
t
) = 0,
u
(5,
t
) = 35,
t
>
0.
Thus, the initialboundary value problem is
∂
u
∂
t
=
k
∂
2
u
∂
x
2
0
≤
x
≤
5,
t
>
0
u
(0,
t
) = 0,
u
(5,
t
) = 35,
t
>
0
u
(
x
, 0) = 100,
0
<
x
≤
5.
You've reached the end of your free preview.
Want to read all 9 pages?
 Winter '16
 Differential Equations, Equations, Partial Differential Equations, Limits, Partial differential equation, uββ