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Solutions to Final Exam
May 24, 2011
problem 1
a
At steady state, we have
T
xx
+
C
= 0
,
for 0
< x < L
T
(
x
= 0) = 0
, T
x
(
x
=
L
) = 0
The general solution to the ODE is
T
=

C
2
x
2
+
ax
+
b
The boundary conditions give us
b
= 0,
a
=
CL
, so we Fnally have
T
S
(
x
) =
Cx
(
L

x
2
)
which plots as the left half of a downfacing parabola. Note that the hottest
section of the rod, at
x
=
L
, is at a temperature of
CL
2
/
2.
b
DeFning
T
T
≡
T

T
S
, where we have
T
S
from (a), gives for
T
T
(referred to as
just
T
for simplicity)
T
xx
=
1
α
2
T
t
,
for 0
< x < L, t >
0
T
(
x
= 0
, t
) = 0
, T
x
(
x
=
L, t
) = 0
, T
(
x, t
= 0) =
f
∗
(
x
)
≡
f
(
x
)

Cx
(
L

x
2
)
,
lim
t
→∞
T
(
x, t
) = 0
1
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Assuming solutions of the form
X
(
x
)Θ(
t
), we have from the PDE
X
′′
X
=
1
α
2
Θ
′
Θ
≡ 
k
2
giving the general solutions
X
′′
+
k
2
X
= 0
→
X
=
c
1
sin
kx
+
c
2
cos(
kx
)
Θ
′
+
α
2
k
2
Θ = 0
→
Θ =
de
−
k
2
α
2
t
The boundary conditions at the ends of the rod give us a StormLiouville
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 Spring '11
 GLAGOLA

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