6. Consider the ICBC problem
u
t
(
x, t
) =
ku
xx
(
x, t
) +
q
(
x
)
,
0
< x < L, t >
0
,
u
(0
, t
) =
α,
u
(
L, t
) =
β, t >
0
,
u
(
x,
0) =
f
(
x
)
,
0
< x < L,
where
α, β
are constants. I will call this Problem A.
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(a) Let
u
∞
(
x
) satisfy
u
′′
∞
(
x
) =

1
k
k
(
x
)
,
0
≤
x
≤
L,
u
∞
(0) =
α,
u
∞
(
L
) =
β.
Assuming
u
solves the ﬁrst ICBC problem, show that the function
v
deﬁned by
v
(
x, t
) =
u
(
x, t
)

u
∞
(
x
)
solves the following Problem B:
v
t
(
x, t
) =
kv
xx
(
x, t
)
,
0
< x < L, t >
0
,
v
(0
, t
) = 0
,
v
(
L, t
) = 0
, t >
0
,
v
(
x,
0) =
f
(
x
)

u
∞
(
x
)
,
0
< x < L,
Conversely
, assuming
v
(
x, t
) solves problem B, show that
u
(
x, t
) =
v
(
x, t
) +
u
∞
(
x
) solves problem A.
(b) Solve Problem B (to the extent that it is possible without having a concrete value for
q
(
x
) and
f
(
x
))
and use the solution of Problem B, and its relation to the solution
u
of Problem A to SHOW that
lim
t
→∞
u
(
x, t
) =
u
∞
(
x
).
Believe it or not, this is NOT a hard exercise.
7. For part a of this exercise you will need to use the fact that the total heat energy at time
t
in a bar stretching
from 0 to
L
whose temperature distribution is given by
u
(
x, t
) is proportional to
∫
L
0
u
(
x, t
)
dx.
It also helps to know that under fairly general conditions (which we, in an applied course, may assume hold)
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 Spring '13
 Schonbek
 Partial differential equation, ICBC, ICBC problem

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