It also helps to know that under fairly general conditions (which we, in an applied course, may assume hold)
d
dt
∫
L
0
u
(
x, t
)
dx
=
∫
L
0
∂u
∂t
(
x, t
)
dx.
(1)
With these preliminaries out of the way, consider the ICBC problem
u
t
(
x, t
) =
ku
xx
(
x, t
) +
q
(
x
)
,
0
< x < L, t >
0
,
u
x
(0
, t
) =
α,
u
x
(
L, t
) =
β, t >
0
,
u
(
x,
0) =
f
(
x
)
,
0
< x < L,
where
α, β
are constants. This will be called Problem C.
(a) Show that if a steady state (equilibrium) solution exists it is necessary that
k
(
β

a
) +
∫
L
0
q
(
x
)
dx
= 0
.
(2)
(b) Show that if and only if condition (2) holds, there is a solution
u
∞
of
u
′′
∞
(
x
) =

1
k
k
(
x
)
,
0
≤
x
≤
L,
u
′
∞
(0) =
α,
u
′
∞
(
L
) =
β.
(c) Assuming (2) let
u
∞
be the solution of
u
′′
∞
(
x
) =

1
k
k
(
x
)
,
0
≤
x
≤
L,
u
′
∞
(0) =
α,
u
′
∞
(
L
) =
β
that satisﬁes
∫
L
0
u
∞
(
x
)
dx
=
∫
L
0
f
(
x
)
dx,
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where
f
(
x
) is as in Problem C. Show that if
u
solves Problem
C
, then
v
(
x, t
) =
u
(
x, t
)

u
∞
(
x
) solves
v
t
(
x, t
) =
kv
xx
(
x, t
)
,
0
< x < L, t >
0
,
v
x
(0
, t
) = 0
,
v
x
(
L, t
) = 0
, t >
0
,
v
(
x,
0) =
f
(
x
)

u
∞
(
x
)
,
0
< x < L,
and use this to explain (prove?) that lim
t
→∞
u
(
x, t
) =
u
∞
(
x
).
(d) Calculus question: In equation (1), on the left hand side the derivative is denoted by
d
dt
, on the right
hand side by
∂
∂t
. Why the change in notation?
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 Spring '13
 Schonbek
 Partial differential equation, ICBC, ICBC problem

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