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16.
u
t
=
ku
xx
u
x
(0
,t
)=0
u
x
(
L, t
)=0
u
(
x,
0) =
f
(
x
)
u
(
x, t
)=
∞
X
n
=0
u
n
cos
nπx
L
e
−
k
(
nπ
L
)
2
t
u
(
x,
0) =
∞
X
n
=0
u
n
cos
nπx
L
≡
f
(
x
)
u
n
are the coeﬃcients of expanding
f
(
x
) in terms of Fourier cosine series.
u
0
=
1
L
Z
L
0
f
(
x
)
dx
u
n
=
2
L
Z
L
0
f
(
x
)cos
nπ
L
xdx
b. The equilibrium is when
u
xx
= 0 subject to the same boundary conditions. The solution
is then obtained by integration with respect to
x
u
x
=
K
and
K
= 0 because of the boundary conditions. Now integrate again to get
u
=
C
. This means that the temperature is constant. There is no other condition to ±x this
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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