Solutions to Homework 11
May 23, 2011
9.52
The problem is
w
xx
+
w
yy
=
1
c
2
w
tt
,
for 0
< x < L,
0
< y < L
w
(
x
= 0) =
w
(
x
=
L
) =
w
(
y
= 0) =
w
(
y
=
L
) = 0;
w
bounded
.
Looking for separable solutions by substituting
w
(
x, y, t
) =
X
(
x
)
Y
(
y
)
T
(
t
)
,
we get
X
′′
Y T
+
XY
′′
T
=
1
c
2
XY T
′′
,
for 0
< x < l,
0
< y < l
X
(0) =
X
(
L
) =
Y
(0) =
Y
(
L
) = 0;
X, Y, T
bounded
.
or
X
′′
X
=
T
′′
c
2
T

Y
′′
Y
=

k
2
T
′′
c
2
T
+
k
2
=
Y
′′
Y
=

l
2
giving us StormLiouville problems for
X
and
Y
, with general solutions
X
′′
+
k
2
X
= 0
→
X
=
c
1
sin(
kx
) +
c
2
cos(
kx
)
Y
′′
+
l
2
Y
= 0
→
Y
=
d
1
sin(
ly
) +
d
2
cos(
ly
)
Using the boundary conditions, we need
c
2
= 0
, k
m
=
mπ
L
, m
= 1
,
2
,
3
, . . .
d
2
= 0
, l
n
=
mπ
L
, n
= 1
,
2
,
3
, . . .
1
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For
T
, the general solution is
T
′′
+
c
2
(
k
2
+
l
2
)
T
= 0
→
T
=
e
1
sin(
c
radicalbig
k
2
+
l
2
t
) +
e
2
cos(
c
radicalbig
k
2
+
l
2
t
)
with no boundary conditions applicable.
However, the requirements on
k
and
l
mean that
c
√
k
2
+
l
2
will be in the form
c
radicalBig
m
2
π
2
L
2
+
m
2
π
2
L
2
, or
cπ
L
√
m
2
+
n
2
.
Considering a superposition of these separable solutions across positive integer
m
and
n
and consolidating the coefficients into
a
mn
≡
c
1
m
d
1
n
e
2
mn
,
b
mn
≡
c
1
m
d
1
n
e
1
mn
, we get
w
(
x, y, t
) =
∞
summationdisplay
m
=1
∞
summationdisplay
n
=1
sin(
mπx
L
) sin(
nπy
L
)(
a
mn
cos(
radicalbig
m
2
+
n
2
cπt
L
+
b
mn
sin(
radicalbig
m
2
+
n
2
cπt
L
))
as requested (noting that
c
2
=
T
ρ
).
9.58
(a) The problem for the steadystate case is
T
xx
= 0
T
(0) =
T
1
, hlT
x
(
L
) +
T
(
L
)

T
0
= 0
The general solution is
T
(
x
) =
a
+
bT
, and putting in the boundary condi
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 Spring '11
 GLAGOLA
 Sin, Cos, Trigraph, Boundary conditions, TRR

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