Calculate
A
m
and
B
m
and use this to calculate:
∞
X
m
=1
,m odd
1
m
2
= 1 +
1
9
+
1
25
+
1
49
· · ·
A
0
=
R
1

1
f
(
x
)
R
1

1
1
=
R
1
0
1
2
=
1
2
A
m
=
R
1

1
f
(
x
) cos(
πmx
)
R
1

1
cos
2
(
πmx
)
=
R
1
0
cos(
πmx
)
1
=
sin(
πmx
)
πm
1
0
=0
(We used the fact that
f
≡
0
on
(

1
,
0)
)
B
0
= 0
MATH 54

FINAL EXAM

SOLUTIONS
11
B
m
=
R
1

1
f
(
x
) sin(
πmx
)
R
1

1
sin
2
(
πmx
)
=
R
1
0
cos(
πmx
)
1
=

cos(
πmx
)
πm
1
0
=

1
πm
(cos(
πm
)

1)
=

1
πm
((

1)
m

1)
(We used the fact that
f
≡
0
on
(

1
,
0)
)
Now, using Parseval’s identity, we get:
∞
X
m
=0
A
2
m
+
B
2
m
=
Z
1

1
(
f
(
x
))
2
A
2
0
+
B
2
0
+
∞
X
m
=1
A
2
m
+
B
2
m
=
Z
1
0
1
1
2
2
+ 0
2
+
∞
X
m
=1
0
2
+

1
πm
((

1)
m

1)
=1
∞
X
m
=1
1
π
2
m
2
((

1)
m

1)
2
=1

1
4
=
3
4
∞
X
m
=1
((

1)
m

1)
2
m
2
=
3
π
2
4
And finally, to conclude, notice that
(

1)
m

1 = 0
if
m
is even
and
= 2
if
m
is odd, hence:
∞
X
m
=1
,m odd
2
2
m
2
=
3
π
2
4
∞
X
m
=1
,m odd
1
m
2
=
3
π
2
4(4)
=
3
π
2
16
12
PEYAM RYAN TABRIZIAN
5.
(5 points)
Use separation of variables to find the general solution to
the following PDE:
u
xx
+
u
yy
=
u
u
(0
, y
) =
u
(1
, y
) =0
(where
u
=
u
(
x, y
)
and
0
< x <
1
,
0
< y <
1
)
Hint:
You can do this!!!
Suppose
u
(
x, y
) =
X
(
x
)
Y
(
y
)
.
Then plug this into the above
equation:
(
X
(
x
)
Y
(
y
))
xx
+ (
X
(
x
)
Y
(
y
))
yy
=
X
(
x
)
Y
(
y
)
X
00
(
x
)
Y
(
y
) +
X
(
x
)
Y
00
(
y
) =
X
(
x
)
Y
(
y
)
And divide all the sides by
X
(
x
)
Y
(
y
)
:
X
00
(
x
)
Y
(
y
)
X
(
x
)
Y
(
y
)
+
X
(
x
)
Y
00
(
y
)
X
(
x
)
Y
(
y
)
=1
X
00
(
x
)
X
(
x
)
+
Y
00
(
y
)
Y
(
y
)
=1
X
00
(
x
)
X
(
x
)
=1

Y
00
(
y
)
Y
(
y
)
=
λ
Hence:
X
00
(
x
) =
λX
(
x
)
(and
Y
00
(
y
) = (1

λ
)
Y
(
y
)
):
And as usual, we get that
X
(0) = 0
and
X
(1) = 0
, and if we do
the
3

cases business as usual, we find that:
λ
=

(
πm
)
2
and
X
(
x
) = sin(
πmx
)
(
m
= 1
,
2
,
· · ·
)
Now go back to
Y
00
(
y
) = (1

λ
)
Y
(
y
) = (1 + (
πm
)
2
)
Y
(
y
)
. The
auxiliary equation is
r
2
= 1+(
πm
)
2
, which gives
r
=
±
p
1 + (
πm
)
2
,
and hence:
Y
(
y
) =
f
A
m
e
√
1+(
πm
)
2
y
+
f
B
m
e

√
1+(
πm
)
2
y
And hence:
MATH 54

FINAL EXAM

SOLUTIONS
13
X
(
x
)
Y
(
y
) =
f
A
m
e
√
1+(
πm
)
2
y
+
f
B
m
e

√
1+(
πm
)
2
y
sin(
πmx
)
And finally take linear combinations:
u
(
x, y
) =
∞
X
m
=1
f
A
m
e
√
1+(
πm
)
2
y
+
f
B
m
e

√
1+(
πm
)
2
y
sin(
πmx
)
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 Spring '08
 Chorin
 Math, Differential Equations, Linear Algebra, Algebra, Equations