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Section 5.2 Dirichlet Problems with Symmetry
81
Solutions to Exercises 5.2
1.
This problem is similar to Example 2. Note that
f
is its own Legendre series:
f
(
θ
)=20(
P
1
(cos
θ
)+
P
0
(cos
θ
))
.
So really there is no need to compute the Legendre coeﬃcients using integrals. We
simply have
A
0
= 20 and
A
1
= 20, and the solution is
u
(
r, θ
) = 20 + 20
r
cos
θ.
3.
We have
u
(
r, θ
)=
∞
±
n
=0
A
n
r
n
P
n
(cos
θ
)
,
with
A
n
=
2
n
+1
2
²
π
0
f
(
θ
)
P
n
(cos
θ
) sin
θdθ
=
2
n
2
²
π
2
0
100
P
n
(cos
θ
) sin
+
2
n
2
²
π
π
2
20
P
n
(cos
θ
) sin
θdθ.
Let
x
= cos
θ
,
dx
=

sin
. Then
A
n
=
50(2
n
+1)
²
1
0
P
n
(
x
)
dx
+ 10 (2
n
²
0

1
P
n
(
x
)
dx.
The case
n
= 0 is immediate by using
P
0
(
x
)=1
,
A
0
=50
²
1
0
dx
+10
²
0

1
dx
=60
.
For
n>
0, the integrals are not straightforward and you need to refer to Exercise 10,
Section 5.6, where they are evaluated. Quoting from this exercise, we have
²
1
0
P
2
n
(
x
)
dx
=0
,n
=1
,
2
,...,
and
²
1
0
P
2
n
+1
(
x
)
dx
=
(

1)
n
(2
n
)!
2
2
n
+1
(
n
!)
2
(
n
,
1
,
2
,....
Since
P
2
n
(
x
) is an even function, then, for
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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