Section 4.6
The Helmholtz and Poisson Equations
65
Solutions to Exercises 4.6
1.
Write (1) in polar coordinates:
φ
rr
+
1
r
φ
r
+
1
r
2
φ
θθ
=

kφ
φ
(
a, θ
)=0
.
Consider a product solution
φ
(
r, θ
)=
R
(
r
)Θ(
θ
). Since
θ
is a polar angle, it follows
that
Θ(
θ
+2
π
)=Θ(
θ
);
in other words, Θ is 2
π
periodic. Plugging the product solution into the equation
and simplifying, we Fnd
R
±±
Θ+
1
r
R
±
1
r
2
R
Θ
±±
=

kR
Θ;
(
R
±±
+
1
r
R
±
+
)
Θ=

1
r
2
R
Θ
±±
;
r
2
R
±±
R
+
r
R
±
R
+
kr
2
=

Θ
±±
Θ
;
hence
r
2
R
±±
R
+
r
R
±
R
+
2
=
λ,
and

Θ
±±
Θ
=
λ
⇒
Θ
±±
+
λ
Θ=0
,
where
λ
is a separation constant. Our knowledge of solutions of second order linear
ode’s tells us that the last equation has 2
π
periodic solutions if and only if
λ
=
m
2
,m
=0
,
±
1
,
±
2
,....
This leads to the equations
Θ
±±
+
m
2
,
and
r
2
R
±±
R
+
r
R
±
R
+
2
=
m
2
⇒
r
2
R
±±
+
rR
±
+(
2

m
2
)
R
.
These are equations (3) and (4). Note that the condition
R
(
a
) = 0 follows from
φ
(
a, θ
⇒
R
(
a
)Θ(
θ
⇒
R
(
a
) = 0 in order to avoid the constant 0 solution.
5.
We proceed as in Example 1 and try
u
(
r, θ
∞
±
m
=0
∞
±
n
=1
J
m
(
λ
mn
r
)(
A
mn
cos
mθ
+
B
mn
sin
mθ
∞
±
m
=0
∞
±
n
=1
φ
mn
(
r, θ
)
,
where
φ
mn
(
r, θ
J
m
(
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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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