1
19. The Dirichlet problem for a circle and Fourier series.
The Greens function for the Laplacian and the corresponding solution to the Dirichlet problem for
Poissons' equation obtained in the previous section, is valid for n ≥ 3. An analysis for the case n =
1
2 with the fundamental solution E(x, x) =
log|x – x | would obtain a similar solution for this
2p
case. However we will derive a solution in the special case of W a circle. That is, we seek a solution
to Laplaces' equation on a circle, satisfying Dirichlet boundary conditions. We choose W a circle
centre the origin, radius a, and Laplaces' equation in polar coordinates
Du(r, q) =
∂2 u 1 ∂u 1 ∂2 u
+
+
= 0, r < a, 0 £ q < 2p,
∂r2 r ∂r r2 ∂q2
subject to boundary conditions
u(a, q) = f(q) , 0 £ q < 2p,
where f is a continuous function on 0 £ q < 2p.
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We use the method of separation of variables and assume that
u(r, q) = R(r)Q(q)
and substitute into the differential equation to obtain
R" Q +
1 '
1
R Q + 2 RQ" = 0,
r
r
or separating variables,
Q ¢¢
r2 R" + rR'
=–
= l2
R
Q
which leads to the two families of ordinary differential equations
r 2 R" + rR'†
– l 2 R = 0, Q " + l 2 Q = 0.
2
Consider the case l = 0; then the solutions of the above equations are
Q = A0 + B0 q ,
R = C0 + D 0 logr
Since solutions u must be periodic in q, and bounded for r £ a, we conclude that B0 = 0, D 0 = 0.
In the case l =/ 0 the solutions of the above equations are
Q = Al coslq + Bl sinlq
R = Cl rl + Dl r– l
where the coefficients depend on l. Again because solutions are periodic in q with period 2p, we conclude that l = n
= 1,2,3,.., and because solutions are bounded for r £ a, Dn = 0, n = 1,2,3,...By superposition we can combine these
solutions to obtain
•
Â
u(r,q) =
Cn r
n
( An
cos nq + Bn sin nq )
n =0
We may as well incorporate the constant Cn into A n and Bn , hence
†
†
•
Â
u(r, q ) =
r
n
( An
cos nq + Bn sin nq )
n =0
At the boundary r = a,
†
†
•
f(q) = u(a, q) =
Â
a
n
( An
cos nq + Bn sin nq )
n =0
=
†
•
a0
+†
2
 ( an cos nq
+ bn sin nq )
n =1
where
†1
an =
2†
p
Ú
p 0
f (t ) cos nt dt , b n =
1
2p
Ú f (t ) sin nt dt
p 0
are the Fourier coefficients of the function f. Equating coefficients of cos n q , sin n q , in these two expressions for
f(q), we obtain that
†
An =
a†n
n
a
, Bn =
bn
an
and hence
u(r, q) =
†
†
a0
2
•
+
Â
n =1
Ê r ˆn
Á ˜ ( a n cos nq + bn sin nq )
Ëa¯
3
This series converges uniformly for 0 £ q < 2p, r < a.
We can get a closed form expression for u by substituting the formulae for the Fourier coefficients as follows.
u(r, q) =
2p
1
Ú
2p 0
Ú
p n =1 0
1
•
Â
p n =1
2p
1
=
Â Ú f (t )( cos nt cos nq
f (t )dt +
2p 0
†
†
2p
•
1
2p
1
=
†
f (t )dt +
1
Ú
† f (t )dt +
2p 0
Ê r ˆn 2 p
Á ˜ Ú f (t ) cos n(t - q )dt
Ëa¯ 0
2p
Ú
p 0
+ sin nt sin nq ) dt
Ê r ˆn
Á ˜ cos n(t - q )dt
Ëa¯
•
f (t )
Â
n =1
where the interchange of the order of summation and integration is allowed by the uniform convergence of the series.
†
Therefore
u(r, q) =
† È
2p
1
Ú
•
Â
f(t) Í1 + 2
2p 0
ÍÎ
n =1
˘
Ê r ˆn
Á ˜ cos n(t - q )˙ dt.
Ëa¯
˙˚
Using the exponential form, 2 cosx = eix + e–i x the term in the square brackets reduces to two infinite geometric
serieswhose sum is
†
†
1 + 2
•
Â
n =1
2
2
Ê r ˆn
a - r
Á ˜ cos n(t - q ) = 2
2
Ëa¯
a - 2ar cos(t - q ) + r
(exercise). Therefore
†
u(r, q) =
1
†
2p
Ú
2p 0
a
f(t)
a
2
2
- r
2
- 2ar cos(t - q ) + r
2
dt,
which is Poissons' integral formula for the circle.
†
†
----------------------------------------------------------Exercise. (Hard) Using the Greens function for the Laplacian from the example at the end of the last section, derive
Poissons' integral formula for the Dirichlet problem for the unit ball in Rn
Ê1 - x 2 ˆ
˜f (x ) dSx
u(x) =
Ú Á
w n | x |=1 ÁË x - x n ˜¯
1
†