21.6 EXERCISES
Using plane polar coordinates, the radial (i.e. normal) derivative of this function is given
by
∂G(r,r 0 )
∂ρ=
r
|r|·∇G(r,r 0 )=
r
2 π|r|·
[
r−r 0
|r−r 0 |^2+
q(r−r 1 )
|r−r 1 |^2]
.
Using (21.107), on the perimeter of the circleρ=athe radial derivative takes the form
∂G(r,r 0 )
∂ρ∣∣
∣
∣
ρ=a=
1
2 π|r|[
|r|^2 −r·r 0
|r−r 0 |^2+
q|r|^2 −q(a^2 /|r 0 |^2 )r·r 0
(a^2 /|r 0 |^2 )|r−r 0 |^2]
=
1
2 πa1
|r−r 0 |^2[
|r|^2 +q|r 0 |^2 −(1 +q)r·r 0]
,
where we have set|r|^2 =a^2 in the second term on the RHS, but not in the first. If we take
q= 1, the radial derivative simplifies to
∂G(r,r 0 )
∂ρ∣
∣
∣
∣ρ=a=1
2 πa,
or 1/L,whereLis the circumference, and so (21.108) withq= 1 is the required Neumann
Green’s function.
Sinceρ(r) = 0, the solution to our boundary-value problem is now given by (21.106) as
u(r 0 )=〈u(r)〉C−∫
CG(r,r 0 )f(r)dl(r),where the integral is around the circumference of the circleC. In plane polar coordinates
r=ρcosφi+ρsinφjandr 0 =ρ 0 cosφ 0 i+ρ 0 sinφ 0 j, and again using (21.107) we find
that onCthe Green’s function is given by
G(r,r 0 )|ρ=a=1
2 π[
ln|r−r 0 |+ln(
a
|r 0 ||r−r 0 |)
+c]
=
1
2 π(
ln|r−r 0 |^2 +lna
|r 0 |+c)
=
1
2 π{
ln[
a^2 +ρ^20 − 2 aρ 0 cos(φ−φ 0 )]
+lna
ρ 0+c}
. (21.109)
Sincedl=adφonC, the solution to the problem is given byu(ρ 0 ,φ 0 )=〈u〉C−a
2 π∫ 2 π0f(φ)ln[a^2 +ρ^20 − 2 aρ 0 cos(φ−φ 0 )]dφ.The contributions of the final two terms terms in the Green’s function (21.109) vanish
because
∫ 2 π
0 f(φ)dφ= 0. The average value ofuaround the circumference,〈u〉C, is a freely
specifiable constant as we would expect for a Neumann problem. This result should be
compared with the result (21.104) for the corresponding Dirichlet problem, but it should
be remembered that in the one casef(φ) is a potential, and in the other the gradient of a
potential.
21.6 Exercises21.1 Solve the following first-order partial differential equations by separating the
variables:
(a)
∂u
∂x−x∂u
∂y=0; (b)x∂u
∂x− 2 y∂u
∂y