1.10 Neumann Problem for a Disk 39
u
∗
has a harmonic extension to R
n
∼ (B
∗
−
∪ (B
∗
−
)
r
), where (B
∗
−
)
r
is the
reflection of B
∗
−
across ∂Ω. By simple geometry, it can be seen that a suf-
ficiently small >0 can be chosen so that B
x,ρ+
∼ B
−
x,δ
maps under the
inversion onto a neighborhood of ∂Ω which also contains Ω ∼ B
∗
−
. Restrict-
ing u
∗
to this neighborhood, u
∗
will be harmonic thereon and its transform
will be harmonic on B
x,ρ+
∼ B
−
x,ρ
, thereby extending u harmonically across
∂B
x,ρ
.
1.10 Neumann Problem for a Disk
Consider a nonempty open set Ω ⊂ R
n
having compact closure and a smooth
boundary. Given a real-valued function g on ∂Ω,theNeumann problem is
that of finding a harmonic function u on Ω such that D
n
u(x)=g(x),x∈ ∂Ω.
There is an obvious difficulty with uniqueness of the solution if u satisfies
the above conditions and c is any constant, then u + c satisfies the same
conditions. Also, not every function g can serve as a boundary function for
the Neumann problem. If u ∈ C
2
(Ω
−
) solves the Neumann problem for the
boundary function g,thenbytakingv = 1 in Green’s Identity
0=
∂Ω
D
n
u(z) dσ(z)=
∂Ω
g(z) dσ(z);
and it follows that a necessary condition for the solvability of the Neumann
problem is that the latter integral be zero.
Before getting into the details, the meaning of the statement D
n
u(x)=
g(x),x∈ ∂Ω, should be clarified since the solution u of the Neumann problem
may be defined only on Ω.Ifn(x) is the outer normal to ∂Ω at x, by definition
D
n
u(x) = lim
t→1−
D
n(x)
u(tx), where D
n(x)
u denotes the derivative of u in
the direction n(x).
The Poisson integral solved the Dirichlet problem for a disk of any dimen-
sion n ≥ 2. An analogous integral for the Neumann problem is not available
for all n ≥ 2. The n = 2 case will be considered in this section and the n ≥ 3
case in the next section.
Consider the n = 2 case, using polar coordinates (r, θ) rather than rect-
angular coordinates. Let g be a real-valued function on the boundary of the
disk B = B
y,ρ
which satisfies the condition
2π
0
g(ρ, θ) dθ =0. (1.17)
Fourier series will be used to construct a harmonic function u on B satisfying
the Neumann condition D
n
u = g on ∂B. The use of Fourier series will not
only provide a method of approximating the solution, but will also lead to