2.2. LAPLACE’S EQUATION 35
THEOREM 12 (Representation formula using Green’s function). If
u
C2(
¯
U ) solves problem (29), then
(30) u(x) =
∂U
g(y)
∂G
∂ν
(x, y) dS(y) +
U
f(y)G(x, y) dy (x U).
Here we have a formula for the solution of the boundary-value problem
(29), provided we can construct Green’s function G for the given domain U.
This is in general a difficult matter and can be done only when U has simple
geometry. Subsequent subsections identify some special cases for which an
explicit calculation of G is possible.
Interpreting Green’s function. Fix x U. Then regarding G as a
function of y, we may symbolically write
−ΔG = δx in U
G = 0 on ∂U,
δx denoting the Dirac measure giving unit mass to the point x.
Before moving on to specific examples, let us record the general assertion
that G is symmetric in the variables x and y:
THEOREM 13 (Symmetry of Green’s function). For all x, y U, x = y,
we have
G(y, x) = G(x, y).
Proof. Fix x, y U, x = y. Write
v(z) := G(x, z), w(z) := G(y, z) (z U).
Then Δv(z) = 0 (z = x), Δw(z) = 0 (z = y) and w = v = 0 on
∂U. Thus our applying Green’s identity on V := U [B(x, ε) B(y, ε)] for
sufficiently small ε 0 yields
(31)
∂B(x,ε)
∂v
∂ν
w
∂w
∂ν
v dS(z) =
∂B(y,ε)
∂w
∂ν
v
∂v
∂ν
w dS(z),
ν
denoting the inward pointing unit vector field on ∂B(x, ε)∪∂B(y, ε). Now
w is smooth near x, whence
∂B(x,ε)
∂w
∂ν
v dS
Cεn−1
sup
∂B(x,ε)
|v| = o(1) as ε 0.
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