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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