34 2. FOUR IMPORTANT LINEAR PDE as ε 0. Hence our sending ε 0 in (24) yields the formula (25) u(x) = ∂U Φ(y x) ∂u ∂ν (y) u(y) ∂Φ ∂ν (y x) dS(y) U Φ(y x)Δu(y) dy. This identity is valid for any point x U and any function u C2( ¯ U ). Now formula (25) would permit us to solve for u(x) if we knew the values of Δu within U and the values of u, ∂u/∂ν along ∂U. However, for our application to Poisson’s equation with prescribed boundary values for u, the normal derivative ∂u/∂ν along ∂U is unknown to us. We must therefore somehow modify (25) to remove this term. The idea is now to introduce for fixed x a corrector function φx = φx(y), solving the boundary-value problem (26) Δφx = 0 in U φx = Φ(y x) on ∂U. Let us apply Green’s formula once more, to compute (27) U φx(y)Δu(y) dy = ∂U u(y) ∂φx ∂ν (y) φx(y) ∂u ∂ν (y) dS(y) = ∂U u(y) ∂φx ∂ν (y) Φ(y x) ∂u ∂ν (y) dS(y). We introduce next this DEFINITION. Green’s function for the region U is G(x, y) := Φ(y x) φx(y) (x, y U, x = y). Adopting this terminology and adding (27) to (25), we find (28) u(x) = ∂U u(y) ∂G ∂ν (x, y) dS(y) U G(x, y)Δu(y) dy (x U), where ∂G ∂ν (x, y) = DyG(x, y) · ν(y) is the outer normal derivative of G with respect to the variable y. Observe that the term ∂u/∂ν does not appear in equation (28): we introduced the corrector φx precisely to achieve this. Suppose now u C2( ¯ U ) solves the boundary-value problem (29) −Δu = f in U u = g on ∂U, for given continuous functions f, g. Plugging into (28), we obtain
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