36 2. FOUR IMPORTANT LINEAR PDE On the other hand, v(z) = Φ(z − x) − φx(z), where φx is smooth in U. Thus lim ε→0 ∂B(x,ε) ∂v ∂ν w dS = lim ε→0 ∂B(x,ε) ∂Φ ∂ν (x − z)w(z) dS = w(x), by calculations as in the proof of Theorem 1. Thus the left-hand side of (31) converges to w(x) as ε → 0. Likewise the right-hand side converges to v(y). Consequently G(y, x) = w(x) = v(y) = G(x, y). b. Green’s function for a half-space. In this and the next subsection we will build Green’s functions for two regions with simple geometry, namely the half-space R+ n and the unit ball B(0, 1). Everything depends upon our explicitly solving the corrector problem (26) in these regions, and this in turn depends upon some clever geometric reflection tricks. First let us consider the half-space R+ n = {x = (x1,...,xn) ∈ Rn | xn 0}. Although this region is unbounded, and so the calculations in the previous section do not directly apply, we will attempt nevertheless to build Green’s function using the ideas developed earlier. Later of course we must check directly that the corresponding representation formula is valid. DEFINITION. If x = (x1,...,xn−1,xn) ∈ R+, n its reflection in the plane ∂R+ n is the point ˜ x = (x1,...,xn−1, −xn). We will solve problem (26) for the half-space by setting φx(y) := Φ(y − ˜) x = Φ(y1 − x1,...,yn−1 − xn−1,yn + xn) (x, y ∈ R+).n The idea is that the corrector φx is built from Φ by “reflecting the singular- ity” from x ∈ R+ n to ˜ x / ∈ R+. n We note φx(y) = Φ(y − x) if y ∈ ∂R+,n and thus Δφx = 0 in R+n φx = Φ(y − x) on ∂R+,n as required.

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