40 2. FOUR IMPORTANT LINEAR PDE DEFINITION. Green’s function for the unit ball is (41) G(x, y) := Φ(y − x) − Φ(|x|(y − ˜)) x (x, y ∈ B(0, 1), x = y). The same formula is valid for n = 2 as well. Assume now u solves the boundary-value problem (42) Δu = 0 in B0(0, 1) u = g on ∂B(0, 1). Then using (30), we see (43) u(x) = − ∂B(0,1) g(y) ∂G ∂ν (x, y) dS(y). According to formula (41), Gyi (x, y) = Φyi (y − x) − Φ(|x|(y − ˜))yi x . But Φyi (y − x) = 1 nα(n) xi − yi |x − y|n , and furthermore Φ(|x|(y − ˜))yi x . = −1 nα(n) yi|x|2 − xi (|x||y − ˜|)n x = − 1 nα(n) yi|x|2 − xi |x − y|n if y ∈ ∂B(0, 1). Accordingly ∂G ∂ν (x, y) = n i=1 yiGyi (x, y) = −1 nα(n) 1 |x − y|n n i=1 yi((yi − xi) − yi|x|2 + xi) = −1 nα(n) 1 − |x|2 |x − y|n . Hence formula (43) yields the representation formula u(x) = 1 − |x|2 nα(n) ∂B(0,1) g(y) |x − y|n dS(y). Suppose now instead of (42) u solves the boundary-value problem (44) Δu = 0 in B0(0,r) u = g on ∂B(0,r)
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