38 2. FOUR IMPORTANT LINEAR PDE Proof. 1. For each fixed x, the mapping y G(x, y) is harmonic, except for y = x. As G(x, y) = G(y, x), x G(x, y) is harmonic, except for x = y. Thus x ∂G ∂yn (x, y) = K(x, y) is harmonic for x R+, n y ∂R+.n 2. A direct calculation, the details of which we omit, verifies (34) 1 = ∂R+n K(x, y) dy for each x R+. n As g is bounded, u defined by (33) is likewise bounded. Since x K(x, y) is smooth for x = y, we easily verify as well u C∞(R+),n with Δu(x) = ∂R+n ΔxK(x, y)g(y) dy = 0 (x R+).n 3. Now fix x0 ∂R+, n ε 0. Choose δ 0 so small that (35) |g(y) g(x0)| ε if |y x0| δ, y ∂R+.n Then if |x x0| δ 2 , x R+,n (36) |u(x) g(x0)| = ∂Rn + K(x, y)[g(y) g(x0)] dy ∂R+∩B(x0,δ)n K(x, y)|g(y) g(x0)| dy + ∂R+−B(x0,δ)n K(x, y)|g(y) g(x0)| dy =: I + J. Now (34), (35) imply I ε ∂R+n K(x, y) dy = ε. Furthermore if |x x0| δ 2 and |y x0| δ, we have |y x0| |y x| + δ 2 |y x| + 1 2 |y x0| and so |y x| 1 2 |y x0|. Thus J 2 g L∞ ∂R+−B(x0,δ)n K(x, y) dy 2n+2 g L∞ xn nα(n) ∂R+−B(x0,δ)n |y x0|−n dy 0 as xn 0+.
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