24 2. FOUR IMPORTANT LINEAR PDE As the expression on the right-hand side of (10) is continuous in the variable x, we see u C2(Rn). 2. Since Φ blows up at 0, we will need for subsequent calculations to isolate this singularity inside a small ball. So fix ε 0. Then (11) Δu(x) = B(0,ε) Φ(y)Δxf(x y) dy + Rn−B(0,ε) Φ(y)Δxf(x y) dy =: + Jε. Now (12) |Iε| C D2f L∞(Rn) B(0,ε) |Φ(y)| dy Cε2| log ε| (n = 2) Cε2 (n 3). An integration by parts (see §C.2) yields (13) = Rn−B(0,ε) Φ(y)Δyf(x y) dy = Rn−B(0,ε) DΦ(y) · Dyf(x y) dy + ∂B(0,ε) Φ(y) ∂f ∂ν (x y) dS(y) =: + Lε, ν denoting the inward pointing unit normal along ∂B(0,ε). We readily check (14) |Lε| Df L∞(Rn) ∂B(0,ε) |Φ(y)| dS(y) Cε| log ε| (n = 2) (n 3). 3. We continue by integrating by parts once again in the term Kε, to discover = Rn−B(0,ε) ΔΦ(y)f(x y) dy ∂B(0,ε) ∂Φ ∂ν (y)f(x y) dS(y) = ∂B(0,ε) ∂Φ ∂ν (y)f(x y) dS(y), since Φ is harmonic away from the origin. Now DΦ(y) = −1 nα(n) y |y|n (y = 0) and ν = −y |y| = y ε on ∂B(0,ε). Consequently ∂Φ ∂ν (y) = ν ·DΦ(y) = 1 nα(n)εn−1 on ∂B(0,ε). Since nα(n)εn−1 is the surface area of the sphere ∂B(0,ε), we have (15) = 1 nα(n)εn−1 ∂B(0,ε) f(x y) dS(y) = −− ∂B(x,ε) f(y) dS(y) −f(x) as ε 0.
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