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 =: Iε + 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) Jε = 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) =: Kε + 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) Cε (n ≥ 3). 3. We continue by integrating by parts once again in the term Kε, to discover Kε = 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) Kε = − 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.
Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2010 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.
