SECOND DERIVATIVES OF THE NEWTONIAN POTENTIAL

CRISTIAN E. GUTIERREZ´ NOVEMBER 4, 2009

We assume f : Ω R is locally Holder¨ continuous, that is, there exists 0 < α 1 → ≤ such that for each K Ω compact there is a constant CK > 0 such that ⊂

α f (x) f (y) CK x y , x, y K. | − | ≤ | − | ∀ ∈

We also assume f is bounded in Ω. The goal is to prove the representation formula (1) below for the second deriva- tives of the Newtonian potential. 2 n Assume n 3, let Γ(x) = Cn x − be the fundamental solution, η : R R is ≥ | | → smooth, η(t) = 0 for t 1, 0 η 1, η(t) = 1 for t 2 and 0 η0 2. ≤ ≤ ≤ ≥ ≤ ≤ Let Ω be a domain for which the divergence theorem holds, and Ω Ω , and 0 ⊂ 0 let f¯(x) = f (x) for x Ω and f¯(x) = 0 in Ω Ω. Define for x Ω ∈ 0 \ ∈

Z Z u(x) = (DijΓ)(x y)( f¯(y) f (x)) dy f (x) DiΓ(x y)νj(y) dσ(y). Ω0 − − − ∂Ω0 −

n Since DijΓ(x) Cn x − , f is bounded and locally Holder¨ continuous, the function | | ≤ | | u(x) is well defined for all x Ω. Let  > 0 and define ∈

Z v(x) = DiΓ(x y)η((x y)/) f (y) dy, Ω − −

and let v(x) = Diw(x), where

Z w(x) = Γ(x y) f (y) dy. Ω − 1 2 C. E. GUTIERREZ´ NOVEMBER 4, 2009

If f is integrable in Ω, then by the homework, v C1(Rn), and if in addition f is  ∈ bounded, w C1(Rn). We have ∈ Z
Djv(x) = Dxj DiΓ(x y)η((x y)/) f (y) dy Ω − − Z
¯ = Dxj DiΓ(x y)η((x y)/) f (y) dy Ω0 − − Z
¯ = Dxj DiΓ(x y)η((x y)/) ( f (y) f (x)) dy Ω0 − − − Z
+ f (x) Dxj DiΓ(x y)η((x y)/) dy Ω0 − − Z
¯ = Dxj DiΓ(x y)η((x y)/) ( f (y) f (x)) dy Ω0 − − − Z
f (x) Dyj DiΓ(x y)η((x y)/) dy − Ω0 − − Z
¯ = Dxj DiΓ(x y)η((x y)/) ( f (y) f (x)) dy Ω0 − − − Z f (x) DiΓ(x y)η((x y)/)νj(y) dσ(y) − ∂Ω0 − − from the divergence theorem. Since x Ω, if we take  dist(x, ∂Ω )/2, then ∈ ≤ 0 x y 2 for y ∂Ω and so η((x y)/) = 1. So | − | ≥ ∈ 0 − Z
¯ Djv(x) = Dxj DiΓ(x y)η((x y)/) ( f (y) f (x)) dy Ω0 − − − Z f (x) DiΓ(x y)νj(y) dσ(y). − ∂Ω0 − Then subtracting we get u(x) Djv (x) −  Z  
¯ = DijΓ(x y) Dxj DiΓ(x y)η((x y)/) ( f (y) f (x)) dy Ω0 − − − − − Z 
 ¯ = Dxj 1 η((x y)/) DiΓ(x y) ( f (y) f (x)) dy Ω0 − − − − Z 
 ¯ = Dxj 1 η((x y)/) DiΓ(x y) ( f (y) f (x)) dy x y 2 − − − − Z| − |≤ " #
1 xj yj = DijΓ(x y) 1 η((x y)/) DiΓ(x y) η0( x y /) − ( f¯(y) f (x)) dy. x y 2 − − − − −  | − | x y − | − |≤ − SECOND DERIVATIVES OF THE NEWTONIAN POTENTIAL November 4, 2009 3

Let K Ω be compact, and K0 = y : dist(y, K) dist(K, ∂Ω)/2 . We have K0 is ⊂ { ≤ } compact, K0 Ω, and if  < dist(K, ∂Ω)/2, then B (x) K0 for all x K. Therefore ⊂  ⊂ ∈ estimating the last integral we obtain for x K ∈ Z   C u(x) Djv(x) DijΓ(x y) + DiΓ(x y) f (y) f (x) dy | − | ≤ x y 2 − −  | − | | −Z|≤ " # C C C α α CK + x y dy C0  . 0 n n 1 K0 ≤ x y 2 x y x y −  | − | ≤ | − |≤ | − | | − | Therefore Djv u uniformly on compact subsets of Ω as  0, so u is continuous → n → in Ω. In addition, v v uniformly in R (this follows directly by subtracting). → 2 This implies that w C (Ω) and u(x) = Dijw(x) for x Ω by the fundamental ∈ ∈ theorem of calculus, because we write

Z yj v(x1, , xj 1, yj, xj+1, , xn) = Djv(x1, , xj 1, t, xj+1, , xn) dt + v(x), − − ··· ··· xj ··· ··· and pass to the limit as  0. So we have proved the following representation → formula for the second derivatives of the Newtonian potential w for x Ω: Z Z ∈ (1) Dijw(x) = (DijΓ)(x y)( f¯(y) f (x)) dy f (x) DiΓ(x y)νj(y) dσ(y), Ω0 − − − ∂Ω0 − under the assumption that f is bounded in Ω and locally Holder¨ continuous for some 0 < α 1. ≤ We next prove that ∆w = f in Ω. Let x Ω, take R sufficiently large such that ∈ Ω BR(x), and apply (1) with Ω0 = BR(x), then ⊂ Z Z Diiw(x) = (DijΓ)(x y)( f¯(y) f (x)) dy f (x) DiΓ(x y)νj(y) dσ(y), x y

Department of Mathematics,Temple University,Philadelphia, PA 19122 E-mail address: [email protected]