Partial Differential Equations

we obtain the estimate
|L ε | ≤ n‖Df‖ ∞
∫
∂B(0;ε)
|Φ(y)| dS ∂B(0;ε) (y) ≤
Applying integration by parts again, we find
∫
∫
K ε = ∆ y Φ(y)f(x − y) dy −
∫
= −
R n \B(0;ε)
∂B(0;ε)
∂B(0;ε)
∂Φ
∂ν (y)f(x − y) dS ∂B(0;ε)(y)
1.2 The Laplace Equation 9
{
Cε| log ε| for n = 2,
Cε for n ≥ 3.
∂Φ
∂ν (y)f(x − y) dS ∂B(0;ε)(y)
since Φ is harmonic in R n \ {0}. Note that ν(y) = y ε
for y ∈ ∂B(0; ε). Moreover, we have
0
ε
y
1
Fig. 1.4.
From this, for n ≥ 3, we obtain
DΦ(y) =
{
−
1
2π
y
− 1
nα(n)
|y| 2 ∀y ≠ 0 if n = 2
y
|y| n ∀y ≠ 0 if n ≥ 3.
∂Φ
∂ν (y) = ν(y) · DΦ(y) = − 1
nα(n)ε n−1
∀y ∈ ∂B(0; ε).
The surface area of ∂B(0; ε) is nα(n)ε n−1 . Now, the continuity of f in x gives
K ε = −
1
nα(n)ε n−1 ∫∂B(0;ε)
f(x − y) dS ∂B(0;ε) (y) −→
ε→0
−f(x).
Similarly we find K ε −→
ε→0
−f(x) also for n = 2. Together with the already proven estimations for L ε and
I ε , we finally conclude
∆u(x) = lim
ε→0
(I ε + L ε + K ε ) = −f(x).
Let U ⊆ R n be open and bounded with smooth boundary ∂U. We consider the boundary value
problem which corresponds to the Poisson equation:
{ −∆u = f on U,
(1.9)
u = g on ∂U,
where now f : U → R and g : ∂U → R are given, and u: U × R + → R is the unknown function (U is the
closure of U in R n ).
For an open subset U of R n , let C k (U) be the space of all functions u: U → R whose restriction onto
U lies in C k (U) and whose partial derivatives up to order k are all uniformly continuous on U (thus, they
can be extended onto U).