2.2. LAPLACE’S EQUATION 23
This reasoning might suggest that the convolution
(8)
u(x)=
Rn
Φ(x y)f(y) dy
=

1
R2
log(|x y|)f(y) dy (n = 2)
1
n(n−2)α(n) Rn
f(y)
|x−y|n−2
dy (n 3)
will solve Laplace’s equation (1). However, this is wrong. Indeed, as inti-
mated by estimate (7),
D2Φ(x
y) is not summable near the singularity at
y = x, and so naive differentiation through the integral sign is unjustified
(and incorrect). We must proceed more carefully in calculating Δu.
Let us for simplicity now assume f Cc 2(Rn); that is, f is twice contin-
uously differentiable, with compact support.
THEOREM 1 (Solving Poisson’s equation). Define u by (8). Then
(i) u
C2(Rn)
and
(ii) −Δu = f in
Rn.
We consequently see that (8) provides us with a formula for a solution
of Poisson’s equation (2) in
Rn.
Proof. 1. We have
(9) u(x) =
Rn
Φ(x y)f(y) dy =
Rn
Φ(y)f(x y) dy;
hence
u(x + hei) u(x)
h
=
Rn
Φ(y)
f(x + hei y) f(x y)
h
dy,
where h = 0 and ei = (0,..., 1,..., 0), the 1 in the
ith-slot.
But
f(x + hei y) f(x y)
h
fxi (x y)
uniformly on
Rn
as h 0, and thus
uxi (x) =
Rn
Φ(y)fxi (x y) dy (i = 1,...,n).
Similarly
(10) uxixj (x) =
Rn
Φ(y)fxixj (x y) dy (i, j = 1,...,n).
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