2.2. LAPLACE’S EQUATION 23

This reasoning might suggest that the convolution

(8)

u(x)=

Rn

Φ(x − y)f(y) dy

=

−

1

2π 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).