11Laplace's and Poisson's equations
IA Vector Calculus
11.3 Integral solutions of Poisson’s equations
11.3.1 Statement and informal derivation
We want to find a solution to Poisson’s equations. We start with a discrete case,
and try to generalize it to a continuous case.
If there is a single point source of strength λ at a, the potential ϕ is
ϕ =
λ
4π
1
|r − a|
.
(we have λ = −4πGM for gravitation and Q/ε
0
for electrostatics)
If we have many sources
λ
α
at positions
r
α
, the potential is a sum of terms
ϕ(r) =
X
α
1
4π
λ
α
|r − r
α
|
.
If we have infinitely many of them, having a distribution of
ρ
(
r
) with
ρ
(
r
0
) d
V
0
being the contribution from a small volume at position
r
0
. It would be reasonable
to guess that the solution is what we obtain by replacing the sum with an integral:
Proposition.
The solution to Poisson’s equation
∇
2
ϕ
=
−ρ
, with boundary
conditions |ϕ(r)| = O(1/|r|) and |∇ϕ(r)| = O(1/|r|
2
), is
ϕ(r) =
1
4π
Z
V
0
ρ(r
0
)
|r − r
0
|
dV
0
For
ρ
(
r
0
) non-zero everywhere, but suitably well-behaved as
|r
0
| → ∞
, we can
also take V
0
= R
3
.
Example. Suppose
∇
2
ϕ =
(
−ρ
0
|r| ≤ a
0 |r| > a.
Fix
r
and introduce polar coordinates
r
0
, θ, χ
for
r
0
. We take the
θ
= 0 direction
to be the direction along the line from r
0
to r.
Then
ϕ(r) =
1
4π
Z
V
0
ρ
0
|r − r
0
|
dV
0
.
We have
dV
0
= r
02
sin θ dr
0
dθ dχ.
We also have
|r − r
0
| =
p
r
2
+ r
02
− 2rr
0
cos θ
by the cosine rule (c
2
= a
2
+ b
2
− 2ab cos C). So
ϕ(r) =
1
4π
Z
a
0
dr
0
Z
π
0
dθ
Z
2π
0
dχ
ρ
0
r
02
sin θ
√
r
2
+ r
02
− 2rr
0
cos θ
=
ρ
0
2
Z
a
0
dr
0
r
02
rr
0
h
p
r
2
+ r
02
− rr
0
cos θ
i
θ=π
θ=0
=
ρ
0
2
Z
a
0
dr
0
r
0
r
(|r + r
0
| + |r − r
0
|)
=
ρ
0
2
Z
a
0
"
dr
0
r
0
r
(
2r
0
r > r
0
2r r < r
0
!#
If r > a, then r > r
0
always. So
ϕ(r) = ρ
0
Z
a
0
r
02
r
dr
0
=
r
0
a
3
3r
.
If r < a, then the integral splits into two parts:
ϕ(r) = ρ
0
Z
r
0
dr
0
r
02
r
+
Z
a
r
dr
0
r
0
= ρ
0
−
1
6
r
2
+
a
2
2
.
11.3.2 Point sources and δ-functions*
Recall that
Ψ =
λ
4π|r − a|
is our potential for a point source. When r 6= a, we have
∇Ψ = −
λ
4π
r − a
|r − a|
3
, ∇
2
Ψ = 0.
What about when
r
=
a
? Ψ is singular at this point, but can we say anything
about ∇
2
Ψ?
For any sphere with center a, we have
Z
S
∇Ψ · dS = −λ.
By the divergence theorem, we have
Z
∇
2
Ψ dV = −λ.
for
V
being a solid sphere with
∂V
=
S
. Since
∇
2
Ψ is zero at any point
r 6
=
a
,
we must have
∇
2
Ψ = −λδ(r − a),
where δ is the 3d delta function, which satisfies
Z
V
f(r)δ(r − a) dV = f(a)
for any volume containing a.
In short, we have
∇
2
1
|r − r
0
|
= −4πδ(r − r
0
).
Using these, we can verify that the integral solution of Poisson’s equation we
obtained previously is correct:
∇
2
Ψ(r) = ∇
2
1
4π
Z
V
0
ρ(r
0
)
|r − r
0
|
dV
0
=
1
4π
Z
V
0
ρ(r
0
)∇
2
1
|r − r
0
|
dV
0
= −
Z
V
0
ρ(r
0
)δ(r − r
0
) dV
0
= −ρ(r),
as required.