10Gauss' Law and Poisson's equation
IA Vector Calculus
10.3 Poisson’s Equation and Laplace’s equation
Definition (Poisson’s equation). The Poisson’s equation is
∇
2
ϕ = −ρ,
where ρ is given and ϕ(r) is to be solved.
This is the form of the equations for gravity and electrostatics, with
−
4
πGρ
and ρ/ε
0
in place of ρ respectively.
When ρ = 0, we get
Definition (Laplace’s equation). Laplace’s equation is
∇
2
ϕ = 0.
One example is irrotational and incompressible fluid flow: if the velocity is
u
(
r
), then irrotationality gives
u
=
∇ϕ
for some velocity potential
ϕ
. Since it is
incompressible, ∇ · u = 0 (cf. previous chapters). So ∇
2
ϕ = 0.
The expressions for
∇
2
can be found in non-Cartesian coordinates, but are a
bit complicated.
We’re concerned here mainly with cases exhibiting spherical or cylindrical
symmetry (use
r
for radial coordinate here). i.e. when
ϕ
(
r
) has spherical or
cylindrical symmetry. Write ϕ = ϕ(r). Then
∇ϕ = ϕ
0
(r)
ˆ
r.
Then Laplace’s equation ∇
2
ϕ = 0 becomes an ordinary differential equation.
– For spherical symmetry, using the chain rule, we have
∇
2
ϕ = ϕ
00
+
2
r
ϕ
0
=
1
r
2
(r
2
ϕ
0
)
0
= 0.
Then the general solution is
ϕ =
A
r
+ B.
– For cylindrical symmetry, with r
2
= x
2
1
+ x
2
2
, we have
∇
2
ϕ = ϕ
00
+
1
r
ϕ
0
=
1
r
(rϕ
0
)
0
= 0.
Then
ϕ = A ln r + B.
Then solutions to Poisson’s equations can be obtained in a similar way, i.e. by
integrating the differential equations directly, or by adding particular integrals
to the solutions above.
For example, for a spherically symmetric solution of
∇
2
ϕ
=
−ρ
0
, with
ρ
0
constant, recall that
∇
2
r
α
=
α
(
α
+ 1)
r
α−2
. Taking
α
= 2, we find the particular
integral
ϕ = −
ρ
0
6
r
2
,
So the general solution with spherical symmetry and constant ρ
0
is
ϕ(r) =
A
r
+ B −
1
6
ρ
0
r
2
.
To determine
A, B
, we must specify boundary conditions. If
ϕ
is defined on all
of
R
3
, we often require
ϕ →
0 as
|r| → ∞
. If
ϕ
is defined on a bounded volume
V , then there are two kinds of common boundary conditions on ∂V :
– Specify ϕ on ∂V — a Dirichlet condition
–
Specify
n · ∇ϕ
(sometimes written as
∂ϕ
∂n
): a Neumann condition. (
n
is
the outward normal on ∂V ).
The type of boundary conditions we get depends on the physical content of
the problem. For example, specifying
∂ϕ
∂n
corresponds to specifying the normal
component of g or E.
We can also specify different boundary conditions on different boundary
components.
Example.
We might have a spherically symmetric distribution with constant
ρ
0
, defined in a ≤ r ≤ b, with ϕ(a) = 0 and
∂ϕ
∂n
(b) = 0.
Then the general solution is
ϕ(r) =
A
r
+ B −
1
6
ρ
0
r
2
.
We apply the first boundary condition to obtain
A
a
+ B −
1
6
ρ
0
a
2
= 0.
The second boundary condition gives
n ·∇ϕ = −
A
b
2
−
1
3
ρ
0
b = 0.
These conditions give
A = −
1
3
ρ
0
b
3
, B =
1
5
ρ
0
a
2
+
1
3
ρ
0
b
3
a
.
Example.
We might also be interested with spherically symmetric solution with
∇
2
ϕ =
(
−ρ
0
r ≤ a
0 r > a
with
ϕ
non-singular at
r
= 0 and
ϕ
(
r
)
→
0 as
r → ∞
, and
ϕ, ϕ
0
continuous at
r = a. This models the gravitational potential on a uniform planet.
Then the general solution from above is
ϕ =
(
A
r
+ B −
1
6
ρ
0
r
2
r ≤ a
C
r
+ D r > a.
Since
ϕ
is non-singular at
r
= 0, we have
A
= 0. Since
ϕ →
0 as
r → ∞
,
D
= 0.
So
ϕ =
(
B −
1
6
ρ
0
r
2
r ≤ a
C
r
r > a.
This is the gravitational potential inside and outside a planet of constant density
ρ
0
and radius a. We want ϕ and ϕ
0
to be continuous at r = a. So we have
B +
1
6
4πρ
0
Ga
2
=
C
a
4
3
πGρ
0
a = −
C
a
2
.
The second equation gives
C
=
−GM
. Substituting that into the first equation
to find B, we get
ϕ(r) =
(
GM
2a
r
a
2
− 3
r ≤ a
−
GM
r
r > a
Since g = −ϕ
0
, we have
g(r) =
(
−
GMr
a
3
r ≤ a
−
GM
r
r > a
We can plot the potential energy:
r
ϕ(r)
r = a
We can also plot −g(r), the inward acceleration:
r
−g(r)
r = a
Alternatively, we can apply Gauss’ Law for a flux of
g
=
g
(
r
)
e
r
out of
S
, a
sphere of radius R. For R ≤ a,
Z
S
g · dS = 4πR
2
g(R) = −4πGM
R
a
3
So
g(R) = −
GMR
a
3
.
For R ≥ a, we can simply apply Newton’s law of gravitation.
In general, even if the problem has nothing to do with gravitation or electro-
statics, if we want to solve
∇
2
ϕ
=
−ρ
with
ρ
and
ϕ
sufficiently symmetric, we
can consider the flux of ∇ϕ out of a surface S = ∂V :
Z
S
∇ϕ · dS = −
Z
V
ρ dV,
by divergence theorem. This is called the Gauss Flux method.