7More partial differential equations
IB Methods
7.5 Poisson’s equation
Let φ : R
3
→ R satisfy the Poisson’s equation
∇
2
φ = −F,
where F (x) is a forcing term.
The fundamental solution to this equation is defined to be G
3
(x, y), where
∇
2
G
3
(x, y) = δ
(3)
(x − y).
By rotational symmetry, G
3
(x, y) = G
3
(|x − y|). Integrating over a ball
B
r
= {|x − y| ≤ r, x ∈ R
3
},
we have
1 =
Z
B
r
∇
2
G
3
dV
=
Z
∂B
r
n · ∇G
3
dS
=
Z
S
2
dG
3
dr
r
2
sin θ dθ dφ
= 4πr
2
dG
3
dr
.
So we know
dG
3
dr
=
1
4πr
2
,
and hence
G
3
(x, y) = −
1
4π|x − y|
+ c.
We often set c = 0 such that
lim
|x|→∞
G
3
= 0.
Green’s identities
To make use of this fundamental solution in solving Poisson’s equation, we first
obtain some useful identities.
Suppose
φ, ψ
:
R
3
→ R
are both smooth everywhere in some region Ω
⊆ R
3
with boundary ∂Ω. Then
Z
∂Ω
φn · ∇ψ dS =
Z
Ω
∇ · (φ∇ψ) dV =
Z
Ω
φ∇
2
ψ + (∇φ) ·(∇ψ) dV.
So we get
Proposition (Green’s first identity).
Z
∂Ω
φn · ∇ψ dS =
Z
Ω
φ∇
2
ψ + (∇φ) ·(∇ψ) dV.
Of course, this is just an easy consequence of the divergence theorem, but
when Green first came up with this, divergence theorem hasn’t existed yet.
Similarly, we obtain
Z
∂Ω
ψn · ∇φ dS =
Z
Ω
ψ∇
2
φ + (∇φ) · (∇ψ) dV.
Subtracting these two equations gives
Proposition (Green’s second identity).
Z
Ω
φ∇
2
ψ −ψ∇
2
φ dV =
Z
∂Ω
φn · ∇ψ −ψn · ∇φ dS.
Why is this useful? On the left, we have things like
∇
2
ψ
and
∇
2
φ
. These
are things we are given by Poisson’s or Laplace’s equation. On the right, we
have things on the boundary, and these are often the boundary conditions we
are given. So this can be rather helpful when we have to solve these equations.
Using the Green’s function
We wish to apply this result to the case
ψ
=
G
3
(
|x − y|
). However, recall that
when deriving Green’s identity, we assumed
φ
and
ψ
are smooth everywhere in
our domain. However, our Green’s function is singular at
x
=
y
, but we want to
integrate over this region as well. So we need to do this carefully. Because of
the singularity, we want to take
Ω = B
r
− B
ε
= {x ∈ R
3
: ε ≤ |x − y| ≤ R}.
In other words, we remove a small region of radius
ε
centered on
y
from the
domain.
In this choice of Ω, it is completely safe to use Green’s identity, since our
Green’s function is certainly regular everywhere in this Ω. First note that since
∇
2
G
3
= 0 everywhere except at x = y , we get
Z
Ω
φ∇
2
G
3
− G
3
∇
2
φ dV = −
Z
Ω
G
3
∇
2
φ dV
Then Green’s second identity gives
−
Z
Ω
G
3
∇
2
φ dV =
Z
S
2
r
φ(n · ∇G
3
) − G
3
(n · φ) dS
+
Z
S
2
ε
φ(n · ∇G
3
) − G
3
(n · φ) dS
Note that on the inner boundary, we have n = −
ˆ
r. Also, at S
2
ε
, we have
G
3
|
S
2
ε
= −
1
4πε
,
dG
3
dr
S
2
ε
=
1
4πε
2
.
So the inner boundary terms are
Z
S
2
ε
φ(n · ∇G
3
) − G
3
(n · φ)
ε
2
sin θ dθ dφ
= −
ε
2
4πε
2
Z
S
2
ε
φ sin θ dθ dφ +
ε
2
4πε
Z
S
2
ε
(n · ∇φ) sin θ dθ dφ
Now the final integral is bounded by the assumption that
φ
is everywhere smooth.
So as we take the limit
ε →
0, the final term vanishes. In the first term, the
ε
’s
cancel. So we are left with
= −
1
4π
Z
S
2
ε
φ dΩ
= −
¯
φ
→ −φ(y)
where
¯
φ is the average value of φ on the sphere.
Now suppose ∇
2
φ = −F . Then this gives
Proposition (Green’s third identity).
φ(y) =
Z
∂Ω
φ(n · ∇G
3
) − G
3
(n · ∇φ) dS −
Z
Ω
G
3
(x, y)F (x) d
3
x.
This expresses
φ
at any point
y
in Ω in terms of the fundamental solution
G
3
, the forcing term
F
and boundary data. In particular, if the boundary values
of φ and n · ∇φ vanish as we take r → ∞, then we have
φ(y) = −
Z
R
3
G
3
(x, y)F (x) d
3
x.
So the fundamental solution is the Green’s function for Poisson’s equation on
R
3
.
However, there is a puzzle. Suppose
F
= 0. So
∇
2
φ
= 0. Then Green’s
identity says
φ(y) =
Z
S
2
r
φ
dG
3
dr
− G
3
dφ
dr
dS.
But we know there is a unique solution to Laplace’s equation on every bounded
domain once we specify the boundary value
φ|
∂Ω
, or a unique-up-to-constant
solution if we specify the boundary value of n · ∇φ|
∂Ω
.
However, to get
φ
(
y
) using Green’s identity, we need to know
φ
and and
n · ∇φ on the boundary. This is too much.
Green’s third identity is a valid relation obeyed by solutions to Poisson’s
equation, but it is not constructive. We cannot specify
φ
and
n ·∇φ
freely. What
we would like is a formula of φ given, say, just the value of φ on the boundary.
Dirichlet Green’s function
To overcome this (in the Dirichlet case), we seek to modify G
3
via
G
3
→ G = G
3
+ H(x, y),
where
∇
2
H
= 0 everywhere in Ω,
H
is regular throughout Ω, and
G|
∂Ω
= 0. In
other words, we find some
H
that does not affect the relations on
G
when acted
on by
∇
2
, but now our
G
will have boundary value 0. We will find this
H
later,
but given such an
H
, we replace
G
3
with
G − H
in Green’s third identity, and
see that all the H terms fall out, i.e. G also satisfies Green’s third identity. So
φ(y) =
Z
∂Ω
[φn · ∇G − Gn · ∇φ] dS −
Z
F G dV
=
Z
∂Ω
φn · ∇G dS −
Z
F G dV.
So as long as we find
H
, we can express the value of
φ
(
y
) in terms of the values
of
φ
on the boundary. Similarly, if we are given a Neumann condition, i.e. the
value of
n · ∇φ
on the boundary, we have to find an
H
that kills off
n · ∇G
on
the boundary, and get a similar result.
In general, finding a harmonic ∇
2
H = 0 with
H|
∂Ω
=
1
4π|x − x
0
|
∂Ω
is a difficult problem. However, the method of images allows us to solve this in
some special cases with lots of symmetry.
Example. Suppose
Ω = {(x, y, z) ∈ R
3
: z ≥ 0}.
We wish to find a solution to
∇
2
=
−F
in Ω with
φ →
0 rapidly as
|x| → ∞
with boundary condition φ(x, y, 0) = g(x, y).
The fundamental solution
G
3
(x, x
0
) = −
1
4π
1
|x − x
0
|
obeys all the conditions we need except
G
3
|
z=0
= −
1
4π
1
[(x − x
0
)
2
+ (y −y
0
)
2
+ z
2
0
]
1/2
6= 0.
However, let
x
R
0
be the point (
x
0
, y
0
, −z
0
) . This is the reflection of
x
0
in the
boundary plane
z
= 0. Since the point
x
R
0
is outside our domain,
G
3
(
x, x
R
0
)
obeys
∇
2
G
3
(x, x
R
0
) = 0
for all x ∈ Ω, and also
G
3
(x, x
R
0
)|
z=0
= G
3
(x, x
0
)|
z=0
.
Hence we take
G(x, x
0
) = G
3
(x, x
0
) − G
3
(x, x
R
0
).
The outward pointing normal to Ω at z = 0 is n = −
ˆ
z. Hence we have
n · ∇G|
z=0
=
1
4π
−(z −z
0
)
|x − x
0
|
3
−
−(z + z
0
)
|x − x
R
0
|
3
z=0
=
1
2π
z
0
[(x − x
0
)
2
+ (y −y
0
)
2
+ z
2
0
]
3/2
.
Therefore our solution is
φ(x
0
) =
1
4π
Z
Ω
1
|x − x
0
|
−
1
|x − x
R
0
|
F (x) d
3
x
+
z
0
2π
Z
R
2
g(x, y)
[(x − x
0
)
2
+ (y −y
0
)
2
+ z
2
0
]
3/2
dx dy.
What have we actually done here? The Green’s function
G
3
(
x, x
0
) in some sense
represents a “charge” at
x
0
. We can imagine that the term
G
3
(
x, x
R
0
) represents
the contribution to our solution from a point charge of opposite sign located at
x
R
0
. Then by symmetry, G is zero at z = 0.
z = 0
x
0
x
R
0
Our solution for
φ
(
x
0
) is valid if we extend it to
x
0
∈ R
3
where
φ
(
x
0
) is solved
by choosing two mirror charge distributions. But this is irrelevant for our present
purposes. We are only concerned with finding a solution in the region
z ≥
0.
This is just an artificial device we have in order to solve the problem.
Example.
Suppose a chimney produces smoke such that the density
φ
(
x, t
) of
smoke obeys
∂
t
φ − D∇
2
φ = F (x, t).
The left side is just the heat equation, modelling the diffusion of smoke, while
the right forcing term describes the production of smoke by the chimney.
If this were a problem for x ∈ R
3
, then the solution is
φ(x, t) =
Z
t
0
Z
R
3
F (y, τ )S
3
(x − y, t − τ ) d
3
y dτ,
where
S
3
(x − y, t − τ ) =
1
[4πD(t − τ)]
3/2
exp
−
|x − y|
2
4D(t − τ)
.
This is true only if the smoke can diffuse in all of
R
3
. However, this is not true
for our current circumstances, since smoke does not diffuse into the ground.
To account for this, we should find a Green’s function that obeys
n · ∇G|
z=0
= 0.
This says that there is no smoke diffusing in to the ground.
This is achieved by picking
G(x, t; y, τ ) = Θ(t −τ)[S
3
(x − y, t − τ ) + S
3
(x − y
R
, t − τ)].
We can directly check that this obeys
∂
t
D
2
∇
2
G = δ(t − τ)δ
3
(x − y)
when x ∈ Ω, and also
n · ∇G|z
0
= 0.
Hence the smoke density is given by
φ(x, t) =
Z
t
0
Z
Ω
F (y, τ )[S
3
(x − y, t − τ ) + S
3
(x − y
R
, t − τ)] d
3
y.
We can think of the second term as the contribution from a “mirror chimney”.
Without a mirror chimney, we will have smoke flowing into the ground. With a
mirror chimney, we will have equal amounts of mirror smoke flowing up from
the ground, so there is no net flow. Of course, there are no mirror chimneys in
reality. These are just artifacts we use to find the solution we want.
Example.
Suppose we want to solve the wave equation in the region (
x, t
) such
that x > 0 with boundary conditions
φ(x, 0) = b(x), ∂
t
(x, 0) = 0, ∂
x
φ(0, t) = 0.
On R
1,1
d’Alembert’s solution gives
φ(x, t) =
1
2
[b(x − ct) + b(x + ct)]
This is not what we want, since eventually we will have a wave moving past the
x = 0 line.
x = 0
To compensate for this, we introduce a mirror wave moving in the opposite
direction, such that when as they pass through each other at
x
= 0, there is no
net flow across the boundary.
x = 0
More precisely, we include a mirror initial condition
φ
(
x,
0) =
b
(
x
) +
b
(
−x
),
where we set
b
(
x
) = 0 when
x <
0. In the region
x >
0 we are interested in, only
the
b
(
x
) term will contribute. In the
x <
0 region, only
x >
0 will contribute.
Then the general solution is
φ(x, t) =
1
2
[b(x + ct) + b(x + c) + b(−x − ct) + b(−x + ct)].