6.6. Laplace and Poisson equations in IR 2 and IR 3 .
Laplace equation:
4u = ∂x21 u + ∂x22 u + · · · + ∂x2n u = 0 in
IR n .
It describes the temperature distribution in equilibrium (time-independent). Solutions are any linear functions a1 x1 + a2 x2 + · · · + an xn or quadratic functions
x21 − x22 ,
x1 x2 , or many higher-order polynomials. These solutions are called har-
monic functions. But the only bounded solutions are the constant solutions.
Poisson equation:
4u = f (~x)
in
IR n .
We assume that |f (~x)| → 0 as |x| → ∞, and we look for bounded solutions only.
Try Fourier transform:
−n
2
û(~ω ) = (2π)
Z
IR n
u(x)eiω·x dx.
Then
−|ω|2 û = fˆ,
û = − |ω|1 2 fˆ.
Inversion: In IR 3 only: (skipped because it needs distribution theorey, see next
semester)
(
1
1
.
)ˆ =
4π|x|
|ω|2
Convolution formula yields
1
u = − 4π|x|
∗ f.
1
u(x) = − 4π
R
f (y)
IR 3 |x−y| dy
in
IR 3
only.
Another method:
Motivation: We know that
~ = q,
divE
i.e., the divergence of an electric field is charge density. For the electric potential A
such that
~ = ∇A,
E
the equation
4A = q
holds. Consider the ideal case where q is a Dirac delta
4A = δ(x).
We can look for a spherically symmetric solution
A(~x) = A(|~x|).
We will find the solution, call it k(x). But before finding it, let us see its significance.
We can see by translation that a solution to
4U = δ(x − x0 )
is
U (x) = k(x − x0 ).
Further, a solution to
4W = f (x0 )δ(x − x0 )
(1)
is
W (x) = k(x − x0 )f (x0 ).
Integrating the W equation (1) in x0 , we find
4
R
IR n
W dx0 =
R
IR n
f (x0 )δ(x − x0 ) dx0
= f (x).
Thus
Z
u(x) =
IR n
(2)
f (x0 )k(x − x0 ) dx0
is a solution to
4u = f (x).
Thus the general solution to 4u = f (x) is given by the convolution of f (x) with the
solution k(x) to the simple Poisson equation with a point charge. The solution k(x)
is call the fundamental solution. From the previous Fourier transform approach,
1
we know it has the form k(x) = − 4π|x|
in three space dimensions.
We remark that the integration step (2) is generally known as Superposition.
2