Page 138 - Electromagnetics

to Laplace’s equation in three major coordinate systems for a variety of problems. For
an introduction to numerical techniques the reader is referred to the books by Sadiku
[162], Harrington [82], and Peterson et al. [146]. Solution to Poisson’s equation is often
undertaken using the method of Green’s functions, which we shall address later in this
section. We shall also consider the solution to Laplace’s equation for bodies immersed in
an applied, or “impressed,” field.

Uniqueness of solution to Poisson’s equation. Before attempting any solutions,
we must ask two very important questions. How do we know that solving the second-order
differential equation produces the same values for E =−∇ as solving the first-order
equations directly for E? And, if these solutions are the same, what are the conditions
for uniqueness of solution to Poisson’s and Laplace’s equations? To answer the first
question, a sufficient condition is to have twice differentiable. We shall not attempt to
prove this, but shall instead show that the condition for uniqueness of the second-order
equations is the same as that for the first-order equations.
Consider a region of space V surrounded by a surface S. Static charge may be located
entirely or partially within V , or entirely outside V , and produces a field within V . This
region may also contain any arrangement of conductors or other materials. Now, assume
that 1 and 2 represent solutions to the static field equations within V with source
ρ(r). We wish to find conditions under which 1 = 2 .
Since we have

∇· [ (r)∇ 1 (r)] =−ρ(r), ∇· [ (r)∇ 2 (r)] =−ρ(r),
the difference field 0 = 2 − 1 obeys

∇· [ (r)∇ 0 (r)] = 0. (3.51)

That is, 0 obeys Laplace’s equation. Now consider the quantity
2
∇· ( 0 ∇ 0 ) = |∇ 0 | + 0 ∇· ( ∇ 0 ).
Integration over V and use of the divergence theorem and (3.51)gives

2
0 (r) [ (r)∇ 0 (r)] · dS = (r)|∇ 0 (r)| dV.
S V
As with the first order equations, we see that specifying either (r) or (r)∇ (r)· ˆ n over
S results in 0 (r) = 0 throughout V , hence 1 = 2 . As before, specifying (r)∇ (r) · ˆ n
for a conducting surface is equivalent to specifying the surface charge on S.

Integral solution to Poisson’s equation: the static Green’s function. The
method of Green’s functions is one of the most useful techniques for solving Poisson’s
equation. We seek a solution for a single point source, then use Green’s second identity
to write the solution for an arbitrary charge distribution in terms of a superposition
integral.
We seek the solution to Poisson’s equation for a region of space V as shown in Figure
3.5. The region is assumed homogeneous with permittivity , and its surface is multiply-
connected, consisting of a bounding surface S B and any number of closed surfaces internal
to V . We denote by S the composite surface consisting of S B and the N internal surfaces
S n , n = 1,..., N. The internal surfaces are used to exclude material bodies, such as the

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