Page 489 - Electromagnetics

so that ψ 1 (x) = Kψ 2 (x) for some constant K. So under homogeneous boundary condi-
tions, every eigenvalue is associated with a unique eigenfunction.
This is false for the periodic boundary conditions (A.82). Eigenfunction expansion then
becomes difficult, as we can no longer assume eigenfunction orthogonality. However, the
Gram–Schmidt algorithm may be used to construct orthogonal eigenfunctions. We refer
the interested reader to Haberman [79].

The harmonic differential equation. The ordinary differential equation
2
d ψ(x) 2
=−k ψ(x) (A.86)
dx 2
2
is Sturm–Liouville with p ≡ 1, q ≡ 0, σ ≡ 1, and λ = k . Suppose we take [a, b] = [0, L]
and adopt the homogeneous boundary conditions
ψ(0) = 0 and ψ(L) = 0. (A.87)
Since p(x)> 0 and σ(x)> 0 on [0, L], equations (A.86) and (A.87) form a regular Sturm–
Liouville problem. Thus we should have an infinite number of discrete eigenvalues. A
power series technique yields the two independent solutions

ψ a (x) = A a sin kx, ψ b (x) = A b cos kx,
to (A.86); hence by linearity the most general solution is
ψ(x) = A a sin kx + A b cos kx. (A.88)

The condition at x = 0 gives A a sin 0 + A b cos 0 = 0, hence A b = 0. The other condition
then requires

A a sin kL = 0. (A.89)
Since A a = 0 would give ψ ≡ 0, we satisfy (A.89) by choosing k = k n = nπ/L for
2
n = 1, 2,.... Because λ = k , the eigenvalues are
λ n = (nπ/L) 2

with corresponding eigenfunctions
ψ n (x) = sin k n x.

Note that λ = 0 is not an eigenvalue; eigenfunctions are nontrivial by definition, and
sin(0πx/L) ≡ 0. Likewise, the differential equation associated with λ = 0 can be solved
easily, but only its trivial solution can fit homogeneous boundary conditions: with k = 0,
2
2
(A.86) becomes d ψ(x)/dx = 0, giving ψ(x) = ax + b; this can satisfy (A.87) only with
a = b = 0.
These “eigensolutions” obey the properties outlined earlier. In particular the ψ n are
orthogonal,
L nπx mπx L
sin sin dx = δ mn ,
0 L L 2
and the eigenfunction expansion of a piecewise continuous function f is given by
∞

nπx
f (x) = c n sin
L
n=1

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