Page 501 - Electromagnetics

Because of the condition at y = 0 let us use
V (x, y) = (A x e −k y x + B x e k y x )(A y sin k y y + B y cos k y y).

The solution form
(x, y) = B(k y )e −k y x sin k y y
V k y
satisfies the finiteness condition and the homogeneous condition at y = 0. The remaining
condition can be satisfied by a continuous superposition of solutions:
∞

V (x, y) = B(k y )e −k y x sin k y ydk y .
0
We must have
∞

V 0 e −ay = B(k y ) sin k y ydk y .
0
Use of the orthogonality relationship (A.115) yields the amplitude spectrum B(k y ), and
we find that
2 ∞ −k y x k y
V (x, y) = e sin k y ydk y . (A.116)
2
π 0 a + k 2 y
As a final example in rectangular coordinates let us consider a problem in which ψ
depends on all three variables:
2
2
∇ ψ(x, y, z) + k ψ(x, y, z) = 0
for
0 ≤ x ≤ L 1 , 0 ≤ y ≤ L 2 , 0 ≤ z ≤ L 3 ,
subject to

ψ(0, y, z) = ψ(L 1 , y, z) = 0,
ψ(x, 0, z) = ψ(x, L 2 , z) = 0,
ψ(x, y, 0) = ψ(x, y, L 3 ) = 0.

Here k = 0 is a constant. This is a three-dimensional eigenvalue problem as described
2
in § A.4, where λ = k are the eigenvalues and the closed surface is a rectangular
box. Physically, the wave function ψ represents the so-called eigenvalue or normal mode
2
2
2
2
solutions for the “TM modes” of a rectangular cavity. Since k + k + k = k ,we
x y z
might have one or two separation constants equal to zero, but not all three. We find,
however, that the only solution with a zero separation constant that can fit the boundary
conditions is the trivial solution. In light of the boundary conditions and because we
expect standing waves in the box, we take
ψ(x, y, z) = [A x sin(k x x) + B x cos(k x x)] ·
· [A y sin(k y y) + B y cos(k y y)] ·
· [A z sin(k z z) + B z cos(k z z)].

The conditions ψ(0, y, z) = ψ(x, 0, z) = ψ(x, y, 0) = 0 give B x = B y = B z = 0. The
conditions at x = L 1 , y = L 2 , and z = L 3 require the separation constants to assume
= pπ/L 3 ,
the discrete values k x = k x m = mπ/L 1 , k y = k y n = nπ/L 2 , and k z = k z p
where k 2 + k 2 + k 2 = k 2 and m, n, p = 1, 2,.... Associated with each of these
x m y n z p mnp

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