Page 159 - Maxwell House

POYNTING's THEOREM 139

3. The following mixed boundary conditions (3.54) are fulfilled at each point of : the
1
′
projection (, )| ∈ 1 ′ tangential to the some part of is given and the projection

1
1
′
′′
′′
(, )| ∈ 1 ′′ tangential to the rest part of is given while = ∪ .
1
1
1

1
1
Evidently, the initial conditions are not required anymore since all analysis shifted to the
frequency domain. Now, the proof of this variant of the uniqueness theorem is based on
Poynting’s theorem (3.48) in the frequency domain and only slightly diverges from the previous
one. From the above discussion surrounding equation (3.59), we know that the surface integral
with Poynting’s vector vanishes. Then the power balance (3.48) for the difference fields
{ (, ), (, )} can be written as
3 3
2
2
′′
′′
0 = ∫ [ (, )| | + (, )| | ]/2 � (3.62)
0
0
3
3
1
0 = () − ()
3
3
It follows from the top equation (3.62) that = 0 if (, ) ≠ 0, (, ) = 0 and = 0
′′
′′

3

3
′′
′′
if (, ) = 0, (, ) ≠ 0. Then the bottom equation (3.62) can be satisfied if only =
3
0 when = 0 and vice versa that finally is proved the uniqueness theorem in frequency
3
domain. Therefore, the fields produced by sources within a lossy region are unique as long as
the tangential components satisfy prescribed conditions at the bounding surface. Clearly, the
energy dissipation inside can be infinitesimal but finite.
1
3.2.4 Cavity Resonators
In the case of loss-free medium, when (, ) = 0, (, ) = 0 the top integral in (3.62)
′′
′′

vanishes not telling us anything about the magnitude of the fields { (, ), (, )}. If so,
3
3
the bottom integral defines the frequencies called resonance frequencies at which the
0
uniqueness theorem is not valid
( ) = ( ) (3.63)
3
0
0
3
In the case of inhomogeneous filling (, ) = = ., (, ) = = . and
′
′
′
′

the electric and magnetic energy defined by (3.53) can be then rewritten as
′
0
( ) = ∫ | | ⁄
2
3

0
2 1 � (3.64)
′
0

( ) = ∫ | | ⁄
2

3
0
2 1
Plugging Maxwell’s equation − x = into the first integral in (3.64), we obtain
′

0 0

from (3.63) and (3.64)

| x 3 |
∫
1
2
= 1 (3.65)
0
′

0 0 ′ ∫ | |
1