Page 157 - Maxwell House

POYNTING's THEOREM 137

To verify the theorem let recourse the proof by contradiction method assuming that we managed
to get two different and correct solutions { (, ), (, )} and { (, ), (, )} of

Maxwell’s equations such as

x = + + 1 , x = + + 2

� (3.55)

− x = , − x =

0
0

and satisfied the identical
Both solutions are induced by the same field sources =
1 2
initial and boundary conditions (3.54). Evidently, the difference between these two solutions
(, ) = (, ) − (, ) ≠ 0 � (3.56)

(, ) = (, ) − (, ) ≠ 0

is also the solution of Maxwell’s equations. Since we assumed that = , = −
1
3
1
2
≡ 0 elsewhere and thereby this difference solution must be solution (3.55) without sources

2

x = +

� (3.57)

− x =

0

Apparently, the solution of homogeneous (no sources) Maxwell’s equations (3.57) must satisfy
the null initial and boundary conditions
(0, ) = 0, (0, ) = 0 � (3.58)
3
3
′′
′
3 = (, ) x = 0, 3 = (, ) x = 0
3

Now, let us look at the power balance applying (3.9) to the solution { , }
3

⁄
0 = () + 3 () + ∯ x ∘ (3.59)

3
1
As a matter of fact, according to (3.18) the net power flow () defined by the last term in
Σ
(3.59) depends only on the tangential to the surface component of electric and magnetic
1
fields. The null boundary conditions (3.58) tell us that this power flow does not exist and,
therefore, this term vanishes and (3.59) is cut to
⁄
3 () = − () (3.60)
Using the fact that the transferred into heat power () ≥ 0 for any moment of time ≥ 0

we can come to conclusion that according to (3.60)
′
3 ( )
′
∫ = 3 () − 3 (0) ≤ 0 (3.61)
0 ′
The internal variable t in (3.61) was replaced by to avoid confusion. Now we can exploit the
′
initial conditions (the first line in (3.58)), not yet used, that tell us that 3 (0) = 0