Page 39 - Electromagnetics

Once Maxwell’s equations are in definite form, standard methods for partial differential
equations can be used to determine whether the electromagnetic model is well-posed. In
a nutshell, the system (2.1)–(2.2) of hyperbolic differential equations is well-posed if and
only if we specify E and H throughout a volume region V at some time instant and also
specify, at all subsequent times,

1. the tangential component of E over all of the boundary surface S,or
2. the tangential component of H over all of S,or
3. the tangential component of E over part of S, and the tangential component of H
over the remainder of S.

Proof of all three of the conditions of well-posedness is quite tedious, but a simplified
uniqueness proof is often given in textbooks on electromagnetics. The procedure used
by Stratton [187] is reproduced below. The interested reader should refer to Hansen [81]
for a discussion of the existence of solutions to Maxwell’s equations.

2.2.1 Uniqueness of solutions to Maxwell’sequations
Consider a simply connected region of space V bounded by a surface S, where both
V and S contain only ordinary points. The fields within V are associated with a current
distribution J, which may be internal to V (entirely or in part). By the initial conditions
that imply the auxiliary Maxwell’s equations, we know there is a time, say t = 0, prior
to which the current is zero for all time, and thus by causality the fields throughout V
are identically zero for all times t < 0. We next assume that the fields are specified
throughout V at some time t 0 > 0, and seek conditions under which they are determined
uniquely for all t > t 0 .
Let the field set (E 1 , D 1 , B 1 , H 1 ) be a solution to Maxwell’s equations (2.1)–(2.2)
associated with the current J (along with an appropriate set of constitutive relations),
and let (E 2 , D 2 , B 2 , H 2 ) be a second solution associated with J. To determine the con-
ditions for uniqueness of the fields, we look for a situation that results in E 1 = E 2 ,
B 1 = B 2 , and so on. The electromagnetic fields must obey
∂B 1
∇× E 1 =− ,
∂t
∂D 1
∇× H 1 = J + ,
∂t
∂B 2
∇× E 2 =− ,
∂t
∂D 2
∇× H 2 = J + .
∂t
Subtracting, we have
∂(B 1 − B 2 )
∇× (E 1 − E 2 ) =− , (2.9)
∂t
∂(D 1 − D 2 )
∇× (H 1 − H 2 ) = , (2.10)
∂t
hence defining E 0 = E 1 − E 2 , B 0 = B 1 − B 2 , and so on, we have
∂D 0
E 0 · (∇× H 0 ) = E 0 · , (2.11)
∂t
∂B 0
H 0 · (∇× E 0 ) =−H 0 · . (2.12)
∂t

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