Page 583 - Engineering Electromagnetics, 8th Edition

D

APPENDIX

The Uniqueness

Theorem

Let us assume that we have two solutions of Laplace’s equation, V 1 and V 2 , both
general functions of the coordinates used. Therefore
2
∇ V 1 = 0
and

2
∇ V 2 = 0
from which

2
∇ (V 1 − V 2 ) = 0
Each solution must also satisfy the boundary conditions, and if we represent the
given potential values on the boundaries by V b , then the value of V 1 on the boundary
V 1b and the value of V 2 on the boundary V 2b must both be identical to V b ,

V 1b = V 2b = V b
or
V 1b − V 2b = 0

In Section 4.8, Eq. (43), we made use of a vector identity,
∇ · (V D) ≡ V (∇ · D) + D · (∇V )

which holds for any scalar V and any vector D.For the present application we shall
select V 1 − V 2 as the scalar and ∇(V 1 − V 2 )as the vector, giving

∇ · [(V 1 − V 2 )∇(V 1 − V 2 )] ≡ (V 1 − V 2 )[∇ · ∇(V 1 − V 2 )]
+∇(V 1 − V 2 ) · ∇(V 1 − V 2 )

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