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Question 5.56: Prove the following uniqueness theorem: If the current densi...

Prove the following uniqueness theorem: If the current density J is specified throughout a volume ν, and either the potential A or the magnetic field B is specified on the surface S bounding ν , then the magnetic field itself is uniquely determined throughout ν . [Hint: First use the divergence theorem to show that

\int\{(\nabla \times U ) \cdot(\nabla \times V )- U \cdot[\nabla \times(\nabla \times V )]\} d \tau=\oint[ U \times(\nabla \times V )] \cdot d a ,

for arbitrary vector functions U and V.]

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Apply the divergence theorem to the function [U × (∇ × V)], noting (from the product rule) that

\nabla \cdot[ U \times( \nabla \times V )]=( \nabla \times V ) \cdot( \nabla \times U )- U \cdot[ \nabla \times( \nabla \times V )] :

\int \nabla \cdot[ U \times( \nabla \times V )] d \tau=\int\{( \nabla \times V ) \cdot( \nabla \times U )- U \cdot[ \nabla \times( \nabla \times V )]\} d \tau=\oint[ U \times( \nabla \times V )] \cdot d a .

As always, suppose we have two solutions, B _{1}\left(\text { and } A _{1}\right) \text { and } B _{2}\left(\text { and } A _{2}\right) \text {. Define } B _{3} \equiv B _{2}- B _{1} (and A _{3} \equiv A _{2}- A _{1} ), so that \nabla \times A _{3}= B _{3} \text { and } \nabla \times B _{3}= \nabla \times B _{1}-\nabla \times B _{2}=\mu_{0} J -\mu_{0} J = 0 . \text { Set } U = V = A _{3} in the above identity:

\int\left\{\left( \nabla \times A _{3}\right) \cdot\left( \nabla \times A _{3}\right)- A _{3} \cdot\left[ \nabla \times\left( \nabla \times A _{3}\right)\right]\right\} d \tau=\int\left\{\left( B _{3}\right) \cdot\left( B _{3}\right)- A _{3} \cdot\left[ \nabla \times B _{3}\right]\right\} d \tau=\int\left(B_{3}\right)^{2} d \tau

 

=\oint\left[ A _{3} \times\left( \nabla \times A _{3}\right)\right] \cdot d a =\oint\left( A _{3} \times B _{3}\right) \cdot d a . But either A is specified (in which case A _{3}= 0 , or else B is

specified (in which case B _{3}= 0 ), at the surface. In either case \oint\left( A _{3} \times B _{3}\right) \cdot d a =0 . \text { So } \int\left(B_{3}\right)^{2} d \tau=0 , and hence  B _{1}= B _{2} . qed

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