Question 4.38: Prove the following uniqueness theorem: A volume ν contains ...

Given two solutions,  V_{1}\left(\text { and } E _{1}=- \nabla V_{1}, D _{1}=\epsilon E _{1}\right) \text { and } V_{2}\left( E _{2}=-\nabla V_{2}, D _{2}=\epsilon E _{2}\right), \text { define } V_{3} \equiv V_{2}-V_{1}\left( E _{3}= E _{2}- E _{1}, D _{3}= D _{2}- D _{1}\right) .

\int_{ \nu } \nabla \cdot\left(V_{3} D _{3}\right) d \tau=\int_{ S } V_{3} D _{3} \cdot d a =0,\left(V_{3}=0 \text { on } S \right), \text { so } \int\left( \nabla V_{3}\right) \cdot D _{3} d \tau+\int V_{3}\left( \nabla \cdot D _{3}\right) d \tau=0 .

But \nabla \cdot D _{3}= \nabla \cdot D _{2}- \nabla \cdot D _{1}=\rho_{f}-\rho_{f}=0, \text { and } \nabla V_{3}= \nabla V_{2}-\nabla V_{1}=- E _{2}+ E _{1}=- E _{3}, \text { so } \int E _{3} \cdot D _{3} d \tau=0 .

But  D _{3}= D _{2}- D _{1}=\epsilon E _{2}-\epsilon E _{1}=\epsilon E _{3}, \text { so } \int \epsilon\left(E_{3}\right)^{2} d \tau=0 . \text { But } \epsilon>0, \text { so } E _{3}=0, \text { so } V_{2}-V_{1}=\text { constant } .

But at surface,  V_{2}=V_{1}, \text { so } V_{2}=V_{1} everywhere.   qed