 Solution

Use symmetric positive conductivity, for example $\alpha I⪯a(x)⪯MI$ with $\alpha >0$. Let $v$ have the same Dirichlet boundary data as $u$, and write $v=u+\varphi$ with $\varphi \in H_0^1(\Omega )$. Symmetry givesThe middle integral vanishes by the weak field equation; equivalently integrate by parts, use $\varphi =0$ on the boundary and $\nabla \cdot (a\nabla u)=0$. HenceThereforeEquality forces $\nabla \varphi =0$, and its zero boundary trace gives $\varphi =0$, proving uniqueness. This is the Dirichlet principle for the conductivity equation.

The positivity is essential to the physical minimum-energy interpretation; symmetry by itself would give only stationarity. For example, take $a=-I$, $\lambda =0$ and $u=0$. The field equation holds, but every nonzero smooth compactly supported $v$ has $J(v)=-{\int }_{\Omega }|\nabla v{|}^2<0$, and scaling $v$ makes the energy unbounded below. Thus if the printed word “conductivity” were not understood to include positivity, the stated minimum claim would require that additional hypothesis.