= Solution
Use symmetric positive conductivity, for example $\alpha I\preceq a(x)\preceq MI$ with $\alpha>0$. Let $v$ have the same <Dirichlet boundary data> as $u$, and write $v=u+\phi$ with $\phi\in H_0^1(\Omega)$. <Symmetry> gives
$$
J(v)=J(u)+2\int_\Omega\nabla\phi\cdot a\nabla u\,dx
+\int_\Omega\nabla\phi\cdot a\nabla\phi\,dx.
$$
The middle integral vanishes by the weak field equation; equivalently integrate by parts, use $\phi=0$ on the boundary and $\nabla\cdot(a\nabla u)=0$. Hence
$$
J(v)-J(u)=\int_\Omega\nabla\phi\cdot a\nabla\phi\,dx
\ge\alpha\int_\Omega|\nabla\phi|^2\,dx\ge0.
$$
Therefore
$$
\boxed{J(u)=\min_{\ v|_{\partial\Omega}=\lambda\cdot x}J(v).}
$$
Equality forces $\nabla\phi=0$, and its zero boundary trace gives $\phi=0$, 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.
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