= Solution
{c}
Suppose $T^{(1)}$ and $T^{(2)}$ are two sufficiently regular <steady state> solutions with the same prescribed <temperature> on the boundary. Their difference $w=T^{(1)}-T^{(2)}$ satisfies
$$
\nabla\cdot(k\nabla w)=0\quad\hbox{in }R,\qquad w=0\quad\hbox{on }\partial R.
$$
The identical <PDE source terms> cancel. Apply the <product rule> and the <divergence theorem> to $wk\nabla w$; this is the weighted form of <Green's first identity>:
$$
\begin{aligned}
0&=\iiint_R w\nabla\cdot(k\nabla w)\,dV\\
&=\iint_{\partial R}wk\nabla w\cdot\mathbf n\,dA-\iiint_Rk\|\nabla w\|^2\,dV
=-\iiint_Rk\|\nabla w\|^2\,dV.
\end{aligned}
$$
Since $k>0$, the nonnegative integrand must vanish, so $\nabla w=0$. Hence $w$ is constant on each connected component of $R$, and its zero <Dirichlet boundary condition> makes each constant zero. Therefore $\boxed{T^{(1)}=T^{(2)}}$: \b[the solution, if it exists, is unique]. This proves <Dirichlet uniqueness for variable-conductivity heat conduction>; it does not require a constant <thermal conductivity>.
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