Dirichlet principle

Among functions with fixed boundary values, the solution of $\nabla \cdot (\kappa \nabla \varphi )=0$ with $\kappa >0$ uniquely minimizes the weighted energy

Let $V$ be a closed linear subspace of a real Hilbert space, and let $Q(v)=a(v,v)$ for a symmetric bounded bilinear form that is nonnegative on the ambient space and coercive on $V$. Then $Q$ has a unique minimizer on every closed affine subspace $g+V$. If $Q(v_j)\to m={\inf }_{g+V}Q$, the midpoint belongs to $g+V$ and the parallelogram law gives $Q(v_j-v_k)\le 2Q(v_j)+2Q(v_k)-4m\to 0$. Coercivity on $V$ makes the minimizing sequence Cauchy; completeness and continuity give a minimizer. The vanishing first variation is $a(u,v)=0$ for all $v\in V$. Two minimizers differ by $w\in V$ with $Q(w)=0$, so coercivity gives uniqueness. This proves the symmetric existence theorem directly, without invoking the Lax-Milgram theorem.

For $g(1)=1$ and finite displayed energy, the angular mode $g(r)\cos \theta$ has Dirichlet energy equal to $\pi$ times that integral. The harmonic function $r\cos \theta$ minimizes the energy with the same boundary data by the Dirichlet principle, giving the bound. Finite energy forces $g(r)\to 0$ as $r\to 0$: Cauchy–Schwarz bounds changes of $g^2$ by twice the product of the two energy tails, so $g^2$ has a limit, and a nonzero limit would make $\int g^2/r$ divergent. Consequently completing the square gives the exact identity ${\int }_0^1[r{g}^{\prime 2}+g^2/r]=1+{\int }_0^{1}r(g^{\prime }-g/r{)}^2$. Equality occurs exactly for $g(r)=r$. This identity also covers finite-energy functions whose polar expression is not classically smooth at the centre.