Past exam of the methods of applied functional analysis course of the University of Oxford / 2010 / A / 1 / b

Write $a(v,w)={\int }_{\Omega }a\nabla v\cdot \nabla w$ and $A[v]=a(v,v)$. The admissible set is the closed affine subspace $H_g^1=g+H_0^1$. Symmetry of the coefficient matrix givesIf $u$ minimizes the energy, its first variation vanishes: differentiating at $t=0$ gives $a(u,\phi )=0$. Hence $u$ is a weak solution.

Conversely, if $u$ is a weak solution, every admissible $v$ has $v-u\in H_0^1$ and thereforeby uniform ellipticity. This is the Dirichlet principle for the stated symmetric energy.

We prove existence directly through quadratic energy minimization by the parallelogram law. Let $m={\inf }_{v\in H_g^1}A[v]$. The set contains $g$, and uniform ellipticity gives $0\le m\le A[g]<\infty$. Choose a minimizing sequence $v_j$ with $A[v_j]\to m$. Since $(v_j+v_k)/2\in H_g^1$, the parallelogram law givesThe differences belong to $H_0^1$, so uniform ellipticity and the Poincaré inequality implyThus $(v_j)$ is a Cauchy sequence. Completeness of $H_0^1$, applied to $v_j-g$, produces $u\in H_g^1$ with $v_j\to u$ in $H^1$. The bounded coefficient matrix makes $a$ a bounded bilinear form, so $A[v_j]\to A[u]$. Consequently $A[u]=m$, and part (i) makes $u$ a weak solution.

For uniqueness, the difference $w$ of two weak solutions lies in $H_0^1$ and satisfies $a(w,w)=0$. Uniform ellipticity gives $\nabla w=0$ almost everywhere, and the Poincaré inequality gives $w=0$. The argument proves existence and uniqueness without assuming the Lax-Milgram theorem.