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

Here uniform ellipticity means that there is a constant $\lambda >0$, independent of $x$, such thatThe assumption that the entries are essentially bounded functions also gives an upper bound $|a(x)\xi |\le \Lambda |\xi |$ with a finite constant $\Lambda$. For this symmetric divergence-form operator, the lower bound makes its energy control the squared Euclidean norm of the gradient.

The space of test functions is $\boxed{C_0^{\infty }(\Omega )=\{\phi \in C^{\infty }(\Omega ):\mathrm{supp}\phi \text{ is a compact subset of }\Omega \}}$. The Sobolev space iswhere $D_{j}u$ denotes a weak derivative, and its Sobolev norm is $\Vert u{\Vert }_{H^1}^2=\Vert u{\Vert }_2^2+{\sum }_j\Vert D_{j}u{\Vert }_2^2$. The zero-boundary Sobolev space is $\boxed{H_0^1(\Omega )={\overline{C_0^{\infty }(\Omega )}}^{H^1}}$. This closure defines the boundary condition even when the boundary has insufficient regularity for a classical boundary value.

A weak solution is a function $u\in H_0^1(\Omega )$ satisfying the weak formulationMultiplication by a test function and integration by parts give this identity from the classical equation. Conversely it is the equation in the sense of distributions, with the zero boundary condition encoded by the zero-boundary Sobolev space. Testing only against $C_0^{\infty }(\Omega )$ is equivalent, by its density in $H_0^1$ and continuity of both sides.

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.