Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 13 / 3

For every $v\in L^2(\Omega )$, the existence result in Question 2 applies to the identity coefficient matrix: $a_{ij}={\delta }_{ij}$ is bounded, symmetric and uniformly elliptic. It gives a unique $u\in W_0^{1,2}(\Omega )$ satisfyingThus every input has exactly one output and $\boxed{T\text{ is well defined}}$. The weak Dirichlet theorem used here requires only a bounded open set; no smoothness of its boundary is needed.

If $u_1=T{v}_1$ and $u_2=T{v}_2$, linearity of the weak equation makes $\alpha u_1+\beta u_2$ solve the equation with forcing $\alpha v_1+\beta v_2$. Uniqueness therefore gives $T(\alpha v_1+\beta v_2)=\alpha T{v}_1+\beta T{v}_2$.

Test the equation for $u=Tv$ with $u$ itself. The Poincaré inequality yieldsIf the gradient norm is nonzero, divide by it; if it is zero, Poincare gives $u=0$ and the estimate is automatic. In either caseHence $T:L^2(\Omega )\to W_0^{1,2}(\Omega )$ is a bounded linear operator. Notice that the energy identity has a minus sign, so $⟨Tv,v⟩=-\Vert D(Tv){\Vert }_2^2\le 0$. Testing the equations for $Tv$ and $Tz$ against each other gives $⟨Tv,z⟩=-\int D(Tv)\cdot D(Tz)=⟨v,Tz⟩$. Thus the bounded Dirichlet inverse is a negative self-adjoint operator on $L^2$.

For finite $p>n$, chooseSince $n\ge 3$ and $p>n$, we have $2<q<2^{\ast }=2n/(n-2)$. The zero-boundary versions of the Sobolev embedding theorem and Rellich-Kondrachov compactness theorem say that $W_0^{1,2}(\Omega )$ embeds continuously into $L^{2^{\ast }}$ and compactly into $L^q$ for $q<2^{\ast }$ on a bounded open set. These versions hold on arbitrary bounded open sets by zero extension; a boundary extension operator is not needed.

If $(v_j)$ is bounded in $L^2$, then $(T{v}_j)$ is bounded in $W_0^{1,2}$ by the preceding part. A subsequence consequently converges in $L^q$ to some $z$. Hölder's inequality givesMultiplication by $w$ is bounded from $L^q$ to $L^2$, and the same estimate proves that $K$ is bounded and well defined. Every bounded input sequence has a subsequence whose images converge in $L^2$, so $\boxed{K\text{ is compact}}$. If $p=\infty$ is included, take $q=2$ and use the same argument. This is the compact L2 potential perturbation of the Dirichlet Laplacian mechanism.

First, multiplication by $w$ maps $W_0^{1,2}$ into $L^2$ by the Hölder and Sobolev estimates just given. Thus a weak solution $u$ of the perturbed equation has $v=\Delta u=f+wu\in L^2$. Uniqueness for the unperturbed Dirichlet problem gives $u=Tv$. Its equation becomesConversely, if this equation has a solution $v\in L^2$, then $u=Tv$ belongs to $W_0^{1,2}$ and satisfies $\Delta u=v=f+wTv=f+wu$ weakly. These constructions are inverse to one another.

The Fredholm alternative for a compact operator states that, on a Banach space, $I-K$ for compact $K$ is bijective if and only if its kernel is zero; in that case its inverse is bounded. Apply it on $L^2$. Kernels correspond under $T$: a nonzero $v$ with $v=wTv$ gives a nonzero $u=Tv$, since $\Delta Tv=v$; a nonzero homogeneous solution $u$ gives $v=wu$, which cannot be zero because a zero-boundary harmonic weak solution is zero. Therefore triviality of the homogeneous PDE kernel is equivalent to triviality of $\ker (I-K)$ and hence to unique solvability for every $f\in L^2$. The reverse implication also follows immediately by taking $f=0$. No sign assumption on $w$ is required.