= Solution
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 becomes
$$
\boxed{(I-K)v=f.}
$$
Conversely, 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.
Back to article page
