Second Existence theorem for Weak Solution in Evans

Question

I have some questions about the proof of second existence theorem (Theorem 4 in 6.2) of weak solutions in Evans’ PDE:

At the last part of step 4, it says

we recall further from D.5 that the dimension of the space $N$ of the solutions of (21) is finite and equals the dimension of the space $N^*$ of solutions of $$v-K^*v=0.$$ We readily check however that (21) holds if and only if $u$ is a weak solution of (11) and that (22) holds if and only if $v$ is weak solution of (12).

I know the weak solution set of (11) is $$S=\{u\in H^1_0(U)|B[u,v]=\langle 0,v \rangle,\forall v\in H^1_0(U)\}=\{u\in H^1_0(U)| u-Ku=0 \}$$ That is, $S=H^1_0(U)\cap N(I-K)=H^1_0\cap N$. Then how we can get $S=N$?(Note that $I-K$ is a operator on $L^2(U)$ but not $H^1_0(U)$) and similarly question for the adjoint $N^*$.

Answer 1

How to get $S=N$? Note that $I-K$ isn't an operator on $H_0^1(U)$.

Recall that, for $g\in L^2(U)$, the notation $L^{-1}_\gamma g$ stands for the unique weak solution $u\in H_0^1(U)$ of $L_\gamma u=g$.

Therefore, the notation $Ku:=\gamma L^{-1}_\gamma u$ means that $Kf$ is defined for each $f\in L^2$ by $$Kf=\text{unique weak solution }u\in H_0^1(U)\text{ of } Lu+\gamma u=\gamma f.$$

Thus, the image of the operator $K:L^2(U)\to L^2(U)$ is indeed a subset of $H_0^1(U)$. Hence, any solution of $u-Ku=0$ is in $H_0^1(U)$ and satisfies $Lu=0$, that is, is a weak solution of $(11)$. In other words: $N\subset S$.

Answer 2

The previous answer indeed covers the first case, but there's a little bit more to do for the adjoint equation.

We see that $v$ is a weak solution of $$\begin{cases}L^*v=0&\text{ in } U\\ v=0&\text{ on }\partial U\end{cases}$$ if and only if $B_\gamma^*[v,\phi]:=B^*[v,\phi]+\gamma(v,\phi)=\gamma(v,\phi)$ for all $v\in H_0^1(U)$. By the first existence theorem in the book one gets in terms of the solution map $v=(L_\gamma^*)^{-1}(\gamma v)$, i.e. $$v-\gamma (L_\gamma^*)^{-1}v=0,$$ so all we need to do in order to finish the equivalence claim is to show that ${L_\gamma^*}^{-1}=(L_\gamma^{-1})^*$. (Note we cannot use eg. the formula $({A^*})^{-1}=(A^{-1})^*$ since the RHS features only a formal adjoint.) For this end calculate for $f,g\in L^2$ (denoting $u=L_\gamma^{-1}f$ and $v=(L_\gamma^*)^{-1}g$) $$ (u,g)=(g,u)=B^*_\gamma [v,u]=B_\gamma [u,v]=(f,v) $$ where we used the definition of weak solution and the fact that $(L_\gamma^*)^{-1},L_\gamma^{-1}$ are maps $L^2(U)\to H^1_0(U)$.