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victorballester7 committed Jan 21, 2024
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\begin{equation*}
{\langle \grad v,\grad w\rangle}_2+{\langle \vf{b}\cdot \grad v,w\rangle}_2=0\quad \forall w\in H_0^1(\Omega)
\end{equation*}
has a finite dimensional solution space $W_0$, as well as the space $V_0$ of solutions of $\mathcal{D}_0$, and $\dim W_0=\dim V_0$. Moreover, if $f\in L^2(\Omega)$, $\mathcal{D}_f$ is solvable if and only if $\langle f,v\rangle=0$ for all $v\in W_0$.
has a finite dimensional solution space $W_0$, the space $V_0$ of solutions of $\mathcal{D}_0$ has also finite dimension and $\dim W_0=\dim V_0$. Moreover, if $f\in L^2(\Omega)$, $\mathcal{D}_f$ is solvable if and only if $\langle f,v\rangle=0$ for all $v\in W_0$.
\end{proposition}
\begin{proof}
We saw in \mcref{INEPDE:Lmu} that for $\mu\geq \mu_0>0$, $L_\mu$ is an isomorphism. Now we want to solve $L_0u=f$. Consider the change of variables $u={L_{\mu_0}}^{-1}w$, with $w=L_{\mu_0}u\in H^{-1}(\Omega)$. Thus, the equation becomes:
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$$
\id-K^*={(\id-K)}^*={\left(L_0{L_{\mu_0}}^{-1}\right)}^*={({L_{\mu_0}}^*)}^{-1}{L_0}^{*}
$$
By \mcref{INEPDE:fredholm}, we have that $(\id-K)w=f$ has a solution if and only if $(\id-K^*)h=0\implies \langle f,h\rangle_{L^2} =0$ for all $h\in L^2$. But ${L_{\mu_0}}^*$ is an isomorphism, so $(\id-K^*)h=0\iff {L_0}^*h=0$.
By \mnameref{INEPDE:fredholm}, we have that $(\id-K)w=f$ has a solution if and only if $(\id-K^*)h=0\implies \langle f,h\rangle_{L^2} =0$ for all $h\in L^2$. But ${L_{\mu_0}}^*$ is an isomorphism, so $(\id-K^*)h=0\iff {L_0}^*h=0$.
\end{proof}
\begin{definition}
We define the following problem:
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