## Fredholm alternative and second order elliptic equation

$\textbf{Thm:}$ (Fredholm alternative) Suppose $T:H\to H$ is a compact mapping of Hilbert space $H$ into itself. Then there exists a countable set $\Lambda\subset\mathbb{R}$ having limit points except possible $\lambda=0$ such that

(1)If $\lambda\not\in \Lambda$, then

$\displaystyle \lambda x-Tx=y$ and $\displaystyle \lambda x-T^\ast x=y$

is uniquely solvable for any $y\in H$. And the inverse mappings $(\lambda I-T)^{-1}$, $(\lambda I-T^\ast)^{-1}$ are bounded.

(2)If $\lambda\in \Lambda$, the the $\text{Ker}(\lambda I-T)$ has positive finite dimension and $\lambda x-Tx=y$ is solvable if and only if $y$ is orthogonal to the $\text{Ker}(\lambda I-T^\ast)$. And vice versa.

For a generalized second order elliptic equation

$\displaystyle Lu=(a^{ij}u_{i})_j+b^iu_i+cu=f$

Usually we require $(a^{ij})$ is strictly elliptic, $b^i, c\in L^\infty(\Omega)$. Since the

$\displaystyle \theta||u||_{H^1_0}\leq (-Lu,u)+\gamma ||u||_{L^2}$

then $L_\sigma=L-\sigma$ is invertible for $\sigma$ big enough. So $Lu=f$ is equivalent to

$\displaystyle L_\sigma u+\sigma u=f$

$\displaystyle u+\sigma L_\sigma^{-1}u=L_{\sigma}^{-1}f$

Let $T=-\sigma IL_\sigma^{-1}: L^2\to L^2$, since $I(u):H_0^1\to L^2$ is compact, then $T$ is compact. Then any $u$ satisfies the above equation is a weak solution in $H^1_0(\Omega)$

$\displaystyle T^\ast=-\sigma I(L^\ast)_\sigma^{-1}$ with $L^\ast v=(a^{ij}v_j)_i-b^iv_i+(c-\partial_ib^i )v$

In the general case to solve $Lu=\mu u+f$, this is equivalent to

$\displaystyle u+(\sigma-\mu )L_\sigma^{-1}u=L_\sigma^{-1}f$

$\displaystyle (Tu,v)_{L^2}=(u,T^*v)_{L^2}$

So there exists a series of $\mu$ such that $\displaystyle Lu=\mu u$ has nontrivial solutions. And the above equation is solvable if and only if

$\displaystyle 0=(L_\sigma^{-1} f,v)_{L^2}=(f,(L^\ast)^{-1}v)_{L^2}=(f, \frac{1}{\mu-\sigma}v)_{L^2}=\frac{1}{\mu-\sigma}(f,v)_{L^2}$.

for any $v\in \text{Ker}(id+(\sigma-\mu)(L^\ast)_\sigma^{-1})$

For instance, consider the following equation

$\displaystyle (1)\quad \begin{cases}\Delta u+u=\sin x\text{ on }\Omega\\u=0\text{ on }\partial \Omega\end{cases}$

If

$(2)\quad \displaystyle \begin{cases}\Delta u+u=0\text{ on }\Omega\\u=0\text{ on }\partial \Omega\end{cases}$

has no nontrivial solution, then $(1)$ is uniquely solvable.

Otherwise $(1)$ is solvable if and only if $\displaystyle \int_\Omega \sin x v=0$ for any $v$ solves $(2)$. The question is when $(2)$ has only the trivial solution.

This has much relation with the poincare inequality, since if $u$ satisfies $(2)$, we will have $\displaystyle \int_\Omega |\nabla u|^2= \int_\Omega u^2$($\partial \Omega$ must have some regularity to justify this).

$\textbf{Lemma1:}$ For $u\in H^1_0(\Omega)$, we have

$\displaystyle ||u||_{L^2}\leq \left(\frac{|\Omega|}{w_n}\right)^{1/n}||\nabla u||_{L^2}$

where $w_n$ is the volume of unit ball in $\mathbb{R}^n$.

$\textbf{Lemma2:}$ If $\Omega$ can be bounded by an n-dimensional strip with width $d$, we have

$\displaystyle ||u||_{L^2}\leq \left(\frac{1}{d}\right)^{2}||\nabla u||_{L^2}$

So if $|\Omega|$ is small enough, or $\Omega$ is thin enough, $(2)$ will only have the trivial solution.

Remark: Why the $T$ is an operator on $L^2(\Omega)$ instead of the $H^1_0(\Omega)$? Pay attention to the “normal to the $Ker(\lambda I- T^*)$” in fredholm alternative.