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}


(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.

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