1
$\begingroup$

Let $\Omega\subset\mathbb R^n$ be a bounded domain with a sufficiently smooth boundary $\partial\Omega$. We assume $\lambda_1\in\mathbb R$ is the principle eigenvalue of the operator $$ -\Delta:\ H^2(\Omega)\to L^2(\Omega), $$ taken with zero Dirichlet boundary condition. Now for any $\lambda\in[0,\lambda_1)$ so that $\lambda$ is not an eigenvalue of $L$, the following boundary value problem \begin{equation} \left\{ \begin{alignedat}{2} &-\Delta u = \lambda u + f &\quad& \mbox{in $\Omega$,} \\ &u=0 &\quad& \mbox{on $\partial\Omega$,} \end{alignedat} \right. \end{equation} where $f$ is assumed to be a $L^2(\Omega)$-function, admits a unique weak solution $u_\lambda\in H_0^1(\Omega)$ via Fredholm alternative principle. Moreover, we have the following estimate $$ \|u_\lambda\|_{L^2(\Omega)} \le C(\lambda) \|f\|_{L^2(\Omega)}. $$

This constant will blow up when $\lambda$ approaches the principle eigenvalue $\lambda_1$.

I am interested to see that what kind of contradiction will occur if we assume $C$ is independent of $\lambda\in[0,\lambda_1]$ for a fixed $f\in L^2(\Omega)$.

$\endgroup$
  • $\begingroup$ Wouldn't it lead to an a priory estimate in $W_2^2$ from which by the method of continuity would follow that the problem is solvable for $\lambda=\lambda_1$ for any $f\in L_2$? $\endgroup$ – Andrew Jul 20 '17 at 15:32
  • 1
    $\begingroup$ Since $u=(-\Delta-\lambda)^{-1}f$, the bound with a uniform $C$ and for all $f\in L^2$ would give the same uniform bound on $\|(-\Delta-\lambda)^{-1}\|$, which is impossible because the operator norm of the resolvent diverges when $\lambda$ approaches the spectrum (easy to show, from the spectral theorem). (The bound with a uniform $C$ for only specific $f$'s might conceivably hold.) $\endgroup$ – Christian Remling Jul 20 '17 at 20:24
  • $\begingroup$ Can the same argument work for the operator $-\Delta + \vec B\cdot \nabla$? $\endgroup$ – CooLee Jul 21 '17 at 1:46
  • $\begingroup$ Here I assume $f$ is a fixed $L^2(\Omega)$-function. $\endgroup$ – CooLee Aug 11 '17 at 11:00
  • 1
    $\begingroup$ The same argument works near any isolated point of the spectrum of an operator $\endgroup$ – Piero D'Ancona Aug 11 '17 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.