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Let $-\Delta f=-f'' $ be the Laplace operator on $[0,l]$ with domain consisting of functions on $[0,l]$ which have are in $H^2([0,l])$ and satisfy $f'(0)=f'(l)=0$.

Then $-\Delta$ is self-adjoint and unbounded.

Now for every $k \in \mathbb{N}$: $$-\Delta \cos( \frac{k \pi}{l})=(\frac{k \pi}{l})^2 \cos(\frac{k \pi}{l})$$

meaning $ \{(\frac{k \pi}{l})^2 : k \in \mathbb{N} \} \subset \sigma(-\Delta)$.

It seems that we have $ \{(\frac{k \pi}{l})^2 : k \in \mathbb{N} \} = \sigma(-\Delta)$ - but can one prove this, i.e. how can we prove that the spectrum is discrete and consists solely of eigenvalues of the form $(\frac{k \pi}{l})^2$ ? Is this even true?

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The functions $$ \varphi_{\lambda}(x)= \cos(\sqrt{\lambda}x),\;\;\;\psi_{\lambda}(x)=\cos(\sqrt{\lambda}(x-l)) $$ satisfy the differential equation $-f''+\lambda f = 0$ subject to the conditions $$ \varphi_{\lambda}(0)=1,\;\varphi_{\lambda}'(0)=0,\;\;\;\;\;\;\psi_{\lambda}(l)=0,\;\psi_{\lambda}'(l)=0. $$ These solutions are linearly independent except when the following Wronskian is $0$: \begin{align} W(\varphi_{\lambda},\psi_{\lambda}) & =\varphi_{\lambda}\psi_{\lambda}'-\varphi_{\lambda}'\psi_{\lambda} \\ & = \sqrt{\lambda}\{\cos(\sqrt{\lambda}x)\sin(\sqrt{\lambda}(x-l)) -\sin(\sqrt{\lambda}x)\cos(\sqrt{\lambda}(x-l))\} \\ & = \sqrt{\lambda}\sin(\sqrt{\lambda}(x-l-x))\\ & =-\sqrt{\lambda}\sin(\sqrt{\lambda}l) \end{align} The resolvent $R(\lambda)f=(-\Delta-\lambda I)^{-1}f$ is constructed for any $\lambda$ for which $W(\varphi_{\lambda},\psi_{\lambda})\ne 0$ as $$ R(\lambda)f = -\frac{\psi_{\lambda}(x)}{W(\varphi_{\lambda},\psi_{\lambda})} \int_{0}^{x}f(t)\varphi_{\lambda}(t)dt -\frac{\varphi_{\lambda}(x)}{W(\varphi_{\lambda},\psi_{\lambda})}\int_{x}^{l}f(t)\psi_{\lambda}(t)dt. $$ You can check that $R(\lambda)f$ satisfies the required conditions to be in the domain of $-\Delta$. In any event, you can verify that $-\Delta-\lambda I$ is invertible if $W(\varphi_{\lambda},\psi_{\lambda})\ne 0$, and that the inverse is the above operator. On other hand, you know that if $W(\varphi_{\lambda},\psi_{\lambda})=0$, then the operator is not invertible because $\varphi_{\lambda}$ is an eigenfunction for such $\lambda$. So that pins down the spectrum.

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