1
$\begingroup$

Let $L$ be the Sturm-Liouville operator with domain $$ D(L)=\Bigg\{\psi\in C^\infty([a,b])\cap L^2([a,b]) : \begin{cases} \alpha\,\psi(a)+\beta\, \psi'(a)&=0\\ \gamma\,\psi(b)+\delta \,\psi'(b)&=0 \end{cases} \Bigg\}\,.$$ and action given by $$ L\psi=-(p\psi')'+q\psi\,, $$ where $p,q\in C^\infty([a,b])$ and $p>0$ on $[a,b]$.

I'm wondering whether $L$ is self-adjoint (in the sense of abstract functional analysis). $L$ is an unbounded linear operator so it isn't enough to check that $L$ is symmetric.

I suspect that $L$ is essentially self-adjoint (i.e. its closure is self-adjoint) but I don't know how to prove it.

Can you give me a hint? Thanks in advance.

$\endgroup$
4
  • $\begingroup$ I don't think it's self adjoint. The domain consists only of smooth functions, which is too small. You ought to be able to approximate more general $H^2$ functions by functions from $D(L)$. $\endgroup$ May 16, 2022 at 19:21
  • $\begingroup$ In order to show that the operator is essentially self-adjoint you need to prove that the defect spaces are trivial, i.e. $\ker(L^*\pm iI)=\{0\}.$ $\endgroup$ May 16, 2022 at 20:27
  • $\begingroup$ @RyszardSzwarc According to you, $D(L^\ast)=\{\psi\in H^2([a,b]): \text{same boundary condition}\}$ and $L^\ast$ acts in the same ways of $L$ but with weak derivative (as Nate Eldredge was saying)? $\endgroup$ May 16, 2022 at 20:39
  • $\begingroup$ @ParcoMacelli Perhaps you are right. I have no much experience concerning differential operators. Another approach could be: show that any function in $L^2$ orthogonal to the range of $L-iI$ is trivial. The same for $L+iI.$ When you write down the condition for orthogonality, perhaps some integration by parts would be possible to get some useful form. $\endgroup$ May 16, 2022 at 20:52

1 Answer 1

0
$\begingroup$

$L$ is essentially self-adjoint in the regular case, as you noted. You can show this by exhibiting explicit bounded inverses $(L\pm iI)^{-1}$ on the domain $\mathcal{D}(L)$ consisting of twice absolutely continuous functions $f$ on $[a,b]$ that satisfy the required endpoint conditions at both $x=a$ and $x=b$. This is done using a Green function solution for the resolvent operator $R(\lambda)=(L-\lambda I)^{-1}$ of the form $$ R(\lambda)f=\frac{\varphi_{\lambda}(x)}{\omega(\lambda)}\int_a^xf(x')\psi_{\lambda}(x')dx+\frac{\psi_{\lambda}(x)}{\omega(\lambda)}\int_x^bf(x')\varphi_{\lambda}(x')dx' $$ where $\psi_{\lambda},\varphi_{\lambda}$ are non-trivial classical solutions of $(L-\lambda I)f=0$ which satisfy $$ \alpha\varphi_{\lambda}(a)+\beta\varphi_{\lambda}'(a)=0 \\ \gamma \psi_{\lambda}(b)+\delta\psi_{\lambda}'(b)=0. $$ A simple normalization that eliminates some complexity can be specified by requiring $$ \varphi_{\lambda}(a)=\beta,\;\;\varphi_{\lambda}'(a)=-\alpha \\ \psi_{\lambda}(b)=\delta,\;\;\psi_{\lambda}'(b)=-\gamma. $$ Then the Wronskian $\omega(\lambda)$ of these solutions is a function that does not depend on $x$: $$ \omega(\lambda)=(p\varphi_{\lambda}')\psi_{\lambda}-\varphi_{\lambda}(p\psi_{\lambda}') $$ The Wronskian either vanishes at no $x\in [a,b]$ or it vanishes identically, because it depends only on $\lambda$. The Wronskian vanishes iff $\{ \varphi_{\lambda},\psi_{\lambda} \}$ is a dependent set of functions of $x$, which is precisely when both functions satisfy $(L-\lambda I)h=0$ as well as the specified conditions at $x=a$ and $x=b$ (in other words, $\lambda$ is an eigenvalue of $L$.) This cannot happen for non-real $\lambda$, and it must happen for an infinite sequence of real values of $\lambda$ that has its only cluster point at $\infty$ (otherwise, the entire function $\omega(\lambda)$ would vanish identically, making every $\lambda$ an eigenvalue of $L$, which is impossible.)

$\endgroup$
3
  • $\begingroup$ Could you give me some references about the green's function for the Sturm-Liouville operator ? $\endgroup$ May 17, 2022 at 17:40
  • $\begingroup$ @ParcoMacelli : There is an old classic text by Titchmarsh about eigenfunction expansions associated with 2nd order ODEs. The exposition is terse, and you'll learn a great deal of what you want to know on this topic within the first 10-20 pages. From Amazon: Eigenfunction expansions associated with second-order differential equations Hardcover – January 1, 1946 by E. C Titchmarsh $\endgroup$ May 17, 2022 at 20:34
  • $\begingroup$ NOTE: I made a correction to my post to add $p$ into the Wronskian. There are good references at the bottom of this wikipedia page: en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory . Look at Zettl and Teschl especially. $\endgroup$ May 17, 2022 at 20:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .