# Is the Sturm-Liouville operator self-adjoint?

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.

• 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 $D(L)$. May 16, 2022 at 19:21
• In order to show that the operator is essentially self-adjoint you need to prove that the defect spaces are trivial, i.e. $\ker(L^*\pm iI)=\{0\}.$ May 16, 2022 at 20:27
• @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)? May 16, 2022 at 20:39
• @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. May 16, 2022 at 20:52

$$L$$ 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.)
• 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. May 17, 2022 at 20:40