I am attempting to show the following operator, which acts on even $2 \pi$-periodic functions, is self-adjoint and find its eigenfunctions and eigenvalues.

$Ly=\frac{d^2y}{dx^2}, \:\: -\pi\leq x \leq \pi$.

I have already done this in the simpler case of $\frac{d^2 y}{dx^2} + \lambda y=0$ with some BCs. In this case, I was able to show that the operator is self-adjoint because the boundary conditions can be used to remove the boundary term when integrating by parts to show that the operator is self-adjoint and the boundary conditions are used to find the eigenfunctions and eigenvalues. I am unsure how to apply it in this case. I know there is significance in the fact that it acts on functions which are $2 \pi$-periodic. Does this mean that $-\pi\leq x \leq \pi$ becomes an implicit boundary condition of sorts? I can guess that the eigenfunctions will be $\cos nx$ but obviously need to prove this.


Operators with an even domain are not going to be selfadjoint because they're not densely-defined, which is a requirement to have an adjoint.

There are many selfadjoint versions of second-differentiation. Let $L_Mf=f''$ be the operator with "maximal" domain consisting of all $f\in L^2$ that are equal a.e. to twice absolutely continuous functions on $[-\pi,\pi]$ with $f'' \in L^2$. The minimal operator $L_m$ is the restriction of $L_M$ to the domain consisting of $f \in \mathcal{D}(L_M)$ such that $f(-\pi)=f'(-\pi)=0$ and $f(\pi)=f'(\pi)=0$. These operators are closed, densely-defined, and adjoints of each other, meaning $L_M=L_m^*$ and $L_m=L_M^*$. Neither is selfadjoint. $L_m$ is symmetric because $$ \langle L_m f,g \rangle = \langle f,L_m g\rangle,\;\; f,g\in\mathcal{D}(L_m). $$ $L_M$ is not symmetric because the following is non-zero for some $f,g\in\mathcal{D}(L_M)$: $$ \langle L_M f,g\rangle-\langle f,L_Mg \rangle=\int_{-\pi}^{\pi}f''\overline{g}-f\overline{g''}dx= (f'\overline{g}-f\overline{g'})|_{-\pi}^{\pi}. $$ The codimension of $\mathcal{D}(L_m)$ in $\mathcal{D}(L_M)$ is $4$. So there are 2 conditions required on $\mathcal{D}(L_M)$ in order to end up with a selfadjoint operator $L_s$ with $\mathcal{D}(L_m)\subset\mathcal{D}(L_s)\subset\mathcal{D}(L_M)$. Common sets of conditions are of the separated type $$ \cos\alpha f(-\pi)+\sin\alpha f'(-\pi)=0,\\ \cos\beta f(\pi)+\sin\beta f'(\pi)=0, $$ or of the periodic type $$ f(-\pi)=f(\pi),\;\; f'(-\pi)=f'(\pi). $$ Either type results in a selfadjoint $L_s$ with $\mathcal{D}(L_m)\subset\mathcal{D}(L_s)\subset\mathcal{D}(L_M)$. You'll end up with a basis of eigenfunctions in all these cases, but the basis changes with the endpoint conditions.

  • Periodic $f(-\pi)=f(\pi),f'(-\pi)=f'(\pi)$. The eigenfunctions are $$ 1,\cos(nx),\sin(nx),\;\; n=1,2,3,\cdots. $$
  • Zero Endpoint $f(-\pi)=0=f(\pi)$. The eigenfunctions are $$ \sin(nx/2),\;\;\; n=1,2,3,\cdots. $$
  • Zero Endpoint Derivatives $f'(-\pi)=0=f'(\pi)$. The eigenfucnctions are $$ 1,\cos(nx/2),\;\;\; n=1,2,3,\cdots. $$
  • Zero left endpoint, Zero right derivative, etc.

The eigenvalues and eigenfunctions change with the endpoint conditions, but they'll form an orthogonal basis of $L^2[-\pi,\pi]$ anyway. Most of the general conditions result in transcendental eigenvalue equations.

  • $\begingroup$ I have now submitted an answer to this question, but I think I came to the same conclusion that the boundary conditions would have to be periodic and that this would produce simple trigonometric eigenfunctions plus one constant eigenfunction. Many thanks for your in-depth answer. $\endgroup$ – Tom May 12 '18 at 17:39

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.