Self-Adjoint Operator on Even Periodic Functions

Question

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.

Answer 1

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.