# Mathieu differential equation

Given the operator $T (\psi)(x):= \psi''(x)-2q \cos(2x)\psi(x)$ with $T : D(T) \subset L^2[0,2\pi]$

I was wondering: What is the right domain $D(T)$ for this operator if we want to solve the Mathieu equation Mathieu equation given by $$T(\psi) = \lambda_n \psi,$$ where $\lambda_n$ is the n-th eigenvalue of the Sturm-Liouville problem with boundary conditions:

$$\psi(0)= \psi(2 \pi),\psi'(0)= \psi'(2 \pi).$$

For sure, we want to take the domain in such a way, that the operator is self-adjoint and all possible solutions are in this domain. Therefore, I was wondering whether there is a canonical domain for this problem?

By the way: I would also be interested in the argument. Why this operator is self-adjoint on this domain. Probably, there is one standard argument why this is the case for all periodic sturm-liouville problems.

• Probably $H^2_{\mathrm{per}}$, the space of $H^2$ functions with $2\pi$-periodicity together with the first derivative. – Siminore Jun 30 '14 at 13:58
• No, you need to embed the periodicity of $f$ and $f'$. – Siminore Jun 30 '14 at 14:37

The operator $T\psi = \psi''-2q\cos(2x)\psi$ is a regular Sturm-Liouville operator because the coefficient of the highest order term does not vanish on the interval, and the other coefficients are nice on the interval. The domain $\mathcal{D}(T)$ is almost suggested in the comments. The Sobolev space consists of all twice-absolutely continuous functions on $[0,2\pi]$, and you need to add periodic conditions: $$\mathcal{D}(T)=\{ f \in L^{2} : f, f' \in \mathcal{AC}[0,2\pi],\; f(0)=f(2\pi),\; f'(0)=f'(2\pi),\;\; f'' \in L^{2}. \}.$$ For $f,g \in \mathcal{D}(T)$ it is easy to verify that $(Tf,g)=(f,Tg)$ where $(\cdot,\cdot)$ is the usual inner-product on $L^{2}[0,2\pi]$. It's not quite as obvious that $T$ is selfadjoint, and not just symmetric. However, this can be proved by showing that $(T\pm iI)$ are surjective. Equivalently, assume $g \in L^{2}[0,2\pi]$, and you must show that $$Tf_{1}-if_{1}=g,\;\; Tf_{2}+if_{2}=g$$ have solutions $f_{1},f_{2}\in \mathcal{D}(T)$. Such a result can be established in a classical way using variation of parameters on eigenfunctions with eigenvalues $\pm i$.
To carry out this plan using classical ODE theory, suppose $\phi_{i}$, $\psi_{i}$ are linearly independent classical solutions of $$-w''+2q\cos(2x)w-iw=0.$$ Then the Wronskian $W(\phi_{i},\psi_{i})=\phi_{i}\psi_{i}'-\psi_{i}\phi_{i}'$ is constant and non-zero. Rescale the solutions, if necessary, to obtain $W\equiv 1$ on $[0,2\pi]$. Then $$f=\phi_{i}(x)\int_{0}^{x} \psi_{i}(u)g(u)\,du+\psi_{i}(x)\int_{x}^{2\pi}\phi_{i}(u)g(u)\,du$$ is a solution of $-f''+2q\cos(2x)f-if = g$ because \begin{align} f' & =\phi_{i}'\int_{0}^{x}\psi g\,du+\psi_{i}'\int_{x}^{2\pi}\phi_{i}g\,du,\\ f'' & =\phi_{i}''\int_{0}^{x}\psi g\,du+\psi_{i}''\int_{x}^{2\pi}\phi_{i}g\,du + \phi_{i}'\psi g-\psi_{i}'\phi g \\ & =\phi_{i}''\int_{0}^{x}\psi g\,du+\psi_{i}''\int_{x}^{2\pi}\phi_{i}g\,du-g \end{align} The general solution of $-f''+2q\cos(2x)f-if=g$ is then $$f = A\phi_{i}+B\psi_{i}+\phi_{i}(x)\int_{0}^{x} \psi_{i}(u)g(u)\,du+\psi_{i}(x)\int_{x}^{2\pi}\phi_{i}(u)g(u)\,du.$$ What remains is to show that there exist constants $A$ and $B$ such that $f(0)=f(2\pi)$ and $f'(0)=f'(2\pi)$. For such $A$, $B$, one has $f \in \mathcal{D}(T)$ and $(T-iI)f=g$. It then follows that $T=T^{\star}$ on the stated domain $\mathcal{D}(T)$.
• NOTE: Showing that $(T\pm iI)$ are surjective also implies that $T$ is densely-defined, which saves a lot of headaches. – Disintegrating By Parts Jun 30 '14 at 21:43
• Sturm-Liouville theory is messy, and a thorough treatment never made it into any standard text, which is a shame. That's why I had to dig it out on my own. Classical existence/uniqueness theorems of ODES is what you want for this bit, though, which is a standard topic. If you have a homogenous ODE $a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_{0}y=0$, then there are $n$ linearly-independent $C^{n}$ solutions on $[a,b]$ if the coefficients functions $a_{j}(x)$ are continuous on $[a,b]$ and if $a_{n}(x)\ne 0$ on $[a,b]$. Such a theorem includes the case you want. – Disintegrating By Parts Jun 30 '14 at 22:43
• Note: For the ODE I just mentioned, there is a unique $y$ such that $y^{(k)}(x_{0})=c_{k}$ are specified for $0 \le k < n$. That's the correspondence with n-dimensional space. – Disintegrating By Parts Jun 30 '14 at 22:48
• You can use any two linearly-independent solutions. For example, choose $y_{1},y_{2}$ in your case so that $y_{1}(0)=1,y_{1}'(0)=0$ and $y_{2}(0)=0,y_{2}'(0)=1$. Such are independent because the vectors $(y(0),y'(0))$ are independent. For the argument I gave, the specifics of the values of $\phi_{i},\psi_{i}$ and their first derivatives at $0$ (or at $2\pi$) are not important, only the scaling so that the Wronskian is identically $1$. The Wronskian is constant in this case because of the missing $y'$ in the equation. For the more general case, the Wronskian(x) isn't constant. – Disintegrating By Parts Jun 30 '14 at 23:03