Boundary conditions for spherical harmonics

Question

How does the constraint that the solution to

$ \\ $ $$\left((1-x^2)y'\right)' - \frac{m^2}{1-x^2}y = \lambda y$$

$ \\ $

be square integrable on $[-1,1]$, force the solution to be bounded at $\pm 1$?

Answer 1

Take the case where $m=0$, which is the ordinary Legendre equation. The solutions for $\lambda=0$ are obtained by solving $((1-x^2)y')'=0$: $$ (1-x^2)y' = C \\ y' = \frac{C}{1-x^2}= \frac{C}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right) \\ y = \frac{C}{2}\ln\left(\frac{1+x}{1-x}\right)+D. $$ All solutions are in $L^2(-1,1)$, but $y$ is not bounded near $\pm 1$ for $C\ne 0$. So, your assertion is not correct for $m=0$. An endpoint condition is required in order for this Sturm-Liouville equation to be selfadjoint, and it turns out that boundedness near $\pm 1$ are endpoint conditions that lead to a selfadjoint problem. This equation is in the limit circle case at both endpoints.

The cases where $m=1,2,3,\cdots$ are different. In these cases, the equation is already essentially selfadjoint without the need to impose endpoint conditions; only $L^2(-1,1)$ is required as a condition. These equations are in the limit point case, where no endpoint conditions are needed for a selfadjoint problem. The only eigenfunctions in $L^2(-1,1)$ turn out to be bounded.

So why do we want a selfadjoint problem? The reason for having a full-blown selfadjoint problem is that you can expand every function in the eigenfunctions; in general, such an expansion may take the form of a Fourier series, a Fourier type integral, or a combination of both. But the important thing is that everything can be built from the pure states of the system in an orthogonal expansion, which is basically a requirement for an axiomatic foundation of Quantum. The Hamiltonian for the Hydrogen isotope requires both discrete and continuous components for a full expansion. The discrete states are bound states, while the continuous states, which require integrals to "sum" them, are unbound states.

In the case of the Associated Legendre equation for $m=1,2,3,\cdots$, it turns out that there are no endpoint conditions that are needed for a selfadjoint problem. The unconstrained problem is already selfadjoint. The only qualification for a classical eigenfunction to be an actual eigenfunction is that the classical eigenfunction be in $L^2(-1,1)$, and all such eigenfunctions turn out to be bounded. But boundedness is not required; it is a consequence of the equation and the requirement that the solution be in $L^2(-1,1)$. For a fixed $m=1,2,3,\cdots$, the set of such eigenfunctions is a complete orthogonal basis of $L^2(-1,1)$ that happens to consist of bounded functions on $(-1,1)$.

To show boundedness of eigenfunctions: To get technical about boundedness, suppose $m=1,2,3,\cdots$ and define $L_m$ by $$ L_m f = -\frac{d}{dx}\left((1-x^2)\frac{df}{dx}\right)+\frac{m^2}{1-x^2}f. $$ Suppose $L_mf = \lambda f$ for some real $\lambda$, and suppose $f\in L^2(-1,1)$. You can assume $f$ is real because both the real and imaginary parts will satisfy the same equation $L_mf=\lambda f$. Then, for $-1 < y < 1$, the following is a uniformly bounded function of $y$: \begin{align} \lambda \int_{0}^{y}f(x)^2dx & = \int_{0}^{y}(L_mf) f dx \\ & = -\int_{0}^{y}((1-x^2)f')'f dx+\int_{0}^{y}\frac{m^2f^2}{1-x^2}dx \\ & = -(1-x^2)f'f|_{0}^{y}+\int_{0}^{y}(1-x^2)f'^2+\frac{m^2f^2}{1-x^2}dx \end{align} It's not easy to argue, but it can be shown that the integral term on the far right must remain bounded as $y\rightarrow\pm 1$. (In fact, it can be shown the evaluation terms on the right must be $0$ as $y\rightarrow\pm 1$.) Therefore any $L^2$ eigenfunction remains uniformly bounded by \begin{align} \left|\frac{f^2(y)}{2}\right| & = \left|\frac{f^2(0)}{2}+\int_{0}^{y}f'fdx\right| \\ & \le \left|\frac{f^2(0)}{2}\right|+\left|\int_{0}^{y} \sqrt{1-x^2}f'\frac{f}{\sqrt{1-x^2}}dx\right| \\ & \le \left|\frac{f^2(0)}{2}\right|+\frac{1}{2}\int_{0}^{y}(1-x^2)f'^2+\frac{f^2}{1-x^2}dx \\ & \le \left|\frac{f^2(0)}{2}\right|+\frac{1}{2}\int_{-1}^{1}(1-x^2)f'^2+\frac{m^2f^2}{1-x^2}dx \\ & \le \frac{|f(0)|^2}{2}+\frac{1}{2}\|\sqrt{1-x^2}f'\|_{L^2}^2+\frac{1}{2}\|mf/\sqrt{1-x^2}\|_{L^2}^2. \end{align}