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I began with the Laplace's equation in the context of spherical harmonics.

From wikipedia, one reads.

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So far I have followed, but in the sequel is stated that

$m \in \Bbb{R}$ since $\Phi$ is periodic. Then assume $Y(\theta,\varphi)$ is regular at the poles of the sphere ($\theta = 0,\pi$) this implies that $\lambda = l (l+1)$ for some integer $l \geq |m|$.

Why is this so?

I tried to work with $$\lim_{\theta \to 0}\lambda \sin^2 \theta + \frac{\sin \theta}{\Theta} \frac{d}{d\theta}\bigg(\sin\theta \frac{d\Theta}{d\theta}\bigg) = m^2 $$

But could only arrive at

$$\lim_{\theta \to 0}\lambda \sin^2 \theta \bigg(\lambda + \frac{\Theta''}{\Theta} \bigg)+ \sin \theta \cos \theta\frac{\Theta'}{\Theta} = m^2 $$

What is the way to go?

Community
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Conrado Costa
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2 Answers2

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Let's have a look at the radial equation that resulted from the separation of variables. That equation is

$$\frac{d}{dr}\left(r^2\frac{dR(r)}{dr}\right)-\lambda R(r)=0 \tag 1$$

Now, using the Frobenius Method we write a series solution for $R(r)$ of the form

$$R(r)=\sum_{n=0}^{\infty}a_nr^{n+\alpha} \tag 2$$

Using $(2)$ in $(1)$ reveals that $\lambda =\alpha(\alpha+1)$ and $a_n=0$ for all $n\ne 0$.

Therefore, the differential equation for $\Theta$ becomes

$$\frac{d}{d x} \left[ (1 - x^2) \frac{d}{d x} \bar\Theta(x) \right] + \left[ \alpha (\alpha + 1) - \frac{m^2}{1 - x^2} \right] \bar\Theta(x) = 0 \tag 3$$

where $x=\cos \theta$ and $\Theta(\theta)=\bar\Theta(\cos \theta)$. It can be shown that solution to $(3)$ is given by

$$\bar\Theta(x) = \frac{1}{\Gamma(1-m)} \left[\frac{1+x}{1-x}\right]^{m/2} \,_2F_1 (-\alpha, \alpha+1; 1-m; \frac{1-x}{2}) \tag 4$$

where $\Gamma$ is the Gamma Function and $_2F_1$ is the Hypergeometric Function.

The expression is $(4)$ can be shown to have a singularity at $x=-1$ unless $\alpha$ is an integer, say $\ell$ and $m$ is restricted by $-\ell \le m\le \ell$. In that case, solutions to $(4)$ are polynomials called the Associated Legendre Polynomials.

Mark Viola
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  • As always a good presentation and set of comments. There is, indeed, a restriction on $\ell$. – Leucippus Jul 21 '15 at 19:29
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    As a side comment it should read $-m \leq \ell \leq m$ in the last sentences. – Leucippus Jul 21 '15 at 19:31
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    @Leucippus Thank you! And nice catch! +1 It is that restriction that is the source of the OP's question, I believe. The ODE has solutions for non-integer $\ell$, but those have undesirable singularities. Another important point to note that neither of us raised is that the Spherical Harmonics form a complete orthogonal set. That is vital to the general expansion. Hope that you're doing well. I am still at home in recovery - hopefully just 1 more week on the wound vac device! – Mark Viola Jul 21 '15 at 19:34
  • How can $(4)$ be shown? – Conrado Costa Jul 21 '15 at 20:08
  • @ConradoCosta Substitute the solution into $(4)$ and see that you get the ODE for $_2F_1$. – Mark Viola Jul 21 '15 at 20:12
  • @Dr.MV, Thank you for your kindness. I am studying legendre polynomials, and I can see that they solve for $\lambda = l(l+1)$ with $l \in Bbb{Z}$. I am not pursuing Frobenius method yet. Since the Differential operator that defines Legendre polynomials is self-adjoint. I guess it remains to show that the solutions of that form (the legendre polynomials) are a complete orthonormal set. This will show that for no other $\alpha$ the above problem admits solution. I might be a while on this matter. – Conrado Costa Jul 23 '15 at 16:44
  • I believe so, still It will take me some time to understand all the way until the conclusion that if $\alpha \in \Bbb{Z}$ then we obtain a solution (as you mentioned there is a singularity at $x = -1$) could you explain why this singularity is an issue? – Conrado Costa Jul 23 '15 at 21:45
  • The singularity is only an issue if the points $\theta =0$ and $\theta =\pi$ are in the domain of the problem. If those points are in the domain, and if we seek non-singular solutions, then we retain only solutions for which $\alpha$ is an integer – Mark Viola Jul 23 '15 at 22:23
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\begin{align} \lambda \, \sin^{2}\theta + \frac{\sin\theta}{F} \frac{d}{d\theta}\left(\sin\theta \, \frac{dF}{d\theta}\right) = m^{2} \end{align} leads to \begin{align} \frac{1}{\sin\theta} \, \frac{d}{d\theta}\left(\sin\theta \, \frac{dF}{d\theta}\right) + \left[ \lambda - \frac{m^{2}}{\sin^{2}\theta} \right] \, F = 0 \end{align} or \begin{align} F'' + \frac{\cos\theta}{\sin\theta} \, F' + \left[ \lambda - \frac{m^{2}}{\sin^{2}\theta} \right] \, F = 0. \end{align} Since $\lambda$ is an arbitrary separation constant let $\lambda = l(l+1)$. This leads to the form of the associated Legendre differential equation and has solution \begin{align} F(\theta) = A \, P_{l}^{m}(\cos\theta) + B \, Q_{l}^{m}(\cos\theta). \end{align} Since $\theta \to 0$ is a required component of the problem then $B=0$ to keep the solution bounded.

Leucippus
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  • but why the boundary conditions imply that $\lambda = l (l+1)$? – Conrado Costa Jul 21 '15 at 14:15
  • @ConradoCosta 1) The form of the Associated Legendre Polynomials requires it to be that way; 2) The boundary conditions require a solution to be bounded as $\theta \to 0$ and $F(\theta+2\pi) = F(\theta)$, ie being continuous from one full rotation to the next. – Leucippus Jul 21 '15 at 14:23
  • Could you explain 1) in greater detail? that is indeed my question. – Conrado Costa Jul 21 '15 at 14:27
  • @ConradoCosta It is defied by the $m^{th}$ derivative of $P_{l}(x)$ multiplied by a factor as seen in the Associated Legendre Wiki page https://en.wikipedia.org/wiki/Associated_Legendre_polynomials. – Leucippus Jul 21 '15 at 14:44
  • Sorry I wasn't clear, the question is why $\lambda = l (l+1)$, in your answer you choose $\lambda = l(l+1)$ but this is a necessary form of $\lambda$? apparently yes, but this is what I fail to see. – Conrado Costa Jul 21 '15 at 14:57
  • @ConradoCosta It was clear in what was being asked. In general $\lambda$ is an arbitrary constant which is subject to change. There are two reasons why it changes to $\lambda = n(n+1)$ for $n$ being an integer. The reasons are: 1) the boundary conditions are $\theta \to 0$ and $F(\theta + 2 \pi) = F(\theta)$ and 2) the form of the differential equation has been defined as "known" with particular parameters. In this case it is the Associated Legendre polynomial. It kind of falls under "known forms" because they have been defined for many years and in today's view are considered as "accepted". – Leucippus Jul 21 '15 at 16:08
  • I believe your answer is correct and complete, I just am unable to learn from it since I don't see clearly why $\lambda = l (l+1)$. I am reading about Legendre polynomials and Sturm-Liouville problems. Maybe after this excursion I will be able to reach the answer. – Conrado Costa Jul 21 '15 at 16:15
  • @ConradoCosta It does take some understanding and acceptence. Books by T. S. Chihara, "Orthogonal Polynomials", and E. D. Rainville, "Special functions", start by these constants of differential equations being known. – Leucippus Jul 21 '15 at 16:46
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    @Leucippus If $\ell$ is not an integer with $|m|\le \ell$, then the solutions to the ODE have a singularity at $\theta = \pi$. Thus, in order for solutions to be non-singular, we must have $\ell$ an integer and we also must restrict $m$. Interesting stuff. – Mark Viola Jul 21 '15 at 19:10