# Solve: $P_n^m(\cos\theta_w)=0$ for $\theta_w$. Zeros of Associated Legendre Polynomial

I am working on boundary value problems with the associated Legendre Polynomial and have the condition that:

$$\frac{P_n^m(\cos\theta_w)}{\sin(\theta_w)}=0$$

I am trying to solve this equation for $$\theta_w$$ for any (positive integer) mode numbers ($$n,m$$).

I started by noting that the denominator is bounded as: $$-1<\sin(\theta_w)<1$$, and won't contribute to the roots.

This simplifies the solution that I need to solve to:

$$P_n^m(\cos\theta_w)=0.$$

I then began considering the definitions of associated Legendre polynomials (note the chain rule when reparameterizing in terms of angles),

$$P_n^m(\cos\theta_w)=\frac{(-1)^m}{2^nn!}(1-\cos^2\theta_w)^{m/2}\left(\frac{1}{-\sin\theta_w}\right)^{m+n}\frac{d^{n+m}}{d\theta_w^{n+m}}(\cos^2\theta_w-1)^n.$$

Now the factorial terms are constants and don't affect the roots, while the $$(1-\cos^2\theta_w)^{m/2}$$ gives zeros for $$\theta_w=0,\pi,2\pi,3\pi...$$, physically this is not realizable for the physical scenario I am modeling so this can be ignored. With the same bounds on $$\sin\theta$$ from above the $$\sin\theta$$ term can be ignored too. This reduces the equation that needs to be solved to:

$$0=\frac{d^{n+m}}{d\theta_w^{n+m}}(\cos^2\theta_w-1)^n.$$

Integrating this $$m+n$$ times gives:

$$(\cos^2\theta-1)^n=\sum_{i=0}^{m+n-1}\frac{c_i\theta_w^{i}}{i!}$$

which is simplified to:

$$-\sin^{2n}\theta_w=\sum_{i=0}^{m+n-1}\frac{c_i(\theta_w)^{i}}{i!}.$$

Here $$c_i$$'s are constants of integration. I think this is a transcendental equation, and am wondering if there are any analytical solutions known for this equation.

I had also looked at the definitions of associated Legendre polynomials involving hypergeometric functions, but to no avail.

An example for a specific mode with the $$m=0,n=2$$ gives:

$$0=\frac{P_2^0(\cos\theta_w)}{\sin(\theta_w)}=\frac{(3\cos^2\theta_w-1)}{2\sin(\theta_w)}$$

This gives the solution that $$\theta_w=\cos^{-1}(1/9)$$. Knowing this root, I found that the transcendental equation would have coefficients $$c_1=0,c_0=4/9$$, so that it reads $$-\sin^{2n}\theta_w=c_0$$. I'm sure if this offers any hints at generalizing to any mode numbers ($$n,m$$).

• If $m$ and $n$ are non-negative integers and $m \leq n$ then $\frac{{P_n^m (\cos \theta )}}{{\sin ^m \theta }}$ is a polynomial of degree $n-m$ in $\cos\theta$. I do not think there are explicit expressions for the zeros of that polynomial in general. If $m>n$ then the Legendre function is identically zero. A good reference for the Legendre functions is the DLMF: dlmf.nist.gov/14
– Gary
Commented Sep 22, 2020 at 15:55
• Thanks for the reference! I had noticed the conditions on $m>n$ in the numerical solutions, but it is nice to find a mathematical basis for them.
Commented Sep 22, 2020 at 16:44
• In F. W. J. Olver's book Asymptotics and Special Functions, p. 469, Ex. 12.5, there is an asymptotic approximation for the $r$th zero for large $n$ and fixed non-negative $m$.
– Gary
Commented Sep 22, 2020 at 17:22

I've recently been looking at that particular problem myself, and I think I've convinced myself that for some arbitrary angle ($$\theta_{w}$$ in your case) there is no general solution. However, the Associated Legendre Functions might. If you let your order be real-valued ($$P_{n}^{m} \rightarrow P_{\nu}^{m}$$ for $$\nu \in \mathbb{R}$$), then you can likely find values of $$\nu$$ which work (I'm currently on this path myself).
• @user3491364 I just discovered the complex-step approximation to derivatives: $$f'(x) \approx \Im \dfrac{f(x+ih)}{h} + O(h^{2})$$ This could be used with Newton's method to find roots to find the specific values of $\nu$ so that $P_{\nu}^{m}(x) = 0$ for a fixed $x$.