I am looking to find the Legendre polynomial for the following standard Associated Legendre equation:
$$\left(\frac{d}{dx}[1-x^2]\frac{d}{dx}+\lambda-\frac{m^2}{1-x^2}\right)\Theta(x)=0$$
I have found the indicial equation for m=0, but not for m$\neq$0. How do I go about solving this case?
Best Answer
What you can show is that, if $y$ is a solution of $$ \frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right)+\mu y =0, $$ then $$ w=(1-x^2)^{m/2}\frac{d^my}{dx^m} $$ is a solution of $$ \frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right)-\frac{m^2}{1-x^2}y+\mu y = 0. \tag{$\dagger$} $$ Therefore, if $P_n(x)$ is the Ordinary Legendre Polynomial, which is the polynomial solution of $((1-x^2)y')'+n(n+1)y=0$, then $$ P_{n,m}(x) = (1-x^2)^{m/2}\frac{d^{m}}{dx^{m}}P_{n}(x) $$ is a solution of the Associated Legendre Equation $(\dagger)$. This holds for a fixed $m$ and for all $n \ge m$ (for $m > n$, the above is identically $0$.) So the Associated Legendre equation has eigenvalues $$ \mu_n=n(n+1),\;\;\; n=m,m+1,m+2,\cdots. $$ For example: $x$ is a solution of the ordinary equation with $\mu=(1)(1+1)=2$. Therefore $$ \sqrt{1-x^2}\frac{d}{dx}x = \sqrt{1-x^2} $$ is a solution of the associated equation with $\mu=(2)(1)$. To check this, $$ \frac{d}{dx}\left((1-x^2)\frac{d}{dx}\sqrt{1-x^2}\right)-\frac{1}{1-x^2}\sqrt{1-x^2} \\ = -\frac{d}{dx}\left((1-x^2)\frac{x}{\sqrt{1-x^2}}\right)-\frac{1^2}{\sqrt{1-x^2}} \\ = -\frac{d}{dx}(x\sqrt{1-x^2})-\frac{1}{\sqrt{1-x^2}} \\ = -\sqrt{1-x^2}+x\frac{x}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \\ = -\sqrt{1-x^2}+\frac{x^2-1+1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \\ = -2\sqrt{1-x^2} $$ Therefore, as expected, $f=\sqrt{1-x^2}$ is a solution of $$ \frac{d}{dx}\left((1-x^2)\frac{df}{dx}\right)-\frac{1^2}{1-x^2}f+(2)(1)f=0. $$ Using the Method of Frobenius: Start by rewriting the equation as $$ \frac{d^2f}{dx^2}-\frac{2x}{1-x^2}\frac{df}{dx}-\frac{m^2}{(1-x^2)^2}f+\frac{\lambda}{1-x^2}f = 0. $$ In order to consider the equation at the singular point $x=-1$, which is a regular singular point, you have to expand in powers of $(x+1)$. For $x=-1$, you only need the coefficients $A$, $B$, $C$ shown below $$ \frac{d^2f}{dx^2}+\left[\frac{A}{x+1}+\cdots\right]\frac{df}{dx}+\left[\frac{B}{(x+1)^2}+\cdots\right]f=0. $$ You can easily determine $A$ and $B$. For example, $$ -\frac{2x}{1-x^2}=\frac{1}{x+1}\left[\frac{2x}{x-1}\right] \implies A = \left.\frac{2x}{x-1}\right|_{x=-1} = 1. \\ -\frac{m^2}{(1-x^2)^2}=\frac{1}{(x+1)^2}\left[-\frac{m^2}{(x-1)^2}\right] \implies B=-\left.\frac{m^2}{(x-1)^2}\right|_{x=-1}=-\frac{m^2}{4}. $$ So the indicial equation comes from solving Euler's equation $$ (x+1)^2f''+A(x+1)f'+Bf = 0,\;\;\; f=(x+1)^r \\ r(r-1)+Ar+B=0 \\ r^2-r+r-\frac{m^2}{4}=0 \\ r^2=\frac{m^2}{2^2} \\ r = \pm \frac{m}{2}. $$ Only the root with the largest real part--in this case $\frac{m}{2}$--is guaranteed to give a solution $(x+1)^{m/2}\sum_{n=0}^{\infty}a_n(x+1)^n$. If you want to take out the singularity at both endpoints $x=\pm 1$, then you can substitute $y=(1-x^2)^{m/2}w$ into the original equation and then solve by a series. In other words, the method of Frobenius leads to the same place in terms of the substitution.