14 - Legendre Function

Motivation of Legendre Equation

Solving Helmholtz Equation using spherical polars $(r,\theta ,\varphi )$ so Laplacian is given by

$${\nabla }^{2}u=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}(ru)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial u}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^{2}u}{\partial {\varphi }^2}=-k^{2}u$$

Separating variables seek solution in form

$$u(r,\theta ,\varphi )=R(r)\Theta (\theta )\Phi (\varphi )$$

Rearranging equation to

$$\frac{r(rR(r){)}^{\prime \prime }}{R(r)}+k^2{r}^2=-\frac{(\sin \theta {\Theta }^{\prime }(\theta ){)}^{\prime }}{\sin \theta \Theta (\theta )}-\frac{{\Phi }^{\prime \prime }(\varphi )}{{\sin }^2\theta \Phi (\varphi )}$$

By usual argument then it must be equal to constant $\lambda$ so

$$-\frac{{\Phi }^{\prime \prime }(\varphi )}{\Phi (\varphi )}=\frac{\sin \theta (\sin \theta {\Theta }^{\prime }(\theta ){)}^{\prime }}{\Theta (\theta )}+\lambda {\sin }^2\theta$$

As $\Phi$ is $2\pi$-periodic function the constant is in form $m^2$ where $m\ge 0$ is an integer
Hence $\Phi =\text{const}$ if

$$m=0\text{ or }\Phi (\varphi )=\cos (m\varphi +\alpha )\text{ if }m>0$$

Hence results in following linear ODE for $\Theta (\theta )$

$$\frac{1}{\sin \theta }\frac{d}{d\theta }\left(\sin \theta \frac{d\Theta }{d\theta }\right)+\left(\lambda -\frac{m^2}{{\sin }^2\theta }\right)\Theta =0$$

As $\theta =0$ and $\theta =\pi$ are both regular singular points then

$$\Theta (\theta )\text{ is sufficiently well-behaved as }\theta \to 0,\pi$$

Hence using change of variables $\cos \theta =x$ and $\Theta (\theta )=y(x)$ with $\frac{d}{d\theta }=-\sin \theta \frac{d}{dx}$ so

$$\frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right)+\left(\lambda -\frac{m^2}{1-x^2}\right)y=0$$

which is known as the associated Legendre Equation for $y(x)$

Associated Legendre Equation

$$\frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right)+\left(\lambda -\frac{m^2}{1-x^2}\right)y=0$$

where $m,\lambda$ can take any complex value generally

When $m=0$ then it is known as the Legendre Equation and Functions

Note that generally focusing on $m,l\in {\mathbb{Z}}^{>0}$ and $\lambda =l(l+1)$

Properties of Legendre Functions

1. $x=\pm 1$ and $x=\infty$ are regular singular points of Associated Legendre Equation
   Indicial Exponents for $x=\pm 1$ are $-\frac{m}{2}$ and $\frac{m}{2}$
   Hence local expansion gives one bounded and unbounded solution as $x\to 1$ and $x\to -1$
   Note that if $m=0$, there is repeated root so solution is of order $\log (x\mp 1)$ as $x\to \pm 1$

2. Consider bounded solutions on $-1<x<1$ then
   Boundedness imposes two conditions, one at either end of interval
   Hence get posed singular Sturm-Liouville problem

$$\begin{array}{rlrl}-((1-x^2)y^{\prime }(x){)}^{\prime }+\frac{m^2}{1-x^2}y(x)& =\lambda y(x)& & \text{ for }-1<x<1\\ y(x)& \text{ bounded}& & \text{ as }x\to \pm 1\end{array}$$

3. Eigenvalues of Sturm-Liouville Form given by $\lambda =l(l+1)$ with integer $l\ge m\ge 0$
   Eigenfunctions are the corresponding associated Legendre functions denoted by

$$y(x)=P_l^m(x)$$

From Sturm-Liouville Theory then orthogonality relation is

$${\int }_{-1}^1{P}_k^m(x)P_l^m(x)dx=0\text{ for }k\ne l$$

4. For $m=0$ and integer $l\ge 0$ then
   Legendre functions are denoted by $P_l(x)$
   Hence $P_l$ is a polynomial of degree $l$ so if solution is as a power series expansion at $x=0$

$$y(x)=\sum _{k=0}^{\infty }{a}_k{x}^k$$

then series terminates with $a_k\equiv 0$ for $k>l$

Hence resulting Legendre Polynomials are given by Rodrigues’ Formula

$$P_l(x)=\frac{1}{2^{l}l!}\frac{d^l}{d{x}^l}[(x^2-1{)}^l]$$

5. Second, linearly independent solution of Legendre Equation $m=0$ is given by
   Legendre Function of Second Kind, denoted by $Q_l(x)$
   Solutions are unbounded as $x\to \pm 1$
   For case $l=0$ then solution $Q_0$ is

$$Q_0(x)=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$$

6. General Case of nonzero $m\le l$ then
   Associated Legendre Functions of First and Second Kind are given by

$$\begin{array}{rl}{P}_l^m(x)& =(-1{)}^m(1-x^2{)}^{m/2}\frac{d^m{P}_l(x)}{d{x}^m}\\ Q_l^m(x)& =(-1{)}^m(1-x^2{)}^{m/2}\frac{d^m{Q}_l(x)}{d{x}^m}\end{array}$$

So associated Legendre Function $P_l^m$ is a polynomial if and only if $m$ is even