Trigonometric Form of the Legendre Differential Equation
Definition
The associated Legendre differential equation in the form of a trigonometric function is as follows.
$$ \begin{align} \frac{ d^{2} y}{ d \theta^{2} }+\cot \theta \frac{ d y}{ d \theta}+ \left( l(l+1) -\frac{m^{2}}{\sin ^{2 }\theta} \right)y=0 \\ \mathrm{or} \quad\frac{1}{\sin \theta}\left(\sin \theta \frac{dy}{d\theta} \right)+ \left(l(l+1) -\frac{ m^{2}}{\sin ^{2} \theta} \right)y=0 \end{align} $$
Explanation
Useful for solving spherical coordinate Laplace’s equation in electromagnetics, quantum mechanics, etc. The solutions are as follows.
$$ \begin{align*} y &= P_{l}^{m}(\cos \theta) \\ &= (1-\cos ^{2}\theta)^{\frac{|m|}{2}} \frac{ d^{|m|} }{ dx^{|m|} } P_{l}(\cos\theta) \end{align*} $$
$P_{l}^{m}(x)$ is called the associated Legendre polynomial, and $P_{l}(x)$ is called the Legendre polynomial.
$$ P_{l}(x)=\dfrac{1}{2^l l!} \dfrac{d^l}{dx^l}(x^2-1)^l $$
Derivation
The associated Legendre differential equation is as follows.
$$ (1-x^{2})\frac{ d^{2}y }{ dx^{2} }-2x \frac{dy}{dx}+\left( \frac{-m^{2}}{1-x^{2}}+l(l+1)\right)y=0 \tag{3} $$
By substituting $x=\cos \theta$ into $dx=-\sin \theta d\theta$, the following is obtained.
$$ \frac{dy}{dx}=\frac{dy}{d\theta}\frac{ d \theta }{ dx }=-\frac{1}{\sin \theta}\frac{ dy }{ d\theta } $$
And it is calculated as follows.
$$ \begin{align*} \frac{ d ^{2} y}{ d^{2}x } &= \frac{ d }{ dx }\left( -\frac{1}{\sin \theta} \frac{ d y}{ d\theta }\right) \\ &= \frac{ d }{ d\theta }\left( -\frac{1}{\sin \theta} \frac{ d y}{ d\theta }\right)\frac{ d \theta }{ dx } \\ &= \left( \frac{\cos \theta}{\sin^{2} \theta} \frac{ d y}{ d\theta } -\frac{1}{\sin \theta}\frac{ d^{2}y }{ d\theta^{2} }\right)\left( -\frac{1}{\sin \theta} \right) \\ &= \frac{1}{\sin ^{2} \theta} \left( \frac{ d ^{2}y }{ d \theta^{2} }-\cot\theta \frac{ d y}{ d\theta }\right) \end{align*} $$
Therefore, substituting this into $(3)$ results in the following.
$$ (1-\cos ^{2 \theta})\left( \frac{1}{\sin ^{2}\theta}\left(\frac{ d ^{2}y }{ d\theta ^{2} }-\cot \theta \frac{ d y}{ d\theta }\right) \right)+2\frac{\cos \theta}{\sin \theta}\frac{ d y}{ d \theta }+\left( \frac{-m^{2}}{1-\cos ^{2}\theta} +l(l+1)\right)y=0 $$
Organizing gives $(1)$.
$$ \begin{align*} &&\frac{ d ^{2}y}{ d \theta^{2} }-\cot \theta \frac{ d y}{ d\theta } + 2\cot \theta \frac{ d y}{ d\theta } +\left( \frac{-m^{2}}{\sin^{2} \theta} +l(l+1)\right)y=0 \\ \implies& &\frac{ d ^{2}y}{ d \theta^{2} }+\cot \theta \frac{ d y}{ d\theta } +\left( \frac{-m^{2}}{\sin ^{2}\theta} +l(l+1)\right)y=0 \end{align*} $$
Combining the second and first terms yields $(2)$.
$$ \frac{1}{\sin \theta} \left( \sin \theta \frac{ d y}{ d\theta } \right)+\left( \frac{-m^{2}}{\sin ^{2}\theta} +l(l+1)\right)y=0 $$