📂Odinary Differential Equations

Associated Legendre Differential Equations and Polynomials

Definition¹

The differential equation given below is called the associated Legendre differential equation.

$$ \begin{equation} \begin{aligned} &&(1-x^{2})\frac{ d^{2}y }{ dx^{2} }-2x \frac{dy}{dx}+\left[ +l(l+1)-\frac{m^{2}}{1-x^{2}} \right]y =&\ 0 \\ \mathrm{or} && \frac{ d }{ dx } \left[ (1-x^{2})y^{\prime} \right] +\left[ l(l+1)-\frac{m^{2}}{1-x^{2}} \right]y =&\ 0 \end{aligned} \label{1} \end{equation} $$

The solution to the associated Legendre differential equation is denoted as $P_{l}^{m}(x)$, and this is called the associated Legendre polynomial or the generalized Legendre polynomial.

$$ \begin{align*} P_{l}^{m}(x)&= (1-x ^{2})^{\frac{|m|}{2}} \frac{ d^{|m|} }{ dx^{|m|} } P_{l}(x) \\ &=(1-x ^{2})^{\frac{|m|}{2}} \frac{ d^{|m|} }{ dx^{|m|} }\left[ \dfrac{1}{2^{l} l!} \dfrac{d^{l}}{dx^{l}}(x^2-1)^{l}\right] \end{align*} $$

Here, $P_{l}(x)$ is the Legendre polynomial. In cases distinguished by the sign of $m$,

$$ P_{l}^{m}(x) = (1-x ^{2})^{\frac{m}{2}} \dfrac{1}{2^{l} l!} \dfrac{d^{l+m}}{dx^{l+m}}(x^2-1)^{l} $$

$$ P_{l}^{-m}=(-1)^{m}\frac{(l-m)!}{(l+m)!}P_{l}^{m}(x) $$

The associated Legendre polynomials appear when solving the Laplace equation in spherical coordinates. Here, the constants $l$, $m$ are related to the quantum numbers in quantum mechanics.

Solution

For $m=0$, it’s the Legendre differential equation. Based on the solution for this case, we can also find a solution for when it is $m\ne 0$. Initially, since the solution to the associated Legendre differential equation is determined by the constants $l$, $m$, let’s denote it as follows.

$$ y=P_{l}^{m}(x) $$

Substitute this into $\eqref{1}$ and organize it as below.

$$ \begin{equation} \frac{ d }{ dx }\left[ (1-x^{2})\frac{ d P_{l}^{m}(x)}{ dx } \right]+\left[ l(l+1)-\frac{m^{2}}{1-x^{2}} \right]P_{l}^{m}(x)=0 \label{2} \end{equation} $$

And let’s assume that the solution has the form below.

$$ P_{l}^{m}(x)=(1-x^{2})^{\frac{|m|}{2}}u(x) $$

Differentiating $x$ once gives

$$ \frac{ d P_{l}^{m}(x)}{ d x }=-|m|x(1-x^{2})^{\frac{|m|}{2}-1}u(x)+(1-x^{2})^{\frac{|m|}{2}}u^{\prime}(x) $$

Substituting this into the first term of $\eqref{2}$ and organizing gives the following.

$$ \begin{align*} \frac{ d }{ dx }\left[ (1-x^{2})\frac{ d P_{l}^{m}(x)}{ dx } \right] =&\ \frac{ d }{ dx }\left[ -|m|x(1-x^{2})^{\frac{|m|}{2}}u(x)+(1-x^{2})^{\frac{|m|}{2}+1}u^{\prime}(x) \right] \\ =&\ -|m|(1-x^{2})^{\frac{|m|}{2}}u(x)+|m|^{2}x^{2}(1-x^{2})^{\frac{|m|}{2}-1}u(x) \\ & -|m|x(1-x^{2})^{\frac{|m|}{2}}u^{\prime}(x)-(|m|+2)x(1-x^{2})^{\frac{|m|}{2}}u^{\prime}(x) \\ & +(1-x^{2})^{\frac{|m|}{2}+1}u^{\prime \prime}(x) \\ =&\ (1-x^{2})^{\frac{|m|}{2}+1}u^{\prime \prime}(x)-2(|m|+1)(1-x^{2})^{\frac{|m|}{2}}u^{\prime}(x) \\ & -[|m|(|m|+1)x^{2}-|m|] (1-x^{2})^{\frac{|m|}{2}-1}u(x) \end{align*} $$

Multiplying both sides by $\dfrac{1}{(1-x^{2})^{|m|/2}}$ gives

$$ \begin{align*} &\frac{1}{(1-x^{2})^{|m|/2}}\frac{ d }{ dx }\left[ (1-x^{2})\frac{ d P_{l}^{m}(x)}{ dx } \right] \\ =&\ (1-x^{2})u^{\prime \prime}(x)-2(|m|+1)xu^{\prime}(x) -[|m|(|m|+1)x^{2}-|m|] (1-x^{2})^{-1}u(x) \end{align*} $$

Therefore, multiplying both sides of $\eqref{2}$ by $\dfrac{1}{(1-x^{2})^{|m|/2}}$ gives

$$ \begin{equation} \begin{aligned} &(1-x^{2})u^{\prime \prime}(x)-2(|m|+1)xu^{\prime}(x) \\ &-\left( \frac{|m|(|m|+1)x^{2}-|m|}{1-x^{2}}+l(l+1)-\frac{m^{2}}{1-x^{2}}\right)u(x)=0 \end{aligned} \label{1} \end{equation} $$

Organizing the coefficients of $u(x)$ gives the following.

$$ \begin{align*} &\frac{|m|(|m|+1)x^{2}-|m|}{1-x^{2}}+l(l+1)-\frac{m^{2}}{1-x^{2}} \\ =&\ \frac{|m|(|m|+1)x^{2}-|m|+l(l+1)(1-x^{2})-m^{2}}{1-x^{2}} \\ =&\ \frac{-m^{2}(1-x^{2})-|m|(1-x^{2})+l(l+1)(1-x^{2})}{1-x^{2}} \\ =&\ l(l+1)-m^{2}-|m| \\ =&\ l(l+1)-|m|(|m|+1) \end{align*} $$

Thus, $\eqref{3}$ is organized into the form below.

$$ \begin{equation} (1-x^{2})\frac{ d^{2} u }{ d x^{2} }-2(|m|+1)x\frac{ d u}{ dx }+[l(l+1)-|m|(|m|+1)]u=0 \label{4} \end{equation} $$

If it is $m=0$, it indeed becomes the Legendre differential equation. Therefore, the solution for when it is $|m|=0$ is $P_{l}^{0}(x)=P_{l}(x)$. Now, let’s differentiate $(4)$ relative to $x$ again. Organizing the coefficients gives the following equation.

$$ \begin{equation} (1-x^{2}) \frac{ d^{3} u }{ d x^{3} } -2[(|m|+1)+1]x\frac{ d^{2} u}{ dx^{2} }+[l(l+1)-(|m|+1)(|m|+2)]\frac{ d u}{ d x}=0 \label{5} \end{equation} $$

Differentiating $\eqref{5}$ relative to $x$ again and organizing the coefficients gives the following.

$$ \begin{equation} (1-x^{2}) \frac{ d^{4} u }{ d x^{4} } -2[(|m|+2)+1]x\frac{ d^{3} u}{ dx^{3} }+[l(l+1)-(|m|+2)(|m|+3)]\frac{ d^{2} u}{ d x^{2}}=0 \label{6} \end{equation} $$

Examining the equations above shows that when $\eqref{4}$ is substituted with $u$ into $\dfrac{ d u}{ d x }$, and $|m|$ is substituted with $|m|+1$ into $\eqref{5}$, we can obtain $\eqref{5}$. Using the same method of substitution for $\eqref{5}$ gives $\eqref{6}$. From this, we can learn that the solution when it is $|m|=0$ is $P_{l}(x)$, when it is $|m|=1$ it is $\dfrac{ d }{ d x }P_{l}(x)$, and when it is $|m|=2$ it is $\dfrac{d^{2}}{dx^{2}}P_{l}(x)$. Therefore, generalizing this gives the following.

$$ u(x)=\frac{d^{|m|}}{dx^{{|m|}}}P_{l}(x) $$

Therefore, the associated Legendre polynomials are as follows.

$$ \begin{align*} P_{l}^{m}(x)&= (1-x ^{2})^{\frac{|m|}{2}} \frac{ d^{|m|} }{ dx^{|m|} } P_{l}(x) \\ &=(1-x ^{2})^{\frac{|m|}{2}} \frac{ d^{|m|} }{ dx^{|m|} }\left[ \dfrac{1}{2^{l} l!} \dfrac{d^{l}}{dx^{l}}(x^2-1)^{l}\right] \end{align*} $$

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