User:Dnessett/Legendre/Associated Legendre Functions Orthogonality for fixed m

[This page is deprecated in favor of User:Dnessett/Legendre/Associated Legendre Functions Orthonormality for fixed m, since the theorem not only shows the orthogonality of the Associated Legendre Functions, but also provides the normalization constant. All future modifications will be made to the referenced page.]

This article provides a proof that the associated Legendre functions are orthogonal for fixed m.

Theorem

$\int \limits _{-1}^{1}P_{\ell }^{m}\left(x\right)P_{k}^{m}\left(x\right)dx={\frac {2}{2\ell +1}}{\frac {\left(\ell +m\right)!}{\left(\ell -m\right)!}}\delta _{\ell k}.$

[Note: This article uses the more common $P_{\ell }^{m}$ notation, rather than $P_{\ell }^{\left(m\right)}$]

Where: $P_{\ell }^{m}\left(x\right)={\frac {\left(-1\right)^{m}}{2^{\ell }\ell !}}\left(1-x^{2}\right)^{\frac {m}{2}}{\frac {d^{\ell +m}}{dx^{\ell +m}}}\left[\left(x^{2}-1\right)^{\ell }\right],$ $0\leq m\leq \ell .$

Proof

The Associated Legendre Functions are regular solutions to the general Legendre equation: $\left(\left[1-x^{2}\right]y^{'}\right)^{'}+\left(\ell \left[\ell +1\right]-{\frac {m^{2}}{1-x^{2}}}\right)y=0$ , where $z^{'}={\frac {dz}{dx}}.$

This equation is an example of a more general class of equations known as the Sturm-Liouville equations. Using Sturm-Liouville theory, one can show that $K_{k\ell }^{m}=\int \limits _{-1}^{1}P_{k}^{m}\left(x\right)P_{\ell }^{m}\left(x\right)dx$ vanishes when $k\neq \ell .$ However, one can find $K_{k\ell }^{m}$ directly from the above definition, whether or not $k=\ell :$

$K_{k\ell }^{m}={\frac {1}{2^{k+\ell }\left(k!\right)\left(\ell !\right)}}\int \limits _{-1}^{1}\left\{\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right\}\left\{{\frac {d^{\ell +m}}{dx^{\ell +m}}}\left[\left(x^{2}-1\right)^{\ell }\right]\right\}dx.$

Since k and ℓ occur symmetrically, one can without loss of generality assume that $\ell \geq k.$ Integrate by parts $\ell +m$ times, where the curly brackets in the integral indicate the factors, the first being $u$ and the second $v'.$ For each of the first $m$ integrations by parts, $u$ in the $\left.uv\right|_{-1}^{1}$ term contains the factor $\left(1-x^{2}\right)$; so the term vanishes. For each of the remaining ℓ integrations, $v$ in that term contains the factor $\left(x^{2}-1\right)$; so the term also vanishes. This means:

$K_{kl}^{m}={\frac {\left(-1\right)^{l+m}}{2^{k+l}\left(k!\right)\left(l!\right)}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}{\frac {d^{l+m}}{dx^{l+m}}}\left[\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right]dx.$

Expand the second factor using Leibnitz' rule:

${\frac {d^{l+m}}{dx^{l+m}}}\left[\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right]=\sum \limits _{r=0}^{l+m}{\frac {\left(l+m\right)!}{r!\left(l+m-r\right)!}}{\frac {d^{r}}{dx^{r}}}\left[\left(1-x^{2}\right)^{m}\right]{\frac {d^{l+k+2m-r}}{dx^{l+k+2m-r}}}\left[\left(x^{2}-1\right)^{k}\right].$

The leftmost derivative in the sum is non-zero only when $r\leq 2m$ (remembering that $m\leq l$ ). The other derivative is non-zero only when $k+l+2m-r\leq 2k$, that is, when $r\geq 2m+(l-k).$ Because $l\geq k$ these two conditions imply that the only non-zero term in the sum occurs when $r=2m$ and $l=k.$ So:

$K_{kl}^{m}={\frac {\left(-1\right)^{l+m}}{2^{2l}\left(l!\right)^{2}}}{\frac {\left(l+m\right)!}{\left(2m\right)!\left(l-m\right)!}}\delta _{kl}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}{\frac {d^{2m}}{dx^{2m}}}\left[\left(1-x^{2}\right)^{m}\right]{\frac {d^{2l}}{dx^{2l}}}\left[\left(1-x^{2}\right)^{l}\right]dx.$

To evaluate the differentiated factors, expand $\left(1-x^{2}\right)^{k}$ using the binomial theorem: $\left(1-x^{2}\right)^{k}=\sum \limits _{j=0}^{k}\left({\begin{array}{c}k\\j\end{array}}\right)\left(-1\right)^{k-j}x^{2\left(k-j\right)}.$ The only thing that survives differentiation $2k$ times is the $x^{2k}$ term, which (after differentiation) equals: $\left(-1\right)^{k}\left({\begin{array}{c}k\\0\end{array}}\right)\left(2k\right)!=\left(-1\right)^{k}\left(2k\right)!$. Therefore:

$K_{kl}^{m}={\frac {1}{2^{2l}\left(l!\right)^{2}}}{\frac {\left(2l\right)!\left(l+m\right)!}{\left(l-m\right)!}}\delta _{kl}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx$ ................................................. (1)

Evaluate $\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}\,dx$ by a change of variable:

$x=\cos \theta \Rightarrow dx=-\sin \theta \,d\theta {\text{ and }}1-x^{2}=\sin \theta .$

Thus, $\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx=\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta .$ [To eliminate the negative sign on the second integral, the limits are switched from $\pi \rightarrow 0\;$ to $\;0\rightarrow \pi$ , recalling that $\;-1=\cos(\pi )\;$ and $\;1=\cos(0)\;$].

A table of standard trigonometric integrals shows: $\int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left.-\sin \theta \cos \theta \right|_{0}^{\pi }}{n}}+{\frac {\left(n-1\right)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta .$ Since $\left.-\sin \theta \cos \theta \right|_{0}^{\pi }=0,$ $\int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left(n-1\right)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta$ for $n\geq 2.$ Applying this result to $\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta$ and changing the variable back to $x$ yields: $\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx={\frac {2\left(l+1\right)}{2l+1}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l-1}dx$ for $l\geq 1.$ Using this recursively:

$\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx={\frac {2\left(l+1\right)}{2l+1}}{\frac {2\left(l\right)}{2l-1}}{\frac {2\left(l-1\right)}{2l-3}}\cdots {\frac {2\left(2\right)}{3}}\left(2\right)={\frac {2^{l+1}l!}{\frac {\left(2l+1\right)!}{2^{l}l!}}}={\frac {2^{2l+1}\left(l!\right)^{2}}{\left(2l+1\right)!}}.$

Applying this result to (1):

$K_{kl}^{m}={\frac {1}{2^{2l}\left(l!\right)^{2}}}{\frac {\left(2l\right)!\left(l+m\right)!}{\left(l-m\right)!}}{\frac {2^{2l+1}\left(l!\right)^{2}}{\left(2l+1\right)!}}\delta _{kl}={\frac {2}{2l+1}}{\frac {\left(l+m\right)!}{\left(l-m\right)!}}\delta _{kl}.$ QED.

See also

• Associated Legendre function
• Spherical harmonics

References

• Kenneth Franklin Riley, Michael Paul Hobson, Stephen John Bence, "Mathematical methods for physics and engineering", (2006) 3$^{rd}$ Edition, Cambridge University Press, ISBN 0-521-67971-0.