User:Dan Nessett/Sandboxes/Sandbox 2

An Associated Legendre Function ${P_\ell}^m$ is related to a Legendre polynomial $P_\ell$ by the following equation:
 * $\map { {P_\ell}^m} x = \paren {1 - x^2}^{m/2} \dfrac {\d^m \map {P_\ell} x} {\d x^m}, \qquad 0 \le m \le \ell$

Although extensions are possible, in this article $\ell$ and $m$ are restricted to integer numbers. For even $m$ the associated Legendre function is a polynomial, for odd $m$ the function contains the factor $\paren {1 - x^2}^{\frac 1 2}$ and hence is not a polynomial.

The associated Legendre functions are important in quantum mechanics and potential theory.

According to Ferrers the polynomials were named "Associated Legendre functions" by the British mathematician  in 1875, where "associated function" is Todhunter's translation of the German term zugeordnete Function, coined in 1861 by  Heine, and "Legendre"  is in honor of the French mathematician  (1752–1833), who was the first to introduce and study the functions.

Differential equation
Define:
 * $\map { {\Pi_\ell}^m} x \equiv \dfrac {\d^m \map {P_\ell} x} {\d x^m}$

where $\map {P_\ell} x$ is a Legendre polynomial.

Differentiating the Legendre differential equation:
 * $\paren {1 - x^2} \dfrac {\d^2 \map { {\Pi_\ell}^0} x} {\d x^2} - 2 x \dfrac {\d \map { {\Pi_\ell}^0} x} {\d x} + \ell \paren {\ell + 1} \map { {\Pi_\ell}^0} x = 0$

$m$ times gives an equation for ${\Pi_\ell}^m$:


 * $\paren {1 - x^2} \dfrac {\d^2 \map { {\Pi_\ell}^m} x} {\d x^2} - 2 \paren {m + 1} x \dfrac {\d \map { {\Pi_\ell}^m} x} {\d x} + \paren {\ell \paren {\ell + 1} - m \paren {m + 1} } \map { {\Pi_\ell}^m} x = 0$

After substitution of:
 * $\map { {\Pi_\ell}^m} x = \paren {1 - x^2}^{-m/2} \map { {P_\ell}^m} x$

and after multiplying through with $\paren {1 - x^2}^{m/2}$, we find the associated Legendre differential equation:
 * $\paren {1 - x^2} \dfrac {\d^2 \map { {P_\ell}^m} x} {\d x^2} - 2 x \dfrac {\d \map { {P_\ell}^m} x} {\d x} + \paren {\ell \paren {\ell + 1} - \dfrac {m^2} {1 - x^2} } \map { {P_\ell}^m} x = 0$

One often finds the equation written in the following equivalent way:
 * $\paren {\paren {1 - x^2} y'}' + \paren {\ell \paren {\ell + 1} - \dfrac {m^2} {1 - x^2} } y = 0$

where the primes indicate differentiation with respect to $x$.

In physical applications it is usually the case that $x = \cos \theta$, then the associated Legendre differential equation takes the form:
 * $\dfrac 1 {\sin \theta} \dfrac \d {\d \theta} \sin \theta \dfrac \d {\d \theta} {P_\ell}^m + \paren {\ell \paren {\ell + 1} - \dfrac {m^2} {\sin^2 \theta} } {P_\ell}^m = 0$

Extension to negative $m$
By the Rodrigues formula, one obtains:


 * $\map { {P_\ell}^m} x = \dfrac 1 {2^\ell \ell!} \paren {1 - x^2}^{m/2} \dfrac {\d^{\ell + m} } {\d x^{\ell + m} } \paren {x^2 - 1}^\ell$

This equation allows extension of the range of $m$ to $-m \le \ell \le m$.

Since the associated Legendre equation is invariant under the substitution $m \to -m$, the equations for ${P_\ell}^{\pm m}$, resulting from this expression, are proportional.

To obtain the proportionality constant consider:
 * $\paren {1 - x^2}^{-m/2} \dfrac {\d^{\ell - m} } {\d x^{\ell - m} } \paren {x^2 - 1}^\ell = c_{l m} \paren {1 - x^2}^{m/2} \dfrac {\d^{\ell + m} } {\d x^{\ell + m} } \paren {x^2 - 1}^\ell, \qquad 0 \le m \le \ell$

and bring the factor $\paren {1 - x^2}^{-m/2}$ to the other side.

Equate the coefficient of the highest power of x on the and  of:
 * $\dfrac {\d^{\ell - m} } {\d x^{\ell - m} } \paren {x^2 - 1}^\ell = c_{l m} \paren {1 - x^2}^m \dfrac {\d^{\ell + m} } {\d x^{\ell + m} } \paren {x^2 - 1}^\ell, \qquad 0 \le m \le \ell$

and it follows that the proportionality constant is:
 * $c_{l m} = \paren {-1}^m \dfrac {\paren {\ell - m}!} {\paren {\ell + m}!}, \qquad 0 \le m \le \ell$

so that the associated Legendre functions of same $\size m$ are related to each other by:
 * $\map { {P_\ell}^{-\size m} } x = \paren {-1}^m \dfrac {\paren {\ell - \size m}!} {\paren {\ell + \size m}!} \map { {P_\ell}^{\size m} } x$

Note that the phase factor $\paren {-1}^m$ arising in this expression is not due to some arbitrary phase convention, but arises from expansion of $\paren {1 - x^2}^m$.

Orthogonality relations
Important integral relations are:
 * $\ds \int_{-1}^1 \map { {P_l}^m} x \map { {P_k}^m} x \rd x = \dfrac 2 {2 l + 1} \dfrac {\paren {l + m} !} {\paren {l - m}!} \delta_{l k}$

and:
 * $\ds \int_{-1}^1 \map { {P_\ell}^m} x \map { {P_\ell}^n} x \dfrac {\d x} {1 - x^2} = \dfrac {\delta_{m n} \paren {\ell + m}!} {m \paren {\ell - m}!}, \qquad m \ne 0$

The latter integral for $n = m = 0$:
 * $\ds \int_{-1}^1 \map { {P_\ell}^0} x \map { {P_\ell}^0} x \dfrac {\d x} {1 - x^2}$

is undetermined (infinite). (see Orthogonality Proofs for details.)

Recurrence relations
The functions satisfy the following difference equations, which are taken from Edmonds.


 * $\paren {\ell - m + 1} \map { {P_{\ell + 1} }^m} x - \paren {2 \ell + 1} x \map { {P_\ell}^m} x + \paren {\ell + m} \map { {P_{\ell - 1} }^m} x = 0$


 * $x \map { {P_\ell}^m} x - \paren {\ell - m + 1} \paren {1 - x^2}^{1/2} \map { {P_\ell}^{m - 1} } x - \map { {P_{\ell - 1} }^m} x = 0$


 * $\map { {P_{\ell + 1} }^m} x - x \map { {P_\ell}^m} x - \paren {\ell + m} \paren {1 - x^2}^{1/2} \map { {P_\ell}^{m - 1} } x = 0$


 * $\paren {\ell - m + 1} \map { {P_{\ell + 1} }^m} x + \paren {1 - x^2}^{1/2} \map { {P_\ell}^{m + 1} } x - \paren {\ell + m + 1} x \map { {P_\ell}^m} x = 0 $


 * $\paren {1 - x^2}^{1/2} \map { {P_\ell}^{m + 1} } x - 2 m x \map { {P_\ell}^m} x + \paren {\ell + m} \paren {\ell - m + 1} \paren {1 - x^2}^{1/2} \map { {P_\ell}^{m - 1} } x = 0 $