Legendre Differential Equation

The second-order Ordinary Differential Equation

$$(1-x^2) {d^2y\over dx^2} - 2x {dy\over dx} + l(l+1)y = 0,$$ (1)

which can be rewritten

$${d\over dx} \left[{(1-x^2) {dy\over dx}}\right]+ l(l+1) y = 0.$$ (2)

The above form is a special case of the associated Legendre differential equation with $m=0$. The Legendre differential equation has Regular Singular Points at $-1$, 1, and $\infty$. It can be solved using a series expansion,

$$y = \sum_{n=0}^\infty a_nx^n$$ (3)

$$y' = \sum_{n=0}^\infty na_nx^{n-1}$$ (4)

$$y'' = \sum_{n=0}^\infty n(n-1)a_nx^{n-2}.$$ (5)

Plugging in,

$$(1-x^2)\sum_{n=0}^\infty n(n-1)a_nx^{n-2}-2x\sum_{n=0}^\infty na_nx^{n-1}+l(l+1)\sum_{n=0}^\infty a_nx^n=0$$ (6)

$\sum_{n=0}^\infty n(n-1)a_nx^{n-2}-\sum_{n=0}^\infty n(n-1)a_nx^n$
$-2x\sum_{n=0}^\infty na_nx^{n-1}+l(l+1)\sum_{n=0}^\infty a_nx^n=0\quad$	(7)

$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-\sum_{n=0}^\infty n(n-1)a_nx^n$
$-2\sum_{n=0}^\infty na_nx^n+ l(l+1)\sum_{n=0}^\infty a_nx^n=0\quad$	(8)

$\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^\infty n(n-1)a_nx^n$
$-2\sum_{n=0}^\infty na_nx^n+ l(l+1)\sum_{n=0}^\infty a_nx^n=0\quad$	(9)

$$\sum_{n=0}^\infty \{(n+1)(n+2)a_{n+2}+[-n(n-1)-2n+l(l+1)]a_n\} = 0,$$ (10)

so each term must vanish and

$$(n+1)(n+2)a_{n+2}-n(n+1)+l(l+1)]a_n = 0$$ (11)

$$a_{n+2} = {n(n+1)-l(l+1)\over (n+1)(n+2)} \,a_n$$

$$= - {[l+(n+1)](l-n)\over (n+1)(n+2)}\, a_n.$$ (12)

Therefore,

$$a_2 = -{l(l+1)\over 1\cdot 2} a_0$$ (13)

$$a_4 = -{(l-2)(l+3)\over 3\cdot 4} a_2$$

$$= (-1)^2 {[(l-2)l][(l+1)(l+3)]\over 1\cdot 2\cdot 3\cdot 4 } a_0$$ (14)

$$a_6 = -{(l-4)(l+5)\over 5\cdot 6} a_4$$

$$= (-1)^3 {[(l-4)(l-2)l][(l+1)(l+3)(l+5)]\over 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6} a_0,$$ (15)

so the Even solution is

$$y_1(x) = 1+\sum_{n=1}^\infty (-1)^n {[(l-2n+2)\cdots(l-2)l][(l+1)(l+3)\cdots (l+2n-1)] \over (2n)!} x^{2n}.$$ (16)

Similarly, the Odd solution is

$$y_2(x)= x+\sum_{n=1}^\infty(-1)^n{[(l-2n+1)\cdots(l-3)(l-1)][(l+2)(l+4)\cdots(l+2n)\over (2n+1)!}x^{2m+1}.$$ (17)

If $l$ is an Even Integer, the series $y_1$ reduces to a Polynomial of degree $l$ with only Even Powers of $x$ and the series $y_2$ diverges. If $l$ is an Odd Integer, the series $y_2$ reduces to a Polynomial of degree $l$ with only Odd Powers of $x$ and the series $y_1$ diverges. The general solution for an Integer $l$ is given by the Legendre Polynomials

$$P_n(x) =c_n\cases{ y_1(x) & for $l$\ even\cr y_2(x) & for $l$\ odd,\cr}$$ (18)

where $c_n$ is chosen so that $P_n(1) = 1$. If the variable $x$ is replaced by $\cos \theta$, then the Legendre differential equation becomes

$${d^2y\over d\theta^2} + {\cos\theta\over\sin\theta}{dy\over dx} + l(l+1)y = 0,$$ (19)

as is derived for the associated Legendre differential equation with $m=0$.

The associated Legendre differential equation is

$${d\over dx} \left[{(1-x^2) {dy\over dx}}\right]+ \left[{l(l+1) - {m^2\over 1-x^2}}\right]y = 0$$ (20)

$$(1-x^2) {d^2y\over dx^2} - 2x {dy\over dx}+\left[{l(l+1) - {m^2\over 1-x^2}}\right]y = 0.$$ (21)

The solutions to this equation are called the associated Legendre polynomials. Writing $x\equiv\cos\theta$, first establish the identities

$${dy\over dx} = {dy\over d(\cos\theta)} = - {1\over\sin\theta}{dy\over d\theta}$$ (22)

$$x{dy\over dx} = -{\cos\theta\over\sin\theta} {dy\over d\theta},$$ (23)

$${d^2y\over dx^2} = {1\over\sin\theta} {d\over d\theta}\left({{1\over\sin\theta}{dy\over d\theta}}\right)$$

$$= {1\over\sin\theta}\left({-\cos\theta\over\sin^2\theta}\right){dy\over d\theta}+{1\over\sin^2\theta} {d^2y\over d\theta^2},$$ (24)

and

$$1-x^2 = 1-\cos^2\theta = \sin^2\theta.$$ (25)

Therefore,

$$(1-x^2){d^2y\over dx^2} = \sin^2\theta {1\over\sin\theta} \left({-\cos\theta\over\sin^2\theta}\right){dy\over d\theta}+ {1\over\sin^2\theta} {d^2y\over d\theta^2}$$

$$= {d^2y\over d\theta^2}-{\cos\theta\over\sin\theta} {dy\over d\theta}.$$ (26)

Plugging (22) into (26) and the result back into (21) gives

$$\left({{d^2y\over d\theta^2}-{\cos\theta\over\sin\theta} {dy... ...r d\theta}+\left[{l(l+1) - {m^2\over\sin^2\theta}}\right]y = 0$$ (27)

$${d^2y\over d\theta^2} + {\cos\theta\over\sin\theta} {dy\over dx} + \left[{l(l+1) - {m^2\over\sin^2\theta}}\right]y = 0.$$ (28)

References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 332, 1972.