A closer look

Let us take a careful look at the equation,

$$(1-x^2)y'' - 2xy' + \Delta y=0$$ (3.2)

We note that the x derivative of $(1-x^2)$ is $-2x$ . This is highly suggestive that the kernel of our function is $(1-x^2)$ itself. The multiple derivatives suggest something a la $w=(1-x^2)^n$ . So let us explore this kernel. Its first derivative is $w'=-2xn(1-x^2)^{n-1}$ or $w'(1-x^2) + 2nxw = 0$ Define $D_n$ to be the nth derivative of this latter equation. Then we have the following formulation:

$$D_0: (1-x^2)w' + 2nxw = 0$$

$$D_1: (1-x^2)w'' -2xw' + 2nxw' + 2nw$$

$$=(1-x^2)w'' + 2x(n-1)w' + 2nw = 0$$

$$D_2: (1-x^2)w''' -2xw'' + 2x(n-1)w'' + 2(N-1)w' + 2nw'$$

$$= (1-x^2)w''' + 2x(n-2)w'' + (4n-2)w' = 0$$

$$D_3: (1-x^2)w^{(4)} - 2xw^{(3)} + 2x(n-2)w^{(3)} + 2(n-2)w{(2)} + (rn-2)w^{(2)}$$

$$=(1-x^2)w^{(4)} + 2x(n-3)w^{(3)} + (6n-6)w^{(2)} = 0$$

If we have instead $w=(x^2-1)^n$ :

$$D_0: (x^2 -1)w' -2nxw = 0$$

$$D_1: (x^2 -1)w'' - 2(n-1)xw' -2nxw = 0$$

$$D_2: (x^2 -1)w^{(3)} -2(n-2)xw^{(2)} -2(2n-1)w' = 0$$

$$D_3: (x^2 -1)w^{(4)} - 2(n-3)xw^{(3)} - 2(3n-3)w^{(2)} = 0$$

$$\vdots \vdots = \vdots$$

$$D_{(n+1)}: (x^2-1)w''^{(n)} + 2xw'^{(n)} -2\left(n^2-\sum_{m=0}^n(m-1)\right)w^n = 0$$

$$(x^2-1)w''^{(n)} + 2xw'^{(n)} -2\left(n^2-\frac{n(n-1)}{2}\right)w^n = 0$$

$$(x^2-1)w''^{(n)} + 2xw'^{(n)} -n(n+1)w^n = 0$$

This latter equation is in the same form as our original equation. That is,

$$y=\frac{d^2}{dx^n}(x^2-1)^n$$ (3.3)

Or, if we so choose,

$$y=\frac{1}{2^n n!}\frac{d^2}{dx^n}(x^2-1)^n$$ (3.4)

This is the so called Legendre Polynomial, denoted by $P_n(x)$ We so assign $\Delta=l(l+1)$ . But what of the $m \neq 0$ cases? In such a situation, our equation of interest can be cast as:

$$(1-x^2)y'' - 2xy' + \left(l(l+1)-\frac{m^2}{1-x^2}\right)y=0$$ (3.5)

Let us continue the derivation series from above; Recall that

$$D_{(n+1)}: (x^2-1)w''^{(n)} + 2xw'^{(n)} -n(n+1)w^n=0$$ (3.6)

Continuing the process we get,

$$D_{k,n+1}: (1-x^2)y^{(2+k)}-x(2+2k)y^{(k+1)} + \left(l(l+1)-k(k+1)\right)y^{(k)} =0$$ (3.7)

Or, in the spirit of what we have done thus far, write:

$$z=\left(\frac{d^k}{dx^k}\right)y \rightarrow (1-x^2)z''-2x(k+1)z'+\left(l(l+1)-k(k+1)\right)z=0$$ (3.8)

This does not yield the correct formulation. When we recast the original equation as:

$$(1-x)^2y'' -2x(1-x^2)y' + \left(l(l+1)(1-x^2)-m^2\right)y=0$$ (3.9)

This easily suggests the following form for our final function:

$$\zeta = (1-x^2)^a\left(\frac{d}{dx}\right)^b y_n$$ (3.10)

To find what form a and b take, we consider $y_1 = x$ with $b=1$ since clearly b must be be a positive integer:

$$\zeta = (1-x^2)^a\frac{d}{dx}(x) \rightarrow \zeta=(1-x^2)^a$$

$$\zeta' = -2ax(1-x^2)^{a-1}$$

$$\zeta''&= -2a(1-x^2)^{a-1} + a(a-1)(4x^2)(1-x^2)^{a-2}$$

$$\ni (1-x^2)\zeta'' -2x\zeta' + \left(l(l+1)-\frac{m^2}{(1-x^2)}\right)\zeta$$

$$= -2a(1-x^2)^a + a(a-1)(4x^2)(1-x^2)^{a-1} + yx^2a(1-x^2)^{a-1}$$

$$+l(l+1)(1-x^2)^a-m^2(1-x^2)^{a-1}=0$$ (3.11)

Upon the expansion of the latter equation for x, we can match each term with its coefficients. These coefficients must equal zero to satisfy the equation. It is easiest to take the highest coefficients of x:

$$a(a-1)4x^2(1-x^2)^{a-1} + 4x^2a(1-x^2)^{a-1} -m^2(1-x^2)^{a-1}=0$$

$$a^2=\frac{m^2}{4} \rightarrow a=\frac{\vert m\vert}{2}$$ (3.12)

Here m=1, but we assume the ansazts of generalization whereby, sans a normalization constant, the full solution of our angular equation is the associated Legendre Function,

$$P^m_l(x)=(1-x^2)^{\frac{\vert m\vert}{2}}\left(\frac{d}{dx}\right)^{\vert m\vert}\left(\frac{1}{2^ll!}\right)\left(\frac{d}{dx}\right)^l(x^2-1)^l$$ (3.13)