Given the Rodrigues' formula for Legendre's polynomials, show that it satisfies the ODE.

Question

The Rodrigues Formula for Legendre's Polynomials is $P_{l}(x)=\frac{1}{2^{l}l!}\frac{d^{l}}{dx^{l}}(x^2-1)^l$.

I wrote $P_{l}(x)=\frac{1}{2^{l}l!}\frac{d^l}{dx^l}\sum_{k=0}^l(-1)^{k-l}\frac{l!}{k!(l-k)!}x^{2k}=\sum_{k=0}^l\frac{(-1)^{k-l}}{2^l}\frac{(2k)!}{k!(l-k)!(2k-l)!}x^{2k-l}$ and put it into the ordinary Legendre's Equation :$\frac{d}{dx}[(1-x^2)\frac{d}{dx}P_{l}(x)]+l(l+1)P_{l}(x)=0$.

I got $\frac{d}{dx}[(1-x^2)\frac{d}{dx}P_{l}(x)]+l(l+1)P_{l}(x)=\sum_{k=0}^l\frac{(-1)^{k-l}}{2^l}\frac{(2k)!}{k!(l-k)!(2k-l-1)!}[(2k-l-1)x^{2k-l-2}+(2k-l+1)x^{2k-l}+\frac{l(l+1)}{2k-l}x^{2k-l}]$

But I can't see why it equals to zero.

Is there other way to do it?

Thanks

Answer 1

This is another approach that does not explicitly use integration. Use $ D $ to stand for $ d/dx$.

First, apply Leibniz rule for the $n+2$ derivative of a product,

$$ \begin{align} D^{n+2}(x^2-1)^{n+1} &= (x^2-1) D^{n+2}(x^2-1)^n \\ & \quad + \left( n+2 \atop 1 \right) 2x D^{n+1}(x^2-1)^n \\ &\quad\quad + 2 \left(n+2 \atop 2\right) D^n (x^2-1)^n \tag{1} \end{align} $$

Second, evaluate the same expression, but take one derivative first

$$ \begin{aligned} D^{n+2}(x^2-1)^{n+1} &= D^{n+1} \Big( D(x^2-1)^{n+1} \Big) \\ &= D^{n+1} \Big( (n+1) (x^2-1)^n \cdot 2x \Big) \end{aligned} $$

And apply Leibniz again, $$ \begin{align} D^{n+2}(x^2-1)^{n+1} &= 2x(n+1) D^{n+1}(x^2-1)^n \\ &\quad + 2(n+1)\left(n+1 \atop 1 \right) D^n(x^2-1)^n \tag{2} \end{align} $$

Now divide through by $2^n n!$, subtract (1) from (2) and use Rodrigues's formula to obtain, $$ \begin{align} 0 &= -(x^2-1) P''_n - 2 (n+2) x P'_n - (n+2)(n+1) P_n \\ &\quad \quad + 2(n+1)x P'_n + 2(n+1)(n+1) P_n \\ &= (1-x^2)P''_n -2x P'_n +n(n+1) P_n \end{align} $$ which is exactly what we wanted.

Answer 2

The function \begin{align*} g_n(x)& :=\frac{\text{d}}{\text{d}x}\left\{\left(1-x^2\right)P_n^{'}(x)\right\}\\ & =\left(1-x^2\right)P_n^{''}(x)-2xP_n^{'}(x)\tag{1} \end{align*} is clearly a polynomial of degree $n$. Then $$g_n(x)=\sum_{s=0}^n c_{n,s}P_s(x).\tag{2}$$ To find the $c_{n,s}$, we multiply by $P_s(x)$, integrate from $-1$ to $1$ and use the following formula $$\int_{-1}^1\left\{P_n(x)\right\}^2\text{d}x=\frac{2}{2n+1}.$$ Two partial integrations can obtain \begin{align*} \frac{2c_{n,s}}{2n+1}& =\int_{-1}^1 P_s(x)g_n(x)\text{d}x\\ & =\int_{-1}^1 P_n(x)g_s(x)\text{d}x. \end{align*} Since $P_n(x)$ is orthogonal to all polynomials of lower degree, we claim that $c_{n,s}=0(s<n)$.

Comparing coefficients of $x^n$ in (1) and (2) yields $$c_{n,n}=-n(n+1).$$ Thus the desired differential equation is $$\left(1-x^2\right)P_n^{''}(x)-2xP_n^{'}(x)+(n^2+n)P_n(x)=0.$$