Show that $\phi_{n}$ satisfies $((1-x^{2})y')'+n(n+1)y=0$.

Question

Show that $\phi_{n}$ satisfies $((1-x^{2})y')'+n(n+1)y=0$, where $\phi_{n}$ is the Legendre polynomial of degree $n$.

Here is what I did so far:

$\begin{align*} ((1-x^{2})y')'+n(n+1)y&=0\\ ((1-x^{2})\phi_{n}')'+n(n+1)\phi_{n}&=0\\ (1-x^{2})\phi_{n}''-2x\phi_{n}'+n(n+1)\phi_{n}&=0 \end{align*}$

I wanted to use the property $\phi_{n}'=x\phi_{n-1}'+n\phi_{n-1}$ to get the following:

$\begin{align*} 0&=(1-x^{2})\phi_{n}''-2x\phi_{n}'+n(n+1)\phi_{n}\\ &=(1-x^{2})\phi_{n}''-2x^{2}\phi_{n-1}'-2xn\phi_{n-1}+n(n+1)\phi_{n} \end{align*}$

I am not sure what to do after this. I wanted to continue using other properties of Legendre polynomials, but I am still stuck.

Answer 1

We have to show that the Legendre polynomials of degree $n$, $ \phi_n(x)$ satisfied the Legendre Equation $$((1-x^{2})y')'+n(n+1)y=0~.$$ i.e., $$\left((1-x^{2})\phi_n(x)'\right)'+n(n+1)\phi_n(x)=0~.$$

${}$

Here \begin{equation} \left((1-x^{2})\phi'_n(x)\right)'+n(n+1)\phi_n(x)\\ =\left(n\phi_{n-1}(x)-nx\phi_n(x)\right)'+n(n+1)\phi_n(x)\\ =n\phi'_{n-1}(x)-nx\phi'_n(x)-n\phi_n(x)+n(n+1)\phi_n(x)\\ =-n^2\phi_{n}(x)-n\phi_n(x)+n(n+1)\phi_n(x)\\ =0 \end{equation} Hence $\phi_{n}$ satisfies $~((1-x^{2})y')'+n(n+1)y=0~$, where $\phi_{n}$ is the Legendre polynomial of degree $n$.

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• in the second line we use the formula $$(1-x^{2})\phi'_n(x)=n\left[\phi_{n-1}(x)-x\phi_n(x)\right]$$
• in the forth line we use the formula $$x\phi'_n(x)-\phi'_{n-1}(x)=n\phi_n(x)$$

For the above mentioned formula, see https://nptel.ac.in/content/storage2/courses/122104018/node86.html