Orthogonality of Legendre polynomial #1

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$$(1-x^{2})\frac{d^{2}P_{n}(x)}{dx^{2}}-2x\frac{dP_{n}(x)}{dx}+n(n+1)P_{n}(x)=0$$

$$(1-x^{2})\frac{d^{2}P_{n}(x)}{dx^{2}}+(-2x)\frac{dP_{n}(x)}{dx}+n(n+1)P_{n}(x)=0$$

$$(1-x^{2})\frac{d}{dx}\frac{dP_{n}(x)}{dx}+\frac{d}{dx}(1-x^{2})\frac{dP_{n}(x)}{dx}+n(n+1)P_{n}(x)=0$$

$$\frac{d}{dx}\left[(1-x^{2})\frac{dP_{n}(x)}{dx}\right]+n(n+1)P_{n}(x)\cdots(a)$$

 

First, substitute \(n\rightarrow m\) of equation \((a)\)

$$\frac{d}{dx}\left[(1-x^{2})\frac{dP_{m}(x)}{dx}\right]+m(m+1)P_{m}(x)=0\cdots(b)$$

 

Second, multiply \(P_{n}(x)\) on equation \((b)\)

$$\frac{d}{dx}\left[(1-x^{2})\frac{dP_{m}(x)}{dx}\right]{\color{blue}P_{n}(x)}+m(m+1)P_{m}(x){\color{blue}P_{n}(x)}=0$$

 

Third, multiply \(P_{m}(x)\) on equation \((a)\)

$$\frac{d}{dx}\left[(1-x^{2})\frac{dP_{n}(x)}{dx}\right]{\color{red}P_{m}(x)}+n(n+1)P_{n}(x){\color{red}P_{m}(x)}=0\cdots(c)$$

 

Fourth, subtract \((b)\) from \((a)\)

But as it is quite complex, let's divide them into first term :

$$\frac{d}{dx}\left[(1-x^{2})P_{m}^{\prime}(x)\right]P_{n}(x)-\frac{d}{dx}\left[(1-x^{2})P_{n}^{\prime}(x)\right]P_{m}(x)$$

 

and second term

$$m(m+1)P_{m}(x)P_{n}(x)-n(n+1)P_{n}(x)P_{m}(x)$$

 

Let's rearrange the first term first.

$$\frac{d}{dx}\left[(1-x^{2})P_{m}^{\prime}(x)\right]P_{n}(x)-\frac{d}{dx}\left[(1-x^{2})P_{n}^{\prime}(x)\right]P_{m}(x)$$

 

$$=\frac{d}{dx}(1-x^{2})P_{m}^{\prime}(x)P_{n}(x)+(1-x^{2})\frac{d^{2}P_{m}(x)}{dx^{2}}P_{n}(x)$$

$$~~~-\frac{d}{dx}(1-x^{2})P_{n}^{\prime}(x)P_{m}(x)-(1-x^{2})\frac{d^{2}P_{n}(x)}{dx^{2}}P_{m}(x)$$

 

$$=\frac{d}{dx}(1-x^{2})\left(P_{m}^{\prime}(x)P_{n}(x)-P_{m}(x)P_{n}^{\prime}(x)\right)$$

$$~~~+(1-x^{2})\left(\frac{d^{2}P_{m}(x)}{dx^{2}}P_{n}(x)-P_{m}(x)\frac{d^{2}P_{n}(x)}{dx^{2}}\right)\cdots(d)$$

 

Equation \((d)\) can be neatly re-rearranged like below:

$$=\frac{d}{dx}\left[(1-x^{2})\left\{P_{m}^{\prime}(x)P_{n}(x)-P_{m}(x)P_{n}^{\prime}(x)\right\}\right]$$

 

* Proof:

$$=\frac{d}{dx}(1-x^{2})\left(P_{m}^{\prime}(x)P_{n}(x)-P_{m}(x)P_{n}^{\prime}(x)\right)$$

$$+(1-x^{2})\frac{d}{dx}\left(P_{m}^{\prime}(x)P_{n}(x)-P_{m}(x)P_{n}^{\prime}(x)\right)$$

 

$$=\frac{d}{dx}(1-x^{2})\left(P_{m}^{\prime}(x)P_{n}(x)-P_{m}(x)P_{n}^{\prime}(x)\right)$$

$$+(1-x^{2})\left[\frac{d^{2}P_{m}(x)}{dx^{2}}P_{n}(x)+P_{m}^{\prime}P_{n}^{\prime}(x)-P_{m}^{\prime}(x)P_{n}^{\prime}(x)-P_{m}(x)\frac{d^{2}P_{n}(x)}{dx^{2}}\right]$$

\(\blacksquare\)

 

Then rearrange the second term.

$$m(m+1)P_{m}(x)P_{n}(x)-n(n+1)P_{n}(x)P_{m}(x)$$

$$=\left(m^{2}+m-n^{2}-n\right)P_{m}(x)P_{n}(x)$$

$$=\left[(m+n)(m-n)+(m-n)\right]P_{m}(x)P_{n}(x)$$

$$=(m+n+1)(m-n)P_{m}(x)P_{n}(x)$$

\(\blacksquare\)

 

Now, let's sum up two terms we rearranged.

$$\frac{d}{dx}\left[(1-x^{2})\left\{P_{m}^{\prime}(x)P_{n}(x)-P_{n}^{\prime}(x)P_{m}(x)\right\}\right]+(m-n)(m+n+1)P_{m}(x)P_{n}(x)=0\cdots(e)$$

 

Integral \((e)\) from -1 to 1,

Then first term of equation \((e)\) will be eliminated, and only the second term will be left.

$$(m-n)(m+n+1)\int_{-1}^{1}{P_{m}(x)P_{n}(x)dx}=0$$

$$\int_{-1}^{1}{P_{m}(x)P_{n}(x)dx}=0, ~~~(m\neq n)$$

 

Reference

Mathematics stack exchange, Proving that Legendre Polynomial is orthogonal