Orthogonality of Legendre polynomial #1

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$$(1-x^2)\frac{d^2{P}_n(x)}{d{x}^2}-2x\frac{d{P}_n(x)}{dx}+n(n+1)P_n(x)=0$$

$$(1-x^2)\frac{d^2{P}_n(x)}{d{x}^2}+(-2x)\frac{d{P}_n(x)}{dx}+n(n+1)P_n(x)=0$$

$$(1-x^2)\frac{d}{dx}\frac{d{P}_n(x)}{dx}+\frac{d}{dx}(1-x^2)\frac{d{P}_n(x)}{dx}+n(n+1)P_n(x)=0$$

$$\frac{d}{dx}[(1-x^2)\frac{d{P}_n(x)}{dx}]+n(n+1)P_n(x)\cdots (a)$$

 

First, substitute $n\to m$ of equation $(a)$

$$\frac{d}{dx}[(1-x^2)\frac{d{P}_m(x)}{dx}]+m(m+1)P_m(x)=0\cdots (b)$$

 

Second, multiply $P_n(x)$ on equation $(b)$

$$\frac{d}{dx}[(1-x^2)\frac{d{P}_m(x)}{dx}]P_n(x)+m(m+1)P_m(x)P_n(x)=0$$

 

Third, multiply $P_m(x)$ on equation $(a)$

$$\frac{d}{dx}[(1-x^2)\frac{d{P}_n(x)}{dx}]P_m(x)+n(n+1)P_n(x)P_m(x)=0\cdots (c)$$

 

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

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

$$\frac{d}{dx}[(1-x^2)P_m^{\prime }(x)]P_n(x)-\frac{d}{dx}[(1-x^2)P_n^{\prime }(x)]P_m(x)$$

 

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.

$$\frac{d}{dx}[(1-x^2)P_m^{\prime }(x)]P_n(x)-\frac{d}{dx}[(1-x^2)P_n^{\prime }(x)]P_m(x)$$

 

$$=\frac{d}{dx}(1-x^2)P_m^{\prime }(x)P_n(x)+(1-x^2)\frac{d^2{P}_m(x)}{d{x}^2}{P}_n(x)$$

$$-\frac{d}{dx}(1-x^2)P_n^{\prime }(x)P_m(x)-(1-x^2)\frac{d^2{P}_n(x)}{d{x}^2}{P}_m(x)$$

 

$$=\frac{d}{dx}(1-x^2)(P_m^{\prime }(x)P_n(x)-P_m(x)P_n^{\prime }(x))$$

$$+(1-x^2)(\frac{d^2{P}_m(x)}{d{x}^2}{P}_n(x)-P_m(x)\frac{d^2{P}_n(x)}{d{x}^2})\cdots (d)$$

 

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

$$=\frac{d}{dx}[(1-x^2)\{P_m^{\prime }(x)P_n(x)-P_m(x)P_n^{\prime }(x)\}]$$

 

* Proof:

$$=\frac{d}{dx}(1-x^2)(P_m^{\prime }(x)P_n(x)-P_m(x)P_n^{\prime }(x))$$

$$+(1-x^2)\frac{d}{dx}(P_m^{\prime }(x)P_n(x)-P_m(x)P_n^{\prime }(x))$$

 

$$=\frac{d}{dx}(1-x^2)(P_m^{\prime }(x)P_n(x)-P_m(x)P_n^{\prime }(x))$$

$$+(1-x^2)[\frac{d^2{P}_m(x)}{d{x}^2}{P}_n(x)+P_m^{\prime }{P}_n^{\prime }(x)-P_m^{\prime }(x)P_n^{\prime }(x)-P_m(x)\frac{d^2{P}_n(x)}{d{x}^2}]$$

$◼$

 

Then rearrange the second term.

$$m(m+1)P_m(x)P_n(x)-n(n+1)P_n(x)P_m(x)$$

$$=(m^2+m-n^2-n)P_m(x)P_n(x)$$

$$=[(m+n)(m-n)+(m-n)]P_m(x)P_n(x)$$

$$=(m+n+1)(m-n)P_m(x)P_n(x)$$

$◼$

 

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

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

 

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

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

$$(m-n)(m+n+1){\int }_{-1}^1{P}_m(x)P_n(x)dx=0$$

$${\int }_{-1}^1{P}_m(x)P_n(x)dx=0,(m\ne n)$$

 

Reference

Mathematics stack exchange, Proving that Legendre Polynomial is orthogonal