Orthogonality of Legendre differential equation and Associate Legendre differential equation

Orthogonality of Legendre:

We know, the legendre diff equation as,

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

Since, \(y=P_n(x)\). So,

$$\frac{d}{dx}\biggl[(1-x^2) \frac{dP_n(x)}{dx}\biggr] + n(n+1) P_n(x)=0\dotsm(a)$$

For \(m^{th}\) order, we get,

$$\frac{d}{dx}\biggl[(1-x^2)\frac{dP_n(x)}{dx}\biggr] - P_n(x)\frac{d}{dx}\biggl[(1-x^2)\frac{dP_m(x)}{dx}\biggr]+[n(n+1)-m(m+1)]P_n(x)\cdot P_m(x)=0$$

$$\frac{d}{dx}\biggl[P_m(x)(1-x^2) \frac{dP_n(x)}{dx}-P_n(x) (1-x^2)\frac{dP_m(x)}{dx}\biggr]+[n(n+1)-m(m+1)] P_n(x)P_m(x)=0$$

Now, integrating both sides with respect to n from -1 to +1, we get

$$\int_{-1}^{+1} \frac{d}{dx}\biggl[P_n(x)(1-x^2) \frac{dP_n(x)}{dx}- P_n(x)\cdot (1-x^2) \frac{dP_m(x)}{dx}\biggr] dx+ [n(n+1)-m(m+1)]\int_{-1}^{+1} P_n(x) P_m(x)dx=0$$

$$or,\;\\;0+n(n+1)- m(m+1)\int_{-1}^{+1} P_n(x) P_m(x) dx=0$$

$$or,\;\; n(n+1)-m(m+1)\int_{-1}^{+1} P_n(x) P_m(x) dx=0$$

Case I

for \(m\neq\)

$$\int_{-1}^{+1} P_n(x) P_n(x) dx=0\dotsm(c)$$

Case II

For \(m=n\)

$$\int_{-1}^{+1} P_n(x) P_n(x) dx\ne0$$

We know, the generating function of Legendre differential polynomial is,

$$\sum_{n=0}^\infty t^n P_n(x)= ( 1-2xt + t^2)^{-\frac12}$$

Squaring both side we get

$$\sum_o^\infty t^{2n} P_n(x) P_n(x) = \frac{1}{(1-2xt + t^2)}$$

Integrating with respect to x from -1 to +1 all cross term vanishes so we only write perfect term. Now, integrating both sides with respect to x from -1 to +1 we get,

$$\sum_{n=0}^\infty t^{2n}\int_{-1}^{+1} P_n(x) P_n(x) dx=\int_{-1}^{+1} \frac{1}{1-2xt+t^2} dx$$

$$=\biggl|\frac{log(1-2xt+t^2)}{-2t}\biggr|_{-1}^{+1}$$

$$=\frac{-1}{2t}[ log (1-t)^2 - log(1+t)^2]$$

$$=\frac{1}{t}[log(1+t)-log(1-t)]$$

$$=\frac1t\biggl[t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\dotsm \biggl( -t-\frac{t^2}{2}-\frac{t^3}{3} - \frac{t^4}{4}+\dotsm\biggr)\bigg]$$

$$=\frac2t\biggl[t+\frac{t^3}{3}+\frac{t^5}{5}+\dotsm+\frac{t^{2n+1}}{2n+1}+\dotsm\biggr]$$

$$=2\biggl[1+\frac{t^2}{3}+\frac{t^4}{5}+\dotsm+\frac{t^{2n}}{2^{n+1}}+\dotsm\biggr]$$

Now, collecting the coefficient of \(t^{2n}\) n both sides, we get,

$$or,\;\;\int_{-1}^{+1} P_n(x)P_n(x)dx=\frac{2}{2n+1}$$

Now, from (c) and (d), we get

$$\int_{-1}^{+1} P_n(x)P_n(x)dx=\frac{2}{2n+1}\delta_{nm}$$

Where, \(\delta_{mn}=1\) For m=n

and\(\delta_{mn}=0\) For m\(\ne\)n

This is the orthogonality relationship of Legendre polynomial.

Associate Legendre differential equation:

The associated Legendre differential equation has a form

$$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+\biggl[ n(n+1)-\frac{m^2}{1-x^2}\biggl]y=0\dotsm(1)$$

For a special case i.e. m=0 equation(1) reduces to Legendre differential equation which has teh form

$$(1-x^2) \frac{d^2y}{dx^2}-2x\frac{dy}{dx}+n(n+1)y=0\dotsm(2)$$

Now differentiating equation(2) with respect to x for m time using Leibnitz theorem.

$$(1-x^2)\frac{d^{2+m_y}}{d^{x^2+m}}+m(-2x)\frac{d^{1+m_y}}{dx^{1+m}}+\frac{m(m-1)}{2}(-2) \frac{d^{my}}{dx^m}-2x\frac{d^{1+m_y}}{dx^{1+m}}-m\cdot2\cdot1\frac{d^{my}}{dx^m}+n(n+1)\frac{d^{m_y}}{dx^m}=0$$

Put \(\frac{dm_y}{dx^m}=z\)

$$(1-x^2)\frac{d^2z}{dx^2}-2x(1+m)\frac{dz}{dx}+[n(n+1)-m(m+1)]z=0$$

Now substitute \(Z=(1-x^2)^{-\frac{m}{2u}}\)

$$\frac{dz}{dx}=(1-x^2)^{-\frac{m}{2}}\frac{dy}{dx}+(-\frac m2 ) (1-x^2)^{-\frac m2 -1} \times (-2x)u$$

$$=\biggl[\frac{dy}{dx}+ \frac{mxu}{1-x^2}\biggr] (1-x^2)^{-\frac m2}$$

Again,

$$\frac{d^2}{dx^2}=\frac{d^2u}{dx^2}+\frac{mu}{1-x^2}\frac{du}{dx}+mxu\times -1 (1-x)^2\times (-2x)(1-x^2)^{\frac{-m}{2}}+\biggl[\frac{du}{dx}+\frac{mxu}{1-x^2}\biggr]\times\frac{-m}{2} (1-x^2)^{-\frac m2 -1}\times(-2x)$$

$$=\frac{d^2u}{dx^2}+\frac{2mx}{1-x^2}\frac{du}{dx}+\frac{mu}{1-x^2}+\frac{2mx^2u+m^2x^2u}{(1-x^2)^2}\biggr](1-x^2)^{-\frac m2}$$

on substituting

$$(1-x^2)\frac{d^2u}{dx^2}-2x\frac{du}{dx}+[n(n+1)-\frac{m^2}{(1-x^2)}\biggr]u=0$$

This is associated differential equation whose solution is y

we have, \(z=(1+x^2)^{\frac{-m}{2u}}\)

$$P_n^m=(1-x^2)^{\frac m2} \frac{d^mP_n(x)}{dx^n}$$

$$=(1-x^2)^{\frac m2} \frac{d^my}{dx^m}$$

$$=(1-x)^{\frac m2} \frac{d_mP_n(x)}{dx^m}$$

This is the solution for associated Legendre.

Reference:

1. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.
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3. Butkov, Eugene.Mathematical Physics. Reading, MA: Addison-Wesley Pub., 1968.
4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
5. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.