Series solution of Legendre differential equation

Series solution of Legendre differential equation :

While solving Helmhotz equation in spherical polar co-ordinate we encounter with a spherical kind of differential equation of the form:

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

$$or,\;\; \frac{d}{dx}\biggl[(1-x^2)\frac{dy}{dx}\biggr]+n(n+1)y=0$$

This equation is called Legendre differential equation. the solution of equation (1) can be found power series method which can be written as,

$$y=\sum_{r=0}^\infty a_r a^{k+r}, a_0\ne0\dotsm(2)$$

since equation (2) is the solution of equation (1). So it must satisfy equation (1)

$$y'=\sum_{r=0}^\infty a_r(k+r) x^{k+r-1}$$

and

$$y''=\sum_{r=0}^\infty a_r (k+r)(k+r-1) x^{k+r-2}$$

Then,

$$(1-x^2) \sum_{r=0}^\infty a_r(k+r)(k+r-1) x^{k+r-2}- 2x\sum_{r=0}^\infty a_r(k+r)x^{k+r-1}+n(n+1)\sum_{r=0}^\infty a_r x^{k+r}=0$$

$$or,\;\; \sum_{r=0}^\infty a_r(k+r)(k+r-1)x^{k+r-2}=\sum_{r=0}^\infty a_r(k+r)(k+r-1)x^{k+r}-2\sum_{r=0}^\infty a_r(k+r)x^{k+r}+ n(n+1)\sum_{r=0}^\infty a_r x^{k+r}=0$$

$$or,\;\;\sum_{r=0}^\infty a_r(k+r)(k+r-1)x^{k+r-2}+\sum_{r=0}^\infty a_r[n(n+1)-(k+r)(k+r-1)-2(k+r)] x^{k+r}=0$$

$$or,\;\;\sum_{r=0}^\infty a_r(k-r)(k-r-1)x^{k-r-2}+\sum_{r=0}^\infty a_r[n(n+1)-(k-r)(k-r+1)]x^{k-r}=0$$

Now, equating the coefficient of \(x^k\) put r=0 (highest term) to zero

$$ a_0[n(n+1) -k(k+1)]=0$$

Since \(a_0\ne0\)

\( n(n+1)-k(k+1)=0\)

or, \(n^2+n-k^2-k=0\)

or, \( (n-k)(n+k)+(n-k)=0\)

or, \( (n-k)(n+k+1)=0\)

Either, k=n or k=-n-1

Equating the coefficient of \(x^{k-1}\) (put r=1) to zero, we get

$$ a_1[n(n+1)-k(k-1)]=0$$

While substituting k=+n and k=-n-1 in equation, the bracketed term never vanishes. So, \(a_1\) must be vanished. So series solution of Legendre differential equation contain no odd term. Now, comparing the coefficient of \(x^{k-r}\) on both sides, we get,

$$or, \;\; a_{r-2}(k-r+2) (k-r+1) + a_r[n(n+1) - (k-r) (k-r+1)]=0$$

$$or, a_r=\frac{-(k-r+1)(k-r+2) a_{r-2}}{n(n+1)-(k-r)(k-r+1)}$$

This is the recurrence relation.

Case I:

For \( k=+n, a_0\ne0, a_1=a_3=a_5=a_7=a_{11}=a_{13}=0\)

$$a_r=\frac{-(n-r+1)(n-r+2)}{n(n+1)-(n-r)(n-r+1)}a_{r-2}$$

$$a_r=\frac{-(n-r+1) (n-r+2)}{r(2n-r+1)}a_{r-2}$$

Put r=2

$$a_2=\frac{-(n-2+1)(n-2+2)}{2(2n-2+1)}a_0$$

$$= \frac{-(n-1)n}{2(2n-1)}a_0$$

Put, r=4

$$a_4=\frac{-(n-3)(n-2)}{4(2n-3)}a_2$$

$$=\frac{-(n-3)(n-2)}{4(2n-3)}\cdot \frac{-(n-1)n}{2(2n-1)}a_0$$

$$=\frac{(-1)^2nn(n-1)(n-2)(n-3)}{2\cdot 4(2n-1)(2n-3)}a_0$$

Put r=2j

$$ a_{2j}=\frac{(-1)^j n(n-1) (n-3)\cdot (n-2j+1) a_0}{2.4...2j(2n-1)(2n-3)\cdot (2n-2j+1)}$$

$$a_{2j}=\frac{(-1)^J n_! a_0}{2^J J!(n-2j)! (2n-1)(2n-3)\cdot (2n-2j+1)}$$

We know the solution of Legendre differential equation for k=n is

$$ y_1=\sum_{r=0}^\infty a_r k^{k-r}$$

$$=\sum_{r=0}^\infty a_r x^{n-r}$$

$$=\sum_{j=0}^\infty a_{2j} x^{n-2j}$$

$$=\sum_{j=0}^\infty \frac{(-1)^j n! a_0 x^{n-2j}}{2^j J! (2n-1) (2n-3)\cdot (2n-2j+1)}$$

$$=\sum_{j=0}^\infty\frac{(-1)^j n(n-1)(n-2)(n-3)\cdot(n-2j+1)(n-2j)!}{2^J j!(2n-1)(2n-3)\cdot (2n-2j+1)}$$

Case II:

\(k=-(n+1\;\;\; a_0\ne0, a_1=a_3=a_5=a_7=0\)

we have,

$$a_r= \frac{-(k-r+1) (k-r+2) a_{r-2}}{[n(n+1)-(k-r)(k-r+1)]}$$

$$=\frac{-(n-1-r+1)(n-1-r+2)a_{r-2}}{n(n+1)-(-n-1-r)(n-1-r+1)}$$

$$=\frac{-(n+r)(n+r-1)}{r(2n+r+1)} a_{r-2}$$

$$=\frac{(n+r)(n+r-1)}{r(2n+r+1)}a_{r-2}$$

Put r=2,

$$a_2=\frac{(n+2)(n+1)}{2(2n+3)}a_0$$

Put r=4,

$$ a_4=\frac{(n+4)(n+3)}{4(2n+5)}a_2$$

$$=\frac{(n+4)(n+3)}{4(2n+5)}\cdot \frac{(n+2)(n+1)}{2(2n+3)}a_0$$

$$=\frac{(n+1)(n+2)(n+3)(n+4)}{2\cdot 4 (2n+3)(2n+5)}a_0$$

$$or\;\; a_{2j}=\frac{(n+1)(n+2)(n+3)\cdot (n+2j)}{2\cdot4\dotsm 2j(2n+3)(2n+5)\cdot (2n+2j+1)}a_0$$

We know the solution is

$$y=\sum_{r=0}^\infty a_r x^{k-r}$$

$$\sum_{r=0}^\infty a_1 x^{-(n+1)-r}$$

$$=\sum_{j=0}^\infty a_{2j} x^{-(n+1)-2j}$$

$$=\sum_{j=0}^\infty \frac{(n+1)(n+2)(n+3)\dotsm(n+2j)a_0 x^{-(n+1-2j)}}{2\cdot4\dotsm 2j(2n+3)(2n+5)\dotsm (2n+2j+1)}$$

$$=\sum_{j=0}^\infty \frac{(n+1)(n+2)\dotsm (n+2j) a_0 x^{-(n+1)-2j}}{2^j J! (2n+3) (2n+5)\dotsm (2n+2j+1)}\dotsm(4)$$

The complete solution is

$$y=AY_1+ BY_2$$

Reference:

1. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.
2. Vaughn, Michael T.Introduction to Mathematical Physics. Weinheim: Wiley-VCH, 2007
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.