Series solution of Laguerre's Differential Eqution

Series Solution of Laguerre's Differential Equation:

Laguerre's differential equation may be written as

$$xy''+(1-x)y'+\lambda y=0$$

Where, \(\lambda\)= constant

The equation has a singularity at x=0. But the singularity is non-essential of removable and hence the method of series integration for solving this equation. For this purpose, we take \(y=\sum_{l=0}^\infty a_l x^{k+l}\) (where k is constant) as the solution of given differential equation. Thus,

$$y'=\sum_l (k+l) a_l x^{k+l-1}$$

and$$ y''=\sum_l (k+l) (k+l-1) a_l x^{k+l-2}$$

Substituting these values in equation (1) , we have

$$\sum_l(k+l)^2a_l x^{k+l-1}- \sum_l a_l(k+l-\lambda) x^{k+l}=0\dotsm(2)$$

This equation is true for all the values of x and hence the coefficients of all the powers of x are identically zero. As equating to zero the coefficient of the lowest power of x, i.e. of \(x^{k-l}\), we have

$$ k^2a_0= 0\dotsm(3)$$

In this identical equation \(a_0\) is taken as arbitrary constant; so equation (3) holds good only if k=0. Then, we have

$$\sum l^2 a_l x^{l-1}- \sum a_l(l-\lambda)x^1=0\dotsm(4)$$

Equating the coefficient of \(x^j\) to zero, we have

$$ a^{j+l}=\frac{j-\lambda}{(j+l)^2}a_j$$

This is the recurrence relation for the coefficients.

Thus,$$ a_l= -\lambda a_0 = (-1) a_0$$

$$a_2=\frac{1-\lambda}{2^2}\cdot (-\lambda a_0$$

$$= (-1)^2 \frac{\lambda (\lambda _1)}{(2!)^2}a_0$$

$$a_3= \frac{2-\lambda}{3^2}a_0$$

$$=-\frac{\lambda(\lambda-1)(\lambda -2)}{(3!)^2}a_0$$

$$=(-1)^3\frac{\lambda(\lambda-1)(\lambda -2)}{(3!)^2}a_0$$

similarly...

$$a_r=\frac{r-1-\lambda}{r^2}a_{r-1}$$

$$=(-1)^r\frac{\lambda(\lambda-1)\dotsm (\lambda-r+1)}{(r!)^2}a_0$$

So $$ y=\sum_{l=0}^\infty a_l x^l= a_0+ a_1 x+ a_2 x^2+ a_3 x^3+ \dotsm+ a^r x^r+ \dotsm$$

$$=a_0\biggl[1-\lambda x+ \frac{\lambda(\lambda-1)}{(2!)^2}x^2-\dotsm + (-1)^r \frac{\lambda(\lambda-1)\dotsm(\lambda-r+1)}{(r!)^2}x^r+\dotsm\biggr]\dotsm(5)$$

If \(\lambda=n\), a positive integer and if we put \(a_0=n!\), then the solution for y becomes the Laguerre

$$L_n(x)=(-1)^n\biggl[x^n-\frac{n^2}{1!}x^{n-1}+\frac{n^2(n-1)^2}{2!}x^{n-2}+\dotsm+(-1)^nn!\biggr]\dotsm(6)$$

This is the expression for Laguerre's Polynomial.

Equation (6) gives

\(L_n(0)= n!\), when\( n=0, L_0(x)=1 \)

When \(n=1, L_1(x)=1-x\)

When \(n=2, L_2(x)= x^2-4x+2\)

When \(n=3, L_3(x)= -x^3+9x^2-36x+6\)

When \(n=4, L_4(x)= x^4-16x^3+72x^2-96x+48\) and so on.

Thus, Laguerre's polynomial is the solution of equation.

$$xL_n''(x)+(1-x)L_n'(x_n)+nL_n(x)=0\dotsm(7)$$

Associated Laguerre's polynomial:

Differentiating equation (7) p times, we have

$$ x\frac{d^{p+2}L}{dx^{p+2}}+(p+1-x) \frac{d^{p+1}L}{dx^{p-1}}+(\lambda-p) \frac{d^p L}{dx^p}\dotsm(8)=0$$

If we substitute \(y=\frac{d^pL_n(x)}{dx^p}\) in this equation, then we get

$$xy''(p+1-x)y'+(\lambda-p)y=0\dotsm(9)$$

Where, p is positive integer.

Thus, the solution of equation (9) is

$$y=\frac{d^p}{dx^p}L_n(x)= L_n^p(x)\dotsm(10)$$

This is called the associated Laguerre polynomial of degree (n-l).

Let be satisfying the differential equation

$$xv''+(p+1-x) v'+ (n-p)v=0\dotsm(11)$$

be related with another function y by the relation

$$y= e^{-x/2} X^{(p-1)/2}v$$

$$v=y.e^\frac x2 x^{-(p-1)/2}$$

Substituting this expression of v in equation (11), we get

$$xy''+2y'+\biggl[n-\frac{p-1}{2}-\frac{4}{x}-\frac{p^2-1}{4x}\biggr]y=0\dotsm(12)$$

As seen from equation (10), \(v=L_n^k(x)\)

So, solution of equation(12) is

$$y=e^{\frac{-x}{2}} x^{\frac{(p-1}{2}} L_n^k(x)=y_{n,k}$$

This function y is called an associated Laguerre Function.

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.