Laguerre Differential Equation

$\begin{displaymath} xy''+(1-x)y'+\lambda y = 0. \end{displaymath}$

(1)

The Laguerre differential equation is a special case of the more general ``associated Laguerre differential equation''

$\begin{displaymath} xy''+(\nu+1-x)y'+\lambda y = 0 \end{displaymath}$

(2)

with $\nu=0$ . Note that if $\lambda=0$ , then the solution to the associated Laguerre differential equation is of the form

$\begin{displaymath} y''(x)+P(x)y'(x)=0, \end{displaymath}$

(3)

and the solution can be found using an Integrating Factor

$\displaystyle \mu$	$\textstyle =$	$\displaystyle \mathop{\rm exp}\nolimits \left({\int P(x)\,dx}\right)=\mathop{\rm exp}\nolimits \left({\int {\nu+1-x\over x}\,dx}\right)$
	$\textstyle =$	$\displaystyle \mathop{\rm exp}\nolimits [(\nu+1)\ln x-x]=x^{\nu+1}e^{-x},$	(4)

$\begin{displaymath} y=C_1\int {dx\over\mu}+C_2= C_1\int{e^x\over x^{\nu+1}}\,dx+C_2. \end{displaymath}$

(5)

The associated Laguerre differential equation has a Regular Singular Point at 0 and an Irregular Singularity at $\infty$ . It can be solved using a series expansion,

$x \sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ (\nu+1)\sum_{n=1}^\infty na_nx^{n-1}$

$- x \sum_{n=1}^\infty na_nx^{n-1}+ \lambda \sum_{n=0}^\infty a_nx^n = 0\quad$ (6)

$\sum_{n=2}^\infty n(n-1)a_nx^{n-1}+ (\nu+1)\sum_{n=1}^\infty na_nx^{n-1}$

$- \sum_{n=1}^\infty na_nx^n + \lambda \sum_{n=0}^\infty a_nx^n = 0\quad$ (7)

$\sum_{n=1}^\infty (n+1)na_{n+1}x^n + (\nu+1)\sum_{n=0}^\infty (n+1)a_{n+1}x^n$

$- \sum_{n=1}^\infty na_nx^n + \lambda \sum_{n=0}^\infty a_nx^n = 0\quad$ (8)

$\begin{displaymath}[(n+1)a_1+\lambda a_0]+ \sum_{n=1}^\infty \{[(n+1)n+(\nu+1)(n+1)]a_{n+1}-na_n+\lambda a_n\}x^n = 0 \end{displaymath}$

(9)

$\begin{displaymath}[(n+1)a_1+\lambda a_0]+ \sum_{n=1}^\infty [(n+1)(n+\nu+1)a_{n+1}+(\lambda-n)a_n]x^n = 0. \end{displaymath}$

(10)

This requires

$\displaystyle a_1$	$\textstyle =$	$\displaystyle - {\lambda\over \nu+1}a_0$	(11)
$\displaystyle a_{n+1}$	$\textstyle =$	$\displaystyle {n-\lambda\over (n+1)(n+\nu+1)}a_n$	(12)

for

. Therefore,

$\begin{displaymath} a_{n+1}= {n-\lambda\over (n+1)(n+\nu+1)}a_n \end{displaymath}$

(13)

for

, 2, ..., so

$\begin{displaymath} y=a_0\left[{1-{\lambda \over \nu+1}x-{\lambda (1-\lambda )\o... ...-\lambda )\over 2\cdot 3(\nu+1)(\nu+2)(\nu+3)}+\cdots}\right]. \end{displaymath}$

(14)

If $\lambda$ is a Positive Integer, then the series terminates and the solution is a Polynomial, known as an associated Laguerre Polynomial (or, if $\nu=0$ , simply a Laguerre Polynomial).

See also Laguerre Polynomial

$x \sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ (\nu+1)\sum_{n=1}^\infty na_nx^{n-1}$
$- x \sum_{n=1}^\infty na_nx^{n-1}+ \lambda \sum_{n=0}^\infty a_nx^n = 0\quad$	(6)

$\sum_{n=2}^\infty n(n-1)a_nx^{n-1}+ (\nu+1)\sum_{n=1}^\infty na_nx^{n-1}$
$- \sum_{n=1}^\infty na_nx^n + \lambda \sum_{n=0}^\infty a_nx^n = 0\quad$	(7)

$\sum_{n=1}^\infty (n+1)na_{n+1}x^n + (\nu+1)\sum_{n=0}^\infty (n+1)a_{n+1}x^n$
$- \sum_{n=1}^\infty na_nx^n + \lambda \sum_{n=0}^\infty a_nx^n = 0\quad$	(8)