info prev up next book cdrom email home

Laguerre Differential Equation

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''
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
\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)

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 $n>1$. Therefore,
a_{n+1}= {n-\lambda\over (n+1)(n+\nu+1)}a_n
\end{displaymath} (13)

for $n=1$, 2, ..., so

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

info prev up next book cdrom email home

© 1996-9 Eric W. Weisstein