Series Solution of Laguerre Differential Equation
Definition
The following differential equation is referred to as the Laguerre differential equation.
$$ xy^{\prime \prime}+(1-x)y^{\prime}+ny=0,\quad n=0,1,2,\cdots $$
Description
The solution to the Laguerre differential equation is called Laguerre polynomials, and the first few Laguerre polynomials are as follows.
$$ \begin{align*} L_{0}(x) &= 1 \\ L_{1}(x) &= -x+1 \\ L_{2}(x) &= \frac{1}{2}\left( x^{2}-4x+2 \right) \\ L_{3}(x) &= \frac{1}{6}\left( -x^{3}+9x^{2}-18x+6 \right) \\ \vdots & \end{align*} $$
Examining the equation to solve it, when $x=0$, the coefficient $P(x)=x$ in $y^{\prime \prime}$ is $0$, so $x=0$ is a singular point, and since the following equation is satisfied, $x=0$ is a regular singular point.
$$ \lim \limits_{x\rightarrow 0} x \frac{1-x}{x}=1<\infty,\quad \lim\limits_{x\rightarrow 0}x^{2}\frac{n}{x}=0 < \infty $$
Therefore, Frobenius method is used.
Solution
$$ xy^{\prime \prime}+(1-x)y^{\prime}+\lambda y=0 $$
Let’s assume the solution of the Laguerre differential equation as the following series.
$$ y= \sum \limits _{n=0}^{\infty} a_{n}x^{n+r} $$
To substitute into the differential equation, calculating $y^{\prime}$ and $y^{\prime \prime}$ gives us the following respectively.
$$ \begin{align*} y^{\prime} &= \sum \limits _{n=0}^{\infty} (n+r)a_{n}x^{n+r-1} \\ y^{\prime \prime}&= \sum \limits _{n=0}^{\infty}(n+r)(n+r-1)a_{n}x^{n+r-2} \end{align*} $$
Substituting these into the differential equation gives $$ \sum \limits _{n=0}^{\infty}(n+r)(n+r-1)a_{n}x^{n+r-1}+ \sum \limits _{n=0}^{\infty} (n+r)a_{n}x^{n+r-1} - \sum \limits _{n=0}^{\infty} (n+r)a_{n}x^{n+r} +\lambda \sum \limits _{n=0}^{\infty} a_{n}x^{n+r}=0 $$
To match the order of $x$ to $n+r$, changing the first two series indices gives us the following.
$$ \sum \limits _{n=-1}^{\infty}(n+r+1)(n+r)a_{n+1}x^{n+r}+ \sum \limits _{n=-1}^{\infty} (n+r+1)a_{n+1}x^{n+r} - \sum \limits _{n=0}^{\infty} (n+r)a_{n}x^{n+r} +\lambda \sum \limits _{n=0}^{\infty} a_{n}x^{n+r}=0 $$
Now, taking out the terms where $n=-1$ and combining the series gives us
$$ r(r-1)a_{0}+ra_{0}+\sum \limits _{n=0}^{\infty} \left[ (n+r+1)(n+r)a_{n+1}+(n+r+1)a_{n+1}-(n+r)a_{n}+\lambda a_{n} \right]x^{n+r}=0 $$
For the above equation to hold, all terms must have a coefficient of $0$, so we get the following condition.
$$ \begin{align*} && r(r-1)a_{0}+ra_{0} &= 0 \\ \implies && r^{2}a_{0} &= 0 \\ \implies && r &= 0 \end{align*} $$
The coefficients inside the series must also be $0$, so substituting $r=0$ gives us the following recursion formula.
$$ \begin{align*} && (n+1)na_{n+1}+(n+1)a_{n+1}-na_{n}+\lambda a_{n} &= 0 \\ \implies && (n+1)^{2}a_{n+1} &= (n-\lambda)a_{n} \\ \implies && a_{n+1} &= \frac{n-\lambda}{(n+1)^{2}}a_{n} \end{align*} $$
Therefore, the coefficients from $n=1$ onwards can all be represented as $a_{0}$. Calculating in order gives us the following.
$$ \begin{align*} a_{1} &= -\lambda a_{0} \\ a_{2} &= \frac{1-\lambda}{2^{2}}a_{1}=\frac{\lambda (\lambda - 1)}{2^{2}}a_{0} \\ a_{3} &= \frac{2-\lambda}{3^{2}}a_{2}=-\frac{\lambda (\lambda-1)(\lambda-2)}{3^{2}\cdot2^{2}}a_{0} \\ \vdots & \\ a_{n} &= (-1)^{n}\frac{\lambda (\lambda -1)\cdots (\lambda-n+2)(\lambda-n+1)}{n^{2}(n-1)^{2}\cdots 2^{2}}a_{0} \end{align*} $$
Therefore, the solution to the differential equation is expressed as the following series.
$$ y= a_{0}\left[ 1-\lambda x +\frac{\lambda (\lambda-1)}{4}x^{2} -\frac{\lambda (\lambda-1)(\lambda-2)}{36}x^{3}+\cdots\right] $$
If the constant $\lambda$ is a non-negative integer, then the series solution becomes a polynomial with a finite number of terms. Since our goal is to find solutions that do not diverge, let’s denote the solution to the Laguerre differential equation for non-negative $\lambda$ as $L_{\lambda}(x)$. Then, the solution for each $\lambda$ is as follows.
$$ \begin{align*} L_{1}(x) &= a_{0} \\ L_{1}(x) &= a_{0}(1-x) \\ L_{2}(x) &= a_{0}\left( 1-2x + \frac{1}{2}x^{2} \right) \\ L_{3}(x) &= a_{0} \left( 1-3x+\frac{3}{2}x^{2}-\frac{1}{6}x^{3}\right) \\ \vdots & \end{align*} $$
Setting $a_{0}$ as $1$ and organizing it so that the coefficient of the highest term is $\pm 1$ gives us
$$ \begin{align*} L_{1}(x) &= 1 \\ L_{1}(x) &= -x+1 \\ L_{2}(x) &= \frac{1}{2}\left( x^{2}-4x+2 \right) \\ L_{3}(x) &= \frac{1}{6}\left( -x^{3}+9x^{2}-18x+6 \right) \\ \vdots & \end{align*} $$
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