Question

For the differential equation 3x^2y''+2xy'+x^2y=0 show that the point x = 0 is a regular singular point (either by using the limit definition or by computing the associated euler equation). compute the recursion formula for the series solution corresponding to the larger root of the indicial equation. with a0 = 1, compute the first three nonzero terms of the series.

Answer 1

Given an ODE of the form

$y''(x)+p(x)y'(x)+q(x)y(x)=f(x)$

a regular singular point $x=c$ is one such that $p(x)$ or $q(x)$ diverge as $x\to c$, but the limits of $(x-c)p(x)$ and $(x-c)^2q(x)$ as $x\to c$ exist.

We have for $x\neq0$,

$3x^2y''+2xy'+x^2y=0\implies y''+\dfrac2{3x}y'+\dfrac13y=0$

and as $x\to0$, we have $x\cdot\dfrac2{3x}\to\dfrac23$ and $x^2\cdot\dfrac13\to0$, so indeed $x=0$ is a regular singular point.

We then look for a series solution about the regular singular point $x=0$ of the form

$y=\displaystyle\sum_{n\ge0}a_nx^{n+k}$

Substituting into the ODE gives

$\displaystyle3x^2\sum_{n\ge0}a_n(n+k)(n+k-1)x^{n+k-2}+2x\sum_{n\ge0}a_n(n+k)x^{n+k-1}+x^2\sum_{n\ge0}a_nx^{n+k}=0$

$\displaystyle3\sum_{n\ge2}a_n(n+k)(n+k-1)x^{n+k}+3a_1k(k+1)x^{k+1}+3a_0k(k-1)x^k$
$\displaystyle+2\sum_{n\ge2}a_n(n+k)x^{n+k}+2a_1(k+1)x^{k+1}+2a_0kx^k$
$\displaystyle+\sum_{n\ge2}a_{n-2}x^{n+k}=0$

From this we find the indicial equation to be

$(3(k-1)+2)ka_0=0\implies k=0,\,k=\dfrac13$

Taking $k=\dfrac13$, and in the $x^{k+1}$ term above we find $a_1=0$. So we have

$\begin{cases}a_0=1\\a_1=0\\\\a_n=-\dfrac{a_{n-2}}{n(3n+1)}&\text{for }n\ge2\end{cases}$

Since $a_1=0$, all coefficients with an odd index will also vanish.

So the first three terms of the series expansion of this solution are

$\displaystyle\sum_{n\ge0}a_nx^{n+1/3}=a_0x^{1/3}+a_2x^{7/3}+a_4x^{13/3}$

with $a_0=1$, $a_2=-\dfrac1{14}$, and $a_4=\dfrac1{728}$.