Frobenius series

Question

Consider the differential equation

\begin{equation} (1+x^3)y''+4xy'+y =0 \end{equation}

I need to find a lower bound on the radius of convergence for above equation at $x=0\ \&\ x=2$.

I wrote the solution in Frobenius form \begin{equation} y(x) =\sum_{k=0}^{\infty}a_k x^{r+k} \end{equation}

Substituting back into the equation I got the following equations:

\begin{align} a_0 r(r-1) &= 0\\ a_1 r(r+1) &= 0\\ a_2 (r+2)(r+1) +a_0(4r+1) &=0\\ a_{k+3}(k+r+3)(k+r+2) + a_k (k+r-1)(k+r) + a_{k+1}(4k+4r+5) &=0 \quad \forall k\ge0 \end{align}

I try to use the fact that the necessary criterion for convergence of Frobenius series is

\begin{equation} \left|\frac{xa_{k}}{a_{k-1}}\right|<1 \implies \left|x\right|<\left|\frac{a_{k-1}}{a_{k}}\right| \end{equation}

For $r=0$, i get the recurrence relations as above. But can proceed on the bound.

Same thing for $r=1$, but I find only one parameter arbitrary instead of two even though this a second order ODE.

For $r=-1$, the solution is trivial. I'm stuck.

Answer 1

Since $a_0$ is the first non-zero term, then we must have $r=0$ or $r=1.$ The second equation then means that $r=0$ or $a_1=0,$ and does not imply that $r=-1$ is a possibility. In particular, $a_1=0$ in the $r=1$ case, which is why only one of the parameters is arbitrary. See here for another example where this happens, as well as an explanation of why we can (and should) assume that $a_0$ is the first non-zero term.

Answer 2

The critical piece of information is the first coefficient (and that all the other coefficients are polynomials), as its roots are the points where the ODE is singular as ODE.

It is a well known fact that an explicit analytical ODE (with leading coefficient $1$) has an analytical or holomorphic solution in any disk where the coefficient functions are analytical. This means that you get a converging power series on any disk that has not the roots of $0=1+x^3$ as interior points, as the the coefficient functions $\frac{4x}{1+x^3}$ and $\frac{1}{1+x^3}$ of the normalized equation are analytical inside such a disk.

The roots of the leading coefficient polynomial are $-1$ and $\frac{1\pm\sqrt{3}i}2$. The the distance from the first center $z_1=0$ to the singularities of the ODE is equally $\rho_1=1$. For the second center $z_2=2$ the minimal distance to the singularities is $\rho_2=|2-\frac{1\pm\sqrt{3}i}2|=\frac{\sqrt{9+3}}2=\sqrt3$.

Answer 3

Let $x=t-1$ ,

Then $(1+(t-1)^3)y''+4(t-1)y'+y=0$

$(1+t^3-3t^2+3t-1)y''+4(t-1)y'+y=0$

$(t^3-3t^2+3t)y''+4(t-1)y'+y=0$

$t(t^2-3t+3)y''+4(t-1)y'+y=0$

Not that this belongs to a Heun's Equation and don't expect simple form of series solution.