# Convergence radius of the solution of an ODE.

Sorry for the vague title... Here's the problem: Consider the ODE $$y'' + \frac{1}{\sqrt{t}}y=0.$$ Given a solution $y$ such that $y(t_0)=0$ for a fixed $t_0>0$ and $y'(t_0)\neq 0.$ What can i say about the convergence radius of the Taylor series of the solution?

Thanks!

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Yes, the Taylor expansion at $t_0.$ Any reason to think that? – Rppacheco Apr 24 '12 at 16:10

We have the following result about differential equation of order $2$ with analytical coefficients:

Consider the equation $y''+ay'+by=0$, where $a$ and $b$ are analytical in a neighborhood of $t_0$, with respective radius of convergence $R_1$ and $R_2$. Then any solution of this equations is analytical in a neighborhood of $t_0$, with radius of convergence $\geq \min(R_1,R_2)$.

(to prove this, we use the definition of radius of convergence of $\sum_n a_nx^n$: it's the supremum of the $M$ such that $\{a_nM^n\}$ is bounded)

Since $\frac 1{\sqrt t}=\frac 1{\sqrt{t_0+t-t_0}}=\frac 1{\sqrt{t_0}}\frac 1{\sqrt{1+\frac{t-t_0}{t_0}}}$, and since the power series of $\frac 1{\sqrt{1+u}}$ has a radius of convergence $1$, $\frac 1{\sqrt{t_0}}$ has a radius of convergence of $t_0$. So the solution of $y''(t)+\frac 1{\sqrt t}y(t)=0$ which satisfies $y(t_0)=0$ and $y'(t_0)\neq 0$ has a radius of convergence $\geq t_0$.

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Thank you very much! Do you any bibliography for this theorem? – Rppacheco Apr 25 '12 at 20:01
I saw it in a French book, "Analyse pour l'agrégation", which is at my library. Maybe it's a classical result about ODE, anyway I will try to go there tomorrow. I remember it's a quite technical proof. – Davide Giraudo Apr 25 '12 at 20:02
I found a bibliography: Birkhoff and Rota's Differential Equations book. Anyway, thank you again! – Rppacheco Apr 28 '12 at 2:34
I guess the proof in the book is quite similar. – Davide Giraudo Apr 29 '12 at 12:03