Consider an explicit ODE Y'=F(Y) where F is holomorphic. At an initial value (z_0,Y_0) where F is analytic, there is a unique solution z |-> Y(z). The question is; is the radius of convergence of this series equal to the modulus of the closest movable singularity of this particular solution?
Dear Professor Gabriel Thomas,
Consider the following ODE: $\dot{x}(t)=[x(t)-1]^{2}$. Its general solution reads as:
\begin{displaymath}
x(t)=\frac{x(0)-1}{1-t[x(0)-1]}+1, \qquad t\geq 0 \enspace .
\end{displaymath}
Its right maximal interval of existence is $[0,t(x_{0}))$ where
\begin{displaymath}
t(x_{0}):=\left\ \begin{array}{rcl} \frac{1}{x(0)-1} & \text{if} & x(0>1 \\ \infty & \text{if} & x(0)\leq 1 \end{array}\right\} \enspace .
\end{displaymath}
If $x(0)=0$ then the solution is $x(t)=-\frac{1}{1+t}+1$ and it exists for all $t\geq 0$ whilst the series expressing this solution converges only for $[0,1)$ (geometric series).
This shows that your claim is false until $t(x_{0})=\infty $. However, we still do not know what is going on when $t(x_{0})
Hello Dr Grabowski,
I have read your counter-example. In the complex field, it is still coherent with my assumption.
The solution for $x(0)=x_0$ is $x(z)=\frac{x_0-1}{1-z[x_0-1]}+1$, related to the geometric series ; it is divergent when $| z (x_0-1)| >1$ so the convergence radius is : $R(x_0)=|\frac{1}{x_0-1}|$. This is what I am claiming.
However in most directions the solutions can by extended to $|z|=\infty$ with the analytic continuation process, but not with the same series as this started at $z=0, x(0)=x_0$.
Thanks for your answer, the question remains open,
Best Regards,
Gabriel THOMAS, Ph D
Dear Gabriel,
Did you tray to use the fact that the radius of convergence of the power series is equal to the distance of its center (here $z=0$) to the nearest singularity - see
https://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions
Bestr regards,
Piotr
Dear Piotr,
your second answer must give the right point of view. As the series is convegent and the solution is unique at $z_0=0$, the solution and the formal power series coïncide in an open set so the singular point of the series (distant of a radius of convergence from the "center") is a so-called movable singularity of the solution map.
This means that e.g. for $z \mapsto tan(z)$, solution at $0$ of $y'=1+y^2$, the radius of convergence $\pi/2$ is related to the coefficients of the power series of tan(z).
Thanks and Best Regards,
Gabriel
Dear Gabriel,
Yes, exactly. a solution can be prolongated until it reach the boundary of the domain of holomorphicity (analyticity) of the RHS. $x'(z)=\frac{1}{a-x}$, $a>0$. Here $x(z)=a-\sqrt([a-x(0)]^{2}-2z}$ and $x'$ becomes unbounded at $z=a^2}/2$.
Best regards,
Piotr
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