On the radius of convergence of solutions of analytic ODE's

Question

Consider the following analytic ODE: $$\frac{d~x}{d~t} = M(t)\cdot x + u(t),\quad x(0)=v $$ where $t$ is a complex variable, $v\in\mathbb{C}^d$, $M(t)$ (resp. $u(t)$) an $d\times d$ matrix (resp. $d\times 1$ matrix) with entries holomorphic functions in an open neighborhood of the origin $0$. By the power series extending of these entries at $0$, one can immediately obtain a formal solution $$f(t)=\sum_n u_nt^n\in \mathbb{C}^d[[t]]$$ of the given ODE.

My question is as follows: suppose that all entries of $M(t)$ and $u(t)$ converge in the disc $D_\varepsilon$ of radius $\varepsilon$ centered at the origin, then does $f(t)$ converge in the disc $D_\varepsilon$?

Answer 1

The solution is given by $$ x(t)=e^{M(t)}\,v, $$ so that the answer is yes.

Edit: as noted by Robert Lewis, this is wrong, unless $M$ is a constant matrix.

Answer 2

For any $r<ε$ the matrix function n $M$ is bounded on $D_r$, the bound is a global Lipschitz constant so that there is a solution for all $|t|\le r$. This means that the solution can be extended to $D_ε$, there are no singularities inside $D_ε$, and its power series must thus have a radius of convergence $\ge ε$.

---

To be more specific, by general properties of a radius of convergence, for any $r<ε$ there is a constant $C>0$ so that $$\|M_k\| r^k<C ~\text{ and }~ \|u_k\|r^k<C$$ for all $k$ for the power series coefficients of $M(t)$ and $u(t)$.

Now consider the differential equation $$ a'(t)=\frac{C(a(t)+1)}{1-\frac{t}r} ~~\text{with solution}~~ a(t)=(a(0)+1)\left(1-\frac{t}r\right)^{-Cr} -1 $$ The series expansion of the solution has a radius of convergence of $r$. The coefficients satisfy the recurrence equation $$ t^k:~~(k+1)a_{k+1}=Cr^{-k}\left(\sum_{j=0}^ka_jr^{j}+1\right) $$

In the given equation the series coefficients for $x(t)$ can be formally constructed via $$ t^k:~~(k+1)x_{k+1}=\sum_{j=0}^kM_jx_{k-j} + u_k $$ The norms of these formal coefficients can now be bounded by the recursive inequality $$ (k+1)\|x_{k+1}\|\le Cr^{-k}\left(\sum_{j=0}^k\|x_{j}\|r^j+1\right) $$ It follows by induction that if $\|x_j\|\le a_j$ for $j=0,...,k$, then also $\|x_{k+1}\|\le a_{k+1}$, which is insured starting with $a_0=\|v\|=\|x_0\|$. By comparison test, the region of convergence includes the disk of radius $r$.