Let $I$ be an open interval and consider a function $f: I\times \mathbb{R}\to \mathbb{R}$, which is analytic in both its variables on the whole domain. Let $u_0:I\mapsto I$ be an analytic function on the whole $I$. Consider the Cauchy problem \begin{align} &\dot u(x,t) = f(t,u(x,t))\,;\\ &u(x,0) = u_0(x)\,, \end{align} where the dot is the derivative wrt $t$.
Assume that the solution $u(x,t)$ of the problem is such that $u(x,t)\in I$ for all $(x,t)\in I\times \mathbb{R}$.
Can I claim that, for all $t$ in some open neighbourhood of $0$, the map $x\mapsto u(t,x)$ is analytic on $I$?
I know that I can use the Cauchy-Kovalevskaya theorem to say that for all $(t,x)\mapsto u(t,x)$ is analytic in some open set of $\mathbb{R}\times I$ containing $\{0\}\times I$, but this does not tell me that for a non-zero $t$, the analyticity wrt $x$ will be on the whole set $I$.