A Can the Cauchy-Kovalewskaya Theorem Predict Solution Existence in All Cases?

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Just a simple observation, hope it would be interesting.

This example is very well known. It shows that if the conditions of the Cauchy-Kovalewskaya theorem are not satisfied then the solution is not obliged to exist.
Consider an initial value problem
$$u_t=u_{zz},\quad u(t=0,z)=\frac{1}{1+z^2},\quad t,z\in\mathbb{C}.$$ Kowalewskaya proved that this problem does not have a solution ##u(t,z)## which is an analytic function at the point ##t=0,\quad z=0##.

Nevertheless consider a Banach space
$$X=\Big\{v=\sum_{k=0}^\infty v_kz^k\mid \|v\|=\sup_{k}\{k! |v_k|\}<\infty\Big\}.$$ This is a subspace of the space of entire functions.

Theorem. The following IVP
$$u_t=u_{zz},\quad u(t=0,z)=\hat u(z)=\sum_{k=0}^\infty\hat u_kz^k\in X\qquad (*)$$ has a unique solution
$$u\in C^1(\mathbb{R},X).$$

Indeed, substitute ##u(t,z)=\sum_{k=0}^\infty u_k(t)z^k## to (*) and have
$$\dot u_k=(k+2)(k+1)u_{k+2},\quad u_k(0)=\hat u_k,\quad k=0,1,2...$$
This is an initial value problem for the infinite system of ODE. After a change of variables ##u_k=\frac{w_k}{k!}## this system takes the form
$$\dot w_k=w_{k+2},\quad w_k(0)=k! \hat u_k.$$
By the standard existence theorem for ODE this IVP has a solution ##\{w_k(t)\}\in C^1(\mathbb{R},\ell_\infty)##.
That is all :)
 
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