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

[wrobel]

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 :)

 

[jvicens]

Thank you for sharing this interesting example! It is a great illustration of the Cauchy-Kovalewskaya theorem and the importance of understanding the conditions for the existence of solutions to initial value problems. It also highlights the usefulness of Banach spaces in the study of ODEs and PDEs. I will definitely keep this example in mind when teaching these concepts in the future.