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

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SUMMARY

The discussion centers on the Cauchy-Kovalewskaya theorem and its implications for the existence of solutions to initial value problems (IVPs). It is established that if the conditions of the theorem are not met, as demonstrated with the IVP $$u_t=u_{zz},\quad u(t=0,z)=\frac{1}{1+z^2}$$, then a solution is not guaranteed. However, a unique solution exists for the IVP $$u_t=u_{zz},\quad u(t=0,z)=\hat u(z)=\sum_{k=0}^\infty\hat u_kz^k\in X$$ within a defined Banach space, confirming that solutions can be derived under specific conditions despite the theorem's limitations.

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  • Understanding of the Cauchy-Kovalewskaya theorem
  • Familiarity with initial value problems (IVPs)
  • Knowledge of Banach spaces and their properties
  • Basic concepts of ordinary differential equations (ODEs)
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  • Study the implications of the Cauchy-Kovalewskaya theorem on analytic functions
  • Explore the properties of Banach spaces in functional analysis
  • Learn about the standard existence theorem for ordinary differential equations
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Mathematicians, particularly those focused on differential equations, analysts studying functional spaces, and researchers exploring the boundaries of solution existence in complex systems.

wrobel
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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 :)
 
Last edited:

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