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How do I see whether or not these ODE's have a global solution:

$x'=t^2+x^2$

$x'=t^2+x$

Why?

2 Answers 2

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Applying the weaker form of the Cauchy theorem (probably known to many as Picard–Lindelöf theorem) to $x'=f(t,x),$ $\partial f/\partial x$ is required to be bounded (this implies it is Lipschitz with respect to $x$, for any $t$), and this is true for the second equation where such derivatives gives $1$, but not for the first, where it gives $2x$.

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Here is a paper relating to this matter: Ref.