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Consider the following IVP: $x'=f(t,x)$ and $x(0)=x_0$, where $x\in \mathbb{R}^n$ and $t\in \mathbb{R}$.

Suppose that for all $(t,x)\in\mathbb{R}^{n+1}$, $|f(t,x)|\leq b(t) |x|^2$.

In order for the solution to the IVP to exist for all $t>=0$, what conditions should we impose on $b(t)$ and $x_0$?

I know that if the function $f: \mathbb{R}^{n+1}\to \mathbb{R}^n$ is continuous, then there is a theorem (Peano Local existence Theorem) that guarantees that the solution to the IVP exists, but I can't seem to find a way to prove the continuity of $f$. I wonder if anyone of you knows another way on how to solve this problem. Thanks!

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  • $\begingroup$ There are two parts to your question: (a) is the local existence, which is essentially addressed by Peano. (b) is the semi-global existence (solution exists for all $t\geq 0$), this is where conditions on $b(t)$ and $x_0$ seems to be more relevant. Can you please clarify your question? $\endgroup$
    – Willie Wong
    Commented Dec 10, 2014 at 16:07

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Your problem is reminiscent of the Osgood's criterion, which can be found on MO here. Local integrability near $0$ for $b$ is only what you need.

A proof of this criterion may be found for instance here, page 55.

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There is no such non-trivial condition. Consider the simple example $$x'=1+x^2$$ It has solutions $x(t)=\tan t$ which blow up at finite time.

No multiple $b(t)(1+x^2)$ will help. To guarantee a solution for all $t>0$ you need something line $|f(t,x)|\leq b(t)|x|$, $|x|$ in the first power.

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  • $\begingroup$ If we change the problem to $|f(t,x)\leq b(t) |x|$, what would be the conditions on $x_0$ and $b(t)$ to guarantee the existence of a solution? $\endgroup$
    – User267845467
    Commented Dec 10, 2014 at 6:33

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