# ODE , locally Lipschitz

I am working on the following problem from Gerald Teschl's book on ODE's and am at a loss of how to proceed.

Suppose $f:\mathbb{R}^n→\mathbb{R}^n$ local Lipschitz. Show that,

if $\lim sup_{|x|\rightarrow \infty}\frac{<x,f(x)>}{|x|^2}<\infty$, then all solution of $x′=f(x)$ are global defined.

Any help would be appreciated. Thanks!

• Almost identical with [Let f:Rm→Rm locally Lipschitz. Show that if lim|x|→∞supf(x).x|x|2 thenLet f:Rm→Rm locally Lipschitz. Show that if lim|x|→∞supf(x).x|x|2 then [closed]](math.stackexchange.com/questions/2715955/…) (closed as off-topic). – user539887 Apr 7 '18 at 22:11

Confirm that $\dfrac{\langle x,f(x)\rangle}{|x|^2}$ is not only continuous but also bounded outside some disk of radius $R$, that is for $|x|\ge R$. Additionally, $\langle x,f(x)\rangle$ is continuous and thus bounded on $|x|\le R$. Let $M$ be a common bound for both cases.
Consider $V(x)=|x|^2$. Then $$\frac{d}{dt}V(x(t))=2\langle x,f(x)\rangle\le 2M\,(1+V(x(t)))$$ which gives an exponential bound on $|x(t)|$ preventing divergence to infinity in finite time.
• How u take this inequality? I just see, $... \leq 2MV(x(t))$ – Mário Bezerra Apr 7 '18 at 17:53
• Let $S$ be the value of the limsup. Then there is some $R$ so that the fraction is smaller than $2S+1$ for $|x|>R$, and only guaranteed there. Note that the fraction has a singularity at $x=0$. Take the supremum of the scalar product over $|x|\le R$. Let $M$ be the maximum of both bounds. – LutzL Apr 7 '18 at 18:03