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Let $X=(X_1,\dots, X_n)\in L^\infty$ be a smooth (non Lipschitz) vector field such that

\begin{equation}\tag{1} X_n \ge c|(X_1, \dots, X_{n-1})|\,. \end{equation}

Does the cone condition (1) imply that the flow $\Phi(t,x)$ is globally defined? I would say so, as it seems that forces integral curves to be confined.

Is this true?

Take $\bar x$ and given the maximal interval $(0, t_{\bar x})$, I argue by contradiction that $t_{\bar x}$ is finite. Take a sequence of times $t_n$ converging to $t_{\bar x}$ and let $x_n$ be the points corresponding to the flow $\Phi(t_n,\bar x)$. If I am able to prove that the sequence $x_n$ is bounded, then it must converge and thus I would be able to extend the flow at $t_{\bar x}$ by setting it equal to $\lim_n x_n$.

How do I prove that these $x_n$ stay bounded, given (1)?

Paolo Intuito
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1 Answers1

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You don't even need the assumption $(1)$.

Local existence and uniqueness is ensured because $X$ is locally Lipschitz. You have that $$ |\Phi(t,x)-x| = \left|\int_0^t X(\Phi(s,x))ds\right| \leq \int_0^t |X(\Phi(s,x))|ds \leq \|X\|_\infty t. $$ This means that solutions stay in a compact set in finite time, which implies that they can be extended to $[0,\infty)$.

Federico
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  • Great thanks. I somehow got distracted by the assumtpion (1) which apparently was not needed. Maybe I can infer something more from it, such that the integral curves will pass through every hyperplane of fixed $n$-th coordinate, $H_h:={x_n = h}$? – Paolo Intuito Nov 26 '18 at 09:17
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    Nope: try $X\equiv 0$. – Federico Nov 26 '18 at 17:16
  • Besides the trivial case, i.e. if $|X|_\infty>0$? – Paolo Intuito Nov 27 '18 at 11:10
  • Neither. A simple counterexample is a non-zero compactly supported vector field – Federico Nov 27 '18 at 13:32
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    For that you would $X_n>0$. – Federico Nov 27 '18 at 13:36
  • How would I go proving something like that? God, these flows give me headaches. Do you have any good references for a nice introduction to flows? – Paolo Intuito Nov 27 '18 at 23:12
  • What is it exactly that you want to prove? I answered your original question. Regarding the property of intersecting every plane ${x_n=h}$, you need $X_n>0$. What are you missing? Maybe ask your final goal that you want to obtain, being aware of https://en.wikipedia.org/wiki/XY_problem – Federico Nov 28 '18 at 13:44
  • How do I prove that if $X_n>0$ holds along with (1), then any integral curve intersect any hyperplane of fixed $n$-th coordinate? – Paolo Intuito Nov 28 '18 at 14:01
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    Just consider a trajectory $x(t)$. Assume $M=\sup_t x_n(t)=\lim_{t\to\infty} x_n(t)<\infty$. Then by $(1)$ you get that $x^=\lim_{t\to\infty} x(t)$ exists. But then $x^$ needs to be a stationary point, giving $X_n(x^*)=0$, contradiction. – Federico Nov 28 '18 at 16:13