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Let $x:\mathbb{R}^+\to \mathbb{R}^n$ such that $\lim_{t\to +\infty} x(t)=0$. Under which conditions can I say that there exists a Lipschitz-Continuous function $f$ such that $\frac{d}{dt}x(t)=f(x)$ and such that $f$ has a unique equilibrium point in $x=0$?

In other words, given a function $x(t)$ (or a sequence of functions $x_n(t)$) how can I prove that it is (they are) generated by an ODE of the form $\dot x = f(x)$?

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  • $\begingroup$ Certainly the question in your second paragraph is not a rewording of the question in the first paragraph at all (as long as I can understand it). If you have a $C^1$ function $x\colon[0,\infty)\to\mathbb{R}^n$ then it is a solution of the ODE $\dot{x}=f(t)$, where $f(t)$ is defined as the derivative of $x$ at $t$. Did you mean that? $\endgroup$
    – user539887
    Jul 10 '18 at 20:17
  • $\begingroup$ You are both right. My explanation was not correct.I apologize. I have updated both title and body. The point is that I want to prove that $x(t)$ is solution of an ODE of the form $\dot x=f(x)$ (not $\dot x=f(t)!!!$). Otherwise I agree that $x(t)$ is $C^1$ is enough. $\endgroup$ Jul 11 '18 at 8:17
  • $\begingroup$ When $n=1$, if $x\colon(-\infty,\infty)\to(0,\infty)$ has negative first derivative and has limit $0$ at infinity, it is a solution of $\dot{x}=f(x)$, where $f(\xi)$ is defined as the derivative of $x$ at that (unique) $t$ at which $x$ takes value $\xi$. Such $f$ is defined on the range $(0,a)$ of $x$, and under some assumptions it can be extended in a Lipschitz way to $[0,a)$ (or even to $[0,a]$). For $n>1$, I don't know. $\endgroup$
    – user539887
    Jul 11 '18 at 8:51
  • $\begingroup$ You are asking if you can complete the curve $(x(t),\dot x(t))$ in $\Bbb R^{2n}$ to a (possibly unique) graph of a function, that is a $n$-dimensional sub-manifold. There seems to be a large gap in dimensionality, lots of information to fill in. $\endgroup$ Jul 11 '18 at 10:51

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