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For $\ x(0)\equiv x_0>0\ $ and a system governed by $$\dot x(t)=-k\ x(t),$$ I find that $$x(t)>0\ \ \ \forall\ t.$$ (Because the solution is $x(t)=e^{-kt}x_0$.)

For which $f$ and $$\dot x(t)=f(x(t)),$$ does this property hold?

I don't know how difficult this is, but the question might be generalized to functions $f(x(t),t)$ and/or higher order ODEs.

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I'm guessing $\lvert f(x) \rvert = O(x)$ as $x \to 0$ is necessary, because (as a negative example) any solution to $dx/dt = -k \sqrt x$ reaches zero in finite time. – Rahul Jan 16 '12 at 9:17
@Rahul Narain: Thanks for the remark. Turns out that the solution of $\dot x(t)=(-\frac \alpha T)\ x(t)^{1-\frac{1}{\alpha}}$ is $x(t)=(1-\frac t T)^\alpha$ for all finite $\alpha$ and these are zero for $t=T$. – NikolajK Jan 16 '12 at 9:43
up vote 3 down vote accepted

If uniqueness holds, then $f(0)=0$ is enough, since $x(t)\equiv0$ is a solution, and two solutions cannot cross.

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And the uniqueness doesn't hold for the equations in my comment? And are there differential equations with other $f$, which whos solutions still don't cross zero? – NikolajK Jan 16 '12 at 14:38
Uniqueness does not hold for $x'=x^k$ for $0<k<1$ ($x^k$ is not Lipstchitz at $x=0$.) If $f(0)\ne0$, solutions such that $x(t_0)=0$ are not identically $0$ and cross de $x$-axis. – Julián Aguirre Jan 16 '12 at 17:41
@Aguirre: Okay, then we know about solutions which cross the axis $x(t_0)=0$ for some $t_0$. But are there $f$ with $f(0)\ne 0$ and solutions that don't cross it? – NikolajK Jan 17 '12 at 8:20
Consider the equation $x'=-x+1$, whose solution is $x=C\,e^{-t}+1$. If $C<0$, then $x$ crosses the horizontal axis at $t=\ln|C|$; if $C\ge0$, then $x>0$ for all $t\in\mathbb{R}$. – Julián Aguirre Jan 17 '12 at 9:12
@Aguirre: Thanks for the examples. Do you think there is a classification in terms of $f$, which says that a differential equation will admit solution which don't cross zero? – NikolajK Jan 17 '12 at 9:46

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