1
$\begingroup$

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.

  • 0
    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.2012-01-16
  • 0
    @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 $t=T$.2012-01-16

1 Answers 1

3

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

  • 0
    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?2012-01-16
  • 0
    Uniqueness does not hold for $x'=x^k$ for $0 ($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.2012-01-16
  • 0
    @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?2012-01-17
  • 0
    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 $t\in\mathbb{R}$.2012-01-17
  • 0
    @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?2012-01-17
  • 0
    Assume existence and uniqueness holds for the equation $x'=f(x)$. If $f(0)=0$, then there is only one solution taking the value $0$, namely $x(t)\equiv0$. If $f(0)\ne0$, there are allways solutions taking the value $0$. In a previous comment I gave an example of an equation with solutions taking the value $0$ and never vanishing solutions. Consider now the equation $x'=1$; all solutions vanish at some point.2012-01-17
  • 0
    Thanks, this is an example for an $f$ for which there is no solution with the desired property. I don't know if you already implied the answer and I maybe just don't understand, but we have no general answer yet to the question "which are (all) the $f$ which admit such a solution (with no crossing)", right?2012-01-17
  • 0
    One possibility (again assuming existence and uniqueness) is "There exists $a$ such that $f(a)=0$". If $a\ge0$ and $x_0>a$, then the solution of $x'=f(x)$, $x(0)=x_0$ is such that $x(t)>a\ge0$ for all $t$. Similarly, if $a<0$ and $x_0, then the solution of $x'=f(x)$, $x(0)=x_0$ is such that $x(t) for all $t$.2012-01-17