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

Consider a first-order autonomous equation in $\mathbb{R}^1$ with $f(x)$ Lipschitz. Suposse $f(0)=f(1)= 0$. Show that solutions starting in $[0,1]$ cannot leave this interval. What is the maximal interval of definition $(T_{-},T_{+})$ for solutions starting in $[0,1]$? Does such a solution have a limit as $t\rightarrow T_{+}$ or $t\rightarrow T_{-}$ ?

Any help would be appreciated. Thanks!

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2 Answers 2

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Since $f$ is Lipschitz, the IVP $x'=f(x)$, $x(0)=x_0$ has a unique solution defined on some maximal interval $(T_-,T_+)$. Since $f(0)=f(1)=0$, $x(t)\equiv0$ and $x(t)\equiv1$ are solutions if $x_0=0$ and $x_0=1$ respectively. Suppose now that $x_0\in(0,1)$. By uniqueness of solution, the graph of the solution cannot cross the lines $x=0$ and $x=1$ (otherwise we would have two different solutions through the same point.) Then $0<x(t)<1$ for all $t$ in the interval of existence of the solution. Moreover, this implies that the solution does not blow-up, and that the interval of existence is $(-\infty,\infty)$.

Edit

I realized after reading orangeskid comment that the equation asked about the limits at the extremes of the interval of existence. This will depend on $f$. If $f$ is positive on $(0,1)$, then $x$ is increasing and $\lim_{t\to-\infty}x(t)=0$, $\lim_{t\to+\infty}x(t)=1$. If $f$ is negative on $(0,1)$, then $x$ is decreasing and $\lim_{t\to-\infty}x(t)=1$, $\lim_{t\to+\infty}x(t)=0$. In general, the limits could be any $\xi\in[0,1]$ such that $f(\xi)=0$.

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  • $\begingroup$ the limits at $\pm \infty$ should be straightforward $\endgroup$
    – orangeskid
    Apr 9, 2018 at 18:15
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Assume $x_0 \in (0,1)$. The solution of $\dot x=f(x)$, $x(t_0)=x_0$ satisfies $$\int_{x_0}^{x(t)} \frac{dx}{f(x)} = t-t_0$$ Since $f(x)>0$ on $(0,1)$ we have $t\mapsto x(t)$ strictly increasing. Now, since $f$ is Lipschitz and $f(0)=f(1)=0$ we have $f(x)\le kx$, $f(1-x)\le k(1-x)$ for some $k$. Therefore, we have $\int_{t_0}^0 \frac{dx}{f(x)}=-\infty$ and $\int_{t_0}^1\frac{dx}{f(x)}=\infty$. It follows that $x(t)$ is defined on $(-\infty, \infty)$ and $\lim_{t\to -\infty} x(t)= 0$ and $\lim_{t\to \infty} x(t)=1$.

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    $\begingroup$ @Robert Lewis: You are correct, I don't know that. I assumed that $0$ and $1$ are consecutive zeroes of $f$, so $f$ has a constant sign on $(0,1)$. I believe we can pass from solutions of $\dot x = f(x)$ to solutions of $\dot x = - f(x)$. So, I just assumed $f>0$ on $(0,1)$. Let me see if I can just eliminate this assumption. Thanks $\endgroup$
    – orangeskid
    Apr 9, 2018 at 18:28
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    $\begingroup$ @Robert Lewis: Yes, it will be the smallest $0$ of $f$ larger than $x_0$. $\endgroup$
    – orangeskid
    Apr 9, 2018 at 18:45
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    $\begingroup$ @Robert Lewis: Yes. Moreover, whenever $x(t)$ approaches a finite limit at say $+\infty$ in the domain of $f$, then $\dot x = f(x)$ also approaches a finite limit. But that limit of $\dot x$ should be zero. So $x$ approaches a zero of $f$. $\endgroup$
    – orangeskid
    Apr 9, 2018 at 19:05
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    $\begingroup$ Yes, that's how I see it. So then $x(t)$ approaches a zero of $f(x)$; the question is whether there is such a zero 'twixt $x_0$ and the endpoints of $(0, 1)$, on either side of $x_0$, right? $\endgroup$ Apr 9, 2018 at 19:14
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    $\begingroup$ @Robert Lewis: Yes, that's correct. If I were the OP I would do a bunch of examples too. $\endgroup$
    – orangeskid
    Apr 9, 2018 at 19:41

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