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I was reading a book of ODE and make a comment using the uniqueness theorem on the behavior of the solutions of an autonomous differential equation $$ \frac{dy}{dt} = f \left(y \right) $$ Suppose $f$ satisfies the hypotheses of the theorem of existence and uniqueness. Take for example $$ f\left( y \right) = \left( {y - 2} \right)\left( {y + 1} \right) $$

Then we know from the theorem that the solutions do not intersect (the uniqueness). Suppose you have a solution satisfying $ y (0) = 1 / 2 $. Clearly $y(t)$ is between -1 and 2, is also decreasing, and bounded, so must tend asymptotically to a horizontal line, which should be $y =c$ for some $c\ge-1$. Why actually happens it tends to the line of phase? That is, why do we actually get $c=-1$? Do you know any good books where you can learn these things? an introductory book?

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  • $\begingroup$ There are some serious formatting problems. I corrected your first equation, please do the same for the rest of your post. $\endgroup$
    – Did
    Commented Aug 29, 2011 at 0:57
  • $\begingroup$ Ok ,I´ll explain again my question. I can know that a solution function is between 2 horizontal lines, which are lines of phase, and also do not intersect. We even know that the function is strictly monotone. At infinity the solution is strictly monotone and bounded must converge to a value. But this value a priori, need not be the value of an equilibrium point (but it happens) my question is why that happens $\endgroup$
    – August
    Commented Aug 29, 2011 at 2:11
  • $\begingroup$ I've taken the liberty of editing the question into something more closely resembling fluent English. August, I hope I have captured your intent. $\endgroup$ Commented Aug 29, 2011 at 3:42

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Suppose $\lim_{t \to \infty} y(t) = L$. For any fixed $T$, $\left|\frac{y(T+N) - y(T)}{N}\right| \to 0$ as $N \to \infty$. By the Mean Value Theorem, $\frac{y(T+N) - y(T)}{N} = y'(t)$ for some $t \in (T,T+N)$. But the differential equation says $y'(t) = f(y(t))$, and by continuity of $f$ we have $\lim_{t \to \infty} f(y(t)) = f(L)$. So we must have $f(L) = 0$.

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  • $\begingroup$ It's been a long while. But I was wondering if you have time to clarify this a bit. Should we say, "For any fixed $N$, $\mid \frac{y(T+N)-y(T)}{N}\mid\to 0$ as $T\to \infty$" instead? As it currently stands, the $t\in (T, T+N)$ may not tends to infinity as $N\to \infty$. $\endgroup$
    – syeh_106
    Commented Nov 22, 2018 at 7:43
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    $\begingroup$ The original is correct. If $f(L) \ne 0$, there would be some $\epsilon > 0$ and $\delta > 0$ such that $|f(y)| > \epsilon$ for $|y - L| < \delta$. Take $T$ large enough that $|y(t) - L| < \delta$ for $t \ge T$. $\endgroup$ Commented Nov 22, 2018 at 15:42
  • $\begingroup$ Indeed. Thanks a lot! I was trying to prove it without using contradiction. I.e. Fix N. $\lim_{T\to \infty}y'(t_T)=0$ where $t_T\in (T, T+N)$. Hence $0=\lim_{T\to \infty} f(y(t_T))=f(L)$. But this isn't necessary. $\endgroup$
    – syeh_106
    Commented Nov 23, 2018 at 0:27

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