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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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    I've taken the liberty of editing the question into something more closely resembling fluent English. August, I hope I have captured your intent.2011-08-29

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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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    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.2018-11-23