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Suppose that $u$ is a monotonic (we can suppose increasing) solution of the ODE $$\ddot{u}+f(u)=0, \text{ in all } \mathbf{R},$$ such that $$\lim_{x\rightarrow\infty}u(x)=M=\sup u.$$ Then $$f(M)=0.$$

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What do you mean by "classical"? –  Pedro Tamaroff Oct 22 '12 at 15:24

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We must assume $f$ to be continuous, otherwise there may be no connection between $f(M)$ and the values of $f(y)$ for $y < M$.

Suppose $f(M) > 0$. By continuity, there exist $\delta > 0$ and $\epsilon > 0$ such that $f(y) > \epsilon$ for $M - \delta \le y < M$. Thus if $u(T) = M - \delta$ we have $\ddot{u}(t) > \epsilon$ for $t \ge T$. Since $\dot{u}(T) \ge 0$, that implies $\dot{u}(t) \ge \epsilon(t - T)$ and $u(t) > M - \delta + \dfrac{\epsilon}{2}(t-T)^2$ for $t \ge T$; in particular this would say $u(t) \to +\infty$ as $t \to \infty$, contradicting the assumption that $\lim_{x \to \infty} u(x) = M$.

Similarly, if $f(M) < 0$ we would get $u(t) \le M - \delta + u'(T) (t-T) - \dfrac{\epsilon}{2} (t-T)^2 \to -\infty$ as $t \to \infty$.

EDIT: perhaps an interesting counterexample without the continuity assumption is $u(t) = -\dfrac{4}{t} + \dfrac{\sin(t^3)}{t^4}$. This function increases monotonically to $0$ as $t \to \infty$, so $\ddot{u}$ is a function of $u$ on $(-\infty,0)$, but this function has no limit as $u \to 0-$: in fact $$\ddot{u} = -\frac{8}{t^3}+20\,{\frac {\sin \left( {t}^{3} \right) }{{t}^{6}}}-18\,{ \frac {\cos \left( {t}^{3} \right) }{{t}^{3}}}-9\,\sin \left( {t}^{3} \right) $$ so the lim sup and lim inf are $+9$ and $-9$.

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