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I've got another question to pose to you. I am given the differential equation x''(t)+a(t)f(x(t))=0, with $a(t)\geq 1$, $f\geq 0$, $ \int_0^{+\infty} f(y)\mathrm d y=+\infty$ and $a,f\in C^0(\mathbb R).$ Then set $I=(t_0,t_1)$ be the maximal interval of definition of the solution; I am then asked to prove that $x(t)$ is bounded above as $t\to t_1^-$. Thanks in advance for your courtesy.

Regards

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    Following Harald comment, one obtains that $\frac{\mathrm d}{\mathrm d t}\left(\frac{x(t)'^2}{2}+\int_0^{x(t)} f(y)\mathrm d y\right)\leq 0.$ But then this function would be bounded by a constant on some interval $[\xi, t_1)$, and if $x$ were not bounded above as $t\to t_1^-$ then you would have that eventually the quantity $\frac{x'^2}{2}$ would become negative, which is impossible i think.2012-03-06

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First of all, observe that x''=-a(t)f(x(t))\le0. Then x' is decreasing and $x$ concave. Deduce that without loss of generality you may assume that $x$ is defined on $[t_0,\infty)$ and x'\ge0 for all $t\ge t_0$.

Next, since $a\ge1$ and $f\ge0$, x''+f(x)\le x''+a\,f(x)=0. Multiply by x' (which is non-negative) to obtain x'x''+x'f(x)\le0\quad\forall t\ge t_0. Let $F$ be a primitive of $f$. The equation above can be written as \Bigl(\frac12(x')^2+F(x)\Bigr)'\le0. Thus \frac12(x')^2+F(x) is decreasing and \frac12(x(t)')^2+F(x(t))\le\frac12(x'(t_0))^2+F(x(t_0))=C\quad\forall t\ge t_0. This implies that $F(x(t))\le C$ for all $t\ge t_0$. There exists $y>0$ such that $F(y)\ge C$ (why?). Finally, since $F$ is increasing (why?), $x(t)\le y$ for all $t\ge t_0$.

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    The solution is concave. This implies that the graph of the solution is below the tangent at any point, that is $x(t)\le A\,t+B$ for constants $A,B\in\mathbb{R}$. Then $x$ is bounded above on $(t_0,t_1)$ whenever t_1<+\infty.2012-03-07