# How to prove the boundedness of the solutions of a nonlinear differential equation

I have the following differential equation: $$\ddot{x} = -\log(x) - 1$$ and I need to prove that every solution of this equation is a bounded function. From the phase plane portrait, it is obvious that this is true:

How can I construct a formal proof for this?

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$x_1=x$, $x_2=\dot{x}$

$\dot{x}_1=x_2$

$\dot{x}_2=-\log(x_1)-1$

we know that to have the solution for the ODE we need $x_1>0$

consider the following function $$V=(x_1\log(x_1)+0.5*x_2^2)$$ the derivative of this function along the system trajectories is $$\dot{V}=x_2\log(x_1)+x_2+x_2(-\log(x_1)-1)=0$$ Therefore, the solutions are bounded!

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Probably worth adding that $V(x(t))=V(x_0)$ for all $t$ ($V(x(t))=0$ for all $t$). In other words $x(t)\in A(x_0)=\{x:V(x)=V(x_0)\}$ for all $t$. $A(x_0)$ defines a closed orbit in the right hand plane. Thus, this is why you see the closed orbits you do on the phase plot (one per initial condition). In addition, the fact that $\dot{x}_1>0$ if $x_2>0$ and $\dot{x}_1<0$ if $x_2<0$ imply that the solutions follow the orbits in a clockwise fashion. – jkn Aug 30 '12 at 0:53
You still need to prove that all level curves of $V$ are bounded. If $V$ was $x_1 + x_2^2$ instead, for example, the conclusion would not follow. – Rahul Aug 30 '12 at 4:38
Good point, actually what is it exactly about $V$ that ensures that each level set is a single closed orbit? Is it that $V(x)\rightarrow +\infty$ as $|x|\rightarrow +\infty$ and that it is strictly increasing with distance from the point $(\frac{1}{e},0)$? – jkn Aug 30 '12 at 20:33
This answer is incomplete, further properties of V are needed. – Did Dec 26 '14 at 12:21

Note that $$\frac{1}{e} + x \log x \geq 0$$ for all positive $x,$ and gives $0$ only at $x = \frac{1}{e}.$ So, when we write $$\frac{\dot{x}^2}{2} + \frac{1}{e} + x \log x = \mbox{constant}$$ we know that the constant is nonnegative. One may differentiate the equation to check it, using the original ODE. If the constant is $0,$ we have $\dot{x}=0, \; x = \frac{1}{e}.$ If the constant is positive, we have $$x \log x \leq \mbox{constant} - \frac{1}{e}.$$

There is an oddity that happens because $$\lim_{x \rightarrow 0^+} x \log x = 0.$$ If we start with $$x(0) = \frac{1}{e}, \; \; \dot{x}(0) = -\sqrt{\frac{2}{e}},$$ then $x$ continues to decrease forever but never quite reaches $0.$

If, Instead, $$x(0) = \frac{1}{e}, \; \; \dot{x}(0) < -\sqrt{\frac{2}{e}},$$ then $x$ reaches $0$ in finite time and with $\dot{x}$ nonzero.

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