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?
 A: 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.
A: $x_1=x$, $x_2=\dot{x}$
$\dot{x}_1=x_2$
$\dot{x}_2=-\log(x_1)-1$
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!
