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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:

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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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    This answer is incomplete, further properties of$V$are needed.2014-12-26
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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.