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Let $x$ be a positive scalar variable whose time derivative satisfies $$|\dot{x}(t)|\leq \exp \left\{\left(-\int_{0}^{t}\frac{1}{x(\tau)} \mathrm{d} \tau \right)\right\},$$ where $|\cdot|$ denotes the absolute value. Does the above inequality imply that $x(t)$ is a bounded function? (Note that we do not need to know about the explicit upper bound for $x(t)$ ). Thanks.

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up vote 6 down vote accepted

Integrate a few times:

Using that $x > 0$ we have that $|\dot{x}(t)| \leq \exp (0) = 1$ which implies that $x(t) \leq M + t$ for some $M > 0$. Plug it back in we have

$$ |\dot{x}(t)| \leq \exp \left( - \int_0^t \frac{1}{M+\tau} \mathrm{d}\tau \right) = \exp \left( - \ln (M+t) + \ln M\right) = \frac{M}{M+t} $$

this implies that

$$ x(t) \leq M + M \left[\ln (M+t) - \ln M\right] $$

for which we can very, very roughly estimate by

$$ x(t) \leq 2\sqrt{M' + t} $$

for a sufficiently large $M'$. Plugging this back in we have that

$$ |\dot{x}(t)| \leq \exp \left( - \int_0^t \frac{1}{2\sqrt{M'+\tau}} \mathrm{d}\tau\right) = \exp \left( \sqrt{M'} - \sqrt{M' + t}\right) $$

Using that $e^{-\sqrt{t}}$ is integrable from $0$ to $\infty$ to some constant, we have that $x$ must be bounded.

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That’s brilliant. Thanks a lot! –  Shiyu Sep 8 '12 at 3:38
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