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## Does this variable have an upper bound?

Let $x$ be a positive scalar variable. Its time derivative satisfies $$|\dot{x}(t)|\le \exp {\left(-\int_{0}^t\frac{1}{x(\tau)}\mathrm{d} \tau\right)}$$ where $|\cdot|$ denotes the absolute value. From the above inequality, can we say $x(t)$ with $t\in[0,+\infty)$ has a finite upper bound? No need to compute an exact upper bound. Is it possible to show there exists a finite upper bound? Thanks.

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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 at 3:38