MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
    
That’s brilliant. Thanks a lot! – Shiyu Sep 8 '12 at 3:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.