Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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 $u:\mathbb{R}\to\mathbb{R}^3$ where $u(t)=(u_1(t),u_2(t), u_3(t))$ be a function that satisfies $$\frac{d}{dt}|u(t)|^2+|u|^2\le 1,\tag{1}$$where $|\cdot|$ is the Euclidean norm. According to Temam's book paragraph 2.2 on page 32 number (2.10), inequality (1) implies $$|u(t)|^2\le|u(0)|^2\exp(-t)+1-\exp(-t),\tag{2}$$but I do not understand why (1) implies (2).

share|cite|improve this question

The basic argument would go like this. Go ahead and let $f(t) = |u(t)|^2$, so that equation (1) says $f'(t) + f(t) \leq 1$. We can rewrite this as $$\frac{f'(t)}{1-f(t)}\leq 1.$$ Let $g(t) = \log(1 - f(t))$. Then this inequality is exactly that $$-g'(t)\leq 1.$$ It follows that $$g(t) = g(0) + \int_0^t g'(s)\,ds\geq g(0) - t.$$ Plugging in $\log(1 - f(t))$ for $g$, we see that $$\log(1-f(t)) \geq \log(1 - f(0)) - t.$$ Exponentiating both sides gives $$1 - f(t) \geq e^{-t}(1 - f(0)),$$ which is exactly the inequality you are looking for.

Edit: In the case when $f(t)>1$, the argument above doesn't apply because $g(t)$ is not defined. Instead, take $g(t) = \log(f(t) -1)$, so that $g'(t)\leq -1$. It follows that $g(t) \leq g(0) - t$ (at least for small enough $t$ that $g$ remains defined), which upon exponentiating again gives the desired inequality.

share|cite|improve this answer
What happens for $1-f(t) \leq 0$? – Fabian May 12 '12 at 4:35
@Fabian: Edited to include this case. – froggie May 12 '12 at 11:04

Your Answer


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.