differential inequality involving the square of the function It is written in a book, (Bertozzi- Majda, vorticity and incompressible flow page 106) that given a differential inequality of the following type:
$ \frac{d}{dt}\|u^{\epsilon}(t)\| \leq c\|u^{\epsilon}(t)\|^2$ we can conclude that $\sup_{t\in [0;T]}\|u^{\epsilon}(t)\| \leq \frac{\|u^{\epsilon}(0)\|}{1-cT\|u^{\epsilon}(0)\|}$ for every $ T < \frac{1}{c\|u^{\epsilon}(0)\|}$.
Can someone explain to me why this is true ?
Thankfully.
 A: Let $v$ be a solution of the Cauchy problem
$$
\frac{d v}{d t} = c v^2, \quad v(0) = \|u_\epsilon(0)\|.
$$
The following statement is more-less evident
Proposition 1. $u(t) \leq v(t)$ for all $t$, where $u(t)$ and $v(t)$ are defined.
($\|u_\epsilon\|$ and $v$ start from equal initial condition $\|u_\epsilon(0)\|$, and the growth rate (derivative) of $\|u_\epsilon\|$ is always less or equal the growth rate of $v$). However, for more rigorous proof, you can proceed with Gronwall's inequality. Subtracting equation for $v$ from inequality for $\|u_\epsilon\|$, and denoting $w := \|u_\epsilon\| - v$ we get
$$
\frac{d w}{d t} = c (\|u_\epsilon\|^2 - v^2) = c (\|u_\epsilon\| + v) (\|u_\epsilon\| - v) = c \beta(t) w(t), \quad w(0) = 0,
$$
where $\beta(t) = (\|u_\epsilon\| + v)$. Applying now the Gronwall inequality, we conclude that $w \leq 0$, and hence $u \leq v$.
At the same time, problem for $v$ has an exact solution by separation of variables:
$$
v(t) = \frac{\|u_\epsilon(0)\|}{1 - c t \|u_\epsilon(0)\|}.
$$
It is easy to see that $v(t)$ is monotonically increase, and hence
$$
\sup_{t \in [0, T]} v(t) \leq \frac{\|u_\epsilon(0)\|}{1 - c T \|u_\epsilon(0)\|}.
$$
Using now Proposition 1, we conclude that
$$
\sup_{t \in [0, T]} u(t) \leq \sup_{t \in [0, T]} v(t) \leq \frac{\|u_\epsilon(0)\|}{1 - c T \|u_\epsilon(0)\|},
$$
which is the desired result.
