Inequality of ODE Suppose $u(t)>0$ satisfies following inequality, how do we solve $u$?
$$u^{'}(t)\leq a \cdot u^{\frac{n+2}{n}}(t)$$
for $t\geq 0$. This seems a little different from Gronwall's inequality, how do we find inequality of $u(t)$, you can add some conditions to $u$ if you want. Thanks for any help.
 A: We first take note of a few simple facts concerning $1 \le n \in \Bbb N$:
$\dfrac{n + 2}{n} = 1 + \dfrac{2}{n}; \tag 1$
set
$k = \dfrac{2}{n}; \tag 2$
then
$\dfrac{n + 2}{n} = 1 + k; \tag 3$
$u'(t) \le au^{1 + k}; \tag 3$
$u^{-k - 1} u'(t) \le a; \tag 4$
$(-\dfrac{1}{k} u^{-k})' \le a; \tag 5$
we next integrate 'twixt $t_0$ and $t > t_0$:
$-\dfrac{1}{k}u^{-k}(t) + \dfrac{1}{k}u^{-k}(t_0) = \displaystyle \int_{t_0}^t (-\dfrac{1}{k} u^{-k}(s))' \; ds \le a(t - t_0); \tag 6$
$-\dfrac{1}{k}u^{-k}(t) + \dfrac{1}{k}u^{-k}(t_0) \le a(t - t_0); \tag 7$
$-\dfrac{1}{k}u^{-k}(t) \le a(t - t_0) - \dfrac{1}{k}u^{-k}(t_0)  ; \tag 8$
$u^{-k}(t) \ge u^{-k}(t_0) -ka(t - t_0); \tag 9$
since
$u(t) > 0, \tag{10}$
it follows that the both sides of (9) are positive for $t$ sufficiently close to $t_0$, and always if $t > t_0$ provided $a \le 0$; hence for such $t$,
$0 < u^k(t) \le (u^{-k}(t_0) -ka(t - t_0))^{-1} = \dfrac{1}{u^{-k}(t_0) -ka(t - t_0)} = \dfrac{u^k(t_0)  }{1 -ku^k(t_0)a(t - t_0)}, \tag{11}$
whence
$0 < u(t) \le ((u^{-k}(t_0) -ka(t - t_0))^{-1/k} = \dfrac{1}{\sqrt[k]{u^{-k}(t_0) -ka(t - t_0)}} = \dfrac{u(t_0)  }{\sqrt[k]{1 - ku^k(t_0)a(t - t_0)}}. \tag{12}$
Inspection of (12) reveals that 
$\displaystyle \lim_{t \to \infty} u(t) = 0, \; a < 0, \tag{13}$
whereas the limit, if one exists, is not determined from (12) if $a \ge 0$.  In fact, a singularity is encountered as
$t \to t_0 + \dfrac{1}{k u^k(t_0)a} \tag{14}$
from below.  The situation as $t \to -\infty$ is similar, with the cases $a < 0$, $a \ge 0$ interchanged.
