A general version of Gronwall's inequality For the following
$$|u(t)|^p\le C_1 \int_0^t |u(s)|^p\,ds+C_2$$
using Gronwall inequality, we have
$$|u(t)|^p\le C_2(1+C_1 te^{C_1 t})$$
Now, my question is, for
$$|u(t)|^p\le K_1 \int_0^t(1+|u(s)|^2) |u(s)|^{p-2}\,ds+K_2$$
What kind of Gronwall inequality will allow us to get the explicit bound like in general?
 A: We assume that $p>2$. Note that
\begin{align*}
|u(t)|^p &\le K_1 \int_0^t \left(1+|u(s)|^2\right)|u(s)|^{p-2} ds + K_2 \\
&=K_1 \int_0^t |u(s)|^{p-2} ds + K_1 \int_0^t |u(s)|^{p} ds + K_2.
\end{align*}
Let $\alpha = (p-2)/p$, and 
\begin{align*}
v(t) = K_1 \int_0^t |u(s)|^{p-2} ds + K_1 \int_0^t |u(s)|^{p} ds + K_2.
\end{align*}
Then,
\begin{align*}
\frac{dv(t)}{dt} &= K_1 |u(t)|^{p-2} + K_1 |u(t)|^{p}\\
&\le K_1 \left(v(t)^{\alpha} + v(t)\right),
\end{align*}
and
\begin{align*}
\frac{dv(t)^{1-\alpha}}{dt} &= (1-\alpha) v(t)^{-\alpha} \frac{dv(t)}{dt}\\
&\le (1-\alpha) K_1 \left(1+ v(t)^{1-\alpha}\right).
\end{align*}
In other words,
\begin{align*}
\frac{d\Big(\ln \big(1+v(t)^{1-\alpha}\big) - (1-\alpha) K_1 t\Big)}{dt} \le 0.
\end{align*}
That is, the function $\ln \big(1+v(t)^{1-\alpha}\big) - (1-\alpha) K_1 t$ is decreasing. Therefore,
\begin{align*}
\ln \big(1+v(t)^{1-\alpha}\big) - (1-\alpha) K_1 t &\le \ln \big(1+v(0)^{1-\alpha}\big)\\
&=\ln\big(1+K_2^{1-\alpha}\big).
\end{align*}
Consequently,
\begin{align*}
v(t) &\le \left[e^{(1-\alpha)K_1 t}\left(1+K_2^{1-\alpha}\right)-1\right]^{\frac{1}{1-\alpha}}\\
&=\left[e^{\frac{2}{p}K_1 t}\Big(1+K_2^{\frac{2}{p}}\Big)-1\right]^{\frac{p}{2}}.
\end{align*}
