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?
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*}