Applying Gronwall inequality for $\frac{d}{dt}\|u(t)\|_{L^{q}}^q \leq f(t) + C\|u(t)\|_{L^{q}}^{q-2}$ Suppose we have a function $u(x,t)$, where $x \in \mathbb R^n$ and $t&gt;0$ denotes time. And suppose $q&gt;2$, then how to apply Gronwall to the inequality below?
$$\frac{d}{dt}\|u(t)\|_{L^{q}}^q \leq f(t) + C\|u(t)\|_{L^{q}}^{q-2}.$$
Here $f(t)$ is any function and $C$ is a constant, both of which do not matter here since my question is how to use Gronwall when we have a power.
I just found the answer myself but the idea was inspired by Alex Francisco who gave his answer above. So we just use Young's inequality, i.e. we consider $\|u(·, t)\|_q^{q-2}=1\times\|u(·, t)\|_q^{q-2} \leq C + C_1\|u(·, t)\|_q^{q}$, where $C$ and $C_1$ are constants.
$\def\peq{\mathrel{\phantom{=}}{}}\def\d{\mathrm{d}}$Note that for any $a, b, x &gt; 0$, by AM-GM,\begin{align*}
&amp;\peq ax^2 + \frac{b}{x^{q - 2}} = (q - 2) · \frac{ax^2}{q - 2} + 2 · \frac{b}{2x^{q - 2}}\\
&amp;\geqslant q \left( \frac{ax^2}{q - 2} \right)^{\frac{q - 2}{q}} \left( \frac{b}{2x^{q - 2}} \right)^{\frac{2}{q}} = q \left( \frac{a}{q - 2} \right)^{\frac{q - 2}{q}} \left( \frac{b}{2} \right)^{\frac{2}{q}}. \tag{1}
\end{align*}
By setting $\left( \dfrac{q - 2}{a} \right)^{\frac{q - 2}{q}} = q \left( \dfrac{b}{2} \right)^{\frac{2}{q}} = c^{-\frac{2(q - 2)}{q}}$, where $c &gt; 0$ is a constant, (1) becomes$$
(q - 2) c^2 x^2 + \frac{2}{q^{\frac{q}{2}} c^{q - 2} x^{q - 2}} \geqslant 1.
$$
Thus for any $c &gt; 0$,$$
x^{q - 2} \leqslant (q - 2) c^2 x^q + \frac{2}{q^{\frac{q}{2}} c^{q - 2}}. \quad \forall x \geqslant 0 \tag{2}
$$

Because $\dfrac{\d}{\d t} (\|u(·, t)\|_q^q) \leqslant f(t) + C \|u(·, t)\|_q^{q - 2}$, then for any $c &gt; 0$,$$
\frac{\d}{\d t} (\|u(·, t)\|_q^q) \leqslant f(t) + C \|u(·, t)\|_q^{q - 2} \leqslant f(t) + \frac{2C}{q^{\frac{q}{2}} c^{q - 2}} + (q - 2) c^2 C \|u(·, t)\|_q^q.
$$
Now Gronwall's inequality can be applied.