A fractional Gronwall lemma I have an energy estimate of the form
$$\dot{u} + \lambda u^{1+ \delta} \le C (1+t)^{-\sigma},$$
where $u=u(t)$ is positive, $\lambda, \delta > 0$ and one can assume $\sigma > 1$ large. I expect the solution to satisfy a bound of the form
$$u \le M\Big(C + u(0)\Big)(1+t)^{-1 / \delta}$$
for some universal $M > 0$.
The only answer I found online is Proposition 2.2 of New nonlinear inequalities and application to control systems by Ghrissi and Hammami. However, they use a lemma which corresponds to the kind of Gronwall inequality I need but for $\lambda \le 0$,
and I did not manage to adapt the proof.
Do you have any ideas?
 A: I figured out how to prove the bound I was looking for. I drew inspiration from the alternative proof of Lemma 3.2 taken from S. Mischler's course on evolution PDEs (first chapter, dedicated to Gronwall's lemma).
The key trick is the following inequality valid for any $u, \ell, \delta > 0$:
$$u \le u \left( \frac{u}{\ell} \right)^{\delta} + \ell,$$
which in turn implies for any $a > 0$ (simply consider $a := \ell^{\delta}$)
$$a u \le u^{1+\delta} + a^{1+1 / \delta}.$$
Note that this holds for any $a$ and therefore any function $a=a(t) > 0$, thus we can plug this lower bound in the differential inequality as to obtain a linear one with a source term which we can partly choose and then optimize!
\begin{align}
\dot{u} + \lambda a u &\le C (1+ t )^{-\sigma} + \lambda a^{1+1/\delta}
\end{align}
$$u \le e^{-\lambda A(t)} u(0) + e^{-\lambda A(t)} \int_0^t e^{\lambda A(t')} \Big\{ C (1+t')^{-\sigma} + \lambda a(t')^{1+1/\delta} \Big\} d t',$$
where we denoted $A(t)=\int_0^t a(t') d t'$. Choosing $a(t)=\frac{\delta + 1}{\delta \lambda (1 + t)}$ we get
\begin{align}
u &\le (1 + t)^{-1-1/\delta} u(0) + (1+t)^{-1-1/\delta} \int_0^t \left\{C (1+t')^{1+1/\delta - \sigma} + \lambda \left(\frac{1+\delta}{\delta \lambda}\right)^{1+1/\delta} (1+t')^{-1-1/\delta} \right\} d t'\\
& \le (1 + t)^{-1-1/\delta} u(0) + \frac{C \delta}{1+\delta(2-\sigma)} (1+t)^{1-\sigma} + \lambda\delta \left(\frac{1+\delta}{\lambda}\right)^{1+1/\delta} (1+t)^{-1/\delta}.
\end{align}
In particular, the optimal decay rate is obtained for $\sigma \ge 1+1/\delta$ and we have
$$u \le (M + u(0)) (1+t)^{-1-1/\delta}, ~ M := \frac{C \delta}{1+\delta(2-\sigma)} + \lambda\delta \left(\frac{1+\delta}{\lambda}\right)^{1+1/\delta}.$$
N.B. In my case, $\lambda \approx u(0)^{-\delta}$ so when $\sigma \ge 1+1/\delta$, there holds indeed
$$u \lesssim (C + u(0))(1+t)^{-1/\delta}.$$
