Nonlinear Grönwall inequality Let $T>0$, $\alpha,\beta>0$ and consider a non-negative continuous function $x$ on $[0,T]$ such that for all $t \in [0,T]$ one has
$$x(t) \leq \alpha+\beta\left(\int_0^t x(s)\,\mathrm ds \right)^{1/2}.$$
Does anyone knows what kind of Grönwall inequality I can get from this ? It would be fantastic if I can get something like $x(t) \leq Ct$.
 A: The inequality implies 
$$
x(t) \le \alpha + \epsilon \int_0^tx(s) ds + \frac{\beta}{4 \epsilon}
$$
for any $\epsilon > 0$. By the usual Gronwall inequality, this implies
$$
x(t) \le \left(\alpha + \frac{\beta}{4 \epsilon} \right) e^{\epsilon t}
$$
for any $t > 0,\epsilon > 0$. Now set $\epsilon = t^{-1}$ to obtain
$$
x(t) \le \left(\alpha + \frac{\beta}{4} t \right)e
$$
You may be able to get better constants by choose $\epsilon = \delta t^{-1}$ with the right $\delta > 0$.
A: By an appropriate change of variables we may assume without loss of generality that $\alpha=\beta=1$. I will also asume that $x(t)\ge0$. Let $u(t)=\int_0^t x(s)\,ds$. Then
$$
u'(t)\le1+\sqrt u\implies\frac{u'}{1+\sqrt u}\le1.
$$
Integrating and using $u(0)=0$ we get
$$
2\sqrt{u}-2\log(1+\sqrt u)\le t.
$$
Let $f(u)$ be the function on the right hand side of the above inequality. It is a concave function, and satisfies the following inequalities:
$$
f(u)\ge\begin{cases}
2(1-\log2)u & \text{if }0\le u\le1,\\
1-2\log2+\sqrt u &\text{if }u>1.
\end{cases}
$$
From here we deduce that $u(t)\le C_1\,t$ where $0\le u\le1$ and $u(t)\le C_2\,t^2$ where $u>1$. This means that for $x$ we have bounds of the form
$$
x(t)\le\begin{cases}
1+A\,\sqrt t & \text{if $t$ small,}\\
B\,t &\text{if $t$ large,}
\end{cases}
$$
for some constants $A$ and $B$.
