Bound on a process satisfying certain integral and differential inequalities. Suppose a non-negative process $x$ satisfies the following integral and differential inequalities:
$$
x_t+C\int_0^tx_s^2ds\,\,<\,\,x_0+\delta+bt,
$$
$$
\dot{x}_t\,\,<\,\,Kx_t^2+a,
$$
where $0<\delta \ll 1$, and $a$, $b$, $C$, $K$ and  are all positive. Is it possible to obtain some constant bound on the process $x$?
Comment:
Note that first equation alone is not enough. $x$ can stay close to zero for a long time and then increase to a large value abruptly so that $x_t$ is large and $\int x_s^2$ is near zero and both add up close to $x_0+\delta+bt$. This way we only get that $x_t<x_0+\delta+bt$. The second equation does not allow such a rapid rise of $x$.
Gronwall kind of inequalities are not helpful because they assume $C<0$. 
Motivation: 
Consider $y\geq 0$ satisfying
$$
\dot{y}_t=-Cy_t^2+b.
$$
Then it is easy to see that $y_t\leq \max\{y_0,\sqrt{b/C}\}$. 
Instead, if $y$ satisfied 
$$
\dot{y}_t \leq -Cy_t^2+b.
$$
then a comparison principle would give same answer $y_t\leq \max\{y_0,\sqrt{b/C}\}$. 
Now consider a process $z\geq 0$ satisfying
$$
\dot{z}_t \leq -Cz_t^2+b + G(z_t,t/\epsilon),
$$
where $\epsilon\ll 1$ and $G$ as a function of its second argument is periodic with mean zero, and as a function of its first argument has atmost quadratic dependence. 
Using the fact that $G$ has atmost quadratic dependence in $z$ we get the second equation of the question. The effect of $G$ on $\dot{z}$ is a very rapid mean zero oscillation. So if we consider integral form of the above equation and approximate the effect of $\int G$ (and assume some other things) then we get the form of the first equation of the question.
 A: I doubt you can obtain such a bound. For simplicity, consider $C=K=b=1$ and $a=2$ (you can adapt what follow to other constants)
Take the familly of functions 
$$f_n(t) = \left\lbrace 
\begin{array} .(t-n) & \text{if} & t \in [n,n+\sqrt[3]{n}] \\ 
(n+2\sqrt[3]{n}-t) & \text{if} & t \in [n+\sqrt[3]{n}, n+2\sqrt[3]{n}] \\
0 & & \text{elsewhere}\end{array} \right.$$
You have that for $t > n$
$$f_n( t ) + \int_0^t f_n^2( s ) ds \leq \sqrt[3]{n} + 2 \int_n^{n+\sqrt[3]{n}} f_n^2( s ) ds  $$
$$ \leq \sqrt[3]{n} + 2 \int_0^{\sqrt[3]{n}} s^2 ds  \leq \sqrt[3]{n} + \frac{2}{3} n $$ 
And for n big enough, you get 
$$ f_n( t ) + \int_0^t f_n^2( s ) ds \leq n < t+\delta$$
So for $n$ big enough, the first equality is verified for $t>n$. The first inequality is also trivialy verified for $t\leq n$.
Now about the second inequality : our functions are not really differentiable, at least in the classical sense. But our function is in $H^1(\mathbb{R})$, so we know that we can approximate it by a smooth function for the $H^1(\mathbb{R})$ convergence. We then use the fact that convergence in $L^2$ imply convergence pointwise for a subsequence to extract a subsequence that converge pointwise, then as the derivative of this subsequence converge in $L^2$ to the derivative of $f_n$, we extract from the previous subsequence a sub-subsequence that converge also pointwise for the derivative.
So for each $f_n$ you can obtain a smooth $\tilde{f_n}$ that verify the inequalities and $\tilde{f_n}(n+\sqrt[3]{n}) > \frac{\sqrt[3]{n}}{2}$, so the familly $(\tilde{f_n})$ is not uniformly bounded. 
