Estimate on a function $u$ satisfying a Riccati equation Suppose you have a smooth function $u:[0,\epsilon] \to \mathbb{R}$ satisying the Riccati equation
$$
u'(x)=f(x)u^2(x)+ g(x)u(x) + h(x)
$$
and
$$
u(0)=0
$$
on $[0,\epsilon]$, where $f,g,h$ are smooth functions with
$$
|f(x)|,|g(x)| \leq A \cdot |x|, \quad \text{and} \quad |h(x)|\leq A.
$$
for some $A>1$.
I'm looking now estimate $u'(x)$ and $u(x)$ in terms of $A$ on a $\textbf{"reasonable good interval"}$ . At the moment I'm having trouble finding that $\textbf{"reasonable good interval"}$.
So far I've got the following. Since $u(0)=0$ there is an $r_0>0$ such that $u(x) \leq 1$ on $[0,r_0]$. Then by applying a Taylor expansion
$$
u'(x) \leq f(x)u(x)+ g(x)u(x) + h(x) \leq 2 A \cdot x \cdot u(x) + A \\
\leq 2 A x^2 \max u'(x) + A.
$$
So in particular for $x \leq (2 A )^{-1}$
$$
 \max u'(x) \leq  2 A
$$
and
$$
\max u(x) \leq x \max u'(x) \leq x 2 A \leq 1.
$$
On the interval $[0, \min(r_0 , (2 A )^{-1})]$. So I recover the assumption $u(x) \leq 1$ again.
However I don't find a way to specify the $r_0$ on which this estimate holds. Is there some trick to justify that I can assume that $$r_0=(2 A )^{-1}?$$
 A: In a similar way to your estimates, you can get
$$
|u'|\le \frac{A\max(4,|x|)}4(4|u|^2+4|u|+1)=\frac{A\max(4,|x|)}4(2|u|+1)^2
$$
which results by Grönwall-like arguments in
$$
1-\frac1{2|u|+1}\le \frac{A}{2}\int_0^{|x|}\max(4,s)\,ds
=\frac{A}{2}\int_0^{|x|}(4+(s-4)_+)\,ds
\\
2|u|+1\le\frac1{1-\frac{A}{4}(8|x|+(|x|-4)_+{}^2)}
\\
|u(x)|\le\frac{\frac A2(8|x|+(|x|-4)_+{}^2)}{4-A(8|x|+(|x|-4)_+{}^2)}
$$
using the notation $X_+=\max(0,X)$. This is valid for $|x|<\frac1{2A}$ for $A\ge \frac18$ and $|x|<\sqrt{\frac4A-16}$ for $A<\frac18$. Meaning this bound is suitable for small values of $A$.

Let's try that again with a different split at $|x|=\frac1A$ so that then
$$
|u'|\le \max(1,A|x|)(|u|^2+|u|+A)\le\max(1,A|x|)((|u|+\tfrac12)^2+A)
\\~\\
\implies
\frac{|u(x)|+\tfrac12}{\sqrt{A}}\le \tan\left(\frac{|x|+\frac1{2A}(A|x|-1)_+{}^2}{\sqrt{A}}+\arctan\frac1{2\sqrt{A}}\right)
=\frac{\tan(v(x))+\frac1{2\sqrt A}}{1-\frac1{2\sqrt A}\tan(v(x))}
\\
u(x)\le \frac{(2A+\frac12)\tan(v(x))}{2\sqrt A-\tan(v(x))}
$$
as
$$
\sqrt{A}v(x)=\int_0^{|x|}\max(1,As)\,ds=\int_0^{|x|}(1+(As-1)_+)\,ds=|x|+\frac1{2A}(A|x|-1)_+{}^2.
$$
This is an upper bound as long as the argument of the tangent remains smaller than $\frac\pi2$ or $v(|x|)<\arctan(2\sqrt A)$. As far as I can see, this should give a good upper bound for larger values of $A$.

This should cover the main variants for bounds on $u$.
