<s>[INST] Showing that $b \leq \frac{b}{\frac{1}{4}+b^2} \Leftrightarrow b^2-2b+\frac{1}{4} \leq 0$ For an ODE where the solution should be on the interval $[-\frac{1}{2},\frac{1}{2}]$ it has to be $b \leq \frac{b}{\frac{1}{4}+b^2}$.
The task is:
Let $f: [-\frac{1}{2}, \frac{1}{2}]\times[-b,b]\to \mathbb{R}$ with $f(x,y):=x^2+y^2$. Find $b>0$ with the theorem of Picard-Lindelöf such that the IVP $y'=f(x,y),\ y(0)=0$ has an unique solution for all $|x|\leq\frac{1}{2}$.
The solution says
$b \leq \frac{b}{\frac{1}{4}+b^2} \Leftrightarrow b^2-2b+\frac{1}{4} \leq 0$.
So $b \in [1-\frac{\sqrt{3}}{2}, 1+\frac{\sqrt3}{2}]$.
I get $b^3-\frac{3}{4}b \leq 0$, so $b\leq \frac{-\sqrt{3}}{2}$ or $b \in [0, \frac{\sqrt3}{2}]$.
How to get $b \leq \frac{b}{\frac{1}{4}+b^2} \Leftrightarrow b^2-2b+\frac{1}{4} \leq 0$?
 [/INST] The "butterfly" (double cone) condition that you have to check is that
$$
Ma\le b,
$$
with $a,b$ the sizes of the box under consideration, $a=\frac12$, and $M=a^2+b^2=\frac14+b^2$ the maximum of the function value inside the box. Inserted this gives
$$
\frac12(\frac14+b^2)\le b\iff \frac14-2b+b^2\le 0.
$$

To compute a non-constant upper bound, one can check that for the differential inequality
$$
|y'|\le \frac14+y^2
$$
one gets
$$
|y(x)|\le\frac12\tan(\frac{|x|}2),
$$
so that in principle the height $b=\frac12\tan(\frac14)$ is sufficient to contain the solution inside the box.
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