Proof of the existence of a solution on IVP with a given interval for $x$ Prove that the solution $y(x)$ exists on the given interval for $x$:
$$y'=y^2 + \cos x^2$$
on $0\leq x\leq 1/2$ with $y(0)=0$.
I think that I have to use Picard's Theorem. But I don't know why I am given an interval for $x$. And the left bound of $x$ is the initial value. Any help?
 A: The task asks if $[0,\frac12]$ is a subset of the domain of the (or a) maximal solution of the IVP. This is a non-trivial question as the solution will diverge to infinity in finite time as soon as it reaches a value larger than $1$, as $y'>y^2-1$ gives an exploding lower bound.

First variant, $y'\le y^2+1$ gives that, as long as the solution exists, $y(x)\le \tan(x)$. In the consequence, the solution remains bounded above for $x<\frac\pi2$. As long as $\cos(x^2)\ge 0$, that is, $x\le\sqrt{\frac\pi2}$, it is also clear that $y$ is increasing and thus positive. Both together give that, for instance, $[0,1]$ is a subset of the maximal domain.

Second variant, if you want to go by the Picard iteration, you first fix a rectangle $R$ via $|x|\le a$, $|y|\le b$. Then the right side function is bounded by $|f(x,y)|\le M=b^2+1$ on $R$. Solutions are guaranteed by this bound to stay inside this rectangle for $|x|\le h$ if $Mh\le b$. Next for the convergence one needs that $q=Lh<1$ for the contraction factor of the Picard iteration, where $L=2b$ is an $y$-Lipschitz constant. Both conditions can be satisfied if $h$ is chosen small enough. As the bounds do not depend on $a$, one can set $a=h$. With $b=1$ these conditions are satisfied for any $h<\frac12$, it follows that the solution exists at least on $[0,1]$, with a limit at $x=\frac12$ due to boundedness.
A: Observe that the unique solution of the IVP
$$
y'=1+y^2, \quad y(0)=0,
$$
is $y(x)=\tan x$, and its maximal interval is $(-\pi/2,\pi/2)$.
Next, define the Picard sequence $y_n(x)=\int_0^x \big(\cos^2t+y_{n-1}^2(t)\big)\,dt$ of
$$
y'=\cos^2(x)+y^2, \quad y(0)=0,
$$
and prove inductively that,
$$
0= y_0(x)\le y_1(x)\le \cdots\le y_n(x)\le y_{n+1}(x)\le \tan x, \quad x\in [0,\pi).
$$
Hence, $y_n$ converges to some $y(x)\le \tan x$, for $x\in [0,\pi/2)$.
Also, for $x\in [0,1/2]$, we have $½<\pi/6$, and hence $\tan(1/2)<\tan(\pi/6)=\frac{\sqrt{3}}{3}$.
$$
0\le y_{n+1}(x)-y_n(x)=\int_0^x \big(y_n^2(t)-y_{n-1}^2(t)\big)\,dt
=\int_0^x \big(y_n(t)+y_{n-1}(t)\big)\big(y_n(t)-y_{n-1}(t)\big)\,dt
\\ \le 2\tan x \int_0^x \big(y_n(t)-y_{n-1}(t)\big)\,dt \le
\frac{2\sqrt{3}}{3}\int_0^x \big(y_n(t)-y_{n-1}(t)\big)\,dt
$$
In particular, $y_1(x)-y_0(x)=\int_0^x \cos^2t\,dt \le x$,
$$
y_2(x)-y_1(x)\le \frac{2\sqrt{3}}{3}\int_0^x \big(y_1(t)-y_0(t)\big)\,dt \le 
\frac{2\sqrt{3}}{3}\int_0^x t\,dt \le \frac{2\sqrt{3}}{3}\cdot \frac{x^2}{2} 
$$
$$
y_3(x)-y_2(x)\le \frac{2\sqrt{3}}{3}\int_0^x \big(y_2(t)-y_1(t)\big)\,dt \le 
\frac{2\sqrt{3}}{3}\int_0^x \frac{2\sqrt{3}}{3}\cdot \frac{t^2}{2}\,dt \le \left(\frac{2\sqrt{3}}{3}\right)^2\cdot \frac{x^3}{3!} 
$$
and inductively
$$
0 \le y_{n}(x)-y_{n-1}(x)\le  \left(\frac{2\sqrt{3}}{3}\right)^{n-1}\cdot \frac{x^{n}}{n!} \le \frac{1}{n!} , \quad x\in [0,1/2],
$$
Hence $\{y_n(x)\}$ converges uniformly in $[0,1/2]$ (Weierstrass M-test) and the limit $y(x)$ satisfies
$$
y(x)=\int_0^x \big(\cos^2t+y^2(t)\big)\,dt, \quad x\in [0,1/2]
$$
and hence $y$ satisfies the IVP.
Note. The interval $[0,1/2]$ could be replaced by $[-a,a]$, for any $0<a<\pi/2$.
