Showing that a sequence of Picard iterates converges I have a sequence of functions:
$$y_{n}(x) = 1 + \int \limits_0^x 1 + t^2 + y_{n-1}^2(t)\,\mathrm dt$$
With $y_0 = 1$.
I'm trying to show that this converges in a box $-1 \le x \le 1$ and $-10 \le y \le 10$. However when I try to show this converges in this box, the bound on the integral gives $|y-1| \le xM$ where $M$ is the maximum of $1 + t^2 + y^2$ in the box, however this maximum is $102$ so the iterates only converge in the box when $|y-1| \le 102$ but this isn't true?
Am I missing something obvious here?
Thanks
 A: The solution of the IVP associated to this Picard iteration, $y'(x)=1+x^2+y(x)^2$, $y(0)=1$, has a lower bound by
$$
y'(x)\ge 1+y(x)^2\implies y(x)\ge\tan(\frac\pi4+x).
$$
This lower bound has a singularity at $x=\frac\pi4<1$, which means that the actual solution has to diverge at some earlier point. The claim is not feasible from the start, solutions can not be confined to the given box.
The box $|y-y_0|\le1$, $|x|\le\frac1{10}$ could work.
A: A theorem for finding an interval of convergence for a Picard iteration scheme is usually stated as:
Given the IVP
\begin{equation*}
y^{\prime} = f(x,y), \quad y(x_{0}) = y_{0},
\end{equation*}
suppose $f$ and $df/dy$ are continuous in the rectangle $R$ given by
\begin{equation*}
x_{0} \le x \le x_{0}+a, \quad \left| y-y_{0}\right| \le b
\end{equation*}
compute the values
\begin{equation*}
M = \max_{(x,y)\in R}\left|f(x,y)\right|, \quad \alpha = \min\left(a, \frac{b}{M}\right).
\end{equation*}
Then the Picard iterates converge to a unique solution on the interval $[x_{0},x_{0}+\alpha]$.
(Note, the interval given by this theorem may be smaller that the actual interval of convergence).  
In your case we have that,
\begin{alignat*}{2}
x_{0} &= 0 \\
a &= 1, \\
b &= 10,\\
M &= \max_{R}\left| 1 + x^{2} + y^{2}\right| \\
&= 1 + a^{2} + b^{2} \\
&= 102 &&\Rightarrow \\
\alpha &= \min\left(1, \frac{10}{102}\right)\\
&\le 1 \\
&= a &&\Rightarrow
\end{alignat*}
which falls short of the desired result.  
Alternatively, we can try looking at the iterates to see if we can learn more about their radius of convergence.
For $y_{n}(x) = 1 + \int_{0}^{x}1+t^{2}+y^{2}_{n-1}(t)dt$, assume the differential equation has the form
\begin{alignat*}{2}
y(x) = 1 + \int_{0}^{x}f(t,y)dt &&\Rightarrow \\
f(t,y) = 1 + t^{2}+y^{2}(t)
\end{alignat*}
and 
\begin{equation*}
y(0) = 1 = y_{0}
\end{equation*}
Now,
\begin{alignat*}{2}
y_{1}(x) &= 1 + \int_{0}^{x}f(t,y_{0})dt \\
&= 1 + \int_{0}^{x}1+t^{2}+1 dt \\
&= 1 + 2x + \frac{1}{2}x^{2},
\end{alignat*}
\begin{alignat*}{2}
y_{2}(x) &= 1 + \int_{0}^{x}f(t,y_{1})dt \\
&= 1 + \int_{0}^{x}1+t^{2}+(1 + 2t + \frac{1}{2}t^{2})^{2} dt \\
&= 1 + 2x + 2x^{2}+2x^{3}+\frac{1}{2}x^{4}+\frac{1}{20}x^{5}
\end{alignat*}
and for $y_{3}$ I get something like:
\begin{equation*}
y_{3}(x) = 1 + 2x + 2x^{2} + \frac{7}{3}x^{3} + 3x^{4} + \frac{13}{5}x^{5}+\frac{202}{120}x^{6} + \frac{52}{70}x^{7}+\frac{22}{80}x^{8}+\cdots
\end{equation*}
So, it seems (loosely) for some constant $M$ large enough
\begin{alignat*}{2}
M\sum_{k=0}^{\infty}x^{k} &\ge \lim_{k\to\infty}y_{k}(x) &&\ge 1+\sum_{k=1}^{\infty}\frac{x^{k}}{k} \Rightarrow\\
M\frac{1}{1-x} &\ge \lim_{k\to\infty}y_{k}(x) &&\ge 1-\log(1-x)
\end{alignat*}
both of which are valid for $\left|x\right| < 1$, but not for the closed interval.  
