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 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.