IVP in which Picard iterates don't converge

Question

I need to exhibit an example of a IVP problem in which the Picard iterates do not converge. I was thinking about something involving a function who is not integrable, or trying to set $y(0)=y_0=0$ in the initial value problem of the type $$\begin{cases} f(t.y)=y'\\ y(t_0)=y_0 \end{cases}$$ and somehow define $f$ so that the quantity $$g_1(t)=\int_0^t f(s,0)ds = \infty$$ where $g_1(t)$ is the first Picard iterate. Thank you in advance!

Answer 1

Try $y'=y^2$ with $y(0) = 1$ on the interval $[0,1]$.

A small bit of work shows that if $y_0(x) =1$, and $y_{n+1}(x) = 1+ \int_0^x y_n(t)^2 dt $, then $y_n(x) \ge 1+ x + \cdots + x^n$, for $x \in [0,1]$.

In fact, if we let $y(x)= {1 \over 1-x}$ for $x \in [0,1)$, it is straightforward to show that if $y_n(x) \le y(x)$ for $x \in [0,1)$ then $y_{n+1}(x)\le y(x)$ for $x \in [0,1)$.

We see that for $x \in [0,1)$ we have $y_n(x) \to y(x)$, and $y_n(1) \to \infty$.