Convergence of Picard Iteration Consider an ODE $\frac{dx}{dt}=f(t,x)$. If $f$ is continuous and there's a unique solution for the ODE in some interval, can we guarantee that the Picard iteration converges to that solution?
My guess for the answer is no, since in Picard-Lindelöf theorem, $f(t, x)$ should also be Lipschitz in $x$, but I can't come up with an counter-example; can anyone give any hints about this question?
 A: Consider the continuous function $f:[0,1] \times \mathbb{R}  \to  \mathbb{R}$ defined as
$$
f(t,x)=\left\{ \begin{array}{cc} 
2t,  & x \le -t^2 \\
-2x/t, & |x|< t^2 \\
-2t, & x \ge t^2
\end{array} \right.,
$$
and the IVP $x'(t)=f(t,x(t))$, $x(0)=0$. Starting Picard-iteration with $u_0(t)=t^2$ yields
$$
u_1(t)= \int_0^t f(s,u_0(s)) ds = \int_0^t f(s,s^2) ds = \int_0^t -2s ds =-t^2,
$$
$$
u_2(t)= \int_0^t f(s,u_1(s)) ds = \int_0^t f(s,-s^2) ds = \int_0^t 2s ds =t^2.
$$
Thus, the iteration sequence $(u_n)$ is periodic and it is divergent in each $t \not=0$. The (unique) solution of the IVP is $x(t)=0$.
To prove that the IVP is uniquely solvable assume that $t \mapsto x(t)$ is a nonzero solution. W.l.o.g. assume $x(t_1) > 0$ for some $t_1>0$ (the case $x(t_1) < 0$ can be done the same way). Now go left until $x$ becomes zero the first time, that is, there is some $t_0 \in [0,t_1)$ such that $x(t_0)=0$ and $x(t)> 0$ $(t \in (t_0,t_1])$. But then $x'(t)=f(t,x(t)) < 0$ $(t \in (t_0,t_1])$ since we are in the second or third case in the definition of $f$. Thus $x$ is strictly decreasing on $[t_0,t_1]$, in contradiction to $0=x(t_0)< x(t_1)$.
