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Consider the Cauchy problem $$ \begin{cases} y'= \cos(y)=f(x,y)\\ y(0)=0 \end{cases} $$ The question is: Does the Picard sequence converge?

My attempt: We have that

$y_0=0$, $y_1(x)= x$, $y_2(x)= \sin(x)$,

$y_3(x)= y_0+ \displaystyle\int_0^x f(s,y_2(s)) ds= \displaystyle\int_0^x \cos(\sin(s)) ds$

We remark that the Picard sequence is not convergent. But my question is how do we prove that the Picard sequence is not convergent without calculating it? What are the conditions of convergence of the Picard sequence?

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Your $f$ is globally Lipschitz with constant $L=1$, the Picard iteration converges on every bounded interval uniformly, or globally on $\Bbb R$ in the norm $\|y\|_L=\sup e^{-2L|x|}|y(x)|$.

We know $|f(x,y)|\le 1$, thus $|y(x)|\le x$.

We know that $(n+\frac12)\pi$, $n\in\Bbb Z$, are stationary points, thus the solution of the IVP $y(0)=0$ is restricted to $-\frac\pi2<y(x)<\frac\pi2$

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