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Given a first order ODE $y'(x)=f(x,y)$ with the initial condition $y(x_0)=x_0$ such that it satisfies Picard thoerem of existence and uniquness, one can compute the solution by Picard iterations :

$$ y_0 (x) \equiv y_0 \, ,$$ and $$ y_{n+1}(x) = y_0 + \int\limits_{x_0}^x f(t,y_n (t)) \, dt \, .$$

Computationally speaking, how fast is this method? Do we have a theorem about its convergence rate? Do we have particulary "bad" examples for its convergence?

Generally speaking, I know this method is not the usual go-to for numerical solutions of ODE's, but why?

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    $\begingroup$ Specifically, how do you intend to evaluate that integral? If I understand your notation correctly, you have simply stated a formal definition of what it means to solve that ODE. Specific numerical methods make certain choices about evaluating the integral. If I am misunderstanding, please clarify your notation and the specific algorithm you have in mind. $\endgroup$ – Bill Greene Apr 5 '17 at 18:14
  • $\begingroup$ @BillGreene , At each step both $f$ and $y_n$ are given. Assume, for the sake of this discussion, that we can use an arbitrarily accurate numerical integration scheme. Even if we had it analytically, we still know nothing about how these iterations converge. $\endgroup$ – Amir Sagiv Apr 6 '17 at 9:50
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    $\begingroup$ That said, Picard's iteration converges linearly. $\endgroup$ – Wolfgang Bangerth Apr 8 '17 at 12:32
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    $\begingroup$ @WolfgangBangerth ,Thanks! Can you refer to the proof or show it, as an answer? $\endgroup$ – Amir Sagiv Apr 9 '17 at 5:19
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    $\begingroup$ Picard iteration is a fixed point iteration. If you can show that it's a contraction, then linear convergence follows immediately based on a theorem that you will find in most introductory textbooks on numerical analysis (e.g. Kincaid and Cheney). $\endgroup$ – Wolfgang Bangerth Apr 10 '17 at 0:23

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