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We have the IVP

\begin{align*} y'(t) &= f(t,y(t))\\ y(t_0) &= y_0. \end{align*}

We consider the Picard iterates $ y^0(t) = y_0 $ and
$$y^k(t) = y_0 + \int_{t_0}^t f(s,y^{k-1}(s))\, ds$$ for $k \in \mathbb{N}$.

I was able to show that $y^k$ converges uniformly under specific circumstances like Lipschitz continuity in the second argument of $f$. Let us say that $y$ is the limit of $y^k$. Next I want to show that $y$ solves the IVP above. I am unsure with my argumentation for $ y'(t) = f(t,y(t)) $:

$$y'(t) = \frac{d}{dt} \lim_{k \to \infty} y^k(t) = \lim_{k \to \infty} \frac{d}{dt} y^k(t) = \lim_{k \to \infty} f(t,y^{k-1}(t)) = f(t,y(t)).$$

The last equation follows from the fact that we assume (Lipschitz) continuity in the second argument, this is not the problem. I am just unsure about the interchange of the limit and differentiation. I would argue that this follows from the fact that $y^k$ converges uniformly to $y$. Could someone confirm if this is true?

Any help is very appreciated.

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The limit is a solution of the Picard integral equation.

The Picard integral equation is equivalent to the initial value problem, which is an almost trivial consequence of the fundamental theorem of calculus. This is true independent of any sequence that converges to the solution.

There is nothing more to show there.

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  • $\begingroup$ Hey, thanks for your reply! The thing is that I need to prove that the limit solves the IVP indeed, sorry if I was unclear. My problem is juat that I don't know if one can interchange $/frac{d}{dt} $ and $/lim$ in this case. Could you please tell me if this step is alright? $\endgroup$ – Diglett Apr 23 '17 at 7:35
  • $\begingroup$ You do not need to consider this exchange of limits if you use the equivalence of solutions of the Picard equation and the IVP. And usually it is not possible to conclude from the convergence of a function sequence to the convergence of the derivatives. $\endgroup$ – LutzL Apr 23 '17 at 7:56
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Okay, now I have found an answer to my problem.

We can interchange $\frac{d}{dt}$ and $\lim$ if the derivative $\frac{d}{dt} y^k(t)$ converges uniformly. If we assume Lipschitz continuity in the second argument, the sequence $\frac{d}{dt} y^k(t) = f(t,y^{k-1}(t))$ converges uniformly indeed because we know that $y^k(t)$ converges uniformly and because of Lipschitz.

Therefore, the interchange is allowed.

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