Limit of Picard iteration solves IVP: Interchange of limit and differentiation

Question

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.

Answer 1

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.

Answer 2

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.