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I'm studying the proof of Peano's existence theorem on this paper.

At page 5 it is said that the problem $$\begin{cases} y(t) = y_0 & \forall t ∈ [t_0, t_0 + c/k] \\ y'(t) = f(t − c/k, y(t − c/k)) & \forall t ∈ (t_0 + c/k, t_0 + c] \end{cases} $$ has a unique solution.

Can you explain me why? Thank you.

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The proof proceeds stepwise. At the outset, the solution is given for $t\in[t_0,t_0+c/k]$. Note carefully that then the right hand side of the second equation $y'(t)=f(t-c/k,y(t-c/k))$ is known for $t\in[t_0+c/k,t_0+2c/k]$, so you can integrate: $$y(t)=y(t_0+c/k)+\int_{t_0+c/k}^\tau f(\tau-c/k,y(\tau-c/k))\,d\tau,\qquad t\in[t_0+c/k,t_0+2c/k].$$ Now repeat the same idea, gaining a unique solution for $t$ in the next interval $[t_0+2c/k,t_0+3c/k]$, and so on.

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The problem is simply a quadrature because $f(t − c/k, y(t − c/k))$ is known for $t \in (t_0 + c/k, t_0 + 2c/k]$. $$t \in (t_0 + c/k, t_0 + 2c/k]\implies t-c/k\in\cdots$$ Same argument for the next intervals.

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