Let $n,d \in \mathbb N$ und $\tau > 0$. Sei $f: [0,n\tau] \times \mathbb R^d \times \mathbb R^d \rightarrow \mathbb R^d, (t,x,y) \rightarrow f(t,x,y)$ continuous with the condition: There exists a $L > 0$ so that $|f(t,x_1,y)−f(t,x_2,y)| \leq L|x_1 −x_2| \forall t \in [0,n\tau],x_1,x_2,y \in \mathbb R^d.$ Furthermore, let $\phi \in C^1([−\tau,0],\mathbb R^d).$ \begin{cases} y′(t) = f(t,y(t),y(t−τ)), \text{ } t\in [0,n\tau],\\ y(t) = \phi(t),\text{ } t \in [−\tau,0] \end{cases} with the condition $\phi'(0) = f(0,\phi(0),\phi(−\tau))$. Show, that this problem has exactly one solution $y \in C^1([−\tau,n\tau],\mathbb R^d)$. Tip: Look at the functions $y_k: [0,\tau] → \mathbb R^d, y_k(t) := y((k − 1)\tau + t)$ for $k=1, \dots,n-1$.
Can anyone give me a tip?
Iterating on the intervals of length $τ$, on each sub-interval $[kτ,(k+1)τ]$, the second argument $y(t)=x(t-τ)$ is a completely known function by the iteration assumption, so the DDE $x'=f(t,x(t),x(t-τ))$ can be considered as ODE $x'(t)=F(t,x(t))$, $F(t,x)=f(t,x,y(t))$, and the existence-and-uniqueness theorem for ODE can be applied.