Delay differential equation when the delay tends to zero.

Question

Consider a delay differential equation with initial condition: $$ x_\tau'(t)=f(t,x_\tau(t),x_\tau(t-\tau)),\,t>0;\quad x_\tau(t)=g(t),\,t\in [-\tau,0], $$ where $\tau>0$ is the delay. My question is under which conditions $x_\tau$ tends to $x$ as $\tau\rightarrow0$, where $x(t)$ solves the problem for $\tau=0$: $$ x'(t)=f(t,x(t),x(t)),\,t>0;\quad x(0)=g(0). $$ Is this a standard result in the literature? I thought of using Gronwall's inequality under a Lipschitz condition for $f$: $|f(t,x_1,y_1)-f(t,x_2,y_2)|\leq K(|x_1-y_1|+|x_2-y_2|)$. Then we derive the inequality $|x_\tau(t)-x(t)|\leq K\int_0^t |x_\tau(s)-x(s)|ds+K\int_0^t |x_\tau(s-\tau)-x(s)|ds$, but I do not know how to work with the second integral $\int_0^t |x_\tau(s-\tau)-x(s)|ds$.

Answer 1

To work with the integral inequality that you derived, $$ |x_\tau(t)-x(t)|\leq K\int_0^t |x_\tau(s)-x(s)|ds+K\int_0^t |x_\tau(s-\tau)-x(s)|ds $$ you can proceed as follows:

First show a uniform in $\tau$ bound for $\|x_\tau(s)\|$ on some interval $[0,T]$. Clearly this has to be possible if the convergence result that you want to prove is true.

This implies a uniform bound for $\dot{x_\tau}(s)$, say $\|\dot{x_\tau}(s)\| \le C_0$ for all $\tau$ and all $s \in [0,T]$.

Assuming also $\|\dot{g}(s)\| \le C_0$ for $s \le 0$ (which is natural), it then follows that $\|x_\tau(s-\tau) - x_\tau(s)\| \le \tau C_0$ for all such $\tau$ and $s \le T$ and therefore $$\begin{aligned} |x_\tau(t)-x(t)|& \leq K\int_0^t |x_\tau(s)-x(s)|ds+K\int_0^t |x_\tau(s-\tau)-x(s)|ds \\ & \leq 2 K\int_0^t |x_\tau(s)-x(s)|ds + KtC_0 \tau \end{aligned} $$ Apply Gronwall's inequality and uniform convergence of $x_\tau$ to $x$ follows on $[0,T]$.