Method of steps

Definition

Let $\tau > 0$ and let $f \colon \R \times \R^n \times \R^n \to \R^n$ be continuous. Consider the delay differential equation $x'(t) = f(t, x(t), x(t - \tau))$ for $t \ge 0$, with a continuous history $x(\theta) = \phi(\theta)$ for $\theta \in [-\tau, 0]$. The method of steps constructs the solution on the intervals $I_j = [j\tau, (j+1)\tau]$, $j = 0, 1, 2, \ldots$, in order.

1. On $I_0$, $x(t - \tau) = \phi(t - \tau)$ is known, so $x$ solves the initial value problem $x'(t) = f(t, x(t), \phi(t - \tau))$, $x(0) = \phi(0)$.
2. Suppose $x$ is known on $[-\tau, j\tau]$. On $I_j$ the term $x(t - \tau)$ is then known, and $x$ solves $x'(t) = f(t, x(t), x(t - \tau))$ with the value $x(j\tau)$ from the previous step.

Let $f$ be locally Lipschitz in its second argument. Then each step has a unique local solution, so the solution is unique as long as it exists.