Single-step autonomous delay differential equation

Definition

Form of the differential equation

This is a particular type of first-order first-degree autonomous delay differential equation, given explicitly as:

$\frac{d(x(t))}{dt}=f(x(t),x(t-\tau ))$

where $f$ is a known function and $\tau >0$ is also known.

Nature of initial value specification

The initial value specification for this type of delay differential equation is a description of $x$ as a function of $t$ on an interval of length $\tau$, typically the left-most such interval in our domain.

Solution method: moving forward

The solution method is called the method of steps. The idea is that, if the function is known on an interval of the form $[a-\tau ,a]$, we can figure out what it is on $[a,a+\tau ]$, and then repeat the process to determine what the function is on $[a+\tau ,a+2\tau ]$, and continue to proceed in this way to determine the function everywhere on $[a,\mathrm{\infty })$.

Let us say that we know that $x(t)=\phi (t)$ on the interval $[a-\tau ,a]$. Then, $x(t)$ is the solution $\psi (t)$ to the following equation on $[a,a+\tau ]$ subject to the condition $\psi (a)=\phi (a)$:

$\frac{d}{dt}(\psi (t))=f(\psi (t),\phi (t-\tau ))$

This is an ordinary first-order first-degree differential equation in $\psi$ with an initial-value specification, so we expect it to have a unique solution.

Solution method: moving backward

We can also do a similar process to move backward. Explicitly, suppose $x(t)=\psi (t)$ on an interval of the form $[a,a+\tau ]$. We want to find out what it looks like on $[a-\tau ,a]$. We set $x(t)=\phi (t)$ on this interval, and we want to solve the following for $t\in [a-\tau ,a]$ subject to the initial value condition $\phi (a)=\psi (a)$:

${\psi }^{\prime }(t+\tau )=f(\psi (t+\tau ),\phi (t))$

Note that this is just an equation in $\phi$ without derivatives, i.e., it is an ordinary equation (a zeroth-order differential equation). However, depending on the nature of $f$, $\psi$, and ${\psi }^{\prime }$, we may have difficulty getting an explicit functional form for $\phi$, and it may be far from unique. Thus, unlike forward motion, which we expect to be uniquely determined by the initial value specification, backward motion may not be uniquely determined.