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

If we are looking for solutions that are once differentiable, and are not insisting on higher order differentiability, the initial value specification is as follows: it is a description of $x$ as a continuous function of $t$ on an interval of length $\tau$, chosen as the left-most interval of the domain where we want the function to be defined. Note that if the left-most interval is $[a-\tau ,a]$, then the delay differential equation becomes active only on $[a,\infty )$.

Further, the initial value specification must satisfy the additional condition that the left-hand derivative of $x$ at $a$ is $f(x(a),x(a-\tau ))$.

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 $[a,\infty )$.

Let us say that we know that $x(t)=\varphi (t)$ 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 $\psi (a)=\varphi (a)$:

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

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 $x(t)=\varphi (t)$ on this interval, and we want to solve the following for $t\in [a-\tau ,a]$ subject to the initial value condition $\varphi (a)=\psi (a)$:

$\psi '(t+\tau )=f(\psi (t+\tau ),\varphi (t))$

Note that this is just an equation in $\varphi$ without derivatives, i.e., it is an ordinary equation (a zeroth-order differential equation). However, depending on the nature of $f$, $\psi$, and $\psi '$, we may have difficulty getting an explicit functional form for $\varphi$, 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.

Facts

Expect piecewise definitions for solutions

Even if the initial value specification is an infinitely differentiable function, it is likely that when we extend it using the method of steps, the solution will have nice differentiability properties within each interval of length $\tau$, but not at the endpoints shared by the interval.

Examples

Consider the delay differential equation:

${\frac {d}{dt}}(x(t))=x(t)+x(t-1)$

Suppose that we are given that $x(t)=t$ on $[0,1]$. We note that this initial value specification is consistent because the derivative $x'(1)$ equals $x(1)+x(1-1)$.

We use the method of steps. Our first goal is to determine $x$ on $[1,2]$. Explicitly, we are trying to find a function $\psi$ on $[1,2]$ such that:

${\frac {d}{dt}}(\psi (t))=\psi (t)+(t-1)$

The differential equation with dependent variable $\psi$ and independent variable $t$ is:

$\psi '-\psi =t-1$

This is a linear differential equation. The general solution would be:

$\psi (t)=-t+Ce^{t}$

We must choose $C$ such that $\psi (1)=\varphi (1)=1$, so $C=2/e$. We get:

$\psi (t)=-t+2e^{t-1}$.

Thus, we have:

$x(t)=-t+2e^{t-1},\qquad t\in [1,2]$

We can now do a similar procedure to find what $x(t)$ looks like in $[2,3]$. Note that we will still get a linear differential equation but with a new particular solution:

$\psi '-\psi =1-t+2e^{t-2}$

The general solution is:

$\psi (t)=t+2te^{t-2}+Ce^{t}$

Plugging in that $\psi (2)=x(2)=2e-2$, we get $C=(2e-8)/e^{2}$, so we get:

$\psi (t)=t+(2t-8)e^{t-2}+2e^{t-1}$

So:

$\psi (t)=t+(2t-8)e^{t-2}+2e^{t-1},\qquad t\in [2,3]$

Overall, we have:

$x(t)=\left\lbrace {\begin{array}{rl}t,&t\in [0,1]\\-t+2e^{t-1},&t\in [1,2]\\t+(2t-8)e^{t-2}+2e^{t-1},&t\in [2,3]\end{array}}\right.$

Note that the function has a piecewise definition, and at each transition point, it is $C^{1}$ (i.e., continuously differentiable) but not twice differentiable.