Delay Differential Equations

Definition A delay differential equation with a constant delay $\tau > 0$ has the form

so the rate of change at time $t$ depends on the state at the earlier time $t - \tau$, and possibly on the present state too. The expression $y(t - \tau)$ is the value of $y$ at the time $t - \tau$, not the number $y(t) - \tau$. An equation of this kind, in which the derivative appears only at the present time, is called retarded. Equations with several delays, with a delay that depends on the state, or with delayed derivatives also occur, and they need more careful treatment.

Definition For $y'(t) = f(t, y(t), y(t - \tau))$ with initial time $t_0$, a history function is a continuous function $\varphi$ on $[t_0 - \tau, t_0]$, and the initial condition is $y(t) = \varphi(t)$ for $t_0 - \tau \le t \le t_0$. A solution on $[t_0 - \tau, t_0 + \beta)$ with $\beta > 0$ is a continuous function that equals $\varphi$ on the history interval, is differentiable on $(t_0, t_0 + \beta)$ with a right derivative at $t_0$, and satisfies the equation for $t_0 \le t < t_0 + \beta$. The value $\varphi(t_0)$ alone is not enough data, because for $t_0 \le t < t_0 + \tau$ the equation uses the values of $y$ at the times of $[t_0 - \tau, t_0)$. The solution is usually not differentiable at $t_0$ itself, since its right derivative comes from the equation and its left derivative from the history.

Definition Let $\tau > 0$ be constant and $\varphi$ continuous on $[t_0 - \tau, t_0]$. For $t_0 \le t \le t_0 + \tau$ the delayed time $t - \tau$ lies in $[t_0 - \tau, t_0]$, where $y = \varphi$ is known, so there the delay equation reduces to the ordinary initial-value problem

Once $y$ is known on $[t_0, t_0 + \tau]$, the same reasoning gives an ordinary problem on $[t_0 + \tau, t_0 + 2\tau]$, whose initial value $y(t_0 + \tau)$ comes from the previous step, and so on forward in time. If $f$ is continuous and locally Lipschitz in its second argument, each step has a unique solution near its starting time, so the delay problem has a unique solution on some interval $[t_0, t_0 + \beta)$. For a linear equation $y'(t) = p(t)y(t) + r(t)y(t - \tau) + g(t)$ with continuous coefficients, each step is a linear equation that can be solved on the whole step, so the solution exists for all $t \ge t_0$. When $f$ does not depend on $y(t)$, each step is a single integration. For $y'(t) = -a\,y(t - \tau)$ with $a \ne 0$, a jump in $y'$ at $t_0$ reappears as a jump in $y''$ at $t_0 + \tau$, in $y'''$ at $t_0 + 2\tau$, and so on, while the lower derivatives are continuous there.

Definition For a delay equation with constant delay $\tau > 0$, the state at time $t$ is the segment $y_t$ of the solution over the last delay interval, defined by $y_t(\theta) = y(t + \theta)$ for $-\tau \le \theta \le 0$. With a continuous history, every segment belongs to the space $C[-\tau, 0]$ of continuous functions on $[-\tau, 0]$. Under the hypotheses of the method of steps, the segment at any time determines the solution at all later times for which it exists, just as the history did at $t_0$. The space $C[-\tau, 0]$ is infinite-dimensional, since the functions $1, \theta, \theta^2, \dots$ on it are linearly independent. For a system of $n$ first-order ordinary differential equations the state is a point of $\mathbb{R}^n$, but for a delay equation no finite list of numbers can play that role in general.

Definition For $y'(t) = -a\,y(t - \tau)$ with real constants $a \ne 0$ and $\tau > 0$, substituting $y = e^{\lambda t}$ with complex $\lambda$ and canceling the nonzero factor $e^{\lambda t}$ gives the characteristic equation

Each root $\lambda$ gives the solution $e^{\lambda t}$, and for a complex root its real and imaginary parts are real solutions. Since $\lambda$ also appears in an exponent, the left side is not a polynomial. The equation has infinitely many complex roots, because multiplying by $\tau e^{\lambda\tau}$ turns it into $z e^{z} = -a\tau$ with $z = \lambda\tau$, whose solutions are the values at $-a\tau$ of the infinitely many branches of the Lambert W function. Only finitely many roots lie in any half-plane $\operatorname{Re}\lambda \ge \gamma$, because such a root satisfies $\lvert \lambda \rvert = \lvert a \rvert e^{-\tau \operatorname{Re}\lambda} \le \lvert a \rvert e^{-\gamma\tau}$. The stability theorem for linear retarded equations with constant coefficients states that if every root has $\operatorname{Re}\lambda < 0$, then every solution, for every continuous history, tends to $0$ exponentially, and that if some root has $\operatorname{Re}\lambda > 0$, some solutions grow exponentially. For $a = 0$ the equation is $y' = 0$, and $\lambda = 0$ is its only root, so the condition $a \ne 0$ matters.

Definition Delay-induced instability is the loss of stability of an equilibrium when a feedback that is stabilizing without delay acts with a large enough delay. For $y'(t) = -a\,y(t - \tau)$ with $a > 0$, the equation without delay is $y' = -ay$, whose solutions decay. With a delay $\tau > 0$, the zero solution is asymptotically stable exactly when $a\tau < \pi/2$. If $0 < a\tau < \pi/2$, no root has $\operatorname{Re}\lambda \ge 0$. Indeed, a root $\lambda = p + iq$ with $p \ge 0$ would satisfy $\lvert \lambda \rvert = a e^{-p\tau} \le a$, so $\lvert q\tau \rvert \le a\tau < \pi/2$ and $\cos(q\tau) > 0$, while the real part of the characteristic equation says

a contradiction. By the stability theorem every solution then decays exponentially. A root $\lambda = i\omega$ on the imaginary axis needs $\cos(\omega\tau) = 0$ and $\omega = a\sin(\omega\tau)$, which first happens at $\omega = \pm a$ and $a\tau = \pi/2$. Differentiating the characteristic equation with respect to $\tau$ gives

whose real part at $\lambda = ia$, $a\tau = \pi/2$ is $a^2/(1 + \pi^2/4) > 0$, so the pair $\pm ia$ moves into the right half-plane as $\tau$ increases. Further crossings occur only at $a\tau = \pi/2 + 2k\pi$, all of them to the right, and a root with $\operatorname{Re}\lambda \ge 0$ always satisfies $\lvert \lambda \rvert \le a$, so no root can arrive in that half-plane from infinity. Hence some root has $\operatorname{Re}\lambda > 0$ for every $a\tau > \pi/2$.