Delay differential equation

Solving DDEs

DDEs are mostly solved in a stepwise fashion with a principle called the method of steps. For instance, consider the DDE with a single delay $${\displaystyle \frac{d}{dt}x(t)=f(x(t),x(t-\tau ))}$$

with given initial condition ${\displaystyle \varphi :[-\tau ,0]\to {\mathbb{R}}^n}$. Then the solution on the interval ${\displaystyle [0,\tau ]}$ is given by ${\displaystyle \psi (t)}$ which is the solution to the inhomogeneous initial value problem $${\displaystyle \frac{d}{dt}\psi (t)=f(\psi (t),\varphi (t-\tau )),}$$ with ${\displaystyle \psi (0)=\varphi (0)}$. This can be continued for the successive intervals by using the solution to the previous interval as inhomogeneous term. In practice, the initial value problem is often solved numerically.

Example

Suppose ${\displaystyle f(x(t),x(t-\tau ))=ax(t-\tau )}$ and ${\displaystyle \varphi (t)=1}$. Then the initial value problem can be solved with integration,

$${\displaystyle x(t)=x(0)+{\int }_{s=0}^t\frac{d}{dt}x(s)ds=1+a{\int }_{s=0}^t\varphi (s-\tau )ds,}$$

i.e., ${\displaystyle x(t)=at+1}$, where the initial condition is given by ${\displaystyle x(0)=\varphi (0)=1}$. Similarly, for the interval ${\displaystyle t\in [\tau ,2\tau ]}$ we integrate and fit the initial condition,

$${\displaystyle \begin{array}{cc}x(t)=x(\tau )+{\int }_{s=\tau }^t\frac{d}{dt}x(s)ds& =(a\tau +1)+a{\int }_{s=\tau }^t\left(a(s-\tau )+1\right)ds\\ & =(a\tau +1)+a{\int }_{s=0}^{t-\tau }\left(as+1\right)ds,\end{array}}$$

i.e., $x(t)=(a\tau +1)+a(t-\tau )\left(\frac{1}{2}a(t-\tau )+1\right).$

Reduction to ODE

In some cases, differential equations can be represented in a format that looks like delay differential equations.

• Example 1 Consider an equation $${\displaystyle \frac{d}{dt}x(t)=f\left(t,x(t),{\int }_{-\mathrm{\infty }}^{0}x(t+\tau )e^{\lambda \tau }d\tau \right).}$$ Introduce ${\displaystyle y(t)={\int }_{-\mathrm{\infty }}^{0}x(t+\tau )e^{\lambda \tau }d\tau }$ to get a system of ODEs $${\displaystyle \frac{d}{dt}x(t)=f(t,x,y),\frac{d}{dt}y(t)=x-\lambda y.}$$
• Example 2 An equation $${\displaystyle \frac{d}{dt}x(t)=f\left(t,x(t),{\int }_{-\mathrm{\infty }}^{0}x(t+\tau )\cos (\alpha \tau +\beta )d\tau \right)}$$ is equivalent to $${\displaystyle \frac{d}{dt}x(t)=f(t,x,y),\frac{d}{dt}y(t)=\cos (\beta )x+\alpha z,\frac{d}{dt}z(t)=\sin (\beta )x-\alpha y,}$$ where $${\displaystyle y={\int }_{-\mathrm{\infty }}^{0}x(t+\tau )\cos (\alpha \tau +\beta )d\tau ,z={\int }_{-\mathrm{\infty }}^{0}x(t+\tau )\sin (\alpha \tau +\beta )d\tau .}$$

The characteristic equation

Similar to ODEs, many properties of linear DDEs can be characterized and analyzed using the characteristic equation. The characteristic equation associated with the linear DDE with discrete delays $${\displaystyle \frac{d}{dt}x(t)=A_{0}x(t)+A_{1}x(t-{\tau }_1)+\cdots +A_{m}x(t-{\tau }_m)}$$ is the exponential polynomial given by $${\displaystyle det(-\lambda I+A_0+A_1{e}^{-{\tau }_1\lambda }+\cdots +A_m{e}^{-{\tau }_m\lambda })=0.}$$

The roots λ of the characteristic equation are called characteristic roots or eigenvalues and the solution set is often referred to as the spectrum. Because of the exponential in the characteristic equation, the DDE has, unlike the ODE case, an infinite number of eigenvalues, making a spectral analysis more involved. The spectrum does however have some properties which can be exploited in the analysis. For instance, even though there are an infinite number of eigenvalues, there are only a finite number of eigenvalues in any vertical strip of the complex plane.

This characteristic equation is a nonlinear eigenproblem and there are many methods to compute the spectrum numerically. In some special situations it is possible to solve the characteristic equation explicitly. Consider, for example, the following DDE: $${\displaystyle \frac{d}{dt}x(t)=-x(t-1).}$$ The characteristic equation is $${\displaystyle -\lambda -e^{-\lambda }=0.}$$ There are an infinite number of solutions to this equation for complex λ. They are given by $${\displaystyle \lambda =W_k(-1),}$$ where W_k is the kth branch of the Lambert W function, so: $${\displaystyle x(t)=x(0)e^{W_k(-1)\cdot t}.}$$

Another example

The following DDE: $${\displaystyle \frac{d}{dt}u(t)=2u(2t+1)-2u(2t-1).}$$

Have as solution in ${\displaystyle \mathbb{R}}$ the function: $${\displaystyle u(t)=\{\begin{array}{ll}F(t+1),& |t|<1\\ 0,& |t|\ge 1\end{array}}$$ with ${\displaystyle F(t)}$ the Fabius function, known as Rvachëv up function.

Applications

• Dynamics of diabetes
• Epidemiology
• Population dynamics
• Classical electrodynamics