Linearizing a delay differential equation at an equilibrium point.

Question

I am confused about the general procedure to linearize a delay differential equation (DDE) at an equilibrium point. I was given the following two examples but I do not know how to get from DDE to linearization at an equilibrium point, $\bar{x}$.

Example 1

Let $$x'(t)=bx(t)[1-x(t-r)] $$ where $b,r$ are constants. The linearization is \begin{align} x'(t)&=bx(t) & \text{ at } \bar{x}=0\\ x'(t)&=-bx(t-r) & \text{ at } \bar{x}=1 \end{align}

Example 2

Let \begin{align} x'_{1}(t)&=-b_1x_1(t)+a_{11}f_1(x_1(t-r_{11}))+a_{12}f_2(x_2(t-r_{12})) \\ x'_{2}(t)&=-b_2x_2(t)+a_{21}f_1(x_1(t-r_{21}))+a_{22}f_2(x_2(t-r_{22})) \end{align} where all $a_{nn}$,$r_{nn}$,$b_{n}$ are constants. Assume that $f_1,f_2$ are differentiable and $f_1(0)=f_2(0)=0$, so that $(x_1,x_2)=(0,0)$ is an equilibrium. The linearization at $\bar{x}=(x_1,x_2)=(0,0)$ is \begin{align} x'_{1}(t)&=-b_1x_1(t)+a_{11}f'_1(0)x_1(t-r_{11})+a_{12}f'_2(0)x_2(t-r_{12}) \\ x'_{2}(t)&=-b_2x_2(t)+a_{21}f'_1(0)x_1(t-r_{21})+a_{22}f'_2(0)x_2(t-r_{22}) \end{align}

Notes

I understand that the Frechet derivative at the equilibrium point is equivalent to the linearization. Put I am not clear on how to use that fact.

In the answer to this question I am looking for a technique that allows me to linearize any delay differential equation, and connects to larger theory of the Frechet derivative.

Answer 1

Informally, if $\dot{x}(t) = f(x(t),x(t-r))$ then we have (using $f(x_1+h_1,x_2+h_2) \approx {\partial f(x_1,x_2) \over \partial x_1} h_1 + {\partial f(x_1,x_2) \over \partial x_2} h_2 $) \begin{eqnarray} \dot{(x+\delta)}(t) &=& f(x(t)+\delta(t),x(t-r)+\delta(t-r)) \\ &\approx& f(x(t),x(t-r))+{\partial f(x(t),x(t-r)) \over \partial x_1} \delta(t) + {\partial f(x(t),x(t-r)) \over \partial x_2} \delta(t-r) \\ \end{eqnarray} From which we get $\dot{\delta}(t) = {\partial f(x(t),x(t-r)) \over \partial x_1} \delta(t) + {\partial f(x(t),x(t-r)) \over \partial x_2} \delta(t-r)$.

Substituting values, we have $\dot{\delta}(t) = b(1-x(t-r)) \delta(t) - b x(t) \delta(t-r)$.

Substituting $x(t) = 0$ or $x(t) = 1$ yields the desired results.