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The delay differential equation given by, $x'(t)=x(t-T)e^{3-x(t-T)}-x(t)$, how to find stability conditions?

So here's my attempt: to find equilibria we have, $x^*e^{3-x^*}-x^*=0$ which leads us to $x^*_1=0$ and $x^*_2=3$.

Linearizing I get, $\frac{\partial f}{\partial x(t)}=-1$ and $\frac{\partial f}{\partial x(t-T)}=e^{3-x(t-T)}(1-x(t-T))$

So for $x^*_2=3$ we have $f(3,3)=0$, and $\frac{\partial f}{\partial x(t)}|_{(3,3)}=-1$ and $\frac{\partial f}{\partial x(t-T)}|_{(3,3)}=-2$ which gives us the linearization of the form $\frac{dy}{dt}=f(3,3)+\frac{\partial f}{\partial x(t)}|_{(3,3)}(y-3)+\frac{\partial f}{\partial x(t-T)}|_{(3,3)}(y(t-T)-3)\approx-y(t)-2y(t-T)$

For $x^*_1=0$, we have $f(0,0)=0$ and $\frac{\partial f}{\partial x(t)}|_{(0,0)}=-1$ and $\frac{\partial f}{\partial x(t-T)}|_{(0,0)}=e^3$ which gives us the linearization of the form $\frac{dy}{dt}=f(0,0)+\frac{\partial f}{\partial x(t)}|_{(0,0)}(y-0)+\frac{\partial f}{\partial x(t-T)}|_{(0,0)}(y(t-T)-0)\approx-y(t)+e^3y(t-T)$

Now, we want solutions of the form $y'=\lambda y_0e^{\lambda t}$ which mean after substitution and a bit of algebra, that $\lambda_{x^*_1}=-1+e^{3-\lambda T}$ and $\lambda_{x^*_2}=-1-2e^{-\lambda T}$.

From here I'm stuck, the back of the books says that $x=0$ is unstable $\forall T$ and that $x=3$ is asymptotically stable if $\sec z<-2$, where $z=-T\tan z$, how did they get here? Am I really off the mark?

user5896534
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1 Answers1

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Here is a general methodology to solve such problems. Assume that the characteristic equation of a time-delay system is given by $C(s,h):=P(s)+e^{-sh}Q(s)$.

We have a result that says that $C(s,0)$ is Hurwitz, then the $C(s,h)$ is Hurwitz for any sufficiently small delay $h$. This can be proven using a perturbation argument. This means that if the delay free system is stable, then the system will be stable for small delays. This is an intrinsic robustness property of this class of time-delay systems.

The idea then is to find the values for the delay for which we have roots on the imaginary axis. So, we look for $(\omega,h)\in\mathbb{R}\times \mathbb{R}_{\ge0}$, such that $C(j\omega,h)=0$. This can be reformulated as $P(j\omega)=-e^{-j\omega h}P(j\omega)$ and taking the modulus squared on both sides yields the expression

$$|P(j\omega)|^2-|Q(j\omega)|^2=0,$$

which is a polynomial in $\omega$ which can be solved analytically in certain very simple cases, but can always be solved numerically.

We then have two cases:

  1. If there no real solutions to this polynomial, then there is no crossing of the imaginary axis by the characteristic roots for any value of the delay. Therefore, if the system with zero delay is stable, then it will be stable for all delays $h\ge0$.

  2. If there are real solutions, then for each solution $\omega_i>0$, we can solve for $C(j\omega_i,h_i)=0$ where $h_i$ will be the critical delay associated with the crossing frequency $\omega_i>0$. To do that, one can rewrite this expression as

$$e^{-\omega_ih_i}=-\dfrac{P(\omega_i)}{Q(\omega_i)}$$

where we have assumed that $Q(\omega_i)\ne 0$, otherwise we can consider instead

$$e^{\omega_ih_i}=-\dfrac{Q(\omega_i)}{P(\omega_i)}.$$

Then, one can use any trigonometric result to solve that. For instance, we have that

$$h_i=\dfrac{1}{\omega_i}\arg\left(-\dfrac{Q(\omega_i)}{P(\omega_i)}\right),$$

with the caveat that the exponential function is periodic. So, we need to find the smallest value for $h_i>0$.

Now let's apply that to your problem. For the equilibrium $x^*=0$, the linearized dynamics is given by $\dot y(t)=-y(t)+e^3y(t-h)$. We have that $P(s)=s+1$ and $Q(s)=-e^3$.

Since $e^3>1$, then we have that $C(s,0)$ is unstable. Then, we have that

$$|P(j\omega)|^2-|Q(j\omega)|^2=\omega^2+1-e^6,$$

which has the positive root $\omega_c=\sqrt{e^6-1}\approx 20.0606 $ and we have that

$$h_c=-\dfrac{1}{\omega_c}\arctan(\omega_c)+\dfrac{2\pi}{\omega_c}\approx0.5506.$$

There it can be a bit tricky, because we need to check what happens to the stability properties in that case, but it can be shown that the system does transition to stability. This can be verified by computing the dominant root $r$ for a delay value larger than 0.5506 using the expression

$$r=\dfrac{1}{h}W_0(e^{3+h}h)-1,$$

where $W_0$ is the principal branch of the Lambert W function. In the present case, let $h=0.6$ and we get $r=2.7826>0$.

For the equilibrium $x^*=3$, the linearized dynamics is given by $\dot y(t)=-y(t)-2y(t-h)$. We have that $P(s)=s+1$, $Q(s)=2$, and that $C(s,0)$ is Hurwitz stable.

Then, we have that

$$|P(j\omega)|^2-|Q(j\omega)|^2=\omega^2-3,$$

and we have that $\omega_c=\sqrt{3}$. Now we consider the expression

$$e^{-\omega_ch_c}=-\dfrac{P(\omega_c)}{Q(\omega_c)}$$

and consider the real and imaginary parts to get

$$1+2\cos(\omega_ch_c)=0,\ \mathrm{and}\ \omega_c-2\sin(\omega_ch_c)$$.

The first one is easy to solve and we get $$h_c=\dfrac{2\pi}{3\sqrt{3}}=\dfrac{2\sqrt{3}\pi}{9}.$$

We can verify that this also solves the second equation as

$$\omega_c-2\sin(\omega_ch_c)=\sqrt{3}-2\sin(2\pi/3)=0.$$

We can also verify using the Lambert-W function formula that

$$r=\dfrac{1}{h_c}W_0(-2e^{h_c}h_c-1=1.7321i,$$

so we are on the imaginary axis and taking $h$ a bit larger than the critical one yields

$$r=\dfrac{1}{h_c+0.01}W_0(-2e^{h_c+0.01}(h_c+0.01)-1=0.0032 + 1.7212i,$$

which illustrates the transition to instability.

As a result, the equilibrium point $x^*=3$ is locally exponentially stable if and only if $h<\dfrac{2\sqrt{3}\pi}{9}$. This certainly correspond to the conditions you have even though that one is much nicer as expressed in closed-form.

KBS
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