0
$\begingroup$

I am attempting to work through "Shahverdiev, Sivaprakasam, and Shore (2002) Lag synchronization in time-delayed systems", but I'm missing something basic up front.

The problem is to take a unidirectionally coupled dynamical system,

$\dot{x} =-\alpha x + (m_1+m_3) \sin x_{\tau_1}$

$\dot{y} =-\alpha y + m_2 \sin y_{\tau_1} + m_3 \sin x_{\tau_2}$

where subscript $\tau$ denotes a time-delay, e.g. $x_{\tau_1} = x(t-\tau_1)$, and determine its stability manifold. The synchronization error is defined to be $\Delta = x_{\tau_2 - \tau_1} - y$, but we are interested in the error dynamics, $\dot{\Delta}$, which the authors list as

$\dot{\Delta}=-\alpha \Delta + m_2 \cos x_{\tau_1} \Delta_{\tau_1}$

My question is about this last step. To differentiate $\Delta$, it seems to me that one could use the expressions for $\dot{x}$ and $\dot{y}$ directly, i.e.

$\dot{\Delta} = \dot{x}(t-\tau_2+\tau_1) - \dot{y}(t)$.

Doing that, however, leaves me with

$\dot{\Delta} = -\alpha \Delta + m_2 \left( \sin x_{\tau_2} - \sin y_{\tau_1} \right)$

I feel as though I'm missing a simple step here. Could anyone point me in the right direction?

Here's a second example, this time from "Senthilkumar, Kurths, and Lakshmanan (2009) Stability of synchronization in coupled time-delay systems using Krasovskii-Lyapunov theory":

The system:

$\dot{x} = -a x(t) + b f(x(t-\tau))$

$\dot{y} = -a y(t) + b f(y(t-\tau)) + K(t)(x(t)-y(t))$

The synchronization error (for small values, the authors state):

$\dot{\Delta} = -(a + K(t))\Delta + b f^\prime(y(t-\tau))\Delta(t-\tau)$

Again, I am unsure of the source of the $f^\prime$ term and why that term isn't simply $b \left( f(x(t-\tau)) - f(y(t-\tau)) \right)$. In both examples, it seems a derivative is being taken, but I don't see why that would be.

$\endgroup$
4
  • $\begingroup$ Did you read Krasovskii book on equations with delay, Krasovskiĭ, N. N. Stability of motion. Applications of Lyapunov's second method to differential systems and equations with delay. Translated by J. L. Brenner Stanford University Press, Stanford, Calif. 1963 vi+188 pp. $\endgroup$
    – user6976
    Commented May 24, 2013 at 23:34
  • $\begingroup$ I've only read several papers on the topic, but I'll try to find the book. Thank you for replying. $\endgroup$
    – lomendil
    Commented May 25, 2013 at 0:16
  • $\begingroup$ You are 100% correct, of course, but your equations are exact and theirs are just linear approximations good for small perturbations. You have seen the formula $f(x)-f(y)\approx f'(y)(x-y)$ (or $f'(x)(x-y)$, if you prefer that one) before, haven't you? Here it is in action. $\endgroup$
    – fedja
    Commented May 25, 2013 at 3:18
  • $\begingroup$ fedja, that's the undergraduate-type thing that I was forgetting! Thank you. I consider that the answer. $\endgroup$
    – lomendil
    Commented May 25, 2013 at 4:13

0

You must log in to answer this question.

Browse other questions tagged .