Estimation of the Boundary of the Limit Cycle of Brusselator Oscillators by the Renormalization Group Method

Abstract

In this paper, a method to explore the limit cycle region in the parameter space of a 2D nonlinear chemical oscillator was provided. Applying the renormalization group method, we derive that the boundaries of the Brusselator oscillator’s limit cycles at various points in parameter space can be predicted. The validity of our method is verified by numerical simulations.

Appendix

Expression for P in Sect. 3.2 Substituting perturbation solution (3.1) and (3.2) into Eqs. (2.10) and (2.11) in the form of

$$\begin{aligned} X&=X_0+\lambda X_1+\lambda ^2 X_2+O(\lambda ^3),\\ Y&=Y_0+\lambda Y_1+\lambda ^2 Y_2+O(\lambda ^3), \end{aligned}$$

we have

$$\begin{aligned}&[\ddot{X_0}+\lambda \ddot{X_1}+\lambda ^2\ddot{X_2}+O(\lambda ^3)]+\omega ^2[X_0+\lambda X_1+\lambda ^2 X_2+O(\lambda ^3)]\\&\quad =\lambda [a^2\delta [X_0+\lambda X_1+\lambda ^2 X_2+O(\lambda ^3)]+\delta [\dot{X_0}+\lambda \dot{X_1}+\lambda ^2 \dot{X_2}+O(\lambda ^3)]\\&\qquad +\frac{2b}{a}[X_0+\lambda X_1+\lambda ^2 X_2+O(\lambda ^3)][\dot{X_0}+\lambda \dot{X_1}+\lambda ^2 \dot{X_2}+O(\lambda ^3)]\\&\qquad +2a([\dot{X_0}+\lambda \dot{X_1}+\lambda ^2 \dot{X_2}+O(\lambda ^3)][Y_0+\lambda Y_1+\lambda ^2 Y_2+O(\lambda ^3)]\\&\qquad +[X_0+\lambda X_1+\lambda ^2 X_2+O(\lambda ^3)][\dot{Y_0}+\lambda \dot{Y_1}+\lambda ^2 \dot{Y_2}+O(\lambda ^3)])\\&\qquad +2[X_0+\lambda X_1+\lambda ^2 X_2+O(\lambda ^3)][\dot{X_0}+\lambda \dot{X_1}+\lambda ^2 \dot{X_2}+O(\lambda ^3)]\nonumber \\&\qquad [Y_0+\lambda Y_1+\lambda ^2 Y_2+O(\lambda ^3)]\\&\qquad +[X_0+\lambda X_1+\lambda ^2 X_2+O(\lambda ^3)]^2[\dot{Y_0}+\lambda \dot{Y_1}+\lambda ^2 \dot{Y_2}+O(\lambda ^3)]]. \end{aligned}$$

To make it easier to find the second order perturbation equation for X to converge, we write the above equation as

$$\begin{aligned}&({\ddot{X}}_0+\lambda {\ddot{X}}_1+\lambda ^2{\ddot{X}}_2)+\omega ^2(X_0+\lambda X_1+\lambda ^2 X_2)+O(\lambda ^3)\nonumber \\&\quad =\lambda [a^2\delta (X_0+\lambda X_1)+\delta ({\dot{X}}_0+\lambda {\dot{X}}_1)+\frac{2b}{a}(X_0+\lambda X_1)({\dot{X}}_0+\lambda {\dot{X}}_1)\nonumber \\&\qquad +2a[({\dot{X}}_0+\lambda {\dot{X}}_1)(Y_0+\lambda Y_1) +(X_0+\lambda X_1)({\dot{Y}}_0+\lambda {\dot{Y}}_1)]\nonumber \\&\qquad +2(X_0+\lambda X_1)({\dot{X}}_0+\lambda {\dot{X}}_1)(Y_0+\lambda Y_1) +(X_0+\lambda X_1)^2({\dot{Y}}_0+\lambda {\dot{Y}}_1)]\nonumber \\&\quad =\lambda [a^2\delta X_0+\delta {\dot{X}}_0+\frac{2b}{a}X_0{\dot{X}}_0+2a({\dot{X}}_0Y_0+X_0{\dot{Y}}_0)+2X_0{\dot{X}}_0Y_0+X_0^2{\dot{Y}}_0]\nonumber \\&\qquad +\lambda ^2[a^2\delta X_1+\delta {\dot{X}}_1+\frac{2b}{a}(X_0{\dot{X}}_1+X_1{\dot{X}}_0)\nonumber \\&\qquad +2a[(\dot{X}_0Y_1+\dot{X}_1Y_0)+(X_0\dot{Y}_1+X_1\dot{Y}_0)]\nonumber \\&\qquad +2(X_0\dot{X}_0Y_1+X_1\dot{X}_0Y_0+X_0\dot{X}_1Y_0)+X_0^2\dot{Y}_1+2X_0X_1\dot{Y}_0]. \end{aligned}$$

(5.1)

Comparing the coefficients of $\lambda ^0$, $\lambda ^1$ and $\lambda ^2$ of both sides of Eq. (5.1), we can obtain the zeroth order, first order and second order perturbation equations for X respectively

$$\begin{aligned} \lambda ^0:\quad \ddot{X_0}+\omega ^2 X_0&=0, \end{aligned}$$

(5.2)

$$\begin{aligned} \lambda ^1:\quad {\ddot{X}}_1+\omega ^2X_1&=a^2\delta X_0+\delta {\dot{X}}_0+\frac{2b}{a}X_0{\dot{X}}_0\nonumber \\&\quad +2a({\dot{X}}_0Y_0+X_0{\dot{Y}}_0)+2X_0{\dot{X}}_0Y_0+X_0^2{\dot{Y}}_0,\end{aligned}$$

(5.3)

$$\begin{aligned} \lambda ^2:\quad {\ddot{X}}_2+\omega ^2X_2&=a^2\delta X_1+\delta {\dot{X}}_1+\frac{2b}{a}(X_0{\dot{X}}_1+X_1{\dot{X}}_0)\end{aligned}$$

(5.4)

$$\begin{aligned}&\quad +2a[(\dot{X}_0Y_1+\dot{X}_1Y_0)+(X_0\dot{Y}_1+X_1\dot{Y}_0)]\nonumber \\&\quad +2(X_0\dot{X}_0Y_1+X_1\dot{X}_0Y_0+X_0\dot{X}_1Y_0)+X_0^2\dot{Y}_1+2X_0X_1\dot{Y}_0. \end{aligned}$$

(5.5)

Taking direct derivatives for (3.3), (3.4), (3.10) and (3.18), we obtain ${\dot{X}}_0$, ${\dot{Y}}_0$, ${\dot{X}}_1$ and ${\dot{Y}}_1$, whose expressions are

$$\begin{aligned} X_0&=A\cos \alpha ,\\ {\dot{X}}_0&=-\omega A\sin \alpha , \\ Y_0&=-\frac{\omega A}{a^2}\sin \alpha -A\cos \alpha ,\\ {\dot{Y}}_0&=-\frac{\omega ^2A}{a^2}\cos \alpha +\omega A\sin \alpha , \\ X_1&=\frac{B_1}{2\omega }t\sin \alpha -\frac{A_1}{2\omega }t\cos \alpha +\frac{A_1}{4\omega ^2}\sin \alpha +\frac{B_1}{4\omega ^2}\cos \alpha \\&\quad -\frac{C_1}{3\omega ^2}\sin 2\alpha -\frac{D_1}{3\omega ^2}\cos 2\alpha -\frac{E_1}{8\omega ^2}\sin 3\alpha -\frac{F_1}{8\omega ^2}\cos 3\alpha ,\\ {\dot{X}}_1&=\frac{B_1}{2\omega }\sin \alpha -\frac{A_1}{2\omega }\cos \alpha +\frac{B_1}{2}t\cos \alpha +\frac{A_1}{2}t\sin \alpha \\&\quad -\frac{2C_1}{3\omega }\cos 2\alpha +\frac{2D_1}{3\omega }\sin 2\alpha -\frac{3E_1}{8\omega }\cos 3\alpha +\frac{3F_1}{8\omega }\sin 3\alpha ,\\ Y_1&=\frac{B_2}{2\omega }t\sin \alpha -\frac{A_2}{2\omega }t\cos \alpha \frac{A_2}{4\omega ^2}\sin \alpha +\frac{B_2}{4\omega ^2}\cos \alpha \\&\quad -\frac{C_2}{3\omega ^2}\sin 2\alpha -\frac{D_2}{3\omega ^2}\cos 2\alpha -\frac{E_2}{8\omega ^2}\sin 3\alpha -\frac{F_2}{8\omega ^2}\cos 3\alpha +\frac{G_2}{\omega ^2},\\ {\dot{Y}}_1&=\frac{B_2}{2\omega }\sin \alpha -\frac{A_2}{2\omega }\cos \alpha +\frac{B_2}{2}t\cos \alpha +\frac{A_2}{2}t\sin \alpha \\&\quad -\frac{2C_2}{3\omega }\cos 2\alpha +\frac{2D_2}{3\omega }\sin 2\alpha -\frac{3E_2}{8\omega }\cos 3\alpha +\frac{3F_2}{8\omega }\sin 3\alpha ,\\ \end{aligned}$$

wherein $\alpha =\omega t+\theta$. Substituting $X_0,\,{\dot{X}}_0,\,Y_0,\,{\dot{Y}}_0,\,X_1,\,{\dot{X}}_1,\,Y_1,\,{\dot{Y}}_1$ into the second order perturbation Eq. (5.5) i.e. (3.20) and simplifying with Maple yields the coefficient P of $\sin \alpha$, see (3.22).