Design of coupled Andronov–Hopf oscillators with desired strange attractors

Abstract

This paper develops a design method for the interconnections of a network of Andronov–Hopf oscillators such that the system exhibits a desired strange attractor. Because of the structure of the oscillators, the desired behavior can be achieved via weak linear coupling, which destabilizes the oscillators’ phase difference. First, a set of sufficient conditions are established that result in phase destabilization, and thus instability, of a desired periodic solution. Then, an additional condition is determined to ensure that all harmonic periodic orbits will be unstable. Finally, additional numerical properties are assessed, where tuning of a small parameter can result in chaos.

Appendix

1.1 Proof of Lemma 1

Proof

Consider the following coordinate transformation $(q, p)\leftrightarrow (r,\theta )$, defined by

$$\begin{aligned}&q = C_\theta r, \qquad p = S_\theta r. \end{aligned}$$

With the new state variables $(r,\theta )$, system (3) with coupling (5) can be expressed as

$$\begin{aligned} \begin{bmatrix} {\dot{r}} \\ R{\dot{\theta }} \end{bmatrix} = \begin{bmatrix} I-R^2 \\ I \end{bmatrix} r + {{\varepsilon }}\begin{bmatrix} C_\theta &{} S_\theta \\ -S_\theta &{} C_\theta \end{bmatrix} H \begin{bmatrix}C_\theta \\ S_\theta \end{bmatrix} r , \end{aligned}$$

(19)

where $R:=\text {diag}(r)$. Let x(t), $t\ge 0$, be an arbitrary trajectory starting at a point in ${\mathbb {S}}_\delta$. Suppose x(t) hits the boundary of ${\mathbb {S}}_\delta$ at $t=t_1 \ge 0$ for the first time. Then, there exists a subset of ${\mathbb {I}}_n$, denoted by ${\mathbb {I}}_{\mathrm{hit}}$, such that $r_k(t_1)^2=1+\delta$ or $r_k(t_1)^2=1-\delta$ for $k\in {\mathbb {I}}_{\mathrm{hit}}$, while $|r_i(t_1)^2-1|<\delta$ for $i\in {\mathbb {I}}_n\backslash {\mathbb {I}}_{\mathrm{hit}}$. When $r_k(t_1)^2=1+\delta$, using (19), the derivative of $r_k(t)^2/2$ at $t=t_1$ is given by

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{r_k^2}{2}\right) =r_k{\dot{r}}_k= (1-r_k^2)r_k^2\\&\quad +\,{\varepsilon }r_k\alpha ^{\mathsf{\mathrm{T}}}r=-\delta (1+\delta )+{\varepsilon }r_k\alpha ^{\mathsf{\mathrm{T}}}r, \end{aligned}$$

for some vector $\alpha \in {\mathbb {R}}^n$ dependent on H and $\theta (t_1)$. Since $|r_k\alpha ^{\mathsf{\mathrm{T}}}r|$ is bounded by a number that depends only on H and $\delta$, the second term can be made smaller in magnitude than the first term by a choice of ${\varepsilon }$. Thus, the derivative is negative, and $r_k^2$ decreases from $1+\delta$. By a similar argument, we see that the value of $r_k^2$ increases when $r_k(t_1)^2=1-\delta$. Hence, x cannot go across the boundary of ${\mathbb {S}}_\delta$, proving the invariance.

To show that ${\mathbb {S}}_\delta$ is locally attractive, suppose x(t) is outside, but not too far from, the boundary of ${\mathbb {S}}_\delta$. That is, there exist $\rho \in {\mathbb {R}}$ and a subset of ${\mathbb {I}}_n$, denoted by ${\mathbb {I}}_{\mathrm{out}}$, such that $\delta \le |r_k(t)^2-1|<\rho <1/2$ for $k\in {\mathbb {I}}_{\mathrm{out}}$. The derivative of $r_k(t)^2/2$ is then bounded by

$$\begin{aligned} \begin{array}{lll} (\mathrm{d}/\mathrm{d}t)(r_k^2/2) \le -\delta (1+\delta )+{\varepsilon }r_k\alpha ^{\mathsf{\mathrm{T}}}r &{} \text{ if } &{} 1+\delta \le r_k(t)^2<1+\rho , \\ (\mathrm{d}/\mathrm{d}t)(r_k^2/2) \ge ~~\delta (1-\delta )+{\varepsilon }r_k\alpha ^{\mathsf{\mathrm{T}}}r &{} \text{ if } &{} 1-\rho < r_k(t)^2 \le 1-\delta . \end{array} \end{aligned}$$

Following a similar argument as before, for a small enough choice of ${\varepsilon }$, the bound on the derivative is sign definite until $r_k(t)$ hits the boundary of ${\mathbb {S}}_\delta$ to take value $1+\delta$ or $1-\delta$. Thus, x will enter ${\mathbb {S}}_\delta$. $\square$

1.2 Proof of Lemma 2

Proof

First, note from (8) that

$$\begin{aligned} \lambda = \sup {\mu } \qquad \text {such that} \qquad \lim \limits _{t\rightarrow \infty } e^{-{ 2}\mu t}||\varPhi (t)||{^2} \rightarrow \infty . \end{aligned}$$

Based on matrix norm properties, the following inequalities hold:

$$\begin{aligned} ||\varPhi (t)||{^2}\ge & {} \dfrac{1}{{2n}}||\varPhi (t)||_F^2\\\ge & {} \dfrac{1}{2n{\Vert P(t)\Vert }}\mathrm{tr}\left( \varPhi (t)^{\mathsf{\mathrm{T}}}P(t)\varPhi (t)\right) \\\ge & {} \dfrac{{ \varsigma }}{{ 2n}}\rho (t), \end{aligned}$$

where $\Vert \cdot \Vert _F$ is the Frobenius norm, and $\varsigma$ is a constant defined such that $||P(t)||<1/\varsigma$ for all $t\ge 0$. Let $\tilde{\mu }_\delta \triangleq \tilde{\mu }-\delta$ for sufficiently small $\delta >0$. Then,

$$\begin{aligned} \lim _{t\rightarrow \infty }e^{-2\tilde{\mu }_\delta t}||\varPhi (t)||{^2} \ge \lim _{t\rightarrow \infty } \dfrac{\varsigma }{2n}e^{-2\tilde{\mu }_\delta t}\rho (t) \rightarrow \infty . \end{aligned}$$

Since $\tilde{\mu }_\delta$ can make the left-hand side go to infinity, we conclude that $\lambda \ge \tilde{\mu }$. $\square$

1.3 Proof of Lemma 5

Proof

By Lemma 4, all possible harmonic solutions have the form (14). Substituting $\xi (t)$ into (3) gives

where and $\varGamma$ denote diagonal matrices of $\omega$ and $\gamma$, respectively. Noting that

$$\begin{aligned} \int _0^{2\pi /\omega } \xi (t)\xi (t)^{\mathsf{\mathrm{T}}}\mathrm{d}t=\frac{1}{2} \varOmega _\varphi VV^{\mathsf{\mathrm{T}}}\varOmega _\varphi ^{\mathsf{\mathrm{T}}}, \qquad V:=\left[ \begin{array}{cc} \gamma &{} 0 \\ 0 &{} \gamma \end{array}\right] , \end{aligned}$$

we see that $\xi (t)$ is a solution of the dynamical system if and only if

This equation is further equivalent to

where we noted that and $\varOmega _\varphi$ commute. The structure of H in (12) implies ${{\bar{H}}}_{11}={{\bar{H}}}_{22}$ and ${{\bar{H}}}_{12}=-{{\bar{H}}}_{21}$, and hence, the second and fourth equalities are satisfied. The first equality is expressed as

$$\begin{aligned} \eta ({\varepsilon },\gamma ):=(\varGamma ^2-I)\gamma -{\varepsilon }{{\bar{H}}}_{11}\gamma =0. \end{aligned}$$

Clearly, $\gamma _i$ satisfying this condition for a small $|{\varepsilon }|$ has to be close to 1, $-1$, or 0. All the harmonic solutions near can be captured by those near due to the freedom in $\varphi$. By the implicit function theorem, $\eta ({\varepsilon },\gamma )=0$ is solvable for $\gamma$ when $|{\varepsilon }|$ is sufficiently small since and hold. In particular, the solution in the neighborhood of is expressed as $\gamma =g({\varepsilon })$ for a continuously differentiable function g such that , and the derivative is given by

The formula for $\gamma$ now follows as the Taylor series and that for $\omega$ is then obtained by solving the third equality. $\square$