Synchronization Conditions of a Mixed Kuramoto Ensemble in Attractive and Repulsive Couplings

Abstract

We study the emergent dynamics of a mixed Kuramoto ensemble in the presence of both attractive and repulsive coupling strengths. To be precise, we consider coupled Kuramoto-type systems consisting of two ensembles in which the oscillators in the same group interact attractively with a positive intra-group coupling strength, whereas the oscillators in the different group communicate repulsively with a negative inter-group coupling strength. For the modeling perspective of the Kuramoto model (KM for brevity), two types of systems are treated in this paper: an attractive–repulsive inertial KM–KM and an attractive–repulsive KM–KM. For these two models, we provide sufficient frameworks leading to complete bi-polar synchronization in which asymptotic configuration tends to a bi-polar state. Our estimates mainly rely on the gradient-like flow formulation and the energy estimate.

A Proof of Theorem 4

Recall that our goal is to establish

$$\begin{aligned} \lim _{t \rightarrow \infty }\Big \{ \Vert {\dot{\varXi }}(t)\Vert + \Vert \varXi (t)-\varXi ^\text {e}\Vert \Big \} = 0. \end{aligned}$$

However, since we have already established the complete synchronization in Proposition 1, we have

$$\begin{aligned} \lim _{t \rightarrow \infty } \Vert {\dot{\varXi }}(t)\Vert = 0. \end{aligned}$$

Hence, it suffices to show that there exists $\varXi ^\text {e} \in {\mathcal {S}}$ such that

$$\begin{aligned} \lim _{t \rightarrow \infty } \Vert \varXi (t)-\varXi ^\text {e}\Vert = 0. \end{aligned}$$

(41)

The proof of (41) will split into two parts. Let $(\varXi _0,\varUpsilon _0) \in {\mathbb {R}}^{2N} \times {\mathbb {R}}^{2N}$ be the initial data such that the solution of (8) is global and bounded, and we define the $\omega$-limit set of $(\varXi _0,\varUpsilon _0)$ by

$$\begin{aligned} \omega (\varXi _0,\varUpsilon _0) := \{a \in {\mathbb {R}}^{2 N}:~\exists t_n \rightarrow \infty ~\text{ such } \text{ that }~\varXi (t_n) \rightarrow a~\text{ as }~n \rightarrow \infty \}. \end{aligned}$$

As a consequence of zero convergence for $\Vert {\dot{\varXi }}\Vert$, we have

$$\begin{aligned} \omega (\varXi _0,\varUpsilon _0) \in {\mathcal {S}}. \end{aligned}$$

Let $\varXi ^\text {e} \in \omega (\varXi _0,\varOmega _0)$, and we may assume that

$$\begin{aligned} \varXi ^\text {e} = 0, \quad V(0)=0 \quad \text{ and } \quad \nabla V_\varXi (0) = 0. \end{aligned}$$

Moreover, we set

$$\begin{aligned} {{\widetilde{\varXi }}} = \varXi -\varXi ^\text {e} \quad \text{ and } \quad \widetilde{V}({{\widetilde{\varXi }}}) = V({{\widetilde{\varXi }}} + \varXi ^\text {e}) - V(\varXi ^\text {e}). \end{aligned}$$

$\bullet$ (Step A: zero convergence of E): we define an energy-like function E(t) as follows:

$$\begin{aligned} E(t) := \frac{1}{2} \Vert \sqrt{M} {\dot{\varXi }}(t)\Vert ^2 + V(\varXi (t)) + \varepsilon \langle \nabla V(\varXi (t)), \sqrt{M} {\dot{\varXi }}(t) \rangle , \end{aligned}$$

where $\varepsilon >0$ is a positive constant to be determined later. In what follows, we simply write $\nabla V$ for $\nabla _\varXi V$ and define the maximum and minimum for $\{m_j\}$:

$$\begin{aligned} m^* := \max \{m_1, \ldots , m_N\}, \quad m_* := \min \{m_1, \ldots , m_N\}, \quad {\widetilde{M}} = \text {diag}\{m_1, \ldots , m_N\}. \end{aligned}$$

Then, we observe

$$\begin{aligned} \frac{\text {d}E(t)}{\text {d}t}&= \langle \sqrt{M}{\dot{\varXi }}(t), \sqrt{M}{\dot{\varXi }}(t) \rangle + \langle \nabla V(\varXi (t)), {\dot{\varXi }}(t) \rangle + \varepsilon \langle \nabla V(\varXi (t)), \sqrt{M} {\ddot{\varXi }}(t) \rangle \\&\quad + \varepsilon \langle \sqrt{M} \nabla ^2 V(\varXi (t)) {\dot{\varXi }}(t), {\dot{\varXi }}(t) \rangle \\&= \langle {\dot{\varXi }}(t), M {\dot{\varXi }}(t) \rangle + \langle {\dot{\varXi }}(t),\nabla V(\varXi (t)) \rangle + \varepsilon \langle \nabla _\varTheta V(\varXi (t)), \sqrt{{\widetilde{M}}} {\ddot{\varTheta }}(t) \rangle \\&\quad + \varepsilon \langle \sqrt{M} \nabla ^2 V(\varXi (t)) {\dot{\varXi }}(t), {\dot{\varXi }}(t) \rangle \\&= \langle \varXi (t), - {\dot{\varXi }}(t) \rangle + \varepsilon \langle \sqrt{{\widetilde{M}}}^{-1}\nabla _\varTheta V(\varXi (t)), {\widetilde{M}} {\ddot{\varTheta }}(t) \rangle + \varepsilon \langle \sqrt{M} \nabla ^2 V(\varXi (t)) {\dot{\varXi }}(t), {\dot{\varXi }}(t) \rangle \\&=-\Vert {\dot{\varXi }}(t)\Vert ^2 - \varepsilon \langle \sqrt{\widetilde{M}}^{-1}\nabla _\varTheta V(\varXi (t)), {\dot{\varTheta }}(t) + \nabla _\varTheta V(\varXi (t)) \rangle \\&\quad + \varepsilon \langle \sqrt{M} \nabla ^2 V(\varXi (t)) {\dot{\varXi }}(t), {\dot{\varXi }}(t) \rangle . \end{aligned}$$

We choose $\varepsilon$ sufficiently small so that

$$\begin{aligned} \varepsilon \sqrt{m^*} \Vert \nabla ^2 V(\varXi )\Vert _\infty < \frac{1}{2}. \end{aligned}$$

Then, we have

$$\begin{aligned} \frac{\text {d}E(t)}{\text {d}t}&\le -\frac{1}{2} \Vert {\dot{\varXi }}(t)\Vert ^2 - \frac{\varepsilon }{\sqrt{m^*}}\Vert \nabla _\varTheta V(\varXi (t)\Vert ^2 -\varepsilon \langle \sqrt{{\widetilde{M}}}^{-1} \nabla _\varTheta V(\varXi (t)), {\dot{\varTheta }}(t) \rangle \\&\le -\frac{1}{2} \Vert {\dot{\varXi }}(t)\Vert ^2 - \frac{\varepsilon }{\sqrt{m^*}}\Vert \nabla _\varTheta V(\varXi (t)\Vert ^2 + \frac{ \varepsilon }{\sqrt{m_*}} \Vert \nabla _\varTheta V(\varXi (t))\Vert \Vert {\dot{\varTheta }}(t)\Vert \\&= -\frac{1}{2} \Vert {\dot{\varTheta }}(t)\Vert ^2 -\frac{1}{2} \Vert \nabla _{\overline{\varPhi }}V(\varXi (t))\Vert ^2- \frac{\varepsilon }{\sqrt{m^*}}\Vert \nabla _\varTheta V(\varXi (t))\Vert ^2 \\&\quad + \frac{ \varepsilon }{\sqrt{m_*}} \Vert \nabla _\varTheta V(\varXi (t))\Vert \Vert {\dot{\varTheta }}(t)\Vert \\&\le -\frac{1}{2} \Vert {\dot{\varTheta }}(t)\Vert ^2 - \frac{\varepsilon }{\sqrt{m^*}}\Vert \nabla V(\varXi (t))\Vert ^2 + \frac{ \varepsilon }{\sqrt{m_*}} \Vert \nabla V(\varXi (t))\Vert \Vert {\dot{\varTheta }}(t)\Vert . \end{aligned}$$

It is easy to see that there exists $\varepsilon _0$ such that for all $\varepsilon \in [0,\varepsilon _0)$,

$$\begin{aligned} -\frac{\text {d}E(t)}{\text {d}t} \ge \varepsilon (\Vert {\dot{\varTheta }}\Vert + \Vert \nabla V(\varXi )\Vert )^2,\quad t>0. \end{aligned}$$

(42)

Thus, we see

$$\begin{aligned} \frac{\text {d}E(t)}{\text {d}t} \le 0, \end{aligned}$$

and since $0 \in \omega (\varXi _0,\varUpsilon _0)$, we can derive

$$\begin{aligned} E(t) \downarrow 0 \quad \text{ as }~ t \rightarrow \infty . \end{aligned}$$

(43)

$\bullet$ (Step B: finite length of the phase vector $\varXi$): Let $\rho$ be the Lojasiewicz exponent of V in Theorem 2. We use the Cauchy-Schwarz inequality to have

$$\begin{aligned} \begin{aligned} E(t)^{1-\rho }&= \left( \frac{1}{2} \Vert \sqrt{M}{\dot{\varXi }}\Vert ^2 + V(\varXi ) + \varepsilon \langle \nabla V(\varXi ), \sqrt{M}{\dot{\varXi }} \rangle \right) ^{1-\rho } \\&= \left( \frac{1}{2} \Vert \sqrt{{\widetilde{M}}}{\dot{\varTheta }}\Vert ^2 + V(\varXi ) + \varepsilon \langle \nabla _\varTheta V(\varXi ), \sqrt{{\widetilde{M}}}{\dot{\varTheta }} \rangle \right) ^{1-\rho }\\&\le \left( \frac{m^*}{2}\right) ^{1-\rho }\Vert {\dot{\varTheta }}\Vert ^{2(1-\rho )} + |V(\varXi )|^{1-\rho } + (m^*)^{\frac{1-\rho }{2}} \Vert \nabla _\varTheta V(\varXi )\Vert ^{1-\rho }\Vert {\dot{\varTheta }}\Vert ^{1-\rho }. \end{aligned} \end{aligned}$$

(44)

We now apply Young’s inequality to find

$$\begin{aligned}&\Vert \nabla _\varTheta V(\varXi )\Vert ^{1-\rho }\Vert {\dot{\varTheta }}\Vert ^{1-\rho } \le (1-\rho )^{1-\rho }\Vert \nabla _\varTheta V(\varXi )\Vert \\&+ \rho ^{1-\rho } \Vert {\dot{\varTheta }}\Vert ^{\frac{1-\rho }{\rho }} \le \Vert \nabla _\varTheta V(\varXi )\Vert + \Vert {\dot{\varTheta }}\Vert ^{\frac{1-\rho }{\rho }} . \end{aligned}$$

Thus, (44) becomes

$$\begin{aligned} E(t)^{1-\rho } \le \left( \frac{m^*}{2}\right) ^{1-\rho } \Vert {\dot{\varTheta }}\Vert ^{2(1-\rho )} + |V(\varXi )|^{1-\rho } + (m^*)^{\frac{1-\rho }{2}} (\Vert \nabla _\varTheta V(\varXi )\Vert + \Vert {\dot{\varTheta }}\Vert ^{\frac{1-\rho }{\rho }} ). \end{aligned}$$

(45)

Since $0= \varXi ^\text {e} \in \omega (\varXi _0,\varUpsilon _0)$, we can choose a sequence $\{t_n\}$ such that $\varXi (t_n) \rightarrow 0$ as $n \rightarrow \infty$. Then, zero convergence of $\Vert {\dot{\varXi }}(t)\Vert$ and (43) yield that for any $r>0$, there exists $n_0 >0$ such that

$$\begin{aligned}&\Vert \varXi (t_n)\Vert< \frac{r}{2}, \quad \frac{C}{\rho \varepsilon }E(t_n)^\rho < \frac{r}{2} \quad \text{ for }~n \ge n_0, \end{aligned}$$

(46)

$$\begin{aligned}&\left( \frac{m^*}{2}\right) ^{1-\rho }\Vert {\dot{\varXi }}(t)\Vert ^{1-2\rho } + (m^*)^{\frac{1-\rho }{2}}\Vert {\dot{\varXi }}(t)\Vert ^{\frac{1}{\rho }-2} \le 2, \quad \forall t \ge t_{n_0}, \end{aligned}$$

(47)

where C is a positive constant:

$$\begin{aligned} C:= \max \{2, (1+(m^*)^{\frac{1-\rho }{2}}\}. \end{aligned}$$

Note that (46) follows from the zero convergences of $\Vert \varXi (t_n)\Vert$ and energy-like function E, and (47) follows from the zero convergence of $\Vert {\dot{\varXi }}\Vert$. We set

$$\begin{aligned} T := \sup \{t\ge t_{n_0}:~\Vert \varXi (s)\Vert < r, \quad \forall s \in [t_{n_0},t]\}. \end{aligned}$$

(48)

Our claim is to prove that $T=\infty$. By using Lojasiewicz’s inequality in Theorem 2, we obtain from (45) that

$$\begin{aligned} \begin{aligned} E(t)^{1-\rho }&\le \left( \frac{m^*}{2}\right) ^{1-\rho }\Vert {\dot{\varTheta }}\Vert ^{2(1-\rho )} + (m^*)^{\frac{1-\rho }{2}}\Vert {\dot{\varTheta }}\Vert ^{\frac{1-\rho }{\rho }} + \left( 1 + (m^*)^{\frac{1-\rho }{2}} \right) \Vert \nabla V(\varXi )\Vert \\&\le 2\Vert {\dot{\varTheta }}\Vert + \left( 1 + (m^*)^{\frac{1-\rho }{2}} \right) \Vert \nabla V(\varXi )\Vert , \qquad t \ge t_{n_0}, \end{aligned} \end{aligned}$$

(49)

where we used (47) for the last inequality. Together with (42) and (49), we find

$$\begin{aligned}&-\frac{\text {d}[E(t)^\rho ]}{\text {d}t} = -\rho {\dot{E}}(t) E(t)^{\rho -1} \ge \frac{\rho \varepsilon (\Vert {\dot{\varTheta }}\Vert + \Vert \nabla V(\varXi )\Vert )^2}{2\Vert {\dot{\varTheta }}\Vert + \left( 1 + (m^*)^{\frac{1-\rho }{2}} \right) \Vert \nabla V(\varXi )\Vert }\nonumber \\&\ge \frac{\rho \varepsilon }{C}(\Vert {\dot{\varTheta }}\Vert + \Vert \nabla V(\varXi )\Vert ). \end{aligned}$$

(50)

We integrate (50) on the time-interval $(t_{n_0},T)$ to get

$$\begin{aligned} \int _{t_{n_0}}^T \Vert {\dot{\varTheta }}(s)\Vert \,\text {d}s \le \frac{C}{\rho \varepsilon } [E(t_{n_0})]^\rho . \end{aligned}$$

(51)

If $T<\infty$, we have

$$\begin{aligned} \Vert \varTheta (T)\Vert \le \int _{t_{n_0}}^T \Vert {\dot{\varTheta }}(s)\Vert \text {d}s + \Vert \varTheta (t_{n_0})\Vert . \end{aligned}$$

Then, it follows from (46) and (51) that $\Vert \varTheta (T)\Vert <r$. This contradicts (48). Hence, we finally have

$$\begin{aligned} T=\infty . \end{aligned}$$

The relation (51) with $T=\infty$ implies that the trajectory $\varTheta (t)$ has a finite length. Hence, we have that $\varTheta (t)$ is convergent and indeed, $\varTheta _e$ is the limit. On the other hand, (50) easily gives

$$\begin{aligned} -\frac{\text {d}[E(t)^\rho ]}{\text {d}t} \ge \frac{\rho \varepsilon }{C}\Vert \nabla _{{{\overline{\varPhi }}}}V(\varXi )\Vert = \frac{\rho \varepsilon }{C} \Vert \dot{{{\overline{\varPhi }}}}\Vert . \end{aligned}$$

(52)

Similarly, we integrate (52) over $(t_{n_0},T=\infty )$ as in (51) to find ${{\overline{\varPhi }}}(t)$ also have a finite length and it converges to ${{\overline{\varPhi }}}_e$. Hence, we conclude that $\varXi ^\text {e} = (\varTheta _e, {{\overline{\varPhi }}}_e)$ is the limit point of the trajectory $\varXi (t) = (\varTheta (t),{{\overline{\varPhi }}}(t))$.