Asymptotic Phase-Locking Dynamics and Critical Coupling Strength for the Kuramoto Model

Abstract

We study the asymptotic clustering (phase-locking) dynamics for the Kuramoto model. For the analysis of emergent asymptotic patterns in the Kuramoto flow, we introduce the pathwise critical coupling strength which yields a sharp transition from partial phase-locking to complete phase-locking, and provide nontrivial upper bounds for the pathwise critical coupling strength. Numerical simulations suggest that multi- and mono-clusters can emerge asymptotically in the Kuramoto flow depending on the relative magnitude of the coupling strength compared to the sizes of natural frequencies. However, theoretical and rigorous analysis for such phase-locking dynamics of the Kuramoto flow still lacks a complete understanding, although there were some recent progress on the complete synchronization of the Kuramoto model in a sufficiently large coupling strength regime (Dörfler and Bullo in SIAM J Appl Dyn Syst 10:1070–1099, 2011, Automatica 50:1539–1564, 2014; Ha et al. in Commun Math Sci 4:1073–1091, 2016). In this paper, we present sufficient frameworks for partial phase-locking of a majority ensemble and the complete phase-locking in terms of the initial phase configuration, coupling strength and natural frequencies. As a by-product of our analysis, we obtain nontrivial upper bounds for the pathwise critical coupling strength in terms of the diameter of natural frequencies, initial Kuramoto order parameter and the system size N. We also show that phase-locked states whose order parameters are less than $N^{-\frac{1}{2}}$ are linearly unstable.

Appendix A. Proof of Lemma 5.3

In this appendix, we provide a proof of Lemma 5.3. The five conditions we must check are as follows.

$$\begin{aligned} \begin{aligned}&(a)~\gamma \in \left( \frac{1}{2},1\right] , \\&(b)~ \beta \in \left( 0,\cos ^{-1}\Big [\frac{1}{\gamma }-1\Big ]\right) , \\&(c)~\kappa >\frac{D(\Omega )}{2\sin \beta (\gamma \cos \beta -(1-\gamma ))}, \\&(d)~\text{ either }\quad R_0\ge \gamma +(1-\gamma )\cos \beta \quad \text{ or }\quad 2\gamma +\frac{D(\Omega )^2}{4\kappa ^2R_0^2}\frac{1}{1-\cos \beta }\le 1+R_0,\\&(e)~\frac{D(\Omega )}{\kappa }< \frac{(2\gamma -1)^{3/2}}{\sqrt{2\gamma }}\frac{2-\gamma }{\sqrt{\gamma /2}+(1-\gamma )}. \end{aligned} \end{aligned}$$

(A.1)

Recall our choice for $\gamma$ and $\beta$:

$$\begin{aligned}&\gamma (R_0)= {\left\{ \begin{array}{ll} 0.5+\frac{0.35}{0.94}R_0&{} 0<R_0\le 0.94,\\ 1-\frac{5}{2}(1-R_0)&{} 0.94<R_0\le 1, \end{array}\right. }\\&\quad \text{ and } \quad \cos \beta (R_0)= {\left\{ \begin{array}{ll} 1-\frac{0.4}{0.94}R_0&{}0<R_0\le 0.94, \\ 0.6&{}0.94<R_0\le 1. \end{array}\right. } \end{aligned}$$

We verify conditions (A.1) for the two cases separately

$$\begin{aligned} R_0\le 0.94 \quad \text{ and } \quad R_0> 0.94. \end{aligned}$$

1.1 A.1 Case A$(R_0\le 0.94)$

We verify the conditions in (A.1) one by one.

• Step 1 (Verification of relation (A.1)-(a)): Since $\gamma (R_0)$ is a linear function of $R_0$, we have

  $$\begin{aligned} 0.5=\gamma (0)< \gamma (R_0)\le \gamma (0.94)=0.85. \end{aligned}$$

• Step 2 (Verification of relation (A.1)-(b)): First, note the equivalence:

  $$\begin{aligned} \text{(A.1)-(b) } \quad \Longleftrightarrow \quad \cos \beta (R_0)>\frac{1}{\gamma (R_0)}-1 \quad \Longleftrightarrow \quad \gamma (R_0)(1+\cos \beta (R_0))-1>0. \end{aligned}$$

  On the other hand, we use the choices for $\gamma (R_0)$ and $\cos \beta (R_0)$ to see

  $$\begin{aligned} \gamma (R_0)(1+\cos \beta (R_0))-1&=\left( 0.5+\frac{0.35}{0.94}R_0\right) \left( 2-\frac{0.4}{0.94}R_0\right) -1\\&=\frac{0.5}{0.94}R_0-\frac{0.14}{0.94^2}R_0^2\\&=\frac{0.36}{0.94}R_0+\frac{0.14}{0.94}R_0\left( 1-\frac{R_0}{0.94}\right) >0, \end{aligned}$$

  where the final statement holds because of $0<R_0\le 0.94$.

• Step 3 (Verification of relation (A.1)-(c)): Note that

  $$\begin{aligned} \begin{aligned} \text{(A.1) }-(c)\quad \Longleftarrow&\quad 1.6\frac{D(\Omega )}{R_0^2}\ge \frac{D(\Omega )}{2\sin \beta (R_0)(\gamma (R_0)\cos \beta (R_0)-(1-\gamma (R_0)))}\\ \Longleftrightarrow&\quad 1.6\frac{1}{R_0^2}\ge \frac{1}{2\sqrt{1-\cos \beta (R_0)}\sqrt{1+\cos \beta (R_0)}(\gamma (R_0)(1+\cos \beta (R_0))-1)}\\ \Longleftrightarrow&\quad 1.6\frac{1}{R_0^2}\ge \frac{1}{2\sqrt{\frac{0.4}{0.94}R_0}\sqrt{2-\frac{0.4}{0.94}R_0}(\frac{0.5}{0.94}R_0-\frac{0.14}{0.94^2}R_0^2)}\\ \Longleftrightarrow&\quad 3.2\frac{1}{R_0^2}\sqrt{\frac{0.4}{0.94}R_0}\sqrt{2-\frac{0.4}{0.94}R_0}(\frac{0.5}{0.94}R_0-\frac{0.14}{0.94^2}R_0^2)\ge 1\\ \Longleftrightarrow&\quad 3.2\sqrt{\frac{0.4}{0.94}}\sqrt{2-\frac{0.4}{0.94}R_0}\left( \frac{0.5}{0.94\sqrt{R_0}}-\frac{0.14}{0.94^2}\sqrt{R_0}\right) \ge 1. \end{aligned} \end{aligned}$$

  In the final inequality, each and every term on the LHS is a positive decreasing function of $R_0$ on (0, 0.94], so the LHS as a whole is a decreasing function of $R_0$ on (0, 0.94]. At $R_0=0.94$, the LHS is 1.0430..., which is larger than 1. We thus conclude that (A.1)-(c) is true.

• Step 4 (Verification of relation (A.1)-(d)): We use the second condition of (A.1)-(d).

  $$\begin{aligned} \text{(A.1) }-(d)\quad \Longleftarrow&\quad 2\gamma (R_0)+\frac{D(\Omega )^2}{4\kappa ^2R_0^2}\frac{1}{1-\cos \beta (R_0)}\le 1+R_0\\ \Longleftarrow&\quad 2\gamma (R_0)+\frac{R_0^2}{4\cdot 1.6^2}\frac{1}{1-\cos \beta (R_0)}\le 1+R_0\\ \Longleftrightarrow&\quad 2\left( 0.5+\frac{0.35}{0.94}R_0\right) +\frac{R_0^2}{4\cdot 1.6^2}\frac{1}{\frac{0.4}{0.94}R_0}\le 1+R_0\\ \Longleftrightarrow&\quad \frac{0.7}{0.94}R_0+\frac{0.94}{4\cdot 1.6^2\cdot 0.4}R_0\le R_0\\ \Longleftrightarrow&\quad \frac{0.7}{0.94}+\frac{0.94}{4\cdot 1.6^2\cdot 0.4}=0.9742\cdots \le 1. \end{aligned}$$

  Hence (A.1)-(d) is true.

• Step 5 (Verification of relation (A.1)-(e)): We have

  $$\begin{aligned} \text{(A.1) }-(e)\quad \Longleftarrow&\quad \frac{1}{1.6}R_0^2\le \frac{(2\gamma (R_0)-1)^{3/2}}{\sqrt{2\gamma (R_0)}}\frac{2-\gamma (R_0)}{\sqrt{\gamma (R_0)/2}+(1-\gamma (R_0))}\\ \Longleftrightarrow&\quad \frac{1}{1.6}R_0^2\le \frac{\left( \frac{0.7}{0.94}R_0\right) ^{3/2}}{\sqrt{1+\frac{0.7}{0.94}R_0}}\frac{1.5-\frac{0.35}{0.94}R_0}{\sqrt{0.25+\frac{0.35}{1.88}R_0}+0.5-\frac{0.35}{0.94}R_0}\\ \Longleftrightarrow&\quad \frac{1}{1.6}\left( \frac{0.94}{0.7}\right) ^{3/2}\frac{\sqrt{R_0}\sqrt{1+\frac{0.7}{0.94}R_0}\left( \sqrt{0.25+\frac{0.35}{1.88}R_0}+0.5-\frac{0.35}{0.94}R_0\right) }{1.5-\frac{0.35}{0.94}R_0}\le 1. \end{aligned}$$

  The LHS of the final inequality has logarithmic derivative

  $$\begin{aligned}&\frac{1}{2R_0}+\frac{\frac{0.7}{0.94}}{2(1+\frac{0.7}{0.94}R_0)}+\frac{\frac{\frac{0.35}{1.88}}{2\sqrt{0.25+\frac{0.35}{1.88}R_0}}-\frac{0.35}{0.94}}{\sqrt{0.25+\frac{0.35}{1.88}R_0}+0.5-\frac{0.35}{0.94}R_0}+\frac{\frac{0.35}{0.94}}{1.5-\frac{0.35}{0.94}R_0}\\&\quad \ge \frac{1}{2\cdot 0.94}+0+\frac{\frac{\frac{0.35}{1.88}}{2\sqrt{0.25+\frac{0.35}{1.88}\cdot 0.94}}-\frac{0.35}{0.94}}{\sqrt{0.25+\frac{0.35}{1.88}R_0}+0.5-\frac{0.35}{0.94}R_0}+0\\&\quad = \frac{1}{2\cdot 0.94}-\frac{0.35}{0.94}\cdot \frac{1-\frac{1}{4\sqrt{0.425}}}{\sqrt{0.25+\frac{0.35}{1.88}R_0}+0.5-\frac{0.35}{0.94}R_0}\\&\quad \ge \frac{1}{2\cdot 0.94}-\frac{0.35}{0.94}\cdot \frac{1-\frac{1}{4\sqrt{0.425}}}{\sqrt{0.25+\frac{0.35}{1.88}\cdot 0}+0.5-\frac{0.35}{0.94}\cdot 0.94}\\&\quad =\frac{1}{2\cdot 0.94}-\frac{0.35}{0.94}\cdot \frac{1-\frac{1}{4\sqrt{0.425}}}{0.65}\\&\quad =0.178\cdots >0, \end{aligned}$$

  and thus is an increasing function of $R_0$ on (0, 0.94]. At $R_0=0.94$, this LHS has value 0.857..., which is less than 1. Hence (A.1)-(e) is true.

1.2 A.2 Case B $(R_0 > 0.94)$

Again, we verify each relation one by one.

• Step 1 (Verification of relation (A.1)-(a)): Since $\gamma (R_0)$ is a linear function of $R_0$,

  $$\begin{aligned} 0.85=\gamma (0.94)< \gamma (R_0)\le \gamma (1)=1. \end{aligned}$$

• Step 2 (Verification of relation (A.1)-(b)): Note the equivalence

  $$\begin{aligned} \text{(A.1) }-(b)\Longleftrightarrow \cos \beta (R_0)>\frac{1}{\gamma (R_0)}-1 \Longleftrightarrow \quad \gamma (R_0)(1+\cos \beta (R_0))-1>0, \end{aligned}$$

  and we can calculate

  $$\begin{aligned} \gamma (R_0)(1+\cos \beta (R_0))-1&=\left( 1-2.5(1-R_0)\right) \cdot 1.6-1=0.6-4(1-R_0)\\&>0.6-4\cdot 0.06=0.36>0, \end{aligned}$$

  where we have used $0.94<R_0\le 1$ in the first inequality.

• Step 3 (Verification of relation (A.1)-(c)): We have

  $$\begin{aligned} \text{(A.1) }-(c)\quad \Longleftarrow&\quad 1.6\frac{D(\Omega )}{R_0^2}\ge \frac{D(\Omega )}{2\sin \beta (R_0)(\gamma (R_0)\cos \beta (R_0)-(1-\gamma (R_0)))}\\ \Longleftrightarrow&\quad 1.6\frac{1}{R_0^2}\ge \frac{1}{2\cdot 0.8\cdot (0.6-4(1-R_0))}\\ \Longleftrightarrow&\quad 2.56\frac{4R_0-3.4}{R_0^2}\ge 1. \end{aligned}$$

  In the final inequality, the LHS has derivative

  $$\begin{aligned} 2.56\left( -\frac{4}{R_0^2}+\frac{6.8}{R_0^3}\right)>2.56\left( -\frac{4}{0.94^2}+6.8\right) =2.2731\cdots >0, \end{aligned}$$

  and thus is an increasing function of $R_0$ on (0.94, 1]. At $R_0=0.94$, the LHS is 1.0430..., which is larger than 1. We thus conclude that (A.1)-(c) is true.

• Step 4 (Verification of relation (A.1)-(d)): This time, we verify the first statement of (A.1)-(d), which is

  $$\begin{aligned} \text{(A.1) }-(d)-(i)\quad \Longleftrightarrow&\quad R_0\ge \gamma (R_0)+(1-\gamma (R_0))\cos \beta (R_0)\\ \Longleftrightarrow&\quad R_0\ge (1-2.5(1-R_0))+2.5(1-R_0)\cdot 0.6\\ \Longleftrightarrow&\quad 0\ge 0. \end{aligned}$$

• Step 5 (Verification of relation (A.1)-(e)): We have

  $$\begin{aligned} \text{(A.1) }-(e)\quad \Longleftarrow&\quad \frac{1}{1.6}R_0^2\le \frac{(2\gamma (R_0)-1)^{3/2}}{\sqrt{2\gamma (R_0)}}\frac{2-\gamma (R_0)}{\sqrt{\gamma (R_0)/2}+(1-\gamma (R_0))}\\ \Longleftrightarrow&\quad \frac{1}{1.6}R_0^2\le \frac{\left( 5R_0-4\right) ^{3/2}}{\sqrt{5R_0-3}}\frac{3.5-2.5R_0}{\sqrt{1.25R_0-0.75}+2.5(1-R_0)}\\ \Longleftrightarrow&\quad \frac{1}{1.6}\frac{R_0^2\sqrt{5R_0-3}(\sqrt{1.25R_0-0.75}+2.5(1-R_0))}{\left( 5R_0-4\right) ^{3/2}(3.5-2.5R_0)}\le 1. \end{aligned}$$

  The LHS of the final inequality has, on (0.94, 1), logarithmic derivative

  $$\begin{aligned}&\frac{2}{R_0}+\frac{5}{2(5R_0-3)}+\frac{\frac{1.25}{2\sqrt{1.25R_0-0.75}}-2.5}{\sqrt{1.25R_0-0.75}+2.5(1-R_0)}-\frac{3}{2}\cdot \frac{5}{5R_0-4}+\frac{2.5}{3.5-2.5R_0}\\&\quad \le \frac{2}{0.94}+\frac{5}{2(5\cdot 0.94-3)}+\frac{\frac{1.25}{2\sqrt{1.25\cdot 0.94-0.75}}-2.5}{\sqrt{1.25R_0-0.75}+2.5(1-R_0)}-\frac{3}{2}\cdot \frac{5}{5\cdot 1-4}\\&\qquad +\frac{2.5}{3.5-2.5\cdot 1}\\&\quad = \frac{2}{0.94}+\frac{5}{3.4}-\frac{2.5-\frac{1.25}{2\sqrt{0.425}}}{\sqrt{1.25R_0-0.75}+2.5(1-R_0)}-7.5+2.5\\&\quad \le \frac{2}{0.94}+\frac{5}{3.4}-\frac{2.5-\frac{1.25}{2\sqrt{0.425}}}{\sqrt{1.25\cdot 1-0.75}+2.5(1-0.94)}-5\\&\quad \le \frac{2}{0.94}+\frac{5}{3.4}-\frac{2.5-\frac{1.25}{2\sqrt{0.425}}}{\sqrt{0.5}+0.15}-5\\&\quad =-3.20\cdots <0, \end{aligned}$$

  and thus is a decreasing function of $R_0$ on (0.94, 1]. At $R_0=0.94$, this LHS has value 0.857..., which is less than 1. Hence (A.1)-(e) is true.