On the Relaxation Dynamics of Lohe Oscillators on Some Riemannian Manifolds

Abstract

We study the collective relaxation dynamics appearing in weakly coupled Lohe oscillators in a large coupling regime. The Lohe models on the unit sphere and unitary group were proposed as a nonabelian generalization of the Kuramoto model on the unit circle and their emergent dynamics has been extensively studied in previous literature for some restricted class of initial data based on the Lyapunov functional approach and order parameter approach. In this paper, we extend the previous partial results to cover a generic initial configuration via the detailed analysis on the order parameter measuring the modulus of the centroid. In particular, we present a detailed relaxation dynamics and structure of the resulting asymptotic states for the Lohe sphere model. We also present new gradient flow formulations for the Lohe matrix models with the same one-body Hamiltonians on some group manifolds. As a direct application of this new formulation, we show that every bounded Lohe flow which originated from any initial configuration converges asymptotically.

Appendices

Appendix A: Verification of Conditions (4.14)

In this section, we show that the choices explicitly verify that the choices (4.22) and (4.23) satisfy the conditions (4.14), that is the conditions

$$\begin{aligned}&\frac{2p}{\left( 2 p-1\right) ^2}D_2(\Omega ) < \kappa _0, \end{aligned}$$

(A.1)

$$\begin{aligned}&\quad 1-2\alpha _0>\frac{1}{2p}, \end{aligned}$$

(A.2)

$$\begin{aligned}&\quad p > \frac{1}{2}+\frac{\rho ^0}{6}, \end{aligned}$$

(A.3)

$$\begin{aligned}&\frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}-\frac{1}{N}\left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}\right\rfloor >\frac{n_0-1}{N}. \end{aligned}$$

(A.4)

1.1 A.1 N is even

In this case, recall that we choose

$$\begin{aligned} \begin{aligned}&\frac{2m_0}{N}<\rho ^0\le \frac{2(m_0+1)}{N}, \quad n_0=\left\lfloor \frac{m_0}{2}\right\rfloor +\frac{N}{2}+1, \\&\kappa _0= \frac{1152}{7} \frac{D_2(\Omega )}{(\rho ^0)^3}, \quad \alpha _0={\left\{ \begin{array}{ll} \rho ^0/8, &{}m_0\le 1,\\ \frac{m_0}{4N}, &{}m_0\ge 2. \end{array}\right. } \end{aligned} \end{aligned}$$

(A.5)

Here, $n_0$ was roughly chosen so that $p=\frac{n_0}{N}\approx \frac{1}{2}+\frac{m_0}{2N}\approx \frac{1}{2}+\frac{\rho ^0}{4}$ and hence that (A.3) would hold in most cases. On the other hand, (A.2) suggests that

$$\begin{aligned} \alpha _0<\frac{1}{2}-\frac{1}{4p}\approx \frac{1}{2}-\frac{1}{4}\frac{1}{1+(p-1)}\approx \frac{1}{2}-\frac{1}{4}(1-(p-1))=\frac{p}{4}=\frac{m_0}{4N}. \end{aligned}$$

When $m_0\ge 2$, this happens to give $\frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}>\frac{n_0-1}{N}$, hence we can finally choose $\kappa _0$ large enough so as to make (A.1) and (A.4) true. On the other hand, when $m_0\le 1$, we need to choose a smaller $\alpha _0$, because then $n_0=\frac{N}{2}+1$ and so

$$\begin{aligned} \frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}>\frac{n_0-1}{N}=\frac{1}{2}\quad \Leftrightarrow \quad \alpha _0<2\rho ^0, \end{aligned}$$

while $\rho _0$ could be very small. The denominator 8 was chosen because $\frac{m_0}{4N}\le \frac{1}{4\cdot 2}=\frac{1}{8}$. The funny coefficient $\frac{1152}{7}$ in the choice for $\kappa _0$ will be explained in the demonstration of (A.4) when $m_0\ge 2$; the condition (A.1) is weaker than (A.4).

• Case A $(m_0\le 1)$: We verify the relations (A.1)–(A.4) one by one. Note that by our choice, we have

  $$\begin{aligned} n_0=\frac{N}{2}+1,\quad \rho ^0\le \frac{4}{N},\quad \alpha _0=\frac{\rho ^0}{8}. \end{aligned}$$

• (Verification of (A.1)) We have

  $$\begin{aligned} p=\frac{n_0}{N}=\frac{1}{2}+\frac{1}{N}, \end{aligned}$$

  so

  $$\begin{aligned} \frac{2p}{(2p-1)^2}=\frac{1+(2/N)}{(2/N)^2}=\frac{N^2}{16}\left( 4+\frac{8}{N}\right) \le \frac{1}{(\rho ^0)^2}\left( 4+4\right) =\frac{8}{(\rho ^0)^2}\le \frac{1152}{7(\rho ^0)^3}, \end{aligned}$$

  hence

  $$\begin{aligned} \frac{2p}{\left( 2p-1\right) ^2}D_2(\Omega )<\frac{1152}{7}\frac{D_2(\Omega )}{(\rho ^0)^3} = \kappa _0. \end{aligned}$$

• (Verification of (A.2)) We have

  $$\begin{aligned} (1-2\alpha _0)\frac{2n_0}{N} = (1-\frac{\rho ^0}{4})(1+\frac{2}{N}) \ge (1-\frac{1}{N})(1+\frac{2}{N})> 1, \quad \text{ i.e., } \quad 1-2\alpha _0>\frac{1}{2p}, \end{aligned}$$

  where we use the fact that $N \ge 3$.

• (Verification of (A.3)) We use (A.5) and $\left\lfloor \frac{m_0}{2}\right\rfloor = 0$ to find

  $$\begin{aligned} p = \frac{n_0}{N}= \frac{1}{2}+\frac{1}{N}\ge \frac{1}{2}+\frac{\rho ^0}{4}>\frac{1}{2}+\frac{\rho ^0}{6}. \end{aligned}$$

• (Verification of (A.4)) By the choice of $\alpha _0$ and $\kappa _0$ in (A.5), we have

  $$\begin{aligned} 0< \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}=\frac{N}{\rho ^0/8}\cdot \frac{7(\rho ^0)^2}{576}=\frac{7N\rho ^0}{72}\le \frac{28}{72}<1, \quad \text{ i.e., } \quad \left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0} \right\rfloor = 0. \end{aligned}$$

  Hence, we have

  $$\begin{aligned} \frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}-\frac{1}{N}\left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}\right\rfloor =\frac{1+7\rho ^0/8}{2-\rho ^0/8}-0>\frac{1}{2}=\frac{n_0-1}{N}. \end{aligned}$$

• Case B $(m_0 \ge 2)$: Similar to Case A, we check the relations (A.1)–(A.4) one by one.

• (Verification of (A.1)): Since

  $$\begin{aligned} n_0=\left\lfloor \frac{m_0}{2}\right\rfloor +\frac{N}{2}+1 > \frac{m_0}{2} + \frac{N}{2}, \end{aligned}$$

  we have

  $$\begin{aligned} 2 p - 1= \frac{2n_0}{N} -1> \frac{m_0}{N}. \end{aligned}$$

  (A.6)

  This yields

  $$\begin{aligned} \frac{2p}{(2p - 1)^2}&= \frac{1}{2p-1}\left( \frac{1}{2p-1} + 1 \right) <\frac{N}{m_0}\left( \frac{N}{m_0}+1\right) \\&=\frac{N}{2(m_0+1)}\cdot \frac{2(m_0+1)}{m_0}\left( \frac{N}{2(m_0+1)}\cdot \frac{2(m_0+1)}{m_0}+1\right) \\&\le \frac{1}{\rho ^0}\cdot 4 \left( \frac{1}{\rho ^0}\cdot 4 +1\right) \le \frac{4}{\rho ^0}\left( \frac{5}{\rho ^0}\right) \le \frac{1152}{7(\rho ^0)^3}, \end{aligned}$$

  Hence

  $$\begin{aligned} \frac{2p}{\left( 2p-1\right) ^2}D_2(\Omega )<\frac{1152}{7}\frac{D_2(\Omega )}{(\rho ^0)^3} = \kappa _0. \end{aligned}$$

• (Verification of (A.2)) We use $\alpha _0=\frac{m_0}{4N}$ and (A.6) to obtain

  $$\begin{aligned} (1-2\alpha _0) p = (1-2\alpha _0)\frac{2n_0}{N}> \Big (1-\frac{m_0}{2N} \Big ) \Big (1+\frac{m_0}{N} \Big )=1+\frac{m_0}{2N} \Big (1-\frac{m_0}{N} \Big )>1, \end{aligned}$$

  i.e., we have

  $$\begin{aligned} 1-2\alpha _0>\frac{1}{2p}. \end{aligned}$$

• (Verification of (A.3)) We again use the relation (A.6) to get

  $$\begin{aligned} p = \frac{n_0}{N}> \frac{1}{2}+\frac{m_0}{2N}=\frac{1}{2}+\frac{2(m_0+1)}{N}\cdot \frac{m_0}{4(m_0+1)}\ge \frac{1}{2}+\rho ^0\cdot \frac{2}{4\cdot 3}=\frac{1}{2}+\frac{\rho ^0}{6}. \end{aligned}$$

• (Verification of (A.4)) By the choice of $\alpha _0 = \frac{m_0}{4N}$ and $\kappa _0$, we have

  $$\begin{aligned} \begin{aligned} \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}&=\frac{4N^2}{m_0}\cdot \frac{7(\rho ^0)^2}{576}=\frac{7N^2(\rho ^0)^2}{144m_0}\le \frac{7\cdot 4(m_0+1)^2}{144m_0} \\&= \frac{7m_0}{36}\left( 1+\frac{1}{m_0}\right) ^2. \end{aligned} \end{aligned}$$

  (A.7)

  On the other hand, we have

  $$\begin{aligned} \begin{aligned} \frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}-\frac{n_0-1}{N}&>\frac{\frac{2m_0}{N}+1-\frac{m_0}{4N}}{1+1-\frac{m_0}{4N}}-\left( \frac{m_0}{2N}+\frac{1}{2}\right) =\frac{4N+7m_0}{8N-m_0}-\frac{N+m_0}{2N} \\&=\frac{m_0(7N+m_0)}{2N(8N-m_0)}. \end{aligned} \end{aligned}$$

  (A.8)

  Thus, we use $m_0 \ge 2$, (A.7) and (A.8) to find

  $$\begin{aligned} \begin{aligned} \frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}&-\frac{n_0-1}{N}-\frac{1}{N}\left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{K_0\rho ^0}\right\rfloor \\&>\frac{m_0(7N+m_0)}{2N(8N-m_0)}-7\frac{(m_0+1)^2}{36Nm_0}\\&=\frac{14N(5m_0^2-8m_0-4)+25m_0^3+14m_0^2+7m_0}{36Nm_0(8N-m_0)}\\&=\frac{14N(m_0-2)(5m_0+2)+25m_0^3+14m_0^2+7m_0}{36Nm_0(8N-m_0)} > 0. \end{aligned} \end{aligned}$$

  (A.9)

  i.e. we have

  $$\begin{aligned} \frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}-\frac{1}{N}\left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{K_0\rho ^0}\right\rfloor >\frac{n_0-1}{N}. \end{aligned}$$

  The role of the coefficient $\frac{1152}{7}$ appears in the last line of (A.9). For any coefficient $\zeta$ replacing $\frac{1152}{7}$, the last line of (A.9) would appear as

  $$\begin{aligned} \frac{Na(m_0)+b(m_0)}{Nm_0(8N-m_0)}, \end{aligned}$$

  where $a(m_0)$ is a quadratic in $m_0$ and $b(m_0)$ is a cubic in $m_0$, depending on $\zeta$. If $a(2)<0$, then we may choose N sufficiently large so that this rational function takes a negative value. Hence we must have $a(2)\ge 0$, which forces us to choose $\zeta \ge \frac{1152}{7}$. This choice of $\zeta$ happens to make the rational function positive for all valid values of $m_0$ and N.

1.2 A.2 N is odd

Recall that we choose $m_0$ to satisfy

$$\begin{aligned} \frac{2m_0-1}{N}\le \rho ^0< \frac{2m_0+1}{N}. \end{aligned}$$

(A.10)

Then, for such $m_0$, we set

$$\begin{aligned} n_0=\left\lfloor \frac{m_0}{2}\right\rfloor +\frac{N+1}{2}, \quad \kappa _0= \frac{640}{3} \frac{D_2(\Omega )}{(\rho ^0)^3}, \quad \alpha _0={\left\{ \begin{array}{ll} \frac{1}{4N}, &{}m_0\le 1,\\ \frac{m_0}{8N}, &{}m_0\ge 2. \end{array}\right. } \end{aligned}$$

(A.11)

We again claim that the choices (A.10)–(A.11) satisfy (A.1)–(A.4). The rationale for choosing $n_0$, and $\alpha _0$ when $m_0\ge 2$, is similar to A.1. However, when $m_0\le 1$, we have $n_0=\frac{N+1}{2}$ and $\frac{n_0-1}{N}=\frac{N-1}{2N}$, which is less than $\frac{1}{2}$, so we can be a little more lenient with our choice of $\alpha _0$ so as to make $\frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}>\frac{n_0-1}{N}$ even when $\rho ^0\approx 0$. Again, the coefficient $\frac{640}{3}$ in the choice of $\kappa _0$ is necessitated by (A.4) when $m_0\ge 2$; the condition (A.1) is weaker than (A.4).

• Case C $(m_0\le 1)$: Note that the relations (A.11) yield

  $$\begin{aligned} \alpha _0=\frac{1}{4N}, \quad n_0=\frac{N+1}{2} \quad \text{ and } \quad \rho ^0<\frac{3}{N}. \end{aligned}$$

  (A.12)

• (Verification of (A.1)) We use the relations (A.12) to obtain

  $$\begin{aligned} \frac{2p}{(2p - 1)^2} = \frac{2\frac{n_0}{N}}{\left( 2\frac{n_0}{N}-1\right) ^2}=N(N+1)=\frac{9(N+1)}{N}\cdot \frac{N^2}{9}< 12\cdot \frac{1}{(\rho ^0)^2}<\frac{640}{3(\rho ^0)^3}. \end{aligned}$$

  Thus, we have

  $$\begin{aligned} \frac{2p}{(2p - 1)^2} D_2(\Omega )<\frac{640}{3}\frac{D_2(\Omega )}{(\rho ^0)^3}= \kappa _0. \end{aligned}$$

• (Verification of (A.2)) We again use (A.12) to get

  $$\begin{aligned} (1-2\alpha _0) 2p = (1-2\alpha _0)\frac{2n_0}{N}=(1-\frac{1}{2N})\cdot \frac{N+1}{N}=1+\frac{1}{2N}\left( 1-\frac{1}{N}\right) >1. \end{aligned}$$

• (Verification of (A.3)) We use (A.12) to get

  $$\begin{aligned} p = \frac{n_0}{N}= \frac{1}{2}+\frac{1}{2N}> \frac{1}{2}+\frac{\rho ^0}{6}. \end{aligned}$$

• (Verification of (A.4)) Note that

  $$\begin{aligned} 0< \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}=4N^2 \cdot \frac{3(\rho ^0)^2}{320}=\frac{3N^2(\rho ^0)^2}{80}<\frac{27}{80}<1, \quad \text{ i.e., }\\ \left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}\right\rfloor = 0. \end{aligned}$$

  This yields

  $$\begin{aligned}&\frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}-\frac{1}{N}\left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}\right\rfloor -\frac{n_0-1}{N} \\&\quad \ge \frac{1-\frac{1}{4N}}{2-\frac{1}{4N}}-0-\frac{N-1}{2N} =\frac{4N-1}{8N-1}-\frac{N-1}{2N}=\frac{7N-1}{2N(8N-1)} >0, \end{aligned}$$

  i.e.,

  $$\begin{aligned} \frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}-\frac{1}{N}\left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}\right\rfloor>\frac{n_0-1}{N}>0. \end{aligned}$$

• Case D $(m_0\ge 2)$: By definition of $n_0$ in (A.11), we have

  $$\begin{aligned} \begin{aligned}&n_0=\left\lfloor \frac{m_0}{2}\right\rfloor +\frac{N+1}{2} > \frac{m_0}{2} -1 + \frac{N+1}{2} = \frac{m_0-1}{2} + \frac{N}{2}, \\&n_0=\left\lfloor \frac{m_0}{2}\right\rfloor +\frac{N+1}{2} \le \frac{m_0}{2} + \frac{N+1}{2}. \end{aligned} \end{aligned}$$

  (A.13)

  These yield

  $$\begin{aligned} 2p = \frac{2n_0}{N}>\frac{m_0-1}{N}+1 \quad \text{ and } \quad 2p < \frac{m_0 + 1}{N} + 1. \end{aligned}$$

  (A.14)

• (Verification of (A.1)) It can easily be seen that $x\mapsto \frac{2x}{(2x-1)^2}=\frac{1}{2x-1}\cdot (\frac{1}{2x-1}+1)$ is decreasing for $x>\frac{1}{2}$. Thus we can use (A.14)$_1$ to see We use (A.10) and (A.14) to see

  $$\begin{aligned} \begin{aligned} \frac{2p}{\left( 2p-1\right) ^2}&<\frac{\frac{m_0-1}{N}+1}{(\frac{m_0-1}{N})^2}<\frac{N(m_0 - 1+N)}{(m_0-1)^2}< \frac{N(m_0 - 1+N)}{(m_0-1)^2}\cdot \underbrace{ \left( \frac{2m_0+1}{N}\cdot \frac{1}{\rho ^0}\right) ^2 }_{> 1} \\&\le \frac{2N^2}{(m_0-1)^2}\cdot \left( \frac{2(m_0-1)+3}{N}\cdot \frac{1}{\rho ^0}\right) ^2 \\&= 2\cdot \left( \frac{2(m_0-1)+3}{m_0-1}\right) ^2\cdot \frac{1}{(\rho ^0)^2}\\&\le 2\cdot 5^2\cdot \frac{1}{(\rho ^0)^2}<\frac{640}{3(\rho ^0)^3}, \end{aligned} \end{aligned}$$

  where in the third inequality, we used (A.10) to find the relation:

  $$\begin{aligned} \frac{2m_0-1}{N}\le \rho ^0\le 1\quad \Rightarrow \quad m_0-1<2m_0-1\le N. \end{aligned}$$

  Thus, we have

  $$\begin{aligned} \frac{2p}{\left( 2p-1\right) ^2}D_2(\Omega )<\frac{640}{3}\frac{D_2(\Omega )}{(\rho ^0)^3}=\kappa _0. \end{aligned}$$

• (Verification of (A.2)) We use $\alpha _0=\frac{m_0}{8N}$ and (A.14) to obtain

  $$\begin{aligned} \begin{aligned} (1-2\alpha _0) 2p&=(1-2\alpha _0)\frac{2n_0}{N}> (1 - \frac{m_0}{4N})(1+\frac{m_0-1}{N})=1+\frac{3m_0-1}{4N}-\frac{m_0(m_0-1)}{4N^2}\\&\ge 1+\frac{m_0}{2N}-\frac{m_0(m_0-1)}{4N^2}=1+\frac{m_0}{4N}\left( 2-\frac{m_0-1}{N}\right) >1, \end{aligned} \end{aligned}$$

  where in the last inequality, we also used the following relation from (A.14):

  $$\begin{aligned} 2 \ge 2p> \frac{m_0-1}{N}+1 \quad \Longrightarrow \quad N > m_0 - 1. \end{aligned}$$

  Hence, we have

  $$\begin{aligned} 1-2\alpha _0>\frac{1}{2p}. \end{aligned}$$

• (Verification of (A.3)) We use $\lfloor \frac{m_0}{2}\rfloor \ge \frac{m_0-1}{2}$ and (A.11) to find

  $$\begin{aligned} n_0\ge \frac{m_0+N}{2}. \end{aligned}$$

  This yields

  $$\begin{aligned} p = \frac{n_0}{N}\ge \frac{1}{2}+\frac{m_0}{2N}=\frac{1}{2}+\frac{2m_0+1}{N}\cdot \frac{m_0}{2(2m_0+1)}> \frac{1}{2}+\rho ^0\cdot \frac{2}{2\cdot 5}>\frac{1}{2}+\frac{\rho ^0}{6}. \end{aligned}$$

• (Verification of (A.4)) We use $\alpha _0=\frac{m_0}{8N}$ and $\kappa _0=\frac{640}{3} \frac{D_2(\Omega )}{(\rho ^0)^3}$ to obtain

  $$\begin{aligned} \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}=\frac{8N^2}{m_0}\cdot \frac{3(\rho ^0)^2}{320}=\frac{3N^2(\rho ^0)^2}{40m_0}< \frac{3(2m_0+1)^2}{40m_0}. \end{aligned}$$

  (A.15)

  On the other hand, we have

  $$\begin{aligned} \begin{aligned} \frac{\rho ^0+(1-\alpha _0)}{1+(1-\alpha _0)}-\frac{n_0-1}{N}&\ge \frac{\frac{2m_0-1}{N}+1-\frac{m_0}{8N}}{1+1-\frac{m_0}{8N}}-\frac{m_0-1+N}{2N}\\&=\frac{8N+15m_0-8}{16N-m_0}-\frac{m_0-1+N}{2N}\\&=\frac{m_0(15N+m_0-1)}{2N(16N-m_0)}. \end{aligned} \end{aligned}$$

  (A.16)

  Thus, it follows from (A.15), (A.16) and $m_0 \ge 2$ that we have

  $$\begin{aligned} \begin{aligned} \frac{\rho _0+(1-\alpha _0)}{1+(1-\alpha _0)}&-\frac{n_0-1}{N}-\frac{1}{N}\left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{\kappa _0\rho ^0}\right\rfloor \\&>\frac{m_0(15N+m_0-1)}{2N(16N-m_0)}-\frac{3(2m_0+1)^2}{40Nm_0}\\&=\frac{12N(9m_0^2-16m_0-4)+32m_0^3-8m_0^2+3m_0}{40m_0N(16N-m_0)}\\&=\frac{12N(m_0-2)(9m_0+2)+8m_0^2(4m_0-1)+3m_0}{40m_0N(16N-m_0)}\\&>0. \end{aligned} \end{aligned}$$

  (A.17)

  This yields

  $$\begin{aligned} \frac{\rho _0+(1-\alpha _0)}{1+(1-\alpha _0)}-\frac{1}{N}\left\lfloor \frac{N}{\alpha _0}\cdot \frac{2D_2(\Omega )}{K_0\rho _0}\right\rfloor >\frac{n_0-1}{N}. \end{aligned}$$

  The role of the coefficient $\frac{640}{3}$ appears in the penultimate term in (A.17), and the reason for this choice is the same as Case B.

Appendix B: Linear Instability of the Bipolar State

In this section, we discuss the linear stability of the bipolar state for the Lohe model with identical oscillators:

$$\begin{aligned} {\dot{x}}_i=\Omega x_i +\frac{\kappa }{N}\sum _{k=1}^N\left( x_k-\frac{\langle x_i,x_k\rangle }{\langle x_i,x_i\rangle }x_i\right) ,\quad i=1,\ldots ,N. \end{aligned}$$

(B.1)

By replacing the variables $x_i$ by $e^{-\Omega t}x_i$, we may assume $\Omega =0$, so that

$$\begin{aligned} {\dot{x}}_i=\frac{\kappa }{N}\sum _{k=1}^N\left( x_k-\frac{\langle x_i,x_k\rangle }{\langle x_i,x_i\rangle }x_i\right) ,\quad i=1,\ldots ,N. \end{aligned}$$

(B.2)

By the O(d)-symmetry of the above model, i.e., by replacing $y_i$ by $Ay_i$ for some $A\in O(d)$, and by permuting the indices, we may assume the bipolar state $(y_1,\ldots ,y_N)$ in question consists of the following:

$$\begin{aligned} y_i={\left\{ \begin{array}{ll} (0,\ldots ,0,1), &{} 1\le i\le r,\\ (0,\ldots ,0,-1), &{} r<i\le N, \end{array}\right. } \end{aligned}$$

where $r=0,\ldots ,N$ describes the number of oscillators on the north pole. We consider a perturbation $\{x_i\}$ of $\{y_i\}$:

$$\begin{aligned} x_i={\left\{ \begin{array}{ll} (x_i^1,\ldots ,x_i^d,\sqrt{1-(x_i^1)^2-\ldots -(x_i^d)^2}), &{} 1\le i\le r,\\ (x_i^1,\ldots ,x_i^d,-\sqrt{1-(x_i^1)^2\ldots -(x_i^d)^2}), &{} r<i\le N, \end{array}\right. } \end{aligned}$$

where $|x_i^a|\ll 1/\sqrt{d}$ for $i=1,\ldots ,N$, $a=0,\ldots ,d-1$. Discarding quadratic terms, we can see that

$$\begin{aligned} \langle x_i,x_j\rangle = {\left\{ \begin{array}{ll} 1+O(\Vert {\mathbf {x}}\Vert ^2), &{} 1\le i,j\le r \text { or } r<i,j\le N,\\ -1+O(\Vert {\mathbf {x}}\Vert ^2), &{} 1\le i\le r<j\le N \text { or }1\le j\le r<i\le N, \end{array}\right. } \end{aligned}$$

where $\Vert {\mathbf {x}}\Vert ^2:=\sum _{i=1}^N\sum _{a=0}^{d-1}(x_i^a)^2$. Hence, if we project (B.2) onto the first d variables, we obtain the system of equations

$$\begin{aligned} {\dot{x}}_i^a= & {} \frac{\kappa }{N}\sum _{k=1}^N x_k^a-\frac{\kappa (2r-N)}{N}x_i^a+O(\Vert {\mathbf {x}}\Vert ^3),\quad i=1,\ldots ,r, \quad a=0,\ldots ,d-1. \end{aligned}$$

(B.3)

$$\begin{aligned} {\dot{x}}_i^a= & {} \frac{\kappa }{N}\sum _{k=1}^N x_k^a-\frac{\kappa (N-2r)}{N}x_i^a+O(\Vert {\mathbf {x}}\Vert ^3),\quad i=r+1,\ldots ,N, \quad a=0,\ldots ,d-1.\nonumber \\ \end{aligned}$$

(B.4)

Note that we do not need the $(d+1)$th component, since it is determined by the first d components. Linearizing (B.3), (B.4) yields

$$\begin{aligned} {\dot{z}}_i^a= & {} \frac{\kappa }{N}\sum _{k=1}^N z_k^a-\frac{\kappa (2r-N)}{N}z_i^a,\quad i=1,\ldots ,r, \quad a=0,\ldots ,d-1. \end{aligned}$$

(B.5)

$$\begin{aligned} {\dot{z}}_i^a= & {} \frac{\kappa }{N}\sum _{k=1}^N z_k^a-\frac{\kappa (N-2r)}{N}z_i^a,\quad i=r+1,\ldots ,N, \quad a=0,\ldots ,d-1. \end{aligned}$$

(B.6)

Then our result on linear stability is as follows:

Theorem B.1

Let (B.5), (B.6) be the linear approximation to (2.7) at the bipolar state $(y_1,\ldots ,y_N)$.

• (1) When $r=0$ or $r=N$, i.e., near the completely synchronized state, the linear system (B.5), (B.6) is diagonalizable with eigenvalue 0 with multiplicity d, and $-1$ with multiplicity $d(N-1)$. Hence the linear stability is indeterminate.

• (2) When $1\le r\le N-1$, i.e., near a bipolar state that is not the completely synchronized state, the linear system (B.5), (B.6) is diagonalizable with eigenvalue 0 with multiplicity d, 1 with multiplicity d, $\frac{2r-N}{N}$ with multiplicity $d(N-r-1)$, $\frac{N-2r}{N}$ with multiplicity $d(r-1)$; hence it is linearly unstable.

Remark B.1

In both cases, 0 is an eigenvalue with multiplicity at least d. This follows from the fact that (2.7) possesses an O(d)-symmetry, and from the fact that the orbit of the corresponding action of O(d) on the bipolar states in $(S^d)^N$ has dimension d.

Proof

This system is just the linear ODE

$$\begin{aligned} {\mathbf {z}}=\kappa \Lambda {\mathbf {z}}, \end{aligned}$$

where

$$\begin{aligned} {\mathbf {z}}^T=(z_1^0,\ldots ,z_N^0,\ldots ,z_1^{d-1},\ldots ,z_N^{d-1}), \end{aligned}$$

and $\Lambda$ is the block matrix

$$\begin{aligned} \Lambda =diag(\underbrace{\Lambda _0,\ldots ,\Lambda _0}_\text {d copies}). \end{aligned}$$

Here, $\Lambda _0$ is defined as

$$\begin{aligned} \Lambda _0:=\frac{1}{N}A-\frac{N-2r}{N}I_N-2\frac{2r-N}{N}B \end{aligned}$$

where A is the $N\times N$-matrix all of whose entries are 1, and B is the $N\times N$-matrix in block form

$$\begin{aligned} B=\begin{pmatrix} I_r &{}0\\ 0&{}0 \end{pmatrix} \end{aligned}$$

where $I_r$ is the $r\times r$ unit matrix, and the 0’s represent zero matrices of the appropriate dimension. According to Corollary C.1 in the next section, $\Lambda$ is diagonalizable, with eigenvalues

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\text { with multiplicity } d, ~-1 \text { with multiplicity }d(N-1) &{} \text {if}\;r=0,N,\\ 0\text { with multiplicity } d, ~1 \text { with multiplicity }d, &{} \\ \frac{2r-N}{N}\text { with multiplicity } d(N-r-1), ~\frac{N-2r}{N} \text { with multiplicity }d(r-1)&{}\text {if}\;1\le r\le N-1. \end{array}\right. } \end{aligned}$$

$\square$

Appendix C: Calculation of Some Eigenvalues

Lemma C.1

Let k be a field of characteristic zero. Let $N\in \mathbb N$ , $m\in \mathbb Z$ such that $0\le r\le N$. Let A be the $N\times N$-matrix all of whose entries are $1\in k$, and let B be the $N\times N$-matrix in the block form

$$\begin{aligned} B=\begin{pmatrix} I_r &{}0\\ 0&{}0 \end{pmatrix} \end{aligned}$$

where $I_r$ is the $r\times r$ unit matrix over k, and the 0’s represent zero matrices of the appropriate dimension. Then for any $\mu \in k$, the matrix $A-\mu B$ has characteristic polynomial

$$\begin{aligned} \phi (t)= {\left\{ \begin{array}{ll} t^{N-1}(t-N)&{} \text {if}\; r=0,\\ t^{N-r-1}(t+\mu )^{r-1}(t^2+(\mu -N)t+\mu (r-N))&{} \text {if}\; 1\le r\le N-1,\\ (t+\mu )^{N-1}(t+(\mu -N))&{} \text {if}\; r=N, \end{array}\right. } \end{aligned}$$

(C.1)

over k.

Proof

• Case A $(r=0)$: Then $B=0$, so we are searching for the characteristic polynomial of A. If $N=1$, then it is immediate that the characteristic polynomial is $\phi (t)=t-1$, as asserted above. So we assume $N\ge 2$. Then it is easy to verify that the relation

  $$\begin{aligned} A^2-NA=0, \end{aligned}$$

  holds,as well as the relations

  $$\begin{aligned} A\ne 0,\quad A-NI_N\ne 0. \end{aligned}$$

  Hence the minimal polynomial m(t) of A is $m(t)=t^2-Nt=t(t-N)$, and since the minimal polynomial and the characteristic polynomial must share the same irreducible factors, we conclude that $\phi (t)$ must be of the form

  $$\begin{aligned} \phi (t)=t^{N-a}(t-N)^a, \end{aligned}$$

  where $a\in \mathbb Z$ is an integer $1\le a\le N-1$. This allows us to calculate, using the binomial theorem,

  $$\begin{aligned} \text {(coefficient of } t^{N-1}\text { in }\phi (t))=-aN, \end{aligned}$$

  but we can also directly calculate

  $$\begin{aligned} \text {(coefficient of } t^{N-1}\text { in }\phi (t))=-\text{ tr }(A)=-N. \end{aligned}$$

  Hence we must have $a=1$, i.e., $\phi (t)=t^{N-1}(t-N)$, as asserted.

• Case B $(r=N)$: Then $B=I_N$, the $N \times N$ identity matrix, so

  $$\begin{aligned} \phi (t)=det(tI_N-(A-\mu B))=det((t+\mu )I_N-A)=(t+\mu )^{N-1}(t+\mu -N), \end{aligned}$$

  by using the result of Case A. This agrees with our assertion (C.1).

• Case C $(1\le r\le N-1)$: Without loss of generality, we may consider $\mu$ to be a variable over k, i.e., that the field of rational functions $k(\mu )$ is a transcendental extension of k, and consider $A-\mu B$ as a matrix over $k(\mu )$. Then we must verify the relation (C.1) over the polynomial ring $k(\mu )[t]$. First, it is easy to verify the relations

  $$\begin{aligned} A^2=NA,\quad B^2=B,\quad ABA=rA. \end{aligned}$$

  (C.2)

  Using these relations, we proceed to verify the following by direct expansion:

  $$\begin{aligned} (A-\mu B)^2&=NA+\mu ^2 B-\mu AB-\mu BA,\\ (A-\mu B)^3&=(N^2-\mu r)A-\mu ^3 B+(\mu ^2-\mu N)AB+(\mu ^2-\mu N)BA+\mu ^2 BAB,\\ (A-\mu B)^4&=(N^3-2N\mu r+r\mu ^2)A+\mu ^4 B+(-\mu ^3+\mu ^2N-\mu N^2+\mu ^2 r)AB\\&\quad +(-\mu ^3+\mu ^2N-\mu N^2+\mu ^2 r)BA+(-2\mu ^3+\mu ^2 N)BAB. \end{aligned}$$

  This permits us to verify the relation

  $$\begin{aligned}&(A-\mu B)^4+(2\mu -N)(A-\mu B)^3\\&\quad +\,(\mu ^2-2\mu N+\mu r)(A-\mu B)^2-\mu ^2(N-r)(A-\mu B)=0. \end{aligned}$$

  (The rationale for the preceding computation is that, the relations (C.2) force the powers of $A-\mu B$ to be linear combinations of the five terms $A, B, AB, BA, \textit{BAB}$ only, and hence that five powers of $A-\mu B$ suffice to give a linear relation. Luckily, we needed only four.) Hence the minimal polynomial m(t) of $A-\mu B$ over $k(\mu )$ is a divisor of

  $$\begin{aligned}&t^4+(2\mu -N)t^3+(\mu ^2-2\mu N+\mu r)t^2-\mu ^2(N-r)t \\&=t(t+\mu )(t^2+(\mu -N)t+\mu (r-N)). \end{aligned}$$

  But $t^2+(\mu -N)t+\mu (r-N)$ is irreducible over $k(\mu )$ if and only if it factors into linear factors, that is to say, if its discriminant (as a quadratic polynomial in t)

  $$\begin{aligned} (\mu -N)^2-4\mu (r-N)=\mu ^2-2(2r-N)\mu +N^2 \end{aligned}$$

  is a square in $k(\mu )$, which holds if and only if its discriminant (as a quadratic polynomial in $\mu$)

  $$\begin{aligned} 4(2r-N)^2-4N^2=16r(r-N)=0 \end{aligned}$$

  is zero in k. Since k has characteristic zero, this is equivalent to

  $$\begin{aligned} r=0~ \text {or}~ r=N\quad \text {in } \mathbb Z. \end{aligned}$$

  These cases were already excluded. Hence $t^2+(\mu -N)t+\mu (r-N)$ is irreducible over $k(\mu )$. Since m(t) and $\phi (t)$ share the same irreducible factors, we conclude that

  $$\begin{aligned} \phi (t)=t^{N-a-2b}(t+\mu )^a(t^2+(\mu -N)t+\mu (r-N))^b, \end{aligned}$$

  where a, b are nonnegative integers such that $a+2b\le N$. As in Case A, we use the binomial theorem to calculate

  $$\begin{aligned} \text {(coefficient of } t^{N-1}\text { in }\phi (t))=a\mu +b(\mu -N), \end{aligned}$$

  and compare this with

  $$\begin{aligned} \text {(coefficient of } t^{N-1}\text { in }\phi (t))=-\text{ tr }(A)=r\mu -N. \end{aligned}$$

  Thus

  $$\begin{aligned} a\mu +b(\mu -N)=r\mu -N \end{aligned}$$

  holds in $k(\mu )$, and since k has characteristic zero, we must have

  $$\begin{aligned} a=r-1,~b=1 \quad \text {in } \mathbb Z, \end{aligned}$$

  i.e.,

  $$\begin{aligned} \phi (t)=t^{N-r-1}(t+\mu )^{r-1}(t^2+(\mu -N)t+\mu (r-N)), \end{aligned}$$

  as asserted. $\square$

Remark C.1

We will only be interested in the case $\mu =4r-2N$. The reason for introducing $\mu$ as a variable is that, in the case $1\le r\le N-1$, the calculation of the characteristic polynomial becomes easier once we consider $\mu$ as a transcendental element over k.

Corollary C.1

The matrix

$$\begin{aligned} \Lambda _0:=\frac{1}{N}A-\frac{N-2r}{N}I_N-2\frac{2r-N}{N}B \end{aligned}$$

over $\mathbb R$ is diagonalizable, with eigenvalues

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\text { with multiplicity } 1, ~-1 \text { with multiplicity }N-1 &{} \text {if}\;r=0,N,\\ 0\text { with multiplicity } 1, ~1 \text { with multiplicity }1, &{} \\ \frac{2r-N}{N}\text { with multiplicity } N-r-1, ~\frac{N-2r}{N} \text { with multiplicity }r-1&{}\text {if}\;1\le r\le N-1. \end{array}\right. } \end{aligned}$$

Proof

Diagonalizability follows from the symmetry of $\Lambda _0$ and the spectral theorem for real symmetric matrices. Let $\mu =2(2r-N)$. Then we can see that

$$\begin{aligned} t^2+(\mu -N)t+\mu (r-N)&=t^2+(4r-3N)t+2(2r-N)(r-N)\\&=(t+2r-N)(t+2r-2N), \end{aligned}$$

so (C.1) becomes

$$\begin{aligned} \phi (t)= {\left\{ \begin{array}{ll} t^{N-1}(t-N)&{} \text {if}\; r=0,\\ t^{N-r-1}(t+4r-2N)^{r-1}(t+2r-N)(t+2r-2N)&{} \text {if}\; 1\le r\le N-1,\\ (t+2N)^{N-1}(t+N)&{} \text {if}\; r=N, \end{array}\right. } \end{aligned}$$

(C.3)

(we used $\mu =2N$ when $r=N$). Thus, the desired characteristic polynomial is

$$\begin{aligned} det&\left[ tI_N-\left( \frac{1}{N}A-\frac{N-2r}{N}I_N-2\frac{2r-N}{N}B\right) \right] \\&=\frac{1}{N^N}det\left[ NtI_N-\left( A-(N-2r)I_N-2(2r-N)B\right) \right] \\&=\frac{1}{N^N}det\left[ (Nt+N-2r)I_N-\left( A-\mu B\right) \right] \\&=\frac{1}{N^N}\phi (Nt+N-2r)\\&={\left\{ \begin{array}{ll} \frac{1}{N^N}(Nt+N-2\cdot 0)^{N-1}(Nt+N-2\cdot 0-N)&{} \text {if}\; r=0,\\ \frac{1}{N^N}(Nt+N-2r)^{N-r-1}((Nt+N-2r)+(4r-2N))^{r-1}&{}\\ \cdot ((Nt+N-2r)+2r-N)((Nt+N-2r)+(2r-2N))&{} \text {if}\; 1\le r\le N-1,\\ \frac{1}{N^N}(Nt+N-2\cdot N+2N)^{N-1}((Nt+N-2\cdot N)+N)&{} \text { if } r=N, \end{array}\right. }\\&={\left\{ \begin{array}{ll} t(t+1)^{N-1}&{} \text {if}\; r=0,\\ t(t-1)(t+\frac{N-2r}{N})^{N-r-1}((t+\frac{2r-N}{N}))^{r-1}&{} \text {if}\; 1\le r\le N-1,\\ t(t+1)^{N-1}&{} \text {if}\; r=N. \end{array}\right. } \end{aligned}$$

This completes the proof.

(The diagonalizability of $\Lambda _0$, which is equivalent to the diagonalizability of $A-\mu B$, can also be shown as follows. If $r=0$ or N, we use the diagonalizability of A. If $1\le r\le N-1$, we use the fact that the minimal polynomial of $A-\mu B$ divides $t(t+\mu )(t^2+(\mu -N)t+\mu (r-N))=t(t+4r-2N)(t+2r-N)(t+2r-2N)$, which consists of distinct linear factors, and hence the minimal polynomial of $A-\mu B$ must factor into distinct linear factors.) $\square$