Emergent Dynamics of a Generalized Lohe Model on Some Class of Lie Groups

Abstract

We introduce a Lohe group which is a new class of matrix Lie groups and present a continuous dynamical system for the synchronization of group elements in a Lohe group. The Lohe group includes classical Lie groups such as the orthogonal, unitary, and symplectic groups, and since Lohe groups need not be compact, global existence of ODEs may fail. The proposed dynamical system generalizes the Lohe model (Lohe in J Phys A 43:465301, 2010; Lohe in J Phys A 42:395101–395126, 2009) itself a nonabelian generalization of the Kuramoto model, and alongside we also generalize the analytical framework (Ha and Ryoo in J Stat Phys 163:411–439, 2016) of emergent and unique phase-locked states. For the construction of the phase-locked states, we introduce Lyapunov functions measuring the ensemble diameter and the dissimilarity between two Lohe flows, and derive Gronwall-type differential inequalities for them. The global existence of solutions then become a consequence of the boundedness of these Lyapunov functions. Our sufficient framework for the emergent dynamics is formulated in terms of coupling strength and initial states, and it leads to the global existence of solutions and the formation and uniqueness of a phase-locked asymptotic state. As a concrete example, we demonstrate how our theory can show emergent phenomenon on the Heisenberg group, where all initial configurations tend to a unique phase-locked state exponentially fast.

Appendices

Appendix 1: Proof of Proposition 2.4

Lemma 7.1

Let ${\mathcal X} = (X_1, \ldots , X_N)$ be a solution to (1.3). Then, we have

$$\begin{aligned} \frac{d}{dt}X_i X_j^{-1}&=~H_i X_i X_j^{-1}-X_i X_j^{-1} H_j\\&\quad +\frac{K}{2N}\sum _{k=1}^N\left[ X_k X_j^{-1} - X_i X_k^{-1} X_i X_j^{-1}-X_iX_j^{-1}X_k X_j^{-1} + X_i X_k^{-1}\right] . \end{aligned}$$

Proof

Recall that $X_i$ satisfies

$$\begin{aligned} \dot{X}_i X_i^{-1}=H_i+\frac{K}{2N}\sum _{k=1}^N\left[ X_k X_i^{-1} - X_i X_k^{-1}\right] , \end{aligned}$$

(7.1)

Then, we use

$$\begin{aligned} \dot{X}_i X_i^{-1} +X_i \dot{X}_i^{-1}=\frac{d}{dt}X_i X_i^{-1}=\frac{d}{dt}I_d=0 \end{aligned}$$

to obtain

$$\begin{aligned} X_i \dot{X}_i^{-1}=-H_i+\frac{K}{2N}\sum _{k=1}^N\left[ -X_k X_i^{-1} + X_i X_k^{-1}\right] . \end{aligned}$$

(7.2)

Hence, the relations (7.1) and (7.2) yield

$$\begin{aligned}&\dot{X}_i =H_i X_i+\frac{K}{2N}\sum _{k=1}^N\left[ X_k - X_i X_k^{-1} X_i\right] , \nonumber \\&\dot{X}_i^{-1}=-X_i^{-1} H_i+\frac{K}{2N}\sum _{k=1}^N\left[ -X_i^{-1}X_k X_i^{-1} + X_k^{-1}\right] . \end{aligned}$$

(7.3)

Then, we use (7.3) to obtain the desired estimate:

$$\begin{aligned} \frac{d}{dt}X_i X_j^{-1}&=~\dot{X}_i X_j^{-1}+X_i \dot{X}_j^{-1}\\&=~H_i X_i X_j^{-1}+\frac{K}{2N}\sum _{k=1}^N\left[ X_k X_j^{-1} - X_i X_k^{-1} X_i X_j^{-1}\right] \\&\quad -X_i X_j^{-1} H_j+\frac{K}{2N}\sum _{k=1}^N\left[ -X_iX_j^{-1}X_k X_j^{-1} + X_i X_k^{-1}\right] \\&=~H_i X_i X_j^{-1}-X_i X_j^{-1} H_j\\&\quad +\frac{K}{2N}\sum _{k=1}^N\left[ X_k X_j^{-1} - X_i X_k^{-1} X_i X_j^{-1}-X_iX_j^{-1}X_k X_j^{-1} + X_i X_k^{-1}\right] . \end{aligned}$$

$\square$

We now present the proof of Proposition 2.4. For this, we split its proof into several steps.

$\bullet$ Step A: We claim that solutions of the form (2.5):

$$\begin{aligned} X_i(t)=X_i^\infty e^{\Lambda t},\quad i=1,\ldots ,N, \end{aligned}$$

where $X_i^\infty \in G$ and $\Lambda \in \mathfrak {g}$ satisfy (2.6):

$$\begin{aligned} X_i^\infty \Lambda (X_i^\infty )^{-1}=H_i+\frac{K}{2N}\sum _{k=1}^N\left[ X_k^\infty (X_i^\infty )^{-1}-X_i^\infty (X_k^\infty )^{-1}\right] ,\quad i=1,\ldots ,N, \end{aligned}$$

are phase-locked states of (1.3).

Proof of claim

We multiply $X_i^{\infty }(X_j^{\infty })^{-1}$ on the right of (2.6) for i to obtain

$$\begin{aligned} X_i^\infty \Lambda (X_j^\infty )^{-1}=H_i X_i^{\infty } (X_j^{\infty })^{-1}+\frac{K}{2N}\sum _{k=1}^N\left[ X_k^{\infty } (X_j^{\infty })^{-1} - X_i^{\infty } (X_k^{\infty })^{-1} X_i^{\infty } (X_j^{\infty })^{-1}\right] . \end{aligned}$$

(7.4)

We again multiply $X_i^{\infty }(X_j^{\infty })^{-1}$ on the left of (2.6) for j to obtain

$$\begin{aligned} X_i^\infty \Lambda (X_j^\infty )^{-1}=X_i^{\infty } (X_j^{\infty })^{-1} H_j+\frac{K}{2N}\sum _{k=1}^N\left[ X_i^{\infty }(X_j^{\infty })^{-1}X_k^{\infty } (X_j^{\infty })^{-1} - X_i^{\infty } (X_k^{\infty })^{-1}\right] . \end{aligned}$$

(7.5)

Subtraction (7.5) from (7.4) gives

$$\begin{aligned} 0&=~H_i X_i^{\infty } (X_j^{\infty })^{-1}-X_i^{\infty } (X_j^{\infty })^{-1} H_j\nonumber \\&\quad +\frac{K}{2N}\sum _{k=1}^N\left[ X_k^{\infty } (X_j^{\infty })^{-1} - X_i^{\infty } (X_k^{\infty })^{-1} X_i^{\infty } (X_j^{\infty })^{-1}\right. \nonumber \\&\left. \quad -X_i^{\infty }(X_j^{\infty })^{-1}X_k^{\infty } (X_j^{\infty })^{-1} + X_i^{\infty } (X_k^{\infty })^{-1}\right] \end{aligned}$$

(7.6)

Thus, it follows from Lemma 7.1 that we have

$$\begin{aligned} \frac{d}{dt}X_i X_j^{-1}&=~H_i X_i X_j^{-1}-X_i X_j^{-1} H_j\\&\quad +\frac{K}{2N}\sum _{k=1}^N\left[ X_k X_j^{-1} - X_i X_k^{-1} X_i X_j^{-1}-X_iX_j^{-1}X_k X_j^{-1} + X_i X_k^{-1}\right] \\&=~H_i X_i^{\infty } (X_j^{\infty })^{-1}-X_i^{\infty } (X_j^{\infty })^{-1} H_j\\&\quad +\frac{K}{2N}\sum _{k=1}^N\left[ X_k^{\infty } (X_j^{\infty })^{-1} - X_i^{\infty } (X_k^{\infty })^{-1} X_i^{\infty } (X_j^{\infty })^{-1}\right. \\&\left. \quad -X_i^{\infty }(X_j^{\infty })^{-1}X_k^{\infty } (X_j^{\infty })^{-1} + X_i^{\infty } (X_k^{\infty })^{-1}\right] \\&=~0. \end{aligned}$$

We used (7.6) at the final equation. This tells us that $X_i X_j^{-1}$ is constant, so ${\mathcal X}$ is a phase-locked state by definition. $\square$

• Step B: We claim that if a solution ${\mathcal X}$ to (2.4) exhibits asymptotic entrainment, then there is a phase-locked state ${\mathcal Y}$ of the form (2.5), (2.6) so that

  $$\begin{aligned} \lim _{t\rightarrow \infty }X_i(t)X_j^{-1}(t)=Y_iY_j^{-1},\quad i,j=1,\ldots ,N. \end{aligned}$$

Proof

We set

$$\begin{aligned} V^{\infty }_i := \lim _{t\rightarrow \infty }(X_iX_1^{-1})(t), \quad i =1, \ldots , N. \end{aligned}$$

Then, by definition of asymptotic entrainment, we have

$$\begin{aligned} V_i^\infty \in G. \end{aligned}$$

On the other hand, we recall Lemma 7.1:

$$\begin{aligned} \frac{d}{dt}X_i X_j^{-1}= & {} H_i X_i X_j^{-1}-X_i X_j^{-1} H_j\nonumber \\&+\frac{K}{2N}\sum _{k=1}^N\left[ X_k X_j^{-1} - X_i X_k^{-1} X_i X_j^{-1}-X_iX_j^{-1}X_k X_j^{-1} + X_i X_k^{-1}\right] .\qquad \end{aligned}$$

(7.7)

Since ${\mathcal X}$ exhibits asymptotic entrainment, the right-hand side of (7.7) attains a limit value, hence the left-hand side $\frac{d}{dt}X_i X_j^{-1}$ must attain a limit value. On the other hand, $X_i X_j^{-1}$ itself also converges. Thus, the limit value of $\frac{d}{dt}X_i X_j^{-1}$ must be 0.

On the other hand, we observe that $X_i X_j^{-1}=(X_iX_1^{-1})(X_jX_1^{-1})$ converges to $V_iV_j^{-1}$. Thus, (7.7) implies

$$\begin{aligned} 0= & {} H_i V^{\infty }_i (V_j^{\infty })^{-1} - V^{\infty }_i (V_j^{\infty })^{-1} H_j \nonumber \\&+\frac{K}{2N}\sum _{k=1}^{N} \left[ V^{\infty }_k(V_j^{\infty })^{-1}- V^{\infty }_i (V_k^{\infty })^{-1} V^{\infty }_i (V_j^{\infty })^{-1} +V^{\infty }_i (V_k^{\infty })^{-1} \right. \nonumber \\&-\left. V^{\infty }_i (V_j^{\infty })^{-1} V^{\infty }_k (V_j^{\infty })^{-1} \right] . \end{aligned}$$

(7.8)

Left-multiplying by $(V_i^{\infty })^{-1}$, right-multiplying by $V^{\infty }_j$, and rearranging terms in (7.8) yield

$$\begin{aligned}&(V_i^{\infty })^{-1} H_i V^{\infty }_i +\frac{K}{2N}\sum _{k = 1}^{N} \left[ (V_i^{\infty })^{-1} V^{\infty }_k- (V_k^{\infty })^{-1} V^{\infty }_i\right] \\&\quad = (V_j^{\infty })^{-1} H_j V^{\infty }_j +\frac{K}{2N}\sum _{k = 1}^{N} \left[ (V_j^{\infty })^{-1} V^{\infty }_k- (V_k^{\infty })^{-1} V^{\infty }_j\right] , \quad 1 \le i, j \le N. \end{aligned}$$

Note that the left-hand side depends only on the index i, and that the right-hand side depends only on the index j. Hence, we introduce an index-independent matrix $\Lambda$ that satisfies

$$\begin{aligned} \Lambda =(V_i^{\infty })^{-1} H_i V^{\infty }_i +\frac{K}{2N}\sum _{k = 1}^{N} \left[ (V_i^{\infty })^{-1} V^{\infty }_k- (V_k^{\infty })^{-1} V^{\infty }_i\right] , \quad 1 \le i, j \le N. \end{aligned}$$

Since $\mathfrak {g}$ is closed under conjugation by elements of G and (2.3) holds, we see that $\Lambda \in \mathfrak {g}$. Upon conjugating (2.6) by $V_i^\infty$, we obtain (2.6):

$$\begin{aligned} V_i^\infty \Lambda (V_i^\infty )^{-1}=H_i+\frac{K}{2N}\sum _{j=1}^N\left[ V_j^\infty (V_i^\infty )^{-1}-V_i^\infty (V_j^\infty )^{-1}\right] ,\quad i=1,\ldots ,N. \end{aligned}$$

By Step A, $\{Y_i\} :=\{V^{\infty }_i e^{-{\mathrm i} \Lambda t}\}$ is a phase-locked state for (2.4). Moreover, $Y_i$ satisfies

$$\begin{aligned} Y_i Y_j^{-1}=V^{\infty }_i (V_j^{\infty })^{-1}=\lim _{t\rightarrow \infty } (X_i X_j^{-1})(t), \end{aligned}$$

as claimed.

• Step C: We claim that if ${\mathcal Y}$ is a phase-locked state, then there is a phase-locked state ${\mathcal Z}$ of the form (2.5), (2.6) so that

  $$\begin{aligned} Y_iY_j^{-1}=Z_iZ_j^{-1},\quad i,j=1,\ldots ,N. \end{aligned}$$

$\square$

Proof of claim

If ${\mathcal Y}$ is a phase-locked state, then ${\mathcal Y}$ exhibits asymptotic entrainment. Thus, we are done by Step B.

• Step D: We claim that every phase-locked state ${\mathcal Y}$ must be of the form (2.5), (2.6). Also, part (ii) of Proposition 2.4 holds.$\square$

Proof of claim

By Step C, there exists a phase-locked state $\{Z_i=V_i^\infty e^{\Lambda t}\}$ of the form (2.5), (2.6), satisfying

$$\begin{aligned} Y_iY_j^{-1}=Z_iZ_j^{-1}=V_i^\infty (V_j^\infty )^{-1},\quad i,j=1,\ldots ,N. \end{aligned}$$

Then, we have

$$\begin{aligned} \dot{Y}_i Y_i^{-1}&=H_i+\frac{K}{2N}\sum _j\left[ Y_j Y_i^{-1} - Y_i Y_j^{-1}\right] \\&= H_i+\frac{K}{2N}\sum _j\left[ V_j^\infty (V_i^\infty )^{-1} - V_i^\infty (V_j^\infty )^{-1}\right] \\&= V_i^\infty \Lambda (V_i^\infty )^{-1}. \end{aligned}$$

Thus, we have

$$\begin{aligned} \frac{d}{dt}\left[ \exp \left( -V_i^\infty \Lambda (V_i^\infty )^{-1}t\right) Y_i\right] =&~\exp \left( -V_i^\infty \Lambda (V_i^\infty )^{-1}t\right) \left[ -V_i^\infty \Lambda (V_i^\infty )^{-1}Y_i+\dot{Y}_i\right] =0, \end{aligned}$$

so that $\exp \left( -V_i^\infty \Lambda (V_i^\infty )^{-1}t\right) Y_i$ is constant, maintaining its value at $t=0$:

$$\begin{aligned} \exp \left( -V_i^\infty \Lambda (V_i^\infty )^{-1}t\right) Y_i=Y_i^0. \end{aligned}$$

(7.9)

On the other hand, we have

$$\begin{aligned} Y_i^0(Y_j^0)^{-1}=V_i^\infty (V_j^\infty )^{-1}\quad \Longrightarrow \quad (V_i^\infty )^{-1}Y_i^0=(V_j^\infty )^{-1}Y_j^0. \end{aligned}$$

Therefore, there is an index-independent matrix $R\in G$ such that

$$\begin{aligned} R=(V_i^\infty )^{-1}Y_i^0,\quad i=1,\ldots ,N. \end{aligned}$$

We insert $Y_i^0=V_i^\infty R$ into (7.9):

$$\begin{aligned} Y_i =\exp \left( V_i^\infty \Lambda (V_i^\infty )^{-1}t\right) Y_i^0 =\exp \left( V_i^\infty \Lambda (V_i^\infty )^{-1}t\right) V_i^\infty R =V_i^\infty \exp (\Lambda t)R. \end{aligned}$$

This implies that (ii) of Proposition 2.4 holds. Manipulating the above equation further, we have

$$\begin{aligned} Y_i= V_i^\infty R\exp (R^{-1}\Lambda Rt). \end{aligned}$$

It is clear that $V_i^\infty R\in G$ and $R^{-1}\Lambda R\in \mathfrak {g}$ satisfy (2.6). Hence $\{Y_i\}$ is of the form (2.5), (2.6). $\square$

Appendix 2: Proof of Lemma 3.1

We first need a simple lemma that relates the behavior of a matrix-valued function to the dynamics of its norm.

Lemma 7.2

Let A(t) and B(t) be $d\times d$-matrix valued functions satisfying the following ODE:

$$\begin{aligned} \frac{d}{dt}A(t)=-KA(t)+B(t). \end{aligned}$$

Then, we have

$$\begin{aligned} \left| \frac{d}{dt}\Vert A\Vert +K\Vert A\Vert \right| \le \Vert B\Vert \end{aligned}$$

whenever $\Vert A\Vert >0$.

Proof

Note that

$$\begin{aligned} \frac{d}{dt}\Vert A\Vert ^2&=\frac{d}{dt} tr(AA^{\dagger }) =tr(\dot{A}A^{\dagger })+tr(A\dot{A}^{\dagger }) =2{\text {Re}} (tr(\dot{A}A^{\dagger }))\\&=2{\text {Re}} (tr(-KAA^{\dagger }+BA^{\dagger })) =-2K\Vert A\Vert ^2+2{\text {Re}} (tr(BA^{\dagger })) \end{aligned}$$

This yields

$$\begin{aligned} \left| \frac{d}{dt}\Vert A\Vert ^2+2K\Vert A\Vert ^2\right| \le 2|{\text {Re}} (tr(BA^{\dagger }))|\le 2\Vert B\Vert \Vert A\Vert . \end{aligned}$$

If $\Vert A\Vert >0$, we may divide both sides by $2\Vert A\Vert$ to obtain the statement. $\square$

We are now ready to present the proof of Lemma 3.1.

Proof of Lemma 3.1

Given a time t, choose indices i and j so that

$$\begin{aligned} D(X(t))=\Vert X_iX_j^{-1}-I_d\Vert . \end{aligned}$$

We approximate the term in the brackets in Lemma 7.1:

$$\begin{aligned}&X_k X_j^{-1} - X_i X_k^{-1} X_i X_j^{-1}-X_iX_j^{-1}X_k X_j^{-1} + X_i X_k^{-1}\\&\quad = -(X_i X_k^{-1}-I_d)\left( X_i X_j^{-1}-I_d\right) -\left( X_iX_j^{-1}-I_d\right) \left( X_k X_j^{-1}-I_d\right) -2\left( X_i X_j^{-1}-I_d\right) . \end{aligned}$$

We now use Lemma 7.1 to rewrite

$$\begin{aligned}&\frac{d}{dt}\left( X_i X_j^{-1}-I_d\right) =-K(X_i X_j^{-1}-I_d) +H_i-H_j+H_i\left( X_iX_j^{-1}-I_d\right) -\left( X_iX_j^{-1}-I_d\right) H_j\nonumber \\&\quad +\frac{K}{2N}\sum _{k=1}^N \left[ -\left( X_i X_k^{-1}-I_d)(X_i X_j^{-1}-I_d\right) -\left( X_iX_j^{-1}-I_d\right) \left( X_k X_j^{-1}-I_d\right) \right] .\nonumber \\ \end{aligned}$$

(7.10)

Finally, we use Lemma 7.2 to obtain

$$\begin{aligned}&\left| \frac{d}{dt}\Vert X_iX_j^{-1}-I_d\Vert +K\Vert X_iX_j^{-1}-I_d\Vert \right| \\&\quad \le \Vert H_i-H_j\Vert +\left\| H_i\left( X_iX_j^{-1}-I_d\right) \right\| +\left\| \left( X_iX_j^{-1}-I_d\right) H_j\right\| \\&\qquad +\frac{K}{2N}\sum _{k=1}^N \left[ \left\| \left( X_i X_k^{-1}-I_d\right) \left( X_i X_j^{-1}-I_d\right) \right\| +\left\| \left( X_iX_j^{-1}-I_d\right) \left( X_k X_j^{-1}-I_d\right) \right\| \right] \\&\quad \le \Vert H_i-H_j\Vert +\Vert H_i\Vert \left\| \left( X_iX_j^{-1}-I_d\right) \right\| +\left\| \left( X_iX_j^{-1}-I_d\right) \right\| \Vert H_j\Vert \\&\qquad +\frac{K}{2N}\sum _{k=1}^N \left[ \left\| \left( X_i X_k^{-1}-I_d\right) \right\| \left\| \left( X_i X_j^{-1}-I_d\right) \right\| +\left\| \left( X_iX_j^{-1}-I_d\right) \right\| \left\| \left( X_k X_j^{-1}-I_d\right) \right\| \right] \\&\quad \le D({\mathcal H})+2\Vert {\mathcal H}\Vert _\infty D({\mathcal X}(t))+KD({\mathcal X}(t))^2, \end{aligned}$$

whenever $\Vert X_iX_j^{-1}-I_d\Vert >0$. By our choice of i and j, we obtain the statement of the lemma. $\square$

Appendix 3: Proof of Lemma 5.1

We start with (7.10) in the proof of Lemma 3.1:

$$\begin{aligned}&\frac{d}{dt}\left( X_i X_j^{-1}-I_d\right) \nonumber \\&\quad =-K\left( X_i X_j^{-1}-I_d\right) +H_i-H_j+H_i\left( X_iX_j^{-1}-I_d\right) -(X_iX_j^{-1}-I_d)H_j\nonumber \\&\qquad +\frac{K}{2N}\sum _{k=1}^N \left[ -\left( X_i X_k^{-1}-I_d\right) \left( X_i X_j^{-1}-I_d\right) -\left( X_iX_j^{-1}-I_d\right) \left( X_k X_j^{-1}-I_d\right) \right] .\nonumber \\ \end{aligned}$$

(7.11)

Similarly, we obtain

$$\begin{aligned}&\frac{d}{dt}(\tilde{X}_i \tilde{X}_j^{-1}-I_d) \nonumber \\&\quad =-K(\tilde{X}_i \tilde{X}_j^{-1}-I_d) +H_i-H_j+H_i\left( \tilde{X}_i\tilde{X}_j^{-1}-I_d\right) -\left( \tilde{X}_i\tilde{X}_j^{-1}-I_d\right) H_j\nonumber \\&\qquad +\frac{K}{2N}\sum _{k=1}^N \left[ -\left( \tilde{X}_i \tilde{X}_k^{-1}-I_d\right) (\tilde{X}_i \tilde{X}_j^{-1}-I_d)-\left( \tilde{X}_i\tilde{X}_j^{-1}-I_d\right) \left( \tilde{X}_k \tilde{X}_j^{-1}-I_d\right) \right] .\nonumber \\ \end{aligned}$$

(7.12)

Subtracting these two equations (7.11) and (7.12) gives

$$\begin{aligned} \frac{d}{dt}(X_i X_j^{-1}-\tilde{X}_i \tilde{X}_j^{-1})= -K(X_i X_j^{-1}-\tilde{X}_i \tilde{X}_j^{-1}) +\mathcal {P}_0+\frac{K}{2N}\sum _{k=1}^N \mathcal {P}_k, \end{aligned}$$

where

$$\begin{aligned} \mathcal {P}_0&:=\left[ H_i-H_j+H_i\left( X_iX_j^{-1}-I_d\right) -(X_iX_j^{-1}-I_d)H_j\right] \\&\quad -\left[ H_i-H_j+H_i\left( \tilde{X}_i\tilde{X}_j^{-1}-I_d\right) -(\tilde{X}_i\tilde{X}_j^{-1}-I_d)H_j\right] \\&= ~H_i(X_iX_j^{-1}-\tilde{X}_i\tilde{X}_j^{-1})-(X_iX_j^{-1}-\tilde{X}_i\tilde{X}_j^{-1})H_j \end{aligned}$$

and

$$\begin{aligned} \mathcal {P}_k&:=\left[ -\left( X_i X_k^{-1}-I_d\right) \left( X_i X_j^{-1}-I_d\right) -\left( X_iX_j^{-1}-I_d\right) \left( X_k X_j^{-1}-I_d\right) \right] \\&\quad -\left[ -\left( \tilde{X}_i \tilde{X}_k^{-1}-I_d)(\tilde{X}_i \tilde{X}_j^{-1}-I_d\right) -\left( \tilde{X}_i\tilde{X}_j^{-1}-I_d\right) \left( \tilde{X}_k \tilde{X}_j^{-1}-I_d\right) \right] \\&=\left[ -\left( X_i X_k^{-1}-I_d\right) \left( X_i X_j^{-1}-I_d\right) +\left( \tilde{X}_i \tilde{X}_k^{-1}-I_d\right) (\tilde{X}_i \tilde{X}_j^{-1}-I_d)\right] \\&\quad -\left[ \left( X_iX_j^{-1}-I_d\right) \left( X_k X_j^{-1}-I_d\right) -\left( \tilde{X}_i\tilde{X}_j^{-1}-I_d\right) \left( \tilde{X}_k \tilde{X}_j^{-1}-I_d\right) \right] \\&=\left[ -\left( X_i X_k^{-1}-\tilde{X}_i \tilde{X}_k^{-1}\right) \left( X_i X_j^{-1}-I_d\right) +\left( \tilde{X}_i \tilde{X}_k^{-1}-I_d\right) \left( \tilde{X}_i\tilde{X}_j^{-1}-X_iX_j^{-1}\right) \right] \\&\quad -\left[ (X_iX_j^{-1}-\tilde{X}_i\tilde{X}_j^{-1})\left( X_k X_j^{-1}-I_d\right) -\left( \tilde{X}_i\tilde{X}_j^{-1}-I_d\right) \left( \tilde{X}_k \tilde{X}_j^{-1}-X_k X_j^{-1}\right) \right] . \end{aligned}$$

Then, we have

$$\begin{aligned} \Vert \mathcal {P}_0\Vert \le ~ \Vert H_i\Vert \Vert X_iX_j^{-1}-\tilde{X}_i\tilde{X}_j^{-1}\Vert +\Vert X_iX_j^{-1}-\tilde{X}_i\tilde{X}_j^{-1}\Vert \Vert H_j\Vert \le ~ 2\Vert {\mathcal H} \Vert _\infty d({\mathcal X},\tilde{{\mathcal X}}) \end{aligned}$$

and

$$\begin{aligned} \Vert \mathcal {P}_k\Vert&\le \left[ \Vert X_i X_k^{-1}-\tilde{X}_i \tilde{X}_k^{-1}\Vert \Vert X_i X_j^{-1}-I_d\Vert +\Vert \tilde{X}_i \tilde{X}_k^{-1}-I_d\Vert \Vert \tilde{X}_i\tilde{X}_j^{-1}-X_iX_j^{-1}\Vert \right] \\&\quad +\left[ \Vert X_iX_j^{-1}-\tilde{X}_i\tilde{X}_j^{-1}\Vert \Vert X_k X_j^{-1}-I_d\Vert +\Vert \tilde{X}_i\tilde{X}_j^{-1}-I_d\Vert \Vert \tilde{X}_k \tilde{X}_j^{-1}-X_k X_j^{-1}\Vert \right] \\&\le ~ 4\max \{D({\mathcal X}),D(\tilde{{\mathcal X}})\}d({\mathcal X},\tilde{{\mathcal X}}). \end{aligned}$$

Finally, we apply Lemma 7.2 to obtain

$$\begin{aligned}&\left| \frac{d}{dt}\Vert X_i X_j^{-1}-\tilde{X}_i \tilde{X}_j^{-1}\Vert +K\Vert X_i X_j^{-1}-\tilde{X}_i \tilde{X}_j^{-1}\Vert \right| \\& \le \Vert \mathcal {P}_0\Vert +\frac{K}{2N}\sum _{k=1}^N \Vert \mathcal {P}_k\Vert \\& \le 2\Vert {\mathcal H} \Vert _\infty d({\mathcal X},\tilde{{\mathcal X}})+\frac{K}{2}\cdot 4\max \{D({\mathcal X}),D(\tilde{{\mathcal X}})\}d({\mathcal X},\tilde{{\mathcal X}})\\& \le 2(\Vert {\mathcal H}\Vert _\infty +K\max \{D({\mathcal X}),D(\tilde{{\mathcal X}})\})d({\mathcal X},\tilde{{\mathcal X}}), \end{aligned}$$

whenever $\Vert X_i X_j^{-1}-\tilde{X}_i \tilde{X}_j^{-1}\Vert >0$. For each time $t \ge 0$, we choose indices i, j so that

$$\begin{aligned} d({\mathcal X}(t),\tilde{{\mathcal X}}(t))=\Vert X_i X_j^{-1}-\tilde{X}_i \tilde{X}_j^{-1}\Vert (t), \end{aligned}$$

to obtain the asserted inequality for $d({\mathcal X}(t),\tilde{{\mathcal X}}(t))$.

Appendix 4: Proof of Lemma 5.2

We next provide the proof of Lemma 5.2. It follows from (1.3) that we have

$$\begin{aligned} \Vert X_i\dot{X}_i^{-1}\Vert= & {} \Vert \dot{X}_iX_i^{-1}\Vert = \Big \Vert H_i+\frac{K}{2N}\sum _j\left( X_jX_i^{-1}-X_iX_j^{-1}\right) \Big \Vert \\= & {} \Big \Vert H_i+\frac{K}{2N}\sum _j\left[ \left( X_jX_i^{-1}-I_d\right) +\left( I_d-X_iX_j^{-1}\right) \right] \Big \Vert \\\le & {} \Vert H\Vert _\infty +KD(\mathcal {X}) \le \Vert H\Vert _\infty +K\alpha _1. \end{aligned}$$

Thus, we have

$$\begin{aligned} \frac{d}{dt}\Vert X_i\Vert ^2&=2{\text {Re}}tr\left( \dot{X}_iX_i^\dagger \right) =2{\text {Re}}tr\left[ \left( \dot{X}_iX_i^{-1}\right) \left( X_iX_i^\dagger \right) \right] \\&\le 2\left\| \dot{X}_iX_i^{-1}\right\| \left\| X_iX_i^\dagger \right\| \le 2\left\| \dot{X}_iX_i^{-1}\right\| \left\| X_i\right\| ^2 \\&\le 2(\Vert H\Vert _\infty +K\alpha _1)\Vert X_i\Vert ^2, \end{aligned}$$

and similarly

$$\begin{aligned} \frac{d}{dt}\Vert X_i^{-1}\Vert ^2&=2{\text {Re}}tr\left( \dot{X}_i^{-1}\left( X_i^{-1}\right) ^\dagger \right) =2{\text {Re}}tr\left[ \left( X_i\dot{X}_i^{-1}\right) \left( \left( X_i^{-1}\right) ^\dagger X_i^{-1}\right) \right] \\&\le 2(\Vert H\Vert _\infty +K\alpha _1)\Vert X_i^{-1}\Vert ^2. \end{aligned}$$

This yields

$$\begin{aligned}&\max _{1\le i \le N} \Vert X_i\Vert \le e^{(\Vert H\Vert _\infty +K\alpha _1)t}\max _{1 \le i \le N} \Vert X_i(0)\Vert , \\&\max _{1\le i \le N} \Vert X_i^{-1}\Vert \le e^{(\Vert H\Vert _\infty +K\alpha _1)t}\max _{1 \le i \le N} \Vert X_i^{-1}(0)\Vert . \end{aligned}$$

Appendix 5: The Explicit Forms of (6.6)

Below, we provide the estimates for three relations appearing in Sect. 6.

• (Estimate of (6.6)$_1$): The term (6.6)$_1$ is given by

  $$\begin{aligned} a_i-a_j=\frac{u_i-u_j}{K}(1-e^{-Kt})+\left( a_i^0-a_j^0\right) e^{-Kt}. \end{aligned}$$

• (Estimate of (6.6)$_3$): The term (6.6)$_3$ is given by

  $$\begin{aligned} c_i-c_j=\frac{w_i-w_j}{K}(1-e^{-Kt})+\left( c_i^0-c_j^0\right) e^{-Kt}. \end{aligned}$$

• (Estimate of (6.6)$_2$): By direct calculation, we have

  $$\begin{aligned}&b_i-b_j+(a_j-a_i)c_j =\left( b_i^0-b_j^0\right) e^{-Kt}+\frac{v_i-v_j}{K}(1-e^{-Kt}) \\&\qquad +\frac{1}{2K^2}(u_iw_i+u_jw_j-2u_iw_j)(1-e^{-Kt})^2 +\frac{1}{2K}(u_i-u_j)\frac{1}{N}c_k^0(-Kte^{-Kt}+e^{-Kt}-e^{-2Kt}) \\&\qquad +\frac{1}{2K}(u_i-u_j)\frac{1}{N}w_k\left( 2te^{-Kt}-\frac{1}{K}+\frac{1}{K}e^{-2Kt}\right) \\&\qquad +\frac{1}{2}\left( u_ic_i^0-u_jc_j^0\right) \left( te^{-Kt}+\frac{1}{K}e^{-Kt}-\frac{1}{K}e^{-2Kt}\right) \\&\qquad +\frac{1}{K}(u_j-u_i)c_j^0(e^{-Kt}-e^{-2Kt}) +\frac{1}{2}(w_i-w_j)\frac{1}{N}\sum _{k=1}^N a_k^0\left( te^{-Kt}-\frac{1}{K}e^{-Kt}+\frac{1}{K}e^{-2Kt}\right) \\&\qquad +\frac{K}{2}\left( a_i^0c_i^0-a_j^0c_j^0\right) \left( -\frac{1}{K}e^{-Kt}+\frac{1}{K}e^{-2Kt}\right) +\left( a_j^0c_j^0-a_i^0c_j^0\right) e^{-2Kt} \\&\qquad +\frac{1}{2K}(w_i-w_j)\frac{1}{N}\sum _{k=1}^Nu_k\left( -2te^{-Kt}+\frac{1}{K}-\frac{1}{K}e^{-2Kt}\right) \\&\qquad +\frac{1}{2}\left( w_ia_i^0-w_ja_j^0\right) \left( -te^{-Kt}+\frac{1}{2}e^{-Kt}-\frac{1}{K}e^{-2Kt}\right) +\frac{1}{K}(a_j^0-a_i^0)w_j(e^{-Kt}-e^{-2Kt}) \\&\qquad +\frac{1}{2}\left( c_i^0-c_j^0\right) \frac{1}{N}\sum _{k=1}^N a_k^0(e^{-Kt}-e^{-2Kt})\\&\qquad +\frac{1}{2}\left( c_i^0-c_j^0\right) \frac{1}{N}\sum _{k=1}^N u_k\left( te^{-Kt}-\frac{1}{K}e^{-Kt}+\frac{1}{K}e^{-2Kt}\right) \\&\qquad +\frac{1}{2}\left( a_i^0-a_j^0\right) \frac{1}{N}\sum _{k=1}^N c_k^0(-e^{-Kt}+e^{-2Kt})\\&\qquad +\frac{1}{2}\left( a_i^0-a_j^0\right) \frac{1}{N}\sum _{k=1}^N w_k\left( -te^{-Kt}+\frac{1}{K}e^{-Kt}-\frac{1}{K}e^{-2Kt}\right) . \end{aligned}$$