Flocking Dynamics of the Inertial Spin Model with a Multiplicative Communication Weight

Abstract

In this paper, we study a flocking dynamics of the deterministic inertial spin (IS) model. The IS model was introduced for the collective dynamics of active particles with an internal angular momentum, or spin. When the generalized moment of inertia becomes negligible compared to spin dissipation (overdamped limit) and mutual communication weight is a function of a relative distance between interacting particles, the deterministic inertial spin model formally reduces to the Cucker–Smale (CS) model with constant speed constraint whose emergent dynamics has been extensively studied in the previous literature. We present several sufficient frameworks leading to the asymptotic mono-cluster flocking, in which spins and relative velocities tend to zero asymptotically. We also provide several numerical simulations for the decoupled and coupled inertial spin models to see the effect of the C–S velocity flocking and compare them with our analytical results.

Appendix A: Proof of Proposition 4.3

First, we define $\mathcal {E}^\infty$ and $\bar{w}_c^\infty$ as the limits of $\mathcal {E}$ and $|{\bar{w}}_c|$, respectively.

• Step A [convergence of (V, S)] By letting $t\rightarrow \infty$ in Proposition 4.2, we have

  $$\begin{aligned} {\mathcal {E}}^\infty + \frac{4\gamma }{\chi ^2 \kappa } \int _0^\infty {\mathcal {S}}(u)\mathrm{d}u = {\mathcal {E}}(t) + \frac{2}{\chi \kappa }{\mathcal {S}}(t) + \frac{4\gamma }{\chi ^2\kappa }\int _0^t {\mathcal {S}}(u)\mathrm{d}u. \end{aligned}$$

This can be rewritten as

$$\begin{aligned}&2\left( \frac{1}{N}\sum _{i=1}^Np_i^2-({\bar{w}}_c^\infty )^2\right) + \frac{4\gamma }{\chi ^2\kappa } \int _t^\infty {\mathcal {S}}(u)\mathrm{d}u \\&\quad = 2\left( \frac{1}{N}\sum _{i=1}^Np_i^2-|{\bar{w}}_c|^2\right) + \frac{2}{\chi \kappa }{\mathcal {S}}(t). \end{aligned}$$

Hence, we have

$$\begin{aligned} ({\bar{w}}_c^\infty )^2 = \frac{2\gamma }{\chi ^2\kappa }\int _t^\infty {\mathcal {S}}(u)\mathrm{d}u - \frac{1}{\chi \kappa } {\mathcal {S}}(t) + |{\bar{w}}_c|^2 \ge \frac{2\gamma }{\chi ^2\kappa }\int _t^\infty {\mathcal {S}}(u)\mathrm{d}u -\frac{1}{\chi \kappa }{\mathcal {S}}(t). \end{aligned}$$

Since we assume ${\bar{w}}_c^\infty =0$ in Proposition 4.4, we have

$$\begin{aligned} \frac{2\gamma }{\chi ^2\kappa } \int _{t}^\infty {\mathcal {S}}(u) \mathrm{d}u - \frac{1}{\chi \kappa } {\mathcal {S}}(t) \le 0, \quad \forall t\in \mathbb {R}. \end{aligned}$$

(A.1)

If we define

$$\begin{aligned} g(t) := \int _t^\infty {\mathcal {S}}(u)\mathrm{d}u, \end{aligned}$$

then (A.1) becomes

$$\begin{aligned} \quad g^\prime (t) + \frac{2\gamma }{\chi } g(t) \le 0. \end{aligned}$$

This implies

$$\begin{aligned} 0\le g(t) \le g(0)e^{-\frac{2\gamma }{\chi } t},\quad \forall t\in \mathbb {R}. \end{aligned}$$

On the other hand, since the following relation holds:

$$\begin{aligned} \left( -g(t)e^{-\frac{\gamma }{\chi } t} \right) ^\prime = {\mathcal {S}}(t) e^{\frac{\gamma }{\chi } t} - \frac{\gamma }{\chi } g(t)e^{\frac{\gamma }{\chi } t}, \end{aligned}$$

we obtain

$$\begin{aligned} \int _0^t {\mathcal {S}}(u) e^{\frac{\gamma }{\chi } u } \mathrm{d}u&= \int _0^t \frac{\gamma }{\chi } g(u) e^{\frac{\gamma }{\chi }u} + g(0) - g(t) e^{\frac{\gamma }{m}t} \\&\le \frac{\gamma }{\chi } g(0) \int _0^t e^{-\frac{\gamma }{\chi } u } \mathrm{d}u + g(0) \\&\le g(0) \left( 1- e^{-\frac{\gamma }{\chi }t } \right) + g(0)= 2g(0). \end{aligned}$$

Thus, we have

$$\begin{aligned} \int _0^\infty {\mathcal {S}}(u) e^{\frac{\gamma }{m}u}\mathrm{d}u< \infty , \quad \text {i.e.,} \quad \int _0^\infty |s_i(u)|^2 e^{\frac{\gamma }{m} u } \mathrm{d}u <\infty , \quad i=1,\ldots ,N. \end{aligned}$$

From the relation $\chi |\dot{w}_i| = p_i|s_i|$ in (4.21)$_1$, this is equivalent with

$$\begin{aligned} \int _0^\infty |\dot{w}_i(u)|^2 e^{\frac{\gamma }{\chi }u}\mathrm{d}u <\infty . \end{aligned}$$

We use Hölder’s inequality to obtain

$$\begin{aligned} \int _0^\infty | \dot{w}_i| \mathrm{d}u&= \int _0^\infty |\dot{w}_i(u)| e^{ \frac{\gamma }{2\chi } u } e^{ -\frac{\gamma }{2\chi } u } \mathrm{d}u \\&\le \left( \int _0^\infty |\dot{w}_i(u)|^2 e^{\frac{\gamma }{\chi } u}du \right) ^\frac{1}{2} \left( \int _0^\infty e^{-\frac{\gamma }{\chi }u}\mathrm{d}u \right) ^\frac{1}{2} <\infty . \end{aligned}$$

Hence, for each i, there exists $(w_i^k)^\infty \in \mathbb {R}$ such that

$$\begin{aligned} \lim _{t\rightarrow \infty } w_i^k(t) = (w_i^k)^\infty , \quad i = 1, \ldots , N,\quad k=1,2,3. \end{aligned}$$

Thus, (V, S) converges to $(V^\infty ,S^\infty ):=(\frac{w_1^\infty }{p_1},\ldots ,\frac{w_N^\infty }{p_N},0,\ldots ,0)$.

•  Step B (linear stability analysis): Now, we show that $(V^\infty ,S^\infty )$ is linearly unstable. For this, we write

  $$\begin{aligned} {\mathcal {I}}= & {} (v_1,\ldots ,v_N, s_1,\ldots ,s_N) \in \mathbb {R}^{6N}~~\text {and}~~ w_i=(w_i^1,w_i^2,w_i^3) \in \mathbb {R}^3, ~~ \\ s_i= & {} (s_i^1,s_i^2,s_i^3) \in \mathbb {R}^3. \end{aligned}$$

We construct $6N\times 6N$ Jacobian matrix ${\mathbf {X}}$ for ${\mathcal {I}}$

$$\begin{aligned} {\mathbf {X}}:= \frac{\partial \dot{{\mathcal {I}}}}{\partial {\mathcal {I}}} = \begin{pmatrix} A &{}\quad B \\ C &{}\quad D \end{pmatrix}, \end{aligned}$$

where A, B, C and D are $3N\times 3N$ matrices whose forms are as follows:

$$\begin{aligned} A= \frac{\partial \dot{w}_j }{\partial w_i}, \quad B= \frac{\partial \dot{w}_j }{\partial s_i},\quad C= \frac{\partial \dot{s}_j }{\partial w_i}, \quad D= \frac{\partial \dot{s}_j }{\partial s_i}. \end{aligned}$$

$\diamond$ (Case of A): We use (4.1) and $s_i^\infty =0$ to find

$$\begin{aligned} A&= \frac{\partial \dot{w}_j }{\partial w_i} = \frac{\partial }{\partial w_i} \left( \frac{1}{\chi } s_j\times w_j\right) \\&=\frac{1}{\chi } \frac{\partial }{\partial w_i }\Big ( s_j^2w_j^3 - s_j^3w_j^2, ~s_j^3w_j^1-s_j^1w_j^3,~s_j^1w_j^2-s_j^2w_j^1\Big ) \\&=\frac{1}{m} {\left\{ \begin{array}{ll} 0 \qquad (i\ne j) \\ \begin{pmatrix} 0 &{}\quad -s_i^3 &{}\quad s_i^2 \\ s_i^3 &{}\quad 0 &{}\quad -s_i^1 \\ -s_i^2 &{}\quad s_i^1 &{}\quad 0 \end{pmatrix} \qquad (i=j) \end{array}\right. } \\&=0. \end{aligned}$$

$\diamond$ (Case of B): We also use (4.1) and $s_i^\infty = 0$ to find

$$\begin{aligned} B= \frac{\partial \dot{w}_j }{\partial s_i}&= \frac{1}{\chi } {\left\{ \begin{array}{ll} 0 \qquad (i\ne j) \\ \begin{pmatrix} 0 &{}\quad w_i^3 &{}\quad -w_i^2 \\ -w_i^3 &{}\quad 0 &{}\quad w_i^1 \\ w_i^2 &{}\quad -w_i^1 &{}\quad 0 \end{pmatrix} \qquad (i=j). \end{array}\right. } \end{aligned}$$

$\diamond$ (Case of C): We use (4.3) and ${\bar{w}}_c^\infty =0$ to obtain

$$\begin{aligned} C&= \frac{\partial \dot{s}_j }{\partial w_i} = \frac{\partial }{\partial w_i} \left( \kappa (w_j\times w_c) - \frac{\gamma }{\chi } s_j\right) \\&= \kappa \frac{\partial }{\partial w_i} \Big ( w_j^2 {\bar{w}}_c^3 - w_j^3 {\bar{w}}_c^2,~w_j^3 {\bar{w}}_c^1-w_j^1{\bar{w}}_c^3,~w_j^1{\bar{w}}_c^2-w_j^2 {\bar{w}}_c^1 \Big ) \\&= {\left\{ \begin{array}{ll} \frac{\kappa }{N} \begin{pmatrix} 0 &{}\quad -w_j^3 &{}\quad w_j^2 \\ w_j^3 &{}\quad 0 &{}\quad -w_j^1 \\ -w_j^2 &{}\quad w_j^1 &{}\quad 0 \end{pmatrix} \qquad (i\ne j) \\ \frac{\kappa }{N} \begin{pmatrix} 0 &{}\quad -w_j^3 &{}\quad w_j^2 \\ w_j^3 &{}\quad 0 &{}\quad -w_j^1 \\ -w_j^2 &{}\quad w_j^1 &{}\quad 0 \end{pmatrix} + \kappa \begin{pmatrix} 0 &{}\quad {\bar{w}}_c^3 &{}\quad -{\bar{w}}_c^2 \\ -{\bar{w}}_c^3 &{}\quad 0 &{}\quad {\bar{w}}_c^1\\ {\bar{w}}_c^2 &{}\quad -{\bar{w}}_c^1 &{}\quad 0 \end{pmatrix} \qquad (i=j) \end{array}\right. } \\&= \frac{\kappa }{N} \begin{pmatrix} 0 &{}\quad -w_j^3 &{}\quad w_j^2 \\ w_j^3 &{}\quad 0 &{}\quad -w_j^1 \\ -w_j^2 &{}\quad w_j^1 &{}\quad 0 \end{pmatrix}. \end{aligned}$$

$\diamond$ (Case of D): We also obtain

$$\begin{aligned} D= \frac{\partial \dot{s}_j }{\partial s_i} = \frac{\partial }{\partial s_i} \left( \kappa (w_j\times w_c) - \frac{\gamma }{\chi } s_j \right) = -\frac{\gamma }{\chi } \delta _{ij}, \end{aligned}$$

where $\delta _{ij}$ is the Kronecker delta.

For the handy notation, we write $3\times 3$ matrix $W_i$ as

$$\begin{aligned} W_i = \begin{pmatrix} 0 &{}\quad w_i^3 &{}\quad -w_i^2 \\ -w_i^3 &{}\quad 0 &{}\quad w_i^1 \\ w_i^2 &{}\quad -w_i^1 &{}\quad 0 \end{pmatrix}, \quad i=1,\ldots ,N. \end{aligned}$$

Then, we can write the matrix B and C as

$$\begin{aligned} B = \frac{1}{\chi } \text {diag} ( W_1,\ldots ,W_N) \quad \text {and} \quad C = -\frac{\kappa }{N} \begin{pmatrix} W_1 &{}\quad \ldots &{}\quad W_1 \\ \ldots &{}\quad \ddots &{}\quad \ldots \\ W_N &{}\quad \ldots &{}\quad W_N \end{pmatrix}. \end{aligned}$$

Note that

$$\begin{aligned} \text {tr} \left( -W_i^2\right) = 2( w_i^1)^2 + 2(w_i^2)^2 + 2(w_i^3)^2 = 2p_i^2. \end{aligned}$$

(A.2)

Then, we compute the characteristic polynomial of ${\mathbf {X}}$ as follows: for each eigenvalue $\lambda$,

$$\begin{aligned} 0&=\text {det} \Big ({\mathbf {X}} - \lambda I_{6N}\Big ) = \text {det} \begin{pmatrix} -\lambda I_{3N} &{} B \\ C &{} \left( - \lambda - \frac{\gamma }{\chi } \right) I_{3N} \end{pmatrix} \\&=\text {det} \left( \lambda \left( \lambda + \frac{\gamma }{\chi } \right) I_{3N} - BC \right) \\&=\text {det} \left( \lambda \left( \lambda + \frac{\gamma }{\chi } \right) I_{3N} - \frac{\kappa }{\chi N} \begin{pmatrix} -W_1^2 &{}\quad \ldots &{}\quad -W_1^2 \\ \ldots &{}\quad \ddots &{}\quad \ldots \\ -W_N^2 &{}\quad \ldots &{}\quad -W_N^2 \end{pmatrix} \right) \\&=: \text {det}\left( \lambda \left( \lambda + \frac{\gamma }{\chi } \right) I_{3N} - {\mathcal {M}} \right) . \end{aligned}$$

It follows from (A.2) that $3N\times 3N$ matrix ${\mathcal {M}}$ has positive trace

$$\begin{aligned} \text {tr}{\mathcal {M}} = \frac{\kappa }{ \chi N} \sum _{i=1}^N \text {tr}\left( -W_i^2 \right) = \frac{2\kappa }{N\chi }\sum _{i=1}^Np_i^2 >0, \end{aligned}$$

and therefore ${\mathcal {M}}$ has an eigenvalue $\lambda _0$ whose real part is positive, i.e., $\text {Re}\lambda _0>0$. For this $\lambda _0$, every $\lambda \in \mathbb {C}$ satisfying

$$\begin{aligned} \lambda \left( \lambda + \frac{\gamma }{m} \right) = \lambda _0 =: c + \mathrm {i}d, \end{aligned}$$

is an eigenvalue of ${\mathbf {X}}$. Now if we write $\lambda$ by $\lambda = a + \mathrm {i}b$, $(a,b,c,d)\in \mathbb {R}^4$, then the above relation implies

$$\begin{aligned} a^2 - b^2 + \frac{\gamma }{\chi } a = \text {Re} \lambda _0=c, \quad 2ab + \frac{\gamma }{\chi } b = \text {Im}\lambda _0=d. \end{aligned}$$

Then,

$$\begin{aligned} c = a^2 + \frac{\gamma }{\chi } a - \frac{\mathrm{d}^2 }{\left( 2a+ \frac{\gamma }{\chi } \right) ^2} = a^2 + \frac{\gamma }{\chi } a - \frac{\mathrm{d}^2 }{4a^2 + \frac{4\gamma }{\chi }a + \frac{\gamma ^2}{\chi ^2}} \end{aligned}$$

and if we write $Z := a^2 + \frac{\gamma }{\chi }a$, we have

$$\begin{aligned} \quad 4Z^2 + \left( \frac{\gamma ^2}{\chi ^2} -4c\right) Z -d^2 -\frac{\gamma ^2}{\chi ^2} c =0. \end{aligned}$$

Since $c=\text {Re}\lambda _0>0$, the above quadratic equation attains two distinct real roots $z_1< 0 <z_2$ and $Z=a^2 + \frac{\gamma }{\chi }a=b^2+c$ has to be positive. Hence, we have

$$\begin{aligned} a^2 + \frac{\gamma }{\chi } a = z_2, \end{aligned}$$

and the above equation always attains one positive real root. In other words, the matrix ${\mathbf {X}}$ has at least one eigenvalue whose real part is positive. Therefore, we can conclude that the equilibrium is linearly unstable.