A Matrix-Valued Kuramoto Model

Abstract

The Kuramoto model is an important model for studying the onset of phase-locking in an ensemble of nonlinearly coupled phase oscillators. Each oscillator has a natural frequency (often taken to be random) and interacts with the other oscillators through the phase difference. It is known that, as the coupling strength is increased, there is a bifurcation in which the incoherent state becomes unstable and a stable phase-locked solution is born. Beginning with the work of Lohe there have been a number of paper that have generalized the Kuramoto model for phase-locking to a non-commuting setting. Here we propose and analyze another such model. We consider a collection of matrix-valued variables that evolve in such a way as to try to align their eigenvector frames. The phase-locked state is one where the eigenframes all align, and thus the matrices all commute. We analyze the stability of the equal frequency phase-locked state and show that it is stable. We also analyze (for the case of symmetric matrices) a dynamic analog of the twist states arising in the standard Kuramoto model, and show that these twist states are dynamically unstable.

Appendices

Appendix A: Derivation of the Gradient Flow

Recall that a complex gradient flow takes the form

$$\begin{aligned} \frac{d \varvec{z}}{dt} = - \nabla _{\bar{\varvec{z}}} E(\varvec{z},\bar{\varvec{z}}) \end{aligned}$$

where $E(\varvec{z},\bar{\varvec{z}})$ is the energy function and $\bar{\varvec{z}}$ denotes the complex conjugate variable.

The energy function defined in Sect. 1,

$$\begin{aligned} E= \sum \limits _{i,j} a_{ij} {{\,\mathrm{Tr}\,}}\left( [\mathbf{M}_i,\mathbf{M}_j]^\dagger [\mathbf{M}_i,\mathbf{M}_j]\right) \end{aligned}$$

The energy E is a scalar valued function of N, $k\times k$ matrices $\mathbf{M}_i$ and $\frac{\partial E}{\partial \bar{\varvec{M}}_i}$ is a $k\times k$ matrix whose entries are the derivatives of E with respect to the corresponding entry of $\bar{\varvec{M}}_i$. Individual components of a matrix will be denoted $M_{i;\alpha \beta }$ with Greek indices for the matrix components. We will use the summation convention that repeated Greek indices are to be summed over, and that repeated Latin indices are to be summed over, with an additional (multiplicative) weight of $a_{ij}$ for the $i,j^{th}$ term.

The energy can be written as

$$\begin{aligned} E= & {} {\bar{M}}_{j;\beta \alpha } {\bar{M}}_{i;\gamma \beta } M_{i;\gamma \delta } M_{j;\delta \alpha } + {\bar{M}}_{i;\beta \alpha } {\bar{M}}_{j;\gamma \beta } M_{j;\gamma \delta } M_{i;\delta \alpha } \\&- {\bar{M}}_{j;\beta \alpha } {\bar{M}}_{i;\gamma \beta } M_{j;\gamma \delta } M_{i;\delta \alpha } -{\bar{M}}_{i;\beta \alpha } {\bar{M}}_{j;\gamma \beta } M_{i;\gamma \delta } M_{j;\delta \alpha }, \end{aligned}$$

again with the summation convention noted in the preceeding paragraph in force, and $M_{i;\alpha ,\beta }$ denoting the $(\alpha ,\beta )$ component of $\mathbf{M}_i$. Differentiating with respect to ${\bar{M}}_{i;\tau \phi }$ gives

$$\begin{aligned} \frac{\partial E}{\partial {\bar{M}}_{i;\tau \phi }}= & {} {\bar{M}}_{j;\phi \alpha } M_{i;\tau \gamma } M_{j;\delta \alpha } + {\bar{M}}_{j;\gamma \tau } M_{j;\gamma \delta } M_{i;\delta \phi } \\&- {\bar{M}}_{j;\phi \alpha } M_{j;\tau \delta } M_{i;\delta \alpha }-{\bar{M}}_{j;\gamma \tau } M_{i;\gamma \delta } M_{j;\delta \phi }. \end{aligned}$$

Note that upon unraveling the summation convention the derivative above can be recognized as the $\tau ,\delta$ entry of a sum of three-fold commutators,

$$\begin{aligned} \frac{\partial E}{\partial \bar{\varvec{M}}_i} = \sum _j a_{ij} [\mathbf{M}_j^\dagger ,[M_{j},M_{i}]]. \end{aligned}$$

Thus the gradient flow for the energy E is the matrix Kuramoto flow (1.7)

$$\begin{aligned} \frac{\partial \mathbf{M}_i}{\partial t}&= -\frac{\partial E}{\partial \mathbf{M}_{i}}\\&=\sum _j a_{ij}[\mathbf{M}_j^\dagger ,[\mathbf{M}_i,\mathbf{M}_j]]. \end{aligned}$$

Appendix B: Computation of the Eigenvalues and Eigenvectors of the Linearization About a Twist State

In order to apply Eq. (2.15), for each $\alpha \in \{0,1,2,\ldots ,N-1\}$ we need to determine the associated eigenvalue for each block of $\mathbf{J}$, Eq. (2.12). Since the first entry of $\mathbf{v}^{(\alpha )}$ is 1 the associated eigenvalue is given by the dot product of the first row of each block with the vector $\mathbf{v}^{(\alpha )}$. The following properties of the discrete Fourier transform will be useful here:

$$\begin{aligned} \sum \limits _{j=0}^{n-1}e^{\frac{2 \pi ijk}{n}} \cos \left( \frac{2 \pi j}{n}\right)&= {\left\{ \begin{array}{ll} 0, &{} k \ne \pm 1 \\ \frac{n}{2}, &{} k= \pm 1 \mod n\end{array}\right. } \end{aligned}$$

(B.1)

$$\begin{aligned} \sum \limits _{j=0}^{n-1} e^{\frac{2 \pi ijk}{n}} \sin \left( \frac{2 \pi j}{n}\right)&= {\left\{ \begin{array}{ll} 0, &{} k \ne \pm 1 \\ \frac{n}{2i}, &{} k=1 \mod n \\ \frac{-n}{2i}, &{} k=-1 \mod n \end{array}\right. } \end{aligned}$$

(B.2)

$$\begin{aligned} \sum \limits _{j=0}^{n-1} e^{\frac{2 \pi ijk}{n}}&= {\left\{ \begin{array}{ll} n, &{} k=0 \mod n \\ 0, &{} k \ne 0 \mod n \end{array}\right. }. \end{aligned}$$

(B.3)

We begin by looking at the $\frac{\partial g}{\partial \theta }$ block. The top row of this block is

$$\begin{aligned} \begin{bmatrix} 1&\cos \left( \frac{2 \pi }{N}\right)&\cos \left( \frac{4 \pi }{N}\right)&\cos \left( \frac{6 \pi }{N}\right)&\cdots&\cos \left( \frac{2 \pi (N-1)}{N}\right) \end{bmatrix}. \end{aligned}$$

To find the $k\mathrm{{th}}$ eigenvalue $a_k$ we take the dot product of this row with $\mathbf{v}^{(k)}$ yielding

$$\begin{aligned} a_k&=1+ e^{2 \pi i k/N} \cos \left( \frac{2 \pi }{N}\right) + e^{4 \pi i k/N} \cos \left( \frac{4 \pi }{N}\right) +\cdots + e^{2 \pi i k (N-1)/N}\cos \left( \frac{2 \pi (N-1)}{N}\right) \\&= \sum _{j=0}^{N-1} e^{\frac{2 \pi i j k}{N}} \cos \left( \frac{2 \pi j}{N}\right) . \end{aligned}$$

By the given identities this sum is equal to N / 2 if $k=1$ or $k=N-1$ and is zero otherwise.

The eigenvalues for the other two blocks follow in the same way. The first row of the $\frac{\partial g}{\partial \varDelta \lambda }$ block is given by

$$\begin{aligned} \begin{bmatrix} 0&\frac{1}{2}\sin \left( \frac{2 \pi }{N}\right)&\frac{1}{2}\sin \left( \frac{4 \pi }{N}\right)&...&\frac{1}{2}\sin \left( \frac{2 \pi (N-1)}{N}\right) \end{bmatrix}. \end{aligned}$$

It is clear that since the entries of the row are given by $\sin \frac{2 \pi j}{N}$ the only non-zero Fourier coefficients will be $k=1$ and $k=N-1$ giving $b_1 = - i N/4$ and $b_{N-1} = i N/4$, with all of the remaining eigenvalues equal to zero.

Finally we consider the block $\frac{\partial f}{\partial \varDelta \theta }$. The first row of this block looks like

$$\begin{aligned} \begin{bmatrix} 0&-1+ \cos \left( \frac{2 \pi }{N}\right)&-1 + \cos \left( \frac{4 \pi }{N}\right)&\cdots&-1+ \left( \cos \left( \frac{2 \pi (N-1)}{N}\right) \right) \end{bmatrix}. \end{aligned}$$

Since this term is the sum of a constant and a cosine term we will have three non-zero Fourier coefficients and thus three non-zero eigenvalues. Applying the above identities gives $c_0 = -N, c_1 = N/2, c_{N-1}= {N}/{2}$ with all of the remaining eigenvalues zero.