Kuramoto模型在Python和MATLAB中的简单实现

haidaowang

事情是这样的，我最近在研究团队编组及内部模式对发挥团队能力的影响，以及如何正确编组让团队能力发挥实现最大化，别问我为什么研究这个，反正稀里糊涂就研究上了。我发现在描述团队编组间及内部同步能力的时候，人们对Kuramoto模型（藏本模型）作了大量的研究,其中包括模型达到完全相位同步的充分条件、耦合强度对于同步的影响、一定条件下振子的收敛速率等。但具体实现一般都在MATLAB中，且网上代码过于复杂（我运行了一遍一堆报错），这里我使用Python和MATLAB对Kuramoto模型简单模拟。
  X5 l% G. @" _- z模拟的话还是一遍举个栗子，边分析边测试效果最好，百度学术上有一篇关于Kuramoto模型的简单论文，我们就用它来实现模拟。

2 K9 d- Z. \. h5 j2 `好吧，那我们开始，首先是Python实现：
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" ~0 H9 L& O/ Y4 x. q) }" O, T. S+ IN维Kuramoto模型的数学描述如下：
; G9 l! _5 a3 u" C% X: `6 k
& y7 m2 y- D! X# @7 l. b
没错，是讨厌的数学公式，没事，它可以改写成这样：

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5 O: x: Q0 D# l  Q* h好像还是有点长，那我们在改写一下：
* p: v: p# z; E- Y  p3 e+ y9 Y
6 q6 d1 _* ~* Q3 Y. }
看着好多了，那我就来说说式子中参数的意义，Kij为耦合矩阵，是为了便于描述不同振子间耦合程度不同的情形。最下面那个式子的r就是我们的目标，反应振子间的相关性，这个相关性就可以描述我们想要的编组内部同步能力。
' v3 M3 ^, `: Y( n( y2 V
哎呦，这个式子看起来好简单，这里要补充一下知识点：同步能力可不是一下子各组该怎么同步直接确定的了，它是一个从开始到稳定的阶段，也就是说随时间变化，最终反映在各组的同步能力才会确定，那么最后图像是什么样子才算同步能力好呢？

同步能力好，是指随着时间的推移，各组的同步能力r逐渐稳定，波动现象消失或固定在某一个小范围内。需要注意的是这和各组r值之间的差距没有关系，我们要的是一个平稳的状态。那怎么办找r和t的关系呢？

( q  n6 i1 Q. T/ A$ S注意看最上面那两个式子，相位（第一项，等号左边那个）上面有个点，这样他可就不简简单单是个相位了，它代表的是相位的变化值，是一个微小的微分值，好吧具体意思就是，那个式子左边展开之后是这样的：

8 {3 _# v! ^+ T  f* n
. n( w) J+ ]" W哎呀，t出现了，其实\theta 与t有关，这里你可能有点绕，因为它们之间的关系是一一对应的，就是说每个时间的t对应了一个\theta ，我下面带入具体数值的时候你就知道了。

5 _  c! |, a9 ^7 w* S# M) ~组间同步能力与时间t的关系出现了！
6 z7 [: p5 ^9 g9 Q5 X6 `" c
也就是说我先用上面的那个公式4计算出来\theta 的值，在带入到公式5，那么t-r关系就可以明确下来了，那现在我们再回过头来看看文章中已经给的例子，看看还有没有未知量。

8 h8 v/ s# k' u) B# M& z栗子是这样的栗子：
/ j0 g8 s9 h, u( ^8 Q9 c
假设某机构内部有 4 个编组，每个编组包括 5 个节点(其中 1 个节点为领导节点) 。另外，将上级领导作为一个独立的编组，且只包含一个节点。假设在领导机关增加4名信息传递人员。当以独立编组模式编组时，指定1名信息传递人员为指挥者，其指挥关系与其他编组一致; 当分散编组时，信息传递力量节点的关系与所在编组其他节点指挥关系一 致。其中，完全分散编组模式时，各信息传递力量节点之间无信息共享通道; 不完全分散编组时，在各信息传递人员节点之间建立一条信息共享通道。各编组模式及其拓扑结构如下图所示：

4 I& `, M: w( Z( {/ D独立编组模式


完全分散编组
; Y/ J4 J1 @# L5 K. R6 p! n0 p5 W! l: L

% s8 H0 y$ X$ d6 E
不完全分散编组



6 T( u* T7 Z, J, @/ m参数数据
8 I( M) g) A3 u! @" d3 l0 g; A



参数确定一下有没有未知量：
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) `2 l( l, |& E7 Q. O/ P& o首先N，数据数目已知，这个有了。
1 G' ]7 g" t; U6 \8 E# q! A7 \
K值是分组内的连接强度，这个是看实际情况，由甲方提供或者自己看着给的，这里就是甲方给的编组图，i与j点的链接强度一目了然，这个有了。

\omega _{i} 是振子i的固有频率，也称自然频率，甲方会给，没法自己估计，这个有了。
; y' ?8 ^- p% p
9 }2 }3 I5 x$ d! E5 A\theta ，\theta 怎么办，初始的\theta 会给，自己也能测的出来，但那么多\theta {j} -\theta{i} 得多少不知道啊，这里通过翻看文章，我发现其实文章是有一个特殊条件的，不然的话是需要研究耦合因子求三种约束条件解情况的，特殊条件就在这：

假设编组内节点的初始相位差为π/2，且编号最小者为0，随编号增大而增大。

' r& q/ N( U' ?# O哦，初始相位差知道了，你还告诉了我各个初始相位，那么\theta _{j} -\theta _{i} 的值就在一个范围内的几个固定值里面啊！

好的，没有未知量了，就是找K的时候麻烦点，没办法，这个决定了编组的不同，写脚本算一下吧：
% L- S) i/ C- u* D
#codingutf-8

##ScriptName:KuramotoSimulation.py
6 V& J& P$ ~* M; }3 B; I
3 p+ }0 c& S* b  v( I0 J! P2 Pimport matplotlib.pyplot as plt

; s! Q8 n7 ]5 {! ?2 M. S2 P+ l7 `. Xfrom pylab import

from sympy import
( \3 ^! q' W8 b3 V9 ]- _$ I
* i& I) Y5 I7 [+ `) X: Ifrom matplotlib.ticker import MultipleLocator, FormatStRFormatter

import math

5 `$ h3 t1 x0 ~  M& F+ l( ^: F1 nimport numpy as np

N = 31      #总节点数
# U7 }+ `% ?  D0 o( O4 O4 b
c=[0,0,math.pi  2,math.pi,3  math.pi 2,0,0,0,math.pi  2,math.pi,3  math.pi 2,0,0,0,math.pi  2,math.pi,3  math.pi 2,0,0,0,math.pi  2,math.pi,3  math.pi 2,0,0,0,math.pi  2,math.pi,3  math.pi 2,0,0]
6 C; z& h3 L% D5 {1 A# t
- F. t$ f: f- H0 s1 Iw = [4,3,3,2,2,1,4,3,3,2,2,1,4,3,3,2,2,1,4,3,3,2,2,1,4,3,3,2,2,1,5]
; [3 V4 f& x) k! w
5 q' i: q( [6 `9 h# O( l/ ik1 = [
) M1 |5 J! X$ M6 f  b9 S9 L
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
# k7 r- l: z; K7 ]' i
1 c. P* c* n3 x! W8 a' |( `[1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2],
2 e; M! Q5 G2 d3 y$ k! Y
[1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

) N! ?% u' Q  c, {/ m7 Z[1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

[1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

[1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

* y! q/ \6 A7 v; }- W' Z& a: t[1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

[0,0,0,0,0,0,1,0.9,0.9,0.9,0.9,0.9,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2],
, j' ?2 o, k0 s! G. _- C
[0,0,0,0,0,0,0.9,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

[0,0,0,0,0,0,0.9,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

" m) p+ k) {( B& d1 `2 b, v6 s[0,0,0,0,0,0,0.9,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
+ W1 I5 Y3 [0 D
; o1 J. M4 H3 D$ F" R  {2 m; m" f[0,0,0,0,0,0,0.9,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

8 Y$ |* w3 p$ a  K$ n[0,0,0,0,0,0,0.9,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
8 N% G( x+ O6 p/ p7 j: L
: n. c8 O% z6 _- H8 f[0,0,0,0,0,0,0,0,0,0,0,0,1,0.8,0.8,0.8,0.8,0.8,0,0,0,0,0,0,0,0,0,0,0,0,2],
4 B, A9 X# o6 B
[0,0,0,0,0,0,0,0,0,0,0,0,0.8,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
: I) m, z6 e) F9 T6 o) H
4 [6 J8 |; G/ D9 e[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

2 s/ {  d! F; `! T[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
, W# X" G9 h. }
[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0],
4 @9 `: L/ t/ x3 ~: N6 g3 \
" a7 v/ g$ u7 q+ h/ Q# T5 G[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0.7,0.7,0.7,0.7,0.7,0,0,0,0,0,0,2],
/ V3 y( l) z/ S2 }! l$ s" }. j) I
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,1,0,0,0,0,0,0,0,0,0,0,0],

, Y9 {8 h) R+ j' _2 ?% E[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,1,0,0,0,0,0,0,0,0,0,0],

[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,1,0,0,0,0,0,0,0,0,0],

5 i8 f7 F1 ?9 `[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,0,1,0,0,0,0,0,0,0,0],
8 M% \0 C! _3 `. \& K7 _& m: b
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,0,0,1,0,0,0,0,0,0,0],
  L; M  w. B! A9 Z/ g
6 I: d$ W: p$ {/ W1 L, s! Q- h" `* ^[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0.6,0.6,0.6,0.6,0.6,2],
  S  }$ Z2 ~/ I# S, [
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,1,0,0,0,0,0],
0 t+ n- L. a4 q
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,1,0,0,0,0],
3 u! n. M0 A5 W' ^% e4 z6 B
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,1,0,0,0],
( W) q4 N9 B% B
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,1,0,0],

[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,0,1,0],
( t0 J" P# c4 E  m0 m
[2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,1]]   

: H+ T+ N) p1 W+ n, pn = [i + 1 for i in range(22)]                #目标划分,24个值,1-24

. j* b. K7 k2 t& U' `7 bt = [j  for j in range(1000)]

ci = 0

C = []

C.append(c)
. g, c& j0 F. x2 n' T
for d in range(1100)

    for i in n

6 B6 o3 J; S# y5 }# x        for j in range(22)

- q6 u8 w/ k. \8 m3 L. p. d            cj = c[j + 1] - c
0 c+ ^+ G7 Q5 _- i- ~$ V" ^* I) T
  \. e; e1 g# z$ _' `1 e( d/ x            ci += k1[j + 1]  math.sin(cj)

5 L# o1 \+ Q0 W2 U        cii = ci + w

7 r5 o: ~7 O$ V* ?+ l" e        h = [0.01  cii + z for z in c]
+ X2 w& O# V9 N1 b9 g0 I, ?+ c
        C.append(h)

        c = h
+ ^8 a# r( |2 I/ {
def r_function(u)
0 f: y9 W% x3 B( j2 R% o6 d
    y1 = 0
0 D& w) H* J& @, f
    y2 = 0

2 o! n- q+ K& b7 L1 |! E    y12 = 0
. W# a/ _* K& {( [( J( d2 m. ]$ E- J$ l2 k# V
- X9 h% k) p( a' s    y22 = 0
7 D. Y. J/ Y: G9 e: T- I+ e6 C0 b
, g2 Z: i: }$ k& v9 C    r = []

    for x in range(1000)

  F: A4 D5 p. }3 X& p5 s4 G        for ul in u

            y1 += math.cos(C[x][ul])
/ g; \9 H+ S+ I, E4 i% _
6 _2 G0 Y9 k2 [3 |9 W7 i% A            y2 += math.sin(C[x][ul])
! k9 L4 o' l1 U4 ]) [
* i- F) M/ B2 q1 Z0 m( E9 a: e        y12 = y1  2
0 r: Q3 O0 j4 t/ @4 I
; Y9 O+ N. M" v3 X) z* p        y22 = y2  2

* H. e* ~! D8 B- Y- R- `2 L7 [        r.append((float(1)  N )  ((y12 + y22)  0.5))

    return r

  Q% y0 J7 S/ \4 Z9 {8 \r1 = r_function([1,2,3,4,5,6])
0 [/ e( ~. C3 V; F  [
r2 = r_function([7,8,9,10,11,12])
9 r9 h' R! j2 q4 {
9 y$ m; J. H3 u9 ir3 = r_function([13,14,15,16,17,18])
8 p6 Q( k( R- x, q' V5 w
) q+ q+ S' F1 ?  f7 }* Rr4 = r_function([19,20,21,22,23,24])
' {/ A8 U( e  E1 \
r5 = r_function([25,26,27,28,29,30])
! y! w' {5 T! s! j" N
#r6 = r_function([31])
" n8 |) z, T7 s/ J+ O; g
. K; @- D! D( ^

# P" ?4 L! L. S$ Hax = subplot(111) #注意一般都在ax中设置,不再plot中设置

4 u- x9 m0 ~. m  OymajorLocator  = MultipleLocator(0.1) #将y轴主刻度标签设置为0.5的倍数
# X, {, v% \, R9 U2 c/ c
+ k. w6 s! ?( W! n0 H) p6 pax.yaxis.set_major_locator(ymajorLocator)
% F2 @) r. H' K: s0 [
plt.plot(t, r1, marker='o', color='green', label='1')

plt.plot(t, r2, marker='D',color='red', label='2')

! Q! B# Q  a' g7 zplt.plot(t, r3, marker='+',color='skyblue', label='3')

: L: p- ^7 R# Z! H$ o+ _+ }plt.plot(t, r4, marker='h', color='blue', label='4')
& M) `: R& J8 a5 c7 T8 d( z
plt.plot(t, r5, marker='_',color='yellow', label='5')

' s2 C# `& z1 X! @/ v& _4 U0 F#plt.plot(t, r6, color='red', label='6')


' X6 F+ T# X) M* v
: n" ]( V4 ?' A( e2 V. {+ b3 d( v+ oplt.legend() # 显示图例
+ `+ z9 d$ R' O9 F' Y* c
3 d, y, G5 r# ]7 E/ T& x6 tplt.xlabel('iteration times')

plt.ylabel('r')

+ l% c9 e8 F' S, I3 p* |5 ]. j( Mplt.show()
3 Z# D7 V) p4 z: V. V9 N, T

独立编组结果如图：
* y, Z1 a" p+ O" v/ F2 }% |- h, x
4 b2 j/ a4 t* E) z

k1独立编组
好的，从图像我们来看看Kuramoto模型在描述这个编组的时候，5组最终稳定，我们说这个团队编组还算科学，但我们改变一下K的值，换成分散编组：

8 W+ I5 f, C8 |# I0 n, Gk2 = [
! O0 l! }. z, H' r5 O3 [6 H0 [( _
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
+ J6 G8 }+ O0 G' ~
% V$ t8 W3 w% w$ _) l6 u& t- s[1,1,0.9,0.8,0.7,0.6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2],

5 i/ X4 o. G6 {; v[1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
( B/ K2 I' {! V4 k, b- S% O! t
[0.9,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

9 _* j! ~5 O- I[0.8,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

[0.7,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
8 g! G2 `# p7 [' m. Y
8 J- b+ W# f$ ]: j: a: B[0.6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

[0,0,0,0,0,0,1,1,0.9,0.8,0.7,0.6,0,0,0,0,0,0,0,0,0,0,0,2],

) ]' o5 W" W: O4 k  a( e) G! `[0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

" h( m- ]: u9 t1 b5 r3 g[0,0,0,0,0,0,0.9,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

[0,0,0,0,0,0,0.8,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
, f; w: j0 w7 Y' r& y3 E* p9 F
[0,0,0,0,0,0,0.7,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0],

" N9 i5 G$ n$ h/ r6 O[0,0,0,0,0,0,0.6,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0],
* O' D$ n& h/ r1 u
7 N1 l' H: h1 E2 R+ f% X0 R[0,0,0,0,0,0,0,0,0,0,0,0,1,0.9,0.8,0.7,0.6,0,0,0,0,0,0,2],

[0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0],
# t; F0 d3 }% U. v
[0,0,0,0,0,0,0,0,0,0,0,0,0.9,0,1,0,0,0,0,0,0,0,0,0],

[0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,1,0,0,0,0,0,0,0,0,0],

, w/ P! H" y, [/ d[0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,0,1,0,0,0,0,0,0,0,0],

[0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,0,1,0,0,0,0,0,0,0],

[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0.9,0.8,0.7,0.6,2],
* b" E# A1 m1 I  D& ^" l
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0],

[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.9,0,1,0,0,0,0],

1 n" W( }" d, E' g- ]8 G2 j4 c[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.8,0,0,1,0,0,0],
: m8 w  l8 w/ ~& R+ i
3 P2 h: C8 i1 Y$ h2 C[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.7,0,0,0,1,0,0],
+ a2 t8 v, t# M' X
& u% `) o5 n- v+ Q/ k[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.6,0,0,0,0,1,0],
( A3 u$ S# W  B
0 e' x9 S3 Y  \' s[2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,2,0,0,0,0,0,1]
5 r6 i. |3 T) ~) D- W
- {; r: D0 j1 V+ p7 S9 y
& b: [5 A$ m! q, n" X
k2独立编组



. x& F3 W( Z1 p7 T# }+ \% j$ H4 k这两幅图像都在开始阶段大幅波动，而后在一定范围内趋于稳定，那么到底哪个分组模式最符合实际，最能突出编组能力呢？

  Y8 P. Q9 O8 f5 O8 P% D这里还有一个公式，来解决这个问题，编组同步能力的量化：

8 |/ x0 V" i" l6 F, n
; F& ]( Z" c9 C: s2 `* s. n1 i1 ^M_{s} 就可以描述某编组的同步效果，r_{s} 是达到稳定状态后序参量的均值，β∈(0,1)是调节因子。我们可以用M_{s} 比较编组内部的好坏。那编组间能力的好坏怎样比较呢？

5 N# B- n  i3 y2 `8 P5 n5 ]这个Kuramoto模型同样有所考虑，它有一个描述整个系统编组能力的公式：
3 }7 t2 {0 C0 V# X- ?9 ~+ ?
' o7 h- K' Y: L4 O& x0 H: P- o
其中，P是编组的数量，M_{i} 是第i个编组的同步能力，\omega _{i} 是编组在整个系统中的权重，\psi {e} 是各编组平均 相位的均值，\Delta \psi 是各编组平均相位的标准差。具体的计算不是这篇文章的重点，就不在计算M{s} 和M的值来比较上述例子独立编组和分散编组的好坏了，本篇文章主要是讲下Kuramoto模型的解决思路，尤其是上面解决\theta 值的方法可以套用在其他Kuramoto模型中，做一个目标估计绰绰有余的。

6 T0 g  X% g. @: P: U; J3 l下面是解决Kuramoto模型常用的MATLAB编程方法，具体思路与上述基本一致，这里不再赘述，K的值我们给另一种编组模式：不完全分散编组模式，也是现在实际上最长用的编组方式，直接上代码：
/ z7 c( M" ]' _( r, P, V6 ]7 z
clc;
( L' G3 B8 Q, h# w  j
5 ]( E3 B1 [& \4 j2 [clear all;
9 v$ a' ~: m9 h1 Q) h$ U- m  K
7 ~" s& t  _0 b$ h5 Xk=[0  1  1  1  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5;

  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   

  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   
3 ~& A+ K6 h3 m, k6 q
: y0 {2 D' d7 z$ Y  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   
, j4 J4 f. K- ?+ g5 t, |
  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   

  0  0  0  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5;

  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   

  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   
( S1 R% {! U& ]2 E1 V/ R. ]
2 R& x- W8 r: Y9 U) i4 y' i  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   
6 L+ G1 n- \. A
- L5 ^: s' G9 E3 E" I0 w  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   

3 Q9 X$ Q. @2 m8 ~  0  0  0  0  0  0  0  0  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  0  0.5;

' @& {' h  U- d9 X7 t  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   

  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   

7 t, t2 c, P) w5 k: L  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   
9 j8 g( n. M3 t6 [) {! |5 K; L; V
8 A7 \! L3 G1 F2 m  C8 v  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0;   
4 i' d% q: L* o/ i4 l* z3 Y
  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  1  1  1  0  0  0  0  0.5;
0 n; |2 ^% v0 r& i/ Z, g
$ Q7 T/ g! h  N. x8 T3 \  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0;   
9 ~0 J2 z4 c1 T# j" W4 g) R  h: v
5 y* c* ^# r4 w' Y8 z3 d9 o) l. a  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0;   

  p4 A( l( [+ E0 v1 Q  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0;   
7 |4 m* M9 M2 s# f
  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0;   

  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  2  2  2  0.5;

  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  2  0  0  0  0;   

6 t4 m* ~5 K+ ^# w- q  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  2  0  0  0  0;   

  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  2  0  0  0  0;   

9 w4 {/ X. P1 E2 W7 n- W  0.5 0 0  0  0  0.5 0 0  0  0 0.5 0  0  0  0  0.5 0 0  0  0  0.5 0 0  0  0;   

  ]     

N = 25;

Step= 10000;
* {7 R5 P/ F6 L! k! w: s% f
Theta=zeros(Step,N);
: J. p; q2 x( `" ~* j2 v
Omega =zeros(Step,N);
; N  k! w$ n+ \1 E$ u
DeltaT=0.01
; c  i3 I+ F% T( r
7 E5 @( ^% N0 i: n; H3 f4 M: qTheta(1,：）=[0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0 pi/2 pi 3*pi/2 0];

Omega(1,：D=2*pi*[2  3  3  4      4 2    3  3      4 4    2  3      3 4    4  2      3 3    4  4      2 2    3  4      1];
+ P8 c+ W0 n7 V0 [8 {. U+ V) J
%  求解微分方程

& W$ M" l+ s3 t! \0 w* afor n = 1:Step
* F, Y& E7 @6 K1 e# [
    for i = 1:N
, G$ g5 m8 _8 i5 x9 k8 {* N
            Sigma = 0;

  _' K* [( z6 Z! _: j7 g' t        for j = 1:N
; R0 c& D9 C# P# b1 J# y$ s, H5 ^
            %%判断k对同步效果的影响
! S. |1 ~+ L" H8 z9 }
= Sigma + k(i,j) * sin( Theta(n,i) - Theta(n,j) );

* L" F4 R) u$ ~- G6 `! l" k        end

! l- j4 V6 |5 C) U+ I            Omega(n+1,i) = Omega(n,i) - Sigma;

            Theta(n+1,i) = Omega(n+1,i)*DeltaT + Theta(n,i);
7 Z3 k2 |$ _, }% S7 |5 Y
& Y: R6 R" X3 g& R0 D    end
/ E" ?0 l3 I" A. R6 z/ t
end
$ B# K) \7 t* b2 H2 ?- a
%  求解序参量r(t)

groupN = 5;

Step = 1000;
8 J( R) ?* c: \; b: U/ w+ h
0 y) \8 _4 w$ `6 `% \- D0 [for i = 1:Step
4 O& x: g" z+ t- Z+ R, M
8 I) C+ y. K3 q+ e/ Y- A5 a    sigma1 = 0;
9 g, ]# b& g; H' W! w$ F' W( g
    sigma2 = 0;
6 i" i+ p7 e4 h( l
: |2 M9 q& i8 X: c! _' k( V6 M/ D! e    sumTheta = 0;

    for j =6:10  %表示相应的群组
8 k. R% |& b, j- H/ ]
+ _2 _+ Z; |5 I. a6 h! W        sigma1=sigma1+cos(Theta(i,j));

        sigma2=sigma2+sin(Theta(i,j));
9 f! P, C9 Q8 Z3 S
    end

$ i5 b& m" I# S9 j- d2 }    r(i)= sqrt( sigma1^2+sigma2^2 )/groupN;

& i) x. c( t& ^end
  G( c: r) u9 _& e
( r2 M  k) F* C3 r0 C+ tx= 0.01：DeltaT：DeltaT*Step;
8 ?$ ~3 |+ S6 s5 j1 O, C
plot(x,r);
. U) g: u6 f& b' M/ yMATLAB版不完全分散编组结果如下图：

regngfpcb

Kuramoto模型在Python和MATLAB中的简单实现