how to calculate synchronization error diagrams

% 定义系统参数

parameters = [1; 1; 4.92; 1; 1; 0; 1183; 303900]

% 定义初始条件

X0 = [0.1; 0.2; 0.3; 1.5; -1; 2]

% 定义时间跨度

tspan = [0 10];

% 定义系统方程（4.3）

[time, Y] = ode45(@(t, Y) error_system(t, Y, parameters), tspan, X0);

function dydt = error_system(t, X, Y, parameters)

% 系统参数

a = parameters(1);

b = parameters(2);

h = parameters(3);

v1 = parameters(4);

v2 = parameters(5);

k1 = parameters(6);

k2 = parameters(7);

k3 = parameters(8);

% 从 Y 中提取误差项

Y(1)=X(4) - X(1);

Y(2)=X(5) - X(2);

Y(3)=X(6) - X(3);

e1 = Y(1);

e2 = Y(2);

e3 = Y(3);

% 计算导数

dydt = [-v1*e1 + h*e2 - a*X(2)*X(3) - k1*e1;

h*e1 - v2*e2 + a*X(4)*X(6) - a*X(1)*X(3) - k2*e2;

a*X(4)*X(5) -a*X(1)*X(2) - b*e3 - k3*e3];

end

Answer 1

[Arnav]

Hi @凯 何,

I understand that you are trying to solve the following system of equations.

The first system is shown below:

The second system needs to be rewritten in the following form as it is currently under-determined in the code:

After the first system is solved, the 2nd system would no longer be under-determined. Assuming the initial conditions for the variables to be , the initial condition for the error becomes .

Since the first system is independent of the second, it can be solved first as shown:

% Solve the first system 
[t1, sol1] = ode45(@(t, y) firstSystem(t, y, a, b, h, v1, v2), tspan, [0.1 0.2 0.3]); 
% Function for the first system of equations 
function dydt = firstSystem(~, y, a, b, h, v1, v2) 
    x1 = y(1); y1 = y(2); z1 = y(3); 
    dydt = [    h*y1 - v1*x1 - a*y1*z1; 
                h*x1 - v2*y1 + a*x1*z1; 
                a*x1*y1 - b*z1 ]; 
end 

Then the second system can be solved by using the result of first system. This can be done as shown:

% Solve the second system 
[t2, sol2] = ode45(@(t, e) secondSystem(t, e, sol1, t1, a, b, h, v1, v2, k1, k2, k3), tspan, [1.4 -1.2 1.7]); % Last argument is initial conditions 
function dedt = secondSystem(t, e, sol1, t1, a, b, h, v1, v2, k1, k2, k3) 
    % Interpolate x1, y1, z1 from the first system's solution 
    x1 = interp1(t1, sol1(:, 1), t); 
    y1 = interp1(t1, sol1(:, 2), t); 
    z1 = interp1(t1, sol1(:, 3), t); 
    ex = e(1); ey = e(2); ez = e(3); 
    dedt = [    -v1*ex + h*ey + a*y1*z1 - a*(ey+y1)*(ez+z1) - k1*ex; 
                h*ex - v2*ey + a*(ex+x1)*(ez+z1) - a*x1*z1 - k2*ey; 
                a*(ex+x1)*(ey+y1) - a*x1*y1 - b*ez - k3*ez  ]; 
end 

The error can be plotted against time using plot(t2, sol2) command.

The interp1 function is used to interpolate values from the solution of the first system. You can learn more about interp1 function by referring to its documentation page:

https://www.mathworks.com/help/releases/R2024a/matlab/ref/interp1.html

Hope it helps.