In this work, I am trying to connect two systems by injecting a delayed wave of the first system into the second system. In this system, there are several problems, including:

function [sol1] = two_system

clc

close all

clear all

global z; %%problems

%%1- The indices for the two systems cannot be equal.

%%2- The global function is only valid for one value of (MD) ...

%%(the last value), while I need all values from the first system.

z= 0.05;

tau = 0.1;

tspan = linspace(0, 5, 300);

opt = odeset('RelTol',1e-3);

sol1 = dde23(@system1,tau ,[0],tspan, opt);

[x1, y1] = ode23(@system2, tspan, [0], opt);

figure(1)

plot(sol1.x, sol1.y(1,:),'--r','LineWidth',1.5)

hold on

plot(x1, y1(:,1),'b','LineWidth',1.5)

L2=length(x1)

L1=length(sol1.x)

end

function dydt = system1(t,y,D)

global z MD;

M1 = y(1);

MD = D(1);

s1 = MD %for checking

F_M1= 2;

theta1=0*pi/180;

A1 = 0.01;

%%%%%%% equation_system1 %%%%%%%

dMdt=(A1*(2*pi*(F_M1))*cos(2*pi*(F_M1)*(t)+theta1));

dydt = [dMdt];

end

function dydt = system2(t,y)

global z MD;

s2=MD %% for checking

M2 = y(1);

F_M2= 2;

theta2=0*pi/180;

A2 = 0.01;

%%%%%%% equation_system2 %%%%%%%

dMdt=(A2*(2*pi*(F_M2))*cos(2*pi*(F_M2)*(t)+theta2))+MD*z;

dydt = [dMdt];

end

[mohammed]

In this system, there are several problems, including:

1- The indices for the two systems cannot be equal.

2- The global function is only valid for one value of (MD) (the last value), while I need all values from the first system.

[Torsten]

Please explain what you finally want to achieve with your code.

[mohammed]

[Torsten]

I don't understand why you don't solve both systems simultaneously in dde23 (it's not obligatory that both equations are delay differential equations for this solver). In this case, the times are automatically synchronized.

[mohammed]

@Torsten This system is hypothetical in order to be useful in solving a more complex system involving synchronization between two chaotic systems.

When solving both systems simultaneously, there will be a problem, which is that the output of the first chaotic system changes every time the delay time (tau) changes, even if the value of the coupling strength is equal to zero (in this example here it is z = 0), while it should remain Fixed because the first system does not contain a delay term. On the other hand, the effect of tau is only on the output of the second system because it contains a delay term.

[Torsten]

I can't follow your description. Maybe I understand better if you write down the equations for the two systems you try to solve in a mathematical way.

Which value for MD from the solution of the first system do you want to insert for the second system in the line

dMdt=(A2*(2*pi*(F_M2))*cos(2*pi*(F_M2)*(t)+theta2))+MD*z;

?

If it is the solution of the first equation, evaluated at t-tau, I don't see the problem solving both equations simultaneously. If you don't use this term in the definition of first equation, the output of the first system won't change with tau, either.

@Sam Chak did exactly this in the code he suggested.

Answer 1

[Sam Chak]

Hi @mohammed

Based on the information you have provided, it appears the two systems could potentially be combined and then solved using the dde23 solver. However, I am uncertain about the specific approach you would like to use to calculate the correlation between and .

%% call dde23 solver
tau     = 0.1;
tspan   = [0 1];
sol     = dde23(@ddefun, tau, @history, tspan);
t       = sol.x;
M       = sol.y;

%% Check the lengths of solutions M1 and M2
size(M)

ans = 1x2

2 52

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%% Plot results
subplot(211)
plot(t, M(1,:), '-o', 'linewidth', 1.5, 'color', '#F09494'), grid on
xlabel('Time t'), ylabel('Amplitude'), title('Solution M1')

subplot(212)
plot(t, M(2,:), '-o', 'linewidth', 1.5, 'color', '#AAE2A8'), grid on
xlabel('Time t'), ylabel('Amplitude'), title('Solution M2')

%% Two-system DDE
function dMdt = ddefun(t, M, Z)
    M1lag   = Z(:,1);   % delayed state M1
    
    F_M1    = 2;
    theta1  = 0*pi/180;
    A1      = 0.01; 
    
    F_M2    = 2;
    theta2  = 0*pi/180;
    A2      = 0.01;
    z       = 0.05;
    
    dMdt    = [A1*2*pi*F_M1*cos(2*pi*F_M1*t + theta1);
               A2*2*pi*F_M2*cos(2*pi*F_M2*t + theta2) + M1lag(1)*z];
end

%% historical data
function s = history(t)
    s = [0; 
         0];
end

[mohammed]

@Sam Chak Thank you for wasting your time with me

I'm aware of this solution, but it doesn't achieve what I want.

Answer 2

[Sam Chak]

Hi @mohammed

Both @Torsten and I have been trying to understand the problem you are attempting to solve. However, it appears the selected value for the factor z is too small, causing the product term 'M1lag(1)*z' to become insignificant in your toy problem.

Would it be possible to instead run a simulation on the chaotic van der Pol oscillator, a double system? I can provide an initial setup with a constant delay of 20 seconds and a factor of 1. You can then adjust the parameters as needed and explain the specific observations you would like to make.

Exploring the chaotic van der Pol oscillator system may provide more meaningful insights than the current approach.

%% call dde23 solver
tau     = 20;       % time delay
tspan   = [0 40];
sol     = dde23(@ddefun, tau, @history, tspan);
t       = sol.x;
x       = sol.y;

%% Plot results
tL  = tiledlayout(2, 2, 'TileSpacing', 'Compact');

nexttile
plot(t, x(1,:),      'linewidth', 1.5, 'color', '#115AD0'), grid on, ylim([-3 3])
xlabel t, ylabel Amplitude,  title('Response, x1(t)')

nexttile
plot(t, x(3,:),      'linewidth', 1.5, 'color', '#EC3114'), grid on, ylim([-3 3])
xlabel t, ylabel Amplitude,  title('Response, x3(t)')

nexttile
plot(x(1,:), x(2,:), 'linewidth', 1.5, 'color', '#F1BC5F'), grid on, axis([-3 3 -5 5])
xlabel x_{1}, ylabel x_{2}

nexttile
plot(x(3,:), x(4,:), 'linewidth', 1.5, 'color', '#3A6A1E'), grid on, axis([-3 3 -5 5])
xlabel x_{3}, ylabel x_{4}

%% Double van der Pol oscillator
function dxdt = ddefun(t, x, Z)
    x1Lag   = Z(:,1);   % delayed state x1
    factor  = 1;        % similar to z in your System 2
    
    % ODEs
    dxdt    = [x(2);
             - x(1) + (1 - x(1)^2)*x(2);
               x(4);
             - x(3) + (1 - x(3)^2)*x(4) + x1Lag(1)*factor];
end

%% historical data
function s = history(t)
    s = [1;     % initial value
         0;
         1;
         0];
end

[mohammed]

@Sam Chak Hi, I really appreciate your interest

After reviewing the method of solving two systems at once by dde23, I noticed that this method was successful for the first model and was not the same for the second model.

In Figure of Model 2, we notice a mismatch in the output of system 1 (which acts as a transmitter) at different values of the delay time (tau), which is not written in the equations of system 1, but rather written in the system 2 (similar to what you discussed in the chaotic van der Pol oscillator 'M1lag(1)*z').

While in Figure of Model 1, the solution was successful using dde23.

In this case, I don't know where the error is