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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:

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function [sol1] = two_system
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);
plot(sol1.x, sol1.y(1,:),'--r','LineWidth',1.5)
hold on
plot(x1, y1(:,1),'b','LineWidth',1.5)
function dydt = system1(t,y,D)
global z MD;
M1 = y(1);
MD = D(1);
s1 = MD %for checking
F_M1= 2;
A1 = 0.01;
%%%%%%% equation_system1 %%%%%%%
dydt = [dMdt];
function dydt = system2(t,y)
global z MD;
s2=MD %% for checking
M2 = y(1);
F_M2= 2;
A2 = 0.01;
%%%%%%% equation_system2 %%%%%%%
dydt = [dMdt];
mohammed on 20 May 2024 at 18:58
@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 on 20 May 2024 at 20:02
Edited: Torsten on 20 May 2024 at 20:49
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
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.

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Accepted Answer

Sam Chak
Sam Chak on 20 May 2024 at 16:44
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
ans = 1x2
2 52
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%% Plot results
plot(t, M(1,:), '-o', 'linewidth', 1.5, 'color', '#F09494'), grid on
xlabel('Time t'), ylabel('Amplitude'), title('Solution M1')
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];
%% historical data
function s = history(t)
s = [0;

More Answers (1)

Sam Chak
Sam Chak on 21 May 2024 at 12:02
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');
plot(t, x(1,:), 'linewidth', 1.5, 'color', '#115AD0'), grid on, ylim([-3 3])
xlabel t, ylabel Amplitude, title('Response, x1(t)')
plot(t, x(3,:), 'linewidth', 1.5, 'color', '#EC3114'), grid on, ylim([-3 3])
xlabel t, ylabel Amplitude, title('Response, x3(t)')
plot(x(1,:), x(2,:), 'linewidth', 1.5, 'color', '#F1BC5F'), grid on, axis([-3 3 -5 5])
xlabel x_{1}, ylabel x_{2}
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(3) + (1 - x(3)^2)*x(4) + x1Lag(1)*factor];
%% historical data
function s = history(t)
s = [1; % initial value
  1 Comment
mohammed on 22 May 2024 at 7:21
Edited: Torsten on 24 May 2024 at 22:47
@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

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