how to find lyapunov exponent for sine function sin(x) and how to plot it? if i have a code for logistic map? instead of logistic map lyapunov i want just sine lyapunov?

clc

clear all

close all

r = 1:0.01:4;

for j=1:length(r)

x(1)=0.1;

lyap = 0;

for i = 1:length(r)

x(i+1) = r(j)*x(i)*(1-x(i)); %logistic map equation

lyap=lyap+log(abs(r(j)*(1-2*x(i)))); % after abs there is dericative of the logistic map

end

lamda(j) = lyap/1000;

end

figure(1)

plot(r,lamda)

grid on

xlabel('control parameter')

ylabel('lyapunov exponent')

Answer 1

[Paras Gupta]

Hi Zubair,

I understand that you want to plot the Lyapunov exponent for the sine function.

The following changes can be made in the provided code to achieve the same:

• Instead of using the logistic map equation, the sine function equation x(i+1) = r(j)*sin(x(i)) is used
• The Lyapunov calculation has been updated to lyap = lyap + log(abs(r(j)*cos(x(i)))) based on the derivative of the sine function.

The modified code is given below:

clc;
clear all;
close all;
r = 1:0.01:4;
for j=1:length(r)
    x(1)=0.1;
    lyap = 0;
  
    for i = 1:length(r)
        x(i+1) = r(j)*sin(x(i));  % Sine function equation
        lyap = lyap + log(abs(r(j)*cos(x(i))));  % Lyapunov calculation for sine function
    end

    lamda(j) = lyap/1000;
end

figure(1)
plot(r,lamda)
grid on
xlabel('control parameter')
ylabel('lyapunov exponent')

You can refer to the following documentations for more information on the function used in the code above:

• sin: https://www.mathworks.com/help/matlab/ref/sin.html
• cos: https://www.mathworks.com/help/matlab/ref/double.cos.html

I hope the above modifications will help to get the required results.