Lyapunov Exponent diagram of one variable out of 3D chaotic map

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tanveer haq 2019-4-9

1
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/455213-lyapunov-exponent-diagram-of-one-variable-out-of-3d-chaotic-map

编辑： tanveer haq 2019-4-9

I am using the code following code to find the lyapunov exponent of variable x:

clear all; clc
rStart = 3.3;
rStep = 0.001;
rEnd = 3.8;
rValues = rStart:rStep:rEnd;  
nPoints = length( rValues );
nIterations = 1000; % number of iterations
LE      = zeros( 1, nPoints );
x       = zeros( 1, nIterations + 1 );
x(1) = 0.1;
y(1) = 0.002;
z(1) = 0.022;
lamda = 3.79;
beta = 0.01;
alpha = 0.0112;
for k = 1:nPoints
  sum = 0;
  for i = 1:nIterations
      
        x(i+1)=k*(lamda*x(i)*(1-x(i))+beta*((y(i)).^2)*x(i)+alpha*((z(i)).^3)); % compute and save next x
        y(i+1)=(lamda*y(i)*(1-y(i))+beta*((z(i)).^2)*y(i)+alpha*((x(i)).^3)); % compute and save next x
        z(i+1)=(lamda*z(i)*(1-z(i))+beta*((z(i)).^2)*z(i)+alpha*((y(i)).^3)); % compute and save next x
        sum = sum + log( abs( rValues( k ) - 2*rValues( k )*x( i ) ) );   
  end
  LE( k ) = sum / nIterations;     
end
figure; plot( rStart:rStep:rEnd, LE, 'k.-' ); 