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using runge kutta 4th order to integrate three different dynamic equation

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Samson
Samson 2019-7-18
编辑： Jan 2019-7-19
% silmulating the rossler and lorenz attractor
% rossler is the driver
%lorenze as the response
% dx = [-(x(2)+ x(3));x(1) + a*x(2);b + x(3)*(x(1)-c)];
%ds = [sigma*(s(2)- s(1)); s(1) *( rho-s(3))-s(2); s(1)*s(2)-beta*s(3)];
% dz = [sigma*(z(2)- z(1));z(1) *( rho-z(3))-z(2);z(1)*z(2)-beta*z(3)];
a= 0.2; b= 0.2;c= 5.7; sigma= 16; beta= 4; rho = 45.92;g = 0;
% Define function handles
dx = [
-(x(2)+ x(3));
x(1) + a*x(2);
b + x(3)*(x(1)-c);
sigma*(s(2)- y(1))-g*(s(1)-x(1));
s(1) *( rho-s(3))-s(2);
s(1)*s(2)-beta*s(3);
sigma*(z(2)- z(1))-g*(z(1)-x(1));
z(1) *( rho-z(3))-z(2);
z(1)*z(2)-beta*z(3);
];
% setting initial conditions
epsu = [0.06;0.01;1];
epsv = [0.85;0.85;1];
epsw = [0.05;0.01;1];
dt = 0.01;
tspan = (dt:dt:1000);
x = zeros(3,tspan(end)/dt);
s = zeros(3,tspan(end)/dt);
z = zeros(3,tspan(end)/dt);
for i = 1: tspan(end)/dt
time = i*dt;
k1x = fx(epsu(i), x(i), a,b,c);
k1y = fx(epsv(i), x(i),s(i),z(i));
k1z = fx(epsw(i), x(i),s(i),z(i));
k2x = fx(x(i)+(dt/2),x(i)+(dt/2)*k1x,s(i)+(dt/2)*k1y,z(i)+(dt/2)*k1z);
k2y = fy(e(x)+(dt/2),x(i)+(dt/2)*k1x,s(i)+(dt/2)*k1y,z(i)+(dt/2)*k1z);
k2z = fz(e(x)+(dt/2),x(i)+(dt/2)*k1x,s(i)+(dt/2)*k1y,z(i)+(dt/2)*k1z);
k3x = fx(e(x)+(dt/2),x(i)+(dt/2)*k2x,s(i)+(dt/2)*k2y,z(i)+(dt/2)*k2z);
k3y = fy(e(x)+(dt/2),x(i)+(dt/2)*k2x,s(i)+(dt/2)*k2y,z(i)+(dt/2)*k2z);
k3x = fz(e(x)+(dt/2),x(i)+(dt/2)*k2x,s(i)+(dt/2)*k2y,z(i)+(dt/2)*k2z);
k4x = fx(e(x)+(dt),x(i)+(dt)*k3x,s(i)+(dt)*k3y,z(i)+(dt)*kz);
k4y = fy(e(x)+(dt),x(i)+(dt)*k3x,s(i)+(dt)*k3y,z(i)+(dt)*kz);
k4z = fz(e(x)+(dt),x(i)+(dt)*k3x,s(i)+(dt)*k3y,z(i)+(dt)*kz);
x(i+1)= e(x)+(dt/6) *(k1x +2*k2x +2*k3x +k4x);
s(i+1)= e(y)+(dt/6) *(k1y +2*k2y +2*k3y +k4y);
z(i+1)= e(z)+(dt/6) *(k1z +2*k2z +2*k3z +k4z);
end
COMMAND WINDOW: ERROR MESSAGE BELOW
Undefined function or variable 's'.
Error in silmu (line 15)
sigma*(s(2)- y(1))-g*(s(1)-x(1));
Kindly tell me where am wrong please

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回答（1 个）

Jakob
Jakob 2019-7-18
In the code you shared, when you call
sigma * (s (2) - and (1)) - g * (s (1) -x (1));
s (1) * (rho-s (3)) - s (2);
s (1) * s (2) -beta * s (3);
s is not defined.

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