beginconditions should give 0, Lorentz system - matlab

Question

I want to plot the Lorentz system (s=10, r=28, b=8/3) for a start condition which should give 0 since it's a start condition in the plane spanned by v1 and v2 and trough the critical point (0,0,0) (equation of the plane : -x + (-9- \sqrt(1201))/56 * y = 0) When I use the Runge-Kutta method to plot the solution with as start conditions ((-9-sqrt(1201))/56,1,10) my graph doesn't converge to 0 and I don't know why.
[X,Y,Z,T] = Runge(T0,(-9-sqrt(1201))/56,1,10,h,1000);
plot(T,X);
plot(T,Y);
plot(T,Z);
I expect a solution in which the lines in the plots go to 0. But I get some random function which goes up and down all the time.
Could this be caused by approximations?
Thanks in advance
This is the function Runge
function [X,Y,Z,T] = Runge(t0,x0,y0,z0,h,n)
X=[x0];
Y=[y0];
Z=[z0];
T=[t0];
k1x = 0;
k1y = 0;
k1z = 0;
k2x=0;
k2y=0;
k2z=0;
k3x=0;
k3y=0;
k3z=0;
k4x=0;
k4y=0;
k4z=0;
for k = 1:n
T(k+1)= T(k) +h;
k1x = F(X(k),Y(k),Z(k));
k1y = G(X(k),Y(k),Z(k));
k1z = H(X(k),Y(k),Z(k));
k2x = F(X(k)+h/2*k1x,Y(k) + h/2*k1y, Z(k) + h/2*k1z);
k2y= G(X(k)+h/2*k1x,Y(k) + h/2*k1y, Z(k) + h/2*k1z);
k2z= H(X(k)+h/2*k1x,Y(k) + h/2*k1y, Z(k) + h/2*k1z);
k3x= F(X(k)+h/2*k2x,Y(k)+h/2*k2y,Z(k) + h/2*k2z);
k3y=G(X(k)+h/2*k2x,Y(k)+h/2*k2y,Z(k) + h/2*k2z);
k3z=H(X(k)+h/2*k2x,Y(k)+h/2*k2y,Z(k) + h/2*k2z);
k4x= F(X(k)+h*k3x,Y(k)+h*k3y,Z(k)+h*k3z);
k4y=G(X(k)+h*k3x,Y(k)+h*k3y,Z(k)+h*k3z);
k4z=H(X(k)+h*k3x,Y(k)+h*k3y,Z(k)+h*k3z);
X(k+1) = X(k) + h/6 * (k1x + 2*k2x + 2*k3x + k4x);
Y(k+1) = Y(k) + h/6 * (k1y + 2*k2y + 2*k3y + k4y);
Z(k+1) = Z(k) + h/6 * (k1z + 2*k2z + 2*k3z + k4z);
end
end

Answer 1

There is no such plane for the non-linear system
Why do you think this plane is invariant?
In the system
F=#(x,y,z) sigma*(y-x);
G=#(x,y,z) x*(rho-z)-y;
H=#(x,y,z) x*y-beta*z;
no non-trivial linear combination of the first two equations is independent of z, which would be required to get an invariant expression in only x and y.
The invariant sub-spaces of the linearization are close to the stable and unstable manifolds, but they are not identical to them. So if you are on some invariant subspace of the linearization, you are in fact some distance away from any curved sub-manifolds related to the non-linear system, and numerical as well as exact solutions will move towards the chaotic attractor.
The Lorentz system is chaotic
Apart from that your observation is correct that floating point errors accumulate to drive the numerical solution away from the exact solution and any properties of it.