Statement of problem:
Taken's theorem shows that we can project a version of the stable attractor for the Lorenz system by looking at a time series form. (The theory is not so important in this case, I'm more concerned with the algorithm I'm implementing on matlab and making it work.)
The Lorenz system of course is given by $\begin{equation}\begin{cases}\frac{dx}{dt}=\sigma y-\sigma x\\ \frac{dy}{dt}=x\rho-y-xz\\ \frac{dz}{dt}=xy-\beta z\end{cases}\end{equation}$.
I'm only considering the case for parameters given by $\sigma=10,\,\rho=28,\,\beta=8/3$.
So if $(x(t),y(t),z(t))$ is a solution, then the mapping has the following simplified form $$\phi_{\tau}(x(t),y(t),z(t))=\langle x(t),x(t-\tau),x(t-2\tau),...,x(t-K\tau)\rangle.$$
Here $\tau>0$. So I'm trying to implement the time delay mapping on matlab for values $K=1$ and $K=2$ and subsequently find the value $\tau$ that will give me the right version of the attractor. By the way, I used euler's method to solve the Lorenz system in this case.
function lorenz
sigma=10;
rho=28;
beta=8/3;
h=0.1;
n=10;
tau=1;
x=[0;-4;-1];%initial conditions%
for i=1:n
z=x(:,i)+h*euler(x(:,i),sigma,rho,beta);
x=[x z];
if i>=3
x(2,:)=x(1,i-tau);
x(3,:)=x(1,i-2*tau);
end
end
figure(1)
plot3(x(1,:),x(2,:),x(3,:));
figure(2)
plot(x(1,:));
function x=euler(x_old,sigma,rho,beta)
x=[0;0;0];
x(1,1)= sigmax_old(2,1)-sigmax_old(1,1);
x(2,1)= rhox_old(1,1)-x_old(2,1)-x_old(1,1)x_old(3,1);
x(3,1)= x_old(1,1)x_old(2,1)-betax_old(3,1);
So am I doing this right? I'm getting it to graph but it's not the same as the Lorenz attractor