How to make the function of the planes from the lorenz attractor

Question

How can I make the equation for the planes of the wings of the lorenzattractor, I know that the critical point should be used.

But I don't know how I should make this plane, the figure shows in yellow which plane I want.

Answer 1

The "wings" don't lie in a plane; the predominantly blue portion on the right of your image seems to indicate that clearly. You can linearize the system at the unstable fixed points to figure out how the system behaves like a linear system near those points, though. The plane that I suspect you are talking about is somewhat akin to a tangent plane to a surface.

To do this, first define

$$ \vec{F}(x,y,z) = \left(\begin{matrix} u(x,y,z) \\ v(x,y,z) \\ w(x,y,z) \end{matrix}\right) = \left(\begin{matrix} \sigma(y-x) \\ x(\rho-z)-y \\ xy-\beta z \end{matrix}\right), $$ then compute the Jacobian of this function $$ \frac{\partial (u,v,w)}{\partial (x,y,z)} = \left(\begin{matrix} \partial u/\partial x & \partial u/\partial y & \partial u/\partial z \\ \partial v/\partial x & \partial v/\partial y & \partial v/\partial z \\ \partial w/\partial x & \partial w/\partial y & \partial w/\partial z \end{matrix}\right) = \left( \begin{matrix} -\sigma & \rho-z & y \\ \sigma & -1 & x \\ 0 & -x & -\beta \\ \end{matrix} \right). $$ Next, plug one of the fixed points $(x,y,z)=\left(\pm\sqrt{\beta(\rho-1)},\pm\sqrt{\beta(\rho-1)},\rho-1\right)$ and the parameters you're using into the Jacobian. Finally, compute the eigen system of the evaluated Jacobian matrix. If all goes well, one eigenvalue should be real and negative, while the other two should be complex conjugates. The real and imaginary parts of the complex eigenvector should be parallel to the plane of rotation of the linearized system. Taking the cross product of those vectors, we can generate the vector shown in the image below:

I emphasize again, though, that the "wing" of the attractor is not contained in a plane.