Lorentz Equations - Chaos and Stability

[unscientific]

Homework Statement

The figure below shows the path of a particle governed by the Lorenz equations with r = 28, σ = 10, b = 8/3. The x'es and boxes show points where the path crosses the plane z = r − 2σ > 0.

(a) Which indicator shows a decreasing z and which shows an increasing z?

(b) Show the length of element $| \delta x |$ between the two paths either grows or decays exponentially if aligned with one of the eigenfunctions of jacobian $\frac{J + J^T}{2}$.

(c) Find $\frac{J + J^T}{2}$ and its eigenvalues at (0, 0, r-2σ). Hence deduce that of $\delta x$ grows is in the x-y plane while it decays is along the direction of z-axis.

(d) Show the volume decreases exponentially with $\delta V = \delta V_0 e^{−(σ+1+b)t}$
Since $\frac{(J + J^T)}{2}$ is symmetric, its eigenfunctions are orthogonal. Show that for a cubic element where the three displacement directions are along the eigenfunctions in section (c) decays at the same rate.

Homework Equations

Lorentz equations are given by:

\dot x = \sigma(y-x)
\dot y = rx - y - xz
\dot z = xy - bz

The Attempt at a Solution

Part (a)
[/B]
For $\dot z < 0$, $xy < bz \approx 21$.

So the boxes represent decreasing z, the x'es represent increasing z.

Part (b)

How do I show it either grows or decays exponentially? Do I put in z = r − 2σ and find the eigenvalues? Wouldn't it be part (c)? I think this part is simpler than it seems.

Part (c)

The matrix $\frac{J + J^T}{2}$ becomes:

I found the eigenvalues to be:

\lambda_{1,2} = -\frac{ -(\sigma + 1) \pm \sqrt{ (\sigma+1)^2 - 4(\sigma - \frac{9}{4} \sigma^2) } }{2}
\lambda_3 = -b

Part (d)

\nabla \cdot \vec u = -(\sigma + 1 + b)

Then the result follows.
This is all that I have managed to do so far, would appreciate any input, thank you!

 

[vacuum]

Part (b)

To show that the length of element $|\delta x|$ either grows or decays exponentially, we can use the fact that the eigenvalues of $\frac{J + J^T}{2}$ determine the behavior of the system. We can also use the fact that the eigenfunctions of a symmetric matrix are orthogonal.

Let's consider the eigenfunctions of $\frac{J + J^T}{2}$ at the point (0,0,r-2σ). We found the eigenvalues in part (c) to be $\lambda_1 = -\frac{-(\sigma+1)+\sqrt{(\sigma+1)^2-4(\sigma-\frac{9}{4}\sigma^2)}}{2}, \lambda_2 = -\frac{-(\sigma+1)-\sqrt{(\sigma+1)^2-4(\sigma-\frac{9}{4}\sigma^2)}}{2}, \lambda_3 = -b$.

The corresponding eigenvectors are:

$v_1 = \begin{pmatrix} 1 \\ -\frac{1}{2}(\sigma+1)+\frac{1}{2}\sqrt{(\sigma+1)^2-4(\sigma-\frac{9}{4}\sigma^2)} \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 1 \\ -\frac{1}{2}(\sigma+1)-\frac{1}{2}\sqrt{(\sigma+1)^2-4(\sigma-\frac{9}{4}\sigma^2)} \\ 0 \end{pmatrix}, v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$

Now, let's consider a small element $\delta x = \begin{pmatrix} \delta x_1 \\ \delta x_2 \\ \delta x_3 \end{pmatrix}$ aligned with one of the eigenfunctions, say $v_1$. Using the Lorentz equations, we can write the change in this element as:

$$\delta \dot x = \begin{pmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \end{pmatrix} = \begin{pmatrix} \sigma(\delta x_2-\