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The Lorenz equation:

$\dot{x}=\sigma\left ( y-x \right )$

$\dot{y}=rx-y-xz$

$\dot{z}=-bz+xy$

The fixed points are $\left ( 0,0,0 \right )$ and $\left ( \pm \sqrt{\left ( r-1 \right )b}, \pm \sqrt{\left ( r-1 \right )b},r-1 \right ), \forall r>1$.

To compute the stability at the origin, Strogaz recommends removing the non-linear terms and ignoring the decoupled equation so as to perform a linearisation.

Thus,

$\dot{x}=\sigma\left ( y-x \right )$

$\dot{y}=rx-y-xz$

$\dot{z}=-bz+xy$

is reduced to

$\dot{x}=\sigma\left ( y-x \right )$

$\dot{y}=rx-y$

Now, we have:

$\begin{bmatrix} \dot{x}\\ \dot{y} \end{bmatrix}$ =$\begin{bmatrix} -\sigma &\sigma \\ r&-1 \end{bmatrix}$ $\begin{bmatrix} x\\ y \end{bmatrix}$

which from the matrix

$\begin{bmatrix} -\sigma &\sigma \\ r&-1 \end{bmatrix}$

one can directly compute the trace and determinant to predict the eigenvalues and therefore the stability.

Am I in the right direction thus far?

Now, if the control parameter $r \mapsto 1$:

The non-zero fixed point for this Lorenz equation tends to $\left ( 0,0,0 \right )$. It coalesce at the origin.

I would like to analyse and understand what the local bifurcation is for r=1. To my knowledge, it should be a saddle-node bifurcation, since $\left ( \pm \sqrt{\left ( r-1 \right )b}, \pm \sqrt{\left ( r-1 \right )b},r-1 \right ), \forall r>1$ and it only coalesce at the origin for r=1.

Would someone illuminate my understanding?

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To compute the stability at the origin, Strogatz recommends removing the non-linear terms and ignoring the decoupled equation so as to perform a linearization.

Well, it could be done this way, but I don't agree that it's easier. If you want to find a linearized system for an equilibrium at the origin, it is it: $$ \dot{x}=\sigma\left ( y-x \right ), $$ $$ \dot{y}=rx-y, $$ $$ \dot{z}=-bz ,$$ just dropping nonlinear terms. The linearization matrix has block-diagonal structure, and it's really easy to compute its characteristic equation and its roots. The last equation is decoupled from others and it gives this $-b$ root of characteristic equation instantaneously, and since the first two equations doesn't include $z$, you can consider them separately. I don't agree that the stability could be determined only by them unless we know something about $b$. It is true for $b > 0$, but completely false for $b < 0$ — in that case equilibrium at the origin has an eigenvalue with positive real part and it's definitely unstable.

I would like to analyse and understand what the local bifurcation is for $r=1$.

Saddle-node bifurcation is not the only bifurcation when equilibria coalesce. In your case three equilibria coalesce simultaneously. As it was explained in one of the previous posts, Lorenz system is symmetric and thus this is a pitchfork bifurcation.

ADDED LATER: Of course, the right way to say here is that "most likely it is a pitchfork bifurcation". In case of symmetric systems this is a codimension-1 bifurcation and so it's most likely to happen. But to be completely sure you need to calculate some additional quantities. Kuznetsov book is a nice source for looking at this subject deeper, and it has many recipes that can be straightforwardly applied to generic systems, not necessarily in normal form already.

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  • $\begingroup$ If for a Lorentz system, there exists 2 negative eigenvalues and one positive eigenvalues for an origin, how should I describe the stability? It feels as though there should be a saddle and an 'incoming' trajectory along an axis. I take a negative and positive eigenvalue corresponding to the x and y component of a coordinate in 3-dim. Plot it on the xy-plane. Then include the z-axis through the xy-plane with trajectories tending towards the origin. $\endgroup$ – Mathematicing Jul 31 '17 at 1:40
  • $\begingroup$ It's unstable if you have eigenvalues with positive real part. What reference book do you use? $\endgroup$ – Evgeny Jul 31 '17 at 13:05

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