### Stability

The jacobian matrix of the system is:

\begin{align} J^{\ast} = \begin{pmatrix} -a & a & 0 \\ c-Z & -1 & -X \\ Y & X & -b \end{pmatrix} \end{align}

Besides, to find the characteristic equation, we let:

\begin{align} det(J^{\ast}- \lambda I) = 0 \Longleftrightarrow \begin{vmatrix} -a-\lambda & a & 0 \\ c-Z & -1-\lambda & -X \\ Y & X & -b-\lambda \end{vmatrix} \end{align} = 0.

In $$\Big(0,0,0\Big)$$ this gives:

\begin{align} \lambda^3 + \lambda^2(1+a+b) + \lambda(a+ab+b-ac) + ab(1-c) = 0 \end{align}.

Applying the Routh-Hurwitz criterion in $$R^3$$, we see that $$\Big(0,0,0\Big)$$ is stable if and only if $$c \leq 1$$.

### Bifurcations

A super-critical pitchfork bifurcation occurs depending on the value of c. If $$0<c\leq 1$$, there is only $$\Big(0,0,0\Big)$$ which is stable. If $$c >1$$, there are 3 equilibrium points as shown in the second box. The two symetric points are then stable. Furthermore, a Hopf-bifurcation is expected when $$c = a\frac{a+b+3}{a-b-1}$$ and chaotic behavior happens when $$c > a\frac{a+b+3}{a-b-1}$$. See more here.

### Infos

In this app you can:

1. Change parameter values
2. Choose initial conditions
3. Change solver options
4. Display phase plane projections (X vs Y, X vs Z or Y vs Z)

These are the equations behind the Lorenz model

\left\{ \begin{align} \frac{dX}{dt} & = a(Y-X),\\ \frac{dY}{dt} & = X(c-Z) - Y,\\ \frac{dZ}{dt} & = XY - bZ, \end{align} \right.

where $$a$$ is the Prandtl number. See my previous App for further explanations.

At steady-state we know that:

\left\{ \begin{align} \frac{dX}{dt} & = 0,\\ \frac{dY}{dt} & = 0,\\ \frac{dZ}{dt} & = 0. \end{align} \right.

This leads to 3 equilibrium points: $$\Big(0,0,0\Big)$$, $$\Big(\sqrt{b(c-1)},\sqrt{b(c-1)}, c-1\Big)$$ and $$\Big(-\sqrt{b(c-1)},-\sqrt{b(c-1)}, c-1\Big)$$.