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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 \(c \leq 1\).

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
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here.
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In this app you can:

- Change parameter values
- Choose initial conditions
- Change solver options
- 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 \(\Big(-\sqrt{b(c-1)},-\sqrt{b(c-1)}, c-1\Big)\).