Setting \(\dot{x} = \dot{y} = \dot{z} = 0\), there are three solutions. One solution which is always present is \( x^* = y^* = z^* = 0 \) Linearize about this solution obtain \( \frac{d}{dt} \begin{pmatrix} \delta x \\ \delta y \\ \delta z \end{pmatrix} = \begin{pmatrix} -\sigma & \sigma & 0 \\ r & -1 & 0 \\ 0 & 0 & -b \end{pmatrix} \begin{pmatrix} \delta x \\ \delta y \\ \delta z \end{pmatrix}. \) The eigenvalues of the linearized dynamics are \( \lambda_{1,2} = -\frac{1}{2}(1 + \sigma) \pm \frac{1}{2} \sqrt{(1 + \sigma)^2 + 4\sigma(r - 1)}, \quad \lambda_3 = -b, \) and thus if \(0 < r < 1\), all three eigenvalues are negative, and the fixed point is a stable node. If \(r > 1\), then \(\lambda_2 > 0\), and the fixed point is attractive in two directions but repulsive in a third, corresponding to a three-dimensional version of a saddle point. For \(r > 1\), a new pair of solutions emerges, with \( x^* = y^* = \pm \sqrt{b(r - 1)}, \quad z^* = r - 1. \) Linearizing about either one of these fixed points, find \( \frac{d}{dt} \begin{pmatrix} \delta x \\ \delta y \\ \delta z \end{pmatrix} = \begin{pmatrix} -\sigma & \sigma & 0 \\ 1 & -1 & -x^* \\ x^* & x^* & -b \end{pmatrix} \begin{pmatrix} \delta x \\ \delta y \\ \delta z \end{pmatrix}. \) The characteristic polynomial of the linearized map is \( P(\lambda) = \lambda^3 + (b + \sigma + 1)\lambda^2 + b(\sigma + r)\lambda + 2b(r - 1). \)
Since \(b, \sigma, r\) are all positive, \(P'(\lambda) > 0\) for all \(\lambda \geq 0\). Since \(P(0) = 2b(r - 1) > 0\), there is always at least one eigenvalue \(\lambda_1\) which is real and negative. The remaining two eigenvalues are either both real and negative, or else they occur as a complex conjugate pair: \(\lambda_{2,3} = \alpha \pm i \beta\). The fixed point is stable provided \(\alpha < 0\). The stability boundary lies at \(\alpha = 0\). Thus, set \( P(i\beta) = \left[ 2b(r - 1) - (b + \sigma + 1)\beta^2 \right] + i \left[ b(\sigma + r) - \beta^2 \right] \beta = 0, \) which results in two equations. Solving these two equations for \(r(\sigma, b)\), find \( r_c = \frac{\sigma(\sigma + b + 3)}{\sigma - b - 1}. \) The fixed point is stable for \(r \in [1, r_c]\). These fixed points correspond to steady convection.
The Lorenz system has commonly been studied with \(\sigma = 10\) and \(b = \frac{8}{3}\). This means that the volume collapse is very rapid, since \( -\nabla \cdot \vec{u} = -\frac{41}{3} \approx -13.67, \) leading to a volume contraction of \( e^{-41/3} \approx 1.16 \times 10^{-6} \) per unit time. For these parameters, one also has \( r_c = \frac{470}{19} \approx 24.74 \)