Consider the Lorenz equations:
$$x' = f(x,y,z) = \sigma(y-x)$$
$$y' = g(x,y,z) = \rho x - y -xz$$
$$z' = h(x,y,z)= -\beta z + xy$$
where $\sigma$ and $\beta$ are viewed as fixed positive constants and $\rho$ is a parameter.
Here is a guide to work through given that the Wiki describes what happens at each of the critical points. The outline of how to derive those is as follows:
- $(1)$ Find the critical points of the system, that is, where you simultaneously have $x' = y' = z' = 0$.
Clearly, a critical point is given by $(x, y, z) = (0,0,0)$.
- $(2)$ Find the Jacobian matrix of the system of equations.
The Jacobian matrix (using partial derivatives) is given by:
$$\tag 2 \displaystyle J = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}& \frac{\partial f}{\partial z} \\\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}& \frac{\partial g}{\partial z}\\ \frac{\partial h}{\partial x} & \frac{\partial h}{\partial y}& \frac{\partial h}{\partial z}\end{bmatrix} = \begin{bmatrix}-\sigma & \sigma & 0\\\rho - z & -1 & -x\\ y & x & -\beta\end{bmatrix}$$
- $(3)$ Lastly, evaluate the eigenvalues of the Jacobian evaluated at each of the critical points for stability analysis. So, evaluating $(2)$ at the CP $(0,0,0)$, yields the matrix:
$$\displaystyle J_{0,0,0} = \begin{bmatrix}-\sigma & \sigma & 0\\\rho & -1 & 0\\ 0 & 0 & -\beta\end{bmatrix}$$
To find the eigenvalues, we set-up and solve for the characteristic polynomial using:
$$| A - \lambda I| = 0 \rightarrow -\beta \lambda^2-\beta \lambda+\beta r \sigma-\beta \lambda \sigma-\beta \sigma-\lambda^3-\lambda^2+\lambda \rho \sigma-\lambda^2 \sigma-\lambda \sigma = 0$$
This leads to the three eigenvalues (note, we could have written these straight off given the special form of the diagonal matrix above) and their corresponding eigenvectors as:
$\displaystyle \lambda_1 = -\beta, v_1 = (0, 0, 1)$
$\displaystyle \lambda_2 = \frac{1}{2} \left(-\sqrt{4 \rho \sigma+\sigma^2-2 \sigma+1)}-\sigma-1\right), v_2 = -\frac{-1+\sigma+\sqrt{1-2 \sigma+4 \rho \sigma+\sigma^2}}{2 \rho}, 1, 0)$
$\displaystyle \lambda_3 = \frac{1}{2}\left(\sqrt{4 \rho \sigma+\sigma^2-2 \sigma+1)}-\sigma-1\right), v_3 = -\frac{-1+\sigma-\sqrt{1-2 \sigma+4 \rho \sigma+\sigma^2}}{2 \rho}, 1, 0)$
Of course those $\lambda$ are the eigenvalues we use to do a stability analysis.
To study the stability of this, we need some complex center manifold theory, so I am going to stay away from that. It leads to a pitchfork bifurcation at the origin.
It is worth noting that there are two other pairs of fixed points at:
- $x = y = \pm \sqrt{\beta(\rho - 1)}, z = \rho -1$
To find this, notice that the first equation gives us $y=x$, substituting this into the second equation gives $x(\rho - 1) - xz = 0 \rightarrow z = \rho-1$, and we substitute these previous two values into the third and have $x^2 = \beta z \rightarrow x = \pm \sqrt{\beta(\rho - 1)}$.
To simplify matters, lets linearize the system by removing the nonlinear terms and analyze the simpler system.
The linearization (eliminate the $xy$ and $xz$ nonlinear terms from the system) is given by:
$$x' = \sigma(y-x)$$
$$y' = \rho x - y$$
$$z' = -\beta z$$
If you look at the equation for $z'$, you notice that it is decoupled from the other two equations and it can be solved outright as $z(t) = c_1 e^{-\beta t}$ and $z(t) \rightarrow 0$ exponentially fast. The other two directions are governed by the system:
$$\begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}-\sigma & \sigma\\\rho & -1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$$
You should be able to analyze this system from all the data provided now.
As you can see, these systems are very complex and include a rich set of mathematics to deal with, so it takes time to get there.
Here is another guide with some of the details for you to work out.
Best Answer
The computation of the equilibrium points is not correct. Since you do not give any assumptions on the parameters, let us suppose that $\alpha,\beta,\gamma$ are non zero. you need to solve simultaneously
\begin{equation} \alpha x-\beta xy=0; \qquad \beta xy-\gamma y=0. \end{equation}
The point $(0,0)$ is clearly an equilibrium point. Observe that $x=0$ implies $\gamma y=0$ and that $y=0$ implies $\alpha x=0$. Thus, your last two equilibrium points are not correct.
Now, suppose $(x,y)$ is not the origin. Then $(x,y)=(\gamma/\beta,\alpha/\beta)$ is the other equilibrium point.
The Jacobian is the matrix
\begin{equation} J=\begin{bmatrix}\alpha-\beta y & -\beta x\\ \beta y & \beta x-\gamma \end{bmatrix}, \end{equation}
then
\begin{equation} J|_{(0,0)}=\begin{bmatrix}\alpha& 0\\ 0 & -\gamma \end{bmatrix}; \qquad J|_{(\gamma/\beta,\alpha/\beta)}=\begin{bmatrix}0 & -\gamma\\ \alpha & 0 \end{bmatrix} \end{equation}
You could construct the following table, depending on the eigenvalues of $J$ at each point. Let $p_1=(0,0)$, $p_2=(\gamma/\beta,\alpha/\beta)$
\begin{array}{|c|c|c|c|} \hline \alpha & \beta & \gamma & p_1 & p_2 \\ \hline + & + & + & saddle & center \\ \hline + & + & - & source & saddle \\ \hline + & - & + & saddle & center \\ \hline + & - & - & source & saddle \\ \hline - & + & + & sink & saddle \\ \hline - & + & - & saddle & center \\ \hline - & - & + & sink & saddle \\ \hline - & - & - & saddle & center \\ \hline \end{array}
And then, for example, have the following phase portraits (at least for the first 2 cases, you could try to do the rest)