## 2 Materials and Methods
### Lorenz Equations  
The Lorenz system of differential equations is shown in Equations (1)-(3) [26], where \(\sigma,r\), \(b\in\mathbb{R}^{+}\).  
\[\frac{dx}{dt}=\sigma(-x+y) \tag{1}\]  
\[\frac{dy}{dt}=rx-y-xy \tag{2}\]  
\[\frac{dz}{dt}=-bx+xy \tag{3}\]  
If the Lorenz Equations (1)-(3) are equal to zero, the critical points are found. In this case, the critical points are \(P_{1}=(0,0,0)\), \(P_{2}=(\sqrt{b(r-1)},\sqrt{b(r-1)},r-1)\), and \(P_{3}=(-\sqrt{b(r-1)},-\sqrt{b(r-1)},r-1)\).  
The Lorenz equations describe the convection phenomenon in the Earth's atmosphere, in this case, \(\sigma=10\) and \(b=\frac{8}{3}\). Moreover, the solution to the Lorenz system of equations follows the form \(\vec{X}=\vec{\xi}e^{\lambda t}\) where \(\vec{\xi}\) represent the eigenvectors, and \(\lambda\) the eigenvalues. On the other hand, for the calculation of the solutions in the neighborhood of the point \(P_{2}\), we start from \(X^{\prime}=AX\), where the matrices \(A\), \(X\), and \(X^{\prime}\) are described in Equations (4)-(6).  
\[A=\begin{pmatrix}10&10&0\\ r&-1&-\sqrt{8/3(r-1)}\\ \sqrt{8/3(r-1)}&\sqrt{8/3(r-1)}&-8/3\end{pmatrix} \tag{4}\]  
\[X=\begin{pmatrix}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{pmatrix} \tag{5}\]  
\[X^{\prime}=\begin{pmatrix}x^{\prime}\\ y^{\prime}\\ z^{\prime}\end{pmatrix} \tag{6}\]  
The eigenvalues come from the characteristic polynomial and are obtained from Equation (7).  
\[|A-\lambda I|=0 \tag{7}\]  
Equation (8) expresses the characteristic polynomial with \(r=28\).  
\[3\lambda^{3}+41\lambda^{2}-50\lambda+2160=0 \tag{8}\]  
From Equation (8), one real root and two complex ones are obtained; these are written in Equations (9)-(11).  
\[\lambda_{1}=-22.558424 \tag{9}\]  
\[\lambda_{2}=4.445878+3.485904i \tag{10}\]  
\[\lambda_{3}=4.445878-3.485904i \tag{11}\]  
Regarding the eigenvectors, it is only necessary to generate two of them to obtain the general solution. Equations (12) and (13) show the eigenvectors \(\vec{\xi}_{1}\) and \(\vec{\xi}_{2}\), respectively.  
\[\vec{\xi}_{1}=\begin{pmatrix}9.163288\\ -11.507650\\ 1\end{pmatrix} \tag{12}\]\[\vec{\xi}_{2}=\begin{pmatrix}0.359510+0.116796i\\ 0.478680+0.294040i\\ 1+0i\end{pmatrix} \tag{13}\]  
\[u(\vec{t})=\begin{pmatrix}0.3595\cos(3.4859)t-0.1167\sin(3.4859)t\\ 0.4786\cos(3.4859)t-0.2940\sin(3.4859)t\\ \cos(3.4859)t\end{pmatrix}e^{dt} \tag{14}\]  
\[v(\vec{t})=\begin{pmatrix}0.1167\cos(3.4859)t-0.3595\sin(3.4859)t\\ 0.2940\cos(3.4859)t-0.4786\sin(3.4859)t\end{pmatrix}e^{dt} \tag{15}\]  
Note that \(d=4.445878\), and the solution \(\vec{\xi}_{2}e^{(d+3.485904i)t}\) contains a real part and a complex part. The real part is presented as \(\vec{u}\) and the complex one as \(\vec{v}\times i\), where \(\vec{u}\) and \(\vec{v}\) are expressed in Equations (14) and (15). Furthermore, if we set \(\vec{w}=\vec{\xi}_{1}e^{-22.5584t}\), the general solution is then written in Equation (16).  
\[\vec{X}(t)=e^{-22.5584t}C_{1}\vec{\xi}_{1}+C_{2}\vec{u}(t)+C_{3}\vec{v}(t)\times i \tag{16}\]  
The function \(\varphi_{2}(t)\) is composed of the third coordinate of the vectors in Equation (16). On the other hand, in this development, \(C_{1}=0\) and \(t_{0}=1/(4.445878)\). The result is given in Equation (17).  
\[\varphi_{y}(t_{0})=(0.999924)C_{2}e+(0.012315)C_{3}e \tag{17}\]  
In addition, the units of the argument \(\theta\) in \(\sin\theta\) and \(\cos\theta\) functions are degrees.