Lyapunov function for a second order system involving trigonometric functions I am studying the stability of the following system:
\begin{aligned}
 \dot{x}_{1} &= -x_{1}^{2} - \sin x_{2}\\
 \dot{x}_{2} &= x_{1} - \frac{\cos x_{2}}{x_{1}}\\
\end{aligned}
The system itself has 2 equilibrium points:
\begin{equation}
  \begin{aligned}
 x_{1} &= -\frac{1}{\sqrt[4]{2}}\\
 {x}_{2} &= -\frac{\pi}{4}\\
\end{aligned}
\qquad
\qquad
\text{and}
\qquad
\qquad
\begin{aligned}
 x_{1} &= \frac{1}{\sqrt[4]{2}}\\
 {x}_{2} &= -\frac{\pi}{4}\\
\end{aligned}
\end{equation}
Which I confirmed by analyzing the phase portrait of the system:

Through linearization, I determined that the left equilibrium point is an unstable focus and the right one is a stable focus according to the eigenvalues of the Jacobian matrix of the system evaluated in each equilibrium point.
\begin{equation}
  \begin{aligned}
  \lambda_{1} &= 1.2613 + 1.1124i\\
  \lambda_{2} &= 1.2613 - 1.1124i\\
\end{aligned}
\qquad
\qquad
\text{,}
\qquad
\qquad
\begin{aligned}
  \lambda_{1} &= -1.2613 + 1.1124i\\
  \lambda_{2} &= -1.2613 - 1.1124i\\
\end{aligned}
\end{equation}
As far as I know, it is possible to make a change of coordinates and translate the stable equilibrium point to the origin. So, my goal consists in finding an appropriate Lyapunov function for determining the stability of the origin (after the change of coordinates), employing Lyapunov's second method.
Among others, here are some candidate functions I've tried so far without satisfactory results:
\begin{equation}
  \begin{aligned}
  V(x) &= x_{1}\sin x_{1} + \frac{1}{2}x_{2}^{2}\\
V(x) &= \frac{1}{2}x_{1}^{2} + (1-\cos x_{2})\\
V(x) &= x_{1}\sin x_{1} + (1-\cos x_{2})\\
\end{aligned}
\end{equation}
I got stuck trying to find a Lyapunov function due to the trigonometric functions involved in the original system. Could someone give me a suggestion or hint?
 A: From the stable equilibrium point $\left(\frac{1}{\sqrt[4]{2}}, - \frac{\pi}{4}\right)$, the state variable transformation can be constructed as
$$
\begin{aligned}
 y_{1} &= x_{1} - \frac{1}{\sqrt[4]{2}} \\
 y_{2} &= x_{2} + \frac{\pi}{4} \\
\end{aligned}
$$
for which there is an inverse
$$
\begin{aligned}
 x_{1} &= y_{1} + \frac{1}{\sqrt[4]{2}} \\
 x_{2} &= y_{2} - \frac{\pi}{4} \\
\end{aligned}
$$
such that system model
$$
\begin{aligned}
 \dot{x}_{1} &= - x_{1}^{2} - \sin(x_{2}) \\
 \dot{x}_{2} &= x_{1} - \frac{\cos(x_{2})}{x_{1}} \\
\end{aligned}
$$
in the new coordinates has the form
$$
\begin{aligned}
 \dot{y}_{1} &= - \left(y_{1} + \frac{1}{\sqrt[4]{2}}\right)^{2} - \sin\left(y_{2} - \frac{\pi}{4}\right) \\
 \dot{y}_{2} &= y_{1} + \frac{1}{\sqrt[4]{2}} - \frac{\cos\left(y_{2} - \frac{\pi}{4}\right)}{y_{1} + \frac{1}{\sqrt[4]{2}}} \\
\end{aligned}
$$
that can be expanded to
$$
\begin{aligned}
 \dot{y}_{1} &= - y_{1}^{2} - 2^{\frac{3}{4}} y_{1} - \frac{1}{\sqrt{2}} - \sin\left(y_{2}\right) \cos\left(\frac{\pi}{4}\right) + \cos\left(y_{2}\right) \sin\left(\frac{\pi}{4}\right) \\
 \dot{y}_{2} &= y_{1} + \frac{1}{\sqrt[4]{2}} - \frac{\cos\left(y_{2}\right) \cos\left(\frac{\pi}{4}\right) + \sin\left(y_{2}\right) \sin\left(\frac{\pi}{4}\right)}{y_{1} + \frac{1}{\sqrt[4]{2}}} \\
\end{aligned}
$$
and then be simplified to
$$
\begin{aligned}
 \dot{y}_{1} &= - \left(y_{1} + 2^{\frac{3}{4}}\right) y_{1} - \left[\cos\left(\frac{\pi}{4}\right) \frac{\sin\left(y_{2}\right)}{y_{2}}\right] y_{2} + \sin\left(\frac{\pi}{4}\right) \cos\left(y_{2}\right) - \frac{1}{\sqrt{2}} \\
 \dot{y}_{2} &= y_{1} - \left[\frac{\sin\left(\frac{\pi}{4}\right)}{y_{1} + \frac{1}{\sqrt[4]{2}}} \frac{\sin\left(y_{2}\right)}{y_{2}}\right] y_{2} - \frac{\cos\left(\frac{\pi}{4}\right) \cos\left(y_{2}\right)}{y_{1} + \frac{1}{\sqrt[4]{2}}} + \frac{1}{\sqrt[4]{2}} . \\
\end{aligned}
$$
Next, the above modeling equations is represented in the nonlinear state-space format
$$
\begin{aligned}
 \dot{\mathbf{y}} &= \mathbf{F}(\mathbf{y}) \mathbf{y} + \mathbf{B} \mathbf{u}(\mathbf{y}) \\
 \begin{bmatrix}
\dot{y}_{1} \\
\dot{y}_{2} \\ 
\end{bmatrix} &= \begin{bmatrix}
- \left(y_{1} + 2^{\frac{3}{4}}\right) & - \frac{1}{\sqrt{2}} \frac{\sin\left(y_{2}\right)}{y_{2}} \\
1 & - \frac{1}{\sqrt{2}} \frac{1}{y_{1} + \frac{1}{\sqrt[4]{2}}} \frac{\sin\left(y_{2}\right)}{y_{2}} \\
\end{bmatrix} \begin{bmatrix}
y_{1} \\
y_{2} \\ 
\end{bmatrix} + \begin{bmatrix}
1 & 0 \\
0 & 1 \\ 
\end{bmatrix} \begin{bmatrix}
\frac{1}{\sqrt{2}} \cos\left(y_{2}\right) - \frac{1}{\sqrt{2}} \\
- \frac{1}{\sqrt{2}} \frac{1}{y_{1} + \frac{1}{\sqrt[4]{2}}} \cos\left(y_{2}\right) + \frac{1}{\sqrt[4]{2}} \\ 
\end{bmatrix} . \\
\end{aligned}
$$
Taking the limit of $\dot{\mathbf{y}}$ as $y_{1} \rightarrow 0$ and $y_{2} \rightarrow 0$, we have
$$
\begin{aligned}
 \begin{bmatrix}
\dot{y}_{1} \\
\dot{y}_{2} \\ 
\end{bmatrix} &= \begin{bmatrix}
- 2^{\frac{3}{4}} & - \frac{1}{\sqrt{2}} \\
1 & - \frac{1}{\sqrt[4]{2}} \\
\end{bmatrix} \begin{bmatrix}
y_{1} \\
y_{2} \\ 
\end{bmatrix} \\
 \begin{bmatrix}
\dot{y}_{1} \\
\dot{y}_{2} \\ 
\end{bmatrix} &= \begin{bmatrix}
- a & - b \\
1 & - c \\
\end{bmatrix} \begin{bmatrix}
y_{1} \\
y_{2} \\ 
\end{bmatrix} \\
 \dot{\mathbf{y}} &= \mathbf{A} \mathbf{y} . \\
\end{aligned}
$$
Since this is a linear system, we postulate there exists a quadratic Lyapunov function $V(\mathbf{y}) = \mathbf{y}^{T} \mathbf{P} \mathbf{y}$, where $\mathbf{P}$ is a real symmetric positive-definite matrix. The derivative of V along the trajectories of $\dot{\mathbf{y}} = \mathbf{A} \mathbf{y}$ is given by
$$
\dot{V}(\mathbf{y}) = \mathbf{y}^{T} \left(\mathbf{P} \mathbf{A} + \mathbf{A}^{T} \mathbf{P}\right) \mathbf{y} = - \mathbf{y}^{T} \mathbf{Q} \mathbf{y}
$$
where $\mathbf{Q}$ is also a symmetric positive-definite matrix defined by
$$
\begin{aligned}
\mathbf{Q} &= 2 (a + c) (a c + b) \begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix} \\
\end{aligned}
$$
and $2 (a + c) (a c + b) = 9 \sqrt[4]{2} > 0$. Solving the Lyapunov equation
$$
\mathbf{P} \mathbf{A} + \mathbf{A}^{T} \mathbf{P} = - \mathbf{Q}
$$
then the unique solution, $\mathbf{P}$ is obtained
$$
\begin{aligned}
\mathbf{P} &= \begin{bmatrix}
c^2 + a c + b + 1 & a - b c \\
a - b c & a^2 + a c + b^2 + b \\
\end{bmatrix} \\
\end{aligned}.
$$
This approach can be used to establish the local stability for the stable equilibrium point.
