I'm analysing the stability of a double physical pendulum, and have determined the dimensionless system of equations to be \begin{align} \label{eq:dimensionless} \begin{aligned} & \ddot{\phi} \cos{(\theta - \phi)} + \dot{\phi}^2 \sin{(\theta - \phi)} + \alpha \ddot{\theta} + \beta \sin{\theta} = 0 \end{aligned} \\[5pt] \begin{aligned} & \ddot{\theta} \cos{(\theta - \phi)} - \dot{\theta}^2 \sin{(\theta - \phi)} + \gamma \ddot{\phi} + \sin{\phi} = 0 \end{aligned} \end{align} where $\alpha$, $\beta$ and $\gamma$ are defined to be \begin{align} \label{eq:dim_parameters} \alpha &= \frac{\big ( I_1 + (m_1 + m_2) L_{1}^2 \big )}{m_2 L_{1} L_{2}} \\[5pt] \beta &= \frac{L_1 (m_1 + m_2)}{m_2 L_{2}} \\[5pt] \gamma &= \frac{(I_2 + m_2 L_{2}^2)}{m_2 L_{1} L_{2}} \, . \end{align} Setting $\ddot{\phi} = \ddot{\theta} = \dot{\phi} = \dot{\theta} = 0$, we find that the stationary points of the system are at $(0, 0), \, (0, \pi), \, (\pi, 0), \, (\pi, \pi)$.
I want to find the stability of the system at these points. Here's how I attempted to find them, but am slightly unsure if my working is correct:
I linearised the system of equations \begin{align} \begin{aligned} & \ddot{\phi} + \alpha \ddot{\theta} + \beta \theta = 0 \\[5pt] & \ddot{\theta} + \phi + \gamma \ddot{\phi} = 0 \, . \end{aligned} \end{align} Rearranging and substituting the equations into each other \begin{align} \begin{aligned} \nonumber & \ddot{\theta} = \frac{\beta \gamma}{1 - \alpha \gamma} \theta - \frac{1}{1 - \alpha \gamma} \phi \end{aligned} \\[5pt] \begin{aligned} \nonumber & \ddot{\phi} = \frac{\alpha}{1 - \alpha \gamma} \phi - \frac{\beta}{1 - \alpha \gamma} \theta \, . \end{aligned} \end{align} Expressing this as a matrix equation \begin{align} \begin{aligned} \label{eq:matrix} \frac{d}{dt} \begin{bmatrix} \theta \\ \dot{\theta} \\ \phi \\ \dot{\phi} \end{bmatrix} &= \begin{bmatrix} 0 & 1 & 0 & 0 \\ \frac{\beta \gamma}{1 - \alpha \gamma} & 0 & -\frac{1}{1 - \alpha \gamma} & 0 \\ 0 & 0 & 0 & 1 \\ -\frac{\beta}{1 - \alpha \gamma} & 0 & \frac{\alpha}{1 - \alpha \gamma} & 0 \end{bmatrix} \begin{bmatrix} \theta \\ \dot{\theta} \\ \phi \\ \dot{\phi} \end{bmatrix} \, . \end{aligned} \end{align} Finding the characteristic equation for the eigenvalues of the matrix \begin{align} \begin{aligned} \nonumber & \text{det}(A - \lambda I) = \begin{vmatrix} -\lambda & 1 & 0 & 0 \\ \frac{\beta \gamma}{1 - \alpha \gamma} & -\lambda & -\frac{1}{1 - \alpha \gamma} & 0 \\ 0 & 0 & -\lambda & 1 \\ -\frac{\beta}{1 - \alpha \gamma} & 0 & \frac{\alpha}{1 - \alpha \gamma} & -\lambda \end{vmatrix} = 0 \\[5pt] & \implies \lambda^4 - \frac{\alpha + \beta \gamma}{1 - \alpha \gamma} \lambda^2 + \frac{\alpha \beta \gamma - \beta}{(1 - \alpha \gamma)^2} = 0 \\[5pt] & \implies \lambda^4 - \frac{\alpha + \beta \gamma}{1 - \alpha \gamma} \lambda^2 - \frac{\beta}{1 - \alpha \gamma} = 0 \end{aligned} \end{align} Solving for the eigenvalues using the quadratic formula gives \begin{align} \begin{aligned} \lambda^2 &= \frac{(\alpha + \beta \gamma) \pm \sqrt{(-(\alpha + \beta \gamma))^2 - 4 (1 - \alpha \gamma) (-\beta)}}{2 (1 - \alpha \gamma)} \\[5pt] &= \frac{(\alpha + \beta \gamma) \pm \sqrt{\alpha^2 - 2 \alpha \beta \gamma + \beta^2 \gamma^2 + 4 \beta}}{2 (1 - \alpha \gamma)} \\[5pt] &= \frac{(\alpha + \beta \gamma) \pm \sqrt{(\alpha - \beta \gamma)^2 + 4 \beta}}{2 (1 - \alpha \gamma)} \end{aligned} \end{align}
I'm not sure about this working. The stability at these points should not depend on the parameters, surely. I feel that $(0, 0)$ should always be stable.
Have I not found the Jacobian correctly? Or am I, perhaps, supposed to use the fact that the parameters are positive? Any help or clarification would be much appreciated.
