I have the below pair of coupled non-linear second-order ODEs for $x$ and $y$ and wish to analyse the stability of the system.

$\gamma_1 \dot{x} - \gamma_2 ( \dot{x} - \dot{y} ) = (m_1 + m_2) d^2 \ddot{x} + I \ddot{x} + m_2 d^2 \ddot{y} + d(m_1 + m_2) g x,$

$\gamma_2 (\dot{x}- \dot{y}) = m_2 d^2 \ddot{y} + I \ddot{y} + m_2 d^2 \ddot{x} + dm_2g y, $

where $\gamma_1$, $\gamma_2$, $m_1$, $m_2$, $g$, $d$ and $I$ are parameters of the system.

What would be the easiest way to analyse stability? I have tried to make substitutions $u = \dot{x}$, $v= \dot{y}$ to convert the system into four first order ODEs, but I cannot write it in the form

$\begin{align} y &= \begin{bmatrix} \dot{u} \\ \dot{v} \end{bmatrix} = \textbf{A} \begin{bmatrix} u \\ v \end{bmatrix} \end{align}$,

where $\textbf{A} $ is the Jacobian, because $\dot{u}$ and $\dot{v}$ have coefficients which cannot be removed by dividing through, can someone advise?


1 Answer 1


First reduce it to the form

$$ \dot X = f(X) $$

with $X = (x_1, x_2, y_1, y_2)^{\dagger}$

After that, verify the equilibrium set by solving $X_0 = \arg\{ f(X) = 0\}$ following with the analysis of $J(X_0) = \nabla_X f(X_0)$ eigenvalues.


Calling $x=x_1, \dot x_1 = x_2, y = y_1, \dot y_1 = y_2$ and solving for $\dot x_1, \dot x_2,\dot y_1, \dot y_2$ we arrive at

$$ \begin{array}{rcl} \dot x_1 &=& x_2\\ \dot x_2 &=& \frac{d^3 g m_2^2 y_1-d g x_1 (m_1+m_2) \left(d^2 m_2+I\right)-\gamma_2 x_2 \left(2 d^2m_2+I\right)+\gamma_2 y_2 \left(2 d^2 m_2+I\right)}{d^4 m_1 m_2-\gamma_1 \left(d^2m_2+I\right)+d^2 I (m_1+2 m_2)+I^2}\\ \dot y_1 &=& y_2\\ \dot y_2 &=& \frac{d^3 g m_2 x_1 (m_1+m_2)-d g m_2 \text{y1} \left(d^2 (m_1+m_2)-\gamma_1+I\right)+\gamma_2 x_2\left(d^2 (m_1+2 m_2)-\gamma_1+I\right)-\gamma_2 y_2 \left(d^2 (m_1+2 m_2)-\gamma_1+I\right)}{d^4m_1 m_2-\gamma_1 \left(d^2 m_2+I\right)+d^2 I (m_1+2 m_2)+I^2} \end{array} $$

  • $\begingroup$ What do you mean with $\dot X = f(X)$, with $X = (x_1, x_2, y_1, y_2)^{\dagger}$, could you give an example? $\endgroup$
    – Tom
    Dec 27, 2022 at 12:28
  • $\begingroup$ Please. See attached note. $\endgroup$
    – Cesareo
    Dec 27, 2022 at 13:51
  • $\begingroup$ If you calculate the Jacobian matrix for the corresponding system of 4 first order equations, it seems to have only constant entries (since $x_1$, $x_2$, $y_1$ or $y_2$ either vanish or become 1 after differentiating), is that a problem ie. the Jacobian will be the same whatever the equilibrium points $X_0$? $\endgroup$
    – Tom
    Dec 27, 2022 at 18:27
  • $\begingroup$ The system is linear and the equilibrium point is at $x_1=x_2=y_1=y_2 = 0$. $\endgroup$
    – Cesareo
    Dec 27, 2022 at 18:36
  • $\begingroup$ Don't you need to evaluate the Jacobian and then substitute in the value for the equilibrium point? But in this case the Jacobian is a constant matrix. Or do you mean that the point at $0$ is automatically stable? $\endgroup$
    – Tom
    Dec 27, 2022 at 18:49

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