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In this paper the authors have the dynamical system

$\begin{align} T_f \dot{y}_f & = -y_f + (1-\alpha(v))\varphi(z,d) \\ T_r \dot{y}_r & = -y_r + \alpha(v) \varphi(z,d) \\ \dot{z} & = -\varphi(z,d) + y_r + u \end{align}$

and they state in eqns (8-10) that the eigenvalues of the linearization at the equilibrium points $(\overline{y}_f, \overline{y}_r, \overline{z})$ are

$\begin{align} \lambda_1 & = -T_f^{-1} \\ \lambda_2 + \lambda_3 & = -\varphi_z(\overline{z},d) - T_r^{-1} \\ \lambda_2 \lambda_3 & = T_r^{-1} \phi_z(\overline{z},d)(1-\alpha(\overline{v})) \end{align}$

Can someone explain to me how these are derived?

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    @Chris Taylor, thanks a lot!2011-06-06

1 Answers 1

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Do you know how to linearize a dynamical system around an equilibrium?

The idea is that you have $x\in \mathbb{R}^n$ and $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ and you define the system $\dot{x}=f(x)$.

Now to find the linearization of the system you have to expand to a Taylor polynomial around the equilibrium and keep only the linear terms. Practically you find the Jacobian matrix of $f$. In most cases you move the equilibrium to $0$ and you end up with $\dot x=Jx$ with $J$ the Jacobian of $f$. For example consider $\dot{x_1}=x_1+x_1 x_2$ $\dot{x_2}=2x_1+x_1^2-x_2$

The equilibrium is $(0,0)$ already and the Jacobian of $f$ here is $\left[ \begin{array}{cc} 1+x_2 & x_1 \\ 2+2x_1 & -1 \end{array}\right]$ at $(0,0)$ this becomes $\left[ \begin{array}{cc} 1 & 0 \\ 2 & -1 \end{array}\right]$ and the linearization of the system is $\left[ \begin{array}{c} \dot x_1 \\ \dot x_2 \end{array}\right]=\left[ \begin{array}{cc} 1 & 0 \\ 2 & -1 \end{array}\right]\left[ \begin{array}{c} x_1 \\ x_2 \end{array}\right].$

The eigenvalues of the equilibrium are the eigenvalues of the Jacobian at the equilibrium.

I hope this helps.