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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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You'll get a better response if you type out the dynamical equations and the eigenvalues here, rather than just linking to a paper and saying "How did they derive that?" Alternatively, you could email the authors of the paper (who will be experts on that paper, unlike people here) but it seems that your main problem is that you don't understand how to linearize a dynamical system around a fixed point and calculate its stability there. You could try reading eg these lecture notes. –  Chris Taylor Jun 1 '11 at 14:25
    
@Chris Taylor I tried to paste an image, but my rate<10. I can't wait the answer from authors. Thanks for link. –  mike_price Jun 2 '11 at 13:39
    
I edited to include the equations. Please go and read the notes at that link (they give a general explanation of what the authors are doing in that paper) and then feel free to ask here if you still have questions. –  Chris Taylor Jun 2 '11 at 13:53
    
@Chris Taylor, thanks a lot! –  mike_price Jun 6 '11 at 11:10
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1 Answer

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.

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