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In this paper the authors have the dynamical system

$$\begin{align} T_f \dot{y}_f & = -y_f + (1-\alpha(v))\varphi(z,d) &(1)\\ T_r \dot{y}_r & = -y_r + \alpha(v) \varphi(z,d) &(2)\\ \dot{z} & = -\varphi(z,d) + y_r + u &(3) \end{align}$$

and they state that the eigenvalues of the linearization(1-3) at the equilibrium points $(\overline{y}_f, \overline{y}_r, \overline{z})$ are

$$\begin{align} \lambda_1 & = -T_f^{-1} &(4)\\ \lambda_2 + \lambda_3 & = -\varphi_z(\overline{z},d) - T_r^{-1} &(5)\\ \lambda_2 \lambda_3 & = T_r^{-1} \phi_z(\overline{z},d)(1-\alpha(\overline{v})) &(6)\\ \end{align}$$

I try linearize it by myself

$$\left( \begin{array}{ccc} -T_f^{-1}-\lambda & 0 & T_f^{-1}(1-\alpha (v))\varphi _z(z,d) \\ 0 & -T_r^{-1}-\lambda & T_r^{-1}\alpha (v)\varphi _z(z,d) \\ 0 & 1 & -\varphi _z(z,d)-\lambda \end{array} \right)$$

Disclosure determinant:

$$\begin{align} (-T_f^{-1}-\lambda)(-T_r^{-1}-\lambda)(-\varphi _z(z,d)-\lambda)-T_r^{-1}\alpha (v)\varphi _z(z,d)(-T_f^{-1}-\lambda)=0 \end{align}$$

$$\begin{align} -(T_f^{-1}+\lambda)(T_r^{-1}+\lambda)(\varphi _z(z,d)+\lambda)+T_r^{-1}\alpha (v)\varphi _z(z,d)(T_f^{-1}+\lambda)=0 \end{align}$$

$$\begin{align} -(T_f^{-1}+\lambda)((T_r^{-1}+\lambda)(\varphi _z(z,d)+\lambda)+T_r^{-1}\alpha (v)\varphi _z(z,d))=0 \end{align}$$

$$\begin{align} T_f^{-1}+\lambda=0 \end{align}$$

$$\begin{align} \lambda _1=-T_f^{-1} &(4^{*}) \end{align}$$

$$\begin{align} (T_r^{-1}+\lambda)(\varphi _z(z,d)+\lambda)+T_r^{-1}\alpha (v)\varphi _z(z,d)=0 \end{align}$$

remove brackets

$$\begin{align} T_r^{-1}\varphi _z(z,d)+T_r^{-1}\lambda+\varphi _z(z,d)\lambda+\lambda^{2}+T_r^{-1}\alpha (v)\varphi _z(z,d)=0 \end{align}$$

$$\begin{align} (1+\alpha (v))T_r^{-1}\varphi _z(z,d)+T_r^{-1}\lambda+\varphi _z(z,d)\lambda+\lambda^{2}=0 &(7) \end{align}$$

Can someone explain to me how they got (5) and (6) from (7)?

UPD Mathematica has solved like this

$$\begin{align} -\frac{1}{T_f} \\ \frac{-1-T_r \varphi _z-\sqrt{1+4 T_r \alpha \varphi _z-2 T_r \varphi _z+T_r^2 \varphi _z^2}}{2 T_r} \\ \frac{-1-T_r \varphi _z+\sqrt{1+4 T_r \alpha \varphi _z-2 T_r \varphi _z+T_r^2 \varphi _z^2}}{2 T_r} \end{align}$$

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up vote 2 down vote accepted

You are doing the wrong thing to find the eigenvalues. To find an eigenvalue $\lambda$ of a matrix $A$, you need to subtract $\lambda$ times the identity matrix from $A$, and then compute the determinant. You are subtracting the matrix $\mathrm{diag}(\lambda_1, \lambda_2, \lambda_3)$ from $A$, whereas you want to be subtracting $\mathrm{diag}(\lambda, \lambda, \lambda)$.

This gives you a polynomial whose order is the dimension of $A$ (in your case $A$ has dimension three, so the polynomial will be a cubic). The roots of this polynomial are the eigenvalues.

In your case there will be one easily computed root of the polynomial, which is $\lambda=-T_f^{-1}$. You can then factor this out and be left with a quadratic in $\lambda$, which you can solve with the standard method.

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Thanks, I understand. Sometimes i feel myself like dumb) Is my linearizing correct without regard for lambda? – mike_price Jul 11 '11 at 14:02

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