Question 5.1: Analyze the stability and the nature of the nearby trajector...

We begin with the calculation of the steady states of the system. Setting the derivative to zero we get from the second equation
y_1(1 − y_1) = 0                         (5.3.20)
implying that at steady state y_1 = 1  or  y_1 = 0. From the first equation, it follows that the corresponding steady state value of y_2  is  y_2 = 0 for both cases. Thus we have two steady states, y_{ss1} = (0, 0)  and  y_{ss2} = (1, 0).

The Jacobian of this system of equations is

J=\begin{bmatrix} 1-3y_1^2 &1 \\ 1-2y_1 & 0 \end{bmatrix}                                            (5.3.21)
There is a nice formula for the eigenvalues of a 2 × 2 system that looks very similar to the quadratic equation,

\lambda =\frac{\tau \pm \sqrt{\tau ^2-4\Delta } }{2}                              (5.3.22)
where τ is the trace of J and Δ is the determinant of J. To prove this formula, consider the general case

A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}                                                 (5.3.23)
The trace of this matrix is τ = a+d and its determinant is Δ= ad−bc. The eigenvalue equation is

det\begin{bmatrix} a − λ & b \\ c & d − λ \end{bmatrix} =0                                                             (5.3.24)

which gives the polynomial equation
λ^2 − (a + d)λ + ad − bc = 0                                (5.3.25)
If we substitute for the trace and determinant, we have
λ^2 − τ λ +Δ = 0                                               (5.3.26)

Using the quadratic formula gives Eq. (5.3.22).
For the steady state at (0, 0), we have a Jacobian

J_{ss1}=\begin{bmatrix} 1 & 1 \\1 &0 \end{bmatrix}                                           (5.3.27)
The trace is τ = 1 and the determinant is Δ= −1, giving eigenvalues

\lambda =\frac{1\pm \sqrt{5} }{2}                                       (5.3.28)
Thus, this steady state is an unstable saddle point.
For the steady state at (1, 0), we have a Jacobian

J_{ss2}=\begin{bmatrix} -2 & 1 \\-1 & 0 \end{bmatrix}                                           (5.3.29)

The trace is τ = −2 and the determinant is Δ= 1, giving eigenvalues

\lambda =\frac{-2\pm \sqrt{0} }{2}=-1                                       (5.3.30)

This is stable node with a repeated eigenvalue.
To investigate the dynamics of this system more thoroughly, Fig. 5.5 shows the phase plane for this problem using RK4 to integrate from a number of different initial conditions. As we can see, the analysis based on the linearized problem is accurate.