Question 5.1: Analyze the stability and the nature of the nearby trajector...
Analyze the stability and the nature of the nearby trajectories for the steady states of
\frac{dy_1}{dt} = y_1 − y^3_1 + y_2 (5.3.18)
\frac{dy_2}{dt} = y_1 − y^2_1 (5.3.19)
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.
