Question 5.3: Analyze the phase plane trajectories for the system dx/dt = ...

To find the steady state solutions, let’s set the time derivatives to zero, multiply the first equation by y, the second by x, and subtract the resulting steady state equations to obtain x^2 + y^2 = 0. This suggests that the unique steady state is (0, 0); see Fig. 5.8

The Jacobian at this steady state is

J_{ss}=\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}                            (5.4.14)
and its eigenvalues are λ_1 = 1+ı  and  λ_2 = 1−ı. Therefore, the equilibrium point is an unstable focus. The question here is what happens as the trajectories spiral away from the steady state.

To determine this, it is helpful to use the polar transformations that we used in Example 5.2. Following a similar procedure like in that example, the original equations can be converted to

\frac{dr}{dt} =r(1 − r^2)                                                          (5.4.15)

\frac{d\theta }{dt} =1                         (5.4.16)
As was the case in Example 5.2, Eq. (5.4.16) suggests that the phase angle θ increases linearly with time. However, in contrast to what we saw in Example 5.2, Eq. (5.4.15) suggests two equilibrium states, one at r = 0 that corresponds to our steady state y_{ss} =(0, 0) and a second one at r = 1 that corresponds to a circle of radius 1, in the y_1, y_2 plane. This periodic (equilibrium) solution was hidden and we were able to uncover it using polar coordinates!

Can we assess the stability of the steady state corresponding to r = 1? Well, if r < 1 in Eq. (5.4.15), then we have \dot{r} > 0 and the radial position increases with time. Conversely, if r > 1 in Eq. (5.4.15), then \dot{r} < 0 and the radial position decreases in time. Since r = 1 is an equilibrium state, this implies that initial conditions outside or inside this closed trajectory will converge to it asymptotically!