Question 5.2: 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{dx}{dt} = −y + ax(x^2 + y^2) (5.3.31)
\frac{dy}{dt} = x + ay(x^2 + y^2) (5.3.32)
where a is a constant.
We start with the calculation of the steady states for this system. Setting the derivatives to zero, multiplying the first equation by x, the second by y, and adding them, we obtain that at steady state (x^2 + y^2) = 0. Therefore, the only steady state for this system is (0, 0). The Jacobian for this system is
J=\begin{bmatrix} 3ax^2 & −1 + 2axy \\ 1 + 2axy & 3ay^2 \end{bmatrix} (5.3.33)
At the steady state (0, 0) we have
J_{ss}=\begin{bmatrix} 0 & -1 \\1 &0 \end{bmatrix} (5.3.34)
The eigenvalue equation is
det\begin{bmatrix} −λ & −1 \\ 1 & −λ \end{bmatrix} =0 (5.3.35)
which gives the polynomial equation
λ^2 + 1 = 0 (5.3.36)
The eigenvalues are purely imaginary, λ = ±ı. This is a case of a neutrally stable steady state. Although the linearized analysis predicts a center, further analysis is required to determine the nature of the trajectories of the nonlinear system close to the steady state.
In some cases, important conclusions can be drawn by switching to polar coordinates,
x = r \cos θ (5.3.37)
y = r \sin θ (5.3.38)
Our aim is to determine the resulting differential equations in terms of θ and r. To get the radial position equation, we can take advantage of the identity
x^2 + y^2 = r^2 (5.3.39)
which upon differentiation with respect to time becomes
x\dot{x} +y\dot{y} =r\dot{r} (5.3.40)
In the latter, we use the dot to denote differentiation with respect to time. Substituting from the original differential equation gives
r\dot{r} =x\left[-y+ax(x^2+y^2)\right] +y\left[x+ay(x^2+y^2)\right] = a(x^2+y^2)^2=ar^4 (5.3.41)
As a result, the radial position differential equation is
\dot{r}=ar^3 (5.3.42)
To get the equation for the angular position, we differentiate the relationships
x = r \cos θ (5.3.43)
y = r \sin θ (5.3.44)
with respect to time,
\dot{x} = \dot{r} \cos θ − (r\sin θ)\dot{θ} = \dot{r} \cos θ − y\dot{θ} (5.3.45)
\dot{y} = \dot{r} \sin θ − (r\cos θ)\dot{θ} = \dot{r} \sin θ − x\dot{θ} (5.3.46)
It then follows that
x\dot{y}-y\dot{x}=\left(x\dot{r}\sin \theta + x^2\dot{\theta } \right) – \left(y\dot{r}\cos \theta – y^2\dot{\theta } \right) = r^2\dot{\theta }+\dot{r}(r\cos\theta \sin\theta – r \sin\theta \cos\theta ) =r^2\dot{\theta } (5.3.47)
from which we can solve for \dot{\theta }
\dot{\theta }=\frac{x\dot{y}-y\dot{x} }{r^2} (5.3.48)
Substituting \dot{x } and \dot{y} from the original differential equations, we get
\dot{\theta }=\frac{x(x + ayr^2) − y(−y + axr^2)}{r^2}=\frac{x^2+y^2}{r^2} (5.3.49)
As a result, the differential equation for the angular velocity becomes
\dot{\theta }=1 (5.3.50)
These equations confirm that the steady state of the system corresponds to r = 0, i.e. the origin (0, 0) in the original coordinates. The stability of this steady state depends on a. When a = 0, the original system in Eqs. (5.3.31) and (5.3.32) is a linear one, and indeed the steady state is a center. When a > 0, the differential equation for the radial position shows that a small positive perturbation of r will grow, with constant angular velocity (increasing phase angle). Thus the solution will rotate counter-clockwise away from the origin, indicating an unstable focus. On the other hand, when a < 0, the differential equation for the radial position shows that a small positive perturbation of r will decay, again with constant angular velocity. In this case the steady state behaves like a stable focus. These features of the solution are shown in Fig. 5.6. The take home message here is that in the case of a center in the linearized equations, nonlinear effects can change the stability of the steady state dramatically.
