Differential Equation (Solutions Attraction)

Question

What does it mean in part D when it states that this solution has a periodic solution that attracts other solutions?

Answer 1

Given that

$\dot x = \mu x + y - x(x^2 + y^2), \tag1$

$\dot y = \mu y - x - y(x^2 + y^2); \tag 2$

we have

$x \dot x = \mu x^2 + xy - x^2(x^2 + y^2), \tag 3$

$y \dot y = \mu y^2 - xy - y^2(x^2 + y^2); \tag 4$

whence,

$r \dot r = x \dot x + y \dot y = \mu x^2 - x^2(x^2 + y^2) + \mu y^2 - y^2(x^2 + y^2)$ $= (\mu - (x^2 + y^2))(x^2 + y^2) = (\mu - r^2)r^2; \tag 5$

thus, with $r \ne 0$,

$\dot r = (\mu - r^2)r; \tag 6$

also,

$y \dot x = \mu xy + y^2 - xy(x^2+ y^2); \tag 7$

$x \dot y = -x^2 + \mu xy - xy(x^2 + y^2); \tag 8$

$x \dot y - y \dot x = -(x^2 + y^2) = -r^2; \tag 9$

$\dot \theta = \dfrac{r^2}{r^2} = 1; \tag{10}$

(a) is (6), (10).

(b) follows directly from (a).

Using $xy$-coordinates, the Jacobean at the origin is

$J = \begin{bmatrix} \mu & 1 \\ -1 & \mu \end{bmatrix}; \tag{11}$

The eigenvalues of $J$ are $\mu \pm i$; thus the origin is a stable spiral for $\mu < 0$; an unstable spiral for $\mu > 0$; for $\mu = 0$ the eigenvalues are insufficient to determine stability and we must look deeper; however, from (4) we have

$\dot r = -r^3, \; \mu = 0; \tag{12}$

thus $\dot r < 0$ when $r > 0$ and we conclude (0) is stable when $\mu = 0$.

When $\mu > 0$, we note from (4) that

$\dot r = 0 \; \text{when} \; r = \sqrt \mu; \tag{13}$

we see that

$r > \sqrt \mu \Longrightarrow \dot r < 0; \tag{14}$

$0 < r < \sqrt m \Longrightarrow \dot r > 0; \tag{15}$

The system spirals toward the periodic orbit at $r = \sqrt \mu$; this is what it means for the solution $r = \sqrt \mu$ to attract other solutions.