Disprove/Prove Existence of Periodic Solution for Autonomous ODE

Question

Consider the system $\dot x = x^2 + y^2 -1$ and $\dot y = y - 2xy$. I am new in this field. I draw the vector filed and I saw that there is no obvious periodic solution. How can I prove/disprove the existence of periodic solution for this system?

Answer 1

The simplest way I can see to show that the system

$\dot x = x^2 + y^2 - 1, \tag 1$

$\dot y = y - 2xy, \tag 2$

has no periodic solution is the invoke the Benixson-Dulac theorem which asserts that a two-dimensional vector field $\mathbf X(x, y)$ has no periodic orbits in a region where $\nabla \cdot \mathbf X(x, y)$ has the same sign almost everywhere; the intuition behind this is easy: if

$\mathbf X = \begin{pmatrix} X_x(x, y) \\ X_y(x, y) \end{pmatrix} \tag 3$

has a closed trajectory $\gamma(t)$, we may use divergence theorem to affirm that

$\displaystyle \int_\Omega \nabla \cdot \mathbf X \; dA = \int_\gamma \mathbf X \cdot \mathbf n \; ds, \tag 4$

where $\Omega$ is the region surrounded by $\gamma(t)$, and $\mathbf n$ is the outward pointing normal with respect to $\Omega$ along $\gamma(t)$; since

$\dot \gamma(t) = \mathbf X(\gamma(t)), \tag 5$

we have

$\mathbf X(\gamma(t)) \cdot \mathbf n(\gamma(t)) = 0 \tag 6$

thus the right-hand integral in (4) vanishes; but the left-hand integral cannot vanish if $\nabla \cdot \mathbf X(x, y)$ has the same sign almost everywhere.

In the present case, we take

$\mathbf X(x, y) = \begin{pmatrix} X_x(x, y) \\ X_y(x, y) \end{pmatrix} = \begin{pmatrix} x^2 + y^2 - 1 \\ y - 2xy \end{pmatrix}, \tag 7$

and we see that

$\nabla \cdot \mathbf X(x, y) = \dfrac{\partial X_x(x, y)}{\partial x} + \dfrac{\partial X_y(x, y)}{\partial y} = 2x + 1 - 2x = 1; \tag 8$

since $\nabla \cdot \mathbf X(x, y) = 1$ everywhere, Bendixson-Dulac precludes the existence of a periodic orbit.