Consider the planar system
$\dot x_1 = x_2 – x_1^3$
$\dot x_2 = -x_1$
Prove that there exists no periodic orbit in this system. I tried to use the Bendixson criteria. The divergence is equal to $-3x_1^2$ which is equal to zero on the $x_2$ axis and therefore we cannot use the Bendixson criteria.
Best Answer
Given the system
$\dot x_1 = x_2 - x_1^3, \tag 1$
$\dot x_2 = -x_1, \tag 2$
we see that the divergence of the vector field $(x_2 - x_1^3, -x_1)^T$ is
$\dfrac{\partial (x_2 - x_1^3)}{\partial x_1} + \dfrac{\partial (-x_1)}{\partial x_2} = -3x_1^2; \tag 3$
since $-3x_1^2$ vanishes only on the set $\{(x_1, x_2) \in \Bbb R^2 \mid x_1 = 0 \}$, which being the $x_2$-axis is of measure $0$, and is negative everywhere else, it follows from the Bendixson-Dulac theorem that (1)-(2) has no periodic orbits.