How should I show that this system has no limit cycle?

[Math100]

Homework Statement

Show that the system $\dot{x}=y(y-x), \dot{y}=x^2$ has no limit cycle.

Relevant Equations

Bendixson's negative criterion states that there are no closed paths in a simply connected region of the phase plane on which $\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}$ is of one sign.

Dulac's test states that there are no closed paths in a simply connected region in which $\frac{\partial(\rho X)}{\partial x}+\frac{\partial(\rho Y)}{\partial y}$ is of one sign, where $\rho(x, y)$ is any function having continuous first partial derivatives.

Proof:

Consider the system $\dot{x}=y(y-x), \dot{y}=x^2$.
By theorem, Bendixson's negative criterion states that there are no closed paths in a simply connected region of the phase plane on which $\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}$ is of one sign.
Let $\dot{x}=X(x, y)=y(y-x)$ and $\dot{y}=Y(x, y)=x^2$.
Then $\frac{\partial X}{\partial x}=-y$ and $\frac{\partial Y}{\partial y}=0$, which gives $\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}=-y+0=-y$.

From here, I don't think applying the Bendixson's negative criterion works in this case. By applying the Dulac's test, I would have to use a function but what function should I use in this problem? I was thinking to use the function $\rho(x, y)=\frac{1}{y}$ but I don't know if this is the right function to use, because I've got $\rho X=\frac{1}{y}\cdot y(y-x)=y-x$ and $\rho Y=\frac{x^2}{y}$, so $\frac{\partial\rho X}{\partial x}+\frac{\partial\rho Y}{\partial y}=-1-\frac{x^2}{y^2}=-1-(\frac{x}{y})^2<0, \forall x, y$. Is this correct? Also, how to determine when to use/apply the Bendixson's negative criterion or the Dulac's test? In this problem, for example, is it okay to apply/use the Dulac's test straight ahead without first applying/using the Bendixson's negative criterion?

 

[pasmith]

It is not necessary to apply Dulac (Bendixon is just Dulac with $\rho \equiv 1$) here.

You know that $\dot y = x^2 > 0$. Is a periodic orbit possible when $\dot y$ cannot change sign?

 

[Math100]

It is not necessary to apply Dulac (Bendixon is just Dulac with $\rho \equiv 1$) here.

You know that $\dot y = x^2 > 0$. Is a periodic orbit possible when $\dot y$ cannot change sign?

No.