How should I show that these systems have no periodic solutions?

[Math100]

Homework Statement

Show that the following systems have no periodic solutions.
a) $\dot{x}=x+x^3-y^2, \dot{y}=x^2-x^4+y^5$
b) $\dot{x}=yx^2-x^3-3xy^2-2y+y^2, \dot{y}=2xy^2-x^2+4x^3-2y$

Relevant Equations

Bendixson's negative criterion: 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.

a) Proof:

Consider the system $\dot{x}=x+x^3-y^2$ and $\dot{y}=x^2-x^4+y^5$.
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 $\dot{x}=X(x, y)=x+x^3-y^2$ and $\dot{y}=Y(x, y)=x^2-x^4+y^5$.
Then $\frac{\partial X}{\partial x}=3x^2+1$ and $\frac{\partial Y}{\partial y}=5y^4$.
This gives $\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}=5y^4+3x^2+1>0, \forall x, y$.
Since $\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}$ is always positive $\forall x, y$ and is of one sign,
it follows that there are no closed paths in a simply connected region of the phase plane.
Therefore, the system $\dot{x}=x+x^3-y^2$ and $\dot{y}=x^2-x^4+y^5$ has no periodic solutions.

b) Proof:

Consider the system $\dot{x}=yx^2-x^3-3xy^2-2y+y^2$ and $\dot{y}=2xy^2-x^2+4x^3-2y$.
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 $\dot{x}=X(x, y)=yx^2-x^3-3xy^2-2y+y^2$ and $\dot{y}=Y(x, y)=2xy^2-x^2+4x^3-2y$.
Then $\frac{\partial X}{\partial x}=2xy-3x^2-3y^2$ and $\frac{\partial Y}{\partial y}=4xy-2$.
This gives $\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}=6xy-3x^2-3y^2-2=6xy-3(x^2+y^2)-2<0, \forall x, y$.
Since $\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}$ is always negative $\forall x, y$ and is of one sign,
it follows that there are no closed paths in a simply connected region of the phase plane.
Therefore, the system $\dot{x}=yx^2-x^3-3xy^2-2y+y^2$ and $\dot{y}=2xy^2-x^2+4x^3-2y$ has no periodic solutions.

Above is my work for these two proofs. May anyone please check/verify/confirm to see if they are correct and accurate?

 

[pasmith]

You need to do more to show that $6xy - 3(x^2 + y^2) - 2$ does not change sign. For example, complete the square to show that $$6xy - 3(x^2 + y^2) - 2 = -3 (x^2 - 2xy + y^2) - 2 = -3(x - y)^2 - 2 < 0.$$

 

[Math100]

You need to do more to show that $6xy - 3(x^2 + y^2) - 2$ does not change sign. For example, complete the square to show that $$6xy - 3(x^2 + y^2) - 2 = -3 (x^2 - 2xy + y^2) - 2 = -3(x - y)^2 - 2 < 0.$$

Thank you so much, that was very helpful!