- #1
Math100
- 802
- 222
- 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?
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?
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