Show that the system has no closed orbits by finding a Lyapunov

1. Feb 21, 2013

Jamin2112

Show that the system has no closed orbits by finding a Lyapunov ....

1. The problem statement, all variables and given/known data

I'm at the point in the problem where I need constants a and b satisfying

ax2(y-x3) + by2(-x-y3) < 0
and ax2+bx2 > 0

for all (x,y)≠(0,0).

2. Relevant equations

Just in case you're wondering, this is to satisfy the V(x,y)=ax2+by2 > 0 and ΔV(x,y)•<y-x3, -x-y3> < 0 so I can apply that one theorem to my problem.

3. The attempt at a solution

Well, it seems reasonable to choose a,b>0 to ensure ax2+bx2 > 0, but I'm having trouble figuring out how to make ax2(y-x3) + by2(-x-y3) < 0 simultaneously.

2. Feb 21, 2013

Staff: Mentor

Re: Show that the system has no closed orbits by finding a Lyapunov ..

For x=y, this can be simplified to $(a-b)x^3 - (a+b)x^5 <0$, which cannot be true both for positive and negative small x.

3. Feb 21, 2013

pasmith

Re: Show that the system has no closed orbits by finding a Lyapunov ..

You won't be able to do that: on the assumption that $V = ax^2 + by^2$ and $\dot V = ax^2(y - x^3) - by^2(x + y^3)$ your system is
$$\dot x = \textstyle\frac12 x(y - x^3) \\ \dot y = -\textstyle\frac12 y(x + y^3).$$
The fixed point at the origin is some kind of non-hyperbolic saddle: x = 0 and y = 0 are invariant. On x = 0, $\dot y < 0$ for $y \neq 0$ and on y = 0, $\dot x < 0$ for $x \neq 0$. Thus there are points arbitrarily close to the origin where motion is unambiguously away from the origin, and at these points $\dot V > 0$.

However, you now know that no periodic orbit can cross the coordinate axes, so the origin cannot be inside a periodic orbit. Since the origin is the only fixed point (any other fixed point must satisfy $x^8 = -1$), there can be no periodic orbits (because in a 2D system every periodic orbit must enclose at least one fixed point).