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Michael Dagg
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 Unique Limit Cycle?   « on: Nov 12th, 2007, 3:01pm » Quote Modify

Show that the autonomous system

x' = x - y - x^3
y' = x + y - y^3

as a unique limit cycle.
 « Last Edit: Nov 12th, 2007, 3:01pm by Michael Dagg » IP Logged

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Sameer
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 Re: Unique Limit Cycle?   « Reply #1 on: Nov 12th, 2007, 6:08pm » Quote Modify

What is unique limit cycle? Never heard of it.
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towr
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 Re: Unique Limit Cycle?   « Reply #2 on: Nov 13th, 2007, 12:03am » Quote Modify

on Nov 12th, 2007, 6:08pm, Sameer wrote:
 What is unique limit cycle? Never heard of it.
It's a closed path that is the limit of the paths that satisfy the system of differential equations (except those paths that converge on a limit point, and those that disappear to infinity). So a particle that starts off anywhere in the plane will eventually be caught up in the cycle or move off to a limit point (or infinity; at least I think that's also allowed).
In physics you often encounter it as a stable periodic solution, if you perturb the system, it moves back to the cycle again. (If there are multiple limit cycles, then a sufficiently large perturbation can switch from one to the other; as long as you can't get flung out to infinity).

If you draw the graphs for x'=0 and y'=0, you can see there's only two places a cycle might exist. And the intersection of the two graphs gives the limit points, although they're not necessarily stable.
 « Last Edit: Nov 13th, 2007, 7:05am by towr » IP Logged

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towr
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 Re: Unique Limit Cycle?   dif_eq1.jpg « Reply #3 on: Nov 13th, 2007, 9:26am » Quote Modify

Here's a picture of the vector field, the green line is the limit cycle. Red is the curve x = y3-y (from y'=0) and blue is the curve y = x - x3 (from x'=0)
It should just about be visible that all the vectors outside the limit circle point inwards, and all vectors inside the limit cycle point outwards. Where the blue and red curve cross we have a stationary point (x'=y'=0, so the point has no inclination to move; but it is not stable)

None of that is an immediate help to solve the problem, though. Even though we can see there is just one limit cycle, that doesn't make a formal proof.
 « Last Edit: Nov 13th, 2007, 9:32am by towr » IP Logged

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Sameer
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 Re: Unique Limit Cycle?   « Reply #4 on: Nov 13th, 2007, 12:35pm » Quote Modify

Isn't there another name for this? Divergence maybe? I will refer my book when I get home.
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 Re: Unique Limit Cycle?   « Reply #5 on: Nov 13th, 2007, 12:56pm » Quote Modify

on Nov 13th, 2007, 12:35pm, Sameer wrote:
 Isn't there another name for this? Divergence maybe? I will refer my book when I get home.
I don't see any alternate names on the wikipage ( http://en.wikipedia.org/wiki/Limit_cycle )
 « Last Edit: Nov 13th, 2007, 12:58pm by towr » IP Logged

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Sameer
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 Re: Unique Limit Cycle?   « Reply #6 on: Nov 13th, 2007, 1:42pm » Quote Modify

on Nov 13th, 2007, 12:56pm, towr wrote:
 I don't see any alternate names on the wikipage ( http://en.wikipedia.org/wiki/Limit_cycle )

I was thinking of divergence and curl in Vector Calculus. I don't know if they are related or not.

http://mathworld.wolfram.com/Divergence.html

http://mathworld.wolfram.com/Curl.html
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ThudnBlunder
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 Re: Unique Limit Cycle?   « Reply #7 on: Nov 13th, 2007, 3:22pm » Quote Modify

Isn't the limit cycle also known as a strange attractor?
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 Re: Unique Limit Cycle?   « Reply #8 on: Nov 14th, 2007, 12:11am » Quote Modify

on Nov 13th, 2007, 3:22pm, ThudanBlunder wrote:
 Isn't the limit cycle also known as a strange attractor?
A limit cycle can be an attractor, but it isn't a strange one.
A limit cycle is a closed trajectory and if it's attractive all neighbouring trajectories spiral into it as t \to  \infty.
A strange attractor on the other hand is just really strange. It doesn't consist of a closed trajectory, and there's no telling where neighbouring trajectories end up.
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Michael Dagg
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 Re: Unique Limit Cycle?   « Reply #9 on: Dec 2nd, 2007, 7:54pm » Quote Modify

That is a good plot towr. The blue and red space
curves reveal something about how the trajectories
behave as they get near the attractor. If you plot
some of the trajectories you will see this.

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Michael Dagg
william wu

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 Re: Unique Limit Cycle?   uniquelimitcycle.png « Reply #10 on: May 20th, 2008, 11:31pm » Quote Modify

Some initial analysis for this problem. BTW, out of curiosity, is there a physical motivation for this particular set of differential equations?

begin analysis:
--------------------
Part I: equilibria

Define
f1(x,y) = x - y - x^3
f2(x,y) = x + y - y^3

We then have the system
x' = f1(x,y)
y' = f2(x,y).

First we find the equilibria; that is, find x* and y* such that the following two equations are satisfied:
(1): 0 = f1(x*,y*)
(2): 0 = f2(x*,y*).

From Equation (1) we have, after some factoring,
(3): y = x(1-x)(1+x).

Plugging this into the second equation yields
0 = x + x(1-x)(1+x)( 1 - x(1-x)(1+x) )( 1 + x(1-x)(1+x) )
= -x(2 - 2x^2 + 3x^4 - 3x^6 + x^8)

Using a computer to find the roots, we see that the only real solution is given by x* = 0, which implies from (3) that y* = 0. Hence the only equilibrium point is at the origin.

Next, we determine the nature of the equilibrium point. The Jacobian of f = (f1,f2) is

Df =
[ 1-3x^2  -1   ]
[ 1  1-3y^2]

which, when evaluated at (0,0), yields the matrix
[ 1 -1]
[ 1 1 ].

The eigenvalues of this matrix are 1+i and 1-i. Since the real parts for both eigenvalues are positive, the equilibrium is unstable -- so trajectories will spiral out from it.

Part II: Closed Orbits

The divergence of f is
div(f) = 1-3x^2 + 1-3y^2
= 2 - 3(x^2 + y^2).

By Bendixson's theorem, any closed orbit must intersect the locus of points such that div(f) = 0; that locus is simply the circle given by the equation

x^2 + y^2 = 2/3.

So, if there is a limit cycle, then it must intersect this circle, since all limit cycles are closed orbits. (However, not all closed orbits are limit cycles.)

However, I'm a little concerned about this statement since it's not so clear from the simulation plots that the cycle intersects the circle of radius sqrt(2/3) ~= 0.816.

 « Last Edit: May 21st, 2008, 12:02pm by william wu » IP Logged

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Eigenray
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 Re: Unique Limit Cycle?   « Reply #11 on: May 21st, 2008, 12:47am » Quote Modify

on May 20th, 2008, 11:31pm, william wu wrote:
 By Bendixson's theorem, any closed orbit must intersect the locus of points such that div(f) = 0

Bendixson's theorem follows from that fact that the flux around a closed orbit is 0, so by Green's theorem the integral of the divergence over the region bounded by the orbit is 0.  So it is this latter region which must contain points both inside and outside the circle div f = 0.

But that's pretty much all I know about closed orbits, and only because it was a homework problem in the class I taught last semester.
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william wu

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 Re: Unique Limit Cycle?   « Reply #12 on: May 21st, 2008, 1:10am » Quote Modify

Mmm, forgive me, I can't tell if you are just explaining the theorem, or correcting something I said. Is it wrong to say that any closed orbit must intersect the circle?

My reasoning:

Say we have a closed orbit that lies completely outside the circle. Then div(f) is always negative in the region encircled by that orbit. Applying Green's theorem then yields a contradiction, as we get 0 = div(f) dx dy < 0.

Similarly, say we have a closed orbit that lies completely inside the circle. Then div(f) is always positive in the region encircled by that orbit. Green's theorem again yields a contradiction, as we get 0 = div(f) dx dy > 0.

Thus, a closed orbit must intersect the circle.
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 Re: Unique Limit Cycle?   « Reply #13 on: May 21st, 2008, 1:25am » Quote Modify

on May 21st, 2008, 1:10am, william wu wrote:
 Say we have a closed orbit that lies completely outside the circle. Then div(f) is always negative in the region encircled by that orbit.

That's true if the region bounded by the orbit lies outside the circle.  But the orbit itself is the boundary of this region.

For example if the orbit were C, the circle centered at the origin of radius 2/3, then C doesn't intersect the circle of radius (2/3), but the integral of div f over the region bounded by C is 0.
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 Re: Unique Limit Cycle?   limitcycle-x2y2eqx4y1.gif « Reply #14 on: May 21st, 2008, 1:50am » Quote Modify

Here are some ideas: Suppose we know that a limit cycle C exists.  Let r2 = x2+y2.  Then

2rr' = (r2)'
= (x2+y2)'
= 2xx' + 2yy'
= 2(x2 + y2 - x4 - y4),

which is positive for r < 1.  It follows that a closed cycle can never enter the circle of radius 1.  In particular, C lies outside the region where div f > 0.

Suppose C' is another cycle.  Then it too lies outside the region div f > 0.  But then the integral of div f over the region bounded by C' is positive if C' is inside C, and negative if C' is outside C.  Since distinct cycles are disjoint, the only possibility is C' = C.

We should be able to show something like the following: Once you go outside the curve x2+y2 = x4+y4, you are pushed back towards the origin, and once inside, you are pushed away again.  So you just go around and around.
 « Last Edit: May 21st, 2008, 1:03pm by Eigenray » IP Logged

william wu

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 Re: Unique Limit Cycle?   « Reply #15 on: May 21st, 2008, 2:57am » Quote Modify

on May 21st, 2008, 1:25am, Eigenray wrote:
 That's true if the region bounded by the orbit lies outside the circle.  But the orbit itself is the boundary of this region.   For example if the orbit were C, the circle centered at the origin of radius 2/3, then C doesn't intersect the circle of radius (2/3), but the integral of div f over the region bounded by C is 0.

OK I see. Actually when I was writing "completely outside the circle", in my mind, I was thinking "region bounded by the orbit" lies outside the circle. But my conclusion was still wrong, as illustrated by your example. So perhaps the correct statement is:

Let denote the path over which div(f) = 0. Then, if there is a closed orbit, the region bounded by that orbit must either (1) cross , or (2) contain (as in your example).

 « Last Edit: May 21st, 2008, 3:22am by william wu » IP Logged

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william wu

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 Re: Unique Limit Cycle?   « Reply #16 on: May 21st, 2008, 3:35am » Quote Modify

Proof that any trajectory must rotate counterclockwise (if y is considered vertical, and x is considered horizontal)

= tan (y/x)
' = (1/(1 + (y/x)^2)) ( y'x - x'y)/(x^2)
= (x^2/(x^2 + y^2)) ( y'x - x'y)/(x^2)

Substituting our equations for x' and y' yields:

' = (x^2/(x^2 + y^2)) ( (x+y-y^3)x - y(x - y - x^3) )/(x^2)
= (x^2/(x^2 + y^2)) ( x^2 + y^2)/(x^2)
= 1.

(I can't do algebra)
 « Last Edit: May 21st, 2008, 4:31am by william wu » IP Logged

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 Re: Unique Limit Cycle?   « Reply #17 on: May 21st, 2008, 4:14am » Quote Modify

I get that

' = (x2 + y2 + x3y - xy3)/(x2 + y2) = 1 + r2 sin(4)/4.

So it is counterclockwise at least for r < 2.  (And once r < 2, r stays < 2, since r' < 0 for r > 2.)
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william wu

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 Re: Unique Limit Cycle?   « Reply #18 on: May 21st, 2008, 4:58am » Quote Modify

You'll find that I make a lot of mistakes

New idea: Define the annulus A = { r ei |  1 r M }, where M is some large number (e.g., 100, although actually "1.5" would suffice.). This is a (1) non-empty compact set. Furthermore, it is (2) positively invariant: if we start somewhere in the annulus, then we can never enter into the inner circle r<1, and we can never exit into the outer region r>M since

(r^2)' = 2((x^2 - x^4) + (y^2 - y^4))

is clearly negative when x and y are both large.

By the Poincare-Bendixson Theorem, conditions (1) and (2) imply that A must contain an equilibrium point or a closed orbit. From my initial analysis, the only equilibrium point present in this sytem is at the origin, which does not lie in A. Hence, A must contain a closed orbit.

So we have shown
(1) there can be no more than one closed orbit (using Bendixson's theorem),
(2) there indeed exists one closed orbit (using the Poincare-Bendixson theorem).

It remains to argue that the closed orbit is actually a limit cycle.
 « Last Edit: May 21st, 2008, 1:50pm by william wu » IP Logged

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