

Title: Unique Limit Cycle? Post by Michael_Dagg on Nov 12^{th}, 2007, 3:01pm Show that the autonomous system x' = x  y  x^3 y' = x + y  y^3 as a unique limit cycle. 

Title: Re: Unique Limit Cycle? Post by Sameer on Nov 12^{th}, 2007, 6:08pm What is unique limit cycle? Never heard of it. 

Title: Re: Unique Limit Cycle? Post by towr on Nov 13^{th}, 2007, 12:03am on 11/12/07 at 18:08:38, Sameer wrote:
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). [hide]If you draw the graphs for x'=0 and y'=0, 

Title: Re: Unique Limit Cycle? Post by towr on Nov 13^{th}, 2007, 9:26am Here's a picture of the vector field, the green line is the limit cycle. Red is the curve x = y^{3}y (from y'=0) and blue is the curve y = x  x^{3} (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. 

Title: Re: Unique Limit Cycle? Post by Sameer on Nov 13^{th}, 2007, 12:35pm Isn't there another name for this? Divergence maybe? I will refer my book when I get home. 

Title: Re: Unique Limit Cycle? Post by towr on Nov 13^{th}, 2007, 12:56pm on 11/13/07 at 12:35:58, Sameer wrote:


Title: Re: Unique Limit Cycle? Post by Sameer on Nov 13^{th}, 2007, 1:42pm on 11/13/07 at 12:56:30, towr wrote:
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 

Title: Re: Unique Limit Cycle? Post by ThudanBlunder on Nov 13^{th}, 2007, 3:22pm Isn't the limit cycle also known as a strange attractor? 

Title: Re: Unique Limit Cycle? Post by towr on Nov 14^{th}, 2007, 12:11am on 11/13/07 at 15:22:56, ThudanBlunder wrote:
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. 

Title: Re: Unique Limit Cycle? Post by Michael_Dagg on Dec 2^{nd}, 2007, 7:54pm 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. 

Title: Re: Unique Limit Cycle? Post by william wu on May 20^{th}, 2008, 11:31pm 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 f_{1}(x,y) = x  y  x^3 f_{2}(x,y) = x + y  y^3 We then have the system x' = f_{1}(x,y) y' = f_{2}(x,y). First we find the equilibria; that is, find x* and y* such that the following two equations are satisfied: (1): 0 = f_{1}(x*,y*) (2): 0 = f_{2}(x*,y*). From Equation (1) we have, after some factoring, (3): y = x(1x)(1+x). Plugging this into the second equation yields 0 = x + x(1x)(1+x)( 1  x(1x)(1+x) )( 1 + x(1x)(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 = (f_{1},f_{2}) is Df = [ 13x^2 1 ] [ 1 13y^2] which, when evaluated at (0,0), yields the matrix [ 1 1] [ 1 1 ]. The eigenvalues of this matrix are 1+i and 1i. 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) = 13x^2 + 13y^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. 

Title: Re: Unique Limit Cycle? Post by Eigenray on May 21^{st}, 2008, 12:47am on 05/20/08 at 23:31:13, william wu wrote:
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. 

Title: Re: Unique Limit Cycle? Post by william wu on May 21^{st}, 2008, 1:10am 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 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdiv(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 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifdiv(f) dx dy > 0. Thus, a closed orbit must intersect the circle. 

Title: Re: Unique Limit Cycle? Post by Eigenray on May 21^{st}, 2008, 1:25am on 05/21/08 at 01:10:36, william wu 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/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3, then C doesn't intersect the circle of radius http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(2/3), but the integral of div f over the region bounded by C is 0. 

Title: Re: Unique Limit Cycle? Post by Eigenray on May 21^{st}, 2008, 1:50am Here are some ideas: Suppose we know that a limit cycle C exists. Let r^{2} = x^{2}+y^{2}. Then 2rr' = (r^{2})' = (x^{2}+y^{2})' = 2xx' + 2yy' = 2(x^{2} + y^{2}  x^{4}  y^{4}), 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 x^{2}+y^{2} = x^{4}+y^{4}, you are pushed back towards the origin, and once inside, you are pushed away again. So you just go around and around. 

Title: Re: Unique Limit Cycle? Post by william wu on May 21^{st}, 2008, 2:57am on 05/21/08 at 01:25:05, Eigenray wrote:
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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cgamma.gif 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cgamma.gif, or (2) contain http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cgamma.gif (as in your example). I like your latest argument! 

Title: Re: Unique Limit Cycle? Post by william wu on May 21^{st}, 2008, 3:35am Proof that any trajectory must rotate counterclockwise (if y is considered vertical, and x is considered horizontal) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif= tan (y/x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif' = (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^2 + y^2)/(x^2) = 1. (I can't do algebra) 

Title: Re: Unique Limit Cycle? Post by Eigenray on May 21^{st}, 2008, 4:14am I get that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif' = (x^{2} + y^{2} + x^{3}y  xy^{3})/(x^{2} + y^{2}) = 1 + r^{2} sin(4http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif)/4. So it is counterclockwise at least for r < 2. (And once r < 2, r stays < 2, since r' < 0 for r > http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2.) 

Title: Re: Unique Limit Cycle? Post by william wu on May 21^{st}, 2008, 4:58am You'll find that I make a lot of mistakes :[ New idea: Define the annulus A = { r e^{i http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif}  1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gifr http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gifM }, where M is some large number (e.g., 100, although actually "1.5" would suffice.). This is a (1) nonempty 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 PoincareBendixson 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 PoincareBendixson theorem). It remains to argue that the closed orbit is actually a limit cycle. 

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