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Math Help - Showing there is a limit cycle

  1. #1
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    Showing there is a limit cycle

    Planar system:

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

    only has a critical point at origin.

    The question asks to show that this system has a periodic orbit.

    The hint for this says to convert to polar coordinates and show that for all \epsilon>0, r'<0 on the circle r=sqrt(2)+ \epsilon and r'>0 on the circle r=1- \epsilon. Then show that this implies that there is a limit cycle in the annulus: {x \inR^2 : 1<= |x| <= sqrt(2)}.

    I'm not sure how to even convert this to polar coordinates. I know that x=rcos( \theta) and y=rsin( \theta). But after i plug it in, I just get a equation and I don't know how to get r' or \theta' from it.
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  2. #2
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    Quote Originally Posted by Borkborkmath View Post
    Planar system:

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

    only has a critical point at origin.

    The question asks to show that this system has a periodic orbit.

    The hint for this says to convert to polar coordinates and show that for all \epsilon>0, r'<0 on the circle r=sqrt(2)+ \epsilon and r'>0 on the circle r=1- \epsilon. Then show that this implies that there is a limit cycle in the annulus: {x \inR^2 : 1<= |x| <= sqrt(2)}.

    I'm not sure how to even convert this to polar coordinates. I know that x=rcos( \theta) and y=rsin( \theta). But after i plug it in, I just get a equation and I don't know how to get r' or \theta' from it.
    Use the formula r^2=x^2+y^2 and take the derivative with respect to time

    2r\dot{r}=2x\dot{x}+2y\dot{y} \iff \dot{r}=\frac{1}{r}(x\dot{x}+y\dot{y}) and

    \theta=\tan^{-1}\left( \frac{y}{x} \right)

    \dot{\theta}=\frac{1}{1+\left( \frac{y}{x}\right)^2}(\frac{\dot{y}}{x}-\frac{y}{x^2}\dot{x})=\frac{1}{x^2+\left( y\right)^2}(x\dot{y}-y\dot{x})=\frac{1}{r^2}(x\dot{y}-y\dot{x})

    Now we need to compute the "inside" of each of the above

    x\dot{x}+y\dot{y}=x(x-y-x^3)+y(x+y-y^3)=x^2+y^2-(x^4+y^4) = x^2+y^2-(x^2+y^2)^2+2x^2y^2=r^2-r^4+2r^2\sin^2(\theta)\cos^2(\theta)

    x\dot{y}-y\dot{x}=x(x+y-y^3)-y(x-y-x^3)=x^2+y^2-xy^3+x^3y=x^2+y^2-xy(y^2-x^2)=r^2-r^4\sin(\theta)\cos(\theta)[\cos^2(\theta)-\sin^2(\theta)]

    This gives the system of ODE's

    \frac{dr}{dt}=r-r^3+2r\sin^2(\theta)\cos^2(\theta)

    \frac{d\theta}{dt}=1-r^2\sin(\theta)\cos(\theta)[\cos^2(\theta)-\sin^2(\theta)]

    either I have made a mistake (very possible) or I can't see how to simplify and solve this system. I will leave this to you at this point.
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  3. #3
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    From what TheEmptySet said

    \dfrac{dr}{dt} = r\left( 1 - r^2 + r \dfrac{\sin^2 2 \theta}{2}\right)

    the following identity might help

    1 - r^2 \le 1 - r^2 + r \dfrac{\sin^2 2 \theta}{2} \le 1 - r^2 + \dfrac{r}{2}
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  4. #4
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    Thanks for your replies.

    When I expand xx'+yy', I get x^2+y^2-x^4-y^4 and just leave it that way. Then i sub in rcos and rsin which leads to: r-r^3(cos^4(\theta)+sin^4(\theta)). Is the way I did it wrong?
    Since you simplified x^4-y^4 to "-(x^2 + y^2)^2+2x^2y^2"
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  5. #5
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    Quote Originally Posted by Borkborkmath View Post
    Thanks for your replies.

    When I expand xx'+yy', I get x^2+y^2-x^4-y^4 and just leave it that way. Then i sub in rcos and rsin which leads to: r-r^3(cos^4(\theta)+sin^4(\theta)). Is the way I did it wrong?
    Since you simplified x^4-y^4 to "-(x^2 + y^2)^2+2x^2y^2"
    No the two expressions are equivalent by trig identities. When I did this I was trying to find a closed form solution, but you can bound this just like Danny did by finding min and max of the function with respect to theta.
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  6. #6
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    Ah ic.

    So now I should just plug in the two given values for r and show the r'<0 and r'>0 respectively.
    Then this leads to the annulus having a limit cycle because: "A compact set that is positively or negatively invariant contains either a limit cycle or an equilibrium point", and since the only critical point in at origin we know this to be a limit cycle?

    Hm, but I seem to run into a problem that when r=sqrt(2)-\epsilon the equation isn't negative.
    Last edited by Borkborkmath; June 6th 2011 at 06:09 PM.
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    Quote Originally Posted by Borkborkmath View Post
    Ah ic.

    So now I should just plug in the two given values for r and show the r'<0 and r'>0 respectively.
    Then this leads to the annulus having a limit cycle because: "A compact set that is positively or negatively invariant contains either a limit cycle or an equilibrium point", and since the only critical point in at origin we know this to be a limit cycle?

    Hm, but I seem to run into a problem that when r=sqrt(2)-\epsilon the equation isn't negative.
    Using your expression we have

    \dot{r}=r-a \cdot r^3, a \in \left[\frac{1}{2},1 \right]

    If you solve this ODE you get

    r(t)=\pm \frac{1}{\sqrt{a+ce^{-t}}} for since r > 0

    r(t)= \frac{1}{\sqrt{a+ce^{-t}}} \lim_{t \to \infty}r(t)=\frac{1}{\sqrt{a}}
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  8. #8
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    Sorry for that. I was doing my math wrong, but I got it and just forgot to re-edit my post.
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