using Poincaré-Bendixon to prove periodic solution existence












2












$begingroup$


I got the system:
$$dot{x} = x-y-y^3-2x(x^2+y^2)$$
$$dot{y}= x+y-2y(x^2+y^2)$$
And it's given that the origin is the only fixed point.
I've converted it to Polar using $dot{r}r=dot{x}x+dot{y}y$ and ended up with:
$$rdot{r}=r^2-2r^3-r^4costheta sin^3theta $$
Which I have simplified to the following but I do not trust my trig within this.
$$dot{r}=r-2r^2-frac{r^3}{4}(sin2theta-frac{1}{2}sin4theta)$$
I'm not very sure if this was the right way to go because when it comes to classifying $dot{r}$ now I have to deal with $sin2theta$ and $sin4theta$. So I guess my question right now is did I end up with the correct $dot{r}$? If yes how do I end up trapping the region in this situation?



I know that I have to check for when $dot{r}>0$ for the outward flow and $dot{r}<0$ for the inward flow. But I do not know how to inspect that with both $sin{2theta}$ and $sin{4theta}$ present at the same time.



My brain right now is a scattered mess and I do not trust anything that comes out of it so I have resorted to share this with you and hopefully someone can put my mind to rest and let me know where I am going wrong because I have tried to solve this question 4 times now and every time I get a different $dot{r}$.






ordinary-differential-equations systems-of-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The term $-2r^3$ looks wrong, since in $x dot x + y dot y$ every term is either of degree 2 or degree 4.
    $endgroup$
    – Hans Lundmark
    Dec 31 '18 at 13:20










  • $begingroup$
    yup, you're right $2r^2(x^2+y^2)=2r^4$ thanks for pointing it out!
    $endgroup$
    – Bluedog
    Dec 31 '18 at 13:43
















2












$begingroup$


I got the system:
$$dot{x} = x-y-y^3-2x(x^2+y^2)$$
$$dot{y}= x+y-2y(x^2+y^2)$$
And it's given that the origin is the only fixed point.
I've converted it to Polar using $dot{r}r=dot{x}x+dot{y}y$ and ended up with:
$$rdot{r}=r^2-2r^3-r^4costheta sin^3theta $$
Which I have simplified to the following but I do not trust my trig within this.
$$dot{r}=r-2r^2-frac{r^3}{4}(sin2theta-frac{1}{2}sin4theta)$$
I'm not very sure if this was the right way to go because when it comes to classifying $dot{r}$ now I have to deal with $sin2theta$ and $sin4theta$. So I guess my question right now is did I end up with the correct $dot{r}$? If yes how do I end up trapping the region in this situation?



I know that I have to check for when $dot{r}>0$ for the outward flow and $dot{r}<0$ for the inward flow. But I do not know how to inspect that with both $sin{2theta}$ and $sin{4theta}$ present at the same time.



My brain right now is a scattered mess and I do not trust anything that comes out of it so I have resorted to share this with you and hopefully someone can put my mind to rest and let me know where I am going wrong because I have tried to solve this question 4 times now and every time I get a different $dot{r}$.






ordinary-differential-equations systems-of-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The term $-2r^3$ looks wrong, since in $x dot x + y dot y$ every term is either of degree 2 or degree 4.
    $endgroup$
    – Hans Lundmark
    Dec 31 '18 at 13:20










  • $begingroup$
    yup, you're right $2r^2(x^2+y^2)=2r^4$ thanks for pointing it out!
    $endgroup$
    – Bluedog
    Dec 31 '18 at 13:43














2












2








2





$begingroup$


I got the system:
$$dot{x} = x-y-y^3-2x(x^2+y^2)$$
$$dot{y}= x+y-2y(x^2+y^2)$$
And it's given that the origin is the only fixed point.
I've converted it to Polar using $dot{r}r=dot{x}x+dot{y}y$ and ended up with:
$$rdot{r}=r^2-2r^3-r^4costheta sin^3theta $$
Which I have simplified to the following but I do not trust my trig within this.
$$dot{r}=r-2r^2-frac{r^3}{4}(sin2theta-frac{1}{2}sin4theta)$$
I'm not very sure if this was the right way to go because when it comes to classifying $dot{r}$ now I have to deal with $sin2theta$ and $sin4theta$. So I guess my question right now is did I end up with the correct $dot{r}$? If yes how do I end up trapping the region in this situation?



I know that I have to check for when $dot{r}>0$ for the outward flow and $dot{r}<0$ for the inward flow. But I do not know how to inspect that with both $sin{2theta}$ and $sin{4theta}$ present at the same time.



My brain right now is a scattered mess and I do not trust anything that comes out of it so I have resorted to share this with you and hopefully someone can put my mind to rest and let me know where I am going wrong because I have tried to solve this question 4 times now and every time I get a different $dot{r}$.






ordinary-differential-equations systems-of-equations







share|cite|improve this question









$endgroup$




I got the system:
$$dot{x} = x-y-y^3-2x(x^2+y^2)$$
$$dot{y}= x+y-2y(x^2+y^2)$$
And it's given that the origin is the only fixed point.
I've converted it to Polar using $dot{r}r=dot{x}x+dot{y}y$ and ended up with:
$$rdot{r}=r^2-2r^3-r^4costheta sin^3theta $$
Which I have simplified to the following but I do not trust my trig within this.
$$dot{r}=r-2r^2-frac{r^3}{4}(sin2theta-frac{1}{2}sin4theta)$$
I'm not very sure if this was the right way to go because when it comes to classifying $dot{r}$ now I have to deal with $sin2theta$ and $sin4theta$. So I guess my question right now is did I end up with the correct $dot{r}$? If yes how do I end up trapping the region in this situation?



I know that I have to check for when $dot{r}>0$ for the outward flow and $dot{r}<0$ for the inward flow. But I do not know how to inspect that with both $sin{2theta}$ and $sin{4theta}$ present at the same time.



My brain right now is a scattered mess and I do not trust anything that comes out of it so I have resorted to share this with you and hopefully someone can put my mind to rest and let me know where I am going wrong because I have tried to solve this question 4 times now and every time I get a different $dot{r}$.






ordinary-differential-equations systems-of-equations




ordinary-differential-equations systems-of-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 12:26









BluedogBluedog

257




257








  • 1




    $begingroup$
    The term $-2r^3$ looks wrong, since in $x dot x + y dot y$ every term is either of degree 2 or degree 4.
    $endgroup$
    – Hans Lundmark
    Dec 31 '18 at 13:20










  • $begingroup$
    yup, you're right $2r^2(x^2+y^2)=2r^4$ thanks for pointing it out!
    $endgroup$
    – Bluedog
    Dec 31 '18 at 13:43














  • 1




    $begingroup$
    The term $-2r^3$ looks wrong, since in $x dot x + y dot y$ every term is either of degree 2 or degree 4.
    $endgroup$
    – Hans Lundmark
    Dec 31 '18 at 13:20










  • $begingroup$
    yup, you're right $2r^2(x^2+y^2)=2r^4$ thanks for pointing it out!
    $endgroup$
    – Bluedog
    Dec 31 '18 at 13:43








1




1




$begingroup$
The term $-2r^3$ looks wrong, since in $x dot x + y dot y$ every term is either of degree 2 or degree 4.
$endgroup$
– Hans Lundmark
Dec 31 '18 at 13:20




$begingroup$
The term $-2r^3$ looks wrong, since in $x dot x + y dot y$ every term is either of degree 2 or degree 4.
$endgroup$
– Hans Lundmark
Dec 31 '18 at 13:20












$begingroup$
yup, you're right $2r^2(x^2+y^2)=2r^4$ thanks for pointing it out!
$endgroup$
– Bluedog
Dec 31 '18 at 13:43




$begingroup$
yup, you're right $2r^2(x^2+y^2)=2r^4$ thanks for pointing it out!
$endgroup$
– Bluedog
Dec 31 '18 at 13:43










1 Answer
1






active

oldest

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2












$begingroup$

You missed a power in combining the last terms of the equations, you should have got
$$
rdot r = r^2-2r^4-xy^3=r^2-(2+sin^3θcosθ)r^4
$$
so that because of the mean inequality of the geometric and quadratic mean $$sqrt[4]{|sqrt3cs^3|}le sqrt{frac{3c^2+s^2+s^2+s^2}4}=sqrt{frac34}$$ we get bounds for the derivative of the radius in terms of the radius only as
$$
r-ar^3le dot rle r-br^3 ~~text{ with }~~ a,b=2pm frac{3sqrt3}{16}.
$$
This means that the radius of a solution decreases in time where the right side is negative, which is for $r>sqrt{1/b}approx 0.773$. The radius increases in time where the leftmost term is positive, that is for $0<r<sqrt{1/a}approx 0.656$.



This gives an invariant annulus or trapping region. Per the given claim of no other stationary points outside the origin one concludes the existence of a limit cycle.



The plot shows clearly the attracting limit cycle in the trapping region $0.6<r<0.8$, visually it is inside the smaller annulus $0.7lessapprox rlessapprox 0.75$.
plot of the dynamics






share|cite|improve this answer











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    $begingroup$

    You missed a power in combining the last terms of the equations, you should have got
    $$
    rdot r = r^2-2r^4-xy^3=r^2-(2+sin^3θcosθ)r^4
    $$
    so that because of the mean inequality of the geometric and quadratic mean $$sqrt[4]{|sqrt3cs^3|}le sqrt{frac{3c^2+s^2+s^2+s^2}4}=sqrt{frac34}$$ we get bounds for the derivative of the radius in terms of the radius only as
    $$
    r-ar^3le dot rle r-br^3 ~~text{ with }~~ a,b=2pm frac{3sqrt3}{16}.
    $$
    This means that the radius of a solution decreases in time where the right side is negative, which is for $r>sqrt{1/b}approx 0.773$. The radius increases in time where the leftmost term is positive, that is for $0<r<sqrt{1/a}approx 0.656$.



    This gives an invariant annulus or trapping region. Per the given claim of no other stationary points outside the origin one concludes the existence of a limit cycle.



    The plot shows clearly the attracting limit cycle in the trapping region $0.6<r<0.8$, visually it is inside the smaller annulus $0.7lessapprox rlessapprox 0.75$.
    plot of the dynamics






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      You missed a power in combining the last terms of the equations, you should have got
      $$
      rdot r = r^2-2r^4-xy^3=r^2-(2+sin^3θcosθ)r^4
      $$
      so that because of the mean inequality of the geometric and quadratic mean $$sqrt[4]{|sqrt3cs^3|}le sqrt{frac{3c^2+s^2+s^2+s^2}4}=sqrt{frac34}$$ we get bounds for the derivative of the radius in terms of the radius only as
      $$
      r-ar^3le dot rle r-br^3 ~~text{ with }~~ a,b=2pm frac{3sqrt3}{16}.
      $$
      This means that the radius of a solution decreases in time where the right side is negative, which is for $r>sqrt{1/b}approx 0.773$. The radius increases in time where the leftmost term is positive, that is for $0<r<sqrt{1/a}approx 0.656$.



      This gives an invariant annulus or trapping region. Per the given claim of no other stationary points outside the origin one concludes the existence of a limit cycle.



      The plot shows clearly the attracting limit cycle in the trapping region $0.6<r<0.8$, visually it is inside the smaller annulus $0.7lessapprox rlessapprox 0.75$.
      plot of the dynamics






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        You missed a power in combining the last terms of the equations, you should have got
        $$
        rdot r = r^2-2r^4-xy^3=r^2-(2+sin^3θcosθ)r^4
        $$
        so that because of the mean inequality of the geometric and quadratic mean $$sqrt[4]{|sqrt3cs^3|}le sqrt{frac{3c^2+s^2+s^2+s^2}4}=sqrt{frac34}$$ we get bounds for the derivative of the radius in terms of the radius only as
        $$
        r-ar^3le dot rle r-br^3 ~~text{ with }~~ a,b=2pm frac{3sqrt3}{16}.
        $$
        This means that the radius of a solution decreases in time where the right side is negative, which is for $r>sqrt{1/b}approx 0.773$. The radius increases in time where the leftmost term is positive, that is for $0<r<sqrt{1/a}approx 0.656$.



        This gives an invariant annulus or trapping region. Per the given claim of no other stationary points outside the origin one concludes the existence of a limit cycle.



        The plot shows clearly the attracting limit cycle in the trapping region $0.6<r<0.8$, visually it is inside the smaller annulus $0.7lessapprox rlessapprox 0.75$.
        plot of the dynamics






        share|cite|improve this answer











        $endgroup$



        You missed a power in combining the last terms of the equations, you should have got
        $$
        rdot r = r^2-2r^4-xy^3=r^2-(2+sin^3θcosθ)r^4
        $$
        so that because of the mean inequality of the geometric and quadratic mean $$sqrt[4]{|sqrt3cs^3|}le sqrt{frac{3c^2+s^2+s^2+s^2}4}=sqrt{frac34}$$ we get bounds for the derivative of the radius in terms of the radius only as
        $$
        r-ar^3le dot rle r-br^3 ~~text{ with }~~ a,b=2pm frac{3sqrt3}{16}.
        $$
        This means that the radius of a solution decreases in time where the right side is negative, which is for $r>sqrt{1/b}approx 0.773$. The radius increases in time where the leftmost term is positive, that is for $0<r<sqrt{1/a}approx 0.656$.



        This gives an invariant annulus or trapping region. Per the given claim of no other stationary points outside the origin one concludes the existence of a limit cycle.



        The plot shows clearly the attracting limit cycle in the trapping region $0.6<r<0.8$, visually it is inside the smaller annulus $0.7lessapprox rlessapprox 0.75$.
        plot of the dynamics







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 16:11

























        answered Dec 31 '18 at 13:27









        LutzLLutzL

        59.9k42057




        59.9k42057






























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