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Using a Lyapunov function to classify stability and sketching a phase portrait

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Using a Lyapunov function to classify stability and sketching a phase portrait


Lyapunov stability question from Arnold's triviumNon linear phase portraitNonlinear phase portrait and linearizationSystem of differential equations, phase portraitDynamical Systems- Plotting Phase PortraitPhase portrait of ODE in polar coordinatesQuestions about stability in the sense of LyapunovLinearization method or Lyapunov function - examplestability using linearization instead of Lyapunov failsLyapunov function instead of linearization













3












$begingroup$



Consider the system
$$x' = -x^3-xy^2k$$
$$y' = -y^3-x^2ky$$
Where $k$ is a given positive integer.



a.) Find and classify according to stability the equilibrium solutions.



$itHint:$ Let $V(x,y) = x^2 + y^2$



b.) Sketch a phase portrait when $k = 1$



$itHint:$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?




a.)
Using $V$, we get $fracddtV=2xx'+2yy'$



Plugging in our system , we get:



$$fracddtV=2x(-x^3-xy^2k)+2y(-y^3-x^2ky)$$
$$=-(x^4+y^4)-x^2y^2k-x^2ky^2<0$$
I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
$$y^2k=-x^2$$
Which only works for $x=y=0$



Therefore our system is asymptotically stable at the origin.



I am having trouble with b.), mostly because the hint is confusing me.



Let $y=ax$, then our system becomes
$$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
$$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.










share|cite|improve this question











$endgroup$
















    3












    $begingroup$



    Consider the system
    $$x' = -x^3-xy^2k$$
    $$y' = -y^3-x^2ky$$
    Where $k$ is a given positive integer.



    a.) Find and classify according to stability the equilibrium solutions.



    $itHint:$ Let $V(x,y) = x^2 + y^2$



    b.) Sketch a phase portrait when $k = 1$



    $itHint:$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?




    a.)
    Using $V$, we get $fracddtV=2xx'+2yy'$



    Plugging in our system , we get:



    $$fracddtV=2x(-x^3-xy^2k)+2y(-y^3-x^2ky)$$
    $$=-(x^4+y^4)-x^2y^2k-x^2ky^2<0$$
    I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
    $$y^2k=-x^2$$
    Which only works for $x=y=0$



    Therefore our system is asymptotically stable at the origin.



    I am having trouble with b.), mostly because the hint is confusing me.



    Let $y=ax$, then our system becomes
    $$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
    $$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
    I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.










    share|cite|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$



      Consider the system
      $$x' = -x^3-xy^2k$$
      $$y' = -y^3-x^2ky$$
      Where $k$ is a given positive integer.



      a.) Find and classify according to stability the equilibrium solutions.



      $itHint:$ Let $V(x,y) = x^2 + y^2$



      b.) Sketch a phase portrait when $k = 1$



      $itHint:$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?




      a.)
      Using $V$, we get $fracddtV=2xx'+2yy'$



      Plugging in our system , we get:



      $$fracddtV=2x(-x^3-xy^2k)+2y(-y^3-x^2ky)$$
      $$=-(x^4+y^4)-x^2y^2k-x^2ky^2<0$$
      I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
      $$y^2k=-x^2$$
      Which only works for $x=y=0$



      Therefore our system is asymptotically stable at the origin.



      I am having trouble with b.), mostly because the hint is confusing me.



      Let $y=ax$, then our system becomes
      $$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
      $$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
      I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.










      share|cite|improve this question











      $endgroup$





      Consider the system
      $$x' = -x^3-xy^2k$$
      $$y' = -y^3-x^2ky$$
      Where $k$ is a given positive integer.



      a.) Find and classify according to stability the equilibrium solutions.



      $itHint:$ Let $V(x,y) = x^2 + y^2$



      b.) Sketch a phase portrait when $k = 1$



      $itHint:$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?




      a.)
      Using $V$, we get $fracddtV=2xx'+2yy'$



      Plugging in our system , we get:



      $$fracddtV=2x(-x^3-xy^2k)+2y(-y^3-x^2ky)$$
      $$=-(x^4+y^4)-x^2y^2k-x^2ky^2<0$$
      I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
      $$y^2k=-x^2$$
      Which only works for $x=y=0$



      Therefore our system is asymptotically stable at the origin.



      I am having trouble with b.), mostly because the hint is confusing me.



      Let $y=ax$, then our system becomes
      $$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
      $$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
      I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.







      ordinary-differential-equations stability-in-odes lyapunov-functions






      share|cite|improve this question















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      edited 3 hours ago







      hkj447

















      asked 4 hours ago









      hkj447hkj447

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          $begingroup$

          Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.



          So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.






          share|cite|improve this answer









          $endgroup$




















            2












            $begingroup$

            Phase portraits - a partial offering



            Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $doty=0$ and $doty=0$.



            $k = 1$



            The linear system is



            $$beginalign
            beginsplit
            dotx &= -x^3 - xy^2 = -x left( x^2 + y^2 right) \
            doty &= -y^3 - x^2y = -y left( x^2 + y^2 right)
            endsplit
            endalign$$



            $$ dotr = fracx dotx + y dotyr = -r^3 $$



            The lone critical point is the origin.



            When $y = a x$, $ainmathbbR$, we have
            $$beginalign
            beginsplit
            dotx &= -x^3left( 1 + a^2 right) \
            doty &= -a y^3left( 1 + a^2 right)
            endsplit
            endalign$$



            k=1



            $k = 2$



            $$beginalign
            beginsplit
            dotx &= -x^3 - xy^4 = -x left( x^2 + y^4 right) \
            doty &= -y^3 - x^4y = -y left( x^2 + y^2 right)
            endsplit
            endalign$$



            $$ dotr = tfrac18 r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



            The bounding curves for $dotr$ are when $cos 4theta = 1$



            $$dotr = -r^3$$



            and when $cos 4theta = -1$



            $$dotr = -tfrac14 r^3 left(r^2+2right)$$



            The bounding curves cross at $r=sqrt2$. At no point is $dotr$ ever positive.



            k=2k=5






            share|cite|improve this answer











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              2 Answers
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              2












              $begingroup$

              Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.



              So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.



                So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.



                  So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.






                  share|cite|improve this answer









                  $endgroup$



                  Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.



                  So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  aghostinthefiguresaghostinthefigures

                  1,4391318




                  1,4391318





















                      2












                      $begingroup$

                      Phase portraits - a partial offering



                      Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $doty=0$ and $doty=0$.



                      $k = 1$



                      The linear system is



                      $$beginalign
                      beginsplit
                      dotx &= -x^3 - xy^2 = -x left( x^2 + y^2 right) \
                      doty &= -y^3 - x^2y = -y left( x^2 + y^2 right)
                      endsplit
                      endalign$$



                      $$ dotr = fracx dotx + y dotyr = -r^3 $$



                      The lone critical point is the origin.



                      When $y = a x$, $ainmathbbR$, we have
                      $$beginalign
                      beginsplit
                      dotx &= -x^3left( 1 + a^2 right) \
                      doty &= -a y^3left( 1 + a^2 right)
                      endsplit
                      endalign$$



                      k=1



                      $k = 2$



                      $$beginalign
                      beginsplit
                      dotx &= -x^3 - xy^4 = -x left( x^2 + y^4 right) \
                      doty &= -y^3 - x^4y = -y left( x^2 + y^2 right)
                      endsplit
                      endalign$$



                      $$ dotr = tfrac18 r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



                      The bounding curves for $dotr$ are when $cos 4theta = 1$



                      $$dotr = -r^3$$



                      and when $cos 4theta = -1$



                      $$dotr = -tfrac14 r^3 left(r^2+2right)$$



                      The bounding curves cross at $r=sqrt2$. At no point is $dotr$ ever positive.



                      k=2k=5






                      share|cite|improve this answer











                      $endgroup$

















                        2












                        $begingroup$

                        Phase portraits - a partial offering



                        Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $doty=0$ and $doty=0$.



                        $k = 1$



                        The linear system is



                        $$beginalign
                        beginsplit
                        dotx &= -x^3 - xy^2 = -x left( x^2 + y^2 right) \
                        doty &= -y^3 - x^2y = -y left( x^2 + y^2 right)
                        endsplit
                        endalign$$



                        $$ dotr = fracx dotx + y dotyr = -r^3 $$



                        The lone critical point is the origin.



                        When $y = a x$, $ainmathbbR$, we have
                        $$beginalign
                        beginsplit
                        dotx &= -x^3left( 1 + a^2 right) \
                        doty &= -a y^3left( 1 + a^2 right)
                        endsplit
                        endalign$$



                        k=1



                        $k = 2$



                        $$beginalign
                        beginsplit
                        dotx &= -x^3 - xy^4 = -x left( x^2 + y^4 right) \
                        doty &= -y^3 - x^4y = -y left( x^2 + y^2 right)
                        endsplit
                        endalign$$



                        $$ dotr = tfrac18 r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



                        The bounding curves for $dotr$ are when $cos 4theta = 1$



                        $$dotr = -r^3$$



                        and when $cos 4theta = -1$



                        $$dotr = -tfrac14 r^3 left(r^2+2right)$$



                        The bounding curves cross at $r=sqrt2$. At no point is $dotr$ ever positive.



                        k=2k=5






                        share|cite|improve this answer











                        $endgroup$















                          2












                          2








                          2





                          $begingroup$

                          Phase portraits - a partial offering



                          Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $doty=0$ and $doty=0$.



                          $k = 1$



                          The linear system is



                          $$beginalign
                          beginsplit
                          dotx &= -x^3 - xy^2 = -x left( x^2 + y^2 right) \
                          doty &= -y^3 - x^2y = -y left( x^2 + y^2 right)
                          endsplit
                          endalign$$



                          $$ dotr = fracx dotx + y dotyr = -r^3 $$



                          The lone critical point is the origin.



                          When $y = a x$, $ainmathbbR$, we have
                          $$beginalign
                          beginsplit
                          dotx &= -x^3left( 1 + a^2 right) \
                          doty &= -a y^3left( 1 + a^2 right)
                          endsplit
                          endalign$$



                          k=1



                          $k = 2$



                          $$beginalign
                          beginsplit
                          dotx &= -x^3 - xy^4 = -x left( x^2 + y^4 right) \
                          doty &= -y^3 - x^4y = -y left( x^2 + y^2 right)
                          endsplit
                          endalign$$



                          $$ dotr = tfrac18 r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



                          The bounding curves for $dotr$ are when $cos 4theta = 1$



                          $$dotr = -r^3$$



                          and when $cos 4theta = -1$



                          $$dotr = -tfrac14 r^3 left(r^2+2right)$$



                          The bounding curves cross at $r=sqrt2$. At no point is $dotr$ ever positive.



                          k=2k=5






                          share|cite|improve this answer











                          $endgroup$



                          Phase portraits - a partial offering



                          Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $doty=0$ and $doty=0$.



                          $k = 1$



                          The linear system is



                          $$beginalign
                          beginsplit
                          dotx &= -x^3 - xy^2 = -x left( x^2 + y^2 right) \
                          doty &= -y^3 - x^2y = -y left( x^2 + y^2 right)
                          endsplit
                          endalign$$



                          $$ dotr = fracx dotx + y dotyr = -r^3 $$



                          The lone critical point is the origin.



                          When $y = a x$, $ainmathbbR$, we have
                          $$beginalign
                          beginsplit
                          dotx &= -x^3left( 1 + a^2 right) \
                          doty &= -a y^3left( 1 + a^2 right)
                          endsplit
                          endalign$$



                          k=1



                          $k = 2$



                          $$beginalign
                          beginsplit
                          dotx &= -x^3 - xy^4 = -x left( x^2 + y^4 right) \
                          doty &= -y^3 - x^4y = -y left( x^2 + y^2 right)
                          endsplit
                          endalign$$



                          $$ dotr = tfrac18 r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



                          The bounding curves for $dotr$ are when $cos 4theta = 1$



                          $$dotr = -r^3$$



                          and when $cos 4theta = -1$



                          $$dotr = -tfrac14 r^3 left(r^2+2right)$$



                          The bounding curves cross at $r=sqrt2$. At no point is $dotr$ ever positive.



                          k=2k=5







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 2 hours ago

























                          answered 3 hours ago









                          dantopadantopa

                          6,76442345




                          6,76442345



























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