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












3












$begingroup$



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



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



$it{Hint:}$ Let $V(x,y) = x^2 + y^2$



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



$it{Hint:}$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?




a.)
Using $V$, we get $frac{d}{dt}V=2xx'+2yy'$



Plugging in our system , we get:



$$frac{d}{dt}V=2x(-x^3-xy^{2k})+2y(-y^3-x^{2k}y)$$
$$=-(x^4+y^4)-x^2y^{2k}-x^{2k}y^2<0$$
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
    $$x' = -x^3-xy^{2k}$$
    $$y' = -y^3-x^{2k}y$$
    Where $k$ is a given positive integer.



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



    $it{Hint:}$ Let $V(x,y) = x^2 + y^2$



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



    $it{Hint:}$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?




    a.)
    Using $V$, we get $frac{d}{dt}V=2xx'+2yy'$



    Plugging in our system , we get:



    $$frac{d}{dt}V=2x(-x^3-xy^{2k})+2y(-y^3-x^{2k}y)$$
    $$=-(x^4+y^4)-x^2y^{2k}-x^{2k}y^2<0$$
    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
      $$x' = -x^3-xy^{2k}$$
      $$y' = -y^3-x^{2k}y$$
      Where $k$ is a given positive integer.



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



      $it{Hint:}$ Let $V(x,y) = x^2 + y^2$



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



      $it{Hint:}$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?




      a.)
      Using $V$, we get $frac{d}{dt}V=2xx'+2yy'$



      Plugging in our system , we get:



      $$frac{d}{dt}V=2x(-x^3-xy^{2k})+2y(-y^3-x^{2k}y)$$
      $$=-(x^4+y^4)-x^2y^{2k}-x^{2k}y^2<0$$
      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
      $$x' = -x^3-xy^{2k}$$
      $$y' = -y^3-x^{2k}y$$
      Where $k$ is a given positive integer.



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



      $it{Hint:}$ Let $V(x,y) = x^2 + y^2$



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



      $it{Hint:}$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?




      a.)
      Using $V$, we get $frac{d}{dt}V=2xx'+2yy'$



      Plugging in our system , we get:



      $$frac{d}{dt}V=2x(-x^3-xy^{2k})+2y(-y^3-x^{2k}y)$$
      $$=-(x^4+y^4)-x^2y^{2k}-x^{2k}y^2<0$$
      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






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      edited 1 hour ago







      hkj447

















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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 $dot{y}=0$ and $dot{y}=0$.



            $k = 1$



            The linear system is



            $$begin{align}
            begin{split}
            dot{x} &= -x^{3} - xy^{2} = -x left( x^{2} + y^{2} right) \
            dot{y} &= -y^{3} - x^{2}y = -y left( x^{2} + y^{2} right)
            end{split}
            end{align}$$



            $$ dot{r} = frac{x dot{x} + y dot{y}}{r} = -r^{3} $$



            The lone critical point is the origin.



            When $y = a x$, $ainmathbb{R}$, we have
            $$begin{align}
            begin{split}
            dot{x} &= -x^{3}left( 1 + a^{2} right) \
            dot{y} &= -a y^{3}left( 1 + a^{2} right)
            end{split}
            end{align}$$



            k=1



            $k = 2$



            $$begin{align}
            begin{split}
            dot{x} &= -x^{3} - xy^{4} = -x left( x^{2} + y^{4} right) \
            dot{y} &= -y^{3} - x^{4}y = -y left( x^{2} + y^{2} right)
            end{split}
            end{align}$$



            $$ dot{r} = tfrac{1}{8} r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



            The bounding curves for $dot{r}$ are when $cos 4theta = 1$



            $$dot{r} = -r^{3}$$



            and when $cos 4theta = -1$



            $$dot{r} = -tfrac{1}{4} r^3 left(r^2+2right)$$



            The bounding curves cross at $r=sqrt{2}$. At no point is $dot{r}$ ever positive.



            k=2k=5






            share|cite|improve this answer











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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












                $begingroup$

                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 2 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 $dot{y}=0$ and $dot{y}=0$.



                      $k = 1$



                      The linear system is



                      $$begin{align}
                      begin{split}
                      dot{x} &= -x^{3} - xy^{2} = -x left( x^{2} + y^{2} right) \
                      dot{y} &= -y^{3} - x^{2}y = -y left( x^{2} + y^{2} right)
                      end{split}
                      end{align}$$



                      $$ dot{r} = frac{x dot{x} + y dot{y}}{r} = -r^{3} $$



                      The lone critical point is the origin.



                      When $y = a x$, $ainmathbb{R}$, we have
                      $$begin{align}
                      begin{split}
                      dot{x} &= -x^{3}left( 1 + a^{2} right) \
                      dot{y} &= -a y^{3}left( 1 + a^{2} right)
                      end{split}
                      end{align}$$



                      k=1



                      $k = 2$



                      $$begin{align}
                      begin{split}
                      dot{x} &= -x^{3} - xy^{4} = -x left( x^{2} + y^{4} right) \
                      dot{y} &= -y^{3} - x^{4}y = -y left( x^{2} + y^{2} right)
                      end{split}
                      end{align}$$



                      $$ dot{r} = tfrac{1}{8} r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



                      The bounding curves for $dot{r}$ are when $cos 4theta = 1$



                      $$dot{r} = -r^{3}$$



                      and when $cos 4theta = -1$



                      $$dot{r} = -tfrac{1}{4} r^3 left(r^2+2right)$$



                      The bounding curves cross at $r=sqrt{2}$. At no point is $dot{r}$ 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 $dot{y}=0$ and $dot{y}=0$.



                        $k = 1$



                        The linear system is



                        $$begin{align}
                        begin{split}
                        dot{x} &= -x^{3} - xy^{2} = -x left( x^{2} + y^{2} right) \
                        dot{y} &= -y^{3} - x^{2}y = -y left( x^{2} + y^{2} right)
                        end{split}
                        end{align}$$



                        $$ dot{r} = frac{x dot{x} + y dot{y}}{r} = -r^{3} $$



                        The lone critical point is the origin.



                        When $y = a x$, $ainmathbb{R}$, we have
                        $$begin{align}
                        begin{split}
                        dot{x} &= -x^{3}left( 1 + a^{2} right) \
                        dot{y} &= -a y^{3}left( 1 + a^{2} right)
                        end{split}
                        end{align}$$



                        k=1



                        $k = 2$



                        $$begin{align}
                        begin{split}
                        dot{x} &= -x^{3} - xy^{4} = -x left( x^{2} + y^{4} right) \
                        dot{y} &= -y^{3} - x^{4}y = -y left( x^{2} + y^{2} right)
                        end{split}
                        end{align}$$



                        $$ dot{r} = tfrac{1}{8} r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



                        The bounding curves for $dot{r}$ are when $cos 4theta = 1$



                        $$dot{r} = -r^{3}$$



                        and when $cos 4theta = -1$



                        $$dot{r} = -tfrac{1}{4} r^3 left(r^2+2right)$$



                        The bounding curves cross at $r=sqrt{2}$. At no point is $dot{r}$ 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 $dot{y}=0$ and $dot{y}=0$.



                          $k = 1$



                          The linear system is



                          $$begin{align}
                          begin{split}
                          dot{x} &= -x^{3} - xy^{2} = -x left( x^{2} + y^{2} right) \
                          dot{y} &= -y^{3} - x^{2}y = -y left( x^{2} + y^{2} right)
                          end{split}
                          end{align}$$



                          $$ dot{r} = frac{x dot{x} + y dot{y}}{r} = -r^{3} $$



                          The lone critical point is the origin.



                          When $y = a x$, $ainmathbb{R}$, we have
                          $$begin{align}
                          begin{split}
                          dot{x} &= -x^{3}left( 1 + a^{2} right) \
                          dot{y} &= -a y^{3}left( 1 + a^{2} right)
                          end{split}
                          end{align}$$



                          k=1



                          $k = 2$



                          $$begin{align}
                          begin{split}
                          dot{x} &= -x^{3} - xy^{4} = -x left( x^{2} + y^{4} right) \
                          dot{y} &= -y^{3} - x^{4}y = -y left( x^{2} + y^{2} right)
                          end{split}
                          end{align}$$



                          $$ dot{r} = tfrac{1}{8} r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



                          The bounding curves for $dot{r}$ are when $cos 4theta = 1$



                          $$dot{r} = -r^{3}$$



                          and when $cos 4theta = -1$



                          $$dot{r} = -tfrac{1}{4} r^3 left(r^2+2right)$$



                          The bounding curves cross at $r=sqrt{2}$. At no point is $dot{r}$ 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 $dot{y}=0$ and $dot{y}=0$.



                          $k = 1$



                          The linear system is



                          $$begin{align}
                          begin{split}
                          dot{x} &= -x^{3} - xy^{2} = -x left( x^{2} + y^{2} right) \
                          dot{y} &= -y^{3} - x^{2}y = -y left( x^{2} + y^{2} right)
                          end{split}
                          end{align}$$



                          $$ dot{r} = frac{x dot{x} + y dot{y}}{r} = -r^{3} $$



                          The lone critical point is the origin.



                          When $y = a x$, $ainmathbb{R}$, we have
                          $$begin{align}
                          begin{split}
                          dot{x} &= -x^{3}left( 1 + a^{2} right) \
                          dot{y} &= -a y^{3}left( 1 + a^{2} right)
                          end{split}
                          end{align}$$



                          k=1



                          $k = 2$



                          $$begin{align}
                          begin{split}
                          dot{x} &= -x^{3} - xy^{4} = -x left( x^{2} + y^{4} right) \
                          dot{y} &= -y^{3} - x^{4}y = -y left( x^{2} + y^{2} right)
                          end{split}
                          end{align}$$



                          $$ dot{r} = tfrac{1}{8} r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$



                          The bounding curves for $dot{r}$ are when $cos 4theta = 1$



                          $$dot{r} = -r^{3}$$



                          and when $cos 4theta = -1$



                          $$dot{r} = -tfrac{1}{4} r^3 left(r^2+2right)$$



                          The bounding curves cross at $r=sqrt{2}$. At no point is $dot{r}$ ever positive.



                          k=2k=5







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 1 hour ago

























                          answered 2 hours ago









                          dantopadantopa

                          6,76442345




                          6,76442345






























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