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I am given a system

$\dot x = f(x,y) = (x^2 + y^2)(x^3 + y^2x -2y - x) \\ \dot y = g(x,y) = (x^2 + y^2)(y^3 + x^2y +2x - y) $

and I am asked if the fixed point at $(0,0)$ is hyperbolic or asymptotically stable. I am also asked whether the system is a Hamiltonian.

So I know that a hyperbolic fixed point is a fixed point that does not have a centre manifold, so this makes me think that it is a hyperbolic fixed point but I'm not really sure.

Then onto asymptotically stable, my understanding of this is that we require the system to be both Liapounov stable and quasi-asymptotically stable, but I really dont know how to go about this.

Finally about it being a Hamiltonian system. I know in order to be a Hamiltonian the system can be written in the form

$H(q_i,p_i)$ such that $\dot q_i= \frac{\partial H}{\partial p_i} \\ \dot p_i= -\frac{\partial H}{\partial q_i}$

But I really dont know how to do this for this DS.

Also i it helps I converted the DS into polar coords, giving me

$\dot r = r^5 - r^3 \\ \dot \theta = 2r^2$

Gragbow
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  • Isn't $r'$ just $r' = r^5$ based on what you wrote? – Moo Jan 24 '18 at 21:50
  • @Moo Oh Yeah, let me just edit that – Gragbow Jan 24 '18 at 22:01
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    Regarding the first question: The definition of hyperbolic fixed point is a fixed point where the Jacobian matrix has no eigenvalues with real part zero, so you just compute the matrix and check. – Hans Lundmark Jan 25 '18 at 07:13
  • @HansLundmark How do I compute the Jacobian matrix though? Edit: I know how to do that actually. Anything on the asympotically stable part? – Gragbow Jan 25 '18 at 11:22
  • @HansLundmark When I compute the Jacobian matrix and at the point (0,0) the Jacobian matrix is just 0 so surely this doesnt help? – Gragbow Jan 25 '18 at 11:45
  • Yes, the Jacobian at the origin is identically zero (not surprisingly, since you have only terms of degree three and higher in your system), and this shows that the origin is not a hyperbolic fixed point. (So you can't use the Jacobian to investigate stability, you have to argue in some other way. Hint: the polar form of the system is very helpful!) – Hans Lundmark Jan 25 '18 at 13:20
  • @HansLundmark Oh I see. My thinking is that I know that the point is not stable as it is not a sink and that both eigenvalues are not less than 0, but I am unsure if this suffices. But asymptotically stable, is it something to do because the polar coords dont depend on $\theta$? – Gragbow Jan 25 '18 at 14:51
  • No, you can't draw any conclusion at all from the eigenvalues, so you can't say yet that the point is unstable. (As I said above, the Jacobian can only be used to investigate stability of hyperbolic fixed point.) But what does $\dot r=r^5$ tell you if $r>0$? – Hans Lundmark Jan 25 '18 at 15:22
  • @HansLundmark Ah ok, my bad.So yeah if $r > 0$ then we can say that $\dot r >0$ and the similarly for $r < 0$. So surely that leads me to being able to say that it is unstable and its a source – Gragbow Jan 25 '18 at 16:09
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    If system is Hamiltonian, it has first integral $H(x, y)$. First integrals exclude the presence of asymptotically stable or completely unstable equilibria. I think you have all other puzzle pieces at hands to draw the conclusion :) – Evgeny Jan 25 '18 at 16:12
  • @Evgeny I knew a Hamiltonian has to be a first integral, but I am still not sure about this stable/asymptotically stable point thing so I'm struggling – Gragbow Jan 25 '18 at 16:16
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    Wait, you've proved that the origin is a source. Maybe non-hyperbolic, but still a source. That tells you everything about a stability of the equilibrium! – Evgeny Jan 25 '18 at 16:26
  • @Evgeny This is the streamplot i end up with, so it looks like a source http://m.wolframalpha.com/input/?i=streamplot%5B%7B%28x%5E2+%2B+y%5E2+%29%28x%5E3+%2B+y%5E2x+%E2%88%92+2y+%E2%88%92+x%29%2C+%28x%5E2+%2B+y%5E2+%29%28y%5E3+%2B+x%5E2+y+%2B+2x+%E2%88%92+y%29%7D%2C%7Bx%2C-6%2C6%7D%2C%7By%2C-6%2C6%7D%5D – Gragbow Jan 25 '18 at 16:40
  • That is not the correct phase portrait because there is a sign issue in the equations from what you wrote above. – Moo Jan 25 '18 at 16:45
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    @Moo Oh god. Thank you for pointing this out, this changes a lot – Gragbow Jan 25 '18 at 16:48
  • @Evgeny It seems like I dont know (0,0) is a source, but more a saddle point. This makes me think that it could be stable? But I dont know how to say it's stable? – Gragbow Jan 25 '18 at 17:54
  • If $r' = r^5$ is correct then no way it's a saddle. – Evgeny Jan 25 '18 at 18:32
  • @Evgeny Im having a nightmare here, basically I'd typed the DS wrong at the start, I've just edited this. – Gragbow Jan 25 '18 at 19:29
  • Are your calculations for $r'$ and the like still correct? – Moo Jan 25 '18 at 19:30
  • @Moo yeah everything now is correct, and that phase flow I posted is correct. Im still struggling on the asymptotically stable and if it is a Hamiltonian though – Gragbow Jan 25 '18 at 19:57
  • @Moo No, they aren't. It should be $r' = 2 r^2(r-1)$ if $r = x^2+y^2$. And that now agrees with a phase plot. The origin is a sink at the phase plot. – Evgeny Jan 25 '18 at 20:23
  • If you want an explanation why Hamiltonian or any conservative system can't have Lyapunov asymptotically stable or completely unstable equilibria check this and look at the part about first integral. – Evgeny Jan 25 '18 at 20:30
  • @Evgeny How did you get that. I've just redone it and ended up with $\dot r = r^4 - r^2$ – Gragbow Jan 25 '18 at 20:30
  • I don't like square roots so I take $r = x^2 + y^2$. I've wrote that explicitly. In that case $\dot{(x^2 + y^2)} = 2x \dot{x} + 2y \dot{y} = 2(x^2+y^2) (x^4 + 2 x^2 y^2 + y^4 - 2xy + 2xy - x^2 -y^2) = 2r (r^2 -r)$. – Evgeny Jan 25 '18 at 20:31
  • @Evgeny so i get $\dot r = r^5 - r^3$ by taking $r^2 = x^2 + y^2$ – Gragbow Jan 25 '18 at 20:34
  • @Gragbow: The case $r<0$ is irrelevant, since $r$ is the distance to the origin, hence nonnegative. – Hans Lundmark Jan 25 '18 at 21:33
  • Try V=y^2+x^2 as a Lyapunov function candidate – Carlos Jan 25 '18 at 21:38
  • @HansLundmark But I dont see how this helps me sorry, I'm struggling with this quite a bit. – Gragbow Jan 25 '18 at 21:49
  • @Carlos Ah I'll try that now, thank you, not really sure how though – Gragbow Jan 25 '18 at 21:51
  • It was just a remark on your comment “and similarly for $r<0$” above. Your argument for $r>0$ is correct, and that's all that's relevant here. – Hans Lundmark Jan 26 '18 at 05:07
  • Oops, I just noticed that you've edited the equations. Have a look at the discussion here then, and see if that helps: https://math.stackexchange.com/questions/2606916/existence-of-unique-limit-cycle-for-r-r%ce%bc-r2-space-%ce%b8-%cf%81r2 – Hans Lundmark Jan 26 '18 at 05:10
  • @HansLundmark Yeah sorry, I should have made it clearer. Basically I had mistyped, so needed to correct it. – Gragbow Jan 26 '18 at 13:21

1 Answers1

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This is about the stability part:

Asymptotic stability of the equilibrium means that it is stable and attractive. The origin $(0,0)$ is an equilibrium. Consider the Lyapunov function candidate $V(x,y) = \frac{1}{2}x^2 + \frac{1}{2}y^2$. The derivative along the trajectories of $\dot x = f(x,y)$ and $\dot y = g(x,y)$ yields

\begin{align} \dot V = x\,\dot x + y\,\dot y &= x(x^2+y^2)(x^3+y^2x-2y-x) + y(x^2+y^2)(y^3+x^2y+2x-y) \\ & = (x^2+y^2)(x^4+y^2x^2-2yx-x^2) + (x^2+y^2)(y^4+x^2y^2+2xy-y^2) \\ & = (x^2+y^2)(x^4+y^4) + (x^2+y^2)(2y^2x^2) - (x^2+y^2)^2 \\ & = (x^2+y^2)\left((x^4+y^4) +(2y^2x^2) \right) - (x^2+y^2)^2 \\ & = (x^2+y^2)(x^2+y^2)^2 - (x^2+y^2)^2 \end{align} Insert $2V = x^2+y^2$: \begin{align} \dot V & = 8\,V^3 - 4 V^2 = -8\,V^2\left(\frac{1}{2}-V\right) \end{align} (Note: this is qualitatively similar to that what could be derived from the representation in polar coordinates)

Accordingly, $\dot V$ is only negative in the set $\{(x,y)\in\mathbb R^2: x^2+y^2<1 \}$ which contains the origin. That is, the origin is (locally) asymptotically stable. The unit disc defines also the region of attraction.

Carlos
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