Consider the following two-dimensional dynamical system,

$$\dot{x} = y -y^3$$ $$\dot{y}= -x-y^2$$

where as usual the dot represents a time derivative. (This system is found on page 165 of Strogatz book 'Nonlinear Dynamics and Chaos'.)

I am trying to find a conserved quantity for this system, apparently it does exist and there is a simple way to find it. Strogatz shows that the system is reversible with a nonlinear center at the origin. The phase portrait looks like:

enter image description here

Does this system have a conserved quantity? Something like $x^2+y^2 = \text{constant}$, for example (this does not work though...). If so how would I go about finding it?

I have tried playing around with the system but have not been able to find one, however the phase portrait leads me to believe that something should be conserved.

  • $\begingroup$ Just a follow-up comment and sort of an advice. In principle reversibility gives you here everything you need for a qualitative analysis of system - no need for seeking first integral really. Note that not all reversible systems are conservative - any system that has symmetric source or sink is an example. $\endgroup$
    – Evgeny
    Jul 17, 2017 at 12:18
  • $\begingroup$ I dont think reverisable implies conservative. link.springer.com/chapter/10.1007/978-3-642-84570-3_27. the author seems to state this in the 3rd paragraph, assuming his conservative definition is the same, which he doesnt state i dont think. $\endgroup$
    – H_1317
    Nov 1, 2017 at 4:11
  • $\begingroup$ my bad, totally misread Evgeny's comment... Why do you think there is a simple way to find it? where does Strogatz say that? $\endgroup$
    – H_1317
    Nov 1, 2017 at 4:13

2 Answers 2


Follow up answer to @lhf's post

The equation

$$(y-y^3) \, \partial_x H - (x+y^2) \, \partial_y H = 0, $$ as proposed by @lhf, is a linear first-order PDE for $H$. The method of characteristics then states that

$$ \frac{\mathrm{d}x}{y-y^3} = \frac{\mathrm{d}y}{-x-y^2} = \frac{\mathrm{d}H}{0}, $$

where the last fraction indicates that $H = c_1$ is a constant along the characteristic curve defined by the first equals sign:

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{x+y^2}{y(1-y^2)}$$

which defines $y$ as a function of $x$. The solution to this (homogeneous?) equation is given by Mathematica as:

$$ \small{ \left(\sqrt{3}+3\right) \log \left[\left(\sqrt{3}-1\right) y^2-2 x-\sqrt{3}-1\right] -\left(\sqrt{3}-3\right) \log \left[\left(\sqrt{3}+1\right) y^2+2 x-\sqrt{3}+1\right]=c_2 } $$ where I have absorbed a factor into the constant of integration. Put now $c_1$ as a function of $c_2$ to have $H = f(c_2)$, where $ f $ is an arbitrary function of its argument. Take for instance $f(\square) = \square$ as the identity function and you will thus find:

$$ \small{ H = \left(\sqrt{3}+3\right) \log \left[\left(\sqrt{3}-1\right) y^2-2 x-\sqrt{3}-1\right] -\left(\sqrt{3}-3\right) \log \left[\left(\sqrt{3}+1\right) y^2+2 x-\sqrt{3}+1\right] }$$

which, if I didn't make any mistake on the transcription, satisfies the original PDE.

Hope you find this useful!

Check with Mathematica: enter image description here

  • $\begingroup$ I just checked on wolfram and indeed it seems to satisfy the original equation! Do you mind double checking in mathematica? $\endgroup$
    – fosho
    Jul 14, 2017 at 19:05
  • $\begingroup$ Please see my edit @Dman $\endgroup$
    – Dmoreno
    Jul 14, 2017 at 19:10
  • $\begingroup$ great! Thanks a lot. $\endgroup$
    – fosho
    Jul 14, 2017 at 19:10
  • $\begingroup$ @Dmoreno how exactly did you use mathematica to derive the parametric equation? I can't figure it out. $\endgroup$
    – Alex
    Jun 27, 2018 at 23:29
  • $\begingroup$ Hi @Alex, what do you mean by 'the parametric equation'? $\endgroup$
    – Dmoreno
    Jun 29, 2018 at 8:25

A conserved quantity $H(x,y)$ must be such that $H(x,y)$ is constant on the integral curves. This happens iff $H_x \dot x + H_y \dot y =0$. So you want to solve $$ \frac{\partial H}{\partial x}( y -y^3) + \frac{\partial H}{\partial y}( -x-y^2) = 0 $$ This suggests that you may try a polynomial for $H$, perhaps with an integrating factor.

  • $\begingroup$ Thanks for the answer. How does one go about solving this PDE? $\endgroup$
    – fosho
    Jul 14, 2017 at 13:03
  • $\begingroup$ @Dman, I don't know the solution. I tried the obvious $\dot x= -H_y, \dot y = H_x$, but it didn't work out. $\endgroup$
    – lhf
    Jul 14, 2017 at 13:13
  • $\begingroup$ @Dman, see math.stackexchange.com/questions/252788/…. $\endgroup$
    – lhf
    Jul 14, 2017 at 13:27
  • $\begingroup$ Considering that close to the origin the system looks like a harmonic oscillator, thought it was worth trying $H = x^2 + y^2 + \Phi(x,y)$, where $\Phi $ is a polynomial going quickly to zero towards the origin, but no luck $\endgroup$ Jul 14, 2017 at 16:36
  • 1
    $\begingroup$ Note that $$\small{H = \left(\sqrt{3}+3\right) \log \left(-2 x+\left(\sqrt{3}-1\right) y^2-\sqrt{3}-1\right)-\left(\sqrt{3}-3\right) \log \left(2 x+\left(\sqrt{3}+1\right) y^2-\sqrt{3}+1\right)}$$ is a solution of the PDE proposed by @lhf (I used the method of characteristics and Mathematica) $\endgroup$
    – Dmoreno
    Jul 14, 2017 at 18:25

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