Consider the nonlinear system \begin{align*} \frac{d}{dt} \begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix} = \begin{pmatrix} x_2(t) \\ -4x_1(t) + x_1^2(t) \end{pmatrix}, \end{align*} which admits periodic orbits near the origin. Consider two initial states $x(0) = x_0$ and $z(0)=z_0$ where $x_0$ and $z_0$ are nonzero and satisfy $x_2(0) = - z_2(0)$. We can think of the situation as putting two particles in the vector field and watch them evolve.

I am trying to show that for any combination of initial states satisfying the above requirements, at some point in time it must happend that $x_2(t) + z_2(t) \ne 0$. This is due to the specific nonlinearity of the periodic orbits. But it seems to be very difficult to show by just looking at the equations \begin{align*} x_2(t) + z_2(t) \equiv 0, \frac{d}{dt} (x_2(t) + z_2(t)) \equiv 0, \dots \end{align*} So I was hoping that there might be a more succinct argument involving the geometry of the periodic orbits. Any hints in the direction of a simpler argument are very welcome.

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    $\begingroup$ One idea: work in extended phase space of two copies of the original system (i.e. state space being $x_1,x_2,z_1,z_2$. Then do a coordinate transformation $y_1=x_2+z_2, y_2=x_2-z_2$ and look at fixed points of this 4D system. $\endgroup$ Oct 24, 2015 at 3:43
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    $\begingroup$ @ seno44: your system describes an anharmonic oscillator (x,y = dx/dt) with the potential $U = 2 x^2 - \frac{1}{3} x^3$ which can be treated with standard methods. Energy conservation is an elliptic curve. The separatrix between finite and unbounded orbits is given by $y^2=\frac{2 x^3}{3}-4 x^2+\frac{64}{3}$ $\endgroup$ Oct 24, 2015 at 21:32
  • $\begingroup$ Thanks for your efforts. I know that the potential is given by $U(x) = \frac{1}{2}x_2^2 + 2 x_1^2 - \frac{1}{3} x_1^3$. The separatrix I am actually not interested in, only the periodic orbits close to the origin. $\endgroup$
    – user45183
    Oct 24, 2015 at 22:38
  • $\begingroup$ @seno44 I would like to add a related question: is there a periodic orbit $\gamma$ with period $T$ such that $\gamma$ start from $(0,y))\;\;y>0$ and arrive at $(0,-y)$ at time $T/2$?If there is a periodic orbit with this property, it would be a counter example to your question. $\endgroup$ Oct 26, 2015 at 11:34
  • $\begingroup$ @AliTaghavi Thanks for your answer. I am not sure what you mean exactly with counter example. I am not trying to find one example which satisfies the property but I am trying to show that all solutions starting with $x_2(t) + z_2(t) = 0$ (starting near the origin) will have $x_2(t) + z_2(t) \ne 0$ at some point in time. $\endgroup$
    – user45183
    Oct 26, 2015 at 12:10

2 Answers 2


As we're interested in two particles, it seems natural to introduce the coordinates $y = x + z$ and $w = x - z$. In these coordinates, we obtain the system \begin{align} \dot{y}_1 &= y_2,\\ \dot{y}_2 &= -4 y_1+\frac{1}{2}y_1^2+\frac{1}{2}w_1^2,\\ \dot{w}_1 &= w_2,\\ \dot{w}_2 &= -4 w_1 + w_1 y_1. \end{align} If I understand you correctly, the question is if, given an initial condition of the form $(y_1,y_2,w_1,w_2) = (a,0,c,d)$, there will be a time $t$ when $y_2(t) \neq 0$. The negation of this statement is equivalent with stating that $y_2(t) = 0$ for all time $t$, i.e. that the set $(a,0,c,d)$ is invariant under the above flow. However, for such an initial condition, we see that $\dot{y}_2(0) = -4a+\frac{1}{2}a^2 + \frac{1}{2}c^2 \neq 0$ for general $a,c$. Hence, the vector field at $(a,0,c,d)$ is not parallel to $(a,0,c,d)$, so the set $(a,0,c,d)$ is not invariant under the flow. Hence, there exists a time $t>0$ such that $y_2(t) \neq 0$, even when $y_2(0) = 0$.

  • $\begingroup$ Thanks for your detailed answer. I have one question: You say that $\dot{y}_2(0) = \dots \ne$ for general $a,c$, but looking closer at it, the set for which $\dot{y}_2(0) = 0$ has infinitely many solutions, namely the set of $(a,c)$ defines a circle centered at 4 with radius 4. But maybe $x_1 + z_1 = a$ and $x_1 - z_1 = c$ does not have a solution for such $(a,c)$? $\endgroup$
    – user45183
    Oct 26, 2015 at 11:40
  • $\begingroup$ Well, not really a circle, but yes: there are choices for $a,c$ such that $\dot{y}_2(0)$ is indeed zero: choose $c = \pm \sqrt{a(8-a)}$, i.e. $x_1(0) = 2 \pm \sqrt{8-(z_1(0)-2)^2}$. $\endgroup$ Oct 26, 2015 at 12:06
  • $\begingroup$ But $-8a + a^2 + c^2 = 0$ gives you a circle. Complete the square in $a$ to get $(a-4)^2 + c^2 = 4^2$. $\endgroup$
    – user45183
    Oct 26, 2015 at 12:11
  • $\begingroup$ Furthermore, for the situation $y_2(t) \neq 0$ not to occur, we want $\dot{y}_2 = 0$ for all time $t$, i.e. that for all time, we have $w_1(t)^2 = y_1(t)(8-y_1(t))$. Taking the time derivative of this identity and using the dynamical system, you obtain a new nontrivial relation between $y_{1,2}(t),w_{1,2}(t)$. Repetition of this process then would yield a number of relations between the dynamical variables which can only be satisfied when $y_{1,2}(t) \equiv 0$ and $w_{1,2} \equiv 0$. $\endgroup$ Oct 26, 2015 at 12:15
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    $\begingroup$ Right, that's essentially the same. Transforming the system might streamline the argument a bit though, but that aspect depends more on personal taste. $\endgroup$ Oct 26, 2015 at 19:21

See EDIT #1 below for a generalization to $N > 2$ particles.

Let us restate the problem for ease of reference. The equations are

$\dot{y}(t)= - 4 x(t) + y(t)^2$

Physically they describe an anharmonic oscillator with spatial coordinate $x(t)$ and velocity $y(t)$ in the potential

$U = 2 x^2 - \frac{1}{3} x^3$

and the energy

$E = \frac{1}{2} y^2 + U(x)$

The problem is then: given two solutions

$s_{1}(t) = (x_{1}(t), y_{1}(t))$
$s_{2}(t) = (x_{2}(t), y_{2}(t))$

corresponding to the initial conditions

$s_{1}(0) = (x_{10}, y_{10})$
$s_{2}(0) = (x_{20}, y_{20})$

Assuming that $y_{10}$ and $y_{20}$ do not vanish show that the quantity

$d(t) = y_{1}(t) + y_{2}(t)$

becomes $\neq 0$ for times $t>0$ even if it is $= 0$ for $t = 0$, i.e.

$y_{10} + y_{20} = 0$.

We prove it indirectly, assuming $d(t) = 0$ for all times.

The idea is to expand the solution $y(t)$ into a power series in $t$.

$y(t) = y(0) + t \dot{y}(0) + t^2 \frac{1}{2} \ddot{y}(0) + t^3 \frac{1}{6} \frac{\partial ^3y(0)}{\partial t^3} + ...$

In order to have $d(t) = 0$, all coefficients must vanish. These can be expressed through the initial values, and the first coefficients are

$c(0) = (y_{10} + y_{20}) $
$c(1) = - 4 (x_{10} + x_{20}) + (x_{10}^2 + x_{20}^2) $
$c(2) = - 4 (y_{10} + y_{20}) + 2 ( x_{10} y_{10} + x_{20} y_{20} ) $
$c(3) = 2 (y_{10}^2 + y_{20}^2) + 2 ( x_{10}^3+x_{20}^3 ) -12 (x_{10}^2 + x_{20}^2) + 16 (x_{10} + x_{20})) $

Since we have $y_{10} + y_{20} = 0$, $d(t) = 0$ requires

$c(1) = 0 = - 4 (x_{10} + x_{20}) + (x_{10}^2 + x_{20}^2) $
$c(2) = 0 = 2 y_{10} (x_{10} - x_{20}) $
$c(3) = 0 = 4 y_{10}^2 + 2 ( x_{10}^3 + x_{20}^3 ) -12 (x_{10}^2 + x_{20}^2) + 16 (x_{10} + x_{20})) $

From $c(2) = 0$ and $y_{10} \neq 0$ we find $x_{20} = x_{10}$. Hence

$c(1) = 0 = - 4 x_{10} + x_{10}^2 $
$c(3) = 0 = y_{10}^2 + x_{10}(x_{10}-2)(x_{10}-4) $

Both solutions of $c(1) = 0$ give $y_{10} = 0$. The contradiction proves the assertion of the OP.


1) In my first attempt I thought I needed the equality of the time period of the oscillation, which is equivalent to stating equality of the energies. But in fact it is not needed, as the developments above show.

2) If we would drop the term $x^2$ in the second equation we can have $d(t) = 0$ by taking $x_{10} = - x_{20}$

3) For a symmetric potential (e.g. $2 x^2 - x^4/4$) we have $d = 0$ iff $x_{10}+x_{20} =0$ and $y_{10}+y_{20} =0$.

EDIT #1 (unfortunately this reasoning is not a proof)

This is an attempt to answer to a comment made by seno44 which reads

"The problem is that I want to generalize this consideration to $N$ particles rather than $N=2$ . In that respect the Taylor expansion approach does not generalize nicely in my opinion."

Assuming that the generalization to $N$ particles is such that each particle obeys the same ODEs and that the quantity $d$ in question is the sum of the velocities, the Taylor method is easily generalized.

As far as I can see it leads to an infinite set of equations of the form $(vx^k ,vy)=0$ $k=0,1,2,...$ where $vy=(y_ {10} ,...,y_{N0} )$ , $vx^k =(x_{10}^k ,...,x_{N0}^k)$ , and $(a,b)$ is the scalar product of the vectors $a$ and $b$ .

This statement is not true. There are other equations from higher derivatives than the ones mentioned, e.g. $(vy^2, vy) = 4 (vx^3, vy)$, and I can't see anymore that the conclusion that there is only the trivial solution for $vy$ follows.

  • $\begingroup$ the starting point is not correct. It should be $x^2(t)$ in the $\dot{y}$ equation! $\endgroup$
    – user45183
    Oct 24, 2015 at 20:25
  • $\begingroup$ @seno 44: Thanks for the hint to my trivial error. I have given a complete solution of the OP in the meantime. $\endgroup$ Oct 27, 2015 at 9:53
  • $\begingroup$ I think you have chosen the route that I was explicitly not interested in as I mentioned in the question. $\endgroup$
    – user45183
    Oct 27, 2015 at 12:32
  • $\begingroup$ @ seno44: I agree, if you consider the Taylor Expansion "very dificult" which, honestly, I wouldn't. But, you are right, there should be some simple symmetry argument to give the proof (even simpler than in the answer of Fritz Veeman) because the contrary holds for a symmetric potential, i.e. you can have $d=0$ iff $x_{10}+x_{20}=0$ and $y_{10}+y_{20}=0$ $\endgroup$ Oct 27, 2015 at 13:35
  • $\begingroup$ The problem is that I want to generalize this consideration to $N$ particles rather than $N=2$. In that respect the Taylor expansion approach does not generalize nicely in my opinion. $\endgroup$
    – user45183
    Oct 27, 2015 at 15:07

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