Noninvariance for a specific nonlinear oscillator

Question

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.

Answer 1

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 $y_2(0) = 0$.

Answer 2

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

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

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

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.

Observations

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.