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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 for higher derivatives than the one mentioned, e.g. $(vy^2, vy) = 4 (vx^3, vy)$. Hence the conclusion that there is only the trivial solution for $vy$ does not follow.

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.

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

Ths is the 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 thta 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 $b$ . As the equations must hold for all $k$, the only solution for $vy$ is the trivial one. QED.

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 for higher derivatives than the one mentioned, e.g. $(vy^2, vy) = 4 (vx^3, vy)$. Hence the conclusion that there is only the trivial solution for $vy$ does not follow.

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

If the generalization I proposed in my last comment is correct, the Taylor method is easily generalized. As far as I can see it leads to an infinite set of equations of the form $(vx^k ,vv)=0$ $k=0,1,2,...$ where $vv=(v_ {10} ,...,v_{N0} )$ , $x^k =(x_{10}^k ,...,x_{N0}^k)$ , and $(a,b)$ is the scalar product of the vectors $a$ and $b$ . As the equations must hold for all $k$, the only solution for $vv$ is the trivial one. QED.

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

Ths is the 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 thta 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 $b$ . As the equations must hold for all $k$, the only solution for $vy$ is the trivial one. QED.