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Deterministic chaos problem: nonperiodic behaviour

  1. Jan 15, 2007 #1

    The following problem is from Halliday and Resnick's Physics, Vol. I, 1967 edition, page 379, Problem 35. It has defeated me for the last month.

    "A particle undergoes motion in the x-y plane. It's equation of motion is given by

    [tex]x = A_x cos(w_x t)[/tex]
    [tex]y = A_y cos(w_y t)[/tex]

    (i)If the ratio of the two angular frequencies is a rational number, then prove that the motion is periodic.

    (ii)If the ratio of the two angular frequencies is irrational, then prove that the particle will pass through every point in the rectangle given by [itex]x = \pm A_x[/itex] and [itex]y = \pm A_y[/itex], but will never pass through a given point with the same velocity twice."

    (i) is elementary, it's (ii) that's giving me the creeps. What is effectively being asked is a proof that the motion will exhibit deterministic chaos, taking a non-repeating but constrained trajectory through the 4D phase space, something like strange attractor behaviour.

    They don't teach non-linear dynamics in our high-school syllabus, so I'm not aware of any standard theorems or tools that can be used to tackle such problems. As a first analysis, I've identified four components to the proposition:

    a) The motion is constrained inside the rectangle. This is obvious.
    b) The motion is non-periodic. This is easily proved by reductio ad absurdum.
    c)Given any value of (x,y) lying in the specified rectangle, There exists some value of t at which the particle will be at (x,y).
    d)The particle does not return to the same point in the phase space at two distinct instants of time.

    Since the equation of motion is deterministic, presumably (b) implies (d). So that leaves me with (c) to prove. How do I do it? Thanks for your help.

    I've also attached my attempt at a proof of (b) below:

    "Let us assume that the motion is periodic, with period T. As the only periods of the real cosine function are integral multiples of [itex]2\pi[/itex], we have

    [tex]T = \frac {2n\pi}{w_x} = \frac {2m\pi}{w_y}
    \implies \frac {w_x}{w_y} = \frac {m}{n}[/tex]

    This implies that the ratio of the angular frequencies is equal to the ratio of two integers, a rational number. But this contradicts the given condition that the said ratio is irrational. Therefore our assumption can not be right. Therefore the motion is non-periodic."

    Last edited: Jan 15, 2007
  2. jcsd
  3. Jan 15, 2007 #2


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    I'm feeling a little uneasy about the question. The motion is periodic in x, so for a fixed X, the times when x = X are two sets of points separated by intervals of 2 pi/w_x. (Two sets, because the X=x twice per cycle). This set of times are all separated by finite intervals, so they don't have any limit points. Doesn't that mean they form a countable set?

    If that's true, then it's false that the particle passes through every point (X,y) because the number of those points is uncountable.

    It seems likely that it passes arbitrarily close to any point (X,y) though.

    The converse of (ii) is obviously true: if it does pass through the same point with the same velocity, then the motion is periodic and the frequency ratio is rational.
  4. Jan 15, 2007 #3
    I don't get your reasoning. "This set of times are all separated by finite intervals, so they don't have any limit points. Doesn't that mean they form a countable set?" Can you make your argument a little more precise here? Who form a countable set?

    It is obvious that the set of all positions of the particle is uncountable (of cardinality aleph-1 to be exact) since a subset is in bijection with an uncountable subset of the time continnum, so they can be put in a bijection with the points of the rectangle.

    The question in (c) is whether the motion of the particle provides a surjective mapping onto the rectangle. In this language, (b)/(d) is equivalent to the condition that the motion provides an injective mapping from the time continnum to the phase space (but it's not necessarily injective to the space part of the phase space, the rectangle).


  5. Jan 15, 2007 #4


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    From what I see, you don't have chaotic motion but quasiperiodic motion, ie. noncommensurate frequencies.

    (If the frequencies are rational, you have a resonant condition - hence the periodic motion.)
  6. Jan 15, 2007 #5


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    Hmm... I dunno why I was trying to make it so difficult before - this is a simpler argument.

    Considering the times when x = X. Write T for the period 2 pi / w_x.

    In the general case there are two times t_1 and t_2 in the interval [0,T] when x = X.

    All the other times when x = X are of the form t_1 + nT and t_2 + nT where n is an integer. So there is a countable set of times when the particle is at a point [X,y] for any y. But there are an uncountable set of points (X,y) with |y| <= A_y, so the particle does not pass through every point in the rectangle.

    FWIW I would think the solution passes arbitarily close to any point. There should be a proof along these lines:

    There is a frequency w'_y very close to w_y where w'_y/w_x is rational and the fraction has a very large denominator when reduced to its lowest terms. This means there is a periodic approximation to the motion which passes through a very large set points (X,y) and the y values are more or less evenly spaced in the interval [-A_y, A_y] (since trig functions are nicely behaved).

    Reading "arbitarily large" for "very large", and turning this into a formal epsilon-delta proof is left as an exercise - I've got a maths degree but I've spent the last 30 years being an engineer not a mathematician....
  7. Jan 18, 2007 #6
    So Halliday and Resnick gave a wrong problem? That seems very strange to me! These books are usually very accurate, at least in the non-theoretical parts.

    I don't get this part, "...the fraction has a very large denominator when reduced to its lowest terms."

    Thanks for the help.

  8. Jan 18, 2007 #7


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    In what sense?
  9. Jan 18, 2007 #8


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    If the periods had a rational ratio p/q where p and q have no common factors, then the motion repeats after q oscillations in X and p oscillations in Y.

    The basic idea behind my argument is that for any irrational frequency ratio r say, there is a ratio p/q close to r, where p and q have no common factors and p and q are both large. And given any ratio p/q, you can always find another ratio p'/q' which is even closer to r, and with p' > p and q' > q.

    So, with an arbitrarily small perturbation of the frequency ratio, the motion passes through an arbitrarily large number of different points on a line across the square. So this type of argument could be used to prove that the motion passes arbitrarily close to ANY point on the line.

    I don't have Halliday and Resnick, but for physicsts (as opposed to mathematicians) the difference between "passes through every point" and "passes arbitrarily close to every point" is not of much practical significance, I think.
  10. Jan 20, 2007 #9
    Well, if it were a case where the measure oif points the particle does not pass through was zero, then one could say the particle passes through every point. But in this case, exactly the opposite is true. The measure of points the particle does not pass through is the same as the measure of the whole rectangle. The particle passes through almost no point of the rectangle.

    In such a case, should a book ask for a proof that the particle passes through every point of the rectangle without any qualifications? I think it's a simple case of mistake on the part of the authors. Thanks.

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