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Singular Points of DE

  1. Nov 2, 2013 #1

    FeDeX_LaTeX

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    The problem statement, all variables and given/known data
    The equation of motion of a particle moving in a straight line is

    ##x'' - x + 2x^3 = 0##

    and ##x = \frac{1}{\sqrt{2}}, x' = u > 0## at ##t = 0##. Identify the singular points in the phase plane and sketch the phase trajectories. Describe the possible motions of the particle, indicating the ranges of u for which these motions occur.

    The attempt at a solution

    I really can't seem to get started on this question -- where are the singular points here, and how might I identify them? The co-efficient of x'' is 1, so it doesn't seem like I can divide by anything useful. I can't set a first derivative equal to zero, because there aren't any in this equation. Is there a form in which I have to rewrite this to get something useful out of it?
     
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  3. Nov 2, 2013 #2

    tiny-tim

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    Hi FeDeX_LaTeX! :smile:
    Multiply throughout by x' and then integrate? :wink:
     
  4. Nov 2, 2013 #3

    haruspex

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    Not an area I've previously worked in, but my reading of http://www.ucl.ac.uk/~ucahdrb/MATH3401/phaseplane.pdf is that the phase plane for a second order ODE is the plane (x', x). The 2nd order ODE is written as a first order ODE in terms of x and y where y = x'. Singular points refer to that context.
     
  5. Nov 6, 2013 #4

    FeDeX_LaTeX

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    Okay, with tiny-tim's suggestion, I had:

    [tex]\frac{dx}{dt} \frac{d^{2}x}{dt^{2}} = (x - 2x^3) \frac{dx}{dt}[/tex]

    [tex]\implies \frac{1}{2} \frac{d}{dt} \left(\frac{dx}{dt}\right)^2 = (x - 2x^3) \frac{dx}{dt}[/tex]

    [tex]\implies \left(\frac{dx}{dt}\right)^2 = x^2 - x^4 + u^2 - \frac{1}{4}[/tex]

    after plugging in our initial conditions.

    We must have ##\left(\frac{dx}{dt}\right)^2 \geq 0##, so we have singularities whenever ##x^2 - x^4 + u^2 - \frac{1}{4} < 0##, i.e. whenever ##(x^2 - \frac{1}{2})^2 > u^2##. That'd give me something of the form x' = f(x) for which I can map a phase-plane plot -- all that remains is to re-arrange and integrate to get an equation in t and x, from which I could probably find the possible motions of the particle, correct?
     
  6. Nov 6, 2013 #5

    HallsofIvy

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    This is just a matter of applying definitions. First, because this asks about "singular poinnts in the phase plane", we have to convert to the "phase plane"! Let v= x' so that x''= v' and the equation becomes [itex]v'- x+ 2x^3= 0[/itex] or [itex]v'= x- 2x^3[/itex].

    So in the "phase plane" we have the pair of equatons [itex]x'= v[/itex], [itex]v'= x- 2x^3[/itex]. A "singular point[/b] is where both x' and v' are 0. That is, where v= 0 and [itex]x- 2x^3= 0[/itex].
     
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