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Nonlinear DE fro CM

  1. Apr 9, 2006 #1
    I'm stuck in a classical mechanics problem. I have to solve this equation (actually I need only [tex]\ddot{x}[/tex]) to conclude it:

    [tex]x \frac{d^2x}{dt^2} + (\frac{dx}{dt})^2 -gx = 0[/tex]

    The book offers a solution [tex]At^n[/tex] and it turns out that it's satisfied for n=2 and A=g/6 :surprised . Well, how am I supposed to guess that solution? It doesn't look Euler equation, or anything else I've seen before. Best I could do was to use brute force and expand it to series.

    Question is, without this guesswork (and brute force), how can I solve/see it?
    Last edited: Apr 9, 2006
  2. jcsd
  3. Apr 9, 2006 #2
    I can give you a way to guess n=2. Consider:
    Which is a variation on [tex]v^2=v_0^2+2a(x-x_0)[/tex], which is only true for constant a, which implies [tex]x=1/2at^2[/tex].

    I can't say, however, that this solution is general.

  4. Apr 9, 2006 #3
    In a word, what I have to say is uuuuugly!

    I'll give you the sketch and you can fill in the details.

    Generally your equation is a mess since it's non-linear. So I transformed it to make it linear. We may show that:
    [tex]a = v \frac{dv}{dx}[/tex]

    So: [tex]xa + v^2 - gx = 0[/tex] becomes
    [tex]xv \frac{dv}{dx}+v^2-gx = 0[/tex]

    [tex]\frac{1}{2} x \frac{d(v^2)}{dx}+ v^2 -gx = 0[/tex]

    Now define [tex]f=v^2[/tex] to get
    [tex]\frac{1}{2} x \frac{df}{dx} + f - gx = 0[/tex]
    which is linear.

    Use [tex]x = e^y[/tex] to get

    which has a solution:
    [tex]f(y)=\frac{2}{3}ge^y+Ae^{-2y}[/tex], where A is a constant.

    So [tex]f(x)=\frac{2}{3}gx+\frac{A}{x^2}[/tex]

    So [tex] \left ( \frac{dx}{dt} \right ) ^2 = \frac{2}{3}gx + \frac{A}{x^2}[/tex]

    and after separating variables I get:

    [tex]t -t_0= \pm \int_{x_0}^x \frac{dx}{\sqrt{\frac{2}{3}gx + \frac{A}{x^2}}}[/tex]

    Which I can't figure out how to integrate. But I find integration (even if it's possible no closed form exists) more tractable than nonlinear differential equations. :smile:

    (By the way, [tex]x = 1/6gt^2[/tex] can be got by using A = 0 in the integral.)

    Last edited: Apr 9, 2006
  5. Apr 9, 2006 #4


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    The equation is non linear. Unfortunately there is no general method for solving non linear equations (except some of them). Mankind need to discover more things in this field.

    Usual techniques when dealing with at first sight non solvable non linear equations (see for example Benders&Orzsag) is to assume solutions. Before nothing, try to test if easy solutions such as exponentials, powers, logarithms or combinations of those powered to something are general solutions. Usually, when an engineer finds that these are not solutions of the differential equations, he/she will try to work out asymptotic solutions by means of perturbation/multiple scale methods, or to use Frobenius/Taylor expansions conveniently for a series solution.

    To sum up: when dealing with non linearity, you are on your own.
  6. Apr 9, 2006 #5


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    There's another curiosity: have you applied initial conditions?? As you see, you've obtained a general solution without any information of the initial conditions!!!!. That's characteristic of non linear diff equations. Sure there are more solutions out there of your equations which incorporate info of the initial conditions. Maybe you can work out some of them assuming an asymptotic behavior and making a dominant balance.....

    Good luck.
  7. Apr 9, 2006 #6
    Yikes! That was what I was afraid of!
    The initial conditions are [tex]x_0 = h>0[/tex], [tex]v_0 = 0[/tex], [tex]a_0 = 0[/tex]

    topsquark, I tried the integral at http://integrals.wolfram.com with this input
    Code (Text):
    1/( (2/3)*g*x + a/(x^2) )^(1/2)
    and it gave a BIG monster!

    Thanks for the replies!
  8. Apr 10, 2006 #7


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    You mean this one right here. It seldom happens that nonlinear ODE-s have solutions which can be expressed in terms of known special functions. But in your case, you got lucky.


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  9. Apr 10, 2006 #8
    Really? I'm not seeing it. Doessn't that in fact say a=va, or v=1 and so x is a line?
  10. Apr 12, 2006 #9


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    [itex] v=\frac{dx}{dt} [/itex] and so

    [tex] \left(a=\right)\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt} [/tex].

  11. Apr 12, 2006 #10
    Gah! I was reading it as a =v(dv/dt). Thanks
  12. Apr 16, 2006 #11
    That's pretty lucky, I'd say.
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