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Second Order DE: Nonlinear Homogeneous

  1. Oct 5, 2007 #1
    I am sure most of you are familiar with the equation: m(x)''+c(x)'+k(x) = 0. Then, we create an auxillary equation that looks like this: mr^2+cr+k = 0. And, then we find the roots of this auxillary equation, calling them r1 and r2. And, if the roots are r1,r2>0 we consider the system to be overdamped and we develop the following general form of the equation to be: x=A*e^(r1*t)+B*e^(r2*t) etc...

    I have an inverted pendulum with damping and understand the equation to be in the form of a second order nonlinear (homogenous) differential equation.

    The second order nonlinear homegenous differential equation is: m(x)''+c(x)'-k*sin(x)=0.
    I have tried everything I know and haven't had much luck... I mean how do you find the roots to something like this? I am sure there is a method to this madness. I would think there is some type of variation of parameters or substitution involved but do not know how to apply substitution etc... to this application.

    I would appreciate any help or advice that might lead me in the right direction...

    Thanks in advance, John
  2. jcsd
  3. Oct 5, 2007 #2


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    There are no generally effective methods to gain the exact solution(s) of non-linear equations, differential or algebraic!

    The "method to this madness" in order to get approximate solutions is to make a guess, giving you a wrong answer, then make a second guess, giving you (hopefully!) a less wrong answer, and so on ad infinitum.

    Such systematized guess-working techniques are very much in use, and to improve those techniques is one of the main branches of applied mathematics areas like computational fluid dynamics and suchlike fields.
  4. Oct 6, 2007 #3


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    Here, for example, since sin(x)= x- (1/6)x3+ ..., a first approximation would be to replace sin(x) by x: mx"+ cx'- kx= 0.

    A slightly more sophisticated method is "quadrature": Let u= x' so that x"= u'. By the chain rule, u'= (du/dx)(dx/dt)= u u' so the equation becomes u u'+ cu- kx= 0. That is now a first order equation for u as a function of x. The problem typically is that even after you have found u, integrating x'= u, to find x as a function of t, in closed form may be impossible (without the damping, this gives elliptic integrals).
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