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Homework Help: 3rd order non linear differential equation

  1. Nov 25, 2012 #1
    Q. Find SP of:

    [tex]\dddot{x} + \ddot{x} + \dot{x} = x^3 -2x^2 - 31x -28[/tex]

    And determine of the solutions as stable or unstable.

    OK, not seen one like this before. I've done it with 2nd order derivatives and wondered if it was the same. by setting the derivatives to zero and solving the RHS. Then turn it into a DE in another variable, linearize, solve and investigate the solutions and their stability.

    Can you set the 3rd derivative just to zero??

    Thanks for your help in advance. Peter
    Last edited: Nov 25, 2012
  2. jcsd
  3. Nov 25, 2012 #2


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    Well the first thing you have to do is determine your solutions. The method is completely analogous to that of a second order equation. Solve the homogeneous system, then solve the non-homogeneous one.
  4. Nov 25, 2012 #3


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    Not necessary to solve the full system if the OP only needs to evaluate the stability of the fixed points.

    The concept of finding fixed points (whether stable or unstable) is that you are searching for solutions of the form ##x(t) = \mbox{const}## which solve your equation. Since your assumed solution is a constant, all derivatives vanish, and you need only find the value of ##x## for which the RHS vanishes.

    Now, I assume you know the rest but I'll run over it just to be sure: these constant solutions obviously will not satisfy all possible initial conditions you could be given; however, if the full solution ever hits one of these fixed points, the system will stay there. So, you want to see which ones are stable or unstable, as that will tell you whether or not the solutions will converge to or diverge from the fixed points you've found. So, you let ##x(t) = x^\ast + \epsilon(t)##, where ##x^\ast## is the fixed point you want to find the stability of and ##\epsilon(t)## is a small perturbation from the fixed point. Since it is small, you can drop terms nonlinear in ##\epsilon(t)## and solve the linearized equation. If ##\epsilon(t)## grows with time you have an unstable fixed point, if it decays with time you have a stable fixed point.
  5. Nov 25, 2012 #4
    Thanks for confirming this Mute. i run in to a problem further down the line. I get solns at the Stationary Point as x=-4, x = 7, x=-1 but after applying [tex]x(t) = x^\ast + \epsilon(t)[/tex], then expanding and linearizing I end up with a DE in [tex]\epsilon[/tex] that I can't solve.

    For example. x* = -1 I'm trying to solve [tex](\epsilon ^3 +\epsilon ^2 +\epsilon + 24) = 0[/tex] which I can't sovle easily if at all. The same problem happends for x* = -4

    What are your thoughts here??
  6. Nov 26, 2012 #5


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    That doesn't look right. Do you mean ##\dddot{\epsilon}## instead of ##\epsilon^3##, etc? Your linearized DE should look something like

    $$\dddot{\epsilon} + \ddot{\epsilon} + \dot{\epsilon} = A + B\epsilon,$$
    for some constants A and B.

    This is just a linear DE that you can solve by assuming a trial solution for the homogeneous equation (i.e., the DE with A = 0) of the form ##\epsilon = \exp(rt)##, from which you get the characteristic polynomial which you solve for r. You would then seek a particular solution to add to the homogeneous solution to take care of a non-zero A term. (If any roots are double roots you also need to seek solutions of the form ##t \exp(st)##, where s is one of the double roots). Do you remember how to do this? It's not really all that much different for a third order DE than a second order equation, the main difference is just that you need to solve a cubic equation for r. You may need to solve for the roots numerically.
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