Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Nonlinear ODE Question

  1. Jun 9, 2006 #1
    I have a naive question about nonlinear ODEs, supose I have one of such equations, and that it is homogeneous

    NL(y(x))=0, (1)

    and I know the solution. Then I want to solve the "associated" non-homogeneous equation

    NL(y(x))=f(x) (2)

    There is a method to attack this kind of problem using the fact that I know the solution of (1)?

    I know that for linear ODEs I can use variation of parameters or Green functions, but what about nonlinear ones?

    I want to check analytical alternatives before surrender to the temptation of numerical methods.
  2. jcsd
  3. Jun 9, 2006 #2
    non-linear equations are quite difficult to handle. there is no straightforward analytic tool.

    there are a few options however:

    -if the equation you are trying to solve is a model of a physical phenomena, you might try to make some assumptions to simplify it. i.e. you could linearize the equation locally about a point if you are only interested in a certain region
    -search to see if your equation has been solved before
    -guess. as long as you can satisfy the equation and the bc's your solution should be unique. I mean all diff eq's are in essence solved by guessing, so if you can get an idea of what kind of function will fit the equation, plug it in and give it a go.
  4. Jun 10, 2006 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Remember WHY we could split up our solution procedure for linear diff.eq's by looking at the particular problem and the homogenous problem separately:
    If Ly=f(x) is a linear diff. eq., then if Y(x) and Z(x) are two sulitions, then the difference function, Y-Z is a solution of the homogenous problem, due to the linearity of L.

    Thus, there is no reason to assume that you can develop something similar for non-linear diff.eqs.
  5. Jun 11, 2006 #4


    User Avatar
    Science Advisor

    The whole point of linearity is that we can split a problem into parts. If y= ax and I replace x with u+ v, then y= a(u+ v)= au+ av, exactly the same as finding y(u) and y(v) separately and then adding them. If y= x2, they y(u+v)= (u+ v)2= u2+ 2uv+ v2= y(u)+ y(v)+ 2uv so knowing y(u) and y(v) does not help me find y(u+v).

    If a d.e. is not linear, then knowing a solution to part of the problem doesn't really help!
  6. Jun 12, 2006 #5
    Thanks to you all.
    Yep, I understand that the "superposition principle" holds just for linear differential operators. What I'm looking for is a similar (or different) procedure that may be used for nonlinear differential equations. I've heard that there are "nonlinear superposition principles" but I have no idea about how they arise or how I could use them.

    I have a lane-emden type ode , which depends on a parameter n.
    My equation has a perturbation term, f(x), which depends solely on the independen variable. When n=0,1 the lane-endem ode is a linear homegeneous ode and every thing is nice and easy. When the perturbation is added I still can solve the resulting ode by standard methods.

    The point is that I know the solution of the homogenous case for other values of n for which the equation is nonlinear. In this cases the lane emden equation looks like L(y(x))=my(x)^n where L is a linear operator (self-adjoint? , have to check that) and m a constant. It seems like a "generalized" eigen-value problem. But here I have no clue about how to obtain the solution when f(x) is added: L(y(x))=my^n+ f(x)
    Maybe I will have to guess or to "cook" an approximate solution :bugeye: as some of you suggested. If it doesn't work then i will try numerically :uhh:
    Last edited: Jun 12, 2006
  7. Jun 12, 2006 #6

    The simplest way in which I had thought to attack the problem was to assume that the superpostion principle holds anyway, in the following sense:
    My Ansatz is


    where y0 is the solution to

    L(y(x))=my(x)^n and ( n not 1 nor 0, sad)

    y1 to


    This will be a solution, won't it? Maybe not a general one, maybe it won't satisfy some physical restrictions imposed by the problem; but maybe it would work.... Is this a valid approach? are there some issues that maybe I've overseen? :surprised


    EDIT: Forget this: when y=y0+y1 is replaced on y^n, i would need to do some "suitable" approximations or constraints, of course.
    Last edited: Jun 13, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook