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Non linear ODE: y'' = a y + b y^3

  1. Jan 12, 2013 #1
    I would like to solve the non linear ODE
    [itex] \frac{d}{dx}f(x)=a f(x)+ b f^3 (x)[/itex]
    with the boundary
    [itex]f(0)=0\quad f(+\infty)=f_0 [/itex]

    How to find analitical solution?
  2. jcsd
  3. Jan 12, 2013 #2
    If it is f ' = a f+b f^3 then integrate dx = df/(a f+b f^3)
    If it is y'' = a y+b y^3 then the first step is the integration of :
    y'' y' = (a y+b y^3) y'
    y'² = a y² + b (y^4)/2 +c
    dx = dy/sqrt(a y² + b (y^4)/2 +c)
    The integration will lead to x(y) on the form of elliptic functions. Very complicated to invert in order to obtain y(x)
    Last edited: Jan 12, 2013
  4. Jan 13, 2013 #3
    Pardon me, I write the uncorrect differential equation: the problem is at the second order

    [itex] \frac{d^2}{dx^2}f(x)=a f(x) + b f^3 (x) [/itex]
    with the boundary
    [itex] f(0)=0,\ f(+\infty)=f_0[/itex]
  5. Jan 13, 2013 #4
    In my opinion, [itex]x(y)=g(y,c_1,c_2)[/itex] is an analytical expression for the solution but I think first, just scrap the a and b and look at:


    then do what Jacq said and get the expression in terms of:


    then try and solve simultaneously the expressions:



    for the constants [itex]c_1,c_2[/itex] and if need be, do so numerically for them just for starters.
  6. Jan 13, 2013 #5
    Hello !

    It is a difficult problem on the analytical viewpoint.
    A difficulty is that elliptic functions are involved. But the major difficulty is due to the condition f(x=+infinity)=finite constant.
    I am not quite sure that the solution given in attachment is a correct answer to the problem, so it should be carefully verified.

    Attached Files:

    • ODE.JPG
      File size:
      33.4 KB
  7. Jan 14, 2013 #6
    In case of some complementary conditions specified in attachment, y(x) can be fully explicited thanks to a Jacobi elliptic function.

    Attached Files:

  8. Jan 14, 2013 #7
    :approve:Nicely done Jacquelin and beautiful too. Thanks.
  9. Feb 9, 2013 #8
    Thank you so much!
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