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Solving an ODE using shooting method

  1. Dec 29, 2014 #1
    Hi, I am trying to solve the following ODE for my maths project:

    ## y'' = \frac{\alpha}{2}y^3 - \frac{3}{2}y^2 + y - \frac{3}{x} y'##

    under the following boundary conditions:

    ## y'(0) = 0 ##
    ## y(x) \rightarrow y \_ \equiv 0\ \text{as}\ x \rightarrow \infty ##



    As a first step, I converted this problem into a set of coupled ODEs:

    ## \frac{dy}{dx} = z##
    ## \frac{dz}{dx} = \frac{\alpha}{2}y^3 - \frac{3}{2}y^2 + y - \frac{3}{x} z##

    under the following boundary conditions:

    ## z(0) = 0 ##
    ## y(x) \rightarrow y \_ \equiv 0\ \text{as}\ x \rightarrow \infty ##


    Next, my source tells me to use the shooting method to convert the BVP into an IVP, which means that I have to use two initial guesses of ## y(0) ## to be able to use the secant method to find the correct value of ## y(0) ##.

    Now, my question is, according to my source, I can avoid the singularity at x = 0 using Taylor expansion as follows:

    ## y(r_{0}) = y_{0} + \frac{1}{16} r_{0}^{2} (2y_{0} - 3y_{0}^{2} + \alpha y_{0}^{3}) ##

    I see how you can estimate ## y(r_{0}) ## where ## r_{0} ## is a tiny distance away from the origin, but I don't really see how they dervied this expression. Could anyone help me out?
     
  2. jcsd
  3. Dec 29, 2014 #2
    Just set y(0)'=0 and y(0)=y_0 in the original DE.
    Then use this as your second derivative in your taylor series.

    EDIT: Tried it myself, something weird is going on. The numerical factors come out different for me.
    Maybe an error? I don't see where the factor of 1/16 comes from. It comes out as 1/4 for me.
     

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    Last edited: Dec 29, 2014
  4. Dec 29, 2014 #3
    Start out by representing y as a function of x at small values of y by y(x) = y(0)+bx2. Then

    y' = 2bx
    y''=2b

    3y'/x = 6b

    So ##8b=\frac{\alpha}{2}y(0)^3-\frac{3}{2}y(0)^2+y(0)##

    So, ##y(r_0)=y(0)+br_0^2=y(0)+\frac{1}{8}r_0^2(\frac{\alpha}{2}y(0)^3-\frac{3}{2}y(0)^2+y(0))##

    Chet
     
  5. Jan 2, 2015 #4
    Thanks for the solution. I also came to the conclusion, but I used ## \frac{y''(0)}{2} ## in place of ## b ##. I have written the code that solves the ODE in my original post, but it's giving me unexpected results. I was wondering if anyone in Physics Forums might want to have a look at it.
     
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