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Solve equation perturbatively

  1. Jul 8, 2010 #1
    Hi all! I want to perturbatively solve this equation in [tex]\beta[/tex] at second order in [tex]\alpha[/tex]

    [tex]\frac{\beta^{2}}{4}-\frac{3}{8}\beta^{4}\alpha+\frac{2}{3}\beta^{6}\alpha^{2}=\frac{1}{4}[/tex]

    I rewrite this formula in this way

    [tex]\beta^{2}=1+\frac{3}{2}\beta^{4}\alpha-\frac{8}{3}\beta^{6}\alpha^{2}[/tex]

    When I try to solve it perturbatively, I obtain

    [tex]\beta^{2}=1+\frac{3}{2}1^{2}\alpha-\frac{8}{3}\alpha^{2}\left(1+\frac{3}{2}1^{2}\alpha\right)^{3}[/tex]

    The result is

    [tex]\beta^{2}=1+\frac{3}{2}\alpha-\frac{8}{3}\alpha^{2}[/tex]

    The square root of which gives

    [tex]\beta=1+\frac{3}{4}\alpha-\frac{155}{96}\alpha^{2}[/tex]

    I know that the correct result is

    [tex]\beta=1+\frac{3}{4}\alpha+\frac{61}{96}\alpha^{2}[/tex]

    Clearly, there is something wrong in the second order of my calculus.

    Can anyone please tell me where I'm mistaking? Thanks a lot.
     
  2. jcsd
  3. Jul 8, 2010 #2

    arildno

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    Hmm..
    Not quite sure what you are doing here, I'm afraid.
    Here's how I would do it.
    Assume a power series expansion as the following:
    [tex]\beta=1+k_{1}\alpha+k_{2}\alpha^{2}+++[/tex]
    We are to determine the k's in orders of alpha.
    Inserting in your equation, and retaining terms of order 0 and 1, we get:
    [tex]1+2k_{1}\alpha=1+\frac{3}{2}*1\to{k}_{1}=\frac{3}{4}[/tex]

    To order 2, we have:
    [tex]k_{1}^{2}\alpha_{2}+2k_{2}\alpha^{2}=\frac{3}{2}*4k_{1}\alpha^{2}-\frac{8}{3}*1*\alpha^{2}[/tex]
    Or, we get the equation for k_2:
    [tex]\frac{9}{16}+2k_{2}=\frac{9}{2}-\frac{8}{3}[/tex]
    Which yields:
    [tex]2k_{2}=\frac{216-128-27}{48}=\frac{61}{48}[/tex]
    whereby the result follows.
     
  4. Jul 8, 2010 #3
    You are right. That's the right way to do it. Thank you very much!
     
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