Is My Perturbative Solution of this Equation Correct?

[norman86]

Hi all! I want to perturbatively solve this equation in \beta at second order in \alpha

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

I rewrite this formula in this way

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

When I try to solve it perturbatively, I obtain

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

The result is

\beta^{2}=1+\frac{3}{2}\alpha-\frac{8}{3}\alpha^{2}

The square root of which gives

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

I know that the correct result is

\beta=1+\frac{3}{4}\alpha+\frac{61}{96}\alpha^{2}

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

Can anyone please tell me where I'm mistaking? Thanks a lot.

 

[arildno]

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:
\beta=1+k_{1}\alpha+k_{2}\alpha^{2}+++
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:
1+2k_{1}\alpha=1+\frac{3}{2}*1\to{k}_{1}=\frac{3}{4}

To order 2, we have:
k_{1}^{2}\alpha_{2}+2k_{2}\alpha^{2}=\frac{3}{2}*4k_{1}\alpha^{2}-\frac{8}{3}*1*\alpha^{2}
Or, we get the equation for k_2:
\frac{9}{16}+2k_{2}=\frac{9}{2}-\frac{8}{3}
Which yields:
2k_{2}=\frac{216-128-27}{48}=\frac{61}{48}
whereby the result follows.

 

[norman86]

You are right. That's the right way to do it. Thank you very much!