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Perturbation theory

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Poirot

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Feb 15, 2012
250
Question: obtain 2-term expansions for the roots of x^3+x^2-w=0 , 0<w<<1.

I assumed an expansion of the form x=a+bw+.... and from this obtained a=-1, b=1 as one solution. How do I work out the form of the other 2 expansions?

Thanks.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Question: obtain 2-term expansions for the roots of x^3+x^2-w=0 , 0<w<<1.

I assumed an expansion of the form x=a+bw+.... and from this obtained a=-1, b=1 as one solution. How do I work out the form of the other 2 expansions?

Thanks.
There are two roots near 0 and one root near 1, you have already dealt with than near 1. Then for the roots near 0 guess a trial solution: \(x=aw^k\) (which is a two term logarithmic expansion: \(\log(x)=A+B\log(w)\) ).

CB
 
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Poirot

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Feb 15, 2012
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sorry Captain Black, are you familiar with perturbation expansions? That is what I'm doing, I should have said the above solution as
x= -1+w+....... as it is an infinite (perturbation) series. I want to know how to arrive at the correct asymptotic sequence.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
sorry Captain Black, are you familiar with perturbation expansions? That is what I'm doing, I should have said the above solution as
x= -1+w+....... as it is an infinite (perturbation) series. I want to know how to arrive at the correct asymptotic sequence.
Sorry Poirot, are you familiar with singular perturbation expansions?

When you find the correct exponent for the trial solution you can use a perturbation expansion for \(a=a_0+a_1w^{k}+...\), where \(a_0\) is one or other of the two zeroth-order coefficients found using the trial solution.

CB
 
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Poirot

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Feb 15, 2012
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I take it you thought I was trying to be funny. Anyway, is there any way to arrive at the correct answer without just guessing?
 

CaptainBlack

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Jan 26, 2012
890
I take it you thought I was trying to be funny. Anyway, is there any way to arrive at the correct answer without just guessing?
It is not a guess. There can be no constant term since the root is going to zero as \(w\) goes to zero, so the simplest candidate is a multiple of some power of \(w\). Try the candidate , find the coefficient and exponent for an initial approximate solution and take it from there.

CB
 
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