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Given a differential equation of the form

$\epsilon \frac{d^ny}{dx^n} + \sum_{k=0}^{n-1} a_k(x)\frac{d^ky}{dx^k}=0$

Then the WKB method says to choose the ansatz $y\sim exp({\frac{i\phi(x)}{\epsilon}})A(x,\epsilon)$

where $A(x,\epsilon) = \sum_{k=0}^{\infty} A_k(x)\epsilon^n$

I wondered what is the motivation for this educated guess.

Thanks in advance.

2 Answers 2

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Look at the introduction to Chapter 10 in Bender & Orszag. ("Click to look inside", then search for "WKB" and go to p.484 in the search results.)

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    Short version: because the Airy function is the simplest function with a "turning point" that is "well-studied" enough, that we look at how we can (locally) approximate solutions to more complicated differential equations with Airy.2010-10-02
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    That is great. Thanks!2010-10-03
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You can get a rough intuitive understanding by looking at the second order ODE $\epsilon y'' + q(x)y=0$, where $0 < \epsilon \ll 1$ and $q(x)$ does not change sign on the interval of interest.

If $q$ was just a real number, the solution would be the linear combination of two exponential functions having arguments with different signs. The second observation you might make is that the multiplication of the highest order term by a small parameter generally implies we have to use a singular perturbation series. If you look at the argument of the exponential function used, you will see it is a singular perturbation series. So in a sense, the form for the WKB solution is the marriage of two things you expect in the solution: we put the singular perturbation series into the argument of the exponential functions.

Putting this form for the solution into the original ODE, letting the small parameter in the singular perturbation series be a variable, solving for it using dominant balance, and then obtaining a series of ODEs (for the different orders of $\epsilon$) you can obtain a solution.