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I'm learning about the WKB method, and I'm applying it to an assignment.

The assignment question asks to find the "leading order" WKB expansion for the particular equation.

For WKB you make the substitution: $$y \sim \exp \left[\frac{1}{\delta}\sum_{n=0}^{\infty} \delta ^n S_n(x)\right], \delta \to 0$$ I'm fine with this, I can manipulate it, solve for $S_n(x)$ etc.but I'm unsure as to what "leading order" means in this context.

For other asymptotic methods when you have $y=y_0+\epsilon y_1+\dots$, it's obvious, you just take the terms multiplying $\epsilon^0$, but it doesn't happen the same in WKB because the $\epsilon$s are in the exponential.

So, what does "leading order" mean for WKB?

Bamboo
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  • First rewrite the equation in terms of new function $f=\ln y$, then the meaning will become clear. – Start wearing purple May 23 '15 at 05:58
  • so you're saying that leading order means that you just take the function $S_n(x)$ that is multiplied by $\epsilon^0$ and raise that to the exponential to get your y to leading order? – Bamboo May 23 '15 at 06:06
  • I am not quite sure about what you mean and what is your problem exactly, so let me consider an example. Suppose we have an equation $\delta^2 y''+ V(x)y =0$ where $\delta$ is a small parameter. Substituting $y=e^f$, the equation becomes $\delta^2 f''+\delta^2 f'^2+ V(x)=0$. Now substitute into this equation the ansatz $f=\frac{S_0}{ \delta }+ S_1+ S_2\delta+ \ldots$ and computing the coefficient of $\delta^0$, we get $S_0'^2+V(x)=0$. – Start wearing purple May 23 '15 at 06:45
  • Sorry, just noticed I'm mixing my terms a bit. In the dominant balance we find that $\delta=\epsilon$. Anyway, I've found the solutions for $S_0$ etc. using the method you outline - that's no trouble at all. I guess my uncertainty is that in the question I'm trying to answer it says "what is the leading order solution", which I interpret to mean "what is $y$ to leading order". It's not entirely clear what this means because I'm used to truncating a summation at a certain term, I don't know how this extends to something that is effectively a product – Bamboo May 23 '15 at 08:12
  • This may indeed depend on interpretation. I am thinking of leading order of $y$ as corresponding to two first terms in $S$, since then we can write $$ y= \ln S_1\cdot e^{S_0/\delta} \Bigl[1+O(\delta)\Bigr].$$ In this case the neglected terms are small as compared to computed ones in the usual "additive" sense. – Start wearing purple May 23 '15 at 08:27
  • Ok, that makes sense. It also matches what felt right in the context of the question, so I'm a bit more comfortable with what I've done now. Thanks for your help – Bamboo May 23 '15 at 09:15
  • You are welcome. I have of course made a misprint in my last comment: $\ln S_1 \to e^{S_1}$. – Start wearing purple May 23 '15 at 09:16

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