Question:
Consider $y'' + \frac{1-x}{x^3} y = 0$
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Derive the first three terms in Liouville Green approximation.
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Hence obtain corresponding asymptotic approximations as x → ∞ of two linearly independent solutions $y(x)$.
Background (i.e. what I know): Let $y(x)$ satisfy the differential equation:
$$y''- q(x)y = 0$$
To approximate the solution around an irregular singular point $x_0$, suppose $y(x) = \exp(S(x))$ where $S(x) = S_0(x) + S_1(x) + …$ where $(S_n(x))$ form an asymptotic sequence as $x\to x_0$. Then substitute our guess to the equation and get:
$$S''+S'^2 = q$$
To derive a first order solution, only include $S_0$ and then $S_0 ''+S_0'^2 = q$. Ignore $S_0''$, we have $S_0'= \pm \sqrt q$, thus:
$$S_0' = \exp\left(\pm \int \sqrt q dx\right)$$
Similarly, to get first two orders, include $S_0, S_1$, then:
$$ S_0 ''+S_1 ''+ S_0'^2 +2S_0'S_1'+S_1''^2 = q$$
Using $S_0'^2 = q$ and ignoring $S_1'',S_1'^2$, we have $2S_0'S_1'=-S_0''$, from which we could deduce $S_1$.
Repeat doing this, we should get all orders…
My question: I only know how to derive the first two terms. I got stuck because I do not know why we could ignore certain terms during the derivation. Of course we could substitute what we got for $S_0,S_1$ back into the equations and see they are indeed much smaller than other terms, but if I wanted to derive a higher order e.g. the 3rd order term, I have no clue which terms I should throw away and which should be kept.
Could someone help me out? Any help is greatly appreciated!
Best Answer
(I have figured out how to do this problem and I think it will be useful to attach the answer for perhaps further reference. Other answers are still very welcome and please kindly point out if the following answer is found to consist of any error: this answer hasn't been checked so could contain calculation errors.)
LG approximations are well-known for its first 2 orders approximations, and further orders should be derived based on the specific choice of $q(z)$, in this case, $\frac{1-x}{x^3}$. Using LG approximations, we get $S_0, S_1$ immediately:
$$S_0 = \exp\left(\int (-q)^{1/2} dx\right),\quad S_0' = (-q)^{1/2}, \quad S_0'' = \frac{-q'}{2(-q)^{1/2}}$$
$$ S_1 = \frac{-1}{4} \ln|q|,\quad S_1' = \frac{-q'}{4q},\quad S_1'' = \frac{q'^2-q''q}{4q^2}$$
At $S_2$ level, have $(S_0' + S_1' + S_2')^2 + (S_0''+S_1''+S_2'') = -q$. Using $S_0'^2 =-q, 2S_0'S_1' = -S_0''$, we have:
$$S_1''+S_2''+S_1'^2 + S_2'^2 + 2S_0'S_2' + 2S_1'S_2' = 0$$
To determine which terms should be kept, compute their orders (for $x\gg1$):
$$q = O(x^{-3}),\quad q' = O(x^{-4}),\quad q'' = O(x^{-5})$$
$$S_0' = O(x^{-3/2})$$
$$S_1' = O(x^{-1}),\quad S_1'' = O(x^{-2})$$
Then keeping only leading terms and assuming $S_2'', S_2'^2 \ll S_1'', S_1'^2$ $(*)$, we should have:
$$S_1''+S_1'^2+2S_1'S_2' = 0$$
Substituting $S_1 = \frac{-1}{4} \ln|q|$ into the formula and keeping only leading terms, we have derived a leading order approximation for $S_2$. We should then substitute this $S_2$ back to $(*)$ to verify the assumptions are indeed satisfied.