Looking at the legendre equation, I want to find the asymptotic behavior as x goes to infinity (leading order, $\nu$~O(1)) $$y''-\frac{2x}{1-x^2}y'+\frac{\nu(\nu+1)}{1-x^2}y=0$$ So I let $y=\exp(S)$ and I rearrange to get $$S''+S'^2-x^2S'^2-x^2S''-2xS'+\nu(\nu+1)=0$$ I get lost from here. Wikipedia introduces $S_0,S_i,...$. They do not really explain where these come from. I believe I want to look at the $x^2$ terms because they will be largest at x goes to infinity, but I do not see how I will get a solution.
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2The usual WKB method, or at least the version described in the Wikipedia article, is not really used to find the $x \to \infty$ behavior of the solution to a differential equation. As indicated in the "An example" section of the Wikipedia article that you mention, it approximates a solution to a differential equation when a parameter in the equation is very small. So, for your equation it may be possible to give an approximation of $y$ when $v$ is very large. Perhaps you should seek an alternate method. – Antonio Vargas Nov 02 '15 at 01:17
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I failed to mention that $\nu ~O(1)$. Does that make a difference – yankeefan11 Nov 02 '15 at 01:18
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I have come across method of dominant balance. Is that more appropriate? – yankeefan11 Nov 02 '15 at 01:21
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1the first thing what comes in my mind is , that one can neglect the ones in the denominators as $x$ gets big. this should simplify things. Indeed we get an Euler ODE which is standard to solve. – tired Nov 02 '15 at 01:58
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So looking at http://mathworld.wolfram.com/EulerDifferentialEquation.html I see that this will result in Bessel Equations. So should the asymptotics at infinity be similar to those at infinity? – yankeefan11 Nov 02 '15 at 14:22
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Or can I just use Frobenius's method? I see that it is valid in [-1,1] but will it suffice at infinity? – yankeefan11 Nov 02 '15 at 14:33
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Sometimes a version of the Frobenius method works at $\infty$, see Dr. MV's answer here. – Antonio Vargas Nov 06 '15 at 03:08