I would like to derive the the asymptotic expansion from the WKB approximation. I thought that is standard and could be found in textbooks, but I did not find any useful materials.
To make the question clear, let us start with an example, say, I have a Bessel's equation, $$ z^2 y''+z y' +(z^2-\nu^2)y=0, $$ whose solution is Bessel's functions $J_{\pm\nu}(z)$. Everyone is familiar with the leading term of asymptotic expansion as $z\to\infty$, that is $$ J_\nu(z)\sim \sqrt{\frac{2}{\pi z}}\cos\left(z-\frac{\nu \pi}{2}-\frac{\pi}{4}\right). $$ My task is to derive this formula from Bessel's equation by WKB approximation.
To use the WKB method, I prefer to make a transformation first $y\to y/\sqrt{z}$, which gives $$ y''+Q(z) y=0,\qquad Q(z)=1+\frac{1/4-\nu^2}{z^2}. $$ The WKB phase can be calculated analytically $$ I=\int^z dz \sqrt{Q(z)} $$ and as $z\to \infty$, one will have $$ I\sim z+\frac{ \pi}{4} \sqrt{4 \nu ^2-1} $$ but the additional constant here is not the phase shift in above cosine function.
Could someone give me a clue of solution, or full derivation, or any relevant references? Thanks in advance.
The similar question for first order differential equation can be found here.
Your equation may be written in the form $$ \frac{{\mathrm{d}^2 y(z)}}{{\mathrm{d}z^2 }} = ( - f(z) + g(z))y(z) $$ with $$ f(z) = 1,\quad g(z) = \frac{{\nu ^2 - 1/4}}{{z^2 }}. $$ This equation has solutions $y_\pm(z)$ satisfying $$ y_ \pm (z) = \exp \left( { \pm \mathrm{i}\int_{}^z {\sqrt {f(t)} \,\mathrm{d}t} } \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{z}} \right)} \right) = C_ \pm \exp ( \pm \mathrm{i}z)\left( {1 + \mathcal{O}\!\left( {\frac{1}{z}} \right)} \right) $$ as $z\to +\infty$, with arbitrary constants $C_\pm$. This can be established from Theorem $2.2$ on p. $196$ of F. W. J. Olver's Asymptotics and Special Functions. The Bessel function $\sqrt{z}J_\nu(z)$ will be an appropriate linear combination of these two solutions.