I'm studying from Applied Partial Differential Equations (Haberman) and got stuck on the last question (c) shown below. How do I find the coefficients $a_1, a_2, b_1$ and $b_2$?
Problem
My Work
So from questions (a) and (b), I found $$J_{1/2}(z)=z^{-1/2}(a_1 cos(z)+a_2sin(z))$$ and $$Y_{1/2}=z^{-1/2}(b_1cos(z)+b_2sin(z))$$
I know that as $z\to0$ (small z):$$J_m(z)=\begin{cases}1 & m=0 \\ \frac{1}{2^mm!}z^m & m\gt 0 \end{cases}$$ and $$Y_m(z)=\begin{cases}\frac{2}{\pi}ln(z) & m=0 \\ \frac{-2^m(m-1)!}{\pi}z^{-m} & m\gt 0 \end{cases}$$
How do I apply these asymptotic functions?
I do one for you, and you will get going, I guess.
First, $(1/2)!=\Gamma(3/2)=\sqrt{\pi}/2$, so, as $z\approx 0$, $$ J_{1/2}(z)\approx \frac{1}{\sqrt{2}\sqrt{\pi}/2}z^{1/2}=\frac{\sqrt{2}}{\sqrt{\pi}}z^{1/2}. $$ I guess you know that $\cos z\approx 1$ and that $\sin z\approx z$ for small $z$. Thus $$ z^{-1/2}(a_1\cos z+a_2\sin z)\approx z^{-1/2}(a_1+a_2z)=\frac{a_1}{z^{1/2}}+a_2z^{1/2}. $$ By comparison, we find that here $a_1=0$ and $a_2=\sqrt{2}/\sqrt{\pi}$.
Now try the rest for yourself.