Bessel equation for half integer

Question

Consider the Bessel differential equation $$x^2y''+xy'+(x^2-\mu^2)y=0$$. The indicial equation gives $r=\pm \mu$, then we have $a_1(1+2r)=0$, and then the recurrence is $$ a_k= \dfrac{a_{k-2}}{(k+r+\mu)(k+r-\mu)}$$. From Frobenius, there will always be a Linearly independent solution for the larger root which in this case is $\mu$. The issue is with the smaller root. Suppose as an example $\mu=-5/2$, then we have $a_1=a_3=0$ and the recurrence is now $$ a_k= \dfrac{a_{k-2}}{(k)(k-5)}$$ so in this case $a_5$ is indeterminate of the form $0/0$. So now we can start the recurrence from $a_5$ for the odd powers $x^5,x^7,.....$. Will this solution be different than $J_{-5/2}(x)$? Can somebody check?

Answer 1

Let $y^\pm$ be the solutions for $r = \pm 5/2$, and let $a^\pm_k$ be their coefficients. From the recursion relation, we have $$ a_k^- = \frac{a^-_{k-2}}{(k-5)k}\Longrightarrow a^-_{k+5} = \frac{a^-_{(k+5)-2}}{k(k+5)}. $$ This means that $a^-_{k+5}$ satisfies the same recursion as $a^+_k$, so their values will be the same up to a constant factor. Of course, $a^+_k = 0$ for every odd $k$, so this doesn't constrain the even-index coefficients $a^-_k$ at all. The odd-index $a^-_k$ coefficients will satisfy $a^-_{k+5} = ca^+_{k}$ for some constant $c$. Since we also have $a^-_1 = a^-_3 = 0$, if we write out the series for $y^-(x)$, we get $$ y^-(x) = \sum_{k=0}^\infty a_k^-x^{k-5/2} = \sum_{k\;\mathrm{even}}^{\infty}a_k^-x^{k-5/2} + a_1x^{-3/2} + a_3 x^{1/2} + \sum_{k\;\mathrm{even}}^\infty a_{k+5} x^{k+5/2} \\= \sum_{k\;\mathrm{even}}^{\infty}a_k^-x^{k-5/2} + c \sum_{k\;\mathrm{even}}^\infty a_k^+ x^{k+5/2} =\sum_{k\;\mathrm{even}}^{\infty}a_k^-x^{k-5/2} + c y^+(x). $$ So the recursion that starts at $a_5$ indeed gives the other linearly independent solution.