= Frobenius resonance for an Euler equation with monomial potential
{c}
{title2=$\beta\mid 2a$}
For $a>0$, $b\ne0$, and a positive integer $\beta$, consider $x^2y''+xy'+(-a^2+bx^\beta)y=0$ near $x=0$. Its <indicial roots> are $a,-a$. The first <Frobenius solution>, normalized to leading coefficient one, is $y_+=x^a\sum_{k\geq0}c_{k\beta}x^{k\beta}$, with $c_{k\beta}=-bc_{(k-1)\beta}/[k\beta(k\beta+2a)]$. If $N=2a$ is a positive integer, a second pure <Frobenius solution> with leading coefficient one would require
$$
n(n-N)d_n=-bd_{n-\beta},\qquad d_0=1,\qquad d_j=0\ (j<0).
$$
Before $n=N$, induction makes $d_n=0$ for indices not divisible by $\beta$. If $\beta$ divides $N$, the coefficient $d_{N-\beta}$ is nonzero, and the exceptional equation $0=-bd_{N-\beta}$ is impossible. If $\beta$ does not divide $N$, the right-hand side is zero and $d_N$ is free. Thus, in the <Frobenius case classification>, a positive integer gap gives Case IIIa exactly when $\beta$ divides $2a$, and Case IIIb otherwise.
The coefficient of the necessary <logarithm> can also be calculated. Let $L=x^2d^2/dx^2+x\,d/dx-a^2+bx^\beta$. The <product rule> gives $L(y_+\log x)=2xy_+'$. Therefore a second solution $x^{-a}\sum d_nx^n+Ky_+\log x$, with $d_0=1$, has
$$
K=-\frac{bd_{N-\beta}}{N}.
$$
For $N=m\beta$ the preceding <recurrence relation> gives $d_{(m-1)\beta}=b^{m-1}/[\beta^{2m-2}((m-1)!)^2]$, so
$$
K=-\frac{b^m}{m\beta^{2m-1}((m-1)!)^2}\ne0.
$$
This proves that the <logarithm> cannot disappear for a special nonzero value of $b$. For $a=0$, <reduction of order> instead produces the repeated-root <Logarithmic Frobenius solution> directly.
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