 Frobenius resonance for an Euler equation with monomial potential

For $a>0$, $b\ne 0$, and a positive integer $\beta$, consider $x^2{y}^{\prime \prime }+x{y}^{\prime }+(-a^2+b{x}^{\beta })y=0$ near $x=0$. Its indicial roots are $a,-a$. The first Frobenius solution, normalized to leading coefficient one, is $y_{+}=x^a{\sum }_{k\ge 0}{c}_{k\beta }{x}^{k\beta }$, with $c_{k\beta }=-b{c}_{(k-1)\beta }/[k\beta (k\beta +2a)]$. If $N=2a$ is a positive integer, a second pure Frobenius solution with leading coefficient one would requireBefore $n=N$, 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 $0=-b{d}_{N-\beta }$ 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 $L=x^2{d}^2/d{x}^2+xd/dx-a^2+b{x}^{\beta }$. The product rule gives $L(y_{+}\log x)=2x{y}_{+}^{\prime }$. Therefore a second solution $x^{-a}\sum d_n{x}^n+K{y}_{+}\log x$, with $d_0=1$, hasFor $N=m\beta$ the preceding recurrence relation gives $d_{(m-1)\beta }=b^{m-1}/[{\beta }^{2m-2}((m-1)!{)}^2]$, soThis 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.