Entire functions of finite order as solutions to certain complex linear differential equations (Q2845048)

It is known that an entire function \(f\) is a solution of the homogeneous linear differential equation NEWLINE\[NEWLINE y''(z)= u(z) y'(z)+ v(z) y(z), \quad z\in {\mathbb C}, \tag{1} NEWLINE\]NEWLINE for some entire functions \(u\) and \(v\) if and only if all the zeros of \(f\) are simple [\textit{J. Nikolaus}, Math. Z. 103, 30--36 (1968; Zbl 0161.27704)].NEWLINENEWLINEOn the other hand, all solutions of (1) are entire functions of finite order if and only if \(u\) and \(v\) are polynomials [\textit{H. Wittich}, Neuere Untersuchungen über eindeutige analytische Funktionen. Berlin etc.: Springer (1955; Zbl 0067.05501)]. Therefore, it is natural to ask when a given entire function of finite order is a solution of the equation (1) with polynomial coefficients.NEWLINENEWLINE{ Theorem A.} Assume that an entire function \(f\) has the form NEWLINE\[NEWLINEf(z)=z^\lambda e^{q(z)} \prod_{k} E_m\Bigl(\frac{z}{a_k}\Bigr), NEWLINE\]NEWLINE where \(q\) is a polynomial, \(\lambda\) is nonnegative integer, \(E_m\) is the Weierstrass primary factor, \(m\) is the genus of the sequence of the zeros of \(f\). Then \(f\) is a solution to (1) with polynomial coefficients \(u\) and \(v\) if and only if all zeros \((a_k)\) of \(f\) are simple and there is some polynomial \(p(z)\) such that the following relations hold: NEWLINE\[NEWLINE p'(a_k)=\sum_{j\neq k} \Bigl( \frac{a_k}{a_j}\Bigr)^m \frac1{a_j-a_k} -\frac{m+\lambda}{a_k}, \quad k=1, 2, \dots,NEWLINE\]NEWLINE and \( \lambda p'(0)=\lambda \sum_j \frac1{a_j}\) if \(m=0\) and \((a_j)\) is non-empty, and \( \lambda p'(0)=0\), otherwise. Moreover, \(p(z)=q(z)- \frac 12 \int u(z)dz\), and the order NEWLINE\[NEWLINE\rho(e^{-\frac 12 \int u} f) =\frac {\deg (\frac 12 u^2+v)+2}{2}.NEWLINE\]NEWLINE In the particular case \(u(z)\equiv 0\), \(p(z)=q(z)\), and \(\rho(f)=\frac{\deg(v)+2}2\), Theorem B gives another description for the case \(u(z)\equiv 0\).