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This article is cited in 2 scientific papers (total in 2 papers) One advance in the proof of the conjecture on meromorphic solutions of Briot–Bouquet type equations A. Ya. Yanchenko National Research University "Moscow Power Engineering Institute"
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     § 1. Introduction. Statement of the main theoremThe paper is concerned with the study of entire solutions of one class of algebraic differential equations. Namely, we consider the so-called autonomous Briot–Bouquet type equations of the form $P(y,y^{(n)})=0$ (where $P$ is an irreducible polynomial of degree $\geqslant 2$ with complex coefficients, $n$ is a natural number). This problem is a part of the more general problem of description of meromorphic solutions of such equations. Corresponding studies have shown that the class of functions $W=\{R(z);\, R(e^{\alpha z}), \, \alpha\in\mathbb{C}; \, \rho(z)\}$ consisting of rational functions and rational functions of exponentials and elliptic functions is important for this problem. One of the first results here is due to C. A. A. Briot, and J. C Bouquet, who showed that, for first-order equations $P(y,y')=0$, all meromorphic solutions lie in the class $W$. The next was Picard [1], who verified that all meromorphic solutions of equations of the form $P(y,y'')=0$ also lie in $W$. However, this result was forgotten and rediscovered only after a hundred years by Hille (see [2]), who proved it under some additional assumptions. Later, Picard’s result was completely rediscovered by Bank and Kaufman [3]. Another proof of this Picard’s theorem was given by A. E. Eremenko (who did not publish his proof learning of this Picard’s paper). The subsequent history of the subject is as follows. In 1980s, A. E. Eremenko put forward the conjecture that, for each natural $n$, any meromorphic solution of the equation $P(y,y^{(n)})= 0$ should lie in the function class $W$. (Of course, it is naturally assumed here that $P$ is an irreducible polynomial of degree $\geqslant 2$, since for a linear equation the corresponding result is not generally true for $n\geqslant 3$.) Eremenko (see [4], [5]) proved his conjecture under some additional assumptions (in his proof, he used one idea of E. Hille). The further advances here are due to Eremenko, Liao, and Wai Ng [6], who showed that, for any $n$, any meromorphic solution of the equation $P(y,y^{(n)})=0$ with at least one pole lies in the class $W$. Thus, the conjecture was verified for all meromorphic solutions save for entire solutions. In the present paper, the above conjecture is proved for an arbitrary $n$ for the case of entire solutions under certain nondegeneracy assumption on the polynomial $P$. In what follows, as usual, $\mathbb{N}$ denotes the set of natural numbers, $\mathbb{C}[\omega_1,\dots,\omega_n]$ is the ring of polynomials $\omega_1,\dots,\omega_n$ over the field of complex numbers $\mathbb{C}$. Given an entire function $f\colon \mathbb{C}\rightarrow\mathbb{C}$, we set $M_f(R)=\max_{|z|\leqslant R} |f(z)|$ for each $R>0$; by $N_f(R)$, we denote the number of zeros of $f(z)$ in the annulus $|z|\leqslant R$ (counting multiplicity). The order $\rho$ of an entire function is defined by the equality $\rho=\varlimsup_{R\to+\infty} (\ln\ln M_f(R)/\ln R)$. If $\rho<+\infty$, then $f(z)$ is said to be an entire function of finite order (see, for example, [7; Ch. 1]). If an entire function $f(z)$ is representable by its Taylor series $f(z)=\sum_{k=0}^{\infty} a_k z^k$, then, for each $R>0$, the maximal term $m(R)$ is defined by $m_f(R)=\max_{k} |a_k|R^{k}$; by definition, the index $\nu_f(R)$ is the greatest $k$ such that $m(R)=|a_k|R^{k}$ (see [3]). The main result of the present paper is as follows. Theorem. Let $n,d\in\mathbb{N}$, $d\geqslant 2$. Let $P$ be an irreducible polynomial from $\mathbb{C}[\omega_1,\omega_2]$ such that: a) $P=\sum_{l=0}^{d} P_l$, where $P_l$ is a homogeneous polynomial of degree $l$ (for each $l=0,\dots,d$); b) $P_d=\prod_{j=1}^{d} (\omega_2-\alpha_j\omega_1)$, where $\alpha_1,\dots,\alpha_d\in\mathbb{C}$ and $\alpha_i\ne \alpha_j$ for $i\ne j$. Let $y=f(z)$ be an entire transcendental function satisfying $P(y,y^{(n)})=0$. Then there exist $a\in\mathbb{C}\setminus \{0\}$, $L\in\mathbb{N}\cup\{0\}$, and $Q\in\mathbb{C}[\omega]$ such that $f(z)=e^{-Laz}Q(e^{az})$. § 2. Auxiliary resultsLemma 1 (see [8], Theorem C). Let $P\in \mathbb{C}[\omega_1,\omega_2]$, let $n$ be a natural number, and let $y=f(z)$ be an entire transcendental function satisfying $P(y,y^{(n)})=0$. Then $f(z)$ is an entire function of finite order. Remark 1. In the actual fact, with the help of the Wiman–Waliron theory (see [9]), it can be shown, under the hypotheses of Lemma 1, that $f(z)$ is an entire function of first-order of normal type. Lemma 2 (see [5], Theorem 1). Let $A$ be an irreducible polynomial from the ring $\mathbb{C}[\omega_1,\omega_2]$ and let $n\in\mathbb{N}$; $\varphi(z)$ be an entire transcendental function satisfying the equation $A(\varphi(z),\varphi^{(n)}(z))=0$. Then the algebraic surface $F=\{(y,s)\colon A(y,s)=0\}$ is a genus-zero surface. Lemma 3. Let $n,M\in\mathbb{N}$, $\sigma_1,\dots,\sigma_M\in\mathbb{C}\setminus\{0\}$; $\varphi(z)=\sum_{j=1}^{M} C_j e^{\sigma_j z}+q(z)$, where $C_j\in\mathbb{C}$, $q(z)\in\mathbb{C}[z]$. Also let $\varphi(z)\notin\mathbb{C}[z]$ and $\{\varphi(z),\varphi^{(n)}(z)\}$ be algebraically dependent over $\mathbb{C}$. Then: a) either there exist numbers $a\in\mathbb{C}\setminus\{0\}$, $K\in\mathbb{N}\cup\{0\}$ and a polynomial $B\in\mathbb{C}[\omega]$ such that $\varphi(z)=e^{-Kaz}B(e^{az})$; b) or there exist a number $\alpha\in\mathbb{C}\setminus\{0\}$ and numbers $D_0,D_1,\dots,D_n\in\mathbb{C}$ such that $\varphi(z)=D_0+\sum_{j=1}^{n}D_j e^{\xi_jz}$ (where all $\xi_j$ are different and $\xi_j^n=\alpha$ for all $j$). Remark 2. If assertion b) of Lemma 3 holds, then the function $y=\varphi(z)$ satisfies the equation $y^{(n)}=\gamma_1y+\gamma_2$ for some $\gamma_1,\gamma_2\in\mathbb{C}$. Proof of Lemma 3. By $\mathbb{Z}$ we denote the modulus $\{\mathbb{Z}\sigma_1\,+\,\dotsb{}+\,\mathbb{Z}\sigma_M\}$. Let $\{\beta_1,\dots,\beta_r\}$ be its basis. Then the functions $\{z,e^{\beta_1z},\dots,e^{\beta_rz}\}$ are algebraically independent over $\mathbb{C}$ (see, for example, [10]). Let
$$
\begin{equation*}
\varphi(z)=\sum_{j_1,\dots,j_r=L_1}^{L_2} A_{\overline{j}} \, e^{(j_1\beta_1+\dots+j_r\beta_r)z}+q(z)
\end{equation*}
\notag
$$
for some $A_{\overline{j}}=A_{j_1\cdots j_r}\in\mathbb{C}$, $L_1,L_2\in\mathbb{Z}$ and $L_1\leqslant L_2$. Consider the polynomial Setting we have $Q(z,e^{\beta_1z},\dots,e^{\beta_rz})=\varphi^{(n)}(z)$. The functions $\{\varphi(z);\,\varphi^{(n)}(z)\}$ are algebraically dependent over $\mathbb{C}$ by the assumption. Hence (in view of the algebraic dependence of $\{z,e^{\beta_1z},\dots,e^{\beta_rz}\}$ over $\mathbb{C}$), $\{P(z,\theta_1,\dots,\theta_r);\, Q(z,\theta_1,\dots,\theta_r)\}$ are algebraically dependent over $\mathbb{C}$, that is, there exists an irreducible polynomial $R\in\mathbb{C}[\omega_1,\omega_2]$ such that $R(P,Q)\equiv 0$.
For $i=1,2$, consider the polynomial $A_i=\partial R(P,Q)/\partial\omega_i$. Differentiating the equality$R(P,Q)=0$, we get the system
$$
\begin{equation}
\begin{aligned} \, &A_1q'(z)+A_2q^{(n+1)}(z)=0, \\ &A_1\,\frac{\partial P}{\partial\theta_1}+A_2\,\frac{\partial Q}{\partial\theta_1}=0, \\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ &A_1\,\frac{\partial P}{\partial\theta_r}+A_2\,\frac{\partial Q}{\partial\theta_r}=0. \end{aligned}
\end{equation}
\tag{1}
$$
Note that at least one of the polynomials $A_1$, $A_2$ is not identically zero. Indeed, since $R$ is not a constant, it follows that either ${\partial R}/{\partial\omega_1}$, or ${\partial R}/{\partial\omega_2}$ is not zero. Suppose, for example, that ${\partial R}/{\partial\omega_1}\ne 0$. Considering the resultant $T(\omega_2)=\operatorname{Res}_{\omega_1}(R,{\partial R}/{\partial\omega_1})$, we find that $T\not\equiv 0$ and $T(Q(z,\theta_1,\dots,\theta_r))\equiv 0$. As a result, we have $Q(z,\theta_1,\dots,\theta_r)\equiv \gamma_0$ for some $\gamma_0\in\mathbb{C}$. But then $P(z,\theta_1,\dots,\theta_r)$ is a constant, which contradicts the assumptions of the lemma. By the assumption, ${\partial P}/{\partial\theta_l}\ne 0$ for all $l$. If $q(z)$ were not in $\mathbb{C}$, then from (1) we would have
$$
\begin{equation*}
A_1q'(z)+A_2q^{(n+1)}(z)=0,\qquad A_1\frac{\partial P}{\partial\theta_1}+A_2\frac{\partial Q}{\partial\theta_1}=0.
\end{equation*}
\notag
$$
Since $(A_1,A_2)\ne(0,0)$, we would have
$$
\begin{equation*}
\begin{vmatrix} q' & q^{(n+1)} \\ \dfrac{\partial P}{\partial\theta_1} &\dfrac{\partial Q}{\partial\theta_1} \end{vmatrix}=0,
\end{equation*}
\notag
$$
and hence,
$$
\begin{equation*}
\frac{q^{(n+1)}}{q'}=\frac{\partial P/\partial\theta_1}{\partial Q/\partial\theta_1}.
\end{equation*}
\notag
$$
By definition of the polynomial $Q$, we have $\partial Q/\partial\theta_1\ne 0$. In the last equality, the left-hand side depends only on $z$, and the (nonzero) right-hand side, only on $\theta_1,\dots,\theta_r$. Hence, there exists a constant $\gamma_1\in\mathbb{C}$ such that $\gamma_1\ne 0$ and $q^{(n+1)}(z)=\gamma_1 q'(z)$, which is impossible for $q(z)\in\mathbb{C}[z]\setminus\mathbb{C}$. Hence $q(z)\in\mathbb{C}$. So, the polynomials $P$, $Q$ may depend only on $\theta_1,\dots,\theta_r$ and are independent of $z$. In addition, $R(P,Q)\equiv 0$. If $r=1$, then the conclusion of the lemma clearly holds. Let $r\geqslant 2$. By Lemma 2, the surface $F=\{(y,s)\colon R(y,s)=0\}$ is birationally equivalent to the sphere. Hence there exists a rational function $t=G(y,s)$ which is a one-to-one mapping of $F$ onto the sphere $\overline{\mathbb{C}}$. Hence, for $t(z)=G(\varphi(z),\varphi^{(n)}(z))$, we have $\varphi(z)=H_1(t(z))$, $\varphi^{(n)}(z)=H_2(t(z))$ for some rational functions $H_1(\omega)$, $H_2(\omega)$. In view of the form of $\varphi(z)$, we have
$$
\begin{equation*}
t(z)=\frac{A(e^{\beta_1z},\dots,e^{\beta_rz})}{B(e^{\beta_1z},\dots,e^{\beta_rz})}
\end{equation*}
\notag
$$
for some relatively prime polynomials $A,B\in\mathbb{C}[\theta_1,\dots,\theta_r]$. In view of the algebraic independence of the functions $e^{\beta_1z},\dots,e^{\beta_rz}$ over $\mathbb{C}$), we have the equalities of the rational functions $P=H_1(A/B)$ and $Q=H_2(A/B)$ for some $H_1,H_2\in\mathbb{C}(\omega)$.
Let $P=P_1/\theta^{\overline{M}}$, where $\theta^{\overline{M}}=\theta_1^{m_1}\cdots \theta_r^{m_r}$ and all $m_i\in\mathbb{N}\cup\{0\}$ and $\{P_1,\theta^{\overline{M}}\}$ are relatively prime polynomials from $\mathbb{C}[\theta_1,\dots,\theta_r]$. In view of the relation between $\varphi(z)$ and $\varphi^{(n)}(z)$), we have $Q=Q_1/\theta^{\overline{M}}$ for relatively prime polynomials $\{Q_1,\theta^{\overline{M}}\}$. Hence
$$
\begin{equation}
\frac{P_1}{\theta^{\overline{M}}}= \frac{\prod_{j=1}^{K_1} (A/B -\xi_{1,j})^{s_{1,j}}}{\prod_{j=1}^{K_2} (A/B-\eta_{1,j})^{g_{1,j}}},\qquad \frac{Q_1}{\theta^{\overline{M}}}= \frac{\prod_{j=1}^{L_1} (A/B -\xi_{2,j})^{s_{2,j}}}{\prod_{j=1}^{L_2} (A/B-\eta_{2,j})^{g_{2,j}}}.
\end{equation}
\tag{2}
$$
In these equalities, all the numbers in each of the sets $\{\xi_{1,j}\}$, $\{\eta_{1,j}\}$, $\{\xi_{2,j}\}$, $\{\eta_{2,j}\}$ are distinct, $s_{l,j},g_{l,j}\in\mathbb{N}$, $\{\xi_{1,j}\}\cap \{\eta_{1,j}\}=\varnothing$, and $\{\xi_{2,j}\}\cap \{\eta_{2,j}\}=\varnothing$.
We first consider the first relation in (2). If its denominator contains at least one bracket (for example, $(A/B-\eta_{1,1})^{g_{1,1}}$), then we have $A/B-\eta_{1,1}=\theta^{\overline{N}}/B_1$ (since all the zeros of the denominator of the left-hand side are determined by the zeros of $\theta^{\overline{M}}$), where $\theta^{\overline{N}}=\theta_1^{N_1}\cdots \theta_r^{N_r}$ (with some $N_1,\dots,N_r\in\mathbb{N}\cup\{0\}$). Next, $B_1$ is a divisor of $B$ ($B_1$, $\theta^{\overline{N}}$ are relatively prime). But in this case, we can change the parameter, by taking $t=B_1/\theta^{\overline{N}}$ as a new parameter, and then consider equalities (2) with $t=B_1/\theta^{\overline{N}}$. If the denominator in the right-hand side (2) contains at least one bracket (for example, $B_1/\theta^{\overline{N}}-\gamma$), then all the zeros (counting multiplicity) of the polynomial $B_1-\gamma\theta^{\overline{N}}$ should lie among those of $\theta^{\overline{M}}$, that is, $B_1-\gamma\theta^{\overline{N}}$ is a divisor of the polynomial $\theta^{\overline{M}}$. Hence $B_1/\theta^{\overline{N}}=\gamma+\theta_1^{p_1}\cdots \theta_r^{p_r}$ for some $p_1\cdots p_r\in\mathbb{Z}$. But then $\varphi(z)$ is a rational function of $e^{az}$ (where $a=p_1\beta_1+\dots+p_r\beta_r$), which has the form required in the lemma, because $\varphi(z)$ is an entire function. So, it can be assumed that there is no denominator in the first equality in (2). As a result, we readily have $A/B=A_1/\theta^{\overline{q}}$ (for some $\theta^{\overline{q}}=\theta_1^{q_1}\cdots \theta_r^{q_r}$, $q_i\in\mathbb{N}\cup\{0\}$). If there is a denominator in the second equality in (2), then by proceeding as in the second equality in (2) and taking into account that if $\varphi^{(n)}(z)$ is a rational function $e^{az}$ for some $a\in\mathbb{C}\setminus\{0\}$, then $\varphi(z)$ has the form required in the lemma (here, we used the form of $\varphi(z)$ in the condition of the lemma), we obtain filially that the right-hand side of equalities (2) involve polynomials of $t=A_1(\theta_1,\dots,\theta_r)/\theta_1^{q_1}\cdots \theta_r^{q_r}$ (for relatively primes $A_1,\theta_1^{q_1}\cdots \theta_r^{q_r}$), that is, we have
$$
\begin{equation}
P\equiv \frac{P_1}{\theta^{\overline{M}}}=a_dt^d+\dots+a_0, \qquad Q\equiv \frac{Q_1}{\theta^{\overline{M}}}=b_{d_1}t^{d_1}+\dots+b_0,
\end{equation}
\tag{3}
$$
where $a_i,b_j\in\mathbb{C}$ for all $i$, $j$ and $a_d\ne 0$, $b_{d_1}\ne 0$. Moreover, in view of the relation between $P$ and $Q$, we have $d=d_1$.
Let $t=\sum_{l=1}^{D} \lambda_l I_l$, where for each $l$ $I_l=\theta_1^{l_1}\cdots \theta_r^{l_r}$ (all $l_i\in\mathbb{Z}$), all the monomials $\{I_l\}$ are distinct, and all $\lambda_l\ne 0$. If $d=1$, then from the first equality in (3) we have $P=a_1\bigl(\sum_{l=1}^{D} \lambda_l I_l\bigr)+ a_0$. But then $Q\,{=}\,a_1\bigl(\sum_{l=1}^{D} \lambda_l\delta_l^{n} I_l\bigr)$, where $I_l\,{=}\,\theta_1^{l_1}\cdots \theta_r^{l_r}$. Hence $\delta_l\,{=}\,l_1\beta_1+\dotsb+l_r\beta_r$. On the other hand, considering the second equality in (3), we get $Q=b_1\bigl(\sum_{l=1}^{D} \lambda_l I_l\bigr)+b_0$, which implies that all $\delta_l^{n}$ are the same. This proves assertion b) of Lemma 3 in this case. Assume now that $d\geqslant 2$. Let $\mathcal{L}\subset\mathbb{Z}^r$ be the set consisting of all the vectors $(l_1,\dots,l_r)$ such that the monomial $I_l=\theta_1^{l_1}\cdots \theta_r^{l_r}$ is contained among the monomials from $t=\sum_{l=1}^{D} \lambda_l I_l$. If $\mathcal{L}$ is a nonzero singleton, then by (3) the conclusion of Lemma 3 holds. Assume that $\mathcal{L}$ contains at least two nonzero elements. We choose some hyperplane $\pi$ defined by $a_1x_1+\dots+a_rx_r=T$ such that $T=\max_{\overline{x}\in\mathcal{L}}(a_1x_1+\dots+a_rx_r)$ (that is, all points from $\pi$ on the same side of the plane $\pi$) and $\pi$ contains two points $M_1$ and $M_2$ ($M_1\ne M_2$). Let $l$ be the straight line generated by $ M_1$ and $M_2$. Assume that $\mathcal{L}\cap l=(m_1,\dots,m_r)+t_k(n_1,\dots,n_r)$, $k=0,\dots,L$, $t_0=0$, $t_1\leqslant\dots\leqslant t_L$, and all $t_i\in\mathbb{N}$. We set $I_0\,{=}\,\theta_1^{m_1}\cdots \theta_r^{m_r}$, $I_1\,{=}\,\theta_1^{n_1}\cdots \theta_r^{n_r}$. Then $t\,{=}\,\sum_{k=0}^{L} \lambda_k I_0I_1^{t_k}+B$, where $B$ is a linear combination of the monomials $I_l\,{=}\,\theta_1^{l_1}\cdots \theta_r^{l_r}$ such that $a_1l_1+\dots+a_rl_r\,{<}\,T$. From (3) we have
$$
\begin{equation*}
P=a_d\biggl( \sum_{k=1}^{L} \lambda_k I_0I_1^{t_k}\biggr)^d+B_1,
\end{equation*}
\notag
$$
where $B_1$ is a linear combination of the monomials $I_l=\theta_1^{l_1}\cdots \theta_r^{l_r}$ such that $a_1l_1+\dots+a_rl_r<dT$. We have
$$
\begin{equation*}
P=a_dI_0^{d}(\gamma_0+\gamma_1 I_1^{t_1}+\dots+ \gamma_2 I_1^{dt_L})^d+B_1,
\end{equation*}
\notag
$$
where all $\gamma_i\ne 0$.
The expression $Q$ consists of the same monomials as $P$, but here if $P$ contains the term $\lambda e \theta_1^{l_1}\cdots \theta_r^{l_r}$, then $Q$ involves $\lambda_l \delta_l^{n} \theta_1^{l_1}\cdots\theta_r^{l_r}$ at the same place, where $\delta_l=l_1\beta_1+\dots+l_r\beta_r$. Hence
$$
\begin{equation*}
Q=a_dI_0^{d}\bigl(\gamma_0(d\delta_0)^{n}+\gamma_1(d\delta_0+t_1\delta_1)^n I_1^{t_1}+\dots+ \gamma_2(d\delta_0+dt_L \delta_L)^n I_1^{dt_L}\bigr)+C_1,
\end{equation*}
\notag
$$
where $C_1$ is the same linear combination of the monomials as $B_1$.
On the other hand, using the second of (3), we find that
$$
\begin{equation*}
Q=b_dI_0^{d}(\gamma_0+\gamma_1 I_1^{t_1}+\dots+ \gamma_2 I_1^{dt_L})^d+\widetilde{B}_1.
\end{equation*}
\notag
$$
Comparing the coefficients in leading terms in both equalities for $Q$ (and taking into account that $a_d\ne 0$, $b_d\ne 0$), we get
$$
\begin{equation*}
(d\delta_0+t_1\delta_1)^n =(d\beta_0)^n,\qquad (d\delta_0+dt_L\delta_1)^n =(d\beta_0)^n.
\end{equation*}
\notag
$$
Dividing both equalities by $(d\beta_0)^n$ and setting $\eta=\delta_1/\delta_0$, we find that
$$
\begin{equation*}
1+\frac{t_1}{d}\eta =\varepsilon_1, \qquad 1+t_L\eta =\varepsilon_2,
\end{equation*}
\notag
$$
where $\varepsilon_1$, $\varepsilon_2$ are the $n$th roots of 1. Since $\eta\ne 0$, we have $(\varepsilon_1-1)=(p/q)(\varepsilon_2-1)$ with $p/q=t_1/dt_L$. Next, $0<t_1\leqslant t_L$ and $d\geqslant 2$, and so $0<p/q<1$ (here we also assume that $p$ and $q$ are relatively prime).
Note that for $n= 2$, the equality = $(\varepsilon_1-1)=(p/q)(\varepsilon_2\,{-}\,1)$ is possible only for $\varepsilon_1=\varepsilon_2=1$ (since $a < p/q < 1$). Therefore, $\eta=0$, which is not true. Let $n\geqslant 3$. Let $K$ be the minimal extension of the field $\mathbb{Q}$ containing all $n$th roots of unity. Let $N_k(\alpha)$ denote the norm of a number $\alpha$ in the field $K$ (see, for example, [10]). We have $n\geqslant 3$, and hence, for at least one conjugate $\mathcal{J}(\varepsilon_2-1)=\mathcal{J}(\varepsilon_2)-1$ we have $|\mathcal{J}(\varepsilon_2-1)|<2$ ($\mathcal{J}$ is an embedding $K$ over $Q$). Moreover, if $\varepsilon$ is an $n$th root of unity, then $|\varepsilon-1|\leqslant 2$. Consider the equalities
$$
\begin{equation*}
N_k(\varepsilon_1-1)=N_k\biggl(\frac{p}{q}(\varepsilon_2-1)\biggr)\quad \Longleftrightarrow\quad \prod_{i=1}^{h} (\mathcal{J}_i(\varepsilon_1)-1)=\biggl(\frac{p}{q}\biggr)^h \prod_{i=1}^{h}(\mathcal{J}_i\varepsilon_2-1)
\end{equation*}
\notag
$$
and ($h$, $A$, $B$ are natural numbers). Since the fraction $p/q$ is irreducible and $0<p/q<1$, we have $q\geqslant 2$. On the other hand, $B<2^h$ due to estimates of the conjugates, which contradicts equality (4). This verifies the claim of Lemma 3 in the case $d\geqslant 2$.
Lemma 3 is proved. Lemma 4 (see [11], Lemma 3). Let $h(z)$ be an entire function of finite order $\rho$. Then, for each $\varepsilon>0$, there exist $R_0>0$ and $\delta>0$ such that, for any $R>R_0$ and $H>0$, the annulus $C_R=\{2R\leqslant |z|\leqslant 3R\}$ contains a finite family (of cardinality $\leqslant N_h(4R)$) $B_R$ of discs with the sum of the radii $\leqslant 2H$ so that, for any $z\in C_R\setminus B_R$, Corollary 1. Let $\varphi(z)$ be an entire function of finite order $\rho$. Then, for each $j\in\mathbb{N}$ and $\varepsilon>0$, there exist $R_j$ and $\delta_j$ such that, for any $R>R_j$ and $H>0$, the annulus $C_R=\{2R\leqslant|z|\leqslant 3R\}$ contains a finite family (of cardinality $\leqslant \sum_{l=0}^{j} N_{\varphi(l)}(4R)$) $B_{j,R}$ of discs with the sum of the radii $\leqslant 2jH$ so that, for each $z\in C_R\setminus B_{j,R}$, Proof. Note that for any $l\in\mathbb{N}$, $\varphi^{(l)}(z)$ is an entire function of order $\rho$ (see, for example, [7], Ch. 1). We apply Lemma 3 to each function $\varphi^{(l)}(z)$ for all $l=0,1,\dots,j-1$.
Let $B_{j,R}$ be the union of families of the exceptional discs for all $l=0,1,\dots,j-1$ and let $\delta_0,\dots,\delta_{j-1}$ be the corresponding constants from inequality (5). Then the sum of the radii of all discs from $B_{j,R}$ is at most $2jH$, the number of discs is at most $(N_f(3R)+N_{f'}(3R)+\dots+N_{f^{(j)}}(3R))$, and moreover, for all $z\in C_R\setminus B_{j,R}$,
$$
\begin{equation*}
\biggl|\frac{\varphi^{(j)}(z)}{\varphi(z)}\biggr|=\biggl|\frac{\varphi'(z)}{\varphi(z)}\biggr|\, \biggl|\frac{\varphi''(z)}{\varphi'(z)}\biggr|\cdots \biggl|\frac{\varphi^{(j)}(z)}{\varphi^{(j-1)}(z)}\biggr|\leqslant \delta_0\delta_1\cdots\delta_{j-1} \biggl( 1+R^{\rho+\varepsilon-1}+\frac{R^{\rho+\varepsilon}}{H}\biggr)^j.
\end{equation*}
\notag
$$
Now to complete the proof, it suffices to put $\delta_j=\delta_0\cdots\delta_{j-1}$. Corollary 2. Let $r\in\mathbb{N}$, $\varphi_1(z),\dots,\varphi_r(z)$ be entire functions of finite order $\rho$. Then, for each $s\in\mathbb{N}$ and $\varepsilon>0$, there exist $R_{s,r}$ and $\lambda_{s,r}$ such that, for any $R>R_{s,r}$ and $H>0$, the annulus $C_R=\{2R\leqslant|z|\leqslant 3R\}$ contains a finite family (of cardinality $\leqslant \sum_{k=1}^{r}\sum_{l=0}^{s} N_{\varphi_k^{(l)}}(4R)$) $B_{r,s,R}$ of discs with the sum of the radii $\leqslant 2rsH$ so that, for each $z\in C_R\setminus B_{r,s,R}$ for any $k=1,\dots,r$ and $j=1,\dots,s$, Remark 3. For a proof of Corollary 2 it suffices to apply Corollary 1 to each of the functions $\varphi_1(z),\dots,\varphi_r(z)$, and, by uniting the discs thrown away in each case, take the maximum of the right-hand sides of all inequalities of the form (6) as the right-hand side of the required inequality. Lemma 5. Let $N\in\mathbb{N}$, $R>0$, $H\in(0,R/2)$, $C_R=\{2R\leqslant|z|\leqslant 3R\}$, let $B_R$ be a union of $N$ closed discs with the sum of the radii smaller than $H$, and $B_R\subset C_R$. Then there exists a number $R_1\in[2R,3R]$ such that $D_R\subset C_R$ and $D_R\cap B_R=\varnothing$, where $D_R$ is the annulus Proof. Let $B_R=\bigcup_{j=1}^{N} A_j$, where $A_j=\{|z-z_j|\leqslant\rho_j\}$ and $|z_1|\leqslant\dots\leqslant |z_N|$. Since $B_R\subset C_R$, we find that $|z_j|\pm \rho_j\subset [2R,3R]$ for all $j=1,\dots,N$. We also note that, for all $j$, $A_j\subset K_j$, where $K_j=\{z\colon |z_j|-\rho_j\leqslant |z|\leqslant |z_j|+\rho_j\}$, and hence, $B_R\subset \bigcup_{j=1}^{N} K_j$.
For each $j=1,\dots,N$, let $\delta_j:=[|z_j|-\rho_j,\, |z_j|+\rho_j]$. Let $\Delta=\bigcup_{j=1}^{N} \delta_j$. Then $\Delta\subset [2R,3R]$. Setting $E=[2R,3R]\setminus \Delta$, we have $E=\bigcup_{k=1}^{M} (\alpha_k,\beta_k)$, where $M\leqslant 2N+2$, $2R\leqslant\alpha_k<\beta_k\leqslant 3R$ for all $k$ and $(\alpha_m,\beta_m)\cap(\alpha_n,\beta_n)=\varnothing$ for all $m\ne n$. For the length of the set $E$, we have the estimate
$$
\begin{equation*}
l(E)\geqslant R-\sum_{j=1}^{N} l(\delta_j)\geqslant R-2\sum_{j=1}^{N} \rho_j\geqslant R-2H>0.
\end{equation*}
\notag
$$
Since $E$ is composed of at most $2N+2$ intervals, there is an interval $(\alpha_{n_0},\beta_{n_0})\subset E$ such that $\beta_{n_0}-\alpha_{n_0}>(R-2H)/(2N+2)$.
We set $R_1=(\alpha_{n_0}+\beta_{n_0})/2$. Then the annulus $D_R=\{ R_1-(R-2H)/(4N+4)\leqslant |z|\leqslant R_1+(R-2H)/(4N+4)\}$ is such that $D_R\subset C_R$ and $D_R\cap \bigl(\bigcup_{j=1}^{N} K_j\bigr)=\varnothing$. Hence $D_R\cap B_R=\varnothing$. Lemma 5 is proved. § 3. Proof of the theoremBy the assumption, where $P_d=\prod_{j=1}^{d} (\omega_2-\alpha_j\omega_1)$, $\deg Q_d<d$.Next, by $\gamma_i$, $i=1,2,\dots$, we will denote positive constants depending only on the function $f(z)$ and the polynomial $P$ (and independent of the numbers $R$ that will be defined below). Let $\rho$ be the order of $f(z)$. Setting $K=1000(nd)^{2}$, $\rho_1=1+\rho$, we have $\rho_1\geqslant 1$. By $\mathcal{L}$ we denote the set of functions
$$
\begin{equation*}
\mathcal{L}=\{f(z);\, f^{(n)}(z)-\alpha_1f(z);\,\dots;\, f^{(n)}(z)-\alpha_d f(z)\}.
\end{equation*}
\notag
$$
Since any function from the set $\mathcal{L}$ is of order $\rho$, there exists $\gamma_1$ such that, for any $R>\gamma_1$, any $l=0,1,\dots,K$, and any $\varphi(z)\in\mathcal{L}$,
$$
\begin{equation}
M_{\varphi^{(l)}}(4R)\leqslant e^{R^{\rho_1}}, \qquad N_{\varphi^{(l)}}(4R)\leqslant R^{\rho_1}.
\end{equation}
\tag{8}
$$
Let $R$ be a sufficiently large number. By Corollary 2 to Lemma 4 to the family of functions from $\mathcal{L}$, there exists a finite family of discs $G_R$ (in view of (8), this family contains at most $\gamma_3 R^{\rho_1+1/2}$ discs) such that a) $G_R\subset \{2R\leqslant|z|\leqslant 3R\}$; b) the sum of the radii of all discs from $G_R$ is at most $\gamma_4 R^{1/2}$; c) for each $z\in\{2R\leqslant |z|\leqslant 3R\}\setminus G_R$, for any function $\varphi(z)\in\mathcal{L}$ and each $l=1,\dots,3nd$,
$$
\begin{equation}
\biggl| \frac{\varphi^{(l)}(z)}{\varphi(z)}\biggr| \leqslant \gamma_5 R^{5nd\rho_1}.
\end{equation}
\tag{9}
$$
Applying Lemma 5 and using estimates (8), it follows that there exist numbers $R_1\subset (2R,3R)$ and $\delta\geqslant \gamma_6R^{-\rho_1+1/2}$ such that, for the ring $D_R=\{R_1-\delta\leqslant |z|\leqslant R_1+\delta\}$, we have a) $D_R\subset \{2R\leqslant |z|\leqslant 3R\}$; b) $D_R\cap G_R=\varnothing$. But then by (9), for each $z\in D_R$, for any function $\varphi(z)\in\mathcal{L}$, for each $l=1,\dots,3nd$, we have
$$
\begin{equation}
\biggl| \frac{\varphi^{(l)}(z)}{\varphi(z)}\biggr| \leqslant \gamma_5 R^{5nd\rho_1}.
\end{equation}
\tag{10}
$$
Hence from (7) for each $z\in D_R$, we have, in view of (10),
$$
\begin{equation*}
|f(z)|^d \prod_{j=1}^{d} \biggl| \frac{f^{(n)}(z)}{f(z)}-\alpha_j\biggr| \leqslant \gamma_7 (1+|f(z)|)^{d-1} R^{2K\rho_1(d-1)}.
\end{equation*}
\notag
$$
Let $z_0\in D_R$ and $|f(z)|>R^{2,5K\rho_1d}$. Then
$$
\begin{equation}
\prod_{j=1}^{d} \biggl| \frac{f^{(n)}(z_0)}{f(z_0)}-\alpha_j\biggr| \leqslant \gamma_8 \frac{R^{2K\rho_1(d-1)}}{|f(z)|}.
\end{equation}
\tag{11}
$$
By the hypotheses of the theorem, $\alpha_i\ne\alpha_j$ for $i\ne j$. Let $j_0$ be such that
$$
\begin{equation*}
\min_{j=1,\dots,d} \biggl| \frac{f^{(n)}(z_0)}{f(z_0)}-\alpha_j\biggr|= \biggl| \frac{f^{(n)}(z_0)}{f(z_0)}-\alpha_{j_0}\biggr|.
\end{equation*}
\notag
$$
For any $j\ne j_0$, we have
$$
\begin{equation*}
\begin{aligned} \, \biggl| \frac{f^{(n)}(z_0)}{f(z_0)}-\alpha_j\biggr|&=\biggl| \frac{f^{(n)}(z_0)}{f(z_0)}-\alpha_{j_0}+\alpha_{j_0}-\alpha_j\biggr|\geqslant |\alpha_{j_0}-\alpha_{j}|- \biggl| \frac{f^{(n)}(z_0)}{f(z_0)}-\alpha_{j_0}\biggr| \\ &\geqslant |\alpha_{j_0}-\alpha_{j}|-\gamma_8 \biggl(\frac{R^{2K\rho_1(d-1)}}{R^{2,5K\rho_1d}}\biggr)^{1/d}\geqslant \gamma_9>0. \end{aligned}
\end{equation*}
\notag
$$
Now an application of (11) gives
$$
\begin{equation*}
\biggl| \frac{f^{(n)}(z_0)}{f(z_0)}-\alpha_{j_0}\biggr|\leqslant \gamma_5 \frac{1}{\gamma_9^{d-1}} \, \frac{R^{2K\rho_1(d-1)}}{|f(z_0)|}
\end{equation*}
\notag
$$
which implies
$$
\begin{equation*}
|f^{(n)}(z_0)-\alpha_{j_0}f(z_0)|\leqslant \gamma_{10} R^{2K\rho_1(d-1)}.
\end{equation*}
\notag
$$
If $z_0\in D_R$ is such that $|f(z_0)|\leqslant R^{2,5K\rho_1d}$, then by from (10) we get
$$
\begin{equation*}
|f^{(n)}(z_0)-\alpha_{j_0}f(z_0)|\leqslant R^{2,6K\rho_1d}.
\end{equation*}
\notag
$$
Hence, for each point $z\in D_R$, there exists a number $i(z)\in\{1,\dots,d\}$ such that
$$
\begin{equation}
|f^{(n)}(z)-\alpha_{i(z)}f(z)|\leqslant \gamma_{11} R^{3K\rho_1d}.
\end{equation}
\tag{12}
$$
Therefore, by (10), for all $l=1,\dots,2nd$,
$$
\begin{equation}
|f^{(n+l)}(z)-\alpha_{i(z)}f^{(l)}(z)|\leqslant \gamma_{12} R^{5K\rho_1d}.
\end{equation}
\tag{13}
$$
We claim that, for all $s=1,\dots,d$,
$$
\begin{equation}
|f^{(ns)}(z)-\alpha_{i(z)}^s f(z)|\leqslant (1+|\alpha_{i(z)}|)^s \gamma_{12} R^{5K\rho_1d}.
\end{equation}
\tag{14}
$$
We verify (14) by induction. 1) For $s=1$, inequality (14) is secured by (12). 2) If the inequality holds for (14) $s=t$, then, for $s=t+1$, using (13) we have
$$
\begin{equation*}
\begin{aligned} \, &|f^{(n(t+1))}(z)\,{-}\,\alpha_{i(z)}^{t+1}f(z)|\leqslant |f^{(n(t+1))}(z)\,{-}\,\alpha_{i(z)}f^{(nt)}(z)|\,{+}\, |\alpha_{i(z)}f^{(nt)}(z)\,{-}\,\alpha_{i(z)}^{t+1}f(z)| \\ &\qquad\leqslant \gamma_{12}R^{5K\rho_1d}+|\alpha_{i(z)}|(1+|\alpha_{i(z)}|)^t \gamma_{12}R^{5K\rho_1d} \leqslant (1+|\alpha_{i(z)}|)^{t+1}\gamma_{12}R^{5K\rho_1d}. \end{aligned}
\end{equation*}
\notag
$$
This proves (14). So, for each $z\in D_R$, there exists a number $i(z)\in\{1,\dots,d\}$ such that, for all $s=1,\dots,d$,
$$
\begin{equation}
|f^{(ns)}(z)-\alpha_{i(z)}^s f(z)|\leqslant \gamma_{13} R^{5K\rho_1d}.
\end{equation}
\tag{15}
$$
Let $Q(t)=\prod_{j=1}^{d} (t-\alpha_j)=t^d+a_{d-1}t^{d-1}+\dots+a_0$. We set
$$
\begin{equation*}
L(f)=f^{(dn)}(z)+a_{d-1}f^{((d-1)n)}(z)+\dots+ a_1f^{(n)}(z)+a_0f(z).
\end{equation*}
\notag
$$
Hence, using (14) and since $Q(\alpha_{i(z)})=0$ for all $i(z)$, we have, for any $z\in D_R$,
$$
\begin{equation*}
\begin{aligned} \, &|L(f(z))|=|L(f(z))-Q(\alpha_{i(z)})f(z)|\leqslant |f^{(dn)}(z)-\alpha_{i(z)}^d f(z)| \\ &\qquad+|a_{d-1}|\, |f^{((d-1)n)}(z)-\alpha_{i(z)}^{d-1}f(z)|+\dots+|a_0|\, |f(z)-f(z)| \leqslant \gamma_{14} R^{5K\rho_1d}. \end{aligned}
\end{equation*}
\notag
$$
Therefore, for each sufficiently large $R$, for the entire function $L(f(z))$, we have the estimate
$$
\begin{equation*}
\max_{|z|=R} |L(f(z))|\leqslant \max_{z\in D_R} |L(f(z))|\leqslant \gamma_{15} R^{5K\rho_1d}.
\end{equation*}
\notag
$$
But then by Liouville’s theorem (see, for example, [7], Ch. 1), this function is a polynomial, that is, $L(f(z))=q(z)$ for some $q_1\in\mathbb{C}[z]$. We set $y=f(z)$. Then, for $y$, we have the linear differential equation
$$
\begin{equation*}
y^{(dn)}+a_{d-1}y^{((d-1)n)}+\dots+a_1y^{(n)}+a_0y=q_1(z).
\end{equation*}
\notag
$$
The characteristic polynomial of this differential equation has the roots $\{\sqrt[n]{\alpha_j}\}$. But then $f(z)=\sum_{j=1}^{M} B_j e^{\delta_jz}+q(z)$, where $B_j\in\mathbb{C}$, $q(z)\in\mathbb{C}[z]$, and each $\delta_j$ is an $n$th root of some $\alpha_m$. The proof of the theorem concludes by application of Lemma 3 and the fact that degree of the irreducible polynomial $P$ is at least two. The author is deeply grateful to A. E. Eremenko for many valuable remarks. 
Citation in format AMSBIB
\Bibitem{Yan22} |
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