RETRACTED ARTICLE: Growth properties at infinity for solutions of modified Laplace equations

Abstract

Let $\mathscr{F}$ be a family of solutions of Laplace equations in a domain D and for each $f\in\mathscr{F}$, f has only zeros of multiplicity at least k. Let n be a positive integer and such that $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$. Let a be a complex number such that $a\neq0$. If for each pair of functions f and g in $\mathscr{F}$, $f^{n}f^{(k)}$ and $g^{n}g^{(k)}$ share a value in D, then $\mathscr{F}$ is normal in D.

1 Introduction

Let D be a domain in $\mathbb{C}$. Let $\mathscr{F}$ be a solution of certain Laplace equations defined in the domain D. $\mathscr{F}$ is said to be normal in D, in the sense of Montel, if for any sequence $\{f_{n}\}\subset\mathscr{F}$, there exists a subsequence $\{f_{n_{j}}\}$ such that $f_{n_{j}}$ converges spherically locally uniformly in D to a meromorphic function or ∞.

Let $g(z)$ be a solution of certain Laplace equations and a be a finite complex number. If $f(z)$ and $g(z)$ have the same zeros, then we say that they share a IM (ignoring multiplicity) (see [1]).

In 1998, Wang and Fang [2] proved the following result.

Theorem A

Let f be a transcendental meromorphic function in the complex plane. Let n and k be two positive integers such that $n\geq k+1$, then $(f^{n})^{(k)}$ assumes every finite non-zero value infinitely often.

Corresponding to Theorem A, there are the following theorems about normal families in [3].

Theorem B

Let $\mathscr{F}$ be a family of meromorphic functions in D, n, k be two positive integers such that $n\geq k+3$. If $(f^{n})^{(k)}\neq1$ for each function $f\in\mathscr{F}$, then $\mathscr{F}$ is normal in D.

Recently, corresponding to Theorem B, Yang [4] proved the following result.

Theorem C

Let $\mathscr{F}$ be a family of meromorphic functions in D. Let n, k be two positive integers such that $n\geq k+2$. Let $a\neq0$ be a finite complex number. If $(f^{n})^{(k)}$ and $(g^{n})^{(k)}$ share a in D for each pair of functions f and g in $\mathscr{F}$, then $\mathscr{F}$ is normal in D.

Recently, Zhang and Li [5] proved the following theorem.

Theorem D

Let f be a transcendental meromorphic function in the complex plane. Let k be a positive integer. Let $L[f]=a_{k}f^{(k)}+a_{k-1}f^{(k-1)}+\cdots+a_{0}f$, where $a_{0}, a_{1},\ldots, a_{k}$ are small functions and $a_{j}$ (≢0) ($j=1,2,\ldots,k$). For $c\neq0, \infty$, let $F=f^{n}L[f]-c$, where n is a positive integer. Then, for $n\geq2$, $F=f^{n}L[f]-c$ has infinitely many zeros.

From Theorem D, we immediately obtain the following result.

Corollary D

Let f be a transcendental meromorphic function in the complex plane. Let c be a finite complex number such that $c\neq0$. Let n, k be two positive integers. Then, for $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$, $f^{n}f^{(k)}-c$ has infinitely many zeros.

It is natural to ask whether Corollary D can be improved by the idea of sharing values similarly with Theorem C. In this paper we investigate the problem and obtain the following result.

Theorem 1

Let $\mathscr{F}$ be a family of meromorphic functions in D. Let n, k be two positive integers such that $n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$. Let a be a complex number such that $a\neq0$. For each $f\in\mathscr{F}$, f has only zeros of multiplicity at least k. If $f^{n}f^{(k)}$ and $g^{n}g^{(k)}$ share a in D for every pair of functions $f, g\in\mathscr{F}$, then $\mathscr{F}$ is normal in D.

Remark 1

From Theorem 1, it is easy to see $\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}\geq2$ for any positive integer k.

Example 1

Let $D=\{z:|z|<1\}$, $n, k\in N$ with $n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$ and n be a positive integer; for $k=2$, let

$$\mathscr{F}=\bigl\{ f_{m}(z)=mz^{k-1}, z\in D, m=1,2,\ldots \bigr\} . $$

Obviously, for any functions $f_{m}$ and $g_{m}$ in $\mathscr{F}$, we have $f_{m}^{n}f^{(k)}_{m}=0$, obviously $f_{m}^{n}f^{(k)}_{m}$ and $g_{m}^{n}g^{(k)}_{m}$ share any $a\neq0$ in D. But $\mathscr{F}$ is not normal in D.

Example 2

Let $D=\{z:|z|<1\}$, $n, k\in N$ with $n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$ and n is a positive integer, and let

$$\mathscr{F}=\bigl\{ f_{m}(z)=e^{mz}, z\in D, m=1,2,\ldots \bigr\} . $$

Obviously, for any $f_{m}$ and $g_{m}$ in $\mathscr{F}$, we have $f_{m}^{n}f^{(k)}_{m}=m^{k}e^{(mn+m)z}$, obviously $f_{m}^{n}f^{(k)}_{m}$ and $g_{m}^{n}g^{(k)}_{m}$ share 0 in D. But $\mathscr{F}$ is not normal in D.

Example 3

Let $D=\{z:|z|<1\}$, $n, k\in N$ with $n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$, and n be a positive integer, let

$$\mathscr{F}=\biggl\{ f_{m}(z)=\sqrt{m}\biggl(z+\frac{1}{m} \biggr), z\in D, m=1,2,\ldots\biggr\} . $$

For functions $f_{m}$ and $g_{m}$ in $\mathscr{F}$, we have $f_{m}f'_{m}=mz+1$. Obviously $f_{m}f'_{m}$ and $g_{m}g'_{m}$ share 1 in D. But $\mathscr{F}$ is not normal in D.

Remark 2

Example 1 shows that the condition that f has only zeros of multiplicity at least k is necessary in Theorem 1. Example 2 shows that the condition $a\neq0$ in Theorem 1 is inevitable. Example 3 shows that Theorem 1 is not true for $n=1$.

2 Lemmas

In order to prove our theorem, we need the following lemmas.

Lemma 2.1

(Zalcman’s lemma, see [6])

Let $\mathcal{F}$ be a family of meromorphic functions in the unit disc △ with the property that, for each $f\in\mathcal{F}$, all zeros of multiplicity are at least k. Suppose that there exists a number $A\geq1$ such that $|f^{(k)}(z)|\leq A$ whenever $f\in\mathcal{F}$ and $f=0$. If $\mathcal{F}$ is not normal in Δ, then for $0\leq\alpha\leq k$, there exist:

• 1. a number $r\in(0,1)$;

• 2. a sequence of complex numbers $z_{n}$, $|z_{n}|< r$;

• 3. a sequence of functions $f_{n}\in\mathcal{F}$;

• 4. a sequence of positive numbers $\rho_{n}\rightarrow0^{+}$

such that $g_{n}(\xi)=\rho_{n}^{-\alpha}f_{n}(z_{n}+\rho_{n}\xi)$ locally uniformly converges (with respect to the spherical metric) to a non-constant meromorphic function $g(\xi)$ on $\mathbb{C}$, and, moreover, the zeros of $g(\xi)$ are of multiplicity at least k, $g^{\sharp}(\xi)\leq g^{\sharp}(0)=kA+1$. In particular, g has order at most 2.

Lemma 2.2

Let n, k be two positive integers such that $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$, and let $a\neq0$ be a finite complex number. If f is a rational but not a polynomial meromorphic function and f has only zeros of multiplicity at least k, then $f^{n}f^{(k)}-a$ has at least two distinct zeros.

Proof

If $f^{n}f^{(k)}-a$ has zeros and has exactly one zero.

We set

$$ f=\frac{A(z-\alpha_{1})^{m_{1}}(z-\alpha_{2})^{m_{2}}\cdots (z-\alpha_{s})^{m_{s}}}{(z-\beta_{1})^{n_{1}}(z-\beta_{2})^{n_{2}}\cdots (z-\beta_{t})^{n_{t}}}, $$

(2.1)

where A is a non-zero constant. Because the zeros of f are at least k, we obtain $m_{i}\geq k$ ($i=1,2,\ldots,s$), $n_{j}\geq 1$ ($j=1,2,\ldots,t$).

For simplicity, we denote

$$\begin{aligned}& m_{1}+m_{2}+\cdots+m_{s}=m\geq ks, \end{aligned}$$

(2.2)

$$\begin{aligned}& n_{1}+n_{2}+\cdots+n_{t}=n\geq t. \end{aligned}$$

(2.3)

From (2.1), we obtain

$$ f^{(k)}=\frac{(z-\alpha_{1})^{m_{1}-k}(z-\alpha_{2})^{m_{2}-k}\cdots (z-\alpha_{s})^{m_{s}-k}g(z)}{(z-\beta_{1})^{n_{1}+k}(z-\beta_{2})^{n_{2}+k}\cdots (z-\beta_{t})^{n_{t}+k}}, $$

(2.4)

where g is a polynomial of degree at most $k(s+t-1)$.

From (2.1) and (2.4), we obtain

$$ f^{n}f^{(k)}=\frac{A^{n}(z-\alpha_{1})^{M_{1}}(z-\alpha_{2})^{M_{2}}\cdots (z-\alpha_{s})^{M_{s}}g(z)}{(z-\beta_{1})^{N_{1}}(z-\beta_{2})^{N_{2}}\cdots (z-\beta_{t})^{N_{t}}}=\frac{p}{q}. $$

(2.5)

Here p and q are polynomials of degree M and N, respectively. Also p and q have no common factor, where $M_{i}=(n+1)m_{i}-k$ and $N_{j}=(n+1)n_{j}+k$. By (2.2) and (2.3), we deduce $M_{i}=(n+1)m_{i}-k\geq k(n+1)-k=nk$, $N_{j}=(n+1)n_{j}+k\geq n+k+1$. For simplicity, we denote

$$\begin{aligned}& \deg P=M=\sum_{i=1}^{s} M_{i}+ \deg(g)\geq nks+ k(s+t-1) \\ & \hphantom{\deg P}=(nks+ks)+k(t-1)\geq(nk+k)s, \end{aligned}$$

(2.6)

$$\begin{aligned}& \deg q=N=\sum_{j=1}^{t} N_{j} \geq(k+1+n)t. \end{aligned}$$

(2.7)

Since $f^{n}f^{(k)}-a=0$ has just a unique zero $z_{0}$, from (2.5) we obtain

$$ f^{n}f^{(k)}=a+\frac{B(z-z_{0})^{l}}{(z-\beta_{1})^{N_{1}}(z-\beta_{2})^{N_{2}}\cdots (z-\beta_{t})^{N_{t}}}=\frac{p}{q}. $$

(2.8)

By $a\neq0$, we obtain $z_{0}\neq\alpha_{i}$ ($i=1,\ldots,s$), where B is a non-zero constant.

From (2.5), we obtain

$$ \bigl[f^{n}f^{(k)}\bigr]'=\frac{(z-\alpha_{1})^{M_{1}-1} (z-\alpha_{2})^{M_{2}-1}\cdots(z-\alpha_{s})^{M_{s}-1}g_{1}(\xi)}{(z-\beta _{1})^{N_{1}+1}\cdots(z-\beta_{t})^{N_{t}+1}}, $$

(2.9)

where $g_{1}(\xi)$ is a polynomial of degree at most $(k+1)(s+t-1)$.

From (2.8), we obtain

$$ \bigl[f^{n}f^{(k)}\bigr]'=\frac{(z-z_{0})^{l-1}g_{2}(z)}{(z-\beta_{1})^{N_{1}+1}\cdots +(z-\beta_{t})^{N_{t}+1}}, $$

(2.10)

where $g_{2}(\xi)=B(l-N)z^{t}+B_{1}z^{t-1}+\cdots+B_{t}$ is a polynomial ($B_{1},\ldots, B_{t}$ are constants).

Now we distinguish two cases.

Case 1. If $l\neq N$, by (2.8), then we obtain $\deg p\geq\deg q$. So $M\geq N$. By (2.9) and (2.10), we obtain $\sum_{i=1}^{s}(M_{i}-1)\leq\deg g_{2}=t$. So $M-s-\deg(g)\leq t$, and $M\leq s+t+\deg(g)\leq(k+1)(s+t)-k<(k+1)(s+t)$. By (2.6) and (2.7), we obtain

$$ M < (k+1) (s+t)\leq(k+1)\biggl[\frac{M}{nk+k}+\frac{N}{n+k+1}\biggr] \leq (k+1)\biggl[\frac{1}{nk+k}+\frac{1}{n+k+1}\biggr]M. $$

By $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$, we deduce $M< M$, which is impossible.

Case 2. If $l= N$, then we distinguish two subcases.

Subcase 2.1. If $M\geq N$, by (2.9) and (2.10), we obtain $\sum_{i=1}^{s}(M_{i}-1)\leq\deg g_{2}=t$. So $M-s-\deg(g)\leq t$, and $M\leq s+t+\deg(g)\leq(k+1)(s+t)-k<(k+1)(s+t)$, then we can proceed similarly to Case 1. This is impossible.

Subcase 2.2. If $M< N$, by (2.9) and (2.10), we obtain $l-1\leq\deg g_{1}\leq(s+t-1)(k+1)$, and then

$$\begin{aligned} N=l \leq&\deg g_{1}+1\leq(k+1) (s+t)-k< (k+1) (s+t) \\ \leq&(k+1)\biggl[\frac{1}{nk+k}+\frac{1}{n+k+1}\biggr]N\leq N. \end{aligned}$$

By $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$, we deduce $N< N$. This is impossible.

If $f^{n}f^{(k)}-a\neq0$ and we know f is rational but not a polynomial, then $f^{n}f^{(k)}$ also is rational but not a polynomial. At this moment, $l=0$ for (2.8), and proceeding as in Case 1, we have a contradiction.

Lemma 2.2 is proved. □

3 Proof of Theorem 1

We may assume that $D=\{|z|<1\}$. Suppose that $\mathscr{F}$ is not normal in D. Without loss of generality, we assume that $\mathscr{F}$ is not normal at $z_{0}=0$. Then, by Lemma 2.1, there exist:

• 1. a number $r\in(0,1)$;

• 2. a sequence of complex numbers $z_{j}$, $z_{j}\rightarrow0$ ($j\rightarrow\infty$);

• 3. a sequence of functions $f_{j}\in\mathcal{F}$;

• 4. a sequence of positive numbers $\rho_{j}\rightarrow0^{+}$

such that $g_{j}(\xi)=\rho_{j}^{-\frac{k}{n+1}}f_{j}(z_{j}+\rho_{j}\xi)$ converges uniformly with respect to the spherical metric to a non-constant meromorphic function $g(\xi)$ in C. Moreover, $g(\xi)$ is of order at most 2.

By Hurwitz’s theorem, the zeros of $g(\xi)$ are at least k multiple.

On every compact subset of $\mathbb{C}$ which contains no poles of g, we see that

$$ f^{n}_{j}(z_{j}+\rho_{j} \xi)f^{(k)}_{j}(z_{j}+\rho_{j} \xi)-a=g_{j}^{n}(\xi) \bigl(g^{(k)}_{j}( \xi )\bigr)-a $$

(3.1)

converges uniformly with respect to the spherical metric to $g^{n}(\xi)(g^{(k)}(\xi))-a$.

If $g^{n}(\xi)(g^{(k)}(\xi))\equiv a$ ($a\neq0$) and g has only zeros of multiplicity at least k, then g has no zeros. From the $g^{n}g^{(k)}$ having no zeros and the $g^{n}(\xi)(g^{(k)}(\xi))\equiv a$, we know g has no poles. Because the $g(\xi)$ is a non-constant meromorphic function in $\mathbb{C}$ and g has order at most 2. We obtain $g(\xi)=e^{d\xi^{2}+h\xi+c}$, where d, h, c are constants and $dh\neq0$. So $g^{n}(\xi)(g^{(k)}(\xi))\not\equiv a$, which is a contradiction.

When $g^{n}(\xi)(g^{(k)}(\xi))-a\neq0$, ($a\neq0$), we distinguish three cases.

Case 1. If g is a transcendental meromorphic function, by Corollary D, this is a contradiction.

Case 2. If g is a polynomial and the zeros of $g(\xi)$ are at least k multiple, and $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$, then $g^{n}(\xi)(g^{(k)}(\xi))-a=0$ must have zeros, which is a contradiction.

Case 3. If g is a non-polynomial ration function, by Lemma 2.2, this is a contradiction.

Next we will prove that $g^{n}g^{(k)}-a$ has just a unique zero. To the contrary, let $\xi_{0}$ and $\xi_{0}^{\ast}$ be two distinct solutions of $g^{n}g^{(k)}-a$, and choose δ (>0) small enough such that $D(\xi_{0}, \delta)\cap D(\xi_{0}^{\ast}, \delta)=\emptyset$ where $D(\xi_{0}, \delta)=\{\xi:|\xi-\xi_{0}|<\delta\}$ and $D(\xi_{0}^{\ast}, \delta)=\{\xi:|\xi-\xi_{0}^{*}|<\delta\}$. From (3.1), by Hurwitz’s theorem, there exist points $\xi_{j}\in D(\xi_{0}, \delta)$, $\xi_{j}^{*}\in D(\xi_{0}^{*}, \delta)$ such that for sufficiently large j,

$$\begin{aligned}& f^{n}_{j}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{j}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0, \\& f^{n}_{j}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{j}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0. \end{aligned}$$

By the hypothesis that for each pair of functions f and g in $\mathscr{F}$, $f^{n}f^{(k)}$ and $g^{n}g^{(k)}$ share a in D, we know that for any positive integer m

$$\begin{aligned}& f^{n}_{m}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{m}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0, \\& f^{n}_{m}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{m}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0. \end{aligned}$$

Fix m, take $j\rightarrow\infty$, and note $z_{j}+\rho_{j}\xi_{j}\rightarrow0$, $z_{j}+\rho_{j}\xi_{j}^{*}\rightarrow0$, then

$$f_{m}^{n}(0) \bigl(f^{(k)}_{m}(0) \bigr)-a=0. $$

Since the zeros of $f_{m}^{n}(0)(f^{(k)}_{m}(0))-a$ have no accumulation point, so $z_{j}+\rho_{j}\xi_{j}= 0$, $z_{j}+ \rho_{j}\xi_{j}^{*}= 0$.

Hence

$$\xi_{j}=-\frac{z_{j}}{\rho_{j}},\quad\quad \xi_{j}^{*}=- \frac{z_{j}}{\rho_{j}}. $$

This contradicts with $\xi_{j}\in D(\xi_{0},\delta)$, $\xi_{j}^{*}\in D(\xi_{0}^{*},\delta)$, and $D(\xi_{0},\delta)\cap D(\xi_{0}^{*},\delta)=\emptyset$. So $g^{n}g^{(k)}-a$ has just a unique zero, which can be denoted by $\xi_{0}$.

From the above, we know $g^{n}g^{(k)}-a$ has just a unique zero. If g is a transcendental meromorphic function, by Corollary D, then $g^{n}g^{(k)}-a= 0$ has infinitely many solutions, which is a contradiction.

From the above, we know $g^{n}g^{(k)}-a$ has just a unique zero. If g is a polynomial, then we set $g^{n}g^{(k)}-a=K(z-z_{0})^{l}$, where K is a non-zero constant, l is a positive integer. Because the zeros of $g(\xi)$ are at least k multiple, and $n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$, we obtain $l\geq3$. Then $[g^{n}g^{(k)}]'=Kl(z-z_{0})^{l-1}$ ($l-1\geq2$). But $[g^{n}g^{(k)}]'$ has exactly one zero, so $g^{n}g^{(k)}$ has the same zero $z_{0}$ too. Hence $g^{n}g^{(k)}(z_{0})=0$, which contradicts $g^{n}g^{(k)}(z_{0})=a\neq0$.

If g is a rational function but not a polynomial, by Lemma 2.2, then $g^{n}g^{(k)}-a=0$ at least has two distinct zeros, which is a contradiction.

Theorem 1 is proved.

4 Discussion

In 2013, Yang and Nevo [4] has proved the following.

Theorem E

Let $\mathscr{F}$ be a family of meromorphic functions in D, n be a positive integer and a, b be two constants such that $a\neq0,\infty$ and $b\neq\infty$. If $n\geq3$ and for each function $f\in\mathscr{F}$, $f'-af^{n}\neq b$, then $\mathscr{F}$ is normal in D.

Recently, Zhang improved Theorem E by the idea of shared values. Meanwhile, Zhang [7] has proved the following.

Theorem F

Let $\mathscr{F}$ be a family of meromorphic functions in D, n be a positive integer and a, b be two constants such that $a\neq0,\infty$ and $b\neq\infty$. If $n\geq4$ and for each pair of functions f and g in $\mathscr{F}$, $f'-af^{n}$ and $g'-ag^{n}$ share the value b, then $\mathscr{F}$ is normal in D.

By Theorem 1, we immediately obtain the following result.

Corollary 1

Let $\mathscr{F}$ be a family of meromorphic functions in a domain D and each f has only zeros of multiplicity at least $k+1$. Let n, k be positive integers and $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$ and let $a\neq0, \infty$ be a complex number. If $f^{(k)}-af^{-n}$ and $g^{(k)}-ag^{-n}$ share 0 for each pair function of f and g in $\mathscr{F}$, then $\mathscr{F}$ is normal in D.

Remark 4.1

Obviously, for $k=1$ and $b=0$, Corollary 1 occasionally investigates the situation when the power of f is negative in Theorem F.

Recently, Zhang [8] proved the following.

Theorem G

Let $\mathscr{F}$ be a family of meromorphic functions in the plane domain D. Let n, be a positive integer such that $n\geq2$. Let a be a finite complex number such that $a\neq0$. If $f^{n}f'$ and $g^{n}g'$ share a in D for every pair of functions $f, g\in\mathscr{F}$, then $\mathscr{F}$ is normal in D.

Question 1

It is natural to ask if the conclusion of Theorems G and 1 still holds for $n\geq1$.