RETRACTED ARTICLE: Normal families and asymptotic behaviors for solutions of certain Laplace equations

Abstract

In this paper, we consider some problems of normal families for solutions of certain Laplace with their derivatives that share a constant. We prove some results which are improvements of some earlier related theorems. Meanwhile, asymptotic behaviors of them are also obtained.

Introduction and results

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, 2]).

In 2009, Schiff [3] proved the following result.

Theorem A

Letfbe a transcendental meromorphic function in the complex plane. Letn, kbe 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 [4].

Theorem B

Let $\mathscr{F}$be a family of meromorphic functions inD. Letn, kbe 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 inD.

Recently, corresponding to Theorem B, Xue [5] proved the following result.

Theorem C

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

Lei, Yang and Fang [6] proved the following theorem.

Theorem D

Letfbe a transcendental meromorphic function in the complex plane. Letkbe 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 $F=f^{n}L[f]-c$, wherenis 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

Letfbe a transcendental meromorphic function in the complex plane. Letcbe a finite complex number such that $c\neq0$. Letn, kbe 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.

From Corollary D, 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 inD. Letn, kbe two positive integers such that $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$. Letabe a finite complex number such that $a\neq0$. If, for each $f\in\mathscr{F}$, fhas only zeros of multiplicity at leastk. If $f^{n}f^{(k)}$and $g^{n}g^{(k)}$shareainDfor every pair of functions $f, g\in\mathscr{F}$, then $\mathscr{F}$is normal inD.

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 $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$ and n be a positive integer, for $k=2$, let

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

For any $f_{m}$ 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 $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$ and n be a positive integer, and let

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

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 $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$, 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 any $f_{m}$ 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 f has only zeros of multiplicity at least k is necessary in Theorem 1. Example 2 shows that $a\neq0$ in Theorem 1 is inevitable. Example 3 shows that Theorem 1 is not true for $n=1$.

Lemmas

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

Lemma 2.1

Zalcman’s lemma (see [7, 8])

Let $\mathscr{F}$be a family of meromorphic functions inDwith the property that for each $f\in\mathscr{F}$, all zeros are of multiplicity at leastk. Suppose that there exists a number $A\geq1$such that $|f^{(k)}(z)|\leq A$whenever $f\in\mathscr{F}$and $f=0$. If $\mathscr{F}$is not normal inD, 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 $f_{n}\in\mathscr{F}$;

• (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 (with respect to the spherical metric) converges to a non-constant meromorphic function $g(\xi)$on $\mathbb{C}$, and moreover, the zeros of $g(\xi)$are of multiplicity at leastk, $g^{\sharp}(\xi)\leq g^{\sharp}(0)=kA+1$, where $g^{\sharp}(z)=\frac{|g'(z)|}{1+|g(z)|^{2}}$. In particular, ghas order at most 2.

Lemma 2.2

Letn, kbe 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. Iffis a rational but not a polynomial meromorphic function andfhas only zeros of multiplicity at leastk, 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

$$ m_{1}+m_{2}+\cdots+m_{s}=m\geq ks $$

(2.2)

and

$$ n_{1}+n_{2}+\cdots+n_{t}=n\geq t. $$

(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)

where 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 $M_{i}=(n+1)m_{i}-k\geq k(n+1)-k=nk$ and $N_{j}=(n+1)n_{j}+k\geq n+k+1$. For simplicity, we denote

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

(2.6)

and

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

(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 $g_{1}(z)$ 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 $g_{2}(z)=B(l-N)z^{t}+B_{1}z^{t-1}+\cdots+B_{t}$ 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

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

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 this is impossible, which is similar to Case 1.

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

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

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 $f_{j}\in\mathscr{F}$;

• (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 $\mathbb {C}$. 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 have

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

(3.1)

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

If $g^{n}(\xi)(g^{(k)}(\xi))\equiv a$ (≠0) and g has only zeros of multiplicity at least k, then g has no zeros. From $g^{n}g^{(k)}$ having no zeros and $g^{n}(\xi)(g^{(k)}(\xi))\equiv a$, we know that g has no poles. Because $g(\xi)$ 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, 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 rational function, by Lemma 2.2, which 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,

$$f^{n}_{j}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{j}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0 $$

and

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

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 for any positive integer m

$$f^{n}_{m}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{m}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0 $$

and

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

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

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

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

Hence

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

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 and l is a positive integer. Because the zeros of $g(\xi)$ are at least k multiple and $n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$, then 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 redcontradicts with $g^{n}g^{(k)}(z_{0})=a\neq0$.

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.

Discussion

In 2013, Ren [9] proved the following theorem.

Theorem E

Let $\mathscr{F}$be a family of meromorphic functions inD, nbe a positive integer anda, bbe 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 inD.

Recently, Ren and Yang [4] improved Theorem E by the idea of shared values. Meanwhile, Yang and Ren [10] also proved the following theorem with some new ideas.

Theorem F

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

By Theorem 1, we immediately obtain the following result.

Corollary 1

Let $\mathscr{F}$be a family of meromorphic functions in a domainDand eachfhas only zeros of multiplicity at least $k+1$. Letn, kbe 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 of functionsfandgin $\mathscr {F}$, then $\mathscr{F}$is normal inD.

Remark 3

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

Recently, Yang and Ren [10] proved the following result.

Theorem G

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

Remark 4

Obviously, our result which has the more extensive form improves Theorems C and G in some sense.

Remark 5

For further study, we pose a question.

Question 1

Does the conclusion of Theorem 1 still hold for $n\geq2$?