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All meromorphic solutions of Fermat-type functional equations

Feng Lü111Corresponding author, Email: [email protected](F. Lü).
College of Science, China University of Petroleum, Qingdao
Shandong, 266580, P.R. China
Abstract

In this paper, by making use of properties of elliptic functions, we describe meromorphic solutions of Fermat-type functional equations f(z)n+f(L(z))m=1f(z)^{n}+f(L(z))^{m}=1 over the complex plane \mathbb{C}, where L(z)L(z) is a nonconstant entire function, mm and nn are two positive integers. As applications, we also consider meromorphic solutions of Fermat-type difference and qq-difference equations.

MSC 2010: 30D30, 33E05, 39A10, 39B32.

Keywords and phrases: Fermat-type functional equation; Meromorphic solution; Weierstrass \wp-function, Residue, Picard’s little theorem.

1 Introduction and main results

In 1637, Fermat [12] stated the conjecture (which is known as Fermat’s last theorem) that the equation xm+ym=1x^{m}+y^{m}=1 cannot have positive rational solutions if m>2m>2. Since then, the equation has been a subject of intense and often heated discussions. In 1995, Wiles [40, 42] proved the profound conjecture.

In 1927, Montel in [38] initially considered the functional equations

fm(z)+gm(z)=1,f^{m}(z)+g^{m}(z)=1, (1.1)

which can be regarded as the analogous of Fermat diophantine equations xm+ym=1x^{m}+y^{m}=1 over function fields. He showed that all the entire solutions f(z)f(z) and g(z)g(z) of (1.1) must be constant if m3m\geq 3, see also Jategaonka [25]. The follow-up works were given by Baker in [1] and Gross in [15], respectively, they generalized Montel’s result by proving that (1.1) does not have nonconstant meromorphic solutions when m>4m>4 and described nonconstant meromorphic solutions for n=2,3n=2,~{}3. In 1970, Yang [45] considered the more general functional equations

fn(z)+gm(z)=1,f^{n}(z)+g^{m}(z)=1, (1.2)

and derived that fm(z)+gn(z)=1f^{m}(z)+g^{n}(z)=1 does not admit nonconstant entire solutions if 1n+1m<1\frac{1}{n}+\frac{1}{m}<1. Since then, (1.2) has been studied in various settings, see [14, 30, 31, 45]. For the convenience, some results can be stated as follows. (see e.g., [8, Proposition 1], [1, 16, 44]).

Theorem A. Suppose f(z)f(z) and g(z)g(z) are nonconstant meromorphic solutions of the functional equation (1.2).

(i) If n=m=3n=m=3, then f(z)=12{1+(h(z))3}(h(z))f(z)=\frac{\frac{1}{2}\left\{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}\right\}}{\wp(h(z))} and g(z)=η2{1(h(z))3}(h(z))g(z)=\frac{\frac{\eta}{2}\left\{1-\frac{\wp^{\prime}(h(z))}{\sqrt{3}}\right\}}{\wp(h(z))} for nonconstant entire function hh, where η3=1\eta^{3}=1 and \wp denotes the Weierstrass \wp-function satisfying ()2431(\wp^{\prime})^{2}\equiv 4\wp^{3}-1 after appropriately choosing its periods.

(ii) If n=2n=2 and m=3m=3, then f(z)=i(h(z))f(z)=i\wp^{\prime}(h(z)) and g(z)=η43(h(z)),g(z)=\eta\sqrt[3]{4}\wp(h(z)), where h(z)h(z) is any nonconstant entire function.

(iii)If m=2m=2 and n=4n=4 , then f(z)=sn(h(z))f(z)=sn^{\prime}(h(z)) and g(z)=sn(h(z))g(z)=sn(h(z)), where h(z)h(z) is any nonconstant entire function and snsn is Jacobi elliptic function satisfying sn2=1sn4{sn^{\prime}}^{2}=1-sn^{4}.

In 1989, Yanagihara [44] considered the existence of meromorphic solutions of Fermat-type functional equation in another direction. In fact, Yanagihara obtained that f3(z)+f3(z+c)=1f^{3}(z)+f^{3}(z+c)=1 does not admit nonconstant meromorphic functions of finite order. This result was also derived by Lü-Han in [22] by making use of the difference analogue of the logarithmic derivative lemma of finite order meromorphic functions, which was established by Halburd and Korhonen [19], Chiang and Feng [9], independently. In 2014, this difference analogue was improved by Halburd, Korhonen and Tohge [21] to meromorphic functions of hyper-order strictly less than 1. Here, the order and hyper-order of a meromorphic function ff are defined as

ρ(f):=lim supr+logT(r,f)logr,ρ2(f)=lim supr+loglogT(r,f)logr,\rho(f):=\limsup\limits_{r\to+\infty}\frac{\log T(r,f)}{\log r},~{}~{}\rho_{2}(f)={\limsup_{r\to+\infty}}\frac{\log\log T(r,f)}{\log r},

where T(r,f)T(r,f) denotes the Nevanlinna characteristic function of ff.

With the help of these results, Korhonen and Zhang in [27] generalize Yanagihara’s theorem to meromorphic solutions of hyper-order strictly less than 1 as follows.

Theorem B. The functional equation

f3(z)+f3(z+c)=1f^{3}(z)+f^{3}(z+c)=1

does not admit nonconstant meromorphic solutions of hyper-order strictly less than 1.

Recently, Lü and his co-workers [7, 22, 36] also considered this kind of problems, and their results can be stated as follows.

Theorem C. Let P(z)P(z) be a polynomial and f(z)f(z) be a nonconstant meromorphic function of hyper-order strictly less than 1. If f(z)f(z) is a solution of

f3(z)+f3(z+c)=eP(z),f^{3}(z)+f^{3}(z+c)=e^{P(z)}, (1.3)

then, P(z)=αz+βP(z)=\alpha z+\beta is a linear function and f(z)=deαz+β3f(z)=de^{\frac{\alpha z+\beta}{3}}, where d(0)d(\neq 0) is constant and d3(1+eαc)=1d^{3}(1+e^{\alpha c})=1.

In recent years, Fermat-type difference as well as differential-difference equations have been studied extensively. As a result, successively a lot of investigations have done by many scholars in this direction, see e.g., [33, 34]. We note that most of above results, including Theorems B and C, were obtained under the condition that the solutions is of hyper-order strictly less than 1, since the difference analogue of the logarithmic derivative lemma is needed in proofs of them. So, it is natural to ask what will happen if the hyper-order condition is omitted. We find that the conclusions of some previous theorems maybe invalid. In [22], Han offered an example to show this point as follows.

Example 1. Let c=πic=\pi i and α,β\alpha,\beta be fixed constants satisfying eαc=1e^{\alpha c}=1. Consider f(z)=121+(h(z))3(h(z))eαz+β3f(z)=\frac{1}{2}\frac{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))}e^{\frac{\alpha z+\beta}{3}}, where h(z)=ezh(z)=e^{z}. Then, a routine computation leads to f(z+c)=η21(h(z))3(h(z))eαz+β3f(z+c)=\frac{\eta}{2}\frac{1-\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))}e^{\frac{\alpha z+\beta}{3}}, where η=eαc3\eta=e^{\frac{\alpha c}{3}}. Further, one has

f3(z)+f3(z+c)=eαz+β,f^{3}(z)+f^{3}(z+c)=e^{\alpha z+\beta},

which implies that the above equation admits meromorphic solution f(z)f(z). Obviously, f(z)f(z) is not the form deαz+β3de^{\frac{\alpha z+\beta}{3}} and the hyper-order of ff is ρ2(f)=1\rho_{2}(f)=1.

Therefore, if the hyper-order condition is omitted, the Fermat-type difference equation may admit the other type of meromorphic solutions. In this paper, we pay attention to above question. Due to properties of elliptic functions, we describe the forms of all meromorphic solutions of some Fermat-type difference equations.

Before giving our main results, we introduce the Weierstrass \wp-function.

The Weierstrass \wp-function is elliptic (also doubly periodic) function with periods ω1\omega_{1} and ω2\omega_{2} (𝖨𝗆ω1ω20{\sf Im}\frac{\omega_{1}}{\omega_{2}}\neq 0) which is defined as

(z)=(z;ω1,ω2):=1z2+μ,ν𝐙;μ2+ν20{1(z+μω1+νω2)21(μω1+νω2)2},\wp(z)=\wp\left(z;\omega_{1},\omega_{2}\right):=\frac{1}{z^{2}}+\sum_{\mu,\nu\in\mathbf{Z};\mu^{2}+\nu^{2}\neq 0}\left\{\frac{1}{\left(z+\mu\omega_{1}+\nu\omega_{2}\right)^{2}}-\frac{1}{\left(\mu\omega_{1}+\nu\omega_{2}\right)^{2}}\right\},

and satisfies, after appropriately choosing ω1\omega_{1} and ω2\omega_{2},

()2=431.\left(\wp^{\prime}\right)^{2}=4\wp^{3}-1.

The period of \wp span the lattice 𝕃=ω1ω2\mathbb{L}=\omega_{1}\mathbb{Z}\oplus\omega_{2}\mathbb{Z}. We also denote

𝕃={μω1+νω2:μ,ν=0,±1,±2,}.\mathbb{L}=\{\mu\omega_{1}+\nu\omega_{2}:~{}~{}\mu,~{}\nu=0,\pm 1,\pm 2,...\}.

Obviously, all the points in 𝕃\mathbb{L} are the poles and periods of \wp. Suppose that DD is the parallelogram with vertices at 0, ω1\omega_{1}, ω2\omega_{2}, ω1+ω2\omega_{1}+\omega_{2}. Note that the order of \wp is 2. Here, the order is the number of poles of \wp or the number of zeros of a\wp-a (aa\in\mathbb{C}) in the parallelogram. Together with ()2=431(\wp^{\prime})^{2}=4\wp^{3}-1, one gets that \wp has two distinct zeros in DD, say θ1\theta_{1} and θ2\theta_{2}. In view of (z)=0(z)=±i\wp(z)=0\Leftrightarrow\wp^{\prime}(z)=\pm i, without loss of generality, throughout the paper, we assume that (θ1)=i\wp^{\prime}(\theta_{1})=-i and (θ2)=i\wp^{\prime}(\theta_{2})=i.

Here, for two meromorphic functions f,gf,~{}g and two points a,b{}a,~{}b\in\mathbb{C}\cup\{\infty\}, the notation f(z)a=0g(z)b=0f(z)-a=0\Rightarrow g(z)-b=0 means that all the zeros of faf-a are the zeros of g(z)bg(z)-b. And the notation f(z)a=0g(z)b=0f(z)-a=0\Leftrightarrow g(z)-b=0 means that faf-a and gbg-b have the same zeros.

In the present paper, more generally, we characterize meromorphic solutions of Fermat-type functional equations as follows.

Proposition 1. Suppose that L(z)L(z) is a nonconstant entire function. Then, the nonconstant meromorphic function f(z)f(z) is a solution of

f3(z)+f3(L(z))=1f^{3}(z)+f^{3}(L(z))=1 (1.4)

if and only if L(z)=qz+cL(z)=qz+c is a linear function with |q|=1|q|=1, and f(z)=12{1+(h(z))3}/(h(z))f(z)=\frac{1}{2}\left\{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}\right\}/\wp(h(z)), where h(z)h(z) is a nonconstant entire function satisfies one of the following equations.

(1). h(L(z))=Ah(z)+θ1+τ1h(L(z))=Ah(z)+\theta_{1}+\tau_{1} with τ1𝕃\tau_{1}\in\mathbb{L};

(2). h(L(z))=Ah(z)+θ2+τ2h(L(z))=Ah(z)+\theta_{2}+\tau_{2} with τ2𝕃\tau_{2}\in\mathbb{L};

(3). h(L(z))=Ah(z)+τ3h(L(z))=Ah(z)+\tau_{3} with τ3𝕃\tau_{3}\in\mathbb{L}, where AA is a constant with A3=1A^{3}=-1 and Aτ𝕃A\tau\in\mathbb{L} for any period of (z)\wp(z).

Proposition 2. Suppose that L(z)L(z) is a nonconstant entire function, suppose that g(z)g(z) is an entire function and f(z)f(z) is a nonconstant meromorphic function, and suppose that n(2),m(3)n(\geq 2),~{}m(\geq 3) are two integers such that (n,m)(3,3)(n,m)\neq(3,3). Then, f(z)f(z) is a solution of

fn(z)+fm(L(z))=eg(z)f^{n}(z)+f^{m}(L(z))=e^{g(z)} (1.5)

if and only if L(z)=qz+cL(z)=qz+c is a linear function, f(z)=Aeg(z)nf(z)=Ae^{\frac{g(z)}{n}} and

g(L(z))=nmg(z)+nLnBA,g(L(z))=\frac{n}{m}g(z)+nLn\frac{B}{A}, (1.6)

where A,BA,~{}B are two nonzero constants with An+Bm=1A^{n}+B^{m}=1. In particularly, if m=nm=n, then |q|=1|q|=1.

It is pointed out that if mnm\neq n in Proposition 2, the conclusion |q|=1|q|=1 maybe invalid, as shown by the following example.

Suppose mnm\neq n and q2=mnq^{2}=\frac{m}{n}. Consider g(z)=z2+ag(z)=z^{2}+a with a constant aa satisfying (1mn)a=nLnBA(1-\frac{m}{n})a=nLn\frac{B}{A}. Then, a calculation yields g(qz)=mng(z)+nLnBAg(qz)=\frac{m}{n}g(z)+nLn\frac{B}{A} and f(z)=Aeg(z)nf(z)=Ae^{\frac{g(z)}{n}} is a solution of (1.5). But |q|1|q|\neq 1.

Let’s turn back to the Fermat-type difference equations. Suppose L(z)=z+cL(z)=z+c with c0c\neq 0. Then, by Proposition 1, we obtain the following result.

Theorem 1. The nonconstant meromorphic function f(z)f(z) is a solution of

f3(z)+f3(z+c)=1f^{3}(z)+f^{3}(z+c)=1 (1.7)

if and only if f(z)=12{1+(h(z))3}/(h(z))f(z)=\frac{1}{2}\left\{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}\right\}/\wp(h(z)), where h(z)h(z) is a nonconstant entire function satisfies one of the following equations.

(1). h(z+c)=Ah(z)+θ1+τ1h(z+c)=Ah(z)+\theta_{1}+\tau_{1} with τ1𝕃\tau_{1}\in\mathbb{L};

(2). h(z+c)=Ah(z)+θ2+τ2h(z+c)=Ah(z)+\theta_{2}+\tau_{2} with τ2𝕃\tau_{2}\in\mathbb{L};

(3). h(z+c)=Ah(z)+τ3h(z+c)=Ah(z)+\tau_{3} with τ3𝕃\tau_{3}\in\mathbb{L}, where AA is a constant with A3=1A^{3}=-1 and Aτ𝕃A\tau\in\mathbb{L} for any period of (z)\wp(z).

Remark 1. It is known that if h(z)h(z) satisfies one of (1)-(3) in Theorem 1, then the order ρ(h(z))1\rho(h(z))\geq 1 and hyper-order ρ2(f)1\rho_{2}(f)\geq 1. (The fact can be found in [14]). So, if ρ2(f)<1\rho_{2}(f)<1, then (1.4) does not admit nonconstant meromorphic solutions. This is the conclusion of Theorem B. Below, we offer example to show that there exist entire function h(z)h(z) with arbitrary order ρ(h(z))(1)\rho(h(z))(\geq 1) satisfying one of (1)-(3).

Example 2. In [39, Theorem 3], Ozawa derived that there exists periodic entire function of arbitrary order σ(1)\sigma(\geq 1). So, there exists an entire function g(z)g(z) such that ρ(g(z))=σ\rho(g(z))=\sigma and g(z+c)=g(z)g(z+c)=g(z). Set h(z)=eazg(z)+bh(z)=e^{az}g(z)+b, where aa, bb are constants with eac=Ae^{ac}=A and (1A)b=θ1+τ1(1-A)b=\theta_{1}+\tau_{1} or θ2+τ2\theta_{2}+\tau_{2} or τ3\tau_{3} with A3=1A^{3}=-1. Then, ρ(h(z))=ρ(g(z))=σ\rho(h(z))=\rho(g(z))=\sigma and a calculation yields that

h(z+c)=Ah(z)+(1A)b.h(z+c)=Ah(z)+(1-A)b.

Thus, h(z)h(z) satisfies one of (1)-(3) in Theorem 1.

By Theorem 1 and Theorem C, we can get the following result.

Theorem 2. Suppose that f(z)f(z) is a nonconstant meromorphic function and P(z)P(z) is a polynomial. Then f(z)f(z) is a solution of the functional equation

f3(z)+f3(z+c)=eP(z)f^{3}(z)+f^{3}(z+c)=e^{P(z)} (1.8)

if and only if P(z)=αz+βP(z)=\alpha z+\beta with two constants α,β\alpha,~{}\beta and f(z)f(z) satisfies one of the following cases.

(a). f(z)=deαz+β3f(z)=de^{\frac{\alpha z+\beta}{3}}, where dd is a nonzero constant and d3(1+eαc)=1d^{3}(1+e^{\alpha c})=1;

(b). f(z)=eαz+β312{1+(h(z))3}/(h(z))f(z)=e^{\frac{\alpha z+\beta}{3}}\frac{1}{2}\left\{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}\right\}/\wp(h(z)), where eαc=1e^{\alpha c}=1 and h(z)h(z) is a nonconstant entire function satisfies one of (1)-(3) in Theorem 1.

With Proposition 2, we derive a theorem as follows.

Theorem 3. Suppose that g(z)g(z) is an entire function and f(z)f(z) is a nonconstant meromorphic function. Suppose that n(2),m(3)n(\geq 2),~{}m(\geq 3) are two integers such that (n,m)(3,3)(n,m)\neq(3,3). Then, f(z)f(z) is a solution of

fn(z)+fm(z+c)=eg(z)f^{n}(z)+f^{m}(z+c)=e^{g(z)} (1.9)

if and only if f(z)=Aeg(z)nf(z)=Ae^{\frac{g(z)}{n}} and g(z+c)=nmg(z)+nLnBAg(z+c)=\frac{n}{m}g(z)+nLn\frac{B}{A}, where A,BA,~{}B are two nonzero constants with An+Bm=1A^{n}+B^{m}=1. In particularly, if the order of g(z)g(z) is less than 1, then (1.9) admits nonconstant meromorphic solutions if and only if n=mn=m and g(z)=αz+βg(z)=\alpha z+\beta, f(z)=Aeαz+βnf(z)=Ae^{\frac{\alpha z+\beta}{n}} with constant α(0)\alpha(\neq 0) such that An(1+eαc)=1A^{n}(1+e^{\alpha c})=1.

Remark 2. We point out that equation (1.9) may have nonconstant meromorphic functions if ρ(g(z))1\rho(g(z))\geq 1, as shown by the following examples.

Example 3. Suppose that mnm\neq n, g(z)=eαz+β+ag(z)=e^{\alpha z+\beta}+a with constants α(0),β,a\alpha(\neq 0),~{}\beta,~{}a. Consider f(z)=Aeg(z)nf(z)=Ae^{\frac{g(z)}{n}} with eαc=nme^{\alpha c}=\frac{n}{m} and (1nm)a=nLnBA(1-\frac{n}{m})a=nLn\frac{B}{A}, where A,BA,~{}B are nonzero constants with An+Bm=1A^{n}+B^{m}=1. By a calculation, one gets g(z+c)=ng(z)m+nLnBAg(z+c)=\frac{ng(z)}{m}+nLn\frac{B}{A} and

fn(z)+fm(z+c)=eg(z).f^{n}(z)+f^{m}(z+c)=e^{g(z)}.

Obviously, ρ(g(z))=1\rho(g(z))=1.

Example 4. Suppose that m=nm=n, and hh is any nonconstant entire periodic function with period cc. Let g(z)=h(z)+azg(z)=h(z)+az with ac=nLnBAac=nLn\frac{B}{A}, where A,BA,~{}B are nonzero constants with An+Bn=1A^{n}+B^{n}=1. Then, we have g(z+c)=g(z)+nLnBAg(z+c)=g(z)+nLn\frac{B}{A}, and f(z)=Aeg(z)nf(z)=Ae^{\frac{g(z)}{n}} is a nonconstant meromorphic solution of

fn(z)+fn(z+c)=eg(z).f^{n}(z)+f^{n}(z+c)=e^{g(z)}.

Obviously, ρ(g(z))1\rho(g(z))\geq 1.

Remark 3. From Theorem 3, we see that the equation (1.9) do not admit meromorphic solutions with poles. It is pointed out that the same argument in Theorems 3 can deal with the Fermat-type difference equations fm(z)+fn(z+c)=eg(z)f^{m}(z)+f^{n}(z+c)=e^{g(z)} with n(2),m(3)n(\geq 2),~{}m(\geq 3) and (m,n)(3,3)(m,n)\neq(3,3), we omit the details here.

The following corollary follows immediately from Theorem 3.

Corollary 1. Suppose that P(z)P(z) is a polynomial and f(z)f(z) is a nonconstant meromorphic function. Suppose that n(2),m(2)n(\geq 2),~{}m(\geq 2) are two integers with m+n5m+n\geq 5 and (n,m)(3,3)(n,m)\neq(3,3). Then, f(z)f(z) is a solution of

fn(z)+fm(z+c)=eP(z)f^{n}(z)+f^{m}(z+c)=e^{P(z)} (1.10)

if and only n=mn=m, f(z)=AeP(z)nf(z)=Ae^{\frac{P(z)}{n}} and P=αz+βP=\alpha z+\beta with two constants α(0),β\alpha(\neq 0),~{}\beta.

From Theorem 3 and Corollary 1, one can easily get the following result.

Corollary 2. Suppose that f(z)f(z) is a nonconstant meromorphic function, and n(2),m(2)n(\geq 2),~{}m(\geq 2) are two integers with m+n5m+n\geq 5. Then, f(z)f(z) is a solution of

fn(z)+fm(z+c)=1f^{n}(z)+f^{m}(z+c)=1 (1.11)

if and only if n=m=3n=m=3.

Finally, as applications of Propositions 1 and 2, we consider meromorphic solutions of Fermat-type qq-difference functional equations. As early as 1952, Valiron in [41] showed that the non-autonomous Schröder qq-difference equation

f(qz)=R(z,f(z))f(qz)=R(z,f(z)) (1.12)

where R(z,f(z))R(z,f(z)) is rational in both arguments, admits a one parameter family of meromorphic solutions, provided that qq\in\mathbb{C} is chosen appropriately. Later, Gundersen et al [18] proved that if |q|>1|q|>1 and the qq-difference equation (1.12) admits a meromorphic solution of order zero, then (1.12) reduces to a qq-difference Riccati equation, i.e. degfR=1\deg_{f}R=1. Some scholars, such as Bergweiler-Hayman [5], Bergweiler-Ishizaki- Yanagihara [6], Eremenko-Sodin [13], Ishizaki-Yanagihara [24] also made contributions to meromorphic solutions of qq-difference functional equations. In 2007, Barneet-Halburd-Korhonen-Morgan in [3] derived the qq-difference analogues of some well-known results in Nevanlinna theory, including the lemma of the logarithmic derivative, the Clunie’s lemma and the second main theorem. With the help of results, meromorphic solutions to qq-difference functional equations were further studied, see [26, 32].

Observe that in Propositions 1 and 2, L(z)L(z) reduce to a linear function, say qz+cqz+c with q0q\neq 0. Therefore, we have described meromorphic solutions of Fermat-type qq-difference functional equations f3(z)+f3(qz+c)=1f^{3}(z)+f^{3}(qz+c)=1 and fn(z)+fn(qz+c)=eg(z)f^{n}(z)+f^{n}(qz+c)=e^{g(z)}, where n,mn,~{}m satisfy some condition and g(z)g(z) is an entire function.

Remark 4. From the above theorems, we see that the case (1,n),(m,1),(2,2)(1,n),~{}(m,1),~{}(2,2) are left. Unfortunately, we cannot deal with these cases and thus leave them for further study.

For the proofs, we will assume that the reader is familiar with basic elements in Nevanlinna theory of meromorphic functions in \mathbb{C} (see e.g. [21, 23, 28, 46]), such as the first and second main theorems, the characteristic function T(r,f)T(r,f), the proximity function m(r,f)m(r,f), the counting function N(r,f)N(r,f) and the reduced counting function N¯(r,f)\overline{N}(r,f). We also need the following notation.

The lower order of a meromorphic function ff is defined as

μ(f):=lim infr+logT(r,f)logr.\mu(f):=\liminf\limits_{r\to+\infty}\frac{\log T(r,f)}{\log r}.

2 Proofs of main results

In this section, we firstly give the proof of Proposition 1.

Proof of Proposition 1.

Firstly, we will prove the necessity. Assume that h(z)h(z) satisfies one of (1)-(3), we will prove that f(z)f(z) is a meromorphic solution of (1.4). Suppose that h(z)h(z) satisfies (1). Then h(L(z))=Ah(z)+θ1+τ1h(L(z))=Ah(z)+\theta_{1}+\tau_{1} with τ1𝕃\tau_{1}\in\mathbb{L}. The assumption that Aτ𝕃A\tau\in\mathbb{L} for any period of (z)\wp(z) defines a mapping as

A:𝕃𝕃,A:τAτ.A:\mathbb{L}\rightarrow\mathbb{L},~{}~{}A:\tau\rightarrow A\tau.

We claim that the mapping A:A: is a bijection.

Observe that A=1A=-1, or eπ3ie^{\frac{\pi}{3}i}, or eπ3ie^{-\frac{\pi}{3}i}. If A=1A=-1, clearly, the claim is valid. Suppose that A=eπ3iA=e^{\frac{\pi}{3}i}. Among all the periods of (ω)\wp(\omega), we list the periods which may take the smallest modulus as follows: ω1\omega_{1}, ω1+ω2\omega_{1}+\omega_{2}, ω2\omega_{2}, ω2ω1\omega_{2}-\omega_{1}, ω1-\omega_{1}, ω2ω1-\omega_{2}-\omega_{1}, ω2-\omega_{2} and ω1ω2\omega_{1}-\omega_{2}. Then, in view of that Aτ=τeπ3iA\tau=\tau e^{\frac{\pi}{3}i} is also a period of \wp, below, we have two possible cases (see Figures 1-2 below).

Refer to caption
Figure 1: (i)
Refer to caption
Figure 2: (ii)

(i). Aω1=ω2A\omega_{1}=\omega_{2} and Aω2=ω2ω1A\omega_{2}=\omega_{2}-\omega_{1} and A(ω2ω1)=ω1A(\omega_{2}-\omega_{1})=-\omega_{1}.

(ii). Aω1=ω1+ω2A\omega_{1}=\omega_{1}+\omega_{2} and A(ω1+ω2)=ω2A(\omega_{1}+\omega_{2})=\omega_{2} and Aω2=ω1A\omega_{2}=-\omega_{1}.

Either (i) or (ii) holds, a simple analysis yields the claim is right. The same argument implies that the claim holds for A=eπ3iA=e^{-\frac{\pi}{3}i}. Therefore, the claim is proved.

Observe that A3=1A^{3}=-1, the claim and the definition of (z)\wp(z) as follows

(z)=(z;ω1,ω2):=1z2+μ,ν𝐙;μ2+ν20{1(z+μω1+νω2)21(μω1+νω2)2}.\wp(z)=\wp\left(z;\omega_{1},\omega_{2}\right):=\frac{1}{z^{2}}+\sum_{\mu,\nu\in\mathbf{Z};\mu^{2}+\nu^{2}\neq 0}\left\{\frac{1}{\left(z+\mu\omega_{1}+\nu\omega_{2}\right)^{2}}-\frac{1}{\left(\mu\omega_{1}+\nu\omega_{2}\right)^{2}}\right\}.

A calculation yields that

(Az)\displaystyle\wp(Az) =1(Az)2+μ,ν𝐙;μ2+ν20{1(Az+μω1+νω2)21(μω1+νω2)2}\displaystyle=\frac{1}{(Az)^{2}}+\sum_{\mu,\nu\in\mathbf{Z};\mu^{2}+\nu^{2}\neq 0}\left\{\frac{1}{\left(Az+\mu\omega_{1}+\nu\omega_{2}\right)^{2}}-\frac{1}{\left(\mu\omega_{1}+\nu\omega_{2}\right)^{2}}\right\} (2.1)
=1A2{1z2+μ,ν𝐙;μ2+ν20{1{z+1A(μω1+νω2)}21{1A(μω1+νω2)}2}}\displaystyle=\frac{1}{A^{2}}\left\{\frac{1}{z^{2}}+\sum_{\mu,\nu\in\mathbf{Z};\mu^{2}+\nu^{2}\neq 0}\left\{\frac{1}{\{z+\frac{1}{A}(\mu\omega_{1}+\nu\omega_{2})\}^{2}}-\frac{1}{\{\frac{1}{A}(\mu\omega_{1}+\nu\omega_{2})\}^{2}}\right\}\right\}
=A{1z2+μ,ν𝐙;μ2+ν20{1{zA2(μω1+νω2)}21{A2(μω1+νω2)}2}}\displaystyle=-A\left\{\frac{1}{z^{2}}+\sum_{\mu,\nu\in\mathbf{Z};\mu^{2}+\nu^{2}\neq 0}\left\{\frac{1}{\{z-A^{2}(\mu\omega_{1}+\nu\omega_{2})\}^{2}}-\frac{1}{\{-A^{2}(\mu\omega_{1}+\nu\omega_{2})\}^{2}}\right\}\right\}
=A{1z2+μ,ν𝐙;μ2+ν20{1(z+μω1+νω2)21(μω1+νω2)2}}\displaystyle=-A\left\{\frac{1}{z^{2}}+\sum_{\mu,\nu\in\mathbf{Z};\mu^{2}+\nu^{2}\neq 0}\left\{\frac{1}{\left(z+\mu\omega_{1}+\nu\omega_{2}\right)^{2}}-\frac{1}{\left(\mu\omega_{1}+\nu\omega_{2}\right)^{2}}\right\}\right\}
=A(z),\displaystyle=-A\wp(z),

which implies that (Az)=(z)\wp^{\prime}(Az)=-\wp^{\prime}(z). It is known that

(w+c)=14[(w)(c)(w)(c)]2(w)(c),\wp(w+c)=\frac{1}{4}[\frac{\wp^{\prime}(w)-\wp^{\prime}(c)}{\wp(w)-\wp(c)}]^{2}-\wp(w)-\wp(c),

for any ww and cc. Thus,

(Aw+θ1)\displaystyle\wp(Aw+\theta_{1}) =14[(Aw)(θ1)(Aw)(θ1)]2(Aw)(θ1)=14[(Aw)+i(Aw)]2(Aw)\displaystyle=\frac{1}{4}[\frac{\wp^{\prime}(Aw)-\wp^{\prime}(\theta_{1})}{\wp(Aw)-\wp(\theta_{1})}]^{2}-\wp(Aw)-\wp(\theta_{1})=\frac{1}{4}[\frac{\wp^{\prime}(Aw)+i}{\wp(Aw)}]^{2}-\wp(Aw) (2.2)
=14((Aw)+i)2(Aw)(Aw)3(Aw)=((Aw)+i)2(Aw)(Aw)2+1(Aw)\displaystyle=\frac{1}{4}\frac{(\wp^{\prime}(Aw)+i)^{2}\wp(Aw)}{\wp(Aw)^{3}}-\wp(Aw)=\frac{(\wp^{\prime}(Aw)+i)^{2}\wp(Aw)}{\wp^{\prime}(Aw)^{2}+1}-\wp(Aw)
=((Aw)+i)2(Aw)((Aw)+i)((Aw)i)(Aw)=2i(Aw)(Aw)i\displaystyle=\frac{(\wp^{\prime}(Aw)+i)^{2}\wp(Aw)}{(\wp^{\prime}(Aw)+i)(\wp^{\prime}(Aw)-i)}-\wp(Aw)=\frac{2i\wp(Aw)}{\wp^{\prime}(Aw)-i}
=A2i(w)(w)i=A2i(w)(w)+i.\displaystyle=\frac{-A2i\wp(w)}{-\wp^{\prime}(w)-i}=\frac{A2i\wp(w)}{\wp^{\prime}(w)+i}.
(Aw+θ1)\displaystyle\wp^{\prime}(Aw+\theta_{1}) =(2i(w)(w)+i)=2i(w)((w)+i)(w)′′(w)((w)+i)2\displaystyle=(\frac{2i\wp(w)}{\wp^{\prime}(w)+i})^{\prime}=2i\frac{\wp^{\prime}(w)(\wp^{\prime}(w)+i)-\wp(w)\wp^{\prime\prime}(w)}{(\wp^{\prime}(w)+i)^{2}} (2.3)
=2i(w)2+i(w)(w)′′(w)((w)+i)2=2i(w)2+i(w)32((w)2+1)((w)+i)2\displaystyle=2i\frac{\wp^{\prime}(w)^{2}+i\wp^{\prime}(w)-\wp(w)\wp^{\prime\prime}(w)}{(\wp^{\prime}(w)+i)^{2}}=2i\frac{\wp^{\prime}(w)^{2}+i\wp^{\prime}(w)-\frac{3}{2}(\wp^{\prime}(w)^{2}+1)}{(\wp^{\prime}(w)+i)^{2}}
=i(w)3i(w)+i.\displaystyle=-i\frac{\wp^{\prime}(w)-3i}{\wp^{\prime}(w)+i}.

Set h(z)=ωh(z)=\omega. Then h(L(z))=Ah(z)+θ1+τ1=Aw+θ1+τ1h(L(z))=Ah(z)+\theta_{1}+\tau_{1}=Aw+\theta_{1}+\tau_{1}. Note that both \wp and \wp^{\prime} are elliptic functions with periods ω1\omega_{1} and ω2\omega_{2}, we have

(h(L(z)))=(Aw+θ1+τ1)=(Aw+θ1),\displaystyle\wp(h(L(z)))=\wp(Aw+\theta_{1}+\tau_{1})=\wp(Aw+\theta_{1}),
(h(L(z)))=(Aw+θ1+τ1)=(Aw+θ1).\displaystyle\wp^{\prime}(h(L(z)))=\wp^{\prime}(Aw+\theta_{1}+\tau_{1})=\wp^{\prime}(Aw+\theta_{1}).

Observe that f(z)=121+(h(z))3(h(z))f(z)=\frac{1}{2}\frac{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))}. So

f(L(z))\displaystyle f(L(z)) =121+(h(L(z)))3(h(L(z)))=121+(Aw+θ1)3(Aw+θ1)=12113i(w)3i(w)+iA2i(w)(w)+i\displaystyle=\frac{1}{2}\frac{1+\frac{\wp^{\prime}(h(L(z)))}{\sqrt{3}}}{\wp(h(L(z)))}=\frac{1}{2}\frac{1+\frac{\wp^{\prime}(Aw+\theta_{1})}{\sqrt{3}}}{\wp(Aw+\theta_{1})}=\frac{1}{2}\frac{1-\frac{1}{\sqrt{3}}i\frac{\wp^{\prime}(w)-3i}{\wp^{\prime}(w)+i}}{\frac{A2i\wp(w)}{\wp^{\prime}(w)+i}} (2.4)
=12((w)+i)13i((w)3i)A2i(w)=1213i21A1(w)3(w)=1Aη21(w)3(w),\displaystyle=\frac{1}{2}\frac{(\wp^{\prime}(w)+i)-\frac{1}{\sqrt{3}}i(\wp^{\prime}(w)-3i)}{A2i\wp(w)}=\frac{1}{2}\frac{-1-\sqrt{3}i}{2}\frac{1}{-A}\frac{1-\frac{\wp^{\prime}(w)}{\sqrt{3}}}{\wp(w)}=\frac{1}{-A}\frac{\eta}{2}\frac{1-\frac{\wp^{\prime}(w)}{\sqrt{3}}}{\wp(w)},

where η=13i2\eta=\frac{-1-\sqrt{3}i}{2} and η3=1\eta^{3}=1. Thus, by (i) of Theorem A, one can easily derive that f(z)3+f(L(z))3=1f(z)^{3}+f(L(z))^{3}=1 and f(z)f(z) is a solution of (1.4). If h(z)h(z) satisfies (2) or (3), then the same argument yields that f(z)f(z) is also a solution of (1.4).

Conversely, we will prove the sufficiency. For convenience, we below denote L(z)L(z) by z¯\overline{z}, which means that z¯=L(z)\overline{z}=L(z) and a point a¯=L(a)\overline{a}=L(a).

Suppose that f(z)f(z) is a nonconstant meromorphic solution of (1.4). Then, Via (i) of Theorem A, one has

f(z)=121+(h(z))3(h(z))andf(z¯)=η21(h(z))3(h(z)),f(z)=\frac{1}{2}\frac{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))}\qquad and\qquad f(\overline{z})=\frac{\eta}{2}\frac{1-\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))}, (2.5)

where hh is a nonconstant entire function over \mathbb{C}. Next, we will prove that h(z)h(z) satisfies one of (1)-(3).

Obviously, T(r,f(z))=T(r,f(z¯))+S(r,f)T(r,f(z))=T(r,f(\overline{z}))+S(r,f). Rewrite the form of f(z)f(z) as

2f(z)(h(z))1=(h(z))3.2f(z)\wp(h(z))-1=\frac{\wp^{\prime}(h(z))}{\sqrt{3}}.

Then,

[2f(z)(h(z))1]2=((h(z))3)2=13[4(h(z))31],[2f(z)\wp(h(z))-1]^{2}=(\frac{\wp^{\prime}(h(z))}{\sqrt{3}})^{2}=\frac{1}{3}[4\wp(h(z))^{3}-1], (2.6)

We rewrite (2.6) as

f(z)[f(z)(h(z))1]=13[(h(z))31](h(z)),f(z)[f(z)\wp(h(z))-1]=\frac{1}{3}\frac{[\wp(h(z))^{3}-1]}{\wp(h(z))}, (2.7)

which implies that

3T(r,(h(z)))\displaystyle 3T(r,\wp(h(z))) =T(r,13[(h(z))31](h(z)))+O(1)=T(r,f(z)[f(z)(h(z))1])+O(1)\displaystyle=T(r,\frac{1}{3}\frac{[\wp(h(z))^{3}-1]}{\wp(h(z))})+O(1)=T(r,f(z)[f(z)\wp(h(z))-1])+O(1)
T(r,f(z))+T(r,f(z)(h(z)))+O(1)\displaystyle\leq T(r,f(z))+T(r,f(z)\wp(h(z)))+O(1)
2T(r,f(z))+T(r,(h(z)))+O(1).\displaystyle\leq 2T(r,f(z))+T(r,\wp(h(z)))+O(1).

Thus, we derive that

T(r,(h(z)))T(r,f(z))+O(1).T(r,\wp(h(z)))\leq T(r,f(z))+O(1). (2.8)

The form of f(z)f(z) yields that f(z¯)=121+(h(z¯))3(h(z¯))f(\overline{z})=\frac{1}{2}\frac{1+\frac{\wp^{\prime}(h(\overline{z}))}{\sqrt{3}}}{\wp(h(\overline{z}))}. Then, the same argument leads to

T(r,(h(z¯)))T(r,f(z¯))+O(1).T(r,\wp(h(\overline{z})))\leq T(r,f(\overline{z}))+O(1). (2.9)

Rewrite (1.4) as f(z¯)3=[f(z)31]=(f(z)1)(f(z)ς)(f(z)ς2),(ς1,ς3=1)f(\overline{z})^{3}=-[f(z)^{3}-1]=-(f(z)-1)(f(z)-\varsigma)(f(z)-\varsigma^{2}),~{}(\varsigma\neq 1,~{}\varsigma^{3}=1), which implies that the zeros of f(z)1f(z)-1, f(z)ςf(z)-\varsigma and f(z)ς2f(z)-\varsigma^{2} are of multiplicities at least 3. Applying Nevanlinna’s first and second theorems to the function f(z)f(z) yields that

2T(r,f(z))m=02N¯(r,1f(z)ςm)+N¯(r,f(z))+S(r,f(z)),\displaystyle 2T(r,f(z))\leq\sum_{m=0}^{2}\overline{N}(r,\frac{1}{f(z)-\varsigma^{m}})+\overline{N}(r,f(z))+S(r,f(z)),
13m=02N(r,1f(z)ςm)+N(r,f(z))+S(r,f(z))\displaystyle\leq\frac{1}{3}\sum_{m=0}^{2}N(r,\frac{1}{f(z)-\varsigma^{m}})+N(r,f(z))+S(r,f(z))
2T(r,f(z))+S(r,f(z)).\displaystyle\leq 2T(r,f(z))+S(r,f(z)).

Therefore

T(r,f(z))=N¯(r,f(z))+S(r,f(z))=N(r,f(z))+S(r,f(z)).T(r,f(z))=\overline{N}(r,f(z))+S(r,f(z))=N(r,f(z))+S(r,f(z)). (2.10)

Further, in view of that f(z)=121+(h(z))3(h(z))f(z)=\frac{1}{2}\frac{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))} and \wp only has multiple poles, one derives that

T(r,f(z))\displaystyle T(r,f(z)) =N¯(r,f(z))+S(r,f(z))=N¯(r,121+(h(z))3(h(z)))+S(r,f(z))\displaystyle=\overline{N}(r,f(z))+S(r,f(z))=\overline{N}(r,\frac{1}{2}\frac{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))})+S(r,f(z)) (2.11)
N¯(r,1(h(z)))+N¯(r,(h(z)))+S(r,f(z))\displaystyle\leq\overline{N}(r,\frac{1}{\wp(h(z))})+\overline{N}(r,\wp(h(z)))+S(r,f(z))
N¯(r,1(h(z)))+12N(r,(h(z)))+S(r,f(z))\displaystyle\leq\overline{N}(r,\frac{1}{\wp(h(z))})+\frac{1}{2}N(r,\wp(h(z)))+S(r,f(z))
32T(r,(h(z)))+S(r,f(z)).\displaystyle\leq\frac{3}{2}T(r,\wp(h(z)))+S(r,f(z)).

The equation f(z)3+f(z¯)3=1f(z)^{3}+f(\overline{z})^{3}=1 yields that N¯(r,f(z¯))=N¯(r,f(z))\overline{N}(r,f(\overline{z}))=\overline{N}(r,f(z)). Then, by (2.10), one has

T(r,f(z¯))\displaystyle T(r,f(\overline{z})) =T(r,f(z))+S(r,f(z))=N¯(r,f(z))+S(r,f(z))\displaystyle=T(r,f(z))+S(r,f(z))=\overline{N}(r,f(z))+S(r,f(z)) (2.12)
=N¯(r,f(z¯))+S(r,f(z)).\displaystyle=\overline{N}(r,f(\overline{z}))+S(r,f(z)).

Then, the same argument as in (2.11) yields that

T(r,f(z¯))32T(r,(h(z¯)))+S(r,f(z)).T(r,f(\overline{z}))\leq\frac{3}{2}T(r,\wp(h(\overline{z})))+S(r,f(z)). (2.13)

All the above discussion yields that

S(r,f(z))=S(r,f(z¯))=S(r,(h(z)))=S(r,(h(z¯))).S(r,f(z))=S(r,f(\overline{z}))=S(r,\wp(h(z)))=S(r,\wp(h(\overline{z}))).

For simplicity, we write

S(r)=S(r,f(z))=S(r,f(z¯))=S(r,(h(z)))=S(r,(h(z¯))).S(r)=S(r,f(z))=S(r,f(\overline{z}))=S(r,\wp(h(z)))=S(r,\wp(h(\overline{z}))).

By (2.5) and a routine computation, we get that

η(1(h(z))3)(h(z¯))=(1+(h(z¯))3)(h(z)).\eta(1-\frac{\wp^{\prime}(h(z))}{\sqrt{3}})\wp(h(\overline{z}))=(1+\frac{\wp^{\prime}(h(\overline{z}))}{\sqrt{3}})\wp(h(z)). (2.14)

We employ the method in [22, 37, 43] to prove this theorem. For a set GG, define the function N(r,G)N(r,G) as

N(r,G)=0rn(t,G)n(0,G)t𝑑t+n(0,G)logr,N(r,G)=\int_{0}^{r}\frac{n(t,G)-n(0,G)}{t}dt+n(0,G)\log r,

where the notation n(r,G)n(r,G) is the number of points in G{|z|<r}G\cap\{|z|<r\}, ignoring multiplicities. We define a set S1S_{1} as

S1={z|(h(z))=0and(h(z¯))=0}.S_{1}=\{z|\wp(h(z))=0~{}~{}and~{}~{}\wp(h(\overline{z}))=0\}.

Arrange S1S_{1} as S1={as}s=1S_{1}=\{a_{s}\}_{s=1}^{\infty} and asa_{s}\rightarrow\infty as ss\rightarrow\infty. Notice, when (h(as))=0\wp(h(a_{s}))=0, then [(h(as))]2=1[\wp^{\prime}(h(a_{s}))]^{2}=-1. Further, when (h(as¯))=0\wp(h(\overline{a_{s}}))=0, then [(h(as¯))]2=1[\wp^{\prime}(h(\overline{a_{s}}))]^{2}=-1. Differentiate (2.14) and apply substitution to observe that

η(1(h(as))3)(h(as¯))[h(z¯)](as)=(1+(h(as¯))3)(h(as)))h(as).\displaystyle\begin{aligned} &\quad\eta(1-\frac{\wp^{\prime}(h(a_{s}))}{\sqrt{3}})\wp^{\prime}(h(\overline{a_{s}}))[h(\overline{z})]^{\prime}(a_{s})\\ &=(1+\frac{\wp^{\prime}(h(\overline{a_{s}}))}{\sqrt{3}})\wp^{\prime}(h(a_{s})))h^{\prime}(a_{s}).\end{aligned}

From which we have that one and the only one of the following situations occurs

{g1(as)=η(1i33)[h(z¯)](as)(1+i33)h(as)=0,g2(as)=η[h(z¯)](as)+h(as)=0,g3(as)=η(1+i33)[h(z¯)](as)(1i33)h(as)=0.\left\{\begin{aligned} &g_{1}(a_{s})=\eta(1-i\frac{\sqrt{3}}{3})[h(\overline{z})]^{\prime}(a_{s})-(1+i\frac{\sqrt{3}}{3})h^{\prime}(a_{s})=0,\\ &g_{2}(a_{s})=\eta[h(\overline{z})]^{\prime}(a_{s})+h^{\prime}(a_{s})=0,\\ &g_{3}(a_{s})=\eta(1+i\frac{\sqrt{3}}{3})[h(\overline{z})]^{\prime}(a_{s})-(1-i\frac{\sqrt{3}}{3})h^{\prime}(a_{s})=0.\end{aligned}\right.

Below, we consider two cases.

Case 1. gi(z)0g_{i}(z)\not\equiv 0 for any i=1,2,3i=1,~{}2,~{}3.

Here, we employ a result of Clunie [11], which can be stated as follows.

Lemma 1. Let ff be a nonconstant entire function and let gg be a transcendental meromorphic function in the complex plane, then T(r,f)=S(r,g(f)))T(r,f)=S(r,g(f))) as rr\rightarrow\infty.

Note that \wp is transcendental. By Lemma 1, we have that T(r,h(z))=S(r,(h(z)))=S(r)T(r,h(z))=S(r,\wp(h(z)))=S(r) and T(r,h(z¯))=S(r,(h(z¯)))=S(r)T(r,h(\overline{z}))=S(r,\wp(h(\overline{z})))=S(r). We also have

T(r,h(z))=O(T(r,h(z))=S(r),T(r,[h(z¯)])=O(T(r,h(z¯)))=S(r).T(r,h^{\prime}(z))=O(T(r,h(z))=S(r),~{}~{}T(r,[h(\overline{z})]^{\prime})=O(T(r,h(\overline{z})))=S(r).

So, T(r,gi)=S(r)T(r,g_{i})=S(r). Further, we have

N(r,S1)\displaystyle N(r,S_{1}) iN(r,1gi)iT(r,1gi)\displaystyle\leq\sum_{i}N(r,\frac{1}{g_{i}})\leq\sum_{i}T(r,\frac{1}{g_{i}}) (2.15)
=iT(r,gi)+O(1)=S(r).\displaystyle=\sum_{i}T(r,g_{i})+O(1)=S(r).

Assume that EE is the set of all zeros of (h)\wp(h), that is E={z|(h(z))=0}E=\{z|\wp(h(z))=0\}. Obviously, S1ES_{1}\subseteq E. Put S2=E\S1S_{2}=E\backslash S_{1}. For any bS2b\in S_{2}, we see that (h(b))=0\wp(h(b))=0 and (h(b¯))0\wp(h(\overline{b}))\neq 0. In view of [(h(b))]2=1[\wp^{\prime}(h(b))]^{2}=-1 and (2.14), one has (h(b¯))=\wp(h(\overline{b}))=\infty. All the above discussions yields that

N¯(r,1(h(z)))\displaystyle\overline{N}(r,\frac{1}{\wp(h(z))}) =N¯(r,E)=N(r,S1)+N¯(r,S2)\displaystyle=\overline{N}(r,E)=N(r,S_{1})+\overline{N}(r,S_{2}) (2.16)
N¯(r,(h(z¯)))+S(r).\displaystyle\leq\overline{N}(r,\wp(h(\overline{z})))+S(r).

Suppose that z0z_{0} is a zero of (h(z))\wp(h(z)) with multiplicity pp. Then z0z_{0} is a zero of ((h(z)))=(h(z))h(z)(\wp(h(z)))^{\prime}=\wp^{\prime}(h(z))h^{\prime}(z) with multiplicity p1p-1. The fact (h(z0))=±i\wp^{\prime}(h(z_{0}))=\pm i yields that z0z_{0} is a zero of hh^{\prime} with multiplicity p1p-1. Thus,

N(r,1(h(z)))=N¯(r,1(h(z)))+N(r,1h(z))=N¯(r,1(h(z)))+S(r).N(r,\frac{1}{\wp(h(z))})=\overline{N}(r,\frac{1}{\wp(h(z))})+N(r,\frac{1}{h^{\prime}(z)})=\overline{N}(r,\frac{1}{\wp(h(z))})+S(r). (2.17)

The equation (2.10) yields that m(r,f(z))=S(r)m(r,f(z))=S(r), since T(r,f(z))=m(r,f(z))+N(r,f(z))T(r,f(z))=m(r,f(z))+N(r,f(z)). Rewrite the form of f(z)f(z) as

1(h(z))=2f(z)(h(z))3(h(z))=2f(z)(h(z))h(z)3(h(z))1h(z).\frac{1}{\wp(h(z))}=2f(z)-\frac{\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))}=2f(z)-\frac{\frac{\wp^{\prime}(h(z))h^{\prime}(z)}{\sqrt{3}}}{\wp(h(z))}\frac{1}{h^{\prime}(z)}.

Then, applying the logarithmic derivative lemma, we have

m(r,1(h(z)))\displaystyle m(r,\frac{1}{\wp(h(z))}) m(r,f(z))+m(r,((h(z)))(h(z)))+m(r,1h(z))+O(1)\displaystyle\leq m(r,f(z))+m(r,\frac{(\wp(h(z)))^{\prime}}{\wp(h(z))})+m(r,\frac{1}{h^{\prime}(z)})+O(1) (2.18)
S(r)+S(r,(h(z)))+S(r)=S(r).\displaystyle\leq S(r)+S(r,\wp(h(z)))+S(r)=S(r).

Combining (2.17) and (2.18) yields that

T(r,(h(z)))\displaystyle T(r,\wp(h(z))) =T(r,1(h(z)))+O(1)=N(r,1(h(z)))+m(r,1(h(z)))+O(1)\displaystyle=T(r,\frac{1}{\wp(h(z))})+O(1)=N(r,\frac{1}{\wp(h(z))})+m(r,\frac{1}{\wp(h(z))})+O(1) (2.19)
=N¯(r,1(h(z)))+m(r,1(h(z)))+O(1)\displaystyle=\overline{N}(r,\frac{1}{\wp(h(z))})+m(r,\frac{1}{\wp(h(z))})+O(1)
=N¯(r,1(h(z)))+S(r).\displaystyle=\overline{N}(r,\frac{1}{\wp(h(z))})+S(r).

We also know that [(h(z¯))]2=4(h(z¯))31[\wp^{\prime}(h(\overline{z}))]^{2}=4\wp(h(\overline{z}))^{3}-1. Then,

[(h(z¯))]2=[h(z¯)]2(h(z¯))2=4[h(z¯)]2(h(z¯))3[h(z¯)]2.[\wp(h(\overline{z}))]^{\prime 2}=[h(\overline{z})]^{\prime 2}\wp^{\prime}(h(\overline{z}))^{2}=4[h(\overline{z})]^{\prime 2}\wp(h(\overline{z}))^{3}-[h(\overline{z})]^{\prime 2}.

Rewrite it as

(h(z¯))2(h(z¯))=(h(z¯))3=14[h(z¯)]2[(h(z¯))]214.\wp(h(\overline{z}))^{2}\wp(h(\overline{z}))=\wp(h(\overline{z}))^{3}=\frac{1}{4[h(\overline{z})]^{\prime 2}}[\wp(h(\overline{z}))]^{\prime 2}-\frac{1}{4}.

By T(r,14[h(z¯)]2)=S(r)T(r,\frac{1}{4[h(\overline{z})]^{\prime 2}})=S(r) and Clunie’s lemma (see [10]), we obtain that m(r,(h(z¯)))=S(r)m(r,\wp(h(\overline{z})))=S(r) and

T(r,(h(z¯)))=m(r,(h(z¯)))+N(r,(h(z¯)))=N(r,(h(z¯)))+S(r).T(r,\wp(h(\overline{z})))=m(r,\wp(h(\overline{z})))+N(r,\wp(h(\overline{z})))=N(r,\wp(h(\overline{z})))+S(r). (2.20)

Note that (h(z¯))\wp(h(\overline{z})) only has multiple poles. So, N¯(r,(h(z¯)))12N(r,(h(z¯)))\overline{N}(r,\wp(h(\overline{z})))\leq\frac{1}{2}N(r,\wp(h(\overline{z}))). Further, combining (2.11), (2.13), (2.16), (2.19) and (2.20), we have that

23T(r,f(z))\displaystyle\frac{2}{3}T(r,f(z)) T(r,(h(z)))+S(r)=N¯(r,1(h(z)))+S(r)N¯(r,(h(z¯)))+S(r)\displaystyle\leq T(r,\wp(h(z)))+S(r)=\overline{N}(r,\frac{1}{\wp(h(z))})+S(r)\leq\overline{N}(r,\wp(h(\overline{z})))+S(r) (2.21)
12N(r,(h(z¯)))+S(r)12T(r,(h(z¯)))+S(r)\displaystyle\leq\frac{1}{2}N(r,\wp(h(\overline{z})))+S(r)\leq\frac{1}{2}T(r,\wp(h(\overline{z})))+S(r)
12T(r,f(z¯))+S(r)=12T(r,f(z))+S(r),\displaystyle\leq\frac{1}{2}T(r,f(\overline{z}))+S(r)=\frac{1}{2}T(r,f(z))+S(r),

which is a contradiction. Thus, the case cannot occur.

Case 2. gi(z)0g_{i}(z)\equiv 0, for some i{1,2,3}i\in\{1,2,3\}.

Firstly, we assume that g1(z)0g_{1}(z)\equiv 0. Then, [h(z¯)]=Ah(z)[h(\overline{z})]^{\prime}=Ah^{\prime}(z), where A=(1+i33)η(1i33)A=\frac{(1+i\frac{\sqrt{3}}{3})}{\eta(1-i\frac{\sqrt{3}}{3})}. Integrating this equation yields h(z¯)=Ah(z)+Bh(\overline{z})=Ah(z)+B, where BB is a fixed constant.

We know that (h(z))\wp(h(z)) has infinitely many poles. Suppose that (h(b0))=\wp(h(b_{0}))=\infty. The equation (2.14) yields that (h(b0¯))=0\wp(h(\overline{b_{0}}))=0 or (h(b0¯))=\wp(h(\overline{b_{0}}))=\infty. Assume (h(b0¯))=\wp(h(\overline{b_{0}}))=\infty. We rewrite (2.14) as

1+(h(z¯))3(h(z¯))[h(z¯)]=η(1(h(z))3)(h(z))[h(z¯)]=η(1(h(z))3)(h(z))Ah(z).\frac{1+\frac{\wp^{\prime}(h(\overline{z}))}{\sqrt{3}}}{\wp(h(\overline{z}))}[h(\overline{z})]^{\prime}=\frac{\eta(1-\frac{\wp^{\prime}(h(z))}{\sqrt{3}})}{\wp(h(z))}[h(\overline{z})]^{\prime}=\frac{\eta(1-\frac{\wp^{\prime}(h(z))}{\sqrt{3}})}{\wp(h(z))}Ah^{\prime}(z). (2.22)

Further, we rewrite (2.22) as

[h(z¯)](h(z¯))+13[(h(z¯))](h(z¯))=Aη[h(z)(h(z))13[(h(z))](h(z))].\displaystyle\frac{[h(\overline{z})]^{\prime}}{\wp(h(\overline{z}))}+\frac{1}{\sqrt{3}}\frac{[\wp(h(\overline{z}))]^{\prime}}{\wp(h(\overline{z}))}=A\eta[\frac{h^{\prime}(z)}{\wp(h(z))}-\frac{1}{\sqrt{3}}\frac{[\wp(h(z))]^{\prime}}{\wp(h(z))}]. (2.23)

Note that (h(z))\wp(h(z)) only has multiple poles. Then, assume that b0b_{0} is a pole of (h(z))\wp(h(z)) and (h(z¯))\wp(h(\overline{z})) with multiplicities m(2)m(\geq 2) and k(2)k(\geq 2), respectively. By taking the residue of both sides of (2.23) at b0b_{0}, we have

k13=Aηm3,-k\frac{1}{\sqrt{3}}=A\eta\frac{m}{\sqrt{3}},

which implies that Aη=kmA\eta=-\frac{k}{m}. It contradicts with Aη=1+i33(1i33)=3+i(3i)A\eta=\frac{1+i\frac{\sqrt{3}}{3}}{(1-i\frac{\sqrt{3}}{3})}=\frac{\sqrt{3}+i}{(\sqrt{3}-i)}.

So, (h(b0¯))=0\wp(h(\overline{b_{0}}))=0 and (h(b0¯))=±i\wp^{\prime}(h(\overline{b_{0}}))=\pm i. Suppose that (h(b0¯))=i\wp^{\prime}(h(\overline{b_{0}}))=i. Assume that b0b_{0} is a zero of (h(z¯))\wp(h(\overline{z})) with multiplicity ss. We rewrite (2.23) as

1(h(z¯))[(h(z¯))](h(z¯))+13[(h(z¯))](h(z¯))\displaystyle\frac{1}{\wp^{\prime}(h(\overline{z}))}\frac{[\wp(h(\overline{z}))]^{\prime}}{\wp(h(\overline{z}))}+\frac{1}{\sqrt{3}}\frac{[\wp(h(\overline{z}))]^{\prime}}{\wp(h(\overline{z}))} =[h(z¯)](h(z¯))+13[(h(z¯))](h(z¯))\displaystyle=\frac{[h(\overline{z})]^{\prime}}{\wp(h(\overline{z}))}+\frac{1}{\sqrt{3}}\frac{[\wp(h(\overline{z}))]^{\prime}}{\wp(h(\overline{z}))} (2.24)
=Aη[h(z)(h(z))13[(h(z))](h(z))].\displaystyle=A\eta[\frac{h^{\prime}(z)}{\wp(h(z))}-\frac{1}{\sqrt{3}}\frac{[\wp(h(z))]^{\prime}}{\wp(h(z))}].

Again by taking the residue of both sides of (2.24) at b0b_{0}, we have

s[1i+13]=Aηm3,s[\frac{1}{i}+\frac{1}{\sqrt{3}}]=A\eta\frac{m}{\sqrt{3}},

which implies that Aη=sm(1i3)A\eta=\frac{s}{m}(1-i\sqrt{3}). Combining Aη=3+i(3i)A\eta=\frac{\sqrt{3}+i}{(\sqrt{3}-i)} yields that

sm=3+i(3i)(1i3)=1+3i4,\frac{s}{m}=\frac{\sqrt{3}+i}{(\sqrt{3}-i)(1-i\sqrt{3})}=\frac{-1+\sqrt{3}i}{4},

which is a contradiction. Therefore, (h(b0¯))=i\wp^{\prime}(h(\overline{b_{0}}))=-i. Thus, all the poles of (h(z))\wp(h(z)) must be the zero of (h(z¯))+i\wp^{\prime}(h(\overline{z}))+i. Suppose that (h(z0))=0\wp(h(z_{0}))=0 and (h(z0))=i\wp^{\prime}(h(z_{0}))=i. The same argument also yields (h(z0¯))=0\wp(h(\overline{z_{0}}))=0 and (h(z0¯))=i\wp^{\prime}(h(\overline{z_{0}}))=i, which means that all the zeros (h(z))i\wp^{\prime}(h(z))-i must be the zero of (h(z¯))i\wp^{\prime}(h(\overline{z}))-i. We denote the facts as follows.

(h(z))=(h(z¯))+i=0,(h(z))i=0(h(z¯))i=0.\wp(h(z))=\infty\Rightarrow\wp^{\prime}(h(\overline{z}))+i=0,~{}~{}\wp^{\prime}(h(z))-i=0\Rightarrow\wp^{\prime}(h(\overline{z}))-i=0. (2.25)

Suppose that 0 is a Picard value of h(z)h(z). Then, we can assume that h(z)=eα(z)h(z)=e^{\alpha(z)}, where α(z)\alpha(z) is an entire function. The equation h(z¯)=Ah(z)+Bh(\overline{z})=Ah(z)+B yields eα(z¯)=Aeα(z)+B=A[eα(z)(BA)]e^{\alpha(\overline{z})}=Ae^{\alpha(z)}+B=A[e^{\alpha(z)}-(-\frac{B}{A})]. So, BA-\frac{B}{A} is also a Picard value of h(z)=eα(z)h(z)=e^{\alpha(z)}, and Picard’s little theorem tells us B=0B=0. Thus, h(z¯)=Ah(z)h(\overline{z})=Ah(z).

Suppose that τ(0)𝕃\tau(\neq 0)\in\mathbb{L}, which means that τ\tau is any fixed period of \wp. Note that τ\tau is not Picard value of h(z)h(z). Assume that h(u0)=τh(u_{0})=\tau. Then, (h(u0))=(τ)=\wp(h(u_{0}))=\wp(\tau)=\infty. From (2.25), one has (h(u0¯))=0\wp(h(\overline{u_{0}}))=0 and (h(u0¯))=i\wp^{\prime}(h(\overline{u_{0}}))=-i. Without loss of generality, we assume that

h(u0¯)=Ah(u0)=Aτ=θ1+ξ1,h(\overline{u_{0}})=Ah(u_{0})=A\tau=\theta_{1}+\xi_{1}, (2.26)

where ξ1𝕃\xi_{1}\in\mathbb{L}. We also know that θ2\theta_{2} is not Picard value of h(z)h(z). Assume h(e0)=θ2h(e_{0})=\theta_{2}. Thus, (h(e0))=(θ2)=0\wp(h(e_{0}))=\wp(\theta_{2})=0 and (h(e0))=(θ2)=i\wp^{\prime}(h(e_{0}))=\wp^{\prime}(\theta_{2})=i. Then, (2.25) yields (h(e0¯))=0\wp(h(\overline{e_{0}}))=0 and (h(e0¯))=i\wp^{\prime}(h(\overline{e_{0}}))=i. Without loss of generality, we assume that

h(e0¯)=Ah(e0)=Aθ2=θ2+ξ2,h(\overline{e_{0}})=Ah(e_{0})=A\theta_{2}=\theta_{2}+\xi_{2}, (2.27)

where ξ2𝕃\xi_{2}\in\mathbb{L}. Clearly, θ2+τ\theta_{2}+\tau is also not Picard value of h(z)h(z). Assume h(t0)=θ2+τh(t_{0})=\theta_{2}+\tau. Then, (h(t0))=(θ2+τ)=0\wp(h(t_{0}))=\wp(\theta_{2}+\tau)=0 and (h(t0))=(θ2+τ)=i\wp^{\prime}(h(t_{0}))=\wp^{\prime}(\theta_{2}+\tau)=i. The same argument as above yields that

h(t0¯)=Ah(t0)=A(θ2+τ)=θ2+ξ3,h(\overline{t_{0}})=Ah(t_{0})=A(\theta_{2}+\tau)=\theta_{2}+\xi_{3}, (2.28)

where ξ3𝕃\xi_{3}\in\mathbb{L}. Combining (2.27) and (2.28) yields that Aτ=ξ3ξ2𝕃A\tau=\xi_{3}-\xi_{2}\in\mathbb{L}. Together with (2.26) yields that θ1𝕃\theta_{1}\in\mathbb{L} is a period of \wp, a contradiction. Thus, 0 is not a Picard value of h(z)h(z).

Below, for any period τ\tau of \wp, we will prove that Aτ𝕃A\tau\in\mathbb{L}, which means that AτA\tau is also a period of \wp.

Note that 0 is not a Picard value of h(z)h(z). Assume that h(d0)=0h(d_{0})=0. Thus, (h(d0))=(0)=\wp(h(d_{0}))=\wp(0)=\infty. Then, (2.25) yields (h(d0¯))=0\wp(h(\overline{d_{0}}))=0 and (h(d0¯))=i\wp^{\prime}(h(\overline{d_{0}}))=-i. Without loss of generality, we assume that

h(d0¯)=Ah(d0)+B=B=θ1+ξ5.h(\overline{d_{0}})=Ah(d_{0})+B=B=\theta_{1}+\xi_{5}. (2.29)

where ξ5𝕃\xi_{5}\in\mathbb{L}. Observe that h(z)h(z) has one finite Picard value at most. So, one of ω1\omega_{1} and ω2\omega_{2} is not a Picard value of h(z)h(z). Without loss of generality, we assume ω1\omega_{1} is not a Picard value of h(z)h(z) and h(c0)=ω1h(c_{0})=\omega_{1}. Then, (h(c0))=(ω1)=(0)=\wp(h(c_{0}))=\wp(\omega_{1})=\wp(0)=\infty, which plus (2.25) implies that (h(c0¯))=0\wp(h(\overline{c_{0}}))=0 and (h(c0¯))=i\wp^{\prime}(h(\overline{c_{0}}))=-i. So, we can set h(c0¯)=θ1+ξ6h(\overline{c_{0}})=\theta_{1}+\xi_{6} with ξ6𝕃\xi_{6}\in\mathbb{L}. Further,

h(c0¯)=Ah(c0)+B=Ah(c0)+θ1+ξ5=Aω1+θ1+ξ5=θ1+ξ6,h(\overline{c_{0}})=Ah(c_{0})+B=Ah(c_{0})+\theta_{1}+\xi_{5}=A\omega_{1}+\theta_{1}+\xi_{5}=\theta_{1}+\xi_{6}, (2.30)

which implies that Aω1=ξ6ξ5𝕃A\omega_{1}=\xi_{6}-\xi_{5}\in\mathbb{L}.

Clearly, there exists an integer NN such that Nω1+τN\omega_{1}+\tau is not a Picard value of h(z)h(z). Suppose that h(t1)=Nω1+τh(t_{1})=N\omega_{1}+\tau. Then, (h(t1))=(Nω1+τ)=(0)=\wp(h(t_{1}))=\wp(N\omega_{1}+\tau)=\wp(0)=\infty, which also implies that (h(t1¯))=0\wp(h(\overline{t_{1}}))=0 and (h(t1¯))=i\wp^{\prime}(h(\overline{t_{1}}))=-i. We can set h(t1¯)=θ1+ξ7h(\overline{t_{1}})=\theta_{1}+\xi_{7} with ξ7𝕃\xi_{7}\in\mathbb{L}. Further,

h(c1¯)=Ah(c1)+θ1+ξ5=A(Nω1+τ)+θ1+ξ5=θ1+ξ7,h(\overline{c_{1}})=Ah(c_{1})+\theta_{1}+\xi_{5}=A(N\omega_{1}+\tau)+\theta_{1}+\xi_{5}=\theta_{1}+\xi_{7}, (2.31)

which leads to

Aτ=ξ7ξ5NAω1𝕃.A\tau=\xi_{7}-\xi_{5}-NA\omega_{1}\in\mathbb{L}.

Thus, we prove that AτA\tau is also a period of \wp.

Observe that A=(1+i33)η(1i33)A=\frac{(1+i\frac{\sqrt{3}}{3})}{\eta(1-i\frac{\sqrt{3}}{3})} and η3=1\eta^{3}=1. A calculation yields A3=1A^{3}=-1 and

A=12+32i=eπ3i,or1232i=eπ3i,or1.A=\frac{1}{2}+\frac{\sqrt{3}}{2}i=e^{\frac{\pi}{3}i},~{}~{}or~{}~{}\frac{1}{2}-\frac{\sqrt{3}}{2}i=e^{-\frac{\pi}{3}i},~{}~{}or~{}~{}-1.

Meanwhile, (2.29) yields that

h(z¯)=Ah(z)+B=Ah(z)+θ1+ξ5,h(\overline{z})=Ah(z)+B=Ah(z)+\theta_{1}+\xi_{5},

which is (1) with τ1=ξ5\tau_{1}=\xi_{5}.

Next, we consider g3(z)0g_{3}(z)\equiv 0 or g2(z)0g_{2}(z)\equiv 0.

If g3(z)0g_{3}(z)\equiv 0, as above discussion, then, one can derive

(h(z))=(h(z¯))i=0,(h(z))+i=0(h(z¯))+i=0.\wp(h(z))=\infty\Rightarrow\wp^{\prime}(h(\overline{z}))-i=0,~{}~{}\wp^{\prime}(h(z))+i=0\Rightarrow\wp^{\prime}(h(\overline{z}))+i=0.

If g2(z)0g_{2}(z)\equiv 0, as above discussion, then, one can derive

(h(z))=(h(z¯))=,(h(z))i=0(h(z¯))+i=0.\wp(h(z))=\infty\Rightarrow\wp(h(\overline{z}))=\infty,~{}~{}\wp^{\prime}(h(z))-i=0\Rightarrow\wp^{\prime}(h(\overline{z}))+i=0.

Further, with the same argument, we can get the conclusions (2) and (3) if g3(z)0g_{3}(z)\equiv 0 and g2(z)0g_{2}(z)\equiv 0, respectively. Here, we omit the details.

Next, we will prove that L(z)L(z) is a linear function with two constant a(0),ba(\neq 0),~{}b. Without loss of generality, we assume that

h(L(z))=Ah(z)+C,h(L(z))=Ah(z)+C, (2.32)

where A3=1A^{3}=-1 and CC is a constant. Suppose that hh is transcendental. We recall the following result (see [11, Theorem 2] and [17, p. 370]).

Lemma 2. If ff (meromorphic) and gg (entire) are transcendental, then

lim suprE,rT(r,f(g))T(r,f)=,\limsup_{r\not\in E,~{}r\rightarrow\infty}\frac{T(r,f(g))}{T(r,f)}=\infty,

where EE is a set of finite Lebesgue measure.

If LL is transcendental, by Lemma 2, we have that

=lim suprE,rT(r,h(L(z)))T(r,h(z))=lim suprE,rT(r,Ah(z)+C)T(r,h(z))=1,\infty=\limsup_{r\not\in E,~{}r\rightarrow\infty}\frac{T(r,h(L(z)))}{T(r,h(z))}=\limsup_{r\not\in E,~{}r\rightarrow\infty}\frac{T(r,Ah(z)+C)}{T(r,h(z))}=1,

a contradiction. Thus, LL is a polynomial. Without loss of generality, we assume L(z)=amzm++a1z+a0L(z)=a_{m}z^{m}+\cdots+a_{1}z+a_{0} with m1m\geq 1 and am0a_{m}\neq 0. We need the following result, which can be seen in [17, (19)]) and [29, (2.7)]), respectively.

Lemma 3. Suppose that g(z)=amzm+am1zm1++a1z+a0g(z)=a_{m}z^{m}+a_{m_{1}}z^{m-1}+...+a_{1}z+a_{0} (am0a_{m}\neq 0) is a non-constant polynomial, then for any ϵ1>0\epsilon_{1}>0 and ϵ2>0\epsilon_{2}>0,

T(r,f(g))(1ϵ2)T(am2rm,f),T(r,f(g))\geq(1-\epsilon_{2})T(\frac{a_{m}}{2}r^{m},f),
T(r,f)1m(1+ϵ1)T(am2rm,f),T(r,f)\leq\frac{1}{m}(1+\epsilon_{1})T(\frac{a_{m}}{2}r^{m},f),

for large rr outside possibly a set of finite Lebesgue measure.

Applying Lemma 3 to the function h(L(z))h(L(z)), we obtain

T(r,h(z))\displaystyle T(r,h(z)) 1m(1+ϵ1)T(am2rm,h(z))1m1+ϵ11ϵ2T(r,h(L(z)))\displaystyle\leq\frac{1}{m}(1+\epsilon_{1})T(\frac{a_{m}}{2}r^{m},h(z))\leq\frac{1}{m}\frac{1+\epsilon_{1}}{1-\epsilon_{2}}T(r,h(L(z))) (2.33)
=1m1+ϵ11ϵ2T(r,Ah(z)+C))=1m1+ϵ11ϵ2T(r,h(z))+O(1),\displaystyle=\frac{1}{m}\frac{1+\epsilon_{1}}{1-\epsilon_{2}}T(r,Ah(z)+C))=\frac{1}{m}\frac{1+\epsilon_{1}}{1-\epsilon_{2}}T(r,h(z))+O(1),

which implies m=1m=1, since ϵ1\epsilon_{1} and ϵ2\epsilon_{2} can be chosen small enough. Thus, L(z)L(z) is a linear function.

Suppose that hh is a polynomial with degree nn. Then, (2.32) yields that h(L(z))h(L(z)) is also a polynomial, which implies that L(z)L(z) is also a polynomial. Further, (2.32) leads to that L(z)L(z) is a linear function.

Thus, the above discussion yields that L(z)=qz+cL(z)=qz+c with q0q\neq 0 and cc is a constant. So, h(qz+c)=Ah(z)+Ch(qz+c)=Ah(z)+C, where CC is a constant.

At the end, we will prove |q|=1|q|=1 by an elementary method. Differentiating the above equation leads to

qh(qz+c)=Ah(z).qh^{\prime}(qz+c)=Ah^{\prime}(z). (2.34)

If h(z)h^{\prime}(z) is a constant, then, q=Aq=A and |q|=|A|=1|q|=|A|=1. Next, we assume that h(z)h^{\prime}(z) is nonconstant. Suppose that |q|<1|q|<1. For each tt^{*}\in\mathbb{C}, by (2.34), we get

h(t)=qAh(qt+c)=(qA)2h[q(qt+c)+c]==(qA)nh[qnt+ck=0n1qk].h^{\prime}(t^{*})=\frac{q}{A}h^{\prime}(qt^{*}+c)=(\frac{q}{A})^{2}h^{\prime}[q(qt^{*}+c)+c]=\cdots=(\frac{q}{A})^{n}h^{\prime}[q^{n}t^{*}+c\sum_{k=0}^{n-1}q^{k}].

Note that |q|<1|q|<1 and qnt+ck=0n1qkc11qq^{n}t^{*}+c\sum_{k=0}^{n-1}q^{k}\rightarrow c\frac{1}{1-q} as nn\rightarrow\infty. Then, let nn\rightarrow\infty, one has that

h(t)=h(c11q)limn(qA)n=0,h^{\prime}(t^{*})=h^{\prime}(c\frac{1}{1-q})\lim_{n\rightarrow\infty}(\frac{q}{A})^{n}=0,

which means that h(z)0h^{\prime}(z)\equiv 0, a contradiction. If |q|>1|q|>1, then we rewrite (2.34) as h(z)=Aph(pz+c)h^{\prime}(z)=Aph^{\prime}(pz+c^{\prime}) with p=1qp=\frac{1}{q} and c=cqc^{\prime}=-\frac{c}{q}. Note that |p|=|1q|<1|p|=|\frac{1}{q}|<1. The same argument as above yields a contradiction. So, we obtain |q|=1|q|=1.

Thus, we finish the proof of this result.

Proof of Proposition 2.

The necessity is obvious. Below, we prove the sufficiency. We rewrite (1.5) as

[f(z)eg(z)n]n+[f(L(z))eg(z)m]m=1.[\frac{f(z)}{e^{\frac{g(z)}{n}}}]^{n}+[\frac{f(L(z))}{e^{\frac{g(z)}{m}}}]^{m}=1.

We will prove that both f(z)eg(z)n\frac{f(z)}{e^{\frac{g(z)}{n}}} and f(L(z))eg(z)m\frac{f(L(z))}{e^{\frac{g(z)}{m}}} are constant.

Observe that Yang’s Theorem yields that both f(z)eg(z)n\frac{f(z)}{e^{\frac{g(z)}{n}}} and f(L(z))eg(z)m\frac{f(L(z))}{e^{\frac{g(z)}{m}}} are constant if 2m+1n<1\frac{2}{m}+\frac{1}{n}<1. So, it is suffice to consider the case (n,m)=(2,3)(n,m)=(2,3) and (2,4)(2,4).

Case 1. (n,m)=(2,3)(n,m)=(2,3).

Suppose that one of f(z)eg(z)2\frac{f(z)}{e^{\frac{g(z)}{2}}} and f(L(z))eg(z)3\frac{f(L(z))}{e^{\frac{g(z)}{3}}} is not constant. Then neither of them is constant. Via (ii) of Theorem A, we have that f(z)eg(z)2=i(h(z))\frac{f(z)}{e^{\frac{g(z)}{2}}}=i\wp^{\prime}(h(z)) and f(L(z))eg(z)3=η43(h(z))\frac{f(L(z))}{e^{\frac{g(z)}{3}}}=\eta\sqrt[3]{4}\wp(h(z)), where h(z)h(z) is any nonconstant entire function. So,

f(L(z))=i(h(L(z)))eg(L(z))2=η43(h(z))eg(z)3.f(L(z))=i\wp^{\prime}(h(L(z)))e^{\frac{g(L(z))}{2}}=\eta\sqrt[3]{4}\wp(h(z))e^{\frac{g(z)}{3}}. (2.35)

Note that wn=nω1w_{n}=n\omega_{1} (n=0,1,2,)(n=0,~{}1,~{}2,...) is a pole of (z)\wp(z) with multiplicity 2. It is known that a nonconstant meromorphic function has four complete multiple values at most. (Here, the constant aa is a complete multiple value of ff if faf-a only has multiple zeros). So, there exists a point wNw_{N} such that h(z)wNh(z)-w_{N} has simple zero, say z0z_{0}. Then, h(z0)=wNh(z_{0})=w_{N} and z0z_{0} is a pole of the function η43(h(z))eg(z)3\eta\sqrt[3]{4}\wp(h(z))e^{\frac{g(z)}{3}} with multiplicity 2. On the other hand, in view of that all the poles of (w)\wp^{\prime}(w) is 3, we derive that z0z_{0} is a pole of i(h(L(z)))eg(L(z))2i\wp^{\prime}(h(L(z)))e^{\frac{g(L(z))}{2}} at least 3 or z0z_{0} is not a pole of i(h(L(z)))eg(L(z))2i\wp^{\prime}(h(L(z)))e^{\frac{g(L(z))}{2}}. Then comparing the multiplicities of both sides of (2.35) at pole-point z0z_{0}, we have a contradiction.

Case 2. (n,m)=(2,4)(n,m)=(2,4).

we assume that one of f(z)eg(z)2\frac{f(z)}{e^{\frac{g(z)}{2}}} and f(L(z))eg(z)4\frac{f(L(z))}{e^{\frac{g(z)}{4}}} is not a constant. Via (iii) of Theorem A, we have that f(z)eg(z)2=sn(h(z))\frac{f(z)}{e^{\frac{g(z)}{2}}}=sn^{\prime}(h(z)) and f(L(z))eg(z)4=sn(h(z))\frac{f(L(z))}{e^{\frac{g(z)}{4}}}=sn(h(z)), where h(z)h(z) is any nonconstant entire function and snsn is Jacobi elliptic function satisfying sn2=1sn4{sn^{\prime}}^{2}=1-sn^{4}. So,

f(L(z))=sn(h(L(z)))eg(L(z))2=sn(h(z))eg(z)4.f(L(z))=sn^{\prime}(h(L(z)))e^{\frac{g(L(z))}{2}}=sn(h(z))e^{\frac{g(z)}{4}}. (2.36)

Suppose that wnw_{n} (n=0,1,2,)(n=0,~{}1,~{}2,...) is a simple pole of sn(z)sn(z). Similarly as above case, there exists a point wNw_{N} such that h(z)wNh(z)-w_{N} has simple zero, say z1z_{1}. Then, h(z1)=wNh(z_{1})=w_{N} and z1z_{1} is a pole of the function sn(h(z))eg(z)4sn(h(z))e^{\frac{g(z)}{4}} with multiplicity 1. On the other hand, in view of that all the pole of sn(w)sn^{\prime}(w) is 2, we get that z1z_{1} is a pole of sn(h(z))eg(L(z))2sn^{\prime}(h(z))e^{\frac{g(L(z))}{2}} at least 2 or z1z_{1} is not a pole of i(h(L(z)))eg(L(z))2i\wp^{\prime}(h(L(z)))e^{\frac{g(L(z))}{2}}. Then comparing the multiplicities of both sides of (2.36) at pole-point z1z_{1}, we have a contradiction.

Therefore, we derive that both f(z)eg(z)n\frac{f(z)}{e^{\frac{g(z)}{n}}} and f(L(z))eg(z)m\frac{f(L(z))}{e^{\frac{g(z)}{m}}} are nonzero constants, say AA and BB with An+Bm=1A^{n}+B^{m}=1. Thus, f(z)=Aeg(z)nf(z)=Ae^{\frac{g(z)}{n}} and

f(L(z))=Beg(z)m=Aeg(L(z))n,f(L(z))=Be^{\frac{g(z)}{m}}=Ae^{\frac{g(L(z))}{n}},

which implies that

g(L(z))=nmg(z)+nLnBA.g(L(z))=\frac{n}{m}g(z)+nLn\frac{B}{A}. (2.37)

With the same argument as in Proposition 1, we can derive that L(z)L(z) is a linear function, say L(z)=qz+cL(z)=qz+c with constants q(0)q(\neq 0), cc.

If m=nm=n, then (2.37) reduces to g(qz+c)=g(z)+nLnBAg(qz+c)=g(z)+nLn\frac{B}{A}. Then, the same discussion as in Proposition 1 yields |q|=1|q|=1.

Thus, we finish the proof of this result.

Based on Theorem 1, we give the proof of Theorem 2 as follows.

Proof of Theorem 2.

The necessity is obvious. Below, we prove the sufficiency. Suppose that the hyper-order of f(z)f(z) is less than 1. Then by Theorem B, we get the conclusion (a). Next, we assume that the hyper-order ρ2(f)1\rho_{2}(f)\geq 1. Rewrite (1.8) as

(f(z+c)eP(z)3)3+(f(z)eP(z)3)3=1.(\frac{f(z+c)}{e^{\frac{P(z)}{3}}})^{3}+(\frac{f(z)}{e^{\frac{P(z)}{3}}})^{3}=1.

Via (i) of Theorem A, we have

f(z)=121+(h(z))3(h(z))eP(z)/3f(z)=\frac{1}{2}\frac{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))}e^{P(z)/3} (2.38)

where h(z)h(z) is an entire function. The fact ρ2(f)1\rho_{2}(f)\geq 1 and (2.38) yields that hh is transcendental. Here, we employ a result of Bergweiler in [4, Lemma 1].

Lemma 3. Let ff be meromorphic and let gg be entire and transcendental. If the lower order μ(fg)<,\mu(f\circ g)<\infty, then μ(f)=0\mu(f)=0.

Note that ρ()=μ()=2\rho(\wp)=\mu(\wp)=2, see [2]. It follows from Lemma 3 that μ[121+(h(z))3(h(z))]=\mu[\frac{1}{2}\frac{1+\frac{\wp^{\prime}(h(z))}{\sqrt{3}}}{\wp(h(z))}]=\infty, so is μ(f)\mu(f). Thus ePe^{P} is a small function of f(z)f(z) and T(r,eP)=S(r,f)T(r,e^{P})=S(r,f). Based on the idea of Liu and Ma in [35], we will obtain the desired result.

Set a(z)=eP(z)a(z)=e^{P(z)} and rewrite (1.8) as f3(z+c)=(f3(z)a(z))f^{3}(z+c)=-(f^{3}(z)-a(z)), which implies that the zeros of f3(z)a(z)f^{3}(z)-a(z) are of multiplicities at least 3. Rewrite (1.8) as f3(z)=(f3(z+c)a(z))f^{3}(z)=-(f^{3}(z+c)-a(z)), which implies that the zeros of f3(z+c)a(z)f^{3}(z+c)-a(z) are of multiplicities at least 3. So, the zeros of f3(z)a(zc)f^{3}(z)-a(z-c) are of multiplicities at least 3. Set G(z)=f3(z)G(z)=f^{3}(z). Assume that the functions a(zc)a(z-c) and a(z)a(z) are distinct from each other. Then, applying the second main theorem of Nevanlinna to GG, one gets that

2T(r,G)\displaystyle 2T(r,G) N¯(r,1G(z)a(zc))+N¯(r,1G(z)a(z))+N¯(r,G)+N¯(r,1G)\displaystyle\leq\overline{N}(r,\frac{1}{G(z)-a(z-c)})+\overline{N}(r,\frac{1}{G(z)-a(z)})+\overline{N}(r,G)+\overline{N}(r,\frac{1}{G})
+S(r,G)13[N(r,1G(z)a(zc))+N(r,1G(z)a(z))+N(r,G)\displaystyle+S(r,G)\leq\frac{1}{3}[N(r,\frac{1}{G(z)-a(z-c)})+N(r,\frac{1}{G(z)-a(z)})+N(r,G)
+N(r,1G)]+S(r,G)43T(r,G)+S(r,G),\displaystyle+N(r,\frac{1}{G})]+S(r,G)\leq\frac{4}{3}T(r,G)+S(r,G),

which is a contradiction. Thus, a(z)=a(zc)a(z)=a(z-c), which implies that eP(z+c)=eP(z)e^{P(z+c)}=e^{P(z)} and eP(z+c)P(z)=1e^{P(z+c)-P(z)}=1. In view of that P(z)P(z) is a polynomial, we derive that P(z)=αz+βP(z)=\alpha z+\beta with two constants α,β\alpha,~{}\beta and eαc=1e^{\alpha c}=1. Therefore, we can rewrite (1.8) as

1=[f(z)eP(z)3]3+[f(z+c)eP(z)3]3=[f(z)eP(z)3]3+[f(z)eP(z+c)3]3.1=[\frac{f(z)}{e^{\frac{P(z)}{3}}}]^{3}+[\frac{f(z+c)}{e^{\frac{P(z)}{3}}}]^{3}=[\frac{f(z)}{e^{\frac{P(z)}{3}}}]^{3}+[\frac{f(z)}{e^{\frac{P(z+c)}{3}}}]^{3}.

Set F(z)=f(z)eP(z)3F(z)=\frac{f(z)}{e^{\frac{P(z)}{3}}}. Then, F(z)3+F(z+c)3=1F(z)^{3}+F(z+c)^{3}=1. By Theorem 1, we get the desired result.

Thus, we finish the proof of Theorem 2. ∎

At the end of this section, we give the proof of Theorem 3.

Proof of Theorem 3.

It is suffice to consider the case ρ(g(z))<1\rho(g(z))<1. Suppose that ff is a nonconstant meromorphic solution of (1.9), then gg is nonconstant, since f(z)=Aeg(z)nf(z)=Ae^{\frac{g(z)}{n}}. Differentiating the equation g(z+c)=nmg(z)+nLnBAg(z+c)=\frac{n}{m}g(z)+nLn\frac{B}{A} one time yields that

g(z+c)=nmg(z).g^{\prime}(z+c)=\frac{n}{m}g^{\prime}(z). (2.39)

Assume that gg^{\prime} is not a constant. If 0 is a Picard value of g(z)g^{\prime}(z), then, g(z)=eh(z)g^{\prime}(z)=e^{h(z)}, where h(z)h(z) is a nonconstant entire function. So, the order ρ(g(z))=ρ(g(z))=ρ(eh(z))1\rho(g(z))=\rho(g^{\prime}(z))=\rho(e^{h(z)})\geq 1, a contradiction. Therefore, there exists a point z2z_{2} such that g(z2)=0g^{\prime}(z_{2})=0. The above equation yields that z2+ncz_{2}+nc also zero of gg^{\prime}. So, n(d|c|+|z2|,g)dn(d|c|+|z_{2}|,g^{\prime})\geq d for any dd\in\mathbb{N}, where n(r,g)n(r,g^{\prime}) denotes the number of zeros of gg^{\prime} in {z:|z|<r}\{z:|z|<r\}. Based on the method in [20, Lemma 3.2], we will derive a contradiction as follows.

ρ(g(z))\displaystyle\rho(g(z)) =ρ(g(z))lim suprlogn(r,g)logrlim supdlogn(d|c|+|z2|,g)log(d|c|+|z2|)\displaystyle=\rho(g^{\prime}(z))\geq\limsup_{r\rightarrow\infty}\frac{\log n(r,g^{\prime})}{\log r}\geq\limsup_{d\rightarrow\infty}\frac{\log n(d|c|+|z_{2}|,g^{\prime})}{\log(d|c|+|z_{2}|)}
lim supdlogdlog(d|c|+|z2|)=1,\displaystyle\geq\limsup_{d\rightarrow\infty}\frac{\log d}{\log(d|c|+|z_{2}|)}=1,

a contradiction. Thus, gg^{\prime} is a nonzero constant and (2.39) yields that m=nm=n. Further, g(z)=αz+βg(z)=\alpha z+\beta, and f(z)=Aeαz+βnf(z)=Ae^{\frac{\alpha z+\beta}{n}} with An(1+eαc)=1A^{n}(1+e^{\alpha c})=1.

Thus, we finish the proof of Theorem 3.

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