# Unicity of Meromorphic Solutions of the Pielou Logistic Equation

Authors: Sheng Li; Baoqin Chen
Journal: Discrete Dynamics in Nature and Society
 (2020)

---

## Abstract

This paper mainly considers the unicity of meromorphic solutions of the Pielou logistic equationyz+1=Rzyz/Qz+Pzyz, where Pz,Qz, and Rz are nonzero polynomials. It shows that the finite order transcendental meromorphic solution of the Pielou logistic equation is mainly determined by its poles and 1-value points. Examples are given for the sharpness of our result.


---

## Body

## 1. Introduction

For a meromorphic functionfz, we use standard notations of the Nevanlinna theory, such as Tr,f, mr,f, and Nr,f (see, e.g., [1–3]). Let Sr,f denote any quantity that satisfies Sr,f=oTr,f as r⟶∞ possibly outside of an exceptional set of finite logarithmic measure. And we define the order of growth of fz by(1)ρf≔limsupr⟶∞logTr,flogr.Also we know that the unicity of solutions of a given equation is always one of its most essential properties. This paper is to discuss the unicity of meromorphic solutions of the Pielou logistic equation(2)yz+1=RzyzQz+Pzyz,where Pz,Qz, and Rz are nonzero polynomials. Equation (2) is an important equation generalized from the famous Verhulst-Pearl equation, which is the most popular continuous model of growth of a population:(3)x′t=xta−bxt,a,b>0.By denotingfz=1/yz, we can get from (2) that(4)Rzfz+1−Qzfz=Pz,which is a linear difference equation.On the growth, zeros, and poles of meromorphic solutions of (2) and (4), Chen proved numbers of significant results in [4]. Then, Cui and Chen [5, 6] began to consider the unicity of meromorphic solutions concerning their zeros, 1-value points, and poles and proved.Theorem 1.
(see [5]). Let fz be a finite order transcendental meromorphic solution of the equation(5)P1zfz+1+P2zfz=0,where P1zandP2z are nonzero polynomials such that P1z+P2z≡0. If a meromorphic function gz shares 0,1,∞ CM with fz, then either fz≡gz or fzgz≡1.Theorem 2.
(see [6]). Let fz be a finite order transcendental meromorphic solution of the equation(6)P1zfz+1+P2zfz=P3z,where P1z,P2z,andP3z are nonzero polynomials such that P1z+P2z≡0. If a meromorphic function gz shares 0,1,∞ CM with fz, then one of the following cases holds:(i)
fz≡gz,(ii)
fz+gz=fzgz,(iii)
There exists a polynomialβz=az+b0 and a constant a0 satisfying ea0≠eb0 such that(7)fz=1−eβzeβzea0−b0−1,gz=1−eβz1−eb0−a0,where a0≠0,b0 are constants.Here and in the following,fz and gz are said to share the value a CM (IM), provided that fz−a and gz−a have the same zeros counting multiplicities (ignoring multiplicities). And fz and gz are said to share the value ∞ CM (IM), provided that fz and gz have the same poles with the same multiplicities (ignoring multiplicities).Cui and Chen’s work is a natural product of generalization work (see, e.g., [1, 3, 7–11]) on the famous Nevanlinna’s 5 IM (4 CM) Theorem (see, e.g., [3, 12]) during the past, about 90 years, especially of the hot research studies on the complex differences and complex difference equations (see, e.g., [1, 4, 8–10, 13–15]) recently. They have given examples to show that all cases of Theorem A and Theorem B can happen, and the numbers of shared values cannot be reduced. Li and Chen [16] turned to consider the following question: What can we say about the unicity of finite order transcendental meromorphic solutions of the equation(8)R1zfz+1+R2zfz=R3z,where R1z≡0,R2z,R3z are rational functions? And we proved some interesting results and also provided some examples for sharpness of them. Two of those results read are as follows.Theorem 3.
(see [16]). Let fz and gz be two finite order transcendental meromorphic solutions of equation (8), where R3z≡0. Suppose that fz and gz share 0,∞ CM. Then, either fz≡gz or(9)fz=R3z2R2zea1z+a0+1,gz=R3z2R2ze−a1z−a0+1,where a1,a0 are constants such that e−a1=ea1=−1, and the coefficients of (8) satisfy R1zR3z+1≡R3zR2z+1.Theorem 4.
(see [16]). Let fz and gz be two finite order transcendental meromorphic solutions of equation (8), where(10)R1z+R2z≡R3z,R1zR3z+1−R1z+1≡R3z−R2zR2z+1.Iffz and gz share 1,∞ CM, then fz≡gz.Remark 1.
Notice thatfz and gz share 0 CM if and only if 1/fz and 1/gz share ∞ CM; fz and gz share ∞ CM if and only if 1/fz and 1/gz share 0 CM; and fz and gz share 1 CM if and only if 1/fz and 1/gz share 1 CM. As a result, for the unicity of finite order transcendental meromorphic solutions equation (2), we only need to consider the case that two CM shared values are 1,∞. Indeed, we prove the following Theorem 5, whose proof is different from that in [5, 6, 16].Theorem 5.
Letxz and yz be two finite order transcendental meromorphic solutions of equation (2). If xz and yz share 1,∞ CM and one of the following cases holds:(i)
Rz≡Pz≡−Qz(ii)
Rz≡Pz and xz has infinitely many poles of multiplicity ≥2(iii)
Rz≡Pz, ρx is not an integer, and xz has at most finitely many simple poles, then xz≡yzWe give some examples for the sharpness of Theorem5 as follows.Example 1.
(1)
x
z
=
2
/
e
π
i
z
+
1 and yz=2/e−πiz+1 satisfy the equation
(11)
y
z
+
1
=
y
z
−
1
+
y
z
.
Here,xz and yz share 1,∞ CM such that they have infinitely many poles and ρx=ρy=1 and Rz≡Pz≡−Qz≡1. This example shows that Theorem 5 may not hold for the case Rz≡Pz≡−Qz.
(2)
x
z
=
1
/
e
π
i
z
+
1 and yz=1/e−πiz+1 satisfy the equation
(12)
y
z
+
1
=
y
z
−
1
+
2
y
z
.
Here,xz and yz share 1,∞ CM such that they have infinitely many simple poles and ρx=ρy=1 and Pz≡2≡Rz≡−Qz≡1. This example shows that Theorem 5 may not hold for the case Rz≡Pz if most (except finitely many) poles of xz are simple or ρx is an integer.Remark 2.
It is interesting to ask a question: whether the shared condition “CM” is replaced by “IM” in Theorem5. We have tried hard but failed to find some negative examples for this question. We conjecture that the conclusions in Theorem 5 still hold when the shared condition “CM” is replaced by “IM.”


## 2. Proof of Theorem 5

To prove Theorem5, we need the following lemma of Clunie (see, e.g., [1, 2]).Lemma 1.
(see [1, 2]). Let fz be a transcendental meromorphic solution of the equation(13)fnPz,f=Qz,f,where Pz,f and Qz,f are polynomials in f and its derivatives with meromorphic coefficients, say aλλ∈I, such that mr,aλ=Sr,f for all λ∈I. If the total degree of Qz,f as a polynomial in f and its derivatives is ≤n, then(14)mr,Pz,f=Sr,f.Proof of Theorem 5.
Sincexz and yz are finite order transcendental solutions of equation (2) and share 1,∞ CM, without loss of generality, assume that ρx≥ρy, and we get(15)xz+1=RzxzQz+Pzxz≔xzAz+Bzxz,(16)yz+1=RzyzQz+Pzyz≔yzAz+Bzyz,(17)yz−1xz−1=ehz,where hz is a polynomial such that deghz≤ρx, and Az=Qz/Rz,Bz=Pz/Rz are rational functions.
Ifeh≡1, then our conclusion holds.
Ifeh≡1, then eh¯≡1, and from (17), we have(18)y=ehx+1−eh,y¯=eh¯x¯+1−eh¯.
Here and in the following, we use the notations(19)f¯=fz+1,f¯¯=fz+2,for any given meromorphic function fz for convenience.
Submitting (18) into (16), we have(20)eh¯x¯=Beh¯−B+1ehx+CBehx+A+B1−eh,where(21)C=Aeh¯−1+1−eh+B1−eheh¯−1.
From (15) and (20), we obtain(22)Beh¯−B+1ehx+CBehx+A+B1−eh=eh¯x¯=eh¯xA+Bx,or equally,(23)BB−1eh¯−1ehx2+BC+ABeh¯−B+1eh−A+B1−eheh¯x+AC=0.
Next, we discuss three cases.Case 1.
R
z
≡
P
z
≡
−
Q
z
. Then, Az≡−1, Bz≡1, and(24)C=Aeh¯−1+eh¯1−eh.
Thus, (23) is of the form(25)1−eh¯+hx=Aeh¯−1+eh¯1−eh.
We claim thateh¯+h≡1. Otherwise, eh¯+h≡1, and then hz≡c1 and (25) yields that Az≡A=ec1:=c2∉0,±1.
If there is a pointz1 such that xz1=1, then yz1=1. We can easily deduce from (2), (16), and (18) that(26)1c2+1=yz1c2+yz1=yz1+1=c2xz1+1+1−c2=c2c2+1+1−c2,which gives c2∈0,1, a contradiction to the fact that c2∉0,±1.
If 1 is a Picard exceptional value ofxz, then 1 is also a Picard exceptional value of yz. What is more, from (2) and (16), we see that 1/1+c2 is a Picard exceptional value of xz and yz. Since 1/1+c2≠1, xz has no other Picard exceptional value. Choose a point z2 such that xz2=c2/1+c2, then(27)yz2=c2xz2+1−c2=11+c2.
This indicates that1/1+c2 is not a Picard exceptional value of yz, a contradiction.
Now, we have proved thateh¯+h≡1. From (25), we get(28)x=eh¯+h−A+1eh¯+Aeh¯+h−1.
Sincexz is transcendental, we see that deghz=n≥1. Set(29)hz=anzn+an−1zn−1+⋯+a1z+a0,where a0,a1,…,an are constants such that an=r1eiθ1≠0.
Substituting (28) into (15), we get(30)eh¯¯+h¯−A¯+1eh¯¯+A¯eh¯¯+h¯−1=x¯=xA+x=eh¯+h−A+1eh¯+AA+1eh¯+h−A+1eh¯,which gives Fz=0, where(31)F=Aeh=+2h¯+h−A¯+1A+1eh=+h¯+h+A¯+1A+1−Aeh=+h¯+A¯+1A+1−Aeh¯+h−A¯+1A+1eh¯+A.
SinceAz≡0 is a rational function, there exist some d>0 and r2>1 such that for all z=reiθ, r>r2, and we have(32)A≥r−d.
Notice that(33)hre−iθ1/n=r1rn1+o1,h¯re−iθ1/n=r1rn1+o1,h¯¯re−iθ1/n=r1rn1+o1,as r⟶+∞. From (31) and (32), we can deduce that(34)limr⟶+∞Fre−iθ1/n=limr⟶+∞e4r1rn1+o11+o1=+∞,a contradiction to the fact that Fz=0.Case 2.
R
z
≡
P
z and xz has infinitely many poles of multiplicity ≥2. From (23), we have(35)x2=Dx+E,where(36)D=−BC+ABeh¯−B+1eh−A+B1−eheh¯BB−1eh¯−1eh,E=−ACBB−1eh¯−1eh.Subcase 1.
h
z is a constant. Then, DzandEz are rational functions and hence have at most finitely many poles. Choose a pole of xz with multiplicity k1≥1, denoted by z3, such that Dz3≠∞,Ez3≠∞. Then, z3 is a pole of x2z with multiplicity 2k1 and a pole of Dzxz+Ez with multiplicity k1. However, from (35), we see that it is impossible.Subcase 2.
h
z is a nonconstant polynomial such that deghz=n≥1. Then, from(37)eh¯−1′=h¯′eh¯,we see that eh¯−1 has at most n zeros of multiplicity ≥2. Then, DzandEz are meromorphic functions which have at most finitely many poles of multiplicity ≥2. Choose a pole of xz with multiplicity k2≥2 denoted by z4 such that z4 is not a pole of Dz,Ez of multiplicity ≥2. Then, z4 is a pole of x2z with multiplicity 2k2≥4 and a pole of Dzxz+Ez with multiplicity at most k2+1. However, from (35) and k2+1<2k2, we find that it is also impossible.Case 3.
R
z
≡
P
z
,
ρ
x is not an integer, and xz has at most finitely many simple poles. Then, deghz<ρx since deghz≤ρx. From Case 2, we can suppose that xz has at most finitely many poles of multiplicity ≥2 and use (35) directly. Now, xz has at most finitely many poles.
On the one hand, we have(38)mr,x=Tr,x−Nr,x=Tr,x+Sr,x.
On the other hand, sincedeghz<ρx, it is easy to find that(39)mr,D≤Tr,D=Sr,x,mr,E≤Tr,E=Sr,x.
Applying Lemma1 to (35), we get(40)mr,x=Sr,x,which contradicts to (38). Our proof of Theorem 5 is thus completed.


## 3. Conclusion

Our result shows that the finite order transcendental meromorphic solution of equation (2) is mainly determined by its poles and 1-value points.



---

*Source: 4253967-2020-03-22.xml*