# Fixed Points of Meromorphic Solutions for Some Difference Equations

Authors: Zong-Xuan Chen; Kwang Ho Shon
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We investigate fixed points of  meromorphic solutionsy(z) for the Pielou logistic equation and obtain some estimates of exponents of convergence of fixed points of  y(z) and its shifts y(z+n), differences Δy(z)=y(z+1)−y(z), and divided differences Δy(z)/y(z).


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## Body

## 1. Introduction and Results

In this paper, we assume the reader is familiar with basic notions of Nevanlinna’s value distribution theory (see [1–3]). In addition, we use the notation σ(f) to denote the order of growth of a meromorphic function f and λ(f) and λ(1/f) to denote, respectively, the exponents of convergence of zeros and poles of f. We also use the notation τ(f) to denote the exponent of convergence of fixed points of f that is defined as
(1)τ(f)=lim¯r→∞logN(r,1/(f(z)-z))logr.Recently, a number of papers (including [4–17]) focus on complex difference equations and difference analogues of Nevanlinna’s theory.The Pielou logistic equation(2)y(z+1)=R(z)y(z)Q(z)+P(z)y(z),
where P(z),Q(z),and R(z) are nonzero polynomials, is an important difference equation, because it is obtained by transform from the well-known Verhulst Pearl equation (see [18, page 99])
(3)x′(t)=x(t)[a-bx(t)](a,b>0),
which is the most popular continuous model of growth of a population.Chen [7] obtained the following theorem.Theorem A.
LetP(z),Q(z),and R(z) be polynomials with P(z)Q(z)R(z)≢0 and let y(z) be a finite order transcendental meromorphic solution of (2). Then
(4)λ(1y)=σ(y)≥1.Example 1.
The functiony(z)=z2z/(2z-1) satisfies the Pielou logistic equation
(5)y(z+1)=2(z+1)y(z)z+y(z),
where y(z) satisfies
(6)(y)=0,λ(1y)=σ(y)=1.
This example shows that the result of Theorem A is sharp.One of the main purposes in this paper is to study fixed points of meromorphic solutions of the Pielou logistic equation (2).The problem of fixed points of meromorphic functions is an important one in the theory of meromorphic functions. Many papers and books (including [18–20]) investigate fixed points of meromorphic functions.Now we consider fixed points of meromorphic functions and their shifts, differences, and divided differences. We see that there are many examples to show that eitherf(z) may have no fixed point, for example, f1(z)=ez+z or the shift f(z+c) of f(z), or the difference Δcf(z)=f(z+c)-f(z) of f(z) may have only finitely many fixed points; for example, for the function f2(z)=ez+z-1, its shift f2(z+1)=eez+z, and its difference Δ2πif2(z)=f2(z+2πi)-f2(z)=2πi have only finitely many fixed points. Even if for a meromorphic function of small growth, Chen and Shon show that there exists a meromorphic function f0 such that σ(f0)<1 and Δcf0(z)=f0(z+c)-f0(z) has only finitely many fixed points (see Theorem 6 of [9]).A divided difference(f(z+c)-f(z))/f(z) may also have only finitely many fixed points; for example, the function f(z)=zez satisfies that its divided difference (f(z+1)-f(z))/f(z)=((z+1)e-z)/z has only finitely many fixed points. Chen and Shon obtained Theorem B.Theorem B (see [9]).
Letc∈ℂ∖{0} be a constant and let f be a transcendental meromorphic function of order of growth σ(f)=σ<1 or of the form f(z)=h(z)eaz, where a≠0 is a constant and h(z) is a transcendental meromorphic function with σ(h)<1. Suppose that p(z) is a nonconstant polynomial. Then
(7)G(z)=f(z+c)-f(z)f(z)-p(z)
has infinitely many zeros.From Theorem B, we easily see that under conditions of Theorem B, the divided differenceG1(z)=(f(z+c)-f(z))/f(z) has infinitely many fixed points. The previous example f(z)=zez shows that result of Theorem B is sharp.However, we discover that the properties on fixed points of meromorphic solutions of (2) are very good. We prove the following theorem.Theorem 2.
LetP(z),Q(z),and R(z) be nonzero polynomials such that
(8)degP(z)≥max{degR(z),degQ(z)},degP(z)≥1.
Set Δy(z)=y(z+1)-y(z). Then every finite order transcendental meromorphic solution y(z) of (2) satisfies the following:(i)
τ(y(z+n))=σ(y(z))≥1(n=0,1,…);(ii)
ifR(z)-(z+1)Q(z)≢0, then τ(Δy(z)/y(z))=σ(y(z));(iii)
if there is a polynomialh(z) satisfying
(9)(-R(z)+Q(z)+zP(z))2-4zP(z)Q(z)=h(z)2,then τ(Δy(z))=σ(y(z)).Remark 3.
Generally,τ(f(z))≠τ(f(z+c)) for a meromorphic function f(z) of finite order. For example, the function f1(z)=ez+z satisfies
(10)τ(f1(z))=0≠τ(f1(z+1))=1.


## 2. Proof of Theorem2

We need the following lemmas for the proof of Theorem2.Lemma 4 (see [12, 17]).
Letw(z) be a nonconstant finite order meromorphic solution of
(11)P(z,w)=0,
where P(z,w) is a difference polynomial in w(z). If P(z,a)≢0 for a meromorphic function a(z) satisfying T(r,a)=S(r,w), then
(12)m(r,1w-a)=S(r,w)
holds for all r outside of a possible exceptional set with finite logarithmic measure.Remark 5.
Using the same method as in the proof of Lemma4 (see [12]), we can prove that in Lemma 4, if all coefficients bλ(z) of P(z,w) satisfy σ(bλ(z))=σ1<σ(w(z))=σ and if P(z,a)≢0 for a meromorphic function a(z) satisfying T(r,a)=S(r,w), then for a given ε(0<ε<σ-σ1),
(13)m(r,1w(z)-a(z))=S(r,w(z))+O(rσ1+ε)
holds for all r outside of a possible exceptional set with finite logarithmic measure.Lemma 6.
Suppose thatR(z),Q(z),and P(z) satisfy the condition (8) in Theorem 2 and that y(z) is a nonconstant meromorphic function. Then
(14)f1(z)=(R(z)-zP(z))y(z)-zQ(z),f2(z)=P(z)y(z)+Q(z)
have at most finitely many common zeros.Proof.
Suppose thatz0 is a common zero of f1(z) and f2(z). Then f2(z0)=P(z0)y(z0)+Q(z0)=0. Thus, y(z0)=-Q(z0)/P(z0). Substituting y(z0)=-Q(z0)/P(z0) into f1(z), we obtain
(15)f1(z0)=-Q(z0)P(z0)(R(z0)-z0P(z0))-z0Q(z0)=-R(z0)Q(z0)P(z0)=0.
Since R(z)Q(z)/P(z) has only finitely many zeros, we see that f1(z) and f2(z) have at most finitely many common zeros.Lemma 7 (see [14]).
Letf(z) be a nonconstant finite order meromorphic function. Then
(16)T(r+1,f(z))=T(r,f(z))+S(r,f(z)).Gol’dberg and Ostrovskii [21, page 66] give that for any constant b,
(17)(1+o(1))T(r-|b|,f(z))≤T(r,f(z+b))≤(1+o(1))T(r+|b|,f(z)).
This and Lemma 7 give the following lemma.Lemma 8.
Letf(z) be a nonconstant finite order meromorphic function. Then
(18)T(r,f(z+1))=T(r,f(z))+S(r,f(z)).Using the same method as in the proof of Lemma6, we can prove Lemmas 9 and 10.Lemma 9.
Suppose thatR(z),Q(z),and P(z) satisfy the condition (8) in Theorem 2 and that y(z) is a nonconstant meromorphic function. Then
(19)f1(z)=R(z)-(z+1)Q(z)-(z+1)P(z)y(z),f2(z)=P(z)y(z)+Q(z)
have at most finitely many common zeros.Lemma 10.
Suppose thatR(z),Q(z),and P(z) satisfy the condition (8) in Theorem 2 and y(z) is a nonconstant meromorphic function. Then
(20)f1(z)=P(z)y(z)2+[-R(z)+Q(z)+zP(z)]y(z)+zQ(z),f2(z)=P(z)y(z)+Q(z)
have at most finitely many common zeros.Proof of Theorem2.
(i) We prove thatτ(y(z+n))=σ(y(z))≥1(n=0,1,…). Suppose that n=0. Set y(z)-z=g(z). So, g(z) is transcendental, T(r,g(z))=T(r,y(z))+O(logr), and S(r,g)=S(r,y). Substituting y(z)=g(z)+z into (2), we obtain
(21)K0(z,g):=P(z)[g(z+1)+z+1][g(z)+z]+Q(z)[g(z+1)+z+1]-R(z)[g(z)+z]=0.
Thus,
(22)K0(z,0)=z(z+1)P(z)+(z+1)Q(z)-zR(z).
By (8) and (22), we see that K0(z,0)≢0. Thus, by Lemma 4 and K0(z,0)≢0, we obtain
(23)N(r,1g(z))=T(r,g(z))+S(r,g(z))
for all r outside of a possible exceptional set with finite logarithmic measure. Thus,
(24)N(r,1y(z)-z)=T(r,y(z))+S(r,y(z))
for all r outside of a possible exceptional set with finite logarithmic measure. So, by Theorem A and (24), we obtain τ(y(z))=σ(y(z))≥1.
Now suppose thatn=1. By (2), we obtain
(25)y(z+1)-z=(R(z)-zP(z))y(z)-zQ(z)Q(z)+P(z)y(z)=(R(z)-zP(z))[y(z)-zQ(z)/(R(z)-zP(z))]Q(z)+P(z)y(z).
By (8), we see that R(z)-zP(z)≢0. Since P(z),Q(z),and R(z) are polynomials, by (25), we see that y(z)-zQ(z)/(R(z)-zP(z)) and Q(z)+P(z)y(z) have the same poles, except possibly finitely many poles. By Lemma 6, we see that (R(z)-zP(z))y(z)-zQ(z) and Q(z)+P(z)y(z) have at most finitely many common zeros. Hence, by (25), we have that
(26)τ(y(z+1))=λ(y(z+1)-z)=λ(y(z)-zQ(z)R(z)-zP(z)).
Suppose that λ(y(z)-zQ(z)/(R(z)-zP(z)))<σ(y(z)). Thus, y(z)-zQ(z)/(R(z)-zP(z)) can be rewritten as the following form:
(27)y(z)-zQ(z)R(z)-zP(z)=zsb0(z)H0(z)eh(z)=b(z)H(z),
where h(z) is a polynomial with degh(z)≤σ(y(z)),  b0(z) and H0(z) are canonical products (b0(z) may be a polynomial) formed by nonzero zeros and poles of y(z)-zQ(z)/(R(z)-zP(z)), respectively, and s is an integer; if s≥0, then b(z)=zsb0(z),H(z)=H0(z)e-h(z); if s<0, then b(z)=b0(z),H(z)=z-sH0(z)e-h(z). Combining Theorem A with properties of canonical product, we see that
(28)λ(b(z))=σ(b(z))=λ(y(z)-zQ(z)R(z)-zP(z))<σ(y(z)),λ(H(z))=σ(H(z))=σ(y(z)).
By (27), we obtain
(29)y(z)=zQ(z)R(z)-zP(z)+b(z)f(z),y(z+1)=(z+1)Q(z+1)R(z+1)-(z+1)P(z+1)+b(z+1)f(z+1),
where f(z)=1/H(z). Thus, by (28) and Lemma 8, we have that
(30)σ(f(z))=σ(H(z))=σ(y(z)),σ(b(z+1))=σ(b(z))<σ(f(z)).
Substituting (29) into (2), we obtain
(31)K1(z,f):={(z+1)Q(z+1)R(z+1)-(z+1)P(z+1)+b(z+1)f(z+1)}·{Q(z)+P(z)[zQ(z)R(z)-zP(z)+b(z)f(z)]}-R(z)[zQ(z)R(z)-zP(z)+b(z)f(z)]=0.
By (31), we obtain
(32)K1(z,0)=[(z+1)Q(z+1)Q(z)R(z)-zQ(z)R(z)[R(z+1)-(z+1)P(z+1)]]×([R(z+1)-(z+1)P(z+1)][R(z)-zP(z)])-1.
By (8), we see that in the numerator of the right side of (32), there exists only one term z(z+1)Q(z)R(z)P(z+1) being of the highest degree. So,
(33)K1(z,0)≢0.
Thus, by (28), (33), Lemma 4, and its Remark 5, we obtain that for any given ε(0<ε<σ(y(z))-σ(b(z)))(34)N(r,1f(z))=T(r,f(z))+S(r,f(z))+O(rσ(b(z))+ε)
holds for all r outside of a possible exceptional set with finite logarithmic measure.
On the other hand, byf(z)=1/H(z) and the fact that H(z) is an entire function, we see that
(35)N(r,1f(z))=0.
Thus, by this and (28), we see that (34) is a contradiction. Hence, λ(y(z)-zQ(z)/(R(z)-zP(z)))=σ(y(z)). By (26), we obtain
(36)τ(y(z+1))=σ(y(z)).
Now suppose thatn=2. By (2), we obtain
(37)g(z+1)=R(z+1)g(z)Q(z+1)+P(z+1)g(z),
where g(z)=y(z+1). By Lemma 8, we have that σ(g(z))=σ(y(z)). By (8), we have
(38)degP(z+1)≥max{degR(z+1),degQ(z+1)},degP(z+1)≥1.
Thus, for (37), applying the conclusion of n=1 above, we obtain
(39)τ(y(z+2))=τ(g(z+1))=σ(g(z))=σ(y(z)).
Continuing to use the same method as above, we can obtain
(40)τ(y(z+n))=σ(y(z))(n=1,2,…).
(ii) Suppose thatR(z)-(z+1)Q(z)≢0. We prove that τ(Δy(z)/y(z))=σ(y(z)). By (2), we obtain
(41)Δy(z)y(z)-z=y(z+1)-y(z)y(z)-z=R(z)-(z+1)Q(z)-(z+1)P(z)y(z)Q(z)+P(z)y(z)=-(z+1)P(z)×(y(z)-(R(z)-(z+1)Q(z))/(z+1)P(z))×(Q(z)+P(z)y(z))-1.
Since y(z)-(R(z)-(z+1)Q(z))/(z+1)P(z) and Q(z)+P(z)y(z) have the same poles, except possibly finitely many poles, by Lemma 9 and (41), we only need to prove that
(42)λ(y(z)-R(z)-(z+1)Q(z)(z+1)P(z))=σ(y(z)).
Set(43)h1(z)=R(z)-(z+1)Q(z),h2(z)=(z+1)P(z).
Suppose that λ(y(z)-h1(z)/h2(z))<σ(y(z)). Using the same method as in the proof of (i), y(z)-h1(z)/h2(z) can be rewritten as the following form:
(44)y(z)=h1(z)h2(z)+b2(z)f2(z),
where f2(z)=1/H2(z) and b2(z) and H2(z) are nonzero entire functions, such that
(45)λ(b2(z))=σ(b2(z))<σ(y(z)),λ(H2(z))=σ(H2(z))=σ(y(z)).
Substituting (44) into (2), we obtain
(46)D2(z,f2(z)):={h1(z+1)h2(z+1)+b2(z+1)f2(z+1)}×{Q(z)+P(z)h1(z)h2(z)+P(z)b2(z)f2(z)}-R(z){h1(z)h2(z)+b2(z)f2(z)}=0,D2(z,0)=h1(z+1)h2(z+1){Q(z)+P(z)h1(z)h2(z)}-R(z)h1(z)h2(z)=h1(z+1)h2(z+1){Q(z)+P(z)R(z)-(z+1)Q(z)(z+1)P(z)}-R(z)h1(z)h2(z)=R(z)h1(z+1)h2(z)-(z+1)h1(z)h2(z+1)(z+1)h2(z+1)h2(z).
Since h1(z) and h2(z) are polynomials, we see that
(47)h1(z+1)h2(z)-(z+1)h1(z)h2(z+1)≢0,
that is,
(48)D2(z,0)≢0.
Using the same method as in the proof of (i), we obtain a contradiction. Hence, (42) holds; that is, τ(Δy(z)/y(z))=σ(y(z)).
(iii) Suppose that there is a polynomialh(z) satisfying
(49)(-R(z)+Q(z)+zP(z))2-4zP(z)Q(z)=h(z)2.
Thus, by (8) and (49), we see that
(50)degh(z)=degP(z)+1.
Now we prove thatτ(Δy(z))=σ(y(z)). By (2), we obtain
(51)y(z+1)-y(z)-z=-P(z)y(z)2+[-R(z)+Q(z)+zP(z)]y(z)+zQ(z)Q(z)+P(z)y(z).
By (49) and (51), we obtain
(52)y(z+1)-y(z)-z=-P(z){y(z)--(zP(z)+Q(z)-R(z))+h(z)2P(z)}×{y(z)--(zP(z)+Q(z)-R(z))-h(z)2P(z)}×(Q(z)+P(z)y(z))-1.
Since P(z),Q(z),and R(z) are polynomials, we see that poles of Q(z)+P(z)y(z) must be poles of P(z)y(z)2+[-R(z)+Q(z)+zP(z)]y(z)+zQ(z). Thus, poles of Q(z)+P(z)y(z) are not zeros of y(z+1)-y(z)-z. By Lemma 10, we see that the numerator and the denominator of the right side of (51) have at most finitely many common zeros. Thus, in order to prove τ(Δy(z))=σ(y(z)), by (52), we only need to prove that
(53)λ(y(z)--(zP(z)+Q(z)-R(z))+h(z)2P(z))=σ(y(z))
or
(54)λ(y(z)--(zP(z)+Q(z)-R(z))-h(z)2P(z))=σ(y(z)).
By (50), we have deg(-zP(z))=degh(z)=degP(z)+1. Combining this with (8), we see that there exists at least one of
(55)-(zP(z)+Q(z)-R(z))+h(z),-(zP(z)+Q(z)-R(z))-h(z),
such that its degree is equal to degP(z)+1. Without loss of generality, we may suppose that
(56)deg(-(zP(z)+Q(z)-R(z))+h(z))=degP(z)+1.
Now we prove that (53) holds. Suppose that
(57)λ(y(z)--(zP(z)+Q(z)-R(z))+h(z)2P(z))<σ(y(z)).
Using a similar method as in the proof of (i), we see that y(z)-(-(zP(z)+Q(z)-R(z))+h(z))/2P(z) can be rewritten as the following form:
(58)y(z)=-(zP(z)+Q(z)-R(z))+h(z)2P(z)+b3(z)f3(z),
where f3(z)=1/H3(z) and  b3(z) and H3(z) are nonzero entire functions, such that
(59)λ(b3(z))=σ(b3(z))<σ(y(z)),λ(H3(z))=σ(H3(z))=σ(y(z)).
Substituting (58) into (2), we obtain
(60)D3(z,f3(z)):={-((z+1)P(z+1)+Q(z+1)-R(z+1))+h(z+1)2P(z+1)+b3(z+1)f3(z+1)-((z+1)P(z+1)+Q(z+1)-R(z+1))+h(z+1)2P(z+1)}·{Q(z)+P(z)-(zP(z)+Q(z)-R(z))+h(z)2P(z)+P(z)b3(z)f3(z)-(zP(z)+Q(z)-R(z))+h(z)2P(z)}-R(z)-(zP(z)+Q(z)-R(z))+h(z)2P(z)-R(z)b3(z)f3(z)=0,D3(z,0):={-(z+1)P(z+1)-Q(z+1)+R(z+1)+h(z+1)2P(z+1)}·{-zP(z)+Q(z)+R(z)+h(z)2}-R(z)-zP(z)-Q(z)+R(z)+h(z)2P(z)=W1(z)-W2(z)4P(z+1)P(z),
where
(61)W1(z)=[-(z+1)P(z+1)-Q(z+1)+R(z+1)+h(z+1)]×[-zP(z)+Q(z)+R(z)+h(z)]P(z),W2(z)=2(-zP(z)-Q(z)+R(z)+h(z))R(z)P(z+1).
By (8), (50), and (56), we see that
(62)deg(-(z+1)P(z+1)-Q(z+1)+R(z+1)+h(z+1))=degP(z)+1,deg(-zP(z)+Q(z)+R(z)+h(z))=degP(z)+1.
Thus, we obtain
(63)degW1(z)=3degP(z)+2,degW2(z)≤3degP(z)+1.
So, by (60) and (63), we see that D3(z,0)≢0.
Using the same method as in the proof of (i), we see that (53) holds.
Thus, Theorem2 is proved.



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