# On the Transcendental Entire Solutions of a Certain Differential Equation and Fixed Points

Authors: Tiantian Shen; Weiran Lü; Guangxue Chu
Journal: ISRN Mathematical Analysis
 (2011)

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## Abstract

We will first show that the following differential equationF(k)−z=eα(F−z) has transcendental entire solutions, where F=fn and α is an entire function, which improve what we have known and answer an open question raised in the work of Zhang and Yang (2009). And then, the examples are discussed.


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## Body

## 1. Introduction and Main Results

In this paper, a meromorphic (resp., entire) function means meromorphic (resp., analytic) in the whole complex plane. We will adopt the standard notations in Nevanlinna's value distribution theory of meromorphic functions, such as the characteristic functionT(r,f),  the counting function of the poles N(r,f),   the proximity function m(r,f), and the reduction counting function N¯(r,f) (see [1–3]). In addition, S(r,f) denotes any quantity satisfying S(r,f)=o{T(r,f)} as r→∞, possibly outside a set of r of finite linear measure.Letf(z) and g(z) be two nonconstant meromorphic functions, a∈C⋃{∞}. We say that f and g share the value a IM if f-a and g-a have the same zeros. If f-a and g-a have the same zeros with the same multiplicities, we say that they share the value a CM. Moreover, if f-z and g-z share 0 CM (resp., IM), we say that f and g share z CM (resp., IM), or we say that f and g have the same fixed-points CM (resp., IM).In 2008, Lei et al. [4] proved the follwing.Theorem A.
Suppose thatf is a nonconstant meromorphic function, n, k are positive integers, and n≥k+5,  a is a nonzero constant. If fn and (fn)(k) share a CM, then f=cewz/n, where c,w are nonzero constants such that wk=1.Recently, Zhang and Yang investigated the power of an entire function sharing one value with its derivative and obtained the following results.Theorem B (see [5]).
Letf be a nonconstant entire function, and let n≥7 be an integer. If fn and (fn)′ share 1 CM, then fn=(fn)′, and f=cez/n, where c is a nonzero constant.Theorem C (see [6]).
Letf be a non-constant entire function, and let  n be a positive integer. If fn and (fn)(k) share 1 CM, and n≥k+1, then fn=(fn)(k), and f=cewz/n, where c,w are nonzero constants such that wk=1.According to the above theorems, one may ask an interesting question: what can be said “iffn and (fn)(k) share z CM”? Our purpose of this paper is to solve this question by giving the transcendental entire solutions of the equation (1.1)F′-z=eα(F-z),
where F=fn, and α is an entire function. As an application, we now use a completely different method from that in [4–6] and give the following results.Theorem 1.1.
Letn(n≥2) be a positive integer, and α be an entire function, then (1.1) has transcendental entire solutions f and
(1.2)f=cez/n,
where c is a nonzero constant.Obviously, from Theorem1.1, we have the following result.Corollary 1.2.
Letf be a transcendental entire function, and let n(n≥2) be a positive integer. If fn and (fn)′ share z CM, then the conclusion of Theorem 1.1 is valid.Obviously, Corollary1.2 improves the results in [6] and answers the open question in [7].In order to illustrate our conditionn≥2 is sharp, we give examples as follows.Example 1.3.
Letf=eez∫0ze-et(1-et)tdt, then f is a non-constant solution of
(1.3)f′-z=ez(f-z),
and f′, f share z CM, while f≢f′.Example 1.4.
Letf=e3z+2z/3+2/9, then f is a non-constant solution of
(1.4)f′-z=3(f-z),
and f, f′ share z CM, but f≢f′.Theorem 1.5.
Letf be a transcendental entire function, n, k positive integers, and n≥k+2; if fn, (fn)(k) share z IM, then f=cewz/n, where c, w are nonzero constants and wk=1.


## 2. Preliminaries

In order to prove the theorems above, we need some lemmas.Lemma 2.1 (see [8, 9]).
Suppose thatf is a meromorphic and transcendental in the whole complex plane and that
(2.1)fnP(f)=Q(f),
where P(f) and Q(f) are differential polynomials in f with functions of small proximity related to f as the coefficients, and the degree of Q(f) is at most n, then
(2.2)m(r,P(f))=S(r,f).Lemma 2.2 (see [1, 10]).
Letf be a transcendental meromorphic function, then
(2.3)m(r,f(l)f)=S(r,f),
for every positive integer l.Lemma 2.3 (see [1, 3]).
Letf be a meromorphic function, and αi(z) with i=1,2,3 are three distinct small functions of f(z), then
(2.4)T(r,f)≤N¯(r,1f-α1)+N¯(r,1f-α2)+N¯(r,1f-α3)+S(r,f).


## 3. Proof of Theorem1.1

By differentiation to (1.1), we have (3.1)F′′-1=α′eα(F-z)+eα(F′-1).
Combining (1.1) and (3.1) yields (3.2)(F′′-1)(F-z)=α′(F′-z)(F-z)+(F′-z)(F′-1),
that is, (3.3)F′′F-α′FF′-(F′)2=(1-α′z)F+zF′′-(α′z+z+1)F′+α′z2.
Substituting F′=nfn-1f′,  F=fn into (3.3) results in (3.4)fnfn-2P=Q,
where (3.5)P=nff′′-nα′ff′-n(f′)2,Q=(1-α′z)fn+z{n(n-1)fn-2(f′)2+nfn-1f′′}-n(α′z+z+1)fn-1f′+α′z2
are differential polynomials in f, and the degree of Q is n. Lemma 2.1 gives m(r,fn-2P)=S(r,f) and (3.6)T(r,fn-2P)=S(r,f).
We now prove P≡0. Suppose that P≢0, and n≥3, then by (3.6) and Lemma 2.1, we obtain T(r,f)=S(r,f), which is impossible. We may now assume that n=2. So, we get (3.7)P=2ff′′-2α′ff′-2(f′)2.
It follows from (3.4) and Lemma 2.2 that T(r,P)=S(r,f), and m(r,P/f2)=S(r,f). These equalities above show that (3.8)m(r,1f)=S(r,f).
In addition, we can see from the expression of P that the multiple zeros of f must be the zeros of P; hence, N(r,1/f)=N¯(r,1/f)+S(r,f). This together with (3.8) and the first theorem (e.g., see [2, Theorem 1.2]) will result in (3.9)T(r,f)=N¯(r,1f)+S(r,f).
By (3.7), we find (3.10)P′=2ff′′′-2α′′ff′-2α′ff′′-2α′(f′)2-2f′f′′.
Let z0 be a simple zero of f; it follows from (3.7) and (3.10) that P(z0)+2{f′(z0)}2=0 and P′(z0)=-2α′(z0){f′(z0)}2-2f′(z0)f′′(z0), which implies that z0 is a zero of Pf′′+(α′P-P′)f′. Let (3.11)g=Pf′′+(α′P-P′)f′f.
Thus, (3.12)T(r,g)=S(r,f).
By (3.11), we obtain (3.13)f′′=α1f+β1f′,
where α1=g/P, β1=(P′-α′P)/P,(3.14)T(r,α1)=S(r,f),T(r,β1)=S(r,f).
Substituting (3.13) into (3.7) will yield (3.15)P=2α1f2+2(β1-α′)ff′-2(f′)2.
Also from (3.13), we find (3.16)f′′′=α2f+β2f′,
where α2=α1′+α1β1, β2=α1+β1′+β12,(3.17)T(r,α2)=S(r,f),T(r,β2)=S(r,f).
It follows from (3.16), (3.13), and (3.10) that (3.18)P′=2(α2-α′α1)f2+2(β2-α′β1-α′′-α1)ff′-2(α′+β1)(f′)2.
From (3.15) and (3.18), we get (3.19)P′=2(α2-α′α1)f2+2(β2-α′β1-α′′-α1)ff′-2(α′+β1){α1f2+(β1-α′)ff′-P2}.
As noted from P′=(α′+β1)P and the definitions of α2 and β2, we can deduce from (3.19) that (3.20)(α1′-2α1α′)f+{β1′-α′′-α′β1+(α′)2}f′=0.
Suppose that α1′-2α1α′≡0, then α1=ce2α, where c is a nonzero constant. If eα≡1, then (1.1) gives f=cez/n, which is the desired result. If eα≢1, this together with (1.1) will lead to N(r,1/f)=S(r,f), which contradicts with (3.9). We now assume that α1′-2α1α′≢0, in the same way, (3.20) shows that N(r,1/f)=S(r,f), which also contradicts with (3.9). Hence, P≡0. Then, it follows from (3.4) that Q≡0 and from (3.3) that (3.21)F′′F-α′F′F-(F′)2=0.
Obviously, we obtain from (3.21) that (3.22)F′′F′=α′+F′F.
By integration, we have F′=dFeα, where d is a nonzero constant. Substituting this into (1.1) will yield (3.23)(d-1)fn=z(1-eα)eα.
If d=1, by this and (1.1), we have F′=F, that is, f=cez/n, where c is a nonzero constant.Ifd≠1, suppose that f has infinitely many zeros, and eα is not a constant, then the zeros of f are the zeros of 1-eα, with multiplicities at least n. It follows from the equation above and the second fundamental theorem (or Lemma 2.3) that (3.24)T(r,eα)≤N¯(r,eα)+N¯(r,1eα)+N¯(r,1eα-1)+S(r,eα),
which with n≥2 may lead to a contradiction. Now, we suppose that f has infinitely many zeros, and eα is a constant. We suppose that eα=B, then we can get a contradiction.Ifd≠1, suppose that f has finitely many zeros, and let (3.25)f=m(z)eβ,
where m(z) is a polynomial and β is an entire function. In this case, (3.23) can be written as (3.26)(1-d)mnzenβ+e-α=1.Applying the second fundamental theorem (or Lemma2.3) to (3.26), we have (3.27)n(T(r,f)=T(r,enβ)+S(r,f)=S(r,f),
which is impossible. This also proves Theorem 1.1.Remark 3.1.
By examining the proof of Theorem1.1 more carefully, here as an extension of Theorem 1.1, we would like to pose the following.Conjecture 3.2.
Letf be a transcendental entire function, and let n be a positive integer. If fn and (fn)(k) share z CM, and n≥k+1, then fn=(fn)(k),
(3.28)f=cewz/n,
where c, w are nonzero constants such that wk=1.


## 4. Proof of Theorem1.5

LetF=fn, if F≢F(k), Lemma 2.3 implies (4.1)T(r,F)≤N¯(r,F)+N¯(r,1F)+N¯(r,1F-z)+S(r,F)≤N¯(r,1f)+N¯(r,1F(k)/F-1)+S(r,f)≤N¯(r,1f)+T(r,F(k)F)+S(r,f)≤N¯(r,1f)+N(r,F(k)F)+S(r,f)≤N¯(r,1f)+kN¯(r,1f)+S(r,f)≤(k+1)T(r,f)+S(r,f),
which contradicts with n≥k+2, and hence F≡F(k), and our desired result follows.Finally, we conclude the paper with the following.Question 1.
LetR be a rational function, and let  α be an entire function, then for two integers n, k such that n≥k+1, what can be said about the transcendental entire solution of (fn)(k)-z=Reα(fn-z), can f be expressed as f=cewz/n for some constants c(≠0), wk=1?Question 2.
What can be said if the condition in Theorem1.5 “n≥k+2” is replaced by “n≥k+1”?



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*Source: 231746-2011-08-15.xml*