Appl. Math. J. Chinese Univ. 2012, 27(1): 94-104 Some results on zeros and uniqueness of difference-differential polynomials LIU Kai LIU Xin-ling CAO Ting-bin Abstract. We consider the zeros distributions of difference-differential polynomials which are the derivatives of difference products of entire functions. We also investigate the uniqueness problems of difference-differential polynomials of entire functions sharing a common value. §1 Introduction and main results A meromorphic function f means meromorphic in the complex plane. If no poles occur, then f reduces to an entire function. Given a meromorphic function f (z ), recall that α(z ) ≡ 0, ∞ is a small function with respect to f (z ), if T (r, α)= S(r, f ), where S(r, f ) is used to denote any quantity satisfying S(r, f )= o(T (r, f )), and r →∞ outside of a possible exceptional set of finite logarithmic measure. The order ρ(f ) is defined by ρ(f ) := lim sup r→∞ log + T (r, f ) log r . In addition, if f − a and g − a have the same zeros, then we say that f and g share the value a IM (ignoring multiplicities). If f − a and g − a have the same zeros with the same multiplicities, then f and g share the value a CM (counting multiplicities). We assume that the reader is familiar with standard symbols and fundamental results of Nevanlinna Theory [10, 12, 22]. The following result is related to the value distribution of differential polynomial, which was firstly given by Hayman [9, Theorem 10] and considered in several papers later, such as [1, 2]. Theorem A. [2, Theorem 1] Let f be a transcendental meromorphic function. If n ≥ 1 is a positive integer, then f n f − 1 has infinitely many zeros. Remark that [f n+1 ] =(n + 1)f n f in Theorem A, Chen [3], Wang and Fang [18, 19] improved Theorem A by proving the following result. Received: 2011-03-29. MR Subject Classification: 30D35, 39A05. Keywords: zero, sharing value, uniqueness, difference-differential polynomial. Digital Object Identifier(DOI): 10.1007/s11766-012-2795-x. This work was partially supported by the NSFC (11026110, 11101201), the NSF of Jiangxi (2010GQS0144).
Transcript
Appl. Math. J. Chinese Univ.2012, 27(1): 94-104
Some results on zeros and uniqueness of
difference-differential polynomials
LIU Kai LIU Xin-ling CAO Ting-bin
Abstract. We consider the zeros distributions of difference-differential polynomials which are
the derivatives of difference products of entire functions. We also investigate the uniqueness
problems of difference-differential polynomials of entire functions sharing a common value.
§1 Introduction and main results
A meromorphic function f means meromorphic in the complex plane. If no poles occur, thenf reduces to an entire function. Given a meromorphic function f(z), recall that α(z) �≡ 0,∞is a small function with respect to f(z), if T (r, α) = S(r, f), where S(r, f) is used to denoteany quantity satisfying S(r, f) = o(T (r, f)), and r → ∞ outside of a possible exceptional set offinite logarithmic measure. The order ρ(f) is defined by
ρ(f) := lim supr→∞
log+ T (r, f)log r
.
In addition, if f − a and g − a have the same zeros, then we say that f and g share the value a
IM (ignoring multiplicities). If f −a and g−a have the same zeros with the same multiplicities,then f and g share the value a CM (counting multiplicities). We assume that the reader isfamiliar with standard symbols and fundamental results of Nevanlinna Theory [10, 12, 22].
The following result is related to the value distribution of differential polynomial, which wasfirstly given by Hayman [9, Theorem 10] and considered in several papers later, such as [1, 2].
Theorem A. [2, Theorem 1] Let f be a transcendental meromorphic function. If n ≥ 1 is apositive integer, then fnf ′ − 1 has infinitely many zeros.
Remark that [fn+1]′ = (n + 1)fnf ′ in Theorem A, Chen [3], Wang and Fang [18, 19]improved Theorem A by proving the following result.
Received: 2011-03-29.MR Subject Classification: 30D35, 39A05.Keywords: zero, sharing value, uniqueness, difference-differential polynomial.Digital Object Identifier(DOI): 10.1007/s11766-012-2795-x.This work was partially supported by the NSFC (11026110, 11101201), the NSF of Jiangxi (2010GQS0144).
LIU Kai, LIU Xin-ling, CAO Ting-bin Some results on difference-differential polynomials 95
Theorem B. Let f be a transcendental entire function, n, k be two positive integers withn ≥ k + 1. Then (fn)(k) − 1 has infinitely many zeros.
Laine and Yang [13] considered the zeros of difference polynomials, and obtained the fol-lowing result.
Theorem C. Let f be a transcendental entire function of finite order and c be nonzero complexconstant. If n ≥ 2, then f(z)nf(z + c) − a has infinitely many zeros, where a ∈ C\{0}.
Recently, Liu and Yang [15, 16] has made some improvements of Theorem C to generaldifference products of meromorphic functions. Some papers [4, 11, 23] are also devoted tothe zeros of difference polynomials of different types. In this paper, we consider Theorem Aand Theorem B for difference implies that f(z)n+1 will be replaced with f(z)nf(z + c) orf(z)n[f(z + c)− f(z)], herein and hereinafter, c is a nonzero complex constant, n, k are positiveintegers, α(z) is a nonzero small function with respect to f . We obtain the following results:
Theorem 1.1. Let f be a transcendental entire function of finite order. If n ≥ k + 2, then thedifference-differential polynomial [f(z)nf(z + c)](k) − α(z) has infinitely many zeros.
Corollary 1.2. Let P (z), Q(z), H(z) be nonzero polynomials. Then the nonlinear difference-differential equation
[f(z)nf(z + c)](k) − P (z) = Q(z)eH(z) (1)has no transcendental entire solution of finite order, provided that n ≥ k + 2. If H(z) is aconstant, then equation (1) has no transcendental entire solution of finite order.
Remark 1.1. Theorem B shows that the zeros multiplicities of f(z)n are at least k + 1 times,but function f(z)nf(z + c) in Theorem 1.1 can have simple zeros. The condition of finite orderin Theorem 1.1 can not be removed, which can be seen by function f(z) = eez
of infinite order,where ec = −n and α(z) is a nonzero polynomial. In addition, the condition of α(z) �≡ 0 cannot be deleted, which can be showed by function f(z) = ez, where c is any complex constant.
Remark 1.2. Theorem 1.1 is an improvement of Theorem C of the case k = 0. If k = 1, thecondition of Theorem 1.1 implies that n ≥ 3. Here, we hope to reduce the condition n ≥ 3 ton ≥ 1. We also hope to reduce the condition n ≥ k+2 to n ≥ k in Theorem 1.1. Unfortunately,we have not succeed in proving this.
Theorem 1.3. Let f be a transcendental entire function of finite order, not a periodic functionwith period c. If n ≥ k + 3, then the difference-differential polynomial [f(z)nΔcf ](k) −α(z) hasinfinitely many zeros.
Corollary 1.4. Let P (z), Q(z), H(z) be nonzero polynomials. Then the nonlinear difference-differential equation
[f(z)nΔcf ](k) − P (z) = Q(z)eH(z) (2)has no transcendental entire solution of finite order, provided that n ≥ k + 3 and H(z) is anonconstant polynomial. If H(z) is a constant, then equation (2) has no transcendental entiresolutions of finite order, except that f is a periodic function with period c.
96 Appl. Math. J. Chinese Univ. Vol. 27, No. 1
Remark 1.3. The condition of α(z) �≡ 0 can not be deleted in Theorem 1.3, which also can beseen by function f(z) = ez, where c is any complex constant.
There are some results about the uniqueness theorem of two differential polynomials sharinga common value. Fang [6, Theorem 1] obtained the following result.
Theorem D. Let f and g be two nonconstant entire functions, n > 2k + 4. If (fn)(k) and(gn)(k) share 1 CM, then either f(z) = c1e
cz, g(z) = c2e−cz, where c1, c2 and c are constants
satisfying (−1)k(c1c2)n(nc)2k = 1 or f = tg where tn = 1.
In this paper, we extend Theorem D to difference-differential polynomials and obtain thefollowing results:
Theorem 1.5. Let f and g be transcendental entire functions of finite order, n ≥ 2k + 6.If [f(z)nf(z + c)](k) and [g(z)ng(z + c)](k) share the value 1 CM, then either f(z) = c1e
Cz,g(z) = c2e
−Cz, where c1, c2 and C are constants satisfying (−1)k(c1c2)n+1[(n + 1)C]2k = 1 orf = tg, where tn+1 = 1.
Theorem 1.6. The conclusion of Theorem 1.5 is also valid, if n ≥ 5k+12 and [f(z)nf(z+c)](k)
and [g(z)ng(z + c)](k) share the value 1 IM.
§2 Some Lemmas
The difference logarithmic derivative lemma, given by Chiang and Feng [5, Corollary 2.5],Halburd and Korhonen [7, Theorem 2.1], [8, Theorem 5.6] independently, plays an importantpart in considering the difference Nevanlinna theory. Here, we state the version of [8, Theorem5.6].
Lemma 2.1. Let f be a transcendental meromorphic function of finite order. Then
m
(r,
f(z + c)f(z)
)= S(r, f). (3)
Lemma 2.2. [5, Theorem 2.1] Let f(z) be a meromorphic function of finite order σ(f). Foreach ε > 0, then
T (r, f(z + c)) = T (r, f) + O(rσ(f)−1+ε) + O(log r). (4)
Thus, if f is a transcendental meromorphic function of finite order σ(f), then we getT (r, f(z + c)) = T (r, f) + S(r, f).
Lemma 2.3. Let f(z) be a transcendental entire function of finite order, and let F = f(z)nf(z+c). Then
T (r, F ) = (n + 1)T (r, f) + S(r, f). (5)
LIU Kai, LIU Xin-ling, CAO Ting-bin Some results on difference-differential polynomials 97
Proof. From Lemma 2.1, Lemma 2.2 and the standard Valiron-Mohon’ko theorem [17], since f
is a transcendental entire function, then we obtain
which is in contradiction with n ≥ k + 4. Hence, we get P (z) ≡ 0, which implies that
f(z)nf(z + c) = g(z)ng(z + c).
Let G = fg , the above equation implies that G(z)n · G(z + c) = 1, then G must be a constant.
Otherwise, we get
nT (r, G) = T (r, G(z + c)) + O(1) = T (r, G) + S(r, G),
thus T (r, G) = S(r, G), a contradiction occurs. Hence, f = tg, where t is a constant such thattn+1 = 1.
Let p be a positive integer and a ∈ C. We denote by Np(r, 1f−a ) the counting function of
the zeros of f − a where an m-fold zero is counted m times if m ≤ p and p times if m > p.
Lemma 2.7. [14, Lemma 2.3] Let f be a nonconstant meromorphic function, and p, k bepositive integers. Then
Np
(r,
1f (k)
)≤ T (r, f (k)) − T (r, f) + Np+k
(r,
1f
)+ S(r, f), (16)
Np
(r,
1f (k)
)≤ kN(r, f) + Np+k
(r,
1f
)+ S(r, f). (17)
LIU Kai, LIU Xin-ling, CAO Ting-bin Some results on difference-differential polynomials 99
Lemma 2.8. [21, Lemma 3] Let F and G be nonconstant meromorphic functions. If F and G
share 1 CM, then one of the following three cases holds:
(i) max{T (r, F ), T (r, G)} ≤ N2(r, 1F ) + N2(r, F ) + N2(r, 1
G) + N2(r, G) + S(r, F ) + S(r, G),
(ii) F = G,
(iii) F · G = 1.
For the proof of Theorem 1.6, we need the following lemma.
Lemma 2.9. [20, Lemma 2.3] Let F and G be nonconstant meromorphic functions sharing thevalue 1 IM. Let
H =F ′′
F ′ − 2F ′
F − 1− G′′
G′ + 2G′
G − 1.
If H �≡ 0, then
T (r, F ) + T (r, G) ≤ 2(
N2(r,1F
) + N2(r, F ) + N2(r,1G
) + N2(r, G))
+ 3(
N(r, F ) + N(r,1F
) + N(r, G) + N(r,1G
))
+ S(r, F ) + S(r, G). (18)
§3 Proofs of Theorem 1.1 and Theorem 1.3
Denote F (z) = f(z)nf(z + c). From Lemma 2.3, then F (z) is not a constant. Assume thatF (z)(k) −α(z) has only finitely many zeros, then from the second main theorem for three smallfunctions [10, Theorem 2.5] and (16), we get
T (r, F (k)) ≤ N(r, F (k)) + N
(r,
1F (k)
)+ N
(r,
1F (k) − α(z)
)+ S(r, F (k))
≤ N1
(r,
1F (k)
)+ N
(r,
1F (k) − α(z)
)+ S(r, F (k))
≤ T (r, F (k)) − T (r, F ) + Nk+1
(r,
1F
)+ S(r, F (k)). (19)
Combining (5) with (19), it implies that
(n + 1)T (r, f) + S(r, f) = T (r, F ) ≤ Nk+1
(r,
1F
)+ S(r, f)
≤ (k + 1)N(
r,1f
)+ N
(r,
1f(z + c)
)+ S(r, f)
≤ (k + 2)T (r, f) + S(r, f), (20)
which is in contradiction with n ≥ k + 2. Thus, Theorem 1.1 is proved. Set G(z) = f(z)nΔcf.
If G(z)(k) −α(z) has only finitely many zeros, using the similar method as above, from Lemma2.4 and the following
T (r, f(z + c) − f(z)) = m(r, f(z + c) − f(z)) ≤ m(r, f) + m
(r,
f(z + c) − f(z)f(z)
)
≤ T (r, f) + S(r, f), (21)
100 Appl. Math. J. Chinese Univ. Vol. 27, No. 1
when f is a transcendental entire function of finite order. Then we get
nT (r, f) + S(r, f) = T (r, G) ≤ Nk+1
(r,
1G
)+ S(r, f)
≤ (k + 1)N(
r,1f
)+ N
(r,
1f(z + c) − f(z)
)+ S(r, f)
≤ (k + 2)T (r, f) + S(r, f), (22)
which is in contradiction with n ≥ k + 3. Thus, we get the proof of Theorem 1.3.
§4 Proofs of Corollary 1.2 and Corollary 1.4
From Theorem 1.1, we get that equation (1) has no transcendental entire solutions of finiteorder if n ≥ k + 2. From Theorem 1.3, equation (2) also has no transcendental entire solutionsof finite order, provided that n ≥ k + 3 and H(z) is a nonconstant polynomial. Hereinafter, wewill prove the case that H(z) is a constant in Corollary 1.2 and Corollary 1.4. The proofs aresimilarly. Here, we just give the proof of Corollary 1.4, when H(z) = A is a constant. Assumethat f is a transcendental entire solution of equation (2), not a periodic function with periodc, then
[f(z)n(f(z + c) − f(z))](k) = P (z) + Q(z)eA.
Integrating above equation k times, we getL(z)
f(z)n+1=
f(z + c) − f(z)f(z)
,
where L(z) is a polynomial. From the standard Valiron-Mohon’ko theorem [17] and f is anentire function, we have
(n + 1)T (r, f) = T
(r,
f(z + c) − f(z)f(z)
)= N
(r,
f(z + c)f(z)
)+ S(r, f)
= T (r, f) + S(r, f),
which is a contradiction . We get the proof of Corollary 1.4.
§5 Proofs of Theorem 1.5 and Theorem 1.6
Let F = [f(z)nf(z + c)](k), G = [g(z)ng(z + c)](k). From Lemma 2.5 and f is an entirefunction, then
T (r, F ) ≤ T (r, f(z)nf(z + c)) + S(r, f(z)nf(z + c)). (23)
Combining (23) with Lemma 2.3, we have S(r, F ) = S(r, f). We also have S(r, G) = S(r, g)from the same reason as above. From (16), we obtain
N2
(r,
1F
)= N2
(r,
1[f(z)nf(z + c)](k)
)
≤ T (r, F ) − T (r, f(z)nf(z + c)) + Nk+2
(r,
1f(z)nf(z + c)
)+ S(r, f). (24)
LIU Kai, LIU Xin-ling, CAO Ting-bin Some results on difference-differential polynomials 101
Thus, from Lemma 2.3 and (24), we get
(n + 1)T (r, f) = T (r, f(z)nf(z + c)) + S(r, f)
≤ T (r, F ) − N2
(r,
1F
)+ Nk+2
(r,
1f(z)nf(z + c)
)+ S(r, f). (25)
From (17), we obtain
N2(r,1F
) ≤ Nk+2
(r,
1f(z)nf(z + c)
)+ S(r, f)
≤ (k + 2)N(
r,1f
)+ N
(r,
1f(z + c)
)+ S(r, f)
≤ (k + 3)T (r, f) + S(r, f). (26)
Similarly as above, we obtain
(n + 1)T (r, g) ≤ T (r, G) − N2
(r,
1G
)+ Nk+2
(r,
1g(z)ng(z + c)
)+ S(r, g). (27)
and
N2(r,1G
) ≤ (k + 3)T (r, g) + S(r, g). (28)
If the (i) of Lemma 2.8 is satisfied,it implies that
which is in contradiction with n ≥ 5k + 12. Thus, we get H ≡ 0. The following proof is trivial,which can be seen in many literatures, the original idea is referred to Yang and Yi [22]. Here,we give the complete proof. By integration for H twice, we obtain
F =(b + 1)G + (a − b − 1)
bG + (a − b), G =
(a − b − 1) − (a − b)FFb − (b + 1)
(32)
which implies that T (r, F ) = T (r, G) + O(1). Since
T (r, F ) ≤ T (r, f(z)nf(z + c)) + S(r, f) = (n + 1)T (r, f) + S(r, f),
then S(r, F ) = S(r, f). So S(r, G) = S(r, g) is. We distinguish into three cases as follows:
Case 1. b �= 0,−1. If a − b − 1 �= 0, then by (32), we get
N(r,1F
) = N
(r,
1G − a−b−1
b+1
). (33)
By the Nevanlinna second main theorem, (16) and (17), we have
which is in contradiction with n ≥ 5k + 12. Thus, a − b − 1 = 0, then
F =(b + 1)GbG + 1
. (36)
Since F is an entire function and (36), then N(r, 1G+ 1
b
) = 0. Using the same method as above,we get
(n + 1)T (r, g) ≤ T (r, G) + Nk
(r,
1gng(z + c)
)− N(r,
1G
) + S(r, g)
≤ Nk
(r,
1gng(z + c)
)+ N
(r,
1G + 1
b
)+ S(r, g)
≤ (k + 1)T (r, g) + S(r, g), (37)
which is a contradiction.
Case 2. b = 0, a �= 1. From (32), we have
F =G + a − 1
a. (38)
Similarly, we also can get a contradiction, Thus, a = 1 follows, it implies that F = G.
Case 3. b = −1, a �= −1. From (32), we obtain
F =a
a + 1 − G. (39)
Similarly, we can get a contradiction, a = −1 follows. Thus, we get F · G = 1.
LIU Kai, LIU Xin-ling, CAO Ting-bin Some results on difference-differential polynomials 103
Finally, if F = G, from Lemma 2.6, then we get f = tg for tn+1 = 1.If F · G = 1, it implies that
[f(z)nf(z + c)](k) · [g(z)ng(z + c)](k) = 1. (40)
Since n ≥ 2k + 6, from (40) and f and g are transcendental entire functions with finite order,we get f and g have no zeros. Thus, f(z) = eb(z) and g(z) = ed(z), where b(z), d(z) are nonzeropolynomials. Thus, substituting f and g into (40), we get nb(z)+b(z+c)+nd(z)+d(z+c) = E,where E is a constant. Thus, b(z) + d(z) = B, where B is a constant. Assume that
b(z) = bpzp + bp−1z
p−1 + . . . + b0
andd(z) = dqz
q + dq−1zq−1 + . . . + b0.
Substitute above into (40), it implies that nb(z) + b(z + c) is a polynomial of degree one. Sob(z) = Cz + D and d(z) = −Cz + B − D, where C is a nonzero constant. Thus, f(z) = c1e
Cz
and g(z) = c2e−Cz, where (−1)k(c1c2)n+1[(n+1)C]2k = 1. Thus, we have completed the proofs.
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Department of Mathematics, Nanchang University, Nanchang 330031, China.
