TU HONG-QIANG AND LIU HUI-FANG
(College of Mathematics and Information Science,Jiangxi Normal University,Nanchang,330022)
Communicated by Ji You-qing
Nevanlinna’s theory has been widely applied to the field of complex difference.Many researchers studied the properties of meromorphic solutions of the following linear difference equation by this theory
wheren∈N,cj(j=1,···,n)are nonzero complex numbers which are different from each other,and obtained lots of results concerning the growth and value distribution of meromorphic solutions of(1.1)(see[3]–[9]).Therein Chiang and Feng[4]considered thecase when there is only one dominating coefficient among all entire coefficients of(1.1),and obtained the following result:
Theorem A[4]Let A0(z),···,An(z)be entire functions.If there exists an integer l(0≤l≤n)such that
When most coefficients of(1.1)have the same order,Qiet al.[9]studied the properties of meromorphic solutions of the following linear difference equation
and obtained the following results:
Theorem B[9]Let Pj(z)and Qj(z)(j=0,1,···,n?1)be polynomials,A(z)be a polynomial of degree k(≥1).If
then each nontrivial meromorphic solution f(z)with finite order of(1.2)satisfies
and so f assumes every nonzero complex value a∈Cinfinitely often.
Theorem C[9]Suppose that the assumptions of Theorem B are satisfied.If f(z)is a nontrivial entire solution with finite order of(1.2)that satisfies λ(f)≤k,then σ(f)=k+1.
Comparing Theorem A with Theorem B and Theorem C,we pose the following questions:
Question 1.2Theorem C provided a criterion which guarantee that each entire solution of(1.2)has the smallest order.Then under the assumptions of Question 1.1,what else condition can guarantee that each meromorphic solution of(1.1)has the smallest order?
In this paper,we investigate the above questions and obtain the following results.
Considering the non-homogeneous linear difference equation
we obtain the following result.
Theorem 1.3Let Aj(z)(j=0,1,···,n)and l satisfy the conditions of Theorem1.1,and let F(z)(/≡0)be a meromorphic function with σ(F)<k+1.Then each meromorphic solution f(z)with finite order of(1.3)satisfies λ(f)=σ(f)≥k+1with at most one possible exceptional solution f0satisfying σ(f0)<k+1.
We need the following lemmas for the proof of the above theorems.
Lemma 2.2[10]Let f(z)be a non-constant meromorphic function.Then for all irreducible rational functions in f
with meromorphic coefficients ai(z),bj(z)(i=0,···,p,j=0,···,q),we have
Lemma 2.3[10]Let g:[0,+∞)→R,h:[0,+∞)→Rbe monotone increasing functions such that g(r)≤h(r)outside of an exceptional set E of finite logarithmic measure.Then,for any α>1,there exists an r0>0such that g(r)≤h(αr)holds for all r>r0.
Lemma 2.4[11]Let fj(z)(j=1,···,n+1,n≥2)be meromorphic functions,gj(z)(j=1,···,n)be entire functions,and satisfy
(ii)when1≤j<k≤n,gj(z)?gk(z)is not a constant;
(iii)when1≤j≤n+1,1≤h<k≤n,
where E?(1,∞)is of finite linear measure or finite logarithmic measure.
Then fj(z)≡0(j=1,···,n+1).
Lemma 2.5[12]Let f(z)be a meromorphic function of order σ(f)=σ<∞.Then for any given ε>0,there exists a set E?(1,∞)of finite linear measure such that for all|z|=r/∈[0,1]∪E,and r sufficiently large,we have
Lemma 2.6[4]Let f be a non-constant meromorphic function with finite order,and η be a nonzero complex number.Then for each ε>0,we have
Lemma 2.7Let Aj(j=0,···,n)be meromorphic functions,and f(/≡0)be a meromorphic solution with finite order of the difference equation
whered1(z)andd2(z)respectively are the canonical products(or polynomials)formed by zeros or poles off(z),such that
andh(z)is a polynomial.By(2.2)and(2.3),we get
Substituting(2.2)into(2.1),we get
By Lemma 2.6 and(2.3)–(2.5),we get
On the other hand,by(2.5),we get
Thus,by(2.7)and(2.8),we get
Then by Lemma 2.4,(2.6),(2.8)and(2.9),we get
This is a contradiction.Lemma 2.7 is thus proved.
Lemma 2.8[8]Let f(z)be a meromorphic solution with finite order of(1.3).If
then λ(f)=σ(f).
Suppose that
Now we discuss the following two cases.
By(3.6)and Lemma 2.3,we getk≤σ?1+ε,which impliesσ(f)=σ≥k+1.
Case 2.If deg(Ql)>deg(Qj)(j=0,1,···,n,j/=l),then we take a suitablezsuch that|z|=randakzk=?|ak|rk.By the similar method as in the proof of Case 1,we can obtainσ(f)≥k+1.
In the following,we prove thatλ(f?φ)=σ(f).
Setg(z)=f(z)?φ(z).Substitutingf(z)=g(z)+φ(z)into(1.1),we obtain
IfH(z)≡0,thenφ(z)is a nonzero meromorphic solution of(1.1).Thus,by the above proof,we get
This is absurd.Hence,
On the other hand,by Lemma 2.2,we get
Since eAis of the regular growth,by(3.9),we getT(r,Aj)=(pj+qj)T(r,eA)+S(r,eA)(j=0,···,n).Hence we get
Then by(3.10)and Lemma 2.6,we get
So by Lemma 2.8,(3.7),(3.8),(3.10)and(3.11),we obtain
Theorem 1.1 is thus proved.
Suppose thatσ(f)>k+1.By(3.10)we get
Then combining with Lemma 2.7 and(3.13),we get
This contradicts the hypothesis of Theorem 1.2.Hence we getσ(f)=k+1.Theorem 1.2 is thus proved.
andf??f0is a solution of the corresponding homogeneous equation(1.1)to(1.3).By Theorem 1.1,we have
a contradiction.Hence(1.3)possesses at most one exceptional solutionf0withσ(f0)<k+1.
Now Suppose thatfis a meromorphic solution of(1.3)withk+1≤σ(f)<∞.Combining(3.10),we have
Hence,by Lemma 2.8,we get
Theorem 1.3 is thus proved.
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Communications in Mathematical Research2018年1期