Some results on transcendental entire solutions of certain nonlinear differential-difference equations
aa r X i v : . [ m a t h . C V ] F e b SOME RESULTS ON TRANSCENDENTAL ENTIRE SOLUTIONSOF CERTAIN NONLINEAR DIFFERENTIAL-DIFFERENCEEQUATIONS
NAN LI, JIACHUAN GENG, AND LIANZHONG YANG
Abstract.
In this paper, we study the transcendental entire solutions for thenonlinear differential-difference equations of the forms: f ( z ) + e ωf ( z ) f ′ ( z ) + q ( z ) e Q ( z ) f ( z + c ) = u ( z ) e v ( z ) , and f n ( z ) + ωf n − ( z ) f ′ ( z ) + q ( z ) e Q ( z ) f ( z + c ) = p e λ z + p e λ z , n ≥ , where ω is a constant, e ω, c, λ , λ , p , p are non-zero constants, q, Q, u, v arepolynomials such that Q, v are not constants and q, u
0. Our results areimprovements and complements of some previous results. Introduction
Let f ( z ) be a transcendental meromorphic function in the complex plane C . Weassume that the reader is familiar with the standard notations and main resultsin Nevanlinna theory (see [10],[12],[20]). Throughout this paper, the term S ( r, f )always has the property that S ( r, f ) = o ( T ( r, f )) as r → ∞ , possibly outside aset E (which is not necessarily the same at each occurrence) of finite logarithmicmeasure. A meromorphic function a ( z ) is said to be a small function with respectto f ( z ) if and only if T ( r, a ) = S ( r, f ).In the past few decades, many scholars have investigated existence, order andvalue distribution of solutions of complex differential or difference equations, see[1, 6, 12, 16] etc.In 1964, Hayman [10] considered the following non-linear differential equation f n ( z ) + Q d ( f ( z )) = g ( z ) , (1.1)where Q d ( f ) is a differential polynomial in f with degree d and obtained the fol-lowing result. Theorem 1.1.
Suppose that f ( z ) is a nonconstant meromorphic function, d ≤ n − , and f, g satisfy N ( r, f ) + N ( r, /g ) = S ( r, f ) in (1.1) . Then we have g ( z ) =( f ( z ) + γ ( z )) n , where γ ( z ) is meromorphic and a small function of f ( z ) . After that, dozens of papers , see [13, 14, 15, 17] etc., focus on the solutions ofthe nonlinear differential equations of the forms f n ( z ) + P ( f ( z )) = h ( z ) , or f n ( z ) f ′ ( z ) + P ( f ( z )) = h ( z ) , Date : February 5, 2021.2010
Mathematics Subject Classification.
Primary 30D35; Secondary 39B32.This work was supported by NNSF of China (No.11801215), and the NSF of ShandongProvince, P. R. China (No.ZR2016AQ20 & No. ZR2018MA021). where P ( f ) denotes a differential polynomial in f of degree at most n −
1, and h is a given meromorphic function such as ue v (where u, v are polynomials), p e α + p e α (where p , p , α , α are polymomials), etc.In 2012, Wen et al. [19] investigated and classified the finite order entire solutionsof the equation f n ( z ) + q ( z ) e Q ( z ) f ( z + c ) = P ( z ) , (1.2)where q, Q, P are polynomials, n ≥ c ∈ C \ { } . Later, Chen[2] replaced P ( z ) in (1.2) by p e λz + p e − λz , where p , p , λ are non-zero constants,and studied its finite order entire solutions when n ≥ f n or f n f ′ on the left-hand side. Hence a natural interesting areaof inquiry is the study of equations which may have two dominated terms on theleft-hand side with the same degree.Motivated by the above equations, Chen, Hu and Wang [4] investigated thefollowing non-linear differential-difference equation(1.3) f n ( z ) + ωf n − ( z ) f ′ ( z ) + q ( z ) e Q ( z ) f ( z + c ) = u ( z ) e v ( z ) , where n is a positive integer, c = 0, ω are constants, q, Q, u, v are polynomials suchthat Q, v are not constants and q, u
0, and obtained the following result.
Theorem 1.2.
Let n be an integer satisfying n ≥ for ω = 0 and n ≥ for ω = 0 .Suppose that f is a non-vanishing transcendental entire solution of finite order of (1.3) . Then every solution f satisfies one of the following results: (1) ρ ( f ) < deg v = deg Q and f = Ce − z/ω , where C is a constant. (2) ρ ( f ) = deg Q ≥ deg v . Then it’s natural to ask: what will happen for the solutions of equation (1.3)when n = 2 and ω = 0? In this paper, we study this problem and obtain thefollowing result, which is a complement of Theorem 1.2. Theorem 1.3.
Let c, e ω = 0 be constants, q, Q, u, v be polynomials such that Q, v are not constants and q, u . Suppose that f is a transcendental entire solutionwith finite order of f ( z ) + e ωf ( z ) f ′ ( z ) + q ( z ) e Q ( z ) f ( z + c ) = u ( z ) e v ( z ) , (1.4) satisfying λ ( f ) < ρ ( f ) , then deg Q = deg v , and one of the following relations holds: (1) σ ( f ) < deg Q = deg v , and f = Ce − z/ e ω (2) σ ( f ) = deg Q = deg v . Two examples are given below to show that our estimates in Theorem 1.3 aresharp.
Example 1.4. f ( z ) = 2 e − z is a transcendental entire solution of the followingdifferential-difference equation f + f f ′ + ze z + z +1 f ( z + 1) = 2 ze z . Here e ω = 1 = 0 , Q = z + z + 1 , v = z and λ ( f ) < σ ( f ) = 1 . Then we have σ ( f ) = 1 < Q = deg v , and f = Ce − z/ e ω , where C = 2 . This shows thatConclusion (1) of Theorem 1.3 occurs. OLUTIONS OF NONLINEAR DIFFERENTIAL-DIFFERENCE EQUATIONS 3
Example 1.5. f ( z ) = e z is a transcendental entire solution of the followingdifferential-difference equation f + f f ′ + e z − z − f ( z + 1) = 2( z + 1) e z . Here e ω = 1 = 0 , Q = z − z − , v = 2 z and λ ( f ) < σ ( f ) = 2 . Then we have σ ( f ) = 2 = deg Q = deg v . This illustrates that Conclusion (2) of Theorem 1.3also exits. It is also interesting to investigate the entire solutions with finite order of thefollowing differential-difference equation f n ( z ) + ωf n − ( z ) f ′ ( z ) + q ( z ) e Q ( z ) f ( z + c ) = p e λ z + p e λ z , (1.5)where n is a positive integer, ω is a constant and c, λ , λ , p , p are non-zero con-stants, q, Q are polynomials such that Q is not a constant and q λ = − λ for the aboveproblem and obtained the following result. Theorem 1.6. If f is a transcendental entire solution with finite order of f n ( z ) + ωf n − ( z ) f ′ ( z ) + q ( z ) e Q ( z ) f ( z + c ) = p e λz + p e − λz , (1.6) then the following conclusions hold: (i) If n ≥ for ω = 0 and n ≥ for ω = 0 , then every solution f satisfies ρ ( f ) = deg Q = 1 . (ii) If n ≥ and f is a solution of (1.6) which belongs to Γ , then f ( z ) = e λz/n + B , Q ( z ) = − n + 1 n λz + b or f ( z ) = e − λz/n + B , Q ( z ) = n + 1 n λz + b where b, B ∈ C , and Γ = { e α ( z ) : α ( z ) is a non-constant polynomial } . For the case λ = λ , equation (1.5) can be reduced to f n ( z ) + ωf n − ( z ) f ′ ( z ) + q ( z ) e Q ( z ) f ( z + c ) = ( p + p ) e λ z , so we can get the order of entire solutions byusing Theorems 1.2 and 1.3.Then it is natural to ask : what will happen for the entire solutions of equation(1.5) when λ = ± λ ? In this paper, we study this problem and obtain thefollowing result. Theorem 1.7. If f is a transcendental entire solution with finite order of (1.5) ,then the following conclusions hold: (1) If n ≥ for ω = 0 and n ≥ for ω = 0 , then every solution f satisfies σ ( f ) = deg Q = 1 . (2) If n ≥ and f is a solution of (1.5) with λ ( f ) < σ ( f ) , then f ( z ) = (cid:18) p nn + ωλ (cid:19) n e λ zn , Q ( z ) = (cid:18) λ − λ n (cid:19) z + b , or f ( z ) = (cid:18) p nn + ωλ (cid:19) n e λ zn , Q ( z ) = (cid:18) λ − λ n (cid:19) z + b , NAN LI, JIACHUAN GENG, AND LIANZHONG YANG where b , b ∈ C satisfy p = q (cid:16) p nn + ωλ (cid:17) n e λ cn + b and p = q (cid:16) p nn + ωλ (cid:17) n e λ cn + b ,respectively. Remark 1.
Obviously, let λ = − λ , we can see that Theorem 1.7(1) is an im-provement of Theorem 1.6 (i). Since any function f which belongs to Γ all satisfythe condition that λ ( f ) < σ ( f ) , thus Theorem 1.7(2) is also an improvement ofTheorem 1.6 (ii). By observing Conclusion (1) of Theorem 1.7, it is natural to ask what will happenfor the entire solutions of equation (1.5) when n = 3 and ω = 0 ? In this paper wealso study this problem and obtain the following result. Theorem 1.8.
Let ω, c, λ , λ , p , p be non-zero constants, q, Q be polynomialssuch that Q is not a constant and q . If f is a transcendental entire solutionwith finite order of f ( z ) + ωf ( z ) f ′ ( z ) + q ( z ) e Q ( z ) f ( z + c ) = p e λ z + p e λ z , (1.7) satisfying N ( r, /f ) < ( κ + o (1) T ( r, f ) , where ≤ κ < and N ( r, /f ) denotesthe counting functions corresponding to simple zeros of f , then σ ( f ) = deg Q = 1 . The two examples below exhibit the sharpness of Theorem 1.8.
Example 1.9. f ( z ) = e z is an transcendental entire solution of the nonlineardifferential-difference equation f + f f ′ + 12 e − z f ( z + log 2) = 2 e z + e − z . Here ω = 1 = 0 , Q = − z , N ( r, /f ) = 0 from the fact that is a Picardexceptional value of f , and σ ( f ) = 1 = deg Q . Example 1.10. f ( z ) = e z − e z is an transcendental entire solution of the non-linear differential-difference equation f − f f ′ − e z f ( z + log 5) = − e z − e z . Here ω = − = 0 , Q = 3 z , N ( r, /f ) = N ( r, /f ) = r/π + o ( r ) < r/π + o ( r ) = T ( r, f ) by using the following Lemma 2.5, and σ ( f ) = 1 = deg Q . Preliminary Lemmas
The following two lemmas play important roles in uniqueness problems of mero-morphic functions.
Lemma 2.1 ([20]) . Let f j ( z ) ( j = 1 , . . . , n ) ( n ≥ be meromorphic functions, andlet g j ( z ) ( j = 1 , . . . , n ) be entire functions satisfying (i) P nj =1 f j ( z ) e g j ( z ) ≡ when ≤ j < k ≤ n, then g j ( z ) − g k ( z ) is not a constant; (iii) when ≤ j ≤ n, ≤ h < k ≤ n , then T ( r, f j ) = o { T ( r, e g h − g k ) } ( r → ∞ , r E ) , where E ⊂ (1 , ∞ ) is of finite linear measure or logarithmic measure.Then, f j ( z ) ≡ j = 1 , . . . , n ) . OLUTIONS OF NONLINEAR DIFFERENTIAL-DIFFERENCE EQUATIONS 5
Lemma 2.2 ([20]) . Let f j ( z ) , j = 1 , , be meromorphic functions and f ( z ) isnot a constant. If X j =1 f j ( z ) ≡ , and X j =1 N (cid:18) r, f j (cid:19) + 2 X j =1 N ( r, f j ) < ( λ + o (1)) T ( r ) , r ∈ I, where λ < , T ( r ) = max ≤ j ≤ { T ( r, f j ) } and I represents a set of r ∈ (0 , ∞ ) withinfinite linear measure. Then f ≡ or f ≡ . The difference analogues of Logarithmic Derivative Lemma (see [3, 7, 8, 9, 11])play important roles in the study of complex difference equations. The followingversion is a special case of [11, Lemma 2.2].
Lemma 2.3. [11] ) Let f be a non-constant meromorphic function, let c, h be twocomplex numbers such that c = h . If he hyper-order of T ( r, f ) i.e. σ ( f ) < , then m (cid:18) r, f ( z + h ) f ( z + c ) (cid:19) = S ( r, f )for all r outside of a set of finite logarithmic measure.The following lemma, which is a special case of [11, Theorem 3.1], gives a rela-tionship for the Nevanlinna characteristics of meromorphic function with its shift. Lemma 2.4. [11]
Let f ( z ) be a meromorphic function with the hyper-order lessthat one, and c ∈ C \ { } . Then we have T ( r, f ( z + c )) = T ( r, f ( z )) + S ( r, f ) . The following lemma gives the Nevanlinna characteristic and counting functionsof a exponential polynomial. For convenience of the readers, let’s first recall somedefinitions and notations concerning exponential polynomial of the form f ( z ) = P ( z ) e Q ( z ) + · · · + P k ( z ) e Q k ( z ) , (2.1)where P j and Q j are polynomials in z for 1 ≤ j ≤ k . Following Steinmetz [18],(2.1) can be written in the normalized form f ( z ) = H ( z ) + H ( z ) e ω z q + · · · + H m ( z ) e ω m z q , (2.2)where H j are either exponential polynomials of order < q or ordinary polynomialsin z , the leading coefficients ω j are pairwise distinct, and m ≤ k . In addition, theconvex hull of a finite set W ⊂ C , denoted by co ( W ), is the intersection of finitelymany closed half-planes each including W , and hence co ( W ) is either a compactpolygon or a line segment. We denote the circumference of co ( W ) by C ( co ( W )).Concerning the exponential polynomial f ( z ) in (2.2), we denote W = { ω , . . . , ω m } and W = W ∪ { } . Lemma 2.5. [18]
Let f ( z ) be given by (2.2) . Then T ( r, f ) = C ( co ( W )) r q π + o ( r q ) . If H ( z ) , then m (cid:18) r, f (cid:19) = o ( r q ) , NAN LI, JIACHUAN GENG, AND LIANZHONG YANG while if H ( z ) ≡ , then N (cid:18) r, f (cid:19) = C ( co ( W )) r q π + o ( r q ) . The following lemma is a revised version of [12, Lemma 2.4.2].
Lemma 2.6.
Let f ( z ) be a transcendental meromorphic solution of the equation: f n P ( z, f ) = Q ( z, f ) , where P ( z, f ) and Q ( z, f ) are polynomials in f and its derivatives with meromorh-phic coefficients, say { a λ | λ ∈ I } , n be a positive integer. If the total degree of Q ( z, f ) as a polynomial in f and its derivatives is at most n , then m ( r, P ( z, f )) ≤ X λ ∈ I m ( r, a λ ) + S ( r, f ) . Lemma 2.7. (the Hadamard factorization theorem [20, Theorem 2.7] or [5, Theo-rem 1.9] ) Let f be a meromorphic function of finite order σ ( f ) . Write f ( z ) = c k z k + c k +1 z k +1 + · · · ( c k = 0) near z = 0 and let { a , a , . . . } and { b , b , . . . } be the zeros and poles of f in C \{ } ,respectively. Then f ( z ) = z k e Q ( z ) P ( z ) P ( z ) , where P ( z ) and P ( z ) are the canonical products of f formed with the non-nullzeros and poles of f ( z ) , respectively, and Q ( z ) is a polynomial of degree ≤ σ ( f ) . Remark 2.
A well known fact about Lemma 2.7 asserts that λ ( f ) = λ ( z k P ) = σ ( z k P ) ≤ σ ( f ) , λ (1 /f ) = λ ( P ) = σ ( P ) ≤ σ ( f ) if k ≥ ; and λ ( f ) = λ ( P ) = σ ( P ) ≤ σ ( f ) , λ (1 /f ) = λ ( z − k P ) = σ ( z − k P ) ≤ σ ( f ) if k < . So we have σ ( f ) = σ ( e Q ) when max { λ ( f ) , λ (1 /f ) } < σ ( f ) . By combining [20, Theorem1.42] with [20, Theorem1.44], we have the followinglemma.
Lemma 2.8 ([20]) . Let f ( z ) be a non-constant meromorphic function in the com-plex plane. If , ∞ are Picard exceptional values of f ( z ) , then f ( z ) = e h ( z ) , where h ( z ) is a non-constant entire function. Moreover, f ( z ) is of normal growth, and (i) if h ( z ) is a polynomial of degree p , then σ ( f ) = p ; (ii) if h ( z ) is a transcendental entire function, then σ ( f ) = ∞ . The following lemma gives a relationship between the growth order of a functionand its derivative.
Lemma 2.9 ([20]) . Suppose that f ( z ) is meromorphic in the complex plane and n is a positive integer. Then f ( z ) and f ( n ) ( z ) have the same order. Proof of Theorem 1.3.
Let f be a transcendental entire solution with finite order of equation (1.4)satisfying λ ( f ) < σ ( f ). Then, by Lemma 2.7 and Remark 2, we can factorize f ( z )as f ( z ) = d ( z ) e h ( z ) , (3.1) OLUTIONS OF NONLINEAR DIFFERENTIAL-DIFFERENCE EQUATIONS 7 where h is a polynomial with deg h = σ ( f ), d is the canonical products formed byzeros of f with σ ( d ) = λ ( f ) < σ ( f ). Obviously, h is a non-constant polynomial,otherwise we’ll have σ ( f ) = σ ( d ) = λ ( f ), a contradiction. Thus we have thatdeg h ≥
1. Let deg h = m ( ≥ h ( z ) = a m z m + a m − z m − + · · · , where a m = 0.We rewrite (1.4) as f + e ωf f ′ + qe Q f c = ue v , (3.2)where f c = f ( z + c ), for simplicity.Obviously, we have σ ( f c ) = σ ( f ) = σ ( f ′ ) by Lemma 2.4 and Lemma 2.9. Sofrom (3.2), by the order property, we getdeg v = σ ( ue v ) ≤ max { σ ( f ′ ) = σ ( f ) = σ ( f c ) , σ ( e Q ) , σ ( q ) } = max { deg h, deg Q } . (3.3)By substituting (3.1) into (3.2), we get (cid:0) d + e ωd · ( d ′ + dh ′ ) (cid:1) e h + qd c e Q + h c = ue v . (3.4) Case 1. σ ( f ) > deg Q , then we have deg h > deg Q ≥
1, and deg v ≤ deg h from(3.3). Subcase 1.1. deg h > deg v . From (3.4) we have (cid:0) d + e ωd · ( d ′ + dh ′ ) (cid:1) e h e a m z m + qd c e h e a m z m = ue v , (3.5)where h = 2 a m − z m − + · · · and h = Q + ( a m mc + a m − ) z m − + · · · are allpolynomials with degree at most m −
1. So, combining with σ ( d ′ ) = σ ( d ) = σ ( d c ) < m , by using Lemma 2.1 to (3.5), we have qd c ≡ , which yields a contradiction. Thus deg h > deg v can not holds. Subcase 1.2. deg h = deg v . Let v ( z ) = v m z m + v m − z m − + · · · , where v m = 0.From (3.4) we have (cid:0) d + e ωd · ( d ′ + dh ′ ) (cid:1) e h e a m z m + qd c e h e a m z m = ue h e v m z m . (3.6)where h = v m − z m − + · · · is a polynomial with degree at most m − h and h are defined as in Subcase 1.1.If v m = 2 a m and v m = a m , combining with σ ( d ′ ) = σ ( d ) = σ ( d c ) < m , by usingLemma 2.1 to (3.6), we get u ≡
0, a contradiction.If v m = 2 a m , then (3.6) can be reduced to (cid:0)(cid:0) d + e ωd · ( d ′ + dh ′ ) (cid:1) e h − ue h (cid:1) e a m z m + qd c e h e a m z m = 0 . Thus, by using Lemma 2.1, we have qd c ≡
0, a contradiction.If v m = a m , then (3.6) can be reduced to (cid:0) d + e ωd · ( d ′ + dh ′ ) (cid:1) e h e a m z m + ( qd c e h − ue h ) e a m z m = 0 . Similarly as above, by Lemma 2.1, we get d + e ωd · ( d ′ + dh ′ ) ≡ . This gives that d = c e − e ω z − h , c ∈ C \ { } . Since deg h > deg Q ≥
1, so we have σ ( d ) = deg h = σ ( f ), which contradicts withthe assumption that σ ( d ) < σ ( f ). NAN LI, JIACHUAN GENG, AND LIANZHONG YANG
Case 2. σ ( f ) = deg Q . Let Q ( z ) = b m z m + b m − z m − + · · · , where b m = 0.Suppose deg v < deg h = deg Q . From (3.4) we have (cid:0) d + e ωd · ( d ′ + dh ′ ) (cid:1) e h e a m z m + ( qd c e f h ) e ( a m + b m ) z m = ue v , (3.7)where f h = ( a m mc + a m − + b m − ) z m − + · · · is a polynomial with degree at most m −
1, and h is defined as in Subcase 1.1.If b m = ± a m , combining with σ ( d ′ ) = σ ( d ) = σ ( d c ) < m , by using Lemma 2.1to (3.7), we get u ≡
0, which yields a contradiction.If b m = a m , then (3.7) can be reduced to (cid:16)(cid:0) d + e ωd · ( d ′ + dh ′ ) (cid:1) e h + qd c e f h (cid:17) e a m z m = ue v . Thus, by using Lemma 2.1, we have u ≡
0, a contradiction.If b m = − a m , then (3.7) can be reduced to (cid:0) d + e ωd · ( d ′ + dh ′ ) (cid:1) e h e a m z m = ue v − qd c e f h . So by Lemma 2.1, we get d + e ωd · ( d ′ + dh ′ ) ≡ . This gives that d = c e − e ω z − h , c ∈ C \ { } . Thus by deg h > deg v ≥
1, we have σ ( d ) = deg h = σ ( f ), which contradicts withthe fact that σ ( d ) < σ ( f ).Therefore, we have deg v = deg h = deg Q from (3.3), which implies that Con-clusion (2) holds. Case 3. σ ( f ) < deg Q , then we have T ( r, f ) = S ( r, e Q ). Thus we get T ( r, f ′ ) = S ( r, e Q ) from Milloux’s theorem and T ( r, f c ) = S ( r, e Q ) from lemma 2.4. Therefore,by (3.2), we have T ( r, e Q ) + S ( r, e Q ) = T ( r, f + e ωf f ′ + qf c e Q )= T ( r, ue v ) = T ( r, e v ) + S ( r, e v ) . Therefore, deg Q = deg v. Differentiating (3.2) yields2 f f ′ + e ω ( f ′ ) + e ωf f ′′ + Ae Q = ( u ′ + uv ′ ) e v , (3.8)with A = q ′ f c + qf ′ c + qf c Q ′ .Eliminating e v from (3.2) and (3.8) to get B e Q + B = 0 , (3.9)where B = uA − qf c ( u ′ + uv ′ ) ,B = u [2 f f ′ + e ω ( f ′ ) + e ωf f ′′ ] − ( f + e ωf f ′ )( u ′ + uv ′ ) . Noticing that σ ( f c ) = σ ( f ) < deg Q , and σ ( f ′′ ) = σ ( f ′ ) = σ ( f ) < deg Q fromLemma 2.9, thus by Lemma 2.1, we get B ≡ B ≡
0. It follows from B ≡ q ′ q + f ′ c f c + Q ′ = u ′ u + v ′ , OLUTIONS OF NONLINEAR DIFFERENTIAL-DIFFERENCE EQUATIONS 9 by integrating, we have qf c e Q = c ue v , where c is a non-zero constant.If c = 1, by substituting qf c e Q = ue v into (3.2), we see that f + e ωf f ′ = 0. Thuswe can easily get f = c e − z/ e ω , where c ∈ C \ { } , which implies that Conclusion(1) holds.If c = 1, we have f = c u − c /q − c e v − c − Q − c . By substituting it into (3.2), we get c u − c q − c c u − c q − c + e ω (cid:18) c u − c q − c (cid:19) ′ + c u − c q − c ( v − c − Q − c ) ′ !! e v − c − Q − c ) = (1 − c ) ue v Since 1 ≤ deg h = σ ( f ) < deg Q = deg v and σ ( f ) = deg( v − c − Q − c ), so byLemma 2.1 we can easily deduce a contradiction by the fact that c = 1 and u Proof of Theorem 1.7.
Suppose that f is a transcendental entire solution with finite order of equation(1.5). We rewrite (1.5) as f n + ωf n − f ′ + qe Q f c = p e λ z + p e λ z , (4.1)where f c = f ( z + c ), for short. From Lemma 2.4, we have σ ( f ) = σ ( f c ).By differentiating both sides of (4.1), we have nf n − f ′ + ω ( n − f n − ( f ′ ) + ωf n − f ′′ + A e Q = p λ e λ z + p λ e λ z , (4.2)where A = q ′ f c + qf ′ c + qf c Q ′ .By eliminating e λ z from equations (4.1) and (4.2), we get λ f n + ( λ ω − n ) f n − f ′ − ω ( n − f n − ( f ′ ) − ωf n − f ′′ + A e Q = p ( λ − λ ) e λ z , (4.3)where A = λ qf c − A .By differentiating (4.3), we have λ nf n − f ′ + ( λ ω − n ) (cid:2) ( n − f n − ( f ′ ) + f n − f ′′ (cid:3) − ω ( n − (cid:2) ( n − f n − ( f ′ ) + f n − f ′ f ′′ (cid:3) − ω ( n − f n − f ′ f ′′ − ωf n − f ′′′ +( A ′ + A Q ′ ) e Q = p ( λ − λ ) λ e λ z . (4.4)By eliminating e λ z from equations (4.3) and (4.4), we obtain λ λ f n + ( λ λ ω − nλ − λ n ) f n − f ′ − ( n −
1) [ λ ω + λ ω − n ] f n − ( f ′ ) − ( λ ω + λ ω − n ) f n − f ′′ + ω ( n − n − f n − ( f ′ ) + 3 ω ( n − f n − f ′ f ′′ + ωf n − f ′′′ + ( λ A − A ′ − A Q ′ ) e Q = 0 . (4.5) Case 1. σ ( f ) <
1. By using logarithmic derivative lemma, Lemma 2.3 andLemma 2.5, from (4.1) we get T (cid:0) r, e Q (cid:1) = m (cid:0) r, e Q (cid:1) = m (cid:18) r, p e λ z + p e λ z − f n − ωf n − f ′ qf c (cid:19) ≤ m (cid:18) r, fqf c (cid:19) + m (cid:18) r, f (cid:19) + m (cid:0) r, p e λ z + p e λ z (cid:1) + m (cid:18) r, f n + ωf n − f ′ f n (cid:19) + m ( r, f n ) + O (1) ≤ ( n + 1) T ( r, f ) + C ( co ( W )) r π + o ( r ) + S ( r, f ) ≤ C ( co ( W )) r π + o ( r ) , where W = { , λ , λ } , thus we have deg Q ≤
1, and noting that deg Q ≥
1, weknow that deg Q = 1. We set Q = az + b , a ∈ C \ { } , b ∈ C .Thus, by using Lemma 2.1 to (4.5), we have λ A − A ′ − A Q ′ = ( λ − a ) A − A ′ ≡ . (4.6) Subcase 1.1. A ≡
0. That is λ qf c − q ′ f c − qf ′ c − qf c a ≡ . This gives that λ − q ′ q − f ′ c f c − a ≡ . By integrating, we have qf c = c e ( λ − a ) z , c ∈ C \ { } . So we have a = λ , and f c = c /q . Otherwise, if a = λ , then we get σ ( f ) = σ ( f c ) = 1, which contradicts with our assumption that σ ( f ) <
1. So we have that f ( z ) = c /q ( z − c ) is a rational function, which contradicts with the assumptionthat f is transcendental. Subcase 1.2. A
0. From (4.6), we get A = c e ( λ − a ) z , c ∈ C \ { } . This gives that( qf c ) ′ + ( a − λ )( qf c ) = − c e ( λ − a ) z , c ∈ C \ { } . (4.7)Since λ = λ , we discuss the following three subcases: Subcase 1.2.1. a = λ . Then (4.7) reduces to ( qf c ) ′ = − c e ( λ − λ ) z . Thus, wehave qf c = c λ − λ e ( λ − λ ) z + c , where c ∈ C . Therefore, by Lemma 2.5 we have T ( r, qf c ) = T (cid:18) r, c λ − λ e ( λ − λ ) z + c (cid:19) = C ( co ( W )) r π + o ( r ) , W = { , λ − λ } . Combining with Lemma 2.4 and the fact that f is transcendental, we get C ( co ( W )) r π + o ( r ) = T ( r, qf c ) ≤ T ( r, f c ) + T ( r, q ) + O (1)= T ( r, f ) + S ( r, f ) , which contradicts with the assumption that σ ( f ) < OLUTIONS OF NONLINEAR DIFFERENTIAL-DIFFERENCE EQUATIONS 11
Subcase 1.2.2. a = λ . Then (4.7) reduces to ( qf c ) ′ + ( λ − λ )( qf c ) = − c .Thus, we have qf c = c λ − λ + c e ( λ − λ ) z , where c ∈ C . We assert that c = 0.Otherwise, if c = 0, we get that f ( z ) = c λ − λ q ( z − c ) is rational, which contradictswith the assumption that f is transcendental. Therefore, since c = 0 and λ = λ ,similarly as in Subcase 1.2.1, by combining with Lemma 2.4, Lemma 2.5, and theassumption that σ ( f ) <
1, we can also get a contradiction.
Subcase 1.2.3. a = λ and a = λ . Then by (4.7), we get that qf c = c λ − λ e ( λ − a ) z + c e ( λ − a ) z , c ∈ C . Therefore, since c = 0, a = λ , and λ = λ , similarly as in Subcase 1.2.1, bycombining with Lemma 2.4, Lemma 2.5, and the assumption that σ ( f ) <
1, wealso get a contradiction.
Case 2. σ ( f ) >
1. Denote P = p e λ z + p e λ z , then we have σ ( P ) = 1 byLemma 2.5, and equation (4.1) can be rewritten as: f n + ωf n − f ′ + ( qf c ) e Q = P. (4.8)Differentiating (4.8) yields nf n − f ′ + ω ( n − f n − ( f ′ ) + ωf n − f ′′ + Le Q = P ′ , (4.9)where L = ( qf c ) ′ + Q ′ ( qf c ). Subcase 2.1. ω = 0 and n ≥
4. Eliminating e Q from (4.8) and (4.9), we have f n − H = P L − P ′ ( qf c ) , (4.10)where H = Lf + ( ωL − nqf c ) f f ′ − ( n − ωqf c ( f ′ ) − ωqf c f f ′′ . Subcase 2.1.1. H
0. From the assumptions that f is entire and q, Q arepolynomials, we have that both H and f H are entire functions. We rewrite P L − P ′ ( qf c ) as P [( qf c ) ′ + Q ′ ( qf c )] − P ′ ( qf c )= P q ( qf c ) ′ qf c f c f · f + ( P Q ′ − P ′ ) q f c f · f = P q (cid:18) q ′ q + f ′ c f c (cid:19) f c f · f + ( P Q ′ − P ′ ) q f c f · f, and rewrite H as q (cid:18) q ′ q + f ′ c f c + Q ′ (cid:19) f c f · f + q (cid:18) ω (cid:18) q ′ q + f ′ c f c + Q ′ (cid:19) − n (cid:19) f c f · f f ′ − ( n − ωq f c f · f ( f ′ ) − ωq f c f · f f ′′ , thus both P L − P ′ ( qf c ) and H are all differential polynomials with meromorphic co-efficients. By logarithmic derivative lemma and Lemma 2.4, we have m ( r, f ′ c /f c ) = S ( r, f c ) = S ( r, f ); by Lemma 2.3, we have m ( r, f c /f ) = S ( r, f ); and by Lemma 2.5,we have m ( r, P ) = O ( r ). Therefore, by using Lemma 2.6 to (4.10) and the factthat n ≥ ω = 0, we obtain that T ( r, H ) = m ( r, H ) = S ( r, f ) + O ( r ) . and T ( r, f H ) = m ( r, f H ) = S ( r, f ) + O ( r ) . Thus, by H T ( r, f ) ≤ T ( r, f H ) + T (cid:18) r, H (cid:19) = S ( r, f ) + O ( r ) , which contradicts with the assumption that σ ( f ) > Subcase 2.1.2. H ≡
0. Then from (4.10), we have
P L − P ′ ( qf c ) = P [( qf c ) ′ + Q ′ ( qf c )] − P ′ ( qf c ) ≡ . This gives that ( qf c ) ′ qf c + Q ′ − P ′ P ≡ . By integration, we see that there exists a c ∈ C \ { } such that qf c = c P e − Q . (4.11)By substituting (4.11) into (4.8), we get that f n + ωf n − f ′ = (1 − c ) P. (4.12)If c = 1, then from (4.12) we have f + ωf ′ = 0. By integration, we get that f = c e − ω z , c ∈ C \ { } , which contradicts with the assumption that σ ( f ) > c = 1. From (4.11), we have f = c P − c q − c e − Q − c , (4.13)and deg Q = deg Q − c = σ ( f ) > σ ( P − c ) = 1 by lemma 2.5.By Substituting (4.13) into (4.12), we have c n − c (cid:18) P − c q − c (cid:19) n + ω (cid:18) P − c q − c (cid:19) n − (cid:18) P − c q − c (cid:19) ′ + P − c q − c ( − Q − c ) ′ !! e − nQ − c = P. Then from deg
Q > σ ( P − c ) = σ ( P ) = 1 and Lemma 2.1, we get that P ( z ) ≡ Subcase 2.2. ω = 0 and n ≥
3. Eliminating e Q from (4.8) and (4.9), we have f n − ( Lf − nqf c f ′ ) = P L − P ′ ( qf c ) . Subcase 2.2.1. Lf − nqf c f ′
0. Since n ≥ ω = 0, similarly as in subcase2.1.1, we have T ( r, Lf − nqf c f ′ ) = m ( r, Lf − nqf c f ′ ) = S ( r, f ) + O ( r ) . and T ( r, f ( Lf − nqf c f ′ )) = m ( r, f ( Lf − nqf c f ′ )) = S ( r, f ) + O ( r ) . Thus, by Lf − nqf c f ′ T ( r, f ) ≤ T ( r, f ( Lf − nqf c f ′ )) + T (cid:18) r, Lf − nqf c f ′ (cid:19) = S ( r, f ) + O ( r ) , which contradicts with the assumption that σ ( f ) > OLUTIONS OF NONLINEAR DIFFERENTIAL-DIFFERENCE EQUATIONS 13
Subcase 2.2.2. Lf − nqf c f ′ ≡
0. Then( qf c ) ′ qf c + Q ′ − n f ′ f ≡ . By integration, we see that there exists a c ∈ C \ { } such that qf c e Q = c f n . (4.14)Substituting (4.14) into (4.8) gives(1 + c ) f n = P. If c = −
1, we have nT ( r, f ) + S ( r, f ) = T ( r, (1 + c ) f n ) = T ( r, P ) = O ( r ),which contradicts with the assumption that σ ( f ) >
1. If c = −
1, then P = p e λ z + p e λ z ≡
0, a contradiction.
Case 3. σ ( f ) = 1. By (4.1), Lemma 2.3, and logarithmic derivative lemma, weobtain T (cid:0) r, e Q (cid:1) = m (cid:0) r, e Q (cid:1) = m (cid:18) r, p e λ z + p e λ z − f n − ωf n − f ′ qf c (cid:19) ≤ m (cid:18) r, qf c (cid:19) + m (cid:0) r, p e λ z + p e λ z (cid:1) + m (cid:0) r, f n + ωf n − f ′ (cid:1) + O (1) ≤ m (cid:18) r, ff c (cid:19) + m (cid:18) r, f (cid:19) + m (cid:18) r, f n + ωf n − f ′ f n (cid:19) + m ( r, f n )+ T (cid:0) r, p e λ z + p e λ z (cid:1) + O (log r ) ≤ ( n + 1) T ( r, f ) + T (cid:0) r, p e λ z + p e λ z (cid:1) + S ( r, f ) . Note that deg Q ≥
1, then by combining with Lemma 2.5, we get1 ≤ deg Q = σ (cid:0) e Q (cid:1) ≤ max { σ ( f ) , σ (cid:0) p e λ z + p e λ z (cid:1) } = 1 , that is σ ( f ) = deg Q = 1. Thus, Conclusion (1) is proved.Next, we prove Conclusion (2). Suppose that f is a transcendental entire solutionwith finite order of equation (1.5) with λ ( f ) < σ ( f ). Then, by Lemma 2.7 andRemark 2, we can factorize f ( z ) as f ( z ) = d ( z ) e h ( z ) , (4.15)where h is a polynomial with deg h = σ ( f ), d is the canonical products formed byzeros of f with σ ( d ) = λ ( f ) < σ ( f ). Similarly as in the proof of Theorem 1.3, wehave σ ( f ) = deg h ≥ d n − ( d + ω ( d ′ + dh ′ )) e nh + qd c e Q + h c = p e λ z + p e λ z . (4.16)Dividing both sides of (4.16) by p e λ z , we obtain f + f + f = 1 , (4.17)where f = − p p e ( λ − λ ) z , f = d n − ( d + ω ( d ′ + dh ′ )) p e nh − λ z , f = qd c p e Q + h c − λ z . Obviously, f is not a constant since λ = λ . Next we discuss two cases: σ ( f ) > σ ( f ) = 1, respectively. We set T ( r ) = max { T ( r, f ) , T ( r, f ) , T ( r, f ) } . If σ ( f ) >
1, then d n − ( d + ω ( d ′ + dh ′ )) /p e − λ z is a small function of e h sincemax { σ ( d ′ ) = σ ( d ) , } < deg h . So we have T ( r ) ≥ T ( r, f ) = nT ( r, e h ) + S ( r, e h ) .Thus, by Milloux’s theorem and Lemma 2.4, we get N (cid:16) r, f (cid:17) T ( r ) = N (cid:16) r, d n − ( d + ω ( d ′ + dh ′ )) (cid:17) T ( r ) ≤ T (cid:0) r, d n − ( d + ω ( d ′ + dh ′ )) (cid:1) T ( r )= O ( T ( r, d )) + O (log r ) T ( r, e h ) · T ( r, e h ) T ( r ) → , and N (cid:16) r, f (cid:17) T ( r ) = N (cid:16) r, qd c (cid:17) T ( r ) ≤ T ( r, qd c ) T ( r ) ≤ T ( r, d ) + S ( r, d ) + O (log r ) T ( r, e h ) · T ( r, e h ) T ( r ) → r → ∞ .Therefore, by using Lemma 2.2, we can deduce that f ≡ f ≡ f ≡
1, that is d n − ( d + ω ( d ′ + dh ′ )) e nh − λ z ≡ p . Obviously, d n − ( d + ω ( d ′ + dh ′ ))
0. Otherwise, if d n − ( d + ω ( d ′ + dh ′ )) ≡
0, we’ll have p ≡
0, acontradiction. So, by Lemma 2.5 and Milloux’s theorem, we have S (cid:0) r, e h (cid:1) + nT (cid:0) r, e h (cid:1) = T (cid:0) r, e nh (cid:1) = T (cid:18) r, p e λ z d n − ( d + ω ( d ′ + dh ′ )) (cid:19) ≤ T (cid:0) r, p e λ z (cid:1) + T (cid:0) r, d n − ( d + ω ( d ′ + dh ′ )) (cid:1) + O (1)= O ( r ) + O ( T ( r, d )) , which contradicts with the assumption that deg h = σ ( f ) > max { σ ( d ) , } .If f ≡
1, then by (4.17), we have f + f ≡
0. That is d n − ( d + ω ( d ′ + dh ′ )) e nh = p e λ z . Following the similar reason as above, we can also get a contradiction.If σ ( f ) = 1, then we have σ ( d ) < h = σ ( e ( λ − λ ) z ) and T ( r ) ≥ T ( r, f ) = T (cid:0) r, e ( λ − λ ) z (cid:1) + S (cid:0) r, e ( λ − λ ) z (cid:1) . Thus, by Milloux’s theorem and Lemma 2.4, weget N (cid:16) r, f (cid:17) T ( r ) = O ( T ( r, d )) + O (log r ) T (cid:0) r, e ( λ − λ ) z (cid:1) · T (cid:0) r, e ( λ − λ ) z (cid:1) T ( r ) → , and N (cid:16) r, f (cid:17) T ( r ) ≤ T ( r, d ) + S ( r, d ) + O (log r ) T (cid:0) r, e ( λ − λ ) z (cid:1) · T (cid:0) r, e ( λ − λ ) z (cid:1) T ( r ) → , as r → ∞ .Therefore, by using Lemma 2.2, we can deduce that f ≡ f ≡ f ≡
1, that is d n − ( d + ω ( d ′ + dh ′ )) e nh − λ z = p . (4.18)We assert that h ′ = λ /n . Otherwise, suppose that h ′ = λ /n , then from σ ( d ′ ) = σ ( d ) < nh − λ z ), by using Lemma 2.1 to (4.18), we get p ≡
0, a
OLUTIONS OF NONLINEAR DIFFERENTIAL-DIFFERENCE EQUATIONS 15 contradiction. Thus h ′ = λ /n . We set h = λ z/n + B , where B is a constant. Bysubstituting it into (4.18), we have d n − (cid:18) d + ω (cid:18) d ′ + λ dn (cid:19)(cid:19) = p e − nB . (4.19)Next, we assert that d is a constant. Otherwise, if d is a non-constant entirefunction, then from (4.19) we get that 0 is a Picard exceptional value of d . Thusby Lemma 2.8, we have d = e α , where α is a non-constant polynomial, whichcontradicts with the assumption that σ ( d ) <
1. So we have that d is a non-zeroconstant, and (4.19) reduces to d n e nB (cid:18) ω λ n (cid:19) = p . Therefore, f = de h = de B e λ z/n = (cid:18) p nn + ωλ (cid:19) n e λ zn . Moreover, from f ≡ f + f ≡
0. That is p e λ z = qd c e Q + h c , which implies that Q = (cid:18) λ − λ n (cid:19) z + b , where b satisfies p = q (cid:16) p nn + ωλ (cid:17) n e λ cn + b .If f ≡
1, then from (4.17) we have f + f = 0. This gives that d n − ( d + ω ( d ′ + dh ′ )) e nh − λ z = p . By using the similar methods as in the case f ≡
1, we get f ( z ) = (cid:18) p nn + ωλ (cid:19) n e λ zn . Furthermore, from f ≡
1, we have qd c e Q + h c − λ z ≡ p . Thus we can deduce Q = (cid:18) λ − λ n (cid:19) z + b , where b satisfies p = q (cid:16) p nn + ωλ (cid:17) n e λ cn + b . From the above discussion, the proofof Conclusion (2) is complete.5. Proof of Theorem 1.8.
Suppose that f is a transcendental entire solution with finite order of equation(1.7).If σ ( f ) <
1, then following the similar method as in the proof of Case 1 ofTheorem 1.7, we can get a contradiction.If σ ( f ) >
1. We denote P = p e λ z + p e λ z , and rewrite (1.7) as: f + ωf f ′ + ( qf c ) e Q = P. (5.1) Differentiating (5.1) yields3 f f ′ + ω f ( f ′ ) + ωf f ′′ + Le Q = P ′ , (5.2)where L = ( qf c ) ′ + Q ′ ( qf c ).Eliminating e Q from (5.1) and (5.2), we have f H = P L − P ′ ( qf c ) , (5.3)where H = Lf + ( ωL − nqf c ) f f ′ − ( n − ωqf c ( f ′ ) − ωqf c f f ′′ . If H ≡
0, then from (5.3) we have
P L − P ′ ( qf c ) ≡
0. Similarly as in the proof ofSubcase 2.1.2 of Theorem 1.7, we get a contradiction. Therefore, H
0. From theassumptions that f is entire and q, Q are polynomials, we have that H is an entirefunction. Since both P L − P ′ ( qf c ) and H/f are all differential polynomials withmeromorphic coefficients, similarly as in the proof of Subcase 2.1.1 of Theorem 1.7,by using Lemma 2.6 to (5.3), we obtain that T ( r, H ) = m ( r, H ) = S ( r, f ) + O ( r ) . and m ( r, H/f ) = S ( r, f ) + O ( r ) . By observing we see that the poles of
H/f all arise from the poles of ( f ′ ) /f ,thus might arise from the zeros of f . Suppose that z is a zero of f with multiplicity p , then it is a pole of ( f ′ ) /f with multiplicity 1 when p = 1; and is a zero of ( f ′ ) /f with multiplicity p − p ≥
2. So we have T ( r, H/f ) = m ( r, H/f ) + N ( r, H/f ) = m ( r, H/f ) + N ( r, /f ) < ( κ + o (1)) T ( r, f ) + S ( r, f ) + O ( r ) . Therefore, by H T ( r, f ) = T ( r, /f ) + O (1) ≤ T (cid:18) r, Hf (cid:19) + T (cid:18) r, H (cid:19) + O (1) < ( κ + o (1)) T ( r, f ) + S ( r, f ) + O ( r ) , ≤ κ < . So we have T ( r, f ) = S ( r, f ) + O ( r ) , which contradicts with the assumption that σ ( f ) > σ ( f ) = 1, then following the similar method as in the proof of Case 3 ofTheorem 1.7, we can get that σ ( f ) = deg Q = 1. Acknowledgements.
Thanks are due to the referee for valuable comments andsuggestions.
References [1] Z. X. Chen,
Complex Differences and Difference Equations , Science Press, 2014.[2] M. F. Chen, Z. S. Gao and J. L.Zhang,
Entire solutions of certain type of non-linear differenceequations , Comput. Methods Funct. Theory (1), (2019) 17–36.[3] Y. M. Chiang and S. J. Feng, On the growth of logarithmic differences, difference quotientsand logarithmic derivatives of meromorphic functions , Trans. Amer. Math. Soc. (7)(2009) 3767–3791. 521–523.[4] W. Chen, P. C. Hu and Q. Wang,
Entire Solutions of Two Certain Types of Non-linearDifferential-Difference Equations , Computational Methods and Function Theory, 2020(3–4).[5] J. B. Conway,
Functions of One Complex Variable I , 2nd ed., springer.com: Springer, 1995.
OLUTIONS OF NONLINEAR DIFFERENTIAL-DIFFERENCE EQUATIONS 17 [6] G. G. Gundersen,
Questions on meromorphic functions and complex differential equations ,Comput. Methods Funct. Theory. , (2017), 195–209.[7] R. G. Halburd and R. J. Korhonen, Difference analogue of the lemma on the logarithmicderivative with applications to difference equations , J. Math. Anal. Appl. (2) (2006)477–487.[8] ,
Nevanlinna theory for the difference operator , Ann. Acad. Sci. Fenn. Math. (2006)463–478.[9] R. G. Halburd, R. J. Korhonen and K. Tohge, Holomorphic curves with shift-invariant hy-perplane preimages , Trans. Amer.Math. Soc. (2014) no. 8, 4267–4298.[10] W. K. Hayman,
Meromorphic Functions , Oxford Mathematical Monographs, ClarendonPress, Oxford, 1964.[11] R. Korhonen,
An extension of Picard’s theorem for meromorphic functions of small hyper-order ,J. Math. Anal. Appl. (2009) 244–253.[12] I. Laine,
Nevanlinna theory and complex differential equations , W. de Gruyter, Berlin, 1993.[13] P. Li,
Entire solutions of certain type of differential equations II , J. Math. Anal. Appl. (2011), 310–319.[14] L. W. Liao, C. C. Yang and J. J. Zhang,
On meromorphic solutions of certain type of non-linear differential equations , Ann. Acad. Sci. Fenn. Math. (2013), 581–593.[15] L. Liao, Non-linear differential equations and Hayman’s theorem on differential polynomials ,Complex Var. Elliptic Equ. (6) (2015), 748–756.[16] J. R. Long, J. Heittokangas, and Z. Ye, On the relationship between the lower order ofcoefficients and the growth of solutions of differential equations , J. Math. Anal. Appl. (1)(2016), 153–166.[17] X. Q. Lu, L. W. Liao and J. Wang,
On meromorphic solutions of a certain type of nonlineardifferential equations , Acta Math. Sin.(Engl. Ser.), (12) (2017), 1597–1608.[18] N. Steinmetz, Zur Wertverteilung von Exponentialpolynomen , Manuscr. Math. (1–2)(1978–1979) 155–167.[19] Z. T. Wen, J. Heittokangas and I. Laine, Exponential polynomials as solutions of certainnonlinear difference equations , Acta Math. Sci. Ser. B (7) (2012), 1295–1306.[20] C. C. Yang and H. X. Yi, Uniqueness theory of meromorphic functions , Mathematics and itsApplications, 557. Kluwer Academic Publishers Group, Dordrecht, 2003.
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