International Journal of Analysis and Applications ISSN 2291-8639 Volume 6, Number 1 (2014), 44-52 http://www.etamaths.com ON THE MEROMORPHIC SOLUTIONS OF CERTAIN NONLINEAR DIFFERENCE EQUATIONS VEENA L. PUJARI Abstract. In this article, we investigate the meromorphic solutions of certain non-linear difference equations using Tumura-Clunie theorem and also provide examples which satisfy our results. 1. Introduction and Main Results Meromorphic solutions of complex differential equations and complex difference equations plays a prominent role in the field of Complex analysis. Solutions of such equations admits several ways of approach, but recently solutions of complex differential or difference equations by Nevanlinna theory techniques has become a subject of great interest. The Clunie lemma and Tumura-Clunie type theorems were efficient tool in find- ing the solutions of complex differential or difference equations. In this article, we solve certain complex non-linear difference equations using Tumura-Clunie type theorems. We assume that the reader is familiar with the ba- sic notions of Nevanlinna’s Value distribution theory [see [8],[9]]. In [7], Anupama J.Patil proved the following the result Theorem A. No trancendental meromorphic function f with N(r,f) = S(r,f) will satisfy an equation of the form a1(z)P(f)Π(f) + a2(z)Π(f) + a3(z) ≡ 0 where a1(z)( 6≡ 0), a2(z) and a3(z) are small functions of f, P(f) = bnf n + bn−1f n−1 + . . . + b1f + b0 where n is a positive integer, bn(6≡ 0),bn−1, . . . ,b0 are small functions of f and Π(f) is a differential polynomial in f i.e, Π(f) = n∑ i=1 αi(z)f ni0 (f′)ni1 (f′′)ni2 . . . (f(m))nim In this paper, we obtain two main results by considering difference function f(z +c) and difference polynomial in place of Π(f) in Theorem A. Theorem 1.1 No transcendental meromorphic function f of finite order ρ with N(r,f) = O(rρ−1+�) + S(r,f) will satisfy the non-linear difference equation of the form a1(z)P(f)f(z + c) + a2(z)f(z + c) + a3(z) ≡ 0 2010 Mathematics Subject Classification. 30D35. Key words and phrases. meromorphic function, difference polynomial, difference equation and Tumura-Clunie theorem. c©2014 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 44 NONLINEAR DIFFERENCE EQUATIONS 45 where c ∈ C, a1(z)( 6= 0), a2(z) and a3(z)( 6= 0) are small functions in the sense of T(r,ai) = O(r ρ−1+�) + S(r,f), i = 1, 2, 3 and P(f) = bnf n + bn−1f n−1 + . . . + b1f + b0 where n is a positive integer, bn( 6≡ 0),bn−1, . . . ,b0 are small functions in the sense of T(r,bj) = O(r ρ−1+�) + S(r,f),j = 0, 1, 2, . . . ,n. Theorem 1.2 No transcendental meromorphic function f of finite order ρ with N(r,f) +N(r, 1/f) = O(rρ−1+�) +S(r,f) will satisfy the difference equation of the form a1(z)P(f)Π(f) + a2(z)Π(f) + a3(z) ≡ 0 where n ≥ 1, a1(z)( 6= 0), a2(z) and a3(z)( 6= 0) are small functions in the sense of T(r,ai) = O(r ρ−1+�) + S(r,f), i = 1, 2, 3 and Π(f) = ∑ λ aλ(z)f(z) l0f(z + c1) l1 . . .f(z + cλ) lλ is a difference polynomial of degree n where n = max ∑λ j=1 lj and c1,c2, . . . ,cλ are distinct values in C and T(r,aλ) = S(r,f) 2. Some Lemmas Lemma 2.1 [2]. Let f be a meromorphic function of finite order ρ, and suppose that Ψ(z) = an(z)f(z) n + · · · + a0(z) has small meromorphic coefficients aj(z), an 6= 0 in the sense of T(r,aj) = O(rρ−1+�)+ S(r,f). Moreover , assume that N(r, 1 Ψ ) + N(r,f) = O(rρ−1+�) + S(r,f). Then Ψ = an ( f + an−1 nan )n . Lemma 2.2 [8]. Suppose f(z) is a meromorphic function in the complex plane and P(z) = a0f n +a1f n−1 +· · ·+an, where a0(6≡ 0),a1, · · · ,an are small functions of f(z). Then T(r,P(f)) = nT(r,f) + S(r,f) Lemma 2.3 [1]. Let f be a meromorphic function with exponent of convergence of poles λ( 1 f ) = λ < +∞, η 6= 0 be fixed, then for each � > 0, N(r,f(z + η)) = N(r,f) + O(rρ−1+�) + O(logr). Lemma 2.4 [3,4]. Let f(z) be a meromorphic function of finite order σ and let c be a fixed non-zero complex constant. Then for each � > 0, we have m ( r, f(z + c) f(z) ) + m ( r, f(z) f(z + c) ) = O(rσ−1+�) Lemma 2.5 [1]. Let f(z) be a meromorphic function of finite order σ, and let η be a fixed non-zero complex number, then for each � > 0, we have T(r,f(z + η)) = T(r,f) + O(rσ−1+�) + O(logr). 46 VEENA L. PUJARI Lemma 2.6 [3]. Let f(z) be a non-constant meromorphic solution of f(z)nP(z,f) = Q(z,f), where P(z,f) and Q(z,f) are difference polynomials in f(z), and let δ < 1 and � > 0. If the degree of Q(z,f) as a polynomial in f(z) and its shifts is at most n, then m(r,P(z,f)) = o ( T(r + |c|,f)1+� rδ ) + o(T(r,f)) for all r outside of a possible exceptional set with finite logarithmic measure. Proof of Theorems Proof of Theorem 1.1 We prove this theorem by contradiction. We first consider the case n ≥ 2. Suppose there exists a transcendental meromorphic function f(z) of finite order ρ with (1) N(r,f) = O(rρ−1+�) + S(r,f) satisfying (2) a1(z)P(f)f(z + c) + a2(z)f(z + c) + a3(z) ≡ 0 i.e a1 [ bnf n + bn−1f n−1 + · · · + b1f + b0 ] f(z + c) + a2f(z + c) + a3 ≡ 0 (3) =⇒ a1bnfnf(z + c) + P1(f)f(z + c) + a3 ≡ 0 where P1(f) = a1bn−1f n−1 + · · · + a1b1f + a1b0 + a2 By our assumption (1) and Lemma 2.3, we have (4) N(r,f(z + c)) = O(rρ−1+�) + S(r,f) Now (3) can be written as a1bnf n + P1(f) ≡− a3 f(z + c) Consider H(z) ≡ fn + P1(f) a1bn ≡− a3 a1bnf(z + c) (5) From (4) and (5), we write N ( r, 1 H ) ≤ N ( r, 1 H ) = N ( r, −a1bnf(z + c) a3 ) = O(rρ−1+�) + S(r,f) With this and by our assumption, we have N ( r, 1 H ) + N(r,f) = O(rρ−1+�) + S(r,f) Now applying Lemma 2.1, we get H(z) = ( f(z) + bn−1 nbn )n (6) NONLINEAR DIFFERENCE EQUATIONS 47 From (5) and (6), we have( f(z) + bn−1 nbn )n ≡− a3 a1bnf(z + c) =⇒ ( f(z) + bn−1 nbn )n f(z + c) ≡− a3 a1bn Thus T ( r, ( f(z) + bn−1 nbn )n f(z + c) ) = T ( r, a3 a1bn ) =⇒ T ( r, ( f(z) + bn−1 nbn )n f(z + c) ) = O(rρ−1+�) + S(r,f)(7) Using (4), we write N ( r, f(z + c) f(z) ) ≤ N(r,f(z + c)) + N ( r, 1 f(z) ) = N ( r, 1 f(z) ) + O(rρ−1+�) + S(r,f)(8) Now, using Lemma 2.4 and (8), we get T ( r, f(z + c) f(z) ) = m ( r, f(z + c) f(z) ) + N ( r, f(z + c) f(z) ) = N ( r, 1 f(z) ) + O(rρ−1+�) + S(r,f) ≤ T(r,f) + O(rρ−1+�) + S(r,f)(9) Now by the first fundamental theorem of Nevanlinna and from (7) and (9), we have T ( r,f ( f(z) + bn−1 nbn )n) = T  r, 1 f ( f(z) + bn−1 nbn )n   + O(1) ≤ T ( r, f(z + c) f(z) ) + T  r, 1 f(z + c) ( f(z) + bn−1 nbn )n   + O(1) ≤ T(r,f) + O(rρ−1+�) + S(r,f)(10) On the other hand, using Lemma 2.2, we write T ( r,f ( f(z) + bn−1 nbn )n) = (n + 1)T(r,f) + S(r,f)(11) Thus from (10) and (11), we get (n + 1)T(r,f) + S(r,f) ≤ T(r,f) + O(rρ−1+�) + S(r,f) nT(r,f) ≤ O(rρ−1+�) + S(r,f) 48 VEENA L. PUJARI which is contradiction. Thus our assumption is false. Next, we shall consider the case n = 1. If n = 1, then (2) becomes a1(z)(b1(z)f(z) + b0)f(z + c) + a2(z)f(z + c) + a3(z) ≡ 0 =⇒ a1b1f(z)f(z + c) + a1b0f(z + c) + a2f(z + c) ≡−a3 =⇒ [a1b1f(z) + (a1b0 + a2)] f(z + c) ≡−a3 =⇒ [ f(z) + (a1b0 + a2) a1b1 ] f(z + c) ≡− a3 a1b1 (12) Degree of − a3 a1b1 is zero and the degree of the term [ f(z) + (a1b0+a2) a1b1 ] is one. Hence by Lemma 2.6, we get m(r,f(z + c)) = o ( T(r + |c|,f)1+� rδ ) + S(r,f), where δ < 1 and � > 0, which holds for all r outside of a possible exceptional set with finite logarithmic measure. Thus using (4), we write T(r,f(z + c)) ≤ O(rρ−1+�) + S(r,f) Now, we write (12) as[ f(z) + (a1b0 + a2) a1b1 ] ≡− a3 a1b1f(z + c) Thus T ( r,f(z) + (a1b0 + a2) a1b1 ) ≡ T ( r,− a3 a1b1f(z + c) ) T(r,f) ≡ O(rρ−1+�) + S(r,f), which is again a contradiction. Thus our assumption is false. Hence the theorem. Example : Let f(z) = 2z2+1 and ρ(f(z)) = 0 (finite order) with N(r,f) = S(r,f). Consider P(f) = f(z) + 1. Then (2) becomes a1(z)(f(z) + 1)f(z + c) + a2(z)f(z + c)a3(z) ≡ 0 =⇒ f(z) + ( a1(z) + a2(z) a1(z) ) ≡− a3(z) a1(z)f(z + c) =⇒ T ( r,f(z) + ( a1(z) + a2(z) a1(z) )) = T ( r,− a3(z) a1(z)f(z + c) ) Applying Nevanlinna’s first fundamental theorem and using Lemma 2.2 and 2.5, we obtain T(r,f) + S(r,f) = T(r,f(z + c)) + O(rρ−1+�) + S(r,f) =⇒ T(r,f) + S(r,f) = T(r,f) + O(rρ−1+�) + S(r,f) Remarks: 1. In Theorem 1.1, a2(z) may or may not be zero. If a2(z) 6= 0, we can proceed as in the proof of Theorem 1.1. If a2(z) = 0, then (2) becomes a1(z)P(f)f(z+c)+a3(z) ≡ 0 =⇒ a1(z)P(f) ≡ −a3(z) f(z+c) we can proceed as in the proof of Theorem 1.1. So, in both the cases, we obtain a non-transcendental meromorphic solution f(z) NONLINEAR DIFFERENCE EQUATIONS 49 of finite order ρ with N(r,f) = O(rρ−1+�) + S(r,f) will satisfy the non-linear difference equation of the form a1(z)P(f)f(z + c) + a2(z)f(z + c) + a3(z) ≡ 0. 2. In Theorem 1.1, a3(z) 6= 0. If a3(z) = 0, then (2) becomes a1(z)P(f)f(z + c) + a2(z)f(z + c) ≡ 0 =⇒ a1(z)P(f) ≡−a2(z) =⇒ P(f) ≡ −a2(z) a1(z) Thus T(r,P(f)) = T ( r, −a2(z) a1(z) ) Using Lemma 2.2, we get nT(r,f) + S(r,f) = O(rρ−1+�) + S(r,f), which is con- tradiction. Similarly,, if a1(z) = 0 we obtain a contradiction. Hence a1(z) 6= 0. Proof of Theorem 1.2 We prove this theorem also by contradiction method. We first consider the case n ≥ 2. Suppose, there exists a transcendental meromor- phic function f(z) of finite order ρ with (13) N(r,f) = O(rρ−1+�) + S(r,f) satisfying the equation (14) a1(z)P(f)Π(f) + a2(z)Π(f) + a3(z) ≡ 0 i.e a1 [ bnf n + bn−1f n−1 + · · · + b1f + b0 ] Π(f) + a2Π(f) + a3 ≡ 0 (15) =⇒ a1bnfnΠ(f) + P1(f)Π(f) + a3 ≡ 0 where P1(f) = a1bn−1f n−1 + · · · + a1b1f + a1b0 + a2 We have difference polynomial as Π(f) = ∑ λ aλ(z)f(z) l0f(z + c1) l1 . . .f(z + cλ) lλ = f(z)n ∑ λ aλ(z)f(z) l0f(z + c1) l1 . . .f(z + cλ) lλ f(z)n = f(z)n ∑ λ aλ(z) ( f(z + c1) f(z) )l1 (f(z + c2) f(z) )l2 . . . ( f(z + cλ) f(z) )lλ By Lemma 2.3 and (13), we get N ( r, f(z + ci) f(z) ) ≤ N (r,f(z + ci)) + N ( r, 1 f(z) ) , i = 1, 2, . . .λ = O(rρ−1+�) + S(r,f)(16) Combining this with the assumption that T(r,aλ) = S(r,f), we obtain that N (r, Π(f)) = O(rρ−1+�) + S(r,f)(17) 50 VEENA L. PUJARI Now (15) can be written as f(z)n + P1(f) a1bn ≡− a3 a1bnΠ(f) ≡ ψ(z) (say)(18) From (17) and (18), we have N ( r, 1 ψ(z) ) ≡ N ( r,− a1bnΠ(f) a3 ) = O(rρ−1+�) + S(r,f)(19) Since ψ(z) = f(z)n + bn−1 bn f(z)n−1 + bn−2 bn f(z)n−2 + . . . + b0 bn + a2 bn By assumption and (19), we write N ( r, 1 ψ ) + N(r,f) ≤ N ( r, 1 ψ ) + N(r,f) = O(rρ−1+�) + S(r,f) Then applying the Lemma 2.1, we get ψ(z) = [ f(z) + bn−1 nbn ]n (20) From (18) and (20), we have[ f(z) + bn−1 nbn ]n ≡− a3 a1bnΠ(f) =⇒ [ f(z) + bn−1 nbn ]n Π(f) ≡− a3 a1bn Thus T ( r, [ f(z) + bn−1 nbn ]n Π(f) ) = T ( r,− a3 a1bn ) = O(rρ−1+�) + S(r,f)(21) Consider Π(f) f(z)n = ∑ λ aλ [ f(z + c1) f(z) ]l1 [f(z + c2) f(z) ]l2 · · · · · · [ f(z + cλ) f(z) ]lλ So m ( r, Π(f) f(z)n ) = ∑ λ [ m(r,aλ) + λ∑ i=1 lim ( r, f(z + ci) f(z) )] using Lemma 2.4 and m(r,aλ) = S(r,f), we have m ( r, Π(f) f(z)n ) = O(rρ−1+�) + S(r,f) Using (16) with this, we obtain T ( r, Π(f) f(z)n ) = O(rρ−1+�) + S(r,f)(22) NONLINEAR DIFFERENCE EQUATIONS 51 Now, by the first fundamental theorem of Nevanlinna and from (21) and (22), we have T ( r,f(z)n [ f(z) + bn−1 nbn ]n) = T  r, 1 f(z)n [ f(z) + bn−1 nbn ]n   + O(1) ≤ T ( r, Π(f) f(z)n ) + T  r, 1 Π(f) [ f(z) + bn−1 nbn ]n   + O(1) = O(rρ−1+�) + S(r,f)(23) By Lemma 2.2, we have T ( r,f(z)n [ f(z) + bn−1 nbn ]n) = 2nT(r,f) + S(r,f)(24) Thus from (23) and (24), we get 2nT(r,f) + S(r,f) ≤ O(rρ−1+�) + S(r,f) which is contradiction. Hence, our assumption is false. Now we shall consider the case when n = 1. If n = 1, then (14) becomes a1(z)(b1f + b0)Π(f) + a2(z)π(f) + a3(z) ≡ 0[ f(z) + a1b0 + a2 a1b1 ] Π(f) ≡− a3 a1b1 (25) The degree of − a3 a1b1 is zero and the degree of the term [ f(z) + a1b0+a2 a1b1 ] is one. Hence applying Lemma 2.6 to (25), we write m(r, Π(f)) = o ( T(r + |c|,f)1+� rδ ) + S(r,f),(26) where δ < 1 and � > 0, which holds for all r outside of a possible exceptional set with finite logarithmic measure. Thus adding (17) and (26), we write T(r, Π(f)) = O(rρ−1+�) + S(r,f)(27) Now (25)can be written as[ f(z) + (a1b0 + a2) a1b1 ] ≡− a3 a1b1Π(f) Thus by (27), we have T ( r,f(z) + (a1b0 + a2) a1b1 ) ≡ T ( r,− a3 a1b1Π(f) ) =⇒ T(r,f) = O(rρ−1+�) + S(r,f) which is again a contradiction. Thus our assumption is false. Hence the theorem. Example: Let f(z) = 2z2 + z + 1 and ρ(f(z)) = 0 with N(r,f) + N(r, 1/f) = 52 VEENA L. PUJARI S(r,f). Consider P(f) = f2(z) + 1 and Π(f) = f(z)f(z + c). Then (2) becomes a1(z)(f 2(z) + 1)(f(z)f(z + c)) + a2(z)(f(z)f(z + c)) + a3(z) ≡ 0 =⇒ f2(z) + ( a1(z) + a2(z) a1(z) ) ≡− a3(z) a1(z)f(z)f(z + c) =⇒ T ( r,f2(z) + ( a1(z) + a2(z) a1(z) )) = T ( r,− a3(z) a1(z)f(z)f(z + c) ) Applying Nevanlinna’s first fundamental theorem and using Lemma 2.2 and 2.5, we obtain 2T(r,f) + S(r,f) = T(r,f) + T(r,f(z + c)) + O(rρ−1+�) + S(r,f) =⇒ 2T(r,f) + S(r,f) = 2T(r,f) + O(rρ−1+�) + S(r,f) Remarks: 1. In Theorem 1.2, a2(z) may or may not be zero. In both the cases, we obtain a non-transcendental meromorphic solution. 2. If a1(z) = 0 and a3(z) = 0, we obtain a contradiction. Hence a1(z) 6= 0 and a3(z) 6= 0. 3. Lemma 2.3, 2.4 and 2.5 fails for meromorphic function of infinite order. Thus Theorem 1.1 and 1.2 are not true for infinite order. References [1] Yik-Man Chang and Shao-Ji Feng, On the Nevanlinna Characteristic of f(z + η) and differ- ence equations in the complex plane, Ramanujan Journal 16(2008), 105-129. 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