Meromorphic solutions of Painlevé III difference equations with Borel exceptional values

Transcription

1 Zhang Journal of Inequalities and Applications 2014, 2014:330 R E S E A R C H Open Access Meromorphic solutions of Painlevé III difference equations with Borel exceptional values Jilong Zhang * * Correspondence: jilongzhang2007@gmail.com LMIB and School of Mathematics & Systems Science, Beihang University, Beijing, , P.R. China Abstract In this paper, we investigate the properties of meromorphic solutions of Painlevé III difference equations. In particular, the difference equation www(w 1)=μ with μ being a non-zero constant is studied. We show that the rational solutions of the equation assume only one form and the transcendental solutions have at most one Borel exceptional value. We also show that the difference equation ww(w 1) 2 =(w λ) 2 does not have nonconstant rational solution, where λ ( 0,1)is a constant. MSC: 30D35; 39A10 Keywords: meromorphic solution; difference; finite order 1 Introduction Let w be a meromorphic function in the complex plane. The z-dependence is supposed by writing w w(z +1)andw w(z 1). We assume that the reader is familiar with the standard notations and results of Nevanlinna value distribution theory (see, e.g.,[1 3]). ρ(w), λ(w) andλ(1/w) denote the order, the exponents of convergence of zeros and poles of w, respectively. Furthermore, we denote by S(r, w) anyquantitysatisfying S(r, w) =o(t(r, w)) for all r outside of a set with finite logarithmic measure and by S(w)= { α meromorphic : T(r, α)=s(r, w) } the field of small functions with respect to w. A meromorphic solution w of a difference equation is called admissible if all coefficients of the equation are in S(w). For example, if a difference equation has only rational coefficients, then all non-rational meromorphic solutions are admissible; if an admissible solution is rational, then all the coefficients must be constants. Recently, with the development of Nevanlinna value distribution theory on difference expressions [4 6], Halburd and Korhonen [7] gave the full classification of the family including Painlevé I and II difference equations. As for the family including Painlevé III difference equations, we recall the following Zhang; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

2 ZhangJournal of Inequalities and Applications 2014, 2014:330 Page 2 of 12 Theorem A ([8]) Assume that the equation ww = R(z, w) (1.1) has an admissible meromorphic solution w of hyper-order less than one, where R(z, w) is rational and irreducible in w and meromorphic in z, then either w satisfies a difference Riccati equation w = αw + β w + γ, where α, β, γ S(w) are algebroid functions, or equation (1.1) canbetransformedtooneof the following equations: ww = ηw2 λw + μ (w 1)(w ν), (1.2a) ww = ηw2 λw, (1.2b) (w 1) ww = η(w λ) (w 1), (1.2c) ww = hw m. (1.2d) In (1.2a), the coefficients satisfy κ 2 μμ = μ 2, λμ = κλμ, κλλ = κλλ and one of the following: (1) η 1, νν =1, κ = ν; (2) η = η = ν, κ 1. In (1.2b), ηη = 1and λλ= λλ. In (1.2c),the coefficients satisfy one of the following: (1) η 1 and either λ = λλ or λ [3] λ [3] = λλ; (2) λλ = λλ, ηλ = λη, ηη = ηη [3] ; (3) ηη = ηη, λ = η; (4) λ [3] λ [3] = λλλ, ηλ = ηη, where λ [3] λ(z +3)and λ [3] λ(z 3).In (1.2d), h S(w) and m Z, m 2. In 2010, Chen and Shon [9] started the topic of researching the properties of finite-order meromorphic solutions of difference Painlevé I and II equations. In fact, they showed that if w is a transcendental finite-order meromorphic solution of the equation w + w = (a 1z + a 2 )w + a 3 or w + w = a 4z + a 5 + a 1 w 2 6, w where a j (1 j 6) are constants with a 1 a 3 a 4 0,thenw has at most one non-zero finite Borel exceptional value. The present author and Yang [10] improved the above result and verified that wdoes not have any Borel exceptional value. And they also considered the difference Painlevé III equations (1.2d) with the constant coefficients. The difference equations (1.2b)and(1.2c) are studied by the present author in the following.

3 ZhangJournal of Inequalities and Applications 2014, 2014:330 Page 3 of 12 Theorem B ([11]) If w is a transcendental finite-order meromorphic solution of ww(w 1)=ηw or ww(w 1)=w 2 λw, where η ( 0), λ ( 0,1) are constants, then (i) λ(w)=ρ(w); (ii) w has at most one non-zero Borel exceptional value for ρ(w)>0. The purpose of this paper is to study the remaining difference equation (1.2a)withconstant coefficients. As it is complicated to discuss meromorphic solutions of (1.2a) when λμ 0, we will consider the following special cases. Theorem 1.1 Suppose that w is a nonconstant rational solution of www(w 1)=μ, (1.3) where μ is a non-zero constant. Then μ = and w assumes the only form of w(z)= with any constant b. 3(z + b) 2 4(z + b +1)(z + b 1) Remark Equation (1.3) isaspecialcaseof(1.2a) in the option (2) as η = λ = ν =0.From the proof of Theorem 1.1, we shall see that the degrees of numerator and denominator of w(z) must be 2. Here, the coefficient 3 is determined. We think that it is because there is 4 a 1in(1.3). Theorem 1.2 Suppose that w is a transcendental finite-order meromorphic solution of (1.3). Then (i) λ(w)=ρ(w); (ii) w has at most one Borel exceptional value for ρ(w)>0. 3(e Example 1.3 The function w 1 (z)= 2πiz +z) 2 is a solution of the difference equation www(w 1)= 33.Itiseasytoseethatλ(w 4(e 2πiz +z+1)(e 2πiz +z 1) )=ρ(w 1 ). Since w = 3, 4(e 2πiz +z+1)(e 2πiz +z 1) then 3 4 is a Picard exceptional value of w 1. This shows that the conclusions of Theorem 1.2 may occur. Remark In w 1 in Example 1.3, the function e 2πiz can be replaced by any finite-order function with period 1. For example, sin(2πz), tan(πz)andsoon. Theorem 1.4 Suppose that w is a nonconstant meromorphic solution of ww(w 1) 2 =(w λ) 2, (1.4) where λ ( 0,1)is a constant. Then w must be transcendental.

4 ZhangJournal of Inequalities and Applications 2014, 2014:330 Page 4 of 12 Theorem 1.5 Suppose that w is a transcendental finite-order meromorphic solution of (1.4). If a and b are two Borel exceptional values of w, then (i) a + b =2, ab = λ; (ii) w = w; (iii) w satisfies the difference Riccati equation w = w λ w 1. is a solution of both the differ- Example 1.6 The transcendental function w 2 (z)= 3eiπz 1 e iπz +1 ence equation ww(w 1) 2 =(w +3) 2 andthericcatiequationw = w+3 w 1.Notingthatw 2(z) has two Picard exceptional values 3 and 1, we see that the conclusions in Theorem 1.5 may occur. 2 Some lemmas Halburd and Korhonen [5] and Chiang and Feng [4] investigated the value distribution theory of difference expressions, a key result of which is a difference analogue of the logarithmic derivative lemma. With the help of the lemma, the difference analogues of the Clunie and Mohon ko lemmas are obtained. Lemma 2.1 ([6]) Let f be a transcendental meromorphic solution of finite order ρ of a difference equation of the form U(z, f )P(z, f )=Q(z, f ), where U(z, f ), P(z, f ) and Q(z, f ) are difference polynomials such that the total degree deg f U(z, f )=ninf(z) and its shifts, and deg f Q(z, f ) n. If U(z, f ) contains just one term of maximal total degree in f (z) and its shifts, then, for each ε >0, m ( r, P(z, f ) ) = O ( r ρ 1+ε) + S(r, f ), possibly outside of an exceptional set of finite logarithmic measure. Lemma 2.2 ([5, 6]) Let w be a transcendental meromorphic solution of finite order of the difference equation P(z, w)=0, where P(z, w) is a difference polynomial in w(z). If P(z, a) 0 for a meromorphic function a S(w), then ( ) 1 m r, = S(r, w). w a We conclude this section by the following lemma. Lemma 2.3 (see, e.g., [3, pp.79-80]) Let f j (j =1,...,n)(n 2) be meromorphic functions, g j (j =1,...,n) be entire functions. If (i) n j=1 f j(z)e g j(z) 0; (ii) g h (z) g k (z) is not a constant for 1 h < k n; (iii) T(r, f j )=S(r, e g h(z) g k (z) ) for 1 j n and 1 h < k n, then f j (z) 0(j =1,...,n).

5 ZhangJournal of Inequalities and Applications 2014, 2014:330 Page 5 of 12 3 Proofsoftheorems Proof of Theorem 1.1 Suppose that w = P(z),whereP(z)andQ(z) are relatively prime polynomials with degrees p and q respectively. It follows from Q(z) (1.3)that P(z +1) P(z 1) P(z) Q(z +1) Q(z 1) Q(z) P(z) Q(z) Q(z) = μ. (3.1) Without loss of generality, we assume that the coefficients of the highest degree terms of P(z)andQ(z)area and 1 respectively. Let s = p q. If s >0,then P(z) Q(z) = azs (1 + o(1)) as z tends to infinity, where a is a non-zero constant. And (3.1)gives a 3 (z +1) s (z 1) s z s( 1+o(1) )( az s( 1+o(1) ) 1 ) = μ, which is a contradiction since the left-hand side of the above equation goes to infinity as r tends to infinity, while the right-hand side is a constant. If s <0,wehave P(z) P(z±1) = o(1) and = o(1) as z tends to infinity. Equation (3.1)yields Q(z) Q(z±1) o(1) = μ, z, which contradicts μ 0. Thus s =0andp = q. Noting that the zeros of Q(z) are not the zeros of P(z) andp(z) Q(z), we get from (3.1) thatallthezerosofq 2 (z) arethezerosofp(z +1)P(z 1).Asthe degrees of Q 2 (z)andp(z +1)P(z 1)areboth2p,weobtain P(z +1)P(z 1)=a 2 Q 2 (z), (3.2) P(z) ( P(z) Q(z) ) = a(a 1)Q(z +1)Q(z 1). (3.3) Now, we aim to prove that the orders of all the zeros of P(z) are even. Otherwise, assume that z 0 is a zero of P(z) with the order k, andk is an odd integer. Then P(z) hastheterm (z z 0 ) k,andp(z +1)P(z 1)hastheterm (z z 0 +1) k (z z 0 1) k (3.4) caused by z 0.Itmeansthatz 0 1andz are both zeros of P(z +1)P(z 1) with the order at least k. On the other hand, it follows from (3.3)thatQ(z+1)Q(z 1)hastheterm(z z 0 ) k exactly. Suppose that Q(z +1)andQ(z 1)havetheterms(z z 0 ) m and (z z 0 ) l respectively, where m and l are non-negative integers satisfying m + l = k.thenq(z)hasterm(z z 0 1) m (z z 0 +1) l exactly, i.e., Q 2 (z) hastheterm(z z 0 1) 2m (z z 0 +1) 2l.By(3.2), P(z +1)P(z 1) has the term (z z 0 +1) 2l (z z 0 1) 2m. (3.5) Without loss of generality, assume that m < l.obviously,2m < k and 2l > k.thenz 0 +1isa zero of P(z +1)P(z 1) with the order 2m < k, which is a contradiction by (3.4). Therefore, all the zeros of P(z)have even orders.

6 ZhangJournal of Inequalities and Applications 2014, 2014:330 Page 6 of 12 Denote P(z)=ar 2 (z), where r(z)=z n + A n 1 z n 1 + A n 2 z n A 1 z + A 0. (3.6) Noting that the coefficient of the highest degree term of Q(z) is1,itiseasytoseefrom (3.2)and(3.3)thatQ = rr and ar 2 rr =(a 1)rr.Let φ = ar 2 rr (a 1)rr. Then φ(z) 0. If deg r = n 2, substituting r by (3.6) in the right-hand side of the last equation, we have that the coefficients of terms z 2n 2, z 2n 3 and z 2n 4 of φ(z)are B 2n 2 = n(4a 3), B 2n 3 =2(n 1)(4a 3)A n 1, B 2n 4 = (15 16a)C 2 n +2(n 3)(4a 3)A n 2 +(n 1)(4a 3)A 2 n 1. We deduce from B 2n 2 =0thata = 3 4. And then B 2n 4 =3Cn 2 0,thusdeg r =1.Letr = z+b. Then w(z)= P(z) Q(z) = 3(z + b) 2 4(z + b +1)(z + b 1), and μ = Proof of Theorem 1.4 Assume to the contrary that w is a nonconstant rational function. Denote w = P(z),whereP(z)andQ(z) are relatively prime polynomials with degrees p and Q(z) q respectively. It follows from (1.4)that ( ) P(z +1) P(z 1) P(z) 2 ( ) P(z) 2 Q(z +1) Q(z 1) Q(z) 1 = Q(z) λ. (3.7) By the same reasoning as in the proof of Theorem 1.1, wehavep = q. Wealsoassume that the coefficients of the highest degree terms of P(z) andq(z) area and 1 respectively. Let z, it follows from (3.7)that a 2 (a 1) 2 =(a λ) 2. (3.8) Obviously, a / {0,1,λ}. Rewriting (3.7) as P(z +1) P(z 1) Q(z +1) Q(z 1) = ( ) P(z) λq(z) 2, P(z) Q(z) and noting that the degrees of P(z) λq(z)andp(z) Q(z)arebothp,wehave (a λ) 2 P(z +1)P(z 1)=a 2( P(z) λq(z) ) 2, (3.9) (a 1) 2 Q(z +1)Q(z 1)= ( P(z) Q(z) ) 2. (3.10)

7 ZhangJournal of Inequalities and Applications 2014, 2014:330 Page 7 of 12 Suppose that z 0 is a zero of P(z) with the order k, andk is an odd integer. Then z 0 1 is a zero of P(z + 1) with the order k. However, since all the zeros of P(z +1)P(z 1)have even orders by (3.8), z 0 1mustbeazeroofP(z 1)withtheoddorderl.Thus,z 0 2is azeroofp(z) withtheoddorderl. Therefore, z 0 2m are all zeros of P(z) by induction, which is impossible. Then all the zeros of P(z) have even orders. Similarly, all the zeros of Q(z)haveevenorderstoo. Denote P(z)=ar 2 (z)andq(z)=t 2 (z), where r(z)=z n + A n 1 z n 1 + A n 2 z n A 1 z + A 0, (3.11) t(z)=z n + B n 1 z n 1 + B n 2 z n B 1 z + B 0. (3.12) We obtain from (3.9)and(3.10)that (a λ)rr = ar 2 λs 2, (a 1)ss = ar 2 s 2. Substituting r and t by (3.11) and(3.12) respectively in the last two equations and comparing the coefficients of terms z 2n 1 and z 2n 2,wehaveA n 1 = B n 1 and n(a λ)=2λ(a n 2 B n 2 ), n(a 1)=2a(A n 2 B n 2 ). It is easy to see that A n 2 B n 2 by a 1.Then a(a λ)=λ(a 1). Combining the above equation with (3.8),we get that a = 0 or a = 1, a contradiction. Proof of Theorem 1.5 Rewriting the difference equation (1.4)as www 2 =2www ww +(w λ) 2, we have from Lemma 2.1 that m(r, w)=s(r, w), then N(r, w)=t(r, w)+s(r, w) and is not the Borel exceptional value of w.thus,a and b are finite complexnumbers. Let P(z, w)=ww(w 1) 2 (w λ) 2. It is easy to see that P(z,0)= λ 2 0. Lemma 2.2 tells us that ( N r, 1 ) = T(r, w)+s(r, w), w and then ab 0.

8 ZhangJournal of Inequalities and Applications 2014, 2014:330 Page 8 of 12 Set f (z)= w(z) a w(z) b. (3.13) Then ρ(f )=ρ(w), λ(f )=λ(w a) <ρ(f )andλ(1/f )=λ(w b) <ρ(f ). Since f is of finite order, we suppose that f (z)=g(z)e dzn, (3.14) where d ( 0) is a constant, n ( 1) is an integer, g(z) is meromorphicand satisfies ρ(g)<ρ(f )=n. (3.15) Then f (z +1)=g(z +1)g 1 (z)e dzn, f (z 1)=g(z 1)g 2 (z)e dzn, (3.16) where g 1 (z)=e ndzn 1 + +d and g 2 (z)=e ndzn 1 + +( 1) nd. bf a We get from (3.13)thatw =.By(1.4)and(3.16), we have f 1 A(z)e 4dzn + B(z)e 3dzn + C(z)e 2dzn + De dzn + E =0, (3.17) where A(z)= [ b 2 (b 1) 2 (b λ) 2] g 2 gg 1 gg 2, B(z)= [ 2b 2 (b 1)(1 a) 2(b λ)(λ a) ] ggg 1 gg 2 + [ (b λ) 2 ab(b 1) 2] g 2 (gg 1 + gg 2 ), C(z)= [ a 2 (b 1) 2 (b λ) 2] g 2 + [ b 2 (a 1) 2 (a λ) 2] (gg 1 gg 2 ) +2 [ (b λ)(λ a)+ab(b 1)(a 1) ] g(gg 1 + gg 2 ), D(z)=2 [ a 2 (b 1)(1 a) (b λ)(λ a) ] g + [ (λ a) 2 ab(1 a) 2] (gg 1 + gg 2 ), E = a 2 (a 1) 2 (a λ) 2. From (3.15), we apply Lemma 2.3 to (3.17), resulting in all the coefficients vanish. We deduce from A(z)=0andE =0that a 2 (a 1) 2 =(a λ) 2, b 2 (b 1) 2 =(b λ) 2. Thus a λ = ±a(a 1), b λ = ±b(b 1). Define G = g, G 1 = gg 1 and G 2 = gg 2.Wediscussthefollowingthreecases.

9 ZhangJournal of Inequalities and Applications 2014, 2014:330 Page 9 of 12 Case 1. Suppose that a λ = a(a 1), b λ = b(b 1). It followsfromthe above equationsthat a and b are distinct zeros of the equation z 2 2z + λ =0. Then a + b =2, ab = λ. (3.18) From B(z)=0, D(z)=0 and(3.18), we have 2b(b 1)(1 a)g 1 G 2 = b(b 1) 2 G(G 1 + G 2 ), 2a(b 1) 2 (a b)g = a(b 1) 2 (a b)(g 1 + G 2 ). If b =1,wegetfrom(3.18)thata = 1, which is a contradiction to a b.thenb 1.Noting that ab 0, the last two equations yield 2G 1 G 2 = G(G 1 + G 2 ), 2G = G 1 + G 2. Combining with the last two equations, we have 4G 1 G 2 =(G 1 + G 2 ) 2,whichmeans,with 2G = G 1 + G 2,that G 1 = G 2 = G. From (3.14)and(3.16), it is easy to see that f = f = f and w = w.wededucefrom(3.13) that w = bf a f 1 = bf + a b(w a) f +1 = w b w a i.e., w = w λ w 1. Case 2. Suppose that + a w b +1 = w ab w 1, a λ = a(a 1), b λ = b(b 1). Then a 2 = λ and b 2 = λ. Without loss of generality, we assume a = λ, b = λ. (3.19) From B(z)=0, D(z)=0 and(3.19), we have 2b(b 1)(1 a)g 1 G 2 = b(b 1) 2 G(G 1 + G 2 ), 2a(b 1)(1 a)g = a(a 1) 2 (G 1 + G 2 ).

10 Zhang Journal of Inequalities and Applications 2014, 2014:330 Page 10 of 12 If a =1orb =1,wegetfrom(3.19) thatλ = 1, which is a contradiction. Then a 1and b 1.Notingthatab 0, the last two equations mean 2(a 1)G 1 G 2 =(b 1)G(G 1 + G 2 ), 2(b 1)G =(a 1)(G 1 + G 2 ). We deduce from the above two equations that (G 1 G 2 ) 2 =0,whichisG 1 = G 2,andboth are equal to b 1 G by the last equation. From (3.14)and(3.16), we obtain that a 1 f = b 1 b 1 f and f = a 1 a 1 f, which yield (a 1) 2 =(b 1) 2, a contradiction since a + b =0. Case 3. Suppose that a λ = a(a 1), b λ = b(b 1). (3.20) Then b 2 = λ.sinceλ 1,itiseasytoseethata 1andb 1by(3.20). Noting that ab 0, we get from B(z)=0,C(z)=0 and D(z)=0 that 2(1 a)(a + b)g 1 G 2 =(b 1)(a b)g(g 1 + G 2 ), (3.21) ( a 2 b 2)[ (b 1) 2 G 2 (a 1) 2 ] G 1 G 2 =4ab(b 1)(1 a)g(g1 + G 2 ), (3.22) 2(b 1)(a + b)g =(a 1)(a b)(g 1 + G 2 ). (3.23) Combining (3.21)with(3.22)and(3.23) respectively, we obtain Then (a b) 2 (b 1) 2 G 2 =(a 1) 2[ (a b) 2 +8ab ] G 1 G 2, (b 1) 2 G 2 = (a 1) 2 G 1 G 2. (a b) 2 = (a b) 2 8ab, which is a + b =0.Thusa = b = ± λ. Substituting a by ± λ in the first equation of (3.20), we get that λ =0orλ = 1, both are impossible. Proof of Theorem 1.2 Let P(z, w)=www(w 1) μ.thenp(z,0)= μ 0.Wededucefrom Lemma2.2that m(r,1/w) = S(r, w), andthenn(r,1/w) = T(r, w)+s(r, w). Thusλ(w) = ρ(w). Assume to the contrary that w has two Borel exceptional values a and b ( a). Obviously, ab 0byλ(w) = ρ(w).let f begiven by (3.13). Then we still have (3.14)-(3.16). Substituting w = where bf a f 1 in (1.3), we obtain A 1 (z)e 4dzn + B 1 (z)e 3dzn + C 1 (z)e 2dzn + D 1 (z)e dzn + E 1 =0, (3.24) A 1 (z)= [ b 3 (b 1) μ ] g 2 gg 1 gg 2,

11 Zhang Journal of Inequalities and Applications 2014, 2014:330 Page 11 of 12 B 1 (z)= [ b 2 (a + b 2ab)+2μ ] ggg 1 gg 2 + [ ab 2 (1 b)+μ ] g 2 (gg 1 + gg 2 ), C 1 (z)= [ ab 2 (a 1) μ ] gg 1 gg 2 [ ab(a + b 2ab)+2μ ] g(gg 1 + gg 2 ) + [ a 2 b(b 1) μ ] g 2, D 1 (z)= [ μ a 2 b(a 1) ] (gg 1 + gg 2 )+ [ a 2 (a + b 2ab)+2μ ] g, E 1 = a 3 (a 1) μ. Lemma 2.3 tellsusthatallthecoefficientsof(3.24) vanish. By a similar way to the above, we deduce from A 1 (z)=0ande 1 =0that a 3 (a 1)=μ, b 3 (b 1)=μ. (3.25) Denote G = g, G 1 = gg 1 and G 2 = gg 2.FromB 1 (z)=0,c 1 (z)=0and(3.25), and noting that a b,wehave (1 2b)G 1 G 2 =(b 1)G(G 1 + G 2 ), (1 2a)G =(a 1)(G 1 + G 2 ). Since the last two equations are both homogeneous, there exist two non-zero constants α and β such that G 1 = αg and G 2 = βg.then (1 2b)αβ =(b 1)(α + β), (3.26) (1 2a)=(a 1)(α + β). (3.27) If α + β =0,thena = b = 1 by (3.26)and(3.27), which is a contradiction. Thus α + β 0. 2 On the other hand, combining (3.14)with(3.16), we have f = αf, f = βf, which yield αβ =1. Itfollowsfrom (3.26)and(3.27)that 1 2b 1 2a = b 1 a 1, which means a = b, a contradiction. Competing interests The author declares that he has no competing interests. Author s contributions The author drafted the manuscript, read and approved the final manuscript. Acknowledgements The author would like to thank the referee for his or her valuable suggestions to the present paper. This research was supported by the NNSF of China Nos , , and the YWF-14-SXXY-008, YWF-ZY of Beihang University. This research was also supported by the youth talent program of Beijing No Received: 15 April 2014 Accepted: 8 August 2014 Published: 02 Sep 2014

12 Zhang Journal of Inequalities and Applications 2014, 2014:330 Page 12 of 12 References 1. Hayman, WK: Meromorphic Functions. Clarendon Press, Oxford (1964) 2. Laine, I: Nevanlinna Theory and Complex Differential Equations. de Gruyter, Berlin (1993) 3. Yang, CC, Yi, HX: Uniqueness Theory of Meromorphic Functions. Kluwer Academic, Dordrecht (2003) 4. Chiang, YM, Feng, SJ: On the Nevanlinna characteristic of f (z + η) and difference equations in the complex plane. Ramanujan J. 16, (2008) 5. Halburd, RG, Korhonen, RJ: Difference analogue of the lemma on the logarithmic derivative with applications to difference equations. J. Math. Anal. Appl. 314, (2006) 6. Laine, I, Yang, CC: Clunie theorems for difference and q-difference polynomials. J. Lond. Math. Soc. 76, (2007) 7. Halburd, RG, Korhonen, RJ: Finite order solutions and the discrete Painlevé equations. Proc. Lond. Math. Soc. 94, (2007) 8. Ronkainen, O: Meromorphic solutions of difference Painlevé equations. Ann. Acad. Sci. Fenn., Math. Diss. 155, 59pp (2010) 9. Chen, ZX, Shon, KH: Value distribution of meromorphic solutions of certain difference Painlevé equations. J. Math. Anal. Appl. 364, (2010) 10. Zhang, JL, Yang, LZ: Meromorphic solutions of Painlevé III difference equations. Acta Math. Sin. 57, (2014) 11. Zhang, JL, Yi, HX: Properties of meromorphic solutions of Painlevé III difference equations. Adv. Differ. Equ. 2013, 256 (2013) / X Cite this article as: Zhang: Meromorphic solutions of Painlevé III difference equations with Borel exceptional values. Journal of Inequalities and Applications 2014, 2014:330