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A class of discrete equations is considered from three perspectives corresponding to three measures of complexity of the solutions: the (hyper-) order of meromorphic solutions in the sense of Nevanlinna, the degree growth of iterates over a function ﬁeld and the height growth of iterates over the rational numbers. In each case, low complexity implies a form of singularity conﬁnement which results in a known discrete Painlevé equation. Keywords Discrete Painlevé equations · Algebraic entropy · Order of meromorphic solutions · Diophantine integrability Mathematics Subject Classiﬁcation 30D35 · 39A12 · 37J35 1 Introduction In this paper we will study the equation a y + b y + c j j j j y + y = , (1) j +1 j −1 Communicated by Athanassios S. Fokas. Risto Korhonen was partially supported by the Academy of Finland grants (#286877) and (#268009). B Risto Korhonen risto.korhonen@uef.ﬁ Rod Halburd r.halburd@ucl.ac.uk Department of Mathematics, University College London, Gower Street, London WC1E 6BT, UK Department of Physics and Mathematics, University of Eastern Finland, P.O. Box 111, 80101 Joensuu, Finland 123 300 R. Halburd, R. Korhonen where c ≡ 0, from three perspectives. In each approach we will explore a different measure of complexity of the solutions and we will interpret y and the coefﬁcients in a slightly different way. Integrability has long been associated with the slow growth of complexity [21], however the ﬁrst highly sensitive, yet heuristic, test for integrability of equations such as (1) was the idea of singularity conﬁnement [6,18]. In each of the three approaches studied here, we will use an analogue of singularity (non-) conﬁnement, suitably re-interpreted, to get a lower bound on the relevant measure of complexity. In analogy with the Painlevé property for differential equations, the idea behind singularity conﬁnement is to study the behaviour of solutions at singular values of the dependent variable. For Eq. (1) we note that, if the coefﬁcient functions are ﬁnite, the only way that y can become inﬁnite starting from ﬁnite values is if y = 0. In j +1 j order to better understand this situation, we consider the perturbed initial conditions y = κ, where κ is arbitrary, and y = and expand the next few iterates as Laurent j −1 j series about = 0. If c = 0, this gives c b j j y = + + O(1), j +1 j +1 2 3 y = a − + + O( ), j +2 j +1 (2) c b (b a + c ) − + O( ) j j j +2 j +1 j +2 y =− − + + O(1). j +3 2 2 2 3 a − + (b /c ) + O( ) j +1 j +1 j In the limit → 0, y will be inﬁnite unless a = 0. If a = 0wehave j +3 j +1 j +1 c − c b − 2b + b j +2 j j +2 j +1 j y = − + O(1). j +3 Taking the limit → 0 we see that y =∞ unless j +3 a = 0, b − 2b − b = 0 and c − c = 0. (3) j +1 j +2 j +1 j j +2 j If these conditions are satisﬁed the singularity is said to be conﬁned as y remains ﬁnite (at least for the next few iterates). Demanding that all singularities are conﬁned in this way means that conditions (3) must hold for all j. Hence Eq. (1) becomes (α j + β)y + (γ + δ(−1) ) y + y = , (4) j +1 j −1 where α, β, γ and δ are constants. In order to avoid technicalities in some of the approaches that follow, we will restrict ourselves to the case in which a , b and c j j j are rational functions of j. This forces δ = 0in Eq. (4), leaving us with the equation (α j + β)y + γ y + y = . (5) j +1 j −1 123 Three Approaches to Detecting Discrete Integrability 301 Equation (5) is known to have a continuum limit to the ﬁrst Painlevé equation, it is the compatibility condition for a related linear problem and it has been derived from the Schlesinger transformations of the third Painlevé equation [5]. Although singularity conﬁnement has been successfully used to ﬁnd many inte- grable discrete equations, Hietarinta and Viallet [13] gave an example of an equation in which the singularities are conﬁned and yet the dynamics appear to be chaotic. The ﬁrst of the three approaches considered in this paper is to study the growth, in the sense of Nevanlinna, of meromorphic solutions of difference equations. To this end we replace Eq. (1) with its complex analytic version: a(z)y(z) + b(z)y(z) + c(z) y(z + 1) + y(z − 1) = , (6) y(z) where a, b and c are rational functions and y is a non-rational meromorphic function. It was suggested in [2] that the existence of sufﬁciently many ﬁnite-order meromorphic solutions is a natural difference equation analogue of the Painlevé property. In [11]the authors used the existence of an admissible ﬁnite-order meromorphic solution to reduce a class of difference equations to a short list of difference Painlevé-type equations. Here admissible means that the solution grows faster than the coefﬁcients in a precise sense. In the case in which the coefﬁcients are rational functions as considered here, this amounts to saying that the solution is non-rational. In [9] it was shown that the same conclusions remain valid if ﬁnite-order is replaced by hyper-order less than one. The order and hyper-order of a meromorphic function will be deﬁned in Sect. 2 of the present paper and we will prove the following. Theorem 1 Let y be a non-rational meromorphic solution of Eq. (6), where a, b and c ≡ 0 are rational functions. If the hyper-order of y is less than one, then a ≡ 0, b(z) = Az + B and c(z) = C, where A, B and C are complex constants. This is a special case of the classiﬁcation in [9,11]. By restricting ourselves to the case of rational coefﬁcients, we eliminate many technical issues that arise from Nevanlinna theory, which allows us to concentrate on the role played by singularity conﬁnement- type arguments in obtaining a lower bound on the hyper-order of solutions. In Sect. 3 we will consider Eq. (1) as a discrete equation (i.e. j ∈ Z) but where each y is a rational function of an external complex variable z. The natural measure of complexity here is the degree growth of the rational iterates y . This is very close to the idea of algebraic entropy [4,13] in which one considers the degree d of the jth iterate of Eq. (1) as a rational function of y and y . The deﬁnition of the algebraic 0 1 entropy is log d lim . j →∞ j Zero algebraic entropy is associated with integrability. By considering y ≡ y (z) and 0 0 y ≡ y (z) as rational functions of z we can use more elementary arguments based on 1 1 complex analysis of a single variable. It also gives us a more reﬁned tool to consider one-parameter families of solutions, rather than considering the whole solution space 123 302 R. Halburd, R. Korhonen at once. If y and y are degree one rational functions, then the degree of y will be the 0 1 n same as d unless some cancellation has occurred on substitution into the expression for y as a function of y and y . n 0 1 Finally, in Sect. 4 we will consider the case in which the coefﬁcients a , b , c j j j are rational functions of j with rational coefﬁcients and the solution of the discrete Eq. (1), y ∈ Q for all sufﬁciently large j. In this setting the natural measure of complexity is the height. The logarithmic height of p/q, where p and q are co-prime integers, is h(p/q) = log max(| p|, |q|). If the logarithmic height of all solutions grows polynomially, we say that the equation is Diophantine integrable [8]. This idea was suggested by applying the observation of Osgood [15–17] and Vojta [22] that there is a formal similarity between Nevanlinna theory and Diophantine approximation to the ﬁrst approach to discrete integrability above. The logarithmic height can be expressed as a sum over all of the (suitably normalised) absolute values on Q (i.e., the usual absolute value and the p-adic absolute values). We will show how a calculation similar to the singularity conﬁnement sequence (2) can be formulated in which the small quantity is small with respect to an arbitrary absolute value on Q. This then induces a lower bound on the height. We will highlight the similarities between the previous two approaches and the proof of the following theorem, which appears in [7] and the PhD thesis of Will Morgan [14]. Theorem 2 Let a ,b and c ≡ 0 be rational functions with coefﬁcients in Q. Suppose n n n that for sufﬁciently large r ,y ∈ Q solves Eq. (1) for all n ≥ r .If 0 n 0 r r {h(a ) + h(b ) + h(c )}= o h(y ) (7) n n n n n=r n=r 0 0 and h(y ) ≤ Kr , (8) n=r for some positive constants K and ρ, then equation (1) reduces to Eq. (5). A similar result leading to the discrete Painlevé II equation has been derived in [3]. Heights were ﬁrst used in Abarenkova, Anglès d’Auriac, Boukraa, Hassani and Mail- lard [1] to estimate entropy. Heights have also been used to detect low complexity solutions in Silverman [20] and Roberts and Vivaldi [19]. 2 Existence of Meromorphic Solutions of Hyper-Order Less Than One In this section we will use the slow growth rate of meromorphic solutions to detect Painlevé type equations out of a natural class of second-order difference equations. Our aim is to review the method of [11] by going through a simple case requiring as few technical details as possible. We will need a small number of concepts from Nevanlinna theory (see e.g. [12]) such as the deﬁnitions of order, hyper-order and the 123 Three Approaches to Detecting Discrete Integrability 303 counting function. Lemma 1 below allows us to translate simple inequalities about the relative frequencies of zeros and poles into statements about the hyper-order. Let f be a meromorphic function in the complex plane. The counting function n(r , f ) is the number of poles of f in the disc {z ∈ C :|z|≤ r }, each pole counted according to its multiplicity. Moreover, we deﬁne log x = max{log x , 0} for any x ≥ 0. Then dt N r , f = (n(t , f ) − n(0, f )) + n 0, f log r ( ) ( ) is the integrated counting function, 2π dθ + i θ m(r , f ) = log | f (re )| 2π is the proximity function, and T (r , f ) = m(r , f ) + N (r , f ) is the Nevanlinna characteristic function of f . The order of f is log T (r , f ) σ( f ) = lim sup , log r r →∞ and the hyper-order is log log T (r , f ) ς( f ) = lim sup . log r r →∞ Note that by restricting the coefﬁcients of (6) to be rational functions rules out the most general form of the difference Painlevé I equation, where c is a period two function. The general case was recovered in [11]. Proof of Theorem 1 We will ﬁrst show that if |z| is sufﬁciently large, then each zero z ˆ of a non-rational meromorphic solution y(z) of (6) may be uniquely associated with a ﬁnite number of poles and zeros of neighboring iterates y(z ± 1), y(z ± 2), y(z ± 3), y(z ± 4) such that the number of zeros divided by the number of poles, both counting multiplicities, is bounded by 4/5 in each such grouping. Let y(z) be a meromorphic solution of (6) and assume ﬁrst that a(z) is not identically zero. Therefore, a, b and c have ﬁnitely many zeros (unless b ≡ 0) and poles, since they are rational functions of z, and so there exists an r ≥ 0 such that a(z) = 0 and c(z) = 0 for all z satisfying |z|≥ r . Suppose that the solution y(z) has a zero of multiplicity k at some point z ˆ in the complex plane. Then y(z) can be expressed as a Laurent series expansion k k+1 y(z) = α(z −ˆz) + O((z −ˆz) ), α ∈ C\{0}, (9) 123 304 R. Halburd, R. Korhonen in a sufﬁciently small neighborhood of z ˆ. All except ﬁnitely many zeros z ˆ of y(z) satisfy |ˆz|≥ r + 1, and so, for these zeros it follows that c(z ˆ + σ) = 0, where σ =±1. Now it follows by substituting (9)tothe Eq.(6) that y(z + σ) has a pole at z ˆ of order at least 2k for σ = 1or σ =−1. The order of the pole at z ˆ + σ can be strictly greater than 2k only if there is a pole of the same order l > 2k at both points z ˆ + 1 and z ˆ − 1. In this case we can ﬁnd even more poles per zero compared to the case where the order of the pole at z ˆ + σ is equal to 2k. Therefore, without loss of generality, we may assume that y(z + σ) has a pole of order 2k at z ˆ, and so −2k 1−2k y(z + σ) = β (z −ˆz) + O((z −ˆz) ), β ∈ C\{0} (10) σ σ for all z in a small enough neighborhood of z ˆ.Now,byshiftingEq. (6), we obtain a(z + σ)y(z + σ) + b(z + σ)y(z + σ) + c(z + σ) y(z + 2σ) + y(z) = . y(z + σ) (11) By substituting the Laurent series expansions (9) and (10)into(11), we have y(z + 2σ) = a(z + σ) + O((z −ˆz) ) (12) in a neighborhood of z ˆ. By continuing in this way, and summarizing the above, it follows that k k+1 y(z) = α(z −ˆz) + O((z −ˆz) ) −2k 1−2k y(z + σ) = β (z −ˆz) + O((z −ˆz) ) (13) y(z + 2σ) = a(z + σ) + O((z −ˆz) ) −2k 1−2k y(z + 3σ) =−β (z −ˆz) + O((z −ˆz) ) y(z + 4σ) = a(z + 3σ) − a(z + σ) + O((z −ˆz) ), where α and β are non-zero. Therefore, the zero of y(z) of order k may be grouped together with the pole of y(z+σ) of order 2k. Note that even if a(z+3σ)−a(z+σ) = 0 we are free to associate the available pole of y(z + 3σ) with the zero of y(z + 4σ), if the zero is of order k at most. If the zero of y(z + 4σ) is of order l > k, then z ˆ + 4 is a starting point of another sequence of the type (13). Hence the number of zeros divided by the number of poles in the sequence (13) is less than or equal to 1/2 counting multiplicities, provided that a ≡ 0in (6). Suppose now that a ≡ 0 so that equation (6) reduces to b(z)y(z) + c(z) y(z + 1) + y(z − 1) = . (14) y(z) We will again consider the case where y(z) has a zero of order k at z =ˆz assuming ﬁrst that both y(z + σ) and y(z − σ) have a pole at least of order 2k. Then, as before, 123 Three Approaches to Detecting Discrete Integrability 305 we may assume without loss of generality that the order of the pole is exactly 2k, and so, by iterating (14), it follows that k k+1 y(z − 2σ) =−α(z −ˆz) + O((z −ˆz) ) −2k 1−2k y(z − σ) = γ (z −ˆz) + O((z −ˆz) ) k k+1 y(z) = α(z −ˆz) + O((z −ˆz) ) (15) −2k 1−2k y(z + σ) = β (z −ˆz) + O((z −ˆz) ) k k+1 y(z + 2σ) =−α(z −ˆz) + O((z −ˆz) ), where α = 0, β = 0 and γ = 0. In this sequence the number of zeros divided σ σ by the number of poles is less than or equal to 3/4, when multiplicities are taken into account. Suppose now that y(z) has a zero of order k at z =ˆz and y(z − σ) has a pole of order l such that k ≤ l < 2k. Then, by (14), −l 1−l y(z − σ) = β (z −ˆz) + O((z −ˆz) ) k k+1 y(z) = α(z −ˆz) + O((z −ˆz) ) c(z) −2k 1−2k y(z + σ) = (z −ˆz) + O((z −ˆz) ) (16) k k+1 y(z + 2σ) =−α(z −ˆz) + O((z −ˆz) ) c(z + 2σ) − c(z) −2k 1−2k y(z + 3σ) = (z −ˆz) + O((z −ˆz) ), where α = 0 and β = 0. The number of zeros divided by the number of poles of y in the set {ˆz − σ, z ˆ, z ˆ + σ, z ˆ + 2σ } is less than or equal to 2/3. It may happen that there is a zero of y at z ˆ + 3σ,orat z ˆ + 4σ , but then this zero becomes a starting point of another sequence of one of the types (15)or (16), or (17) below. If there is another sequence of the type (16) progressing in the opposite direction from the point z ˆ − σ , then the corresponding zero-pole ratio of the two combined sequences in the set {ˆz − 4σ, . . . , z ˆ + 2σ } is less than or equal to 4/5. We still need to deal with the case where y(z) has a zero of order k and y(z − σ) has a pole of order l < k (or it assumes a ﬁnite value). The iteration of equation (14) yields −2 −1 y(z + σ) = c(z)y(z) + b(z)y(z) − y(z − σ) b(z + σ) b(z + σ)b(z) 2 3 4 y(z + 2σ) =−y(z) + y(z) − y(z) + O(y(z) ) c(z) c(z) (17) −2 y(z + 3σ) = (c(z + 2σ) − c(z))y(z) 2c(z + 2σ)b(z + σ) −1 + −b(z) + − b(z + 2σ) y(z) + y(z − σ) + O(1). c(z) 123 306 R. Halburd, R. Korhonen There are 2k zeros (by taking y(z) into account) and 4k poles in the sequence (17), unless c(z + 2σ) − c(z) = 0 and b(z + 2σ) − 2b(z + σ) + b(z) = 0. (18) (To be exact, there are only 3k poles in (17) if the ﬁrst of the equations in (18) holds and the second one does not. In this case the zero-pole ratio in this sequence is 2/3.) Now unless equations (18) hold for all z at least one of them will fail to hold for all sufﬁciently large |z|. If these equations hold for all z, then (18) become linear difference equations. Solving these equations, and taking into account that the coefﬁcients b(z) and c(z) are rational functions by assumption, it follows that b(z) = Az + B and c(z) = C for complex constants A, B and C. In this case, the proof is complete. Otherwise, we have been able to associate each zero of y(z) with an appropriate number of zeros and poles of “nearby” iterates y(z ± 1), y(z ± 2), y(z ± 3), y(z ± 4) for all sufﬁciently large |z| such that within each grouping the number of zeros divided by the number of poles is at most 4/5. Therefore, 1 4 n r , ≤ n(r + 2, y) + O(1). (19) y 5 Lemma 1 below, which is a special case of [10, Lem. 2.1], applied to (19) with a = 0, s = 2 and τ = 4/5, implies that the hyper-order of y is at least one. Lemma 1 ([10]) Let f (z) be a non-rational meromorphic solution of P(z, f ) = 0, (20) where P(z, f ) is difference polynomial in f (z) with rational coefﬁcients, and let a ∈ C satisfy P(z, a) ≡ 0. If there exists s > 0 and τ ∈ (0, 1) such that n r , ≤ τ n(r + s, f ) + O(1), (21) f − a then the hyper-order ς( f ) of f is at least 1. We conclude that the only possible case when non-rational solutions of hyper-order less than one can exist is when b(z) = Az + B and c(z) = C. 3 Polynomial Degree Growth of Rational Iterates In this section we will prove the following: Theorem 3 Let {y } be a sequence of non-constant rational functions of z solving j j ∈N Eq. (1) where a ,b and c ≡ 0 are rational functions of j. If the degree of {y } j j j j j ∈N grows at most polynomially in j, then a = 0,b = Aj + B and c = C + D(−1) , j j j where A, B, C and D are constants. 123 Three Approaches to Detecting Discrete Integrability 307 The key idea behind the proof is to use the fact that the degree of a rational function is the number of zeros or poles (counting multiplicities) in CP . We will use singularity conﬁnement-type calculations similar to (2) to relate the number of zeros and poles of nearby iterates. Proof of Theorem 3 The ﬁrst part of the proof is largely analogous to the ﬁrst part of the proof of Theorem 1, and it consists of determining the relative zero and pole densities of the solution sequence. We assume ﬁrst that a is not identically zero. Therefore, a , j j b and c have ﬁnitely many zeros (unless b ≡ 0) and poles, since they are rational j j j functions of j, and so there exists a j ≥ 0 such that c = 0 for all j satisfying 0 j | j|≥ j . We will show that if | j | is sufﬁciently large, then each zero of y may be uniquely associated with a ﬁnite number of poles and zeros of neighboring iterates y such that the number of zeros divided by the number of poles, both counting multiplicities, is bounded by 4/5 in each grouping obtained. Suppose that the rational function y has a zero of multiplicity k at some point z ˆ on the complex sphere. By using a Möbius transformation, if necessary, we may assume without loss of generality that z ˆ = 0. In the following, expressions such as “y has j +1 a pole” will refer to a pole at z = 0. Since y has a zero of order k,itfollows from Eq. (1) that y has a pole of order at least 2k for σ = 1or σ =−1. Since we have j +σ taken j to be sufﬁciently large, it follows that a = 0, and so iteration of Eq. (1) j +σ gives k k+1 y = αz + O(z ) −2k 1−2k y = β z + O(z ) j +σ σ (22) y = a + O(z ) j +2σ j +σ −2k 1−2k y =−β z + O(z ) j +3σ σ y = a − a + O(z ), j +4σ j +3σ j +σ where α and β are non-zero. Therefore, the zero of y of order k may be associated σ j with the pole of y of order 2k. Note that even if a − a = 0, then we can j +σ j +3σ j +σ associate the available pole of y with this zero of y (or the zero is a starting j +3σ j +4σ point of a new sequence of iterates of the type (22) in a similar way as in the case (13) of meromorphic solutions). Hence the number of zeros divided by the number of poles in sequence (22) is less than or equal to 1/2 (counting multiplicities) under the assumption that a ≡ 0. Suppose now that a ≡ 0 so that Eq. (1) reduces to b y + c j j j y + y = . (23) j +1 j −1 We will again consider the case where y has a zero of order k at z = 0 assuming ﬁrst that both y and y have a pole at least of order 2k. Then, by iterating (23), it j +σ j −σ follows that 123 308 R. Halburd, R. Korhonen k k+1 y =−αz + O(z ) j −2σ −2k 1−2k y = γ z + O(z ) j −σ σ k k+1 y = αz + O(z ) (24) −2k 1−2k y = β z + O(z ) j +σ σ k k+1 y =−αz + O(z ), j +2σ where α = 0, β = 0 and γ = 0. In this sequence the number of zeros divided σ σ by the number of poles is less than or equal to 3/4, when multiplicities are taken into account. Suppose now that y has a zero of order k at z = 0 and y has a pole of order l j j −σ such that k ≤ l < 2k. Then, by (23), −l 1−l y = β z + O(z ) j −σ σ k k+1 y = αz + O(z ) −2k 1−2k y = z + O(z ) j +σ 2 (25) k k+1 y =−αz + O(z ) j +2σ c − c j +2σ j −2k 1−2k y = z + O(z ), j +3σ where α = 0 and β = 0. The number of zeros divided by the number of poles in (25) is less than or equal to 2/3, provided that y is non-zero. If y vanishes, j +3σ j +3σ the sequence of iterates starting from y of (25) becomes a special case of the j +3σ sequence (26) below. If there are two sequences of the type (25) joined together in a similar way as in the meromorphic case (16), then the number of zeros divided by the number of poles in the combined sequence is less than or equal to 4/5. We still need to deal with the case where y has a zero of order k and y has a j j −σ pole of order l < k (or it assumes a ﬁnite value). The iteration of equation (23) yields −2 −1 y = c y + b y − y j +σ j j j −σ j j b b b j +σ j +σ j 2 3 4 y =−y + y − y + O(y ) j +2σ j j j j c (26) 2c b j +2σ j +σ −2 −1 y = (c − c )y + −b + − b y + y + O(1). j +3σ j +2σ j j j +2σ j −σ j j There are 2k zeros (by taking y into account) and 4k (or 3k) poles in the sequence (26), unless c − c = 0 and b − 2b + b = 0. (27) j +2σ j j +2σ j +σ j Now unless Eq. (27) hold for all j at least one of them will fail to hold for all sufﬁciently large j. If these equations hold for all j then b = Aj + B and c = C + D(−1) j j for constants A, B, C and D. Otherwise, note that deg y is the total number of zeros 123 Three Approaches to Detecting Discrete Integrability 309 of y on the complex sphere counting multiplicities and it is also the total number of poles of y . Since we have been able to associate each zero of y uniquely with j j an appropriate number of zeros and poles of iterates y close to y for all sufﬁciently i j large j such that within each grouping the number of zeros divided by the number of poles is less than or equal to 4/5, we have D (y ) ≤ D (y ) + O(1), n →∞, n j n+s j for some s > 0, where D (y ) := deg (y ). n j j j =−n It follows that 1 5 lim sup log D (y ) ≥ log > 0, n j n 4 n→∞ so the degree growth of {y } is exponential. j j ∈N 4 Diophantine Integrability In this section we will consider the solution (y ) of Eq. (1) to be a sequence of rational numbers and we will explore the growth of the height of such solutions subject to the assumptions of Theorem 2. We denote the usual absolute value on Q by |·| .Any non-trivial absolute value on Q is equivalent to |·| or to one of the p-adic absolute values |·| , for some prime p.Given aprime p, any non-zero x ∈ Q can be written a r as x = p for a, b, r ∈ Z, where p ab. Then the p-adic absolute value of x is −r deﬁned to be |x | = p .The p-adic absolute values are non-Archimedean, which means they satisfy the strong triangle inequality |x + y| ≤ max |x | , |y| , p p p for all x and y ∈ Q. An important identity for our calculations is the following expression for the loga- rithmic height in terms of absolute values h(x ) = log |x | , p≤∞ where the sum is over all primes p as well as the “prime at inﬁnity” p =∞. We begin by ﬁxing a particular absolute value |·| on Q and we use this to determine a certain length scale for the nth iterate. Since the coefﬁcient functions are rational functions of n, for sufﬁciently large n, they are either identically zero or they are ﬁnite 123 310 R. Halburd, R. Korhonen and non-zero. From now on we work with sufﬁciently large n in this sense. For some 0 <δ < 1/2 we deﬁne by −δ −1 = κ max 1, |c | , |b | , |a | , p n n n n p p p (28) −1 −1 |c | , |c | , |b | , |b | , |a | , |a | , n+1 n−1 n+1 n−1 n+1 n−1 p p p p p p where κ = 1 for all p < ∞ and κ = 3. The following lemma, the proof of which is p ∞ elementary and can be found in [7], relates small and large values of y with respect to the given absolute value. Lemma 2 Fix a prime p ≤∞. Let (y ) be a solution to equation (1) where a c ≡ 0. n n n Suppose that for some integer m, |y | < , where is deﬁned by (28). Then either m m m |y | ≥ and |y | ≥ , m+1 m+2 m+2 p p 2−δ |y | or | | | | y ≥ and y ≥ . m−1 m−2 m−2 p p 2−δ |y | Lemma 3 If (y ) is a solution of equation (1) satisfying the assumptions of Theorem 2, then a ≡ 0. Proof Assume that a ≡ 0. Let |·| denote the absolute value corresponding to the prime p. For sufﬁciently large r , deﬁne S (r ) ={n : r ≤ n ≤ r and |y | < } 1 0 n n S (r ) ={n : r ≤ n ≤ r and |y | ≥ }. 2 0 n n Now 1 1 1 + + + log = log + log . (29) |y | |y | |y | k k k k=r k∈S (r ) k∈S (r ) 0 1 2 Using Lemma 2 gives r +1 1 1 + + log ≤ log |y |. (30) y 2 − δ k∈S (r ) k=r −1 1 0 123 Three Approaches to Detecting Discrete Integrability 311 Also + + −1 + −1 log ≤ log ≤ log k k |y | k∈S (r ) k∈S (r ) k=r 2 2 0 + −1 = log κ max{1, |c | , |b | , |a | , |c | , |c | , p k k k k+1 k−1 k=r −1 −1 |b | , |b | , |a | , |a | } k+1 k−1 k+1 k−1 r +1 + + −1 + ≤ (r − r ) log κ + log |c | + 3log |b | 0 p k k k=r −1 + + + −1 + log |a | + 2log |c | + 2log |a | , k k k where if |b | = 0 it is excluded from the list. Recall that 1 1 + + h(x ) = log |x | = h = log . x x p≤∞ p≤∞ So taking the sum over all primes p ≤∞ in equation (29)gives r r +1 h(y ) ≤ h(y ) k k 2 − δ k=r k=r −1 0 0 ⎛ ⎞ r +1 ⎝ ⎠ + 3 (h(a ) + h(b ) + h(c )) + (r − r ) log 3 . (31) k k k 0 k=r −1 So from (7), we have ⎛ ⎞ r r +1 r ⎝ ⎠ h(y ) ≤ h(y ) + o h(y ) , k k k 2 − δ k=r k=r −1 k=r 0 0 0 which is impossible if y satisﬁes (8). We conclude this section by quoting another lemma from [7] which bears a strong similarity to the singularity conﬁnement-type calculation in (2). We do not give the precise deﬁnition of here but merely note that it depends on the absolute values of various combinations of the coefﬁcient functions. 123 312 R. Halburd, R. Korhonen k+3 Lemma 4 Let (y ) ⊆ Q/{0} with k − 1 ≥ r satisfy n 0 n=k−1 c + b y n n n y + y = . (32) n+1 n−1 −1/2 If |y | ≤ |y | and, for sufﬁciently small δ> 0, |y | < , then k−1 k k k c b −1/2 k k 1. y = + + A , where |A | ≤ |y | . k+1 2 k k k y k k+1 2 3−4δ 2. y =−y + y + B , where |B | ≤ |y | k+2 k k k k k+2 b −2 b +b k+2 k+1 k c −c c k+2 k k 3. y = + + C , where k+3 2 k y k+2 k+2 c − c k+2 k 1−δ −1/2 |C | ≤ max |y | , |y | k k+2 k+2 for non-Archimedean absolute values and c − c k+2 k 1−δ −1/2 |C | ≤ 2 |y | + 3 |y | k k+2 k+2 for Archimedean absolute values. 4. |y | = |y | for non-Archimedean absolute values and |y | > |y | > k+2 k k k+2 |y | for Archimedean absolute values. Acknowledgements Open access funding provided by University of Eastern Finland (UEF) including Kuopio University Hospital. 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Computational Methods and Function Theory – Springer Journals
Published: Apr 29, 2019
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