Rational Solutions of the Sasano System of Type A₅⁽²⁾

In this paper, we completely classify the rational solutions of the Sasano system of type A₅⁽²⁾, which is given by the coupled Painlevé III system. This system of differential equations has the affine Weyl group symmetry of type A₅⁽²⁾.

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Date:	2011
Main Author:	Matsuda, K.
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Language:	English
Published:	Інститут математики НАН України 2011
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spelling	irk-123456789-1468052019-05-16T15:48:37Z Rational Solutions of the Sasano System of Type A₅⁽²⁾ Matsuda, K. In this paper, we completely classify the rational solutions of the Sasano system of type A₅⁽²⁾, which is given by the coupled Painlevé III system. This system of differential equations has the affine Weyl group symmetry of type A₅⁽²⁾. 2011 Article Rational Solutions of the Sasano System of Type A₅⁽²⁾ / K. Matsuda // Symmetry, Integrability and Geometry: Methods and Applications. — 2011. — Т. 7. — Бібліогр.: 26 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 33E17; 34M55 DOI:10.3842/SIGMA.2011.030 http://dspace.nbuv.gov.ua/handle/123456789/146805 en Symmetry, Integrability and Geometry: Methods and Applications Інститут математики НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
description	In this paper, we completely classify the rational solutions of the Sasano system of type A₅⁽²⁾, which is given by the coupled Painlevé III system. This system of differential equations has the affine Weyl group symmetry of type A₅⁽²⁾.
format	Article
author	Matsuda, K.
spellingShingle	Matsuda, K. Rational Solutions of the Sasano System of Type A₅⁽²⁾ Symmetry, Integrability and Geometry: Methods and Applications
author_facet	Matsuda, K.
author_sort	Matsuda, K.
title	Rational Solutions of the Sasano System of Type A₅⁽²⁾
title_short	Rational Solutions of the Sasano System of Type A₅⁽²⁾
title_full	Rational Solutions of the Sasano System of Type A₅⁽²⁾
title_fullStr	Rational Solutions of the Sasano System of Type A₅⁽²⁾
title_full_unstemmed	Rational Solutions of the Sasano System of Type A₅⁽²⁾
title_sort	rational solutions of the sasano system of type a₅⁽²⁾
publisher	Інститут математики НАН України
publishDate	2011
url	http://dspace.nbuv.gov.ua/handle/123456789/146805
citation_txt	Rational Solutions of the Sasano System of Type A₅⁽²⁾ / K. Matsuda // Symmetry, Integrability and Geometry: Methods and Applications. — 2011. — Т. 7. — Бібліогр.: 26 назв. — англ.
series	Symmetry, Integrability and Geometry: Methods and Applications
work_keys_str_mv	AT matsudak rationalsolutionsofthesasanosystemoftypea52
first_indexed	2025-07-11T00:38:19Z
last_indexed	2025-07-11T00:38:19Z
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fulltext	Symmetry, Integrability and Geometry: Methods and Applications SIGMA 7 (2011), 030, 20 pages Rational Solutions of the Sasano System of Type A (2) 5 Kazuhide MATSUDA Department of Engineering Science, Niihama National College of Technology, 7-1 Yagumo-chou, Niihama, Ehime, 792-8580, Japan E-mail: matsuda@sci.niihama-nct.ac.jp Received November 5, 2010, in final form March 17, 2011; Published online March 25, 2011 doi:10.3842/SIGMA.2011.030 Abstract. In this paper, we completely classify the rational solutions of the Sasano system of type A (2) 5 , which is given by the coupled Painlevé III system. This system of differential equations has the affine Weyl group symmetry of type A (2) 5 . Key words: affine Weyl group; rational solutions; Sasano system 2010 Mathematics Subject Classification: 33E17; 34M55 1 Introduction Paul Painlevé and his colleagues [21, 4] intended to find new transcendental functions defined by second order nonlinear differential equations. In general, nonlinear differential equations have moving branch points. If a solution has moving branch points, it is too complicated and is not worth considering. Therefore, they determined the second order nonlinear differential equations with rational coefficients which have no moving branch points. As a result, the standard forms of such equations turned out to be given by the following six equations: PI : y′′ = 6y2 + t, PII : y′′ = 2y3 + ty + α, PIII : y′′ = 1 y (y′)2 − 1 t y′ + 1 t (αy2 + β) + γy3 + δ y , PIV : y′′ = 1 2y (y′)2 + 3 2 y3 + 4ty2 + 2(t2 − α)y + β y , PV : y′′ = ( 1 2y + 1 y − 1 ) (y′)2 − 1 t y′ + (y − 1)2 t2 ( αy + β y ) + γ t y + δ y(y + 1) y − 1 , PVI : y′′ = 1 2 ( 1 y + 1 y − 1 + 1 y − t ) (y′)2 − ( 1 t + 1 t− 1 + 1 y − t ) y′ + y(y − 1)(y − t) t2(t− 1)2 ( α+ β t y2 + γ t− 1 (y − 1)2 + δ t(t− 1) (y − t)2 ) , where ′ = d/dt and α, β, γ, δ are all complex parameters. In this article, our concern is with the Bäcklund transformations and special solutions which are given by rational, algebraic functions or classical special functions. Each of PJ (J = II, III, IV,V,VI) has Bäcklund transformations, which transform solutions into other solutions of the same equation with different parameters. It was shown by Okamoto [17, 18, 19, 20] that the Bäcklund transformation groups of the Painlevé equations except for PI are isomorphic to the extended affine Weyl groups. For PII, PIII, PIV, PV, and PVI, the Bäcklund transformation groups correspond to A (1) 1 , A (1) 1 ⊕ A (1) 1 , A (1) 2 , A (3) 3 , and D (1) 4 , respectively. mailto:matsuda@sci.niihama-nct.ac.jp http://dx.doi.org/10.3842/SIGMA.2011.030 2 K. Matsuda While generic solutions of the Painlevé equations are “new transcendental functions”, there are special solutions which are expressible in terms of rational, algebraic, or classical special functions. For example, Airault [1] constructed explicit rational solutions of PII and PIV with their Bäcklund transformations. Milne, Clarkson and Bassom [13] treated PIII, and described their Bäcklund transformations and exact solution hierarchies, which are given by rational, algebraic, or certain Bessel functions. Bassom, Clarkson and Hicks [2] dealt with PIV, and described their Bäcklund transformations and exact solution hierarchies, which are expressed by rational functions, the parabolic cylinder functions or the complementary error functions. Clarkson [3] studied some rational and algebraic solutions of PIII and showed that these solutions are ex- pressible in terms of special polynomials defined by second order, bilinear differential-difference equations which are equivalent to Toda equations. Furthermore, the rational solutions of PJ (J = II, III, IV,V,VI) were classified by Yablonski and Vorobev [25, 24], Gromak [6, 5], Murata [14, 15], Kitaev, Law and McLeod [8], Mazzoco [11] and Yuang and Li [26]. Noumi and Yamada [16] discovered the equation of type A (1) l (l ≥ 2), whose Bäcklund transformation group is isomorphic to the extended affine Weyl group W̃ (A (1) l ). The Noumi and Yamada systems of typesA (1) 2 andA (1) 3 correspond to the fourth and fifth Painlevé equations, respectively. Moreover, we [9, 10] classified the rational solutions of the Noumi and Yamada systems of types A (1) 4 and A (1) 5 . Sasano [22] found the coupled Painlevé V and VI systems which have the affine Weyl group symmetries of types D (1) 5 and D (1) 6 . In addition, he [23] obtained the equation of the affine Weyl group symmetry of type A (2) 5 , which is defined by A (2) 5 (αj)0≤j≤3  tq′1 = 2q21p1 − q21 + (α0 + α1 + α3)q1 − t+ 4tp2 + 2q1q2p2, tp′1 = −2q1p 2 1 + 2q1p1 − (α0 + α1 + α3)p1 + α0 − 2p1q2p2, tq′2 = 2q22p2 − q22 + (α0 + α1 + α3)q2 − t+ 4tp1 + 2q1p1q2, tp′2 = −2q2p 2 2 + 2q2p2 − (α0 + α1 + α3)p2 + α1 − 2q1p1p2, α0 + α1 + 2α2 + α3 = 1/2, where ′ = d/dt. This system of differential equations is also expressed by the Hamiltonian system: t dq1 dt = ∂H ∂p1 , t dp1 dt = −∂H ∂q1 , t dq2 dt = ∂H ∂p2 , t dp2 dt = −∂H ∂q2 , where the Hamiltonian H is given by H = q21p 2 1 − q21p1 + (α0 + α1 + α3)q1p1 − α0q1 − tp1 + q22p 2 2 − q22p2 + (α0 + α1 + α3)q2p2 − α1q2 − tp2 + 4tp1p2 + 2q1p1q2p2. Let us note that Mazzocco and Mo [12] studied the Hamiltonian structure of the PII hierar- chy, and Hone [7] studied the coupled Painlevé systems from the similarity reduction of the Hirota–Satsuma system and another gauge-related system, and presented their Bäcklund trans- formations and special solutions. A (2) 5 (αj)0≤j≤3 has the Bäcklund transformations s0, s1, s2, s3, π, which are given by s0 : (∗)→ ( q1 + α0 p1 , p1, q2, p2, t;−α0, α1, α2 + α0, α3 ) , Rational Solutions of the Sasano System of Type A (2) 5 3 s1 : (∗)→ ( q1, p1, q2 + α1 p2 , p2, t;α0,−α1, α2 + α1, α3 ) , s2 : (∗)→ ( q1, p1 − α2q2 q1q2 + t , q2, p2 + α2q1 q1q2 + t , t;α0 + α2, α1 + α2,−α2, α3 + 2α2 ) , s3 : (∗)→ ( q1 + α3 p1 + p2 − 1 , p1, q2 + α3 p1 + p2 − 1 , p2, t;α0, α1, α2 + α3,−α3 ) , π : (∗)→ (q2, p2, q1, p1, t;α1, α0, α2, α3) , with the notation (∗) = (q1, p1, q2, p2, t;α0, α1, α2, α3). The Bäcklund transformation group 〈s0, s1, s2, s3, π〉 is isomorphic to the affine Weyl group of type A (2) 5 . Our main theorem is as follows: Theorem 1.1. For a rational solution of A (2) 5 (αj)0≤j≤3, by some Bäcklund transformations, the solution and parameters can be transformed so that (q1, p1, q2, p2) = (0, 1/4, 0, 1/4) and (α0, α1, α2, α3) = (α3/2, α3/2, α2, α3) = (α3/2, α3/2, 1/4− α3, α3), respectively. Furthermore, for A (2) 5 (αj)0≤j≤3, there exists a rational solution if and only if one of the following occurs: (1) −2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z, (2) −2α0 + α3 ∈ Z, 2α1 + α3 ∈ Z, (3) 2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z, (4) 2α0 + α3 ∈ Z, 2α1 + α3 ∈ Z, (5) −2α0 + α3 ∈ Z, α3 − 1/2 ∈ Z, (6) −2α1 + α3 ∈ Z, α3 − 1/2 ∈ Z. This paper is organized as follows. In Section 2, for A (2) 5 (αj)0≤j≤3, we determine meromorphic solutions at t = ∞. Then, we find that the constant terms a∞,0, c∞,0 of the Laurent series of q1, q2 at t =∞ are given by a∞,0 := −2α0 + α3, c∞,0 := −2α1 + α3, respectively. In Section 3, for A (2) 5 (αj)0≤j≤3, we determine meromorphic solutions at t = 0. Then, we see that the constant terms a0,0, c0,0 of the Laurent series of q1, q2 at t = 0 are given by the parameters α0, α1, α2, α3. In Section 4, for A (2) 5 (αj)0≤j≤3, we treat meromorphic solutions at t = c ∈ C∗, where in this paper, C∗ means the set of nonzero complex numbers. Then, we observe that q1, q2 have both a pole of order of at most one at t = c and the residues of q1, q2 at t = c are expressed by nc (n ∈ Z). Thus, it follows that a∞,0 − a0,0 ∈ Z, c∞,0 − c0,0 ∈ Z, (1.1) which gives a necessary condition for A (2) 5 (αj)0≤j≤3 to have a rational solution. In Section 5, using the meromorphic solution at t = ∞, 0, we first compute the constant terms of the Laurent series of the Hamiltonian at t = ∞, 0. Furthermore, by the meromorphic solution at = c ∈ C∗, we calculate the residue of H at t = c. 4 K. Matsuda In Section 6, by equation (1.1), we obtain the necessary conditions for A (2) 5 (αj)0≤j≤3 to have rational solutions, which are given in our main theorem. Furthermore, we show that if there exists a rational solution for A (2) 5 (αj)0≤j≤3, the parameters can be transformed so that −2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z. In Section 7, we define shift operators, and for a rational solution of A (2) 5 (αj)0≤j≤3, we transform the parameters to (α0, α1, α2, α3) = (α3/2, α3/2, α2, α3). In Section 8, we determine rational solutions of A (2) 5 (α3/2, α3/2, α2, α3) and prove our main theorem. In Appendix A, using the shift operators, we give examples of rational solutions. 2 Meromorphic solutions at t =∞ In this section, for A (2) 5 (αj)0≤j≤3, we treat meromorphic solutions at t = ∞. For the purpose, in this paper, we define the coefficients of the Laurent series of q1, p1, q2, p2 at t =∞ by a∞,k, b∞,k, c∞,k, d∞,k, k ∈ Z. 2.1 The case where q1, p1, q2, p2 are all holomorphic at t =∞ Proposition 2.1. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p1, q2, p2 are all holomorphic at t =∞. Then, q1 = (−2α0 + α3) + · · · , p1 = 1/4 + (−2α1 + α3)(−2α1 − α3)t −1/4 + · · · , q2 = (−2α1 + α3) + · · · , p2 = 1/4 + (−2α0 + α3)(−2α0 − α3)t −1/4 + · · · . Proposition 2.2. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p1, q2, p2 are all holomorphic at t =∞. Then, it is unique. Proof. We set q1 = a∞,0 + a∞,−1t −1 + · · ·+ a∞,−(k−1)t −(k−1) + a∞,−kt −k + a∞,−(k+1)t −(k+1) + · · · , p1 = 1/4 + b∞,−1t −1 + · · ·+ b∞,−(k−1)t −(k−1) + b∞,−kt −k + b∞,−(k+1)t −(k+1) + · · · , q2 = c∞,0 + c∞,−1t −1 + · · ·+ c∞,−(k−1)t −(k−1) + c∞,−kt −k + c∞,−(k+1)t −(k+1) + · · · , p2 = 1/4 + d∞,−1t −1 + · · ·+ d∞,−(k−1)t −(k−1) + d∞,−kt −k + d∞,−(k+1)t −(k+1) + · · · , where a∞,0, b∞,−1, c∞,0, d∞,−1 all have been determined. Comparing the coefficients of the terms t−k (k ≥ 1) in tp′1 = −2q1p 2 1 + 2q1p1 − (α0 + α1 + α3)p1 + α0 − 2p1q2p2, tp′2 = −2q2p 2 2 + 2q2p2 − (α0 + α1 + α3)p2 + α1 − 2q1p1p2, we have 3a∞,−k/8− c∞,−k/8 = −kb∞,−k + (α0 + α1 + α3)b∞,−k + 2 ∑ a∞,−lb∞,−mb∞,−n − 2 ∑ a∞,−lb∞,−m + 2 ∑ c∞,−lb∞,−md∞,−n, Rational Solutions of the Sasano System of Type A (2) 5 5 −a∞,−k/8 + 3c∞,−k/8 = −kd∞,−k + (α0 + α1 + α3)d∞,−k + 2 ∑ c∞,−ld∞,−md∞,−n − 2 ∑ c∞,−ld∞,−m + 2 ∑ a∞,−lb∞,−md∞,−n, where the first and third sums extend over nonnegative integers l, m, n such that l+m+n = k and 0 ≤ l < k, and the second sums extend over nonnegative integers l, m such that l +m = k and m ≥ 1. Therefore, a∞,−k, c∞,−k are both inductively determined. Comparing the coefficients of the terms t−k (k ≥ 1) in tq′1 = 2q21p1 − q21 + (α0 + α1 + α3)q1 − t+ 4tp2 + 2q1q2p2, tq′2 = 2q22p2 − q22 + (α0 + α1 + α3)q2 − t+ 4tp1 + 2q1p1q2, we obtain 4d∞,−(k+1) = −ka∞,−k − (α0 + α1 + α3)a∞,−k − 2 ∑ a∞,−la∞,−mb∞,−n + ∑ a∞,−la∞,−m − 2 ∑ a∞,−lc∞,−md∞,−n, 4b∞,−(k+1) = −kc∞,−k − (α0 + α1 + α3)c∞,−k − 2 ∑ c∞,−lc∞,−md∞,−n + ∑ c∞,−lc∞,−m − 2 ∑ c∞,−la∞,−mb∞,−n, where the first and third sums extend over nonnegative integers l, m, n such that l+m+n = k, and the second sums extend over nonnegative integers l, m such that l + m = k. Therefore, b∞,−(k+1), d∞,−(k+1) are both inductively determined, which proves the proposition. � 2.2 The case where one of (q1, p1, q2, p2) has a pole at t =∞ In this subsection, we deal with the case in which one of (q1, p1, q2, p2) has a pole at t =∞. For the purpose, by π, we have only to consider the following two cases: (1) q1 has a pole at t =∞ and p1, q2, p2 are all holomorphic at t =∞, (2) p1 has a pole at t =∞ and q1, q2, p2 are all holomorphic at t =∞. 2.2.1 The case where q1 has a pole at t =∞ Proposition 2.3. For A (2) 5 (αj)0≤j≤3, there exists no solution such that q1 has a pole at t =∞ and p1, q2, p2 are all holomorphic at t =∞. 2.2.2 The case where p1 has a pole at t =∞ Proposition 2.4. For A (2) 5 (αj)0≤j≤3, there exists no solution such that p1 has a pole at t =∞ and q1, q2, p2 are all holomorphic at t =∞. 2.3 The case where two of (q1, p1, q2, p2) have a pole at t =∞ In this subsection, we deal with the case in which two of (q1, p1, q2, p2) has a pole at t =∞. For the purpose, by π, we have only to consider the following four cases: (1) q1, p1 have both a pole at t =∞ and q2, p2 are both holomorphic at t =∞, (2) q1, q2 have both a pole at t =∞ and p1, p2 are both holomorphic at t =∞, (3) q1, p2 have both a pole at t =∞ and p1, q2 are both holomorphic at t =∞, (4) p1, p2 have both a pole at t =∞ and q1, q2 are both holomorphic at t =∞. 6 K. Matsuda 2.3.1 The case where q1, p1 have a pole at t =∞ Proposition 2.5. For A (2) 5 (αj)0≤j≤3, there exists no solution such that q1, p1 have both a pole at t =∞ and q2, p2 are both holomorphic at t =∞. 2.3.2 The case where q1, q2 have a pole at t =∞ Proposition 2.6. For A (2) 5 (αj)0≤j≤3, there exists no solution such that q1, q2 have both a pole at t =∞ and p1, p2 are both holomorphic at t =∞. 2.3.3 The case where q1, p2 have a pole at t =∞ By direct calculation, we can obtain the following two lemmas: Lemma 2.1. Suppose that for A (2) 5 (αj)0≤j≤3, q1 ≡ 0. Then, one of the following occurs: (1) α0 = 1 4 , α3 = 1 2 , and (q1, p1, q2, p2) = ( 0, 1 4 + (4α1 − 1)(4α1 + 1) 16t ,−2α1 + 1 2 , 1 4 ) , (2) α0 = α3 2 , α1 = α3 2 , and (q1, p1, q2, p2) = ( 0, 1 4 , 0, 1 4 ) , (3) α0 = α3 2 , α1 = −α3 2 , and (q1, p1, q2, p2) = ( 0, 1 4 , 2α3, 1 4 ) . Lemma 2.2. Suppose that for A (2) 5 (αj)0≤j≤3, q2 ≡ 0. Then, one of the following occurs: (1) α1 = 1 4 , α3 = 1 2 , and (q1, p1, q2, p2) = ( −2α0 + 1 2 , 1 4 , 0, 1 4 + (4α0 − 1)(4α0 + 1) 16t ) , (2) α0 = α3 2 , α1 = α3 2 , and (q1, p1, q2, p2) = ( 0, 1 4 , 0, 1 4 ) , (3) α0 = −α3 2 , α1 = α3 2 , and (q1, p1, q2, p2) = ( 2α3, 1 4 , 0, 1 4 ) . By Lemma 2.2, we find that q2 6≡ 0. Now, let us assume that q1 has a pole of order n0 (n0 ≥ 1) and p2 has a pole of order n3 (n3 ≥ 1). Lemma 2.3. For A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p2 have both a pole at t =∞ and p1, q2 are both holomorphic at t =∞. Then, n0 6= n3. Proof. We suppose that n0 = n3. Especially, we treat the case where n0 = n3 = 1 and show contradiction. If n0 = n3 > 1, we can prove contradiction in the same way. Comparing the coefficients of the terms t2, t in tq′1 = 2q21p1 − q21 + (α0 + α1 + α3)q1 − t+ 4tp2 + 2q1q2p2, tp′1 = −2q1p 2 1 + 2q1p1 − (α0 + α1 + α3)p1 + α0 − 2p1q2p2, we have 2a2∞,1b∞,0 − a2∞,1 + 4d∞,1 + 2a∞,1c∞,0d∞,1 = 0, Rational Solutions of the Sasano System of Type A (2) 5 7 −2a∞,1b 2 ∞,0 + 2a∞,1b∞,0 − 2b∞,0c∞,0d∞,1 = 0, (2.1) respectively. Comparing the coefficients of the terms t, t2 in tq′2 = 2q22p2 − q22 + (α0 + α1 + α3)q2 − t+ 4tp1 + 2q1p1q2, tp′2 = −2q2p 2 2 + 2q2p2 − (α0 + α1 + α3)p2 + α1 − 2q1p1p2, we obtain 2c2∞,0d∞,1 − 1 + 4b∞,0 + 2a∞,1b∞,0c∞,0 = 0, −2c∞,0d 2 ∞,1 − 2a∞,1b∞,0d∞,1 = 0, (2.2) which implies that b∞,0 = 1/4. Furthermore, from the second equation in (2.1) and the first equation in (2.2), it follows that a∞,1 = 0, which is impossible. � Lemma 2.4. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p2 have both a pole at t =∞ and p1, q2 are both holomorphic at t =∞. Then, n0 < n3. Proposition 2.7. For A (2) 5 (αj)0≤j≤3, there exists no solution such that q1, p2 have both a pole at t =∞ and p1, q2 are both holomorphic at t =∞. Proof. We treat the case where (n0, n3) = (1, 2) and show contradiction. The other cases can be proved in the same way. Comparing the coefficients of the terms t3 in tq′1 = 2q21p1 − q21 + (α0 + α1 + α3)q1 − t+ 4tp2 + 2q1q2p2, we have d∞,2 = 0, which is impossible. � 2.3.4 The case where p1, p2 have a pole at t =∞ Proposition 2.8. For A (2) 5 (αj)0≤j≤3, there exists no solution such that p1, p2 have both a pole at t =∞ and q1, q2 are both holomorphic at t =∞. 2.4 The case where three of (q1, p1, q2, p2) have a pole at t =∞ In this subsection, considering π, we treat the following two cases: (1) q1, p1, q2 all have a pole at t =∞ and p2 is holomorphic at t =∞, (2) q1, p1, p2 all have a pole at t =∞ and q2 is holomorphic at t =∞. 2.4.1 The case where q1, p1, q2 have a pole at t =∞ Proposition 2.9. For A (2) 5 (αj)0≤j≤3, there exists no solution such that q1, p1, q2 all have a pole at t =∞ and p2 is holomorphic at t =∞. 8 K. Matsuda 2.4.2 The case where q1, p1, p2 have a pole at t =∞ By Lemma 2.2, let us note that q2 6≡ 0. Lemma 2.5. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p1, p2 all have a pole at t =∞ and q2 is holomorphic at t =∞. Moreover, assume that q1, p1, p2 has a pole of order n0, n1, n3 (n0, n1, n3 ≥ 1) at t =∞, respectively. Then, n3 ≥ n0 + n1. Proof. Considering that tp′1 = −2q1p 2 1 + 2q1p1 − (α0 + α1 + α3)p1 + α0 − 2p1q2p2, we can prove the lemma. � Therefore, we define the nonnegative integer k by n3 = n0 + n1 + k. Lemma 2.6. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p1, p2 all have a pole at t = ∞ and q2 is holomorphic at t = ∞. Then, c∞,0 = c∞,−1 = · · · = c∞,−(k−1) = 0, a∞,n0b∞,1 + c∞,−kd∞,n3 = 0, and n0 − k ≥ 1. Proof. Considering that tp′1 = −2q1p 2 1 + 2q1p1 − (α0 + α1 + α3)p1 + α0 − 2p1q2p2, we find that c∞,0 = c∞,−1 = · · · = c∞,−(k−1) = 0, a∞,n0b∞,1 + c∞,−kd∞,n3 = 0. Furthermore, considering that tq′1 = 2q21p1 − q21 + (α0 + α1 + α3)q1 − t+ 4tp2 + 2q1q2p2, we can show the lemma. � Proposition 2.10. For A (2) 5 (αj)0≤j≤3, there exists no solution such that q1, p1, p2 all have a pole at t =∞ and q2 is holomorphic at t =∞. Proof. We treat the case where n1 = 1. The other cases can be proved in the same way. Comparing the coefficients of the terms tn0+1 in tp′1 = −2q1p 2 1 + 2q1p1 − (α0 + α1 + α3)p1 + α0 − 2p1q2p2, we have −2a∞,n0b∞,0 − 2a∞,n0−1b∞,1 + 2a∞,n0 − 2c∞,−kd∞,n3−1 − 2c∞,−k−1d∞,n3 = 0. If n0 − k ≥ 3, comparing the coefficients of the terms t2n0 in tq′1 = 2q21p1 − q21 + (α0 + α1 + α3)q1 − t+ 4tp2 + 2q1q2p2, we obtain 2b∞,1a∞,n0−1 + 2a∞,n0b∞,0 − a∞,n0 + 2c∞,−kd∞,n3−1 + 2c∞,−k−1d∞,n3 = 0. Then, it follows that a∞,n0 = 0, which is impossible. If n0 − k = 2, comparing the coefficients of the terms t2, t3n0−1 in tq′2 = 2q22p2 − q22 + (α0 + α1 + α3)q2 − t+ 4tp1 + 2q1p1q2, tp′2 = −2q2p 2 2 + 2q2p2 − (α0 + α1 + α3)p2 + α1 − 2q1p1p2, Rational Solutions of the Sasano System of Type A (2) 5 9 we have 2d∞,n3c∞,−kc∞,−k−1 + 2d∞,n3−1c 2 ∞,−k + 4b∞,1 + 2c∞,−ka∞,n0b∞,0 + 2c∞,−ka∞,n0−1b∞,1 = 0, −2c∞,−kd∞,n3−1 − 2c∞,−k−1d∞,n3 − 2a∞,n0b∞,0 − 2a∞,n0−1b∞,1 = 0, respectively. Then, it follows that b∞,1 = 0, which is impossible. If n0 − k = 1, comparing the coefficients of the terms t2 in tq′2 = 2q22p2 − q22 + (α0 + α1 + α3)q2 − t+ 4tp1 + 2q1p1q2, we obtain 2c2∞,−kd∞,n3 + 4b∞,1 + 2a∞,n0b∞,1c∞,−k = 4b∞,1 = 0, which is impossible. � 2.5 The case where all of (q1, p1, q2, p2) have a pole at t =∞ Lemma 2.7. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p1, q2, p2 all have a pole at t =∞. Moreover, assume that q1, p1, q2, p2 have a pole of order n0, n1, n2, n3 (n0, n1, n2, n3 ≥ 1) at t =∞, respectively. Then, n0 + n1 = n2 + n3. Proof. Considering tp′1 = −2q1p 2 1 + 2q1p1 − (α0 + α1 + α3)p1 + α0 − 2p1q2p2, we can show the lemma. � Therefore, we see that n0 + n1 = n2 + n3 ≥ 2. Proposition 2.11. For A (2) 5 (αj)0≤j≤3, there exists no solution such that q1, p1, q2, p2 all have a pole at t =∞. Proof. We treat the case where n0 + n1 = n2 + n3 = 2. The other cases can be proved in the same way. Comparing the coefficients of the term t3 in tp′1 = −2q1p 2 1 + 2q1p1 − (α0 + α1 + α3)p1 + α0 − 2p1q2p2, we have a∞,1b∞,1 + c∞,1d∞,1 = 0. Comparing the coefficients of the term t2 in tq′1 = 2q21p1 − q21 + (α0 + α1 + α3)q1 − t+ 4tp2 + 2q1q2p2, tp′1 = −2q1p 2 1 + 2q1p1 − (α0 + α1 + α3)p1 + α0 − 2p1q2p2, tq′2 = 2q22p2 − q22 + (α0 + α1 + α3)q2 − t+ 4tp1 + 2q1p1q2, tp′2 = −2q2p 2 2 + 2q2p2 − (α0 + α1 + α3)p2 + α1 − 2q1p1p2, we obtain 2a∞,1a∞,0b∞,1 + 2b∞,0a 2 ∞,1 − a2∞,1 + 4d∞,1 + 2a∞,1c∞,1d∞,0 + 2a∞,1c∞,0d∞,1 = 0, −2a∞,1b∞,0 − 2a∞,0b∞,1 + 2a∞,1 − 2c∞,1d∞,0 − 2c∞,0d∞,1 = 0, 2c∞,1c∞,0d∞,1 + 2d∞,0c 2 ∞,1 − c2∞,1 + 4b∞,1 + 2c∞,1a∞,1b∞,0 + 2c∞,1a∞,0b∞,1 = 0, −2c∞,1d∞,0 − 2c∞,0d∞,1 + 2c∞,1 − 2a∞,1b∞,0 − 2a∞,0b∞,1 = 0, (2.3) 10 K. Matsuda respectively. Based on the second and fourth equations of (2.3), we have a∞,1 = c∞,1. From the first and second equations of (2.3), we obtain a2∞,1 + 4d∞,1 = 0. From the third and fourth equations of (2.3), we have c2∞,1 + 4b∞,1 = 0. Therefore, since a∞,1b∞,1 + c∞,1d∞,1 = 0, it follows that a∞,1b∞,1 = 0, which is impos- sible. � 2.6 Summary Proposition 2.12. For A (2) 5 (αj)0≤j≤3, there exists a meromorphic solution at t = ∞. Then, q1, p1, q2, p2 are uniquely expanded as follows: q1 = (−2α0 + α3) + · · · , p1 = 1/4 + (−2α1 + α3)(−2α1 − α3)t −1/4 + · · · , q2 = (−2α1 + α3) + · · · , p2 = 1/4 + (−2α0 + α3)(−2α0 − α3)t −1/4 + · · · . 3 Meromorphic solution at t = 0 In this section, we treat meromorphic solutions at t = 0. Then, in the same way as Proposi- tion 2.12, we can show the following proposition: Proposition 3.1. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a meromorphic solution at t = 0. Then, one of the following occurs: (1) q1, p1, q2, p2 are all holomorphic at t = 0, (2) p1 has a pole of order one at t = 0 and q1, q2, p2 are all holomorphic at t = 0, (3) p2 has a pole of order one at t = 0 and q1, p1, q2 are all holomorphic at t = 0. In this paper, we define the coefficients of the Lauren series of q1, p1, q2, p2 at t = 0 by a0,k, b0,k, c0,k, d0,k, k ∈ Z. In this section, we prove that the constant terms of q1, q2 at t = 0, a0,0, c0,0 are zero, or expressed by the parameters, αj (0 ≤ j ≤ 3). 3.1 The case where q1, p1, q2, p2 are all holomorphic at t = 0 Proposition 3.2. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p1, q2, p2 are all holomorphic at t = 0. Then, one of the following occurs: (1) a0,0 = 0, −(α0 + α1 + α3)b0,0 + α0 = 0, c0,0 = 0, −(α0 + α1 + α3)d0,0 + α1 = 0, (2) a0,0 = 0, (−α0 +α1−α3)b0,0 +α0 = 0, c0,0 = α0−α1 +α3, (−α0 +α1−α3)d0,0−α1 = 0, (3) a0,0 = −α0 + α1 + α3, (α0 − α1 − α3)b0,0 − α0 = 0, c0,0 = 0, (α0 − α1 − α3)d0,0 + α1 = 0, (4) a0,0 = −α0 − α1 + α3, (α0 + α1 − α3)b0,0 − α0 = 0, c0,0 = −α0 − α1 + α3, (α0 + α1 − α3)d0,0 − α1 = 0. 3.2 The case where p1 has a pole at t = 0 Proposition 3.3. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that p1 has a pole at t = 0 and q1, q2, p2 are all holomorphic at t = 0. Then, q1 = (−8α0 − 8α3 + 6)t/{(4α1 − 1)(4α1 + 1)}+ · · · , p1 = (4α1 − 1)(4α1 + 1)t−1/16 + · · · , q2 = (−2α1 + 1/2) + · · · , p2 = 1/4 + · · · . Rational Solutions of the Sasano System of Type A (2) 5 11 3.3 The case where p2 has a pole at t = 0 Proposition 3.4. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that p2 has a pole at t = 0 and q1, p1, q2 are all holomorphic at t = 0. Then, q1 = (−2α0 + 1/2) + · · · , p1 = 1/4 + · · · , q2 = (−8α1 − 8α3 + 6)t/{(4α0 − 1)(4α0 + 1)}+ · · · , p2 = (4α0 − 1)(4α0 + 1)t−1/16 + · · · . 4 Meromorphic solution at t = c ∈ C∗ In this section, we deal with meromorphic solutions at t = c ∈ C∗, where C∗ means the set of nonzero complex numbers. Proposition 4.1. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a meromorphic solution at t = c ∈ C∗ such that some of (q1, p1, q2, p2) have a pole at t = c. Then, one of the following occurs: (1) q1 has a pole at t = c and p1, q2, p2 are all holomorphic at t = c, (2) q2 has a pole at t = c and q1, p1, p2 are all holomorphic at t = c, (3) q1, q2 have both a pole at t = c and p1, p2 are both holomorphic at t = c, (4) q1, p2 have both a pole at t = c and p1, q2 are both holomorphic at t = c, (5) p1, q2 have both a pole at t = c and q1, p2 are both holomorphic at t = c, (6) p1, p2 have both a pole at t = c and q1, q2 are both holomorphic at t = c, (7) q1, p1, q2, p2 all have a pole at t = c. 4.1 The case where q1 has a pole at t = c ∈ C∗ Proposition 4.2. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1 has a pole at t = c ∈ C∗ and p1, q2, p2 are all holomorphic at t = c. Then, either of the following occurs: (1) { q1 = c(t− c)−1 + · · · , p1 = −α0 c (t− c) + · · · , (2)  q1 = −c(t− c)−1 + · · · , p1 = 1 + α1 + α3 c (t− c) + · · · , q2 = O(t− c), p2 = O(t− c). 4.2 The case where q2 has a pole at t = c ∈ C∗ Proposition 4.3. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q2 has a pole at t = c ∈ C∗ and q1, p1, p2 are all holomorphic at t = c. Then, either of the following occurs: (1) { q2 = c(t− c)−1 + · · · , p2 = −α1 c (t− c) + · · · , (2)  q1 = O(t− c), p1 = O(t− c), q2 = −c(t− c)−1 + · · · , p2 = 1 + α0 + α3 c (t− c) + · · · . 12 K. Matsuda 4.3 The case where q1, q2 have a pole at t = c ∈ C∗ Proposition 4.4. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, q2 have both a pole at t = c ∈ C∗ and p1, p2 are both holomorphic at t = c. Then, either of the following occurs: (1)  q1 = −c(t− c)−1 + · · · , p1 = bc,0 + bc,1(t− c) + · · · , q2 = −c(t− c)−1 + · · · , p2 = dc,0 + dc,1(t− c) + · · · , (2)  q1 = c(t− c)−1 + · · · , p1 = −α0 c (t− c) + · · · , q2 = c(t− c)−1 + · · · , p2 = −α1 c (t− c) + · · · , where bc,0 + dc,0 = 1 and bc,1 + dc,1 = α3 c . 4.4 The case where q1, p2 have a pole at t = c ∈ C∗ Proposition 4.5. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p2 have both a pole at t = c ∈ C∗ and p1, q2 are both holomorphic at t = c. Then, one of the following occurs: (1)  q1 = 4c(t− c)−1 + 8/3 + · · · , p1 = 0− α0/{5c} · (t− c) + · · · , q2 = (t− c) + (3α0 − α1 − α3 + 2)/{2c} · (t− c)2 + · · · , p2 = c(t− c)−2 − 4α0/5 · (t− c)−1 + · · · , (2)  q1 = −c(t− c)−1 + (−1/4− α0) + · · · , p1 = 0 + α0/{5c} · (t− c) q2 = (t− c) + (−3α0 − α1 − α3 + 2)/{2c} · (t− c)2 + · · · , p2 = c(t− c)−2 + 4α0/5 · (t− c)−1 + · · · , (3)  q1 = c(t− c)−1 + (3/4− α0) + · · · , p1 = 1/2− 1/{12c} · (t− c) + · · · , q2 = −(t− c) + [(α1 + α3)/c− 3/{4c}](t− c)2 + · · · , p2 = −c/2 · (t− c)−2 − 1/6(t− c)−1 + · · · . 4.5 The case where p1, q2 have a pole at t = c ∈ C∗ Proposition 4.6. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that p1, q2 have both a pole at t = c ∈ C∗ and q1, p2 are both holomorphic at t = c. Then, one of the following occurs: (1)  q1 = (t− c) + (3α1 − α0 − α3 + 2)/{2c} · (t− c)2 + · · · , p1 = c(t− c)−2 − 4α1/5 · (t− c)−1 + · · · , q2 = 4c(t− c)−1 + 8/3 + · · · , p2 = 0− α1/{5c} · (t− c) + · · · , (2)  q1 = (t− c) + (−3α1 − α0 − α3 + 2)/{2c} · (t− c)2 + · · · , p1 = c(t− c)−2 + 4α1/5 · (t− c)−1 + · · · , q2 = −c(t− c)−1 + (−1/4− α1) + · · · , p2 = 0 + α1/{5c} · (t− c) + · · · , Rational Solutions of the Sasano System of Type A (2) 5 13 (3)  q1 = −(t− c) + [(α0 + α3)/c− 3/{4c}](t− c)2 + · · · , p1 = −c/2 · (t− c)−2 − 1/6(t− c)−1 + · · · , q2 = c(t− c)−1 + (3/4− α1) + · · · , p2 = 1/2− 1/{12c} · (t− c) + · · · . 4.6 The case where p1, p2 have a pole at t = c ∈ C∗ Proposition 4.7. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that p1, p2 have both a pole at t = c ∈ C∗ and q1, q2 are both holomorphic at t = c. Then, q1 = (−4dc,−1) + ac,1(t− c) + · · · , p1 = bc,−1(t− c)−1 + (3/8 + 2b2c,−1/c) + · · · , q2 = (−4bc,−1) + cc,1(t− c) + · · · , p2 = dc,−1(t− c)−1 + (3/8 + 2d2c,−1/c) + · · · , where the coefficients satisfy 16bc,−1dc,−1 + c = 0, ac,1bc,−1c+ cc,1dc,−1c+ c 2 (α0 + α1 + α3) = c 2 . 4.7 The case where q1, p1, q2, p2 have a pole at t = c ∈ C∗ Proposition 4.8. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p1, q2, p2 all have a pole at t = c ∈ C∗. Then, q1 = −2c(t− c)−1 + ( √ c− 4/3) + ac,1(t− c) + · · · , p1 = √ c/4 · (t− c)−1 + 1/2 + bc,1(t− c) + · · · , q2 = −2c(t− c)−1 + (− √ c− 4/3) + cc,1(t− c) + · · · , p2 = − √ c/4 · (t− c)−1 + 1/2 + dc,1(t− c) + · · · , where the coefficients satisfy bc,1 + dc,1 = α3/{2c}, ac,1 √ c− cc,1 √ c = 2 + 2α3 − 2α0 − 2α1. 4.8 Summary Proposition 4.9. (1) Suppose that for A (2) 5 (αj)0≤j≤3, there exists a meromorphic solution at t = c ∈ C∗. Then, q1, q2 have both a pole of order at most one at t = c and the residues of q1, q2 at t = c are expressed by nc (n ∈ Z). (2) Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution. Then, a∞,0 − a0,0 ∈ Z, c∞,0 − c0,0 ∈ Z. Proof. Case (1) is obvious. Let us prove case (2). From the discussions in Sections 2, 3 and 4, it follows that q1 = a∞,0 + m1∑ j=1 njcj t− cj , q2 = c∞,0 + m2∑ k=1 n′kc ′ k t− c′k , nj , n ′ k ∈ Z, where m1, m2 are both positive integers and ck ∈ C∗ (1 ≤ k ≤ m1) and c′j ∈ C∗ (1 ≤ j ≤ m2) are poles of q1 and q2, respectively. If q1 or q2 is holomorphic in C∗, then its second sum is considered to be zero. Considering the constant terms of the Taylor series of q1, q2 at t = 0, we can prove the proposition. � 14 K. Matsuda 5 The Laurent series of the Hamiltonian H In this section, for a meromorphic solution at t = ∞, 0, we first compute the constant terms h∞,0, h0,0 of the Laurent series of the Hamiltonian H at t =∞, 0. Moreover, for a meromorphic solution at t = c ∈ C∗, we calculate the residue of H at t = c. 5.1 The Laurent series of H at t =∞ Proposition 5.1. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a meromorphic solution at t =∞. Then, h∞,0 = 3 4 (α0 + α1 + α3) 2 − 1 2 (−2α0 + α3)(−2α1 + α3)− 3(α0 + α1)α3. 5.2 The Laurent series of H at t = 0 5.2.1 The case where q1, p1, q2, p2 are all holomorphic at t = 0 Proposition 5.2. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p1, q2, p2 are all holomorphic at t = 0. Then, h0,0 =  0 if case (1) occurs in Proposition 3.2, −α1(α0 + α3) if case (2) occurs in Proposition 3.2, −α0(α1 + α3) if case (3) occurs in Proposition 3.2, −α3(α0 + α1) if case (4) occurs in Proposition 3.2. 5.2.2 The case where p1 has a pole at t = 0 Proposition 5.3. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that p1 has a pole at t = 0 and q1, q2, p2 are all holomorphic at t = 0. Then, h0,0 = −1 4 (α0 + α1 + α3) 2 + α2 1 + 3 16 . 5.2.3 The case where p2 has a pole at t = 0 Proposition 5.4. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that p2 has a pole at t = 0 and q1, p1, q2 are all holomorphic at t = 0. Then, h0,0 = −1 4 (α0 + α1 + α3) 2 + α2 0 + 3 16 . 5.3 The Laurent series of H at t = c ∈ C∗ 5.3.1 The case where q1 has a pole at t = c ∈ C∗ Proposition 5.5. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1 has a pole at t = c ∈ C∗ and p1, q2, p2 are all holomorphic at t = c. Then, H is holomorphic at t = c. 5.3.2 The case where q2 has a pole at t = c ∈ C∗ Proposition 5.6. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q2 has a pole at t = c ∈ C∗ and q1, p1, p2 are all holomorphic at t = c. Then, H is holomorphic at t = c. Rational Solutions of the Sasano System of Type A (2) 5 15 5.3.3 The case where q1, q2 have a pole at t = c ∈ C∗ Proposition 5.7. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, q2 have both a pole at t = c ∈ C∗ and p1, p2 are both holomorphic at t = c. Then, H is holomorphic at t = c. 5.3.4 The case where q1, p2 have a pole at t = c ∈ C∗ Proposition 5.8. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, q2 have both a pole at t = c ∈ C∗ and p1, p2 are both holomorphic at t = c. Then, H has a pole of order one at t = c and Res t=c H =  c if case (1) occurs in Proposition 4.5, c if case (2) occurs in Proposition 4.5, c/2 if case (3) occurs in Proposition 4.5. 5.3.5 The case where p1, q2 have a pole at t = c ∈ C∗ Proposition 5.9. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that p1, q2 have both a pole at t = c ∈ C∗ and q1, p2 are both holomorphic at t = c. Then, H has a pole of order one at t = c and Res t=c H =  c if case (1) occurs in Proposition 4.6, c if case (2) occurs in Proposition 4.6, c/2 if case (3) occurs in Proposition 4.6. 5.3.6 The case where p1, p2 have a pole at t = c ∈ C∗ Proposition 5.10. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that p1, p2 have both a pole at t = c ∈ C∗ and q1, q2 are both holomorphic at t = c. Then, H has a pole of order one at t = c and Res t=c H = c/4. 5.3.7 The case where q1, p1, q2, p2 have a pole at t = c ∈ C∗ Proposition 5.11. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a solution such that q1, p1, q2, p2 all have a pole at t = c ∈ C∗. Then, H has a pole of order one at t = c and Res t=c H = c/4. 5.4 Summary Proposition 5.12. (1) Suppose that for A (2) 5 (αj)0≤j≤3, there exists a meromorphic solution at t = c ∈ C∗. Then, the residue of H at t = c is expressed by nc/4 (n ∈ Z). (2) Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution. Then, 4(h∞,0− h0,0) ∈ Z. Proof. Case (1) is obvious. Case (2) can be proved in the same way as Proposition 4.9. � 16 K. Matsuda 6 Necessary condition . . . (1) 6.1 The case where q1, p1, q2, p2 are all holomorphic at t = 0 6.1.1 The case where a0,0 = 0, c0,0 = 0 Proposition 6.1. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution such that q1, p1, q2, p2 are all holomorphic at t = 0. Moreover, assuming that a0,0 = 0, c0,0 = 0, then, −2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z. Proof. The proposition follows from Propositions 2.12, 4.9. � 6.1.2 The case where a0,0 = 0, c0,0 6= 0 Proposition 6.2. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution such that q1, p1, q2, p2 are all holomorphic at t = 0. Moreover, assuming that a0,0 = 0, c0,0 6= 0, then, −2α0 + α3 ∈ Z, 2α1 + α3 ∈ Z. Proof. The proposition follows from Propositions 2.12, 3.2 and 4.9. � 6.1.3 The case where a0,0 6= 0, c0,0 = 0 Proposition 6.3. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution such that q1, p1, q2, p2 are all holomorphic at t = 0. Moreover, assuming that a0,0 6= 0, c0,0 = 0, then, 2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z. Proof. The proposition follows from Propositions 2.12, 3.2 and 4.9. � 6.1.4 The case where a0,0 6= 0, c0,0 6= 0 Proposition 6.4. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution such that q1, p1, q2, p2 are all holomorphic at t = 0. Moreover, assuming that a0,0 6= 0, c0,0 6= 0, then, 2α0 + α3 ∈ Z, 2α1 + α3 ∈ Z. Proof. From Propositions 2.12, 3.2 and 4.9, it follows that α0 − α1 ∈ Z. If α0 6= 0, by Proposition 3.2, we find that s0(q1, p1, q2, p2) is a rational solution of A (2) 5 (−α0, α1, α2 + α0, α3) such that all of s0(q1, p1, q2, p2) are holomorphic at t = 0 and a0,0 = 0, c0,0 6= 0. Then, from Proposition 6.2, we obtain the necessary condition. If α1 6= 0, by s1 and Proposi- tion 6.3, we obtain the necessary condition in the same way. If α0 = α1 = 0 and α2 6= 0, by Proposition 3.2, we see that s2(q1, p1, q2, p2) is a rational solution of A (2) 5 (α2, α2,−α2, α3 + 2α2) such that all of s0(q1, p1, q2, p2) are holomorphic at t = 0 and a0,0 6= 0, c0,0 6= 0. Based on the above discussion, considering that α0+α1+2α2+α3 = 1/2, we can obtain the necessary condition. The remaining case is that α0 = α1 = α2 = 0, α3 = 1/2. We prove that for A (2) 5 (0, 0, 0, 1/2), there exists no rational solution such that q1, p1, q2, p2 are all holomorphic at t = 0 and a0,0 6= 0, c0,0 6= 0. If there exists such a rational solution, by Proposition 3.2, we find that b0,0 = d0,0 = 0. Then, s3(q1, p1, q2, p2) is a rational solution of A (2) 5 (0, 0, 1/2,−1/2) such that all of s3(q1, p1, q2, p2) are holomorphic at t = 0 and a0,0 = c0,0 = 0. Therefore, it follows from Proposition 6.2 that −2 · 0 + (−1/2) ∈ Z, which is impossible. � Rational Solutions of the Sasano System of Type A (2) 5 17 6.2 The case where p1 has a pole at t = 0 Proposition 6.5. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution such that p1 has a pole at t = 0 and q1, q2, p2 are all holomorphic at t = 0. Then, −2α0 + α3 ∈ Z, α3 − 1/2 ∈ Z. Proof. The proposition follows from Propositions 2.12, 3.3 and 4.9. � By s1s2, we can prove the following corollary. Corollary 6.1. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution such that p1 has a pole at t = 0 and q1, q2, p2 are all holomorphic at t = 0. Then, by some Bäcklund transformations, the parameters can be transformed so that −2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z. 6.3 The case where p2 has a pole at t = 0 Proposition 6.6. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution such that p2 has a pole at t = 0 and q1, p1, q2 are all holomorphic at t = 0. Then, −2α1 + α3 ∈ Z, α3 − 1/2 ∈ Z. Proof. The proposition follows from Propositions 2.12, 3.4 and 4.9. � By s0s2, we can prove the following corollary. Corollary 6.2. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution such that p2 has a pole at t = 0 and q1, p1, q2 are all holomorphic at t = 0. Then, by some Bäcklund transformations, the parameters can be transformed so that −2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z. 6.4 Summary Proposition 6.7. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution. Then, one of the following occurs: (1) −2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z, (2) −2α0 + α3 ∈ Z, 2α1 + α3 ∈ Z, (3) 2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z, (4) 2α0 + α3 ∈ Z, 2α1 + α3 ∈ Z, (5) −2α0 + α3 ∈ Z, α3 − 1/2 ∈ Z, (6) −2α1 + α3 ∈ Z, α3 − 1/2 ∈ Z. Corollary 6.3. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution. Then, by some Bäcklund transformations, the parameters can be transformed so that −2α0 + α3 ∈ Z, −2α1 + α3 ∈ Z. 7 Necessary condition . . . (2) 7.1 Shift operators In order to transform the parameters to the standard form, let us construct shift operators. 18 K. Matsuda Proposition 7.1. Let the shift operators T0, T1, T2 be defined by T0 = πs2s3s2s1s0, T1 = s0T0s0, T2 = s2T0s2, respectively. Then, T0(α0, α1, α2, α3) = (α0 + 1/2, α1 + 1/2, α2 − 1/2, α3), T1(α0, α1, α2, α3) = (α0 − 1/2, α1 + 1/2, α2, α3), T2(α0, α1, α2, α3) = (α0, α1, α2 + 1/2, α3 − 1), respectively. 7.2 The properties of Bäcklund transformations Proposition 7.2. (1) If p1 ≡ 0 for A (2) 5 (αj)0≤j≤3, then α0 = 0. (2) If p2 ≡ 0 for A (2) 5 (αj)0≤j≤3, then α1 = 0. (3) If q1q2 + t ≡ 0 for A (2) 5 (αj)0≤j≤3, then α2 = 0. (4) If p1 + p2 − 1 ≡ 0 for A (2) 5 (αj)0≤j≤3, then α3 = 0. By this proposition, we can consider s0 as the identical transformation, if p0 ≡ 0. In the same way, we consider each of s1, s2, s3 as the identical transformation, if p2 ≡ 0, or if q1q2 + t ≡ 0, or if p1 + p2 − 1 ≡ 0, respectively. 7.3 Reduction of the parameters to the standard form By Corollary 6.3, using T0, we can transform the parameters to (α0, α1, α2, α3) = (α3/2, α3/2, α2, α3). Proposition 7.3. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution. Then, by some Bäcklund transformations, the parameters can be transformed so that −2α0 + α3 = 0, −2α1 + α3 = 0. 8 Classif ication of rational solutions 8.1 Rational solution of A (2) 5 (α3/2, α3/2, α2, α3) Proposition 8.1. For A (2) 5 (α3/2, α3/2, α2, α3), there exists a rational solution and (q1, p1, q2, p2) = (0, 1/4, 0, 1/4). Moreover, it is unique. Proof. The proposition follows from the direct calculation and Proposition 2.2. � 8.2 Proof of main theorem Let us prove our main theorem. Proof. Suppose that for A (2) 5 (αj)0≤j≤3, there exists a rational solution. Then, from Proposi- tion 6.7, we find that the parameters satisfy one of the conditions in the theorem. Moreover, from Proposition 7.3, we see that the parameters can be transformed so that −2α0 + α3 = −2α1 + α3 = 0. From Proposition 8.1, it follows that for A (2) 5 (α3/2, α3/2, α2, α3), there exists a unique rational solution such that (q1, p1, q2, p2) = (0, 1/4, 0, 1/4), which proves the main theorem. � Rational Solutions of the Sasano System of Type A (2) 5 19 A Examples of rational solutions In this appendix, we give examples of rational solutions of A (2) 5 (αj)0≤j≤3. For the purpose, we use the shift operators, T0, T1, T2, and the seed rational solution, (q1, p1, q2, p2) = (0, 1/4, 0, 1/4) for A (2) 5 (α3/2, α3/2, α2, α3). Then, we obtain the following examples of rational solutions: for A (2) 5 (α3/2 + 1/2, α3/2 + 1/2, α2 − 1/2, α3), (q1, p1, q2, p2) = ( −1, 1 4 + α3 2(t+ 4α2 3) − 4α3 + 1 4(t+ 1) ,−1, 1 4 − α3 2(t+ 4α2 3) + 4α3 + 1 4(t+ 1) , ) ; for A (2) 5 (α3/2− 1/2, α3/2 + 1/2, α2, α3), q1 = 2α3 − 1 1 + α3(1− 2α3) t + − 2α3 + 2 1 + 2α3 + 1 −t+ α3(2α3 + 1)− 1 1 + α3(−2α3 + 1) t , p1 = 1 4 + 2α3 + 1 −4t+ 4α3(2α3 + 1)− 1 1− α3(2α3 − 1) t , q2 = − 1 1− α3(2α3 − 1) t , p2 = 1 4 + α3(2α3 − 1) t − 2α3 + 1 − 4 1 + α3(1− 2α3) t + 4t 2α3 − 1 1 + α3(1− 2α3) t ; for A (2) 5 (α3/2, α3/2, α2 + 1/2, α3 − 1), (q1, p1, q2, p2) = ( − 1, 1 4 − 2α3 − 1 4{t+ (2α3 − 1)2} + 4α3 − 3 2(t+ 1) ,−1, 1 4 + 2α3 − 1 4{t+ (2α3 − 1)2} − 4α3 − 3 2(t+ 1) ) . Acknowledgments The author wishes to express his sincere thanks to Professor Yousuke Ohyama. In addition, he is also indebted the referees for their useful comments. 20 K. Matsuda References [1] Airault H., Rational solutions of Painlevé equation, Stud. Appl. 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Math. 54 (2002), 648–670. http://dx.doi.org/10.1088/0305-4470/36/36/306 http://dx.doi.org/10.1007/BF02393211 http://dx.doi.org/10.1088/0305-4470/34/11/316 http://arxiv.org/abs/0708.2960 http://dx.doi.org/10.1088/0305-4470/34/11/320 http://arxiv.org/abs/nlin.SI/0007036 http://www.ams.org/leavingmsn?url=http://dx.doi.org/10.1088/0951-7715/20/12/006 http://arxiv.org/abs/nlin.SI/0610066 http://dx.doi.org/10.1111/1467-9590.00044 http://arxiv.org/abs/math.QA/9808003 http://dx.doi.org/10.1007/BF01458459 http://dx.doi.org/10.1007/BF01762370 http://dx.doi.org/10.1007/BF02419020 http://arxiv.org/abs/0704.2327 http://dx.doi.org/10.4153/CJM-2002-024-0 1 Introduction 2 Meromorphic solutions at t= 2.1 The case where q1, p1, q2, p2 are all holomorphic at t= 2.2 The case where one of (q1,p1,q2,p2) has a pole at t= 2.2.1 The case where q1 has a pole at t= 2.2.2 The case where p1 has a pole at t= 2.3 The case where two of (q1,p1,q2,p2) have a pole at t= 2.3.1 The case where q1, p1 have a pole at t= 2.3.2 The case where q1, q2 have a pole at t= 2.3.3 The case where q1, p2 have a pole at t= 2.3.4 The case where p1, p2 have a pole at t= 2.4 The case where three of (q1,p1,q2,p2) have a pole at t= 2.4.1 The case where q1, p1, q2 have a pole at t= 2.4.2 The case where q1, p1, p2 have a pole at t= 2.5 The case where all of (q1,p1,q2,p2) have a pole at t= 2.6 Summary 3 Meromorphic solution at t=0 3.1 The case where q1, p1, q2, p2 are all holomorphic at t=0 3.2 The case where p1 has a pole at t=0 3.3 The case where p2 has a pole at t=0 4 Meromorphic solution at t=cC* 4.1 The case where q1 has a pole at t=cC* 4.2 The case where q2 has a pole at t=cC* 4.3 The case where q1, q2 have a pole at t=cC* 4.4 The case where q1, p2 have a pole at t=cC* 4.5 The case where p1, q2 have a pole at t=cC* 4.6 The case where p1, p2 have a pole at t=cC* 4.7 The case where q1, p1, q2, p2 have a pole at t=cC* 4.8 Summary 5 The Laurent series of the Hamiltonian H 5.1 The Laurent series of H at t= 5.2 The Laurent series of H at t=0 5.2.1 The case where q1, p1, q2, p2 are all holomorphic at t=0 5.2.2 The case where p1 has a pole at t=0 5.2.3 The case where p2 has a pole at t=0 5.3 The Laurent series of H at t=cC* 5.3.1 The case where q1 has a pole at t=cC* 5.3.2 The case where q2 has a pole at t=cC* 5.3.3 The case where q1, q2 have a pole at t=cC* 5.3.4 The case where q1, p2 have a pole at t=cC* 5.3.5 The case where p1, q2 have a pole at t=cC* 5.3.6 The case where p1, p2 have a pole at t=cC* 5.3.7 The case where q1, p1, q2, p2 have a pole at t=cC* 5.4 Summary 6 Necessary condition … (1) 6.1 The case where q1, p1, q2, p2 are all holomorphic at t=0 6.1.1 The case where a0,0=0, c0,0=0 6.1.2 The case where a0,0=0, c0,0=0 6.1.3 The case where a0,0=0, c0,0=0 6.1.4 The case where a0,0=0, c0,0=0 6.2 The case where p1 has a pole at t=0 6.3 The case where p2 has a pole at t=0 6.4 Summary 7 Necessary condition … (2) 7.1 Shift operators 7.2 The properties of Bäcklund transformations 7.3 Reduction of the parameters to the standard form 8 Classification of rational solutions 8.1 Rational solution of A5(2)(3/2,3/2,2,3) 8.2 Proof of main theorem A Examples of rational solutions References

Rational Solutions of the Sasano System of Type A₅⁽²⁾

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