First-order Second-degree Equations Related with Painlevé Equations

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1 ( Will be set by the publisher ) ISSN (print), (online) International Journal of Nonlinear Science Vol. xx (2008) No. xx, pp. xxx-xxx First-order Second-degree Equations Related with Painlevé Equations Ayman Sakka 1 and Ugurhan Mugan 2+ 1 Islamic University of Gaa, Department of Mathematics P.O.Box 108, Rimal, Gaa, Palestine 2 Bilkent University, Department of Mathematics, Bilkent, Ankara, Turkey (Received xxx 2008, accepted xxx 2008, will be set by the editor) Abstract. The first-order second-degree equations satisfying the Fuchs theorem concerning the absence of movable critical points, related with Painlevé equations, and one-parameter families of solutions which solve the first-order second-degree equations are investigated. Keywords: Painlevé equations, Fuchs theorem. 1 Introduction Painlevé equations, PI-PVI, which are second order first-degree equations v = F (v, v, ) where F is rational in v, algebraic in v and locally analytic in with the Painlevé property, which were first derived at the beginning of the 20 th century by Painlevé and his school [1]. A differential equation is said to have Painlevé property if all solutions are single valued around all movable singularities. Movable means that the position of the singularities varies as a function of initial values. Painlevé equations may be regarded as the nonlinear counterparts of some classical special functions. They also arise as reductions of solutions of soliton equations solvable by the inverse scattering method and may be studied as the compatibility condition of the isomonodromy deformation problem. Recently, there have been studies of integrable mappings and discrete systems, including discrete analogous of the Painlevé equations. The Riccati equations is the only example for the first-order first-degree equation which has the Painlevé property. By Fuchs theorem, the irreducible form of the first order algebraic differential equation of the second-degree with Painlevé property is given as (v ) 2 = (A 2 v 2 + A 1 v + A 0 )v + B 4 v 4 + B 3 v 3 + B 2 v 2 + B 1 v + B 0, (1) where A j, j = 0, 1, 2 and B k, j = 0, 1, 2, 3, 4 are functions of and set of parameters denoted by α [2]. Higher order (n 3) and second order higher-degree (k 2) with Painleé property were subject to the articles [3, 4, 5] Painlevé equations, PI-PVI, posses a rich internal structure. For example, for certain choice of the parameters PII-PVI admit one parameter families of solutions, rational, algebraic and expressible in terms of the classical transcendental functions: Airy, Bessel, Weber-Hermite, Whitteker, hypergeometric functions respectively. But, all the known one parameter families of solutions appear as the solutions of Riccati equations. In this article, we investigate the one parameter families of solutions of PII-PVI which solves the first-order second-degree equation of the form (1). Let v() be a solution of one of the Painlevé equations v = P 2 (v ) 2 + P 1 v + P 0. (2) where P 0, P 1, P 2 depend on v, and set of parameters α. Differentiating equation (1) and using (2) to replace v and (1) to replace (v ) 2, one gets 2+ Corresponding author. Tel. : ; Fax: address: mugan@fen.bilkent.edu.tr. Φv + Ψ = 0, (3)

2 where Φ = (P 1 2A 2 v A 1 )(A 2 v 2 + A 1 v + A 0 ) + P 2 (A 2 v 2 + A 1 v + A 0 ) 2 + 2P 0 4B 4 v 3 (3B 3 + A 2 )v2 (2B 2 + A 1 )v (B 1 + A 0 ) + 2P 2(B 4 v 4 + B 3 v 3 + B 2 v 2 + B 1 v + B 0 ), Ψ = (B 4 v 4 + B 3 v 3 + B 2 v 2 + B 1 v + B 0 )[P 2 (A 2 v 2 (4) + A 1 v + A 0 ) + 2P 1 2A 2 v A 1 ] P 0 (A 2 v 2 + A 1 v + A 0 ) (B 4 v4 + B 3 v3 + B 2 v2 + B 1 v + B 0 ). One can determine the coefficients A j, j = 0, 1, 2 and B k, j = 0, 1, 2, 3, 4 of (1) by setting Φ = Ψ = 0. Therefore, the Painlevé equation (2) admit one-parameter family of solution characteried by equation (1) if and only if Φ = Ψ = 0. For the sake of the completeness, we will examine all possible cases, and hence, we recover some of the well known one-parameter families of solutions as well as the new ones. 2 Painlevé II Equation Let v be a solution of the PII equation v = 2v 3 + v + α. (5) In this case the equation (3) takes the form of (φ 3 v 3 + φ 2 v 2 + φ 1 v + φ 0 )v + ψ 5 v 5 + ψ 4 v 4 + ψ 3 v 3 + ψ 2 v 2 + ψ 1 v + ψ 0 = 0, (6) where φ 3 = 4(B A2 2 1), φ 2 = A 2 + 3B 3 + 3A 1 A 2, φ 1 = A 1 + 2B 2 + 2A 0 A 2 + A 2 1, φ 0 = A 0 + B 1 + A 0 A 1 2α, ψ 5 = 2A 2 (B 4 + 1), ψ 4 = B 4 + A 1B 4 + 2A 2 B 3 + 2A 1 ψ 3 = B 3 + A 1B 3 + 2A 2 B 2 + 2A 0 + A 2, ψ 2 = B 2 + A 1B 2 + 2A 2 B 1 + A 1 + αa 2, ψ 1 = B 1 + A 1B 1 + 2A 2 B 0 + A 0 + αa 1, ψ 0 = B 0 + A 1B 0 + αa 0. (7) Setting ψ 5 = 0 yields A 2 = 0 or B 4 = 1. If A 2 = 0, then one can not choose A j, j = 0, 1 and B k, k = 0, 1,..., 4 so that φ j = 0, j = 0, 1, 2, 3 and ψ k = 0, k = 0, 1,..., 4. If B 4 = 1, then φ j = 0, j = 0, 1, 2, 3 and ψ k = 0, k = 0, 1,..., 4 if and only if A 2 = ±2, A 1 = 0, A 0 = ±, B 3 = B 1 = 0, B 2 =, B 0 = 1 4 2, and α = ± 1 2. With these choices of A j, B k, equation (1) gives the well known equation 2v = ±(2v 2 + ), (8) which gives the one parameter family of solutions of PII if α = ±1/2, [6]. There is no first-order seconddegree equation related with PII. 3 Painlevé III Equation Let v solve the third Painlevé equation PIII v = 1 v (v ) 2 1 v + γv (αv2 + β) + δ v. (9) Then the equation (3) takes the following form (φ 4 v 4 + φ 3 v 3 + φ 2 v 2 + φ 1 v + φ 0 )v + ψ 6 v 6 + ψ 5 v 5 + ψ 4 v 4 + ψ 3 v 3 + ψ 2 v 2 + ψ 1 v + ψ 0 = 0, (10) where φ 4 = 2γ 2B 4 A 2 2, φ 3 = 2α A 2 B 3 A 1 A 2 1 A 2, φ 2 = (A A 1), φ 1 = 2β + B 1 + A 0 A 1 A 0 1 A 0, φ 0 = A B 0 + 2δ, ψ 6 = A 2 (B 4 + γ), ψ 5 = (B B 4 + A 2 B 3 + γa 1 + α A 2), ψ 4 = A 0 B 4 B 3 2 B 3 γa 0 α A 1 A 2 B 2, ψ 3 = A 0 B 3 B 2 2 B 2 A 2 B 1 β A 2 α A 0, ψ 2 = A 0 B 2 B 1 2 B 1 A 2 B 0 β A 1 δa 2, ψ 1 = A 0 B 1 B 0 2 B 0 β A 0 δa 1, ψ 0 = A 0 (B 0 δ) (11)

3 Setting ψ 0 = ψ 6 = 0 gives A 0 (B 0 δ) = 0, A 2 (B 4 + γ) = 0. (12) Thus, one should consider the following four cases separately. Case 1. A 0 = A 2 = 0: If B 4 = γ, B 3 = 2α, B 2 = K, B 2 1 = 2β, B 0 = δ, and A 1 = 2a 1 then, φ j = 0, j = 0, 1,..., 4 and ψ k = 0, k = 1, 2,...5, where K is arbitrary constant and a 1 is a constant and satisfies γ(a 1 + 1) = 0, α(a 1 + 1) = 0, β(a 1 1) = 0, δ(a 1 1) = 0. (13) Therefore, we have the following subcases: i. a , α = β = γ = δ = 0 ii. a 1 = 1, β = δ = 0 and iii. a 1 = 1, α = γ = 0. When α = β = γ = δ = 0, the general solution of PIII is v(; 0, 0, 0, 0) = c 1 c 2 where c 1, c 2 are constants. When a 1 = 1, β = δ = 0, (1) gives the following first-order second-degree equation for v The transformations v = (v + v) 2 = γ 2 v 4 + 2αv 3 + (K + 1)v 2. (14) w γ 1/2 (w + 1) + α, v = γ 1/2 v 2 + wv, (15) give one-to-one correspondence between solutions v of PIII and solutions w() of the following Riccati equation w w 2 2w + K = 0. (16) β = δ = 0 case for PIII was also considered in [6, 7]. When a 1 = 1, α = γ = 0, v satisfies the following first-order second-degree equation (v v) 2 = (K + 1)v 2 2βv δ 2. (17) Equation (17) was also considered in [5], and the solution of PIII for α = γ = 0 was first given in [6]. Case 2. A 0 0, A 2 0: Equation (12) gives B 4 = γ and B 0 = δ. To set φ j = 0, j = 0, 1,..., 4 and ψ k = 0, k = 1, 2,..., 5, one should choose B 3 = 1 [2α A 2(2a 1 + 1)], B 2 = [ a A 0A 2 ], B 1 = 1 [2β + 2a 1A 0 A 0 ], (18) and A 1 = 2a 1, A2 0 = 4δ, where a 1 = 2α A 2 1 and α, β, γ, δ satisfy Then, equation (1) becomes βγ 1/2 + α( δ) 1/2 + 2γ 1/2 ( δ) 1/2 = 0. (19) v + γ 1/2 v 2 + [αγ 1/2 + 1]v + ( δ) 1/2 = 0. (20) Equation (20) was given in the literature before, see for example [6]. Case 3. A 0 0, A 2 = 0: Equation (12) gives B 0 = δ. One should choose B 4 = γ, B 3 = 2α, B 2 = a 2 1, B 1 = 1 [2β + 2a 1A 0 A 0 ], A 2 0 = 4δ, A 1 = 2a 1, and α = 0 in order to get φ j = 0, j = 0,..., 4, and ψ k = 0, k = 1,..., 5, where a 1 is a constant such that γ(a 1 + 1) = 0, 2β + A 0 (a 1 1) = 0. (21) So, if γ 0, then β = 2( δ) 1/2 and (1) gives the equation (20) with α = 0. If γ = 0 then (1) gives the equation (17) with K = 1 β2 δ. Case 4. A 0 = 0, A 2 0: Equation (12) gives B 4 = γ, and if B 0 = δ, B 3 = 1 [2α A 2(2a 1 + 1)], B 2 = a2 1, B 2 1 = 2β, A2 2 = 4γ, A 1 = 2a 1, and β = 0, then φ j = 0, j = 0,..., 4, ψ k = 0, k = 1,..., 5, where a 1 is a constant and satisfies δ(a 1 1) = 0, 2α A 2 (a 1 + 1) = 0. (22) So, if δ 0, equation (1) gives the equation (20) for β = 0. If δ = 0, then v solves the equation (14) with K = α2 γ 1.

4 4 Painlevé IV Equation Let v be a solution of the PIV equation Then the equation (3) takes the form where v = 1 2v (v ) v3 + 4v 2 + 2( 2 α)v + β v. (23) (φ 4 v 4 + φ 3 v 3 + φ 2 v 2 + φ 1 v + φ 0 )v + ψ 6 v 6 + ψ 5 v 5 + ψ 4 v 4 + ψ 3 v 3 + ψ 2 v 2 + ψ 1 v + ψ 0 = 0, (24) φ 4 = 3(1 B A2 2 ), φ 3 = 8 A 2 2B 3 2A 1 A 2, φ 2 = 4( 2 α) B A2 1 A 0A 2 A 1, φ 1 = A 0, φ 0 = 1 2 A2 0 + B 0 + 2β, ψ 6 = 3 2 A 2(B 4 + 1), ψ 5 = (B A 1B A 2B 3 + 4A A 1), ψ 4 = 1 2 A 0B 4 B A 1B A 2B 2 2( 2 α)a 2 4A A 0, ψ 3 = 1 2 A 0B 3 B A 1B A 2B 1 2( 2 α)a 1 4A 0, ψ 2 = 1 2 A 0B 2 B A 1B A 2B 0 2( 2 α)a 0 βa 2, ψ 1 = 1 2 A 0B 1 B A 1B 0 βa 1, ψ 0 = 1 2 A 0(B 0 2β). (25) Setting ψ 0 = ψ 6 = 0 implies A 2 (B 4 + 1) = 0, A 0 (B 0 2β) = 0. (26) respectively. Therefore, there are four subcases: i. A 0 = A 2 = 0, ii. A 0 = 0, A 2 0, iii. A 0 0, A 2 0, and iv. A 0 0, A 2 = 0. For the first case there are no choice of A j and B k such that Φ = Ψ = 0. In the second and third cases, one should choose A 2 = 2ɛ, A 1 = 4ɛ, A 0 = 2( 2β) 1/2, B 4 = 1, B 3 = 4, B 2 = 4[ 2 + α + ɛ( 2β) 1/2 + ɛ], B 1 = 4ɛ( 2β) 1/2, B 0 = 2β and where ɛ = ±1. Then v satisfies the following Ricatti equation ( 2β) 1/2 + 2ɛα + 2 = 0, (27) v = ɛ(v 2 + v 2α 2ɛ). (28) Equation (28) was also considered in [8, 9]. When A 0 0, A 2 = 0, one has to choose A 0 = 4, A 1 = 0, B 4 = 1, B 3 = 4, B 2 = 4( 2 α), B 1 = 0, B 0 = 4, and β = 2. Therefore, one can state the following theorem Theorem 4.1 The Pianlevé IV equation admits a one-parameter family of solution characteried by if and only if β = 2. (v + 2) 2 v 4 4v 3 4( 2 α)v 2 = 0, (29) (29) was also given in [10]. If f() = + a and g() = a, where a 2 = 4α, then (29) can be written as (v + 2) 2 = v 2 (v f)(v g). (30) If a = 0, that is α = 0, then (29) reduces to the following Riccati equation v = ɛv(v + ) 2, ɛ = ±1. (31) Equation (31) is nothing but the equation (28) with α = 0. If a 0, by using the transformation v = fw2 g w 2 1 [2], (30) can be transformed to the following Riccati equation: By introducing w = 2ɛy fy, x = +a, (32) can be linearied. 2 2w = ɛ(fw 2 g), ɛ = ±1. (32)

5 5 Painlevé V Equation If v solves the PV equation v = 3v 1 2v(v 1) (v ) 2 1 v + α 2 v(v β(v 1)2 1) γ δv(v + 1) v + v v 1, (33) then the equation (3) takes the form (φ 5 v 5 +φ 4 v 4 +φ 3 v 3 +φ 2 v 2 +φ 1 v+φ 0 )v +ψ 7 v 7 +ψ 6 v 6 +ψ 5 v 5 +ψ 4 v 4 +ψ 3 v 3 +ψ 2 v 2 +ψ 1 v+ψ 0 = 0, (34) where φ 5 = 2α B A2 2, φ 4 = 3B A2 2 6α A A 2, φ 3 = 2B 3 + B A A 1A 2 + A 0 A 2 + A A 2 A 1 1 A [3α + β + γ + δ 2 ], 2 φ 2 = 2B 1 + B A A 0A 1 + A 0 A 2 + A A 1 A 0 1 A 0 2 [α + 3β + γ δ 2 ], 2 φ 1 = 3B A β + A A 0, φ 0 = ( 2β + B A2 0 ), ψ 7 = 1 2 A 2(B 4 + 2α ), ψ 2 6 = B 4 ( 3 2 A A 1 2 ) 1 2 A 2B 3 + α (3A 2 2 A 1 ) B 4, ψ 5 = B 4 ( 1 2 A A ) + B 3( 3 2 A A 1 2 ) 1 2 A 2B 2 A 2 (3α + β + γ + δ 2 ) + α (3A A 0 ) + B 4 B 3, ψ 4 = B 3 ( 1 2 A A ) + B 2( 3 2 A A 1 2 ) 1 2 A 2B A 0B 4 A 1 (3α + β + γ + δ 2 ) + A 2 2 (α + 3β + γ δ 2 ) + 3α A B 3 B 2, ψ 3 = B 2 ( 1 2 A A ) + B 1( 3 2 A A 1 2 ) 1 2 A 2B A 0B 3 A 0 (3α + β + γ + δ 2 ) + A 2 1 (α + 3β + γ δ 2 ) 3β A B 2 B 1, ψ 2 = B 1 ( 1 2 A A ) + B 0( 3 2 A A 1 2 ) + A 0 (α + 3β + γ δ 2 ) 2 + β (A 2 2 3A 1 ) + B 1 B 0, ψ 1 = B 0 ( 3 2 A A ) 1 2 A 0B 1 + β (A 2 1 3A 0 ) + B 0, ψ 0 = 1 2 A 0(B 0 2β ). 2 Setting ψ 7 = ψ 0 = 0 implies A 0 (B 0 2β 2 ) = 0, (35) A 2(B 4 + 2α ) = 0. (36) 2 Therefore, one should consider the following four distinct cases. Case 1. A 0 0, A 2 0: In this case (36) implies B 4 = 1 4 A2 2 and B 0 = 1 4 A2 0. In order to get φ j = 0, j = 0, 5, one should choose A 0 = 8β, A = 8α, then φ 2 j, j = 1, 4 are identically ero, and φ j = 0, j = 2, 3 if 2 B B A A 1 A 0 + d d (A 1) 2α 6β 2γ + 2δ 2 = 0, (37) 2 B B A A 1 A 2 d d (A 1) + 6α + 2β + 2γ + 2δ A 0 A 2 = 0, (38) respectively. The equations ψ j = 0, j = 1, 6, give B 1 = 1 2 A 0A 1 and B 3 = 1 2 A 1A 2. Then, the equations (37) and (38) give B 2 = 1 4 [(A 0 + A 1 + A 2 ) 2 + A A 0A 2 + 8δ], and d d (A 1) = 2 (A 2 A 0 )(A 0 + A 1 + A 2 ) + 2γ. (39) With these choices ψ 5 = ψ 2 = 0 identically, and ψ 4 = ψ 3 = 0 if ( (A 0 + A 1 + A 2 ) 3 + 8δ A 1 + 2A 2 2 ) 4γ (A 0 + A 1 + A 2 ) = 0, (40) ( (A 0 + A 1 + A 2 ) 3 + 8δ A 1 + 2A ) + 4γ (A 0 + A 1 + A 2 ) = 0. (41) respectively. Adding the above equations gives (A 0 + A 1 + A 2 )[(A 0 + A 1 + A 2 ) 2 + 8δ] = 0. (42)

6 If A 0 + A 1 + A 2 = 0, then equations (39) and (40) give γ = 0 and δ(a 2 A 0 2 ) = 0. Thus, if then v solves the following Riccati equation γ = 0, δ[(2α) 1/2 ( 2β) 1/2 1] = 0, (43) v (2α) 1/2 v 2 [( 2δ) 1/2 (2α) 1/2 + ( 2β) 1/2 ]v + ( 2β) 1/2 = 0. (44) If A 0 + A 1 + A 2 0, then equations (42) and (40) give ( 2δ) 1/2 [1 ( 2β) 1/2 (2α) 1/2 ] = γ, (45) and the equations (39) and (41) are identically satisfied. Therefore if α, β, γ, and δ satisfy the relation (45), then PV has one-parameter family of solution which is given by the following Riccati equation v (2α) 1/2 v 2 [( 2δ) 1/2 (2α) 1/2 + ( 2β) 1/2 ]v + ( 2β) 1/2 = 0. (46) Equation (46) is the well known one-parameter family of solution of PV [11]. Case 2. A 0 = A 2 = 0: In order to make φ j = 0, j = 1,..., 5 and ψ k = 0, k = 1,..., 6 one should choose B 4 = 2α, B 2 0 = 2β, and A 2 1, B n, n = 1, 2, 3 satisfy the following equations 2B 3 + B A2 1 A 1 1 A 1 + 6α 2 + 2β 2 + 2γ 2B 1 + B A2 1 + A A 1 2α 2 6β 2 2γ + 2δ = 0, (47) + 2δ = 0, (48) B 3 + B 3 ( 1 2 A 1 2 ) + 4α 2 A 1 = 0, (49) B 3 B 2 + B 3 ( 1 2 A ) + B 2( 1 2 A 1 2 ) A 1( 3α 2 + β 2 + γ + δ) = 0, (50) B 2 B 1 + B 2 ( 1 2 A ) + B 1( 1 2 A 1 2 ) + A 1( α 2 + 3β 2 + γ δ) = 0, (51) B 1 + B 1 ( 1 2 A ) 4β 2 A 1 = 0. (52) Adding the equations (49-52) gives A 1 [B 1 + B 2 + B (α β) 2δ] = 0. (53) 2 Thus, two cases A 1 = 0 and A 1 0, should be considered separately. Case 2.a. A 1 = 0: Solving the equations (49-52), gives B j = K j, j = 1, 2, 3, where K 2 j are constants, then (47) and (48) gives γ = δ = 0, 2K 3 + K 2 + 6α + 2β = 0, 2K 1 + K 2 2α 6β = 0. (54) If one lets K 2 = 2K, then equations (54) give K 3 = (K + 3α + β) and K 1 = (K α 3β). Thus, for γ = δ = 0, v() satisfies The transformation 2 (v ) 2 = 2αv 4 (K + 3α + β)v 3 + 2Kv 2 (K α 3β)v 2β. (55) v = (v 1)[w + µ (2µ 1)v], v = [(w + µ )2 + 2β] w (w + µ )2 2β, (56) where (2µ 1) 2 = 8α, give one-to-one correspondence between solutions v of (55) and solutions w of the following equation w = w 2 + 2w + K 3α + 3β 1. (57)

7 The relation between PV with γ = δ = 0 and equation (57) was considered in [12]. Case 2.b. A 1 0: In this case Φ = Ψ = 0 implies that α + β = 0, γ = ( 2δ) 1/2 and v satisfies v = (2α) 1/2 (v 1) 2 + γv. (58) (58) is the special case α + β = 0 of the one-parameter family of PV, see equation (44). Case 3. A 0 = 0, A 2 0: Equation (36) gives B 4 = 2α. Setting φ 2 5 = 0 and ψ 6 = 0 give A 2 2 = 8α 2 B 3 = 1 2 A 1A 2, and equations φ j = 0, j = 0,..., 4 are satisfied if B 2 = 1 [ 2 A A 1 2 A 1 A A 2 1 6α 2β 2γ 2δ2 ], B 1 = 1 [ 2 A A A 1 A 2 4α 4β 2γ], and A 1 satisfies the following equations: B 0 = 2β 2. (59) d2 d 2 (A 1 ) = d d (A 1)[ 5 2 A A 1 1] 10α(2A 1 + A 2 ) 2β(A 1 + A 2 ) γ(3a 2 + A 1 2) 2δ(A 1 + A 2 2) A 2 1 (6A 2 + A 1 ), d2 d 2 (A 1 ) = d d (A 1)[3A 2 + A 1 2] 6α(A 1 + A 2 ) 4β(A 1 + A 2 ) γ(3a 2 + A 1 4) + 2δ(A 1 + 2) A 2 1 (A 2 + A 1 ), (60) (61) d2 d 2 (A 1 ) = d d (A 1)[ 1 2 A A 1 1] + 2αA 1 2β(A 1 + A 2 ) + γ(a 1 + 2) A 2 1 A 2, (62) Solving the equation (60) for d2 d 2 (A 1 ) and using in (61) and (62) give the equations for d d (A 1). Eliminating d d (A 1) between these equations and using α = A 2 2 give (A 1 + A 2 )[(A 1 + A 2 ) 2 + 8δ] = 0. (63) Therefore, if A 1 + A 2 0, then equation (63) implies (A 1 + A 2 ) 2 + 8δ = 0. The equations (60), (61), (62) are satisfied if (A 1 + A 2 )(A 2 2) + 4γ = 0 and β = 0. Thus, if γ + ( 2δ) 1/2 [(2α) 1/2 1] = 0 and β = 0, v solves the following Riccati equation v = v[(2α) 1/2 v + ( 2δ) 1/2 (2α) 1/2 )]. (64) Equation (64) is the special case, β = 0, of (46). If A 1 + A 2 = 0 then equation (60), (61), (62) are satisfied if (A 2 2 )(γ + 2δ) = 0. Therefore, when γ = δ = 0, the second-degree equation (1) reduces to the following Riccati equation v = (2α) 1/2 v(v 1) + ( 2β) 1/2 (v 1). (65) Equation (65) is the particular case, K = α β, of equation (55). If γ and δ are not both ero, then one has A 2 = 2 and hence α = 1 2. Thus we have B 4 = 1, B 2 3 = 2, 2 B 2 = 1 (2δ 2 + 2γ + 2β + 1), B 2 1 = 2 (γ + 2β). Then, for α = 1/2 and γ, δ not both ero, v solves 2 the following first-order second-degree equation [v v(v 1)] 2 + 2(δ 2 + γ + β)v 2 2(γ + 2β)v + 2β = 0. (66) If one lets f() = b + a and g() = c + a, where a 2 = 2β, a(b + c) = 2γ, bc = 2δ, then (66) takes the following form [v v(v 1)] 2 = (fv a)(gv a). (67) If a = 0, that is, if β = γ = 0, then equation (67) is reduced to the following Riccati equation v = v(v 1) + ( 2δ) 1/2 v. (68) Equation (68) is the special case, β = γ = 0 and α = 1 2, of (46). If b = c, that is, if γ2 = 4βδ, then equation (67) is reduced to the following Riccati equation v = v(v 1) + [( 2δ) 1/2 + ( 2β) 1/2 ]v ( 2β) 1/2. (69)

8 Equation (69) is the special case, α = 1 2, of (46). If β 0 and γ 2 4βδ 0, then the solution of equation (66) is given by v = a(w2 +1), where w() fw 2 +g solves the following Riccati equation w = ɛ(fw 2 + g), ɛ = ±1, (70) which can be transformed to the following linear equation by the transformation w = 2ɛy fy : 2 f()y + ay g()f 2 ()y = 0. (71) If b 0, the change of variable = a b x transforms equation (71) to the equation ( 1 ÿ x 1 1 ) ẏ + a2 (x 1)(cx b) x 4bx 2 y = 0. (72) If b = 0, that is δ = 0, then the change of variable = 1 ac x2 transforms equation (71) to the Bessel equation x 2 ÿ + xẏ + (x 2 + a 2 )y = 0. (73) Case 4. A 0 0, A 2 = 0: In this case, equation (36) gives B 0 = 2β. Setting φ 2 0 = 0 and ψ 1 = 0 implies that A 2 0 = 8β and B 2 1 = 1 2 A 0A 1. The conditions φ j = 0 for j = 1,..., 5 are satisfied if and ψ j = 0, j = 2,..., 6 if B 2 = 1 [ 2 A A A 0 A A 2 1 2α 6β 2γ + 2δ2 ], B 3 = 1 [ 2 A A A 2 1 4α 4β 2γ], B 4 = 2α, 2 d2 d 2 (A 1 ) = d d (A 1)[ 5 2 A A 1 + 1] + 2α(A 1 + A 0 ) + 10β(2A 1 + A 0 ) +γ(3a 0 + A 1 + 2) 2δ(A 1 + A 0 + 2) A 2 1 (6A 0 + A 1 ), d2 d 2 (A 1 ) = d d (A 1)[3A 0 + A 1 + 2] + 4α(A 1 + A 0 ) + 6β(A 1 + A 0 ) +γ(3a 0 + A 1 + 4) + 2δ(A 1 2) A 2 1 (A 0 + A 1 ), (74) (75) (76) d2 d 2 (A 1 ) = d d (A 1)[ 1 2 A A 0 1] + 2α(A 0 + A 1 ) 2βA 1 γ(a 1 2) A 2 1 A 0, (77) Solving the equation (75) for d2 (A d 2 1 ) and using in (76) and (77) give the first order differential equations for A 1. Eliminating d d (A 1) between these equations and using β = A 2 0 gives (A 1 + A 0 )[(A 1 + A 0 ) 2 + 8δ] = 0. (78) If A 1 + A 0 0, then one obtains (A 1 + A 0 ) 2 + 8δ = 0. The equations (76) and (77) are satisfied if α = 0 and (A 0 + 2)(A 0 + A 1 ) = 4γ. Thus, when γ = ( 2δ) 1/2 [1 ( β) 1/2 ], v satisfies the following Riccati equation v = [( 2δ) 1/2 + ( 2β) 1/2 ]v ( 2β) 1/2. (79) This is the special case, α = 0, of equation (46). If A 1 + A 0 = 0, equations (75), (76), (77) are satisfied only if (A 0 + 2)(γ 2δ) = 0. Therefore, if A , then one should have γ = δ = 0, and v satisfies (65). If A = 0, then β = 1 2 and the second-degree equation (1) gives [v (v 1)] 2 = 2αv 4 2(γ + 2α)v 3 2(δ 2 γ α)v 2. (80) The Lie-point discrete symmetry v = 1 v, ᾱ = β, β = α, γ = γ, δ = δ of PV [6] transforms solutions v(; α, 1 2, γ, δ) of (80) to solutions v(; 1 2, β, γ, δ) of equation (66).

9 6 Painlevé VI Equation If v solves the sixth Painlevé equation, PVI v = 1 2 { 1 v + 1 v v }(v ) 2 { v }v + v(v 1)(v ) {α + β v 2 + γ( 1) (v 1) 2 + δ( 1) v ) 2 }, (81) then the equation (3) takes the form (φ 6 v 6 + φ 5 v 5 + φ 4 v 4 + φ 3 v 3 + φ 2 v 2 + φ 1 v + φ 0 )v + ψ 8 v 8 + ψ 7 v 7 + ψ 6 v 6 + ψ 5 v 5 + ψ 4 v 4 + ψ 3 v 3 + ψ 2 v 2 + ψ 1 v + ψ 0 = 0, (82) where φ 6 = 2α B A2 2, φ 5 = 2( + 1)B 4 + ( + 1)A 2 2 4α(+1) A 2 ( 1) ( 1) A 2, φ 4 = ( 1) A 2 ( 1) ( 1) (A 1 A 2 ) A2 1 + A 0A 2 + ( + 1)A 1 A A2 2 + B 2 + ( + 1)B 3 3B 4 + ( + 1)A 2 A [α( ) + β + (δ + γ)( 1)], φ 3 = ( 1) (A 1 A 2 ) ( 1) ( 1) (A 0 A 1 ) + 2A 0 A 1 A 1 A 2 + 2B 1 B 3 + ( + 1)A 1 A 0 A 2 4 [(α + β)( + 1) + (γ + δ)( 1)], ( 1) 2 φ 2 = ( 1) (A 0 A 1 ) + ( 1) ( 1) A A2 0 ( + 1)A 0A 1 A 0 A A B 0 ( + 1)B 1 B 2 A 1 + ( + 1)A [α + β( ) + (γ + δ)( 1)], ( 1) 2 φ 1 = [2( + 1)B 0 + ( + 1)A β(+1) + A ( 1) ( 1) A 0], φ 0 = [B A β ], ( 1) 2 ψ 8 = 1 2 A 2[B 4 + 2α ], ψ 7 = B 4 [( + 1)A A 1 2( 1) ( 1) ] 1 2 A α 2B 3 + [2( + 1)A 2 A 1 ] B 4, ψ 6 = B 4 [ 3 2 (A 0 A 2 ) + ( 1) + 2( 1) ( 1) ] + B 3[( + 1)A A 1 2( 1) ( 1) ] 1 2 A 2B 2 + ( + 1)B 4 B 3 + α [2( + 1)A 1 A 0 ] A 2 [α( ) + β + (δ + γ)( 1)], ψ 5 = B 3 [ 3 2 (A 0 A 2 ) + B 4 [ 1 2 A 1 + ( + 1)A 0 + ( 1) + 2( 1) ( 1) ] + B 2[( + 1)A A 1 2( 1) ( 1) ] 1 2 A 2B 1 ( 1) ] + 2α(+1) A 0 A 1 [α( ) + β+ (δ + γ)( 1)] + 2A 2 [(α + β)( + 1) + (γ + δ)( 1)] + ( + 1)B ( 1) 2 3 B 4 B 2, ψ 4 = B 2 [ 3 2 (A 0 A 2 ) + ( 1) + 2( 1) ( 1) ] + B 1[( + 1)A A 1 2( 1) ( 1) ] 1 2 A 2B 0 B 3 [ 1 2 A 1 + ( + 1)A 0 + ( 1) ] A 0B 4 A 0 [α( ) + β + (δ + γ)( 1)] + 2A 1 [(α + β)( + 1) + (γ + δ)( 1)] ( 1) 2 A 2 [α + β( ) + γ( 1) + δ( 1)] + ( + 1)B ( 1) 2 2 B 3 B 1, ψ 3 = B 1 [ 3 2 (A 0 A 2 ) + B 2 [ 1 2 A 1 + ( + 1)A 0 + ( 1) + 2( 1) ( 1) ] + B 0[( + 1)A A 1 2( 1) ( 1) ] A 0B 3 ( 1) ] + 2β(+1) A ( 1) A 0 [(α + β)( + 1) + (γ + δ)( 1)] ( 1) 2 A 1 [α + β( ) + (γ + δ)( 1)] + ( + 1)B ( 1) 2 1 B 2 B 0, ψ 2 = B 0 [ 3 2 (A 0 A 2 ) + ( 1) + 2( 1) ( 1) ] B 1[ 1 2 A 1 + ( + 1)A 0 + B 1 + β [2( + 1)A ( 1) 2 1 A 2 ] A 0 [α + β( ) + (γ + δ)( 1)], ( 1) 2 ψ 1 = β [2( + 1)A ( 1) 2 0 A 1 ] A 0B 1 B 0 [( + 1)A A 1 + ( 1) ] B 0, ψ 0 = 2 A 0[B 0 2β ( 1) 2 ]. Setting ψ 8 = ψ 0 = 0 implies A 0 ( B 0 2β ) ( ( 1) 2 = 0, A 2 B 4 + ( 1) ] A 0B 2 + ( + 1)B 0 (83) ) 2α 2 ( 1) 2 = 0. (84) To solve these equations one may distinguish between the following four cases: Case 1. A 0 0, A 2 0: Equation (84) gives B 4 = 2α, B 0 = 2β. If A ( 1) 2 2 = 2a 2 ( 1), A 0 = 2a 0 ( 1),

10 B 3 = 1 2 A 1A 2, and B 1 = 1 2 A 1A 0 then, φ 0 = φ 6 = ψ 1 = ψ 7 = 0, where a 2 2 = 2α and a2 0 = 2β. Then φ 1 = φ 5 = 0 identically, and φ j = 0, j = 2, 3, 4 if A 1 = 2(λ+µ) ( 1), and B 2 = 2 [(λ 2 + a 0 λ γ β) 2 + (2µλ + a 0 λ + a 0 µ µ + 2a 0 a 2 a 0 α β + γ δ) + µ 2 + µ(a 0 + 1) + a 0 β + δ], (85) where λ and µ are constants such that λ + µ + a 0 + a 2 = 0, λ(a 2 a 0 1) = a 2 α β γ δ, and µ(a 2 a 0 1) = a 0 α β + γ + δ. To set ψ j = 0, j = 2,..., 6, λ and µ should satisfy (λ + a 2 1)[(λ + a 0 ) 2 2γ] = 0 and (µ + a 2 )[(µ + a 2 ) 2 2γ] = 0. The above five conditions on λ and µ are satisfied if (2α) 1/2 ( 2β) 1/2 (2γ) 1/2 (1 2δ) 1/2 = 1. (86) Therefore, if α, β, γ, and δ satisfy the condition (86), then v satisfies the following Riccati equation: ( 1)v = (2α) 1/2 v 2 [(( 2β) 1/2 + (2γ) 1/2 ) + (2α) 1/2 (2γ) 1/2 ]v + ( 2β) 1/2. (87) This is the well known one parameter family of solutions of PVI [6, 13, 14]. Case 2. A 0 = A 2 = 0: In order to set φ j = 0, j = 0,..., 6, one should choose B 4 = 2α, B 0 = 2β, A ( 1) 2 1 = a 1, B 2 = ( + 1)B A2 1 + A 1 + ( 1) ( 1) A 2 1 [α( ) + β + γ( 1) + δ( 1)], B 1 = B ( + 1)A 1 (2 + 1) ( 1) A [(α + β)( + 1) + (γ + δ)( 1)], ( 1) 2 (88) where a 1 is a constant. with these choices ψ 1 = ψ 7 = 0 identically, and ψ 6 = 0, if B 3 satisfies d d [2 ( 1) 2 B 3 ] 1 2 a 1( 1) 2 B 3 2α [a 1( + 1) ] = 0. (89) Therefore, there are three distinct subcases: i. a 1 (a 1 2) 0, ii. a 1 = 0, and iii. a 1 = 2. If a 1 (a 1 2) 0, then one can not choose B 3 and a 1 so that ψ j = 0, j = 2, 3, 4, 5. If a 1 = 0, then equation (89) gives B 3 = 4α+b 3, where b 3 is a constant and ψ j = 0, j = 2, 3, 4, 5 only if b 3 = 2(γ + β) and α = δ = 0. Therefore, if α = δ = 0, one-parameter family of solutions of PVI are given by 2 ( 1) 2 (v ) 2 = 2(v ) 2 [(γ + β)v β]. (90) If γ + β = 0, then equation (90) reduces to the linear equation ( 1)v = ( 2β) 1/2 (v ), (91) which is the special case, α = δ = 0 and γ + β = 0, of the equation (87). If γ + β 0, then the transformation v = 1 {2[( 1)u (γ + β) u + n( 1) + m]2 + β}, (92) where 2m(m 1) = γ and 2n(n 1) = β, transform equation (90) to the hypergeometric equation ( 1)u + [2(n + m + 1) (2n + 1)]u + (n + m)(n + m 1)u = 0. (93) If a 1 = 2, then equation (89) gives B 3 = b 3 4α, where b 3 is a constant, and ψ j = 0, j = 2, 3, 4, 5 only if b 3 = 2 and α = β = 0, γ = δ = 1 2. Therefore, with these values of the parameters, PVI has one-parameter family of solutions given by ( 1) 2 (v v) 2 = v(v 1) 2. (94) The transformation v = [( 1) u u ] 2 transforms equation (94) into the hypergeometric equation ( 1)u + (1 + 2)u + 1 u = 0. (95) 2

11 Case 3. A 0 0, A 2 = 0: (84), and φ 0 = φ 6 = 0 give B 0 = respectively. Then, one can obtain B 1 = 1 2 A 0A 1, 2β ( 1) 2, A 2 0 = B 2 = 1 A 2 0 (+1) A 0 A 1 A A A [α + β( ) + γ( 1) + δ( 1)], B 3 = 1 A 0A 1 1 A (+1) 8β ( 1) 2 and B 4 = 2α A 1 + (2 + 1) 2 ( 1) A 1 2 [(α + β)( + 1) + (γ + δ)( 1)]. (96) by solving ψ 1 = 0, φ j = 0, j = 2, 3 respectively. Then, φ 1 = φ 2 = ψ 5 = ψ 7 = ψ 8 = 0 identically and φ 4 = 0 gives ( 1)A 1 + ( 1)A 1 A 0 = 0. (97) If one lets A 0 = a 0 ( 1), where a2 0 = 8β. Then equation (97) implies that A 1 = a 1 constant. The equation ψ 6 = 0 now gives a 0 ( 1), where a 1 is a (a 1 2)[a a 0 a 1 + 4a 1 8(γ + δ α)] = 0, (a 0 + a 1 )[a a 0 a 1 + 4a 1 8(γ + δ + α)] = 0. (98) So, there are four distinct cases: i. a 1 = 2, a 0 +a 1 0, ii. a 1 2, a 0 +a 1 = 0, iii. a 1 2, a 0 +a 1 0, and iv. a 1 = 2, a 0 + a 1 = 0. If a 1 = 2, a 0 + a 1 0 and a 1 2, a 0 + a 1 = 0, then one can not choose B k, k = 0,..., 4 such that ψ j = 0, j = 2, 3, 4, 5. When a 1 2, a 0 + a 1 0, ψ j = 0, j = 2,..., 5 only if α = δ = 0, γ + β = 0, and then v satisfies (91). If a 1 = 2, a 0 = 2, then β = 1 2, B 0 = 1, B ( 1) 2 1 = 2 and ( 1) 2 B 2 = 1 [2α + 2γ( 1) + 2δ( 1) 2 + 1], B 3 = 1 [(1 2γ 2δ)( 1) 2α( + 1)]. (99) Then ψ j = 0, j = 2,..., 5 identically. Hence, we have the following theorem Theorem 6.1 The Pianlevé VI equation admits a one-parameter family of solution characteried by 2 [( 1)v (v 1)] 2 = v 2 {2αv 2 [(2α + λ) + 2α λ]v + 2γ 2 + (2α + λ 4γ) + 2γ λ}, (100) where λ = 2γ + 2δ 1, if and only if β = 1 2. The equation (100) can be linearied as follows: If α = λ = 0, then equation (100) can be reduced to the following linear equation ( 1)v = [(2γ) 1/2 ( 1) ]v. (101) This is the special case, α = 0, and β = 1 2, of equation (87). If α = 0 and λ 0, then the solution of equation (100) is given by where w is the solution of the following Riccati equation v = 1 λ [( 1)w2 2γ( 1) λ], (102) ( 1)w = ɛ[( 1)w 2 2γ( 1) λ], ɛ = ±1. (103) The transformation w = 2ɛ(y +ny) y, where 4n 2 = 1 2δ, transform equation (103) into the hypergeometric equation: If α 0, then equation (100) can be written as ( 1)y + (2n + 1)( 1)y 1 λy = 0 (104) 4 2 [( 1)v (v 1)] 2 = v 2 (av f)(av g), (105) where f() = b( 1) + a, g() = c( 1) + a, a 2 = 2α, bc = 2γ, a(b + c) = 2α + λ. If b = c, that is, if 2(2α) 1/2 (2γ) 1/2 = 2α + 2γ + 2δ 1, then equation (105) is reduced to the Riccati equation ( 1)v = (2α) 1/2 v 2 [(2γ) 1/2 ( 1) + (2α) 1/2 ]v, (106)

12 which is the special case, β = 1 2, of equation (87). If (2α) 1/2 (2γ) 1/2 2α + 2γ + 2δ 1, then the solution of equation (100) is given by where w is the solution of the following Riccati equation The transformation v = fw2 g a(w 2 1), (107) ( 1)w = ɛ(fw 2 g), ɛ = ±1. (108) w = 2ɛ 1)y [( + n( 1) + m], (109) f y where 4n 2 = (b a)(c a), 4m 2 = a 2, transforms equation (108) into the following linear equation y = [ 2n+1 + 2m+1 1 b b b+a ]y 1 4( 1)(b b+a) [b(2a2 + 8nm ab ac) (b a)(2a 2 + 8nm ab ac) 4m(b a) + 4an]y. (110) Equation (110) is known as Huen s equation [15]. Case 4. A 0 = 0, A 2 0: In this case, equation (84) and φ 6 = φ 0 = 0 give B 4 = 2α A 2 2 = 8α B 0 = 2β ( 1) 2 B 3 = 1 2 A 2A 1, and Then, φ 2 = 0, yields, respectively. Solving the equations ψ 7 = 0, and φ j = 0, j = 3, 4 give B 2 = { 1 4 A 2( ) ( + 1)A 2A 1 A 1 ( 1) ( 1) A A2 1 A [β + γ( 1) + δ( 1)]}, B 1 = 1 2 {( + 1)A 1 + (2 + 1) ( 1) A 1 A 1 A 2 + A ( + 1)A2 2 4 ( 1) 2 [β( + 1) + (γ + δ)( 1)]. (111) ( 1)A 1 + ( 1)A 1 A 2 = 0. (112) and φ 5 = φ 1 = ψ 2 = ψ 8 = ψ 7 = 0 identically. Let A 2 = a 2 ( 1), where a2 2 = 8α, then equation (112) implies that A 1 = a 1 a 2 ( 1), where a 1 is a constant. The equation ψ 2 = 0 now gives a 1 [a 2 2+2a 2 a 1 4a 1 4a 2 +8(γ+δ β)] = 0, (a 2 +a 1 2)[a 2 2+2a 2 a 1 4a 1 4a 2 +8(γ+δ+β)] = 0. (113) So, there are four subcases to be considered: i. a 1 0, a 2 + a 1 2, ii. a 1 = 0, a 2 2, iii. a 1 0, a 1 = 2 a 2, and iv. a 1 = 0, a 2 = 2. If a 1 = 2 a 2, a 1 0, then one can not choose B k, k = 0,..., 4 so that ψ j = 0, j = 2,..., 5. When a 1 = 0, a 2 2, ψ j = 0, j = 2,..., 5 only if α + δ = (2α) 1/2, γ = β = 0. In this case v satisfies the following Ricatti equation ( 1)v = (2α) 1/2 v(v 1), (114) which is the special case, γ = β = 0, of the equation (87). When a 1 0, a 2 + a 1 2 0, then ψ j = 0, j = 2,..., 5 only if β = 0 and (2α) 1/2 (2γ) 1/2 (1 2δ) 1/2 1 = 0. Then v satisfies the following Ricatti equation This is the special case, β = 0, of (87). If a 1 = 0, a 2 = 2, then one has α = 1 2, B 4 = B 2 = B 1 = ( 1)v = (2α) 1/2 v 2 [(2γ) 1/2 ( 1) + (2α) 1/2 ]v. (115) 1 2, B 3 =, 1 2 ( 1) 2 [2β + 2γ( 1) + 2δ( 1) ] (116) 1 [(2γ + 2δ 1)( 1) + 2β( + 1)] ( 1) 2

13 Now the equations ψ j = 0, j = 3, 4, 5, 6 are satisfied identically. Thus, PVI admits a one-parameter family of solution characteried by [( 1)v v(v 1)] 2 +2[λ 2 +(γ +β λ) γ]v 2 [(λ+γ +β) +β γ λ]v +2β 2 = 0, (117) where λ = δ 1 2, if and only if α = 1 2. The Lie point-discrete symmetry v = 1 v, ᾱ = β, β = α, γ = γ, δ = δ, = 1 of PVI [16] transforms solutions v(; 1 2, β, γ, δ) of equation (117) into solutions v( ; ᾱ, 1 2, γ, δ) of equation (100). 7 Conclusion We investigated the first-order, second-degree equations satisfying the Fuchs theorem concerning the absence of movable singular points except the poles, related with the Painlevé equations. By using these first-order, second-degree equations, one parameter families of solutions of PII-PVI are also obtained. For the sake of completeness we examine all possible cases and hence some of the well known results as well as the new ones are obtained. References [1] E.L. Ince, Ordinary Differential Equations, (Dover, New York, 1956). [2] R. Chalkley, J. Diff. Eq. 68 (1987) 72. [3] U. Muğan and F. Jrad, J. Phys. A:Math. Gen. 32 (1999) [4] U. Muğan and A. Sakka, J. Math. Phys. 44 (1999) [5] A. Sakka, J. Phys. A:Math. Gen. 34 (2001) 623. [6] A.S. Fokas and M. J. Ablowit, J. Math. Phys. 23 (1982) [7] N. A. Lukashevich, Diff. Urav. 1 (1965) 561. [8] N.A. Lukashevich, Diff. Urav. 3 (1976) 771. [9] K. Okamoto, Math. Ann. 275 (1986) 221. [10] D. Albrecht, E. Mansfield, A. Milne, J. Phys. A: Math. Gen. 29, (1996) 973. [11] V.I. Gromak, Diff. Urav. 12, No.4 (1976) 740. [12] A. Sakka and U. Muğan, J. Phys. A: Math. Gen. 31 (1998) [13] N. A. Lukashevich and A.I. Yablonskii, Diff. Urav. 3 (1967) 264. [14] U. Muğan and A. Sakka, J. Math. Phys. 36 (1995) [15] E.T. Whittaker and G.N. Watson, A Course of Modern Analysis,(1902),(New York:MacMillan) [16] M. Maocco, arxiv:nlin.si/ Jul 2000.