First-order Second-degree Equations Related with Painlevé Equations

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1 ISSN (print), (online) International Journal of Nonlinear Science Vol.8(2009) No.3,pp First-order Second-degree Equations Related with Painlevé Equations Ayman Sakka 1, Uğurhan Muğan 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 27 August 2008, accepted 15 January 2009) 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 the 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 (IST). Ablowit, Ramani, and Segur [2] showed that all the ordinary differential equations (ODE) obtained by the exact similarity transforms from a partial differential equation (PDE) solvable by IST have the Painlevé property. The Painlevé property confirms the integrability properties of a PDE. Wiess, Tabor and Carnevale [3] defined a Painlevé property for PDE that does not refer to that for ODE s. This method is commonly used to investigate the integrability of a given PDE [4, 5]. Painlevé equations can also be obtained as the compatibility condition of the isomonodromy deformation problem. Recently, there have been studies of integrable mappings and discrete systems, including the discrete analogous of the Painlevé equations. The Riccati equation 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 [6]. Higher order (n 3) and second order higher-degree (k 2) with Painlevé property were subject to the articles [7 9]. Painlevé equations, PI-PVI, possess 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 Corresponding author. Tel. : ; Fax: address: mugan@fen.bilkent.edu.tr Copyright c World Academic Press, World Academic Union IJNS /276

2 260 International Journal of Nonlinear Science,Vol.8(2009),No.3,pp 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 where Φv + Ψ = 0, (3) 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 solutions 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 In this case, equation (3) takes the form of v = 2v 3 + v + α. (5) (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, [10]. 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 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) IJNS for contribution: editor@nonlinearscience.org.uk

3 Ayman Sakka, Uğurhan Muğan: First-order Second-degree Equations Related with 261 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) 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 = 0, 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 [10, 11]. 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 [9], and the solution of PIII for α = γ = 0 was first given in [10]. 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 [10]. Case 3. A 0 = 0, A 2 = 0: Equation (12) gives B 0 = δ. One should choose B 4 = γ, B 3 = 2α, IJNS homepage:

4 262 International Journal of Nonlinear Science,Vol.8(2009),No.3,pp 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 Painlevé IV Equation Let v be a solution of the PIV equation where Then equation (3) takes the form of 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 = 2ε, A 1 = 4ε, A 0 = 2( 2β) 1/2, B 4 = 1, B 3 = 4, B 2 = 4[ 2 + α + ε( 2β) 1/2 + ε], B 1 = 4ε( 2β) 1/2, B 0 = 2β, and where ε = ±1. Then v satisfies the following Riccati equation ( 2β) 1/2 + 2εα + 2 = 0, (27) v = ε(v 2 + v 2α 2ε). (28) Equation (28) was also considered in [12, 13]. 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 IJNS for contribution: editor@nonlinearscience.org.uk

5 Ayman Sakka, Uğurhan Muğan: First-order Second-degree Equations Related with 263 Theorem 1 The Painlevé 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 [14]. 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 = εv(v + ) 2, ε = ±1. (31) Equation (31) is nothing but equation (28) with α = 0. If a = 0, by using the transformation v = fw2 g [6], (30) can be transformed to the following Riccati w 2 1 equation: 2w = ε(fw 2 g), ε = ±1. (32) By introducing w = 2εy fy 5 Painlevé V Equation If v solves the PV equation +a, x = 2, (32) can be linearied. 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 equation (3) takes the form of (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 (35) A 0 (B 0 2β 2 ) = 0, 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 IJNS homepage:

6 264 International Journal of Nonlinear Science,Vol.8(2009),No.3,pp 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. By adding the above equations, one gets (A 0 + A 1 + A 2 )[(A 0 + A 1 + A 2 ) 2 + 8δ] = 0. (42) 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 solutions of PV [15]. 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) IJNS for contribution: editor@nonlinearscience.org.uk

7 Ayman Sakka, Uğurhan Muğan: First-order Second-degree Equations Related with 265 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 equations (49-52), gives B j = K j, j = 1, 2, 3, where K 2 j are constants, then (47) and (48) give 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 equation (54) gives 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) The relation between PV with γ = δ = 0 and equation (57) was considered in [16]. 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) Equation (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 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 that (A 1 + A 2 ) 2 + 8δ = 0. Equations (60), (61), and (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) IJNS homepage:

8 266 International Journal of Nonlinear Science,Vol.8(2009),No.3,pp Equation (64) is a 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, first-order 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, 2 B 3 = 2, B 2 2 = 1 (2δ 2 + 2γ + 2β + 1), B 2 1 = 2 (γ + 2β). Then, for α = 1/2 and γ, δ not both ero, 2 v solves 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 γ 2 = 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) 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 = ε(fw 2 + g), ε = ±1. (70) (70) can be transformed to the following linear equation by the transformation w = 2εy 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 y x 1 1 ) y + 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 y + x y + (x 2 + a 2 )y = 0. (73) Case 4. A 0 = 0, A 2 = 0: In this case, equation (36) gives B 0 = 2β 2. Setting φ 0 = 0 and ψ 1 = 0 implies that A 2 0 = 8β 2, and B 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 (74) 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 ), (75) (76) IJNS for contribution: editor@nonlinearscience.org.uk

9 Ayman Sakka, Uğurhan Muğan: First-order Second-degree Equations Related with 267 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 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 = 0, then one should have γ = δ = 0, and v satisfies (65). If A = 0, then β = 1 2 and first-order 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, α = β, β = α, γ = γ, δ = δ of PV [10] transforms solutions v(; α, 1 2, γ, δ) of (80) to solutions v(; 1 2, β, γ, δ) of equation (66). 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) IJNS homepage:

10 268 International Journal of Nonlinear Science,Vol.8(2009),No.3,pp 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), 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) IJNS for contribution: editor@nonlinearscience.org.uk

11 Ayman Sakka, Uğurhan Muğan: First-order Second-degree Equations Related with 269 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 [10, 17, 18]. 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 the following equation: 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 which is a special case, α = δ = 0 and γ + β = 0, of the equation (87). If γ + β = 0, then the transformation v = ( 1)v = ( 2β) 1/2 (v ), (91) 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 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, and 2β ( 1) 2, A 2 0 = 8β ( 1) 2, and B 4 = 2α 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) A 1 + (2 + 1) 2 ( 1) A 2 1 [(α + β)( + 1) + (γ + δ)( 1)], (96) IJNS homepage:

12 270 International Journal of Nonlinear Science,Vol.8(2009),No.3,pp 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 is a constant. The equation ψ 6 = 0 now gives a 0 ( 1), where a 1 (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 2 The Painlevé 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. 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 a 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 satisfies the following Riccati equation v = 1 λ [( 1)w2 2γ( 1) λ], (102) ( 1)w = ε[( 1)w 2 2γ( 1) λ], ε = ±1. (103) The transformation w = 2ε(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 following Riccati equation ( 1)v = (2α) 1/2 v 2 [(2γ) 1/2 ( 1) + (2α) 1/2 ]v. (106) (106) 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 v = fw2 g a(w 2 1), (107) IJNS for contribution: editor@nonlinearscience.org.uk

13 Ayman Sakka, Uğurhan Muğan: First-order Second-degree Equations Related with 271 where w solves the following Riccati equation The transformation ( 1)w = ε(fw 2 g), ε = ±1. (108) w = 2ε 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 m+1 1 b b b+a ]y 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 [19]. 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 respectively. Solving the equations ψ ( 1) 2 7 = 0, and φ j = 0, j = 3, 4 give B 3 = 1 2 A 2A 1, and Then, φ 2 = 0, yields 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 Riccati 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 Riccati equation: ( 1)v = (2α) 1/2 v 2 [(2γ) 1/2 ( 1) + (2α) 1/2 ]v. (115) Equation (115) is the special case, β = 0, of (87). If a 1 = 0, a 2 = 2, then one has α = 1 2, B 4 = 1 2, B 3 =, B 2 = B 1 = 1 2 ( 1) 2 [2β + 2γ( 1) + 2δ( 1) ], (116) 1 [(2γ + 2δ 1)( 1) + 2β( + 1)]. ( 1) 2 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, α = β, β = α, γ = γ, δ = δ, = 1 of PVI [20] transforms solutions v(; 1 2, β, γ, δ) of equation (117) into solutions v( ; α, 1 2, γ, δ) of equation (100). IJNS homepage:

14 272 International Journal of Nonlinear Science,Vol.8(2009),No.3,pp Conclusion It is well known that for certain choice of parameters the Painlevé equations, PII-PVI, admit one parameter family of solutions, rational, algebraic and expressible in terms of the classical transcendental functions such as Airy, Bessel, Weber-Hermite, Whitteker, hypergeometric functions respectively. All these known one parameter family of solutions satisfy the Riccati equations. In this article, 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 PII-PVI. By using these first-order second-degree equations, one parameter family 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. Acknowledgments The work of UM is partially supported by The Scientific and Technological Research Council of Turkey (TÜBİTAK) under grand number 108T977. References [1] E.L. Ince: Ordinary Differential Equations, Dover: New York(1956) [2] Mark J. Ablowit, Alfred Ramani, Harvey Segur: A connection between nonlinear evolution equations and ordinary differential equations of P-type. I, J. Math. Phys., 21(4), (1980) [3] John Weiss, M. Tabor, George Carnevale: The Painlevé property for partial differential equations, J. Math. Phys., 24(3), (1983) [4] Jiuli Yin: Painlevé integrability, Bäcklund transformation and solitary solutions satability of modified DGH equation, International Journal of Nonlinear Science, 2(3), (2006) [5] Suping Qian: Painlevé analysis and symmetry reductions to the strong dispersive DGH equation, International Journal of Nonlinear Science, 1(2), (2006) [6] Roger Chalkley: New contributions to the related work of Paul Appell, Laarus Fuchs, Georg Hamel, and Paul Painlev on nonlinear differential equations whose solutions are free of movable branch points, J. Diff. Eq., 68, (1987) [7] Uğurhan Muğan, Fahd Jrad: Painlevé test and the first Painlevé hierarchy, J. Phys. A: Math. Gen., 32(45), (1999) [8] Uğurhan Muğan, Ayman Sakka: Second-order Second-degree Painlevé Equations Related with Painlevé I-VI Equations and Fuchsian Type Transformations, J. Math. Phys., 40(7), (1999) [9] Ayman Sakka: Second-order fourth-degree Painlevé-type equations, J. Phys. A: Math. Gen., 34(3), (2001) [10] Athanassios S. Fokas, Mark J. Ablowit: On a unified approach to transformations and elementary solutions of Painlevé equations, J. Math. Phys., 23, (1982) [11] Nikolai A. Lukashevich: Elementary solutions of certain Painlevé equations, Diff. Urav., 1(6), (1965) [12] Nikolai A. Lukashevich: Theory of the fourth Painlevé equation, Diff. Urav., 3(5), (1967) [13] Kauo Okamoto: Studies on the Painlevé equations.3. 2nd and 4th Painlevé equations, PII and PIV, Math. Ann., 275(2), (1986) IJNS for contribution: editor@nonlinearscience.org.uk

15 Ayman Sakka, Uğurhan Muğan: First-order Second-degree Equations Related with 273 [14] David W. Albrecht, Eliabeth L. Mansfield and Alice E. Milne: Algorithms for special integrals of ordinary differential equations, J. Phys. A: Math. Gen., 29(5), (1996) [15] Valerii I. Gromak: Solutions of Painlevé s fifth problem, Diff. Urav., 12(4), (1976) [16] Ayman Sakka, Uğurhan Muğan: Second-order second-degree Painlevé equations related to Painlevé IV,V,VI equations, J. Phys. A: Math. Gen., 31(10), (1998) [17] Nikolai A. Lukashevich, A.I. Yablonskii: On a set of solutions of the sixth Painlevé equation, Diff. Urav., 3(3), (1967) [18] Uğurhan Muğan, Ayman Sakka: Schlesinger Transformations For Painlevé VI Equation J. Math. Phys., 36, (1995) [19] E.T. Whittaker, G.N. Watson: A Course of Modern Analysis, MacMillan: New York (1943) [20] Marta Maocco: Rational solutions of the Painlevé VI equation J. Phys. A: Math. Gen., 34(11), (2001) IJNS homepage: