Second-order second degree Painlevé equations related with Painlevé I, II, III equations

Transcription

1 J. Phys. A: Math. Gen Printed in the UK PII: S Second-order second degree Painlevé equations related with Painlevé I, II, III equations A Sakka and U Muğan Department of Mathematics, Bilkent University, Bilkent, Ankara, Turkey Received 30 December 1996 Abstract. The algorithmic method introduced by Fokas and Ablowit to investigate the transformation properties of Painlevé equations is used to obtain a one-to-one correspondence between the Painlevé I, II and III equations and certain second-order second degree equations of Painlevé type. 1. Introduction Second-order and first degree equations y = F,y,y 1.1 F is rational in y, algebraic in y and locally analytic in with the property that the only movable singularities are poles, that is, the Painlevé property, were classified at the turn of the century by Painlevé and his school [22, 15, 17]. Within the Möbius transformation, they found 50 such equations. Among all these equations, six of them are irreducible and define classical Painlevé transcendents, PI, PII,...,PVI. The remaining 44 equations are either solvable in terms of the known functions or can be transformed into one of the six equations. Besides the physical importance, the Painlevé equations possess a rich internal structure. Some of these properties can be summaried as follows. i For a certain choice of parameters, PII PVI admit a one-parameter family of solutions which are either rational or expressible in terms of the classical transcendental functions. For example, PII admits a one-parameter family of solutions expressible in terms of Airy functions [9]. ii There are transformations Bäcklund or Schlesinger associated with PII PVI, these transformations map the solution of a given Painlevé equation to the solution of the same equation but with different values of parameters [11, 19, 20]. iii PI PV can be obtained from PVI by the process of contraction [17]. It is possible to obtain the associated transformations for PII PV from the transformation for PVI. iv They can be obtained as the similarity reduction of the nonlinear partial differential equations solvable by inverse scattering transform IST. Since the work of Kowalevskaya that was the first connection between the integrability and the Painlevé property. v PI PVI can be considered as the isomonodromic conditions of a suitable linear system of ordinary differential equations with rational coefficients possessing both regular and irregular singularities [18]. Moreover, the initial value problem of PI PVI can be studied by using the inverse monodromy transform IMT [12, 13, 21]. address: mugan@fen.bilkent.edu.tr /97/ $19.50 c 1997 IOP Publishing Ltd 5159

2 5160 A Sakka and U Muğan The Riccati equation is the only example of the first-order first degree equation which has the Painlevé property. Before the work of Painlevé and his school Fuchs [14, 17] considered the equation of the form F,y,y = F is polynomial in y and y and locally analytic in, such that the movable branch points are absent, that is, the generaliation of the Riccati equation. Briot and Bouquet [17] considered the subcase of 1.2, that is, first-order binomial equations of degree m Z + : y m + F,y = F,y is a polynomial of degree at most 2m in y. It was found that there are six types of equation of the form 1.3. But, all these equations are either reducible to a linear equation or solvable by means of elliptic functions [17]. Second-order binomial-type equations of degree m 3 y m + F,y,y = F is polynomial in y and y and locally analytic in, was considered by Cosgrove [4]. It was found that there are nine classes. Only two of these classes have arbitrary degree m and the others have degree three, four and six. As in the case of first-order binomialtype equations, all these nine classes are solvable in terms of the first, second and fourth Painlevé transcendents, elliptic functions or by quadratures. Chay [3], Garnier [16] and Bureau [1] considered the third-order differential equations possessing the Painlevé property of the following form y = F,y,y,y 1.5 F is assumed to be rational in y,y,y and locally analytic in. But, in [1] the special form of F,y,y,y F,y,y,y = f 1, yy + f 2, yy 2 + f 3, yy + f 4, y 1.6 f k, y are polynomials in y of degree k with analytic coefficients in was considered. In this class no new Painlevé transcendents were discovered and all of them are solvable either in terms of the known functions or one of six Painlevé transcendents. Second-order second degree Painlevé-type equations of the following form y 2 = E,y,y y + F,y,y 1.7 E and F are assumed to rational in y, y and locally analytic in was the subject of the articles [2, 8]. In [2] the special case of 1.7 y = M,y,y + N,y,y 1.8 was considered, M and N are polynomials of degree two and four respectively in y, rational in y and locally analytic in. Also, in this classification, no new Painlevé transcendents were found. In [8], the special form, E = 0 and hence F is polynomial in y and y of 1.7 was considered and that six distinct class of equations were obtained by using the so-called α-method. These classes were denoted by SD I,...,SD VI and are solvable in terms of the classical Painlevé transcendents PI,...,PVI, elliptic functions or solutions of the linear equations. Second-order second degree equations of Painlevé type appear in physics [5 7]. Moreover, second degree equations are also important in determining transformation properties of the Painlevé equations [10, 11]. In [11], the aim was to develop an algorithmic method to investigate the transformation properties of the Painlevé equations. But, certain

3 Second degree Painlevé equations 5161 new second degree equations of Painlevé type related with PIII and PVI were also discussed. By using the same notation, the algorithm introduced in [11] can be summaried as follows. Let v be a solution of any of the fifty Painlevé equations, as listed by Gambier [15] and Ince [17], each of which takes the form v = P 1 v 2 + P 2 v + P P 1,P 2,P 3 are functions of v, and a set of parameters α. The transformation, i.e. Lie-point discrete symmetry which preserves the Painlevé property of 1.9 of the form u; ˆα =Fv;α, is the Möbius transformation u; ˆα= a 1v + a a 3 v + a 4 v, α solves 1.9 with the set of parameters α and u; ˆαsolves 1.9 with the set of parameters ˆα. Lie-point discrete symmetry 1.10 can be generalied by involving the v ; α, i.e. the transformation of the form u; ˆα=Fv ; α,v,α,. The only transformation which contains v linearly is the one involving the Riccati equation, i.e. u, ˆα = v + av 2 + bv + c 1.11 dv 2 + ev + f a, b, c, d, e, f are functions of only. The aim is to find a, b, c, d, e, f such that 1.11 defines a one-to-one invertible map between solutions v of 1.9 and solutions u of some second-order equation of the Painlevé type. Let J = dv 2 + ev + f Y = av 2 + bv + c 1.12 then differentiating 1.11 and using 1.9 to replace v and 1.11 to replace v, one obtains: Ju = [P 1 J 2 2dJv ej]u 2 + [ 2P 1 JY +P 2 J +2avJ +bj + 2dvY + ey d v 2 + e v + f ]u + [P 1 Y 2 P 2 Y +P 3 2avY by + a v 2 + b v + c ] There are two distinct cases. I Find a,...,f such that 1.13 reduces to a linear equation for v, Au,u,v+Bu,u,= Having determined a,...,f upon substitution of v = B/A into 1.11 one can obtain the equation for u, which will be one of the fifty Painlevé equations. II Find a,...,f such that 1.13 reduces to a quadratic equation for v, Au,u,v 2 +Bu,u,v+Cu,u,= Then 1.11 yields an equation for u which is quadratic in the second derivative. As mentioned before in [11] the aim is to obtain the transformation properties of PII PVI. Hence, the case I for PII PV, and case II for PVI was investigated. In this article, we investigate the transformation of type II to obtain the one-to-one correspondence between PI, PII, PIII and the second-order second degree Painlevé-type equations. Some of the second degree equations related with PI PIII were obtained in [2, 8] but most of them have not been considered in literature. Instead of having the transformation of the form 1.11 which is linear in v, one may use the appropriate transformations related to m v m + P j, vv m j = 0 m>1 j=1

4 5162 A Sakka and U Muğan P j, v is a polynomial in v, which satisfies the Fuchs theorem concerning the absence of movable critical points [14, 17]. This type of transformations and the transformations of type II for PIV PVI will be published else. 2. Painlevé I Let v be a solution of PI v = 6v Then, for PI equation 1.13 takes the form of [2d 2 u 2 4adu + 2a 2 ]v 3 + [du + 3deu 2 + d 3ae 3bdu a 3ab + 6]v 2 +[eu + 2df + e 2 u 2 + e 2af 2be 2cdu b b 2 2ac]v +[fu +ef u 2 + f bf ecu c bc + ] = Now, the aim is to choose a,b,...,f in such a way that 2.2 becomes a quadratic equation for v. There are two cases: either the coefficient of v 3 is ero or not. Case I. 2d 2 u 2 4adu + 2a 2 = 0. In this case the only possibility is a = d = 0, and one has to consider the two cases separately i e = 0 and ii e 0. Case I.i. e = 0. One can always absorb c and f in u by a proper Möbius transformation, and hence, without loss of generality, one sets c = 0, and f = 1. Then equation 2.2 takes the following form, 6v 2 + b b 2 v u bu = The procedure discussed in the introduction yields the following second-order second degree Painlevé-type equation for u [u + bu b + 2b 2 u b b 2 b bb b 3 2b 1] 2 = [u b bb b 3 ] 2 [24u 24bu + b b ] 2.4 and there exist the following one-to-one correspondence between solutions v and u u = v + bv v = u + bu 2b + b 2 u 2b u + b bb b 3 The change of variable u = pxyx + qx, = x fx=c 1 x+c 2 Rx = exp 5 b d = c 4/5 4 c 4 f 2 R 3/5 dx + c 5 px = 1 24 c 3/5 4 f 1 R 1/5 { [ qx = 1 24 c3/5 4 R 1/5 f R R 1 Ṙ 2 2c 1 5f Ṙ } 24c 4/5 4 f 2 R 2/5 x + c 3 f 2 dx + c 6 ] c j, j = 1, 2,...,6 are constants and Ṙ = dr, transforms 2.4 into the following form, dx ÿ 2 = [Axy + Bx] 2 [c 1 xẏ y + c 2 ẏ + c 3 ] 2.7

5 Second degree Painlevé equations 5163 Ax and Bx are given in terms of fxand Rx. Equation 2.7 was first obtained by Cosgrove and Scoufis [8] and labelled as SD-V.A. Case I.ii. e 0. Without loss of generality one can set b = 0 and e = 1. Hence, equations 1.11 and 2.2 become u = v + c v + f Av 2 + Bv + C = respectively, A = 6 B = u + u 2 C = f u + fu 2 a 1 u a 0 +6f a 1 =c f a 0 =c +6f 2 +. The discriminant of the second equation of 2.8 is = u + u f 2 24a 1 u + a If is not a complete square, that is, a 1 and a 0 are not both ero, then the first equation of 2.8 and v = fu + f u 2a 1 u fu 3 +a 1 u 2 a 1 2a f 2 u c u + uu u 3 12fu 12c define a one-to-one correspondence between a solution v of PI and a solution u of the following second degree equation [2a 1 u + a 0 u 2a 1 u 2 + R 2 uu Q 4 u] 2 = [2a 1 u a 1 u 2 + a 1 2a 0u + a fa 1] R 2 u = a 1 u 2 a 1 4a 0u a fa 1 Q 4 u = a 1 u 4 + a 1 u3 + a fa 1u f a 1 2f a 1 2a1 2 u f a 0 2f a f 2 a 1 2a 0 a 1. Note that if a 1 = f = 0, then y = usolves the following equation y 2yy = 1 2 y y 2 + y + 1 y y The second-order second degree Painlevé-type equation for y was first obtained by Bureau [2]. If is a complete square, that is a 0 = a 1 = 0 then u satisfies PX in [17 p 334]. Case II. 2d 2 u 2 4adu + 2a 2 0. In this case equation 2.2 can be written as v + hav 2 + Bv + C = A = 2d 2 u 2 4adu + 2a 2 B = du + d3e 2dhu 2 + d 3ae 3bd + 4adhu a 3ab + 2a 2 h + 6 C = e dhu + e 2 + 2df 3deh + 2d 2 h 2 u 2 +e hd 2af 2be 2cd + 3aeh + 3bdh 4adh 2 u b b 2 ha 2ac + 3abh 6h 2a 2 h

6 5164 A Sakka and U Muğan and h is a function of. f = he dh and b, c, d, e satisfy the following equations e 2dhh + bh ah 2 c = 0 c bc + = hb ha b 2 2ac + 3abh 6h 2a 2 h One has to distinguish two cases: i d = 0 and ii d 0. Case II.i. d = 0. When d = 0, without loss of generality, one can choose b = 0 and e = 1, then equations 1.11 and 2.2 take the following forms u = v + av 2 + c v + f Av 2 + Bv + C = respectively, A = 2a 2 B = 3au + a + 2a 2 f + 6 C = u + u 2 + af u + fa +2a 2 f +6+2ac Clearly, a should be different than ero, then 2.17 and f = h yield c = f af 2 f + 6f 2 + = Then, for these choices u satisfies the following second degree equation of Painlevé type [8a 3 u + 2a 2 3au 7a + 6a 2 f + 6u Q 3 u] 2 = [2a 2 u R 2 u] = 8a 2 u a 2 u 2 2aa 1 u a 0 Q 3 u = a 3 u 3 + a 2 5a + 6a 2 f + 42u 2 + a[2aa 1 2a 12a 2a 2 f 6 + a 0 ]u +aa 0 a 03a 2a 2 f 6 R 2 u = a 2 u 2 + 2aa + 2a 2 f 12u + 2a 3a 2 12a + 4ca a 1 = 3a + 2a 2 f + 18 a 0 = a 2 4a 2 f 3a 4a 2 4ac + 3a 2 f 2 + 6f+36. Case II.ii. d 0. Without loss of generality, one can set a = 0,d = 1. Then f = he h and the first equation of 2.17 gives e 2hh + bh c = If e = 2h, then f = h 2 and equations 1.11 and 2.2 become u = v + bv + c v + h 2 Av 2 + Bv + C = respectively, A = 2u 2 B = u + 4hu 2 3bu 6 C = hu + 2h 2 u 2 + a 1 3bhu + a 0 6h a 1 = 2h + bh c a 0 = b b 2 12h c bc + + ha 0 6h = 0. The discriminant of the second equation of 2.24 is 2.25 = u 3bu 6 2 8u 2 a 1 u + a

7 Second degree Painlevé equations 5165 If is not a complete square, that is, a 1 and a 0 are not both ero, then u satisfies the following second degree equation [4ua 1 u + a 0 u 32a 1 u + a 0 u 2 R 2 uu + Q 3 u] 2 = [3a 0 u 2a 1 ba 1u 2 2a 0 + ba 0 12a 1 u + 6a 0 ] R 2 u = 2[a 1 3ba 1u 2 + a 0 + 3ba 0 18a 1 u 6a 0 ] Q 3 u = 2[3ba 1 + 2a 1a 0 36h 3b 2 ]u 3 +[12a 1 + 3b2a 0 ba 0 + 4a 0 a 0 36h]u a 0 + 3ba 0u + 36a 0. If e 2h, then c = h + bh, f = he h and equations 1.11 and 2.2 become 2.28 v + bv + c u = v + hv + e h respectively, Av 2 + Bv + C = A = 2u 2 B = u + 3e 2hu 2 3bu 6 C = e hu + ee hu 2 + a 1 3be + 3bhu + a 0 6e h a 1 = e 2h + be 2h a 0 = b b 2 6e h + 6h 2 + = The discriminant of the second equation of 2.29 is = [u e 2hu 2 3bu 6] 2 8u 2 a 1 u + a If is not a complete square, that is, a 1 and a 0 are not both ero, then u satisfies the following second degree equation [4ua 1 u + a 0 u 7a 1 u + 3a 0 u 2 F 2 uu Q 5 u] 2 = [a 1 u 3a 0 u + R 3 u] F 2 u = [a 1 6ba 1 + 3a 0 e 2h]u 2 + a 0 + 3ba 0 24a 1 u 6a 0 Q 5 u = a 1 e 2h 2 u 5 [2e 2ha 1 ba 1 + 4a 2 1 e 2h2 a 0 ]u 4 [3b2a 1 7ba 1 + 4a 1 2a 0 + 6h 21e + 2e 2ha 0 + 5ba 0]u 3 [12a 1 3ba 1 + 3b2a 0 ba a 0 a 0 15e 6h]u 2 12a 0 + 3ba 0 3a 1 u 36a 0 R 3 u = a 1 e 2hu 3 + [2a 1 5ba 1 3a 0 e 2h]u 2 + 2a 0 + ba 0 18a 1 u 6a 0. If is a complete square, that is, a 1 = a 0 = 0, then w = 6/u solves PXXVIII [17, p 340]. 3. Painlevé II In this section we consider the equation PII. Let v be a solution of PII v = 2v 3 + v + α. 3.1

8 5166 A Sakka and U Muğan One finds that P 1 = P 2 = 0 and P 3 = 2v 3 + v + α by comparing 3.1 with 1.9. Then equation 1.13 becomes [2d 2 u 2 4adu + 2a 2 2]v 3 + [du + 3deu 2 + d 3ae 3bdu a 3ab]v 2 +[eu + 2df + e 2 u 2 + e 2af 2be 2cdu b b 2 2ac + ]v +[fu +ef u 2 + f bf ecu c bc + α] = To reduce 3.2 to a quadratic equation for v, there are two cases depending on whether the coefficient of v 3 is ero or not. Case I. 2d 2 u 2 4adu+2a 2 2 = 0. This implies that d = 0,a 2 =1. One has to consider the two cases: i e = 0, and ii e 0 separately. Case I.i. e = 0. With a proper Möbius transformation, one can choose c = 0, and f = 1. Then equations 1.11 and 3.2 take the form of u = v + av 2 + bv Av 2 + Bv + C = respectively, A = 3ab B = 2au + b 0 C = u bu α b 0 = b b When b 0, u satisfies the following second-order second degree Painlevé-type equation: [18b 2 u + 6b2au 2b 0 + 3u Q 3 u] 2 = [4u 2 2ab 0 + 6b 2 3u + 3bb 0 2b b 0 + 6aαb] = [12abu 4u 2 4ab + 2b 2 + u b0 2 12αab] Q 3 u = 8u a2b 2 + u 2 + 6bb 0 b b2 b 0 + 2b 0 + 6b 4 + 4aαbu abb 0 b 0 2ab ab2 0 6αbb αbb 2 +. When b = 0 the discriminant is a complete square, u is a solution of PXXXIV in [17, p 340]. Case I.ii. e 0. Without loss of generality, one can choose b = 0 and e = 1. Hence equations 1.11 and 3.2 become u = v + av 2 + c v + f Av 2 + Bv + C = respectively, A = 3au B = u + u 2 2af u + b 0 C = [fu +fu 2 +a 1 + af 2 u + a 0 + fb 0 ] 3.8 a 1 =f af 2 c a 0 = c + 2af c f + α b 0 = 2ac. The discriminant of the second equation of 3.7 is = u + u 2 + 4af u + 2ac aua 1 u + a If is not a complete square, that is, a 1 and a 0 are not both ero. Then u satisfies the following second degree equation [6ua 1 u + a 0 u 24a 1 u + a 0 u 2 + F 3 uu Q 5 u] 2 = [22a 1 u a 0 u R 3 u]

9 Second degree Painlevé equations 5167 F 3 u = 2a 1 u 3 3a 1 11a af a 1 u 2 3a 0 20af a a 1 b 0 u a 0 b 0 Q 5 u = 2a 1 u 5 + 3a 1 a af a 1 u 4 + [3a af a f 2 4ac a 1 8af a 0 ]u 3 + [b 0 3a af a af a aα + 1a 1 2a 0 20f ac ]u 2 + [b 0 3a 0 + 4af a a 0 2aα a 1 b 2 0 ]u a 0b 2 0 R 3 u = 2a 1 u 3 3a 1 + 4af a 1 5a 0 u 2 3a 0 8af a 0 + 4a 1 b 0 u a 0 b 0. If is a complete square, that is, a 1 = a 0 = 0, then u satisfies PXXXV [17, p 340]. Case II. 2d 2 u 2 4adu + 2a In this case 3.2 can be written as v + hav 2 + Bv + C = A = 2d 2 u 2 4adu + 2a 2 2 B = du + d3e 2dhu 2 + d 3ae 3bd + 4adhu a 3ab + 2a 2 h 2h C = e dhu + e 2 + 2df 3deh + 2d 2 h 2 u 2 +e hd 2af 2be 2cd + 3aeh + 3bdh 4adh 2 u b b 2 ha 2ac + 3abh + 2h 2 2a 2 h h is a function of, f = he dh, and b, c, d, e satisfy the following equations e 2dhh ah 2 + bh c = 0 c bc + α = hb ha b 2 2ac + 3abh + 2h 2 2a 2 h There are two distinct cases: i d = 0 and ii d 0. Case II.i. d = 0. With a proper Möbius transformation, one can set b = 0,e = 1. Therefore equations 1.11 and 3.2 become u = v + av 2 + c v + f Av 2 + Bv + C = respectively, A = 2a 2 1 B = 3au + b 0 C = u + u 2 + af u + c 0 b 0 = a + 2fa 2 1 c 0 = fb 0 +2ac Then h = f and equation 3.14 imply c = f af 2 f = 2f 3 + f α When A 0, u satisfies the following second-order second degree equation of Painlevé type: [8a u + 2aa 2 13au 7a + 6fa 2 6fu Q 3 u] 2 = [2aa 2 1u R 2 u]

10 5168 A Sakka and U Muğan = [8a 2 1u a 2 + 8u 2 2aa 1 u a 0 ] Q 3 u = a 2 + 8a 2 + 2u 3 + a[5a 2 14a + 6fa 2 1a 2 + 4]u 2 +[2aa 2 1a 1 + 2af a 1 2a 1 2a 2 + 1a + a 0 a 2 + 2]u +a 2 1a 0 + 2af a 0 3aa a 0 R 2 u = aa 2 10u 2 + 2[a 2 + 2a + 2fa 2 1a 2 3]u 2a 2 1[a + 2ca a] + 3aa 2 a 1 = 3a + 2fa 2 1 a 0 =a 4fa 2 1a 12f 2 a a 2 12ac Case II.ii. d 0. Without loss of generality we set a = 0,d =1. Then the first equation of 3.14 gives e 2hh + bh c = If e = 2h, then f = h 2 and equations 1.11 and 3.2 become u = v + bv + c v + h 2 Av 2 + Bv + C = respectively, A = 2u 2 1 B = u +4hu 2 3bu + 2h C = hu + 2h 2 u 2 + a 1 3bhu + a 0 + 4h 2 a 1 = 2h + bh c a 0 = b b 2 + 6h c bc + α + ha 0 + 4h 2 = 0. The discriminant of the second equation of 3.21 is = u 3bu + 6h 2 8u 2 1a 1 u + a If is not a complete square, that is, a 1 and a 0 are not both ero, then u satisfies the following second degree equation [4u 2 1a 1 u + a 0 u 32a 1 u 2 + a 0 u a 1 u 2 2F 3 uu + Q 4 u] 2 = [3a 0 u + a 1 u R 3 u] F 3 u = a 1 3ba 1u 3 + a 0 + 3ba ha 1 u 2 a 1 6ha 0u a 0 + 6ba ha 1 Q 4 u = 2[3ba 1 2ba 1 + 2a 1 a h 2 + 3]u 4 [12ha 1 + 2ba 1 4a 1 a 1 + 6c 3b2a 0 ba 0 4a 0 a h 2 + 3]u 3 3[b2a 1 7ba 1 + 4a 1 6h ha 0 + 4a 05bh 2c]u 2 +2[6ha 1 ba a 1 a 1 + 6c 3ba 0 2a 0 a h b 2 ]u + 4[3ha 0 a 0a 1 + 6c 6bh + 9h 2 a 1 ] R 3 u = 2a 1 ba 1u 3 + 2a 0 + ba ha 1 u 2 2a 1 + ba 1 6ha 0 u 2a 0 + 2ba 0 + 3ha 1. If e 2h, then c = h + bh and f = he h and equations 1.11 and 3.2 become u = v + bv + c v + hv + e h Av 2 + Bv + C =

11 respectively, Second degree Painlevé equations 5169 A = 2u 2 1 B = u +3e 2hu 2 3bu 6 C = e hu + ee hu 2 + a 1 3be + 3bhu + a 0 2ee h a 1 = g + bg a 0 = b b e 2 2eh + 2h 2 h 2h 3 h + α = 0 g = e 2h The discriminant of the second equation of 3.26 is = [u gu 2 3bu + 22e h] 2 8u 2 1a 1 u + a If is not a complete square, that is, a 1 and a 0 are not both ero, then by using the linear transformation u = py + q, p and q are solutions of the following equations p 2gpq 3bp = 0 q gq 2 3bq + 22e h = y satisfies the following second degree equation [F 3 yy p 4 Q 2 yy 2 p 3 R 3 yy p 4 S 6 y] 2 = [T 2 yy + G 4 y] 2 [p 2 y pgy 2 2 2F 3 y] 3.30 F 3 y = 4p 2 y 2 + 2pqy + q 2 1c 1 y + c 0 Q 2 y = 7pf 3 y 2 + 3pf 2 + 5qf 3 y + 3pf 1 5qf 2 2qc 1 R 3 y = [c 1 + g3p3 f 2 5qc 1 ]y 3 + [ppf 2 + 2p f 2 + g4p 3 f 1 + 5qc 0 ]y 2 +[ppf 1 + 2p f 1 + 4gp 3 f 0 ]y + ppf 0 + 2p f 0 S 6 y = g 2 p 3 f 3 y 6 [2p 2 gf 3 2g f 3 g 2 3qc 1 pc 0 + 4p 3 f3 2 ]y5 [2p 2 gf 2 2g f 2 + 4f 3 3qc 1 pc 0 g 2 2p 3 f 1 5qc 0 ]y 4 [2p 2 gf 1 2g f 1 + 4f 2 3qc 1 pc 0 + 4f 3 p 3 f 1 5qc 0 ]y 3 [2p 2 gf 0 2g f 0 + 4f 2 p 3 f 1 5qc qc 1 f 1 + 5p 3 f 0 f 3 ]y 2 4[2p 3 f1 2 5qc 0f 1 + f 0 3qc 1 pc 0 ]y 4f 0 2p 3 f 1 5qc 0 T 2 y = p[p 3 f 3 y 2 + 5qc 1 3pc 0 y + 11qc 0 4p 3 f 1 ] G 4 y = gp 5 f 3 y 4 + p 4 [2f 3 f 32gq b + 3gpf 2 ]y p 2 [2p 2 f 2 + 8f 3qp q p + 2gp 3 f 1 +3gqc 0 + 4p 2 f 2 qg + b]y 2 + [4pqc 0 4p q + pq c 0 +2q 2 1c 1 4qq c 1 + 4gp 5 f 0 4p 4 f 1 gq + b]y +2[q 2 1c 0 2qq c 0 + 2p 4 f 0 gq + b] c 1 = pa 1 c 0 = qa 1 + a 0 f 3 = c 1 p 2 f 2 = 2qc 1 + pc 0 p 3 f 1 = c 1q pqc 0 p 4 f 0 = c 0q 2 1 p 4. If is a complete square, then w = u+1 2 is a solution of PXLV [17, p 342].

12 5170 A Sakka and U Muğan 4. Painlevé III Let v be a solution of PIII v = v 2 Then, equation 1.13 takes the form of: [d 2 u 2 2adu + a 2 γ]v 4 + v v + γv3 + α v2 + β + δ v. 4.1 a + a + α [ + fu ef u 2 + [ du + deu 2 + d ae bd + d u ] [ ab v 3 + eu + e + e u b + b ] v 2 f + f + bf + ce u c + c + β ] + bc v [f 2 u 2 2cf u + c 2 + δ] = 0. There are three distinct cases to reduce 4.2 to a quadratic equation in v. Case I. If d 2 u 2 2adu + a 2 γ 0. Then 4.2 can be written as v 2 + hv + gav 2 + Bv + C = A = d 2 u 2 2adu + a 2 γ B = du + de dhu 2 + d + d bd ae + 2adh u a + a + α ab ha 2 γ 4.4 C =e dhu d[dg + he dh]u 2 + [e + e d h + d ] bd ae + 2adh + 2adg u b + b + h a + a + α ab + h 2 a 2 γ ga 2 γ and a, b, c, d, e, f, g, h satisfy ge dh = 0 he dh = f dg dghe dh = f 2 d 2 g 2 dh[dg + he dh] dge dh = ef g [e + e d h + d ] bd ae + 2adh + 2adg = 2cf g [b + b h a + a + α h ] ab h 2 a 2 γ+ga 2 γ =c 2 +δ [e + e h d + d bd ae + 2adh + 2adg h [b + b h a + a + α +g d + d bd ae + 2adh = f + f +g a + a+α ab ] + bf + ce ] ab h 2 a 2 γ+2ga 2 γ = c + c + β + bc

13 Second degree Painlevé equations 5171 Hence there are two subcases: i e dh and ii e = dh. Case I.i. e dh. Then 4.5 implies f = g = h = 0, and c 2 c + β + δ = 0 ce = 0 + bc = Equation 4.6 gives c = β = δ = 0. In the case of β = δ = 0, the transformation [11] transforms PIII into v w = v + γ 1/2 v = w γ 1/2 w + α + γ 1/2 w = 1 ww 4.7 which has the first integral w = 1 2 w2 + w + k, k =constant. There are two subcases which should be considered separately: 1 d = 0 and 2 d 0. Case I.i-1: d = 0. Then without loss of generality one can set b = 0 and e = 1. With these choices equations 1.11 and 4.2 become u = v + av 2 v Av 2 + Bv + C = respectively, A = a 2 γ B =au + a + a + α C = u + u. 4.9 Note that A 0, thus the second-order second degree Painlevé-type equation related with PIII is: [2 2 a 2 γ 2 u F 1 uu Q 3 u] 2 = [aa 2 γu R 2 u] = [4 2 a 2 γu 2 a 2 u 2 2aa a 2 + αa + 2γu a + a + α 2 ] F 1 u = a 2 γ[a 2 4γu+aa 5a 2 3αa + 2γ] Q 3 u = γ 2 a 2 u 3 +γaa + a 4 + αa 3 2γa 2 +2αγa + 4γ 2 u 2 +[ 2 aa 2 γa a 2 a a 2 + αa 2 + γ a a +2αa + γ 2 +γ2a 2 +4αa + α 2 ]u + a + 1 a + 1 α [ 2 a 2 γa 2 aa 2 + a 2 γa +αa 2 + γ+α 2 + γa] R 2 u = γau 2 a 3 αa 2 3γa γαu+a 2 γ a + 2a a + a + α aa 1 αa 1 γ As a special case of 4.10, if a = 0,γ 0 then the transformation u = e x y + α2 4γ x = e x 4.12

14 5172 A Sakka and U Muğan transforms 4.10 to the following second degree equation ÿ = 2y + a 0 ẏ + 2b 1 y + b 0 ẏ 4.13 a 0 = 1 α2 4γ x b2 1 = α2 γ b 0 = b 1 a 0. Equation 4.13 was also obtained in [2]. Case I.i-2. d 0. With a proper Möbius transformation one can set a = 0 and d = 1. Hence, equations 1.11 and 4.2 respectively become u = v + bv v 2 + ev Av 2 + Bv + C = A = u 2 γ B = u + eu 2 b 1 u 1 α C = eu + g 1 be + 1 e u + g 0 1 αe + γe2 g 1 =e +be g 0 = b + b αe + γe The discriminant of the second equation of 4.14 is = [ u eu 2 b 1 u + 2γe 1 ] 2 α 4u 2 γ g 1 u + g If g 1 and g 0 are not both ero, then y = u q, p and q are solutions of the p following equations p 2epq p b 1 = 0 q eq 2 q b 1 + 2γe 1 α = satisfies the following second degree Painlevé equation [pf 3 yy pq 2 yy 2 R 4 yy + p 2 S 5 y] 2 = p 2 [T 2 yy + G 3 y] 2 [y epy 2 2 2p 2 F 3 y] 4.18

15 Second degree Painlevé equations 5173 F 3 y = 2c 1 y + c 0 y 2 + 2a 1 y + a 0 Q 2 y = 3c 1 y 2 + c 0 + 5a 1 c 1 y + a 1 c 0 + 2a 0 c 1 R 4 y = ep 2 c 1 y 4 + [pc 1 + 2p c 1 + 2ep 2 c 0 a 1 c 1 ]y 3 +[pf 2 + 2p f 2 + ep 2 8a 1 c 0 + a 0 c 1 ]y 2 +[pf 1 + 2p f 1 + 4ep 2 f 0 ]y + pf 0 + 2p f 0 S 5 y = [ec 1 2e c 1 + pe 2 c 0 a 1 c 1 ]y 5 +[ef 2 2e f 2 4c 1 c 0 a 1 c 1 + pe 2 a 1 c 0 a 0 c 1 ]y 4 +[ef 1 2e f 1 4pc 1 a 1 c 0 a 0 c 1 4pf 2 c 0 a 1 c 1 ]y 3 +[ef 0 2e f 0 4pf 2 a 1 c 0 a 0 c 1 4pf 1 c 0 a 1 c 1 ]y 2 4p[f 1 a 1 c 0 a 0 c 1 + f 0 c 0 a 1 c 1 ]y 4pf 0 a 1 c 0 a 0 c T 2 y = c 1 y 2 c 0 3a 1 c 1 y + 2a 0 c 1 a 1 c 0 G 3 y = [c 1 + c 1eq 2b + epf 2 ]y 3 +[f 2 4c 1a 1 + ep3a 1c 0 + a 0 c 1 + 2f 2 eq b]y 2 +[f 1 2c 1a 0 4c 0a 1 + 2epf 0 + 2f 1 eq b]y +f 0 2c 0a 0 + 2f 0eq b a 1 = 1 p q a 0 = 1 p 2q2 γ c 1 = 1 p g 1 c 0 = 1 p 2qg 1 + g 0 f 2 = 2a 1 c 1 + c 0 f 1 = 2a 1 c 0 + a 0 c 1 f 0 = a 0 c 0. If g 1 = g 0 = 0 and γ 0 then w = u γ u+ is a solution of PXL in [17, p 341]. If γ g 1 = g 0 = 0 and γ = 0 then w = α is a solution of PXVI in [17, p 335]. It should be u noted that both PXL and PXVI have first integrals [17]. Case I.ii: e = dh. Then the second equation of 4.5 gives f = dg and hence d 0. Without loss of generality one can take a = 0 and d = 1. Thus 4.4 and 4.5 yield respectively A = u 2 γ B = u b 1 u + γe 1 α C = f u 2 + g 1 u + g g 1 = e + be g 0 = b + 1 b α e + γe2 γf and Thus 1.11 becomes fg 1 +2c = 0 eg 1 + c + f + 2bf = 0 fg 0 = c 2 +δ eg 0 = f γe 1 α +c + c+β +bc = u = v + bv + c v 2 + ev + f. 4.22

16 5174 A Sakka and U Muğan The discriminant of Av 2 + Bv + C = 0is = [ u b 1 u+γe 1 ] 2 α +4u 2 γ f u 2 + g 1 u + g If is not a complete square, then one obtains the following the second-order second degree equation related with PIII [2u 2 γ cu g 0 u ucu g 0 u 2 + F 3 uu + Q 5 u] 2 = [ucu g 0 u + R 4 u] F 3 u = u 2 γ3cb 1 u + g 0 2b 1g 0 + cb 0 Q 5 u = 8δu 5 + 2cb 1 + b g 0u 4 2g 0 b 1 2cb 0 b 1g 0 + 4g2 0 + g 0b 2 1 4cb 0b 1 12γδu 3 2g 0 b 0 +2γcb 1 b 0g 0 +16γcg 0 +γcb g 0b 0 b 1 2cb 2 0 u2 +2γg 0 b 1 2cγ b 0 γb 1g 0 +4γg2 0 2cγ b 0b 1 + 4δγ 2 g 0 b0 2 u +γ2g 0 b 0 +b 0g 0 +4cγg 0 cb0 2 R 4 u = 2ceu 4 2eg 0 + bcu 3 + g 0 + 2bg 0 + b 1 g 0 2γ ceu 2 +2cb + 2γeg 0 γcb 1 b 0 g 0 u γg 0 +2bg 0 + cb 0 b 1 = b 1 b 0 =γe 1 α 4.25 if f = 0. If f 0 then by using the linear transformation u = py + q, p and q are given as follows p = 1 exp bs ds 1 [ q = p γ es α ] ds 4.26 ps s y solves the following second degree equation of Painlevé type [F 4 yy Q 3 yy 2 R 4 yy + F 4 ys 3 y] 2 = p 2 [T 2 yy G 5 y] 2 [y 2 + 2pF 4 y] 4.27

17 Second degree Painlevé equations 5175 F 4 y = 2py 2 + 2a 1 y + a 0 fy 2 + 2c 1 y + c 0 Q 3 y = p2fy 3 + 3f 3 y 2 + f 2 y + f 1 R 4 y = pf + 2p fy 4 +pf 3 + 2p f 3 y 3 +pf 2 + 2p f 2 y 2 + pf 1 + 2p f 1 y + pf 0 + 2p f 0 S 3 y = 2p 3 [2fy 3 + 3f 3 y 2 + f 2 y + f 1 ] T 2 y = c p y2 + f 2 2c 0 4a 1 c 1 y + f 1 2a 1 c 0 G 5 y = 2epfy 5 [f 2feq b 4epf 3 ]y 4 2[f 3 2fa 1 2f 3eq b epf 2 ]y 3 [f 2 8c 1a 1 2fa 0 2f 2eq b 4epf 1 ]y 2 2[f 1 2c 0a 1 2c 1a 0 2f 1eq b epf 0 ]y [f 0 2c 0a 0 2f 0eq b] a 0 = 1 p 2 q2 γ a 1 = q c 1 = 1 p p f q c c 0 = 1 p 2 f p2 c1 2 + δ f 3 = c 1 + fa 1 f 2 =c 0 +4a 1 c 1 +fa 0 f 1 =a 1 c 0 +a 0 c 1 f 0 =a 0 c If is a complete square, that is, C = 0, then ec = 0. If e 0 then this case reduces to the case I.i-2 with a 1 = a 0 = 0. If e = 0 and γ = 0, then w = u is a solution of PIII. If e = 0 and γ 0, then wx = u γ u+ γ, 2 = 2x, is a solution of PV with δ = 0 [11]. Case II. d 2 u 2 2adu+a 2 γ = 0 du +deu 2 +d ae bd + d u a + a+α ab 0. Then 4.2 can be written as: v + f Av 2 + Bv + C = A = au + a + α [ B = u + af + 1 u + f ] a + α C = fu 2 f af 2 + c + c + c + β and + a + α f 2 a 2 γ = 0 ff af 2 c = 0 f c + c + β + a + α f 2 = c 2 + δ. If f 0 then 4.31 implies c = f af 2 c + c + β The discriminant of Av 2 + Bv + C = 0 reads [ = u + af + 1 a + α u + f a + α f 2 = 1 f c2 + δ ] 2 4 au + a + α fu 2 2cu + c2 + δ f 4.33

18 5176 A Sakka and U Muğan and equation 1.11 becomes u = v + av 2 + c v + f Let y = u q, p and q are given as p p = 1 [ ] exp a fsds q = a + α [ p 1 ] a 0 a q = α fsds a = 0. Then y is a solution of the following second degree Painlevé-type equation [F 3 yy Q 2 yy 2 R 3 yy F 3 ys 2 y] 2 = [Q 2 yy T 4 y] 2 [y 2 2pF 3 y] 4.37 F 3 y = 2ay + σ fy 2 + 2c 1 y + c 0 Q 2 y = ay + σ fy + c 1 R 3 y = a f + p p f y 3 2 p + f + p f 2 y 2 1 p + f + p f 0 p 2 y + σ c + p c 0 S 2 y = 2p[2af y 2 + 3ac 1 + σf y + ac 0 + σc 1 ] [ ] T 4 y = 2apfy 4 a f + p p f 2pf 2 + aqf [ σ f + p p f 1 p + 2a c + p c 1 [ 1 p 0 p 2σ c + + a c + [ 0 p σ c + p c 0 p c 1 ] 2σqc 0 p c 0 σ = a +α c 1 = 1 p f q c c 0 = 1 p 2 f p2 c δ f 2 = σf + 2ac 1 f 1 = ac 0 + 2σc 1. y 3 ] 2pf 1 + qf 2 y 2 ] 2σpc 0 + qf 1 y If f = 0, then w = u is a solution of the equation w + ρw + σ 2 w 2 = 2 2 w 2 + ρw 2 + σw+τ ρ = 4γ 1/2 δ 1/2 σ = 4[α δ 1/2 βγ 1/2 ] τ = 4α + γ 1/2 [β + δ 1/2 ] Equation 4.39 was first obtained in [11] and also in [8] which was denoted as SD-III. When f = c = 0 then is a complete square, and w = u is a solution of 4.7.

19 Second degree Painlevé equations 5177 Case III. d 2 u 2 2adu + a 2 γ = 0 and du + deu 2 + d ae bd + d u a + a+α ab = 0. Then 1.11 and 4.2 become u = v + av 2 + bv + c Av 2 + Bv + C = ev + f respectively, A = eu + e + 1 e u b + 1 b B = fu ef u 2 + f + 1 f + bf + ce u c + c + β + bc and C = f 2 u 2 2cf u + c 2 + δ. a 2 γ = 0 ae = 0 The discrete Lie-point symmetry of PIII [11] a + α 4.41 ab = v = 1 v ᾱ = β β = α γ = δ δ= γ 4.43 transforms this case to the case I.i. References [1] Bureau F 1964 Ann. di Math 66 1 [2] Bureau F 1972 Ann. di Math [3] Chay J 1911 Acta Math [4] Cosgrove C M 1993 Stud. Appl. Math [5] Cosgrove C M 1977 J. Phys. A: Math. Gen [6] Cosgrove C M 1977 J. Phys. A: Math. Gen [7] Cosgrove C M 1978 J. Phys. A: Math. Gen [8] Cosgrove C M and Scoufis G 1993 Stud. Appl. Math [9] Erugin N P 1958 Dokl. Akad. Nauk 2 [10] Fokas A S and Yortsos Y C 1981 Lett. Nuovo Cimento [11] Fokas A S and Ablowit M J 1982 J. Math. Phys [12] Fokas A S and Zhou X 1992 Commun. Math. Phys [13] Fokas A S, Muğan U and Zhou X 1992 Inverse Problems [14] Fuchs R 1907 Math. Ann [15] Gambier B 1909 Acta. Math [16] Garnier R 1912 Ann. Sci. Ec. Norm. Super [17] Ince E L 1956 Ordinary Differential Equations New York: Dover [18] Jimbo M and Miwa T 1981 Physica 2D 407 Jimbo M and Miwa T 1981 Physica 4D 47 Ueno K 1980 Proc. Japan Acad. A Jimbo M, Miwa T and Ueno K 1981 Physica 2D 306 Jimbo M 1979 Prog. Theor. Phys [19] Muğan U and Fokas A S 1992 J. Math. Phys [20] Muğan U and Sakka A 1995 J. Math. Phys [21] Muğan U and Sakka A 1995 J. Phys. A: Math. Gen [22] Painlevé P 1900 Bull. Soc. Math. France Painlevé P 1912 Acta. Math. 25 1