Second-order second-degree Painlevé equations related to Painlevé IV,V,VI equations

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1 J. Phys. A: Math. Gen ) Printed in the UK PII: S )8567- Second-order second-degree Painlevé equations related to Painlevé IV,V,VI equations A Sakka and U Muğan Department of Mathematics, Bilkent University, Bilkent, Ankara, Turkey Received 4 July 1997 Abstract. The algorithmic method introduced by Fokas and Ablowitz to investigate the transformation properties of Painlevé equations is used to obtain one-to-one correspondence between the Painlevé IV, V and VI equations and the second-order second-degree equations of Painlevé type. 1. Introduction Painlevé and his school addressed a question raised by Picard concerning second-order first-degree ordinary differential equations of the form y = Fz,y,y ) 1.1) F is rational in y, algebraic in y and locally analytic in z, which have the property that singularities other than poles of any of the solutions are fixed, 15, 17. This property is known as the Painlevé property. Within the Möbius transformation, Painlevé and his colleagues found that there are 50 canonical equations of the form 1.1). Among these equations six are irreducible and define classical Painlevé transendents PI PVI. These may be regarded as nonlinear counterparts of some of the classical special functions. For example, PIII has solutions which have similar properties to the Bessel functions. The solutions of the 11 equations of the remaining 44 equations can be expressed in terms of the solutions of the Painlevé equations PI, PII, or PIV and 33 equations are solvable in terms of the linear equations of order two or three, or are solvable in terms of the elliptic functions. Although the Painlevé equations were discovered from strictly mathematical considerations they have appeared in many physical problems. Besides their physical importance, the Painlevé equations possess a rich internal structure. Some of these properties can be summarized as follows. i) For a certain choice of parameters PII PVI admit oneparameter families of solutions which are either rational or expressible in terms of the classical transcendental functions. For example, PVI admits a one-parameter family of solutions expressible in terms of hypergeometric 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, 0. 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 address: mugan@fen.bilkent.edu.tr /98/ $19.50 c 1998 IOP Publishing Ltd 471

2 47 A Sakka and U Muğan the transformation for PVI. iv) They can be obtained as the similarity reduction of the nonlinear partial differential equations solvable by the inverse scattering transform IST). v) PI PVI can be considered as the isomonodromic conditions of 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) 1, 13, 1. The Riccati equation is the only example for the first-order first-degree equation which has the Painlevé property. Before the work of Painlevé and his school, Fuchs 14, 17 considered equations of the form Fz,y,y ) = 0 1.) F is polynomial in y and y, and locally analytic in z, such that the movable branch points are absent, that is, the generalization of the Riccati equation. Briot and Bouquet 17 considered a subcase of 1.), that is, first-order binomial equations of degree m Z + : y ) m + Fz,y) = 0 1.3) Fz,y) is a polynomial of degree at most m in y. It was found that there are six types of equation of the form 1.3). However, 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 + Fz,y,y ) = 0 1.4) F is polynomial in y and y and locally analytic in z, were 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. Chazy 3, Garnier 16 and Bureau 1 considered third-order differential equations possessing the Painlevé property of the following form y = Fz,y,y,y ) 1.5) F is assumed to be rational in y,y,y and locally analytic in z. However, in 1 the special form of Fz,y,y,y ) Fz,y,y,y ) = f 1 z, y)y + f z, y)y ) + f 3 z, y)y + f 4 z, y) 1.6) f k z, y) are polynomials in y of degree k with analytic coefficients in z 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 ) = Ez,y,y )y + Fz,y,y ) 1.7) E and F are assumed to rational in y and y and locally analytic in z were the subject of, 8. In the special case of 1.7) y = Mz,y,y ) + Nz,y,y ) 1.8) was considered, M and N are polynomials of degree and 4 respectively in y, rational in y and locally analytic in z. Also, in this classification, no new Painlevé transcendents were found. In 8, the special form, E = 0 and hence F polynomial in y and y of 1.7) was considered and six distinct classes of equations were obtained by using the so-called α-method. These classes were denoted by SD-I,...,SD-VI and all are

3 Second degree Painlevé equations 473 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. However, certain new second-degree equations of Painlevé type related to PIII and PVI were also discussed. By using the same notation, the algorithm introduced in 11 can be summarized as follows: let vz) be a solution of any of the 50 Painlevé equations, as listed by Gambier 15 and Ince 17, each of which takes the form v = P 1 v ) + P v + P 3 1.9) P 1,P,P 3 are functions of v, z 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 uz; ˆα) =Fvz;α), z) is the Mobius transformation uz; ˆα)= a 1z)v + a z) 1.10) a 3 z)v + a 4 z) vz, α) solves 1.9) with the set of parameters α, and uz; ˆα)solves 1.9) with the set of parameters ˆα. Lie-point discrete symmetry 1.10) can be generalized by involving the v z; α), i.e. the transformation of the form uz; ˆα)=Fv z; α),vz,α),z). The only transformation which contains v linearly is the one involving the Riccati equation, i.e. uz, ˆα) = v + av + bv + c 1.11) dv + ev + f a, b, c, d, e, f are functions of z only. The aim is to find a, b, c, d, e, f such that 1.11) define 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 + ev + f Y = av + bv + c 1.1) then differentiating 1.11) and using 1.9) to replace v and 1.11) to replace v, one obtains, Ju = P 1 J djv eju + P 1 JY +P J +avj + bj +dvy + ey d v + e v + f )u + P 1 Y P Y +P 3 avy by + a v + b v + c. 1.13) There are two distinct cases. 1) Find a,...,f such that 1.13) reduce to a linear equation for v, Au,u,z)v+Bu,u,z)= ) Having determined a,...,f upon substitution of v = B/A in 1.11) one can obtain the equation for u, which will be one of the 50 Painlevé equations. ) Find a,...,f such that 1.13) reduces to a quadratic equation for v, Au,u,z)v +Bu,u,z)v+Cu,u,z)= ) 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, case 1 for PII PV, and case for PVI was investigated. In this article, we investigate the transformation of type to obtain the one-to-one correspondence between PIV, PV, PVI and the second-order second-degree Painlevé-type

4 474 A Sakka and U Muğan equations. Similar work has been carried out for PI, PII, PIII in 3. Some of the seconddegree equations related to PIV PVI has been obtained in, 8 without giving the relation between the Painlevé equations but many of them have not been considered in the 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 z, v)v ) m j m>1 1.16) j=1 P j z, v) is a polynomial in v, which satisfies the Fuchs theorem concerning the absence of movable critical points 14, 17. This type of transformation yields the relation between the Painlevé equations PI PVI and higher-order higher-degree Painlevétype equations. Throughout this article denotes the derivative with respect to z and. denotes the derivative with respect to x.. Painlevé IV Let vz) be a solution of the PIV equation, v = 1 v v ) + 3 v3 + 4zv + z α)v + β v..1) Then, for PIV the equation 1.13) takes the form A 4 v 4 + A 3 v 3 + A v + A 1 v + A 0 = 0.) A 4 = 3d u adu + a 1 A 3 = du + deu + d ae bd)u a ab + 4z) A = eu + df + e )u + e af be cd)u b b ac + 4z 4α) A 1 = f u + f u c ) A 0 = f u cf u + c + β)..3) Now, the aim is to choose a,b,...,f so that.) becomes a quadratic equation for v. There are three cases: 1) A 4 = A 3 = 0, ) A 4 = 0,A 3 0 and 3) A 4 0. Case 1. In this case the only possibility is e = d = 0,a = 1 and b = az. One can always absorb c and f in u by a proper Möbius transformation, and hence, without loss of generality, one can set c = 0, and f = 1. Then equation.) takes the following form, au + a α)v u v + u + β = 0..4) Following the procedure discussed in the introduction yields the following second-order second-degree Painlevé-type equation for uz) u 3au 4a α)u aβ = 4z u au + a α)u + β).5) and there exits the following one-to-one correspondence between solutions vz) of PVI and solutions uz) of the equation.5) u = v + av + azv v = u azu 3au 4a α)u aβ..6) 4azau + a α)

5 Second degree Painlevé equations 475 The change of variable uz) = 4ayx), x = a z transforms equation.5) into the equation ÿ 1 ) Q 3 y) = x ẏ Q 3 y) y.7) Q 3 y) := 4y 3 + a α)y + 1 βy βa α). Equation.7) was first obtained by Bureau, equation 19., p 07 without giving the relation to PV and the special case, α = a, of.7) was solved in terms of the first Painlevé transcendent. However, we have not been able to recover this relation. Case. In this case d = 0, a =1 and to reduce the equation.) to a quadratic equation for v one should take e 0. Since one can always absorb b and e in u by a proper Möbious transformation, without loss of generality one may take b = 0, e = 1. Now equation.) can be written as v + f )Av + Bv + C) = 0.8) A = 4au + z) B = u + u af u + ac + 4zf z + α).9) C = fu f af )u + c + fac+4zf z + α) and c, f satisfy the following equations ff af c) = 0 fc +fac+4zf z + α) = c + β..10) If f 0, then equations.10) give c = f af f = 1 f f ) + 3 f 3 4zf + z α)f + β f..11) Let µ = β) 1/, and uz) = ξz)yx) + ηz), x = ζz), ηz) = 1 f f af + µ) z ξz) = exp {ηẑ) + af ẑ)} dẑ.1) ζz) = 1 z ξẑ)dẑ. Then one obtains the following second-order second-degree equation of Painlevé type for yx) 4y + a 0 )ÿ yẏ) ẏ y ) P y)ẏ y ) Q 3 y) = = ẏ y R y) ẏ y ) S 3 y).13) P y) = p 3p y + p a 0 ah pq)y p a0 a 0 7ah + 8pq) + α aµ a Q 3 y) = 16p y + a 0 )hy + h + µ)y + µ R y) = p 5p y + p3pa 0 + q)y + p a0 +4pq ah)a 0 α + aµ + a S 3 y) = 16ap yy + a 0 )hy + µ)..14)

6 476 A Sakka and U Muğan The coefficients in equations.13) and.14) are given as follows h = µx + ν µ + ν 0 ν=constant px) := ξz) qx) := ηz) a 0 := 1 ξz) ηz) + az..15) The functions px), qx), and a 0 x) satisfy the following equations pṗ pq + ah) = 0 p q + p a0 pq + h)a 0 aµ α) = 0 p a 0 + a0 ) aha.16) 0 α + aµ + a = 0. The one-to-one correspondence between solutions vz) of the PIV and solutions yx) of the equation.13) is given as follows y = v + av ηv µ 4ay + a 0 )v + pẏ y )v + yhy + µ) = 0..17) ξv + f) If f = 0, then equation.10) gives c = β) 1/. Thus C = 0 and equation.8) gives a linear equation for v. Then one obtains the following transformation for the PIV equation 11 v = 1 v v v zv β) 1/.18) ᾱ = 1 4 α + 3 β)1/ β = α + 1 β)1/. Case 3. In this case equation.) can be written as follows v + gv + h)b v + B 1 v + B 0 ) = 0.19) B j,j = 1,,3 are functions of u,u,z and g, h are functions of z only. One may consider the two subcases 3.1) d = 0, and 3.) d 0 separately. Case 3.1. If d = 0, then one must take f = g= h=0 and c = µ, µ := β) 1/. Since e 0 then without loss of generality one can take b = 0,e = 1. The quadratic equation for v becomes 3a 1)v au + a + 4z)v + u + u + µa z + α) = 0..0) One may distinguish between the two cases a + 3 = 0 and a Case If a + 3 = 0, then by using the change of variable u = νy + 1 x),x = νz, aν ν is a nonzero constant, one obtains the following second-order second-degree equation of Painlevé type for yx) λ = aν κ = 1 ν ÿ λxẏ + λ λ x )y κλx = 4y + σ) ẏ +λxy + κ).1) α + aµ + ) a σ = 1 α aµ )..) ν a

7 Second degree Painlevé equations 477 The one-to-one correspondence between solutions vz) of the PIV and solution yx) of the equation.1) is given by y = av 3v zv + aµ aνv 3v + aνy λx)v ν ẏ + y λxy + λ x + κ) = 0..3) Equation.1) was first obtained by Cosgrove 8, and was labelled as SD-IV.A. Case If a + 3 0, let z = rx), az) := sx), and uz) = px)yx) + qx), px), qx) and rx) satisfy the following equations ṡ ) ps + 3)ṙ = 6s 1) s +3)ṗ q) = 4s ṙ + 4r ) ṡ.4) s + 3)p q qṗ + q ) = ṙ + 4r 6s 1)µs r + α). such that s 1)s + 3) 0. Then the equations py + q = v + av + µ Av + Bv + C = 0.5) v A = 3s 1) B = spy + sq + ṡ ) ṙ + 4r C = 1 p s + 3)ẏ + 3p s 1)y 3s 1) +4ps sq + ṡ ) ṙ + 4r y + sq + ṡ ) ṙ.6) give a one-to-one correspondence between solutions vz) of the PIV and solutions yx) of the following second-order second-degree equation of Painlevé type ÿ + y + a 0 )ẏ 4y y + a 0 ) = 16 s + 3) ss 9)y + c 1 y + c 0 ẏ y )..7) The functions a 0,c 1,c 0 are given in terms of the functions px), qx), rx) and sx) as follows a 0 = 1 ps + 3) qs + 3) + ss 9) pṡ + 16rs 6s 1) c 1 = p sqs 9) r5s + 3) + s + 3) 6s 1)ṡ c 0 = s + 3) 1s 1) s + ss + 3)s + 15) ṡ) 18s 1) 1 p sq s 9) 4rq5s + 3) 3sr + 3 p s 1)µs 3) + 4sr αs + 1)..8)

8 478 A Sakka and U Muğan Case 3.. If d 0, then without lose of generality one can take a = 0, d = 1. Equation.) can be written as follows v + gv + f )B v + B 1 v + B 0 ) = 0.9) B = 3u 1) B 1 = u +4e 3 g)u 4bu + 3 g 8z B 0 = e g)u + e g f)u +e + b g be c)u b b + 4z 4α) g3 g 8z) + 3f.30) and b, c, e, f, g satisfy the following equations: ge g) = 0 fe g) = 0 fe +be c) = 0 f + bf = ge + be c) fb b +3e 8ze 3f + 4z 4α) = c + β gb b + 3e 8ze + 4z 4α) = c + f3e 4z)..31) The following three subcases 3..1) f = g=0, 3..) f = 0, g 0, 3..3) f 0 may be considered separately. Case If f = g = 0, then equation.31) gives c = µ, µ + β = 0. Let z = rx), uz) = px)yx)+qx) px), qx) and rx) are solutions of the following equations pṙ = 1 ṗ = t + sq) p q = sq + tq 3s + 4r.3) and tx) and sx) are arbitrary functions. Moreover, defining bz) := tx), ez) := sx) then the quadratic equation for v can be written as 3y + a 1 y + a 0 )v + ẏ sy )v + c 1 y + c 0 ) = 0.33) a 1 = p 1 q a 0 = p q 1) c 1 =ṡ+p 1 ts µ) c 0 = p pqc 1 pṫ t 8rs + 3s + 4r 4α)..34) Assume that c 1 and c 0 are not both zero, then equations.33) and py + q = v + bv + c v + ev.35) give one-to-one correspondence between solutions vz) of the PIV and solutions yx) of the following second-order second-degree Painlevè-type equation P 3 y)ÿ syẏ ṡy ) Q y)ẏ sy ) R 4 y)ẏ sy ) + P 3 y)f y) = F y)ẏ sy ) + G 4 y) ẏ sy ) P 3 y).36)

9 Second degree Painlevé equations 479 P 3 y) = 6c 1 y 3 + g y + g 1 y + g 0 ) Q y) = 5c 1 y + g + a 1 c 1 )y + 3g 1 a 1 c 0 R 4 y) = 33sc 1 y 4 + c 1 + sg )y 3 + g + sg 1 )y + g 1 y + g 0 F y) = c 1 y g 4a 1 c 1 )y + g 1 3a 1 c 0 G 4 y) = 3 sc 1 y g + ṗ c 1 + ṗ p c 1 + 4sa 1 c 1 ) y 3 ) p g 4a 1 c 1 + 3sg 1 sa 1 c 0 y + g 1 + ṗ ) p g 1 a 0 c 1 a 1 c 0 + sg 0 y + g 0 + ṗ p g 0 a 0 c 0.37) and g = 1 c 0 + 4a 1 c 1 ) g 1 = a 0 c 1 +a 1 c 0 g 0 = 1 a 0c 0..38) When c 1 = c 0 = 0, one obtains e + be c = 0 b b +3e 8ze + 4z α) = 0.39) and equation.33) reduces to a linear equation for v. In this case w = u+1 solves PXLII u 1 in 17, p 341. Case 3... If f = 0, g 0, then equation.31) gives g = e, and c = β) 1/ e + be β) 1/ = 0 b b +3e 8ze + 4z α) = 0..40) In this case, B 0 = 0 in equation.30) and equation.9) is linear in v, B v + B 1 = 0. This case is the same as case 3..1 when c 1 = c 0 = 0. Case If f 0, then equation.31) gives g = e and c = 1 e + be) f + bf ce = 0 f c + f3e 8z) = ec + β).41) fb b +3e 8ze 3f + 4z 4α) = c + β. Note that if e = 0, then equations.41) imply c = 0,f = 0 which contradicts the assumption f 0, thus one has to take e 0. Let µ = β) 1/ and let uz) = ξz)yx) + ηz), x = ζz), z f ẑ) + µeẑ) ξz) = exp dẑ fẑ) ζz) = 1 z ξẑ)eẑ) dẑ.4) ηz) = 1 f e + be + µ). Then the equation.9) can be written as 3py + qy + p 1 q 1)v sẏ y )v yhy + µ) = 0.43)

10 480 A Sakka and U Muğan hx) = µx + ν µ + ν 0 ν=constant.44) px) := ξz) qx) := ηz) sx) := ξz)ez). Equation.43) and y = v ηv eη b)v µ.45) ξv + ev + f) give one-to-one correspondence between solutions vz) of the PIV and solutions yx) of the following second-order second-degree Painlevé equation F y)ÿ yẏ) py + q)ẏ y ) P 3 y)ẏ y ) + F y)q 3 y) = py + q)ẏ y ) R 3 y) ẏ y ) + S 4 y).46) F y) = 3py + qy + p 1 q 1) 1 P 3 y) = 3 py 3 + ṗ p ṡ ) s + 3q y + q q ṡ ) s + q 1 y + 1 p p q q 1 ṗ p q 1) p + ṡ ) s Q 3 y) = 4s {4hpy 3 + 7hq + 5µp)y + 8µq + 3hp 1 q.47) 1)y +3µp 1 q 1)} R 3 y) = 3 py 3 + 6qy + 1p qṗ p q + 5q 3)y + 1p q 1)ṗ + q) 1p q q S 4 y) = 1s yhy + µ)py + qy + p 1 q 1). 3. Painlevé V Let vz) be a solution of the fifth Painlevé equation, PV, v = 3v 1 vv 1) v ) 1 z v + α z vv βv 1) 1) + + γ δvv + 1) v + z v z v ) Equation 1.13) becomes a fifth-order polynomial in v as follows A 5 v 5 + A 4 v 4 + A 3 v 3 + A v + A 1 v + A 0 = 0 3.) A 5 = du a) α z A 4 = du + d u a ) 3du a) + 6α du a) + z z A 3 = eu + e u b ) du + d u a ) eu b) + eu b) z du a) eu b) + f u c) + 1 6α z z β z γ z A = f u + f u c ) eu + e u b ) eu b) eu b) z δ 3.3)

11 Second degree Painlevé equations 481 f u c) eu b) + du a) 1 + α z z + 6β z + γ z δ A 1 = f u + f u c ) + 3f u c) + 6β f u c) + z z A 0 = f u c) + β z. Note that if A 5 = 0, then A 4 = 0. Therefore in order to reduce 3.) to a quadratic equation for v one must consider the following three cases: 1) A 5 = A 3 = 0, ) A 5 = 0,A 3 0 and 3) A 5 0. Case 1. If A 5 = A 3 = 0, then d = e = 0. Thus one should take f 0 and hence without loss of generality c = 0, f=1. This gives a = α = 0 and b + b + z b + β z + γ + δ = ) z Let zuz) = px)yx) + µ, z = rx) = exp x 1 dˆx, px) = µx + ν, pˆx) µ = β) 1/, and ν is a constant such that µ + ν 0. Then the quadratic equation for v can be written as yy + µp 1 ) ) 1 ) 1 v 1 + ẏ y ) v 1 y + c 1 y + c 0 ) = 0 3.5) c 0 = 1 p p c 1 + δr ) 3.6) and c 1 x) = 1 µ zbz) satisfies the Riccati equation p Equations 3.5) and c 1 + c1 + γ + δr)rp = ) y = p 1 zv + bv) µ 3.8) give one-to-one correspondence between a solution vz) of PV and a solution yx) of the following second-order second-degree equation: y + c 1 y + c 0 )ÿ yẏ) y + c 1 )ẏ y ) P 3 y)ẏ y ) + Q 5 y) = y + c 1 )ẏ y ) R 3 y) ẏ y ) + 8yy + µp 1 )y + c 1 y + c 0 ) 3.9) P 3 y) = 4y 3 + 3c 1 µp 1 )y 4p µpc 1 + γr)y p 3 p 3 c1 3 +µp c1 + γ + δr)rpc 1 + δ3µ + )r Q 5 y) = 8y + c 1 y + c 0 )3y 3 + 5c 1 + 4µp 1 )y + 3µp 1 c 1 + c 0 )y +µp 1 c 0 R 3 y) = 8y c 1 + µp 1 )y + p 10p c1 + µpc 1 + γ + 6δr)ry +p 3 3p 3 c1 3 + µp c1 + γ + 3δr)rpc 1 + δ3µ + )r. 3.10)

12 48 A Sakka and U Muğan Case. If A 5 = 0,A 3 0, then d = 0,a = α)1/. Ife=0, then 3.) cannot be reduced z to a quadratic equation in v. Let e 0 and without loss of generality let b = 0,e = 1. Then equation 3.) can be reduced to the following quadratic equation for v: A 3 v + A fa 3 )v + A 1 fa +f A 3 =0 3.11) f and c satisfy the following equations ff +1)f af c) = 0 3f + 1)c + f a ) + f f 1)ac + δ) = f + 1) f c + c ) γ z z f β 3.1) f + 1). z One has to consider the following three cases f = 0,f = 1 and ff +1) 0 separately. Case.1. If f = 0, then equation 3.1) gives z c + β = 0. Thus A 0 = A 1 = 0 and equation 3.) reduce to a linear equation A 3 v + A = 0 for v. Therefore one obtains the following transformation for the PV 11 v = 1 δ) 1/ zv zv α) 1/ v + α) 1/ β) 1/ ) + δ) 1/ zv + β) 1/ ᾱ = 1 16δ {γ + δ)1/ 1 α) 1/ β) 1/ } β = 1 16δ {γ δ)1/ 1 α) 1/ β) 1/ } γ = δ) 1/ β) 1/ α) 1/ δ = δ. 3.13) Case.. If f = 1, then equation 3.1) gives a + c) + δ = 0. Assume that γ and δ are not both zero, and let yx) = zu ν) 1 4µ 1), x = z κ, µ = α) 1/ + 1,ν= δ)1/, and ν) 1/ when ν 0 κ = ) when ν = 0. Then the quadratic equation for v can be written as follows ) 1 ) 1 y + µy + µ + β) v 1 + xẏ v 1 + κ x νy ν + γ)= ) The equations 3.15) and y = zv + µ 1)v νz + 4µ 1)v + µ 3.16) v 1) give one-to-one correspondence between solutions vz) of PV and solutions yx) of the following second-order second-degree Painlevé-type equation ÿ 1 Q 3 y) = y x y x ẏ Q 3 y) Q 3 y) := 4y 3 + ν ν4µ 1) + γy + 4 ν νµ µ + β) + µγ y + ν µ + β)γ ν) 3.17)

13 when ν 0, and ÿ 1 Second degree Painlevé equations 483 Q y) = 4 y x y ẏ Q y) Q y) := γy +µy + µ + β) 3.18) when ν = 0. The equations 3.17) and 3.18) were obtained by Bureau, equations 18.6), 0.5), p 06, 09 resp. Note that if γ = δ = 0, then w = zv + 8α) 1/ v 8α) 1/ + 1v + 8α) 1/ ) v 1 is a solution of the following equation zw = ww 3.0) which has the first integral zw = w + w + K, K is the integration constant. Case.3. If ff +1) 0, then equation 3.1) gives c = f α)1/ f z f = 3f + 1 ff +1) f ) 1 z f + α z ff βf + 1) +1) + + γ z f z δf f 1) f +. f ) By using the linear transformation uz) = ξz)yx) + ηz) and the change of variable x = ζz), z µ = β) 1/ ẑf ẑ) + µ + 1)f ẑ) + µ ξz) = exp dẑ ẑf ẑ) ζz) = 1 z ξẑ)fẑ) + 1 dẑ ηz) = 1 3.) zf zf α) 1/ f + µ) one can write the quadratic equation for v as follows ) φ 1 ) 1 φ + 1) yy + a 1) v 1 + ẏ y ) v 1 y + c 1 y + c 0 ) = 0 3.3) hx) = µx + ν a 1 = µ µ + ν 0 h φx) := fz) c 1 = c 0 = c 1 + δr φ h φ + 1) 1 ηf + 1) a c ξf + 1) rx) = exp and φ + φc 1 + α)1/ φ µ h h φ = 0 c 1 + c1 + γrφ h φ + 1) + δr φ φ 1) = 0. h φ + 1) 3 Equations 3.3) and y = φx) hφ + 1) dx 3.4) 3.5) 1 zξv + f) zv + α) 1/ v zηv µ) 3.6)

14 484 A Sakka and U Muğan give a one-to-one correspondence between solutions vz) of the PV and solution yx) of the following equation: y + c 1 y + c 0 )ÿ yẏ) y + c 1 )ẏ y ) P 3 y)ẏ y ) + Q 5 y) = y + c 1 )ẏ y ) R 3 y) ẏ y ) + S 4 y) 3.7) ) φ P 3 y) = 4y 3 + 3c 1 + φφ + 1) a 1 y ) φ + c 1 + c 0 + c 1 φφ + 1) a 1 y ) φ + c 0 +c 0 φφ + 1) a 1 Q 5 y) = 8φ φ + 1 y + c 1 y + c 0 )3y 3 + 5c 1 + 4a 1 )y + 3a 1 c 1 + c 0 )y + a 1 c 0 4φ + 1) φ 11φ + 5) R 3 y) = φ + 1) y3 + a 1 + φφ + 1) φ + 1) c 1 y ) 4φ 5φ + 3) + φ + 1) c 1 + φ + 1) c φ 0 c 1 φφ + 1) a 1 c 1 y ) φ c 0 φφ + 1) a 1 c 0 + 4φc 0c 1 φ + 1 S 4 y) = 8φ φ + 1 yy + a 1)y + c 1 y + c 0 ). 3.8) Case 3. If A 5 0, then equation 3.) can be written as v 3 + gv + hv + k)b v + B 1 v + B 0 ) = 0 3.9) B j,j =0,1, are functions of u,u,z and g, h, k are functions of z only. If k = 0, then one obtains A 0 = 0, and hence f = 0,z c +β = 0. This implies that A 1 = 0, and 3.) takes the form and equation 1.11) becomes A 5 v 3 + A 4 v + A 3 v + A = ) u = v + bv + c v + ev. 3.31) The discrete Lie-point symmetry of PV 11 v = 1 v ᾱ = β β= α, γ = γ δ=δ 3.3) transform this case to case.

15 Second degree Painlevé equations 485 When k 0, one should take γ = 0,d 0, and without loss of generality a = 0, d= 1. The functions e, g, h, k should satisfy e = f +1), g = f +), h = f +1, k = f, b, c, f are solutions of the following equations b + c)f 1) = 0 b + c) c 1 ) = 0 z b + c) + δ = 0 f + bf 1) = ) f b + bz bc + αz ff 1) f + 1)c + β f 1) = 0. z The equations 1.11) and 1.15) become u = v + bv + c v 1)v f) B = u α B 0 = 1 f B v +B 1 v+b 0 =0 3.34) B z 1 = u + f 1)u + z u α f 1) z f u cf u + c + β. z 3.35) The following two subcases 3.1) f = 1 and 3.) f 1 should be considered separately. Case 3.1. If f = 1, then equation 3.33) gives b = µ z + ν, c = µ z, ν = δ)1/, and µ is a constant such that νµ + 1) = 0. Let yx) = izu + µ), x = ln z. Then the equations y = zv + µv + νzv µ iv 1) 3.36) y + iµy µ + α)v iẏv y iµy µ β) = 0 give one-to-one correspondence between solutions vz) of PV and yx) of the following second-order second-degree Painlevé-type equation ÿ 1 Q 4 y) = y iν expx) ẏ Q 4 y) y 3.37) Q 4 y) := y 4 + µ + α β)y 4iµα + β)y + µ α)µ + β). When ν = i equation 3.37) becomes ÿ 1 Q 4 y) = 4y expx) ẏ Q 4 y) y 3.38) Q 4 y) := y 4 + µ + α β)y 4iµα + β)y + µ α)µ + β). Equation 3.38) was also obtained by Bureau, equation 16.13), p 03 but the relation between 3.38) and PV has not been mentioned. Case 3.. If f 1, then 3.33) gives δ = 0,b = c, c = f, and f satisfies PV with f 1 δ = γ = 0 f = 3f 1) ff 1) f ) 1 z f + α z ff 1) + βf 1). 3.39) z f

16 486 A Sakka and U Muğan Let uz) = ξz)yx) + ηz), x = ζz), z µ = β) 1/ ẑf ẑ) + µ + 1)f ẑ) µ ξz) = exp ẑf ẑ) ζz) = 1 z ξẑ)fẑ) 1 dẑ ηz) = and let φx) := fz). Then the equations dẑ 1 zf f 1) zf + µf 1) y = zv zηv + zη + µ)v µ zξv 1)v f) y + a 1 y + a 0 )v φ 1)ẏ y )v φyy + c 1 ) = ) 3.41) hx) = µx + ν c 1 = µ µ + ν 0 hx) a 1 x) = c 1 φ a 0 = a1 φ αφ h 3.4) give one-to-one correspondence between solutions vz) of PV and yx) of the following second-degree equation ÿ yẏ y d 1 )ẏ y ) d 1 = φφ 1) φ + µ d = h 1 d 3 = φφ 1) φ µφ + 1) hφ 1) Q 4 y) = Q 4 y) = d y d 3 ) ẏ y ) + Q 4 y). 3.43) y φ + 1) φ 1) 4φy φ 1) y3 + a 1 + c 1 )y + a 0 + 4a 1 c 1 )y + a 0 c 1 and the function φx) satisfies the following equation 3.44) φ = 3 φ φ) ḣ h φ + 4α h φ3 µ φ. 3.45) h 4. Painlevé VI Let vz) be a solution of PVI 1 v + 1 v v z ) v ) v = 1 vv 1)v z) + α + βz z z 1) then, for PVI the equation 1.13) takes the form 1 z + 1 z ) v v z v + γz 1) v 1) ) δzz 1) + v z) 4.1) A 6 v 6 + A 5 v 5 + A 4 v 4 + A 3 v 3 + A v + A 1 v + A 0 = 0 4.)

17 Second degree Painlevé equations 487 A 6 = du a) α z z 1) A 5 = du + d u a ) z + 1)du a) z 1) 4αz + 1) + du a) + zz 1) z z 1) A 4 = eu + e u b ) z + 1)du + d u a ) eu b) z 1) + eu b) zz 1) z z 1) αz + 4z + 1) + βz + γ + δz)z 1) du a) z + 1)eu b) + f u c) 3zdu a) + z + z 1) zz 1) A 3 = f u + f u c ) z + 1)eu + e u b ) + zdu + d u a ) 4 + α + β)z + 1) + γ + δ)z 1) zz 1) eu b) f u c) zdu a) + z + z 1) zz 1) + z z 1) du a) + f u c) z 1 zz 1) A = zeu + e u b ) z + 1)f u + f u c ) + zeu b) + z eu b) z 1 zz 1) αz + βz + 4z + 1) + γ z + δ)z 1) + f u c) z + 1)eu b) 3f u c) + zdu a) z + z 1) zz 1) A 1 = zf u + f u c ) + z + 1)f u c) + A 0 = z f u c) + β. z 1) z 4βz + 1) f u c) + z 1 z 1) 4.3) Note that if A 6 = 0 then A 5 = 0. Moreover, one cannot find the functions a, b, c, d, e, f such that A 6 = A 5 = A 4 = 0. Thus, to reduce 4.) to a quadratic equation for v one may consider the following two cases 1) A 6 = 0, A 4 0 and ) A 6 0. Case 1. If A 6 = 0, then one obtains d = 0,a = α)1/. Note that when e = 0 equation zz 1) 4.) cannot be reduced to a quadratic equation for v. Therefore one should take e 0, and hence without loss of generality b = 0,e =1. In this case equation 4.) can be written as v + f) B v + B 1 v + B 0 ) = 0 4.4) f and c are solutions of the following equations ff +1)f + z) = 0 zz 1)af + c) + ff +1) +βzf + 1)f + z) + γz 1)f f + z) 4.5) +δ 1)zz 1)f f + 1) = 0. The following three subcases should be considered separately.

18 488 A Sakka and U Muğan Case 1.1. If f = 0, then equation 4.5) implies z 1) c + β = 0. By using the change of variable µ + 1)z yx) = zu z 1) µ 1 x = arcsin z + 1 µ = α) 1/ + β) 1/ z 1) z 1 4.6) one may write the quadratic equation for v as follows ) z 1 ) z 1 A v B v C = 0 4.7) A = y + λy + λ γ 1+sin x B = 1 sin x ẏ 4.8) C = sin x + 1 sin x 1 y λy + λ + δ 1) and λ = 1 α)1/ β) 1/ + 1. The second-degree equation for yx) is ÿ 1 Q 4 y) = 4 tan x y K ẏ Q 4 y) y sin x Q 4 y) := y 4 + δ γ λ 1)y 4.9) + λγ + δ 1)y +λ γ )λ + δ 1) K = µ. The equation 4.9) was obtained by Bureau, equation 16.1), p 0 and also by Fokas and Ablowitz 11. Case 1.. If f = 1,then equation 4.5) gives z a + c) = γ. The quadratic equation for v becomes 1 A v 1 ) 1 + B z v 1 ) + C = ) z A = zu + cu + c + β z 1) B = z 1 zz 1)u + zu + za + c 4.11) 1 C = zz 1) z 1) u z 1)z a + c + 1)u + z a + c + 1) + δ 1. The Lie-point symmetry of PVI 11 v z;ᾱ, β, γ, δ) = 1 vz; α, β, γ, δ) 4.1) z = 1 z ᾱ=α β= γ γ = β δ=δ transforms this case to the case 1.1 Case 1.3. If f = z, then equation 4.5) gives za + c + 1) = 1 δ. The quadratic equation for v now may be written as ) 1 ) 1 A v 1 + B v 1 + C = )

19 Second degree Painlevé equations 489 z u + zcu + c + β z 1) B = u + A = 1 z C = z 1)u a + c) + a + c)u + z 1 γ z z 1) The Lie-point symmetry of PVI 11 v z;ᾱ, β, γ, δ) = 1 1 z)vz; α, β, γ, δ) z = 1 α=ᾱ β = γ γ = δ+ 1 1 z transforms this case to the case z 1) za + c) u + zz 1) zz 1) δ= β ) 4.15) Case. If A 6 0, then to reduce equation 4.) to a quadratic equation for v one should take d 0 and hence without loss of generality one may take a = 0,d =1. Then equation 4.) can be written as v 4 + gv 3 + hv + kv + l)b v + B 1 v + B 0 ) = ) B j,j = 1,,3 are functions of u,u,z and b, c, e, f, g, h, k, l may be chosen such that one of the following two cases are satisfied. i) e = z + 1) f = z g = z + 1) h = z + 4z + 1 k = zz + 1) l = z 1 µ b = µ + ν)z + µ ν c = zz 1) z ) µ and ν are constants such that µ + ν µν = γ + δ) µ + ν = µ + ν) + 4γ δ) 4.18) ii) e = z, g = z, h = z,f = k = l =0 z 1) c + β = 0 zb + c + 1) + δ 1 = ) Case.1. With the choices 4.17), 4.18) equations 1.11) and 1.13) take the following form u = respectively, v + bv + c v 1)v z) A = u Av + Bv + C = 0 4.0) α B = u z 1) + z z 1) zz 1) u C = zu cu + 1 z c + β z z 1). 4.1) Let yx) = zz 1)u 1 z+1 µ, x = arcsin z 1, then yx) is a solution of 4.9) with K = ν 1, and Q 4 y) := y 4 + β α 1 µ )y + µα + β)y µ α) 1 4 µ + β). 4.)

20 490 A Sakka and U Muğan Using the fact that both subcases 1.1 and.1 give the same second-degree equation one obtains the following new Lie-point symmetry for PVI: v = v z ᾱ = γ β = δ 1 γ = α δ = β + 1 v ) Case.. In this case equations 1.11) and 1.13) respectively become u = v + bv + c Av z) + Bv z) + C = 0 4.4) vv z) A = u α B = u z 1) + z z 1) zz 1) u 4.5) C = z 1)u b + c) + b + c)u + z 1 γ z z 1). The Lie-point symmetry 4.1) of PIV transforms this to case.1. References 1 Bureau F 1964 Ann. Math 66 1 Bureau F 197 Ann. Math Chazy J 1911 Acta Math Cosgrove C M 1993 Stud. Appl. Math Cosgrove C M 1977 J. Phys. A: Math. Gen Cosgrove C M 1977 J. Phys. A: Math. Gen Cosgrove C M 1978 J. Phys. A: Math. Gen Cosgrove C M and Scoufis G 1993 Stud. Appl. Math Erugin N P 1958 Dokl. Akad. Nauk. BSSR 10 Fokas A S and Yortsos Y C 1981 Lett. Nuovo Cimento Fokas A S and Ablowitz M J 198 J. Math. Phys Fokas A S and Zhou X 199 Commun. Math. Phys Fokas A S, Muğan U and Zhou X 199 Inverse Problems Fuchs R 1907 Math. Ann Gambier B 1909 Acta. Math Garnier R 191 Ann. Sci. Ec. Norm. Super Ince E L 1956 Ordinary Differential Equations New York: Dover) 18 Jimbo M and Miwa T 1981 Physica D 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 D 306 Jimbo M 1979 Prog. Theor. Phys Muğan U and Fokas A S 199 J. Math. Phys Muğan U and Sakka A 1995 J. Math. Phys Muğan U and Sakka A 1995 J. Phys. A: Math. Gen Painlevé P 1900 Bull. Soc. Math. Fr Painlevé P 191 Acta. Math Sakka A and Muğan U 1997 J. Phys. A: Math. Gen