Painlevé Test and Higher Order Differential Equations

Transcription

1 Journal of Nonlinear Mathematical Physics Volume 9, Number ), Article Painlevé Test and Higher Order Differential Equations Uğurhan MUĞAN and Fahd JRAD SUNY at Buffalo, Department of Mathematics, 305 Mathematics Bldg., Buffalo, NY U.S.A. umugan@acsu.buffalo.edu, Permanent Address: Bilkent University, Department of Mathematics, Bilkent, Ankara, Turkey mugan@fen.bilkent.edu.tr Cankaya University, Department of Mathematics and Computer Sciences, Cankaya, Ankara, Turkey fahd@cankaya.edu.tr Received September 6, 200; Revised November, 200; Accepted April 5, 2002 Abstract Starting from the second Painlevé equation, we obtain Painlevé type equations of higher order by using the singular point analysis. Introduction Painlevé and his school [20, 2, 4, 5] investigated second-order first-degree equations of the form y = F z,y,y ),.) where F is rational in y, algebraic in y and locally analytic in z, and has no movable critical points. This property is known as the Painlevé property and ordinary differential equations ODE), which possess it, are said to be of Painlevé type. They found that, within a Möbius transformation, there exist fifty such equations. Distinguished among these fifty equations are the six Painlevé equations, PI,..., PVI. Any other of the fifty equations either can be integrated in terms of known functions or can be reduced to one of these six equations. Higher order first-degree in the polynomial class and higher order higher degree equations of Painlevé type were investigated by Fuchs [3, 5], Briot and Bouquet [5], Chazy [5], Bureau [3, 4], Exton [], Martynov [7], Cosgrove [8, 9], Kudryashov [6], Clarkson, Joshi and Pickering [6] and also in the articles [22, 23, 9, 8]. The Riccati equation is Copyright c 2002 by U Muğan and F Jrad

2 Painlevé Test and Higher Order Differential Equations 283 the only example for the first-order first-degree equation which has the Painlevé property. The best known third order equation is Chazy s natural-barrier equation y =2yy 3y n 2 6y y 2) 2,.2) The case n = appears in several physical problems. The equation.2) is integrable for all real and complex n and n =. Its solutions are rational for real n in 2 n 5, and have a circular natural barrier for n 7andn =. In this article the second Painlevé hierarchy is investigated by using the Painlevé ODE test or singular point analysis. It is possible to obtain the equation of Painlevé typeofany order, as well as the known ones, starting from the second Painlevé equation. Painlevé ODE test which is an algorithm introduced by Ablowitz, Ramani, Segur ARS) [, 2] tests whether a given ODE satisfies the necessary conditions to be of Painlevé type. The procedure to obtain higher order Painlevé type equations starting from the second Painlevé equation may be summarized as follows: I. Take an n th order Painlevé type differential equation y n) = F z,y,y,...,y n )),.3) where F is analytic in z and rational in its other arguments. If y y 0 z z 0 ) α as z z 0, then α is a negative integer for certain values of y 0. Moreover, the highest derivative term is one of the dominant terms. Then the dominant terms are of order α n. There are n resonances r 0 =,r,r 2,...,r n, with all r i, i =, 2,...,n ) being nonnegative real distinct integers such that Qr j ) = 0, j = 0,, 2,...,n ). The compatibility conditions for the simplified equation that retains only dominant terms of.3) are identically satisfied. Differentiation of the simplified equation with respect to z yields y n+) = G z,y,y,...,y n)),.4) where G contains the terms of order α n, and the resonances of.4) are the roots of Qr j )α + r n) = 0. Hence equation.4) has a resonance r n = n α additional to the resonances of.3). Equation.4) passes the Painlevé test provided that r n r i, i =, 2,...,n ) and is a positive integer. Moreover the compatibility conditions are identically satisfied, that is z 0,y r,...,y rn are arbitrary. II. Add the dominant terms which are not contained in G. Then the resonances of the new equation are the zeros of a polynomial Qr) ofordern +. Find the coefficients of Qr) such that there is at least one principal Painlevé branch, that is, all n+ resonances except r 0 = ) are real positive distinct integers for at least one possible choice of α, y 0 ). The other possible choices of α, y 0 ) may give the secondary Painlevé branch, that is, all the resonances are distinct integers. III. Add the nondominant terms which are the terms of weight less than α n, with coefficients analytic in z. Find the coefficients of the nondominant terms by the use of the compatibility conditions and transformations which preserve the Painlevé property. In this article we consider only the principal branch, that is, all the resonances r i except r 0 = ) are positive real distinct integers and the number of resonances is equal to the

3 284U Muğan and F Jrad order of the differential equation for a possible choice of α, y 0 ). Then, the compatibility conditions give full set of arbitrary integration constants. The other possible choices of α, y 0 ) may give secondary branches which possess several distinct negative integer resonances. Negative but distinct integer resonances give no conditions which contradict integrability [2]. Higher order equations with negative but distinct integer resonances might be investigated separately. In the present work we start with the second Painlevé equation and obtain the third, fourth and some fifth order equations of Painlevé type. All of the third order and some of the fourth order equations are found in the literature, but for the sake of completeness we present the known equations with appropriate references. A similar procedure was used in [8] to obtain the higher order equations of Painlevé type by starting from the first Painlevé equation. The procedure can also be used to obtain the higher order equations by starting from PIII,..., PVI. These results will be published elsewhere. 2 Third order equations: P 3) II The second Painlevé equation, PII, is y =2y 3 + zy + ν. 2.) The Painlevé test gives that there is only one branch and the resonances are, 4). The dominant terms of 2.) are y and 2y 3 which are of order 3 asz z 0. Differentiation of the simplified equation y =2y 3 gives y = ay 2 y, 2.2) where a is a constant which can be introduced by replacing y with λy, such that 6λ 2 = a. Addition of the polynomial type terms of order 4gives the following simplified equation y = yy + a 2 y 2 + a 3 y 2 y + a 4 y 4, 2.3) where a i, i =,...,4are constants. Substitution of y = y 0 z z 0 ) + βz z 0 ) r, 2.4) into the simplified equation 2.3) gives the following equations, Qr) = 0, for the resonance r and for y 0, respectively, Qr) =r +) { r 2 y 0 +7)r [ a 3 y a 2 )y 0 8 ]} =0, a 4 y0 3 a 3 y a 2 )y 0 +6=0. 2.5a) 2.5b) Equation 2.5b) implies that, in general, there are three branches of Painlevé expansions if a 4 0. Now one should determine y 0j, j =, 2, 3, and a i suchthatatleastoneof the branches is the principal branch. There are three cases which should be considered separately. Case I. a 3 = a 4 = 0: In this case y 0 takes a single value, i.e. there is only one branch. Equation 2.5a) implies that r 0 = andr r 2 = 6. Therefore there are the following four

4 Painlevé Test and Higher Order Differential Equations 285 possible cases: a. y 0 = 6/a 2 : r,r 2 )=, 6), =0, b. y 0 = 2/ : r,r 2 )=2, 3), = a 2, c. y 0 = 2/ : r,r 2 )= 2, 3), = 2a 2 /3, d. y 0 = 4/ : r,r 2 )=, 6). 2.6) The case d is not be considered since r = is a double resonance. The compatibility conditions are identically satisfied for the first two cases. To find the canonical form of the third-order equations of Painlevé type, one should add nondominant terms the coefficients of which are analytic functions of z, that is, one should consider the following equation for each case y = yy + a 2 y 2 + A y + A 2 yy + A 3 y 3 + A 4 y + A 5 y 2 + A 6 y + A 7, 2.7) where A k z), k =,...,7, are analytic functions of z. Substitution of y = y 0 z z 0 ) + 6 y j z z 0 ) j, 2.8) j= into equation 2.7) gives the recursion relation for y j. Then one can find A k such that the recursion relation, i.e. the compatibility conditions for j = r,r 2, are identically satisfied and hence y r, y r2 are arbitrary. I.a: The Painlevé property is preserved under the following linear transformation yz) =µz)ut)+νz), t = ρz), 2.9) where µ, ν and ρ are analytic functions of z. By using the transformation 2.9) one can set A 4 = A 5, A =0,anda 2 = 6. The compatibility condition at the resonance r = gives A 2 = A 3. The arbitrariness of y in the recursion relation for j = 6 and the recursion relation yield that A 5 A 2 5 =0, A 6 A 5 A 6 =0, A 7 3 A 5A 7 = 6 A2 6, A 3 =0. 2.0) According to the equation 2.0a) there are three cases should be considered separately. I.a.i: A 5 = 0: From equation 2.0) all coefficients A k can be determined uniquely. The canonical form of the third order equation for this case is y = 6y 2 +c z + c 2 )y + 72 c2 z c c 2 z c2 2z 2 + c 3 z + c 4, 2.) where c i, i =,...,4are constants. If c = c 2 = 0, then 2.) can be written as u =6u 2 c 3 z c 4, 2.2) where u = y. If c 3 = 0, then the solution of 2.2) can be written in terms of elliptic functions. If c 3 0, 2.2) can be transformed into the first Painlevé equation.

5 286 U Muğan and F Jrad If c = 0 and c 2 0, 2.) takes the following form by replacing y by γy and z by δz such that γδ = and c 2 δ 3 =6 y = 6y 2 +6y +3z 2 + c 3 z + c 4, 2.3) where c 3 = c 3 δ 5, c 4 = c 4 δ 4. Equation 2.3) was given in [5] and [3]. If c 0 and c 2 = 0, replacement of y by γy and z by δz in 2.) such that γδ = and c δ 4 = 2 yields y = 6y 2 +2zy +2z 4 + c 3 z + c 4, 2.4) where c 3 = c 3 δ 5, c 4 = c 4 δ 4. Equation 2.4) was given by Chazy [5] and Bureau [3]. It should be noted that 2.) can be reduced to 2.4) by the replacement of z by z c 2 /c ) and then replacing y by γy and z by δz such that γδ =,c δ 4 = 2. I.a.ii: A 5 = 6 : Without loss of generality the integration constant c can be set to z+c) 2 zero. From 2.0) the coefficients A k can be determined and the canonical form of the equation is y = 6y 2 +6z 2 y + y 2) + c z 3 + c 2 z 2) y + c 3 z 2 + c 4 z c2 z c c 2 z ) 4 c2 2z 2, 2.5) where c i, i =,...,4, are constants. If c = c 2 = 0, 2.5) is a special case of the equation given by Chazy [5]. If c = 0 and c 2 0, 2.5) takes the following form by replacing y by γy and z by δz such that γδ = and c 2 δ =24 y = 6y 2 +6z 2 y + y 2 +4y ) + c 3 z 2 + c 4 z +24z 2, 2.6) where c 3 = c 3 δ 6 and c 4 = c 4 δ 3. Equation 2.6) is given in [3]. If c 0 and c 2 = 0, then equation 2.5) takes the form y = 6y z 2 y + y 2 )+8z 3 y + z 8 + c 3 z 2 + c 4 z, 2.7) where c 3, c 4 are constants. Equation 2.7) was given in [3]. I.a.iii: If one replaces A 5 with 6Â5, A 6 with 6Â6 and A 7 with 6Â7, then equations 2.0) yields  5 6Â2 5 =0,  6 6Â5Â6 =0,  7 2Â5Â7 =  ) Integration of 2.8a) once gives  5) 2 =4Â3 5 α, 2.9) where α is an integration constant. Then  5 = Pz;0,α ), 2.20)

6 Painlevé Test and Higher Order Differential Equations 287 where P is the Weierstrass elliptic function. If  6 = 0, 2.8c) implies that Â7 satisfies Lamé s equation. Hence  7 = c E z)+c 2 F z), 2.2) where c and c 2 are constants, E z) andf z) are the Lamé functions of degree one of the first and second kind, respectively. They are given as zζa) σz + a) zζa) σz a) E z) =e, F z) =e, 2.22) σz) σz) where ζ is the ζ-weierstrass function such that ζ = Pz), σ is the σ-weierstrass function such that σ z) σz) = ζz) anda is a parameter such that Pa;0,α ) = 0. Then the equation y = 6y 2 +6Pz;0,α ) y + y 2) zζa) σz + a) zζa) σz a) + c e + c 2 e, 2.23) σz) σz) where c =6c and c 2 =6c 2. Equation 2.23) was considered in [5]. If Â6 0,Pz;0,α ) also solves equation 2.8b). Then 2.8c) implies that Â7 satisfies the nonhomogeneous Lamé s equation. Hence,  7 z) =k z)e z)+k 2 z)f z), 2.24) where k z) =k z P 2 t;0,α ) F t)dt, k 2 z) =k 2 + W t) z P 2 t;0,α ) E t)dt 2.25) W t) with k and k 2 are constants of integration and W z) =E F E F. I.b: The coefficients A k z), k =,...,7, of the nondominant terms can be found by using the linear transformation 2.9) and the compatibility conditions. The linear transformation 2.9) allows one to set a 2 = 2, A z) = 0, A 2 z) = A 3 z) and the compatibility conditions give that A 2 z) =A 6 z) =0andA 4 z) =A 5 z). So the canonical form of the equation is y = 2 yy + y 2) + A 4 y + y 2) + A 7, 2.26) where A 4 and A 7 are arbitrary analytic functions of z. If one lets u = y + y 2, then 2.26) can be reduced to a linear equation for u. Equation 2.26) was given in [3]. I.c: Without loss of generality one can choose = 2. Then the simplified equation is y =2yy 3y 2, 2.27) which was also considered in [5, 3]. Case II. a 4 = 0: In this case y 0 satisfies the quadratic equation a 3 y a 2 )y 0 6= ) Therefore there are two branches corresponding to,y 0j ), j =, 2. The resonances satisfy equation 2.5a). Now one should determine y 0j and a i, i =, 2, 3 such that one of the branches is the principal branch.

7 288 U Muğan and F Jrad If y 0j are the roots of 2.28), by setting P y 0j )= [ a 3 y 2 0j 22 + a 2 )y 0j 8 ], j =, 2, 2.29) and if r j,r j2 ) are the resonances corresponding to y 0j, then one has r j r j2 = P y 0j )=p j, j =, 2, 2.30) where p j are integers such that at least one is positive. Equation 2.28) gives that a 3 = 6 y 0 y 02, 2 + a 2 = a 3 y 0 + y 02 ). 2.3) Then 2.29) can be written as P y 0 )=6 y ) 0, Py 02 )=6 y ) ) y 02 y 0 For p p 2 0, the p j satisfy the following Diophantine equation p + p 2 = ) Now one should determine all finite integer solutions of Diophantine equation. One solution of 2.33) is p,p 2 )=2, 2). The following cases should be considered: i) If p > 0, p 2 > 0andp <p 2, then p > 6andp 2 > 2. ii) If p > 0, p 2 < 0, then p < 6. Based on these observations there are the following nine integer solutions of Diophantine equation, viz p,p 2 )=2, 2), 7, 42), 8, 24), 9, 8), 0, 5), 2, 3), 3, 6), 4, 2), 5, 30). 2.34) For each p,p 2 ), one should write r j,r j2 ) such that r ji are distinct integers and r j r j2 = p j, j =, 2. Then y 0j and a i can be obtained from 2.3) and 2.32) and r j + r j2 = y 0j +7, j =, 2, 2.35) There are the five following cases such that all the resonances are distinct integers for both branches. The resonances and the simplified equations for these cases are II.a: = a 2 =0,y0 2 = 6 :r,r 2 )=3, 4), a 3 y 02 = y 0 : r 2,r 22 )=3, 4), y = a 3 y 2 y, 2.36) II.b: y 0 = : r,r 2 )=2, 4), y 02 = 3 : r 2,r 22 )=4, 6), y = yy +2y 2 +2 y 2 y ), 2.37)

8 Painlevé Test and Higher Order Differential Equations 289 II.c: y 0 = 3 : r,r 2 )=, 3), y 02 = 6 : r 2,r 22 )= 2, 3), a y = yy + y 2 ) 3 y 2 y, 2.38) II.d: y 0 = 2 : r,r 2 )=, 4), y 02 = 6 : r 2,r 22 )= 3, 4), a y = yy +2y 2 ) 2 y 2 y, 2.39) II.e: y 0 = : r,r 2 )=, 5), y 02 = 6 : r 2,r 22 )= 5, 6), y = yy +5y 2 y 2 y ). 2.40) For each case the compatibility conditions for the simplified equations are identically satisfied. To find the canonical form of the third order equations of Painlevé type, one should add nondominant terms with the coefficients which are analytic functions of z. II.a: By using the linear transformation 2.9) one can set 2A + A 3 =0,A 2 = 0 and a 3 = 6. The compatibility conditions at j =3, 4, for the both branches allow one to determine the coefficients A k. The canonical form of the equation for this case is ) y =6y 2 y 2 c2 z 2 c 2 z c 3 y + c y 2 c 2 z c 2 ) y 4 c3 z c c 2 z + 2 c c 3, 2.4) where c, c 2, c 3 are constants. If one replaces z c 2 c 2 γδ = and c δ 2 = 2, then 2.4) yields by z, y by γy and z by δz such that u =6u 2 u +2zuu +4 z 2 + k ) u +4zu +4u 2, 2.42) where u = y z and k is a constant. Equation 2.42) was considered in [5, 3], and its first integral is PIV. If c = c 2 = 0, 2.4) can be solved in terms of elliptic functions. If c = 0 and c 2 0, 2.4) gives y =6y 2 y + c 2 z + c ) 3 y + c 2 y. 2.43) c 2 If one introduces t = z + c 3 c 2, then the first integral of 2.43) is PII. II.b: By using the linear transformation 2.9) one can always choose 2A + A 3 =0, A 2 = 0 and =. Then the compatibility conditions for the both branches, that is the arbitrariness of y 2 and y 4 for the first branch and y 42 and y 62 for the second branch, imply that all the coefficients A k of nondominant terms, are zero. So the canonical form for this case is y = yy 2y 2 +2y 2 y. 2.44) Equation 2.44) was given in [5, 3].

9 290 U Muğan and F Jrad II.c: By using the linear transformation 2.9) one can always set A 3 = A 5 = 0 and = 3. Then the compatibility conditions at j =, 3 give that A = c /2, A 2 = c,c = constant and A 4 = A 6. Then the canonical form of the equation is y = 3yy 3y 2 3y 2 y + 2 c y + c yy + A 4 y + A 4y + A ) The first integral of 2.45) gives that u = 3uu u 3 + B u + B 2, 2.46) where u = y c /6), and B and B 2 are arbitrary analytic functions of z. Equation 2.45) was considered in [5]. II.d: One can always choose A 3 = A 5 = 0 and = 2 by the linear transformation 2.9). The arbitrariness of y and y 4 for the first branch and y 42 for the second branch imply that A = A 2 = A 7 = 0 and A 4 =2A 6. The canonical form is y = 2yy 4y 2 2y 2 y + A 4 y + 2 A 4y. 2.47) The first integral of 2.47) is y = y 2 2y 2yy y3 2 + A 4y + c, c = constant. 2.48) y The equation 2.47) was considered in [5, 3]. II.e: By the linear transformation 2.9) one can choose A = A 3 = 0 and =. Then the compatibility conditions give that A 2 = A 5 =0,A 6 = A 4 /3andA 7 = A 4 /3. After the replacement of y by y and A 4 by 3A 4 the canonical form of the equation for this case is y = yy +5y 2 y 2 y +3A 4 y + A 4y + A ) Equation 2.49) has the first integral y yy y 3 + A 4 y + A 2 8 4) = y y 2) y + y ) 2 A 4 +4 y y 2) 2A 4 y 3 + A 4y + A 4) +4A 2 4 y 2 +4A 4 A 4y +4A c, 2.50) where A 4 is an arbitrary function of z and c is an arbitrary constant of integration. Equation 2.49) was also considered in [5, 3]. Case III. a 4 0: In this case there are three branches corresponding to,y 0j ), j =, 2, 3, where y 0j are the roots of 2.5b). Equation 2.5b) implies that 4 j= y 0j = a 3 a 4, y 0i y 0j = 2 + a 2 ), a 4 i j 3 y 0j = ) a 4 j= If the resonances except r 0 =, which is common for all branches) are r ji, i =, 2, corresponding to y 0j and if one sets P y 0j )= [ a 3 y 2 0j 22 + a 2 )y 0j 8 ], j =, 2, 3, 2.52)

10 Painlevé Test and Higher Order Differential Equations 29 then 2.5a) implies that 2 r ji = P y 0j )=p j, 2.53) i= where p j are integers and in order to have a principal branch at least one of them is positive. Equations 2.5) and 2.52) give p =6 y ) 0 y ) 0, p 2 =6 y ) 02 y ) 02, y 02 y 03 y 0 y 03 p 3 =6 y ) 03 y ) ) y 0 y 02 and hence, the p j satisfy the following Diophantine equation 3 j= p j = ) Moreover equation 2.54) gives that p j = y 0 y 02 y 03 ) 2 y 0 y 02 ) 2 y 0 y 03 ) 2 y 02 y 03 ) 2, 2.56) j= if 0, that is, if p > 0, then either p 2 or p 3 is a negative integer. One should consider the case = 0 separately. III.a: = 0: In this case the sum of the resonances for all three branches is fixed and 2 r ji =7, j =, 2, ) i= Under this condition the solutions of the Diophantine equation 2.55) are p,p 2,p 3 )= 0, 0, 30) and 0, 2, 60). III.a.i: p,p 2,p 3 )=0, 0, 30): Equation 2.54) can be written as where p y 02 y 03 )=ky 0, p 2 y 03 y 0 )=ky 02, p 3 y 0 y 02 )=ky 03, 2.58) k = 6 y 0 y 02 y 03 y 0 y 02 )y 02 y 03 )y 0 y 03 ). 2.59) For k = ±0 5 the system 2.58) has the nontrivial solutions y 0j, j =, 2, 3. For these values of y 0j the resonances and the coefficients a i, i =2, 3, 4,are y 0 = ν ) 5 : r,r 2 )=2, 5), y 02 = ν + ) 5 : r 2,r 22 )=2, 5), y 03 =6ν : r 3,r 32 )= 3, 0), a 2 = 2 ν, a 3 = 2 ν 2, a 4 =, ν = constant, 2.60) 4ν3

11 292 U Muğan and F Jrad for both values of k. For these values of y 0j and a i the simplified equation passes the Painlevé test for all branches. The linear transformation 2.9) and the compatibility conditions at the resonances of the first and second branches are enough to determine all coefficients A k z) of the nondominant terms. The canonical form of the equation for this case is, y =2y 2 +72y 2 y +54y 4 + c, 2.6) where c is an arbitrary constant. Equation 2.6) can be obtained with the choice of ν =/ 5 ) and replacement of y with 6y/ 5 ). Equation 2.6) was given in [5, 8]. III.a.ii: p,p 2,p 3 )=0, 2, 60): For this case equation 2.58) has nontrivial solution y 0j for k = ±20 3. Then y 0j, a i and the corresponding resonances are as follows: y 0 = ± ) 3 3 : r,r 2 )=2, 5), y 02 = ± ν ν : r 2,r 22 )=3, 4), y 03 = 6 ± ) 3 : r 3,r 32 )= 5, 2), ν a 2 =3 7 ± 3 3 ν, a 3 = 40 ± 4 3 ν 2, a 4 = 7 ± 3 3 ν 3, ν = constant. 2.62) By using the linear transformation 2.9) one can choose ν = ± 3andA = A 2 =0. All other coefficients A k of the nondominant terms can be determined from the compatibility conditions at the resonances of the first and second branches. The canonical form for this case is as follows: y = 27 ± c y 2 + y 4) + 20 ± 42 3 y 2 y ± ) 3 y + y ± 43 3 c ) 32 or y =6y 2 y ± 7 ) y 3 + y 2) ) 3 c y ) 3 c y 2 9 ± 7 ) 3 c , 2.64) where c =44c/ ). Equation 2.64) was considered in [8]. III.b: 0: Sincep,p 2 > 0, p 3 < 0, equation 2.56) can be written as p p 2 ˆp 3 =6n 2, 2.65) where n is a constant and ˆp 3 = p 3. Then the Diophantine equation 2.55) yields p p 2 =ˆp 3 p + p 2 ) n ) and, since p p 2 ) 2 0, then p + p 2 ) 2 4ˆp 3 p + p 2 )+4n )

12 Painlevé Test and Higher Order Differential Equations 293 Therefore 0 < ˆp 3 n. Hence one may assume that n is a positive integer. When ˆp 3 = n, equations 2.65) and 2.66) give p,p 2,p 3 )=6,n, n) as the solution of the Diophantine equation. For the case of ˆp 3 <n, if one assumes that p <p 2 if p = p 2, 2.66) implies that p and p 2 are complex numbers), then the Diophantine equation 2.55) implies that p < 2. Equations 2.55) and 2.66) give that p ˆp 3 ) 2 = n 2 [6p 6 p )ˆp 3 ], p p 2 ) 2 = n 2 [6p +6 p )p 2 ] 2.68) for p < 6 and for 6 <p < 2 respectively. Equation 2.68) imply that [6p 6 p )ˆp 3 ] and [6p +6 p )p 2 ] must be squares of integers. By the use of these results, p j the multiplication of the resonances for the branches corresponding the y 0j, j =, 2, 3, are p,p 2,p 3 )=4, 6, 0), 5, 870, 26), 5, 95, 2), 7, 4, 722), 7, 38, 399), 7, 33, 54), 8, 22, 264), 8, 6, 48), 9, 5, 90), 0, 4, 20),, 3, 858). 2.69) For each values of p,p 2,p 3 ) given in 2.69) one should follow the given steps below for 4, 6, 0). When p,p 2,p 3 ) = 4, 6, 0), p = 4implies that possible integral values of r i, i =, 2arer,r 2 )=, 4),, 4). Then r j + r j2 = y 0j +7, j =, 2, 3, 2.70) implies that y 0 = 2/ and y 0 = 2/ for r,r 2 )=, 4),, 4) respectively. On the other hand y 0j satisfies equation 2.58) for k = ±20. For k = 20, y 02 = 9y 0 /4, but the resonance equation for the second branch r 2 2i 7 + y 02 )r 2i + p 2 =0, i =, 2, 2.7) implies that 7 + y 02 be an integer. So, in order to have integer resonances r 2,r 22 )for the second branch, y 02 has to be integral. A similar argument holds for the third branch, but for k = 20, both y 0 and y 02 are not integers. Also for k = 20 the resonances for the second and third branches are not integers. Following the same steps one cannot find the integral resonances for the second and third branches for all other cases of p,p 2,p 3 ) given in 2.69). When p,p 2,p 3 )=6,n, n), equation 2.58) has a nontrivial solution y 0j for k = ±n. We have y 0 =0,y 02 = y 03 for k = n and y 0 =2ν, y 02 = ν6 n), y 03 = ν6 + n) for k = n, where ν is an arbitrary constant. Since y 0 =0fork = n, thiscaseisnotbe considered. For k = n, 2 + a 2, a 3 and a 4 can be determined from equation 2.5) to be 80 n2 2 + a 2 = 2ν36 n 2 ), 2 3 = ν 2 36 n 2 ), 4 = 2ν 3 36 n 2 ). 2.72) Since p = 6, then all possible distinct integral resonances for the first branch are r,r 2 ) =, 6), 2, 3),, 6), 2, 3). Because of the double resonance, r 0 = r =, the case, 6) is not considered. When r,r 2 )=, 6), equation 2.70) implies that = 0. This case was considered in case III.a. For the other possible resonances, 2, 3)

13 294U Muğan and F Jrad and 2, 3), one can obtain the a i, i =, 2, 3, 4, and y 0j, j =, 2, 3. Once the coefficients of the resonance equation 2.5a) are known one should look at the distinct integer resonances for the second and third branches. We have only two cases, such that all the resonances are distinct integers for all branches. The resonances and the corresponding simplified equations are as follows: III.b.i: y 0 = 2 : r,r 2 )= 2, 3), y 02 = 6 n) : r 2,r 22 )=,n), y 03 = 6 + n) : r 3,r 32 )=, n), a [ y = yy n2 ) 236 n 2 ) y 2 2 ] 36 n 2 y 2 y n 2 ) a2 y 4, n, ) It should be noted that as n the simplified equation reduces to 2.27). III.b.ii: y 0 = 2 : r,r 2 )=2, 3), y 02 = n ) : r 2,r 22 )=6,n/6), 6 y 03 = + n ) : r 3,r 32 )=6, n/6), 6 [ y = yy 468 n n 2 y n 2 y 2 y n 2 a2 y 4 ], n 6, ) The canonical form of the equations corresponding to the above cases can be obtained by adding the nondominant terms with the analytic coefficients A k, k =,...,7. III.b.i: By using the transformation 2.9) one can set A 3 = A 4 = 0 and = 2. The compatibility conditions at the resonances imply that all the coefficients are zero except A 6 and A 7 which remain arbitrary for n =2. Forn =3A 7 is arbitrary and all the other coefficients are zero. For n =4, 5 and 6 all coefficients A k are zero. However, it was proved in [7] that for n 4the equation does not admit nondominant terms. The canonical forms of the equations for n = 2 and n = 3 are y =2yy y y2 y + 8 y4 + A 6 y + A 7, 2.75) y =2yy y y2 y y4 + A 7, 2.76) respectively. Equations 2.75) and 2.76) were given in [5] and [8], and both can be linearized by letting y = 2u /u and y = 3u /2u respectively. III.b.ii: The linear transformation 2.9) and the compatibility conditions at the resonances of the first and second branches give the canonical form as y = 2yy 26 2m2 + m 2 y y m 2 + y 2) y 2 + A 5 y + y 2) m2 A 5 ) 48 2 A2 5 + c z + c 2, 2.77)

14 Painlevé Test and Higher Order Differential Equations 295 where m =6/n, m, 6, c and c 2 are arbitrary constants and A 5 is an arbitrary function of z. Equation 2.77) was given in [5] and [8] and is equivalent to y + y 2 = m2 48 A 5 m2 u, u =6u Fourth order equations: P 4) II 4m 2 ) c z + c 2 ). 2.78) Differentiation of 2.3) with respect to z gives the terms y 4), yy, y y, y 2 y, yy 2, y 3 y, all of which are of order 5 forα =, as z z 0. Addition of the term y 5, which is also of order 5, gives the following simplified equation y 4) = yy + a 2 y y + a 3 y 2 y + a 4 yy 2 + a 5 y 3 y + a 6 y 5, 3.) where a i, i =,...,6, are constants. Substitution of 2.4) into 3.) gives the following equations for resonance r and for y 0, respectively, Qr) =r +) { r 3 + y 0 )r 2 [ a 3 y a 2 )y 0 46 ] r a 5 y a 3 + a 4 )y a 2 )y 0 96 } =0, a 6 y0 4 a 5 y0 3 +2a 3 + a 4 )y a 2 )y 0 24= a) 3.2b) Equation 3.2b) implies that in general there are four branches of Painlevé expansion, if a 6 0, corresponding to the roots y 0j, j =, 2, 3, 4. Now one should determine y 0j and a i such that at least one of the branches is the principal branch. Depending on the number of branches there are four cases. Each case should be considered separately. Case I. a 5 = a 6 =0,2a 3 + a 4 = 0: In this case there is only one branch which should be the principal branch. There are following two subcases which will be considered separately. I.a: = 0: In this case the equation 3.2.a) gives that the resonances r,r 2,r 3 ) additional to r 0 = ) satisfy 3 3 r i =, r i = 24. Under these conditions the only i= i= possible distinct positive integer resonances are r,r 2,r 3 )=, 4, 6). Then 3.2) implies that a 3 = 0 and y 0 = 2/a 2. Therefore the simplified equation is y 4) = a 2 y y. 3.3) To obtain the canonical form of the equation, one should add the nondominant terms, viz y, yy, y, y 2, y 2 y, yy, y, y 4, y 3, y 2, y,, that is terms of order greater than 5 as z z 0 with coefficients A k z), k =,...,2, which are analytic functions of z. The coefficients A k can be determined by using the linear transformation 2.9) and the compatibility conditions at the resonances. One can choose a 2 = 2, A 2 =0and2A 3 A 6 + A 9 =0 by the linear transformation 2.9). The compatibility conditions, that is the arbitrariness of y, y 4, y 6, give that A 3 A 2 3 =0, A + A 2 = A 3 /3, A 4 =6A, A 5 = A 8 = A 9 =0, A 6 =2A 3, A 7 A 3 A 7 =2A A 3 A +2A 2 A 3, A 0 = A 3 A A 3, A =A 7 A 0 ) A A 7 A 0 ), A 2 + A A 2 = 6 A 7 A 0 ) )

15 296 U Muğan and F Jrad According to the solution of 3.4a), there are the four following cases: I.a.i: A 3 =0,A = 0: Then the canonical form of the equation is y 4) = 2y y +c z + c 2 )y + c y + c z + c 2 ) 3 + c 3, 3.5) 8c where c i, i =, 2, 3, are arbitrary constants. Integration of 3.5) once gives y = 6y 2 +c z + c 2 )y + 72c 2 c z + c 2 ) 4 + c 3 z + c 4, 3.6) where c 4 is an integration constant. If c 0 and c 2 = 0, then the equation 3.6) takes the form of 2.4). For c = 0 and c ) yields 2.3). I.a.ii: A 3 =0,A =/z c): Without loss of generality one can choose the constant of integration c as zero. Then the canonical form of the equation is y 4) = 2y y + z y + 6 z y 2 +c z c 2 )y + c 2 z y + 24 c2 z 3 9 c c 2 z c2 2z + c 3 z. 3.7) If c = c 2 = 0, then 3.7) is equivalent to u = z u + c 3), y = 6y 2 + u. 3.8) If c = 0 and c 2 0, after replacement of z by γz, y by βy, such that βγ = and c 2 γ 3 =6 equation 3.7) takes the form y 4) = 2y y + z y + 6 y 2 + y ) 6y +3z + c 3 z z, 3.9) where c 3 = c 3 γ 4.Ifc 0 and c 2 = 0, the equation 3.7) takes the form y 4) = 2y y + y +6y 2) +2zy +6z 3 + c 3 z z, 3.0) where c 3 is an arbitrary constant. I.a.iii: A 3 =6/z c) 2 : For simplicity let c = 0. Then the canonical form of the equation is where y 4) = 2y y + A y +6y 2) + 6 z 2 y +2yy ) + A 7 y + A 0 y 2 + A y + A 2, 3.) A = 2c z 3 c 2 z c z 3 + c 2 ), A 7 = c z 3 + c 2 5 c c 3 z c 2c 3 z 3 + c c 4 z 24c + c 2 c 4 z 2 6c 2 z 3 A 0 = 2 z 3 6 c z 3 + c 2 2c c 2 z 3), ),

16 Painlevé Test and Higher Order Differential Equations 297 A = c z 3 + c 2 A 2 = c c z0 + c 2c c 5 z c c c 2c 2 4 c z 4 + c 2 z) 2 [ 24 z z7 + c c 3 c 4 60 z5 + c 2c 3 c 5 ), 5 z2 c 2 c 3 z 0 48c 2 z 8 + 4c c 2 c 3 z 7 +5c 2 5 c 4 z 5 c2 2 c 3 5 z4 +4c c 2 c 4 z 2 42c c 2 z c 2 2c 3 z +6c 2 2z 2 c z 4 + c 2 z ) 6c c 3 5 z6 + 3c 2c 3 5 z3 48c + c c 4 2c 2 c 4 z 2 +6c 2 z 3 )], 3.2) where c i, i =,...,5, are constants. Equations 3.5), 3.7), 3.9), 3.0) and 3.) were considered in [3, 9, 7]. I.a.iv: A 3 =6Pz;0,α ): If one replaces A 3 with 6Â3 in the equation 3.4a), then Weierstrass elliptic function Pz;0,α ), where α is an arbitrary integration constant, is a solution of the resulting equation. If one lets A = à /Ã, then the equation 3.4b) gives Lamé s equation for Ã. Therefore, à = c E z)+c 2 F z), 3.3) where c and c 2 are constants and E z) andf z) are the Lamé functions of degree one and of the first and second kind respectively, which were given in 2.22). Similarly replacement of A with 6Â, A 3 with 6Â3 and A 7 with 6Â7 in the equation for A 7 in 3.4) yields  7 6Pz;0,α )Â7 =72  3  + Â2  3 ). 3.4) So the homogenous solution of the above equation is nothing but the Weierstrass elliptic function Pz;0,α ). Therefore the compatibility conditions 3.4) allow one to determine all the nonzero coefficients A k z) in terms of the Weierstrass elliptic function Pz). I.b: 0: Equation 3.2a) implies that r r 2 r 3 = 24. Under this condition there are four possible cases of r,r 2,r 3 ) such that r i > 0 and distinct integers, but there is only the following case out of the four cases such that the compatibility conditions at the resonances for the simplified equations are identically satisfied and y 0 0 r,r 2,r 3 )=2, 3, 4), y 0 = 2/, a 2 =3, a 3 = a 4 =0. 3.5) By adding the nondominant terms to the simplified equation, using the linear transformation 2.9) and the compatibility conditions one finds the canonical form of the equation y 4) = 2yy 6y y + A y +2yy +2y 2) + A 3 y +2yy )+A 7 y + y 2) + A 2, 3.6) where A, A 3, A 7 and A 2 are arbitrary functions of z. If one lets u = y 2 + y, then the equation 3.6) can be linearized. Equation 3.6) was considered in [3, 9]. Case II. a 5 = a 6 = 0: In this case there are two branches corresponding to,y 0j ), j =, 2, where y 0j are the roots of 3.2.b) and y 0 + y 02 = 23 + a 2 ) 2a 3 + a 4, y 0 y 02 = 24 2a 3 + a )

17 298 U Muğan and F Jrad Let r j,r j2,r j3 ) be the roots additional to r 0 = ) of the resonance equation 3.2a) corresponding to y 0j. When one sets P y 0j )= 22a 3 + a 4 )y 2 0j a 2 )y 0j +96, j =, 2, 3.8) 3.2a) implies that 3 r ji = P y 0j )=p j, j =, 2, 3.9) i= where the p j are integers and at least one of them is a positive integer in order to have the principal branch. Let the branch corresponding to y 0 be the principal branch, that is p > 0. Equations 3.7) and 3.8) give P y 0 )=24 y 0 y 02 ) = p, Py 02 )=24 y ) 02 = p ) y 0 Hence the p j satisfy the following Diophantine equation, if p p 2 0, + = p p ) There are 2 integer solutions p,p 2 ) of 3.2) such that one of the p j is positive. Once p is known, for each p one can write possible distinct positive integers r,r 2,r 3 )such 3 that r i = p. Thenforeachsetofr,r 2,r 3 ), a k, k = 2, 3, 4, and y 0j can be i= determined in terms of by using 3 r ji =+ y 0j, i= r ji r jk = a 3 y0j a 2 )y 0j +46, 3.22) i k for j =, and the equation 3.7). Then for these values of a k and y 0j one should find all cases such that the resonance equation 3.2a) has distinct integral roots r 2i corresponding to y 02. Only for the following cases a) p,p 2 )=2, 24) and b) p,p 2 )=20, 20) are all the resonances distinct integers for both branches, one of which is the principal branch. The resonances and the simplified equations for these cases are as follows: II.a: p,p 2 )=2, 24) : y 0 = 3 : r,r 2,r 3 )=, 3, 4), y 02 = 6 : r 2,r 22,r 23 )= 2, 3, 4), y 4) = yy +3y y 3 y 2 y 2 3 yy 2 ), 3.23) II.b: p,p 2 )=20, 20) : y 0 = : r,r 2,r 3 )=, 4, 5), y 02 = 6 : r 2,r 22,r 23 )= 5, 4, 6), y 4) = yy +y y y 2 y 2 yy 2). 3.24)

18 Painlevé Test and Higher Order Differential Equations 299 For case II.a the compatibility conditions at the resonances of the simplified equation are identically satisfied. For the case II.b the compatibility condition at the resonance r 3 = 5 implies that y 4 = 0 which contradicts with the arbitrariness of y 4. Moreover in the case II.b, if one lets y = λu such that λ =, integration of the simplified equation once yields u = uu +5u 2 u 2 u + c, 3.25) where c is an arbitrary integration constant. Equation 3.25) is not a Painlevé type equation unless c = 0. It was studied in [3, 7]. Hence we consider case II.a. Adding the nondominant terms to the simplified equation and by using the linear transformation 2.9) and the compatibility conditions of the first branch we can determine the coefficients A k z) of the nondominant terms. The canonical form of the equation for the case II.a is y 4) = 3yy 9y y 3y 2 y 6yy 2 + Ry +2R y + R y + A 9 y 3 +3yy + y Ry ) + A 2, 3.26) where Rz) =A 3 z) A 9 z) anda 3 and A 9 are arbitrary analytic functions of z. Ifone lets u = y +3yy + y 3 Ry, 3.27) then equation 3.26) can be reduced to a linear equation for u. Equation 3.26) was considered in [3, 9]. Case III. a 6 = 0: There are three branches corresponding to y 0j, j =, 2, 3, which are the roots of the equation 3.2b). If one lets 3 r ji = p j = P y 0j )=a 5 y0j 3 22a 3 + a 4 )y0j 2 i= a 2 )y 0j +96, j =, 2, 3, 3.28) where p j are integers and at least one of them is positive, by the use of the same procedure as was carried in the previous case the p j satisfy the following Diophantine equation: 3 j= p j = 24, 3.29) if p p 2 p 3 0 and, if 0, p j = y 0 y 02 y 03 ) 2 y 0 y 02 ) 2 y 0 y 03 ) 2 y 02 y 03 ) ) j= Let p,p 2 > 0andp 3 < 0. If r j,r j2,r j3 ) are the resonances corresponding to y 0j respectively, then they satisfy equation 3.22) for j =, 2, 3. There are the following two cases which should be considered separately.

19 300 U Muğan and F Jrad III.a: = 0: Equation 3.22a) for j = implies that there are five possible values of r,r 2,r 3 ) and hence five possible values of p. For each value of p one can solve 3.29) such that p 2 > 0, p 3 < 0 and both are integers. Then for each p,p 2,p 3 ) the equations p =24 y ) 0 y ) 0, p 2 =24 y 02 y 03 p =24 y ) 03 y 03 y 0 y 02 give the equations 2.58) for y 0j for k = y 02 y 0 ) y 02 y 03 ), ), 3.3) 24 y 0 y 02 y 03 y 0 y 02 )y 02 y 03 )y 0 y 03 ). 3.32) The system 2.58) has nontrivial solution if k 2 = p p 2 +p p 3 +p 2 p 3 ). For each value of k one can find y 0j and a i, i =3, 4, 5, in terms of a 2. Once the coefficients of the resonance equation 3.2a) are known for all branches, one should look at the cases such that the roots of 3.2a) are distinct integers for the second and third branches. There is only one case, p,p 2,p 3 )=40, 40, 20), and k =40 5. The y 0j, the resonances and the simplified equation for this case are as follows: y 0 = 4 ) 5 : r,r 2,r 3 )=2, 4, 5), a 2 y 02 = 4 + ) 5 : r 2,r 22,r 23 )=2, 4, 5), a 2 y 03 = 24 : r 3,r 32,r 33 )= 3, 4, 0), a 2 y 4) = a 2 y y + 8 a 2y 2 y + 4 a 2yy 2 + ) 64 a2 2y 3 y. 3.33) The compatibility conditions are identically satisfied for the simplified equation. To obtain the canonical form of the equation one should add the nondominant terms with analytic coefficients A k z), k =,...,2. The linear transformation 2.9) and the compatibility conditions at the resonances of the first and second branches give the following equation y 4) =24y y +72y 2 y +44yy y 3 y ) Integration of 3.34) once gives 2.6). III.b: 0: In this case the resonances r j,r j2,r j3 )andy 0j satisfy 3.22) for j =, 2, 3and 3 y 0j = 2a 3 + a 4 ), a 5 i= y 0j y 0k = a 2 ), a 5 j k 3 i= y 0j = 24 a 5, 3.35) respectively. The p j = n 2 = 3 j= r ji satisfy the Diophantine equation 3.29). If one lets 24 2 y 0 y 02 y 03 ) 2 y 0 y 02 ) 2 y 0 y 03 ) 2 y 02 y 03 ) 2, 3.36)

20 Painlevé Test and Higher Order Differential Equations 30 then 3.30) gives p p 2 ˆp 3 =24n 2, 3.37) where ˆp 3 = p 3 and p < 48. If one follows the procedure given in the previous section, 3.29) and 3.37) give that p ˆp 3 ) 2 = n 2 [24p 24 p )ˆp 3 ], p p 2 ) 2 = n 2 [24p +24 p )p 2 ], 3.38) for p < 24and for 24< p < 48 respectively. So the right hand sides of both equations in 3.38) must be complete squares. Based on these conditions on p i, i =, 2, 3, there are 7 integer solutions p,p 2,p 3 ) of the Diophantine equation 3.29). For each solution p,p 2,p 3 ), one can find y 0j by solving the system of equations 2.58). Then one can write possible resonances r,r 2,r 3 )foreachp provided that 3 y 0 = r i 3.39) i= are all integers. There are the following three cases such that all the resonances of all three branches are distinct integers. III.b.i: p,p 2,p 3 )=5, 60, 24) : y 0 = 2 : r,r 2,r 3 )=, 3, 5), y 02 = 2 : r 2,r 22,r 23 )= 2, 5, 6), y 03 = 8 : r 3,r 32,r 33 )= 4,, 6), a 2 = 2, a 3 = 2 a2, a 4 = 7 4 a2, a 5 = 8 a ) III.b.ii: p,p 2,p 3 )=24,n, n), n > 0, n 24: y 0 = 2 : r,r 2,r 3 )=2, 3, 4), y 02 = n ) : r 2,r 22,r 23 )= 4, 6, n ), y 03 = + n ) :r 3,r 32,r 33 )= 4, 6, n ), n2 a 2 = 576 n 2, a 3 = n 2 a2, a 4 = n 2 a2, a 5 = n 2 a3. 3.4) III.b.iii: p,p 2,p 3 )=24,n, n), n > 0, n 4, 24: y 0 = 2 : r,r 2,r 3 )= 2, 3, 4), y 02 = 6 n 4 ) : r 2,r 22,r 23 )=, 4, n 4 ), y 03 = 6+ n ) : r 3,r 32,r 33 )=, 4, n ), 4 4

21 302 U Muğan and F Jrad a 2 = n2 576 n 2, a 3 = n 2 a2, a 4 = n 2 a2, a 5 = n 2 a ) For all three cases the simplified equations pass the Painlevé test. To obtain the canonical form of the equation one should add the nondominant terms with the coefficients A k z), k =,...,2. The linear transformation 2.9) and the compatibility conditions at the resonances give the following equations: III.b.i: y 4) = 2yy y y 2y 2 y 7yy 2 y 3 y 2 + A 6 y + yy ) + 3 A 6 y 2 +4y ) + 3 A A 6A 6, 3.43) where A 6 is an arbitrary function of z. Equation 3.43) was given in [9]. III.b.ii: The compatibility condition at the resonance r = 6 for the third branch gives A + A 2 = ) So following two subcases should be considered separately. III.b.ii.: A = 0: The canonical form of the equation is y 4) = 2yy 6 [ m 2 m 2 9 ) y y 8y 2 y 6 yy 2 + y 3 y )] + A 3 y +2yy )+A 3 + c ) y + y 2) + A 2, 3.45) where m = n/24, m, 4, 6, A 3 is an arbitrary function of z and A 2 = m2 48 A 3 A 3 A 3 c A 3 +2c 2 z + c 2 ), c,c 2 = constant. 3.46) The result 3.45) was given in [9]. III.b.ii.2: A =/z c): Without loss of generality one can set c = 0. The canonical form of the equation is y 4) = 2yy [ m 2 ) m 2 y y +48y 2 y +96 yy 2 + y 3 y )] + { y +2yy [ 26 2m 2 ) z m 2 y 2 +48y 2 y +24y 4]} ) + A 3 y +2yy )+ A y 3 A 3 z + c z + y 2) + A 2, 3.47) where A 3 is an arbitrary function of z and A 2 = m2 48 A 3 z A 3 A 3 A 3 + 2z A2 3 c za 3 + ) 2 c2 z 3 + c 2 z, 3.48) where c and c 2 are constants. The result 3.47) was given in [9].

22 Painlevé Test and Higher Order Differential Equations 303 is III.b.iii: If we let m = n/4, m, 4, 6, the canonical form of the equation for m =2 If one sets y 4) =2yy +5y y 3 2 y2 y 3yy 2 + [ 2 y3 y + A y 2yy 3 2 y y2 y ] 8 y4 A 7 y + A 7 y + A 7y + A ) u = y 2yy 3 2 y y2 y 8 y4 A 7 y, 3.50) then 3.49) can be reduced to a linear equation for u. It should be noted that 3.50) belongs to P 3) II and was given in 2.75). For m =3 y 4) =2yy y y 6 9 y2 y 32 9 yy y3 y + A y 2yy 7 3 y y2 y 4 27 y4 where A and A 2 are arbitrary functions of z. If one sets ) + A 2, 3.5) u = y 2yy 7 3 y y2 y 4 27 y4, 3.52) 3.5) can be reduced to a linear equation in u. Equation 3.52) belongs to P 3) II and was given in 2.76). Equations 3.49) and 3.5) were given in [9]. It should be noted, that for m 4, integration of the simplified equation once gives the simplified equation of the case given in 2.74) with an additional integration constant c. Thusform 4the simplified equation is not of Painlevé typeifc 0. Case IV. a 6 0: In this case there are four branches corresponding to,y 0j ), 3 j =, 2, 3, 4.Ifr j,r j2,r j3 ) are the resonances corresponding to the branches, r ji = p j such that the p j are integers and at least one of them is positive. Then 3.2a) implies that P y 0j )=a 5 y 3 0j 22a 3 + a 4 )y 2 0j a 2 )y 0j +96=p j, j =, 2, 3, ) On the other hand 3.2b) implies that 4 y 0j = a 5, y 0j y 0i = 2a 3 + a 4, a 6 a 6 j= j i k Then 3.53) yields j i y 0j y 0i y 0k = 23 + a 2 ) a 6, p j = P y 0j )=24 j k 4 j= i= y 0j = 24 a ) y ) 0j, j =, 2, 3, ) y 0k

23 304U Muğan and F Jrad Therefore the p j satisfy the Diophantine equation 4 j= p j = ) To find the simplified equation one should follow the following steps: a) Find all integer solutions p,p 2,p 3,p 4 ) of the Diophantine equation 3.56). b) For each pair p,p 2 )from the solution set of the Diophantine equation, write all possible r j,r j2,r j3 ) such that 3 r ji = p j, j =, 2. c) Determine y 0 and y 02 in terms of,if 0,byusing i= the equation 3.22a) for j =, 2. d) Use 3.55) to find y 03 and y 04 in terms of. e) Eliminate the cases r j,r j2,r j3 ) j =, 2, such that y 0k, k =3, 4, are not integers see the equation 3.22a)). f) Find a i, i =2,...,6, in terms of by using the 3.53) and 3.54). Once all the coefficients of the equation 3.2a) are known, look at the cases such that the roots of 3.2a) are distinct integers for y 03 and y 04. There are four cases such that all the resonances are distinct integers for all branches. These cases and the corresponding simplified equations are as follows: IV.a: p,p 2,p 3,p 4 )=6, 4, 6, 24) : y 0 = 5 : r,r 2,r 3 )=, 2, 3), y 02 = 0 : r 2,r 22,r 23 )= 2,, 2), y 03 = 5 : r 3,r 32,r 33 )= 3, 2, ), y 04 = 20 : r 4,r 42,r 43 )= 4, 3, 2), a y 4) = yy +2y y 2 5 y 2 y 3 5 yy a2 y 3 y ) 625 a3 y 5, 3.57) IV.b: p,p 2,p 3,p 4 )=36, 36, 84, 504) : y 0 = 5 : r,r 2,r 3 )=2, 3, 6), a 2 y 02 = 0 : r 2,r 22,r 23 )=2, 3, 6), a 2 y 03 = 5 : r 3,r 32,r 33 )= 2, 6, 7), a 2 y 04 = 20 : r 4,r 42,r 43 )= 7, 6, 2), a 2 y 4) = a 2 [y y + 5 a 2 y 2 y + yy 2 25 a2 2y 5 IV.c: p,p 2,p 3,p 4 )=36, 36, 44, 44) : y0 2 = 0 : r,r 2,r 3 )=2, 3, 6), a 3 y 02 = y 0 : r 2,r 22,r 23 )=2, 3, 6), )], 3.58)

24 Painlevé Test and Higher Order Differential Equations 305 y03 2 = 40 : r 3,r 32,r 33 )= 3, 6, 8), a 3 y 04 = y 03 : r 4,r 42,r 43 )= 3, 6, 8), y 4) = a 3 y 2 y + yy 2 3 ) 50 a 3y 5, 3.59) IV.d: p,p 2,p 3,p 4 )=20, 20, 60, 60) : y 0 = 2 : r,r 2,r 3 )=, 2, 0), y 02 = 8 : r 2,r 22,r 23 )= 0,, 2), y 03 = 4 : r 3,r 32,r 33 )= 2, 2, 5), y 04 = 6 : r 4,r 42,r 43 )= 3, 2, 0), a y 4) = yy 7 2 y y + 4 y 2 y 5 4 yy a2 y 3 y ) 6 a3 y ) The simplified equation for the case IV.d does not pass the Painlevé test. So this case is not be considered. The canonical forms for the other cases can be obtained by adding the nondominant terms with the coefficients A k z), k =,...,2 to the simplified equations. All coefficients A k can be obtained by using the linear transformation 2.9) and the compatibility conditions at the resonances. The canonical forms are as follows: IV.a: y 4) = 5yy 0 y y + y 2 y + y 3 y ) 5yy 2 y 5 + A y +4yy +3y 2 +6y 2 y + y 4) + A 3 y +3yy + y 3) + A 7 y + y 2) + A y + A ) If one lets y = u /u, 3.6) gives the fifth order linear equation for u. Equation 3.6) was given in [9]. IV.b: y 4) = 5y y +5y 2 y +5yy 2 y 5 +c z + c 2 )y + c 3, 3.62) where c i are constants. The result 3.62) was given in [9]. IV.c: y 4) =0y 2 y +0yy 2 6y 5 + c y 2y 3) +c 2 z + c 3 )y + c 4, 3.63) where c i are constants. The result 3.63) was given in [9, 6]. 4 Fifth order equations: P 5) II Differentiation of 3.) and addition of the term y 6 which is also of order 6 asz z 0 gives the following simplified equation of order five y 5) = yy 4) + a 2 y y + a 3 y 2 + a 4 y 2 y + a 5 yy y + a 6 y 3 + a 7 y 3 y + a 8 y 2 y 2 + a 9 y 4 y + 0 y 6, 4.)

25 306 U Muğan and F Jrad where a i, i =,...,0, are constants. Substitution of 2.4) into 4.) into the equation above gives the following equations for resonance r and for y 0 respectively, { Qr) =r +) r y 0 )r 3 [ a 4 y0 2 + a 2 )y 0 0 ] r 2 [ a 7 y0 3 7a 4 + a 5 )y a 2 +4a 3 )y ] r [ a 9 y0 4 22a 7 + a 8 )y a 4 +2a 5 + a 6 )y a 2 +2a 3 )y ] } =0, 4.2) 0 y 5 0 a 9 y a 7 + a 8 )y 3 0 6a 4 +2a 5 + a 6 )y a 2 +2a 3 )y = ) Equation 4.3) implies that there are five branches if a 6 0. Ifr j,r j2,r j3,r j4 ), j = 4,...,5, are the distinct integer resonances corresponding to,y 0j )andif r ji = p j, where p j are integers and at least one of them is positive, then the p j satisfy the following Diophantine equation, 5 j= = p j 20. i= 4.4) The determination of all integer solutions p,p 2,p 3,p 4,p 5 ) of the Diophantine equation is quite difficult. So, for the sake of completeness, in this section we present special cases such as single, double and triple branch cases. We also give an example for the case of the four branches. Since the procedure to obtain the canonical form of the differential equations is the same as described in the previous sections, we only give the canonical form of the differential equations for each cases. The canonical form of the equation can be obtained by adding the nondominant terms y 4), yy, y y, y 2 y, yy 3, y 3 y, y 5, y, yy, y 2, y 2 y, y 4, y, yy, y 3, y, y 2, y, with coefficients A k z), k =,...,9, which are analytic functions of z, respectively. Case I. If a l =0,l =4,...,0, there is only one branch and there are two cases such that the resonances are distinct positive integers. I.a: y 0 = 2/ : r,r 2,r 3,r 4 )=2, 3, 4, 5), y 5) = 2yy 4) 8y y 6y 2 + A y 4 +2yy +6y y ) + A 8 y +2yy +2y 2) + A 3 y +2yy )+A 6 y + y 2) + A 9, 4.5) where A, A 8, A 3, A 6, A 9 are arbitrary analytic functions of z. Equation 4.5) can be linearized by letting u = y + y 2. I.b: y 0 = 2/a 2 : r,r 2,r 3,r 4 )=, 4, 5, 6). In this case the linear transformation and the compatibility conditions give A i =0,i =,...,7, A = A 2 = A 5 = 0 and A 9 2 A2 9 =0. 4.6)

26 Painlevé Test and Higher Order Differential Equations 307 Depending on the solution of 4.6) there are following two subcases. I.b.i: A 9 = 0. The canonical form of the equation is y 5) = 2y y 2y 2 +c z + c 2 )y +2c y + 6 c z + c 2 ) 2, 4.7) where c and c 2 are constants. If c 0, 4.7) can be reduced to 2.4). If c =0, 4.7) can be reduced to a third order equation which belongs to the hierarchy of the first Painlevé equation, P 3) I [8], by integration once and letting y = u. I.b.ii: A 9 =2/z 2. The canonical form of the equation is y 5) = 2y y 2y ) 3 z 2 2 y + yy +2y 2 + c z 3 + c 2 z 2 24 z 3 48 z 3 yy + 6c z 2 4c 2 z ) z 4 y + 4c + 4c ) 2 z 4 y + 24 z 4 y2 + 6 ) y c z 3 + c 2 z 2 ) 2, 4.8) where c and c 2 are constants. Case II. a 7 = = 0 = 0: In this case there are two branches. The resonances and the canonical form of the equation are y 0 = 3 : r,r 2,r 3,r 4 )=, 3, 4, 5), y 02 = 6 : r 2,r 22,r 23,r 24 )= 2, 3, 4, 5), y 5) = 3yy 4) 2y y 9y 2 8yy y 6y 3 3y 2 y +Ry) + 3 A [ 0 y +3yy +3y 2 y +3y 2 Ry) ] + A 5 y +3yy + y 3 Ry ) + A 9, 4.9) where R = A 8 A 9 /3andA 8, A 9, A 0, A 5 and A 9 are arbitrary analytic functions of z. Equation 4.9) can be linearized if one lets u = y +3yy + y 3 Ry. 4.0) Case III. a 9 = 0 = 0: In this case there are three branches. The resonances and the canonical form of the equations are as follows: III.a: y 0 = 2 : r,r 2,r 3,r 4 )=2, 3, 4, 5), y 02 = n : r 2,r 22,r 23,r 24 )=4, 5, 6,n), y 03 = +n : r 3,r 32,r 33,r 34 )=4, 5, 6, n), y 5) = 2yy 4) + [ 56 8n 2 ) n 2 y y n 2) y 2 +48y 2 y + 288yy y +96 y 3 + y 3 y ) + 288y 2 y 2] + A 8 y +2yy + y 2) +2A 8 + c z + c 2 )y +2yy )+A 8 +2c ) y + y 2) + A 9, 4.)