Department of Mathematics Luleå University of Technology, S Luleå, Sweden. Abstract

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1 Nonlinear Mathematical Physics 1997, V.4, N 4, Transformation Properties of ẍ + f 1 t)ẋ + f t)x + f t)x n = 0 Norbert EULER Department of Mathematics Luleå University of Technology, S Luleå, Sweden Abstract In this paper, we consider a general anharmonic oscillator of the form ẍ + f 1 t)ẋ + f t)x+f t)x n = 0, with n Q. We seek the most general conditions on the functions f 1, f and f, by which the equation may be integrable, as well as conditions for the existence of Lie point symmetries. Time-dependent first integrals are constructed. A nonpoint transformation is introduced by which the equation is linearized. 1 Introduction Recently we have reported some results on the integrability of the nonlinear anharmonic oscillator ẍ + f 1 t)ẋ + f t)x + f t)x n = 0. 1) Here ẋ dx/dt, ẍ d x/dt and n Q. Conditions on the functions f 1, f, and f as well as the constant n were derived for which the equation admits point transformations in integrable equations. The Lie point symmetries were obtained only for the case where f 1, f and f are constants. The Painlevé analysis for special cases of n was performed. For more details, we refer to the papers of Euler et al [6], Duarte et al [] and Duarte et al []. In the present paper, we generalize those results, introduce a nonpoint transformation which linearizes 1), and do a Lie point symmetry classification of 1), whereby conditions for the existence of Lie point symmetries are given on f 1, f and f. Before doing so, we would like to make some literatorical remarks on point transformations, nonpoint transformations, and integrability of ordinary differential equations ODEs), relevant in the present considerations. In being faced with a nonlinear ordinary differential equation NODE), one unsually wants to construct its general solution. If the general solution can be obtained, the equation is said to be integrable. Constructing such solutions for NODEs is in general difficult. In fact, in most cases the general solution of NODEs cannot be obtained in closed form, so that one has to be satisfied by solving the equation numerically or by constructing some special exact solutions. Much attention has been focused on the classification of NODEs as integrable and nonintegrable ones. In the case of second order ODEs, the construction Copyright c 1997 by Mathematical Ukraina Publisher. All rights of reproduction in form reserved.

2 TRANSFORMATION PROPERTIES 11 of a first integral is of fundamental importance. It is desirable to have a simple approach to obtaining time-dependent first integrals of NODEs. Several methods for the identification of integrable ODEs have been proposed. A method dating back to the beginning of the development of differential calculus, is to find a coordinate transformation which transforms a particular differential equation in a differential equation with a known general solution. To find a transformation which transforms a NODE in a linear ODE would certainly be a way in which to solve the NODE in general. In particular, the problem of linearizing second-order ODEs has been of great interest. The utilization of point transformations for the linearization is the usual procedure see, for example Duarte et al [1], Sarlet et al [1], and Moreira [11]). Since the time of Tresse [17], it is known that the most general second-order ODE which may be linearized by a point transformation, is of the form ẍ + Λ x, t)ẋ + Λ x, t)ẋ + Λ 1 x, t)ẋ + Λ 0 x, t) = 0, ) whereby the functions Λ j must satisfy the following conditions: Λ 1xx Λ xt + Λ tt + 6Λ Λ 0x + Λ 0 Λ x Λ Λ 1t Λ 1 Λ t Λ Λ 1x + Λ Λ t = 0, Λ tt Λ 1xt + Λ 0xx 6Λ 0 Λ t Λ Λ 0t + Λ 0 Λ x + Λ Λ 0x + Λ 1 Λ t Λ 1 Λ 1x = 0. ) We use the notation Λ 1x Λ 1 / x, Λ 1xx Λ 1 / x, etc.) In fact, ) is the most general second-order ODE which may be point transformed by the invertible point transformation XT ) = F x, t), in the free particle equation d X dt = 0. T x, t) = Gx, t), T, X) t, x) Transformation 4) is obtained by solving F and G from Λ = G x F xx G xx F x ) 1, Λ = G t F xx + G x F tx F x G tx F t G xx ) 1, Λ 1 = G x F tt + G t F tx F t G tx F x G tt ) 1 Λ 0 = G t F tt G tt F t ) 1. 0, 4) Here G t F x G x F t 0. The compatibility condition of system 6) is given by ). If the point transformation 4) is known, the first integrals, Lie point symmetries, and general solution of 5) may be used to obtain the corresponding ones for ). We use this result in Section 4 in the classification of Lie point symmetries for 1). In particular, the first integral of 5) is I dx dt ) = dx dt, so that the first integral of ) takes the form It, x, ẋ) = F t + F x ẋ G t + G x ẋ ; 5) 6)

3 1 N. EULER which is, in general, a time-dependent first integral. The free particle equation 5) admits eight Lie point symmetry generators forming the sl, R) Lie algebra under the Lie bracket. Those Lie point symmetry generators are G 1 = T, G 6 = T X, G = X, G = T T, G 4 = X X, G 5 = X T, G 7 = T T T + X ), G 8 = X T X T + X ). X This is the maximum number of Lie point symmetries which any second-order ODE might admit. In fact, any nonlinear second order ODE may admit the sl, R) Lie point symmetry algebra provided it admits eight Lie point symmetries. In such a case a point transformation can be found which would linearize the equation, i.e., transform the equation in the free particle equation 5). This leads to the statement: A necessary and sufficient condition for a second-order ODE to be linearizable by a point transformation, is that the equation admits the sl, R) Lie point symmetry algebra. Linearization by point transformations was studied in some detail by several authors see for example the works of Leach [9], Sarlet et al [1], and Duarte et al [1]). An example of a nonlinear second-order ODE that admits the sl, R) Lie point symmetry algebra is the equation Leach [9]) ẍ + αxẋ + α 9 x = 0, 7) where α is an arbitrary real constant. This equation plays an important role in our Lie point symmetry classification of 1) see Section 4). The general solution of 7) was obtained by Duarte et al [1] by the invertible point transformation XT ) = t x 1 6 αt, T x, t) = 1 x 1 αt, which transforms 7) in the free particle equation 5). The general solution of the free particle equation is XT ) = k 1 T + k, so that the general solution of 7) follows: t k 1 xt) = 1 6 αt 1 k 1 αt + k. 8) Here k 1 and k are integrating constants. This result is used to solve some of the equations in Table 1 and Table of Section 4. It is clear that if one is able to find the invertible point transfromation by which a NODE may be linearized, the general solution of the NODE is easily obtained. We refer to the book of Steeb [14]. Since 1) is not linearizable by a point transformation, we aim to find point transformations in other integrable equations Section and Section 5), and to linearize 1) by a nonpoint transformation Section ). If a NODE admits a Lie point symmetry, the symmetry may be used to calculate point transformations which transform the NODE either in an autonomous ODE or an ODE with lower order. A Lie point symmetry classification of 1) is performed in Section 4. For more details on Lie point symmetries, we refer to the books of Olver [1], Fushchych et al [8] and Steeb [15].

4 TRANSFORMATION PROPERTIES 1 The problem of classifying second order ODEs with respect to the singularity structure of their soluitions, was considered by a school of French Mathematicians under the leadership of P. Painlevé in the period from 189 till 190. They classified the equation A 1 x, t)ẍ + A x, t)ẋ + A x, t)ẋ + A 4 x, t) = 0, m j A j x m j = 0, j = 1,..., 4 9) m 1,..., m 4 may be different integers) with respect to the following classification criterion: The critical points of solutions of 9), that are branch points and essential singularities, should be fixed points. Any function which is a solution of an equation of this class of ODEs would, therefore, have only poles as movable singularities. They obtained fifty second-order ODEs. The equations satisfying the above criterion are said to have the Painlevé property. Fourty-four of these fifty equations can be solved by standard functions. The remaining six are known as the Painlevé transcendents; they define transcendental functions. It is important to note that the Painlevé transcendents admit no Lie point symmetry transformations. The classification of 9) was done under the Mobius group of transformations XT ) = ψ 1t)x + ψ t) ψ t)x + ψ 4 t), T = φt), where ψ j and φ are analytic functions of t. ODE, one could ask the question: Given a particular nonlinear second-order Does there exist an invertible point transformation which may transform a given nonlinear ODE in one of the integrable second-order ODEs classified by Painlevé? This is generally a difficult question to answer. In our paper, Euler et al [7], an invertible point transformation was obtained for an anharmonic oscillator of the form 1) by which the equation may be transformed in the second Painlevé transcendent. We discuss this result in Section 5 of the present paper in detail. It is clear that the point transformation 4) preserves the Lie point symmetry stucture as well as the integrability structure of a given ODE. By introducing a nonpoint transformation of the form XT ) = F x, t), dt x, t) = Gx, t)dt, 10) one preserves only the integrability structure and not the symmetry structure of the equation. A transformation of this type was considered by Euler et al 1994) in their calculations of approximate solutions of nonlinear multidimensional heat equations. Duarte et al 1994) made use of transformation 10) and obtained equations which may be nonpoint transformed in the free particle equation 5). They showed, by way of examples, transformation 10) may lead to the linearization of NODEs not linearizable by a point transformation. In Section of the present paper, we utilize this transformation for the linearization of 1).

5 14 N. EULER First integrals by point transformations In this section we consider the problem of constructing invertible point transformations of the form 4), i.e., XT ) = F x, t), T x, t) = Gx, t), for equation 1), i.e., ẍ + f 1 t)ẋ + f t)x + f t)x n = 0. Note that 1) is a special case of ). That is, for Λ = Λ = 0, Λ 1 = αt) equation ) takes the form ẍ + α 1 t)ẋ + Λ 0 x, t) = 0. 11) By condition, ) it follows that 11) may be linearized by a point transformation of the form 4) if and only if Λ 0 is a linear function of x, where α is an arbitrary function of t. This leads to the following result: Equation 1), with n / {0, 1}, cannot be linearized by a point transformation. We now consider the integrable equation d X dt + Xn = 0, which admits the first integral I X, dx ) = 1 dt ) dx + Xn+1 dt n ) By the point transformation 4) equation 1) takes the form where ẍ + A ẋ + A ẋ + A 1 ẋ + A 0 = 0, 1) A = F xx G x G xx F x + G xf n) 1, A = G t F xx + G x F xt F x G xt F t G xx + G t G xf n) 1, A 1 = G x F xt + G t F xt F t G xt F x G tt + G t G x F n) 1, A 0 = G t F tt F t G tt + G t F n) 1 14) and F x G t F t G x 0. In order to obtain an equation of the form 1), we set F x, t) = ft)x, Gx, t) = gt), 15) where f, g are smooth functions, to be determined in terms of the coefficient functions of 1), namely f 1, f and f. System 14) leads to A = A = 0, A 1 = fġ f g, A 0 = ġ f f g x + ġf n 1 x n. fġ fġ

6 TRANSFORMATION PROPERTIES 15 The functions f 1, f and f then take the form f 1 t) = f f ġ g, We can state the following Theorem 1: Equation ẍ + f 1 t)ẋ + f t)x + f t)x n = 0 may be point transformed in the equation d X dt + Xn = 0, by the transformation f t) = f f f g f ġ, f t) = ġ f n 1. 16) XT ) = ft)x, T x, t) = gt) in the following cases: a) For n {, 0, 1} the transformation coefficients are t ) ft) = Cf 1/n+) f 1 ζ) t) exp n + dζ, 17) gt) = t 1/ f ζ) f n 1)/ dζ 18) ζ) with the following conditions on the equation coefficients f = 1 f n + 4 f n + f n + ) f ) + n 1 n + ) b) For n = the transformation coefficients are gt) = f f ) 1 f 1 + f n n + 1 n + ) f 1.19) t ρ ) f ρ) exp φζ) dζ dρ, 0) t ) ft) = exp φζ) dζ, 1) where φ is the solution of the Riccati equation φ = φ f 1 t)φ + f t). ) The condition on the equation coefficients is f 1 t) = 1 f. ) f

7 16 N. EULER To prove Theorem 1 one needs to invert system 16) and integrate to obtain f and g. The compatibility condition of 16) results in the differential relations 19) and ), which provides the condition of existence of an invertible point transformation of 1) in the integrable equation 1). By the point transformation 4) with 15), the first integral of 1) is It, x, ẋ) = 1 f ġ x + ḟ ) g ẋ + 1 n + 1 f n+1 x n+1, n 1), where f and g as well as the corresponding conditions on f 1, f, and f, are given in Theorem 1. Linearization by nonpoint transformation In this section, we make use of the nonpoint transformation 10), i.e., XT ) = F x, t), dt x, t) = Gx, t)dt. Let us pose the following problem: Find functions F and G in transformation 10), by which the general anharmonic oscillator 1) transforms in d X dt + k dx 1 dt + k X p = 0. 4) Here k 1, k are real constants, and p Q. Applying transformation 10) to 4), we obtain where ẍ + A x, t)ẋ + A 1 x, t)ẋ + A 0 x, t) = 0 5) A x, t) = F xx F x G x G, A 0 x, t) = F tt F x F t F x G t G + k 1 A 1x, t) = F xt G t F x G G x F t + k 1, G F x F t k G F p. F x F x In order to obtain an equation of the form 1), we set A = 0, A 1 = f 1 t), A 0 = f t)x + f t)x n. The condition A = 0 leads to the following special form for 10): XT ) = ft)x m, dt x, t) = gt)x m 1 dt, 6) so that f 1 t) = m + 1 f m f ġ g + k 1, f t) = 1 f m f f ) ġ f g + k f 1, f t) = k f m g f p 1. We can now state Theorem : Equation ẍ + f 1 t)ẋ + f t)x + f t)x n = 0

8 TRANSFORMATION PROPERTIES 17 may be nonpoint transformed in the equation d X dt + k dx 1 dt + k X p = 0, k 1, k R, p Q, by transformation 6), with { ft) = f m/n+) m t } m exp f1 ρ)dρ k 1 n + n + t, ) m 1/ gt) = f 1 n+1)/m) k 7) and p = n m n {, 1}, m {0, 1}, p 1, mp + 1), if and only if f = 1 f n + 4 ) f n + f n + ) + n 1 ) f 1 f n + ) f 1 + f f n n + 1 { n + ) f 1 k 1 f + n + ) 4 } n 1)f 1 4k 1. f 8) Remark: Conditions 19) and 8) are identical if k 1 = 0. The nonpoint transformation does, therefore, not identify a wider class of integrable equations of the form 1). Let us now find a nonpoint transformation which linearizes 1). Note that the constant m, in the nonpoint transformation 6), may be chosen arbitrary except for 0 and 1). With the choice m = n + 1, equation 1), for n Q\{, 1, 1}, is linearized in d X dt + k 1 With this value for m, transformation 6) reduces to where dx dt + k = 0, k 0. 9) XT ) = ft)x n+1, dt = ft) = f n+1)/n+) exp { n + 1 k f t)ft) x n dt, 0) n + 1 n + ) t ) } n + 1 f1 ρ)dρ k 1 t. 1) n + Thus, if condition 8) holds, 1) may be linearized by transformation 0). Note also that 9) may be point transformed in the free particle equation. For k 1 = 0, a first integral of 1) takes the form It, x, ẋ) = 1 ) Ft + F x ẋ + F, ) G with F x, t) = ft)x n+1, Gx, t) = n + 1 k f t)ft) x n, and f given by 1) if condition 8) with k 1 = 0) is satisfied.

9 18 N. EULER 4 Lie point symmetry transformations 4.1 Introduction In this section, we obtain continuous transfromations which leave equation 1) invariant, and therefore transform solutions of 1) to solutions of 1). This type of transformations forms a group, namely, the Lie point transfromation group. Let a Lie point transformation be given in the following form: t = ϕx, t, ε), x = ψx, t, ε). ) Here ε is the group parameter, the group identity is the identity transformation at ε = 0, and the group inverse is the inverse transformation. One can define an infinitesimal generator Z for the Lie point transformation group by Z = ξx, t) t + ηx, t) x 4) so that tx, t, ε) = t + εzt + Oε ), xx, t, ε) = x + εzx + Oε ). Integral curves of the generator Z are group orbits of the transformation group; that is by integrating the autonomous system d t dε = ξ x, t), d x dε = η x, t) 5) with the initial conditions xε = 0) = x, tε = 0) = t, we arrive at the finite transformation ). A function Jx, t) is an invariant of the Lie point transformation group invariant under the action of the transformation group) if and only if ZJx, t) = 0. 6) This is known as the invariance condition. Clearly, the invariant functions of a Lie point transformation group are the first integrals of the corresponding autonomous system 5). In order to find a Lie point transformation group which leaves a second order ODE F t, x, ẋ, ẍ) = 0 7) invariant, we need to prolong the infinitesimal generator Z to Z ) = Z + η 1) ẋ + η) ẍ and apply the invariance condition to the ODE at F = 0, i.e., Z ) F = 0. F =0 8) The prolongation coefficients of Z are η r) = dr dt r [ηx, t) ẋξx, t)] + xr+1) ξx, t).

10 TRANSFORMATION PROPERTIES 19 A generator which satisfies condition 8) for a particular ODE is known as a Lie point symmetry generator for that ODE. The corresponding Lie point transformation is known as a Lie point symmetry transformation for the particular ODE. For some ODE, the invariance condition may lead to several Lie point symmetry generators. This set of Lie point symmetry generators form an algebra under the Lie bracket, known as a Lie point symmetry algebra for the equation. It is clear that an invertible point transformation which transforms one ODE in another, will also transform Lie point symmetry generators of one equation in Lie point symmetry generators of other equation. In particular, the sl, R) Lie point symmetry algebra of ) is spanned by the following Lie point symmetry generators G 1 = Q T t + P T x, G = Q X t + P X x, ) G 4 = F Q X t + P X, G 5 = F x Q T t + P T G 7 = G GQ T + F Q X ) t + G GP T + F P X ) x, G 8 = F F Q X + GQ T ) t + F F P X + GP T ) x. G = G Q T x t + P T x ), G 6 = G ), Q X t + P X x This is obtained by applying the point transformation 4) and transforming the Lie point symmetry generators of the free particle equation 5). We denote the inverse transformation by xx, T ) = P X, T ), tx, T ) = QX, T )). The integrable equation 1) admits the following Lie point symmetry generators G 1 = T, G = T ) T X n 1 X, so that Lie point symmetry generators of 1) can be obtained by the point transformations derived in Section if the appropriate conditions are satisfied. The Lie point symmetry generators for 1), obtained by the point transformation of the form 4) with F and G given by 15), are of the form G 1 = Q T t + P T x, { G = gt)q T n 1 ), ) } { ) } 9) xq X t gt)p T ft)xp X n 1 x, whereby the conditions given in Theorem 1 have to be satisfied. This result is contained in our Lie point symmetry classification of 1) see subsection 4.). Lie point symmetries of an ODE may be used to find invertible point transfromations for ODEs. Let 7) admit the Lie point symmetry generator 4). An invertible point transformation of the form 4), which transforms 7) in an ODE of the autonomous form GX, Ẋ, Ẍ) = 0, is obtained by solving the system of first-order PDEs ZT = 1, ZX = 0,

11 0 N. EULER whereas the solution of ZT = 0, ZX = 1 provides the point transformation in an equation of the form HT, Y, Ẏ ) = 0, where ẊT ) = Y T ). Thus, if an ODE admits Lie point symmetries, it may be used to find first integrals of the ODE. This procedure was followed by Leach and Maharaj [10] for an anharmonic oscillator with multiple anharmonicities. 4. Lie point symmetry classification of 1) Our aim in this section is to do a general Lie point symmetry classification of 1), that is, we give the most general conditions on f 1, f and f for which Lie point symmetries of 1) exist. The aim is not to find all possible functions by which 1) admits Lie point symmetries but merely to give theorems of existence. We consider this form of classification useful since particular equations of the form 1) can easily be tested for the existence of Lie point symmetries. On applying the invariance condition 8) on 1), we obtain the following restrictions on the infinitesimal functions ξ and η for generator 4): ξx, t) = h 1 t)x + h t), ηx, t) = ḣ1 f 1 h 1 )x + g t)x + g 1 t). Here h j, g j are smooth functions to be determined by the conditions x n+1 A 1 + x n A + x n 1 A + x A 4 + xa 5 + A 6 = 0, x n B 1 + xb + B = 0. 40) The A s and B s are functions of f 1, f, f, h 1, h, g 1 and g. In particular, A 1 = n)f 1 f + h 1 f + nf ḣ 1, A = n 1)f g + h f + f ḣ, A = nf g 1, A 4 = f 1 f h 1 f 1 h 1 f 1 + d dt f h 1 ) f1 ḣ1 f 1 ḣ 1 h 1 f1 + h ) 1, A 5 = f 1 ġ + h f + f ḣ + g, A 6 = f g 1 + f 1 ġ 1 + g 1, B 1 = f h 1, B = f h 1 d dt h 1f 1 ) + ḧ1, B = ġ + d dt f 1h ) ḧ. In general, one has to consider three cases depending on the nonlinearity: The linear case n {0, 1}, the case n = as well as the case n Q\{0, 1, }. Case 1: The linear case, i.e., n = 0 and n = 1. The equation can be point transformed in the free particle equation. The Lie point symmetry algebra is sl, R), as discussed in the introduction. The Lie point symmetry generators are of the form Z = h 1 t)x + h t)) ) {ḣ1 t + t) f 1 t)h 1 t) x + g 1 t)x + g t)} x, where h 1, h, g 1, and g take on particular functional forms, in terms of f 1, f and f. This case was discussed in detail by Duarte et al [1].

12 TRANSFORMATION PROPERTIES 1 Case : n =. System 40) reduces to the system A 1 = 0, A + A 4 = 0, A + A 5 = 0, A 6 = 0, B 1 = 0, B = 0, B = 0. From the equation B 1 = 0, it follows that h 1 = 0 so that the Lie point symmetry generator takes on the form Z = h t) t + g t)x + g 1 t)) x, 41) where the remaining conditions on g 1, g and h are f g + h f + f ḣ = 0, 4) f g 1 + f 1 ġ + h f + f ḣ + g = 0, 4) f g 1 + f 1 ġ 1 + g 1 = 0, 44) ġ + d dt f 1h ) ḧ = 0. 45) Note that h 1 = 0 for all n. In solving conditions 4) 45) we have to consider two subcases, namely g 1 = 0 and g 1 0. Subcase.1: g 1 = 0. By 4) and 45), we obtain g t) = d dt [ln f ] h ḣ 1 ḣ 1 f 1h, 46) ḣ = c 1 1 d ) 5 dt [ln f ] f 1 h. 47) Inserting 46) and 47) into 4) leads to an expression of the form F 1 f 1, f, f ) h + c 1 F f 1, f, f ) = 0, 48) where F 1 = 1f1 f + 50f 1 f f 80f 1 f f f f + f1 f f 100f f f + 5f f 1 f + 1f 1 f f 84 f 50f f 1 49) 15f 1 f f + 105f f f 5f f ) ) / 15f ), F = 6f1 f + 5f f 10f f 1 f 1 f f + 6f ) ) 5f f / 5f. 50) This leads to the following Theorem : The most general Lie point symmetry generator 4), which the equation ẍ + f 1 t)ẋ + f t)x + f t)x = 0 may admit, is of the form Z = h t) t + g t)x x if and only if f 1, f and f satisfy one of the following conditions:

13 N. EULER a) F = 0, then h is given by h t) = f /5 1 t [ t /5 exp f1 ζ)dζ) c 1 f ρ) exp 1 ρ ] f1 ζ)dζ) dρ + c. 5 5 and g is given by 46). b) F 1 = 0 with c 1 = 0, then h is given by h t) = c f /5 1 t exp f1 ζ)dζ) 5 and g is given by 46). c) F 1 0 and F 0 with c 1 0, then h t) = c 1F F 1 and g is given by 46), whereby the condition on f 1, f and f is F 1 F F F 1 F1 f 5 f 1 1 ) 5 f 1 F 1 F = 0. Here F 1 and F are given by 49) and 50), respectively. A note on the proof of Theorem : If F = 0, it follows that F 1 0. The infinitesimal function h in the symmetry generator 41) is then obtained by integrating 47), whereby g is given by 46). If F 0 and c 1 0, then F 1 must be nonzero, so that h = c 1 F 1 /F has to satisfy 47). Remark: Condition F = 0 is identical to the condition by which 1) is point transformable in the integrable equation 1) and linearizable by the nonpoint transformation 0) with n = ). The Lie point symmetries 9) with n = ) obtained by the point transformations of Section, are those corresponding to Theorem a. The most general Lie point symmetry generator which follows from the conditions of Theorem b, and Theorem c cannot be obtained from the symmetries of the integrable equation 1). Subcase.: g 1 0. Equations 4) and 45) remain the same, therefore, relations 46) and 47) hold also for this subcase. By 4), g 1 is given by g 1 t) = 1 f f 1 ġ + h f + f ḣ + g ), 51) so that 44) leads to the expression F 1 f 1, f, f ) h + c 1 F f 1, f, f ) = 0

14 TRANSFORMATION PROPERTIES where F 1 = 7f 5 1 f 5 150f 1f f f 1 f 5 f f 1 f 5 f 1 500f 1 f 5 f 15f f 5 f 15f 5 f 1 f 70f 4 1 f 4 f + 500f 1 f f 4 f + 500f f 4 f 800f 1 f 4 f 1 f +15f f 4 f 1 f 6875f 4 f 1 f + 115f 1 f 4 f f + 70f 1 f f 115f 1f f f f 1 f f 1 f 500f f f 1485f 1 f f + 500f f f 150f f 1 f 0790f 1 f f f f 1 f 5 f f 5 f 1 f1 1075f 1 f 4 f f f f f 1 565f 1 f 5 f f 4 f f 1100f1 f 4 f + 565f 1 f f 4 f 565f 1 f 4 f 1 f + 975f 4 f f + 600f1 f f f 065f f f f f f 1 f f +00f 1 f f f 10950f f f 15f 4 f 1 f 565f 1 f f + 975f f f +750f 1 f 5f ) 1 475f 4 f f ) 1 15f 5f ) + 15f1 f 4f ) + 15f f 4f ) 865f f f ) + 950f f f ) 975f f f ) + 150f 5f 4) f 1 f 4f 4) 565f f f 4) + 65f 4f 5) F = ) / 650f 6 ), 6f 4 1 f 4 65f f f 1 f 4 f f 4 f 1 150f 1f 4 f 88f 1 f f +150f 1 f f f 160f 1 f f 1 f + 500f f f + 49f 1 f f 500f f f +680f f 1 f f 1 f f 564 f f 1f 4 f 1 950f f f1 150f 4 f 190f1 f f + 150f f f 00f f 1 f 190f 1 f f f +5940f f f 1075f f + 500f 4f ) f 1 f f ) 1600f f f ) + 50f f 4) ) / 65f 5 ). This leads to the following Theorem 4: The most general Lie point symmetry generator 4), which the equation ẍ + f 1 t)ẋ + f t)x + f t)x = 0 may admit, is of the form 5) 5) Z = h t) t + {g t)x + g 1 t)} x if and only if f 1, f and f satisfy one of the following conditions: a) F = 0, then h is given by h t) = f /5 1 t [ t /5 exp f1 ζ)dζ) c 1 f ρ) exp 1 ρ ] f1 ζ)dζ) dρ + c, 5 5 g 1 by 51), and g is given by 46). b) F 1 = 0 with c 1 = 0, then h is given by h t) = c f /5 1 t exp f1 ζ)dζ), 5 g 1 by 51), and g is given by 46).

15 4 N. EULER c) F 1 0 and F 0 with c 1 0, then h t) = c 1F F 1, g 1 by 51), and g is given by 46), whereby the condition on f 1, f and f is F 1 F F F 1 F1 f 5 f 1 1 ) 5 f 1 F 1 F = 0 Here F 1 and F are given by 5) and 5), respectively. Case : n Q\{0, 1, }. This leads to the system A = 0 A = 0, B = 0, A 5 = 0. From the equation A = 0, it follows that g 1 = 0, so that the remaining conditions on h and g are n 1)f g + h f + f ḣ = 0, 54) f 1 ġ + h f + f ḣ + g = 0, 55) ġ + d dt f 1h ) ḧ = 0. 56) To solve this system of equations, we need to consider two subcases: Subcase.1: n =. This leads to Theorem 5: The most general Lie point symmetry generator 4), which the equation ẍ + f 1 t)ẋ + f t)x + f t)x = 0 may admit, is of the form Z = h t) t + g t)x x if and only if f 1 = 1 f, f where h is a solution of with and h ) + 4Γt)ḣ + Γt)h = 0 58) Γt) = 1 16 [ ] d dt ln f ) + 1 d 4 dt ln f ) + f g t) = 1 d 4 dt ln f ) + 1 ḣ. 57)

16 TRANSFORMATION PROPERTIES 5 Note that condition 57) is identical to the condition derived in Section, for an invertible point transformation of 1) with n =. Therefore, all Lie point symmetries of 1) with n =, which follow from Theorem 5, may also be obtained by applying the point transformation, in Section, in the Lie point symmetries of 1) with n = ). Note that 58) may be transformed in the free particle equation 5): Let At) = ḣ h. In this case, 58) reduces to Ä + A A + 4Γt)A + A + Γt) = 0. Comparing this equation with ), we find that condition ) is satisfied. Subcase.: n. By 54) and 56), we obtain ) ) 1 d 1 g t) = h n 1 dt ln f ) ḣ, 59) n 1 ḣ = c 1 ) { n 1 n + n 1 Inserting 59) and 60) into 55) leads to where F 1 f 1, f, f ) h + c 1 F f 1, f, f ) = 0, F 1 = } d dt ln f ) f 1 h. 60) [ 4n 1)f 1 f + n + 5n + n 9)f 1 f f 8nn + )f 1f f 1 +n + 9n + 7n + 7)f f n 6n )f 1 f f 4n + ) f f f n + )n 9)f f 1 f + n 1)n + 5)f 1 f f n + 4)n + 5) f n + ) f f 1 n 1)n + )f 1 f f + n + )n + 5)f f f n + ) f f ) ] / [ n + ) f ], [ F = n + 1)f1 f + n + ) f f n + )f f 1 n 1)f 1 f f + n + 4) f n + )f f ] / [ n + ) f ] 6). This leads to Theorem 6: The most general Lie point symmetry generator 4), which the equation ẍ + f 1 t)ẋ + f t)x + f t)x n = 0, with n Q\{, 0, 1, }, may admit, is of the form 61) Z = h t) t + g t)x x if and only if f 1, f and f satisfy one of the following conditions:

17 6 N. EULER a) F = 0, then h is given by h t) = f /n+) n 1 exp n + and g is given by 59). t f1 ζ)dζ) [ c 1 t f /n+) ρ) exp b) F 1 = 0 with c 1 = 0, then h is given by h t) = c f /n+ n 1 t exp f1 ζ)dζ), n + and g is given by 59). c) F 1 0 and F 0 with c 1 0, then h t) = c 1F F 1, and g is given by 46), whereby the condition on f 1, f and f is F 1 F F F 1 F1 n 1 ) f n + n 1 f 1 f 1 F 1 F = 0. Here F 1 and F are given by 5) and 5), respectively. n 1 ρ ] f1 ζ)dζ) dρ + c, n + Remark: The condition F = 0 is identical to the condition for transforming 1) in the integrable equation 1) and linearizing 1) by the non-oint transformation 0). An important special case of equation 1) is the case where f 1 = f = 0. By applying Theorem to Theorem 6 we calculate the conditions on f for which there exist Lie point symmetries of 1) see Table 1 and Table ). The general form of f can be given in all cases, except where g 1 0, i.e., Theorem 4. For this case, we only list the conditions on f Table 1). Let us view the original determining equations for this case: g 1 = 0, g t) = 1 ḣ 1 c 1, f t) = 1 4 h ) g 1. This set of equations may be converted into a condition on h, namely h h 4) 5 + ġ1 h 1 ) ḣ g 1 c 1 h ) = 0, where h ) 0. We did study solutions of this equation.

18 TRANSFORMATION PROPERTIES 7 Table 1: ẍ + f t) x = 0 Theorem : Lie Symmetry Generator Z = h t) t + g t)x x Theorem a h t) = c 1 k 1 k 1 t + k ) + c k 1 t + k ) g t) = c k 1 k 1 t + k ) c 1 Z 1 = k 1 t + k ) t + k 1k 1 t + k )x x, Z = 1 k 1 k 1 t + k ) t + n + 1 n 1 x x [Z 1, Z ] = Z 1 Condition on f F 5f { 6 f 5f f } = 0 General solution for f f t) = k 1 t + k ) 5, k 1, k R Theorem b h t) = c k /5 1 5 t + ) 5 k 1t + k, g t) = c 5 k /5 t k 1 ) Condition on f F f { 84f + 105f f f 5f f ) } = 0 General solution for f f t) = k 1 5 t + 5 k 1t + k ), k 1, k, k R

19 8 N. EULER Thoerem c Table 1 continued) h t) = k 1 t + k t + k g t) = k 1 t + 1 k c 1 ) Condition on f 1f f 4f f 4f f ) + 6f f f f ) 15f = 0 f ) ) 1f f f 4) + 10f f f 4) Condition on f with the subsititions: At) = f f 1 and BA) = At) 10B A B ) ) B 5B + A B B ) + 1AB B 1B = 0 Note: The equation for B can be linearized by a point transformation. General solution for f { f t) = k 4 k 1 t + k t + k ) 5/ c1 exp t dρ k 1 ρ + k ρ + k }, k j R Theorem 4: Lie Symmetry Generator Z = h t) t + {g 1t) + g t)x} x Theorem 4a { t } h t) = f /5 /5 c 1 f ρ)dρ + c, g t) = 1 5 f f h c 1 g 1 t) = k 1 t + k, k 1, k R Condition on f F 1 65f 5 { 564f f f f 1075f f 1600f f f ) + 50f f 4) } = 0

20 TRANSFORMATION PROPERTIES 9 Table 1 continued) Condition on f with the subsititions: At) = f f 1 and BA) = At) 50B B + 50BB ) 600ABB 5B + 490A B 49A 4 = 0 Note: The equation for B cannot be linearized by a point transformation. Theorem 4b h t) = c f /5, g t) = c f 5 f 7/5 g 1 t) = k 1 t + k, k 1, k R Condition on f 1 F 1 650f 6 { f f f f + 975f f f + 950f f f ) 975f f f ) 565f f f 4) + 65f 4 f 5) } = 0 Condition on f with the subsititions: At) = f f 1 and BA) = At) 65B B + 500B B A)B + 65BB ) 500BAB ) + 15B9A 5B)B +750B A 450BA + 196A 5 = 0 Theorem 4c h t) = F, g t) = 1 f h c 1 F 1 5 f g 1 t) = k 1 t + k, k 1, k R Condition on f 4948 f 6 f + 1 terms + 500f 6 f 4) f 6) = 0

21 0 N. EULER Table : ẍ + f t) x n = 0 Theorem 5: n =. Lie Symmetry Generator Z = h t) t + g t)x x h t) = c 1 t + c t + c, g t) = c 1 t + 1 c, f = constant Z 1 = t t + xt x, Z = t t + 1 x x, Z = t [Z 1, Z ] = Z 1, [Z 1, Z ] = Z, [Z, Z ] = Z Theorem 6: n Q\{, 0, 1, }. Lie Symmetry Generators Z = h t) t + g t)x x Theorem 6a h t) = c 1 k 1 k 1 t + k ) + c k 1 t + k ), g t) = c k 1 k 1 t + k ) c 1 ) n + 1 n 1 Z 1 = k 1 t + k ) t + k 1k 1 t + k )x x, Z = 1 k 1 t + k ) ) n + 1 k 1 t + x n 1 x [Z 1, Z ] = Z 1 Condition on f F n + ) f { n + 4) f n + )f f } = 0 General solution for f f t) = k 1 t + k ) n+), k 1, k R Theorem 6b { h t) = c k /n+) 1 } n + t + k 1 n + t + k, g t) = c n + k /n+) t k 1 )

22 TRANSFORMATION PROPERTIES 1 Table continued) Condition on f 1 F 1 n + ) f { n + 9n + 0) f + n + 8n + 15)f f f n + ) f f ) } = 0 General solution for f { f t) = k 1 } n+)/ n + t + k 1 n + t + k Theorem 6c: h t) = k 1 t + k t + k, g = k 1 t + 1 k c 1 ) Condition on f n + 5) f f 6n + 5)f f 4n + 4) f f ) n)f f f f ) n + )f f ) ) n + 4)f f f 4) + n + )f f f 4) = 0 Condition on f with the subsititions: At) = f f 1 and BA) = At) { n + )B ) A B } { } B n + )B + A B B ) + 1AB B 1B = 0 Note: The equation for B can be linearized by a point transformation. General solution for f { ) n 1 t } dρ f t) = k 4 k 1 t + k t + k ) n+)/ exp c 1 k 1 ρ, k j R + k ρ + k

23 N. EULER 4. Invertible point transformations by Lie point symmetries One can now use the Lie point symmetry generators obtained above to construct time dependent first integrals of 1). We give the point transformation in general form and do an example to illustrate the procedure. In Table 1 and Table we make use of the substitution f f = At), BA) = At), by which we are able to reduce the order of the differential conditions on f by two. It follows that f f = B + A, f ) f = B B + BA + A, f 4) f = B B + B ) B + 4B BA + B + 6BA + A 4, f 5) f = B B + 4B B B + B ) B + 5B B A + 5B ) BA + 10B B +10B BA + 15B A + 10BA + A 5, where B db/da, etc. Let us consider the transformation of 1) in an autonomous form by the use of the most general Lie point symmetry generator for the nonlinear equation 1), namely Z = h t) t + {g 1t) + g t)x} x. The defining equations for the point transformations are XT ) = F x, t), T x, t) = Gx, t) h t) T t + {g 1t) + g t)x} T x = 1, h t) X t + {g 1t) + g t)x} X x = 0. The general solution of this system is t 1 XT ) = ϕ 1 ω), T x, t) = ω = x exp { t g ρ) h ρ) dρ } h ρ) dρ + ϕ ω), t { g 1 ρ) h ρ) exp ρ g ζ) h ζ) dζ } dρ, where ϕ 1 and ϕ are arbitrary functions of ω and may be chosen in a convenient form. To construct a point transformation which may transfrom 1) in a first-order ODE, one needs to solve the system h t) T t + {g 1t) + g t)x} T x = 0, h t) X t + {g 1t) + g t)x} X x = 1. 6)

24 TRANSFORMATION PROPERTIES The general solution of this system is t 1 XT ) = h ρ) dρ + ψ 1ω), { t } g ρ) ω = x exp h ρ) dρ T x, t) = ψ ω), t { g 1 ρ) h ρ) exp ρ g ζ) h ζ) dζ } dρ, where ψ 1 and ψ may be chosen arbitrary. It can be shown that this point transformation reduces 1) in an Abel equation of the first kind, i.e., dy dt + Γ T )Y + Γ T )Y + Γ 1 T )Y + Γ 0 T ) = 0 with Y = dx/dt. Since the known solutions of the Abel equation are restricted to special forms of the functions Γ, it is usually a useless exercise. However, if the equation admits a two-dimensional Lie point symmetry algebra, it is well known that the reduction may be performed such that the reduced equation in this case, the Abel equation) admits a Lie point symmetry. In particular, if Z 1 and Z are symmetry generators of the equation to be reduced, and [Z 1, Z ] = λz 1, then the generator Z 1 should be used to perform the reduction. This will ensure that the symmetry generator Z is preserved by the reduction, i.e., the reduced equation will admit the generator Z in transformed form. This is important to note, since the integrating factor µ of any first-order ODE of the form ẋ = Mx, t)/nx, t), which admits the Lie point symmetry generator Z = ξ / t + η / x, is given by µ = Nη + Mξ) 1. Note also that it is not a simple task to find Lie point symmetry generators for a first order ODE. In particular, the determining equation of a Lie point symmetry generator for the ODE ẋ = fx, t) is ξ f t η f x + η t + f η x f ξ t f ξ x = 0. As an example, we transform ẍ + k 1 n + t + k ) n+)/ 1 n + t + k x n = 0, k 1, k R, 64) for n Q\{, 0, 1}, in an autonomous form. This case corresponds to Theorem 6b given in Table. Note that g = 1 ḣ, h t) = 1 n + t + k 1 n + t + k. The general form of the point transformation is given by 6). We let ϕ 1 = ω, ϕ = 0. It follows that XT ) = h 1/ x, T x, t) = t dρ h ρ). This transformation leads to the autonomous equation d { ) X dt k1 + k } X + c n+)/ X n = 0 n + n +

25 4 N. EULER which has the first integral T X, dx ) = dt ) dx dt A first integral for 64) is then It, x, ẋ) = 1 1 ) h ḣx + h ẋ 1 h { ) k1 + k } ) X + c n+)/ X n+1 = 0. n + n + n n + 1 cn+)/ h n+1)/ x n+1. { ) k1 + k } x n + n + 5 Transforming in the second Painlevé transcendent By using the Lie point symmetry generators classified in the previous section, we are able to construct point transformations by which 1) may be transformed in a second equation. However, this method does not allow for a transformation in an equation without a Lie point symmetry. This can easily be shown: Consider a second equation in the variables X, T ) with no symmetry. To construct a point transformation in this equation by a Lie point symmetry generator Z = ξ / t + η / t of the first equation with variables x, t), we need to solve the system ξ X t + η X x = 0, ξ T t + η T x = 0, which implies that the Jacobian is zero. There is an important class of equations which do not admit Lie point symmetries, but are integrable. These are the six Painlevé transcendents, as discussed in the introduction. We consider the following problem Euler et al 1991): Find the condition on f 1, f, and f for which 1), with n =, i.e., ẍ + f 1 t)ẋ + f t)x + f t)x = 0, may be point transformed in the second Painlevé transcendent d X dt T X X a = 0, a R. 65) By the point transformation XT ) = ft)x, T x, t) = gt) 66) we obtain the following expressions: f 1 t) = f f ġ g, Inverting this system leads to Theorem 7: Equation ẍ + f 1 t)ẋ + f t)x + f t)x = 0 f t) = f f f g f ġ gġ, f t) = fġ).

26 TRANSFORMATION PROPERTIES 5 may be point transformed in d X dt T X X = 0 by the invertible point transformation XT ) = ft)x, T x, t) = gt), where { t } ft) = kf 1/6 1 exp f 1ρ)dρ, gt) = k 18f 8/ 6f f + 7 f f 1 f f 1 f 1 f + 6f f 8f 1 f under the following conditions: ) { t exp f1 ρ)dρ}, 9f 4) f ) 54f f f ) + 18f f f 1 6 f f + 19 f f f 78 f f f f f f f + f f f 1 11 f f f f 1 + 6f f 1 f 7 f f f + 90f f f 7 f f1 f 57f f 1 f f 1 + 7f f f f 1 14f f f 1 54 f f 4 90 f f 4f f ) 1 f f 1 f 4f 1 + 6f 1 f 4 6 f 1 f 4f + 60f 1 f 4f 1 6f 4f f1 + 8f 4f 1 4 = 0, 67) with 6f f + 7 f f 1 f f 1 f 1 f + 6f f 8f 1 f 0. 68) Remark: The l.h.s. of 68) equal to zero is identical to the condition obtained in Section by which 1) may be point transformed in d X/dT + X = 0, and which linearizes 1) by a nonpoint transformation with n = ). It is easy to show that if f 1, f and f are such that the l.h.s. of 68) is equal to zero, then condition 67) is satisfied identically for those functional forms. In Euler et al [6], we showed that 1) passes the Painlevé test if and only if condition 67) is satisfied. We refer to the book of Steeb and Euler [16] for more details on the Painlevé test of nonlinear evolution equations.) Thus, if conditions 67) and 68) are satisfied, equation 1) has the Painlevé property and is therefore integrable; this is true since there exists an invertible point transformation in the second Painlevé transcendent. Moreover, equation 1) with f 1, f and f, which make the l.h.s. of 68) equal to zero, also has the Painlevé property, since the Painlevé test is passed and the equation can invertibly be point transformed in the integrable equation d X/dT + X = 0. Let us finally consider the special case where f 1 = f = 0. By 67), we obtain the condition 9A ) 18ÄA + 1AA 9A A 4 = 0, 69) where At) = f /f. This equation admits three Lie point symmetry generators Z 1 = t 6 t + 1 At + 1 ) A, Z = t t A A, Z = t.

27 6 N. EULER By using these symmetry properties, 69) may be transformed in the following Abel equation du dt = g 1T )U + g T )U + g T )U, where ) dx 1 UT ) =, T u, A) = u exp{ X}, XT ) = ln A, ua) = da dt dt, g 1 T ) = 1 1 T 9 4 ) T + 5T 6T, g T ) = 1 T 7T ), g T ) = 1 T. 6 Conclusion We have seen that if condition 19) with n / {, 1, 0, 1}) is satisfied, equation 1) has the following properties: a) Equation 1) may be point transformed in d X dt + Xn = 0. b) Equation 1) may be linearized in d X dt + k = 0, k R\{0} by a nonpoint transformation. c) Equation 1) admits a two-dimensional Lie point symmetry algebra. By the Lie point symmetry classification, we have observed that 1) admits a Lie symmetry generator with g 1 0 only in the case n =. This is a very complicated case to solve in general. We have given the general conditions on f 1, f and f for the existence of Lie symmetries in this case as well as all other cases of n. Leach and Maharaj [10] calculated some special cases where the Lie point symmetries may be given explicitly. For the case n =, we have given the necessary and sufficient conditions on f 1, f and f by which 1) has the Painlevé property. This includes also condition 19) with n =. A detailed Painlevé analysis of 1) will be the subject of a future paper.

28 TRANSFORMATION PROPERTIES 7 References [1] Duarte L.G.S., Duarte S.E.S. and Moreira I.C., One-dimensional equations with the maximum number of symmetry generators, J. Phys. A: Math. Gen., 1987, V.0, L701 L704. [] Duarte L.G.S., Euler N., Moreira I.C. and Steeb W.-H., Invertible point transformations, Painlevé analysis and anharmonic oscillators, J. Phys. A: Math. Gen., 1990, V., [] Duarte L.G.S., Moreira I.C., Euler N. and Steeb W.-H., Invertible Point transformations, Lie Symmetries and the Painlevé test for the equation ẍ + f 1t)ẋ + f t)x + f t)x n = 0, Physica Scripta, 1991, V.4, [4] Duarte L.G.S., Moreira I.C. and Santos F.C., Linearization under non-point transformations, J. Phys. A: Math. Gen., 1994, V.7, L79 L74. [5] Euler M., Euler N. and Köhler A., On the construction of approximate solutions for a multidimensional nonlinear heat equation, J. Phys. A: Math. Gen., 1994, V.7, [6] Euler N., Steeb W.-H. and Cyrus K., On exact solutions for damped anharmonoc oscillators, J. Phys. A: Math. Gen., 1989, V., L195 L199. [7] Euler N., Steeb W.-H., Duarte L.G.S. and Moreira I.C., Invertible point transformations, Painlevé test, and the second Painlevé transcendent, Int. J. Theor. Phys, 1991, V.0, [8] Fushchych W.I., Shtelen W.M. and Serov N.I., Symmetry Analysis and Exact Solutions of Equations of Nonlinear Mathematical Physics, Kluwer, Dordrecht, 199. [9] Leach P.G.L., First integrals for the modified Emden equation q + αt) q + q n = 0, J. Math. Phys, 1985, V.6, [10] Leach P.G.L. and Maharaj S.D., A first integral for a class of time-dependent anharmonic oscillators with multiple anharmonicities, J. Math. Phys., 199, V., [11] Moreira I.C., Europhys. Lett, 1990, V.1, 91. [1] Olver P.J., Applications of Lie Groups to Differential Equations, Springer, New York, [1] Sarlet W., Mahomed F.M. and Leach P.G.L., Symmetries of non-linear differential equations and linearization, J. Phys. A: Math. Gen., 1987, V.0, [14] Steeb W.-H., Invertible Point Transformations and Nonlinear Differential Equations, World Scientific, Singapore, 199. [15] Steeb W.-H., Continuous Symmetries, Lie Algebras, Differential Equations and Computer Algebra, World Scientific, Singapore, [16] Steeb W.-H. and Euler N., Nonlinear Evolution Equations and Painlevé Test, World Scientific, Singapore, [17] Tresse M.A., Determination des invariants puctuels de l equation differentielle ordinaire de second orde, Hirzel, Leipzig, 1896.