The Riccati equation method with variable expansion coefficients. I. Solving the Burgers equation

The Riccati equation methodwith variable expansion coefficients. I. Solving the Burgers equation Solomon M. Antoniou SKEMSYS Scientific Knowledge Engineering and Management Systems 37 Κoliatsou Street, Corinthos 20100, Greece [email protected] 2 Abstract We introduce the Riccati equation method with variable expansion coefficients. This method is used to find travelling wave solutions to the Burgers equation 0 u a u u u xx x t = − + . The important new feature of the method lies in the fact that the ( − ξ dependent) coefficients A and B of the Riccati equation 2 BY A Y + = ′ satisfy their own nonlinear ODEs, which can be further solved by one of the known methods, like Jacobi's elliptic equation method, the − ′ ) G / G ( expansion method, the projective Riccati equation method, and numerous other methods, like the reduction to Ermakov's equation. We also introduce the − ′ ) G / G ( expansion method with variable expansion coefficients. The second method presupposes Riccati's equation method with variable expansion coefficients. The solutions obtained by the − ′ ) G / G ( expansion method are expressed in terms of hypergeometric function even in the simplest case. More complicated cases might require hypercomputing facilities with symbolic capabilities. Both methods are new, computationally demanding, the use of which reveals a rich number of solutions not known previously. Keywords: Riccati method, nonlinear evolution equations, traveling wave solutions, Burgers equation, exact solutions. 3 1. Introduction. Nonlinear partial differential equations arise in a number of areas of Mathematics and Physics in an attempt to model physical processes, like Chemical Kinetics (Gray and Scott [98]), Fluid Mechanics (Whitham [204]), or biological processes like Population Dynamics (Murray [154]). In the recent past, there is a number of new methods which have been invented in solving these equations. Among the new methods are (1) the inverse scattering transform method (Ablowitz, Kaup, Newell and Segur [17], Ablowitz and Clarkson [18], Ablowitz and Segur [19], Drazin and Johnson [50], Gardner et. al. [90]-[92], Gelfand and Levitan [93], Kay and Moses [121], Lax [132], Marchenko [147], Miura [149]-[152], Novikov, Manakov, Pitaevskii and Zakharov [158], Ramm [170], Wadati [187], Wadati, Sanuki and Konno [188] ) (2) Hirota’s bilinear method (Hirota [111] and [112], Hientarinen [110], Hereman and Zhuang [109] ) (3) the Algebro-Geometric approach (Belokolos, Bobenko, Enolskii, Its and Matveev [32] and references) (4) the Bäcklund transformation method (Miura [149], Rogers and Shadwick [172]) (5) the tanh-coth method (Malfliet [142], Malfliet and Hereman [144] and [145], Bekir and Cevikel [31], Wazwaz [199], Fan [72], Abdel-All, Razek and Seddeek [3]) (6) the sn-cn method (Baldwin et al [27]) (7) the F-expansion method (Wang and Li [192], Abdou [6] and [7]) (8) the Jacobi elliptic function method (Abbott, Parkes and Duffy [2], Chen and Wang [44], Fan and Zhang [82], Liu and Li [135] and [136], Liu, Fu, Liu and Zhao [137] and [138], Lu and Shi [141], Xiang [206], Yan [209] and [210],Yang [214]) 4 (9) the Riccati equation method (Yan and Zhang [213]) (10) the Weierstrass elliptic function method (Kudryashov [124]) (11) the Exp-function method (He and Wu [104], Naher, Abdullah and Akbar [155] and [156], Mohyuddin, Noor and Noor [153], Yildirim and Pinar [218], Aslan [26], Bekir and Boz [30]) (11) the − ′ G / G expansion method (Abazari and Abazari [1], Borhanibar and Moghanlu [37], Borhanibar and Abazari [38], Elagan, Sayed and Hamed [52], Feng, Li and Wan [84], Jabbari, Kheiri and Bekir [118], Naher, Abdullah and Akbar [157], Ozis and Aslan [163], Wang, Li and Zhang [194], Zayed [220], Zayed and Gepreel [222] ) (12) the homogeneous balance method (Fan [71], Wang, Zhou and Li [191], Yan and Zhang [213]) (13) the direct algebraic method (Soliman and Abdou [180]) (14) the basic equation method (Kudryashov [125] ) and its variants, like the simplest equation method (Vitanov [184]-[186], Yefimova [217]) (15) the Cole-Hopf transformation method (Cole [48], Hopf [115], Salas and Gomez [174]) (16) the Adomian decomposition method (Adomian [21] and [22], Abdou [4], Cherruault [46], Cherruault and Adomian [47], El-Wakil, Abdou and Elhanbaly [58], El-Wakil and Abdou [59], Rach, Baghdasarian and Adomian [169], Wazwaz [196] and [197]) (17) the Painleve truncated method (Weiss, Tabor and Carnevale [202] and [203], Zhang, Wu and Lou [223], Estevez and Gordoa [69] and [70]) (18) the homotopy perturbation method (Taghizadeh, Akbari and Ghelichzadeh [183], Yahya et al [207], Liao [134]) (19) the reduced differential transformation method (Keskin and Oturanc [122], Arora, Siddiqui and Singh [25]) (20) the Lie symmetry method (Lie point symmetries, potential symmetries, nonclassical symmetries, the direct method) (Bluman and Kumei [33], 5 Bluman and Anco [34], Bluman and Cole [35], Bluman, Cheviakov and Anco [36], Cantwell [40], Hydon [116], Olver [160], Ovsiannikov [162], Schwarz [175], Steeb [181], Stephani [182]) (21) the variational iteration method (He [102], Abdou [8], Wazwaz [198]) (22) the first integral method (Raslan [171], Feng [85], Feng and Wang [86]) (23) the integral bifurcation method (Rui, Xie, Long and He [173] and references) The implementation of most of these methods was made possible only using Symbolic Languages (Grabmeier, Kaltofen and Weispfenning [97]) like Mathematica (Baldwin et al [28] and [29], Göktaş and Hereman [96]), Macsyma (Hereman [106], Hereman and Takaoka [108], Hereman and Zhuang [109]), Maple, etc. In many cases some special elimination methods (Wang [189]) and computational algorithms were also used (Wu [205]). In this paper we introduce the Riccati equation method with variable expansion coefficients and we find traveling wave solutions of the Burgers equation. The paper is organized as follows: In Section 2 we introduce the basic ingredients of the method used. In Section 3 we consider Burgers equation and Riccati’s equation method of solution where the expansion coefficients are not constants, i.e. they depend on the variable ξ. In this section we consider first the Riccati expansion method and then the extended Riccati method. In the first case we find closed-form solutions for ) ( u ξ expressed in terms of hyperbolic tangent functions. In the second case we establish that A and B are proportional each other. We also find a closed form expression for the function ) ( u ξ expressed in terms of ) ( A ξ . The function ) ( A ξ is proved to satisfy a second order nonlinear ordinary differential equation solved in Appendices A, B and C. In Section 4 we consider the various forms the function ) ( u ξ takes using the various solutions found in the 6 Appendices. In Section 5 we consider the − ′ ) G / G ( expansion method with variable coefficients as a method of solution of the Burgers equation. 2. The Method. We consider an evolution equation of the general form ) , u , u , u ( G u xx x t L = or ) , u , u , u ( G u xx x tt L = (2.1) where u is a sufficiently smooth function. We introduce a new variable ξ given by ) t x ( k ω − = ξ (2.2) where k and ω are constants. Changing variables, since ) ( u ) k ( u t ξ ′ ω − = , ) ( u k u x ξ ′ = , ) ( u k u 2 xx ξ ′ ′ = , … (2.3) equations (2.1) become ordinary nonlinear differential equations | | ¹ | \ | ξ ξ = ξ ω − L , d u d k , d du k , u G d du ) k ( 2 2 2 (2.4) or | | ¹ | \ | ξ ξ = ξ L , d u d k , d du k , u G d u d k 2 2 2 2 2 2 (2.5) Solutions of evolution equations depending on ) t x ( k ω − ≡ ξ are called traveling wave solutions. Equations (2.4) or (2.5) will be solved considering expansions of the form ∑ = = ξ n 0 k k k Y a ) ( u (2.6) or ∑ ∑ = = + = ξ n 0 k k k n 0 k k k Y b Y a ) ( u (2.7) where all the expansion coefficients depend on the variable ξ , ) ( a a k k ξ ≡ , ) ( b b k k ξ ≡ for every n , , 2 , 1 , 0 k L = 7 contrary to the previously considered research where the expansion coefficients were considered as constants. The function ) ( Y Y ξ ≡ satisfy Riccati’s equations 2 BY A ) ( Y + = ξ ′ (2.8) or 2 Y R Y Q P ) ( Y ⋅ + ⋅ + = ξ ′ (2.9) where again all coefficients A, B and P, Q, R depend on the variable ξ . In solving equations (2.4) or (2.5), we consider the expansions (2.6) or (2.7) and then we balance the nonlinear term with the highest derivative term of the function ) ( u ξ , which determines n (the number of the expansion terms). Equating similar powers of the function ) ( Y ξ , we can determine the various coefficients and thus find the solution of the equation considered. Our motivation in using Riccati's method with variable expansion coefficients, was to obtain traveling wave solutions expressed in terms of special functions (Hypergeometric, Bessel, etc). Introducing variable expansion coefficients, we obtain variable coefficients A and B of the Riccati equation (2.8). Converting Riccati equation into a second order linear ordinary differential equation with variable coefficients, its solutions might be expressed in terms of special functions. 3. The Burgers equation and its solutions. Burgers equation (Burgers [39]) was introduced in an attempt to model a theory of turbulence. There is a number of explicit solutions based on Lie symmetries, either classical (Antoniou [24], Liu, Li and Zhang [139], Ouhadan, Mekkaoui and El Kinani [161]), or nonclassical (Gandarias [88], Mansfield [146], Qin, Mei and Xu [168]). Some solutions have also been found using the tanh-method (Hereman and Malfliet [107]) or some of its descendants, like the Exp-Function Method (Ebaid [51]) and the Homotopy perturbation method (Taghizadeh, Akbari and Ghelichzadeh [183]). On the other hand Burgers equation can be reduced to the 8 heat equation through the famous Cole-Hopf transformation (Cole [48], Hopf [115], Olver [160] and Mansfield [146]). We consider next Burgers equation in the form 0 u a u u u xx x t = − + (3.1) and try to find traveling wave solutions of this equation. We introduce a new variable ξ given by ) t x ( k ω − = ξ (3.2) where k and ω are constants. Changing variables, since ) ( u ) k ( u t ξ ′ ω − = , ) ( u k u x ξ ′ = and ) ( u k u 2 xx ξ ′ ′ = equation (3.1) becomes an ordinary nonlinear differential equation 0 ) ( u k a ) ( u ) ( u k ) ( u ) k ( 2 = ξ ′ ′ − ξ ′ ξ + ξ ′ ω − (3.3) Integrating once the above equation, we obtain 0 2 c ) ( u k a ) ( u 2 1 ) ( u ) ( = ξ ′ − ξ + ξ ω − (3.4) where 0 c is a constant. We consider the following cases in solving equation (3.4). 3.1. First Case. The Riccati Method. We consider the solution of equation (3.4) to be of the form ∑ = = ξ n 0 k k k Y a ) ( u (3.5) where ) ( a a k k ξ ≡ for every n , , 2 , 1 , 0 k L = . 3.1.1. Method I. We first consider that ) ( Y Y ξ ≡ satisfies Riccati’s equation 2 BY A ) ( Y + = ξ ′ (3.6) where A and B depend on the variable ξ : ) ( A A ξ ≡ and ) ( B B ξ ≡ . We substitute (3.5) into (3.4) and take into account Riccati’s equation (3.6). We then balance the first order derivative term with that of the nonlinear term. The 9 order of the nonlinear term ) ( u 2 ξ is n 2 and that of the first order derivative term is 1 n 2 ) 1 n ( + = + − . We thus get the equation 1 n n 2 + = from which we obtain 1 n = . Therefore Y a a ) ( u 1 0 + = ξ (3.7) We first calculate ) ( u ξ ′ from the above equation, taking into account Riccati’s equation (3.6) and that the various coefficients 0 a , 1 a , A and B depend on ξ. We find 2 1 1 1 0 Y B a Y a ) A a a ( ) ( u + ′ + + ′ = ξ ′ (3.8) Because of (3.7) and (3.8), equation (3.4) becomes, arranged in powers of Y, Y } a ) a ( a k a { c a k a a a 2 1 A a k a 1 0 1 0 0 0 2 0 1 ω − − ′ + ) ` ¹ ¹ ´ ¦ + | ¹ | \ | ′ − ω − − 0 Y ) B k a 2 a ( a 2 1 2 1 1 = − − (3.9) We now have to determine the coefficients 0 a , 1 a , A and B from the above equation, equating to zero the coefficients of Y. Equating to zero the coefficient of 2 Y , we determine the coefficient 1 a : B k a 2 a 1 = (3.10) Equate to zero the coefficient of Y, 0 a ) a ( a k a 1 0 1 = ω − − ′ we can determine the coefficient 0 a : B B k a a a k a a 1 1 0 ′ + ω = ′ + ω = (3.11) Equate now to zero the constant term: 0 c a k a a a 2 1 A a k a 0 0 0 2 0 1 = + | ¹ | \ | ′ − ω − − (3.12) This equation takes on the form 10 D k a 2 ) 2 a ( a a A a 0 0 0 1 + ω − + ′ − = (3.13) where D is a constant ( k a c D 0 ≡ ). Equation (3.13) - because of (3.10) and (3.11) - can be written as D k a 2 B B k a B B k a B B k a A ) B k a 2 ( + | ¹ | \ | ′ + ω − | ¹ | \ | ′ + ω + ′ | ¹ | \ | ′ − = or M B B 4 1 B B 2 1 AB 2 + | ¹ | \ | ′ + ′ | ¹ | \ | ′ − = (3.14) where M is another constant ( 2 2 2 0 k a 4 c 2 M ω − ≡ ). What we need, in order to determine ) ( u ξ , is 0 a , 1 a and Y (see equation (3.7)). On the other hand, Y is determined by Riccati's equation 2 BY A ) ( Y + = ξ ′ once A and B are known quantities. We thus need four quantities: 0 a , 1 a , A and B. We do have however three equations only, (3.10), (3.11) and (3.14). There is one equation missing. However there is no need for an extra equation, as we shall see below. First of all, Riccati’s equation 2 BY A ) ( Y + = ξ ′ under the substitution ) ( w ) ( w ) ( B 1 ) ( Y ξ ξ ′ ⋅ ξ − = ξ (3.15) takes on the form of a linear second order ordinary differential equation 0 ) ( w B A ) ( w B B ) ( w = ξ + ξ ′ | ¹ | \ | ′ − ξ ′ ′ (3.16) with unknown function ) ( w ξ . We now transform equation (3.16) under the substitution ) ( y ) ( B ) ( w ξ ⋅ ξ = ξ (3.17) 11 The derivatives of the function ) ( u ξ transform as | ¹ | \ | ′ + ′ = ξ ′ y B B y 2 1 B 1 ) ( w (3.18) ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ ′ ′ + ′ ′ + | | ¹ | \ | ′ ⋅ − ′ ′ = ξ ′ ′ y B y B y B ) B ( 4 1 B 2 1 B 1 ) ( w 2 (3.19) Equation (3.16), because of (3.14) and (3.17)-(3.19), takes on the form + | ¹ | \ | ′ + ′ | ¹ | \ | ′ − ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ ′ ′ + ′ ′ + | | ¹ | \ | ′ ⋅ − ′ ′ y B B y 2 1 B B B 1 y B y B y B ) B ( 4 1 B 2 1 B 1 2 0 y B M B B 4 1 B B 2 1 2 = ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ + | ¹ | \ | ′ + ′ | ¹ | \ | ′ − + which gives us upon multiplying by B , the simple equation (some miraculous cancellations take place) 0 ) ( y M ) ( y = ξ ⋅ + ξ ′ ′ (3.20) Considering various forms of the constant M, we can determine ) ( y ξ and then ) ( w ξ from (3.16). For 2 n M = , equation (3.20) admits the general solution ) n sin( C ) n cos( C ) ( y 2 1 ξ + ξ = ξ , whereas for 2 n M − = admits the general solution ) n sinh( C ) n cosh( C ) ( y 2 1 ξ + ξ = ξ . If 0 M = then 2 1 C C ) ( y + ξ = ξ . The function ) ( Y ξ can be determined from (3.15) and then ) ( u ξ from (3.7), calculating first 1 a and 0 a from (3.10) and (3.11) respectively. We thus find that | | ¹ | \ | ξ ξ ′ ⋅ ξ − + | ¹ | \ | ′ + ω = + = ξ ) ( w ) ( w ) ( B 1 B k a 2 B B k a Y a a ) ( u 1 0 or ) ( w ) ( w k a 2 B B k a ) ( u ξ ξ ′ ⋅ − ′ + ω = ξ (3.21) Since ) ( y ) ( B ) ( w ξ ⋅ ξ = ξ , we obtain 12 ) ( y ) ( y ) ( B ) ( B 2 1 ) ( w ) ( w ξ ξ ′ + ξ ξ ′ ⋅ = ξ ξ ′ Using the above expression, we obtain from (3.21) that ) ( y ) ( y k a 2 ) ( u ξ ξ ′ ⋅ − ω = ξ (3.22) This is quite a remarkable result : no matter what the coefficients of Riccati’s equation are, we arrive at the equation (3.22) where ) ( y ξ satisfies equation (3.20). We thus obtain the following three solutions, depending on the values of the constant M (which appears in equation (3.20)) ) n tan( C ) n tan( C 1 n k a 2 ) ( u ξ + ξ ⋅ − ⋅ − ω = ξ for 2 n M = (3.23) ) n tanh( C ) n tanh( C 1 n k a 2 ) ( u ξ + ξ ⋅ + ⋅ − ω = ξ for 2 n M − = (3.24) ξ ⋅ + ⋅ − ω = ξ C 1 C k a 2 ) ( u for 0 M = (3.25) where 2 1 C / C C = , with 1 C and 2 C are the two arbitrary constants of the general solution of the differential equation (3.20). 3.1.2. Method II. We consider instead of (3.6), the equation 2 Y R Y Q P Y ⋅ + ⋅ + = ′ (3.26) where P, Q and R are functions depending on ξ . In this case considering again ( ) ( a a 0 0 ξ = , ) ( a a 1 1 ξ = ) Y a a ) ( u 1 0 + = ξ (3.27) we find 2 1 1 1 1 0 RY a Y ) Q a a ( ) P a a ( ) ( u + + ′ + + ′ = ξ ′ (3.28) taking into account equation (3.26). Equation (3.4) then becomes + + ′ − ω − + − Y )} Q a a ( k a a ) a {( Y ) R k a 2 a ( a 2 1 1 1 1 0 2 1 1 13 0 1 0 0 0 c P a k a a k a ) 2 a ( a 2 1 = − ′ − ω − + (3.29) Equating to zero the coefficients of the powers of Y, we obtain the following system of equations R k a 2 a 1 = (3.30) 0 ) Q a a ( k a a ) a ( 1 1 1 0 = + ′ − ω − (3.31) 0 1 0 0 0 c P a k a a k a ) 2 a ( a 2 1 = − ′ − ω − (3.32) Equation (3.31), because of (3.30), takes on the form | ¹ | \ | ′ + + ω = R R Q k a a 0 (3.33) From (3.32) we obtain D k a 2 ) 2 a ( a a P a 0 0 0 1 + ω − + ′ − = (D is a constant) Using (3.30) and (3.33), we obtain from the above equation M R R Q 4 1 R R Q 2 1 P R 2 + ) ` ¹ ¹ ´ ¦ | ¹ | \ | ′ + + ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ ′ | ¹ | \ | ′ + ′ − = ⋅ (3.34) Using the usual substitution w w R 1 Y ′ ⋅ − = ′ (3.35) equation (3.26) transforms into 0 w P R w R R Q w = + ′ | ¹ | \ | ′ + − ′ ′ (3.36) Substituting the product P R in the above equation given by (3.34), we arrive at 0 w M R R Q 4 1 R R Q 2 1 w R R Q w 2 = ( ( ¸ ( ¸ + ) ` ¹ ¹ ´ ¦ | ¹ | \ | ′ + + ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ ′ | ¹ | \ | ′ + ′ − + ′ | ¹ | \ | ′ + − ′ ′ (3.37) We introduce another function f by 14 | ¹ | \ | ′ + = ′ R R Q f f (3.38) Equation (3.37) then becomes 0 w M f f 4 1 f f 2 1 w f f w 2 = ( ( ¸ ( ¸ + | ¹ | \ | ′ + ′ | ¹ | \ | ′ − + ′ ′ − ′ ′ (3.39) Under the substitution y f w = (3.40) equation (3.39) takes on the form 0 y M y = ⋅ + ′ ′ (3.41) We thus have obtained | ¹ | \ | ′ + + ω = R R Q k a a 0 , R k a 2 a 1 = and w w R 1 Y ′ ⋅ − = where y f w = with | ¹ | \ | ′ + = ′ R R Q f f and 0 y M y = ⋅ + ′ ′ . Therefore | ¹ | \ | ′ ⋅ − + | ¹ | \ | ′ + ω = ξ w w R 1 R k a 2 f f k a ) ( u or w w k a 2 f f k a ) ( u ′ ⋅ − ′ + ω = ξ (3.42) Since, using (3.40), y y f f 2 1 w w ′ + ′ ⋅ = ′ (3.43) we obtain from (3.39) that y y k a 2 ) ( u ′ ⋅ − ω = ξ (3.44) 15 We thus arrive again at the same result: no matter what the coefficients of Riccati’s equation (3.26) are, we arrive at the equation (3.41), where ) ( y ξ satisfies equation (3.41), which are equations (3.22) and (3.20) respectively of Method I. Conclusion. Burgers equation 0 u a u u u xx x t = − + under the substitution ) t x ( k ω − = ξ transforms into 0 2 c ) ( u k a ) ( u 2 1 ) ( u ) ( = ξ ′ − ξ + ξ ω − . Considering the expansion Y a a ) ( u 1 0 + = ξ where Y satisfies either one of the Riccati equations 2 BY A ) ( Y + = ξ ′ or 2 Y R Y Q P Y ⋅ + ⋅ + = ′ , where all the coefficients depend on ξ , we obtain the three solutions (3.23)-(3.25), depending on the values of the constant M. 3.1.3. Method III. Equation (3.4) is a Riccati equation by itself and under the substitution w w ) k a 2 ( u ′ − = transforms into the linear second order equation 0 w k a 2 c w k a w 0 = − ′ ω + ′ ′ (3.45) with constant coefficients. Let 2 2 0 2 k a c k a 2 + ω ≡ ∆ be the discriminant of the auxiliary to (3.45) equation. (I) If 2 0 2 n c k a 2 − = + ω , n real, then | | ¹ | \ | ξ ω − × ) ` ¹ ¹ ´ ¦ | | ¹ | \ | ξ + | | ¹ | \ | ξ = k a 2 exp k a 2 n sin C k a 2 n cos C w 2 1 and thus | | ¹ | \ | ξ + | | ¹ | \ | ξ ⋅ − ⋅ − ω = k a 2 n tan C k a 2 n tan C 1 n u , 2 1 C / C C = . (II) If 2 0 2 n c k a 2 = + ω , n real, then | | ¹ | \ | ξ ω − × ) ` ¹ ¹ ´ ¦ | | ¹ | \ | ξ + | | ¹ | \ | ξ = k a 2 exp k a 2 n sinh C k a 2 n cosh C w 2 1 16 and thus | | ¹ | \ | ξ + | | ¹ | \ | ξ ⋅ + ⋅ − ω = k a 2 n tanh C k a 2 n tanh C 1 n u , 2 1 C / C C = . (III) If 0 c k a 2 0 2 = + ω , then | | ¹ | \ | ξ ω − × ξ + = k a 2 exp ) C C ( w 1 2 and thus ξ ⋅ + ⋅ − ω = C 1 C k a 2 u , 2 1 C / C C = . We thus see that we derive the same set of solutions as in the previous two sections: it suffices to perform the rescaling n ) k a 2 ( n → . The reader might wonder why we have considered all these three cases instead to consider only the last one, since we get identical solutions. The reason is that the first two methods will be used as a prelude to the extended Riccati equation method, considered next, where we obtain quite different solutions. 3.2. Second Case. The Extended Riccati Method. In this case we consider the expansion ∑ ∑ = = + = ξ n 1 k k k n 0 k k k Y b Y a ) ( u and balance the first order derivative term with the second order nonlinear term of (3.4). We then find 1 n = and thus Y b Y a a ) ( u 1 1 0 + + = ξ (3.46) where again require all the coefficients 0 a , 1 a and 1 b depend on ξ, and Y satisfies Riccati’s equation 2 BY A Y + = ′ . From equation (3.42) we obtain (taking into account 2 BY A Y + = ′ ) 2 1 1 2 1 1 1 1 0 Y b A Y b BY a Y a ) B b A a a ( ) ( u − ′ + + ′ + − + ′ = ξ ′ (3.47) Therefore equation (3.4), under the substitution (3.46) and (3.47), becomes + − + ′ − + + ω − ) B b A a a ( k a ) b a 2 a ( 2 1 a ) ( 1 1 0 1 1 2 0 0 17 + − + ′ − ω − + 2 1 1 1 1 0 Y ) B k a 2 a ( a 2 1 Y } a k a a ) a {( 0 2 1 1 1 1 0 c Y ) A k a 2 a ( b 2 1 Y b k a b ) a ( = + + ′ − ω − + (3.48) Equating the coefficients of Y to zero, obtain a system of differential equations from which we can determine the various coefficients. Equating to zero the coefficients of 2 Y and 2 Y / 1 , we find that B k a 2 a 1 = and A k a 2 b 1 − = (3.49) respectively. Equating to zero the coefficients of Y and Y / 1 and taking into account the values of 1 a and 1 b , we find B B k a a 0 ′ + ω = and A A k a a 0 ′ + ω = (3.50) respectively. The above two equations imply that A and B are proportional each other: A s B 2 ⋅ − = (3.51) with 2 s − being the proportionality factor with s real. We do not consider the case A s B 2 ⋅ = , because this choice leads to imaginary type of solutions (see equation (3.58) below). The zeroth order term appearing in (3.48) takes on the form, taking into account the values of 1 a and 1 b : 2 ) 2 a ( a a k a AB ) k a 8 ( 0 0 0 2 2 ω − + ′ − = The above equation, after introducing the value of 0 a expressed in terms of A, takes on the form 2 2 m 2 A A 2 1 A A B A 8 − | ¹ | \ | ′ + ′ | ¹ | \ | ′ − = (3.52) where m is defined by 18 k a 2 1 m ω ⋅ = , 0 m > (3.53) From equation (3.52), since A s B 2 ⋅ − = , we obtain the following second order nonlinear ordinary differential equation 0 ) ( A s 8 ) ( A m 2 )) ( A ( 2 3 ) ( A ) ( A 4 2 2 2 2 = ξ ⋅ − ξ ⋅ + ξ ′ − ξ ′ ′ ⋅ ξ (3.54) Equation (3.54) can be solved using a number of methods (the auxiliary equation method, the − ′ G / G expansion method, the reduction to Ermakov’s equation, the projective Riccati equation expansion method, etc.). This has been done in Appendix A, where we have found a considerable number of different solutions. We now have to determine the function ) ( w ξ from the equation (3.16) 0 ) ( w B A ) ( w B B ) ( w = ξ + ξ ′ | ¹ | \ | ′ − ξ ′ ′ (3.55) This equation written in terms of A, using (3.47), takes on the form 0 ) ( w A s ) ( w A A ) ( w 2 2 = ξ − ξ ′ | ¹ | \ | ′ − ξ ′ ′ (3.56) The previous equation can be written as (see also the Note below, for another method of solution) 0 w A s A A w A w 2 2 = − ′ ′ − ′ ′ which is equivalent to 0 w A s A w 2 = − ′ | ¹ | \ | ′ (3.57) Multiplying by A w′ , we obtain 0 w w s A w A w 2 = ′ − ′ | ¹ | \ | ′ ′ which is equivalent to 0 ) w ( s A w 2 2 2 = ′ − ′ ( ( ¸ ( ¸ | ¹ | \ | ′ and from this by integration 0 w s A w 2 2 2 = − | ¹ | \ | ′ (3.58) where we have put the constant of integration equal to zero. We thus get 19 ) ( A s ) ( w ) ( w ξ ± = ξ ξ ′ (3.59) (Note. Another method of solution of equation (3.57) would be the following: Under the substitution A w′ = υ , since 2 A w A A w ′ ′ − ′ ′ = υ′ , equation (3.57) takes on the form 0 w w s 2 = ⋅ υ ′ ⋅ − υ′ , which is equivalent to 0 ) w ( s ) ( 2 2 2 = ′ ⋅ − ′ υ and by integration (we have put the integration constant equals to zero) 0 w s 2 2 2 = ⋅ − υ and then ) ( w s ) ( ξ ± = ξ υ , i.e. ) ( w s A w ξ ± = ′ , which is (3.59)). Therefore s 1 ) ( w ) ( w ) ( A s 1 ) ( w ) ( w ) ( B 1 ) ( Y 2 ± = ξ ξ ′ ⋅ ξ = ξ ξ ′ ⋅ ξ − = ξ (3.60) We then obtain the following expression for the function ) ( u ξ using (3.46), (3.50), (3.51) and (3.60): ) ( A s k a 4 ) ( A ) ( A k a ) ( u ξ ± ξ ξ ′ + ω = ξ (3.61) We thus arrive at the following Theorem: Theorem. Burgers equation 0 u a u u u xx x t = − + under the substitution ) t x ( k ω − = ξ transforms into 0 2 c ) ( u k a ) ( u 2 1 ) ( u ) ( = ξ ′ − ξ + ξ ω − . Considering the expansion Y b Y a a ) ( u 1 1 0 + + = ξ , where Y satisfies the Riccati equation 2 Y B A ) ( Y + = ξ ′ , with all the coefficients 0 a , 1 a , 1 b and A, B depending on ξ, we obtain that A s B 2 ⋅ − = and ) ( u ξ given by the formula ) ( A s k a 4 ) ( A ) ( A k a ) ( u ξ ± ξ ξ ′ + ω = ξ , where ) ( A ξ satisfies the differential equation 0 ) ( A s 8 ) ( A m 2 )) ( A ( 2 3 ) ( A ) ( A 4 2 2 2 2 = ξ ⋅ − ξ ⋅ + ξ ′ − ξ ′ ′ ⋅ ξ 20 4. The final solutions. Using the various solutions of the function ) ( A ξ calculated in Appendices A, B and C, we are able to find the different forms of the function ) ( u ξ from equation (3.61). 4.I. First Family of Solutions. Using (A.10) for ) ( A ξ , we obtain from (3.61) )] m tanh( ) C m a a ( ) m a C a [( ] ) m tanh( C [ )] m ( tanh 1 [ m a k a ) C 1 ( ) ( u 1 0 1 0 2 2 1 2 ξ − ρ + − ρ ξ + ξ − − + ω = ξ ) ` ¹ ¹ ´ ¦ | | ¹ | \ | ξ + + ξ ⋅ ⋅ ρ − + ± ) m tanh( C 1 ) m tanh( C m a a s k a 4 1 0 where the coefficients 0 a and 1 a satisfy the relations (four different combinations) 2 2 2 1 s 16 a ρ = , 2 2 2 0 s 16 m a = . (4.Ia) For s 4 a 1 ρ = and s 4 m a 0 = , we obtain )] m tanh( 1 [ ] ) m tanh( C [ )] m ( tanh 1 [ m k a ) C 1 ( ) ( u 2 ξ + − ξ + ξ − + + ω = ξ ) ` ¹ ¹ ´ ¦ ξ + + ξ ⋅ − ± ) m tanh( C 1 ) m tanh( C 1 m k a (4.1) (4.Ib) For s 4 a 1 ρ − = and s 4 m a 0 = , we obtain )] m tanh( 1 [ ] ) m tanh( C [ )] m ( tanh 1 [ m k a ) C 1 ( ) ( u 2 ξ + ξ + ξ − − − ω = ξ ) ` ¹ ¹ ´ ¦ ξ + + ξ ⋅ + ± ) m tanh( C 1 ) m tanh( C 1 m k a (4.2) (4.Ic) For s 4 a 1 ρ = and s 4 m a 0 − = , we obtain 21 )] m tanh( 1 [ ] ) m tanh( C [ )] m ( tanh 1 [ m k a ) C 1 ( ) ( u 2 ξ + ξ + ξ − − − ω = ξ ) ` ¹ ¹ ´ ¦ ξ + + ξ ⋅ + ) m tanh( C 1 ) m tanh( C 1 m k a m (4.3) (4.Id) For s 4 a 1 ρ − = and s 4 m a 0 − = , we obtain )] m tanh( 1 [ ] ) m tanh( C [ )] m ( tanh 1 [ m k a ) C 1 ( ) ( u 2 ξ − ξ + ξ − + − ω = ξ ) ` ¹ ¹ ´ ¦ ξ + + ξ ⋅ + − ± ) m tanh( C 1 ) m tanh( C 1 m k a (4.4) 4.II. Second Family of Solutions. Using (A.24) for ) ( A ξ , we obtain from (3.61) ± ) ` ¹ ¹ ´ ¦ ξ ξ + ⋅ ξ ξ − | | ¹ | \ | ξ − ⋅ ξ ⋅ + ξ ⋅ ⋅ + ω = ξ ) ( H ) ( F a a ) ( H a ) ( F D a a a m 4 exp )} ( F a ) ( H m a 4 { m 4 k a ) ( u 1 0 2 0 0 0 1 2 0 2 3 1 2 ) ` ¹ ¹ ´ ¦ ξ ξ + ± ) ( H ) ( F a a s k a 4 1 0 where 2 2 1 s 16 1 a = and 2 2 2 0 s 4 m a = (four different combinations). (4.IIa) For s 4 1 a 1 = and s 2 m a 0 = , we obtain ξ ξ ξ − ξ ) ` ¹ ¹ ´ ¦ ξ ξ + ξ ξ − ) ` ¹ ¹ ´ ¦ ξ ξ + ⋅ + ω = ξ m m 2 m 2 2 e ) ( H ) ( H ) ( F m 2 e ) ( H ) ( F D s 2 e ) ( H ) ( F s 8 m m s k a 2 ) ( u 22 ) ` ¹ ¹ ´ ¦ ξ ξ + ± ) ( H ) ( F m 2 k a (4.5) where ξ − − = ξ m 2 2 e s m D ) ( F and ξ − + ξ + = ξ m 2 e s 2 m D E ) ( H (E, D constants). Since ) m ( tanh D s 2 m E D s 2 m E ) m ( tanh s m D s m D ) ( H ) ( F 2 2 ξ | | ¹ | \ | ξ + − + | | ¹ | \ | ξ + + ξ | | ¹ | \ | + + | | ¹ | \ | − = ξ ξ equation (4.5) can be written as × + ω = ξ s k a 2 ) ( u ) ( Z ) ( Z ) ( X m 2 ) m tanh( ) ( Z ) ( X ) D m 4 ( s 2 m ) ( Z ) ( X ) D m 4 ( s 2 m 2 2 3 2 2 3 ξ ) ` ¹ ¹ ´ ¦ ξ ξ + ξ ) ` ¹ ¹ ´ ¦ ξ ξ + + − ) ` ¹ ¹ ´ ¦ ξ ξ − + × ) ` ¹ ¹ ´ ¦ ξ ξ + ± ) ( Z ) ( X m 2 k a (4.6) where ) m ( tanh s m D s m D ) ( X 2 2 ξ | | ¹ | \ | + + | | ¹ | \ | − = ξ (4.7) ) m ( tanh D s 2 m E D s 2 m E ) ( Z ξ | | ¹ | \ | ξ + − + | | ¹ | \ | ξ + + = ξ (4.8) (4.IIb) For s 4 1 a 1 = and s 2 m a 0 − = , we obtain ξ − ξ − ξ ξ ) ` ¹ ¹ ´ ¦ ξ ξ − ξ ξ + ) ` ¹ ¹ ´ ¦ ξ ξ − ⋅ + ω = ξ m m 2 m 2 2 e ) ( H ) ( H ) ( F m 2 e ) ( H ) ( F D s 2 e ) ( H ) ( F s 8 m m s k a 2 ) ( u 23 ) ` ¹ ¹ ´ ¦ ξ ξ + − ± ) ( H ) ( F m 2 k a (4.9) where ξ − = ξ m 2 2 e s m D ) ( F and ξ − ξ + = ξ m 2 e s 2 m D E ) ( H (E, D constants). Since ) m ( tanh D s 2 m E D s 2 m E ) m ( tanh s m D s m D ) ( H ) ( F 2 2 ξ | | ¹ | \ | ξ + + − | | ¹ | \ | ξ + − ξ | | ¹ | \ | + − | | ¹ | \ | − = ξ ξ equation (4.9) can be written as × + ω = ξ s k a 2 ) ( u ) ( Z ) ( Z ) ( X m 2 ) m tanh( ) ( Z ) ( X ) D m 4 ( s 2 m ) ( Z ) ( X ) D m 4 ( s 2 m 2 2 3 2 2 3 ξ ) ` ¹ ¹ ´ ¦ ξ ξ − ξ ) ` ¹ ¹ ´ ¦ ξ ξ + − + ) ` ¹ ¹ ´ ¦ ξ ξ − − × ) ` ¹ ¹ ´ ¦ ξ ξ + − ± ) ( Z ) ( X m 2 k a (4.10) where ) m ( tanh s m D s m D ) ( X 2 2 ξ | | ¹ | \ | + − | | ¹ | \ | − = ξ (4.11) ) m ( tanh D s 2 m E D s 2 m E ) ( Z ξ | | ¹ | \ | ξ + + − | | ¹ | \ | ξ + − = ξ (4.12) (4.IIc) For s 4 1 a 1 − = and s 2 m a 0 = , we obtain ξ − ξ − ξ ξ ) ` ¹ ¹ ´ ¦ ξ ξ − ξ ξ − ) ` ¹ ¹ ´ ¦ ξ ξ + − ⋅ + ω = ξ m m 2 m 2 2 e ) ( H ) ( H ) ( F m 2 e ) ( H ) ( F D s 2 e ) ( H ) ( F s 8 m m s k a 2 ) ( u 24 ) ` ¹ ¹ ´ ¦ ξ ξ − ± ) ( H ) ( F m 2 k a (4.13) where ξ + = ξ m 2 2 e s m D ) ( F and ξ + ξ + = ξ m 2 e s 2 m D E ) ( H (E, D constants). Since ) m ( tanh D s 2 m E D s 2 m E ) m ( tanh s m D s m D ) ( H ) ( F 2 2 ξ | | ¹ | \ | ξ + − − | | ¹ | \ | ξ + + ξ | | ¹ | \ | − − | | ¹ | \ | + = ξ ξ equation (4.13) can be written as × − ω = ξ s k a 2 ) ( u ) ( Z ) ( Z ) ( X m 2 ) m tanh( ) ( Z ) ( X ) D m 4 ( s 2 m ) ( Z ) ( X ) D m 4 ( s 2 m 2 2 3 2 2 3 ξ ) ` ¹ ¹ ´ ¦ ξ ξ − ξ ) ` ¹ ¹ ´ ¦ ξ ξ + − + ) ` ¹ ¹ ´ ¦ ξ ξ − − × ) ` ¹ ¹ ´ ¦ ξ ξ − ± ) ( Z ) ( X m 2 k a (4.14) where ) m ( tanh s m D s m D ) ( X 2 2 ξ | | ¹ | \ | − − | | ¹ | \ | + = ξ (4.15) ) m ( tanh D s 2 m E D s 2 m E ) ( Z ξ | | ¹ | \ | ξ + − − | | ¹ | \ | ξ + + = ξ (4.16) (4.IId) For s 4 1 a 1 − = and s 2 m a 0 − = , we obtain ξ ξ ξ − ξ ) ` ¹ ¹ ´ ¦ ξ ξ + ξ ξ − ) ` ¹ ¹ ´ ¦ ξ ξ + ⋅ − ω = ξ m m 2 m 2 2 e ) ( H ) ( H ) ( F m 2 e ) ( H ) ( F D s 2 e ) ( H ) ( F s 8 m m s k a 2 ) ( u 25 ) ` ¹ ¹ ´ ¦ ξ ξ + ± ) ( H ) ( F m 2 k a (4.17) where ξ − + = ξ m 2 2 e s m D ) ( F and ξ − − ξ + = ξ m 2 e s 2 m D E ) ( H (E, D constants). Since ) m ( tanh D s 2 m E D s 2 m E ) m ( tanh s m D s m D ) ( H ) ( F 2 2 ξ | | ¹ | \ | ξ + + + | | ¹ | \ | ξ + − ξ | | ¹ | \ | − + | | ¹ | \ | + = ξ ξ equation (4.17) can be written as × − ω = ξ s k a 2 ) ( u ) ( Z ) ( Z ) ( X m 2 ) m tanh( ) ( Z ) ( X ) D m 4 ( s 2 m ) ( Z ) ( X ) D m 4 ( s 2 m 2 2 3 2 2 3 ξ ) ` ¹ ¹ ´ ¦ ξ ξ + ξ ) ` ¹ ¹ ´ ¦ ξ ξ + + − ) ` ¹ ¹ ´ ¦ ξ ξ − + × ) ` ¹ ¹ ´ ¦ ξ ξ + ± ) ( Z ) ( X m 2 k a (4.18) where ) m ( tanh s m D s m D ) ( X 2 2 ξ | | ¹ | \ | − + | | ¹ | \ | + = ξ (4.19) ) m ( tanh D s 2 m E D s 2 m E ) ( Z ξ | | ¹ | \ | ξ + + + | | ¹ | \ | ξ + − = ξ (4.20) 4.III. Third Family of Solutions. Using (A.31) for ) ( A ξ , we obtain from (3.61) ± − ± − − + ω = ξ ξ ξ 2 2 2 m 2 1 2 m 4 2 1 2 2 2 2 s m 16 ) e C C m 2 ( ] e C ) s 4 C ( m 4 [ m 2 k a ) ( u 26 2 2 2 m 2 1 2 m 2 1 2 s m 16 ) e C C m 2 ( e C m 4 s k a 4 − ± ± ξ ξ (4.21) The above expression can also be written as } C m s k a 16 ) ; s , m ( U { ) ; s , m ( V )] m ( tanh 1 [ m k a 2 ) ( u 1 2 2 ± ξ ξ ξ + + ω = ξ (4.22) where ) m 2 tanh( ] s m 16 C ) 1 m 4 [( s m 16 C ) 1 m 4 ( ) ; s , m ( U 2 2 2 1 2 2 2 2 1 2 ξ − + − − − = ξ (4.23) − ξ − ± = ξ 2 1 2 1 2 )} m tanh( ) C C m 2 ( ) C C m 2 ( { ) ; s , m ( V m )] m ( tanh 1 [ )] m 2 ( tanh 1 [ s m 16 2 2 2 ξ + × ξ − − (4.24) 4.IV. Fourth Family of Solutions. The projective Riccati equation method provides us with eight families of solutions, each family containing some subfamilies of solutions. In Appendix A, (Section A.IV) we have found twenty one solutions in total. To save space, the reader can easily write down the various expressions for ) ( u ξ , using the different solutions of ) ( A ξ found in that Section and (3.61). The complete set is to be reported in electronic form somewhere else. 4.V. Fifth Family of Solutions. We now use (A.71)-(A.74) for ) ( A ξ and the general expression (3.61) and we find four additional families of solutions. (4.Va) Using (A.71) and (3.61), we have ξ − ξ ξ ξ − ξ ξ − + ± + + ω = ξ m 2 m 1 m m 2 m 1 m 2 e C e C e s k a 4 e C e C e C m 2 k a ) ( u or ) m tanh( ) C C ( ) C C ( )] m tanh( 1 [ C m k a 2 ) ( u 2 1 2 1 2 ξ − + + ξ − + ω = ξ 27 ) m tanh( ) C C ( ) C C ( ) m tanh( 1 s k a 4 2 1 2 1 ξ − + + ξ + ± (4.25) (4.Vb) Using (A.72) and (3.61), we have ξ − ξ ξ − ξ − ξ ξ − ± − − ω = ξ m m 1 m m m 1 m 1 2 e s 2 e m C e m s k a 4 e s 2 e m C e C m 2 k a ) ( u or ) m tanh( ) s 2 m C ( ) s 2 m C ( )] m tanh( 1 [ C k a m 2 ) ( u 1 1 1 2 ξ + + − ξ + − ω = ξ ) m tanh( ) s 2 m C ( ) s 2 m C ( ) m tanh( 1 s k a m 4 1 1 ξ + + − ξ − ± (4.26) (4.Vc) Using (A.73) and (3.61), we have ξ − ξ ξ − ξ − ξ ξ + ± + − ω = ξ m m 1 m m m 1 m 1 2 e s 2 e m C e m s k a 4 e s 2 e m C e C m 2 k a ) ( u or ) m tanh( ) s 2 m C ( ) s 2 m C ( )] m tanh( 1 [ C k a m 2 ) ( u 1 1 1 2 ξ − + + ξ + − ω = ξ ) m tanh( ) s 2 m C ( ) s 2 m C ( ) m tanh( 1 s k a m 4 1 1 ξ − + + ξ − ± (4.27) (4.Vd) Using (A.74) and (3.61), we have (we find 0 ) ( A = ξ ′ in this case) ξ − ξ ξ − ξ + + + + ± ω = ξ m 2 2 2 m 1 m 2 m 1 2 e C s 4 s 4 e C s m 2 e C s m 2 s m 2 e C m s k a 4 ) ( u or k a m 2 ) ( u ± ω = ξ (4.28) 4.VI. Sixth Family of Solutions. Using the results of Appendix B and (3.61), we have ] ) ( a a [ s k a 4 ) ( a a ) ( a k a ) ( u 1 0 1 0 1 ξ ϕ + ± ξ ϕ + ξ ϕ′ + ω = ξ (4.29) 28 where ) (ξ ϕ is a solution of the differential equation (B.2), the various coefficients corresponding to the Solutions 1-11 of Appendix B. The final formulas are too lengthy to write them down here. 4.VII. Seventh Family of Solutions. Using the results of Appendix C, we obtain ξ ⋅ ξ ⋅ + ξ ⋅ + + = ′ ) s , m ( F ) s , m ( F e D E e ) s , m ( F D G G and ξ ⋅ ξ ⋅ + ξ ⋅ + = ′ ′ ) s , m ( F ) s , m ( F 2 e D E e ) s , m ( F G G We thus have ξ ⋅ ξ ⋅ + ξ ⋅ + + ⋅ ± = | ¹ | \ | ′ + = ξ ) s , m ( F ) s , m ( F 0 1 0 e D E e ) s , m ( F D s 4 1 a G G a a ) ( A and = | ¹ | \ | ′ + ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | ¹ | \ | ′ − | ¹ | \ | ′ ′ + ′ = ξ ξ ′ G G a a G G G G a a ) ( A ) ( A 1 0 2 1 0 ξ ⋅ ξ ⋅ ξ ⋅ ξ ⋅ ξ ⋅ ξ ⋅ + ξ ⋅ + + ⋅ ± ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | ¹ | \ | + ξ ⋅ + + − | | ¹ | \ | + ξ ⋅ + ± ′ = ) s , m ( F ) s , m ( F 0 2 ) s , m ( F ) s , m ( F ) s , m ( F ) s , m ( F 2 0 e D E e ) s , m ( F D s 4 1 a e D E e ) s , m ( F D e D E e ) s , m ( F s 4 1 a The final expression for ) ( u ξ , follows from (3.61) and the above formulas. The coefficient 0 a satisfies equation (C.7), which is actually equation (A.1) and therefore admits all the solutions reported in Appendices A, B and C. 5. Solution of the Burgers equation using the G'/G-expansion. In this section we solve Burgers equation using the − | ¹ | \ | ′ G G expansion. It will be clear by the end of this section that we need the solutions found previously using the Riccati equation methods. Using the expansion 29 | ¹ | \ | ′ + = ξ G G a a ) ( u 1 0 (5.1) assuming that the coefficients 0 a and 1 a are − ξ dependent, Burgers equation (3.4) becomes − ) ` ¹ ¹ ´ ¦ | ¹ | \ | ′ + + ) ` ¹ ¹ ´ ¦ | ¹ | \ | ′ + ω − 2 1 0 1 0 G G a a 2 1 G G a a ) ( 0 2 1 1 0 c G G G G a G G a a k a = ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ ( ( ¸ ( ¸ | ¹ | \ | ′ − ′ ′ + | ¹ | \ | ′ ′ + ′ − (5.2) Upon expanding the above equation and equating the coefficients of G to zero, we obtain Coefficient of 0 G : 0 0 2 0 0 c a k a a 2 1 a ) ( = ′ − + ω − (5.3) Coefficient of 1 G − : 0 G a k a G a k a G a a a ) ( 1 1 1 0 1 = ′ ′ − ′ ′ − ′ + ω − (5.4) Coefficient of 2 G − : 0 ) G ( a k a ) G ( a 2 1 2 1 2 2 1 = ′ + ′ (5.5) From equation (5.3) we conclude that ) ( a a 0 0 ξ ≡ is a solution of the equation (3.4). Assuming that 0 a 1 ≠ we obtain from (5.5) that k a 2 a 1 − = (5.6) showing that 1 a is a constant. Finally from equation (5.4) we obtain 0 G ) a k a ( G k a 0 = ω + ′ − + ′ ′ (5.7) The above equation can be reduced to a linear first order differential equation and can be solved in principle provided that 0 a is a known quantity. However the coefficient 0 a satisfies equation (5.3), which has been solved, using the Riccati 30 equation method with variable expansion coefficients. The solutions of equation (5.3) are actually the various expressions of the function ) ( u ξ of Section 4. The solutions of (5.3) substituted into (5.7) lead to equations with rather complicated solutions, despite the fact that equation (5.7) is a linear differential equation. It is obvious that we could use some other methods in solving equation (5.3), using, say, the Riccati equation method with constant coefficients. Using the expansion Y b b a 1 0 0 + = with 2 Y M L Y ⋅ + = ′ ( 0 b , 1 b , L, and M constants) and substituting into (5.3), after equating to zero the coefficients of Y, we obtain the following set of equations 0 b M k a b 2 1 1 1 = | ¹ | \ | − , 0 ) b ( b 0 1 = ω − , 0 2 0 1 0 c b 2 1 b L k a b = + − ω − From the previous equations we obtain the following two sets of solutions Solution 1. ω = 0 b , M k a 2 b 1 = , M 1 k a 4 c 2 L 2 2 0 2 ⋅ + ω − = Solution 2. ω = 0 b , L 1 k a 2 c 2 b 0 2 1 ⋅ + ω − = , L 1 k a 4 c 2 M 2 2 0 2 ⋅ + ω − = For Solution 1, Riccati's equation 2 Y M L Y ⋅ + = ′ becomes 2 2 2 0 2 Y M M 1 k a 4 c 2 Y ⋅ + ⋅ + ω − = ′ which admits the solution | | | ¹ | \ | + ξ + ω + ω − = ) C ( k a 2 c 2 tanh M k a 2 c 2 Y 1 0 2 0 2 Therefore | | | ¹ | \ | + ξ + ω + ω − ω = ) C ( k a 2 c 2 tanh c 2 a 1 0 2 0 2 0 31 For Solution 2, Riccati's equation 2 Y M L Y ⋅ + = ′ becomes 2 2 2 0 2 Y L 1 k a 4 c 2 L Y ⋅ ⋅ + ω − = ′ which admits the solution | | | ¹ | \ | + ξ + ω + ω = ) C ( k a 2 c 2 tanh c 2 L k a 2 Y 1 0 2 0 2 Therefore | | | ¹ | \ | + ξ + ω + ω − ω = ) C ( k a 2 c 2 tanh c 2 a 1 0 2 0 2 0 (5.8) We thus see that 0 a is given by the same expression in both solutions. Because of the expression (5.8) for 0 a , equation (5.7) becomes K G 2 K tanh K K 1 G − = ′ | ¹ | \ | | ¹ | \ | ξ + − + ′ ′ (5.9) where we have put 0 c 0 = , 0 C 1 = and k a K ω = for simplicity. Equation (5.9) is a rather complicated equation, which admits a closed-form solution expressed in terms of the hypergeometric function. Details of the solution are given in Appendix D. Using the results of that Appendix, we obtain the following expression for the function ) ( u ξ : × − | ¹ | \ | ξ ω − ω = ξ k a 2 2 K tanh ) ( u 2 2 ) K 1 ( 1 2 2 ) K 1 ( 1 2 2 K ) 1 K 3 K 2 ( K 2 ) ; K ( Z e K ) 1 K ( 4 ) ; K ( V K 2 K h sec ] e K ) 1 K 3 K 2 ( K 2 ) ; K ( U K [ + − + ξ ⋅ − + ξ ⋅ | ¹ | \ | ξ × + − + ξ ⋅ × ξ − ξ − (5.10) The functions ) ; K ( U ξ , ) ; K ( V ξ and ) ; K ( Z ξ are defined in Appendix D. The function ) ; K ( Z ξ is expressed in terms of the hypergeometric function. 32 We thus see that even in the simplest case, we obtain very complicated equations. In more complex situations, we might need supercomputing facilities with symbolic capabilities. We also conclude that in order to find solutions using the − ′ ) G / G ( expansion method with variable expansion coefficients, we have to consider the Riccati expansion method with variable coefficients first. Appendix A. In this Appendix we shall solve equation (3.54): 0 ) ( A s 8 ) ( A m 2 )) ( A ( 2 3 ) ( A ) ( A 4 2 2 2 2 = ξ ⋅ − ξ ⋅ + ξ ′ − ξ ′ ′ ⋅ ξ (A.1) A.I) First Method. We consider an expansion of the form ∑ = ξ ϕ = ξ n 0 k k k ) ( a ) ( A (A.2) where ) (ξ ϕ satisfies Jacobi's differential equation 4 2 ) ( d d ϕ ρ + ϕ µ + λ = ξ ϕ ξ (A.3) (In Appendix B we consider a full fourth order polynomial). Upon substitution of (A.2) into (A.1) and balancing 4 A with either A A ′ ′ ⋅ or 2 ) A ( ′ , and taking into account (A.3), we obtain 1 n = . We thus substitute ) ( a a ) ( A 1 0 ξ ϕ + = ξ (A.4) Since 4 2 1 1 a ) ( a ) ( A ϕ ρ + ϕ µ + λ = ξ ϕ′ = ξ ′ and ) 2 ( a ) ( A 3 1 ϕ ρ + ϕ µ = ξ ′ ′ equation (A.1) becomes + ϕ ρ + − + ϕ ρ + − 3 2 1 2 1 0 4 2 1 2 2 1 ) a s 16 ( a a 2 ) 2 1 a s 8 ( a + ϕ µ + + − + ϕ µ − + − + ) 4 1 m a s 8 ( a a 4 ) 12 1 m 3 1 a s 8 ( a 6 2 2 0 2 1 0 2 2 2 0 2 2 1 0 a 2 3 a s 8 a m 2 2 1 4 0 2 2 0 2 = λ − − + (A.5) 33 Equating to zero the coefficients of the different powers of ϕ, we find ρ = 2 2 1 s 16 1 a , 2 2 2 0 m s 16 1 a = , 2 m 2 − = µ , 4 m = ρ λ (A.6) We thus obtain that 4 2 2 2 2 4 m 2 m ) ( ϕ ρ + ϕ − ρ = ξ ϕ′ (A.7) where we have redefined 2 ρ → ρ . The last equation can be written as ρ − ϕ ρ = ξ ϕ′ 2 2 m ) ( (A.8) This is a Riccati equation which can be transformed into a second order ODE, admitting solution ) m tanh( C 1 ) m tanh( C m ) ( ξ + + ξ ⋅ ⋅ ρ − = ξ ϕ (A.9) Therefore ) ( A ξ is given by (A.4) with ) (ξ ϕ given by (A.9): | | ¹ | \ | ξ + + ξ ⋅ ⋅ ρ − + = ξ ) m tanh( C 1 ) m tanh( C m a a ) ( A 1 0 (A.10) where the coefficients 0 a and 1 a satisfy the relations (four different combinations) 2 2 2 1 s 16 a ρ = , 2 2 2 0 s 16 m a = (A.11) A.II) Second Method. We consider the − ′ ) G / G ( expansion of the form k n 0 k k G G a ) ( A ∑ = | ¹ | \ | ′ = ξ (A.12) with k a ( n , , 1 , 0 k L = ) constants and ) ( G G ξ = . (In Appendix C we consider variable expansion coefficients: ) ( a a k k ξ ≡ ). 34 Upon substitution of (A.12) into (A.1) and balancing 4 A with either A A ′ ′ ⋅ or 2 ) A ( ′ , we obtain 1 n = . We thus substitute | ¹ | \ | ′ + = ξ G G a a ) ( A 1 0 (A.13) Notice that 0 a and 1 a have nothing to do with the similar coefficients appearing in (3.7), (3.24) or (3.42). Since ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | ¹ | \ | ′ − ′ ′ = ξ ′ 2 1 G G G G a ) ( A and ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | ¹ | \ | ′ + ′ ′ ⋅ ′ − ′ ′ ′ = ξ ′ ′ 3 2 1 G G 2 G G G 3 G G a ) ( A equation (A.1) becomes ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | ¹ | \ | ′ + ′ ′ ⋅ ′ − ′ ′ ′ ) ` ¹ ¹ ´ ¦ | ¹ | \ | ′ + 3 2 1 0 1 G G 2 G G G 3 G G G G a a a ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | ¹ | \ | ′ + | ¹ | \ | ′ + ⋅ + ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | ¹ | \ | ′ + | ¹ | \ | ′ | ¹ | \ | ′ ′ − | ¹ | \ | ′ ′ − 2 2 1 1 0 2 0 2 4 2 2 2 1 G G a G G a a 2 a m 2 G G G G G G 2 G G a 2 3 0 G G a G G a a 4 G G a a 6 G G a a 4 a s 8 4 4 1 3 3 1 0 2 2 1 2 0 1 3 0 4 0 2 = ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | ¹ | \ | ′ + | ¹ | \ | ′ + | ¹ | \ | ′ + | ¹ | \ | ′ + ⋅ − From the above equation we obtain the following set of coefficients, equating each one of them to zero: Coefficient of 4 G − : 0 ) G ( a s 8 ) G ( a 2 3 ) G ( a 2 4 4 1 2 4 2 1 4 2 1 = ′ − ′ − ′ (A.14) Coefficient of 3 G − : 0 ) G ( a a s 32 ) G ( a a 2 3 3 1 0 2 3 1 0 = ′ − ′ (A.15) Coefficient of 2 G − : 2 2 1 2 2 2 1 2 1 1 0 ) G ( a m 2 ) G ( a 2 3 ) G ( ) G ( a ) G ( ) G ( a a ) 3 ( ′ + ′ ′ − ′ ′ ′ ′ + ′ ′ ′ − 0 ) G ( a a s 48 2 2 1 2 0 2 = ′ − (A.16) Coefficient of 1 G − : 0 G a a s 32 G a a m 4 G a a 1 3 0 2 1 0 2 1 0 = ′ − ′ + ′ ′ ′ (A.17) 35 Coefficient of 0 G : 0 a s 8 a m 2 4 0 2 2 0 2 = − (A.18) The above system (A.14)-(A.18) is equivalent to 2 2 1 s 16 1 a = , 2 2 2 0 s 4 m a = , G m 4 G 2 ′ = ′ ′ ′ , 0 a m 4 G G a 2 G G a 1 2 0 2 1 = + | ¹ | \ | ′ ′ ′ + | ¹ | \ | ′ ′ ′ (A.19) The last equation is a quadratic equation with respect to the ratio G / G ′ ′ ′ and because its discriminant is zero, we obtain 1 0 a a G G − = ′ ′ ′ (A.20) The above equation combined with the equation G m 4 G 2 ′ = ′ ′ ′ , gives us 0 1 2 a a m 4 G G − = ′ ′ ′ ′ ′ (A.21) and by integration | | ¹ | \ | ξ − + ξ + = 0 1 2 3 2 1 a a m 4 exp C C C G (A.22) Therefore | | ¹ | \ | ξ − + ξ + | | ¹ | \ | ξ − | | ¹ | \ | − + = ′ 0 1 2 3 2 1 0 1 2 0 1 2 3 2 a a m 4 exp C C C a a m 4 exp a a m 4 C C G G or, adjusting the above expression and redefining the constants (i.e. 3 2 0 C / C a D = and 3 1 0 C / C a E = ), we obtain | | ¹ | \ | ξ − + ξ + | | ¹ | \ | ξ − − = ′ 0 1 2 0 0 1 2 1 2 a a m 4 exp a D E a a m 4 exp a m 4 D G G We thus have 36 | | | | | ¹ | \ | | | ¹ | \ | ξ − + ξ + | | ¹ | \ | ξ − − + = | ¹ | \ | ′ + = 0 1 2 0 0 1 2 1 2 1 0 1 0 a a m 4 exp a D E a a m 4 exp a m 4 D a a G G a a A (A.23) where 2 2 1 s 16 1 a = and 2 2 2 0 s 4 m a = (four different combinations). Equation (A.23) can also be written as ) ( H ) ( F a a A 1 0 ξ ξ ⋅ + = (A.24) where | | ¹ | \ | ξ − − = ξ 0 1 2 1 2 a a m 4 exp a m 4 D ) ( F (A.25) | | ¹ | \ | ξ − + ξ + = ξ 0 1 2 0 a a m 4 exp a D E ) ( H (A.26) A.III) Third Method. Using the transformation ) ( w 1 ) ( A 2 ξ = ξ (A.27) equation (A.1) transforms into the Ermakov equation (Ermakov [68]) 3 2 2 w s 4 ) ( w m ) ( w − − = ξ − ξ ′ ′ (A.28) Equation 0 ) ( w m ) ( w 2 = ξ − ξ ′ ′ admits two independent solutions ξ = m 1 e w and ξ − = m 2 e w . We consider one of them, ξ − = m 2 e w and introduce the transformation ∫ ξ = η 2 w d , i.e. ξ = η m 2 e m 2 1 , ) m 2 ( ln m 2 1 η = ξ and ) ( w e z m ξ = ξ , ) ( z z η = . Since } ) ( w ) ( w m { e d dz m ξ ′ + ξ = η ξ − and } ) ( w ) ( w m { e d z d 2 m 3 2 2 ξ ′ ′ + ξ − = η ξ − 37 Ermakov’s equation (A.28) transforms into the equation 3 2 2 2 z s 4 d z d − − = η (A.29) Using one of the standard methods, i.e. multiplying both members by η d dz 2 or introducing the variable p by η = d dz p , we arrive at the equation 1 2 2 2 C z s 4 d dz + = | | ¹ | \ | η (A.30) From the above equation we obtain the two equations 1 2 2 C z s 4 d dz + = η and 1 2 2 C z s 4 d dz + − = η The first of the above equations admits the solution 2 1 2 2 1 C C s 4 z C + η = + while the second 2 1 2 2 1 C C s 4 z C + η − = + . We thus have, squaring both the previous two equations and solving with respect to 2 z 1 2 2 2 2 m 2 1 2 2 C m 4 s m 16 ) e C C m 2 ( z − ± = ξ and then, since ) ( w e z m ξ = ξ , ξ − ξ × − ± = ξ m 2 1 2 2 2 2 m 2 1 2 2 e C m 4 s m 16 ) e C C m 2 ( ) ( w We thus obtain, using the above expression and (A.27), that ) ( A ξ is given by 2 2 2 m 2 1 2 m 2 1 2 s m 16 ) e C C m 2 ( e C m 4 ) ( A − ± ⋅ = ξ ξ ξ (A.31) The above expression can be written in terms of ) m tanh( ξ . In fact since 38 2 m 2 m 1 m 2 1 2 ) e s m 4 ( ) e C e C m 2 ( C m 4 ) ( A ξ − ξ ξ − − ± ⋅ = ξ we can easily transform the above expression into 2 2 2 2 1 2 1 2 2 1 2 )] m ( tanh 1 [ s m 16 )] m ( tanh ) C C m 2 ( ) C C m 2 ( [ )] m ( tanh 1 [ C m 4 ) ( A ξ − − ξ − ± ξ − = ξ m A.IV. Fourth Method. We shall now use the projective Riccati equation method (see for example Abdou [5]) in solving equation (A.1). We consider the expansion ] ) ( g b ) ( f a [ ) ( f a ) ( A N 1 i i i 1 i 0 ∑ = − ξ + ξ ξ + = ξ (A.32) where the functions ) ( f ξ and ) ( g ξ satisfy the system ) ( g ) ( f p ) ( f ξ ξ = ξ ′ (A.33) ) ( f r ) ( g p q ) ( g 2 ξ − ξ + = ξ ′ (A.34) ( ( ¸ ( ¸ ξ δ + + ξ − − = ξ ) ( f q r ) ( f r 2 q p 1 ) ( g 2 2 2 (A.35) The system of equations (A.33) and (A.34) admits five families of solutions given by (I) If 2 2 µ − λ = δ and 0 q p < then ξ − ⋅ µ + ξ − ⋅ λ + = ξ q p cosh q p sinh r q ) ( f (A.36) ξ − ⋅ µ + ξ − ⋅ λ + ξ − ⋅ µ + ξ − ⋅ λ ⋅ − = ξ q p cosh q p sinh r q p sinh q p cosh p q p ) ( g (A.37) with ( ( ¸ ( ¸ ξ µ − λ + + ξ − − = ξ ) ( f q r ) ( f r 2 q p 1 ) ( g 2 2 2 2 2 (A.38) (II) If 2 2 µ − λ − = δ and 0 q p > then 39 ξ ⋅ µ + ξ ⋅ λ + = ξ q p cos q p sin r q ) ( f (A.39) ξ ⋅ µ + ξ ⋅ λ + ξ ⋅ µ + ξ ⋅ λ ⋅ = ξ q p cos q p sin r q p sin q p cos p q p ) ( g (A.40) with ( ( ¸ ( ¸ ξ µ − λ − + ξ − − = ξ ) ( f q r ) ( f r 2 q p 1 ) ( g 2 2 2 2 2 (A.41) (III) If 0 q = , then ζ + ξ σ + ξ = ξ 2 2 r p 1 ) ( f (A.42) ζ + ξ σ + ξ σ + ξ ⋅ − = ξ 2 2 r p r p p 1 ) ( g (A.43) with ) ( f p r 2 p ) ( f p r 2 ) ( g 2 2 2 2 ξ ( ( ¸ ( ¸ ζ − σ + ξ = ξ (A.44) where σ and ζ are free parameters. (IV) If 1 p ± = and 2 r − = δ then ) ( r p 2 r 6 q ) ( f ξ ψ + = ξ (A.45) ) ( 12 q ) ( 12 ) ( g ξ ψ′ + ξ ψ′ = ξ (A.46) where ) (ξ ψ satisfies Weirstrass equation 216 q p ) ( 12 q ) ( 4 )) ( ( 3 2 3 2 − ξ ψ − ξ ψ = ξ ψ′ with solution ) ( ) ( ξ ℘ = ξ ψ . The relation between f and g is given by 40 q p f p r 2 g 2 − = (A.47) (V) If 1 p ± = and 25 r 2 − = δ then ) ( 72 q p 5 r 6 q 5 ) ( f 2 ξ ψ + = ξ (A.48) ) ( )] ( 12 q p [ ) ( q ) ( g ξ ψ ξ ψ + ξ ψ′ − = ξ (A.49) with ) ( ) ( ξ ℘ = ξ ψ and | | ¹ | \ | + − − = 2 2 2 f q 25 r 24 f r 2 q p 1 g (A.50) Upon substituting (A. 32) into (A.1) and balancing 4 A with either A A ′ ′ ⋅ or 2 ) A ( ′ , we obtain 1 n = . We thus substitute ) ( g b ) ( f a a ) ( A 1 1 0 ξ + ξ + = ξ (A.51) Since )} ( f r ) ( g p q { b ) ( g ) ( f p a ) ( A 2 1 1 ξ − ξ + + ξ ξ = ξ ′ and − ξ ξ − ξ + ξ = ξ ′ ′ ) ( g ) ( f p r b 3 ) ( g q p b 2 ) ( f q p a ) ( A 1 1 1 ) ( g p b 2 ) ( g ) ( f p a 2 ) ( f r p a 3 2 1 2 2 1 2 1 ξ + ξ ξ + ξ − equation (A.1) becomes × ξ + ξ + )} ( g b ) ( f a a { 1 1 0 − ξ ξ − ξ + ξ × ) ( g ) ( f p r b 3 ) ( g q p b 2 ) ( f q p a { 1 1 1 − ξ + ξ ξ + ξ − )} ( g p b 2 ) ( g ) ( f p a 2 ) ( f r p a 3 2 1 2 2 1 2 1 + ξ − ξ + + ξ ξ − 2 2 1 1 )}] ( f r ) ( g p q { b ) ( g ) ( f p a [ 2 3 0 )} ( g b ) ( f a a { s 8 )} ( g b ) ( f a a { m 2 4 1 1 0 2 2 1 1 0 2 = ξ + ξ + − ξ + ξ + + (A.52) 41 Upon expanding and rearranging, we obtain an equation which contains a constant term and powers of the functions ) ( f ξ and ) ( g ξ . We further substitute ) ( g 2 ξ by (A.35) wherever powers of the function ) ( g ξ enter. We thus obtain finally an equation which will contain powers of g no greater than one. Equating the coefficients of this equation to zero, we obtain the following. Constant coefficient: − − + 2 1 2 2 1 2 0 4 2 0 2 ) b ( m p q 2 ) b ( ) a ( s p q 48 ) a ( m 2 0 ) a ( s 8 ) b ( s p q 8 4 0 4 4 1 4 2 2 = − − (A.53) Coefficient of f: + − + + 2 1 2 0 4 2 1 1 0 4 4 1 4 2 1 0 2 ) b ( ) a ( s p r 96 ) b ( a a s p q 96 ) b ( s p r q 32 a a m 4 0 a a q p ) b ( r q a ) a ( s 32 ) b ( m p r 4 1 0 2 1 1 3 0 4 2 1 2 = − − − + (A.54) Coefficient of g: 0 b a m 4 b ) a ( s 32 ) b ( a s p q 32 1 0 2 1 3 0 4 3 1 0 4 = + − (A.55) Coefficient of 2 f : + + δ − + + δ 2 1 2 2 2 1 2 1 4 2 1 2 0 4 2 ) b ( m q p ) r ( 2 ) b ( ) a ( s p q 48 ) b ( ) a ( s q p ) r ( 48 + δ − − − + 4 1 4 2 4 1 4 2 2 2 1 1 0 4 2 1 2 ) b ( s p 16 ) b ( s p r 48 ) b ( a a s p r 192 ) b ( 2 r 5 0 a a r p 3 ) a ( 2 q p ) a ( ) a ( s 48 ) b ( 2 ) a ( m 2 1 0 2 1 2 1 2 0 4 2 1 2 1 2 = + + − δ + + (A.56) Coefficient of 3 f + + δ − − δ + 1 0 2 2 1 2 1 4 4 1 4 2 2 a a q ) r ( p 2 ) b ( ) a ( s p r 96 ) b ( s q p ) r ( r 32 42 + − + δ − + 3 1 0 4 2 1 2 2 1 1 0 4 ) a ( a s 32 ) b ( q ) r ( r 2 ) b ( a a s q p d 96 0 ) b ( a a s q p r 96 2 1 1 0 4 2 = + (A.57) Coefficient of 4 f + δ + − δ − δ + 2 1 2 2 2 1 2 2 1 2 1 2 1 4 2 ) b ( q 2 ) a ( q 2 r p ) a ( q 2 p ) b ( ) a ( s q p ) r ( 48 − δ − − − + δ + 4 1 4 2 2 2 4 1 4 2 2 4 4 1 4 2 1 2 4 2 1 2 2 ) b ( s q p 8 ) b ( s q p r 8 ) a ( s 8 ) b ( q 2 r ) b ( q r 0 ) b ( s q p r 16 4 1 4 2 2 2 = δ − (A.58) Coefficient of g f + − − − + 3 1 0 4 1 1 1 1 2 0 4 1 0 1 1 2 ) b ( a s p r 64 b a q p b a ) a ( s 96 b a p r b a m 4 0 ) b ( a s p q 32 3 1 1 4 = + (A.59) Coefficient of g f 2 + δ + − + δ 1 0 2 3 1 0 4 2 b a q ) r ( p 2 ) b ( a s q p ) r ( 32 0 b ) a ( a s 96 ) b ( a s p r 64 b a p r 1 2 1 0 4 3 1 1 4 1 1 = − − + (A.60) Coefficient of g f 3 0 b ) a ( s 32 b a q ) r ( p ) b ( a s q p ) r ( 32 1 3 1 4 1 1 2 3 1 1 4 2 = − + δ − + δ (A.61) Solving the system of the nine simultaneous equations (A.53)-(A.61), using any of the known Computer Algebra Systems (Axiom, Macsyma, Maple, Mathematica), we obtain a considerable number of solutions, from which we select only the nontrivial ones: 43 Solution I. 2 0 s 4 m a ± = , m 2 b a 1 1 ρ = , 1 1 b b = , 1 2 b s 8 p = , 1 2 2 b s 2 m q − = where ρ is any root of the equation 2 2 r + δ = ρ . Solution II. 2 0 s 4 m a ± = , m 2 b a 1 1 ρ = , 1 1 b b = , 1 2 b s 8 p − = , 1 2 2 b s 2 m q − = where ρ is any root of the equation 2 2 r + δ = ρ . Solution III. 2 0 s 4 m a ± = , 1 1 a a = , 1 1 b b = , 2 1 2 1 2 2 1 2 ) b ( ) b ( r ) a ( m 4 − = δ , 1 2 b s 8 p = , 1 2 2 b s 2 m q − = Solution IV. 2 0 s 4 m a ± = , 1 1 a a = , 1 1 b b = , 2 1 2 1 2 2 1 2 ) b ( ) b ( r ) a ( m 4 − = δ , 1 2 b s 8 p − = , 1 2 2 b s 2 m q = Solution V. 2 0 s 4 m a ± = , 0 a 1 = , 1 1 b b = , 1 2 b s 4 p = , 1 2 2 b s 4 m q − = Solution VI. 2 0 s 4 m a ± = , 0 a 1 = , 1 1 b b = , 1 2 b s 8 p = , 1 2 2 b s 2 m q − = and r is any root of 0 r 2 = δ + . Solution VII. 2 0 s 4 m a ± = , 0 a 1 = , 1 1 b b = , 1 2 b s 8 p − = , 1 2 2 b s 2 m q = and r is any root of 0 r 2 = δ + . Solution VIII. 0 a 0 = , m s 8 p a 2 1 ρ = , 0 b 1 = , p p = , p m 4 q 2 − = and ρ is any root of 2 2 r + δ = ρ . 44 For each set of coefficients and parameters (Solution I – Solution VIII) there correspond in principle five families of solutions. However not all of them appear in the final solutions, because there is a conflict between the values of the coefficients and the values of the parameters of the various solutions. Family I. This family corresponds to the Solution I 2 0 s 4 m a ± = , m 2 b a 1 1 ρ = , 1 1 b b = , 1 2 b s 8 p = , 1 2 2 b s 2 m q − = where ρ is any root of the equation 2 2 r + δ = ρ . From (A.51) and Solution I, we obtain ) ( g b ) ( f m 2 b s 4 m ) ( A 1 1 2 ξ + ξ ρ + ± = ξ (A.62) with 1 2 b s 8 p = , 1 2 2 b s 2 m q − = and ρ satisfies the equation 2 2 r + δ = ρ . Sub-family Ia. Since 0 m 4 q p 2 > − = , we consider first the choice 2 2 µ − λ = δ . The functions ) ( f ξ and ) ( g ξ in (A.62) are given by (A.36) and (A.37) respectively: ) ( U 1 b s 2 m ) ( f 1 2 2 ξ ⋅ − = ξ and ) ( U ) ( U 2 b s 2 m ) ( g 1 2 2 ξ ξ ′ ξ ⋅ − = ξ where ) | m | 2 ( cosh ) | m | 2 ( sinh r ) ( U ξ ⋅ µ + ξ ⋅ λ + = ξ We thus obtain ) ( U ) ( U m s 4 1 s 4 m ) ( A 2 2 ξ ξ ′ ⋅ ξ − ρ ⋅ − ± = ξ where ρ is any root of the equation 2 2 2 2 r + µ − λ = ρ . 45 Sub-family Ib. If 1 p ± = and 2 r − = δ , we have 0 = ρ and then 0 a 1 = . In this case we have ) ( g b s 4 m ) ( A 1 2 ξ + ± = ξ where ) ( g ξ is given by (A.46). (Ib.a) For 1 p = , we have 2 1 s 8 1 b = and 2 m 4 q − = . Therefore 2 m 4 ) ( 12 ) ( 12 ) ( g − ξ ℘′ ξ ℘′ = ξ and thus 2 2 2 m 4 ) ( 12 ) ( s 2 3 s 4 m ) ( A − ξ ℘′ ξ ℘′ ⋅ + ± = ξ where ) (ξ ℘ is Weirstrass function satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ (Ib.b) For 1 p − = , we have 2 1 s 8 1 b − = and 2 m 4 q = . Therefore 2 m 4 ) ( 12 ) ( 12 ) ( g + ξ ℘′ ξ ℘′ = ξ and thus 2 2 2 m 4 ) ( 12 ) ( s 2 3 s 4 m ) ( A + ξ ℘′ ξ ℘′ ⋅ − ± = ξ Sub-family Ic. If 1 p ± = and 25 r 2 − = δ , ρ is satisfies the equation 25 r 24 2 2 = ρ while ) ( f ξ and ) ( g ξ are given by (A.48) and (A.49) respectively. (Ic.a) For 1 p = , we have 2 1 s 8 1 m 2 a ⋅ ρ = , 2 1 s 8 1 b = and 2 m 4 q − = . Therefore ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ + − = ξ and ) ( ] m 4 ) ( 12 [ ) ( m 4 ) ( g 2 2 ξ ℘ − ξ ℘ ξ ℘′ = ξ We thus have the following expression for ) ( A ξ : ] m 4 ) ( 12 [ ) ( ) ( s 2 m r 1 ) ( 3 m s 24 m 5 s 4 m ) ( A 2 2 2 2 2 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ + | | ¹ | \ | − ξ ℘ ρ + ± = ξ where ) (ξ ℘ is Weirstrass function satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ 46 (Ic.b) For 1 p − = , we have 2 1 s 8 1 m 2 a ⋅ ρ − = , 2 1 s 8 1 b − = and 2 m 4 q = . Therefore ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ − = ξ and ) ( ] m 4 ) ( 12 [ ) ( m 4 ) ( g 2 2 ξ ℘ − ξ ℘ ξ ℘′ − = ξ We thus have the following expression for ) ( A ξ : ] m 4 ) ( 12 [ ) ( ) ( s 2 m r 1 ) ( 3 m s 24 m 5 s 4 m ) ( A 2 2 2 2 2 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ + | | ¹ | \ | − ξ ℘ ρ + ± = ξ Note. If 2 2 µ − λ − = δ and 0 q p > leads to 0 m 2 < . The case 0 q = cannot be considered either, since it leads to 0 m= and then the coefficient of ) ( f ξ in (A.62) becomes infinite. Family II. This family corresponds to the Solution II 2 0 s 4 m a ± = , m 2 b a 1 1 ρ = , 1 1 b b = , 1 2 b s 8 p − = , 1 2 2 b s 2 m q − = where ρ is any root of the equation 2 2 r + δ = ρ . From (A.51) and Solution II, we obtain ) ( g b ) ( f m 2 b s 4 m ) ( A 1 1 2 ξ + ξ ρ + ± = ξ (A.63) with 1 2 b s 8 p − = , 1 2 2 b s 2 m q − = and ρ satisfies the equation 2 2 r + δ = ρ . Sub-family IIa. Since 0 m 4 q p 2 > = , we first consider the case 2 2 µ − λ − = δ . The functions ) ( f ξ and ) ( g ξ are given by (A.39) and (A.40) respectively: ) ( V 1 b s 2 m ) ( f 1 2 2 ξ ⋅ − = ξ and ) ( V ) ( Z b s 4 | m | 2 ) ( g 1 2 ξ ξ ⋅ − = ξ where ) | m | 2 ( sin ) | m | 2 ( cos ) ( Z ξ ⋅ µ + ξ ⋅ λ = ξ ) | m | 2 ( cos ) | m | 2 ( sin r ) ( V ξ ⋅ µ + ξ ⋅ λ + = ξ 47 We thus obtain ) ( V ) ( Z | m | m s 4 1 s 4 m ) ( A 2 2 ξ ξ + ρ ⋅ − ± = ξ where ρ is any root of the equation 2 2 2 2 r µ − λ − = ρ . Sub-family IIb. If 1 p ± = and 2 r − = δ then 0 = ρ and thus 0 a 1 = . In this case we have ) ( g b s 4 m ) ( A 1 2 ξ + ± = ξ where ) ( g ξ is given by (A.46). (IIb.a) For 1 p = , we have 2 1 s 8 1 b − = and 2 m 4 q = . Therefore 2 m 4 ) ( 12 ) ( 12 ) ( g + ξ ℘′ ξ ℘′ = ξ and thus 2 2 2 m 4 ) ( 12 ) ( s 2 3 s 4 m ) ( A + ξ ℘′ ξ ℘′ ⋅ − ± = ξ where ) (ξ ℘ is Weirstrass function satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 − ξ ℘ − ξ ℘ = ξ ℘′ (IIb.b) For 1 p − = , we have 2 1 s 8 1 b = and 2 m 4 q − = . Therefore 2 m 4 ) ( 12 ) ( 12 ) ( g − ξ ℘′ ξ ℘′ = ξ and thus 2 2 2 m 4 ) ( 12 ) ( s 2 3 s 4 m ) ( A − ξ ℘′ ξ ℘′ ⋅ + ± = ξ Sub-family IIc. If 1 p ± = and 25 r 2 − = δ then ρ satisfies the equation 25 r 24 2 2 = ρ . The functions ) ( f ξ and ) ( g ξ are given by (A.48) and (A.49) respectively. (IIc.a) For 1 p = , we have 2 1 s 8 1 b − = , 2 1 s 8 1 m 2 a ⋅ ρ − = and 2 m 4 q = . We then have the following expressions for ) ( f ξ and ) ( g ξ ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ + = ξ and ) ( )] ( 12 m 4 [ ) ( m 4 ) ( g 2 2 ξ ℘ ξ ℘ + ξ ℘′ − = ξ Therefore 48 )] ( 12 m 4 [ ) ( ) ( s 2 m ) ( 3 m r 1 s 24 m 5 s 4 m ) ( A 2 2 2 2 2 2 ξ ℘ + ξ ℘ ξ ℘′ ⋅ + ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ ξ ℘ + ρ − ± = ξ where ) (ξ ℘ is Weirstrass function satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 − ξ ℘ − ξ ℘ = ξ ℘′ (IIc.b) For 1 p − = , we have 2 1 s 8 1 b = , 2 1 s 8 1 m 2 a ⋅ ρ = and 2 m 4 q − = . We then have the following expressions for ) ( f ξ and ) ( g ξ ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ − − = ξ and ) ( )] ( 12 m 4 [ ) ( m 4 ) ( g 2 2 ξ ℘ ξ ℘ + ξ ℘′ = ξ Therefore )] ( 12 m 4 [ ) ( ) ( s 2 m ) ( 3 m r 1 s 24 m 5 s 4 m ) ( A 2 2 2 2 2 2 ξ ℘ + ξ ℘ ξ ℘′ ⋅ + ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ ξ ℘ + ρ − ± = ξ Note. If 2 2 µ − λ = δ and 0 q p < leads to 0 m 2 < . The case 0 q = cannot be considered either, since it leads to 0 m= and then the coefficient of ) ( f ξ in (A.63) becomes infinite. Family III. This family corresponds to the Solution III 2 0 s 4 m a ± = , 1 1 a a = , 1 1 b b = , 2 1 2 1 2 2 1 2 ) b ( ) b ( r ) a ( m 4 − = δ , 1 2 b s 8 p = , 1 2 2 b s 2 m q − = From (A.51) and Solution III, we obtain ) ( g b ) ( f a s 4 m ) ( A 1 1 2 ξ + ξ + ± = ξ (A.64) where 2 1 2 1 2 2 1 2 ) b ( ) b ( r ) a ( m 4 − = δ , 1 2 b s 8 p = , 1 2 2 b s 2 m q − = 49 Sub-family IIIa. Since 0 m 4 q p 2 < − = for 2 2 µ − λ = δ , i.e. 2 1 2 1 2 2 1 2 2 2 ) b ( ) b ( r ) a ( m 4 − = µ − λ the functions ) ( f ξ and ) ( g ξ are given by (A.36) and (A.37) respectively: ) ( U 1 s b 2 m ) ( f 2 1 2 ξ ⋅ − = ξ , ) ( U ) ( U s 4 1 ) ( g 2 ξ ξ ′ ⋅ ξ ⋅ = ξ where ) | m | 2 ( cosh ) | m | 2 ( sinh r ) ( U ξ ⋅ µ + ξ ⋅ λ + = ξ Therefore ) ( U ) ( U s 4 1 ) ( U 1 s b 4 a m 2 s 4 m ) ( A 2 2 1 1 2 2 ξ ξ ′ ⋅ ξ ⋅ + ξ ⋅ − ± = ξ where the following relation holds 2 1 2 2 2 2 1 2 b ) r ( a m 4 + µ − λ = . Sub-family IIIb. If 1 p ± = and 2 r − = δ , i.e. 2 1 2 1 2 2 1 2 2 ) b ( ) b ( r ) a ( m 4 r − = − then 0 a 1 = and ) ( g ξ is given by (A.46). (IIIb.a) If 1 p = , then 2 1 s 8 1 b = , 2 m 4 q − = and 2 m 4 ) ( 12 ) ( 12 ) ( g − ξ ℘′ ξ ℘′ = ξ . Therefore 2 2 2 m ) ( 3 ) ( s 8 3 s 4 m ) ( A − ξ ℘′ ξ ℘′ ⋅ + ± = ξ where ) (ξ ℘ is Weirstrass function satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ (IIIb.b) If 1 p − = , then 2 1 s 8 1 b − = , 2 m 4 q = and 2 m 4 ) ( 12 ) ( 12 ) ( g + ξ ℘′ ξ ℘′ = ξ . Therefore 50 2 2 2 m ) ( 3 ) ( s 8 3 s 4 m ) ( A − ξ ℘′ ξ ℘′ ⋅ − ± = ξ Sub-family IIIc. If 1 p ± = and 25 r 2 − = δ , i.e. 2 1 2 1 2 2 1 2 2 ) b ( ) b ( r ) a ( m 4 25 r − = − then ) ( f ξ and ) ( g ξ are given by (A.48) and (A.49) respectively. (IIIc.a) For 1 p = , we have 2 1 s 8 1 b = and 2 m 4 q − = and thus ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ + − = ξ and ) ( ] m 4 ) ( 12 [ ) ( m 4 ) ( g 2 2 ξ ℘ − ξ ℘ ξ ℘ = ξ Therefore ] m ) ( 3 [ ) ( ) ( s 8 m ) ( 3 m r 1 3 m a 10 s 4 m ) ( A 2 2 2 2 2 1 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ + | | ¹ | \ | ξ ℘ + − + ± = ξ where ) (ξ ℘ is Weirstrass function satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ with 2 1 2 2 1 2 b r 24 a m 100 = . (IIIc.b) For 1 p − = , we have 2 1 s 8 1 b − = and 2 m 4 q = and thus ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ − = ξ and ) ( ] m 4 ) ( 12 [ ) ( m 4 ) ( g 2 2 ξ ℘ − ξ ℘ ξ ℘′ − = ξ Therefore ] m ) ( 3 [ ) ( ) ( s 8 m ) ( 3 m r 1 3 m a 10 s 4 m ) ( A 2 2 2 2 2 1 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ + | | ¹ | \ | ξ ℘ − + ± = ξ Note. If 2 2 µ − λ − = δ and 0 q p > leads to 0 m 2 < . The choice 0 q = leads to 0 m= . Family IV. This family corresponds to the Solution IV 51 2 0 s 4 m a ± = , 1 1 a a = , 1 1 b b = , 2 1 2 1 2 2 1 2 ) b ( ) b ( r ) a ( m 4 − = δ , 1 2 b s 8 p − = , 1 2 2 b s 2 m q = From (A.51) and Solution IV, we obtain ) ( g b ) ( f a s 4 m ) ( A 1 1 2 ξ + ξ + ± = ξ (A.65) where 2 1 2 1 2 2 1 2 b b r a m 4 − = δ , 1 2 b s 8 p − = , 1 2 2 b s 2 m q = Sub-family IVa. Since 0 m 4 q p 2 < − = for 2 2 µ − λ = δ , i.e. 2 1 2 1 2 2 1 2 2 2 b b r a m 4 − = µ − λ , then ) ( f ξ and ) ( g ξ are given by (A.36) and (A.37) respectively: ) ( U 1 b s 2 m ) ( f 1 2 2 ξ ⋅ = ξ and ) ( U ) ( U b s 4 1 ) ( g 1 2 ξ ξ ′ ξ ⋅ − = ξ where ) | m | 2 ( cosh ) | m | 2 ( sinh r ) ( U ξ ⋅ µ + ξ ⋅ λ + = ξ We thus have ) ( U ) ( U s 4 1 ) ( U 1 b s 2 a m s 4 m ) ( A 2 1 2 1 2 2 ξ ξ ′ ⋅ ξ ⋅ − ξ ⋅ + ± = ξ with 2 1 2 2 2 2 1 2 b ) r ( a m 4 + µ − λ = . Sub-family IVb. If 1 p ± = and 2 r − = δ , i.e. 2 1 2 1 2 2 1 2 2 b b r a m 4 r − = − , we have 0 a 1 = . (IVb.a) For 1 p = , we have 2 1 s 8 1 b − = , 2 m 4 q − = and thus from (A.46) we get 2 m 4 ) ( 12 ) ( 12 ) ( g − ξ ℘′ ξ ℘′ = ξ and then 2 2 2 m ) ( 3 ) ( s 8 3 s 4 m ) ( A − ξ ℘′ ξ ℘′ ⋅ − ± = ξ 52 where ) (ξ ℘ is the Weirstrass function, satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ (IVb.b) For 1 p − = , we have 2 1 s 8 1 b = , 2 m 4 q = and thus from (A.46) we get 2 m 4 ) ( 12 ) ( 12 ) ( g + ξ ℘′ ξ ℘′ = ξ and then 2 2 2 m ) ( 3 ) ( s 8 3 s 4 m ) ( A + ξ ℘′ ξ ℘′ ⋅ + ± = ξ Sub-family IVc. If 1 p ± = and 25 r 2 − = δ , i.e. 2 1 2 1 2 2 1 2 2 b b r a m 4 25 r − = − then ) ( f ξ and ) ( g ξ are given by (A.48) and (A.49) respectively. (IVc.a) For 1 p = , we have 2 1 s 8 1 b − = , 2 m 4 q − = and then ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ + − = ξ and ) ( ] m 4 ) ( 12 [ m 4 ) ( g 2 2 ξ ℘ − ξ ℘ = ξ Therefore ] m ) ( 3 [ ) ( ) ( s 8 m ) ( 3 m r 1 3 m a 10 s 4 m ) ( A 2 2 2 2 2 1 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ − | | ¹ | \ | ξ ℘ + − + ± = ξ where ) (ξ ℘ is the Weirstrass function, satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ with 4 2 2 2 1 s m 800 r 3 a = . This last equation comes from equating the two different expressions of δ and using the value of 1 b . (IVc.b) For 1 p − = , we have 2 1 s 8 1 b = , 2 m 4 q = and then ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ − = ξ and ) ( ] m 4 ) ( 12 [ m 4 ) ( g 2 2 ξ ℘ − ξ ℘ − = ξ 53 Therefore ] m ) ( 3 [ ) ( ) ( s 8 m ) ( 3 m r 1 3 m a 10 s 4 m ) ( A 2 2 2 2 2 1 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ − | | ¹ | \ | ξ ℘ − + ± = ξ Note. The case 2 2 µ − λ − = δ and 0 q p > leads to 0 m 2 < . The choice 0 q = leads to 0 m= . Family V. This family corresponds to the Solution V 2 0 s 4 m a ± = , 0 a 1 = , 1 1 b b = , 1 2 b s 4 p = , 1 2 2 b s 4 m q − = From (A.51) and Solution V, we obtain ) ( g b s 4 m ) ( A 1 2 ξ + ± = ξ (A.66) where 1 2 b s 4 p = , 1 2 2 b s 4 m q − = Sub-family Va. Since 0 m 4 q p 2 < − = , the choice 2 2 µ − λ = δ , the function ) ( g ξ is given by (A.37): ) ( X ) ( X 2 b s 4 m ) ( g 1 2 ξ ξ ′ ⋅ ξ ⋅ = ξ where ) | m (| cosh ) | m (| sinh r ) ( X ξ ⋅ µ + ξ ⋅ λ + = ξ Therefore ) ( X ) ( X s 2 m s 4 m ) ( A 2 2 ξ ξ ′ ξ ⋅ + ± = ξ Sub-family Vb. If 1 p ± = and 2 r − = δ , then ) ( g ξ is given by (A.46). (Vb.a) For 1 p = , we have 2 1 s 4 1 b = , 2 m q − = and then 2 m ) ( 12 ) ( 12 ) ( g − ξ ℘′ ξ ℘′ = ξ . Therefore 2 2 2 m ) ( 12 ) ( s 3 s 4 m ) ( A − ξ ℘′ ξ ℘′ ⋅ + ± = ξ 54 where ) (ξ ℘ is Weirstrass function satisfying the equation 216 m ) ( 12 m ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ (Vb.a) For 1 p − = , we have 2 1 s 4 1 b − = , 2 m q = and then 2 m ) ( 12 ) ( 12 ) ( g + ξ ℘′ ξ ℘′ = ξ . Therefore 2 2 2 m ) ( 12 ) ( s 3 s 4 m ) ( A + ξ ℘′ ξ ℘′ ⋅ − ± = ξ Sub-family Vc. If 1 p ± = and 25 r 2 − = δ then ) ( g ξ is given by (A.49). (Vc.a) For 1 p = , we get 2 1 s 4 1 b = , 2 m q − = and then ] m ) ( 12 [ ) ( ) ( m ) ( g 2 2 − ξ ℘ ξ ℘ ξ ℘′ = ξ . Therefore ] m ) ( 12 [ ) ( ) ( s 4 m s 4 m ) ( A 2 2 2 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ + ± = ξ (Vc.b) For 1 p − = , we get 2 1 s 4 1 b − = , 2 m q = and then ] m ) ( 12 [ ) ( ) ( m ) ( g 2 2 − ξ ℘ ξ ℘ ξ ℘′ = ξ . Therefore ] m ) ( 12 [ ) ( ) ( s 4 m s 4 m ) ( A 2 2 2 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ + ± = ξ Note. The case 2 2 µ − λ − = δ and 0 q p > leads to 0 m 2 < . The case 0 q = leads to 0 m= . Family VI. This family corresponds to the Solution VI 2 0 s 4 m a ± = , 0 a 1 = , 1 1 b b = , 1 2 b s 8 p = , 1 2 2 b s 2 m q − = and r is any root of 0 r 2 = δ + . From (A.51) and Solution VI, we obtain ) ( g b s 4 m ) ( A 1 2 ξ + ± = ξ (A.67) 55 where 1 2 b s 8 p = , 1 2 2 b s 4 m q − = and r is any root of 0 r 2 = δ + . Sub-family VIa. Since 0 m 4 q p 2 < − = , the choice 2 2 µ − λ = δ , i.e. 0 r 2 2 2 = µ − λ + and ) ( g ξ is given by (A.37): ) ( U ) ( U b s 4 1 ) ( g 1 2 ξ ξ ′ ξ ⋅ = ξ , where ) | m | 2 ( cosh ) | m | 2 ( sinh r ) ( U ξ ⋅ µ + ξ ⋅ λ + = ξ Therefore ) ( U ) ( U s 4 m s 4 m ) ( A 2 2 ξ ξ ′ ξ ⋅ + ± = ξ where r satisfies the equation 0 r 2 2 2 = µ − λ + . Sub-family VIb. If 1 p ± = and 2 r − = δ (i.e. r is any real) then ) ( g ξ is given by (A.46). (VIb.a) For 1 p = , we get 2 1 s 8 1 b = , 2 m 4 q − = and thus 2 m 4 ) ( 12 ) ( 12 ) ( g − ξ ℘′ ξ ℘′ = ξ . Therefore 2 2 2 m ) ( 3 ) ( s 8 3 s 4 m ) ( A − ξ ℘′ ξ ℘′ ⋅ + ± = ξ , where ) (ξ ℘ is the Weirstrass function, satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ (VIb.b) For 1 p − = , we get 2 1 s 8 1 b − = , 2 m 4 q = and thus 2 m 4 ) ( 12 ) ( 12 ) ( g + ξ ℘′ ξ ℘′ = ξ . Therefore 2 2 2 m ) ( 3 ) ( s 8 3 s 4 m ) ( A + ξ ℘′ ξ ℘′ ⋅ − ± = ξ . Sub-family VIc. If 1 p ± = and 25 r 2 − = δ , i.e. 0 25 r r 2 2 = − then ) ( g ξ is given by (A.49). 56 (VIc.a) For 1 p = , we get 2 1 s 8 1 b = , 2 m 4 q − = and then ] m 4 ) ( 12 [ ) ( ) ( m 4 ) ( g 2 2 − ξ ℘ ξ ℘ ξ ℘′ = ξ .Therefore ] m ) ( 3 [ ) ( ) ( s 8 m s 4 m ) ( A 2 2 2 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ + ± = ξ (VIc.b) For 1 p − = , we get 2 1 s 8 1 b − = , 2 m 4 q = and then ] m 4 ) ( 12 [ ) ( ) ( m 4 ) ( g 2 2 − ξ ℘ ξ ℘ ξ ℘′ − = ξ . Therefore ] m ) ( 3 [ ) ( ) ( s 8 m s 4 m ) ( A 2 2 2 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ + ± = ξ . Note. The case 2 2 µ − λ − = δ and 0 q p > leads to 0 m 2 < . The case 0 q = leads to 0 m= . Family VII. This family corresponds to the Solution VII 2 0 s 4 m a ± = , 0 a 1 = , 1 1 b b = , 1 2 b s 8 p − = , 1 2 2 b s 2 m q = and r is any root of 0 r 2 = δ + . From (A.51) and Solution VII, we obtain ) ( g b s 4 m ) ( A 1 2 ξ + ± = ξ (A.68) where 1 2 b s 8 p − = , 1 2 2 b s 4 m q = and r is any root of 0 r 2 = δ + . Sub-family VIIa. Since 0 m 4 q p 2 < − = , the choice 2 2 µ − λ = δ , i.e. 0 r 2 2 2 = µ − λ + , then ) ( g ξ is given by (A.37): ) ( U ) ( U b s 4 1 ) ( g 1 2 ξ ξ ′ ξ ⋅ − = ξ , where ) | m | 2 ( cosh ) | m | 2 ( sinh r ) ( U ξ ⋅ µ + ξ ⋅ λ + = ξ . 57 Therefore ) ( U ) ( U s 4 1 s 4 m ) ( A 2 2 ξ ξ ′ ξ ⋅ − ± = ξ where r satisfies the equation 0 r 2 2 2 = µ − λ + . Sub-family VIIb. If 1 p ± = and 2 r − = δ (i.e. r any real) then ) ( g ξ is given by (A.46). (VIIb.a) For 1 p = , we have 2 1 s 8 1 b − = , 2 m 4 q − = and thus 2 m 4 ) ( 12 ) ( 12 ) ( g − ξ ℘′ ξ ℘′ = ξ . Therefore 2 2 2 m ) ( 3 ) ( s 8 3 s 4 m ) ( A − ξ ℘′ ξ ℘′ ⋅ − ± = ξ where ) (ξ ℘ is Weirstrass function satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ (VIIb.b) For 1 p − = , we have 2 1 s 8 1 b = , 2 m 4 q = and thus 2 m 4 ) ( 12 ) ( 12 ) ( g + ξ ℘′ ξ ℘′ = ξ . Therefore 2 2 2 m ) ( 3 ) ( s 8 3 s 4 m ) ( A + ξ ℘′ ξ ℘′ ⋅ + ± = ξ Sub-family VIIc. If 1 p ± = and 25 r 2 − = δ , i.e. 0 25 r r 2 2 = − then ) ( g ξ is given by (A.49). (VIIc.a) For 1 p = , we get 2 1 s 8 1 b − = , 2 m 4 q − = and then ] m 4 ) ( 12 [ ) ( ) ( m 4 ) ( g 2 2 − ξ ℘ ξ ℘ ξ ℘′ = ξ .Therefore ] m ) ( 3 [ ) ( ) ( s 8 m s 4 m ) ( A 2 2 2 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ − ± = ξ (VIIc.b) For 1 p − = , we get 2 1 s 8 1 b = , 2 m 4 q = and then ] m 4 ) ( 12 [ ) ( ) ( m 4 ) ( g 2 2 − ξ ℘ ξ ℘ ξ ℘′ − = ξ . Therefore 58 ] m ) ( 3 [ ) ( ) ( s 8 m s 4 m ) ( A 2 2 2 2 − ξ ℘ ξ ℘ ξ ℘′ ⋅ − ± = ξ . Note. The case 2 2 µ − λ − = δ and 0 q p > leads to 0 m 2 < . The case 0 q = leads to 0 m= . Family VIII. This family corresponds to the Solution VIII 0 a 0 = , m s 8 p a 2 1 ρ = , 0 b 1 = , p p = , p m 4 q 2 − = and ρ is any root of 2 2 r + δ = ρ . From (A.51) and Solution VIII, we obtain ) ( f m s 8 p ) ( A 2 ξ ρ = ξ (A.69) where p p = , p m 4 q 2 − = and ρ is any root of the equation 2 2 r + δ = ρ . Sub-family VIIIa. Since 0 m 0 q p 2 > ⇔ < , the choice 2 2 µ − λ = δ , i.e. 2 2 2 2 r + µ − λ = ρ , then ) ( f ξ is given by (A.36): ) ( U 1 p m 4 ) ( f 2 ξ ⋅ − = ξ where ) | m | 2 ( cosh ) | m | 2 ( sinh r ) ( U ξ ⋅ µ + ξ ⋅ λ + = ξ .Therefore ) ( U 1 s 2 m ) ( A 2 ξ ⋅ ρ − = ξ where ρ satisfies the equation 2 2 2 2 r + µ − λ = ρ . Sub-family VIIIb. If 1 p ± = and 25 r 2 − = δ , leads to 2 2 r 25 24 = ρ and then ) ( f ξ is given by (A.48). (VIIIb.a) For 1 p = , we have 2 m 4 q − = and thus ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ + − = ξ . where ) (ξ ℘ is Weirstrass function satisfying the equation 27 m 8 ) ( 3 m 4 ) ( 4 )) ( ( 6 4 3 2 + ξ ℘ − ξ ℘ = ξ ℘′ 59 Therefore | | ¹ | \ | ξ ℘ + − ρ = ξ ) ( 3 m r 1 s 12 m 5 ) ( A 2 2 , and ρ is any root of the equation 25 r 24 2 2 = ρ . (VIIIb.b) For 1 p − = , we have 2 m 4 q = and thus ) ( 9 m 10 r 3 m 10 ) ( f 4 2 ξ ℘ − = ξ . Therefore | | ¹ | \ | ξ ℘ − ρ − = ξ ) ( 3 m r 1 s 12 m 5 ) ( A 2 2 . Note. The case 2 2 µ − λ − = δ and 0 q p > leads to 0 m 2 < . The case 0 q = leads to 0 m= . The case 2 r − = δ cannot be considered since in this case 0 = ρ and then 0 a 1 = (i.e. 0 ) ( A = ξ ). A.V) Fifth Method. We consider now an expression of the form ξ − ξ ξ − ξ + + + + = ξ m 1 0 m 1 m 1 0 m 1 e d c e c e b a e a ) ( A (A.70) and substitute back into (A.1). We then derive an equation which contains powers of the exponentials ξ ±m e . Equating all the coefficients of these exponentials to zero, we obtain the following set of solutions 1 1 a a = , 0 a 0 = , 0 b 1 = , 1 1 c c = , 0 c 0 = , 1 1 d d = 1 1 a a = , 0 0 a a = , 1 1 b b = , m s a 2 c 1 1 − = , m s a 2 c 0 0 − = , m s b 2 d 1 1 − = 0 a 1 = , 0 a 0 = , 1 1 b b = , 1 1 c c = , 0 c 0 = , m s b 2 d 1 1 − = 0 a 1 = , 0 a 0 = , 1 1 b b = , 1 1 c c = , 0 c 0 = , m s b 2 d 1 1 = s 2 c m a 1 1 = , 0 0 a a = , 1 1 b b = , 1 1 c c = , m s a 2 c 0 0 = , m s b 2 d 1 1 = We thus have 60 ξ − ξ ξ + = ξ m 1 m 1 m 1 e d e c e a ) ( A , ξ − ξ ξ − − = ξ m 1 m 1 m 1 e m b s 2 e c e b ) ( A ξ − ξ ξ − + = ξ m 1 m 1 m 1 e m b s 2 e c e b ) ( A , ξ − ξ ξ − ξ + + + + = ξ m 1 0 m 1 m 1 0 m 1 e m b s 2 m a s 2 e c e b a e s 2 c m ) ( A or ξ − ξ ξ + = ξ m 2 m 1 m e C e C e ) ( A , 1 1 1 a c C = , 1 1 2 a d C = (A.71) ξ − ξ ξ − − = ξ m m 1 m e s 2 e C m e m ) ( A , 1 1 1 b c C = (A.72) ξ − ξ ξ − + = ξ m m 1 m e s 2 e C m e m ) ( A , 1 1 1 b c C = (A.73) 2 m 2 2 m 1 m 2 m 1 2 s 4 e C s 4 e C s m 2 s m 2 e C s m 2 e C m ) ( A + + + + = ξ ξ − ξ ξ − ξ , 0 1 1 a c C = , 0 1 2 a b C = (A.74) Appendix B. We consider an expansion of the form ) ( a a ) ( A 1 0 ξ ϕ + = ξ (B.1) where ) (ξ ϕ satisfies Jacobi's differential equation 4 4 3 3 2 2 1 0 n n n n n ) ( d d ϕ + ϕ + ϕ + ϕ + = ξ ϕ ξ (B.2) Contrary to the previously considered case (Appendix A, equation (A.3)) we consider the right hand side of (B.2) to be a complete fourth degree polynomial. Since 4 4 3 3 2 2 1 0 1 1 n n n n n a ) ( a ) ( A ϕ + ϕ + ϕ + ϕ + = ξ ϕ′ = ξ ′ and ) n 4 n 3 n 2 n ( a 2 1 ) ( A 3 4 2 3 2 1 1 ϕ + ϕ + ϕ + = ξ ′ ′ 61 equation (A.1) becomes + ϕ + − + ϕ | ¹ | \ | − 3 4 2 1 2 1 0 4 2 1 2 4 2 1 ) n a s 16 ( a a 2 a s 8 n 2 1 a + ϕ | ¹ | \ | + − − + 2 3 1 0 2 1 2 0 2 2 2 1 2 1 2 n a a 2 3 a a s 48 n a 2 1 a m 2 + ϕ − + − + ) a a s 32 n a a n a a a m 4 ( 1 3 0 2 2 1 0 1 2 1 1 0 2 0 n a a 2 1 n a 2 3 a s 8 a m 2 1 1 0 0 2 1 4 0 2 2 0 2 = + − − + (B.3) Equating to zero all the coefficients, we find 0 a s 8 n 2 1 a 2 1 2 4 2 1 = | ¹ | \ | − 0 ) n a s 16 ( a a 2 4 2 1 2 1 0 = + − 0 n a a 2 3 a a s 48 n a 2 1 a m 2 3 1 0 2 1 2 0 2 2 2 1 2 1 2 = + − − 0 a a s 32 n a a n a a a m 4 1 3 0 2 2 1 0 1 2 1 1 0 2 = − + − 0 n a a 2 1 n a 2 3 a s 8 a m 2 1 1 0 0 2 1 4 0 2 2 0 2 = + − − From the above system we consider only the following set of nontrivial solutions, assuming that 0 a 1 ≠ : Solution 1. 0 0 a a = , 1 1 a a = , 3 1 0 3 1 2 0 2 1 2 2 0 0 a ) a n a a s 48 a m 4 ( a n + − = , 2 1 0 3 1 2 0 2 1 2 0 1 a ) a n 3 a a s 128 a m 8 ( a n + − = , 1 0 3 1 2 0 2 1 2 2 a a n 3 a a s 96 a m 4 n + − = 3 3 n n = , 2 1 2 4 a s 16 n = . Solution 2. 0 0 a a = , 2 2 2 0 2 0 3 1 n m 4 a s 96 a n 3 a + − = , 62 2 3 2 2 2 2 0 2 2 0 2 2 2 0 n 27 ) n m 4 a s 96 ( ) a s 48 n m 8 ( n + − − + = , 3 2 2 2 0 2 2 0 2 2 2 1 n 3 ) n m 4 a s 96 ( ) a s 32 n m 4 ( n + − − + = , 2 2 n n = , 3 3 n n = , 2 2 2 2 0 2 2 3 2 0 2 4 ) n m 4 a s 96 ( n a s 144 n + − = Solution 3. 0 0 a a = , 1 2 0 2 2 2 0 1 n ) a s 32 n m 4 ( a a − + = , 2 2 2 2 0 2 2 1 2 0 2 2 2 0 ) n m 4 a s 32 ( 3 n ) a s 48 n m 8 ( n − − − + = 1 1 n n = , 2 2 n n = , 1 2 0 2 2 2 2 0 2 2 2 3 n 3 ) a s 96 n m 4 ( ) a s 32 n m 4 ( n + + − − + = , 2 1 2 2 2 2 0 2 2 0 2 4 n ) n m 4 a s 32 ( a s 16 n − − = Solution 4. 0 0 a a = , s 4 a 1 ρ = , 2 4 4 2 0 2 4 2 3 0 2 0 2 0 n ) n a s 12 n m n a s ( a s 64 n − + ρ = 4 2 0 2 2 3 0 0 1 n ) a s 32 m 2 n a s 3 ( a s 16 n ρ − ρ + = 4 4 2 0 2 4 2 3 0 2 n ) n a s 24 n m n a s 3 ( 4 n − + ρ = where ρ is any root of the equation 0 n 4 2 = − ρ . Solution 5. 0 0 a a = , s 4 a 1 ρ = , 4 2 0 2 2 2 2 0 2 0 n 3 ) a s 48 n m 8 ( a s 16 n − + = ρ − + = ) a s 32 n m 4 ( a s 4 n 2 0 2 2 2 0 1 , 0 2 0 2 2 2 3 a s 12 ) a s 96 n m 4 ( n + + − ρ = 63 where ρ is any root of the equation 0 n 4 2 = − ρ . Solution 6. 0 0 a a = , s 4 a 1 ρ = , ) a s 64 m a s 16 n ( n a s 3 4 n 3 0 3 2 0 1 4 0 0 − + ρ ⋅ = 0 2 0 1 3 0 3 2 a s 4 m a s 16 n a s 128 n − ρ + = , 2 0 2 4 1 2 0 3 0 3 3 a s 48 n n m a s 32 a s 512 n + ρ − ρ = where ρ is any root of the equation 0 n 4 2 = − ρ . Solution 7. 0 0 a a = , s 4 a 1 ρ = , 0 4 0 2 0 2 2 4 0 4 1 a s 4 n n 3 a m s 64 a s 256 n ρ + − = 2 0 2 4 0 2 0 2 2 4 0 4 2 a s 16 n n 3 a m s 128 a s 768 n + − = , 3 0 3 4 0 2 0 2 2 4 0 4 3 a s 64 ) n n a m s 64 a s 768 ( n + − ρ = where ρ is any root of the equation 0 n 4 2 = − ρ . Solution 8. 0 0 a a = , 0 1 a a ρ = , 2 1 2 0 2 2 m 4 n a s 32 n − ρ + = , ) a n s 16 n m 4 n m 24 n a n s 384 ( n 9 1 n 2 0 1 2 1 2 0 2 2 1 2 0 0 2 0 3 − + ρ − ρ + ρ = ) a s 16 m 4 n ( n a s 3 16 n 2 0 2 2 1 0 2 0 2 4 − + ρ ⋅ = where ρ is any root of the equation 0 m 4 a s 16 n n 3 2 2 0 2 1 2 0 = − + ρ − ρ Solution 9. 0 0 a a = , 0 1 a a ρ = , 3 2 2 0 2 2 0 2 2 1 1 0 n 3 m 8 a s 128 ) a s 16 m 4 n ( n 3 1 n − ρ − ρ − + ρ ⋅ − = 2 1 2 0 2 2 m 4 n a s 32 n − ρ + = , ) n 3 m 8 a s 128 ( n a s 16 n 3 2 2 0 2 1 2 0 2 4 − ρ − ρ − = where ρ is any root of the equation 0 n 3 ) m 8 a s 128 ( n 3 2 2 0 2 2 1 = − ρ − + ρ 64 Solution 10. 0 0 a a = , 0 1 a a ρ = , 2 1 2 0 2 2 m 4 n a s 32 n − ρ + = ) a n s 16 n m 4 n m 24 n a n s 384 ( n 9 1 n 2 0 1 2 1 2 0 2 2 1 2 0 0 2 0 3 − + ρ − ρ + ρ = ) a s 16 m 4 n ( n a s 3 16 n 2 0 2 2 1 0 2 0 2 4 − + ρ ⋅ = where ρ is any root of the equation 0 m 4 a s 16 n n 3 2 2 0 2 1 2 0 = − + ρ − ρ . Solution 11. 0 0 a a = , 0 1 a a ρ = , ρ − + ρ = 2 2 0 2 0 2 1 m 4 a s 16 n 3 n 2 2 0 2 0 2 2 m 8 a s 48 n 3 n − + ρ = , 2 2 0 2 4 a s 16 n ρ = where ρ is any root of the equation 0 n ) m 4 a s 48 ( n 3 2 2 0 2 3 0 = − ρ − + ρ . B.1. Solutions of the differential equation (B.2). For the first three Solutions, we use the following Lemma: Lemma: If (B.2) is expressed as ) D q p ( ) D q p ( ) n ( r ) ( d d − − ϕ + − ϕ + ϕ = ξ ϕ ξ (B.4) then its solution is given by D q 4 e ) p n ( 4 e D q n 4 e ) p n p D q ( 4 e n 2 K r K r 2 2 K r 2 2 K r 2 + + + + − − + − = ϕ ξ − ξ − ξ − ξ − (B.5) where D q ) p n ( K 2 2 − + = (B.6) The proof of the Lemma is based on the following formula × − + − = − − ϕ + − ϕ + ϕ ϕ ∫ D q ) p n ( 1 ) D q p ( ) D q p ( ) n ( d 2 2 ( ¸ ( ¸ − − ϕ ⋅ − + + + + ϕ − − + + ϕ × } D q ) p ( D q ) p n ( ) p n ( ) n ( D q ) p n ( { n 2 ln 2 2 2 2 2 2 65 Integrating equation (B.4) using the above integral, we find (we set the integration constant equals to zero) = − − ϕ ⋅ − + + + + ϕ − − + + ϕ } D q ) p ( D q ) p n ( ) p n ( ) n ( D q ) p n ( { n 2 2 2 2 2 2 2 ) D q ) p n ( r ( exp 2 2 ξ ⋅ − + − = Solving the above equation with respect to ϕ, we obtain (B.5). The solution (B.5) can also be expressed in terms of the hyperbolic tangent function as ) K r ( tanh 1 ) p n ( 4 ) K r tanh( ) D q 4 1 ( ) D q 4 1 ( ) K r ( tanh 1 ) p n p D q ( 4 ) K r tanh( ) D q 4 1 ( n ) D q 4 1 ( n 2 2 2 2 2 2 2 2 ξ − + + ξ − − + ξ − − − + ξ − − + − = ϕ (B.7) For the Solution 1, we have × | | ¹ | \ | + ϕ = ϕ + ϕ + ϕ + ϕ + 2 1 0 2 1 2 4 4 3 3 2 2 1 0 a a a s 16 n n n n n | | ¹ | \ | − − − ϕ | | ¹ | \ | + − − ϕ × 2 1 2 2 1 2 3 2 1 0 2 1 2 2 1 2 3 2 1 0 a s 32 D a s 32 n s a a 32 a s 32 D a s 32 n s a a 32 where 2 1 2 3 2 1 0 ) a s m 16 ( ) n s a a 64 ( D − − ≡ Equation (B.2) then gives by integration either (B.5) or (B.7) where 2 1 2 3 2 1 0 a s 32 n s a a 32 p − = and 2 1 2 a s 32 1 q = For the Solution 2, we have 66 × + − = ϕ + ϕ + ϕ + ϕ + 2 2 2 2 0 2 2 3 2 0 2 4 4 3 3 2 2 1 0 ) n m 4 a s 96 ( n a s 144 n n n n n × | | ¹ | \ | + − + ϕ × 2 3 2 2 2 0 2 n 3 n m 4 a s 96 | | ¹ | \ | + − + − + − − ϕ × 3 2 0 2 2 2 2 0 2 3 2 0 2 2 2 2 2 2 0 2 n a s 288 D ) n m 4 a s 96 ( n a s 288 ) n m 4 ( ) n m 4 a s 96 ( | | ¹ | \ | + − − − + − − ϕ × 3 2 0 2 2 2 2 0 2 3 2 0 2 2 2 2 2 2 0 2 n a s 288 D ) n m 4 a s 96 ( n a s 288 ) n m 4 ( ) n m 4 a s 96 ( where 2 0 2 2 2 2 0 2 ) a s m 48 ( ) n m 4 a s 96 ( D − − + ≡ Equation (B.2) then gives by integration either (B.5) or (B.7) where 3 2 0 2 2 2 2 2 2 0 2 n a s 288 ) n m 4 ( ) n m 4 a s 96 ( p − + − = , 3 2 0 2 2 2 2 0 2 n a s 288 n m 4 a s 96 q + − = For the Solution 3, we have × − − = ϕ + ϕ + ϕ + ϕ + 2 1 2 2 2 2 0 2 2 0 2 4 4 3 3 2 2 1 0 n ) n m 4 a s 32 ( a s 16 n n n n n × | | ¹ | \ | − + + ϕ × 2 2 0 2 2 2 1 a s 32 n m 4 n × | | ¹ | \ | − − + − − − − ϕ × ) n m 4 a s 32 ( a s 96 D n ) n m 4 a s 32 ( a s 96 ) m 4 n ( n 2 2 2 0 2 2 0 2 1 2 2 2 0 2 2 0 2 2 2 1 × | | ¹ | \ | − − − − − − − ϕ × ) n m 4 a s 32 ( a s 96 D n ) n m 4 a s 32 ( a s 96 ) m 4 n ( n 2 2 2 0 2 2 0 2 1 2 2 2 0 2 2 0 2 2 2 1 where 2 0 2 2 2 2 0 2 ) a s m 48 ( ) n m 4 a s 96 ( D − − + ≡ Equation (B.2) then gives by integration either (B.5) or (B.7) where 67 ) n m 4 a s 32 ( a s 96 ) m 4 n ( n p 2 2 2 0 2 2 0 2 2 2 1 − − − = , ) n m 4 a s 32 ( a s 96 n q 2 2 2 0 2 2 0 2 1 − − = For the rest of the Solutions (Solution 4-Solution 11), we may use Section 11 (pp. 209-211), of Chapter 7 of H. D. Davis: "Introduction to Nonlinear Differential and Integral Equations", Dover 1962. In all these cases we shall also need the algorithm of determining the roots of a fourth degree polynomial. The reader may consult my paper "Solving Third, Fourth and Fifth Degree Polynomial Equations" available from www.docstoc.com. The solutions are then expressed in terms of Jacobi's or Weirstrass functions or even in terms of Elliptic Integrals of various kinds. Appendix C. In this Appendix we shall solve equation (A.1) using the − ′ G / G expansion method with variable coefficients. We substitute | ¹ | \ | ′ + = ξ G G a a ) ( A 1 0 (C.2) where 0 a and 1 a are − ξ dependent quantities, ) ( a a 0 0 ξ = and ) ( a a 1 1 ξ = , while ) ( G G ξ = . Since ( ( ¸ ( ¸ | ¹ | \ | ′ − ′ ′ + | ¹ | \ | ′ ′ + ′ = ξ ′ 2 1 1 0 G G G G a G G a a ) ( A and ( ( ¸ ( ¸ | ¹ | \ | ′ + ′ ′ ⋅ ′ − ′ ′ ′ + ( ( ¸ ( ¸ | ¹ | \ | ′ − ′ ′ ′ + | ¹ | \ | ′ ′ ′ + ′ ′ = ξ ′ ′ 3 2 1 2 1 1 0 G G 2 G G G 3 G G a G G G G a 2 G G a a ) ( A equation (A.1) becomes + ¦ ¹ ¦ ´ ¦ ( ( ¸ ( ¸ | ¹ | \ | ′ − ′ ′ ′ + | ¹ | \ | ′ ′ ′ + ′ ′ ) ` ¹ ¹ ´ ¦ | ¹ | \ | ′ + 2 1 1 0 1 0 G G G G a 2 G G a a G G a a 68 ¦ ) ¦ ` ¹ ( ( ¸ ( ¸ | ¹ | \ | ′ + ′ ′ ⋅ ′ − ′ ′ ′ + 3 2 1 G G 2 G G G 3 G G a ¦ ¹ ¦ ´ ¦ + | ¹ | \ | ′ ′ ′ + ( ( ¸ ( ¸ | ¹ | \ | ′ − ′ ′ + | ¹ | \ | ′ ′ + ′ − G G ) a ( ) a ( 2 G G G G ) a ( G G ) a ( ) a ( 2 3 1 0 2 2 2 1 2 2 1 2 0 ¦ ) ¦ ` ¹ ( ( ¸ ( ¸ | ¹ | \ | ′ − ′ ′ | ¹ | \ | ′ ′ + ( ( ¸ ( ¸ | ¹ | \ | ′ − ′ ′ ′ + 2 1 1 2 1 0 G G G G G G ) a ( ) a ( 2 G G G G ) a ( ) a ( 2 ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | ¹ | \ | ′ + | ¹ | \ | ′ + ⋅ + 2 2 1 1 0 2 0 2 G G a G G a a 2 a m 2 0 G G a G G a a 4 G G a a 6 G G a a 4 a s 8 4 4 1 3 3 1 0 2 2 1 2 0 1 3 0 4 0 2 = ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | ¹ | \ | ′ + | ¹ | \ | ′ + | ¹ | \ | ′ + | ¹ | \ | ′ + ⋅ − From the above equation, upon expanding, we obtain the following set of coefficients, equating each one of them to zero: Coefficient of 4 G − : 0 ) G ( a s 8 ) G ( a 2 3 ) G ( a 2 4 4 1 2 4 2 1 4 2 1 = ′ − ′ − ′ (C.3) Coefficient of 3 G − : 0 ) G ( a a s 32 ) G ( a a ) G ( a a 2 3 3 1 0 2 3 1 1 3 1 0 = ′ − ′ ′ + ′ (C.4) Coefficient of 2 G − : + ′ ′ ′ ′ + ′ ′ ′ + ′ ′ ′ − + ′ ′ − ) G )( G ( a a 2 ) G ( a a ) G )( G ( a a ) 3 ( ) G ( a a ) 2 ( 1 1 2 1 1 1 0 2 1 0 + ′ ′ ′ ′ − ′ ′ − ′ ′ − ′ ′ ′ ′ + ) G )( G ( a a 3 ) G ( a 2 3 ) G ( ) a ( 2 3 ) G ( ) G ( a 1 1 2 2 1 2 2 1 2 1 0 ) G ( a a s 48 ) G ( a m 2 ) G ( a a 3 2 2 1 2 0 2 2 2 1 2 2 1 0 = ′ − ′ + ′ ′ + (C.5) Coefficient of 1 G − : + ′ ′ ′ − ′ ′ ′ − ′ ′ ′ + ′ ′ ′ + ′ ′ ′ + ′ ′ ′ G a a 3 G a a 3 G a a G a a G a a 2 G a a 1 0 1 0 0 1 1 0 1 0 1 0 0 G a a s 32 G a a m 4 1 3 0 2 1 0 2 = ′ − ′ + (C.6) 69 Coefficient of 0 G : 0 a s 8 a m 2 ) a ( 2 3 a a 4 0 2 2 0 2 2 0 0 0 = − + ′ − ′ ′ (C.7) From equation (C.3) we obtain 2 2 1 s 16 1 a = (C.8) We thus see that 1 a is a constant. Taking (C.8) into account, equation (C.4) becomes an identity. Equations (C.5) and (C.6) are then equivalent to the equations 0 a a s 48 a m 2 a 3 G G a 2 3 G G a G G a 3 1 2 0 2 1 2 0 2 1 1 0 = − + ′ + | ¹ | \ | ′ ′ ′ − ′ ′ ′ ′ + ′ ′ ′ − (C.9) and 2 0 2 2 0 0 a s 32 m 4 G G a a 3 G G + − ′ ′ ′ ′ = ′ ′ ′ ′ (C.10) respectively. Equation (C.9), because of (C.10), becomes a quadratic equation with respect to the ratio G / G ′ ′ ′ : 0 a a s 16 a m 2 a 3 G G a 3 a a a 3 G G a 2 3 1 2 0 2 1 2 0 0 0 1 0 2 1 = + + ′ − ′ ′ ′ | | ¹ | \ | − ′ − | ¹ | \ | ′ ′ ′ (C.11) The discriminant of the above equation is 2 2 0 1 2 0 2 0 0 2 s 4 m 3 a a 36 a 75 a a s 16 9 D − ′ − + | | ¹ | \ | ′ = (C.12) and thus 1 0 0 1 0 a 3 D a 3 a a a G G ± | | ¹ | \ | − ′ = ′ ′ ′ (C.13) We substitute the above ratio into (C.10) and the divide the resulting expression by (C.13). We then obtain the ratio G / G ′ ′ ′ ′ ′ : 70 ) s , m ( F G G = ′ ′ ′ ′ ′ (C.14) where 1 1 0 0 0 1 0 0 2 0 2 2 1 0 2 0 0 a 3 D a a 3 a a D a a a a s 32 m 4 a a 9 a a 3 ) s , m ( F ± | | ¹ | \ | − ′ ′ ± + − | | ¹ | \ | ′ − | | ¹ | \ | ′ = (C.15) Integrating (C.14), we get ξ + ξ + = F 3 2 1 e C C C G (C.16) Therefore we obtain ξ ⋅ ξ ⋅ + ξ + ⋅ + = ′ ) s , m ( F ) s , m ( F e D E e ) s , m ( F D G G (C.17) where D and E are also arbitrary constants (i.e. 3 2 C / C D = and 3 1 C / C E = ). In all the previous expressions, s 4 1 a 1 ± = while 0 a satisfies differential equation (C.7). Equation (C.7) is essentially equation (A.1). In solving this equation, we can consider again an expansion of the form (C.2), 2 2 1 2 0 0 G G ) a ( ) a ( a | ¹ | \ | ′ + = where 2 0 ) a ( and 2 1 ) a ( are variable coefficients, the subscript 2 denoting the second step in the process. We would then derive equations similar in form to (C.7) and (C.17). This process can continue forever. After say n steps, we derive the equations 2 2 1 2 0 1 0 G G ) a ( ) a ( ) a ( | ¹ | \ | ′ + = 3 3 1 3 0 2 0 G G ) a ( ) a ( ) a ( | ¹ | \ | ′ + = ……………………………… 71 1 n 1 n 1 1 n 0 n 0 G G ) a ( ) a ( ) a ( + + + | ¹ | \ | ′ + = where each one of the k 0 ) a ( ( 1 n , , 2 , 1 k + = L ) satisfy (C.2) and | ¹ | \ | ′ G G are expressed by relations similar to (C.17). It is obvious that the process will terminate by considering one of the known solutions of (A.1) only. Appendix D. We shall solve equation (5.9) K G 2 K tanh K K 1 G − = ′ | ¹ | \ | | ¹ | \ | ξ + − + ′ ′ Using the substitution H G = ′ , equation (5.9) becomes K H 2 K tanh K K 1 H − = | ¹ | \ | | ¹ | \ | ξ + − + ′ (D.1) The above is a linear first order differential equation admitting the function | ¹ | \ | ξ − ξ − 2 K cosh e 2 ) K 1 ( as an integrating factor. We thus obtain | ¹ | \ | ξ = ( ¸ ( ¸ | ¹ | \ | ξ ξ − ξ ξ − ξ − 2 K cosh e K 2 K cosh e ) ( H d d 2 ) K 1 ( 2 ) K 1 ( and by integration × + − = | ¹ | \ | ξ ξ − ξ − ξ − ) K 1 ( 2 2 ) K 1 ( e ) 1 K 3 K 2 ( 2 K 2 K cosh e ) ( H 1 2 K )} K sinh( ) 1 K ( K ) K cosh( ) 1 K ( ) K 2 1 ( { + ξ − + ξ − + − × where 1 K is an arbitrary constant. We thus obtain that × | ¹ | \ | ξ + − = ξ 2 K h sec ) 1 K 3 K 2 ( 2 K ) ( H 2 2 + ξ − + ξ − + − × )} K sinh( ) 1 K ( K ) K cosh( ) 1 K ( ) K 2 1 ( { 2 72 | ¹ | \ | ξ + ξ − 2 K h sec e K 2 ) K 1 ( 1 (D.2) In order to find G, we have to integrate the above expression. We need the following integrals: | ¹ | \ | ξ = ξ | ¹ | \ | ξ ∫ 2 K tanh K 2 d 2 K h sec 2 (D.3) | ¹ | \ | ξ − ξ = ξ | ¹ | \ | ξ ξ ∫ 2 K tanh K 2 2 d 2 K h sec ) K cosh( 2 (D.4) | ¹ | \ | | ¹ | \ | ξ = ξ | ¹ | \ | ξ ξ ∫ 2 K cosh ln K 4 d 2 K h sec ) K sinh( 2 (D.5) × − = ξ | ¹ | \ | ξ ξ − ξ − ∫ ) 1 K ( 2 ) K 1 ( e ) 1 K 2 ( K 2 d 2 K h sec e ( ¸ ( ¸ | ¹ | \ | − − − − − | ¹ | \ | − − − − × ξ ξ ξ K 1 2 K 1 2 K e , K 1 K 2 , K 1 K , 1 F ) 1 K 2 ( e , K 1 K 3 , K 1 K 2 , 1 F e ) 1 K ( | ¹ | \ | ξ − ξ − 2 K tanh e K 2 ) 1 K ( (D.6) Introducing the notation + | ¹ | \ | ξ − ξ − + | ¹ | \ | ξ − = ξ 2 K tanh ) 1 K ( K ) 1 K ( 2 2 K tanh K ) K 2 1 ( 2 ) ; K ( V 2 | ¹ | \ | | ¹ | \ | ξ − + 2 K cosh ln ) 1 K ( 4 (D.7) − | ¹ | \ | − − − − = ξ ξ ξ K 1 2 K e , K 1 K 3 , K 1 K 2 , 1 F e ) 1 K ( ) ; K ( Z | ¹ | \ | ξ − − | ¹ | \ | − − − − − ξ 2 K tanh ) 1 K 2 ( e , K 1 K 2 , K 1 K , 1 F ) 1 K 2 ( K 1 2 (D.8) we find that ( 2 K is an arbitrary constant) 2 1 2 K ) ; K ( Z ) 1 K 2 ( K 2 K ) ; K ( V ) 1 K 3 K 2 ( 2 K ) ( G + ξ − + ξ + − = ξ (D.9) and then 73 2 2 ) K 1 ( 1 2 2 ) K 1 ( 1 2 2 K ) 1 K 3 K 2 ( K 2 ) ; K ( Z e K ) 1 K ( 4 ) ; K ( V K 2 K h sec ] e K ) 1 K 3 K 2 ( K 2 ) ; K ( U K [ G G + − + ξ ⋅ − + ξ ⋅ | ¹ | \ | ξ × + − + ξ ⋅ = ′ ξ − ξ − (D.10) where ) K sinh( ) 1 K ( K ) K cosh( ) 1 K ( ) K 2 1 ( ) ; K ( U 2 ξ − + ξ − + − = ξ (D.11) References [1] R. 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