Computers and Mathematics with Applications

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1 Computers and Mathematics with Applications 60 (00) Contents lists available at ScienceDirect Computers and Mathematics with Applications journal homepage: Symmetry analysis of an integrable Ito coupled system Hassan A. Zedan Department of Mathematics Faculty of Science Kafr El-Sheikh University Egypt a r t i c l e i n f o a b s t r a c t Article history: Received February 00 Accepted 3 October 00 Keywords: Coupled evolution system Truncated Painleve expansion Symbolic computation Weiss Kruskal simplification In this paper we study the invariance analysis integrability properties and P-property of the Ito coupled nonlinear partial differential equations. We explore several new solutions for the Ito system through the Lie symmetry analysis. Moreover this work has been devoted to study the integrability aspects of the Ito system through higher order symmetries. We are also investigating the existence of higher order symmetries for the Ito system. Interestingly our investigations reveal a rich variety of particular solutions which have not been reported in the literature for this model. 00 Elsevier Ltd. All rights reserved.. Introduction For the past two decades the Lie group method has been applied to solve a wide range of problems and to explore many physically interesting solutions of nonlinear phenomena [ 4]. In recent years several extensions and modifications of the classical Lie algorithm have been proposed in order to arrive at new solutions of partial differential equations (PDEs) [5]. Lie symmetry analysis is one of the most powerful methods to obtain particular solutions of differential equations [6]. It is based on the study of their invariance with respect to one-parameter Lie group of point transformations whose infinitesimal generators are represented as vector fields. Once the Lie groups that leave the differential equations invariant are known we can construct an exact solution called a group invariant solution which is invariant under the transformation. In this paper we investigate the invariance analysis and the Painleve analysis to the following Nonlinear Ito coupled system u t = v x v t = (v xxx +3uv x +3vu x ) ww x w t = w xxx +3uw x. Let us consider a one-parameter Lie group of infinitesimal transformations of the form: U = u + εη (x t u v w) V = v + εη (x t u v w) W = w + εη 3 (x t u v w) X = x + εζ (x t u v w) T = t + εζ (x t u v w) ε. The functions η η η 3 ζ and ζ are the infinitesimal of transformations for the variables u v w x and t respectively. In order to find the infinitesimal we need to extend the group to calculate how derivative terms transform. The transformation (.) (.) address: hassanzedan003@yahoo.com. Current Address: Math. Dept. Faculty of Science King Abdul Aziz University P.O. Box 8003 Jeddah 589 Saudi Arabia /$ see front matter 00 Elsevier Ltd. All rights reserved. doi:0.06/j.camwa

2 H.A. Zedan / Computers and Mathematics with Applications 60 (00) (.) together with the transformations for the first second... derivatives are called the first second... extensions. We denote the infinitesimal for u t v t w t u x v x w x u y v y w y u tt by η () t η() t η() 3t η() x η() x η() 3x η() y η() y η() y η() tt respectively. Using these various extensions the infinitesimal criterion for the invariance of (.) under group (.) admits an infinitesimal generator of the form: η () t = η () x η () t = (η (3) xxx + 3η v x + 3η () x u + 3η u x + 3η () x v) η 3w x η () 3x w η () 3t = η (3) 3xxx + 3η w x + 3η () 3x u where η () t η() t η() 3t η() x η() x η() 3x η() y η() y η() y η() tt are extending infinitesimal of transformations given by [ 4]. The invariance of Eq. (.) under the infinitesimal transformations (.) leads to [9]: η = au η = 4av η 3 = 3aw ζ = ax + b ζ = 3at + c (.4) where a b and c are arbitrary constants. The associated Lie vector fields are V = x + 3t u u 4v v 3w w V = V 3 =. The non-zero commutation relation between the vector fields are [V V ] = V [V V ] = 3V 3. Solving the characteristic equation associated with the infinitesimal symmetries (.4) one obtains z = ax + b (3at + c) 3 w = (3at + c) 3 u w = (3at + c) 4 3 v w 3 = (3at + c)w. We wish to note that while deriving the above similarity reductions under similarity transformation (.5) one can reduce the system of PDEs (.) to the system of ODEs of the form: w + zw + w = 0 4w zw + a w + 6w w + 6w w + w 3w 3 = 0 3w 3 + zw 3 + a w 3 3w w 3 = 0 (.6) where prime denotes differentiation with respect to z.. Painleve (P) analysis In order to verify whether the reduced system of ODEs (.6) is integrable or not we apply Ablowitz Ramani Segur (ARS) algorithm [6 8] to the system of ODEs (.6). Since the independent variable appeared explicitly in the above system of ODEs (.6) first let us rewrite Eq. (.6) of the form: w + τw + z 0w + w = 0 4w τw z 0w + a w + 6w w + 6w w + w 3w 3 = 0 3w 3 + τw 3 + z 0w 3 + a w 3 3w w 3 = 0 (.) where τ = z z 0 and z 0 is a movable singular point. Now let us represent the solution to the system of ODEs (.) locally as a Laurent series and let the leading order be of the form w = a 0 τ α w = b 0 τ β w 3 = c 0 τ γ (.) where a 0 b 0 and c 0 are arbitrary constants and α β and γ are integers to be determined. Substituting (.) into the system of ODEs (.) and balancing the dominant terms we obtain α = β β 3 = β + α and β 3 = γ. Then α = β = γ =. Now let us consider the Laurent expansion of the form w = a 0 τ + β τ r w = b 0 τ + β τ r w 3 = c 0 τ + β 3 τ r. (.3) (.3) (.5)

3 3090 H.A. Zedan / Computers and Mathematics with Applications 60 (00) Substituting the system of equations (.3) into the system of ODEs (.) and balancing the most singular terms again we obtain r = 4 6. Let us assume the Laurent series of the form w = a j τ j w = b j τ j w 3 = c j τ j. (.4) Substituting the system of equations (.4) into the system of ODEs (.) and equating various powers of τ n and solving the resultant equations then we obtain α 0 = α = α 4 = α 5 = α 6 = β 0 = β = β 5 = β 6 = γ 0 = γ = γ = γ 3 = γ 4 = γ 5 = γ 6 = 0 α = 3 z 0 α 3 = 3 β = z β 0 3 = z 0 β 4 =. The Laurent series solution (.4) is meromorphic ; consequently the similarity reduced system of ODEs (.) also possesses the Painleve property. Even though the system of ODEs (.6) is integrable in general it is very difficult to integrate it explicitly and obtain a general solution. However one can obtain a number of physically interesting solutions by considering certain special choices of the infinitesimal symmetries (.) which we present few of them in the following sections. Then we obtain the solution of the system of ODEs w = 3 z w = z w 3 = 0 and hence the solution of the system of PDEs (.) are u = (ax + b) 3(3at + c) v = (a x + abx + b ) w = 0. (.5) (3at + c) Generic subcases: Let us recall that in the derivation of the general similarity reductions we made an assumption that a 0 and c 0. Now let us consider another cases and the possible similarity reductions... Case : a = c = 0 b is arbitrary The similarity variables take the form: z = t w = u w = v w 3 = w. The similarity reduced system of ODEs turns out to be w = 0 w = 0 w 3 = 0. Thus the solution obtained will be stationary. (.6).. Case : a = b = 0 c is arbitrary The similarity variables take the form: z = x w = u w = v w 3 = w. The similarity reduced system of ODEs turns out to be w = 0 w + 6w w + 6w w + w 3w 3 = 0 w 3 3w w 3 = 0. From the first equation of the system of ODEs (.8) we obtain w = k and from the second one we obtain w = k 6w 3. 6k When we substitute w into the third equation and integrate it twice we obtain w = 3 k 4 + k 3 w 3 + k w k k 3 w 4 k. 3 (.7) (.8)

4 H.A. Zedan / Computers and Mathematics with Applications 60 (00) Subcase : k = k 3 = 0 k and k 4 are arbitrary we obtain k sn k 4 x I k k k 4 w = and u = v = k. sn k k 4 x k I k4 k... Subcase : k 3 = k 4 = 0 k and k are arbitrary we obtain k k x w = k and u = 5k k + 3x 6k v = k...3. Subcase 3: k 3 = 0 k k and k 4 are arbitrary we obtain and w = sn u = k + k v = k. A x sn 4k k 4 +k +k A k k 4 k A k k 4 A x 4k k 4 +k +k A k k 4 k4 k A 8k k 4 + k. where A = k (k A ) and A k = If we use the Painleve analysis we obtain u = x v = c 0 w = c 0 x. (.9).3. Case 3: b = c = 0 a is arbitrary The similarity variables take the form: z = x t 3 w = t 3 u w = t 4 3 v w 3 = tw. (.0)

5 309 H.A. Zedan / Computers and Mathematics with Applications 60 (00) The similarity reduced system of ODEs turns out to be w + zw + 3w = 0 4w zw + 6w + 8w w + 8w w + 36w 3w 3 = 0 3w 3 + zw 3 + 3w 3 9w w 3 = 0 (.) where prime denotes differentiation with respect to z. In order to verify whether the reduced system of ODEs (.9) is integrable or not we apply Ablowitz Ramani Segur (ARS) algorithm [6 8] to the system of ODEs (.9). Since the independent variable appeared explicitly in the above system of ODEs (.9) first let us rewrite the system of ODEs (.9) of the form: w + τw + z 0w + 3w = 0 4w τw z 0w + 6w + 8w w + 8w w + 36w 3w 3 = 0 3w 3 + τw 3 + z 0w 3 + 3w 3 9w w 3 = 0 (.) where τ = z z 0 and z 0 is a movable singular point. Then we obtain w = 9 z w = 8 z w 3 = 0 and hence the solution of the system of PDEs (.) are u = x 9t v = x 8t w = 0. (.3).4. Case 4: a = 0 b and c are arbitrary (leads travelling wave solution) The similarity variables take the form: z = cx bt w = u w = v w 3 = w. (.4) The similarity reduced system of ODEs turns out to be bw + cw = 0 bw + c3 w + 6cw w + 6cw w + cw 3w 3 = 0 bw 3 + c3 w 3 3cw w 3 = 0 (.5) where prime denotes differentiation with respect to z. Let us assume the Laurent series of the form w = a j τ j w = b j τ j w 3 = c j τ j. (.6) Then we obtain the solution of the system of PDEs (.) are or u = c3 (cx bt) + a bc v = (cx bt) b( 6ca + b) w = c 6c (.7) u = 4c 3 (cx bt) + b 3c v = 4bc (cx bt) b 6c 6bcc 6c w = 6bcc (cx bt) + c. (.8)

6 H.A. Zedan / Computers and Mathematics with Applications 60 (00) Case 5: b = 0 a and c are arbitrary The similarity variables and similarity functions take the form z = ax (3at + c) 3 w = (3at + c) 3 u w = (3at + c) 4 3 v w 3 = (3at + c)w. (.9) We wish to note that while deriving the above similarity reductions under similarity transformation (.9) one can reduce the system of PDEs (.) to the system of ODEs of the form: w + zw + w = 0 4w zw + a w + 6w w + 6w w + w 3w 3 = 0 3w 3 + zw 3 + a w 3 3w w 3 = 0 (.0) where prime denotes differentiation with respect to z. In order to verify whether the reduced system of ODEs (.0) is integrable or not we apply Ablowitz Ramani Segur (ARS) algorithm [6 8] to the system of ODEs (.0). Since the independent variable appeared explicitly in the above system of ODEs (.0) first let us rewrite the system of ODEs (.0) of the form: w + τw + z 0w + w = 0 4w τw z 0w + a w + 6w w + 6w w + w 3w 3 = 0 3w 3 + τw 3 + z 0w 3 + a w 3 3w w 3 = 0 (.) where τ = z z 0 and z 0 is a movable singular point. Let us assume the Laurent series of the form w = a j τ j w = Then we obtain b j τ j w 3 = w = 3 z w = z w 3 = 0 and hence the solution of the system of PDEs (.) are x c j τ j. (.) ax u = 3(3at + c) v = a w = 0. (.3) (3at + c).6. Case 6: c = 0 a and b are arbitrary z = ax + b t 3 w = t 3 u w = t 4 3 v w 3 = tw. (.4) We wish to note that while deriving the above similarity reductions under similarity transformation (.4) one can reduce the system of PDEs (.) to the system of ODEs of the form: w + zw + 3w = 0 4w zw + 6a w + 8w w + 8w w + 36w 3w 3 = 0 3w 3 + zw 3 + a w 3 3w w 3 = 0 (.5) where prime denotes differentiation with respect to z.

7 3094 H.A. Zedan / Computers and Mathematics with Applications 60 (00) In order to verify whether the reduced system of ODEs (.5) is integrable or not we apply Ablowitz Ramani Segur (ARS) algorithm [6 8] to the system of ODEs (.5). Since the independent variable appeared explicitly in the above system of ODEs (.5) then we can rewrite the system of ODEs (.5) in the form: w + τw + z 0w + 3w = 0 4w τw z 0w + 6a w + 8w w + 8w w + 36w 3w 3 = 0 3w 3 + τw 3 + z 0w 3 + a w 3 3w w 3 = 0 (.6) where τ = z z 0 and z 0 is a movable singular point. Let us assume the Laurent series of the form w = a j τ j w = b j τ j w 3 = Then we obtain w = 9 z w = 8 z w 3 = 0 and hence the solution of the system of PDEs (.) are c j τ j. (.7) u = ax + b (ax + b) v = w = 0. (.8) 9t 8t 3. Painleve (P) analysis for the original PDE: Methodology: Consider a system of M polynomial differential equations F i (u(z) u (z) u (z)... u (m i) (z)) = 0 i =... M. (3.) Step (Determine the dominant behavior). It is sufficient to substitute u i (z) = x i g α i (z) i =... M where x i is a constant into (3.) to determine the leading exponents α i. In the resulting polynomial system equating every two or more possible lowest exponents of g(z) in each equation gives a linear system for α i. The linear system is then solved for α i and each solution branch is investigated. The traditional Painleve test requires that all the α i s are integers and that at least one is negative. An alternative approach is to use the weak Painleve test which allows certain rational α i s and resonances; see [0 ] for more information on the weak Painleve test. If one or more exponents α i remain undetermined we assign integer values to the free α i so that every equation in (3.) has at least two different terms with equal lowest exponents. For each solution α i we substitute u i (z) = u i0 (z)g α i (z) i =... M into (3.). We then solve the (typically) nonlinear equation for u i0 (z) which is found by balancing the leading terms. By leading terms we mean those terms with the lowest exponent of g(z). Step (Determine the resonances). For each α i and u i0 (z) we calculate the r r r 3 r m for which u i0 (z) is an arbitrary function in u i (z) = g α i (z) u ik (z)g k (z) i =... M. k=0 To do this we substitute u i (z) = u i0 (z)g α i (z) + uir (z)g α i+r (z) into (3.) and keep only the lowest order terms in g(z) that are linear in u ir. This is carried out by computing the solutions for r of det(q r ) = 0 where the M M matrix Q r satisfies Q r u r = 0 u r = (u r u r u 3r u Mr ) T. If any of the resonances are non-integer then the Laurent series solutions of (3.) have a movable algebraic branch point and the algorithm terminates. If r m is not a positive integer then the algorithm terminates. Step 3 (Find the constants of integration and check compatibility conditions). For the system to possess the Painleve property the arbitrariness of u ir (z) must be verified up to the highest resonance level. This is carried out by substituting u i (z) = g α i (z) r m k=0 u ik (z)g k (z) into (3.) where r m is the largest positive integer resonance. To simplify step 3 we can use Weiss Kruskal simplification.

8 H.A. Zedan / Computers and Mathematics with Applications 60 (00) The manifold defined by g(z) = 0 is noncharacteristic that means g zl (z) 0 for some l on the manifold g(z) = 0. By the implicit function theorem we can then locally solve g(z) = 0 for z l so that g(z) = z l h(z... z l z l+... z N ) for some arbitrary function h. Now we apply the method into our problem (the Ito coupled system). Let u = u 0 (ϕ(x t)) α v = v 0 (ϕ(x t)) α w = w 0 (ϕ(x t)) α 3 (3.) where u 0 v 0 and w 0 are functions in x t then system (.) becomes u 0 (ϕ(x t))α + α u 0 (ϕ(x t)) α ϕ(x t) = v 0 (ϕ(x t))α + α v 0 (ϕ(x t)) α ϕ(x t) (3.3) v 0 (ϕ(x t))α + α v 0 (ϕ(x t)) α ϕ(x t) = 3 v 0 (ϕ(x 3 t))α v 0 6α (ϕ(x ϕ(x t) v t))α 6α 0 (ϕ(x t))α ϕ(x t) v 0 6α (ϕ(x t))α ϕ(x t) v 0 + 6α (ϕ(x t))α ϕ(x t) ϕ(x t) 3 ϕ(x t) v 0 (ϕ(x t)) α 3 α 3 6v 0 (ϕ(x t)) α ϕ(x t) α ϕ(x t) v 0 (ϕ(x t)) α 3 α v 0 (ϕ(x t)) α ϕ(x t) α 3 ϕ(x t) + 6v 0 (ϕ(x t)) α ϕ(x t) ϕ(x t) 3 α 4v 0 (ϕ(x t)) α 3 α 6 u 0 (ϕ(x t))α +α v 0 6u 0 (ϕ(x t)) α +α α 6u 0 (ϕ(x t)) α +α v 0 6u 0(ϕ(x t)) α +α v 0 α ϕ(x t) v 0 ϕ(x t) w 0 (ϕ(x t)) α w 0 3 ϕ(x t) w(ϕ(x 0 t))α3 α 3 w0 (x t) (ϕ(x t)) α 3 + α 3 w 0 (x t)(ϕ(x t)) α 3 ϕ(x t) 3 w 0 (x t) = (ϕ(x t)) α 3 w 0 (x t) + 3α 3 3 (ϕ(x t)) α 3 ϕ(x t) + 3α w0 (x t) 3 (ϕ(x t)) α 3 ϕ(x t) w0 (x t) + 3α 3 (ϕ(x t)) α 3 ϕ(x t) w0 (x t) 3α 3 (ϕ(x t)) α 3 ϕ(x t) + α 3 3 w 0(x t)(ϕ(x t)) α 3 3 ϕ(x t) 3 + 3α w 3 0(x t)(ϕ(x t)) α 3 ϕ(x t) ϕ(x t) 3α w 3 0(x t)(ϕ(x t)) α 3 3 ϕ(x t) + α 3 w 0 (x t)(ϕ(x t)) α 3 3 ϕ(x t) 3α 3 3 w 0 (x t)(ϕ(x t)) α 3 ϕ(x t) ϕ(x t) + α 3 w 0 (x t)(ϕ(x t)) α 3 3 ϕ(x t) 3 3u 0 (x t)(ϕ(x t)) α w0 +α 3 (x t) 3α 3 u 0 (x t)w 0 (x t)(ϕ(x t)) α +α 3 ϕ(x t) 3 (3.4). (3.5) From Eq. (3.3) we obtain α = α and from Eq. (3.4) we obtain α 3 = α + α. Hence α = α =. From Eq. (3.4) we obtain α 3 = α 3. Hence α 3 =.

9 3096 H.A. Zedan / Computers and Mathematics with Applications 60 (00) Substituting α = α = α 3 = into Eqs. (3.3) (3.5) and then requiring the leading terms (of (ϕ(x t)) 5 in Eq. (3.5)) balance give u 0 = ϕ(xt) 4 from leading terms (of (ϕ(x t)) 3 in Eq. (3.3)) balance and substituting about u 0 gives v 0 = ϕ(xt) ϕ(xt) 4 and from leading terms (of (ϕ(x t)) 5 in Eq. (3.3)) balance and substituting about u 0 and v 0 gives w 0 = 3 ϕ(xt) ϕ(xt) ± I. We have two cases first when w 0 = 3 ϕ(xt) ϕ(xt) I and the second one when w 0 = 3 ϕ(xt) ϕ(xt) I we will study these two cases. Substituting ϕ(x t) u(x t) = 4 (ϕ(x t)) + u r (x t)(ϕ(x t)) r ϕ(x t) ϕ(x t) v(x t) = 4 (ϕ(x t)) + v r (x t)(ϕ(x t)) r and w(x t) = ± I ϕ(x t) 3 ϕ(x t) (ϕ(x t)) + w r (x t)(ϕ(x t)) r into (.) and equating the coefficient of (ϕ(x t)) r 5 in Eq. (3.4) we obtain the following characteristic equation for the resonances 5 ϕ(x t) ϕ(x t) 4 8( 8 + r)( 6 + r)( 4 + r)( 3 + r)( + r)( + r)( + r) = 0. ϕ(xt) Assuming ϕ(xt) 0 and 0 then r = r = r = r = 3 r = 4 r = 6 r = 8. We now substitute u = (ϕ(x t)) v = (ϕ(x t)) w = (ϕ(x t)) 8 u k (x t)(ϕ(x t)) k 0 8 v k (x t)(ϕ(x t)) k 0 8 w k (x t)(ϕ(x t)) k 0 into (.) and use the Weiss Kruskal simplification [34] (i.e. ϕ(x t) = x h(t)) we obtain u 0 = 4 v 0 = 4h t w 0 = I h (t) u = 0 v = 0 w = 0 u = 3 h (t) v = c (x t) w = I h (t)((h (t)) + v ) h (t) u 3 = c (x t) v 3 = 3 ( h (t) 3h (t)u 3 ) w 3 = I h (t)(h (t) + h (t)u 3 ) u 4 = c 3 (x t) v 4 = (u 3t h (t)u 4 ) w 4 = I h (t)u 4 6 h (t) u 5 = 68(h (t)) ((h (t)) h (t) + 40h (t)u 4t h (t)v t + 30(h (t)) 3 u 3 + 6h (t)v + 60h (t)u 3 v ) v 5 = 68(h (t)) ((h (t)) h (t) 6h (t)u 4t h (t)v t + 30(h (t)) 3 u 3 + 6h (t)v + 60h (t)u 3 v ) w 5 = I h (t)(5(h (t)) h (t) + 3h (t)u 68 (h 4t + 4h (t)v t + 4(h (t)) 3 u 3 + h (t)v (t)) + 48h (t)u 3 v ) (3.6)

10 H.A. Zedan / Computers and Mathematics with Applications 60 (00) u 6 = c 4 (x t) v 6 = 67(h (t)) 3 ( h (t)) 3 h (t) 30(h (t)) 4 u 3t + 40h (t)h (t)u 4t 40(h (t)) u 4tt 8h (t)h (t)v t + (h (t)) v tt 30(h (t)) 3 h (t)u 3 60(h (t)) v t u (h (t)) 4 u 6 + (h (t)) v 6h (t)h (t)v 60(h (t)) u 3t v + 60h (t)h (t)u 3 v w 6 = I h (t)(h (t)) h (t)h (t) 4(h (t)) u 88 (h 3t 6h (t)h (t)u 3 (t)) + 7(h (t)) u (h (t)) u 6 u 7 = 4 (u 4t u 3 u 4 ) v 7 = 0 ( 5h (t)u 4t + 4u 6t + 60h (t)u 3 u 4 ) w 7 = I h (t)( 30(h (t)) 4 h (t) + 70(h (t)) u 0080 (h 3tt 360(h (t)) 3 u 4t + 344(h (t)) u 6t (t)) 3 45(h (t)) 3 v t + 45(h (t)) 5 u (h (t)) u 3t u 3 560(h (t)) h (t)u 4 680(h (t)) 3 u 3 u 4 30(h (t)) h (t)v + 0h (t)u 4t v 90h (t)v t v + 80(h (t)) 3 u 3 v + 60h (t)v + 80h (t)u 3 v ) u 8 = c 5 (x t) v 8 = 44 ( 44u 8h (t) u 4 u 3t u 3 u 4t + u 4tt ) w 8 = I h (t)79u 8064 (h (t)) v 3 3 (h (t)) + 008u 4 (h (t)) u 8 (h (t)) u 3 (h (t)) u 3 v h (t)h (t) + 30u 3 (h (t)) 3 h (t) + 6v (h (t)) (h (t)) (h (t)) v h (t)h (t) 5(h (t)) 3 h (t) 48v (h (t)) u 3t 4(h (t)) 4 u 3t + 58u 3 (h (t)) u 4t + 36h (t)h (t)u 4t 4u 3 (h (t)) v t h (t)h (t)v t 3(h (t)) u 4tt 4(h (t)) v tt where c (x t) c (x t) c 3 (x t) c 4 (x t) and c 5 (x t) are arbitrary functions then we establish u v w by substituting into (3.6) from the two cases. 4. Conclusion In this paper we have explored only Lie point symmetries. However in recent years several works have been devoted to study the integrability aspects of coupled systems (.) through higher order symmetries [5]. Presently we are also investigating the existence of higher order symmetries for the nonlinear Ito coupled system. References [] P.J. Olver Applications of Lie Groups to Differential Equations Springer New York 986. [] G.W. Bluman S. Kumei Symmetries and Differential Equations Springer New York 989. [3] H. Stephani Differential Equations: Their Solutions Using Symmetries Cambridge University Press Cambridge 990. [4] N.H. Ibrahimov CRC Handbook of Lie Group Analysis of Differential Equations CRC Press Boca Raton 996. [5] P.A. Clarkson Chaos Solitons Fractals 5 (995) 6. [6] M.J. Ablowitz A. Ramani Segur J. Math. Phys. (983) 75. [7] M. Lakshmanan R. Sahadevan Phys. Rep. 4 (993). [8] A. Ramani B. Grammaticos T. Bountis Phys. Rep. 80 (989) l59. [9] A. Head Comput. Phys. Comm. 77 (993) 4. [0] A.N.W. Hone Painleve tests singularity structure and integrability Nonlinear Sci. (005) 34. [] B. Grammaticos B. Dorizzi Math. Comput. Simulation 37 (4 5) (994) [] B. Grammaticos A. Ramani K.M. Tamizhmani T. Tamizhmani A.S. Carstea J. Phys. A 40 (30) (007) F75 F735. [3] M. Jimbo M.D. Krudkal T. Miwa Phys. Lett. 9A (98) [4] M.D. Krudkal P.A. Clarkson Stud. Appl. Math. 86 (87) (99) 65. [5] P.J. Olver V.V. Sokolov Comm. Maths. Phys. 93 (998) 45.