Modified ( G ′ /G) -expansion method for solving nonlinear partial differential equations
MModified ( G (cid:48) /G ) -expansion method for solving nonlinearpartial differential equations Mahouton Norbert Hounkonnou ∗ and Rolland Finangnon Kanfon † International Chair in Mathematical Physics and Applications(ICMPA-UNESCO Chair), University of Abomey-Calavi,072B.P.50, Cotonou, Rep. of Benin
E-mails: ∗ [email protected], † [email protected] Abstract
This paper addresses a modified ( G (cid:48) /G )-expansion method to obtain new classes of solutionsto nonlinear partial differential equations (NPDEs). The cases of Burgers, KdV and Kadomtsev-Petviashvili NPDEs are exhaustively studied. Relevant graphical representations are shown in eachcase. Over the second half of the past century, there was a flurry of activities regarding the investigation, witha solid mathematical background and a wide range of physical observations and practical applications,of nonlinearities in complex phenomena governing our universe. Indeed, nonlinear equations, widelyused to describe these complex phenomena, pervade many branches of mathematics and physics,including fluid mechanics, plasma physics, quantum electrodynamics, solid-state physics, nuclear andatomic physics, plasma waves, biology, and so on. See, for instance, [1, 2, 3] (and references therein) formore details from both theoretical and experimental viewpoints. The knowledge of exact solutions tothese equations constitutes certainly one of the keys for better understanding the nonlinearities. Thisexplains the search for suitable analytical methods for solving nonlinear equations. The most popularones are today the variational iteration method [4], [5], inverse scattering method [6], Hirota bi-linearform[7], Painlev´e analysis [8], direct algebraic method [9], tanh-function method [10], [11], [12], sine-cosine method [13], Darboux transformation [14, 15], homotopy analysis method, ( G (cid:48) /G )-expansionmethod [16], [17] and their extensions.This paper addresses a modified ( G (cid:48) /G )-expansion method to obtain new classes of solutions tononlinear partial differential equations (NPDEs). This approach consists, first, in linearizing theNPDE, and secondly, in integrating the resulting partial differential equation (PDE) with a non-zerointegration constant dependent constant. Then, the ( G (cid:48) /G )-expansion method is applied to deducethe solutions of the nonlinear PDE.The paper is organized as follows. In Section (2) we describe the ( G (cid:48) /G )-method used to solvethe NPDEs. We show that any solution φ of the NPDE can be expanded as polynomials dependingon the new variable G (cid:48) /G , where G (cid:48) is the derivative of G with respect to the coordinate variables.The general form of φ is also given. In Section (3) we explicitly solve the Burgers partial differential1 a r X i v : . [ m a t h - ph ] O c t quation, the Kadomtsev-Petviashvili and KdV equations. Relevant graphical representations areexhibited in each case. Section (4) is devoted to concluding remarks. The nonlinear partial differential equations (NPDEs), although difficult to solve, are object of intensivestudy in recent literature as shown in [1]-[25] and references therein. In this section, we consider thefollowing types of NPDEs: Q ( u, u t , u x , u xx , · · · ) = 0 , x, t ∈ R , (1) P ( u, u t , u x , u y , u tt , u xx , u yy , u tx , u ty , u xy , · · · ) = 0 , (2) u t = ∂u∂t , u x = ∂u∂x , u xx = ∂ u∂x , u tx = ∂ u∂t∂x , · · · ; x, y, t ∈ R , (3)where u = u ( x, t ) ∈ C ∞ ( R ) and v = v ( x, y, t ) ∈ C ∞ ( R ) are solutions of (1) and (2), respectively.Making the transformations u ( x, t ) = u ( ν ) and v ( x, y, t ) = v ( η ) (4)with ν = x − ωt, η = kx + αy + ω (cid:48) t, (5)where ω, ω (cid:48) , k and α ∈ R , the equations (1) and (2) can be rewritten as H ( u, u (cid:48) , u (cid:48)(cid:48) , · · · ) = 0 , (6)where u ( ξ ) = u ( ν ) for equation (1) and u ( ξ ) = v ( η ) for equation (2).Assume that equation (6) is integrable with respect to ξ without cancelling the integrating con-stants. Then introduce the following transformation [10] u ( ξ ) = φ ( ξ ) + c , c ∈ R . (7)Substituting (7) into (6) and setting the constant part equal to zero in the resulting nonlinear ODEin φ and assuming that the function φ and its derivatives have the following asymptotic values: φ ( ξ ) → φ ± as ξ → ±∞ , (8)for n ≥ φ ( n ) → ξ → ±∞ , (9)2nd that φ ± satisfies the algebraic equation in φ [10], we get the value of c .The ( G (cid:48) /G )-expanding method consists in setting φ ( ξ ) = α m (cid:32) G (cid:48) G (cid:33) m + α m − (cid:32) G (cid:48) G (cid:33) m − + · · · (10)where α m (cid:54) = 0 and G = G ( ξ ) satisfies the second order linear ordinary differential equation (LODE)in the form [22, 19]: G (cid:48)(cid:48) + λG (cid:48) + µG = 0 . (11)The parameter m can be found by substituting along with equation (11) into equation (6) and consid-ering a homogeneous balance between the highest order derivative and highest order nonlinear termin equation (6), where k, α, ω, a , a , · · · , a m are to be determined. Substituting (7) and (10)into (4) yields a set of algebraic equations for k, α, a , a , · · · , a m , vanishing all coefficients of( G (cid:48) /G ) i . From these relations, k, α, ω, a , a , · · · , a m can be obtained. Having determined theseparameters, knowing that m is a positive integer in most cases, and using (7) and (10) we obtainanalytical solutions u ( x, t ) and v ( x, y, t ) of (1) and (2), respectively.The solutions of the equation (11) are given as follows: • If λ − µ > G ( ξ ) = e − λ ξ (cid:32) k cosh (cid:16) (cid:112) λ − µ ξ (cid:17) + k sinh (cid:16) (cid:112) λ − µ ξ (cid:17)(cid:33) (12)and G (cid:48) G = − λ (cid:112) λ − µ f ( ξ ; λ, µ ) . (13)where f ( ξ ; λ, µ ) = k sinh (cid:16) √ λ − µ ξ (cid:17) + k cosh (cid:16) √ λ − µ ξ (cid:17) k cosh (cid:16) √ λ − µ ξ (cid:17) + k sinh (cid:16) √ λ − µ ξ (cid:17) (14) • If λ − µ < G ( ξ ) = e − λ ξ (cid:32) k cos (cid:32) (cid:112) µ − λ ξ (cid:33) + k sin (cid:32) (cid:112) µ − λ ξ (cid:33)(cid:33) (15)and G (cid:48) G = − λ (cid:112) µ − λ g ( ξ ; λ, µ ) . (16)where g ( ξ ; λ, µ ) = − k sin (cid:16) √ µ − λ ξ (cid:17) + k cos (cid:16) √ µ − λ ξ (cid:17) k cos (cid:16) √ µ − λ ξ (cid:17) + k sin (cid:16) √ µ − λ ξ (cid:17) , (17)3 If λ − µ = 0, G ( ξ ) = ( k ξ + k ) e − λ ξ (18)and G (cid:48) G = − λ k k ξ + k . (19)The equation (10) can be re-expressed as φ ( ξ ) = m (cid:88) k =1 α k (cid:32) G (cid:48) ( ξ ) G ( ξ ) (cid:33) k + α , α ∈ R , α k ∈ R ∀ k > , (20)giving φ (cid:48) ( ξ ) = − m (cid:88) k =1 kα k µ (cid:32) G (cid:48) G (cid:33) k − + λ (cid:32) G (cid:48) G (cid:33) k + (cid:32) G (cid:48) G (cid:33) k +1 (21)and φ (cid:48)(cid:48) ( ξ ) = m (cid:88) k =1 kα k (cid:40) µ ( k − (cid:32) G (cid:48) G (cid:33) k − + µλ (2 k − (cid:32) G (cid:48) G (cid:33) k − (22)+ k ( λ + 2 µ ) (cid:32) G (cid:48) G (cid:33) k + λ (2 k + 1) (cid:32) G (cid:48) G (cid:33) k +1 + ( k + 1) (cid:32) G (cid:48) G (cid:33) k +2 (cid:41) . In this section, we obtain new classes of solutions to some relevant NPDEs such as the Burgersequations [10], KdV equation [10] and Kadomtsev-Petviashvili equation (also called KP equationin the literature) [21]. In each case the most relevant graphical representations are shown with anappropriate choice of parameters.
Let us consider the nonlinear Burgers equation in two-dimensional space-time given by the followingrelation : u t + αuu x + βu xx = 0 . (23)Using (4), the equation (23) takes the form M ( u, u (cid:48) , u (cid:48)(cid:48) ) = − ωu (cid:48) + αuu (cid:48) + βu (cid:48)(cid:48) = 0 . (24)Equation (24) can be integrated to give the integral form (cid:90) M ( u, u (cid:48) , u (cid:48)(cid:48) ) dξ = − ωu + 12 αu + βu (cid:48) + c C = 0 , c C ∈ R . (25)4uppose that u = φ + c . Then the equation (25) can be transformed into the form αc φ − ωφ + 12 αφ + βφ (cid:48) + c (cid:18) αc + C − ω (cid:19) = 0 . (26)Using the equations (8) and (9), the functions φ ± satisfy the relation αc φ ± − ωφ ± + 12 αφ ± = 0 , (27)with the condition c (cid:18) αc + C − ω (cid:19) = 0 . (28)Two cases, deduced from the equation (28), deserve investigation in order to find explicit solutions tothe equation (23). Proposition 3.1 (Case 1) c = 0 leads to the two following situations:(i) ξ = x − (2 C − β ) t : We have U = βα + βα (cid:20) − λ + √ λ − µ f ( ξ ; λ, µ ) (cid:21) , if λ − µ > , βα + βα (cid:20) − λ + √ µ − λ g ( ξ ; λ, µ ) (cid:21) , if λ − µ < βα + βα (cid:16) − λ + k k ξ + k (cid:17) , if λ − µ = 0 , (29) (ii) ξ = x − (2 C − βµ ) t : We get U = βµα + βα (cid:20) − λ + √ λ − µ f ( ξ ; λ, µ ) (cid:21) , if λ − µ > , βµα + βα (cid:20) − λ + √ µ − λ g ( ξ ; λ, µ ) (cid:21) if λ − µ < βµα + βα (cid:104) − λ + k k ξ + k (cid:105) if λ − µ = 0 , (30) Proof of relations (29) and (30) .Equation (26) becomes − ωφ + 12 αφ + βφ (cid:48) = 0 . (31)Using (10) and balancing φ into φ (cid:48) , we get2 m = m + 1 so , m = 1 . (32)Finally φ ( ξ ) = α (cid:16) G (cid:48) G (cid:17) + α . (33)5igure 1: Solution U for α = 2 , β = 1 , µ =4 , c = 2 , c = 2 , C = 2 Figure 2: Solution U for α = 2 , β = 1 , µ =5 , c = 2 , c = 4 , C = 2Substituting (33) and (21) (for m = 1) in (31), we obtain the following expression − ωα (cid:16) G (cid:48) G (cid:17) − ωα + 12 αα + αα α (cid:16) G (cid:48) G (cid:17) + 12 αα (cid:16) G (cid:48) G (cid:17) + βα − βα µ − βα µ (cid:16) G (cid:48) G (cid:17) − βα (cid:16) G (cid:48) G (cid:17) = 0 . (34)Vanishing the polynomial coefficients yields the system − ωα + αα + βα − βα µ = 0 − ωα + αα α − βα µ = 0 αα − βα = 0 (35)whose solutions are given by ω = 2 C − βα = βα α = βα , or ω = 2 C − βµα = βµα α = βα . (36)The relations (29) and (30) are therefore well satisfied. (cid:4) Proposition 3.2 (Case 2) c = α ( ω − C ) furnishes two situations:(i) ξ = x − (2 C − β + βµ ) tU = − β − βλ +2 C +2 βµα + β √ λ − µα f ( ξ ; λ, µ ) , if λ − µ > , − β − βλ +2 C +2 βµα + β √ µ − λ α g ( ξ ; λ, µ ) , if λ − µ < , − β +2 C − βλ +2 βµα + βα (cid:16) k k ξ + k (cid:17) , if λ − µ = 0 . (37)6 ii) ξ = x − (2 C − βµ ) t : U = − βλ +2 Cα + β √ λ − µα f ( ξ ; λ, µ ) , if λ − µ > , − βλ +2 Cα + β √ µ − λ α g ( ξ ; λ, µ ) , if λ − µ < , − βλ +2 Cα + βα (cid:16) k k ξ + k (cid:17) if λ − µ = 0 . (38)Figure 3: Solution U for α = 2 , β = 1 , λ =5 , µ = 4 , c = 2 , c = 4 , C = 2 Figure 4: Solution U for α = 2 , β = 1 , λ =4 , µ = 5 , c = 2 , c = 4 , C = 2 Proof of relations (37) and (38)) .The equation (26) becomes ωφ − Cφ + 12 αφ + βφ (cid:48) = 0 . (39)Applying ( G (cid:48) /G )-expansion method on (39), using (10) and balancing φ into φ (cid:48) , we get m = 1 and φ ( ξ ) = α (cid:0) G (cid:48) G (cid:1) + α . (40)Using (40), the relation (39) is re-expressed as a polynomial in ( G (cid:48) /G ) and we obtain ωα (cid:16) G (cid:48) G (cid:17) + ωα − Cα (cid:16) G (cid:48) G (cid:17) − Cα + 12 αα + αα α (cid:16) G (cid:48) G (cid:17) + 12 αα (cid:16) G (cid:48) G (cid:17) + βα − βα µ − βα µ (cid:16) G (cid:48) G (cid:17) − βα (cid:16) G (cid:48) G (cid:17) = 0 . (41)Vanishing the polynomial coefficients provides the system ωα − Cα + αα + βα − βα µ = 0 ωα − Cα + αα α − βα µ = 0 αα − βα = 0 (42)which can be solved to give the solutions ω = 2 C − β + βµα = βα α = βα , or ω = 2 C − βµα = βµα α = βα . (43)Similarly as in the previous case, the relations (37) and (38) are well satisfied. (cid:4) .2 KdV nonlinear equation We consider the following KdV equation: u t + αuu x + γu xxx = 0 . (44)To solve it we use the transformation (4) to reduce the equation (44) into an ordinary differentialequation (ODE): − ωu + 12 αu + γu (cid:48)(cid:48) + c C = 0 , (45)where c C is constant. Combining (7) and (45), we get( αc − ω ) φ + 12 αφ + γφ (cid:48)(cid:48) + c ( 12 αc + C − ω ) = 0 . (46)The solutions of (44) are also given in two different cases. Proposition 3.3 (Case 1) c = 0 leads to two situations:(i) If ξ = x − γ (4 µ − λ ) t, we get U = − γµ + γλ α − γ ( λ − µ ) α f ( ξ ; λ ; µ ) , if λ − µ > , − γµ + γλ α − γ ( λ − µ ) α g ( ξ ; λ ; µ ) , if λ − µ < , − γµ +6 γµλ − γλ α + γ ( λ − µ ) α (cid:16) k k ξ + k (cid:17) − γα (cid:16) k k ξ + k (cid:17) , if λ − µ = 0 . (47) (ii) If ξ = x + γ (4 µ − λ ) t , we have: U = γ ( λ − µ ) √ λ − µα f ( ξ ; λ, µ ) − γ ( λ − µ ) α f ( ξ ; λ ; µ )+ − γµ +6 γµλ − γλ α , if λ − µ > , γ ( λ − µ ) √ λ − µα g ( ξ ; λ, µ ) − γ ( λ − µ ) α g ( ξ ; λ ; µ )+ − γµ +6 γµλ − γλ α , if λ − µ < , − γµ +6 γµλ − γλ α + γ ( λ − µ ) α (cid:16) k k ξ + k (cid:17) − γα (cid:16) k k ξ + k (cid:17) , if λ − µ = 0 . (48) Proof of relations (47) and (48) .The equation (46) is reduced to − ωφ + 12 αφ + γφ (cid:48)(cid:48) = 0 . (49)Taking into account (10), and balancing φ into φ (cid:48)(cid:48) , we get m = 2 and arrive at the following system: − ωα + αα + 2 γα µ + γα λµ = 0 − ωα + αα α + 6 γα λµ + 2 γα µ + α µ + α λ γ = 0 − ωα + αα α + αα + 4 γα λ + 3 α γλ + 8 α µγ = 0 αα α + 10 γα λ + 2 α γ = 0 αα + 6 α γ = 0 . (50)8igure 5: Solution U for α = 2 , γ = 1 , λ =5 , µ = 4 , c = 2 , c = 4 , C = Figure 6: Solution of U for α = 2 , γ =1 , λ = 4 , µ = 5 , c = 2 , c = 2 , C = 4which can be solved to give ω = γ (4 µ − λ ) α = − γ (2 µ + λ ) α α = − γλα α = − γα or ω = − γ (4 µ − λ ) α = − γµα α = − γλα α = − γα . (51)The relations (47) and (48) then hold. (cid:4) Proposition 3.4 (Case 2) c = ω − Cα amounts to the solutions of (44) in the two following situa-tions:(i) ξ = x − (2 C + 4 γµ − λ γ ) t gives U = − γµ + γλ +2 Cα − γ ( λ − µ ) α f ( ξ ; λ ; µ ) , if λ − µ > , − γµ + γλ +2 Cα − γ ( λ − µ ) α g ( ξ ; λ ; µ ) , if λ − µ < and − γµ + γλ +2 Cα − γα (cid:16) k k ξ + k (cid:17) , if λ − µ = 0 . (52) (ii) ξ = x − (2 C + 4 γµ + λ γ ) t yields U = − γµ +3 γλ +2 Cα − γ ( λ − µ ) α f ( ξ ; λ ; µ ) , if λ − µ > , − γµ +3 γλ +2 Cα − γ ( λ − µ ) α g ( ξ ; λ ; µ ) , if λ − µ < , − γµ +3 γλ +2 Cα − γα (cid:16) k k ξ + k (cid:17) , if λ − µ = 0 . (53) Proof of relations (52) and (53) .The equation (46) can be re-expressed as( ω − C ) φ + 12 αφ + γφ (cid:48)(cid:48) = 0 . (54)9igure 7: Solution of U for α = 2 , γ =5 , λ = 1 , µ = 1 / , c = 2 , c = 4 , C = 1 , β =1 Figure 8: Soliton solution of U for α =1 / , γ = 1 , λ = 2 , µ = 1 / , c = 1 , c =0 . , C = 0 . φ into φ (cid:48)(cid:48) imply that m = 2 and we arrive at the followingsystem: ωα − Cα + αα + 2 γα µ + γα λµ = 0 ωα − Cα + αα α + 6 γα λµ + 2 γα µ + α µ + α λ γ = 0 ωα − Cα + αα α + αα + 4 γα λ + 3 α γλ + 8 α µγ = 0 αα α + 10 γα λ + 2 α γ = 0 αα + 6 α γ = 0 (55)whose solutions are given by ω = 2 C + 4 γµ − λ γα = − γµα α = − γλα α = − γα or ω = 2 C − γµ + λ γα = γ (2 µ + λ ) α α = − γλα α = − γα . (56)Therefore we simply get relations (52) and (53)). (cid:4) This subsection is devoted to the resolution of the Kadomtsev-Petviashvili NPDE in three dimensionsgiven by ( u t + 6 uu x + u xxx ) x + 3 σ u yy = 0 , σ = ± . (57)Using the transformation (4), the equation (57) takes the form N ( u, u (cid:48) , u (cid:48)(cid:48) , u (cid:48)(cid:48)(cid:48) ) = kωu (cid:48)(cid:48) + 6 k ( u (cid:48) u ) (cid:48) + k u (4) + 3 σ α u (cid:48)(cid:48) = 0 . (58)Integrating this latter equation, we get the following nonlinear ODE (cid:90) N ( u, u (cid:48) , u (cid:48)(cid:48) , u (cid:48)(cid:48)(cid:48) ) dξ = ( kω + 3 σ α ) u (cid:48) + 6 k u (cid:48) u + k u (cid:48)(cid:48)(cid:48) + c C = 0 , (59)10here c C is a constant to be determined. Substituting (7) into (59), we simply get( kω + 3 σ α + 6 c k ) φ + 3 k φ + k φ (cid:48)(cid:48) + c ( kω + 3 σ α + 3 k c + C ) = 0 . (60)The equation (57) can be now solved in different cases to give the next results: Proposition 3.5 (Case 1) c = 0 . The solutions of (57) are:(1) If ξ = x − (4 k µ − σ α − k λ ) t/k : U = − k µ + k λ − k ( λ − µ )2 f ( ξ ; λ, µ ) , if λ − µ > , − k µ + k λ − k (4 µ − λ )2 g ( ξ ; λ, µ ) , if λ − µ < , − k µ + k λ − k (cid:16) c c ξ + C (cid:17) , if λ − µ = 0 . (61) (ii) If ξ = x − (4 k µ − σ α − k λ ) t/k : U = − k µ + k λ − k ( λ − µ )2 f ( ξ ; λ, µ ) , if λ − µ > , − k µ + k λ − k (4 µ − λ )2 g ( ξ ; λ, µ ) , if λ − µ < , − k µ + k λ − k (cid:16) c c ξ + C (cid:17) , if λ − µ = 0 . (62) Proof of relations (61) and (62).The equation (60) takes the form( kω + 3 σ α ) φ + 3 k φ + k φ (cid:48)(cid:48) = 0 . (63)Let us consider (10) and balance φ into φ (cid:48)(cid:48) to get m = 2.Then we arrive at the following system k α + 6 k α = 06 k αα + 10 k α λ + 2 k α = 03 σ α α + kωα + 3 k α + 3 k α λ + 8 k α µ + 6 k α α α + 4 k α λ = 0 kωα + 2 k α µ + αk α α + 3 σ α α + 6 k α λµ + k α λ = 0 kωα + k α λµ + 3 k α + 3 σ α α + 2 k α µ = 0 (64)giving the solutions ω = k µ − σ α − k λ k α = − k µα = − k µα = − k or ω = − k µ +3 σ α − k λ k α = − k µ − k λ α = − k λα = − k . (65)There result the relations (61) and (62). (cid:4) Proposition 3.6 (Case 2) c = − C − σ α − k ω . It leads to the solutions of (57) in two situations: i) ξ = kx + αy − (4 k µ − k C − k λ + 6 k σ α − σ α ) t/k ( − k ) provides U = − k µ +3 k λ − C − σ α − k ω − k ( λ − µ )2 f ( ξ ; λ, µ ) , if λ − µ > , − − k µ +3 k λ − C − σ α − k ω − k (4 µ − λ )2 g ( ξ ; λ, µ ) if λ − µ < , − k µ +3 k λ − C − σ α − k ω − k (cid:16) c c ξ + C (cid:17) if λ − µ = 0 (66) (ii) ξ = kx + αy − ( − k µ + 2 k C + k λ + 6 k σ α − σ α ) t/k ( − k ) gives U = − k µ + k λ − C − σ α − k ω − k ( λ − µ )2 f ( ξ ; λ, µ ) , if λ − µ > , − k µ + k λ − C − σ α − k ω − k (4 µ − λ )2 g ( ξ ; λ, µ ) , if λ − µ < − k µ + k λ − C − σ α − k ω − k (cid:16) c c ξ + C (cid:17) if λ − µ = 0 . (67) Proof of (66) and (67).The equation (60) can be rewritten as( kω + 3 σ α − k C − k σ α − k ω ) φ + 3 k φ + k φ (cid:48)(cid:48) = 0 . (68)Considering (10) and balancing φ into φ (cid:48)(cid:48) , we get m = 2 and obtain the following system k C + 6 k α = 06 k α α + 10 k α λ + 2 k α = 0 − k σ α α − k ωα + 6 k α α − k Cα + 3 k α λ +3 k α + kωα + 8 k α µ + 3 σ α α + 4 k α λ = 02 k α µ − k ωα + 2 k α λ + kωα − k Cα + 6 k α − k σ α α + 6 k α α + 3 σ α α = 0 kωα − k Cα − k σ α α + 3 k α +3 σ α α + 2 k α µ − k ωα + k α λµ = 0 (69)Its solutions are given by ω = − k +6 k σ α +4 k µ − σ α − k λ k ( − k ) α = − k µα = − k λα = − k (70)or (71) ω = − k µ +2 k C + k λ +6 k σ α − σ α k ( − k ) α = − k µ − k λ α = − k λα = − k (72)12alidating the relations (66) and (67). (cid:4) In this work we have modified the ( G (cid:48) /G )-expansion method to find new classes of solutions to knownimportant NPDEs, in addition to solutions obtained by the usual ( G (cid:48) /G )-expansion method. The mostrelevant graphical representations have been drawn to show the pertinence of the obtained analyticalsolutions. Acknowledgment
RK thanks the University of Abomey-Calavi and the Ministry for High Education and ScientificResearch of Benin for financial support. This work is partially supported by the ICTP through theOEA-ICMPA-Prj-15. The ICMPA is in partnership with the Daniel Iagolnitzer Foundation (DIF),France.
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