Successfully reported this slideshow.
Upcoming SlideShare
×

# Ab31169180

200 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

### Ab31169180

1. 1. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180Travelling Wave Solutions of BBM and Modified BBM Equations by Modified F-Expansion Method Priyanka M. Patel1* and V. H. Pradhan1 Department of Applied Mathematics and Humanities, S. V. National Institute of Technology, Surat-395007, IndiaAbstract A new modified F-expansion method is ut  u x   u u x  u x x x  0 (2)introduced to obtain the travelling wavesolutions like soliton and periodic, of Benjamin- Consider the travelling wave solutions of eq. (2),Bona-Mahony (BBM) equation and modified under the transformationBBM equation. The method is convenient,effective and can be applied to other nonlinear u( x, t )  u( ) where   k ( x   t ) (3)partial differential equations in themathematical physics. where k  0 and  are constants that do determined later. By substituting eq. (3) into eq. (2)Keywords : Modified F-expansion method, we obtainBenjamin-Bona-Mahony (BBM) equation,modified BBM equation, travelling wave solutions. k  u  k u  k  u u  k 2 u  01. Introduction  u  u   u u  k u  0 (4) It is well known that nonlinear partialdifferential equations (NLPDEs) are widely used to Integrating eq. (4) with respect to „  ‟ anddescribe many important phenomena and dynamic considering the zero constant for integration weprocesses in physics, mechanics, chemistry and havebiology, etc. There are many methods to solveNLPDEs such as tanh-sech method [2], homotopy u2analysis method [14],sine-cosine method [8], (1   ) u    k u  0 (5) 2G / G  expansion method [4,9],differential where prime denotes differentiation with respect totransform method [7], and so on. In this paper, weapply the modified F-expansion method [3,9,10] to  . By balancing the order of u 2 and u in eq. (5),the generalized BBM equation we have 2 n  n  2 then n  2 So we can see its exact solutions in the formut  ux   u N ux  ux x x  0 with N  1 (1) u ( )  a0  a2 F 2 ( )  a1 F 1 ( )  a1 F ( )  a2 F 2 ( ) with constant parameter  . Wazwaz [2] (6)has obtained the analytical solution of eq.(1) withtanh-sech method and sine-cosine method.Abbasbandy [14] has applied homotopy analysis where a 0 , a1 , a2 , a1 , a 2 are constants to bemethod (HAM) to obtain the analytical solution of determined later. And F ( ) satisfies thegeneralized BBM equation for N  1 and N  2 . following Riccati equationIn the present paper we have applied modified F-expansion method to solve eq.(1) for N  1 and F ( )  A  B F ( )  C F 2 ( ) (7) N  2 . A set of new solutions are obtained otherthan the solutions obtained by previous authors. where A , B , C are constants. Substituting eq. (6)*Corresponding author- Email address :pinu7986@gmail.com into eq. (5) and using eq. (7), the left-hand side of eq. (5) can be converted into a finite series in2. Travelling wave solutions of BBM F p ( ) , ( p = -4, -3 ,-2, -1, 0, 1, 2, 3, 4).equation Equating each coefficient of F ( ) to zero yields pFor N  1, eq.(1) reduces to the BBM equation or the following set of algebraic equations.the regularized long-wave equation (RLW). 169 | P a g e
2. 2. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180 a2  F 4 ( ) : (12 A2 k  a2 )  2  F 3 ( ) : 2 A2 ka1  10 ABka2  a1a2   a  2  F 2 ( ) : Ak (3Ba1  8Ca2 )  1  a2 (1  4 B 2 k    a0 ) 2  F ( ) : a1 (1  B k  2 ACk    a0 )  a2 (6 BCk  a1 ) 1 2   a 2   F 0 ( ) : a0  BCka1  2C 2 ka2  ABka1  2 A2 ka2  a0  0  a1a1  a2 a2  (8) 2  F ( ) : a1 (1  B k  2 ACk    a0 )  a2 (6 ABk  a1 ) 1 2   a12  F ( ) : 3BCka1  4 B ka2  2 2  a1 (1  8 ACk    a0 ) 2   F ( ) : 2C ka1  10 BCka2  a1a2 3 2  1  F 4 ( ) : a2 (12C 2 k  a2 )  2  Solving the above algebraic equations by symbolic computations , we have the following solutions:Case 1: A  0 , we have 12BCk 12C 2 ka0  a0 , a1  0 , a2  0 , a1  , a2  ,   1  B 2 k  a0 (9)  Case 2: B  0 , we have 12C 2 k a0  a0 , a1  0, a2  0, a1  0, a2  ,   1  8 ACk  a0     (10) 12 A k 2 12C k2a0  a0 , a1  0, a2  , a1  0, a2  ,   1  8 ACk  a0     So we can list the solutions of u as follows:When A  0, B  1, C  1 , from appendix and case 1, we have 3k  u1 ( )  a0  sec h 2   (11)  2where   k ( x  (1  k  a0 ) t )When A  0, B  1, C  1 , from appendix and case 1, we have 3k  u2 ( )  a0  csc h 2   (12)  2where   k ( x  (1  k  a0 ) t ) 1 1When A  , B  0, C   , from appendix and case 2, we have 2 2 3k u3 ( )  a0  coth( )  csc h( ) 2 (13)  3k 3k coth( )  csc h( ) coth( )  csc h( )  2u4 ( )  a0   2 (14)   3ku5 ( )  a0   tanh( )  i sec h( ) 2 (15)  170 | P a g e
3. 3. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180 3k 3k  tanh( )  i sec h( )  tanh( )  i sec h( ) 2u6 ( )  a0   2 (16)  where   k  x  1  2k  a0  t When A  1, B  0, C  1 , from appendix and case 2, we have 12ku7 ( )  a0  tanh 2   (17)  12k 12ku8 ( )  a0  tanh 2    tanh 2   (18)   12ku9 ( )  a0  coth 2   (19)  12k 12ku10 ( )  a0  coth 2    coth 2   (20)   where   k ( x  (1  8k  a0 ) t ) 1 1When A  , B  0, C  , from appendix and case 2, we have 2 2 3k u11    a0  sec    tan   2    (21) 3k 2 3k u12    a0  sec    tan    sec    tan   2       (22) 3k u13    a0  csc    cot    2    (23) 3k 2 3ku14    a0  csc    cot    csc    cot   2       (24)where   k  x  (1  2k  a0 ) t  1 1When A   , B  0, C   , from appendix and case 2, we have 2 2 3ku15    a0  sec    tan   2    (25) 3k 2 3ku16    a0  sec    tan     sec    tan   2       (26) 3ku17    a0  csc    cot   2    (27) 3k 2 3ku18    a0  csc    cot     csc    cot   2       (28)where   k  x  (1  2k  a0 ) t When A  1  1 , B  0, C  1 1 , from appendix and case 2, we have 12ku19    a0  tan 2   (29)  12k 12ku20    a0  tan 2    tan 2   (30)   12ku21    a0  cot 2   (31)  171 | P a g e
4. 4. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180 12k 12ku22    a0  cot 2    cot 2   (32)  where   k  x  1  8k  a0  t 2.1 Discussion shape and propagates at constant speed. It is a Travelling wave solutions from eq.(11) to stable solution. The first observation of this kind ofeq.(20) represents the soliton solutions and from wave was made in 1834 by John Scott Russell [5].eq. (21) to eq.(32) represents the periodic If we draw the solution with a constant speed, wesolutions. will get the following figure (A.1) which shows theWe discuss one travelling wave solution given by right moving wave at different values of t .eq. (11). The travelling wave solution (11) is asolitary like solution which does not change its 4 t=0 t=1 t=2 3.5 t=3 3 2.5 u 2 1.5 1 -10 -8 -6 -4 -2 0 2 4 6 8 10 x Fig. A.1. Plot of travelling wave solution eq. (11) with k    a0  1 and t  0,1, 2,3 If we draw the travelling wave solution spread out. If we take negative values of  ((11) with different positive values of  (   1,  1.5,  2 ), we will get the following  1,1.5, 2 ), we will get the following figure figure (A.2-II) which shows the value of negative(A.2-I) which shows the value of positive amplitude. Here negative amplitude means theamplitude in decreasing. In other words the value wave is going to look upside down. From figuresof positive amplitude tends to zero. From this (A.2-I) and (A.2-II) we can say that positive andphysically we can say that the fluid becomes more negative amplitudes are the same. 4 1 alpha=1 alpha=1.5 alpha= -1 alpha=2 alpha= -1.5 3.5 0.5 alpha= -2 3 0 2.5 -0.5 u u 2 -1 1.5 -1.5 1 -2 -10 -8 -6 -4 -2 0 2 4 6 8 10 -10 -8 -6 -4 -2 0 2 4 6 8 10 x x (I) (II) Fig. A.2. Different solitary waves profiles given by eq. (11) at t  0 , a0  k  1 for (I)   1,1.5,2 and (II)   1,  1.5,  2 172 | P a g e
5. 5. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180If we draw the travelling wave solution (11) with amplitude in increasing. The amplitude isdifferent values of k ( k  1, 2,3 ), we will get the proportional to the velocity, which means thatfollowing figure (A.3) which describes the value of larger pulse travel faster than smaller ones. 10 k=1 9 k=2 k=3 8 7 6 u 5 4 3 2 1 -10 -8 -6 -4 -2 0 2 4 6 8 10 xFig. A.3. Different solitary waves profiles given by eq. (11) at t  1 , a0    1 for k  1, 2,3The other solitary and periodic solutions are graphically shown in fig.(A.4) to fig.(A.10).Solitary solutions Fig.A.4 Fig.A.5Fig.A.4. Travelling wave solution given by eq. (17) with values a0    k  1 and Fig.A.5. Travelling wavesolution given by eq.(14) with values a0  2 and   k 1 Fig.A.6Fig.A.6 Travelling wave solution given by absolute of eq. (15) with values a0  2 and   k  1 173 | P a g e
6. 6. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180Trigonometric solutions : Fig.A.7 Fig.A.8Fig.A.7. Travelling wave solution given by eq. (23) and Fig.A.8 travelling wave solution given by eq.(25) withvalues a0    k  1 Fig.A.9 Fig.A.10Fig.A.9. Travelling wave solution given by eq. (29) and Fig.A.10. Travelling wave solution given by eq.(31)with values a0    k  13. Travelling wave solutions of modified BBM equationFor N  2 , eq.(1) reduces to the modified BBM equation.ut  ux   u 2 ux  ux x x  0 (33)Now, we construct the travelling wave solutions of eq. (33), under the transformation v  x, t   v   where   k  x   t  (34)where k  0 and  are constants. Substituting eq. (34) into the eq. (33), we get k  v  k v   k v 2 v  k 3 v  0 (35)Integrating eq. (35) with respect to „  ‟ and considering the zero constant for integration we have 1    v  v3  k 2v  0 (36) 3where prime denotes differentiation with respect to  . By balancing the order of v and v in eq. (36), we have 3 3 m  m  2 then m  1. So we can write v    a0  a1 F 1    a1 F   (37)where a 0 , a1 , a1 are constants to be determined later. And F ( ) satisfies Riccati equation given by eq. (7).Substituting eq. (37) into eq. (36) and using eq. (7), the left-hand side of eq. (36) can be converted into a finite 174 | P a g e
7. 7. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180series in F ( ) , ( p = -3 ,-2,-1, 0, 1, 2, 3). Equating each coefficient of F ( ) to zero yields the following p pset of algebraic equations. a13 F 3   : 2 A2 k 2 a1   3 F   : 3 ABk a1  a0 a12 2 2  F 1   : a1  B 2 k 2 a1  2 ACk 2 a1  a1  a0 2 a1  a12 a1   a03 F   : a0  BCk a1  ABk a1  a0  0 2 2  2a0 a1a1  (38) 3 F 1   : a1  B 2 k 2 a1  2 ACk 2 a1  a1  a0 2 a1  a1a12  F 2   : 3BCk 2 a1  a0 a12   a1  3 F   : 2C k a1  3 2 2 3  Solving the algebraic equations above by symbolic computations, we have the following solutions:Case 3: A  0 , we have 3 6 B2k 2a0   i B k , a1  0 , a1   i C k ,   1  (39) 2  2Case 4: B  0 , we have 6 a0  0, a1  0, a1   i C k ,   1  2 ACk 2     (40) 6 6 2a0  0, a1   i A k , a  iC k ,   1  4 ACk   1  So we can list the solutions of v as follows:When A  0, B  1, C  1 , from appendix and case 3, we have 3  v1 ( )   i k tanh   (41) 2 2   k2  where   k  x   1  t    2 When A  0, B  1, C  1 , from appendix and case 3, we have 3  v2 ( )   i k coth   (42) 2 2   k2  where   k  x   1  t    2  1 1When A  , B  0, C   , from appendix and case 4, we have 2 2 ik 6 v3 ( )    coth( )  csc h( )  (43) 2  ik 6v4 ( )    tanh    i sec h    2    (44) 175 | P a g e
8. 8. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180   k2  where   k  x  1  t    2   ik 6 ik 6 coth( )  csc h( )  coth( )  csc h( )  1v5 ( )   (45) 2  2  ik 6 1 ik 6v6 ( )    tanh    i sec h        tanh    i sec h    46) 2  2   where    k x  1  k 2  t When A  1, B  0, C  1 , from appendix and case 4, we have 6v7 ( )   i k tanh   (47)  6v8 ( )   i k coth   (48) where    k x  1  2 k 2  t  6 6v9 ( )  i k tanh 1    i k tanh   (49)   6 6v10 ( )   i k coth 1    i k coth   (50)   where    k x  1  4 k 2  t  1 1When A  , B  0, C  , from appendix and case 4, we have 2 2 ik 6 v11     sec    tan    2    (51) ik 6 v12      csc    cot    2    (52)   1 2 where   k  x  1  k  t    2   ik 6 1 ik 6 v13     sec    tan       sec    tan    2    (53) 2  ik 6 1 ik 6v14      csc    cot        csc    cot    2    (54) 2 where    k x   1  k 2  t  1 1When A   , B  0, C   , from appendix and case 4, we have 2 2 ik 6v15     sec    tan    2    (55) ik 6v16      csc    cot    2    (56) 176 | P a g e
9. 9. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180   k2  where   k  x  1  t    2  ik 6 1 ik 6v17     sec    tan       sec    tan    2    (57) 2  ik 6 1 ik 6v18      csc    cot        csc    cot    2    (58) 2 where    k x   1  k 2  t When A  1  1 , B  0, C  1 1 , from appendix and case 4, we have 6v19        i k tan   (59)  6v20        i k cot   (60) where    k x  1  2 k 2  t  6 6v21        i k tan 1       i k tan   (61)   6 6v22        i k cot 1       i k cot   (62)  where    k x   1  4 k 2  t The travelling wave solutions given by eq.(41) to eq.(50) represents soliton-like solutions and eq.(51) to eq.(62)represents periodic solutions. Some absolute of solitary and periodic solutions are shown in fig.(B.1) tofig.(B.6).Solitary solutions: Fig.B.1 Fig.B.2 177 | P a g e
10. 10. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180 Fig.B.3Fig.B.1 Travelling wave solution given by eq.(41), Fig.B.2 Travelling wave solution given by eq.(42) andFig.B.3 Travelling wave solution given by eq.(47) with values   k  1Trigonometric solutions: Fig.B.4 Fig.B.5 Fig.B.6Fig.B.4 Travelling wave solution given by eq. (51), Fig.B.5 Travelling solution given by eq.(56) and Fig.B.6Travelling wave solution given by eq.(60) with values   k  14. Conclusion shown that this method is direct, concise, effective In this paper, we have used a modified F- and can be applied to other NLPDEs in theexpansion method to construct more exact mathematical physics. All the obtained solutionssolutions of the nonlinear BBM and modified BBM satisfies BBM and modified BBM equations.equations with the aid of Mathematica. We have 178 | P a g e
11. 11. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180Appendix ARelations between values of ( A, B, C ) and corresponding F   in Riccati equationF    A  B F    C F 2  A B C F   1 1  0 1 1  tanh   2 2 2 1 1  0 1 1  coth   2 2 21 1 0  coth    csch   , tanh    i sec h  2 21 0 1 tanh   , coth  1 1 0 sec    tan   , csc    cot  2 2 1 1 0  sec    tan   , csc    cot   2 21 1 0 1 1 tan   , cot   10 0 0  ( m is an arbitrary constant) C  marbitraryconstant 0 0 Aarbitrary exp  B   Aconstant 0 0 BReferences [1]. A.I.Volpert, Vitaly A. Volpert, Vladimir [5]. J-H. He, “Soliton perturbation”,Selected A. Volpert, “Travelling wave solutions of works of J-H.He, (2009), pp. 8453-8457. parabolic systems”, American [6]. M.D.Abdur Rab, A.S.Mia, T.Akter, Mathematical Society, (1993). “Some travelling wave solutions of Kdv- [2]. A.Wazwaz, “New travelling wave Burger equation”, International Journal of solutions of different physical structures to Mathematical Analysis, (2012), pp.1053- generalized BBM equation”, Elsevier, 1060. (2006), pp.358-362. [7]. M.T.Alquran, “Applying differential [3]. G.Cai, Q.Wang, “A modified F-expansion transform method to nonlinear partial method for solving nonlinear PDEs”, differential equations: A modified Journal of Information and Computing approach”, Applications and Applied Science”, (2007), pp.03-16. Mathematics:An International Journal, [4]. J.Manafianheris, “Exact solutions of the (2012), pp.155-163. BBM and MBBM equations by the [8]. M.T.Alquran, “Solitons and periodic generalized (G‟/G)-expansion method”, solutions to nonlinear partial differential International Journal of Genetic equations by the Sine-Cosine method”, Engineering, (2012), pp.28-32. Applied Mathematics & Information 179 | P a g e
12. 12. Priyanka M. Patel, V. H. Pradhan / International Journal of Engineering Research and Applications (IJERA) ISSN: 2248-9622 www.ijera.com Vol. 3, Issue 1, January -February 2013, pp.169-180 Sciences: An International Journal, [12]. N.Taghizadeh, M.Mirzazadeh, “The (2012), pp.85-88. modified extended tanh method with the[9]. M.T.Darvishi, Maliheh Najafi, Riccati equation for solving nonlinear Mohammad Najafi, “Travelling wave partial differential equations”, solutions for the (3+1)-dimensional Mathematica Aeterna, vol.2, (2012), breaking soliton equation by (G‟/G)- pp.145-153. expansion method and modified F- [13]. P.G.Estevez, S.Kuru, J.Negro and expansion method”, International Journal L.M.Nieto, “Travelling wave solutions of of Computational and Mathematical the generalized Benjamin-Bona-Mahony Sciences 6:2, (2012), pp.64-69. equation”, Cornell University Library,[10]. M.T.Darvishi, Maliheh Najafi, Mohammad Najafi, “Travelling wave (2007), pp.1-19. solutions for foam drainage equation by [14]. S.Abbasbandy, “Homotopy analysis modified F-expansion method”, Food and method for generalized Benjamin-Bona- Public health, (2012), pp. 6-10. Mahony equation”, ZAMP, (2008), pp.51-[11]. M.Wadati, “Introduction to solitons”, 62. Pramana Journal of Physics, (2001), pp.841-847. 180 | P a g e