The tanh–coth method combined with the Riccati equation for solving non-linear equation

Chaos, Solitons and Fractals 40 (2009) 1467–1474 www.elsevier.com/locate/chaos

The tanh–coth method combined with the Riccati equation for solving non-linear equation Ahmet Bekir Dumlupinar University, Art-Science Faculty, Department of Mathematics, Ku¨tahya, Turkey Accepted 10 September 2007

Abstract In this work, we established abundant travelling wave solutions for some non-linear evolution equations. This method was used to construct solitons and traveling wave solutions of non-linear evolution equations. The tanh–coth method combined with Riccati equation presents a wider applicability for handling non-linear wave equations. Ó 2007 Elsevier Ltd. All rights reserved.

1. Introduction The investigation of the travelling wave solutions for non-linear partial differential equations plays an important role in the study of non-linear physical phenomena. Non-linear wave phenomena appears in various scientific and engineering fields, such as fluid mechanics, plasma physics, optical fibers, biology, solid state physics, chemical kinematics, chemical physics and geochemistry. Non-linear wave phenomena of dispersion, dissipation, diffusion, reaction and convection are very important in non-linear wave equations. In recent years, new exact solutions may help to find new phenomena. A variety of powerful methods, such as inverse scattering method [1,11], bilinear transformation [8], the tanh– sech method [9,10,13], extended tanh method [2,5,17], sine–cosine method [16,20,21], homogeneous balance method [4], Exp-function method [6,7] and improved tanh-function [3] method were used to develop non-linear dispersive and dissipative problems. The pioneer work Malfiet [9,10] introduced the powerful tanh method for a reliable treatment of the non-linear wave equations. The useful tanh method is widely used by many such as in [13–15] and by the references therein. Later, the tanh–coth method, developed by Wazwaz [17], is a direct and effective algebraic method for handling non-linear equations. Various extensions of the method were developed as well. Our first interest in the present work is in implementing the tanh–coth method combined with Riccati equation method to stress its power in handling non-linear equations, so that one can apply it to models of various types of non-linearity. The next interest is in the determination of exact travelling wave solutions for Benjamin–Bona–Mahony (BBM) equation. Searching for exact solutions of non-linear problems has attracted a considerable amount of research work where computer symbolic systems facilitate the computational work.

E-mail address: [email protected] 0960-0779/$ - see front matter Ó 2007 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2007.09.029

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2. The tanh–coth method Wazwaz has summarized for using tanh–coth method. A PDE P ðu; ut ; ux ; uxx ; utt ; uxxx ; . . .Þ ¼ 0

ð2:1Þ

can be converted to on ODE QðU ; U 0 ; U 00 ; U 000 ; . . .Þ ¼ 0

ð2:2Þ

upon using a wave variable n ¼ x  bt. Eq. (2.2) is then integrated as long as all terms contain derivatives where integration constants are considered zeros. Introducing a new independent variable Y ¼ tanhðnÞ or Y ¼ cothðnÞ

n ¼ x  bt;

ð2:3Þ

leads to change of derivatives: d d ¼ ð1  Y 2 Þ ; dn dY   d2 d d2 ¼ ð1  Y 2 Þ 2Y þ ð1  Y 2 Þ 2 ; 2 dY dY dn  3 2 3  d d 2 2 2 d 2 2 d : ¼ ð1  Y Þ ð6Y  2Þ Þ þ ð1  Y Þ  6Y ð1  Y dY dY 2 dY 3 dn3

ð2:4Þ

The tanh–coth method [17,18] admits the use of the finite expansion U ðnÞ ¼ SðY Þ ¼

m X

ak Y k þ

k¼0

m X

bk Y k ;

ð2:5Þ

k¼1

where m is a positive integer, for this method, that will be determined. Expansion (2.5) reduces to the standard tanh method [9] for bk ¼ 0; 1 6 k 6 m. The parameter m is usually obtained, as stated before, by balancing the linear terms of the highest order in the resulting equation with the highest order non-linear terms. If m is not an integer, then a transformation formula should be used to overcome this difficulty. Substituting (2.5) into the ODE results is an algebraic system of equations in powers of Y that will lead to the determination of the parameters ak ðk ¼ 0; . . . ; mÞ, bk ðk ¼ 1; . . . ; mÞ and b. The function Y satisfies the Riccati equation Y 0 ¼ A þ BY þ CY 2 ;

ð2:6Þ

where A, B and C are constants [19], and Y0 ¼

dY ðnÞ ; dn

n ¼ x  bt:

ð2:7Þ

3. The Riccati equation method The Riccati equation Y 0 ¼ A þ BY þ CY 2 ;

ð3:1Þ

has specific solutions for B ¼ 0 given in [12] by 1 A¼ ; 2 1 A¼ ; 2 A ¼ 1; A ¼ 1; A ¼ 1; A ¼ 1;

1 n n C ¼  ; Y 1 ¼ tanh ; coth ; 2 2 2 1 n n C ¼ ; Y 2 ¼ tan n  sec n; tan ;  cot ; 2 2 2 C ¼ 1; Y 4 ¼ tanh n; coth n; C ¼ 1; Y 5 ¼ tan n;  cot n; 1 1 C ¼ 4; Y 7 ¼ tanh 2n; coth 2n; 2 2 1 1 C ¼ 4; Y 8 ¼ tan 2n;  cot 2n: 2 2

ð3:2Þ

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Other values for Y can be derived for other arbitrary values for A and C. To show the efficiency of the method described in the previous part, we present some example.

4. Benjamin–Bona–Mahony equation Benjamin–Bona–Mahony equation (BBM) ut þ ux þ uux  uxxt ¼ 0:

ð4:1Þ

Using the wave variable n ¼ x  bt, the system (4.1) is carried to a system of ODEs ð1  bÞu0 þ uu0 þ bu000 ¼ 0;

ð4:2Þ

by once time integrating we find ð1  bÞu þ

u2 þ bu00 ¼ 0: 2

ð4:3Þ

Balancing u00 with u2 in (4.3) gives m þ 2 ¼ 2m;

ð4:4Þ

so that m ¼ 2:

ð4:5Þ

The tanh–coth method admits the use of the finite expansion U ðnÞ ¼ SðY Þ ¼ a0 þ a1 Y 2 þ

b1 : Y2

ð4:6Þ

Substituting Eq. (4.6) in (4.3), collecting the coefficients of Y i ði ¼ 0; . . . ; 8Þ and set it to zero we obtain the system 12ba1 C 2 þ a21 ¼ 0; 10ba1 BC ¼ 0; 4ba1 B2 þ a0 a1 þ a1 þ 8ba1 AC  ba1 ¼ 0; 6ba1 AB ¼ 0; 2a1 b1 þ a20  2ba0 þ 2a0 þ 4bb1 C 2 þ 4ba1 A2 ¼ 0; 6bb1 BC ¼ 0;

ð4:7Þ

4bb1 B2 þ b1 þ a0 b1  bb1 þ 8bb1 AC ¼ 0; 10bb1 AB ¼ 0; b21 þ 12bb1 A2 ¼ 0: Solving this system by Maple gives B ¼ 0 and the following six sets of solutions: (i) The first set: a0 ¼

4AC ; 4AC  1

a1 ¼ 0;

b1 ¼

12A2 ; 1  4AC

b¼

1 : 1  4AC

ð4:8Þ

a1 ¼ 0;

b1 ¼

12A2 ; 1 þ 4AC

b¼

1 : 1 þ 4AC

ð4:9Þ

a1 ¼

12C 2 ; 1  4AC

b1 ¼ 0;

b¼

1 : 1  4AC

ð4:10Þ

a1 ¼

12C 2 ; 1 þ 4AC

b1 ¼ 0;

b¼

1 : 1 þ 4AC

ð4:11Þ

(ii) The second set: a0 ¼

12AC ; 4AC þ 1

(iii) The third set: a0 ¼

4AC ; 4AC  1

(iv) The fourth set: a0 ¼

12AC ; 4AC þ 1

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(v) The fifth set: a0 ¼

8AC ; 1  16AC

a1 ¼

12C 2 ; 1  16AC

b1 ¼

12A2 ; 1  16AC

b¼

1 : 1  16AC

ð4:12Þ

a1 ¼

12C 2 ; 1 þ 16AC

b1 ¼

12A2 ; 1 þ 16AC

b¼

1 : 1 þ 16AC

ð4:13Þ

(vi) The sixth set: a0 ¼

24AC ; 16AC þ 1

This in turn gives the following general set of solutions 4AC 12A2 1  ; Y 2 ðnÞ; b ¼ 4AC  1 1  4AC 1  4AC 12AC 12A2 1 Y 2 ðnÞ; b ¼ uII ¼  ; 4AC þ 1 1 þ 4AC 1 þ 4AC 2 4AC 12C 1 Y 2 ðnÞ; b ¼ uIII ¼  ; 4AC  1 1  4AC 1  4AC 2 12AC 12C 1 Y 2 ðnÞ; b ¼  ; uIV ¼ 4AC þ 1 1 þ 4AC 1 þ 4AC 2 2 8AC 12C 12A 1 Y 2 ðnÞ  Y 2 ðnÞ; b ¼ uV ¼  ; 1  16AC 1  16AC 1  16AC 1  16AC 24AC 12C 2 12A2 1 Y 2 ðnÞ  Y 2 ðnÞ; b ¼  ; uVI ¼ 1 þ 16AC 16AC þ 1 1 þ 16AC 1 þ 16AC

uI ¼

ð4:14Þ ð4:15Þ ð4:16Þ ð4:17Þ ð4:18Þ ð4:19Þ

where A and C are arbitrary constants and Y takes many trigonometric and hyperbolic functions as shown in (3.2). Case I: We first consider uI ðx; tÞ. We use the first result of (4.8). We then apply the related Y functions for this choice of A and C. Using the first case in (3.2) where A ¼ 12 and C ¼  12 gives the solution    1 n u1 ¼ 1 þ 3coth2 ; ð4:20Þ 2 2 and soliton solution    1 n u2 ¼ 1 þ 3tanh2 ; 2 2

ð4:21Þ

where b ¼ 12. For A ¼ 1 and C ¼ 1 we find b ¼ 15, and we therefore obtain the solution 4 u3 ¼ ½1 þ 3coth2 ðnÞ; 5

ð4:22Þ

and the soliton solution 4 u4 ¼ ½1 þ 3tanh2 ðnÞ: 5

ð4:23Þ

For A ¼ 1 and C ¼ 1 we find b ¼  13, and we therefore obtain the solutions 4 u5 ¼ ½1 þ 3cot2 ðnÞ; 3 4 u6 ¼ ½1 þ 3 tan2 ðnÞ: 3

ð4:24Þ ð4:25Þ

For A ¼ 1 and C ¼ 4 we find b ¼ 171 , and we therefore obtain the solution u7 ¼

16 ½1 þ 3coth2 ð2nÞ; 17

ð4:26Þ

and the soliton solution u8 ¼

16 ½1 þ 3tanh2 ð2nÞ: 17

ð4:27Þ

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1471

For A ¼ 1 and C ¼ 4 we find b ¼  151 , and we therefore obtain the solutions 16 ½1 þ 3cot2 ð2nÞ; 15 16 u10 ¼ ½1 þ 3 tan2 ð2nÞ: 15

u9 ¼

ð4:28Þ ð4:29Þ

Case II: We first consider uII ðx; tÞ. We use the second result of (4.9). Using the first case in (3.2) where A ¼ 12 and C ¼ 12 gives the solutions    3 n u11 ¼ 1  cot2 ; ð4:30Þ 2 2    3 n u12 ¼ 1  tan2 ; ð4:31Þ 2 2 " # 3 1 ; ð4:32Þ u13 ¼ 1 2 ðtan n  sec nÞ2 where b ¼ 12. For A ¼ 1 and C ¼ 1 we find b ¼  13, and we therefore obtain the solution u14 ¼ 4½1 þ coth2 ðnÞ:

ð4:33Þ

and the soliton solution u15 ¼ 4½1 þ tanh2 ðnÞ: For A ¼ 1 and C ¼ 1 we find b ¼

ð4:34Þ 1 , 5

and we therefore obtain the solutions

12 u16 ¼ ½1  cot2 ðnÞ; 5 12 u17 ¼ ½1  tan2 ðnÞ: 5

ð4:35Þ ð4:36Þ

For A ¼ 1 and C ¼ 4 we find b ¼  151 , and we therefore obtain the solution u18 ¼

48 ½1 þ coth2 ð2nÞ; 15

ð4:37Þ

and the soliton solution u19 ¼

48 ½1 þ tanh2 ð2nÞ: 15

ð4:38Þ

For A ¼ 1 and C ¼ 4 we find b ¼ 171 , and we therefore obtain the solutions 48 ½1  cot2 ð2nÞ; 17 48 u21 ¼ ½1  tan2 ð2nÞ: 17

u20 ¼

ð4:39Þ ð4:40Þ

Case III: We next consider uIII ðx; tÞ. We use the third result of (4.10). Using the first case in (3.2) where A ¼ 12 and C ¼  12 gives the soliton solution    1 n u22 ¼ 1  3tanh2 ; ð4:41Þ 2 2 and the solution    1 n u23 ¼ 1  3coth2 ; 2 2

ð4:42Þ

where b ¼ 12. For A ¼ 1 and C ¼ 1 we find b ¼ 15, and we therefore obtain the soliton solution 4 u24 ¼ sech2 ðnÞ: 5

ð4:43Þ

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and the travelling wave solution 4 u25 ¼  csch2 ðnÞ: 5

ð4:44Þ

For A ¼ 1 and C ¼ 1 we find b ¼  13, and we therefore obtain the soliton solution u26 ¼

4 sec2 ðnÞ; 3

ð4:45Þ

and the travelling wave solution 4 u27 ¼ csc2 ðnÞ: 3

ð4:46Þ

For A ¼ 1 and C ¼ 4 we find b ¼ 171 , and we therefore obtain the soliton solution u28 ¼

16 ½1  3tanh2 ð2nÞ; 17

ð4:47Þ

and the solution u29 ¼

16 ½1  3coth2 ð2nÞ: 17

ð4:48Þ

For A ¼ 1 and C ¼ 4 we find b ¼  151 , and we therefore obtain the solutions u9 and u10 . Case VI: We next consider uVI ðx; tÞ. We use the fourth result of (4.11). Using the first case in (3.2) where A ¼ 12 and C ¼ 12 gives the solutions u11 and u12 and the solution 3 u30 ¼ ½1  ðtan n  sec nÞ2 Þ; 2

ð4:49Þ

where b ¼ 12. For A ¼ 1 and C ¼ 1 we find b ¼  13, and we therefore obtain the solutions u14 and u15 . For A ¼ 1 and C ¼ 1 we find b ¼ 15, and we therefore obtain the solutions u16 and u17 . For A ¼ 1 and C ¼ 4 we find b ¼  151 , and we therefore obtain the solutions u18 and u19 . For A ¼ 1 and C ¼ 4 we find b ¼ 171 , and we therefore obtain the solutions u20 and u21 . Case V: We next consider uV ðx; tÞ. We use the fifth result of (4.12). Using the first case in (3.2) where A ¼ 12 and C ¼  12 gives the solution      1 n n u31 ¼  4 þ 3tanh2 þ 3coth2 ; ð4:50Þ 5 2 2 where b ¼ 15. For A ¼ 12 and C ¼ 12 we find b ¼  13, and we therefore obtain the solutions      4 n n u32 ¼  1  3 tan2  3cot2 ; 3 2 2 4 u33 ¼  ½1  3ðtan n  sec nÞ2  3ðtan n  sec nÞ2 : 3

ð4:51Þ ð4:52Þ

For A ¼ 1 and C ¼ 1 we find b ¼ 171 , and we therefore obtain the solution u34 ¼ 

4 ½2 þ 3tanh2 ðnÞ þ 3coth2 ðnÞ: 17

ð4:53Þ

For A ¼ 1 and C ¼ 1 we find b ¼  151 , and we therefore obtain the solution u35 ¼ 

4 ½2  3 tan2 ðnÞ  3cot2 ðnÞ: 15

ð4:54Þ

For A ¼ 1 and C ¼ 4 we find b ¼ 651 , and we therefore obtain the solution u36 ¼ 

1 ½32 þ 48tanh2 ð2nÞ þ 3coth2 ð2nÞ; 65

For A ¼ 1 and C ¼ 4 we find b ¼  631 , and we therefore obtain the solution 1 u37 ¼  ½32  48 tan2 ð2nÞ  3cot2 ð2nÞ; 63

ð4:55Þ

ð4:56Þ

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Case VI: We next consider uVI ðx; tÞ. We use the sixth result of (4.13). Using the first case in (3.2) where A ¼ 12 and C ¼  12 gives the solution     n n u38 ¼ 2 þ tanh2 þ coth2 ; ð4:57Þ 2 2 where b ¼  13. For A ¼ 12 and C ¼ 12 we find b ¼ 15, and we therefore obtain the solutions      3 n n u39 ¼  2 þ tan2 þ cot2 ; 5 2 2 3 2 2 u40 ¼  ½2 þ ðtan n  sec nÞ þ ðtan n  sec nÞ : 5

ð4:58Þ ð4:59Þ

For A ¼ 1 and C ¼ 1 we find b ¼  151 , and we therefore obtain the solution u41 ¼ 

12 ½2  tanh2 ðnÞ  coth2 ðnÞ: 15

ð4:60Þ

For A ¼ 1 and C ¼ 1 we find b ¼ 171 , and we therefore obtain the solution u42 ¼ 

12 ½2 þ tan2 ðnÞ þ cot2 ðnÞ: 17

ð4:61Þ

For A ¼ 1 and C ¼ 4 we find b ¼  631 , and we therefore obtain the solution u43 ¼ 

12 ½8  4tanh2 ð2nÞ  coth2 ð2nÞ; 63

ð4:62Þ

For A ¼ 1 and C ¼ 4 we find b ¼ 651 , and we therefore obtain the solution u44 ¼ 

3 ½32 þ 4 tan2 ð2nÞ  cot2 ð2nÞ; 65

ð4:63Þ

5. Conclusion The tanh–coth method combined with Riccati equation was successfully used to establish solitary wave solutions. Many well known non-linear wave equations were handled by this method. The performance of the this method is reliable and effective and gives more solutions. The applied method will be used in further works to establish more entirely new solutions for other kinds of non-linear wave equations. The availability of computer systems like Mathematica or Maple facilitates the tedious algebraic calculations. The method which we have proposed in this letter is also a standard, direct and computerizable method, which allows us to solve complicated and tedious algebraic calculation.

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