The Generalized Kudryashov Method and Its ...

Scientific Journal of Mathematics Research

Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

The Generalized Kudryashov Method and Its Applications for Solving Nonlinear PDEs in Mathematical Physics E. M. E. Zayed1 , G. M. Moatimid2 , Abdul-Ghani Al-Nowehy3 1 2 3

Department of Mathematics, Faculty of Sciences, Zagazig University, Zagazig, Egypt

Department of Mathematics, Faculty of Education, Ain Shams University,Roxy, Hiliopolis, Cairo, Egypt

Department of Mathematics, Faculty of Education and science, Taiz University, Taiz, Yemen 1 3 [email protected]; [email protected]

Abstract- In this paper, we use the generalized Kudryashov method for finding exact solutions of nonlinear partial differential equations (PDEs) in mathematical physics. To illustrate the validity of this method, the generalized Kudryashov method is applied to four nonlinear PDEs namely, the Burgers equation, the modified Benjamin-Bona-Mahony (mBBM) equation, the Potential Kadomtsev-Petviashvili (PKP) equation and the Cahn-Hilliard equation. As a result, many analytical exact solutions are obtained including symmetrical Fibonacci function solutions, hyperbolic function solutions, and rational solutions. Physical explanations for the solutions of the given four nonlinear PDEs are obtained. Keywords- Nonlinear PDEs; Generalized Kudryashov Method; Symmetrical Hyperbolic Fibonacci Function; Exact solutions

I. INTRODUCTION The interest of studying the physical problems is shifted to find the exact solution of the corresponding nonlinear PDE. These equations are mathematical models of complex physical phenomena that arised in engineering, applied mathematics, chemistry, biology, mechanics, physics, etc. Thus, it is very important to reveal exact solutions of nonlinear partial differential equations. Calculating the exact and numerical solutions, in particular traveling wave solutions of nonlinear equations in mathematical physics play an important role in soliton theory [1,2]. The exact solution of a differential equation gives information about the construction of the complex physical phenomena. Therefore, researchers carried out a huge amount of research work to explore the exact traveling wave solutions of nonlinear physical phenomena. Accordingly, they developed many potential and useful methods and techniques, such as the homogeneous balance method [3], the Hirota's bilinear transformation method [4,5], the tanh-function method [6,7], the (G'/G)-expansion method [8-13], the Exp-function method [14-16], the multiple exp-function method [17,18], the symmetry method [19,20] ,the modified simple equation method [21-23], the improved (G'/G)-expansion method [24], a multiple extended trial equation method [25], the Jacobi elliptic function expansion method [26], the Bäcklund transform [27,28], the generalized Riccati equation meyhod [29], the modified extended Fan sub equation method [30], the auxiliary equation method [31,32], the first integral method [33,34], the modified Kudryashov method [35-41] and so on. The objective of this paper is to construct the exact solutions of Burgers equation [42], the modified Benjamin-BonaMahony (mBBM) equation [9,43,44],the Potential Kadomtsev-Petviashvili (PKP) equation [24,44] and the Cahn-Hilliard equation [45-47], by using the generalized Kudryashov method. II. DESCRIPTION OF THE GENERALIZED KUDRYASHOV METHOD Suppose that a nonlinear PDE has the following from: 𝐹 𝑢, 𝑢𝑡 , 𝑢𝑥 , 𝑢𝑡𝑡 , 𝑢𝑥𝑡 , 𝑢𝑥𝑥 , … = 0,

(2.1)

where 𝑢 = 𝑢(𝑥, 𝑡) is an unknown function, 𝐹 is a polynomial in 𝑢 = 𝑢(𝑥, 𝑡) and its partial derivatives, in which the highest order derivatives and nonlinear terms are involved. The main steps of the generalized Kudryashov method are the following: Step 1. First of all, we use the wave transformation:

u( x, t )  U ( ),

  kx  t ,

(2.2)

where 𝑘 and 𝜆 are arbitrary constants with 𝑘, 𝜆 ≠ 0, to reduced Eq. (2.1) into a nonlinear ordinary differential equation (ODE) with respect to the variable  of the form

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Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

H (U ,U ,U ,U ,...)  0, where

(2.3)

H is a polynomial in U ( ) and its total derivatives U , U , U ,... such that U  

dU d

, U  

d 2U d 2

and so on.

Step 2. We assume that the formal solution of the ODE (2.3) can be written in the following rational form:

U ( ) 

where

i in0 ai Q ( ) AQ( )  , j mj0 b j Q ( ) BQ( )

(2.4)

Q  11a , AQ( )  in0 aiQi ( ) and BQ( )  mj0 b j Q j ( ) . The function Q is the solution of the

equation

Q  QQ  1 ln(a),

0  a  1.

(2.5)

Taking into consideration (2.4), we obtain

 AB  AB  U ( )  QQ  1   ln( a), B2  QQ  1 U ( )  (2Q  1)( AB  AB) ln 2 (a) 2 B 2 Q 2 Q  1  B( AB  AB)  2 ABB  2 A( B) 2 ln 2 (a), B3



(2.6)



(2.7)

U ( )  Q 3 Q  1 ln 3 (a) 3

 ( AB  AB  3 AB  3 AB) B  6 B( AB  AB) 6 A( B) 3    B3 B 4    ( AB  AB  2 AB) B  2 A( B) 2  3 2 3Q 2 Q  1 (2Q  1)   ln (a) B3    AB  AB  3 QQ  1(6Q 2  6Q  1)   ln (a), B2  (2.8) and similar for higher order differentiation terms. Step 3. Under the terms of the given method, we suppose that the solution of Eq. (2.3) can be written in the following form:

a0  a1Q  a2Q 2 ··· anQ n U ( )  . b0  b1Q  b2Q 2 ···bmQ m

(2.9)

To calculate the values m and n in (2.9) that is the pole order for the general solution of Eq. (2.3), we progress conformably as in the classical Kudryashov method on balancing the highest order nonlinear terms and the highest order derivatives of U ( ) in Eq. (2.3) and we can determine a formula of 𝑚 and 𝑛. We can receive some values of 𝑚 and 𝑛. Step 4. We substitute (2.4) into Eq. (2.3) to get a polynomial R( Q ) of Q and equate all the coefficients of to zero, to yield a system of algebraic equations for 𝑎𝑖 (𝑖 = 0,1, . . . , 𝑛) and 𝑏 𝑗 𝑗 = 0,1, . . . , 𝑚 .

Q i , (i=0,1,2,...)

Step 5. We solve the algebraic equations obtained in Step 4 by using Mathematica or Maple, to get 𝑘, 𝜆 and the coefficients of 𝑎𝑖 (𝑖 = 0,1, . . . , 𝑛) and 𝑏 𝑗 (𝑗 = 0,1, . . . , 𝑚). In this way, we attain the exact solutions to Eq. (2.3).

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The obtained solutions depend on the symmetrical hyperbolic Fibonacci functions given [48]. The symmetrical Fibonacci sine, cosine, tangent, and cotangent functions are, respectively, defined as:

a x  ax , 5 a x  ax tan Fs ( x)  x , a  ax

a x  ax , 5 a x  ax cot Fs ( x)  x , a  ax

(2.10)

2 ch( x. ln( a)), 5 cot Fs ( x)  coth( x. ln( a)).

(2.11)

sFs ( x) 

cFs ( x) 

2 sh( x. ln( a)), 5 tan Fs ( x)  tanh( x. ln( a)), sFs( x) 

cFs ( x) 

III. APPLICATIONS In this section we construct the exact solutions in terms of the symmetrical hyperbolic Fibonacci functions of the following four nonlinear PDEs by using the generalized Kudryashov method described in Sec. 2 3.1

Example 1. The nonlinear Burgers Equation This equation is well-known [49] and can be written in the form:

ut  uxx  u Pux  0,

(3.1.1)

where  is a nonzero constant and P  1. Eq. (3.1.1), one type of partial differential equations, was first presented by Burgers in 1948 as a model for turbulent phenomena of viscous fluids [50]. The Burgers equation defines the far field of wave propagation in nonlinear dissipative systems. It is well known that this equation is linearizable to the heat equation by using the Cole-Hopf transform. This equation has been considered in a number of fields of implementation such as traffic flows and formation of large clusters in the universe. Let us now solve Eq. (3.1.1) by using the generalized Kudryashov method. To this end, we use the traveling wave transformation

u( x, t )  U ( ), where

  kx  t ,

(3.1.2)

k and  are real constants with k ,   0 , to reduce Eq. (3.1.1) to the following nonlinear ODE:  U   k 2U   kU PU   0.

Integrating Eq. (3.1.3) with respect to

 and vanishing the constant of integration, we get

 U  k 2U   k Balancing

(3.1.3)

U P1  0. P 1

(3.1.4)

U  and U P1 in (3.1.4), then the following relation is attained:

n  m  1  ( P  1)(n  m)  n  m 

1 . P

(3.1.5)

Since the balance number of Eq. (3.1.4) is not integer when 𝑃 > 1, then we take the transformation into consideration 1

U ( )  v P ( ).

(3.1.6)

Substituting (3.1.6) into Eq. (3.1.4) we have the new equation

P( P  1)v  k 2 ( P  1)v  kPv2  0. - 21 -

(3.1.7)


Balancing the

Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

v and v 2 in (3.1.7), then the following relation is obtained: n  m  1  2(n  m)  n  m  1.

(3.1.8)

If we choose 𝑚 = 1 and 𝑛 = 2, then the formal solution of Eq. (3.1.7) has the form:

a0  a1Q  a2Q 2 , b0  b1Q

(3.1.9)

 (a1  2a2Q)(b0  b1Q)  b1 (a0  a1Q  a2Q 2 )   v ( )  QQ  1   ln( a). (b0  b1Q) 2  

(3.1.10)

v( )  and consequently,

Substituting (3.1.9) and (3.1.10) into (3.1.7), collecting the coefficients of each power of of the coefficients to zero, we obtain the following system of algebraic equations:

Q i , (i=0,1,2,3,4) and setting each

Q 4 :  k 2 a2b1 ln(a)  kPa22  k 2 Pa2b1 ln(a)  0, k 2 a2b1 ln( a)  k 2 Pa2b1 ln( a)  2kPa1a2  2k 2 Pa2b0 ln( a)  P 2 a2b1  Pa2b1 3 Q :  2k 2 a2b0 ln( a)  0, 2

Q : Q1 :

 kPa12  P 2 a1b1  2k 2 Pa2b0 ln( a)  2kPa0 a2  2k 2 a2b0 ln( a)  k 2b1a0 ln( a)  Pa1b1  k 2 a1b0 ln( a)  Pa2b0  k 2 Pa1b0 ln( a)  P 2 a2b0  0,  k 2 Pb1a0 ln( a)  k 2 Pa1b0 ln( a)  Pa1b0  k 2 a1b0 ln( a)  Pa0b1  P 2 a1b0  k 2b1a0 ln( a)  2kPa0 a1  P 2 a0b1  0,

Q0 :  kPa02  P 2 a0b0  Pa0b0  0. (3.1.11) Solving the system of algebraic Eqs. (3.1.11) by Maple or Mathematica, we obtain the following sets: Set 1:

a0  0, a1  0, a2  

kb1 ( P  1) ln( a) 1 2k 2 ln( a) , b0   b1 ,   P 2 P

(3.1.12)

k  k , a  a, b1  b1. Substituting (3.1.12) into (3.1.9), we get the following solution:

v( ) 

 2k ( P  1) ln( a) . P 1  a 1  a







(3.1.13)

From (3.1.6) and (3.1.13) we have 1

  2k ( P  1) ln( a)  P U ( )    . 2  P 1  a 





With the help of (2.10) and (2.11) the exact solution of Eq. (3.1.1) has the form:

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(3.1.14)


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39 1

P  k ( P  1) ln( a)   2k 2 ln( a)    u ( x, t )   t   1  , cot Fs  kx  P P      

(3.1.15)

or 1

P  k ( P  1) ln( a)     2k 2 ln( a)  u ( x, t )   t  ln( a)   1  . coth  kx  P P        

(3.1.16)

Set 2:

k (b0  b1 )( P  1) ln( a) k 2 ln( a) ,  , P P k  k , a  a, b0  b0 , b1  b1.

a0  0, a2  0, a1  

(3.1.17)

Consequently, we have the exact solutions of Eq. (3.1.1) in the form:

  k (b0  b1 )( P  1)1  tanh  ln( a)  P u ( x, t )    ,     P  2 b  b 1  tanh  0 1  

(3.1.18)

  k (b0  b1 )( P  1)1  coth  ln( a)  P u ( x, t )    , P 2b0  b1 1  coth   

(3.1.19)

1

or 1

where 



 kx  1 2

k 2 ln(a ) 2P



t ln(a), and b0 ,b1 are arbitrary constants.

Set 3:

kb1 ( P  1) ln( a) kb ( P  1) ln( a) k 2 ln( a) , a2   1 ,   , P P P k  k , a  a, b1  b1.

a0  0, b0  0, a1 

(3.1.20)

Now, we have the exact solutions of Eq. (3.1.1) in the form: 1

P  k ( P  1) ln( a) 1  tanh  , u ( x, t )   2 P  

(3.1.21)

or 1

 k ( P  1) ln( a) P 1  coth  , u ( x, t )   2 P   where 



 kx  1 2

k 2 ln(a ) 2P



t ln(a).

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(3.1.22)


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

Set 4:

kb1 ( P  1) ln( a) a1 P k 2 ln( a) , b0   ,  , P k ( P  1) ln( a) P k  k , a1  a1 , a  a, b1  b1.

a0  0, a2  

(3.1.23)

Consequently, we have the exact solutions of Eq. (3.1.1) in the form: 1

P  k ( P  1) ln(a) tanh  1 , u ( x, t )   2 P  

(3.1.24)

or 1

where 



 kx  1 2

P  k ( P  1) ln( a) coth  1 , u ( x, t )   2 P  

k 2 ln(a ) 2P

(3.1.25)



t ln(a).

Set 5:

a0 

kb0 ( P  1) ln( a) kb ( P  1) ln( a) a2 P , a1  a2  0 , b1   P P k ( P  1) ln( a)

k 2 ln( a) ,   , k  k , a  a, b0  b0 , a2  a2 . P

(3.1.26)

Now, we have the exact solutions of Eq. (3.1.1) in the form: 1

 4k1  2(a2  k1 )1  tanh   a2 1  tanh 2  P u ( x, t )    , 4b0  2 ak21b0 1  tanh   

(3.1.27)

or 1

 4k  2(a  k )1  coth   a 1  coth 2  P 2 1 2 u ( x, t )   1  , 2 a 2 b0 4b0  k1 1  coth    where

k1 

kb0 ( P 1) ln(a ) P

,



  12 kx  k

2

ln(a ) 2P

(3.1.28)



t ln(a), and b0 , a2 are arbitrary constants.

Set 6:

a0 

2kb0 ( P  1) ln( a) 4kb0 ( P  1) ln( a) 2kb0 ( P  1) ln( a) , a1   , a2  , P P P

2k 2 ln( a) b1  2b0 ,    , k  k , a  a, b0  b0 . P Consequently, we have the exact solutions of Eq. (3.1.1) in the form:

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(3.1.29)


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

 k ( P  1)1  tanh 2 ln( a)  P u ( x, t )    , 2 P tanh  

(3.1.30)

 k ( P  1)1  coth 2 ln( a)  P u ( x, t )    , 2 P coth  

(3.1.31)

1

or 1

where 



 kx  1 2

k 2 ln(a ) P



t ln(a).

Set 7:

kb0 ( P  1) ln( a) kb ( P  1) ln( a) k 2 ln( a) , a1   0 ,   , P P P k  k , a  a, b0  b0 , b1  b1.

a2  0, a0 

(3.1.32)

Now, we have the exact solutions of Eq. (3.1.1) in the form:

 kb0 ( P  1)1  tanh  ln( a)  P u ( x, t )    ,  P 2b0  b1 1  tanh  

(3.1.33)

 kb ( P  1)1  coth  ln( a)  P u ( x, t )   0  ,  P 2b0  b1 1  coth  

(3.1.34)

1

or 1

where 



 kx  1 2

k 2 ln(a ) 2P



t ln(a), and b0 ,b1 are arbitrary constants.

On comparing our results (3.1.30), (3.1.31), (3.1.33) and (3.1.34) with [49], where the authors solved Eq. (3.1.1) using the fractional order derivative 0    1. We deduce that all results are equivalent in the special case with a  e and   1, while our results (3.1.16), (3.1.18), (3.1.19), (3.1.21), (3.1.22), (3.1.24), (3.1.25), (3.1.27) and (3.1.28) are new, and not discussed elsewhere. 3.2

Example 2. The modified Benjamin-Bona-Mahony (mBBM) Equation

In this subsection, we apply the given method described in Sec. 2, to solve the modified Benjamin-Bona-Mahony (mBBM) equation [43, 51] in the form:

ut  ux u 2ux  uxxx  0,

(3.2.1)

where 𝜈 is a non zero positive constant. Eq. (3.2.1) was first derived to describe an approximation for surface of long waves in nonlinear dispersive media. It can also characterize the hydromagnetic waves in a cold plasma, acoustic waves in harmonic crystals and acoustic gravity waves in compressible fluids [51]. We proceed by considering the traveling wave transformation

u( x, t )  U ( ),

  kx  ct ,

where 𝑘 and 𝑐 are real constants with 𝑘, 𝑐 ≠ 0. Eq. (3.2.1) can be reduced to the following nonlinear ordinary differential equation:

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(3.2.2)


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

(c  k )U  kU 2U   k 3U   0. Integrating Eq. (3.2.3) with respect to



(3.2.3)

and vanishing the constant of integration, we get

1 (c  k )U  kU 3  k 3U   0. 3 By balancing

(3.2.4)

U  with U 3 , we have:

n  m  2  3(n  m)  n  m  1

(3.2.5)

If we choose 𝑚 = 1 and 𝑛 = 2, then

U ( ) 

a0  a1Q  a2Q 2 , b0  b1Q

(3.2.6)

and consequently, we deduce that

 (a1  2a2Q)(b0  b1Q)  b1 (a0  a1Q  a2Q 2 )   U ( )  QQ  1   ln( a), (b0  b1Q) 2  

U ( ) 

(3.2.7)

QQ  1 (2Q  1)[(a1  2a2Q)(b0  b1Q)  b1 (a0  a1Q  a2Q 2 )] ln 2 (a) (b0  b1Q) 2 Q 2 Q  1 ln 2 (a) (b0  b1Q) 3 2







 2a2 (b0  b1Q) 2  2b1 (a1  2a2Q)(b0  b1Q)  2b12 (a0  a1Q  a2Q 2 ) , (3.2.8)

U ( )  Q 3 Q  1 ln 3 (a) 3

  6a2b1 (b0  b1Q)  6b12 (a1  2a2Q) 6b1 3(a0  a1Q  a2Q 2 )     (b0  b1Q) 3 (b0  b1Q) 4   3Q 2 Q  1 (2Q  1) ln 3 (a) 2

 2a (b  b Q) 2  2b1 (a1  2a2Q)(b0  b1Q)  2b12 (a0  a1Q  a2Q 2 )   2 0 1  (b0  b1Q) 3   2 3 QQ  1(6Q  6Q  1) ln (a)  (a  2a2Q)(b0  b1Q)  b1 (a0  a1Q  a2Q 2 )   1 . 2 ( b  b Q ) 0 1   (3.2.9)

Substituting (3.2.6)-(3.2.8) into (3.2.4), collecting all the terms of the same orders coefficients to zero, we obtain the following set of algebraic equations:

- 26 -

Q i , (i  0,1,..., 6), and setting each


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

Q 6 : 18k 3 a 2 b 0 b 1 ln2 a  3ka 1 a 22  9k 3 a 2 b 21 ln2 a  0, Q 5 : 18k 3 a 2 b 0 b 1 ln2 a  3ka 1 a 22  9k 3 a 2 b 21 ln2 a  0, Q4 :

Q3 :

2

Q :

3ka 21 a 2  18k 3 a 2 b 20 ln2 a  27k 3 a 2 b 0 b 1 ln2 a  3ka 0 a 22  3k 3 a 2 b 21 ln2 a 3ka 2 b 21  3ca 2 b 21  0, 3ka 1 b 21  6ka 0 a 1 a 2  6ka 2 b 0 b 1  3ca 1 b 21  3k 3 a 1 b 0 b 1 ln2 a  ka 31  3k 3 b 21 a 0 ln2 a 30k 3 a 2 b 20 ln2 a  9k 3 a 2 b 0 b 1 ln2 a  6k 3 b 1 a 0 b 0 ln2 a  6k 3 a 1 b 20 ln2 a  6ca 2 b 0 b 1  0, 3ca 0 b 21  3ca 2 b 20  9k 3 a 1 b 20 ln2 a  3ka 0 a 21  3ka 0 b 21  6ka 1 b 0 b 1  3ka 2 b 20  3ka 20 a 2 12k 3 a 2 b 20 ln2 a  3k 3 a 1 b 0 b 1 ln2 a  9k 3 b 1 a 0 b 0 ln2 a  3k 3 b 21 a 0 ln2 a  0,

Q 1 : 3ca 1 b 20  6ca 0 b 0 b 1  3ka 1 b 20  3k 3 a 1 b 20 ln2 a  6ka 0 b 0 b 1  3ka 20 a 1  3k 3 b 1 a 0 b 0 ln2 a  0, Q 0 : 3ca 0 b 20  3ka 0 b 20  ka 30  0.

3.2.10 (3.2.10)

On solving the above set of algebraic Eqs. (3.2.10) with the aid of Maple or Mathematica, we get the following cases: Case 1:

a0  0, a1  2

6



6

kb0 ln( a), a2  2



kb0 ln( a), b1  2b0 ,

(3.2.11)

c  k (k ln (a)  1), k  k , a  a, b0  b0 ,  0. 2

2

Inserting (3.2.10) into (3.2.6), we get the following solution:

 a U ( )  2k ln( a)  2  1  a 6

 . 

(3.2.12)

With the help of (2.10) and (2.11) the exact solution of Eq. (3.2.1) in the form:

u ( x, t )  

6









k ln( a). cosech k x  (k 2 ln 2 (a)  1)t ln( a) .

(3.2.13)

Case 2:

a0  

6



kb0 ln( a), a1  2

6



kb0 ln( a), a2  2

6



kb0 ln( a),

(3.2.14)

b1  2b0 , c  k (2k ln (a)  1), k  k , a  a, b0  b0 ,  0. 2

2

Now, we have the exact solution

u ( x, t )  k

6









ln( a). coth k x  (2k 2 ln 2 (a)  1)t ln( a) .

- 27 -

(3.2.15)


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

Case 3:

a0  

  1 6 6 1 kb0 ln( a), a2  2a1  kb0 ln( a), c  k (k 2 ln 2 (a)  2), 2   2  

(3.2.16)

6  a1   6kb0 ln( a)  b1    , a1  a1 , k  k , a  a, b0  b0 ,  0. 3  k ln( a)  Consequently, we get the exact solutions 2 1  k1  a1 1  tanh   (a1  k1 )1  tanh   u ( x, t )   , 2 b0  k2 1  tanh  

(3.2.17)

or 2 1  k1  a1 1  coth   (a1  k1 )1  coth   u ( x, t )   , 2 b0  k2 1  coth  

where

k1  

6



kb0 ln(a), k2  

1 6



a1   6kb0 ln(a ) k ln(a )

,

  12 k x  12 (k 2 ln 2 (a)  2)t ln(a),

(3.2.18)

and

a1 ,b0 are arbitrary

constants. Case 4:

a0  0, b0  0, a1  

1 6 6 kb1 ln( a), a2  2 kb1 ln( a), 2  

(3.2.19)

1 c  k (k 2 ln 2 (a)  2), k  k , a  a, b1  b1 ,  0. 2 Now, we have the exact solutions

u ( x, t )  

1   1 6 1  k ln( a). tanh  k  x  (k 2 ln( a) 2  2)t  ln( a), 2  2  2  

(3.2.20)

u ( x, t )  

1   1 6 1  k ln( a). coth  k  x  (k 2 ln( a) 2  2)t  ln( a). 2  2  2  

(3.2.21)

or

Case 5:

a0   c

1 6 6  a1   6kb0 ln( a)  kb0 ln( a), a2  0, b1    , 2  3  k ln( a) 

1 k (k 2 ln 2 (a)  2), a1  a1 , k  k , a  a, b0  b0 ,  0. 2

Consequently, we get the exact solutions:

- 28 -

(3.2.22)


u ( x, t ) 



Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

kb0 ln( a)  a1 1  tanh  , 2b0  2k3 1  tanh 

6



(3.2.23)

or

u ( x, t )  where

k3  

1 6



kb0 ln( a)  a1 1  coth  , 2b0  2k3 1  coth 



a1   6kb0 ln(a ) k ln(a )

6



,

  12 k x  12 (k 2 ln 2 (a)  2)t ln(a),

(3.2.24)

and

a1 , b0 are arbitrary constants.

Case 6:

a0  

1 6 1 6 kb0 ln( a), a2  0, a1   k ln( a)(2b0  b1 ), 2  2 

(3.2.25)

1 c  k (k 2 ln 2 (a)  2), k  k , a  a, b0  b0 ,  0. 2 Consequently, we obtain the exact solutions:

u ( x, t ) 









6

k ln( a)2b0 tanh  b1 1  tanh  , 4b0  2b1 1  tanh 

(3.2.26)

k ln( a)2b0 coth  b1 1  coth  , 4b0  2b1 1  coth 

(3.2.27)

or

u ( x, t )  where 



6



 12 k x  12 (k 2 ln 2 (a)  2)t ln(a), and b0 , b1 are arbitrary constants.

On comparing our results (3.2.20) and (3.2.21) with [44], where the authors solved Eq. (3.2.1) using the fractional order derivative 0    1. We deduce that all results are equivalent in the special case with   1, while our results (3.2.13), (3.2.15), (3.2.17), (3.2.18), (3.2.23), (3.2.24), (3.2.26) and (3.2.27) are new, and not discussed elsewhere. 3.3

Example 3. The Potential Kadomstev-Petviashvili (PKP) Equation

In this subsection, we apply the given method to solve the following potential Kadomstev-Petviashvili (PKP) equation [24,44,52]:

1 3 3 u xxxx  u xu xx  u yy  u xt  0. 4 2 4

(3.3.1)

To this end, we use transformation

u( x, y, t )  U ( ),

  kx  ly  ct ,

(3.3.2)

where 𝑘 , 𝑙 and c are arbitrary constants with 𝑘, 𝑙, 𝑐 ≠ 0, to reduce Eq. (3.3.1) to the following nonlinear ordinary differential equation:

1 4  3 3 3 k U  k U U   l 2U   kcU   0. 4 2 4 Integration the Eq. (3.3.3) with respect to



and vanishing the constant of integration, we obtain

- 29 -

(3.3.3)


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

1 4 3 3 k U   k 3U 2  ( l 2  kc)U   0. 4 4 4 By balancing U  with solutions (3.2.6).

(3.3.4)

U 2 , we have n  m  1. If we choose m  1 and n  2, then Eq. (3.3.4) has the same formal

Substituting (3.2.6)-(3.2.9) into (3.3.4), and equating all the coefficients of system of algebraic equations:

Q i , (i  0,1,..., 7) to zero, we obtain a

Q 7 : 6k 4 a 2 b 31 ln2 a  3k 3 a 22 b 21 lna  0, Q 6 : 6k 3 a 22 b 21 lna  12k 3 a 22 b 0 b 1 lna  12k 4 a 2 b 31 ln2 a  24k 4 a 2 b 0 b 21 ln2 a  0, Q5 :

7k 4 a 2 b 31 ln2 a  6k 3 a 1 b 0 a 2 b 1 lna  48k 4 a 2 b 0 b 21 ln2 a  4kca 2 b 31  36k 4 a 2 b 20 b 1 ln2 a 3k 3 a 22 b 21 lna  3l2 a 2 b 31  24k 3 a 2 b 21 a 0 lna  12k 3 a 22 b 20 lna  0, 12l2 a 2 b 0 b 21  12k 3 a 1 b 20 a 2 lna  k 4 a 2 b 31 ln2 a  72k 4 a 2 b 20 b 1 ln2 a  12k 3 a 22 b 0 b 1 lna

Q 4 : 12k 3 a 1 b 0 a 2 b 1 lna  3l2 a 2 b 31  24k 3 a 22 b 20 lna  12k 3 a 2 b 21 a 2 lna  16kca 2 b 0 b 21 12k 3 a 2 b 0 b 1 a 0 lna  4kca 2 b 31  24k 4 a 2 b 30 ln2 a  28k 4 a 2 b 0 b 21 ln2 a  0,

6k 4 b 1 a 0 b 20 ln2 a  3l2 b 31 a 0  3l2 a 1 b 0 b 21  15l2 a 2 b 20 b 1  12l2 a 2 b 0 b 21  4kca 1 b 0 b 21 6k 4 a 1 b 20 b 1 ln2 a  k 4 a 1 b 0 b 21 ln2 a  6k 3 a 1 b 0 b 1 a 0 lna  12k 3 a 22 b 20 lna  4kcb 31 a 0

Q 3 : 54k 4 a 2 b 30 ln2 a  3k 3 a 21 b 20 lna  k 4 b 31 a 0 ln2 a  3k 3 b 21 a 20 lna  6k 4 b 21 a 0 b 0 ln2 a 20kca 2 b 20 b 1  16kca 2 b 0 b 21  6k 4 a 1 b 30 ln2 a  4k 4 a 2 b 0 b 21 ln2 a  6k 3 a 1 b 0 a 2 b 1 lna 41k 4 a 2 b 20 b 1 ln2 a  6k 3 a 2 b 21 a 0 lna  24k 3 a 1 b 20 a 2 lna  24k 3 a 2 b 0 b 1 a 0 lna  0, 12k 4 b 1 a 0 b 20 ln2 a  6l2 a 2 b 30  3l2 b 31 a 0  15l2 a 2 b 20 b 1  6l2 b 21 a 0 b 0  38k 4 a 2 b 30 ln2 a 10k 4 b 21 a 0 b 0 ln2 a  10k 4 a 1 b 20 b 1 ln2 a  k 4 a 1 b 0 b 21 ln2 a  12k 3 a 1 b 0 b 1 a 0 lna

Q 2 : 6k 3 a 21 b 20 lna  k 4 b 31 a 0 ln2 a  6k 3 b 21 a 20 lna  8kca 1 b 20 b 1  4kca 1 b 0 b 21  4kcb 31 a 0 20kca 2 b 20 b 1  8kcb 21 a 0 b 0  12k 3 a 2 b 0 b 1 a 0 lna  12k 4 a 1 b 30 ln2 a  8kca 2 b 30 5k 4 a 2 b 20 b 1 ln2 a  12k 3 a 1 b 20 a 2 lna  6l2 a 1 b 20 b 1  0,

4k 4 a 1 b 20 b 1 ln2 a  8kca 1 b 20 b 1  6l2 a 2 b 30  7k 4 a 1 b 30 ln2 a  3l2 a 1 b 30  4kca 1 b 30

Q1 :

3l2 b 1 a 0 b 20  4kcb 1 a 0 b 20  3k 3 b 21 a 20 lna  8kcb 21 a 0 b 0  8k 4 a 2 b 30 ln2 a 6l2 b 21 a 0 b 0  7k 4 b 1 a 0 b 20 ln2 a  4k 4 b 21 a 0 b 0 ln2 a  6l2 a 1 b 20 b 1  8kca 2 b 30 3k 3 a 21 b 20 lna  6k 3 a 1 b 0 b 1 a 0 lna  0,

Q0 :  3l 2 a1b03  3l 2b1a0b02  4kcb1a0b02  k 4b1a0b02 ln 2 (a)  4kca1b03  k 4 a1b03 ln 2 (a)  0. (3.3.5)

- 30 -


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

Solving the above algebraic Eqs. (3.3.5) with the aid of Maple or Mathematica, the following cases of solutions are obtained: Case 1:

a1  2a0 , a2  4kb0 ln( a), b1  2b0 , c  

1 (4k 4 ln 2 (a)  3l 2 ) 4k

(3.3.6)

, a0  a0 , k  k , a  a, l  l , b0  b0 . Consequently, we have the exact solutions:

u ( x, y, t ) 

a0 tanh  kb0 ln( a).1  tanh  , b0 tanh( )

(3.3.7)

u ( x, y, t ) 

a0 coth  kb0 ln( a).1  coth  , b0 coth

(3.3.8)

2

or 2

where 



1 2

kx  ly 

( 4 k 4 ln 2 ( a )  3l 2 ) t 4k

ln(a), and a ,b 0

0

are arbitrary constants.

Case 2:

a0  0, a1  0, a2  2kb1 ln( a), b0  0, c  

1 4 2 (k ln (a)  3l 2 ), 4k

(3.3.9)

k  k , a  a, l  l , b1  b1. In this case we have the results:

u( x, y, t )  k ln(a).1  tanh ,

(3.3.10)

u( x, y, t )  k ln(a).1  coth ,

(3.3.11)

or

where 



1 2

kx  ly 

( 4 k 4 ln 2 ( a )3l 2 ) t 4k

ln(a).

Case 3:

a0  0, a2  2kb1 ln( a), b0  0, c  

1 4 2 (k ln (a)  3l 2 ), 4k

(3.3.12)

a1  a1 , k  k , a  a, l  l , b1  b1. In this case we get the results:

u ( x, y, t ) 

a1  k ln( a).1  tanh , b1

(3.3.13)

u ( x, y, t ) 

a1  k ln( a).1  coth , b1

(3.3.14)

or

where 



1 2

kx  ly 

( 4 k 4 ln 2 ( a )  3l 2 ) t 4k

ln(a), and a ,b 1

1


- 31 -


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

Case 4:

a2  0, a1  2k (b0  b1 ) ln( a) 

a0b1 1 , c   (k 4 ln 2 (a)  3l 2 ), b0 4k

(3.3.15)

a0  a0 , k  k , a  a, l  l , b0  b0 , b1  b1. Consequently, we have the exact solutions:

u ( x, y, t ) 





2a0  2 k (b0  b1 ) ln( a)  a20bb01 1  tanh  2b0  b1 1  tanh 

,

(3.3.16)

,

(3.3.17)

or

u ( x, y, t )  where 



1 2

kx  ly 





2a0  2 k (b0  b1 ) ln( a)  a20bb01 1  coth 

( 4 k 4 ln 2 ( a )3l 2 ) t 4k

2b0  b1 1  coth 

ln(a), and a , b , b 0

0


1

Case 5:

a1  2kb0 ln( a) 

a0b1 1 , a2  2kb1 ln( a), c   (k 4 ln 2 (a)  3l 2 ), b0 4k

(3.3.18)

a0  a0 , k  k , a  a, l  l , b0  b0 , b1  b1. In this case we have the results:

2a0  2k1 1  tanh   kb1 ln( a).1  tanh  , 2b0  b1 1  tanh 

(3.3.19)

2a  2k1 1  coth   kb1 ln( a).1  coth  u ( x, y, t )  0 , 2b0  b1 1  coth 

(3.3.20)

2

u ( x, y, t )  or

2

kx  ly 

where

k1  kb0 ln(a)  a20bb01 , 

3.4

Example 4. The Cahn-Hilliard Equation

1 2

( 4 k 4 ln 2 ( a )3l 2 ) t 4k

ln(a), and a , b , b are arbitrary constants. 0

0

1

In this subsection, we apply the given method to the Cahn-Hilliard equation [46, 49] in the form:

ut  ux  6uux   (3u 2  1)uxx  uxxxx  0. 2

(3.4.1)

Eq. (3.4.1) was first introduced in 1958 as a model for process of phase separation of a binary alloy under the critical temperature [53]. This equation arose as the modelling equation in numerous other contexts with disparate length scales. For example, the improved Cahn-Hilliard equation is used to represent the evolution of two components of intergalactic material or in ecology in the modeling of the dynamics of two populations or in bio mathematics in modeling the dynamics of the biomass and the solvent components of a bacterial film [54]. Let us solve Eq. (3.4.1) by using the given method of Sec. 2. To this end, we use the traveling wave transformation

u( x, t )  U ( ),

  kx  t ,

- 32 -

(3.4.2)


Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

where 𝑘 and 𝜆 are arbitrary constants with 𝑘, 𝜆 ≠ 0, to reduce Eq. (3.4.2) to the following nonlinear ordinary differential equation:

 (  k )U   3k 2 (U 2U )  k 2U   k 3U   0. Integrating the Eq. (3.4.3) with respect to



(3.4.3)

and vanishing the constant of integration, we obtain

 (  k )U  3k 2U 2U   k 2U   k 3U   0. By balancing U  with solution (3.2.6).

(3.4.4)

U 2U , we get n  m  1. If we choose m  1 and n  2, then we have the same formal

Substituting (3.2.6)-(3.2.9) into (3.4.4), and collecting all the terms with the same power of equating each coefficient to zero, yields a set of algebraic equations:

Q i , (i  0,1,..., 8) together,

Q 8 : 3k 2 a 32 b 1 lna  6k 4 a 2 b 31 ln3 a  0, Q 7 : 6k 2 a 32 b 0 lna  12k 4 a 2 b 31 ln3 a  6k 2 a 1 a 22 b 1 lna  3k 2 a 32 b 1 lna  24k 4 a 2 b 0 b 21 ln3 a  0, 6

Q :

3k 2 a 21 a 2 b 1 lna  6k 2 a 32 b 0 lna  7k 4 a 2 b 31 ln3 a  k 2 a 2 b 31 lna  36k 4 a 2 b 20 b 1 ln3 a 3k 2 a 0 a 22 b 1 lna  6k 2 a 1 a 22 b 1 lna  48k 4 a 2 b 0 b 21 ln3 a  15k 2 a 1 a 22 b 0 lna  0, 4k 2 a 2 b 0 b 21 lna  28k 4 a 2 b 0 b 21 ln3 a  12k 2 a 0 a 22 b 0 lna  ka 2 b 31  12k 2 a 21 a 2 b 0 lna

Q 5 : k 4 a 2 b 31 ln3 a  k 2 a 2 b 31 lna  3k 2 a 21 a 2 b 1 lna  24k 4 a 2 b 30 ln3 a  72k 4 a 2 b 20 b 1 ln3 a a 2 b 31  3k 2 a 0 a 22 b 1 lna  15k 2 a 1 a 22 b 0 lna  0, 6k 4 b 1 a 0 b 20 ln3 a  k 4 a 1 b 0 b 21 ln3 a  5k 2 a 2 b 20 b 1 lna  ka 1 b 31  18k 2 a 0 a 1 a 2 b 0 ln3 a 6k 4 b 21 a 0 b 0 ln3 a  6k 4 a 1 b 30 ln3 a  41k 4 a 2 b 20 b 1 ln3 a  3ka 2 b 0 b 21  4k 2 a 2 b 0 b 21 lna

Q 4 : 3k 2 a 31 b 1 lna  54k 4 a 2 b 30 ln3 a  3a 2 b 0 b 21  12k 2 a 21 a 2 b 0 lna  3k 2 a 21 b 1 a 0 lna a 1 b 31  k 4 b 31 a 0 lna 3  4k 4 a 2 b 0 b 21 lna 3  6k 4 a 1 b 20 b 1 ln3 a  3k 2 a 20 a 2 b 1 lna k 2 b 31 a 0 lna  12k 2 a 0 a 22 b 0  k 2 a 1 b 0 b 21 lna  0,

a 0 b 31  3a 2 b 20 b 1  3a 1 b 0 b 21  3ka 1 b 0 b 21  3ka 2 b 20 b 1  5k 4 a 2 b 20 b 1 ln3 a  k 2 b 31 a 0 lna 10k 4 b 21 a 0 b 0 ln3 a  3k 2 a 31 b 0 lna  12k 4 a 1 b 30 ln3 a  38k 4 a 2 b 30 ln3 a  5k 2 a 2 b 20 b 1 lna

Q 3 : 2k 2 a 2 b 30 lna  12k 4 b 1 a 0 b 20 ln3 a  k 4 a 1 b 0 b 21 ln3 a  18k 2 a 0 a 1 a 2 b 0 lna  k 4 b 31 a 0 ln3 a 3k 2 a 21 b 1 a 0 lna  10k 4 a 1 b 20 b 1 ln3 a  3k 2 a 20 a 2 b 1 lna  k 2 a 1 b 0 b 21 lna  6k 2 a 20 a 1 b 1 lna ka 0 b 31  6k 2 a 20 a 2 b 0 lna  6k 2 a 0 a 21 b 0 lna  2k 2 a 1 b 20 b 1 lna  2k 2 b 21 a 0 b 0 lna  0,

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Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

k 2 b 1 a 0 b 20 lna  3ka 0 b 0 b 21  4k 4 b 21 a 0 b 0 ln3 a  6k 2 a 0 a 21 b 0 lna  7k 4 a 1 b 30 ln3 a 2

Q :

3k 2 a 30 b 1 lna  7k 4 b 1 a 0 b 20 ln3 a  6k 2 a 20 a 1 b 1 lna  6k 2 a 20 a 2 b 0 lna  2k 2 a 2 b 30 lna 3k 2 a 20 a 1 b 0 lna  3ka 1 b 20 b 1  4k 4 a 1 b 20 b 1 ln3 a  8k 4 a 2 b 30 ln3 a  3a 1 b 20 b 1 2k 2 a 1 b 20 b 1 lna  2k 2 b 21 a 0 b 0 lna  a 2 b 30  k 2 a 1 b 30 lna  3a 0 b 0 b 21  ka 2 b 30  0,

Q1 :

3a 0 b 20 b 1  3k 2 a 20 a 1 b 0 lna  3k 2 a 30 b 1 lna  k 4 b 1 a 0 b 20 ln3 a  ka 1 b 30  a 1 b 30 3ka 0 b 20 b 1  k 2 b 1 a 0 b 20 lna  k 4 a 1 b 30 ln3 a  ka 1 b 30 lna  0,

Q 0 : a 0 b 30  ka 0 b 30  0. (3.4.5) On solving the above algebraic Eqs. (3.4.5) with the aid of Maple or Mathematica, the following cases of solutions are obtained: Case 1:

a0  b0 , a1  2b0 , a2  0, b1  0,   

2 2 ,k  , ln( a) ln( a)

(3.4.6)

a  a, b0  b0 . In this case, we have the exact solutions:

 1  u ( x, t )   tanh ( x  t ) ,  2 

(3.4.7)

 1  u ( x, t )   coth ( x  t ) .  2 

(3.4.8)

or

Case 2:

a0  b0 , a1  2b0  b1 , a2  0,   

2 2 ,k  , ln( a) ln( a)

(3.4.9)

a  a, b1  b1 , b0  b0 . In this case, we obtain the following exact solutions:

u ( x, t ) 

 2b0 tanh(

1 2



( x  t ))  b1 1  tanh(



2b0  b1 1  tanh(

1 2

1 2

( x  t ))



( x  t ))

,

(3.4.10)

.

(3.4.11)

or

u ( x, t ) 

 2b0 coth(

1 2



( x  t ))  b1 1  coth(



2b0  b1 1  coth(

1 2

- 34 -

1 2

( x  t ))



( x  t ))


where

Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

b0 ,b1 are arbitrary constants.

IV. PHYSICAL EXPLANATIONS FOR SOME OF OUR SOLUTIONS In this section, we will illustrate the application of the results established above. Exact solutions of the results describe different nonlinear waves. For the established exact kink and anti-kink solutions with symmetrical hyperbolic Fibonacci functions are special kinds of solitary waves. Kink and anti-kink solutions have a remarkable property that keeps its identity upon interacting with other. Kink and anti-kink solutions have particle-like structures, for example, magnetic monopoles, and extended structures, like, domain walls and cosmic strings, that have implications in cosmology of the early universe. Let us now examine Figs. 1 to 4 as it illustrates some of our results obtained in this article. To this end, we select some special values of the parameters obtained, for example, in some of them, kink and anti-kink solutions (3.1.31) and (3.1.33) of the Burgers equation whit 1 < 𝑥 < 10 and 0 < 𝑡 < 1, (3.2.20) and (3.2.21) of the modified Benjamin-Bona-Mahony equation with 𝜈 = 𝑘 = 1, 𝑎 = 𝑒, −10 < 𝑥, 𝑡 < 10, (3.3.10) and (3.3.11) of the Potential Kadomtsev-Petviashvili equation with 𝑎₀ = 𝑏₀ = 𝑘 = 𝑙 = 1, 𝑦 = 1, −10 < 𝑥, 𝑡 < 10, (3.4.8) and (3.4.10) of the Cahn-Hilliard equation when the −10 < 𝑥 < 10 𝑎𝑛𝑑 0 < 𝑡 < 10, respectively. The other figures are ignored for simplicity

3D Plot kink solution (3.1.33) with 𝑃 = 2, 𝑘 = 1, = 3, 𝑎 = 𝑒.

3D Plot anti-kink solution (3.1.31) with 𝑏0 = 𝑃 = 3, 𝑏1 = 𝑘 = 1, 𝛽 = 2, 𝑎=e.

Figs. 1 Symmetrical Fibonacci hyperbolic function solutions of the nonlinear Burgers equation

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Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

3D Plot of the kink solution (3.2.2).

3D Plot of the anti-kink solution (3.2.21).

Figs. 2 Symmetrical Fibonacci hyperbolic function solutions of the mBBM equation

3D Plot of the kink solution (3.3.10) when a = 10.

3D Plot of the anti-kink solution (3.3.11) when a = e.

Figs. 3 Symmetrical Fibonacci hyperbolic function solutions of the PKP equation

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Apr. 2015, Vol. 5 Iss. 2, PP. 19-39

3D Plot of the anti-kink solution (3.4.8).

3D Plot of the kink solution (3.4.10).

Figs. 4 Symmetrical Fibonacci hyperbolic function solutions of the Cahn-Hilliard equation V. CONCLUSIONS In this paper, it is shown that the symmetrical hyperbolic Fibonacci function solutions can be obtained by the general Expafunction by using generalized Kudryashov method. We have extended successfully the generalized Kudryashov method to solve four nonlinear partial differential equations. As applications, we have obtained many new symmetrical hyperbolic Fibonacci function solutions for the Burgers equation, the modified Benjamin-Bona-Mahony (mBBM) equation, the Potential Kadomtsev-Petviashvili (PKP) equation and the Cahn-Hilliard equation. As one can see, the generalized Kudryashov method is powerful, effective and convenient for solving nonlinear PDEs and the physical explanation of some solutions of these equations have been presented in Sec. 4. Finally, the generalized Kudryashov method provides a powerful mathematical tool to obtain more general exact analytical solutions of many nonlinear PDEs in mathematical physics. REFERENCES

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