NEW EXACT SOLUTIONS OF SOME NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS VIA THE IMPROVED EXP-FUNCTION METHOD

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132

NEW EXACT SOLUTIONS OF SOME NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS VIA THE IMPROVED EXP-FUNCTION

METHOD

M. F. El-Sabbagh, R. Zait and R. M. Abdelazeem

Mathematics Department, Faculty of Science, MiniaUniversity, Egypt.

ABSTRACT

In this paper, we establish new exact solutions of some nonlinear partial differential equations (PDEs) of interest such as the Kaup–Kupershmidt, the generalized shallow water, the Boussinesq equations via the improved Exp–

function method. Also the method is used to construct periodic and solitary wave solutions for the considered equations as well.

Keywords: Nonlinear PDEs, Exact solutions, The improved Exp–function method.

1. INTRODUCTION

The nonlinear evolution equations (NLEEs) are widely used as models to describe complex physical phenomena in various field of science, particularly in fluid mechanics, solid state physics, plasma waves and chemical physics.

Nonlinear equations covers also the following subjects: surface wave in compressible fluid, hydro magnetic waves in cold plasma, acoustic waves in un harmonic crystal, ect. . The wide applicability of these equations is the main reason for they have attracted so much attention from mathematicians in the last decades. The investigation of the exact solutions of non linear partial differential equations (PDEs) plays an important role in the study of non-linear physical phenomena. When we want to understand the physical mechanism of phenomena in nature, described by non linear PDEs, exact solutions have to be explored. The study of nonlinear PDEs becomes one of the most important topics in mathematical physics. Recently there are many new methods to obtain exact solutions of nonlinear PDEs such as sine-cosine function method [1-5], tanh function method [6-8], (𝐺^/ )-expansion method 𝐺 [9-13 ], extended Jacobi elliptic function method [14, 15]. He and Wu [16], proposed a straightforward and concise method, called Exp-function method [17-24], to obtain generalized solitary wave solutions of nonlinear PDEs. Ali [25] improved this method and obtained new exp-function solutions and periodic solutions as well.

In this paper, we use the improved Exp-function method [25, 26 ] to search for new solitary wave solutions, compact like solutions and periodic solutions of some nonlinear PDEs, such as the Kaup–Kupershmidt equation [27-35], the generalized shallow water equation [36, 37], and the Boussinesq equation [38-42].

2. THE IMPROVED EXP-FUNCTION METHOD:

We present the improved Exp-function method [25] in the following steps:

1- Consider the following nonlinear PDE with two independent variables 𝑥, 𝑡 and dependant variable 𝑢:

𝑁(𝑢, 𝑢𝑡, 𝑢𝑥, 𝑢𝑥𝑥, 𝑢𝑥𝑡, 𝑢𝑡𝑡, … ) = 0 (1), where 𝑁 is in general a polynomial function of its argument and the subscripts denote the partial derivatives.

2- We seek a traveling wave solution of Eq. (1) in the form

𝑢 𝑥, 𝑡 = 𝑢 𝜉 , 𝜉 = 𝑘𝑥 + 𝜔𝑡 2 , where 𝑘 and 𝜔 are constants to be determined

3- Using the transformation (2), Eq. (1) can be reduced to an ordinary differential equation (ODE):

𝐺 𝑢, 𝑢^′, 𝑢^′′, … = 0 3 , where 𝐺 is a polynomial of 𝑢 and its derivatives.

4- Through this method, we express the solution of the nonlinear PDE (1) in the form:

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133 𝑢 𝜂 = ^𝑚_{𝑗 =0}𝐴𝑗𝑒𝑥𝑝⁡(𝑗𝜉)

𝐵_𝑖𝑒𝑥𝑝⁡(𝑖𝜉)

𝑛𝑖=0

4 ,

where 𝑚 and 𝑛 are positive integers that could be freely chosen.

5- To determine 𝑢 𝑥, 𝑡 explicitly, one may apply the following producer:

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Substitute Eq. (4) into Eq. (3), then the lift hand side of Eq. (3) is converted into a polynomial in 𝑒𝑥𝑝⁡(𝜉). Setting all coefficients of 𝑒𝑥𝑝⁡(𝜉) to zero yield a system of algebraic equations for 𝐴0, 𝐴1, 𝐴2, … . 𝐴𝑚, 𝐵0, 𝐵1, 𝐵2, … . 𝐵𝑛, 𝑘 𝑎𝑛𝑑 𝜔.

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Solve these algebraic equations to obtain A0, A1, A2, … , Am, 𝐵0, 𝐵1, 𝐵2, … , 𝐵𝑛, 𝑘 𝑎𝑛𝑑 𝜔.

3. APPLICATIONS

In order to illustrate the effectiveness of the above method, examples of mathematical and physical interests are chosen as follows:

3.1The Kaup–Kupershmidt equation [27-35]

The Kaup–Kupershmidt equation is the nonlinear fifth-order partial differential equation; It is the first equation in a hierarchy of integrable equations with Lax operator ^𝜕³

𝜕𝑥³+ 2𝑢_𝜕𝑥^𝜕 +^𝜕𝑢_𝜕𝑥. It has properties similar (but not identical) to those of the better known KdV hierarchy in which the Lax operator has order two.

In the present paper we introduce new exact solutions of the Kaup–Kupershmidt equation via the improved Exp- function method as follows:

Consider the Kaup–Kupershmidt equation is given as:

𝑢_𝑡= 𝑢_{𝑥𝑥𝑥𝑥𝑥} − 20𝑢𝑢_𝑥𝑥𝑥 − 50𝑢_𝑥𝑢_𝑥𝑥+ 80𝑢²𝑢_𝑥 (5) Using the transformation:

𝑢 = 𝑢 𝜂 , 𝜂 = 𝑥 − 𝑐𝑡 6 , where 𝑐 is a constant to be determined later. Substituting Eq. (6) into Eq. (5) we get

𝑐𝑢^′− 𝑢^{′′′′′}+ 20𝑢𝑢^′′′+ 50𝑢^′𝑢^′′− 80𝑢²𝑢^′= 0 7 ,

where primes denote derivatives with respect to 𝜂. Now we study the following cases:

𝑪𝒂𝒔𝒆 𝟏: 𝒎 = 𝟐, 𝒏 = 𝟐:

According to the improved Exp-function method, the travelling wave solution of Eq. (5) in this case can be written as:

𝑢 𝑥, 𝑡 = 𝐴₀+ 𝐴₁𝑒𝑥𝑝⁡(𝜂) + 𝐴₂𝑒𝑥𝑝⁡(2𝜂)

𝐵0+ 𝐵1𝑒𝑥𝑝⁡(𝜂) + 𝐵2𝑒𝑥𝑝⁡(2𝜂) (8) In case 𝐵2≠ 0, Eq. (8) can be simplified as:

𝑢 𝑥, 𝑡 = 𝐴0+ 𝐴1𝑒𝑥𝑝⁡(𝜂) + 𝐴2𝑒𝑥𝑝⁡(2𝜂)

𝐵₀+ 𝐵₁𝑒𝑥𝑝⁡(𝜂) + 𝑒𝑥𝑝⁡(2𝜂) (9) Substituting Eq. (9) into Eq. (7), and using the Maple, equating to zero the coefficients of all powers of 𝑒𝑥𝑝⁡(𝜂)yields a set of algebraic equations for 𝐴0, 𝐴1, 𝐴2, 𝐵0, 𝐵1 and c. Solving these system of algebraic equations, with the aid of Maple, we obtain families of solutions as the following:

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134 𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:

𝐵0= 1, 𝐴1= −5, 𝐴0=¹₂, 𝑐 = −11, 𝐴2=¹₂, 𝐵1= 2.

Thus the following solution is

𝑢1 𝑥, 𝑡 =

12 − 5𝑒𝑥𝑝⁡(𝑥 + 11𝑡) +1

2 𝑒𝑥𝑝⁡(2𝑥 + 22𝑡)

1 + 2𝑒𝑥𝑝⁡(𝑥 + 11𝑡) + 𝑒𝑥𝑝⁡(2𝑥 + 22𝑡) (10)

𝑭𝒊𝒈𝒖𝒓𝒆 𝟏. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟓 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏𝟎 . 𝑭𝒂𝒎𝒊𝒍𝒚 𝟐:

𝐵0= 1, 𝐴1=−5

8 , 𝐴0= 1

16, 𝑐 =−1

16, 𝐴2= 1

16, 𝐵1 = 2.

𝑢2 𝑥, 𝑡 = 16 −1 5

8 𝑒𝑥𝑝⁡(𝑥 + 𝑡 16) + 1

16 𝑒𝑥𝑝⁡(2𝑥 +𝑡 8) 1 + 2𝑒𝑥𝑝⁡(𝑥 + 𝑡

16) + 𝑒𝑥𝑝⁡(2𝑥 +𝑡 8)

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𝑭𝒊𝒈𝒖𝒓𝒆 𝟐. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟓 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏𝟏 . 𝑪𝒂𝒔𝒆 𝟐: 𝒎 = 𝟑, 𝒏 = 𝟑:

According to the improved Exp-function method, the travelling wave solution of Eq. (5) in this case can be written as:

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135 𝑢 𝑥, 𝑡 = 𝐴₀+ 𝐴₁𝑒𝑥𝑝⁡(𝜂) + 𝐴₂𝑒𝑥𝑝⁡(2𝜂) + 𝐴₃𝑒𝑥𝑝⁡(3𝜂)

𝐵0+ 𝐵1𝑒𝑥𝑝⁡(𝜂) + 𝐵2𝑒𝑥𝑝⁡(2𝜂) + 𝐵3𝑒𝑥𝑝⁡(3𝜂) (12) In case 𝐵3≠ 0, Eq. (12) can be simplified as:

𝑢 𝑥, 𝑡 = 𝐴0+ 𝐴1𝑒𝑥𝑝⁡(𝜂) + 𝐴2𝑒𝑥𝑝⁡(2𝜂) + 𝐴3𝑒𝑥𝑝⁡(3𝜂)

𝐵₀+ 𝐵₁𝑒𝑥𝑝⁡(𝜂) + 𝐵₂𝑒𝑥𝑝⁡(2𝜂) + 𝑒𝑥𝑝⁡(3𝜂) 13 Substituting Eq. (13) into Eq. (7), and using the Maple, equating to zero the coefficients of all powers of 𝑒𝑥 𝑝 𝜂 yields a set of algebraic equations for 𝐴₀, 𝐴₁, 𝐴₂, 𝐴₃, 𝐵₀, 𝐵₁, 𝐵₂ and c. Solving the system of algebraic equations given above, with the aid of Maple, we obtain:

𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:

𝐴0= 0, 𝐴1=¹₈𝐵₂², 𝐴2=⁻⁵₂ 𝐵2, 𝐴3=¹₂, 𝐵0= 0, 𝐵1 =¹₄𝐵₂², 𝐵2= 𝐵2, 𝑐 = −11.

𝑢5 𝑥, 𝑡 =

14 𝐵²²𝑒𝑥𝑝⁡(𝑥 + 11𝑡) − 5𝐵₂𝑒𝑥𝑝⁡(2𝑥 + 22𝑡) + 𝑒𝑥𝑝⁡(3𝑥 + 33𝑡)

12 𝐵²²𝑒𝑥𝑝⁡(𝑥 + 11𝑡) + 2𝐵2𝑒𝑥𝑝⁡(2𝑥 + 22𝑡) + 2𝑒𝑥𝑝⁡(3𝑥 + 33𝑡) (14)

If 𝐵2= 1, then

𝑢₅ 𝑥, 𝑡 =

14 𝑒𝑥𝑝⁡(𝑥 + 11𝑡) − 5𝑒𝑥𝑝⁡(2𝑥 + 22𝑡) + 𝑒𝑥𝑝⁡(3𝑥 + 33𝑡) 12 𝑒𝑥𝑝⁡(𝑥 + 11𝑡) + 2𝑒𝑥𝑝⁡(2𝑥 + 22𝑡) + 2𝑒𝑥𝑝⁡(3𝑥 + 33𝑡)

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𝑭𝒊𝒈𝒖𝒓𝒆 𝟑. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟓 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏𝟓 , 𝑩𝟐= 𝟏.

𝑭𝒂𝒎𝒊𝒍𝒚 𝟐:

𝐴0= 0, 𝐴1=₆₄¹ 𝐵₂², 𝐴2=⁻⁵₁₆𝐵2, 𝐴3=₁₆¹ , 𝐵0= 0, 𝐵1=¹₄𝐵₂², 𝐵2= 𝐵2, 𝑐 = −₁₆¹.

𝑢6 𝑥, 𝑡 =

14 𝐵²²𝑒𝑥𝑝⁡(𝑥 + 𝑡

16) − 5𝐵²𝑒𝑥𝑝⁡(2𝑥 +𝑡

8) + 𝑒𝑥𝑝⁡(3𝑥 +3𝑡 16) 4(𝐵₂²𝑒𝑥𝑝⁡(𝑥 + 𝑡

16) + 4𝐵²𝑒𝑥𝑝⁡(2𝑥 +𝑡

8) + 4𝑒𝑥𝑝⁡(3𝑥 +3𝑡 16))

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136 If 𝐵2= 1, then

𝑢₆ 𝑥, 𝑡 =

14 𝑒𝑥𝑝⁡(𝑥 + 𝑡

16) − 5𝑒𝑥𝑝⁡(2𝑥 +𝑡

8) + 𝑒𝑥𝑝⁡(3𝑥 +3𝑡 16) 4(𝑒𝑥 𝑝 𝑥 + 𝑡

16 + 4𝑒𝑥 𝑝 2𝑥 +𝑡

8 + 4𝑒𝑥 𝑝 3𝑥 +3𝑡 16 )

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𝑭𝒊𝒈𝒖𝒓𝒆 𝟒. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟓 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏𝟕 , 𝑩𝟐= 𝟏.

3.2 The generalized shallow water equation: [36, 37]

The shallow water wave equations describe the evolution of incompressible flow, neglecting density change along the depth. The shallow water wave equations are applicable to cases where the horizontal scale of the flow is much bigger than the depth of fluid. The shallow water equations have been extensively used for a wide variety of coastal phenomena, such as tide-currents, pollutant- dispersion storm-surges, tsunami-wave propagation, etc..

In the present paper we introduce new exact solutions of the generalized shallow water equation via the improved Exp-function Method as follows:

Consider the generalized shallow water equation:

𝑢_{𝑥𝑥𝑥𝑡} + 𝛼𝑢_𝑥𝑢_𝑥𝑡 + 𝛽𝑢_𝑡𝑢_𝑥𝑥 − 𝑢_𝑥𝑡− 𝑢_𝑥𝑥 = 0 18 , where 𝛼 and 𝛽 are arbitrary nonzero constants.

Using the transformation:

𝑢 = 𝑢 𝜂 , 𝜂 = 𝑥 − 𝑐𝑡 19 , where 𝑐 is a constant to be determined later. Substituting Eq. (19) into Eq. (18) we get

−𝑐𝑢^′′′′− 𝑐𝛼𝑢^′𝑢^′′− 𝑐𝛽𝑢^′𝑢^′′− 𝑐𝑢^′′− 𝑢^′′= 0 (20), where the prime denotes the differential with respect to 𝜂.

Now we study the following cases:

𝑪𝒂𝒔𝒆 𝟏: 𝒎 = 𝟐, 𝒏 = 𝟐:

𝑢 𝑥, 𝑡 = 𝐴₀+ 𝐴₁𝑒𝑥𝑝⁡(𝜂) + 𝐴₂𝑒𝑥𝑝⁡(2𝜂)

𝐵0+ 𝐵1𝑒𝑥𝑝⁡(𝜂) + 𝐵2𝑒𝑥𝑝⁡(2𝜂) (21)

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137 In case 𝐵2≠ 0, Eq.(21) can be simplified as:

𝑢 𝑥, 𝑡 = 𝐴0+ 𝐴1𝑒𝑥𝑝⁡(𝜂) + 𝐴2𝑒𝑥𝑝⁡(2𝜂)

𝐵0+ 𝐵1𝑒𝑥𝑝⁡(𝜂) + 𝑒𝑥𝑝⁡(2𝜂) 22 Substituting Eq. (22) into Eq. (20), and using the Maple, equating to zero the coefficients of all powers of 𝑒𝑥𝑝⁡(𝜂)yields a set of algebraic equations for 𝐴₀, 𝐴₁, 𝐴₂, 𝐵₀, 𝐵₁ and c. Solving the system of algebraic equation given above, with the aid of Maple, we obtain:

𝐴₀=^𝐵⁰^(−24+𝐴_𝛼+𝛽²^{𝛽 +𝐴}²^𝛼), 𝐴₁= 0, 𝐴₂= 𝐴₂, 𝐵₀= 𝐵₀, 𝐵₁= 0, 𝑐 =⁻¹₃.

𝑢1 𝑥, 𝑡 =

𝐵0(−24 + 𝐴2𝛽 + 𝐴2𝛼)

𝛼 + 𝛽 + 𝐴₂𝑒𝑥𝑝⁡(2𝑥 +2𝑡 3 ) 𝐵0+ 𝑒𝑥𝑝⁡(2𝑥 +2𝑡

3 )

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If 𝐴₂= 𝐵₀= 1, 𝛼 = 𝛽 = 2, then

𝑢1 𝑥, 𝑡 =−5 + 𝑒𝑥𝑝⁡(2𝑥 +2𝑡 3 ) 1 + 𝑒𝑥𝑝⁡(2𝑥 +2𝑡

3 )

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𝑭𝒊𝒈𝒖𝒓𝒆 𝟓. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟏𝟖 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟐𝟒 , 𝑨_𝟐= 𝑩_𝟎= 𝟏, 𝜶 = 𝜷 = 𝟐.

𝑪𝒂𝒔𝒆 𝟐: 𝒎 = 𝟑, 𝒏 = 𝟑:

𝑢 𝑥, 𝑡 = 𝐴0+ 𝐴1𝑒𝑥𝑝⁡(𝜂) + 𝐴2𝑒𝑥𝑝⁡(2𝜂) + 𝐴3𝑒𝑥𝑝⁡(3𝜂)

𝐵0+ 𝐵1𝑒𝑥𝑝⁡(𝜂) + 𝐵2𝑒𝑥𝑝⁡(2𝜂) + 𝐵3𝑒𝑥𝑝⁡(3𝜂) (25) In case 𝐵3≠ 0, Eq.(25) can be simplified as:

𝑢 𝑥, 𝑡 = 𝐴₀+ 𝐴₁𝑒𝑥𝑝⁡(𝜂) + 𝐴₂𝑒𝑥𝑝⁡(2𝜂) + 𝐴₃𝑒𝑥𝑝⁡(3𝜂)

𝐵0+ 𝐵1𝑒𝑥𝑝⁡(𝜂) + 𝐵2𝑒𝑥𝑝⁡(2𝜂) + 𝑒𝑥𝑝⁡(3𝜂) 26 Substituting Eq. (26) into Eq. (20), and using the Maple, equating to zero the coefficients of all powers of 𝑒𝑥𝑝⁡(𝜂) yields a set of algebraic equations for𝐴₀, 𝐴₁, 𝐴₂, 𝐴₃, 𝐵₀, 𝐵₁, 𝐵₂ and c. Solving the system of algebraic equation given above, with the aid of Maple, we obtain:

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138 𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:

𝐴0= 0, 𝐴1=𝐵₁(−24 + 𝐴₃𝛽 + 𝐴₃𝛼)

𝛼 + 𝛽 , 𝐴2= 0, 𝐴3= 𝐴3, 𝐵0= 0, 𝐵1= 𝐵1, 𝐵2= 𝐵2, 𝑐 =−1 3 Thus the following solution is

𝑢2 𝑥, 𝑡 =

𝐵1(−24 + 𝐴3𝛽 + 𝐴3𝛼)

𝛼 + 𝛽 𝑒𝑥𝑝⁡(𝑥 +𝑡

3) + 𝐴³𝑒𝑥𝑝⁡(3𝑥 + 𝑡) 𝐵1𝑒𝑥𝑝⁡(𝑥 +𝑡

3) + 𝑒𝑥𝑝⁡(3𝑥 + 𝑡)

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If 𝐵1= 𝐴3= 2, 𝛼 = 𝛽 = 1, then

𝑢₂ 𝑥, 𝑡 =−20𝑒𝑥𝑝⁡(𝑥 +𝑡

3) + 2𝑒𝑥𝑝⁡(3𝑥 + 𝑡) 2𝑒𝑥𝑝⁡(𝑥 +𝑡

3) + 𝑒𝑥𝑝⁡(3𝑥 + 𝑡)

(28)

𝑭𝒊𝒈𝒖𝒓𝒆 𝟔. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟏𝟖 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟐𝟖 , 𝑨𝟑= 𝑩𝟏= 𝟐, 𝜶 = 𝜷 = 𝟏.

𝐴0=𝐵0(𝐴3𝛽 + 𝐴3𝛼 − 36)

𝛼 + 𝛽 , 𝐴1= 0, 𝐴2= 0, 𝐴3= 𝐴3, 𝐵0 = 𝐵0, 𝐵1= 0, 𝐵2= 0, 𝑐 =−1 8 Thus the following solution is

𝑢3 𝑥, 𝑡 =

𝐵0(𝐴3𝛽 + 𝐴3𝛼 − 36)

𝛼 + 𝛽 + 𝐴₃𝑒𝑥𝑝⁡(3𝑥 +3𝑡 8 ) 𝐵0+ 𝑒𝑥𝑝⁡(3𝑥 +3𝑡

8 )

(29)

If 𝐵₀= 𝐴₃= 2, 𝛼 = 𝛽 = 1, then

𝑢3 𝑥, 𝑡 =−32 + 2𝑒𝑥𝑝⁡(3𝑥 +3𝑡 8 ) 2 + 𝑒𝑥𝑝⁡(3𝑥 +3𝑡

8 )

(30)

(8)

139

𝑭𝒊𝒈𝒖𝒓𝒆 𝟕. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟏𝟖 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟑𝟎 , 𝑨𝟑= 𝑩𝟎= 𝟐, 𝜶 = 𝜷 = 𝟏.

3.3 The Boussinesq equation: [38-42]

The Boussinesq-type equations, which include the lowest-order effects of nonlinearity and frequency dispersion as additions to the simplest non-dispersive linear long wave theory, provide a sound and increasingly well-tested basis for the simulation of wave propagation in coastal regions. The standard Boussinesq equations for variable water depth were first derived by Peregrine (1967), who used depth-averaged velocity as a dependent variable.

In the present paper we introduce new exact solutions of the Boussinesq equation via the improved Exp-function method as follows:

Consider the Boussinesq equation:

𝑢𝑡𝑡− 𝑢𝑥𝑥 − 𝑢𝑥𝑥𝑥𝑥 − 6(𝑢𝑥)²− 6𝑢𝑢𝑥𝑥 = 0 (31) Using the transformation:

𝑢 = 𝑢 𝜂 , 𝜉 = 𝑘𝑥 + 𝜔𝑡 32 , where 𝑘, 𝜔 are constants to be determined later. Substituting Eq. (32) into Eq. (31) we get

𝜔²𝑢^′′− 𝑘²𝑢^′′− 𝑘⁴𝑢^′′′′− 6𝑘²(𝑢^′)²− 6𝑘²𝑢𝑢^′′= 0 33 , where the prime denotes the differential with respect to 𝜉. Now we study the following cases:

𝑪𝒂𝒔𝒆 𝟏: 𝒎 = 𝟐, 𝒏 = 𝟑:

𝑢 𝑥, 𝑡 = 𝐴0+ 𝐴1𝑒𝑥𝑝⁡(𝜉) + 𝐴2𝑒𝑥𝑝⁡(2𝜉)

𝐵0+ 𝐵1𝑒𝑥𝑝⁡(𝜉) + 𝐵2𝑒𝑥𝑝⁡(2𝜉) + 𝐵3𝑒𝑥𝑝⁡(3𝜉) (34) In case 𝐵3≠ 0, Eq.(34) can be simplified as:

𝑢 𝑥, 𝑡 = 𝐴₀+ 𝐴₁𝑒𝑥𝑝⁡(𝜉) + 𝐴₂𝑒𝑥𝑝⁡(2𝜉)

𝐵0+ 𝐵1𝑒𝑥𝑝⁡(𝜉) + 𝐵2𝑒𝑥𝑝⁡(2𝜉) + 𝑒𝑥𝑝⁡(3𝜉) 35 Substituting Eq. (35) into Eq. (33), and using the Maple, equating to zero the coefficients of all powers of 𝑒𝑥 𝑝 𝜉 yields a set of algebraic equations for 𝐴₀, 𝐴₁, 𝐴₂, 𝐵₀, 𝐵₁, 𝐵₂, 𝑘 𝑎𝑛𝑑 𝜔. Solving the system of algebraic equation given above, with the aid of Maple, we obtain:

(9)

140 𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:

𝐴₀= 0, 𝐴₁= 0, 𝐴₂= 𝑘²𝐵₂, 𝐵₀= 0, 𝐵₁=𝐵22

4 , 𝐵₂= 𝐵₂, 𝜔 = 𝑘 1 + 𝑘² Thus the following solution is

𝑢₁ 𝑥, 𝑡 = 𝑘²𝐵2𝑒𝑥𝑝⁡(2𝑘𝑥 + 2𝑘𝑡 1 + 𝑘²)

𝐵22

4 𝑒𝑥𝑝⁡(𝑘𝑥 + 𝑘𝑡 1 + 𝑘²) + 𝐵2𝑒𝑥𝑝⁡(2𝑘𝑥 + 2𝑘𝑡 1 + 𝑘²) + 𝑒𝑥𝑝⁡(3𝑘𝑥 + 3𝑘𝑡 1 + 𝑘²)

(36)

If 𝑘 = 1, 𝐵2= 2, then

𝑢1 𝑥, 𝑡 = 2𝑒𝑥𝑝⁡(2𝑥 + 2𝑡 2)

𝑒𝑥𝑝⁡(𝑥 + 𝑡 2) + 2𝑒𝑥𝑝⁡(2𝑥 + 2𝑡 2) + 𝑒𝑥𝑝⁡(3𝑥 + 3𝑡 2) (37)

𝑭𝒊𝒈𝒖𝒓𝒆 𝟖. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟑𝟏 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟑𝟕 , 𝑩𝟐= 𝟐, 𝒌 = 𝟏.

𝐴0= 0, 𝐴1= 0, 𝐴2= 𝑘²𝐵2, 𝐵0= 0, 𝐵1=𝐵₂²

4 , 𝐵2= 𝐵2, 𝜔 = −𝑘 1 + 𝑘² Thus the following solution is

𝑢₂ 𝑥, 𝑡 = 𝑘²𝐵2𝑒𝑥𝑝⁡(2𝑘𝑥 − 2𝑘𝑡 1 + 𝑘²)

𝐵22

4 𝑒𝑥𝑝⁡(𝑘𝑥 − 𝑘𝑡 1 + 𝑘²) + 𝐵2𝑒𝑥𝑝⁡(2𝑘𝑥 − 2𝑘𝑡 1 + 𝑘²) + 𝑒𝑥𝑝⁡(3𝑘𝑥 − 3𝑘𝑡 1 + 𝑘²)

(38)

If 𝐵2= 2, 𝑘 = 1, then

𝑢2 𝑥, 𝑡 = 2𝑒𝑥 𝑝 2𝑥 − 2𝑡 2

𝑒𝑥𝑝⁡(𝑥 − 𝑡 2) + 2𝑒𝑥𝑝⁡(2𝑥 − 2𝑡 2) + 𝑒𝑥𝑝⁡(3𝑥 − 3𝑡 2) (39)

(10)

141

𝑭𝒊𝒈𝒖𝒓𝒆 𝟗. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟑𝟏 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟑𝟗 , 𝑩𝟐= 𝟐, 𝒌 = 𝟏.

𝑪𝒂𝒔𝒆 𝟐: 𝒎 = 𝟐, 𝒏 = 𝟒:

𝑢 𝑥, 𝑡 = 𝐴0+ 𝐴1𝑒𝑥 𝑝 𝜉 + 𝐴2𝑒𝑥 𝑝 2𝜉

𝐵0+ 𝐵1𝑒𝑥 𝑝 𝜉 + 𝐵2𝑒𝑥 𝑝 2𝜉 + 𝐵3𝑒𝑥 𝑝 3𝜉 + 𝐵4𝑒𝑥 𝑝 4𝜉 (40) In case 𝐵₄≠ 0, Eq.(40) can be simplified as:

𝑢 𝑥, 𝑡 = 𝐴0+ 𝐴1𝑒𝑥 𝑝 𝜉 + 𝐴2𝑒𝑥 𝑝 2𝜉

𝐵0+ 𝐵1𝑒𝑥 𝑝 𝜉 + 𝐵2𝑒𝑥 𝑝 2𝜉 + 𝐵3𝑒𝑥 𝑝 3𝜉 + 𝑒𝑥 𝑝 4𝜉 41 Substituting Eq. (41) into Eq. (33), and using the Maple, equating to zero the coefficients of all powers of 𝑒𝑥 𝑝 𝜉 yields a set of algebraic equations for 𝐴₀, 𝐴₁, 𝐴₂, 𝐵₀, 𝐵₁, 𝐵₂, 𝐵₃, 𝑘 and 𝜔. Solving the system of algebraic equation given above, with the aid of Maple, we obtain:

𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:

𝐴₀= 0, 𝐴₁= 0, 𝐴₂= 𝐴₂, 𝐵₀= 𝐴22

64𝑘⁴, 𝐵₁= 0, 𝐵₂= 𝐴2

4𝑘², 𝐵₃= 𝐵₃, 𝜔 = 𝑘 1 + 4𝑘² Thus the following solution is

𝑢3 𝑥, 𝑡 = 𝐴₂𝑒𝑥𝑝⁡(2𝑘𝑥 + 2𝑘𝑡 1 + 4𝑘²)

𝐴₂² 64𝑘⁴+ 𝐴₂

4𝑘²𝑒𝑥𝑝⁡(2𝑘𝑥 + 2𝑘𝑡 1 + 4𝑘²) + 𝑒𝑥𝑝⁡(4𝑘𝑥 + 4𝑘𝑡 1 + 4𝑘²)

(42)

If 𝐴₂= 2, 𝑘 = 1, then

𝑢₃ 𝑥, 𝑡 = 2𝑒𝑥𝑝⁡(2𝑥 + 2𝑡 5)

16 +1 1

2 𝑒𝑥𝑝⁡(2𝑥 + 2𝑡 5) + 𝑒𝑥𝑝⁡(4𝑥 + 4𝑡 5)

(43)

(11)

142

𝑭𝒊𝒈𝒖𝒓𝒆 𝟏𝟎. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟑𝟏 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟒𝟑 , 𝑨𝟐= 𝟐, 𝒌 = 𝟏.

𝐴0== 0, 𝐴1= 0, 𝐴2= 𝐴2, 𝐵0= 𝐴22

64𝑘⁴, 𝐵1= 0, 𝐵2= 𝐴2

4𝑘², 𝐵3= 𝐵3, 𝜔 = −𝑘 1 + 4𝑘² Thus the following solution is

𝑢4 𝑥, 𝑡 = 𝐴₂𝑒𝑥𝑝⁡(2𝑘𝑥 − 2𝑘𝑡 1 + 4𝑘²)

𝐴₂² 64𝑘⁴+ 𝐴₂

4𝑘²𝑒𝑥𝑝⁡(2𝑘𝑥 − 2𝑘𝑡 1 + 4𝑘²) + 𝑒𝑥𝑝⁡(4𝑘𝑥 − 4𝑘𝑡 1 + 4𝑘²)

(44)

If 𝐴2= 2, 𝑘 = 1, then

𝑢₄ 𝑥, 𝑡 = 2𝑒𝑥𝑝⁡(2𝑥 − 2𝑡 5)

16 +1 1

2 𝑒𝑥𝑝⁡(2𝑥 − 2𝑡 5) + 𝑒𝑥𝑝⁡(4𝑥 − 4𝑡 5)

(45)

𝑭𝒊𝒈𝒖𝒓𝒆 𝟏𝟏. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟑𝟑 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟒𝟓 , 𝑨𝟐= 𝟐, 𝒌 = 𝟏.

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143 4. CONCULSION

In this paper, the improved Exp-function method has been successfully applied to obtain new solutions of three nonlinear partial differential equations. Thus, the improved Exp-function method can be extended to solve the problems of nonlinear partial differential equations which arising in the theory of solitons and other areas.

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