Exact solutions of two nonlinear partial differential equations by using the first integral method

  • Hossein Jafari1, 2Email author,

    Affiliated with

    • Rahmat Soltani1,

      Affiliated with

      • Chaudry Masood Khalique2 and

        Affiliated with

        • Dumitru Baleanu3, 4, 5

          Affiliated with

          Boundary Value Problems20132013:117

          DOI: 10.1186/1687-2770-2013-117

          Received: 22 August 2012

          Accepted: 12 April 2013

          Published: 7 May 2013

          Abstract

          In recent years, many approaches have been utilized for finding the exact solutions of nonlinear partial differential equations. One such method is known as the first integral method and was proposed by Feng. In this paper, we utilize this method and obtain exact solutions of two nonlinear partial differential equations, namely double sine-Gordon and Burgers equations. It is found that the method by Feng is a very efficient method which can be used to obtain exact solutions of a large number of nonlinear partial differential equations.

          Keywords

          first integral method double sine-Gordon equation Burgers equation exact solutions

          1 Introduction

          With the availability of symbolic computation packages like Maple or Mathematica, the search for obtaining exact solutions of nonlinear partial differential equations (PDEs) has become more and more stimulating for mathematicians and scientists. Having exact solutions of nonlinear PDEs makes it possible to study nonlinear physical phenomena thoroughly and facilitates testing the numerical solvers as well as aiding the stability analysis of solutions. In recent years, many approaches to solve nonlinear PDEs such as the extended tanh function method [16], the modified extended tanh function method [7, 8], the exp-function method [911], the Weierstrass elliptic function method [12], the Laplace decomposition method [13, 14] and so on have been employed.

          Among these, the first integral method, which is based on the ring theory of commutative algebra, due to Feng [1519] has been applied by many authors to solve different types of nonlinear equations in science and engineering [2023]. Therefore, in the present article, the first integral method is applied to analytic treatment of some important nonlinear of partial differential equations.

          The rest of this article is arranged as follows. In Section 2, the basic ideas of the first integral method are expressed. In Section 3, the method is employed for obtaining the exact solutions of double sine-Gordon (SG) and Burgers equations, and finally conclusions are presented in Section 4.

          2 The first integral method

          Let
          P ( u , u t , u x , u x x , u x t , ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ1_HTML.gif
          (1)
          be a nonlinear partial differential equation (PDE) with u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq1_HTML.gif as its solution. We introduce the transformation
          u ( x , t ) = u ( ξ ) , ξ = x c t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ2_HTML.gif
          (2)
          where c is constant. Then we have
          t ( ) = c ξ ( ) , x ( ) = ξ ( ) , 2 t 2 ( ) = c 2 d 2 d ξ 2 ( ) , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ3_HTML.gif
          (3)
          Thus PDE (1) is then transformed to the ordinary differential equation (ODE)
          Q ( u , u ξ , u ξ ξ , ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ4_HTML.gif
          (4)
          We now introduce a new transformation, namely
          X ( ξ ) = u ( ξ ) , Y ( ξ ) = u ξ ( ξ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ5_HTML.gif
          (5)
          and this gives us the system of ODEs
          { X ξ ( ξ ) = Y ( ξ ) , Y ξ ( ξ ) = F ( X ( ξ ) , Y ( ξ ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ6_HTML.gif
          (6)

          If we can find the integrals of (6) under the same conditions, the qualitative theory of differential equations [24] tells us that the general solutions of (6) can be obtained directly. But in general, it is very difficult even for a single first integral. Since for a plane autonomous system, there is no methodical theory which gives us first integrals, we will therefore apply the division theorem to find one first integral (6), which will reduce (4) to a first-order integral for an ordinary differential equation. By solving this equation, exact solutions of (1) will be obtained. We recall the division theorem.

          Theorem 2.1 (Division theorem, see [25])

          Let P ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq2_HTML.gif and Q ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq3_HTML.gif be polynomials of two variables x and y in C [ x , y ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq4_HTML.gif, and let P ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq2_HTML.gif be irreducible in C [ x , y ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq4_HTML.gif. If Q ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq3_HTML.gif vanishes at all zero points of P ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq2_HTML.gif, then there exists a polynomial G ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq5_HTML.gif in C [ x , y ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq4_HTML.gif such that Q ( x , y ) = P ( x , y ) G ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq6_HTML.gif.

          The division theorem follows immediately from the Hilbert-Nullstellensatz theorem [26].

          Theorem 2.2 (Hilbert-Nullstellensatz theorem)

          Let K be a field and L be an algebraic closure of K. Then:
          1. (i)

            Every ideal γ of K [ X 1 , , X n ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq7_HTML.gif not containing 1 admits at least one zero in L n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq8_HTML.gif.

             
          2. (ii)

            Let x = ( x 1 , , x n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq9_HTML.gif, y = ( y 1 , , y n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq10_HTML.gif be two elements of L n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq8_HTML.gif; for the set of polynomials of K [ X 1 , , X n ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq7_HTML.gif zero at x to be identical with the set of polynomials of K [ X 1 , , X n ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq7_HTML.gif zero at y, it is necessary and sufficient that there exists a K-automorphism S of L such that y i = S ( x i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq11_HTML.gif for 1 i n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq12_HTML.gif.

             
          3. (iii)

            For an ideal α of K [ X 1 , , X n ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq7_HTML.gif to be maximal, it is necessary and sufficient that there exists an x in L n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq8_HTML.gif such that α is the set of polynomials of K [ X 1 , , X n ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq7_HTML.gif zero at x.

             
          4. (iv)

            For a polynomial Q of K [ X 1 , , X n ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq7_HTML.gif to be zero on the set of zeros in L n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq8_HTML.gif of an ideal γ of K [ X 1 , , X n ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq7_HTML.gif, it is necessary and sufficient that there exists an integer m > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq13_HTML.gif such that Q m γ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq14_HTML.gif.

             

          3 Applications

          3.1 Exact solutions to the double sine-Gordon equation

          Consider the double sine-Gordon (SG) equation [27, 28]
          u x t = sin u + sin 2 u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ7_HTML.gif
          (7)
          In order to apply the first integral method described in Section 2, we first introduce the transformations
          sin u = v v 1 2 i , sin 2 u = v 2 v 2 2 i , v = e i u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ8_HTML.gif
          (8)
          Using (2) and (3), Eq. (7) becomes
          c u = v v 1 2 i + v 2 v 2 2 i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ9_HTML.gif
          (9)
          We next use the transformation
          { v = e i u , u = v i v , u = 1 i ( v v ( v ) 2 v 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ10_HTML.gif
          (10)
          We obtain
          2 c v v + 2 c ( v ) 2 v 4 v 3 + v + 1 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ11_HTML.gif
          (11)
          Next, we introduce new independent variables X = v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq15_HTML.gif, Y = d v d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq16_HTML.gif which change (11) to the system of ODEs
          { d X d ξ = Y , d Y d ξ = 2 c Y 2 X 4 X 3 + X + 1 2 c X . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ12_HTML.gif
          (12)
          Assume that d ξ = X d τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq17_HTML.gif, then (12) becomes
          { d X d τ = X Y , d Y d τ = Y 2 a X 4 a X 3 + a X + a , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ13_HTML.gif
          (13)

          where a = 1 2 c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq18_HTML.gif.

          According to the first integral method, we suppose that X = X ( τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq19_HTML.gif and Y = Y ( τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq20_HTML.gif are nontrivial solutions of Eq. (13) and
          P ( X , Y ) = i = 0 m a i ( X ) Y i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equa_HTML.gif
          is an irreducible polynomial in the complex domain C [ X , Y ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq21_HTML.gif such that
          P ( X ( τ ) , Y ( τ ) ) = i = 0 m a i ( X ( τ ) ) Y ( τ ) i = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ14_HTML.gif
          (14)
          where a i ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq22_HTML.gif ( i = 0 , 1 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq23_HTML.gif) are polynomials in X and a m ( X ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq24_HTML.gif. Equation (14) is called the first integral to Eq. (13). Applying the division theorem, one sees that there exists a polynomial H ( X , Y ) = h ( X ) + g ( X ) Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq25_HTML.gif in the complex domain C [ X , Y ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq26_HTML.gif such that
          d P d τ = P X X τ + P Y Y τ = ( h ( X ) + g ( X ) Y ) ( i = 0 m a i ( X ) Y i ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ15_HTML.gif
          (15)
          Suppose that m = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq27_HTML.gif in (14), and then, by comparing with the coefficients of Y i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq28_HTML.gif ( i = 2 , 1 , 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq29_HTML.gif) on both sides of (15), we have
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ16_HTML.gif
          (16)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ17_HTML.gif
          (17)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ18_HTML.gif
          (18)
          Since a 1 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq30_HTML.gif is a polynomial in X, from (16) we conclude that a 1 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq30_HTML.gif is a constant and g ( X ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq31_HTML.gif. For simplicity, we take a 1 ( X ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq32_HTML.gif. Then Eq. (17) indicates that deg h ( X ) deg a 0 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq33_HTML.gif. Thus, from Eq. (18) we conclude that deg h ( X ) = deg a 0 ( X ) = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq34_HTML.gif. Now suppose that
          h ( X ) = A 2 X 2 + A 1 X + A 0 , a 0 ( X ) = B 2 X 2 + B 1 X + B 0 ( A 2 0 , B 2 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ19_HTML.gif
          (19)
          where A 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq35_HTML.gif, A 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq36_HTML.gif, A 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq37_HTML.gif, B 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq38_HTML.gif, B 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq39_HTML.gif, B 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq40_HTML.gif are all constants to be determined. Substituting Eq. (19) into Eq. (17), we obtain
          A 2 = B 2 , A 1 = 0 , A 0 = B 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equb_HTML.gif
          Then
          h ( X ) = B 2 X 2 B 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equc_HTML.gif
          Substituting a 0 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq41_HTML.gif, a 1 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq30_HTML.gif and h ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq42_HTML.gif in (18) and setting all the coefficients of powers X to be zero, we obtain a system of nonlinear algebraic equations, and by solving it, we obtain
          B 2 = B 1 = B 0 = ± a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ20_HTML.gif
          (20)
          Substituting (20) in (14), we obtain
          Y ± a ( X 2 + X + 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ21_HTML.gif
          (21)
          Combining Eq. (21) with (13), second-order differential Eq. (11) can be reduced to
          d v d ξ = a ( v 2 + v + 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ22_HTML.gif
          (22)
          Solving Eq. (22) directly and changing to the original variables, we obtain the exact solutions to Eq. (7):
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ23_HTML.gif
          (23)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ24_HTML.gif
          (24)

          where c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq43_HTML.gif is an arbitrary constant and a = 1 2 c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq18_HTML.gif.

          Therefore, the exact solutions to the double sine-Gordon (SG) equation can be written as
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ25_HTML.gif
          (25)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ26_HTML.gif
          (26)

          where c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq43_HTML.gif is an arbitrary constant.

          3.2 Exact solutions to the Burgers equation

          The Burgers equation [29]
          u t + u u x a u x x = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ27_HTML.gif
          (27)

          is one of the most famous nonlinear diffusion equations. The positive parameter a refers to a dissipative effect.

          Using (2) and (3), Eq. (27) becomes
          c u + u u a u = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ28_HTML.gif
          (28)
          We rewrite (28) as follows:
          u = 1 a u u c a u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ29_HTML.gif
          (29)
          By introducing new variables X = u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq44_HTML.gif and Y = u ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq45_HTML.gif, Eq. (29) changes into a system of ODEs
          { X ( ξ ) = Y , Y ( ξ ) = 1 a X Y c a Y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ30_HTML.gif
          (30)
          Now, the division theorem is employed to seek the first integral to (30). Suppose that X = X ( ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq46_HTML.gif and Y = Y ( ξ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq47_HTML.gif are nontrivial solutions to (30), and P ( X , Y ) = i = 0 m a i ( X ) Y i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq48_HTML.gif is an irreducible polynomial in C [ X , Y ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq49_HTML.gif such that
          P ( X ( ξ ) , Y ( ξ ) ) = i = 0 m a i ( X ) Y i = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ31_HTML.gif
          (31)
          where a i ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq22_HTML.gif ( i = 0 , 1 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq23_HTML.gif) are polynomials in X and a m ( X ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq24_HTML.gif. Equation (31) is called the first integral to Eq. (30). Due to the division theorem, there exists a polynomial H ( X , Y ) = h ( X ) + g ( X ) Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq25_HTML.gif in C [ X , Y ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq21_HTML.gif such that
          d P d ξ = P X X ξ + P Y Y ξ = ( h ( X ) + g ( X ) Y ) ( i = 0 m a i ( X ) Y i ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ32_HTML.gif
          (32)
          Suppose that m = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq27_HTML.gif in (31). By comparing the coefficients of Y i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq28_HTML.gif ( i = 2 , 1 , 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq29_HTML.gif) on both sides of (32), we have
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ33_HTML.gif
          (33)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ34_HTML.gif
          (34)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ35_HTML.gif
          (35)
          Since a 1 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq30_HTML.gif is a polynomial in X, from (33) we conclude that a 1 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq30_HTML.gif is a constant and g ( X ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq50_HTML.gif. For simplicity, we take a 1 ( X ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq32_HTML.gif, and balancing the degrees of h ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq42_HTML.gif and a 0 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq41_HTML.gif, we conclude that deg h ( X ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq51_HTML.gif or 1. If deg h ( X ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq52_HTML.gif, suppose that h ( X ) = A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq53_HTML.gif, then from (34), we find
          a 0 ( X ) = 1 2 a X 2 + ( A + c a ) X + B , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equd_HTML.gif

          where B is an arbitrary integration constant.

          Substituting a 0 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq41_HTML.gif and h ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq42_HTML.gif in (35) and setting all the coefficients of powers X to be zero, we obtain a system of nonlinear algebraic equations, and by solving it, we obtain
          A = 0 , B 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ36_HTML.gif
          (36)
          Using (36) in (31), we obtain
          Y ( ξ ) = 1 2 a X 2 c a X B . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ37_HTML.gif
          (37)
          Combining Eq. (37) with the first part of (30), we obtain the exact solutions of Eq. (29) as follows:
          X 1 ( ξ ) = c + 2 a B + c 2 [ 1 tanh ( 2 a B + c 2 2 a ( ξ + ξ 0 ) ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ38_HTML.gif
          (38)

          where ξ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq54_HTML.gif is an arbitrary constant.

          Therefore, the exact solutions to the Burgers equation can be written as
          u 1 ( x , t ) = c + 2 a B + c 2 [ 1 tanh ( 2 a B + c 2 2 a ( x c t + ξ 0 ) ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ39_HTML.gif
          (39)

          where ξ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq54_HTML.gif is an arbitrary constant and a > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq55_HTML.gif.

          Now suppose that m = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq56_HTML.gif. By an application of the division theorem, we can conclude that there exists a polynomial H ( X , Y ) = h ( X ) + g ( X ) Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq57_HTML.gif in C [ X , Y ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq21_HTML.gif such that
          d P d ξ = P X X ξ + P Y Y ξ = ( h ( X ) + g ( X ) Y ) ( i = 0 2 a i ( X ) Y i ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ40_HTML.gif
          (40)
          Comparing the coefficients of Y i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq28_HTML.gif ( i = 3 , 2 , 1 , 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq58_HTML.gif) of both sides of (40) yields
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ41_HTML.gif
          (41)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ42_HTML.gif
          (42)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ43_HTML.gif
          (43)
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ44_HTML.gif
          (44)
          Since a 2 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq59_HTML.gif is a polynomial of X, from (41) we conclude that a 2 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq59_HTML.gif is a constant and g ( X ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq60_HTML.gif. For simplicity, we take a 2 ( X ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq61_HTML.gif, and balancing the degrees of h ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq42_HTML.gif, a 0 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq41_HTML.gif and a 1 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq30_HTML.gif, we conclude that deg h ( X ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq52_HTML.gif or 1. If deg h ( X ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq52_HTML.gif, suppose that h ( X ) = A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq62_HTML.gif. Then from (42), we find
          a 1 ( X ) = 1 a X 2 + ( A + 2 c a ) X + B , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Eque_HTML.gif
          where B is an arbitrary constant of integration. From (43) we have
          a 0 ( X ) = 1 4 a 2 X 4 ( 2 3 a A + c a 2 ) X 3 + ( A 2 2 + 3 c 2 a A + c 2 a 2 B 2 a ) X 2 + ( A B + c B a ) X + D , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equf_HTML.gif
          where D is an arbitrary constant of integration. Substituting a 0 ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq41_HTML.gif and h ( X ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq42_HTML.gif in (44) and setting all the coefficients of powers of X to zero, we obtain a system of nonlinear algebraic equations. Solving these equations, we obtain
          A = 0 , B 0 , D 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ45_HTML.gif
          (45)
          Now using (45) in (31), we get
          Y ( ξ ) = 1 2 a X 2 c a X B 2 ± B 2 4 D 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ46_HTML.gif
          (46)
          Combining Eq. (46) with the first part of (30), we obtain the exact solutions to Eq. (29) in the form
          X 2 ( ξ ) = c + a B + c 2 a B 2 4 D × [ 1 tanh ( a B + c 2 a B 2 4 D 2 a ( ξ + ξ 0 ) ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ47_HTML.gif
          (47)
          where ξ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq54_HTML.gif is an arbitrary constant of integration. Therefore, the exact solutions to the Burgers equation can be written as
          u 2 ( x , t ) = c + a B + c 2 a B 2 4 D × [ 1 tanh ( a B + c 2 a B 2 4 D 2 a ( x c t + ξ 0 ) ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_Equ48_HTML.gif
          (48)

          where ξ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq54_HTML.gif is an arbitrary constant and a > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-117/MediaObjects/13661_2012_Article_364_IEq55_HTML.gif.

          4 Conclusions

          The first integral method was employed successfully to solve some important nonlinear partial differential equations, including the double sine-Gordon and Burgers equations, analytically. Some exact solutions for these equations were formally obtained by applying the first integral method. Due to the good performance of the first integral method, we feel that it is a powerful technique in handling a wide variety of nonlinear partial differential equations. Also, this method is computerizable, which permits us to accomplish difficult and tiresome algebraic calculations on a computer with ease.

          Declarations

          Authors’ Affiliations

          (1)
          Department of Mathematics, University of Mazandaran
          (2)
          International Institute for Symmetry Analysis and Mathematical Modelling, Department of Mathematical Sciences, North-West University
          (3)
          Department of Mathematics and Computer Sciences, Faculty of Art and Sciences, Çankaya University
          (4)
          Department of Chemical and Materials Engineering, Faculty of Engineering, King Abdulaziz University
          (5)
          Institute of Space Sciences

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          © Jafari et al.; licensee Springer. 2013

          This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.