# Bilinear Equation of the Nonlinear Partial Differential Equation and Its Application

Authors: Xiao-Feng Yang; Yi Wei
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The homogeneous balance of undetermined coefficient method is firstly proposed to derive a more general bilinear equation of the nonlinear partial differential equation (NLPDE). By applying perturbation method, subsidiary ordinary differential equation (sub-ODE) method, and compatible condition to bilinear equation, more exact solutions of NLPDE are obtained. The KdV equation, Burgers equation, Boussinesq equation, and Sawada-Kotera equation are chosen to illustrate the validity of our method. We find that the underlying relation among theG′/G-expansion method, Hirota’s method, and HB method is a bilinear equation. The proposed method is also a standard and computable method, which can be generalized to deal with other types of NLPDE.


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## Body

## 1. Introduction

The nonlinear partial differential equation (NLPDE) is known to describe a wide variety of phenomena not only in physics but also in biology, chemistry, and several other fields [1–3]. The investigation of the exact solutions for NLPDE plays an important role on the study of nonlinear physical phenomena [4–9]. In recent years, many powerful methods are used to obtain the exact solutions of NLPDE, for example, the inverse scattering method [7], Bäcklund and Darboux transformation method [8], homotopy perturbation method [9], first integral method [10–12], variational iteration method [13, 14], sub-ODE method [15–17], Jacobi elliptic function method [18], tanh-sech method [19], G′/G-expansion method [20, 21], Hirota’s method [22–24], and homogeneous balance (HB) method [25–27].As the direct methods, theG′/G-expansion method, Hirota’s method, and HB method are very effective for constructing the exact solutions of NLPDE. Exact traveling wave solutions, N-soliton solutions, and solitary wave solutions of some NLPDE are obtained by using the above three methods [20–30]. Fan improved the HB method to investigate the BT, Lax pairs, symmetries, and exact solutions for some NLPDE [31, 32]. He also showed that there are many links among the HB method, Weiss-Tabor-Carnevale method, and Clarkson-Kruskal method.However, there is no unified direct method which can be used to deal with all types of NLPDE. And also, no literature is available to illustrate the underlying relations among the three direct methods.In the present paper, by improving some key steps in the HB method [26], we propose a new method, HB of undetermined coefficient method, which can be used to derive the bilinear equation of NLPDE. Based on the bilinear equation, by applying the perturbation method, sub-ODE method, and compatible condition, more exact solutions of NLPDE are obtained. We illustrate the real meaning of balance numbers. We show the underlying relations among the G′/G-expansion method, Hirota’s method, and HB method.This paper is organized as follows: the HB of undetermined coefficient method is described in Section2. In Sections 3 and 4, the KdV equation and Burgers equation are chosen as examples to illustrate the method, respectively. In Section 5, the bilinear equations of Boussinesq equation and Sawada-Kotera equation are derived, respectively. Some brief conclusions are given in Section 6.


## 2. Description of the HB of Undetermined Coefficient Method

Let us consider a general NLPDE, say, in two variables(1)Pu,ut,ux,uxx,uxt,⋯=0,where P is a polynomial function of its arguments and the subscripts denote the partial derivatives. The HB of undetermined coefficient method consists of three steps.Step 1.
Suppose that the solution of Equation (1) is of the form
(2)u=amnlnwm,n+∑i,j=0i+j≠0,m+ni=m,j=naijlnwi,j+a00,where u=ux,t, w=wx,t, lnwi,j=∂i+jlnwx,t/∂xi∂tj, m, n (balance numbers), and aiji=0,1,⋯,m,j=0,1,⋯,n (balance coefficients) are constants to be determined later. By balancing the highest nonlinear terms and the highest order partial derivative terms, balance numbers are obtained. Substituting Equation (2) into Equation (1) and balancing the terms with wx/wiwt/wj yield a set of algebraic equations for aiji=1,⋯,m,j=1,⋯,n.Step 2.
Solving the set of algebraic equations and simplifying Equation (1), we can get the bilinear equation of Equation (1) directly or after integrating some times (Generally, integrating times equal to the orders of the lowest partial derivative of Equation (1).) with respect to x,t.Step 3.
Generally, in order to obtain the exact solutions of Equation (1), there are three methods to deal with the bilinear equation of Equation (1):
(i)
Applying the perturbation method to the bilinear equation of Equation (1), N-soliton solution of Equation (1) can be obtained.
(ii)
By using traveling wave transformations(3)wx,t=wξ,ξ=x−Vt,the bilinear equation of Equation (1) satisfies the following ODE:
(4)w″+λw′+µw=0,where the prime denotes the derivation with respect to ξ and λ, μ, and V are constants to be determined later.
Substituting Equations (3) and (4) into Equation (1), it is converted into the following equation:
(5)l1w2+l2ww′+l3w′2=0,where l1, l2, and l3 are polynomial functions of V, λ, and μ.
Settingl1=l2=l3=0 yields a set of algebraic equations for V, λ, and μ. Solving the set of algebraic equations and using the solutions of Equation (4), w can be determined. Substituting w into Equation (2), the exact traveling wave solutions of Equation (1) are obtained.
(iii)
By applying the compatible conditionwxt=wtx to the bilinear equation of Equation (1), it is reduced to an ODE. Solving the ODE, more exact solutions of Equation (1) can be obtained.Next, we choose the KdV equation and Burgers equation to illustrate our method.


## 3. Application to the KdV Equation

Let us consider the celebrated KdV equation in the form(6)ut+uux+δuxxx=0,where δ is a constant. Suppose that the solution of Equation (6) is of the form
(7)u=amnlnwm,n+∑i,j=0i+j≠0,m+ni=m,j=naijlnwi,j+a00,where m, n, and aiji=0,1,⋯,m,j=0,1,⋯,n are constants to be determined later.Balancinguxxx and uux in Equation (6), it is required that m+3=2m+1 and n=2n. Then, Equation (7) can be written as
(8)u=a20lnwxx+a10lnwx+a00.From Equation (8), one can calculate the following derivatives:
(9)u=a20wxxw−wx2w2+a10wxw+a00,ux=a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2,ut=a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2,uxxx=a20wxxxxxw−5wxxxxwx+10wxxxwxxw2+20wxxxwx2+30wxx2wxw3−60wxxwx3w4+24wx5w5+a10wxxxxw−4wxxxwx+3wxx2w2+12wxxwx2w3−6wx4w4.Equating the coefficients ofwx/w5 and wx/w4 on the left-hand side of Equation (6) to zero yields a set of algebraic equations for a20 and a10 as follows:
(10)−2a202+24δa20=0,.3a20a10−6δa10=0.Solving the above algebraic equations, we geta20=12δ and a10=0. Substituting a20 and a10 back into Equation (8), we get
(11)u=12δlnwxx+a00.where a00 is an arbitrary constant. Substituting Equation (11) into Equation (6), we get
(12)12δK1+K2+K3=0,where
(13)K1=wxxtw−2wxwxt+wxxwtw2+2wx2wtw3,K2=a00wxxxw−3wxxwxw2+2wx3w3,K3=δwxxxxxw+2wxxxwxx−5wxxxxwxw2+16wxxxwx2−6wxwxx2w3.Simplifying Equation (12) and integrating once with respect to x, we get
(14)∂∂xwxtw−wxwt+δwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2=0.Equation (14) is identical to
(15)wxtw−wxwt+δwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2−Ctw2=0,where Ct is an arbitrary function of t and a00 is an arbitrary constant.In particular, takingCt as zero in Equation (15), we get the bilinear equation of Equation (6) as follows:
(16)wxtw−wxwt+δwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2=0.Equation (16) can be written concisely in terms of D-operators as
(17)DxDt+δDx4+a00Dx2w·w=0,where
(18)DxmDtna·b=∂x−∂x′m∂t−∂t′nax,tbx′,t′x′=x,t′=t.Remark 1.
Applying Hirota’s method [22–24], the bilinear equation of Equation (6) can be written as
(19)DxDt+δDx4w·w=0.
Equation (19) is obtained by setting a00=0 in Equation (17). Obviously, Equation (19) is a special case of Equation (17).
(i)
Now, we apply the perturbation method to Equation (17) to derive N-soliton solution of Equation (6). Suppose that w can be expanded as follows:
(20)w=1+εw1+ε2w2+⋯εNwN+⋯,where ε is a parameter and wi=wix,ti=1,2,⋯.
Substituting Equation (20) into Equation (17) and arranging it at each order of ε, we get
(21)ε:DxDt+a00Dx+δDx3w1·1+1·w1=0,ε2:DxDt+a00Dx+δDx3w2·1+w1·w1+1·w2=0,ε3:DxDt+a00Dx+δDx3w3·1+w2·w1+w1·w2+1·w3=0,⋯
The order-ε equation can be rewritten as a linear differential equation for w1 as follows:
(22)∂∂x∂∂t+a00∂∂x+δ∂3∂x3w1=0.
Solving Equation (22), we get
(23)w1=eP1x−a00P1+δP13t,where P1 is an arbitrary constant.
The coefficient ofε2 can be rearranged as follows:
(24)2∂∂x∂∂t+a00∂∂x+δ∂3∂x3w2=−DxDt+a00Dx+δDx3w1⋅w1.
Substituting Equation (23) into Equation (24), the right-hand side of Equation (24) equals zero. Therefore, we can choose
(25)w2=0.
Substituting Equations (23) and (25) into Equation (20), we get
(26)w=1+eP1x−a00P1+δP13t+ξ10,where P1, a00, and ξ10 are arbitrary constants.
Substituting Equation (26) into Equation (11), 1-soliton solution of Equation (6) can be obtained. If we choose w1=eP1x−a00P1+δP13t+eP2x−a00P2+δP23t in Equation (22), similar to the process of obtaining 1-soliton solution, we can get 2-soliton solution of Equation (6) as follows:
(27)w=1+eη1+eη2+P1−P22P1−P22eη1+η2,where η1=P1x−a00P1+δP13t+ξ10, η2=P2x−a00P2+δP23t+ξ20, Pi,ξi0i=1,2, and a00 are arbitrary constants.
Substituting Equation (27) into Equation (11), 2-soliton solution of Equation (6) can be obtained. Similarly, we can get N-soliton solution of Equation (6).Remark 2.
Obviously, settinga00=0 in Equations (23) and (27), 1-soliton and 2-soliton solutions of Equation (6) are identical to Hirota’s results [22–24].Remark 3.
By using the properties ofD-operators [22–24], a Bäcklund transformation of Equation (17) can be obtained as follows:
(28)Dt+a00+αDx+δDx3w∗·w=0,Dx2−βDx−αw∗·w=0,where w∗ and w satisfy Equation (17) and α, β, and a00 are arbitrary constants.
(ii)
Now, we discuss Equation (16) by using the sub-ODE method.
Using transformationswx,t=wξ,ξ=x−Vt, Equation (16) is reduced to
(29)a00−Vw″w−w′2+δw″″w−4w′w″′+3w″2=0,where the prime denotes the derivation with respect to ξ and V is a constant to be determined later.
Noticing the bilinear property of Equation (16), suppose that w satisfies the following ODE:
(30)w″+λw′+μw=0,where λ and μ are parameters.
Substituting Equation (30) into Equation (16), we get
(31)l1w2+l2ww′+l3w′2=0,where
(32)l1=μV−a00+δ4μ−λ2,l2=λV−a00+δ4μ−λ2,l3=V−a00+δ4μ−λ2.
Settingl1=l2=l3=0 yields a set of algebraic equations for V, λ, and µ. Solving this set of algebraic equations, we get
(33)V=a00+δλ2−4μ,where λ, µ, and a00 are arbitrary constants.
Substituting Equation (30) into Equation (11), we get
(34)u=12δwxw+λ22+3δλ2−4μ+a00.
Substituting the general solutions of Equation (30) into Equation (34), we get three types of traveling wave solutions of Equation (6) as follows.
Whenλ2−4µ>0,
(35)u1x,t=u1ξ=−3δAC1eA/2ξ−C2e−A/2ξC1eA/2ξ−C2e−A/2ξ2+3δA+a00,where
(36)A=λ2−4µ,V=a00+Aδ,ξ=x−a00+Aδt,and λ, μ, C1, C2, and a00 are arbitrary constants.
TakingC1=C3+C4/2 and C2=C3−C4/2, Equation (35) can be rewritten as
(37)u2x,t=u2ξ=−3δAC3sinhA/2ξ+C4coshA/2ξC3coshA/2ξ+C4sinhA/2ξ2+3δA+a00,where C3, C4, and a00 are arbitrary constants and A, V, and ξ are given by Equation (36).
In particular, ifC4/C3<1, then Equation (37) is reduced to
(38)u3x,t=u3ξ=3δAsech2A2ξ+ξ0+a00,where C3, C4, and a00 are arbitrary constants and A, V, and ξ are given by Equation (36), ξ0=arctanhC4/C3.
Whenλ2−4μ<0,
(39)u4x,t=u4ξ=3δA−C1sin−A/2ξ+C2cos−A/2ξC1cos−A/2ξ+C2sin−A/2ξ2+3δA+a00,where C1, C2, and a00 are arbitrary constants and A, V, and ξ are given by Equation (36).
Obviously, Equation (39) can be written as
(40)u5x,t=u5ξ=3δAsec2−A2ξ+ξ0+a00,where C1, C2, and a00 are arbitrary constants and A, V, and ξ are given by Equation (36), ξ0=−arctanC2/C1.
Whenλ2−4µ=0,
(41)u6x,t=u6ξ=−12δC2C1+C2ξ2+a00,where V=a00, ξ=x−a00t, C1, C2, and a00 are arbitrary constants.
(iii)
Now, we discuss Equation (16) from the compatible condition. Equation (16) can be written as
(42)δwxxxx+wxt+a00wxxw+δ3wxx2−4wxxxwx−a00wx2−wxwt=0.
Noticewx≠0; otherwise, we can only get a trivial solution. Setting the second term of Equation (42) to zero and solving wt yield
(43)wt=δ3wxx2−4wxxxwx−a00wx2wx.
Substituting Equation (43) into Equation (42), we get
(44)wxxxx+wxx3−2wxwxxwxxxwx2=0.
Integrating Equation (44) once with respect to x, we get
(45)wxxx−wxx2wx=bt,where bt is an arbitrary function of t.
Using transformationsY=wxx2 and X=wx, Equation (45) is reduced to
(46)dYdX=2YX+2bt.
Solving the above equation, we get(47)Y=ctX2−2btX,namely,
(48)wxx=ctwx2−2btwx,where bt and ct are arbitrary functions of t.Case 1.
Whenbt=ct=0, from Equation (48), we get
(49)w=c1tx+c2t,where c1t and c2t are arbitrary functions of t.
Substituting the above equation into Equation (48), we get
(50)xdc1tdt+dc2tdt=−a00c1t.
Setting the coefficients ofxii=0,1 to zero in the above equation, we get
(51)dc1tdt=0,dc2tdt=−a00c1t.
Solving the above equations, we get(52)c1t=C1,c2t=C2−a00C1t,where Cii=1,2 are arbitrary constants. Then, we get
(53)w=C1x−a00t+C2.
Substituting Equation (53) into Equation (11), we get an exact solution of Equation (6) as follows:
(54)u7x,t=a00−12δC12C1x−a00t+C22,where Cii=1,2 and a00 are arbitrary constants.Case 2.
Whenbt=0 and ct>0, similar to Case 1, we get
(55)w=C3+C2eC1x−a00+δC12t,and an exact solution of Equation (6) as follows:
(56)u8x,t=12δC12C2C3eC1x−a00+δC12tC3+C2eC1x−a00+δC12t2+a00,where Cii=1,2,3 and a00 are arbitrary constants.Case 3.
Whenbt≠0 and ct=0 similar to Case 1, we get
(57)w=C1x−a00t+C23+12δC1t+C3,and an exact solution of Equation (6) as follows:
(58)u9x,t=a00−36δC1x−a00t+C2C1x−a00t+C23−24δC1t−2C3C1x−a00t+C23−12δC1t+C32,where Cii=1,2,3 and a00 are arbitrary constants.Case 4.
Whenbt≠0 and ct>0, similar to Case 1, we get
(59)w=C4eC1x−a00+δC12t+C3e−C1x−a00+δC12t+C2x−C2a00+3δC12t+C5,and an exact solution of Equation (6) as follows:
(60)u10x,t=12δA1−B1+a00,where
(61)A1=C12C4eC1x−a00+δC12t+C12C3e−C1x−a00+δC12tC4eC1x−a00+δC12t+C3e−C1x−a00+δC12t+C2x−C2a00+3δC12t+C5,B1=C1C4eC1x−a00+δC12t−C1C3e−C1x−a00+δC12t+C22C4eC1x−a00+δC12t+C3e−C1x−a00+δC12t+C2x−C2a00+3δC12t+C52,where C22+4C12C3C4=0, Cii=1,2,3,4,5, and a00 are arbitrary constants.Case 5.
Whenbt≠0 and ct<0, similar to Case 1, we get
(62)w=−C1C2x+C1cosC2x+δC22t−a00t+C3+3δC1C23t−a00C1C2t−C4C23C23,and an exact solution of Equation (6) as follows:
(63)u11x,t=12δA2−B2+a00,where
(64)A2=−C1C22cosC2x+δC22t−a00t+C3C1C2x+C1cosC2x+δC22t−a00t+C3+3δC1C23t−C1C2a00t−C23C4,B2=C12C22−1+sinC2x+δC22t−a00t+C32C1C2x+C1cosC2x+δC22t−a00t+C3+3δC1C23t−a00C1C2t−C23C42,where Cii=1,2,3,4 and a00 are arbitrary constants.So far, based on the bilinear equation which is derived by using the HB of undetermined coefficient method, many exact solutions of the KdV are obtained by applying the perturbation method, sub-ODE method, and compatible condition. Our results can compare with theG′/G-expansion method, Hirota’s method, and HB method [20–30].


## 4. Application to the Burgers Equation

Let us consider the Burgers equation in the form(65)ut+uux+δuxx=0,where δ is a constant.Suppose that the solution of Equation (65) is of the form
(66)u=amnlnwm,n+∑i,j=0i+j≠0,m+ni=m,j=naijlnwi,j+a00,where m, n, and aiji=0,1,⋯,m,j=0,1,⋯,n are constants to be determined later.Balancinguxx and uux in Equation (65), it is required that m+2=2m+1 and n=2n. Then, Equation (66) can be written as
(67)u=a10lnwx+a00.From Equation (67), one can calculate the following derivatives:
(68)u=a10wxw+a00,ux=a10wxxw−wx2w2,ut=a10wxtw−wxwtw2,uxx=a10wxxxw−3wxxwxw2+2wx3w3.Equating the coefficients ofwx/w3 on the left-hand side of Equation (65) to zero yields an algebraic equation for a10 as follows:
(69)−a102+2a10δ=0.Solving the above algebraic equation, we geta10=2δ. Substituting a10 back into Equation (67), we get
(70)u=2δlnwx+a00,where a00 is an arbitrary constant.Substituting Equation (70) into Equation (65), we get
(71)2δK1+K2+K3=0,where
(72)K1=wxtw−wxwtw2,K2=a00wxxw−wx2w2,K3=δwwxxx−wxwxxw2.Simplifying Equation (71), we get
(73)wxtw−wxwt+a00wxxw−wx2+δwwxxx−wxwxx=0.Equation (73) can be written concisely in terms of D-operators as
(74)Dxwt+δwxx+a00wx·w=0.Equation (74) is identical to
(75)wt+δwxx+a00wx−Ctw=0,where Ct is an arbitrary function of t and a00 is an arbitrary constant.In particular, takingCt as constant C in Equation (75), we get
(76)wt+δwxx+a00wx−Cw=0.Remark 4.
Applying Hirota’s method [22–24], the bilinear equation of Equation (74) can be written as
(77)Dxwt+δwxx⋅w=0.
Equation (77) is obtained by setting a00=0 in Equation (74). Obviously, Equation (77) is a special case of Equation (74).Remark 5.
Equations (70) and (76) are general Cole-Hopf transformations. In fact, setting a00=C=0 in Equations (70) and (76), we get the famous Cole-Hopf transformations
(78)u=2δlnwx,wt+δwxx=0.
(i)
Now, we apply the perturbation method to Equation (74) to derive N-soliton solution of Equation (65). Suppose that w can be expanded as follows:
(79)w=1+εw1+ε2w2+⋯εNwN+⋯,where ε is a parameter and wi=wix,ti=1,2,⋯.
Substituting Equation (79) into Equation (74) and arranging it at each order of ε, we get
(80)ε:12Dxt+a00Dx2w1⋅1+1⋅w1+δDxw1xx⋅1=0,ε2:12Dxt+a00Dx2w2⋅1+w1⋅w1+1⋅w2+δDxw1xx⋅w1+w2xx⋅1=0,ε3:12Dxt+a00Dx2w3⋅1+w2⋅w1+w1⋅w2+1⋅w3+δDxw1xx⋅w2+w2xx⋅w1+w3xx⋅1=0,⋯
The order-ε equation can be rewritten as a linear differential equation for w1 as follows:
(81)∂∂x∂∂t+a00∂∂x+δ∂2∂x2w1=0.
Solving Equation (81), we get
(82)w1=eP1x−a00P1+δP12t,where P1 is an arbitrary constant.
The coefficient ofε2 can be rearranged as follows:
(83)∂∂x∂∂t+a00∂∂x+δ∂2∂x2w2=−12Dxt+a00Dx2w1⋅w1−δDxw1xx⋅w1.
Substituting Equation (82) into Equation (83), the right-hand side of Equation (83) equals zero. Therefore, we can choose
(84)w2=0.
Substituting Equations (82) and (84) into Equation (79), we get
(85)w=1+eP1x−a00P1+δP12t+ξ10,where P1, a00, and ξ10 are arbitrary constants.
Substituting Equation (85) into Equation (70), 1-soliton solution of Equation (65) can be obtained.
If we choosew1=eP1x−a00P1+δP12t+eP2x−a00P2+δP22t in Equation (81), Equation (74) has no 2-soliton solution. But there exists a solution as follows:
(86)w=1+∑i=1neη1,where ηi=Pix−a00Pi+δPi2t+ξ10, Pi,ξ10i=1,2,⋯,n, and a00 are arbitrary constants.
(ii)
Now, we discuss Equation (73) by using the sub-ODE method.
Using transformationswx,t=wξ, ξ=x−Vt, Equation (73) is reduced to
(87)a00−Vw″w−w′2+δww″′−w′w″=0,where the prime denotes the derivation with respect to ξ and V is a constant to be determined later.
Noticing the bilinear property of Equation (87), suppose that w satisfies the following ODE:
(88)w″+λw′+μw=0,where λ and μ are parameters.
Substituting Equation (88) into Equation (87), we get
(89)l1w2+l2ww′+l3w′2=0,where
(90)l1=μδλ+V−a00,l2=λδλ+V−a00,l3=δλ+V−a00.
Settingl1=l2=l3=0 yields a set of algebraic equations for V, λ, and μ. Solving the set of algebraic equations, we get
(91)V=a00−δλ,where λ, μ, and a00 are arbitrary constants.
Substituting the general solutions of Equation (88) into Equation (70), we get three types of traveling wave solutions of Equation (65) as follows.
Whenλ2−4µ>0,
(92)u1x,t=u1ξ=δAC1eA/2ξ−C2e−A/2ξC1eA/2ξ+C2e−A/2ξ+a00−δλ,where
(93)A=λ2−4μ,V=a00−δλ,ξ=x−a00−δλt,where C1, C2, λ, µ, and a00 are arbitrary constants.
TakingC1=C3+C4/2 and C2=C3−C4/2, Equation (93) can be written as
(94)u2x,t=u2ξ=δAC3sinhA/2ξ+C4coshA/2ξC3coshA/2ξ+C4sinhA/2ξ+a00−δλ,where C3, C4, and a00 are arbitrary constants and A and ξ are given by Equation (94).
In particular, ifC4/C3<1, then Equation (94) is reduced to
(95)u3x,t=u3ξ=δAtanhA2ξ+ξ0+a00−δλ,where C3, C4, and a00 are arbitrary constants and A and ξ are given by Equation (94), ξ0=arctanhC4/C3.
Whenλ2−4µ<0,
(96)u4x,t=u4ξ=δ−A−C3sin−A/2ξ+C4cos−A/2ξC3cos−A/2ξ+C4sin−A/2ξ+a00−δλ,where C3, C4, and a00 are arbitrary constants and A and ξ are given by Equation (94).
Obviously, Equation (96) can be written as
(97)u5x,t=u5ξ=−δ−Atan−A2ξ−ξ0+a00−δλ,where λ, µ, and a00 are arbitrary constants and A and ξ are given by Equation (94), ξ0=−arctanC4/C3.
Whenλ2−4µ=0,
(98)u6x,t=u6ξ=2δC2C1+C2ξ+a00−δλ,where ξ=x−a00−δλt, C1, C2, λ, µ, and a00 are arbitrary constants.
(iii)
Now, we discuss Equation (73) from the compatible condition.
Using the compatible condition, we can get nothing but Equation (76). Using transformations wx,t=wξ,ξ=x−Vt, Equation (76) is reduced to
(99)δw″+a00−Vw′−Cw=0.
Substituting the general solutions of Equation (99) into Equation (70), we get three types of traveling wave solutions of Equation (65) as follows.
Whena00−V12+4Cδ>0,
(100)u7x,t=u7ξ=ΔC1eΔ/2δξ−C2e−Δ/2δξC1eΔ/2δξ+C2e−Δ/2δξ+V1,where
(101)Δ=a00−V12+4Cδ,ξ=x−V1t,where V1, C, C1, C2, and a00 are arbitrary constants.
TakingC1=C3+C4/2 and C2=C3+C4/2, Equation (101) can be written as
(102)u8x,t=u8ξ=ΔC3sinhΔ/2δξ+C4coshΔ/2δξC3coshΔ/2δξ+C4sinhΔ/2δξ+V1,where V1, C, C3, C4, and a00 are arbitrary constants and Δ and ξ are given by Equation (101).
In particular, ifC4/C3<1, then Equation (102) is reduced to
(103)u9x,t=u9ξ=ΔtanhΔ2δξ+ξ0+V1,where ξ0=arctanhC4/C3, V1, C, C3, C4, and a00 are arbitrary constants and Δ and ξ are given by Equation (101).
Whena00−V22+4Cδ<0,
(104)u10x,t=u10ξ=−Δ−C3sin−Δ/2δξ+C4cos−Δ/2δξC3cos−Δ/2δξ+C4sin−Δ/2δξ+V2,where V2, C, C3, C4, and a00 are arbitrary constants and Δ and ξ are given by Equation (101).
Obviously, Equation (104) can be written as
(105)u11x,t=u11ξ=−−Δtan−Δ2δξ−ξ0+V2,where ξ0=−arctanC4/C3, V2, C, C3, C4, and a00 are arbitrary constants and Δ and ξ are given by Equation (101).
Whena00−V32+4Cδ=0,
(106)u12x,t=u12ξ=2δC2C1+C2ξ+V3,where ξ=x−V3t, V3, C1, C2, and a00 are arbitrary constants.
Moreover, note that Equation (76) is linear, so we can get the solution of Equation (65) as follows:
(107)w=∑i=1n1w1i+∑i=1n2w2i+∑i=1n3w3iu13=2δlnw+a00,where
(108)w1i=C1,1ie−a00−V1i+Δ1i/2ξ1i+C2,1ie−a00−V1i−Δ1i/2ξ1i,w2i=C1,2icos−Δ2iξ2i2+C2,2isin−Δ2iξ2i2e−a00−V2iξ2i/2,w3i=C1,3i+C2,3iξ3ie−a00−V3iξ3i/2,where ξ1i=x−V1it, Δ1i=a00−V1i2+4Cδ>0i=1,⋯,n1; ξ2i=x−V2it, Δ2i=a00−V2i2+4Cδ<0i=1,⋯,n2; ξ3i=x−V3it, Δ3i=a00−V3i2+4Cδ=0i=1,⋯,n3; C1,1i,C2,1ii=1,⋯,n1; C1,2i,C2,2ii=1,⋯,n2; and C1,3i,C2,3ii=1,⋯,n3 are arbitrary constants, and n1, n2, and n3 are arbitrary but finite integers.Remark 6.
We can deal with Equation (76) by using some assumptions. For example, when we suppose that w=−αt+Wx and C=0, we get
(109)w=−αt+αxa00+C1+C2e−a00x/δ,u14x,t=α2δ−ta002+a00x+a002C1−C2e−a00x/δαx−a00t+a00C1+C2e−a00x/δ,where C1, C2, α, and a00 are arbitrary constants.
When we suppose thatw=−αt+Wx and C=a00=0, we get
(110)w=−αt+αx22δ+C1x+C2,u15x,t=4δαx+δC12δC1x+C2−αt+αx2,where C1, C2, and α are arbitrary constants.
Similarly, we can assume thatw=∑i=1npixqit; then, a new solution of Equation (65) can be obtained. Being similar to above process, we omit it.So far, applying the HB of undetermined coefficient method to the Burgers equation, we get the bilinear equation of Burgers equation. Moreover, we reduce the Burgers to a linear equation. Based on them, many exact solutions of the Burgers equation are obtained by applying the perturbation method, sub-ODE method, and compatible condition. Our results can compare with theG′/G-expansion method, Hirota’s method, and HB method [20–30].


## 5. Bilinear Equation of the Boussinesq Equation and Sawada-Kotera Equation

In this section, we derive the bilinear equations of the Boussinesq equation and Sawada-Kotera equation by using the HB of undetermined coefficient method. Being similar to Section4, we omit the process of solving exact solutions.Example 1.
The generalized Boussinesq equation reads(111)utt+2αuxt+α2+βuxx+γuuxx+δuxxxx=0,where α, β, γ, and δ are known constants.
In order to balanceuxxxx and uux in Equation (111), it is required that m+4=2m+2 and n=2n. Then, we suppose that the solution of Equation (111) is of the form
(112)u=a20lnwxx+a10lnwx+a00,where ai0i=0,1,2 are constants to be determined later.
Substituting Equation (112) into Equation (111) and equating the coefficients of wx/w6 and wx/w5 on the left-hand side of Equation (111) to zero yield a set of algebraic equations for a20 and a10. Solving the algebraic equations, we get a20=6δ/γ and a10=0. Substituting a20 and a10 back into Equation (112), we get
(113)u=6δγlnwxx+a00.where a00 is an arbitrary constant.
Substituting Equation (113) into Equation (111), we get
(114)6δγK1+K2+K3=0,where
(115)K1=βwxxxxw−3wxx2+4wxwxxxw2+12wxxwx2w3−6wx4w4+α212wxxwx2w3−3wxx2+4wxwxxxw2+wxxxxw−6wx4w4+α2wxxxtw−6wxxwxt+6wxxtwx+2wtwxxxw2+12wxwtwxx+12wx2wxtw3−12wtwx3w4+wxxttw−2wxwxtt+wxxwtt+2wtwxxt+2wxt2w2+2wx2wtt+2wxxwt2+8wxwtwxtw3−6wx2wt2w4,K2=a002γwxxxxw−8γwxwxxx+6γwxx2w2+24γwxxwx2w3−12γwx4w4,K3=δwxxxxxxw+2wxxx2−6wxwxxxxx−3wxxwxxxxw2+18wx2wxxxx−6wxx3w3+18wx2wxx2−24wx3wxxxw4.
Simplifying Equation (114) and integrating twice with respect to x, we get
(116)∂2∂x2α2+β+2γa00wwxx−wx2w2+2αwwxt−wxwtw2+wwtt−wt2w2+δwwxxxx−4wxwxxx3wxx2w2=0.
Equation (116) is identical to
(117)α2+β+2γa00wwxx−wx2+2αwwxt−wxwt+wwtt−wt2+δwwxxxx−4wxwxxx+3wxx2−C1tx+C2tw2=0,where C1t and C2t are arbitrary functions of t and a00 is an arbitrary constant.
In particular, lettingC1t=C2t=0 in Equation (117), we get the bilinear equation of Equation (111) as follows:
(118)α2+β+2γa00wwxx−wx2+2αwwxt−wxwt+wwtt−wt2+δwwxxxx−4wxwxxx+3wxx2=0.
Equation (118) can be written concisely in terms of D-operators as
(119)2αDxDt+δDx4+α2+β+2γa00Dx2+Dt2w·w=0,where a00 is an arbitrary constant.Example 2.
The Sawada-Kotera equation reads(120)ut+15u3+uuxxx+uxxxxx=0.
In order to balanceuxxxxx and u3 in Equation (120), it is required that m+5=3m+1 and n=3n. Then, we suppose that the solution of Equation (120) is of the form
(121)u=a20lnwxx+a10lnwx+a00,where ai0i=0,1,2 are constants to be determined later.
Substituting Equation (121) into Equation (120) and equating the coefficients of wx/w7 and wx/w6 on the left-hand side of Equation (120) to zero yield a set of algebraic equations for a20 and a10. Solving the algebraic equations, we get a20=2 and a10=0. Substituting a20 and a10 back into Equation (121), we get
(122)u=2lnwxx+a00,where a00 is an arbitrary constant.
Substituting Equation (122) into Equation (120), we get
(123)2K1+K2+K3=0,where
(124)K1=45a002wxxxw−3wxwxxw2+2wx3w3,K2=15a00wxxxxxw+2wxxwxxx−5wxwxxxxw2+8wxxxwx2−6wxwxx2w3,K3=wxxtw−2wxwxt+wxxwtw2+2wx2wtw3+wxxxxxxxw+9wxxwxxxxx−5wxxxwxxxx−7wxwxxxxxxw2+20wxwxxx2−30wxwxxwxxxx+12wx2wxxxxxw3.
Simplifying Equation (123) and integrating once with respect to x, we get
(125)∂∂x45a002wwxx−wx2w2+15a00wwxxxx−4wxwxxx+3wxx2w2+wwxt−wxwtw2+wwxxxxxx−6wxwxxxxx+15wxxwxxxx−10wxxx2w2=0.
Equation (125) is identical to
(126)45a002wwxx−wx2+15a00wwxxxx−4wxwxxx+3wxx2+wwxt−wxwt+wwxxxxxx−6wxwxxxxx+15wxxwxxxx−10wxxx2−Ctw2=0,where Ct is an arbitrary function of t and a00 is an arbitrary constant.
In particular, lettingCt=0 in Equation (126), we get the bilinear equation of Equation (120) as follows:
(127)45a002wwxx−wx2+15a00wwxxxx−4wxwxxx+3wxx2+wwxt−wxwt+wwxxxxxx−6wxwxxxxx+15wxxwxxxx−10wxxx2=0.
Equation (127) can be written concisely in terms of D-operators as
(128)45a002Dx2+15a00Dx4+DxDt+Dx6w⋅w=0,where a00 is an arbitrary constant.


## 6. Conclusions

The HB of undetermined coefficient method is successfully used to establish the bilinear equation of NLPDE. By applying the perturbation method, sub-ODE method, and compatible condition to the bilinear equation, more exact solutions of NLPDE are obtained. We illustrate the real meaning of balance numbers. We show the underlying relations among theG′/G-expansion method, Hirota’s method, and HB method. Many well-known NLPDE can be handled by the HB of undetermined coefficient method. The performance of our method is found to be simple and efficient. The availability of computer systems like Maple facilitates the tedious algebraic calculations. Our method is also a standard and computable method, which allows us to solve complicated and tedious algebraic calculations.



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