Research article

On traveling wave solutions of a class of KdV-Burgers-Kuramoto type equations

  • Received: 21 May 2019 Accepted: 25 August 2019 Published: 20 September 2019
  • MSC : 34A05, 34A34

  • In the paper, the traveling wave solutions of a KdV–Burgers-Kuramoto type equation with arbitrary power nonlinearity are considered. Lie symmetry analysis method on the equation is performed, which shows that the equation possesses traveling wave solutions. By qualitative analysing the equivalent autonomous system of the traveling wave equation of the equation, the existence of the traveling wave solutions of the equation is presented. Through analysing the associated determining system, the non-trivial infinitesimal generator of Lie symmetry admitted by the traveling wave solutions equation under the certain parametric conditions is found. The traveling wave solutions of the KdV–Burgers-Kuramoto type equation by solving the invariant surface condition equation under the certain parametric conditions are obtained.

    Citation: Yanxia Hu, Qian Liu. On traveling wave solutions of a class of KdV-Burgers-Kuramoto type equations[J]. AIMS Mathematics, 2019, 4(5): 1450-1465. doi: 10.3934/math.2019.5.1450

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  • In the paper, the traveling wave solutions of a KdV–Burgers-Kuramoto type equation with arbitrary power nonlinearity are considered. Lie symmetry analysis method on the equation is performed, which shows that the equation possesses traveling wave solutions. By qualitative analysing the equivalent autonomous system of the traveling wave equation of the equation, the existence of the traveling wave solutions of the equation is presented. Through analysing the associated determining system, the non-trivial infinitesimal generator of Lie symmetry admitted by the traveling wave solutions equation under the certain parametric conditions is found. The traveling wave solutions of the KdV–Burgers-Kuramoto type equation by solving the invariant surface condition equation under the certain parametric conditions are obtained.


    Many phenomena in physics, engineering and other subjects can be described by nonlinear partial differential equations. In order to well understand various nonlinear phenomena in nature, the exact solutions for the nonlinear partial differential equations has to be studied. Finding effective methods to solve and analyse these equations has been an interesting subject in the field of differential equations. Sometimes, It is difficult to express various wave solutions of nonlinear partial differential equations explicitly in terms of elementary functions. In many cases it is possible to find and prove the existence of traveling wave solutions by the qualitative theory of differential equations and dynamical systems. From the physical point of view, traveling waves usually describe transition processes. Such as solitons and propagation with a finite speed, and thus they give more insight into the physical aspects of the problems. In order to obtain the traveling wave solutions of nonlinear partial differential equations, several powerful approaches to finding the exact solutions have been proposed, such as Backlund transformation method in [1], Darboux transformation [2,3,4,5,6], Hirota's method [7,8,9], the Fokas method [10], exp-function method in [11,12], tanh-sech method in [13], hyperbolic function method in [14], first integral method in [15,16], homogeneous balance principle in [17], Lie symmetry analysis method in [18,19] and the references therein [20,21,22]. Lie group theory plays an important role in studying the solutions of differential equations, for example, see [23,24,25,26,27,28,29]. In [25], Lie symmetry analysis is performed on the fifth-order KdV types of equations which arise in modelling many physical phenomena, and the exact analytic solutions are obtained.

    In this paper, we consider a class of KdV-Burgers-Kuramoto (KdV-BK) type equation with arbitrary power nonlinearity,

    ut+upux+αuxx+βuxxx+γuxxxx=0, (1.1)

    where α,β,γ0 and p0 are real constants. It contains dispersive terms uxx,uxxx and uxxxx and also the nonlinear term upux. Here the choice of p=1 leads equation (1.1) to the KdV-Burgers-Kuramoto(KBK) equation

    ut+uux+αuxx+βuxxx+γuxxxx=0. (1.2)

    Equation (1.1) is referred to as the generalized KdV-Burgers-Kuramoto(KBK) equation. In [30], equation (1.1) was used for explanation of the origin of persistent wave propagation through medium of reaction-diffusion type. In [31,32], the motion of a viscous incompressible flowing down an inclined plane is studied by the application of equation (1.1). Mathematical model for consideration dissipative waves in plasma physics by means of equation (1.1) was presented in [33]. The equation (1.2) is simultaneously involved in nonlinearity, dissipation, dispersion and instability, and is suggested by Kuramoto [34,35]. The KBK equation is a classical nonlinear partial differential equation, which effectively describes the turbulent motion and other unstable motions. It is well known that there are many works to deal with equation (1.2) in recent years, see [34,35], [36,37,38,39] and the references therein. For example, In [36,37], The travelling wave periodic solutions of equation (1.2) were considered and obtained.

    In this paper, we are mainly interested in equation (1.1). The rest of the paper is organized as follows. In section 2, Lie symmetries for (1.1) are found by differentiating the symmetry condition, and the partial differential equation has traveling wave solutions in this sense is shown. In section 3, the existence of traveling wave solutions of the equation is presented by analysing the corresponding autonomous system. Under the certain parametric conditions, the non-trivial infinitesimal generator of Lie groups admitted by deduced ordinary differential equation is obtained through analysing the determining system in section 4. Based on the Lie symmetry theory, a class of traveling wave solutions of (1.1) is presented by solving the invariant surface condition equation under the certain parametric conditions in section 5. Section 6 is conclusions.

    In this section, we consider Lie symmetry to equation (1.1). Based on Lie symmetry analysis theory [19], the generator X of Lie symmetry admitted by (1.1) has the expression

    X=ξ(x,t)x+τ(x,t)t+η(x,t,u)u,

    where η(x,t,u)=g(x,t)u+h(x,t), and ξ(x,t),τ(x,t),g(x,t),h(x,t) are function need to be determined. The surface u=u(x,t) is invariant, provided that

    ηξuxτut=0, (2.1)

    when u=u(x,t). (2.1) is called the invariant surface condition. The linearized symmetry condition is as follows:

    X(4)Δ=0, (2.2)

    when Δ=0, where

    Δ=uxxxx+1γ(ut+upux+αuxx+βuxxx),
    X(4)=ξx+τt+ηu+ηxux+ηtut+ηxxuxx+ηxxxuxxx+ηxxxxuxxxx,

    and

    ηx=ηx+(ηuξx)uxτxut,ηt=ηt+(ηuτt)utξtux,ηxx=ηxx+(2ηuxξxx)uxτxxut+(ηu2ξx)uxx2τxuxt,ηxxx=ηxxx+(3ηuxxξxxx)uxτxxxut+(3ηux3ξxx)uxx3τxxuxt+(ηu3ξx)uxxx3τxutxx,ηxxxx=ηxxxx+(4ηuxxxξxxxx)uxτxxxxut+(6ηuxx4ξxxx)uxx4τxxxutx+(4ηux6ξxx)uxxx6τxxutxx+(ηu4ξx)uxxxx4τxutxxx. (2.3)

    Substituting (2.3) into (2.2), one can have

    (gu+h)pup1ux+[(gxu+hx)+(gξx)uxτxut]up+α[(gxxu+hxx)+(2gxξxx)uxτxxut+(g2ξx)uxx2τxuxt]+β[gxxxu+hxxx+(3gxxξxxx)uxτxxxut+(3gx3ξxx)uxx3τxxuxt+(g3ξx)uxxx3τxutxx]+γ[gxxxxu+hxxxx+(4gxxxξxxxx)uxτxxxxut+(6gxx4ξxxx)uxx4τxxxutx+(4gx6ξxx)uxxx6τxxutxx+(g4ξx)uxxxx4τxutxxx]+gtu+ht+(gτt)utξtux=0. (2.4)

    The linearized symmetry condition gives us a systematic approach to finding Lie point symmetries. We can obtain uxxxx from the restriction Δ=0 and substitute uxxxx into (2.4). According to their dependence on derivative of u, we can get a linear system of determining equation for ξ,τg,h:

    τx=0;3βτx6γτxx=0;β(g3ξx)+γ(4gx6ξxx)β(g4ξx)=0;2ατx3βτxx4γτxxx=0;γ(6gxx4ξxxx)+β(3gx3ξxx)+α(g2ξx)α(g4ξx)=0;γτxxxxβτxxxατxx+gτt(g4ξx)=0;γ(4gxxxξxxxx)+β(3gxxξxxx)+α(2gxξxx)ξt=0;pg+(gξx)(g4ξx)=0;hp=0;gx=0;hx=0;αgxx+βgxxx+γgxxxx+gt=0;αhxx+βhxxx+γhxxxx+ht=0. (2.5)

    We solve the first equation of (2.5) to obtain

    τ=A(t),

    where A(t) is an arbitrary function.

    Owing to p0, the ninth equation of (2.5) yields

    h=0,

    and the tenth equation of (2.5) yields

    g=B(t),

    where B(t) is an arbitrary function. The twelfth equation of (2.5) tells us

    B(t)=c3,

    where c3 is an arbitrary constant. So,

    g=c3.

    From the eighth equation of (2.5), we have

    ξx=13pc3.

    The sixth equation of (2.5) gives us the result

    τt=43pc3.

    Substituting the expression of ξx into the third equation of (2.5) yields c3=0. Thus, g=0,ξx=0,τt=0. The fifth equation of (2.5) is satisfied apparently. One can get ξt=0 from the seventh equation of (2.5). So, the generator of Lie symmetry is

    X=c1x+c2t,

    where c1,c2 are arbitrary constants.

    The derived above generator X implies that (1.1) has a invariant solution in the form

    u=u(v),v=xct,

    where c is an arbitrary constant.

    Substituting the traveling wave solution of the form u=u(v),v=xct into (1.1), we can get the traveling wave solution equation as follows,

    cu+upu+αu+βu+γu(4)=0. (3.1)

    For (3.1) with p1, performing the integration once, one has the following traveling wave solutions equation

    cu+up+1p+1+αu+βu+γu=k, (3.2)

    where k is an integration constant. Let R=dudv,w=dRdv,k=0, (3.2) can be rewritten as

    {u=RR=ww=1γ(cuup+1p+1αRβw). (3.3)

    As we know, a solitary wave solution of (1.1) corresponds to a heteroclinic orbit of system (3.3). We will prove that the heteroclinic orbit of system (3.3) does exist when the parametric conditions are satisfied correspondingly.

    For convenience, we consider p being a positive integer. When p is an even number and (p+1)c>0, (3.3) has three equilibrium points

    u1=(0,0,0),u2,3=(±p(1+p)c,0,0).

    When p is an odd number, (3.3) has two equilibrium points

    u1=(0,0,0),u2=(p(1+p)c,0,0).

    The coefficient matrix A1 of the linearization system of (3.3) about the equilibrium point u1 is

    A1=[010001cγαγβγ].

    Then, the characteristic equation for A1 is

    f1(λ)=λ3+βγλ2+αγλcγ. (3.4)

    Similarly, one can get the coefficient matrix A2 of the linearization system of (3.3) about the equilibrium points u2,3. The matrix is

    A2=[010001pcγαγβγ].

    Then, the characteristic equation for A2 is

    f2(λ)=λ3+βγλ2+αγλ+pcγ. (3.5)

    We have the following result for (3.3).

    Theorem 3.1. If α,β,γ and c are of same sign, p is a positive integer and pcγαβ<0, then the equilibrium point A1 of (3.3) has a one-dimensional unstable manifold and the equilibrium point A2,3 (or A2) of (3.3) has a three-dimensional stable manifold.

    Proof. The proof is based on Argument Principle. For proving the equilibrium point A1 of (3.3) possessing a one-dimensional unstable manifold, one needs to prove the characteristic equation (3.4) has only one root in the right half complex plane, and for proving the equilibrium point A2,3 (or A2) of (3.3) possessing a three-dimensional stable manifold, one needs to prove the characteristic equation (3.5) has three roots in the left half complex plane.

    Let us to consider f1(λ) firstly. Since f1(λ) is analytic in complex plane, the number of roots in the right half complex plane is

    12πlimRΔCargf1(z), (3.6)

    where C is composed of ΓR:z=Reiθ,π2θπ2, a straight line (Ri,Ri) is on the imaginary axis and ΔCargf1(z) denotes the total change quantity in the argument of f1(z) along C. Apparently, (3.6) equals

    12πlimRΔΓRargf1(z)+12πlimRΔ(Ri,Ri)argf1(z).

    The first part of the above formula is

    12πlimRΔΓRargz3+12πlimRΔΓRarg(1+βγz2+αγzcγz3)=12πlimRΔΓRarg(R3e3θi)=32.

    The second part of that formula is

    12πlimRΔ(R,R)argf1(iy),

    where f1(iy)=(βγy2cγ)+(αγyy3)i, and f1(0)=cγ. Because β,γ and c are of same sign, βγ>0,cγ>0. So Re(f1(iy))<0 and f(0)<0, the image f1(iy) only lies on the left complex plane. For |R|, f1(iy) has the asymptotic behavior,

    Re(f1(iy))βγy2<0,Im(f1(iy))y3.

    So,

    12πlimRΔ(R,R)arg(f1(iy))=π23π22π=12.

    The number of roots of f1(λ)=0 in the right half complex plane is 1. Therefore, the equilibrium point A1 of (3.3) has a one-dimensional unstable manifold.

    Similarly, we consider the number of roots of f2(λ)=0 in the left half complex plane. Since f2(λ) is analytic in complex plane, the number of roots in the left half complex plane is

    12πlimRΔCargf2(z), (3.7)

    where C is composed of ΓR:z=Reiθ,π2θ3π2, a straight line (Ri,Ri) on the imaginary axis and ΔCargf2(z) denotes the total change quantity in the argument of f2(z) along C. Apparently, (3.7) equals

    12πlimRΔΓRargf2(z)+12πlimRΔ(Ri,Ri)argf2(z).

    The first part of the above formula is

    12πlimRΔΓRargz3+12πlimRΔΓRarg(1+βγz2+αγz+pcγz3)=12πlimRΔΓRarg(R3e3θi)=32.

    The second part of that formula is

    12πlimRΔ(R,R)argf2(iy),

    where f2(iy)=(βγy2+pcγ)+(αγyy3)i and f2(0)=pcγ.

    We first compute the quantity Δ(,0)arg(f2(iy)). Note that

    Re(f2(iy))=βγy2+pcγ,Im(f2(iy))=αγyy3,

    it is obvious that as y increases from to 0, Re(f2(iy)) increases monotonously from to pcγ, and Im(f2(iy)) decreases monotonously from + to 2α3γα3γ afterwards increases monotonously to 0. Owing to αγ>pcβ from the assumption pcγαβ<0, as y increases from to 0, the image f2(iy) starts in second quadrant of complex plane, intersects the minus Re-axis, passes through the third quadrant, intersects the Im-axis at a certain point, passes through the forth quadrant and finally ends up the point (pcγ,0) of the positive Re-axis. So, Δ(,0)arg(f2(iy))=3π2. Similarly, we can get Δ(0,+)arg(f2(iy))=3π2. Therefore, The number of roots of f2(λ)=0 in the left half complex plane is 3. Therefore, the equilibrium point A2,3 (or A2) of (3.3) has a three-dimensional stable manifold. This completes the proof.

    The local trajectory in the neighborhood of these equilibrium points of (3.3) are shown in Figures 1 and 2. In Figure 1, we let α=710,β=200,γ=170,p=4,c=1, and the initial value is (0.000001,0.000001,0.000001). In Figure 2, we let α=710,β=200,γ=170,p=4,c=1, and the initial value is (0.000001,0.000001,0.000001). In these Figures, there are the images in coordinate plane (t,u),(t,R),(t,w),(u,R),(u,w),(R,w) and in coordinate space (u,R,w), which show that the equilibrium point (0,0,0) is unstable and the equilibrium points (±p(p+1)c,0,0) is stable.

    Figure 1.  The local trajectory.
    Figure 2.  The local trajectory.

    Theorem 3.2. When α,β,γ,c are of same sign, p is a positive integer and pcγαβ<0, (3.3) has potentially a heteroclinic orbit.

    Proof. Owing to Theorem 3.1, the sum of the dimension of the unstable manifold Wu(A1) and the stable manifold Ws(A2,3) is four. The dimension of the phase plane of (3.3) is three. Therefore, these two manifolds potentially intersect in R3 along one-dimension curve, which is a heteroclinic orbit of (3.3).

    As we know, the heteroclinic orbit of (3.3) corresponds a traveling wave solution of (1.1). Motivated by the above results, we will consider to obtain traveling wave solutions of (1.1) by using Lie symmetry method.

    In the section, we consider to search for Lie symmetries admitted by (3.2).

    We suppose that V=ξ(x,y)x+η(x,y)y is the infinitesimal generator of Lie symmetry admitted by (3.2). Here, for convenience, symbols y and x are in place of u and v, respectively.

    Therefore,

    V(3)=ξ(x,y)x+η(x,y)y+η(1)(x,y,y)y+η(2)(x,y,y,y)y+η(3)(x,y,y,y)y

    is the 3th-extended infinitesimal generator, where

    η(1)(x,y,y)=ηx+(ηyξx)yξyy2,η(2)(x,y,y,y)=ηxx+(2ηxyξxx)y+(ηyy2ξxy)y2ξyyy3+(ηy2ξx3ξyy)y,η(3)(x,y,y,y,y)=ηxxx+(3ηxxyξxxx)y+3(ηxyy2ξxxy)y2+(ηyyy3ξxyy)y3ξyyyy4+3(ηxyξxx+(ηyy3ξxy)y2ξyyy2)y3ξyy2+(ηy3ξx4ξyy)y. (4.1)

    The linearized symmetry condition

    V(3)(yf(x,y,y,y))=0,y=f(x,y,y,y), (4.2)

    where f(x,y,y,y)=yp+1γ(p+1)+cγy+kγαγyβγy. Plugging (4.1) to (4.2) and replacing y by f(x,y,y,y), This yields

    η(ypγcγ)+αγ(ηx+(ηyξx)yξyy2)+βγ(ηxx+(2ηxyξxx)y+(ηyy2ξxy)y2ξyyy3+(ηy2ξx3ξyy)y)+ηxxx+(3ηxxyξxxx)y+3(ηxyy2ξxxy)y2+(ηyyy3ξxyy)y3ξyyyy4+3(ηxyξxx+(ηyy3ξxy)y2ξyyy2)y3ξyy2+(ηy3ξx4ξyy)(yp+1γ(p+1)+cγy+kγαγyβγy)=0. (4.3)

    Both ξ and η are independent of y and y, After setting the coefficients of the powers (y)i(y)j(i,j=0,1,2,3,4) in (4.3) to zero, one can get the determining equations system,

    ξyyy=0, (4.4)
    βγξyy+ηyyy3ξxyy=0, (4.5)
    αγξy+βγ(ηyy2ξxy)+3(ηxyy2ξxxy)+4αγξy=0, (4.6)
    ξy=0, (4.7)
    3βγξy+3(ηyy3ξxy)+4βγξy=0, (4.8)
    ξyy=0, (4.9)
    βγ(ηy2ξx)+3(ηxyξxx)βγ(ηy3ξx)=0, (4.10)
    αγ(ηyξx)+βγ(2ηxyξxx)+(3ηxxyξxxx)αγ(ηy3ξx)=0, (4.11)
    η(ypγcγ)+αγηx+βγηxx+ηxxx+(ηy3ξx)(yp+1(p+1)γ+cγy+kγ)=0. (4.12)

    The equation (4.7) gives

    ξ=ξ(x). (4.13)

    After putting (4.13) into equation (4.8) yields

    η=a1(x)y+a2(x), (4.14)

    where a1(x) and a2(x) are functions of x. Equation (4.4), (4.5), (4.6) and (4.9) are satisfied by (4.13) and (4.14). Substituting (4.13) and (4.14) into (4.10), (4.11) and (4.12), we have the system

    βξ+3γa1(x)3γξ=0,2αξ+2βa1(x)βξ+3γa1(x)γξ=0,(a1(x)y+a2(x))(ypγcγ)+αγ(a1(x)y+a2(x))+βγ(a1(x)y+a2(x))+(a1(x)y+a2(x))+(a1(x)3ξ)(yp+1(p+1)γ+cγy+kγ)=0. (4.15)

    The third equation of (4.15) is a polynomial of y with degree p+1 which is zero if and only if each variable coefficient is set to zero,

    pa1(x)+3ξ=0, (4.16)
    a2(x)=0, (4.17)
    αa1(x)+βa1(x)+γa1(x)3cξ=0, (4.18)
    (a1(x)3ξ)k=0. (4.19)

    So the above set of differential equations of a1(x) and a2(x) can be solved according to the following two cases,

    Case 1: k=0

    Substituting ξ derived from (4.16) into the first equation of (4.15), under the condition p3, one has

    a1(x)=c1eβpx3(γp+3γ)

    and

    ξ=pc13eβpx3(γp+3γ),

    where c1 is an integration constant. Substituting ξ and a1(x) into the second equation of (4.15), one can get a parametric condition

    β2(9+p)p+3β2(6+p)(3+p)18αγ(3+p)2=0. (4.20)

    The other parametric condition can be obtained by pulling a1(x) to (4.18) and it is

    β3(4p2+9p)+9αβγ(3+p)2+27cγ2(3+p)3=0. (4.21)

    Integrating ξ and using a1(x),a2(x), we have

    ξ=c1γ(p+3)βeβpx3γ(p+3)+c2,

    and

    η=c1eβpx3γ(p+3)y,

    where c2 is an arbitrary constant. The infinitesimal generator

    X=[c1γ(p+3)βeβpx3γ(p+3)+c2]x+c1eβpx3γ(p+3)yy. (4.22)

    If p=3, we can get a1(x) and ξ(x) being all zeros form substituting ξ into the first equation of (4.15). So, ξ=c,η=0, where c is an arbitrary constant. The infinitesimal generator is X=cx.

    Case 2: k0,

    In this case, from (4.16) and (4.19), we can have a1(x)=0,ξ(x)=0. Furthermore, (4.15) and (4.18) hold. We can obtain

    ξ=c,η=0,

    where c is an arbitrary constant. The infinitesimal generator is X=cx.

    For equation (3.1) with p=1, performing the integration once, one has

    cu+lnu+αu+βu+γu=k, (4.23)

    Similarly, we can obtain the system satisfied by generators of the Lie group admitted by (4.23) by the linearized symmetry condition (4.2).

    Thus, we can get

    a1(x)=0,a2(x)=0,ξ=c,

    where c is an arbitrary constant. In this case, the infinitesimal generator of Lie symmetry is X=cx.

    Therefore, we can have the following result about the Lie symmetries admitted by (3.2).

    Theorem 4.1.

    Case 1. p1,p3 and k=0.

    when (4.20) and (4.21) are satisfied, (3.2) accepts Lie symmetry with the infinitesimal generator (4.22).

    Case 2. p1,p3 and k0.

    (3.2) accepts Lie symmetry with the infinitesimal generator X=cx, where c is an arbitrary constant.

    Case 3. p=1 or p=3.

    (3.2) accepts Lie symmetry with the infinitesimal generator is X=cx, where c is an arbitrary constant.

    In the section, we consider to obtain the traveling wave solutions of the KdV-Burgers-Kuramoto type equation under the parametric conditions in Section 4 using the invariant curve condition

    Q=ηyξ=0. (5.1)

    If the infinitesimal generator of Lie symmetry admitted by (3.2) is

    X=cx,

    then

    Q=yc=0,

    that is, a trivial solution y=constant is obtained. The trivial solution has no new useful meaning. So, we consider the following case.

    Under the parameter conditions of (4.20) and (4.21), the infinitesimal generator is (4.22). Inserting ξ(x) and η(x,y) of (4.22) into (5.1), one has

    y=yc2c1eβpx3γ(3+p)γ(p+3)β.

    After solving the above equation, we can have

    y=c3|c2c1γ(p+3)βeβpx3γ(3+p)|3p, (5.2)

    where c3 is an integration constant. Using the identity e2x1+e2x=12tanhx+12 and choosing c1c2=βγ(p+3), we obtain the invariant solutions of (1.1),

    u(x,t)=c3(±β2γ(p+3))3p[tanh(βp(xct)6γ(p+3))+1]3p,

    where c3 is an arbitrary constant.

    When p=1, (1.1) becomes (1.2), the parametric conditions (4.20) and (4.21) change to

    47β2=144αγ (5.3)

    and

    13β3+144αβγ+1728cγ2=0. (5.4)

    Substituting (5.3) into (5.4) yields the condition

    5αβ+47γc=0. (5.5)

    Accordingly, under the conditions (5.3) and (5.5), one can get a traveling wave solution of (1.2),

    u(x,t)=c3(±β8γ)3[tanhβ(xct)24γ+1]3,

    where c3 is an arbitrary constant. After using the identity sech2t=1tanh2t, the traveling wave solution can be turn to

    u(x,t)=3c3(β3γ)3sech2β(xct)24γc3(β3γ)3sech2β(xct)24γtanhβ(xct)24γ±4c3(β3γ)3(tanhβ(xct)24γ+1).

    The solution is be equivalent to that in the existing literatures, for example [35,39].

    When p=2, the parametric conditions (4.20) and (4.21) change to

    71β2=225αγ (5.6)

    and

    34β3+225αβγ+27×53cγ2=0. (5.7)

    Substituting (5.6) into (5.7) yields the condition

    7αβ+639γc=0. (5.8)

    Accordingly, under the conditions (5.6) and (5.8), one can get traveling wave solutions of (1.2)

    u(x,t)=c3(|β10γ|)32[tanh2β(xct)30γ+1]32.

    where c3 is an arbitrary constant. The solution is a new solution of (1.1) compared with the results in [40].

    In this paper, we consider the solutions of a KdV-Burgers-Kuramoto type equation with an arbitrary power nonlinearity upux. Firstly, we present the condition of the existence of traveling wave solutions of the equation based on the qualitative theory of differential equations. Then, with p(p1,p3) being an arbitrary constant, when k=0, and the corresponding parametric conditions (4.20) and (4.21) are satisfied, we derive the invariant solutions of the KdV-Burgers-Kuramoto equation by solving the determining system and the invariant curve condition equation. When p=2, the KdV-Burgers-Kuramoto equation with the nonlinearity term u2ux is the equation in [40]. Compared with the results in [40], a new solution is obtained.

    For the case γ=0 in (1.1), the KdV-Burgers-Kuramoto type equation with an arbitrary power nonlinearity can be turn to the KdV-Burgers type equation. One can not directly get the traveling wave solutions from the traveling wave solutions of (1.1) with γ=0 because the equation has singular points. Thus, the dynamical properties and the qualitative analysis to the traveling wave solutions equation have to be studied. These will be investigated in our future work.

    The first author would like to thank Prof. Zhaosheng Feng (University of Texas-Rio Grande Valley) for his discussions and helpful suggestions.

    All authors declare no conflicts of interest in this paper.



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