# Saddle-Node Heteroclinic Orbit and Exact Nontraveling Wave Solutions for (2+1)D KdV-Burgers Equation

Authors: Da-Quan Xian
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We have undertaken the fact that the periodic solution of (2+1)D KdV-Burgers equation
does not exist. The Saddle-node heteroclinic orbit has been obtained. Using the Lie group method, we
get two-(1+1)-dimensional PDE, through symmetric reduction; and by the direct integral method, spread
F-expansion method, and(G′/G)-expansion method, we obtain exact nontraveling wave solutions, for the
(2+1)D KdV Burgers equation, and find out some new strange phenomenons of sympathetic vibration to
evolution of nontraveling wave.


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## Body

## 1. Introduction

We consider the (2+1)-dimensional Korteweg-de Vries Burgers ((2+1)D KdV Burgers) equation(1)(ut+uux-βuxx+αuxxx)x+γuyy=0,
where u:Rx×Ry×Rt+→R, α, β, and γ are real parameters. Equation (1) is model equation for wide class of nonlinear wave models in an elastic tube, liquid with small bubbles, and turbulence [1–3]. Much attention has been put on the study of their exact solutions by some methods [4], such as, a complex line soliton by extended tanh method with symbolic computation [5], exact traveling wave solutions including solitary wave solutions, periodic wave and shock wave solutions by extended mapping method, and homotopy perturbation method [6, 7].It is well known that the investigation of exact solutions of nonlinear evolution equations plays an important role in the study of nonlinear physical phenomena. Many effective methods have been presented [7–22], such as functional variable separation method [8, 9], homotopy perturbation method [12], F-expansion method [7, 13], Lie group method [14, 15], variational iteration method [16], homoclinic test method [17–19], Exp-function method [20, 21], and homogeneous balance method [22]. Practically, there is no unified method that can be used to handle all types of nonlinearity.In this paper, we will discuss the existence of periodic traveling wave solution and seek the Saddle-Node heteroclinic orbit, and further use the Lie group method with the aid of the symbolic computation system Maple to construct the non-traveling wave solutions for (1).


## 2. Existence of Periodic Traveling Wave Solution of (1)

Introducing traveling wave transformation in this form(2)u(x,y,t)=u(ξ),ξ=px+qy-ct
permits us to convert (1) into an ODE for u=u(ξ)(3)p(puuξ-βp2uξξ+αp3uξξξ)ξ-ruξξ=0,
where r=pc-q2γ, Integrating (3) with respect to ξ twice and taking integration constant to A yields
(4)2αp4uξξ-2βp3uξ+p2u2-2ru=A.
Letting uξ=v, thus nonlinear ordinary differential equation (4) is equivalent to the autonomous dynamic system as follows:
(5)dudξ=v,(6)dvdξ=12αp4(2βp3v-p2u2+2ru+A).
The dynamic system (5) has two balance points:
(7)P1(u1,v1)=(r+r2+p2Ap2,0),P2(u2,v2)=(r-r2+p2Ap2,0).
The Jacobi matrixes at the balance points for the right-hand side of (5) are obtained as follows, respectively:
(8)J1=(01-r2+p2Ap4αβpα),J2=(01r2+p2Ap4αβpα).
Their latent equations are expressed, respectively, as,
(9)p3λ(pαλ-β)+r2+p2A=0,p3λ(pαλ-β)-r2+p2A=0.
Relevant latent roots are as follows respectively:
(10)λ1=pβ±p2β2-4αr2+p2A2p2α,λ2=pβ±p2β2+4αr2+p2A2p2α.
Obviously, if p2β2>4αr2+p2A, then λ1 are two positive real roots, therefore P1 is a nonsteady node point. If 0<p2β2<4αr2+p2A, then λ1 are conjugate complex roots and real part is positive, so P1 is a nonsteady focus point. And λ2 is a positive and minus real root, thus P2 is a saddle point. From (5), we know the phase trajectory on the phase plane satisfies
(11)dvdu=2βp3v-p2u2+2ru+A2αp4v.
Integrating (11), we can obtain
(12)H(u,v)=Au+ru2-13p2u3+2βp3uv-αp4v2,
where H(u,v) is a total energy or Hamiliton function of system (4). Apparently
(13)uξ≠-∂H∂v,vξ≠∂H∂u.
Consequently, the system expressed in (12) is not a conservative one, then periodic traveling wave solution of (1) does not exist.We conclude the above analysis in the following theorem.Theorem 1.
Under the traveling wave transformation, the periodic solution of (2+1)-dimensional KdV-Burgers equation does not exist.
But, saddle-node heteroclinic orbits and nontraveling periodic solution do exist, which will be discussed later in this paper.


## 3. Saddle-Node Heteroclinic Orbits of KdV-Burgers Equation

First, we assume the solutions of (4) in the form
(14)u(ξ)=r+r2+p2Ap2+b(1+eaξ)2.
Substituting (14) into (4) yields
(15)2(4αp4a2+r2+p2A+2βp3a)e2aξ-4(αp4a2-r2+p2A-βp3a)eaξ+2r2+p2A+p2b=0.
Then we get
(16)4αp4a2+r2+p2A+2βp3a=0,αp4a2-r2+p2A-βp3a=0,2r2+p2A+p2b=0.
Solving the system (16) gets
(17)a=-β5αp,b=-12β225α,r2+p2A=6p2β225α.
Substituting (17) into (14) obtains
(18)u(ξ)=r+r2+p2Ap2-12β225α1(1+e-(β/5αp)ξ)2=u1-3β225α(1+tanhβ20αξ)2.
Evidently, ξ→-∞⇒u(ξ)→u1, ξ→+∞⇒u(ξ)→u1-(6β2/25α)=u2. Thus (18) is a saddle-node heteroclinic orbit through nonsteady node point P1 and saddle point P2 [23].Ecumenic, taking the Hamiliton functionH(u,v)=B, we obtain
(19)dudξ=v=3pβu±3u[3Aα+3(p2β2+rα)u-p2αu2]-9Bα3αp2,
where B is an arbitrary constant. Integrating (19) with respect to ξ we have
(20)∫u(ξ)3αp23pβs±3s[3Aα+3(p2β2+rα)s-p2αs2]-9Bαds=ξ+ξ0,
where ξ0 is an arbitrary constant. We can see that (4) has the general solution (20) and all partial cases as include above result can be found from the general solution of (20). Example, take αr2+p2A-p2β2=0, 3Bα+Aβ2=0, rα+p2β2=0 in (20), we find a solution of (4) as follows:
(21)u(ξ)=-3β24α[1+tanh(β4pαξ+ξ0)]2.
It is a heteroclinic orbit too.


## 4. Li Symmetry of (1)

This section devotes to Li symmetry of (1) [14, 15]. Let
(22)σ=σ(x,y,t,u,ut,ux,uy,…).
be the Li symmetry of (1). From Lie group theory, σ satisfies the following equation
(23)σxt+2uxσx+uσxx+σuxx-βσx3+ασx4+γσyy=0.
We take the function σ in the form
(24)σ=a1ux+a2uy+a3ut+a4u+a5,
where ai=ai(x,y,t):Rx×Ry×Rt+→R  (i=1,…,5) are functions to be determined later. Substituting (3) into (2) yields
(25)a1=-12γk2′(t)y+k1(t),a2=k2(t),a3=c,a4=0,a5=12γk2′′(t)y-k1′(t),
where kj(t)  (j=1,2) are arbitrary functions of t, c is an arbitrary constant. Substituting (25) into (24), we obtain the Li symmetries of (1) as follows:
(26)σ=[-12γk2′(t)y+k1(t)]ux+k2(t)uy+cut+12γk2′′(t)y-k1′(t).


## 5. Symmetry Reduction and Solutions of (1)

Based on the integrability of reduced equation of symmetry (26), we are to consider the following three cases.Case 1.
Takingk2(t)=0 and c=0 in (26) yields
(27)σ=k1(t)ux-k1′(t).
The solution of the differential equation σ=0 is
(28)u=k1′(t)k1(t)x+F(y,t),F(y,t):Ry×Rt+→R.
Substituting (28) into (1) yields the function F(y,t) which satisfies the following linear PDE:
(29)k1′′k1+γ∂2F∂y2=0.
By integrating both sides, we find out the following result:
(30)F(y,t)=-k1′′2γk1y2+k3(t)y+k4(t),
where k3(t), k4(t) are new arbitrary functions of t. Substituting (30) into (28), we can get the solutions of (1) as follows:
(31)u1(x,y,t)=k1′(t)k1(t)x-k1′′2γk1y2+k3(t)y+k4(t).
(1) Given ki(t)=cn(t,0.95)  (i=1,3,4), x=1, γ=0.6 in (31), the local structure of u1 is obtained (Figure 1). Where cn(t,0.95) is an Jacobian elliptic cosine function.
(2) Givenk1(t)=sech(t), k3(t)=sin(t), k4(t)=cn(t,0.1), y=1, γ=0.6 in (31), the local structure of u1 is obtained (Figure 2).Figure 1
The strange phenomenon which is a sympathetic vibration of periodicity on thet-axis and paraboloid on y-axis for u1(x,y,t) as x=1.Figure 2
The periodic solution which is a periodic nontraveling wave traveling on thet-axis for u1(x,y,t) as y=1.Case 2.
Takek1(t)=t, k2(t)=1 and c=0 in (26), then
(32)σ=tux+uy-1.
Solving the differential equation σ=0, we can get
(33)u=y+F(t,ξ),ξ=x-ty.
Substituting (33) into (1) and integrating once with respect to ξ yield
(34)Ft+FFξ+γt2Fξ-βFξξ+αFξξξ=0.
Again, further using the transformation of dependent variable to (34),
(35)F(t,ξ)=F(θ),θ=k(t-13γt3+ξ).
Substituting (35) into (34) and integrating once with respect to θ yield
(36)2k2αF′′-2kβF′+F2+2F+A=0,
where A is an integration constant, F′=dF(θ)/dθ. We assume that the solution of (36) can be expressed in the form
(37)F(θ)=a0+a1w(θ)+a2w(θ)2,
where ai  (i=0,1,2) are constants to be determined later, w(θ) satisfies the following auxiliary equation
(38)w′=p+qw2.
Substituting (37) and (38) into (36) and equating the coefficients of all powers of w to zero yield a set of algebra equations for a0, a1, a2, and A as follows. 							(39)w4:a2(a2+12αk2q2)=0,w3:-4βka2q+2a1a2+4αk2a1q2=0,w2:a12+16αk2a2qp-2βka1q+2a2+2a2a0=0,w1:2a1a0-4βka2p+2a1+4αk2a1qp=0,w0:2a0+A+4αk2a2p2+a02-2βka1p=0.
Solving the system of function equations with the aid of Maple, we obtain
(40)a0=3β2-25α25α,a1=6β2q25sα,a2=3β2q25αp.
when k=β/10sα, pq<0, A=(625α2-36β4)/625α2, where s=-pq.
It is known that solutions of (38) are as follows [24]:
(41)w(θ)=-stanh(sθ),w(θ)=-scoth(sθ).
Substituting (41), (40), (37), and (35) into (33), we obtain solutions of (1) as follows:
(42)u2(x,y,t)=125α{3β2(β10sα(x-ty+t-γ3t3))-25α-3qβ2×[tanh(β10sα(x-ty+t-γ3t3))-2pqtanh2(β10sα(x-ty+t-γ3t3))]}+y,u3(x,y,t)=125α{3β2(β10sα(x-ty+t-γ3t3))-25α-3qβ2×[coth(β10sα(x-ty+t-γ3t3))-2pqcoth2(β10sα(x-ty+t-γ3t3))]}+y.
(see Figures 3 and 4).Figure 3
Local structure ofu2(x,y,t) is shown as x=1,  α=1,  β=10,  p=-1,  q=1,  and γ=6.Figure 4
Local structure ofu3(x,y,t) is shown as x=1,  α=1,  β=10,  p=-1,  q=1,  γ=6.Remark 2.
If we direct assume that the solution of (34) can be expressed in the form
(43)F(t,ξ)=a0(t)+a1(t)w(θ)+a2(t)w(θ)2,
where θ=f(t)ξ+g(t), f(t), and g(t) are continuous functions of t to be determined later. w(θ) satisfies the auxiliary equation (38). Substituting (43) and (38) into (34), equating the coefficients of all powers of w to zero yields a set of function equations for a0(t), a1(t), a2(t), f(t), and g(t) as follows:
(44)w5:2fa2q(12f2q2α+a2)=0,w4:-3fq(-2a1q2f2α+2qfa2β-a1a2)=0,w3:-2βa1f2q2+2a22fp+40αa2f3q2p+2a2g′q+2a0a2fq+a12fq+2a2f′ξq+2γt2a2fq=0,w2:-8βa2f2pq+a1g′q+a2′+a0a1fq+γt2a1fq+8αa1f3pq2+a1f′ξq+3a1a2fq=0,w1:a12fp+16αa2f3p2q+a1′+2γt2a2fp+2a0a2fp+2a2g′p+2a2f′ξp-2βa1f2pq=0,w0:a1g′p+a1f′ξp+a0a1fp-2βa2f2p2+a0′+2αa1f3p2q+γt2a1fp=0.
Solving the system of function equations, we obtain
(45)a0(t)=3β225α,a1(t)=±6β2q25sα,a2(t)=3β2q25αp,f(t)=±β10sα,g(t)=∓βγ30sαt3.
This result indicate the idea is equivalent to idea of Case 2 above.Case 3.
Takek2(t)=0 and c=1 in (26), then
(46)σ=k1(t)ux+ut-k1(t).
Solving the differential equation σ=0, we obtain
(47)u=k1(t)+F(ξ,y),ξ=x-∫k1(t)dt.
Substituting (47) into (1) yield
(48)αFξξξξ-βFξξξ+FFξξ+Fξ2+γFyy=0.
Using the transformation F(ξ,y)=F(η),η=kξ-cy and integrating the resulting equation with respect to η we have
(49)k2F2+2γc2F+2k4αF′′-2k3βF′+A=0,
where A is an arbitrary constant, F′=dF/dη. Suppose that the solution of ODE (49) can be expressed by a polynomial in (G′/G) as follows:
(50)F(η)=bn(G′G)n+⋯,
where G=G(η) satisfies the second-order LODE in the form [25]
(51)G′′+λG′+μG=0.
Balancing F′′ with F2 in (49) gives n=2. So that
(52)F(η)=b2(G′G)2+b1(G′G)+b0,b2≠0,
where bi  (i=0,1,2) and μ are constants to be determined later. Substituting (52) and (51) into (49). Setting these coefficients of the G′/G to zero, yields a set of algebraic equations as follows:
(53)k2b2(12αk2+b2)=0,2k2(10αk2b2λ+b1b2+2αk2b1+2βkb2)=0,8αk4b2λ2+2βk3b1+k2b12+16αk4b2μ+2k2b2b0+6αk4b1λ+4βk3b2λ+2γc2b2=0,2k2b1b0+4βk3b2μ+4αk4b1μ+2γc2b1+2αk4b1λ2+2βk3b1λ+12αk4b2λμ=0,2γc2b0+2αk4b1λμ+A+2βk3b1μ+4αk4b2μ2+k2b02=0.
Solving the algebraic equations above yields
(54)b0=15k3λα(5kλα+2β)-3k2β2+25c2αγ25k2α,b1=-12k(5kαλ+β)5,b2=-12k2α.
when 25k2α2(4μ-λ2)+β2=0 and 625α2(Ak2-c2γ2)+36k4β4=0. Consequently, we obtain the following solution of (1) for λ2-4μ>0:
(55)u4(x,y,t)=-12k2ατ2×[(C1sinhτ(k(x-∫‍k1(t)dt)-cy)((k(x-∫‍k1(t)dt)-cy))2+C2coshτ(k(x-∫‍k1(t)dt)-cy))×(C1coshτ(k(x-∫‍k1(t)dt)-cy)+C2sinhτ×(k(x-∫‍k1(t)dt)-cy))-1]2+(12k2λατ-12k(5kαλ+β)5)×[(C1sinhτ(k(x-∫‍k1(t)dt)-cy)((k(x-∫‍k1(t)dt)-cy))2+C2coshτ(k(x-∫‍k1(t)dt)-cy))×(C1coshτ(k(x-∫‍k1(t)dt)-cy)+C2sinhτ×(k(x-∫‍k1(t)dt)-cy))-1]2+15k3λα(5kλα+2β)-3k2β2+25c2αγ25k2α+λ24,
where τ=(1/2)λ2-4μ.


## 6. Conclusions

Based on the fact that the periodic solution of (2+1)D KdV-Burgers equation does not exist, we have obtained Saddle-node Heteroclinic Orbits. By applying the Lie group method, we reduce the (2+1)D KdV Burgers equation to (1+1)-dimensional equations including the (1+1)-dimensional linear partial differential equation with constants coefficients (29), (48) and (1+1)-dimensional nonlinear partial differential equation with variable coefficients (34). By solving the equations (29), (34), and (48), we obtain some new exact solutions and discover the strange phenomenon of sympathetic vibration to evolution of nontraveling wave soliton for the (2+1)D KdV Burgers equation. Our results show that the unite of Lie group method with others is effective to search simultaneously exact solutions for nonlinear evolution equations. Other structures of solutions with symmetry (26) are to be further studied.



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