# Analytical Solutions for the Elastic Circular Rod Nonlinear Wave, Boussinesq, and Dispersive Long Wave Equations

Authors: Shi Jing; Yan Xin-li
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

The solving processes of the homogeneous balance method, Jacobi elliptic function expansion method, fixed point method, and modified mapping method are introduced in this paper. By using four different methods, the exact solutions of nonlinear wave equation of a finite deformation elastic circular rod, Boussinesq equations and dispersive long wave equations are studied. In the discussion, the more physical specifications of these nonlinear equations, have been identified and the results indicated that these methods (especially the fixed point method) can be used to solve other similar nonlinear wave equations.


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## Body

## 1. Introduction

Nonlinear partial differential equations are widely used to describe complex phenomena in various fields of sciences such as biology, chemistry, communication, and especially many branches of physics like condensed matter physics, field theory, fluid dynamics, plasma physics, and optics, and so forth. In this paper, we consider nonlinear wave equation of a finite deformation elastic circular rod, Boussinesq equations, and dispersive long wave equations with same physical behavior. The elastic circular rod is one important component in the structures, in which the dynamics of these components are governed by double nonlinearity and double dispersion wave equation [1]. The Boussinesq equation, which was first introduced in 1871, arises in several physical applications. The dynamics of shallow water waves, which are seen in various places like sea beaches, lakes, and rivers, are governed by the Boussinesq equation. In recent years there has been much interest in some variants of the Boussinesq systems [2–6]. These coupled Boussinesq equations [7] arise in shallow water waves for two-layered fluid flow. This situation occurs when there is an accidental oil spill from a ship which results in a layer of oil floating above the layer of water. The (2+1)-dimensional dispersive long wave equations [8, 9] were first derived by Boiti as a compatibility condition for a “weak” Lax pair. A good understanding of the solutions for these equations is very helpful to coastal and civil engineers in applying the nonlinear water model to coastal harbor design.In the field of nonlinear science to find exact solutions for a nonlinear system is one of the most fundamental and significant studies. The evaluation of exact solutions of nonlinear wave has complicated nonlinear defects; such equations are often very difficult to be solved. Although intensive investigations have made significant progress in recent years, many methods have been proposed to construct exact solutions, such as Weierstrass elliptic function method, the homogeneous balance method, sine-cosine method, the nonlinear transformation method, the hyperbolic tangent functions finite expansion, improved mapping approach, and further extended tanh method.The homogeneous balance method is a powerful tool to find solitary wave solutions of nonlinear partial differential equations. Fan introduced the homogeneous balance method into the search for Bäcklund transformations and obtained more solutions [10].The mapping method is a very effective direct method to construct exact solutions of nonlinear equations. Zhang et al. make use of the auxiliary equation and the expanded mapping methods to find the new exact periodic solutions for(2+1)-dimensional dispersive long wave equations [11].The basic idea of the fixed point methods consists in finding an iteration function, which generates successive approximations to the solution [12].Many periodic solutions have been recently expressed in terms of various Jacobi-elliptic functions for a wide class of nonlinear evolution equations, which have been obtained by means of Jacobi elliptic expansion method [13].The longitudinal wave equation of a finite deformation elastic circular rod, Boussinesq equations, and dispersive long wave equations are nonlinear partial differential equations of different scientific field. We find that they have the same characteristics. In this paper, the analytical solutions of the differential equations for the elastic circular rod, Boussinesq equations, and dispersive long wave equations are solved by using homogeneous balance method, Jacobi elliptic function method, fixed point method, and modified mapping method. The more physical specifications of these nonlinear equations have been identified and the results indicated.


## 2. Three Types of Nonlinear Wave Equations

The longitudinal wave equation of a finite deformation elastic circular rod is [1]
(1)utt-c02uxx=[3E2ρu2+E2ρu3+ν2R22(utt-c12uxx)]xx,
where c0=E/ρ is the longitudinal wave velocity for a linear elastic rod and c1=μ/ρ is the shear wave velocity. ρ is the density of the material. ν is the Poisson ratio. R is the diameter of bar. E is Yong’s modulus of material. μ is the elastic shear modulus of material. t is the time variable.Make the traveling wave transformation(2)u(x,t)=u(ξ),ξ=k(x-λt).Substituting (2) into (1) and integrating it with respect to ξ twice, we can obtain
(3)k2u′′+β1u+β2u2+β3u3=N,
where N is the second integral constant. The first integral constant is zero and
(4)β1=2(λ2-c02)ν2R2(λ2-c12),β2=3c02ν2R2(λ2-c12),β3=c02ν2R2(λ2-c12).The variant Boussinesq equations were discussed in [7]. They are coupling wave equations which can be expressed as follows:
(5)Ht+(Hu)x+uxxx=0,ut+Hx+uux=0,
where u(x,t) is the velocity and H(x,t) is the total depth. Make the traveling wave transformation
(6)H(x,t)=H(ξ),u(x,t)=u(ξ),ξ=k(x-λt).Substituting (6) into (5) and integrating with respect to ξ, we have
(7)-λH+Hu+k2u′′=a1,H=λu-12u2+b1,
where a1, b1 are integral constants.Substituting the second formula of (7) into the first one and assuming β1=b1-λ2, β2=(3/2)λ, β3=-(1/2), N=a1+λb1, we can obtain (3). So the solutions of (7) are equivalent to that of (3) and the second formula of (7).(
2
+
1
)-dimension dispersive long wave equations were discussed in [8, 9]. They can be written as follows:
(8)uty+hxx+12(u2)xy=0,ht+(hu+u+uxy)x=0.Suppose that the traveling wave solutions for (8) can be given in the following forms:
(9)u(x,y,t)=u(ξ),h(x,y,t)=h(ξ),ξ=x+ny+ωt.Substituting (9) into (8), we have
(10)nu′′+(1-nω2)u-32nωu2-n2u3=N1,h=-nωu-12nu2,
where N1 is the second integral constant.Obviously, the first formula of (10) and (3) have the same characteristics. On the other hand, the solutions of (10) and (7) are same with each other except for the coefficients. Therefore, in following parts, the solutions of (5) are presented.


## 3. Homogeneous Balance Method

In [14, 15] the homogeneous balance method is used to solve nonlinear wave equations. For example, the following equation was solved in [15]:
(11)ut+6uux+uxxx=0.The exact solution of the following corresponding equation was given by using the same method:(12)ut-Duxx+λ(u3+αu2+βu)=0.To solve (5), the homogeneous balance method was improved in this paper.Now, suppose that(13)u(x,t)=f′ωx+b,H(x,t)=g′ωx2+c,
where f(ω), g(ω), and ω(x,t) are the undetermined functions and b, c are undetermined constants.Substituting (13) into (5), the equations can be rewritten as follows:
(14)(f′′g′+f′g′′+f(4))ωx4+6f′′′ωx2ω2x+f′′(cωx2+3ω2x2+4ωxω3x)+g′′(bωx3+ωx2ωt)+3f′g′ωx2ω2x+f′(cω2x+ω4x)+2g′(bωxω2x+ωxωxt)=0(g′′+f′f′′)ωx3+f′′(ωxωt+bωx2)+f′2ωxω2x+f′(ωxt+bω2x)+2g′ωxω2x=0,
where ω2x=ωxx, ω3x=ωxxx, ω4x=ωxxxx. In (14), let
(15)f′′g′+f′g′′+f(4)=0,g′′+f′f′′=0.Integrating the second formula of (15), we have
(16)g′+12f′2=0.Furthermore, we can obtain(17)g′=-12f′2,g′′=-f′f′′.Substituting (17) into the first formula of (15), we have
(18)f(4)-32f′2f′′=0.Integrating (18), we have
(19)f′′′-12f′3=0.We know the solution of (19) is
(20)f=2ln⁡⁡ω.Obviously, we have(21)f′2=-2f′′,f′f′′=-f′′′,f′g′=-f′′′,f′3=2f′′′.Substituting (17) and (21) into (14), the equations can be rewritten as follows:
(22)ωx2(3ω2x+bωx+ωt)f′′′+(cωx2+3ω2x2+4ωxω3x+2bωxω2x+2ωxωxt)f′′+(cω2x+ω4x)f′=0,ωx(ωt+bωx)f′′+(ωxt+bω2x)f′=0.Letting all coefficients off in (22) be zero, we have
(23)3ω2x+bωx+ωt=0,cωx2+3ω2x2+4ωxω3x+2bωxω2x+2ωxωxt=0,cω2x+ω4x=0,ωt+bωx=0,ωxt+bω2x=0.From first and fifth second formula of (23), it can be obtained that
(24)ω=α(t)x+β(t),
where α(t) is undetermined function and β(t) is arbitrary function.From second formula of (23), we have
(25)α(t)=e-(c/2)t+k,
where k is the integral constant.Therefore, from (13), (20), (24), and (25), we know the solutions of (5) are
(26)u1(x,t)=2α(t)α(t)x+β(t)+b,H1(x,t)=-2α2(t)[α(t)x+β(t)]2+c.


## 4. Jacobi Elliptic Function Method

In [16], the NLS equation and Zakharov equation were studied by using the Jacobi elliptic function method. In this paper, (1) and (5) are discussed by using this method.


### 4.1. Jacobi Elliptic Sine Function Method

Let(27)u=a0+a1snξ.We know(28)dudξ=a1cnξdnξ,d2udξ2=-(1+m2)a1snξ+2m2a1sn3ξ,u2=a02+2a0a1snξ+a12sn2ξ,u3=a03+3a02a1snξ+3a0a12sn2ξ+a13sn3ξ,
where cnξ and dnξ are the Jacobi elliptic cosine function and the third type Jacobi elliptic function, respectively, and m(0<m<1) is module.Substituting (27) into (3), we have
(29)β1a0+β2a02+β3a03-N+a1[-k2(1+m2)+β1+2β2a0+3β3a02]snξ+a12(β2+3β3a0)sn2ξ+a1(2k2m2+β3a12)sn3ξ=0.Let the coefficients of all derivatives ofsnξ be zero and we have
(30)a0=-β23β3,a1=±2km-β3,k2=11+m2(β1+2β2a0+3β3a02),N=β1a0+β2a02+β3a03.Therefore the solution of (3) is
(31)u2(x,t)=a0+a1snk(x-λt),
where a0, a1, and k are denoted by (30), and λ is arbitrary constant.Whenm→1, snξ→tanhξ,
(32)u2(x,t)=a0+a1tanhk(x-λt).Thus, the solutions of (5) are (32) and the following formula:
(33)H2(x,t)=b1+λu2-12u22.Assumingλ=-1, b1=1.5, β1=0.5, β2=-1.5, and β3=-0.5, it can be obtained that a0=-1, a1=2, and k=1. The solutions of (5) are u(x,t)=-1+2tanh⁡(x+t) and H(x,t)=2-2tanh2(x+t). The solitary wave and behavior of the solutions u(x,t)=-1+2tanh⁡(x+t) and H(x,t)=2-2tanh2(x+t) are shown in Figures 1 and 2, respectively, for 0≤t≤1 and -10≤x≤10. The waveform is similar to the result in [17].Figure 1
The solitary wave and behavior of the solutionsu(x,t)=-1+2tanh⁡(x+t) for 0⩽t⩽1 and -10⩽x⩽10.Figure 2
The solitary wave and behavior of the solutionsH(x,t)=2-2tanh2(x+t) for 0⩽t⩽1 and -10⩽x⩽10.


### 4.2. Jacobi Elliptic Cosine Function Method

Let(34)u=a0+a1cnξ.Thus(35)u′=-a1snξdnξ,u′′=(2m2-1)a1cnξ-2m2cn3ξ,u2=a02+2a0a1cnξ+a12cn2ξ,u3=a03+3a02a1cnξ+3a0a12cn2ξ+a13cn3ξ.Substituting (35) into (3), we can obtain
(36)β1a0+β2a02+β3a03-N+a1[k2(2m2-1)+β1+2β2a0+3β3a02]cnξ+a12(β2+3β3a0)cn2ξ+a1(-2k2m2+β3a12)cn3ξ=0.Let the coefficients of all derivatives ofcnξ be zero and we have
(37)a0=-β23β3,a1=±km2β3,k2=β22-3β1β33β3(2m2-1),N=β1a0+β2a02+β3a03.Therefore, the solution of (3) is
(38)u3(x,t)=a0+a1cnk(x-λt),
where a0, a1 and k are denoted by (37), and λ is arbitrary constant.The solutions of (7) are (38) and the following formula:
(39)H3(x,t)=b1+λu3(x,t)-12u32(x,t).Takingm→1, cnξ→sechξ, (38) can be rewritten as follows:
(40)u3(x,t)=a0+a1sechk(x-λt).Assumingλ=1, β1=0.5, β2=-1.5, and β3=0.5, it can be obtained that a0=1, a1=2, and k=1. The solution of (3) is u(x,t)=1+2sech(x-t). The solitary wave and behavior of the solutions are shown in Figure 3, respectively, for 0≤t≤1 and -10≤x≤10.Figure 3
The solitary wave and behavior of the solutionsu(x,t)=1+2sech(x-t) for 0⩽t⩽1 and -10⩽x⩽10.


### 4.3. Third Kind of Jacobi Elliptic Function Method

Let(41)u=a0+a1dnξ.Thus(42)u′=-m2a1snξ,u′′=(2-m2)a1dnξ-2a1dn3ξ,u3=a03+3a02a1dnξ+3a0a12dn2ξ+a13dn3ξ.Substituting them into (3), we can obtain
(43)β1a0+β2a02+β3a03-N+a1[k2(2-m2)+β1+2β2a0+3β3a02]dnξ+a12(β2+3β3a0)dn2ξ+a1(-2k2+a12β3)dn3ξ=0.Letting the coefficients of all derivatives ofdnξ be zero, we have
(44)a0=-β23β3,a1=±k2β3,k2=1m2-2(3β1β3-β223β3),N=β1a0+β2a02+β3a03.Therefore the solution of (3) is
(45)u4(x,t)=a0+a1dnk(x-λt),
where a0, a1, and k are denoted by (44), and λ is arbitrary constant.Thus the solutions of (7) are (45) and the following formula:
(46)H4(x,t)=b1+λu4(x,t)-12u42(x,t).Lettingm→1, then (45) equals (38).


### 4.4. Jacobi Elliptic Functioncsξ Method

Let(47)u=a0+a1csξ,csξ≡cnξsnξ.Then(48)u′=-a1(1+cs2ξ)dnξ,u′′=(2-m2)a1csξ+2a1cs3ξ.Substituting (47) and (48) into (3), we can obtain
(49)D=β1a0+β2a02+β3a03-N+a1[k2(2-m2)+β1+2β2a0+3β3a02]csξ+a12(β2+3β3a0)cs2ξ+a1(2k2+a12β3)cs3ξ.Let the coefficients in above formula be zero, and we have(50)a0=-β23β3,a1=±k22-β3,k2=1m2-2(3β1β3-β223β3),N=β1a0+β2a02+β3a03.According to (47), the solution of (3) is
(51)u5(x,t)=a0+a1csk(x-λt),
where a0, a1, and k are denoted by (50), and λ is arbitrary constant.Therefore, the solutions of (7) are (51) and the following formula:
(52)H5=b1+λu5(x,t)-12u52(x,t).Whenm→1,
(53)u5(x,t)=a0+a1cschk(x-λt).Assumingβ1=(1-nω2)/n, β2=-(3/2)ω, β3=-(1/2), N=-(N1/n), and k=1, we can obtain (3). Choosing n=-(2/3), ω=1, then β1=-(5/2), β2=-(3/2). Substituting β1, β2, β3 into (50), it can be obtained that a0=-1, a1=2, and k=1. The solutions of (8) are u(x,y,t)=-1+2csch(x-(2/3)y+t) and h(x,y,t)=-(1/3)+(4/3)csch2(x-(2/3)y+t). The solitary wave and behavior of the solutions u(x,t)=-1+2csch(x-(2/3)y+t) and h(x,y,t)=-(1/3)+(4/3)csch2(x-(2/3)y+t) are shown in Figures 4 and 5, respectively, for t=0, -4≤x≤4, -4≤y≤4, and x-(2/3)y≠0.Figure 4
The solitary wave and behavior of the solutionsu(x,y,t)=-1+2csch(x-2/3y+t) for t=0, -4⩽x⩽4, and -4⩽y⩽4.Figure 5
The solitary wave and behavior of the solutionsh(x,y,t)=-1/3+4/3csch2(x-2/3y+t) for t=0, -4⩽x⩽4, and -4⩽y⩽4.


## 4.1. Jacobi Elliptic Sine Function Method

Let(27)u=a0+a1snξ.We know(28)dudξ=a1cnξdnξ,d2udξ2=-(1+m2)a1snξ+2m2a1sn3ξ,u2=a02+2a0a1snξ+a12sn2ξ,u3=a03+3a02a1snξ+3a0a12sn2ξ+a13sn3ξ,
where cnξ and dnξ are the Jacobi elliptic cosine function and the third type Jacobi elliptic function, respectively, and m(0<m<1) is module.Substituting (27) into (3), we have
(29)β1a0+β2a02+β3a03-N+a1[-k2(1+m2)+β1+2β2a0+3β3a02]snξ+a12(β2+3β3a0)sn2ξ+a1(2k2m2+β3a12)sn3ξ=0.Let the coefficients of all derivatives ofsnξ be zero and we have
(30)a0=-β23β3,a1=±2km-β3,k2=11+m2(β1+2β2a0+3β3a02),N=β1a0+β2a02+β3a03.Therefore the solution of (3) is
(31)u2(x,t)=a0+a1snk(x-λt),
where a0, a1, and k are denoted by (30), and λ is arbitrary constant.Whenm→1, snξ→tanhξ,
(32)u2(x,t)=a0+a1tanhk(x-λt).Thus, the solutions of (5) are (32) and the following formula:
(33)H2(x,t)=b1+λu2-12u22.Assumingλ=-1, b1=1.5, β1=0.5, β2=-1.5, and β3=-0.5, it can be obtained that a0=-1, a1=2, and k=1. The solutions of (5) are u(x,t)=-1+2tanh⁡(x+t) and H(x,t)=2-2tanh2(x+t). The solitary wave and behavior of the solutions u(x,t)=-1+2tanh⁡(x+t) and H(x,t)=2-2tanh2(x+t) are shown in Figures 1 and 2, respectively, for 0≤t≤1 and -10≤x≤10. The waveform is similar to the result in [17].Figure 1
The solitary wave and behavior of the solutionsu(x,t)=-1+2tanh⁡(x+t) for 0⩽t⩽1 and -10⩽x⩽10.Figure 2
The solitary wave and behavior of the solutionsH(x,t)=2-2tanh2(x+t) for 0⩽t⩽1 and -10⩽x⩽10.


## 4.2. Jacobi Elliptic Cosine Function Method

Let(34)u=a0+a1cnξ.Thus(35)u′=-a1snξdnξ,u′′=(2m2-1)a1cnξ-2m2cn3ξ,u2=a02+2a0a1cnξ+a12cn2ξ,u3=a03+3a02a1cnξ+3a0a12cn2ξ+a13cn3ξ.Substituting (35) into (3), we can obtain
(36)β1a0+β2a02+β3a03-N+a1[k2(2m2-1)+β1+2β2a0+3β3a02]cnξ+a12(β2+3β3a0)cn2ξ+a1(-2k2m2+β3a12)cn3ξ=0.Let the coefficients of all derivatives ofcnξ be zero and we have
(37)a0=-β23β3,a1=±km2β3,k2=β22-3β1β33β3(2m2-1),N=β1a0+β2a02+β3a03.Therefore, the solution of (3) is
(38)u3(x,t)=a0+a1cnk(x-λt),
where a0, a1 and k are denoted by (37), and λ is arbitrary constant.The solutions of (7) are (38) and the following formula:
(39)H3(x,t)=b1+λu3(x,t)-12u32(x,t).Takingm→1, cnξ→sechξ, (38) can be rewritten as follows:
(40)u3(x,t)=a0+a1sechk(x-λt).Assumingλ=1, β1=0.5, β2=-1.5, and β3=0.5, it can be obtained that a0=1, a1=2, and k=1. The solution of (3) is u(x,t)=1+2sech(x-t). The solitary wave and behavior of the solutions are shown in Figure 3, respectively, for 0≤t≤1 and -10≤x≤10.Figure 3
The solitary wave and behavior of the solutionsu(x,t)=1+2sech(x-t) for 0⩽t⩽1 and -10⩽x⩽10.


## 4.3. Third Kind of Jacobi Elliptic Function Method

Let(41)u=a0+a1dnξ.Thus(42)u′=-m2a1snξ,u′′=(2-m2)a1dnξ-2a1dn3ξ,u3=a03+3a02a1dnξ+3a0a12dn2ξ+a13dn3ξ.Substituting them into (3), we can obtain
(43)β1a0+β2a02+β3a03-N+a1[k2(2-m2)+β1+2β2a0+3β3a02]dnξ+a12(β2+3β3a0)dn2ξ+a1(-2k2+a12β3)dn3ξ=0.Letting the coefficients of all derivatives ofdnξ be zero, we have
(44)a0=-β23β3,a1=±k2β3,k2=1m2-2(3β1β3-β223β3),N=β1a0+β2a02+β3a03.Therefore the solution of (3) is
(45)u4(x,t)=a0+a1dnk(x-λt),
where a0, a1, and k are denoted by (44), and λ is arbitrary constant.Thus the solutions of (7) are (45) and the following formula:
(46)H4(x,t)=b1+λu4(x,t)-12u42(x,t).Lettingm→1, then (45) equals (38).


## 4.4. Jacobi Elliptic Functioncsξ Method

Let(47)u=a0+a1csξ,csξ≡cnξsnξ.Then(48)u′=-a1(1+cs2ξ)dnξ,u′′=(2-m2)a1csξ+2a1cs3ξ.Substituting (47) and (48) into (3), we can obtain
(49)D=β1a0+β2a02+β3a03-N+a1[k2(2-m2)+β1+2β2a0+3β3a02]csξ+a12(β2+3β3a0)cs2ξ+a1(2k2+a12β3)cs3ξ.Let the coefficients in above formula be zero, and we have(50)a0=-β23β3,a1=±k22-β3,k2=1m2-2(3β1β3-β223β3),N=β1a0+β2a02+β3a03.According to (47), the solution of (3) is
(51)u5(x,t)=a0+a1csk(x-λt),
where a0, a1, and k are denoted by (50), and λ is arbitrary constant.Therefore, the solutions of (7) are (51) and the following formula:
(52)H5=b1+λu5(x,t)-12u52(x,t).Whenm→1,
(53)u5(x,t)=a0+a1cschk(x-λt).Assumingβ1=(1-nω2)/n, β2=-(3/2)ω, β3=-(1/2), N=-(N1/n), and k=1, we can obtain (3). Choosing n=-(2/3), ω=1, then β1=-(5/2), β2=-(3/2). Substituting β1, β2, β3 into (50), it can be obtained that a0=-1, a1=2, and k=1. The solutions of (8) are u(x,y,t)=-1+2csch(x-(2/3)y+t) and h(x,y,t)=-(1/3)+(4/3)csch2(x-(2/3)y+t). The solitary wave and behavior of the solutions u(x,t)=-1+2csch(x-(2/3)y+t) and h(x,y,t)=-(1/3)+(4/3)csch2(x-(2/3)y+t) are shown in Figures 4 and 5, respectively, for t=0, -4≤x≤4, -4≤y≤4, and x-(2/3)y≠0.Figure 4
The solitary wave and behavior of the solutionsu(x,y,t)=-1+2csch(x-2/3y+t) for t=0, -4⩽x⩽4, and -4⩽y⩽4.Figure 5
The solitary wave and behavior of the solutionsh(x,y,t)=-1/3+4/3csch2(x-2/3y+t) for t=0, -4⩽x⩽4, and -4⩽y⩽4.


## 5. Fixed Point Method

In general, fixed point theory can be divided into two types. The first type is only used to discuss the existence of solution. The second type is used not only to discuss the existence and uniqueness of solution but also to search the fixed point. We are more interested in the second type.Lemma 1 (see [18–20]).
SupposeE is a real Banach space which has normal cone. Consider u0,v0∈E, u0<v0. The operator A:[u0,v0]×[u0,v0]→E is mixed monotone and satisfies (1)
(54)u0≤A(u0,v0),A(v0,u0)≤v0(2)
for allu,v(u0≤u≤v≤v0), ∃ constant α∈(0,1), satisfies
(55)∥A(v,u)-A(u,v)∥≤α∥v-u∥.
Then there exists a uniqueu¯(u¯∈[u0,v0]) which satisfies(56)A(u¯,u¯)=u¯,
where u¯ is called the fixed point of A.
In[u0,v0] for all w0, letting wn=A(wn-1,wn-1), n=1,2,…, we have
(57)u¯=lim⁡n→∞wn.In this paper,E is taken as a continuous function space C[0,1] which is defined in the closed interval [0,1]. Introducing equivalent norm, it is easy to know that C[0,1] is real Banach space which has normal cone. Let
(58)u0=0,v0=1.Integrating (3) with respect to ξ, we can obtain
(59)dudξ=-k-2∫0ξ[β1u(s)+β2u2(s)+β3u3(s)]dξ+Nξ+C1,
where C1 is integral constant.Integrating the above formula with respect toξ and using the formula in [12], we have
(60)u=-k-2∫0ξ(ξ-s)[β1u(s)+β2u2(s)+β3u3(s)]ds+12Nξ2+C1ξ+C2,
where C2 is integral constant.Ifβ2>0, β3>0 (i.e., λ2>c1), and β1<0 (i.e., λ2<c02), suppose
(61)A(u,v)=-k-2∫0ξ(ξ-s)[β1u(s)+β2v2(s)+β3v3(s)]ds+12Nξ2+C1ξ+C2.Theorem 2.
In (61), supposing β2>0, β3>0, β1<0, C1≥0, C2≥0, and k-2≤N/(β2+β3)≤2(1+C1+C2)/(β2+β3-β1), there exists unique fixed point u¯∈[u0,v0] for the operator A which is defined by (61). Then (57) is true.Proof.
Obviously,A:[u0,v0]×[u0,v0]→C[0,1] is mixed monotone. Consider
(62)A(u0,v0)=A(0,1)=12[N-k-2(β2+β3)]ξ2+C1ξ+C2≥0=u0,A(v0,u0)=12(N-β1k-2)ξ2+C1ξ+C2≤12N(1-β1β2+β3)+C1+C2≤1=v0.
ThusA satisfies the term (1) of Lemma 1.
For allu,v∈[u0,v0], u≤v,
(63)A(v,u)-A(u,v)=-k-2∫0ξ(ξ-s)×[β1v(s)+β2u2(s)+β2u3(s)-β1u(s)-β2v2(s)-β3v3(s)]ds=k-2∫0ξ(ξ-s)eMse-Ms[v(s)-u(s)]×{β3[v2(s)+u(s)v(s)+u2(s)]-β1+β2[v(s)+u(s)]+β3[v2(s)+u(s)v(s)+u2(s)]}ds≤(-β1+2β2+3β3)k-2∥v-u∥∫0ξeMSds≤-β1+2β2+3β3k2M∥v-u∥eMξ.
Then, we have(64)∥A(v,u)-A(u,v)∥≤α,
where α=(-β1+2β2+3β3)/(k2M), and when M is large enough, α∈(0,1).
Therefore, operatorA which is defined by (61) satisfies the term (2) of Lemma 1. The proof is completed.Obviously, the fixed point of the operator defined by (61) is equivalent to the solution of (3). We denote
(65)u6(x,t)=u¯(ξ).By using the classical iteration algorithms, such as Mann iteration method, Ishikawa iteration method, and Noor iteration method, the solutions can be obtained. Then, the traveling wave solutions of (5) are (65) and the following formula:
(66)H6(x,t)=b1+λu6(x,t)-12u62(x,t).


## 6. Modified Mapping Method

The modified mapping method was introduced and the nonlinear evolution equation was solved in [21, 22]. In [22], several hundreds of Jacobi elliptic function expansion solutions were obtained. We also use this method in this paper.Let(67)u=A0+A1f+B1f-1,
where f′′=pf+qf3⇔f′2=r+pf2+(1/2)qf4. r, p, and q are constants.Thus(68)u′=A1f′-B1f-2f′,u′′=A1pf+A1qf3+B1pf-1+2B1rf-3,u2=A02+2A1B1+2A0A1f+2A0B1f-1+A12f2+B12f-2,u3=A03+6A0A1B1+3A1(A02+A1B1)f+3A0A12f2+A13f3+3B1(A02+A1B1)f-1+3A0B12f-2+B13f-3.Substituting (68) into (3), we have
(69)β1A0+β2(A02+2A1B1)+β3(A03+6A0A1B1)-N+A1[k2p+β1+β2A0+3β3(A02+A1B1)]f+A12(β2+3β3A0)f2+A1(k2q+β3A12)f3+B1[k2p+β1+2β2A0+3β3(A02+A1B1)]f-1+B12(β2+3β3A0)f-2+B1(2k2r+β3B12)f-3=0.Notice that we get the same algebraic equations when the coefficients off,f-1 and f2,f-2 in (69) are zero. Therefore there are only five algebra equations. Solve them and we have
(70)A0=-β23β3,A1=±k-qβ3,B1=±k-2rβ3,k=±β22-3β1β3-3k2β32qr3β3p.In order to makeA1, B1 be real number, q and r should be different signs with β3.(i) Choosingr=1, p=-2, and q=2, we have f=tanh⁡ξ. Then the solitary wave solutions of (3) are
(71)u7(x,t)=A0+A1tanh⁡ξ+B1tanh-1ξ.Assumingn=-2, ω=1, k=1, and β3=-(1/2), it can be obtained that β1=-(3/2), β2=-(3/2), A0=-1, A1=2, and B1=2. The solutions of (8) are u(x,y,t)=-1+2tanh⁡(x-2y+t)+2tanh-1(x-2y+t) and h(x,y,t)=7+4tanh2(x-2y+t)+4tanh-2(x-2y+t). The solitary wave and behavior of them are shown in Figures 6 and 7, respectively, for t=0, -4≤x≤4, -4≤y≤4, and x-2y≠0.Figure 6
The solitary wave and behavior of the solutionsu(x,y,t)=-1+2tanh⁡(x-2y+t)+2tanh-1(x-2y+t) for t=0, -4⩽x⩽4, and -4⩽y⩽4.Figure 7
The solitary wave and behavior of the solutionsh(x,y,t)=7+4tanh2⁡(x-2y+t)+4tanh-2(x-2y+t) for t=0, -4⩽x⩽4, and -4⩽y⩽4.(ii) Choosingr=1, p=-(1+m2), and q=2m2, thus
(72)f=snξ,
where m(0<m<1) is the module of Jacobi elliptic function.Therefore the expansion solutions of (3) are
(73)u8(x,t)=A0+A1snξ+B1sn-1ξ.Lettingm→1, then (73) equals (71).(iii) Choosingr=1/4, p=(m2-2)/2, and q=m2/2, we have
(74)f=snξ(1+dnξ).Therefore the solutions of (3) are
(75)u9(x,t)=A0+A1snξ1+dnξ+B1(1+dnξ)snξ,
where A0, A1, B1 of ui(i=7,8,9) are denoted by (70).Takingm→1, (75) can be rewritten as follows:
(76)u9(x,t)=A0+A1tanhξ1+sech⁡ξ+B1(1+sech⁡ξ)tanh⁡ξ.Assumingn=-2, ω=1, k=1, and β3=-(1/2), it can be obtained that β1=-(3/2), β2=-(3/2), A0=-1, A1=1, and B1=1. The solutions of (8) are u(x,y,t)=-1+tanh⁡(x-2y+t)/(1+sech⁡(x-2y+t))+(1+sech⁡(x-2y+t))/tanh⁡(x-2y+t) and h(x,y,t)=1+[tanh(x-2y+t)/(1+sech(x-2y+t))]2+[(1+sech(x-2y+t))/tanh(x-2y+t)]2. The solitary wave and behavior of them are shown in Figures 8 and 9, respectively, for t=0, -4≤x≤4, -4≤y≤4, and x-2y≠0.Figure 8
The solitary wave and behavior of the solutionsu(x,y,t)=-1+ tanh⁡(x-2y+t)/(1+sech⁡(x-2y+t))+(1+sech⁡(x-2y+t))/tanh⁡(x-2y+t) for t=0, -4⩽x⩽4, and -4⩽y⩽4.Figure 9
The solitary wave and behavior of the solutionsh(x,y,t)=1+ tanh2⁡(x-2y+t)/(1+sech⁡(x-2y+t))2+(1+sech⁡(x-2y+t))2/tanh2(x-2y+t) for t=0, -4⩽x⩽4, and -4⩽y⩽4.


## 7. Summary and Conclusions

In summary, by using different methods, nonlinear wave equation of a finite deformation elastic circular rod, Boussinesq equations, and dispersive long wave equations are solved in this paper, and several analytical solutions are obtained. Kink soliton solutions are obtained when we use the homogenous balance method. The Jacobi elliptic function method is utilized and periodic solutions are obtained. Whenm→1, the solutions reduce to solitary solution. Their wave forms are the bell type, kink type, exotic type, and peakon type. By using the modified mapping method, peakon solutions of (3), (5), (8) are obtained. On the other hand, after proving the existence and uniqueness of fixed point of operator A, we can use the classical iteration algorithms to get the solutions. The method has wide application and can be used to solve other nonlinear wave equations. Fixed point method may be significant and important for the explanation of some special physical problems whose analytical solution cannot be obtained. The results indicate the following. First, the natural phenomena and the physical properties of the equations are abundant, which should be further discovered. Second, there are many kinds of methods for solving these equations. Third, the arbitrary constant can be determined by the initial and boundary conditions and we can obtain the exact solutions for certain engineering or scientific problems. It is shown that the methods, proposed in this paper for three types of nonlinear wave equations, are feasible for determining exact solutions of other nonlinear equations.



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*Source: 487571-2014-04-22.xml*