The Green-Naghdi system was first derived by Green and Naghdi [1] in 1976 for describing a fully nonlinear shallow water gravity wave with a free surface or an interfacial surface. The system also appears in different physical contexts such as bubbly fluid dynamics and magneto-hydro dynamics. Its various aspects have been studied. For instance, Li [2] showed that the system has no eigenvalues with a positive real part and solitary waves with a small amplitude are linearly stable. Deng, Guo and Wang [3] obtained the exact expressions of its smooth soliton wave solutions, cusp soliton wave solutions, smooth periodic wave solutions and periodic cusp wave solutions and gave some numerical simulations of these solutions.

The Green-Naghdi asymptotic model for small aspect ratio waves

$S_{t}-\frac{1}{k^{2}}S_{xxt}-\frac{1}{2k^{2}}\sqrt{\frac{g}{k}}S_{xxx}+\frac{3}{2}\sqrt{\frac{g}{k}}S_{x}+\sqrt{gk}SS_{x}=\frac{1}{3k}\sqrt{\frac{g}{k}}(5S_{x}S_{xx}+SS_{xxx}),$

(1.1)

which was first derived by Kraenkelb, Leona and Mannaa [4] in 2005, is a $k$-dependent equation that describes the nonlinear deformations of the wave with a given wave number $k$. Here $S(x, t)$ is the free surface deformation, $g$ is the gravitation, $t$ is the time variable, and $x$ is the propagation direction. Thus the parameters $g, k$ are both positive. Taking $\alpha=\sqrt{\frac{g}{k}}$, (1.1) becomes

$k^{2}S_{t}-S_{xxt}-\frac{\alpha}{2}S_{xxx}+\frac{3}{2}\alpha k^{2}S_{x}+k^{3}\alpha SS_{x}-\frac{5}{3}k\alpha S_{x}S_{xx}-\frac{1}{3}k\alpha SS_{xxx}=0.$

(1.2)

Recently, many mathematicians are very interested in traveling wave solutions of many mathematical and physical models. For instance, Huang and Liu investigated new exact traveling wave solutions of Fisher equation and Burgers-Fisher equation by using the method of an auxiliary ODE method in [5]; Rong, Tang and Huang [6] considered bifurcations of traveling solutions for the K$(n, 2n, -n)$ equations; Tang et al. [7] discussed traveling wave solutions for the generalized special type of the Tzitzeica-Dodd-Bullough equation; Zhong and Gu [8] studied the bifurcation of traveling wave solutions for symmetric regularized wave equations and so on.

To the best of our knowledge, there is no result about the traveling wave solutions of (1.2). In this paper, we discuss the bifurcation of the traveling wave system of the equation (1.2) under the different parameter conditions by qualitative analysis methods of planar systems (See, e.g., [9]), give the corresponding phase portraits by using the software Maple and show the exact expressions of smooth periodic wave solutions, kink-like wave solutions, antikink-like wave solutions, compacton-like wave solutions, periodic cusp wave solutions, solitary wave solutions and cusp solitary wave solutions. Furthermore, we simulate them.

This paper is organized as follows. In Section 2, (1.2) is changed into a traveling wave system. Phase portraits are given in Section 3. Section 4 shows the exact expressions of bounded wave solutions and the numerical simulations of these solutions.

Let $\xi=x-ct$, where $c\neq 0$ is the wave speed. Substituting $S(x, t)=u(x-ct)=u(\xi)$ into (1.2) we get the following ordinary differential equation

$(\frac{3}{2}\alpha -c)k^{2}u_{\xi}+(c-\frac{\alpha}{2})u_{\xi\xi\xi}+k^{3}\alpha u u_{\xi}-\frac{5}{3}k\alpha u_{\xi}u_{\xi\xi}-\frac{1}{3}k\alpha uu_{\xi\xi\xi}=0.$

(2.1)

Integrating (2.1) once with respect to $\xi$ yields the traveling wave equation

$\beta+(\frac{3}{2}\alpha -c)k^{2}u+\frac{1}{2}k^{3}\alpha u^{2}+(c-\frac{\alpha}{2}-\frac{1}{3}k\alpha u)u_{\xi\xi}-\frac{2}{3}k\alpha (u_{\xi})^{2}=0,$

(2.2)

where $\beta$ is the constant of integration. Let $v=u_{\xi}$ and $b=\frac{\alpha}{2}-c$, and we have the following traveling wave system from (2.2)

$\left\{ \begin{array}{ll} \frac{d u}{d\xi}=v, \\ \frac{d v}{d\xi}=\frac{1}{\frac{1}{3}k\alpha u+b}\left(-\frac{2}{3}k\alpha v^{2}+\frac{1}{2}k^{3}\alpha u^{2}+(\alpha+b)k^{2}u+\beta\right). \end{array} \right.$

(2.3)

It is not convenient to study the phase portraits of system (2.3) because it has a singular line $ u=-\frac{3b}{k\alpha}.$ Thus, we introduce a transformation

$d\xi=(\frac{1}{3}k\alpha u+b)d\tau.$

(2.4)

Then system (2.3) is changed to

$\left\{ \begin{array}{ll} \frac{d u}{d\tau}=(\frac{1}{3}k\alpha u+b)v, \\ \frac{d v}{d\tau}=-\frac{2}{3}k\alpha v^{2}+\frac{1}{2}k^{3}\alpha u^{2}+(\alpha+b)k^{2}u+\beta. \end{array} \right.$

(2.5)

The first integral of (2.3) and (2.5) is

$\begin{aligned} H(u, v)&=-\frac{1}{270}k^{5}\alpha^{3}(2\alpha+11b)u^{5}-\frac{1}{2}kb^{2}(\alpha kb+kb^{2}-\alpha\beta)u^{2}\\ &\quad +\frac{1}{2}(\frac{1}{3}k\alpha u+b)^{4}v^{2} -\frac{1}{18}k^{2}\alpha b(9kb^{2}+6k\alpha b+2\alpha \beta)u^{3}\\ &\quad -b^{3}\beta u-\frac{1}{216}k^{3}\alpha^{2}(2\alpha\beta+18k\alpha b+45kb^{2})u^{4}-\frac{1}{324}k^{6}\alpha^{4}u^{6}. \end{aligned}$

(2.6)

Obviously, (2.3) and (2.5) have the same topological phase portraits except the singular line $l$: $u=-\frac{3b}{k\alpha}.$ In the following, we focus on system (2.5). Let

$ v_{1}= \sqrt{\frac{3}{2k\alpha}\left[\beta-\frac{k}{2\alpha}3b(2\alpha-b)\right]}, v_{2}= -\sqrt{\frac{3}{2k\alpha}\left[\beta-\frac{k}{2\alpha}3b(2\alpha-b)\right]}, \\ u_{1} = \frac{1}{k^{2}\alpha}\left[-k(\alpha+b)-\sqrt{k^{2}(\alpha+b)^{2}-2\alpha k\beta}\right], \\ u_{2}= \frac{1}{k^{2}\alpha}\left[-k(\alpha+b)+\sqrt{k^{2}(\alpha+b)^{2}-2\alpha k\beta}\right], $

which will be used later.

Let

$f(u)=\frac{1}{2}k^{3}\alpha u^{2}+(\alpha+b)k^{2}u+\beta.$

(3.1)

It is easy to see that

$f^{\prime}(u)=k^{3}\alpha u+(\alpha+b)k^{2}.$

(3.2)

The symmetric axis and the discriminant of $f$ are

$u=-\frac{\alpha+b}{k\alpha}$

(3.3)

and

$\Delta=2\alpha k^{3}[\frac{k}{2\alpha}(\alpha+b)^{2}-\beta].$

(3.4)

The coefficient matrix of the linearized system of (2.5) at an equilibrium $E(u_{e}, v_{e})$ is

$\left( \begin{array}{cc} \frac{1}{3}k\alpha v_{e} \frac{1}{3}k\alpha u_{e}+b \\ f'(u_{e}) -\frac{4}{3}k\alpha v_{e} \\ \end{array} \right),$

(3.5)

whose determinant and trace are

$D(E)=-\frac{4}{9}k^{2}\alpha^{2}v_{e}^{2}-(\frac{1}{3}k\alpha u_{e}+b)f'(u_{e}), T(E)=-k\alpha v_{e}.$

(3.6)

By the qualitative theory of differential equations for an equilibrium of a planar dynamical system [9], we know that the equilibrium $E(u_{e}, v_{e})$ is a saddle point if $D<0$; It is a node if $D>0$ and $T\neq 0$; it is a center if $D>0$ and $T=0$; It is degenerate if $D=0$. Using these, we can obtain the phase portraits of (2.5) under different parameter conditions.

Note that from $c\neq 0$ and $b=\frac{\alpha}{2}-c$ we have

$\frac{k}{2\alpha}(\alpha+b)^{2}-\frac{3}{2\alpha}k b(2\alpha-b)=\frac{k}{2\alpha}(\alpha-2b)^{2}>0, $

which means $\frac{k}{2\alpha}(\alpha+b)^{2}>\frac{3}{2\alpha}k b(2\alpha-b).$ Using these, we discuss the bifurcation of system (2.5) by the relative position of the function $f(u)$ to the singular line $l$ (see Figure).

Theorem 1 If $2b<\alpha$ and $\frac{3}{2\alpha}k b(2\alpha-b)<\beta<\frac{k}{2\alpha}(\alpha+b)^{2}, $ then system (2.5) has two equilibriums $ E_{1}(-\frac{3b}{k\alpha}, v_{1})$ and $E_{2}(-\frac{3b}{k\alpha}, v_{2})$ on the singular line $l$ and two equilibriums $E_{3}(u_{1}, 0)$ and $E_{4}(u_{2}, 0)$ at the left side of $l$. $ E_{1}$, $E_{2}$ and $E_{3}$ are saddle points, and $ E_{4}$ is a center (see (1)-(3) of Figure).

Proof The definition of $f$ yields

$f(-\frac{3b}{k\alpha})=\beta-\frac{3kb}{2\alpha}(2\alpha-b).$

(3.7)

From

$-\frac{2}{3}k\alpha v^{2}+f(-\frac{3b}{k\alpha})=0, $

we have

$v^{2}=\frac{3}{2k\alpha}[\beta-\frac{3}{2\alpha}k b(2\alpha-b)].$

(3.8)

By $\frac{3}{2\alpha}k b(2\alpha-b)<\beta$, we get that system (2.5) has two equilibrium points $ E_{1}(-\frac{3b}{k\alpha}, v_{1})$ and $E_{2}(-\frac{3b}{k\alpha}, v_{2})$. By (3.6) we have $D(E_{1})<0$ and $D(E_{2})<0$. Thus, $ E_{1}(-\frac{3b}{k\alpha}, v_{1})$ and $E_{2}(-\frac{3b}{k\alpha}, v_{2})$ are saddle points.

From $\beta<\frac{k}{2\alpha}(\alpha+b)^{2}$ and (3.4), we get that $\Delta>0.$ Thus, $f(u)$ has two real roots $u_{1}$ and $u_{2}$, and system (2.5) has two equilibriums $E_{3}(u_{1}, 0)$ and $ E_{4}(u_{2}, 0).$ Next, we discriminate the relative position of $E_{3}, E_{4}$ and $l.$ From $2b<\alpha$ we have

$-\frac{3b}{k\alpha}-(-\frac{\alpha+b}{k\alpha})=\frac{\alpha-2b}{k\alpha}>0.$

(3.9)

By (3.3) we obtain that the symmetric axis of $f$ is at the left side of the singular line $l$. By (3.7) and $\frac{3}{2\alpha}k b(2\alpha-b)<\beta, $ we have $f(-\frac{3b}{k\alpha})>0.$ Thus, we get that $E_{3}$ and $ E_{4}$ are at the left side of $l$ (see (a) of Figure 1). Furthermore, it is easy to see that $f'(u_{1})<0$, $f'(u_{2})>0$, $\frac{1}{3}k\alpha u_{i}+b<0$ ($i=1, 2$). By (3.6) we have that $D(E_{3})<0$ and $D(E_{4})>0$. Thus, $E_{3}$ is a saddle point and $E_{4}$ a center.

Theorem 2 If $2b<\alpha$ and $\beta=\frac{3}{2\alpha}k b(2\alpha-b), $ then system (2.5) has two equilibriums $E_{3}(u_{1}, 0)$ and $ E_{4}(u_{2}, 0).$ $E_{3}$ is a saddle point, $ E_{4}$ is a degenerate saddle point (see (4) of Figure).

Proof The proof that $E_{3}$ is a saddle point is similar as one of Theorem 1. We only prove that $ E_{4}$ is a degenerate saddle point.

By $\beta=\frac{k}{2\alpha}3b(2\alpha-b), $ we have $u_{2}= -\frac{3b}{k\alpha}.$ Under the transformation $u=y-\frac{3b}{k\alpha}$ and $ v=z$, system (2.5) can be written as

$\left\{ \begin{array}{ll} \frac{d y}{d\tau}=\frac{1}{3}k\alpha y z, \\ \frac{d z}{d\tau}=k^{2}(\alpha-2b)y-\frac{2}{3}k\alpha z^{2}+\frac{1}{2}k^{3}\alpha y^{2}. \end{array} \right.$

(3.10)

Thus, we focus on $O(0, 0)$ of system (3.10). Let $Y=k^{2}(\alpha-2b)y-\frac{2}{3}k\alpha z^{2}+\frac{1}{2}k^{3}\alpha y^{2}, X=z, $ and we have

$\left\{ \begin{array}{ll} \frac{d X}{d\tau}=Y, \\ \frac{d Y}{d\tau}=\frac{2}{9}k^{2}\alpha^{2} X^{3}-k\alpha XY +(|X|^{3}+|XY|)O(|(X, Y)|). \end{array} \right.$

(3.11)

By Theorem 7.2 in Chapter 2 of [9], we get that $O(0, 0)$ is a degenerate saddle point of (3.11). which means that $O(0, 0)$ is a degenerate saddle point of (3.10). Thus, $ E_{4}$ is a degenerate saddle point of (2.5).

Theorem 3 If $2b<\alpha$ and $\beta=\frac{k}{2\alpha}(\alpha+b)^{2}, $ then system (2.5) has three equilibriums $ E_{1}(-\frac{3b}{k\alpha}, v_{1})$, $E_{2}(-\frac{3b}{k\alpha}, v_{2})$ and $E_{3}(u_{1}, 0)$. $ E_{1}$ and $E_{2}$ are saddle points, and $E_{3}$ is a cusp (see (5) of Figure).

Proof The proof that $E_{1}$ and $E_{2}$ are saddle points is similar as one of Theorem 1. Next we only prove that $E_{3}$ is a cusp. With the transformation $u=m-\frac{\alpha+b}{k\alpha}$ and $ v=y$, system (2.5) becomes

$\left\{ \begin{array}{ll} \frac{d m}{d\tau}=\frac{2b-\alpha}{3}y+\frac{1}{3}k\alpha m y, \\ \frac{d y}{d\tau}=\frac{1}{2}k^{3}\alpha m^{2}-\frac{2}{3}k\alpha y^{2}. \end{array} \right.$

(3.12)

Obviously, the topological structure of system (2.5) near the equilibrium $E_{3}$ is homeomorphic to the one of system (3.12) near $O.$ Let $w=m$ and $z=\frac{2b-\alpha}{3}y+\frac{1}{3}k\alpha m y.$ System (3.12) is changed into

$\left\{ \begin{array}{ll} \frac{d w}{d\tau}=z, \\ \frac{d z}{d\tau}=\frac{1}{2}k^{3}\alpha w^{2}+z^{2}\frac{k\alpha}{\alpha-2b+k\alpha w} . \end{array} \right.$

(3.13)

By Theorem 7.3 in Chapter 2 of [9], we get that the equilibrium $E_{3}$ is a cusp point.

Theorem 4 If $2b<\alpha$ and $\beta<\frac{3}{2\alpha}k b(2\alpha-b), $ then system (2.5) has two equilibriums $E_{3}(u_{1}, 0)$ and $ E_{4}(u_{2}, 0)$ which are both saddles (see (4) of Figure).

The proof is same as one of Theorem 1.

Theorem 5 If $2b<\alpha$ and $\beta>\frac{k}{2\alpha}(\alpha+b)^{2}, $ then system (2.5) has two equilibriums $ E_{1}(-\frac{3b}{k\alpha}, v_{1})$ and $E_{2}(-\frac{3b}{k\alpha}, v_{2})$ which are both saddle points (see (5) of Figure).

If $2b>a$ we have similar results as follows.

Theorem 6 For $2b>\alpha, $

(1) If $\frac{3}{2\alpha}k b(2\alpha-b)<\beta<\frac{k}{2\alpha}(\alpha+b)^{2}, $ then (2.5) has four equilibriums $ E_{1}(-\frac{3b}{k\alpha}, v_{1})$, $E_{2}(-\frac{3b}{k\alpha}, v_{2})$, $E_{3}(u_{1}, 0)$ and $ E_{4}(u_{2}, 0)$. $ E_{1}$, $E_{2}$ and $E_{4}$ are saddle points, and $ E_{3}$ is a center (see (1)-(3) of Figure).

(2) If $\beta=\frac{k}{2\alpha}3b(2\alpha-b), $ then system (2.5) has two equilibriums $E_{3}(-\frac{3b}{k\alpha}, 0)$ and $ E_{4}(u_{2}, 0).$ $E_{4}(u_{2}, 0)$ is a saddle point and $ E_{3}$ is degenerate saddle point (see (4) of Figure).

(3) If $\beta=\frac{k}{2\alpha}(\alpha+b)^{2}.$ then system (2.5) has three equilibriums $ E_{1}(-\frac{3b}{k\alpha}, v_{1})$, $E_{2}(-\frac{3b}{k\alpha}, v_{2})$ and $E_{3}(-\frac{\alpha+b}{k\alpha}, 0)$. $ E_{1}$ and $E_{2}$ are saddle points and $E_{3}$ is a cusp (see (5) of Figure).

(4) If $\beta<\frac{k}{2\alpha}3b(2\alpha-b).$ then system (2.5) has two equilibriums $E_{3}(u_{1}, 0)$ and $E_{4}(u_{2}, 0).$ $E_{3}$ and $E_{4}$ are both saddle points (see (6) of Figure).

(5) If $\beta>\frac{k}{2\alpha}(\alpha+b)^{2}.$ then system (2.5) has two equilibriums $ E_{1}(-\frac{3b}{k\alpha}, v_{1})$ and $E_{2}(-\frac{3b}{k\alpha}, v_{2})$ which are both saddle points (see (7) of Figure).

In this section, we give smooth periodic wave solutions, kink-like wave solutions, antikink-like wave solutions, compacton-like wave solutions, periodic cusp wave solutions, solitary wave solutions and cusp solitary wave solutions and their numerical simulations. Let $u(\xi)$ be a traveling wave solution of (1.2) for $\xi\in(-\infty, +\infty)$ and

$\lim\limits_{\xi\rightarrow-\infty}u(\xi)=A, \lim\limits_{\xi\rightarrow+\infty}u(\xi)=B, $

where $\xi=x-ct$ and $A, B$ are constants. $u(\xi)$ is called a solitary wave solution of (1.2) if $A=B$ and a kink (or antikink) wave solution if $A\neq B.$ Usually, a solitary wave solution of (1.2) corresponds to a homoclinic orbit of system (2.5), a kink (or antikink) wave solution of (1.2) corresponds to a heteroclinic orbit of system (2.5), and a periodic traveling wave solution of (1.2) corresponds to a periodic orbit of system (2.5). In the following, we just consider the cases that $2b<\alpha$ and $\frac{3}{2\alpha}k b(2\alpha-b)<\beta<\frac{k}{2\alpha}(\alpha+b)^{2}$ (see (1)-(3) of Figure). Using the same way, we can discuss other cases. Note that, from (2.4), we have $\xi\rightarrow -\infty$ if $\tau\rightarrow +\infty$ and $u<-\frac{3b}{k\alpha}$.

Theorem 7 (Periodic wave solutions) Suppose that $2b<\alpha$, $\frac {3 }{2\alpha}k b(2\alpha-b)<\beta<\frac{k}{2\alpha}(\alpha+b)^{2}$. Consider the following conditions:

(1) $H(u_{1}, 0)\geq H(-\frac{3b}{k\alpha}, v_{1})$ and $u_{2}<u_{0}<-\frac{3b}{k\alpha}$ (see (1) and (2) of Figure);

(2) $H(u_{1}, 0)< H(-\frac{3b}{k\alpha}, v_{1})$ and $u_{2}<u_{0}<e_{0}$, where $e_{0}\in (u_{2}, -\frac{3b}{k\alpha})$ is the solution of equation $H(u, 0)=H(u_{1}, 0)$ (see (3) of Figure).

If one of the two conditions holds, then (1.2) has a periodic wave solution

$u(\xi ) = \left\{ {\begin{array}{*{20}{l}} {{\phi _1}(\xi - 2n{T_0})}&{{\rm{ for}}\;\xi \in [2n{T_0},(2n + 1){T_0}),}\\ {{\phi _1}( - \xi + 2(n + 1){T_0})}&{{\rm{ for}}\;\xi \in ((2n + 1){T_0},2(n + 1){T_0}),} \end{array}} \right.$

$n=0, \pm 1, \pm 2, \cdots, $ with $u(0)=u_{0}$ and $v(0)=0$ satisfying

$\int_{u_{0}}^{\phi_{1}}-\frac{1}{\sqrt{F(s)}}\mathrm{d}s=\xi,$

(4.1)

where

$\begin{aligned} F(u)&=\frac{2}{(\frac{1}{3}k\alpha u+b)^{4}} \left( \frac{1}{324}k^{6}\alpha^{4}u^{6}+\frac{1}{270}k^{5}\alpha^{3}(2\alpha+11b)u^{5}+b^{3}\beta u\right.\\&\quad +\frac{1}{2}kb^{2}(\alpha kb+kb^{2}+\alpha\beta)u^{2} +\frac{1}{18}k^{2}\alpha b(9kb^{2}+6k\alpha b+2\alpha \beta)u^{3} \\&\left. \quad+\frac{1}{216}k^{3}\alpha^{2}(2\alpha\beta+18k\alpha b+45kb^{2})u^{4}-H(u_{0}, 0)\right), \end{aligned}$

(4.2)

$T_{0}$ is given by

$T_{0}=\int_{u_{0}}^{u_{0}^{-}}-\frac{1}{\sqrt{F(s)}}\mathrm{d} s, $

and $u_{0}^{-}\in(u_{1}, u_{2})$ is a solution of $H(u, 0)=H(u_{0}, 0)$.

Proof By (1) of Theorem 1 system (2.5) has a periodic orbit $\Gamma=\Gamma_{+}\cup \Gamma_{-}$ since the equilibrium $E_{4}$ is a center. Take $(u_{0}, 0)\in \Gamma$. From the definition of $H(u, v)$ in (2.6), $\Gamma$ lies on the curve given by $H(u, v)=H(u_{0}, 0)$. Note that $H(u, 0)=H(u_{0}, 0)$ has a solution $u_{0}^{-}\in(u_{1}, u_{2})$, i.e., $\Gamma$ intersects the $u-$axis at the point $(u_{0}^{-}, 0)$.

$H(u, v)=H(u_{0}, 0)$ yields $v=\pm \sqrt{F(u)}.$ By (2.3) we have

$\mathrm{d} \xi=\frac{\mathrm{d} u}{v}=\frac{\mathrm{d} u}{\pm \sqrt{F(u)}}.$

Integrating the above along $\Gamma$ in clockwise gives (4.1).

Theorem 8 (Kink-like or antikink-like wave solutions) Suppose that $2b<\alpha, $ $\frac{3}{2\alpha}k b(2\alpha-b)<\beta<\frac{k}{2\alpha}(\alpha+b)^{2}$ and $H(u_{1}, 0)> H(-\frac{3b}{k\alpha}, v_{1})$. Let $\Gamma_{1}$ and $\Gamma_{2}$ denote the orbit connecting $E_{3}(u_{1}, 0)$ (see (1) of Figure). Take an initial value $(u_{0}, v_{0})$ with $u_{1}<u_{0}<u_{2}$ and $v_{0}>0$ (or $v_{0}<0$) on the $\Gamma_{1}$ (or $\Gamma_{2}$). Then (1.2) has a kink-like (or antikink-like) wave solution $u=u(\xi)$ satisfying

$\int_{u_{0}}^{u}\frac{1}{\sqrt{F(s)}}\mathrm{d} s=\xi \quad \left(\mbox{or} \;\int_{u_{0}}^{u}-\frac{1}{\sqrt{F(s)}}\mathrm{d} s=\xi\right)$

for $\xi\in(-\infty, T_{1})$ (or $\xi\in(\tilde{T}_{1}, +\infty)$), where $F(u)$ is defined in (4.2) by replacing $H(u_{0}, 0)$ with $H(u_{1}, 0)$ and $T_{1}$ (or $\tilde{T}_{1}$) is given by

$T_{1}=\int_{u_{0}}^{-\frac{3b}{k\alpha}}\frac{1}{\sqrt{F(s)}}\mathrm{d} s \quad \left(\mbox{or} \; \tilde{T}_{1}=\int_{u_{0}}^{-\frac{3b}{k\alpha}}-\frac{1}{\sqrt{F(s)}}\mathrm{d} s \right).$

The proof is similar to the one of Theorem 3.3.

Theorem 9 (Compacton-like wave solutions) Suppose that $2b<\alpha, $ $\frac{3}{2\alpha}k b(2\alpha-b)<\beta<\frac{k}{2\alpha}(\alpha+b)^{2}$ and $H(u_{1}, 0)> H(-\frac{3b}{k\alpha}, v_{1})$. If $u_{1}<e_{1}<e_{0}, $ where $e_{0}\in(u_{1}, u_{2})$ is the solution of equation $H(u, 0)=H(-\frac{3b}{k\alpha}, v_{1})$ (see (1) of Figure), then (1.2) has a compacton-like wave solution

$u(\xi ) = \left\{ {\begin{array}{*{20}{l}} {{\phi _2}(\xi )}&{{\rm{for}}\;\xi \in [0,{T_2}),}\\ {{\phi _2}( - \xi )}&{{\rm{for}}\;\xi \in ( - {T_2},0),} \end{array}} \right.$

with $(u(0), v(0))=(e_{1}, 0)$ satisfying

$\int_{e_{1}}^{\phi_{2}}\frac{1}{\sqrt{F(s)}}\mathrm{d} s=\xi, $

where $F(u)$ is defined in (4.2) by replacing $H(u_{0}, 0)$ with $H(e_{1}, 0)$ and $T_{2}$ is given by

$T_{2}=\int_{e_{1}}^{-\frac{3b}{k\alpha}}\frac{1}{\sqrt{F(s)}}\mathrm{d} s.$

Theorem 10 (Periodic cusp wave solutions) Suppose that $2b<\alpha, $ $\frac{3k}{2\alpha}b(2\alpha-b)<\beta<\frac{k}{2\alpha}(\alpha+b)^{2}$and $H(u_{1}, 0)> H(-\frac{3b}{k\alpha}, v_{1})$. If $u_{1}<e_{0}<u_{2}, $where $e_{0}\in(u_{1}, u_{2})$ is the solution of the equation $H(u, 0)=H(-\frac{3b}{k\alpha}, v_{1})$ (see (1) of Figure), then (1.2) has a periodic cusp wave solution

$u(\xi)=\left\{ \begin{array}{ll} \phi_{3}(\xi-2nT_{3}) \qquad \hbox{ for}\; \xi\in[2nT_{3}, (2n+1)T_{3}), \\ \phi_{3}(-\xi+2nT_{3}) \quad \hbox{for}\; \xi\in[(2n-1)T_{3}, 2nT_{3}), \end{array} \right. $

$ n=0, \pm1, \pm2, \cdots, $ with $u(0)=e_{0}$ and $v(0)=0$ satisfying

$\int_{e_{0}}^{\phi_{3}}\frac{1}{\sqrt{F(s)}}\mathrm{d} s=\xi,$

where $F(u)$ is defined in (4.2) by replacing $H(u_{0}, 0)$ with $H(e_{0}, 0)$, and $T_{3}$ is given by

$T_{3}=\int_{e_{0}}^{-\frac{3b}{k\alpha}}\frac{1}{\sqrt{F(s)}}\mathrm{d} s.$

Theorem 11 (Solitary wave solutions]) Suppose $2b<\alpha$, $\frac{3}{2\alpha}k b(2\alpha-b)<\beta<\frac{k}{2\alpha}(\alpha+b)^{2}$ and $H(u_{1}, 0)<H(-\frac{3b}{k\alpha}, v_{1})$ (see (3) Figure). Then (1.2) has a solitary wave solution

$u(\xi)=\left\{\begin{array}{ll} \phi_{4}(\xi)&\hbox{for}\; \xi\in[0, +\infty), \\ \phi_{4}(-\xi)&\hbox{for}\; \xi\in(-\infty, 0) \end{array} \right.$

with $u(0)=e_{0}$ and $v(0)=0$ satisfying

$\int_{e_{0}}^{\phi_{4}}-\frac{1}{\sqrt{F(s)}}\mathrm{d} s=\xi,$

(4.3)

where $e_{0}\in (u_{2}, -\frac{3b}{k\alpha})$ is the solution of equation $H(u, 0)=H(u_{1}, 0)$ and $F(u)$ is defined in (4.2) by replacing $H(u_{0}, 0)$ with $H(u_{1}, 0).$

Theorem 12 (Cusp solitary wave solutions) Suppose that $2b<\alpha, $ $\frac{3}{2\alpha}k b(2\alpha-b)<\beta<\frac{k}{2\alpha}(\alpha+b)^{2}$ and $H(u_{1}, 0)= H(-\frac{3b}{k\alpha}, v_{1})$. Let $\Gamma_{1}$ (or $\Gamma_{2}$) denote the orbits connecting $(u_{1}, 0)$ and $(-\frac{3b}{k\alpha}, v_{1})$(or $(-\frac{3b}{k\alpha}, -v_{1})$). Take an initial value $(u(0), v(0))=(u_{0}, v_{0})\in \Gamma_{1}$. Then (1.2) has a cusp solitary wave solution

$u(\xi)=\left\{ \begin{array}{ll} \phi_{5}(\xi)&\hbox{for}\; \xi\in(-\infty, T_{4}); \\ \phi_{5}(-\xi+2T_{4})&\hbox{for}\; \xi\in(T_{4}, +\infty), \end{array} \right. $

where $\phi_{5}$ satisfies

$\int_{u_{0}}^{\phi_{5}}\frac{1}{\sqrt{F(s)}}\mathrm{d} s=\xi, $

$F(u)$ is defined in (4.2) by replacing $H(u_{0}, 0)$ with $H(u_{1}, 0)$ and

$T_{4}=\int_{u_{0}}^{-\frac{3b}{k\alpha}}\frac{1}{\sqrt{F(s)}}\mathrm{d} s=\xi.$

First, in order to simulate some bounded wave solutions, we take $b={1\over 2}, $ $\beta={5.3 \over 7}$ and $k=0.2$ which imply $\alpha=7$ for $g=9.8$. After simple calculations, we obtain $u_{1}$=$-6.64483, $ $u_{2}$=$-4.06944, $ $v_{1}=0.70801$, the singular line $l$: $u$=$-1.07143, $ $H(u_{1}, 0)=0.57234$ and $H(-1.07143, 0.70801)=0.02327$ Thus, the conditions in Theorems 4.1-4.4 are satisfied. Solving equation $H(u, 0)=H(-1.07143, 0.70801)$ yields $e_{0}=-5.03733$

The graph of the periodic wave solution $u(\xi)$ of (1.2) with an initial value $(u(0), v(0))=(-3, 0)$ is shown in (1) of Figure. The graph of the kink-like (or antikink-like) wave solution $u(\xi)$ of (1.2) with an initial value

$(u(0), v(0))=(-6, 0.11160)$

(or $(u(0), v(0))=(-6, -0.11160)$) is shown in (2) (or (3)) of Figure. The graph of the compacton-like wave solution $u(\xi)$ of (1.2) with an initial value $(u(0), v(0))=(-6, 0)$ is shown in (4) of Figure. The graph of the periodic cusp wave solution $u(\xi)$ of (1.2) with an initial value $(u(0), v(0))=(-5.03733, 0)$ is shown in (5) of Figure.

Second, in order to simulate solitary wave solutions, we take $b={1\over 2}, $ $\beta={5.55 \over 7}$ and $k=0.2$ which imply $\alpha=7$ for $g=9.8$. After simple calculations, we obtain

$u_{1}=-5.97573, u_{2}=-4.73855, v_{1}=0.73453, $

the singular line $l$: $u$=$-1.07143, $ $H(u_{1}, 0)=-0.13603$ and $H(-1.07143, 0.73453)=0.02446.$ Thus, the conditions in Theorem 4.5 are satisfied. Solving equation $H(u, 0)=H(-5.97573, 0)$ yields $e_{0}=-3.74457.$ The graph of the solitary wave solution $u(\xi)$ of (1.2) with an initial value $(u(0), v(0))$=$(-3.74457, 0)$ is shown in (6) of Figure.

Last, in order to simulate cusp solitary wave solutions, we take $b={1\over 2}, $ $\beta={5.481 \over 7}$ and $k=0.2$ which imply $\alpha=7$ for $g=9.8$. After simple calculations, we obtain

$u_{1}=-6.21428, u_{2}=-4.49999, v_{1}=0.72730, $

the singular line $l$: $u$=$-1.07143, $ $H(u_{1}, 0)=0.02413$ and $H(-1.07143, 0.72730)=0.02413.$ Thus, the conditions in Theorem 4.6 are satisfied. Solving equation

$H(-5, v)=H(-6.21428, 0)$

yields $v=0.17172$ (or $v=-0.17172$). The graph of the cusp solitary wave solution $u(\xi)$ of (1.2) with an initial value $(u(0), v(0))=(-5, 0.17172)$ is shown in (7) of Figure.