1. Introduction
Rogue waves were originally used to describe the extremely destructive water waves on the ocean, which appeared from nowhere and disappeared without a trace [1]. From the point of view of mathematics, rogue waves are a kind of special non-singular rational solutions of nonlinear evolution equations. In 1965, the concept of ‘freak rogue waves' was first proposed by Draper [2]. Since then, the study of rogue waves has attracted widespread attention like those [3–19] and some effective methods have been proposed for constructing rogue wave solutions, such as Darboux transformation [4, 6, 7], Hirota's bilinear method [8, 9], inverse scattering transform [10, 11] and so on. To be more specific, Solli, Ropers, Koonath and Jalali [3] reported the rogue waves observed by them in the optical system and modelled the generation of these rogue waves through the generalized nonlinear Schrödinger (NLS) equation; Akhmediev, Ankiewicz and Soto-Crespo [4] presented a modified Darboux scheme for constructing rogue wave solutions of the standard self-focusing NLS equation; Yan [5] gave the term ‘financial rogue waves' based on the nonlinear option equation; Guo, Ling and Liu [7] proposed the generalized Darboux transformation to construct Nth-order rogue wave solutions of physically interested models and illustrated its feasibility with two concrete examples; Ohta and Yang [8] derived Nth-order rogue wave solutions of the NLS equation by Hirota's bilinear method; Randoux, Suret and El [10] used a local periodization procedure to give inverse scattering transform analysis of rogue wave solutions of the NLS equation; Ma [16] presented the polynomial conjecture related to rogue waves in the Korteweg-de Vries equation.
More recently, some researchers [20–27] have focused on studying rogue wave solutions, especially higher-order rogue wave solutions of (3 + 1)-dimensional nonlinear models. In 2018, Zhaqilao [22] proposed the so-called symbolic computation approach to constructing rogue wave solutions with controllable centers of two high-dimensional nonlinear systems. Compared with the other methods [4, 8, 10], the symbolic computation approach [22] is direct and does not depend on the τ-function and Lax pair analysis which require high skills. More importantly, a pair of parameters embedded in the method [22] has been proven to be able to control the centers of rogue waves in terms of position and quantity. It is worth mentioning that the method [22] without these two parameters is only its initial version [21]. In this paper, we find that in addition to controlling the center of a first-order rogue wave, these two parameters can also control the fission and asymmetry of higher-order rogue waves. However, the findings on fission and asymmetry of higher-order rogue waves are not reported in [22]. Since the high asymmetry is another prominent feature of rogue waves besides steepness and instantaneity, it is of great significance to construct asymmetric rogue waves. Although there are a few studies on asymmetric rogue waves, such as [28–30], these studies either stay on the image display or have no substantive discussion on the regularity of control factors. In this article, we shall use the method [22] to construct higher-order rogue wave solutions with controllable fission and asymmetry of the following (3 + 1)-dimensional generalized Boussinesq (gB) equation [27]:1 ) and obtained its two-soliton solutions based on Bell polynomial theory. Meanwhile, Yan [27] also obtained a breather wave solution and first-order rogue wave solution of equation (1 ) by using the extended homoclinic test method. As far as we know, the higher-order rogue wave solution of equation (1 ) and its fission and asymmetry has not been reported.
$\begin{eqnarray}{u}_{tt}+\alpha {({u}^{2})}_{xx}+\beta {u}_{xxxx}+\gamma {u}_{xx}-{u}_{yy}-{u}_{zz}=0,\end{eqnarray}$
where $\alpha ,$ $\beta $ and $\gamma $ are constants, and the influence laws of two parameters controlling the fission and asymmetry of higher-order rogue wave solutions are discussed. In 2019, Yan [27] bilinearized equation (We arrange the rest of this paper as follows: section 2 describes Zhaqilao's symbolic computation approach [22]; section 3 gives the Painlevé analysis and bilinear form of equation (1 ); in section 4 , the first-, second-, third- and fourth-order rogue wave solutions of equation (1 ) are constructed by the method [22], and then the effects of the two parameters included in the constructed rogue wave solutions on the center, fission and asymmetry of the rogue waves are discussed; and section 5 concludes this paper and discusses the potential physical applications of the results obtained.
2. Description of Zhaqilao's symbolic computation approach
To construct rogue wave solutions of the (3 + 1)-dimensional gB equation (1 ), we first recall Zhaqilao's symbolic computation approach [22]. For a given (3 + 1)-dimensional nonlinear partial differential equation (PDE) assumed as:
$\begin{eqnarray}P(u,{u}_{x},{u}_{y},{u}_{z},{u}_{t},{u}_{xx},{u}_{xy},{u}_{xz},{u}_{xt},\cdot \cdot \cdot )=0,\end{eqnarray}$
where $P$ is usually a polynomial of $u$ and its various partial derivatives with respect to $x,$ $y,$ $z$ and $t,$ or any other expression which can be transformed into such a polynomial, we outline the steps of the method [22] as follows:Based on the Painlevé analysis, we take a transformation:
$\begin{eqnarray}u=T(f),\end{eqnarray}$
where f is an undetermined function of the independent variables $x,$ $y,$ $z$ and $t;$ With the help of equation (
$\begin{eqnarray}R({D}_{\xi },{D}_{z};f)=0,\,\xi =x+ay-ct,\end{eqnarray}$
where $a$ and $c$ are undetermined real constants, ${D}_{\xi }$ and ${D}_{z}$ are Hirota's operators [31]: $\begin{eqnarray}\begin{array}{l}{D}_{\xi }^{m}{D}_{z}^{n}f(\xi ,z)\cdot g(\xi ,z)={({\partial }_{\xi }-{\partial }_{\xi ^{\prime} })}^{m}{({\partial }_{z}-{\partial }_{z^{\prime} })}^{n}\\ \,\times f(\xi ,z)g(\xi ^{\prime} ,z^{\prime} ){| }_{\xi =\xi ^{\prime} ,z=z^{\prime} },\,(m,n\in N);\end{array}\end{eqnarray}$
We suppose that
$\begin{eqnarray}\begin{array}{l}f={\tilde{F}}_{n+1}(\xi ,z;\theta ,\vartheta )={F}_{n+1}(\xi ,z,a)+2\theta z{F}_{n}(\xi ,z,b)\\ \,+2\vartheta \xi {F}_{n}(\xi ,z,c)+({\theta }^{2}+{\vartheta }^{2}){F}_{n-1}(\xi ,z,a),\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{l}{F}_{n}(\xi ,z,\varepsilon )=\displaystyle \sum _{k=0}^{n(n+1)/2}\displaystyle \sum _{i=0}^{k}{\varepsilon }_{n(n+1)-2k,2i}{z}^{2i}{\xi }^{n(n+1)-2k},\,\\ (\varepsilon =a,b,c),\end{array}\end{eqnarray}$
with the initial values $\begin{eqnarray}{F}_{-1}(\xi ,z,a)={F}_{0}(\xi ,z,b)={F}_{0}(\xi ,z,c)=0,\end{eqnarray}$
while ${a}_{j,l},$ ${b}_{j,l}$ and ${c}_{j,l}(j,l\in \{0,2,4,\cdot \cdot \cdot ,n(n+1)\})$ are all real undetermined constants, $\theta $ and $\vartheta $ are two embedded parameters in the rogue wave solutions to be constructed. Substituting equation (
Solving the system of polynomial equations derived above and substituting the obtained values of ${a}_{j,l},$ ${b}_{j,l}$ and ${c}_{j,l}(j,l\in \{0,2,4,\cdot \cdot \cdot ,n(n+1)\})$ into equation (
3. Painlevé analysis and bilinear form
By Painlevé analysis, the following auto-Bäcklund transformation for the (3 + 1)-dimensional gB equation (
$\begin{eqnarray}u=\displaystyle \frac{6\beta }{\alpha }{(\mathrm{ln}\,\phi )}_{xx}+{u}_{0},\end{eqnarray}$
where ${u}_{0}$ is a solution of equation (1 ), while the real function $\phi $ of $x,$ $y,$ $z$ and $t$ satisfies the system of nonlinear PDEs:
$\begin{eqnarray}{\phi }_{t}^{2}-{\phi }_{y}^{2}-{\phi }_{z}^{2}+(\gamma +2\alpha {u}_{0}){\phi }_{x}^{2}-3\beta {\phi }_{xx}^{2}+4\beta {\phi }_{x}{\phi }_{xxx}=0,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\phi }_{x}^{2}({\phi }_{tt}-{\phi }_{yy}-{\phi }_{zz}+6\gamma {\phi }_{xx}+9\beta {\phi }_{xxxx})\\ \,+\,{\phi }_{xx}({\phi }_{t}^{2}-{\phi }_{y}^{2}-{\phi }_{z}^{2}-3\beta {\phi }_{xx}^{2})\\ \,+\,2{\phi }_{x}[{({\phi }_{t}^{2}-{\phi }_{y}^{2}-{\phi }_{z}^{2})}_{x}+4\alpha {u}_{0}{\phi }_{x}^{3}]=0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\phi }_{{xx}}\left({\phi }_{{tt}}-{\phi }_{{yy}}-{\phi }_{{zz}}\right)-2\beta {\phi }_{{xxx}}^{2}+3\beta {\phi }_{{xx}}{\phi }_{{xxxx}}\\ \,+\,2\left({\phi }_{{xt}}^{2}-{\phi }_{{xy}}^{2}-{\phi }_{{xz}}^{2}\right)\\ \,+\,2\alpha {u}_{0,{xx}}{\phi }_{x}^{2}+2{\phi }_{x}\left[{\left({\phi }_{{tt}}-{\phi }_{{yy}}-{\phi }_{{zz}}\right)}_{x}+6\alpha {u}_{0,x}{\phi }_{{xx}}^{2}\right.\\ \,+\,\left.2\left(\gamma +2\alpha {u}_{0}\right){\phi }_{{xxx}}+3\beta {\phi }_{{xxxxx}}\right]=0.\end{array}\end{eqnarray}$
Firstly, we suppose that
$\begin{eqnarray}u=\displaystyle \sum _{j=0}^{\infty }{v}_{j}{\phi }^{j-n},\,({v}_{0}\ne 0),\end{eqnarray}$
where ${v}_{j}(j=0,1,2,\cdot \cdot \cdot )$ are real functions of $x,$ $y,$ $z$ and $t$ to be determined later.Secondly, substituting equation (13 ) into equation (1 ) and analyzing the leading terms of $\alpha {({u}^{2})}_{xx}$ and $\beta {u}_{xxxx}$ yields
$\begin{eqnarray}-2n-2=-n-4,\end{eqnarray}$
$\begin{eqnarray}2\alpha n(2n+1){v}_{0}^{2}{\phi }_{x}^{2}+\beta n(n+1)(n+2)(n+3){v}_{0}{\phi }_{x}^{4}=0,\end{eqnarray}$
which hint $\begin{eqnarray}n=2,\,{v}_{0}=-\displaystyle \frac{6\beta }{\alpha }{\phi }_{x}^{2}.\end{eqnarray}$
Further letting13 ), (16 ) and (17 ) into equation (1 ), then setting all the coefficients of the same powers of ${\phi }^{s}(s=-5,-4,-3,-2,-1,0)$ to be zeros, we have10 )–(12 ).
$\begin{eqnarray}{v}_{j}=0(j\geqslant 3),\,{v}_{2}={u}_{0},\end{eqnarray}$
and inserting equations ( $\begin{eqnarray}{v}_{1}=\displaystyle \frac{6\beta }{\alpha }{\phi }_{xx},\end{eqnarray}$
$\begin{eqnarray}{v}_{2,tt}+\alpha {({v}_{2}^{2})}_{xx}+\beta {v}_{2,xxxx}+\gamma {v}_{2,xx}-{v}_{2,yy}-{v}_{2,zz}=0,\end{eqnarray}$
besides equations (With the help of equations (13 ) and (16 )–(18 ) we finally reach equation (9 ). It is easy to see from equations (17 ) and (19 ) that ${v}_{2}$ or ${u}_{0}$ is a solution of equation (1 ).
The (3 + 1)-dimensional gB equation (
$\begin{eqnarray}(\omega {D}_{\xi }^{2}+\beta {D}_{\xi }^{4}-{D}_{z}^{2})f\cdot f=0,\end{eqnarray}$
where $f$ is a real function of $\xi $ and $z,$ and $\omega $ and $\xi $ always represent $\begin{eqnarray}\omega =-{a}^{2}+{c}^{2}+\gamma ,\end{eqnarray}$
$\begin{eqnarray}\xi =x+ay-ct,\end{eqnarray}$
where $a$ and $c$ are two real constants. Obviously, equations (
$\begin{eqnarray}\omega {u}_{\xi \xi }+3{({u}^{2})}_{\xi \xi }+\beta {u}_{\xi \xi \xi \xi }-{u}_{zz}=0.\end{eqnarray}$
We next introduce $\begin{eqnarray}u={u}_{0}+{q}_{\xi \xi },\end{eqnarray}$
here u0 is assumed as a constant. Equation ( $\begin{eqnarray}\omega {q}_{\xi \xi }+\alpha {({u}_{0}+{q}_{\xi \xi })}^{2}+\beta {q}_{\xi \xi \xi \xi }-{q}_{zz}=0\end{eqnarray}$
by integrating the resulting equation derived from the substitution of equation (In view of equations (9 ) and (24 ), we assume that26 ) into equation (25 ), then the following equivalent equation is derived:27 ) into equation (20 ) by Hirota's operators (5).
$\begin{eqnarray}q=\displaystyle \frac{6\beta }{\alpha }\,\mathrm{ln}\,f,\end{eqnarray}$
and substitute equation ( $\begin{eqnarray}\begin{array}{l}\omega ({f}_{\xi \xi }f-{{f}_{\xi }}^{2})-({f}_{zz}f-{{f}_{z}}^{2})+\beta ({f}_{\xi \xi \xi \xi }f-4{f}_{\xi \xi \xi }{f}_{\xi }+3{{f}_{\xi \xi }}^{2})\\ \,=\,0,\end{array}\end{eqnarray}$
if and only if ${u}_{0}=0.$ We finally transform equation (4. Rogue wave solutions
Based on the bilinear form (27 ), we construct the first-, second-, three- and four-order rogue wave solutions of the (3 + 1)-dimensional gB equation (1 ).
4.1. First-order rogue wave solution
For the first-order rogue wave solution, we consider the case of $n=0$ in equation (6 ) and select28 ) into equation (27 ) and setting all the coefficients of the same powers of ${z}^{p}{\xi }^{q}(p,q\,=0,1,2,\cdot \cdot \cdot )$ to be zeros, we get a set of polynomial equations for ${a}_{2,0},$ ${a}_{0,2}$ and ${a}_{0,0}.$ Solving the polynomial equations, we have
$\begin{eqnarray}f={F}_{1}(\xi ,z,a)={a}_{2,0}{\xi }^{2}+{a}_{0,2}{z}^{2}+{a}_{0,0}.\end{eqnarray}$
Substituting equation ( $\begin{eqnarray}{a}_{2,0}=1,\,{a}_{0,2}=-\omega ,\,{a}_{0,0}=-\displaystyle \frac{3\beta }{\omega },\end{eqnarray}$
and then obtain the first-order rogue wave solution of equation (1 ):28 ) so that $f$ can be taken the following form:30 ) still solves equation (1 ). This is due to equations (28 ) and (31 ) having the same partial derivatives with respect to $\xi $ and $z.$
$\begin{eqnarray}u=-\displaystyle \frac{12\omega \beta [3\beta +\omega {(\xi -\theta )}^{2}+{\omega }^{2}{(z-\vartheta )}^{2}]}{\alpha {[3\beta -\omega {(\xi -\theta )}^{2}+{\omega }^{2}{\left(z-\vartheta \right)}^{2}]}^{2}}.\end{eqnarray}$
Here a pair of parameters $\theta $ and $\vartheta $ have been embedded into equation ( $\begin{eqnarray}f={\tilde{F}}_{1}(\xi ,z;\theta ,\vartheta )={a}_{2,0}{(\xi -\theta )}^{2}+{a}_{0,2}{(z-\vartheta )}^{2}+{a}_{0,0},\end{eqnarray}$
which ensures that equation (In figure 1, the first-order rogue wave solution (30 ) is shown by setting30 ). As can be seen from figures 1 and 2, the first-order rogue wave is symmetric about the lines $\xi =0$ and $z=0,$ and its center is controlled by the parameters $\theta $ and $\vartheta .$ Specifically, when $\theta =\vartheta =0,$ the first-order rogue wave has the extreme value at the location $(0,0),$ but for different $\theta =\vartheta =3,$ the extreme value appears at the point $(3,3).$
$\begin{eqnarray}a=-4.6,\,c=-3.15,\,\alpha =3,\,\beta =1,\,\gamma =7,\end{eqnarray}$
and $\theta =\vartheta =0,$ respectively. At the same time, except for different $\theta =\vartheta =3,$ the selection of other parameters is the same as that in figure 1, we use figure 2 to show the first-order rogue wave solution (
Figure 1. First-order rogue wave localized in ( |
Figure 2. First-order rogue wave localized in ( |
4.2. Second-order rogue wave solution
For the second-order rogue wave solution, we consider equation (6 ) when $n=1,$ that is
$\begin{eqnarray}\begin{array}{l}f={\tilde{F}}_{2}(\xi ,z;\theta ,\vartheta )={F}_{2}(\xi ,z,a)+2\theta z{F}_{1}(\xi ,z,b)\\ \,+\,2\vartheta \xi {F}_{1}(\xi ,z,c)+{\theta }^{2}+{\vartheta }^{2},\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{l}{F}_{2}(\xi ,z,a)=({a}_{4,2}{z}^{2}+{a}_{4,0}){\xi }^{4}+({a}_{2,4}{z}^{4}+{a}_{2,2}{z}^{2}){\xi }^{2}\\ \,+\,{a}_{0,6}{z}^{6}+{a}_{0,4}{z}^{4}+{a}_{6,0}+{F}_{1}(\xi ,z,a),\end{array}\end{eqnarray}$
and ${F}_{1}(\xi ,z,a),$ ${F}_{1}(\xi ,z,b)$ and ${F}_{1}(\xi ,z,c)$ are determined by: $\begin{eqnarray}{F}_{1}(\xi ,z,\varepsilon )={\varepsilon }_{2,0}{\xi }^{2}+{\varepsilon }_{0,2}{z}^{2}+{\varepsilon }_{0,0},\,(\varepsilon =a,b,c).\end{eqnarray}$
Substituting equation (33 ) into equation (27 ) and setting all the coefficients of the same powers of ${z}^{p}{\xi }^{q}(p,q=0,1,2,\cdot \cdot \cdot )$ to be zeros, we get a set of polynomial equations for ${a}_{6,0},$ ${a}_{4,2},$ ${a}_{4,0},$ ${a}_{2,4},$ ${a}_{2,2},$ ${a}_{2,0},$ ${a}_{0,6},$ ${a}_{0,4},$ ${a}_{0,2},$ ${a}_{0,0},$ ${b}_{2,0},$ ${b}_{0,2},$ ${b}_{0,0},$ ${c}_{2,0},$ ${c}_{0,2}$ and ${c}_{0,0}.$ Solving the polynomial equations, we gain
$\begin{eqnarray}\begin{array}{l}{a}_{6,0}=1,\,{a}_{4,2}=-3\omega ,\,{a}_{4,0}=-\displaystyle \frac{25\beta }{\omega },\,{a}_{2,4}=3{\omega }^{2},\\ {a}_{2,2}=90\beta ,\,{a}_{2,0}=-\displaystyle \frac{125{\beta }^{2}}{{\omega }^{2}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{0,6}=-{\omega }^{3},\,{a}_{0,4}=-17\beta \omega ,\,{a}_{0,2}=-\displaystyle \frac{475{\beta }^{2}}{\omega },\\ {a}_{0,0}=-\displaystyle \frac{16875{\beta }^{3}+{\omega }^{2}{\theta }^{2}+9{\omega }^{3}{\theta }^{2}}{9{\omega }^{3}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{2,0}=1,\,{b}_{0,2}=\displaystyle \frac{\omega }{3},\,{b}_{0,0}=-\displaystyle \frac{5\beta }{3\omega },\,{c}_{2,0}=1,\,{c}_{0,2}=3\omega ,\\ {c}_{0,0}=\displaystyle \frac{\beta }{\omega },\end{array}\end{eqnarray}$
and therefore obtain the second-order rogue wave solution of equation (1 ):33 )–(38 ).
$\begin{eqnarray}u=\displaystyle \frac{6\beta }{\alpha }{(\mathrm{ln}\,{\tilde{F}}_{2}(\xi ,z;\theta ,\vartheta ))}_{\xi \xi },\end{eqnarray}$
where ${\tilde{F}}_{2}(\xi ,z;\theta ,\vartheta )$ is determined by equations (In figures 3 and 4, we show the second-order rogue wave solution (39 ). It can be seen from figure 3 that when $\theta =\vartheta =0$ the second-order rogue wave has two interaction wave peaks. However, with the increase of $\theta $ and $\vartheta ,$ the computer simulation shows that these two interaction wave peaks gradually separate and give birth to the third wave peak. For the case of $\theta =\vartheta =200,$ figure 4 shows the second-order rogue wave with three separated wave peaks. In addition, we can see that the second-order rogue wave in figure 3 is symmetric with respect to the lines $\xi =0$ and $z=0,$ but the one in figure 4 is asymmetric.
Figure 3. Second-order rogue wave localized in ( |
Figure 4. Second-order rogue wave localized in ( |
4.3. Third-order rogue wave solution
For the third-order rogue wave solution, we consider equation (6 ) with $n=2:$
$\begin{eqnarray}\begin{array}{l}f={\tilde{F}}_{3}(\xi ,z;\theta ,\vartheta )={F}_{3}(\xi ,z,a)+2\theta z{F}_{2}(\xi ,z,b)\\ \,+\,2\vartheta \xi {F}_{2}(\xi ,z,c)+({\theta }^{2}+{\vartheta }^{2}){F}_{1}(\xi ,z,a),\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{l}{F}_{3}(\xi ,z,a)={a}_{12,0}{\xi }^{12}+({a}_{10,2}{z}^{2}+{a}_{10,0}){\xi }^{10}\\ \,+\,({a}_{8,4}{z}^{4}+{a}_{8,2}{z}^{2}+{a}_{8,0}){\xi }^{8}\\ \,+\,({a}_{6,6}{z}^{6}+{a}_{6,4}{z}^{4}+{a}_{6,2}{z}^{2}){\xi }^{6}\\ \,+\,({a}_{4,8}{z}^{8}+{a}_{4,6}{z}^{6}+{a}_{4,4}{z}^{4}){\xi }^{4}\\ \,+\,({a}_{2,10}{z}^{10}+{a}_{2,8}{z}^{8}+{a}_{2,6}{z}^{6}){\xi }^{2}+{a}_{0,12}{z}^{12}+{a}_{0,10}{z}^{10}\\ \,+{a}_{0,8}{z}^{8}+{F}_{2}(\xi ,z,a),\end{array}\end{eqnarray}$
and ${F}_{2}(\xi ,z,a),$ ${F}_{2}(\xi ,z,b)$ and ${F}_{2}(\xi ,z,c)$ are determined by: $\begin{eqnarray}\begin{array}{l}{F}_{2}(\xi ,z,\varepsilon )=({\varepsilon }_{4,2}{z}^{2}+{\varepsilon }_{4,0}){\xi }^{4}+({\varepsilon }_{2,4}{z}^{4}+{\varepsilon }_{2,2}{z}^{2}){\xi }^{2}\\ \,+{\varepsilon }_{0,6}{z}^{6}+{\varepsilon }_{0,4}{z}^{4}+{\varepsilon }_{6,0}+{F}_{1}(\xi ,z,\varepsilon ),\,(\varepsilon =a,b,c),\end{array}\end{eqnarray}$
respectively.Substituting equation (40 ) into equation (27 ) and setting all the coefficients of the same powers of ${z}^{p}{\xi }^{q}(p,q=0,1,2,\cdot \cdot \cdot )$to be zeros, we get a set of polynomial equations for the undetermined constants in equations (35 ), (41 ) and (42 ), from which we have
$\begin{eqnarray}\begin{array}{l}{a}_{12,0}=1,\,{a}_{10,2}=-6\omega ,\,{a}_{10,0}=-\displaystyle \frac{98\beta }{\omega },\,{a}_{8,4}=15{\omega }^{2},\\ {a}_{8,2}=690\beta ,\,{a}_{8,0}=\displaystyle \frac{735{\beta }^{2}}{{\omega }^{2}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{6,6}=-20{\omega }^{3},\,{a}_{6,4}=-1540\omega \beta ,\,{a}_{6,2}=-\displaystyle \frac{18620{\beta }^{2}}{\omega },\\ {a}_{6,0}=-\displaystyle \frac{75460{\beta }^{3}}{3{\omega }^{3}},\,{a}_{4,8}=15{\omega }^{4},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{4,6}=1460{\omega }^{2}\beta ,\,{a}_{4,4}=37450{\beta }^{2},\,{a}_{4,2}=\displaystyle \frac{220500{\beta }^{3}}{{\omega }^{2}},\\ {a}_{4,0}=-\displaystyle \frac{5187875{\beta }^{4}}{3{\omega }^{4}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{2,10}=-6{\omega }^{5},\,{a}_{2,8}=-570{\omega }^{3}\beta ,\,{a}_{2,6}=-35420\omega {\beta }^{2},\\ {a}_{2,4}=\displaystyle \frac{14700{\beta }^{3}}{\omega },\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{2,2}=-\displaystyle \frac{565950{\beta }^{4}}{{\omega }^{3}},\,\\ {a}_{2,0}=-\displaystyle \frac{3994663750{\beta }^{5}+3{\omega }^{4}{\theta }^{2}-75{\omega }^{5}{\vartheta }^{2}}{75{\omega }^{5}(1+{\theta }^{2}+{\vartheta }^{2})},\\ {a}_{0,12}={\omega }^{6},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{0,10}=58{\omega }^{4}\beta ,\,{a}_{0,8}=4335{\omega }^{2}{\beta }^{2},\,{a}_{0,6}=\displaystyle \frac{798980{\beta }^{3}}{3},\\ {a}_{0,4}=\displaystyle \frac{16391725{\beta }^{4}}{3{\omega }^{2}},\end{array}\end{eqnarray}$
$\begin{eqnarray}{a}_{0,2}=\displaystyle \frac{7522418750{\beta }^{5}+3{\omega }^{4}{\theta }^{2}-75{\omega }^{5}{\vartheta }^{2}}{75{\omega }^{4}(1+{\theta }^{2}+{\vartheta }^{2})},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{0,0}=\displaystyle \frac{21970650625{\beta }^{6}+27{\omega }^{4}\beta {\theta }^{2}-675{\omega }^{5}\beta {\vartheta }^{2}}{225{\omega }^{6}(1+{\theta }^{2}+{\vartheta }^{2})},\,\\ {b}_{6,0}=1,\,{b}_{4,2}=\omega ,\,{b}_{4,0}=-\displaystyle \frac{21\beta }{\omega },\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{2,4}=-\displaystyle \frac{9{\omega }^{2}}{5},\,{b}_{2,2}=-38\beta ,\,{b}_{2,0}=-\displaystyle \frac{133{\beta }^{2}}{{\omega }^{2}},\\ {b}_{0,6}=-\displaystyle \frac{{\omega }^{3}}{5},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{0,4}=\displaystyle \frac{7\omega \beta }{5},\,{b}_{0,2}=\displaystyle \frac{49{\beta }^{2}}{\omega },\,{b}_{0,0}=-\displaystyle \frac{3773{\beta }^{3}}{3{\omega }^{3}},\,{c}_{6,0}=1,\\ {c}_{4,2}=9\omega ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{c}_{4,0}=-\displaystyle \frac{13\beta }{\omega },\,{c}_{2,4}=-5{\omega }^{2},\,{c}_{2,2}=-230\beta ,\,\\ {c}_{2,0}=-\displaystyle \frac{245{\beta }^{2}}{{\omega }^{2}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{c}_{0,6}=-5{\omega }^{3},\,{c}_{0,4}=-45\omega \beta ,\,{c}_{0,2}=-\displaystyle \frac{535{\beta }^{2}}{\omega },\\ {c}_{0,0}=-\displaystyle \frac{12005{\beta }^{3}}{3{\omega }^{3}}\end{array}\end{eqnarray}$
and therefore obtain the third-order rogue wave solution of equation (1 ):40 )–(54 ).
$\begin{eqnarray}u=2{(\mathrm{ln}\,{\tilde{F}}_{3}(\xi ,z;\theta ,\vartheta ))}_{\xi \xi },\end{eqnarray}$
where ${\tilde{F}}_{3}(\xi ,z;\theta ,\vartheta )$ is determined by equations (In figure 5, we show the third-order rogue wave solution (55 ) with three interaction wave peaks, which is symmetric with respect to the lines $\xi =0$ and $z=0.$ The highest wave peak is located at the center $(0,0),$ and the other two entangled second highest wave peaks surround the center. Similar to the second-order rogue wave in figure 3, the three interaction wave peaks in figure 5 gradually separate and give birth to another three wave peaks as long as one or all of $\theta $ and $\vartheta $ is/are given sufficiently large value(s). When either $\theta $ or $\vartheta $ is equal to zero, we show the third-order rogue wave solution (55 ) that is symmetric with respect to the lines $\xi =0$ and $z=0$ in figures 6 and 7, respectively. That is to say that $\theta =0$ makes the third-order rogue wave symmetric with respect to the line $\xi =0,$ while $\vartheta =0$ makes the third-order rogue wave symmetric with respect to the line $z=0.$ However, for the non-zeros $\theta $ and $\vartheta ,$ the third-order rogue wave is asymmetric, see figure 8 for example.
Figure 5. Third-order rogue wave localized in ( |
Figure 6. Third-order rogue wave localized in ( |
Figure 7. Third-order rogue wave localized in ( |
Figure 8. Third-order rogue wave localized in ( |
4.4. Fourth-order rogue wave solutions
For the fourth-order rogue wave solution, we consider equation (6 ) with $n=3:$
$\begin{eqnarray}\begin{array}{l}f={\tilde{F}}_{4}(\xi ,z;\theta ,\vartheta )={F}_{4}(\xi ,z,a)+2\theta z{F}_{3}(\xi ,z,b)\\ \,+\,2\vartheta \xi {F}_{3}(\xi ,z,c)+({\theta }^{2}+{\vartheta }^{2}){F}_{2}(\xi ,z,a),\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{l}{F}_{4}(\xi ,z,a)={a}_{20,0}{\xi }^{20}+({a}_{18,2}{z}^{2}+{a}_{18,0}){\xi }^{18}\\ \,+\,({a}_{16,4}{z}^{4}+{a}_{16,2}{z}^{2}+{a}_{16,0}){\xi }^{16}\\ \,+\,({a}_{14,6}{z}^{6}+{a}_{14,4}{z}^{4}+{a}_{14,2}{z}^{2}+{a}_{14,0}){\xi }^{14}\\ \,+\,({a}_{12,8}{z}^{8}+{a}_{12,6}{z}^{6}+{a}_{12,4}{z}^{4}+{a}_{12,2}{z}^{2}+{a}_{12,0}){\xi }^{12}\\ \,+\,({a}_{10,10}{z}^{10}+{a}_{10,8}{z}^{8}+{a}_{10,6}{z}^{6}+{a}_{10,4}{z}^{4}){\xi }^{10}\\ \,+\,({a}_{8,12}{z}^{12}+{a}_{8,10}{z}^{10}+{a}_{8,8}{z}^{8}+{a}_{8,6}{z}^{6}){\xi }^{8}\\ \,+\,({a}_{6,14}{z}^{14}+{a}_{6,12}{z}^{12}+{a}_{6,10}{z}^{10}+{a}_{6,8}{z}^{8}){\xi }^{6}\\ \,+\,({a}_{4,16}{z}^{16}+{a}_{4,14}{z}^{14}+{a}_{4,12}{z}^{12}+{a}_{4,10}{z}^{10}){\xi }^{4}\\ \,+\,({a}_{2,18}{z}^{18}+{a}_{2,16}{z}^{16}+{a}_{2,14}{z}^{14}+{a}_{2,12}{z}^{12}){\xi }^{2}\\ \,+\,{a}_{0,20}{z}^{20}+{a}_{0,18}{z}^{18}+{a}_{0,16}{z}^{16}+{a}_{0,14}{z}^{14}+{F}_{3}(\xi ,z,a),\end{array}\end{eqnarray}$
and ${F}_{3}(\xi ,z,a),$ ${F}_{3}(\xi ,z,b)$ and ${F}_{3}(\xi ,z,c)$ are determined by: $\begin{eqnarray}\begin{array}{l}{F}_{3}(\xi ,z,\varepsilon )={\varepsilon }_{12,0}{\xi }^{12}+({\varepsilon }_{10,2}{z}^{2}+{\varepsilon }_{10,0}){\xi }^{10}\\ \,+\,({\varepsilon }_{8,4}{z}^{4}+{\varepsilon }_{8,2}{z}^{2}+{\varepsilon }_{8,0}){\xi }^{8}\\ \,+\,({\varepsilon }_{6,6}{z}^{6}+{\varepsilon }_{6,4}{z}^{4}+{\varepsilon }_{6,2}{z}^{2}){\xi }^{6}\\ \,+\,({\varepsilon }_{4,8}{z}^{8}+{\varepsilon }_{4,6}{z}^{6}+{\varepsilon }_{4,4}{z}^{4}){\xi }^{4}\\ \,+\,({\varepsilon }_{2,10}{z}^{10}+{\varepsilon }_{2,8}{z}^{8}+{\varepsilon }_{2,6}{z}^{6}){\xi }^{2}+{\varepsilon }_{0,12}{z}^{12}+{\varepsilon }_{0,10}{z}^{10}\\ \,+\,{\varepsilon }_{0,8}{z}^{8}+{F}_{2}(\xi ,z,\varepsilon ),\,(\varepsilon =a,b,c),\end{array}\end{eqnarray}$
respectively.Substituting equation (56 ) into equation (27 ) and setting all the coefficients of the same powers of ${z}^{p}{\xi }^{q}(p,q=0,1,2,\cdot \cdot \cdot )$ to be zeros, we get a set of polynomial equations, and from which the undetermined constants can be determined as follows:1 ):56 )–(96 ).
$\begin{eqnarray}\begin{array}{l}{a}_{20,0}=1,\,{a}_{18,2}=-10\omega ,\,{a}_{18,0}=-\displaystyle \frac{270}{\omega },\,{a}_{16,4}=45{\omega }^{2},\\ {a}_{16,2}=3030\beta ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{16,0}=\displaystyle \frac{16605{\beta }^{2}}{{\omega }^{2}},\,{a}_{14,6}=-120{\omega }^{3},\,{a}_{14,4}=-13800\omega \beta ,\\ {a}_{14,2}=-\displaystyle \frac{275400{\beta }^{2}}{\omega },\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{14,0}=-\displaystyle \frac{351000{\beta }^{3}}{{\omega }^{3}},\,{a}_{12,8}=210{\omega }^{4},\,{a}_{12,6}=34440{\omega }^{2}\beta ,\\ {a}_{12,4}=1367100{\beta }^{2},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{12,2}=\displaystyle \frac{10621800{\beta }^{3}}{{\omega }^{2}},\,{a}_{12,0}=-\displaystyle \frac{18877950{\beta }^{4}}{{\omega }^{4}},\\ {a}_{10,10}=-252{\omega }^{5},{a}_{10,8}=-52500{\omega }^{3}\beta ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{10,6}=-3308760\omega {\beta }^{2},\,{a}_{10,4}=-\displaystyle \frac{56586600{\beta }^{3}}{\omega },\\ {a}_{10,2}=-\displaystyle \frac{107534700{\beta }^{4}}{{\omega }^{3}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{10,0}=-\displaystyle \frac{2094264900{\beta }^{5}}{{\omega }^{5}},{a}_{8,12}=210{\omega }^{6},\\ {a}_{8,10}=50820{\omega }^{4}\beta ,\,{a}_{8,8}=4513950{\omega }^{2}{\beta }^{2},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{8,6}=135286200{\beta }^{3},\,{a}_{8,4}=\displaystyle \frac{1071960750{\beta }^{4}}{{\omega }^{2}},\\ {a}_{8,2}=\displaystyle \frac{4871002500{\beta }^{6}}{{\omega }^{4}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{8,0}=-\displaystyle \frac{178095030750}{{\omega }^{6}},{a}_{6,14}=-120{\omega }^{7},\\ {a}_{6,12}=-31080{\omega }^{5}\beta ,{a}_{6,10}=-3601080{\omega }^{3},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{6,8}=-151237800\omega {\beta }^{3},{a}_{6,6}=-\displaystyle \frac{3824793000{\beta }^{4}}{\omega },\\ {a}_{6,4}=-\displaystyle \frac{636363000{\beta }^{5}}{{\omega }^{3}}\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{6,2}=\displaystyle \frac{521628471000{\beta }^{6}}{{\omega }^{5}},\,\\ {a}_{6,0}=\displaystyle \frac{341392530943000{\beta }^{7}+{\theta }^{2}{\omega }^{6}-49{\vartheta }^{2}{\omega }^{7}}{49{\omega }^{7}(1+{\theta }^{2}+{\vartheta }^{2})},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{4,16}=45{\omega }^{8},\,{a}_{4,14}=11400{\omega }^{6}\beta ,\,{a}_{4,12}=1619100{\omega }^{4}{\beta }^{2},\\ {a}_{4,10}=94613400{\omega }^{2}{\beta }^{3},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{4,8}=2667498750{\beta }^{4},\,{a}_{4,6}=\displaystyle \frac{45237339000{\beta }^{5}}{{\omega }^{2}},\\ {a}_{4,4}=-\displaystyle \frac{405853402500{\beta }^{6}}{{\omega }^{4}},\end{array}\end{eqnarray}$
$\begin{eqnarray}{a}_{4,2}=\displaystyle \frac{1631039216625000{\beta }^{7}+3{\omega }^{6}{\theta }^{2}-147{\omega }^{7}{\vartheta }^{2}}{49{\omega }^{6}(1+{\theta }^{2}+{\vartheta }^{2})},\end{eqnarray}$
$\begin{eqnarray}{a}_{4,0}=\displaystyle \frac{9338356860958125{\beta }^{8}+25\beta {\omega }^{6}{\theta }^{2}-1225\beta {\omega }^{7}{\vartheta }^{2}}{49{\omega }^{8}(1+{\theta }^{2}+{\vartheta }^{2})},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{2,18}=-10{\omega }^{9},\,{a}_{2,16}=-2190{\omega }^{7}\beta ,\\ {a}_{2,14}=-354600{\omega }^{5}{\beta }^{2},\\ {a}_{2,12}=-35645400{\omega }^{3}{\beta }^{3},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{2,10}=-671510700\omega ,\,{a}_{2,8}=-\displaystyle \frac{31477666500{\beta }^{5}}{\omega },\\ {a}_{2,6}=-\displaystyle \frac{1982064357000{\beta }^{6}}{{\omega }^{3}},\end{array}\end{eqnarray}$
$\begin{eqnarray}{a}_{2,4}=-\displaystyle \frac{4454010668835000{\beta }^{7}+3{\omega }^{6}{\theta }^{2}-147{\omega }^{7}{\vartheta }^{2}}{49{\omega }^{5}(1+{\theta }^{2}+{\vartheta }^{2})},\end{eqnarray}$
$\begin{eqnarray}{a}_{2,2}=-\displaystyle \frac{42646827589616250{\beta }^{8}+90\beta {\omega }^{6}{\theta }^{2}-4410\beta {\omega }^{7}{\theta }^{2}}{49{\omega }^{7}(1+{\theta }^{2}+{\vartheta }^{2})},\end{eqnarray}$
$\begin{eqnarray}{a}_{2,0}=\displaystyle \frac{34112014318541250{\beta }^{9}+125{\beta }^{2}{\omega }^{6}{\theta }^{2}-6125{\beta }^{2}{\omega }^{7}{\theta }^{2}}{49{\omega }^{9}(1+{\theta }^{2}+{\vartheta }^{2})},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{0,20}={\omega }^{10},\,{a}_{0,18}=150{\omega }^{8}\beta ,\,{a}_{0,16}=23085{\omega }^{6}{\beta }^{2},\\ {a}_{0,14}=3299400{\omega }^{4}{\beta }^{3},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{0,12}=360709650{\omega }^{2}{\beta }^{4},\,{a}_{0,10}=21813668100{\beta }^{5},\\ {a}_{0,8}=\displaystyle \frac{1200881855250{\beta }^{6}}{{\omega }^{2}},\end{array}\end{eqnarray}$
$\begin{eqnarray}{a}_{0,6}=\displaystyle \frac{2116777295997000{\beta }^{7}+{\omega }^{6}{\theta }^{2}-49{\omega }^{7}{\vartheta }^{2}}{49{\omega }^{4}(1+{\theta }^{2}+{\vartheta }^{2})},\end{eqnarray}$
$\begin{eqnarray}{a}_{0,4}=\displaystyle \frac{17085505551148125{\beta }^{8}+17\beta {\omega }^{6}{\theta }^{2}-833\beta {\omega }^{7}{\vartheta }^{2}}{49{\omega }^{6}(1+{\theta }^{2}+{\vartheta }^{2})},\end{eqnarray}$
$\begin{eqnarray}{a}_{0,2}=\displaystyle \frac{170251365580773750{\beta }^{9}-19\beta {\omega }^{6}{\theta }^{2}+\left.931\beta {\omega }^{7}{\vartheta }^{2}\right)}{49{\omega }^{8}(1+{\theta }^{2}+{\vartheta }^{2})},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{a}_{0,0}\,=\\ \displaystyle \frac{289952121707600625{\beta }^{10}+1875{\beta }^{3}{\omega }^{6}{\theta }^{2}-91875{\beta }^{3}{\omega }^{7}{\vartheta }^{2}}{49{\omega }^{10}(1+{\theta }^{2}+{\vartheta }^{2})},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{12,0}=1,\,{b}_{10,2}=2\omega ,\,{b}_{10,0}=-\displaystyle \frac{90\beta }{\omega },\,{b}_{8,4}=-9{\omega }^{2},\,\\ {b}_{8,2}=-270\beta ,\,{b}_{8,0}=-\displaystyle \frac{375{\beta }^{2}}{{\omega }^{2}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{6,6}=\displaystyle \frac{36}{7}{\omega }^{3},\,{b}_{6,4}=876\omega \beta ,\,{b}_{6,2}=\displaystyle \frac{6900{\beta }^{2}}{\omega },\\ {b}_{6,0}=-\displaystyle \frac{20700{\beta }^{3}}{{\omega }^{3}},\,{b}_{4,8}=\displaystyle \frac{25}{7}{\omega }^{4},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{4,6}=\displaystyle \frac{780}{7}{\omega }^{2}\beta ,\,{b}_{4,4}=-16950{\beta }^{2},\,{b}_{4,2}=-\displaystyle \frac{117900{\beta }^{3}}{{\omega }^{2}},\\ {b}_{4,0}=-\displaystyle \frac{1371825{\beta }^{4}}{{\omega }^{4}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{2,10}=-\displaystyle \frac{18}{7}{\omega }^{5},\,{b}_{2,8}=-\displaystyle \frac{1310}{7}{\omega }^{3}\beta ,\,{b}_{2,6}=-\displaystyle \frac{78420}{7}\omega {\beta }^{2},\\ {b}_{2,4}=-\displaystyle \frac{188100{\beta }^{3}}{\omega },\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{2,2}=-\displaystyle \frac{2359350{\beta }^{4}}{{\omega }^{3}},\,{b}_{2,0}=-\displaystyle \frac{41985450{\beta }^{5}}{{\omega }^{5}},\\ {b}_{0,12}=-\displaystyle \frac{{\omega }^{4}}{7},{b}_{0,10}=-\displaystyle \frac{10}{7}{\omega }^{4}\beta ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{0,8}=-\displaystyle \frac{225}{7}{\omega }^{2}{\beta }^{2},\,{b}_{0,6}=-\displaystyle \frac{17100}{7}{\beta }^{3},\\ {b}_{0,4}=-\displaystyle \frac{1582425{\beta }^{4}}{{\omega }^{2}},{b}_{0,2}=-\displaystyle \frac{23955750{\beta }^{5}}{{\omega }^{4}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{b}_{0,2}=\displaystyle \frac{42017625{\beta }^{6}}{{\omega }^{6}},\,{b}_{0,0}=\displaystyle \frac{42017625{\beta }^{6}}{{\omega }^{6}},\,{c}_{12,0}=1,\\ {c}_{10,2}=18\omega ,\,{c}_{10,0}=-\displaystyle \frac{74\beta }{\omega },\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{c}_{8,4}=-25{\omega }^{2},\,{c}_{8,2}=-1870\beta ,\,{c}_{8,0}=-\displaystyle \frac{825{\beta }^{2}}{{\omega }^{2}},\\ {c}_{6,6}=-36{\omega }^{3},\,{c}_{6,4}=-1740\omega \beta ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{c}_{6,2}=\displaystyle \frac{26580{\beta }^{2}}{\omega },\,{c}_{6,0}=-\displaystyle \frac{42300{\beta }^{3}}{{\omega }^{3}},\,{c}_{4,8}=63{\omega }^{4},\,\\ {c}_{4,6}=5460{\omega }^{2}\beta ,\,{c}_{4,4}=-7350{\beta }^{2},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{c}_{4,2}=-\displaystyle \frac{{333900}^{3}}{{\omega }^{2}},\,{c}_{4,0}=-\displaystyle \frac{2228625{\beta }^{4}}{{\omega }^{4}},\\ {c}_{2,10}=-14{\omega }^{5},\,{c}_{2,8}=-1890{\omega }^{3}\beta ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{c}_{2,6}=-149100\omega {\beta }^{2},\,{c}_{2,4}=-\displaystyle \frac{1304100{\beta }^{3}}{\omega },\\ {c}_{2,2}=-\displaystyle \frac{16899750{\beta }^{4}}{{\omega }^{3}},\,{c}_{2,0}=-\displaystyle \frac{78255450{\beta }^{5}}{{\omega }^{5}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{c}_{0,12}=-7{\omega }^{6},\,{c}_{0,10}=-294{\omega }^{4}\beta ,\,{c}_{0,8}=-15225{\omega }^{2}{\beta }^{2},\,\\ {c}_{0,6}=-623700{\beta }^{3},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{c}_{0,4}=-\displaystyle \frac{18974025{\beta }^{4}}{{\omega }^{2}},\,{c}_{0,2}=-\displaystyle \frac{395101350{\beta }^{5}}{{\omega }^{4}},\\ {c}_{0,0}=-\displaystyle \frac{44983575{\beta }^{6}}{{\omega }^{6}}.\end{array}\end{eqnarray}$
We, therefore, obtain the fourth-order rogue wave solution of equation ( $\begin{eqnarray}u=2{(\mathrm{ln}\,{\tilde{F}}_{4}(\xi ,z;\theta ,\vartheta ))}_{\xi \xi },\end{eqnarray}$
where ${\tilde{F}}_{4}(\xi ,z;\theta ,\vartheta )$ is determined by equations (Similarly, we show in figure 9 the fourth-order rogue wave solution (97 ) with four interaction wave peaks. The highest two peaks are located at the center $(0,0),$ and the other entangled second highest wave peaks surround the center. Similar to the second- and third-order rogue waves in figures 3 and 5, the four interaction wave peaks in figure 9 gradually separate and give birth to another five wave peaks as long as one or all of $\theta $ and $\vartheta $ is/are given sufficiently large value(s). For example, see figures 10 and 11 for two symmetric fourth-order rogue waves, the one corresponding to $\theta =0$ and $\vartheta =6\times {10}^{7}$ is symmetric with respect to the line $z=0,$ and the other given $\theta =6\times {10}^{7}$ and $\vartheta =0$ is symmetric with respect to the line $\xi =0.$ However, the parameters $\theta =6\times {10}^{7}$ and $\vartheta =6\times {10}^{7}$ lead to an asymmetric fourth-order rogue wave, see figure 12.
Figure 9. Fourth-order rogue wave localized in ( |
Figure 10. Fourth-order rogue wave localized in ( |
Figure 11. Fourth-order rogue wave localized in ( |
Figure 12. Fourth-order rogue wave localized in ( |
5. Conclusion and discussion
In summary, we have constructed first-, second-, third- and fourth-order rogue wave solutions (30 ), (39 ), (55 ) and (97 ) of the (3 + 1)-dimensional gB equation (1 ) through Zhaqilao's symbolic calculation approach [22]. To the best of our knowledge, the high-order rogue wave solutions (39 ), (55 ) and (97 ) are new. This paper shows that the two parameters $\theta $ and $\vartheta $ included in solutions (30 ), (39 ), (55 ) and (97 ) control the following aspects:
| (1) The center $(\theta ,\vartheta )$ of the first-order rogue wave solution ( | |
| (2) The fission giving birth to more wave peaks of the second-, third- and fourth-order rogue wave solutions ( | |
| (3) The symmetry of high-order rogue wave solutions ( |
In view of the controllability of parameters $\theta $ and $\vartheta $ on the center, fission and asymmetry of first-, second-, third- and fourth-order rogue wave solutions (30 ), (39 ), (55 ) and (97 ), we can consider their potential applications in physics, such as the design of adjustable resistors and controllable switching devices under specific environment or with special requirements. Since the gB equation (1 ) or its reduced equation (23 ) is a kind of fluid equation, if the fluid is assumed to be a conductive liquid, we can not only use the vibration of the wave peak to trigger the connected power supply, but also adjust the size of the resistance by adjusting the number of fission wave peaks, so as to achieve the purpose of designing adjustable resistance and controllable switching device. If feasible, we can also consider using the controllability of wave crest position and asymmetry of rogue wave solutions (30 ), (39 ), (55 ) and (97 ) to design adjustable resistors and controllable switching devices that meet other conditions including the horizontal position and vertical height design of the device electric shock point. Nonetheless, we are still looking forward to other physical applications of these obtained rogue wave solutions with controllable center, fission and asymmetry.
For the method [22], it is a novel and meaningful discovery that the fission and asymmetry of higher-order rouge waves can be controlled by the embedded parameters $\theta $ and $\vartheta .$ High asymmetry is an important observation point in the study of rogue waves. Although the rogue wave solutions obtained in this paper have asymmetric controllability, there is still a certain distance from the highly asymmetric rogue waves, which raises a problem worthy of study for our next work. Nonlinear problems are always full of infinite charm and challenges. Some meaningful developments of nonlinear models deserve attention, such as the H2 optimal controller of coupled stochastic algebraic Riccati equations [32], and the influence of higher-order nonlinear effects on optical solitons [33].