西南师范大学学报(自然科学版)   2019, Vol. 44 Issue (10): 16-22.  DOI: 10.13718/j.cnki.xsxb.2019.10.005
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  • 积分微分KP层次方程的精确行波解    [PDF全文]
    宋佳谦 , 刘小华     
    贵州民族大学 数据科学与信息工程学院, 贵阳 550025
    摘要:对KP层次方程进行积分变换和行波变换得到常微分方程,利用扩展试验方程法把求解常微分方程的问题转化为求解代数方程组的问题,根据不同情况得到了KP层次方程的钟状解、三角函数解、双曲函数解和椭圆函数解的精确表达式,这些解的显示表达式是首次求出的.这种方法对于求解非线性偏微分方程十分有效并且能够得到许多新的精确解.
    关键词扩展试验方程法    KP层次方程    行波解    

    非线性偏微分方程在数学、物理、化学以及其它学科和工程领域起着非常重要的作用.很多学者研究了非线性偏微分方程,因此许多求解非线性偏微分方程的方法也发展起来.比如Jacobi椭圆函数展开法[1]、辅助函数法[2]、指数函数法[3]、sine-cosine法[4]G′/G展开法[5-6]、Bäcklund变换法[7]等.扩展试验方程法是一种常用于求解非线性偏微分方程精确解的方法,文献[8]利用扩展试验方程法得到了广义Benjamin方程和Burger-Kdv方程的精确行波解,文献[9]通过扩展试验方程法考虑了非线性耦合Schrodinger Boussinesq偏微分方程的精确行波解.

    本文主要探讨积分微分KP层次方程

    $ {u_t} = \frac{1}{2}{u_{xxy}} + \frac{1}{2}\partial _x^{ - 2}\left( {{u_{yyy}}} \right) + 2{u_x}\partial _x^{ - 1}\left( {{u_y}} \right) + 4u{u_y} $ (1)

    的精确行波解,其中

    $ u = u(x,y,t)\;\;\;\;\partial _x^{ - 1} = \int_{ - \infty }^x {\rm{d}} x $

    KP层次方程在数学物理和工程中有许多重要的应用,是一类非常重要的非线性偏微分方程.文献[10]利用重复齐次平衡法得到了(3+1)维KP方程的孤子解和周期解,文献[11]通过G′/G-展开法得到了(2+1)维KP方程的行波解.文献[12]通过扩展tanh法得到了耦合KP方程的行波解.我们将对方程(1)进行积分变换,之后再对其进行行波变换,然后利用扩展试验方程法对方程(1)的精确行波解进行研究.

    1 方程(1)的行波变换

    在方程(1)中做变换u=vxx,可将积分微分KP层次方程(1)转换成如下的方程

    $ {v_{xxt}} = \frac{1}{2}{v_{xxxxy}} + \frac{1}{2}{v_{yyy}} + 2{v_{xxx}}{v_{xy}} + 4{v_{xx}}{v_{xxy}} $ (2)

    对方程(2)进行行波变换,再对其进行一次积分,令v(xyt)=v(ξ),其中ξ=x+yktk是波速,方程(2)可化为

    $ \left( {k + \frac{1}{2}} \right){v_{\xi \xi }} + \frac{1}{2}{v^{(4)}} + 3{\left( {{v_{\xi \xi }}} \right)^2} + C = 0 $ (3)

    其中$v_{\xi\xi}=\frac{\partial^{2} v}{\partial \xi^{2}}, v^{(4)}=\frac{\partial^{4} v}{\partial \xi^{4}}$C是积分常数.令w=vξξ,方程(3)可化为如下等价的方程

    $ \left( {k + \frac{1}{2}} \right)w + \frac{1}{2}{w_{\xi \xi }} + 3{w^2} + C = 0 $ (4)

    综上推导可知,方程(4)的解为w=vξξ,而方程(1)的解为u=vxx,显然方程(1)的解与方程(4)的解具有相同形式.为此下面只需对方程(4)进行研究.

    2 方程(1)的精确行波解

    上节给出了方程(1)的积分变换和行波变换,本节将利用扩展试验方程法对方程(1)的精确行波解进行研究.首先给出扩展试验方程法的步骤.

    2.1 扩展试验方程法

    考虑如下的常微分方程

    $ P\left( {{w^{\left( m \right)}},{w^{\left( {m - 1} \right)}}, \cdots ,w',w} \right) = 0 $ (5)

    步骤1  假设方程(5)有如下形式的解

    $ w = \sum\limits_{i = 0}^\delta {{\tau _i}} {Y^i} $ (6)

    其中τi(i=0,1,…,δ)为待定系数,Y满足如下方程

    $ {\left( {Y'} \right)^2} = \mathit{\Lambda }\left( Y \right) = \frac{{\mathit{\Phi }\left( Y \right)}}{{\mathit{\Psi }\left( Y \right)}} = \frac{{{\lambda _\theta }{Y^\theta } + {\lambda _{\theta - 1}}{Y^{\theta - 1}} + \cdots + {\lambda _1}Y + {\lambda _0}}}{{{\zeta _\varepsilon }{Y^\varepsilon } + {\zeta _{\varepsilon - 1}}{Y^{\varepsilon - 1}} + \cdots + {\zeta _1}Y + {\zeta _0}}} $ (7)

    τiλiζi是待常数,可以得到

    $ w' = \frac{{\mathit{\Phi '}\left( Y \right)\mathit{\Psi }\left( Y \right) - \mathit{\Phi }\left( Y \right)\mathit{\Psi '}\left( Y \right)}}{{2{\mathit{\Psi }^2}\left( Y \right)}}\left( {\sum\limits_{i = 0}^\delta {i{\tau _i}{Y^{i - 1}}} } \right) + \frac{{\mathit{\Phi }\left( Y \right)}}{{\mathit{\Psi }\left( Y \right)}}\left( {\sum\limits_{i = 0}^\delta {i(i - 1){\tau _i}{Y^{i - 2}}} } \right) $ (8)

    Φ(Y),Ψ(Y)是关于Y的多项式.

    步骤2  平衡最高阶导数项和最高阶非线性项,可得到δζλ的关系.

    步骤3  把方程(6),(7)代入方程(5)得到关于Y的多项式,令Yi的各项系数为0,得到方程组,解方程组得到τiλiζi.

    步骤4  将方程(7)转化为积分形式

    $ \pm \xi = \int {\frac{{{\rm{d}}Y}}{{\sqrt {\mathit{\Lambda }\left( Y \right)} }}} = \int {\sqrt {\frac{{\mathit{\Psi }\left( Y \right)}}{{\mathit{\Phi }\left( Y \right)}}} {\rm{d}}Y} $ (9)

    根据(9)式可以得到方程(5)的行波解.

    2.2 方程(1)的精确行波解

    根据方程(1)与方程(4)解之间的关系,先用扩展试验方程法对方程(4)的精确行波解进行研究.现在应用扩展试验方程法求解方程(4),通过(6),(7),(8)式可以得到

    $ w = {\tau _\delta }{Y^\delta } + {\tau _{\delta - 1}}{Y^{\delta - 1}} + \cdots + {\tau _0} $ (10)
    $ {w^2} = \tau _\delta ^2{Y^{2\delta }} + \cdots + 2{\tau _\delta }{\tau _{\delta - 1}}{Y^{2\delta - 1}} + \cdots + \tau _0^2 $ (11)
    $ w' = \delta {\tau _\delta }{Y^{\delta - 1}}Y' + (\delta - 1){\tau _{\delta - 1}}{Y^{\delta - 2}}Y' + \cdots + {\tau _1}Y' $ (12)
    $ {\left( {w'} \right)^2} = {\left( {\delta {\tau _\delta }{Y^{\delta - 1}} + \cdots + {\tau _1}} \right)^2}\frac{{{\lambda _\theta }{Y^\theta } + {\lambda _{\theta - 1}}{Y^{\theta - 1}} + \cdots + {\lambda _0}}}{{{\zeta _\varepsilon }{Y^\varepsilon } + {\zeta _{\varepsilon - 1}}{Y^{\varepsilon - 1}} + \cdots + {\zeta _0}}} $ (13)
    $ w'' = \frac{{{{\left( {\delta {\tau _\delta }{Y^{\delta - 1}} + \cdots + {\tau _1}} \right)}^2}\frac{{{\lambda _\theta }{Y^\theta } + {\lambda _{\theta - 1}}{Y^{\theta - 1}} + \cdots + {\lambda _0}}}{{{\zeta _\varepsilon }{Y^\varepsilon } + {\zeta _{\varepsilon - 1}}{Y^{\varepsilon - 1}} + \cdots + {\zeta _0}}}}}{{2\delta {\tau _\delta }{Y^{\delta - 1}}Y' + 2(\delta - 1){\tau _{\delta - 1}}{Y^{\delta - 2}}Y' + \cdots + {\tau _1}Y'}} $ (14)

    利用齐次平衡法,可以得到δ-2+θε=2δ,即δ=θε-2.

    假设ε=0,θ=3,则δ=1,于是有

    $ w = {\tau _0} + {\tau _1}Y $ (15)
    $ w'' = \frac{{{\tau _1}}}{{2{\zeta _0}}}\left( {3{\lambda _3}{Y^2} + 2{\lambda _2}Y + {\lambda _1}} \right) $ (16)

    把(15),(16)式代入方程(4),合并同类项,令Ym(m=0,1,2)的系数等于0,得到代数方程组为

    $ \left\{ {\begin{array}{*{20}{l}} {4{\zeta _0}\tau _1^2 + {\lambda _3}{\tau _1} = 0}\\ {{\zeta _0}{\tau _1} + 2{\zeta _0}{\tau _1}k + 12{\zeta _0}{\tau _0}{\tau _1} + {\lambda _2}{\tau _1} = 0}\\ {4{\zeta _0}C + 2{\zeta _0}{\tau _0} + 4{\zeta _0}{\tau _0}k + 12{\zeta _0}\tau _0^2 + {\lambda _1}{\tau _1} = 0} \end{array}} \right. $ (17)

    解得

    $ \left\{ \begin{array}{l} {\tau _1} = - \frac{{{\lambda _3}}}{{4{\zeta _0}}}\\ {\tau _0} = \frac{{{\lambda _2} \pm \sqrt {48\zeta _0^2C - 3{\lambda _1}{\lambda _3} + \lambda _2^2} }}{{ - 12{\zeta _0}}}\\ k = \frac{{ - {\zeta _0} \pm \sqrt {48\zeta _0^2C - 3{\lambda _1}{\lambda _3} + \lambda _2^2} }}{{2{\zeta _0}}} \end{array} \right. $ (18)

    根据方程(9)可知

    $ \pm \xi = \int {\frac{{{\rm{d}}Y}}{{\sqrt {\mathit{\Lambda }\left( Y \right)} }}} = \int {\sqrt {\frac{{\mathit{\Psi }\left( Y \right)}}{{\mathit{\Phi }\left( Y \right)}}} {\rm{d}}Y} = \sqrt {\frac{{{\lambda _3}}}{{{\zeta _0}}}} \int {\frac{{{\rm{d}}Y}}{{\sqrt {{Y^3} + \frac{{{\lambda _2}}}{{{\lambda _3}}}{Y^2} + \frac{{{\lambda _1}}}{{{\lambda _3}}}Y + \frac{{{\lambda _0}}}{{{\lambda _3}}}} }}} $ (19)

    其中λ3ζ0>0,为了对(19)式进行积分,下面将讨论以下几种情况:

    情况1  若

    $ {Y^3} + \frac{{{\lambda _2}}}{{{\lambda _3}}}{Y^2} + \frac{{{\lambda _1}}}{{{\lambda _3}}}Y + \frac{{{\lambda _0}}}{{{\lambda _3}}} = {\left( {Y - \alpha } \right)^3} $ (20)

    α是非零常数,等式两边系数相等,得到

    $ \left\{ {\begin{array}{*{20}{l}} {{\lambda _2} = - 3\alpha {\lambda _3}}\\ {{\lambda _1} = 3{\alpha ^2}{\lambda _3}}\\ {{\lambda _0} = - {\alpha ^3}{\lambda _3}} \end{array}} \right. $ (21)

    把(21)式代入(18)式,得到

    $ \left\{ {\begin{array}{*{20}{l}} {{\tau _1} = - \frac{{{\lambda _3}}}{{4{\zeta _0}}}}\\ {{\tau _0} = \frac{{ - 3\alpha {\lambda _3} \pm 4{\zeta _0}\sqrt {3C} }}{{ - 12{\zeta _0}}}}\\ {k = - \frac{1}{2} \pm 2\sqrt {3C} } \end{array}} \right. $ (22)

    λ3ζ0αC为任意常数

    $ \pm \xi = \sqrt {\frac{{{\lambda _3}}}{{{\zeta _0}}}} \int {\frac{{{\rm{d}}Y}}{{\sqrt {{{(Y - \alpha )}^3}} }}} = - \sqrt {\frac{{{\lambda _3}}}{{{\zeta _0}}}} \frac{2}{{\sqrt {Y - \alpha } }} $ (23)

    得到

    $ Y = \alpha + \frac{{4{\lambda _3}}}{{{\xi ^2}{\zeta _0}}} $ (24)

    把(22),(24)式代入方程(15),得到方程(1)的钟状解u1(图 1),

    图 1 方程(1)的解u1
    $ {u_1} = \mp \frac{{\sqrt {3C} }}{3} + \frac{{\lambda _3^2}}{{{\xi ^2}\zeta _0^2}} $ (25)

    其中

    $ \begin{array}{*{20}{c}} {k = - \frac{1}{2} \pm 2\sqrt {3C} }&{\xi = x + y - kt}&{{\lambda _3}{\zeta _0} > 0} \end{array} $

    情况2  若

    $ {Y^3} + \frac{{{\lambda _2}}}{{{\lambda _3}}}{Y^2} + \frac{{{\lambda _1}}}{{{\lambda _3}}}Y + \frac{{{\lambda _0}}}{{{\lambda _3}}} = {\left( {Y - {\alpha _1}} \right)^2}\left( {Y - {\alpha _2}} \right) $ (26)

    其中α1α2是非零常数,等式两边系数相等,得到

    $ \left\{ {\begin{array}{*{20}{l}} {{\lambda _2} = - 3{\lambda _3}\left( {2{\alpha _1} + {\alpha _2}} \right)}\\ {{\lambda _1} = {\lambda _3}\left( {2{\alpha _1}{\alpha _2} + \alpha _1^2} \right)}\\ {{\lambda _0} = - {\lambda _3}\alpha _1^2{\alpha _2}} \end{array}} \right. $ (27)

    把(27)式代入(18)式,得到

    $ \left\{ {\begin{array}{*{20}{l}} {{\tau _1} = - \frac{{{\lambda _3}}}{{4{\zeta _0}}}}\\ {{\tau _0} = \frac{{{\lambda _3}\left( {2{\alpha _1} + {\alpha _2}} \right) \pm \sqrt {48\zeta _0^2C + \lambda _3^2{{\left( {{\alpha _1} - {\alpha _2}} \right)}^2}} }}{{12{\sigma _0}}}}\\ {k = \frac{{ - {\zeta _0} \pm \sqrt {48\zeta _0^2C + \lambda _3^2{{\left( {{\alpha _1} - {\alpha _2}} \right)}^2}} }}{{2{\zeta _0}}}} \end{array}} \right. $ (28)

    λ3ζ0Cα1α2为任意常数.

    α2>α1时,

    $ \pm \xi = \sqrt {\frac{{{\lambda _3}}}{{{\zeta _0}}}} \int {\frac{{{\rm{d}}Y}}{{\left( {Y - {\alpha _1}} \right)\sqrt {Y - {\alpha _2}} }}} = 2\sqrt {\frac{{{\lambda _3}}}{{{\zeta _0}\left( {{\alpha _2} - {\alpha _1}} \right)}}} \arctan \left( {\frac{{\sqrt {Y - {\alpha _2}} }}{{\sqrt {{\alpha _2} - {\alpha _1}} }}} \right) $ (29)

    得到

    $ {Y_1} = {\alpha _2} + \left( {{\alpha _2} - {\alpha _1}} \right){\tan ^2}\left( {\frac{\xi }{2}\sqrt {\frac{{{\zeta _0}\left( {{\alpha _2} - {\alpha _1}} \right)}}{{{\lambda _3}}}} } \right) $ (30)

    把(28),(30)式代入方程(15),得到方程(1)的三角函数解u2(图 2),

    图 2 方程(1)的解u2
    $ {u_2} = \frac{{{\lambda _3}\left( {2{\alpha _1} + {\alpha _2}} \right) \pm \sqrt {48\zeta _0^2C + \lambda _3^2{{\left( {{\alpha _1} - {\alpha _2}} \right)}^2}} }}{{12{\zeta _0}}} - \frac{{{\lambda _3}}}{{4{\zeta _0}}}\left( {{\alpha _2} + \left( {{\alpha _2} - {\alpha _1}} \right){{\tan }^2}\left( {\frac{\xi }{2}\sqrt {\frac{{{\zeta _0}\left( {{\alpha _2} - {\alpha _1}} \right)}}{{{\lambda _3}}}} } \right)} \right) $ (31)

    其中

    $ \begin{array}{*{20}{c}} {k = \frac{{ - {\zeta _0} \pm \sqrt {48\zeta _0^2C + \lambda _3^2{{\left( {{\alpha _1} - {\alpha _2}} \right)}^2}} }}{{2{\zeta _0}}}}\\ {\begin{array}{*{20}{c}} {{\alpha _2} > {\alpha _1}}&{{\lambda _3}{\zeta _0} > 0}&{\xi = x + y - kt} \end{array}} \end{array} $

    同样地,当α1>α2时,

    $ {Y_2} = {\alpha _1} + \left( {{\alpha _1} - {\alpha _2}} \right)\csc {h^2}\left( {\frac{\xi }{2}\sqrt {\frac{{{\zeta _0}\left( {{\alpha _1} - {\alpha _2}} \right)}}{{{\lambda _3}}}} } \right) $ (32)

    把(28),(32)式代入方程(15),得到方程(1)的双曲函数解u3

    $ {u_3} = \frac{{{\lambda _3}\left( {2{\alpha _1} + {\alpha _2}} \right) \pm \sqrt {48\zeta _0^2C + \lambda _3^2{{\left( {{\alpha _1} - {\alpha _2}} \right)}^2}} }}{{12{\zeta _0}}} - \frac{{{\lambda _3}}}{{4{\zeta _0}}}\left( {{\alpha _1} + \left( {{\alpha _1} - {\alpha _2}} \right)\csc {h^2}\left( {\frac{\xi }{2}\sqrt {\frac{{{\zeta _0}\left( {{\alpha _1} - {\alpha _2}} \right)}}{{{\lambda _3}}}} } \right)} \right) $ (33)

    其中

    $ \begin{array}{*{20}{c}} {k = \frac{{ - {\sigma _0} \pm \sqrt {48\zeta _0^2C + \lambda _3^2{{\left( {{\alpha _1} - {\alpha _2}} \right)}^2}} }}{{2{\zeta _0}}}}\\ {\begin{array}{*{20}{c}} {{\alpha _1} > {\alpha _2}}&{{\lambda _3}{\zeta _0} > 0}&{\xi = x + y - kt} \end{array}} \end{array} $

    情况3  若

    $ {Y^3} + \frac{{{\lambda _2}}}{{{\lambda _3}}}{Y^2} + \frac{{{\lambda _1}}}{{{\lambda _3}}}Y + \frac{{{\lambda _0}}}{{{\lambda _3}}} = \left( {Y - {\alpha _1}} \right)\left( {Y - {\alpha _2}} \right)\left( {Y - {\alpha _3}} \right) $ (34)

    α1α2α3是非零常数,等式两边系数相等,得到

    $ \left\{ {\begin{array}{*{20}{l}} {{\lambda _2} = - {\lambda _3}\left( {{\alpha _1} + {\alpha _2} + {\alpha _3}} \right)}\\ {{\lambda _1} = {\lambda _3}\left( {{\alpha _1}{\alpha _2} + {\alpha _1}{\alpha _3} + {\alpha _3}{\alpha _2}} \right)}\\ {{\lambda _0} = - {\lambda _3}{\alpha _1}{\alpha _2}{\alpha _3}} \end{array}} \right. $ (35)

    把(35)式代入(18)式,得到

    $ \left\{ {\begin{array}{*{20}{l}} {{\tau _1} = - \frac{{{\lambda _3}}}{{4{\zeta _0}}}}\\ {{\tau _0} = \frac{{{\lambda _3}\left( {{\alpha _1} + {\alpha _2} + {\alpha _3}} \right) \pm \sqrt {48\zeta _0^2C - \lambda _3^2\left( {{\alpha _1}{\alpha _2} + {\alpha _1}{\alpha _3} + {\alpha _3}{\alpha _2}} \right) + \lambda _3^2\left( {\alpha _1^2 + \alpha _2^2 + \alpha _3^2} \right)} }}{{ - 12{\zeta _0}}}}\\ {k = \frac{{ - {\zeta _0} \pm \sqrt {48\zeta _0^2C - \lambda _3^2\left( {{\alpha _1}{\alpha _2} + {\alpha _1}{\alpha _3} + {\alpha _3}{\alpha _2}} \right) + \lambda _3^2\left( {\alpha _1^2 + \alpha _2^2 + \alpha _3^2} \right)} }}{{2{\zeta _0}}}} \end{array}} \right. $ (36)

    其中λ3ζ0Cα1α2α3为任意常数,

    $ Y = {\alpha _1} + \left( {{\alpha _2} - {\alpha _1}} \right)s{n^2}\left[ {\frac{{\overline { - {\lambda _3}{\zeta _0}\left( {{\alpha _1} - {\alpha _3}} \right)} \xi }}{{2{\lambda _3}}},\quad \sqrt {\frac{{{\alpha _1} - {\alpha _2}}}{{{\alpha _1} - {\alpha _3}}}} } \right] $ (37)

    把(36),(37)式代入方程(15),得到方程(1)的椭圆函数解u4(图 3),

    图 3 方程(1)的解u4
    $ \begin{array}{l} {u_4} = \frac{{{\lambda _3}\left( {{\alpha _1} + {\alpha _2} + {\alpha _3}} \right) \pm \sqrt {48\zeta _0^2C - \lambda _3^2\left( {{\alpha _1}{\alpha _2} + {\alpha _1}{\alpha _3} + {\alpha _3}{\alpha _2}} \right) + \lambda _3^2\left( {\alpha _1^2 + \alpha _2^2 + \alpha _3^2} \right)} }}{{ - 12{\zeta _0}}} - \\ \;\;\;\;\;\;\frac{{{\lambda _3}}}{{4{\zeta _0}}}\left[ {{\alpha _1} + \left( {{\alpha _2} - {\alpha _1}} \right)s{n^2}\left[ {\frac{{\sqrt { - {\lambda _3}{\zeta _0}\left( {{\alpha _1} - {\alpha _3}} \right)} \xi }}{{2{\lambda _3}}},\quad \sqrt {\frac{{{\alpha _1} - {\alpha _2}}}{{{\alpha _1} - {\alpha _3}}}} } \right]} \right] \end{array} $ (38)

    其中

    $ \begin{array}{*{20}{c}} {k = \frac{{ - {\zeta _0} \pm \sqrt {48\zeta _0^2C - \lambda _3^2\left( {{\alpha _1}{\alpha _2} + {\alpha _1}{\alpha _3} + {\alpha _3}{\alpha _2}} \right) + \lambda _3^2\left( {\alpha _1^2 + \alpha _2^2 + \alpha _3^2} \right)} }}{{2{\zeta _0}}}}\\ {\begin{array}{*{20}{c}} {\xi = x + y - kt}&{{\lambda _3}{\zeta _0} > 0} \end{array}} \end{array} $
    3 总结

    利用扩展试验方程法得到了积分微分KP层次方程的钟状解、三角函数解、双曲函数解和椭圆函数解的精确表达式.通过查阅文献,这些解的显示表达式是首次求出的,可以看出扩展试验方程法是求解非线性偏微分方程的有效方法,可以用于求解数学物理中的非线性偏微分方程.

    参考文献
    [1]
    付遵涛, 刘式适, 刘式达. 非线性波方程求解的新方法[J]. 物理学报, 2004, 53(2): 343-348.
    [2]
    KANGALGIL F, AYAZ F. New Exact Travelling Wave Solutions for the Ostrovsky Equation[J]. Physics Letters A, 2008, 372(11): 1831-1835. DOI:10.1016/j.physleta.2007.10.045
    [3]
    张金华, 丁玉敏. 一族非线性色散偏微分方程的指数函数型和三角函数型精确解[J]. 西南大学学报(自然科学版), 2012, 34(1): 28-33.
    [4]
    WAZWAZ A M. A Class of Nonlinear Fourth Order Variant of a Generalized Camassa-Holm Equation with Compact and Noncompact Solutions[J]. Applied Mathematics and Computation, 2005, 165(2): 485-501. DOI:10.1016/j.amc.2004.04.029
    [5]
    何彩霞, 刘小华. 耦合KdV型方程有界行波解的存在性及其显式表达式[J]. 西南师范大学学报(自然科学版), 2014, 39(7): 26-29.
    [6]
    许丽萍. 利用G'/G展开法构造非线性微分差分方程的精确解[J]. 西南大学学报(自然科学版), 2010, 32(7): 34-40.
    [7]
    范恩贵, 张鸿庆. Whitham-Broer-Kaup浅水波方程的Bäcklund变换和精确解[J]. 应用数学和力学, 1998, 19(8): 667-670.
    [8]
    BELGACEM F, BULUT H, BASKONUS H M, et al. Mathematical Analysis of the Generalized Benjamin and Burger-Kdv Equations via the Extended Trial Equation Method[J]. Journal of the Association of Arab Universities for Basic and Applied Sciences, 2014, 16(1): 91-100. DOI:10.1016/j.jaubas.2013.07.005
    [9]
    GEPREEL K A. Extended Trial Equation Method for Nonlinear Coupled Schrodinger Boussinesq Partial Differential Equations[J]. Journal of the Egyptian Mathematical Society, 2016, 24(3): 381-391. DOI:10.1016/j.joems.2015.08.007
    [10]
    KHALFALLAH M. New Exact Traveling Wave Solutions of the (3+1)-Dimensional Kadomtsev-Petviashvili(KP) Equation[J]. Communications in Nonlinear Science and Numerical Simulation, 2009, 14(4): 1169-1175. DOI:10.1016/j.cnsns.2007.11.010
    [11]
    NEIRAMEH A, EBRAHIMI M, MAHMEIANI A G. The ($ \frac{{G'}}{{G}}$)-Expansion Method for(2+1)-Dimensional Kadomtsev-Petviashvili Equation[J]. Journal of King Saud University-Science, 2011, 23(2): 179-181. DOI:10.1016/j.jksus.2010.07.006
    [12]
    ADEM A R. Symbolic Computation on Exact Solutions of a Coupled Kadomtsev-Petviashvili Equation:Lie Symmetry Analysis and Extended Tanh Method[J]. Computers and Mathematics with Applications, 2017, 74(8): 1897-1902. DOI:10.1016/j.camwa.2017.06.049
    The Exact Traveling Wave Solution of the Integral Differential KP Hierarchy Equation
    SONG Jia-qian , LIU Xiao-hua     
    School of Data Science and Information Engineering, Guizhou Minzu University, Guiyang 550025, China
    Abstract: By means of the integral transformation and traveling wave transformation, the KP hierarchical equation can be turned into the equivalent ordinary differential equation. With the help of the extended trial equation method, the problem of solving ordinary differential equation is transformed into that of solving algebraic equation system. According to the solutions of algebraic equation system and the extended trial equation, we have obtained the exact expressions of bell-shaped solution, triangular solution, hyperbolic solution and elliptic function solution of the KP hierarchical equation. The explicit expressions of these solutions are obtained for the first time. With the extended trial equation method, many new exact solutions of nonlinear partial differential equations can be obtained effectively.
    Key words: extend trial equation method    KP hierarchical equation    travelling wave solution    
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