寻求非线性偏微分方程的精确解一直是解决和研究非线性问题的关键.近年来, 精确解的求法不断涌现, 如Backlund变换法、Hirota变换法、变量分离法、反散射变换法等.最近, 王明亮等提出的$ (G'/G) $展开法^[1-3], 即假设非线性偏微分方程的行波解可用$ (G'/G) $的多项式来表示, 且$ G $满足一类二阶线性常微分方程, 由此得到一个代数方程组, 将求解微分方程的问题转化为求此代数方程组的解.此方法不需要任何初始或边界条件, 可以简洁、有效地求解非线性偏微分方程.目前, 在此方法的基础上, 出现了许多扩展和改进, 这些改进主要是从将$ (G'/G) $展开法的正幂展开推广到正负幂展开^[4]; 改变$ (G'/G) $的展开形式^[5-7]; 改变函数$ G $满足的方程^[8-10]等方面进行了延伸.本文也是以$ (G'/G) $展开法的基本思想为依据, 是将其展开形式改进为$ \left( {\frac{{G - G'}}{{G + G'}}} \right) $的形式, 并首次尝试将函数$ G $满足的常系数方程改进为一类二阶变系数的非线性方程, 以Burgers-KPP方程为例进行了求解, 得到了该方程的多个显式行波解.

Burgers-Kolmogorov-Petrovskii-Piscounov方程

$ \begin{equation} {u_t} - \alpha {u_{xx}} + \lambda u{u_x} + \beta ({u^3} + \gamma {u^2} + \delta u) = 0, \end{equation} $

(1.1)

其中$ \alpha , \;\beta , \lambda , \gamma , \delta $均为常数.该类方程是既包含耗散作用又包含频散作用的非线性演化方程, 它广泛应用于流体力学、热传导、理论物理等领域.当$ (\alpha , \;\beta , \lambda , \gamma , \delta ) $取不同参数时, 它囊括着许多著名的方程.例如广义KPP方程, Huxley方程, 广义Fisher方程, Burgers-Fisher方程, Fitzhugh-Nagumo方程, Newell-Whitehead方程等.文献[11]用Cole-Hopf变换法得到了该方程的孤子解, 并对解的渐进性质进行了论证; 通过tanh函数展开法, 该方程的单孤波解和周期波解由文献[12]得到; 文献[13]通过变系数辅助方程并结合齐次平衡法得到了该方程的行波解.

将非线性偏微分方程

$ \begin{equation} P\left( {u, {u_x}, {u_t}, {u_{xx}}, {u_{xt}}, {u_{tt}}, \cdots } \right) = 0 \end{equation} $

(2.1)

作行波变换.令$ u\left( {x, t} \right) = u(\xi ) $, $ \xi = x - ct $, 其中$ c $表示波速, 是一常数, 则方程(2.1)化为

$ \begin{equation} P\left( {u, u', u'', \cdots } \right) = 0. \end{equation} $

(2.2)

设方程(2.1)的解为

$ \begin{equation} u(\xi ) = \sum\limits_{i = 0}^l {{a_i}(\xi ){{\left( {\frac{{G - G'}}{{G + G'}}} \right)}^i}}, \end{equation} $

(2.3)

这里$ {a_i}(\xi ) $ $ \left( {i = 0, 1, 2, \cdots , l} \right) $为待定的函数, 参数$ l $可通过齐次平衡法确定, $ G = G(\xi ) $满足一类二阶变系数非线性常微分方程

$ \begin{equation} GG'' + p(\xi ){(G')^2} + q(\xi )GG' = 0, \end{equation} $

(2.4)

其中$ p(\xi ), \;q(\xi ) $均为$ \xi $的任意函数.通过借助Mathematica符号计算软件, 可以得到方程(2.4)的解

$ G(\xi ) = {C_2}{\rm{exp}}\left( {\int_1^\xi {{ {{{{\rm{e}}^{ - \int_1^\tau {q(\eta )d\eta } }}} \over {{C_1} + \int_1^\tau {(1 + p(\varsigma )){{\rm{e}}^{ - \int_1^\varsigma {q(\eta )d\eta } }}} {\rm{d}}\varsigma }}}{\rm{d}}\tau } } \right), $

其中$ {C_1}, {C_2} $为积分常数, 同时可得

$ \begin{equation} \left( {\frac{{G - G'}}{{G + G'}}} \right) = 1 - \frac{2}{{{\rm{1}} + {{\rm{e}}^{\int_1^\xi {q(\eta )d\eta } }}\left( {{C_1} + \int_1^\xi {(1 + p(\varsigma )){{\rm{e}}^{ - \int_1^\varsigma {q(\eta )d\eta } }}} {\rm{d}}\varsigma } \right)}}. \end{equation} $

(2.5)

将(2.3)式代入(2.2)式, 并结合(2.4)式, 合并$ {\left( {\frac{{G - G'}}{{G + G'}}} \right)} $的各同幂次项, 并令$ {\left( {\frac{{G - G'}}{{G + G'}}} \right)} $的各次幂的系数为零, 从中求出$ {a_i}(\xi ), \;p(\xi ), q(\xi ) $, 再将求得的$ p(\xi ), \;q(\xi ) $代入(2.5)式, 最后将得到的$ \left( {\frac{{G - G'}}{{G + G'}}} \right) $函数及$ {a_i}(\xi ) $代回到(2.3)式, 即得到方程(2.1)的解.

对方程(1.1)作行波变换.令$ u = u(\xi ) = u(x - ct) $, 从而化为

$ \begin{equation} - cu' - \alpha u'' + \lambda uu' + \beta ({u^3} + \gamma {u^2} + \delta u) = 0. \end{equation} $

(3.1)

设方程(1.1)的解能够表示成多项式$ u(\xi ) = \sum\limits_{i = 0}^l {{a_i}(\xi ){{\left( {\frac{{G - G'}}{{G + G'}}} \right)}^i}} $, 且$ G = G(\xi ) $满足一类二阶变系数非线性常微分方程

$ GG'' + p(\xi ){(G')^2} + q(\xi )GG' = 0, $

其中$ p(\xi ), \;q(\xi ) $均为$ \xi $的任意函数.利用齐次平衡法, 有$ 3l = l + 2 $, 得$ l = 1 $.则方程(1.1)的解表示为

$ \begin{equation} u(\xi ) = {a_1}(\xi )\left( {\frac{{G - G'}}{{G + G'}}} \right) + {a_0}(\xi ), \end{equation} $

(3.2)

由方程(2.4)和(3.2)式可得

$ \begin{equation*} \begin{split} u'\left( \xi \right) = &\frac{1}{2}(1 + p(\xi ) - q(\xi )){a_1}(\xi ){\left( {\frac{{G - G'}}{{G + G'}}} \right)^2} + ( - (1 + p(\xi )){a_1}(\xi )+{a_1}^\prime (\xi ))\left( {\frac{{G - G'}}{{G + G'}}} \right)\\ & + \frac{1}{2} ((1 + p(\xi ) + q(\xi )){a_1}(\xi ) + 2{a_0}^\prime (\xi )), \\ u''\left( \xi \right) = &\frac{1}{2}{(1 + p(\xi ) - q(\xi ))^2}{a_1}(\xi ){\left( {\frac{{G - G'}}{{G + G'}}} \right)^3} + \frac{1}{2}({a_1}(\xi )(3q(\xi ) - 3 - 3p{(\xi )^2} + 3p(\xi )(q(\xi ) - 2 ) \\ & + p'(\xi ) - q'(\xi )) + 2(1 + p(\xi ) - q(\xi )){a_1}^\prime (\xi )){\left( {\frac{{G - G'}}{{G + G'}}} \right)^2} + \frac{1}{2}({a_1}(\xi )(3 + 6p(\xi ) + 3p{(\xi )^2} \\ & - q{(\xi )^2} - 2p'(\xi )) - 4(1 + p(\xi )){a_1}^\prime (\xi ) + 2{a_1}^{\prime \prime }(\xi ))\left( {\frac{{G - G'}}{{G + G'}}} \right) + \frac{1}{2}( - {a_1}(\xi )(1 + p{(\xi )^2} \\ & + q(\xi ) + p(\xi )(2 + q(\xi )) - p'(\xi ) - q'(\xi )) + 2((1 + p(\xi ) + q(\xi )){a_1}^\prime (\xi ) + {a_0}^{\prime \prime }(\xi ))), \\ \end{split} \end{equation*} $

将上面的$ u $及其各阶导数代入(3.1)式, 合并$ {\left( {\frac{{G - G'}}{{G + G'}}} \right)} $的同幂次项并比较方程两端的系数, 化简可得

$ \begin{align} {\left( {\frac{{G - G'}}{{G + G'}}} \right)^3} : &\; {a_1}(\xi )( - \alpha {(1 + p(\xi ) - q(\xi ))^2} + \lambda (1 + p(\xi ) - q(\xi )){a_1}(\xi ) + 2\beta {a_1}{(\xi )^2}) = 0, \end{align} $

(3.3)

$ \begin{align} {\left( {\frac{{G - G'}}{{G + G'}}} \right)^2}: & \; 2(\beta \gamma - \lambda - \lambda p(\xi ) + 3\beta {a_0}(\xi )){a_1}{(\xi )^2} - 2\alpha (1 + p(\xi ) - q(\xi )){a_1}^\prime (\xi ) + {a_1}(\xi )( 3 \alpha{} \\ &{} + 3\alpha p{(\xi )^2} -c+ \lambda {a_0}(\xi ) + q(\xi )(c - 3\alpha - \lambda {a_0}(\xi )) + p(\xi )(- c + 6\alpha - 3\alpha q(\xi ) {}\\ &{} + \lambda {a_0}(\xi )) - \alpha p'(\xi )+ \alpha q'(\xi ) + 2\lambda {a_1}^\prime (\xi )) = 0, \end{align} $

(3.4)

$ \begin{align} {\left( {\frac{{G - G'}}{{G + G'}}} \right)^1} : &\; \lambda (1 + p(\xi ) + q(\xi )){a_1}{(\xi )^2} + {a_1}(\xi )(2c - 3\alpha + 2\beta \delta - 3\alpha p{(\xi )^2} + 2{a_0}(\xi )( 3\beta{a_0}(\xi ) {} \\ &{}+ 2\beta \gamma- \lambda ) + 2p(\xi )(c - 3\alpha - \lambda {a_0}(\xi )) + \alpha (q{(\xi )^2} + 2p'(\xi )) + 2\lambda {a_0}^\prime (\xi )) + (4\alpha {}\\ &{} + 4 \alpha p(\xi )- 2c + 2\lambda {a_0}(\xi )){a_1}^\prime (\xi ) - 2\alpha {a_1}^{\prime \prime }(\xi ) = 0, \end{align} $

(3.5)

$ \begin{align} {\left( {\frac{{G - G'}}{{G + G'}}} \right)^0}: & \; 2\beta \gamma {a_0}{(\xi )^2} + 2\beta {a_0}{(\xi )^3} - (c - \alpha - \alpha p(\xi ))(1 + p(\xi ) + q(\xi )){a_1}(\xi ) - 2c{a_0}^\prime (\xi ) {}\\ &{}+ {a_0}(\xi ) (2\beta \delta + \lambda (1 + p(\xi ) + q(\xi )){a_1}(\xi ) + 2\lambda {a_0}^\prime (\xi )) - \alpha ({a_1}(\xi )(p'(\xi ) + q'(\xi )) {}\\ & {}+ 2((p(\xi ) + q(\xi ) + 1){a_1}^\prime (\xi ) + {a_0}^{\prime \prime }(\xi ))) = 0. \end{align} $

(3.6)

由(3.3)和(3.4)式, 可求得

$ \begin{align} {a_1}(\xi ) = &\frac{1}{{4\beta }}\left( { - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } } \right)(1 + p(\xi ) - q(\xi )), \end{align} $

(3.7)

$ \begin{align} {a_0}(\xi ) = & - \frac{{( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{4\beta (1 + p(\xi ) - q(\xi ))}}((1 + p(\xi ))(1 + p(\xi ) - q(\xi )) - (p'(\xi ) - q'(\xi ))){} \\ & {} + \frac{{ - 12\alpha \beta \gamma + \lambda (c - \gamma \lambda ) \pm (3c + \gamma \lambda )\sqrt {{\lambda ^2} + 8\alpha \beta } }}{{4({\lambda ^2} + 9\alpha \beta )}}. \end{align} $

(3.8)

现令$ p(\xi ) = q(\xi ) + f(\xi ) + {k_1}\xi + {k_2} $, 其中$ f(\xi ) $为$ \xi $的任意函数, $ {k_1}, {k_2} $为任意常数, 将其与(3.7)、(3.8)式代入(3.5)式, 借助Mathematica符号计算软件, 得到.若

$ \begin{align} \Delta = & \frac{1}{{{\lambda ^2} + 4\alpha \beta \mp \lambda \sqrt {{\lambda ^2} + 8\alpha \beta } }}\left( {8\delta {{({\lambda ^2} + 9\alpha \beta )}^2} + c(3c + 2\gamma \lambda )(5{\lambda ^2} + 36\alpha \beta )} \right.{} \\ & {} \left. { - {\gamma ^2}(216{\alpha ^2}{\beta ^2} + 36\alpha \beta {\lambda ^2} + {\lambda ^4}) \pm \lambda {{(3c + \gamma \lambda )}^2}\sqrt {{\lambda ^2} + 8\alpha \beta } } \right). \end{align} $

(3.9)

则有以下情况

(1)  当$ \Delta> 0 $时, 有

$ \begin{eqnarray} &&p(\xi ) = - R\tan \left( {\frac{{R\xi }}{2} \mp {C_1}} \right) + \frac{{f'(\xi ) + {k_1}}}{{1 + f(\xi ) + {k_1}\xi + {k_2}}} + f(\xi ) + {k_1}\xi + {k_2}, \end{eqnarray} $

(3.10)

$ \begin{eqnarray} && q(\xi ) = - R\tan \left( {\frac{{R\xi }}{2} \mp {C_1}} \right) + \frac{{f'(\xi ) + {k_1}}}{{1 + f(\xi ) + {k_1}\xi + {k_2}}}, \end{eqnarray} $

(3.11)

其中$ {C_1} $为积分常数, $ R = \pm \frac{{\beta \sqrt \Delta }}{{({\lambda ^2} + 9\alpha \beta )}} $.

(2)  当$ \Delta < 0 $且$ {C_1} = 0 $时, 有

$ \begin{eqnarray} &&p(\xi ) = R\tanh \left( {\frac{{R\xi }}{2}} \right) + \frac{{f'(\xi ) + {k_1}}}{{1 + f(\xi ) + {k_1}\xi + {k_2}}} + f(\xi ) + {k_1}\xi + {k_2}, \end{eqnarray} $

(3.12)

$ \begin{eqnarray} && q(\xi ) = R\tanh \left( {\frac{{R\xi }}{2}} \right) + \frac{{f'(\xi ) + {k_1}}}{{1 + f(\xi ) + {k_1}\xi + {k_2}}}, \end{eqnarray} $

(3.13)

其中$ R = \pm \frac{{\beta \sqrt { - \Delta } }}{{({\lambda ^2} + 9\alpha \beta )}} $.

将上述得到的(3.7)–(3.13)式代入(3.6)式中, 得到了以下几种情况.

(1)  $ c = - \frac{{\gamma \lambda }}{3} $, $ \delta = \frac{{2{\gamma ^2}}}{9} $, 则此时$ \Delta = - 2\alpha \beta + \frac{1}{2}\lambda ( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } ) $, 有

① 当$ \Delta > 0 $时, 则$ R = \pm \frac{{\gamma \sqrt \Delta }}{{3\alpha }} $, 结合(3.10)、(3.11)式, 代入(3.7)、(3.8)和(2.5)式, 可得

$ \begin{eqnarray} {a_1}(\xi ) & = & \frac{1}{{4\beta }}\left( { - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } } \right)(1 + f(\xi ) + {k_1}\xi + {k_2}), \end{eqnarray} $

(3.14)

$ \begin{eqnarray} {a_0}(\xi ) & = & - \frac{{( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{12\alpha \beta }}\left( {3\alpha (1 + f(\xi ) + {k_1}\xi + {k_2}) - \gamma \sqrt \Delta \tan \left( {\frac{{\gamma \sqrt \Delta \xi }}{{6\alpha }} \mp {C_1}} \right)} \right)\\ && - \frac{\gamma }{3}, \end{eqnarray} $

(3.15)

$ \begin{eqnarray} \left( {\frac{{G - G'}}{{G + G'}}} \right) & = & - \frac{{8\gamma \sqrt \Delta \cos{C_1}}}{{2{C_2}\gamma \sqrt \Delta \cos{{{C_1}}^2} + (12\alpha \mp {C_2}\gamma \sqrt \Delta \sin (2{C_1})) \tan\left( {\frac{{\gamma \sqrt \Delta \xi }}{{6\alpha }}} \right)}}{} \\ && {} \cdot \frac{1}{{\cos\left( {\frac{{\gamma \sqrt \Delta \xi }}{{3\alpha }} \mp {C_1}} \right) + \cos{C_1}}} \cdot \frac{1}{{1 + f(\xi ) + {k_1}\xi + {k_2}}} + 1, \end{eqnarray} $

(3.16)

其中$ {C_1}, {C_2} $为积分常数.再将(3.14)–(3.16)式代入(3.2)式, 则得到了方程(1.1)的解

$ u(\xi ) = \frac{{\gamma ( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )\left( { - 6\alpha \sqrt \Delta \cos \left( {\frac{{\gamma \sqrt \Delta \xi }}{{6\alpha }}} \right) + {C_2}\gamma \Delta \cos{C_1}\sin \left( {\frac{{\gamma \sqrt \Delta \xi }}{{6\alpha }} \mp {C_1}} \right)} \right)}}{{12\alpha \beta \left( {{C_2}\gamma \sqrt \Delta \cos \left( {\frac{{\gamma \sqrt \Delta \xi }}{{6\alpha }} \mp {C_1}} \right)\cos {C_1} + 6\alpha \sin \left( {\frac{{\gamma \sqrt \Delta \xi }}{{6\alpha }}} \right)} \right)}} - \frac{\gamma }{3}. $

② 当$ \Delta < 0 $且$ {C_1} = 0 $时, 则$ R = \pm \frac{{\gamma \sqrt { - \Delta } }}{{3\alpha }} $, 结合(3.12)、(3.13)式, 代入(3.7)、(3.8)和(2.5)式, 可得

$ \begin{eqnarray} {a_1}(\xi ) & = & \frac{1}{{4\beta }}\left( { - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } } \right)(1 + f(\xi ) + {k_1}\xi + {k_2}), \end{eqnarray} $

(3.17)

$ \begin{eqnarray} {a_0}(\xi ) & = & - \frac{{( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{12\alpha \beta }}\left( {3\alpha (1 + f(\xi ) + {k_1}\xi + {k_2}) + \gamma \sqrt { - \Delta } \tanh \left( {\frac{{\gamma \sqrt { - \Delta } \xi }}{{6\alpha }}} \right)} \right)\\ && - \frac{\gamma }{3}, \end{eqnarray} $

(3.18)

$ \begin{eqnarray} \left( {\frac{{G - G'}}{{G + G'}}} \right) & = & 1 - \frac{{4\gamma \sqrt { - \Delta } }}{{{C_2}\gamma \sqrt { - \Delta } + 6\alpha \tanh \left( {\frac{{\gamma \sqrt { - \Delta } \xi }}{{6\alpha }}} \right)}} \cdot \frac{1}{{1 + \cosh\left( {\frac{{\gamma \sqrt { - \Delta } \xi }}{{3\alpha }}} \right)}} \\ &&\cdot \frac{1}{{(1 + f(\xi ) + {k_1}\xi + {k_2})}}, \end{eqnarray} $

(3.19)

其中$ {C_2} $为积分常数.再将(3.17)–(3.19)式代入(3.2)式, 则得到了方程(1.1)的解

$ u(\xi ) = - \frac{{\gamma ( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )\left( {6\alpha \sqrt { - \Delta } - {C_2}\gamma \Delta \tanh \left( {\frac{{\gamma \sqrt { - \Delta } \xi }}{{6\alpha }}} \right)} \right)}}{{12\alpha \beta \left( {{C_2}\gamma \sqrt { - \Delta } + 6\alpha \tanh \left( {\frac{{\gamma \sqrt { - \Delta } \xi }}{{6\alpha }}} \right)} \right)}} - \frac{\gamma }{3}. $

(2)  $ c = \frac{\gamma }{4}( - \lambda + \sqrt {{\lambda ^2} + 8\alpha \beta } ) $, $ \delta = \frac{{{\gamma ^2}(8\alpha \beta (36\alpha \beta + 5{\lambda ^2}) - 2\lambda (6\alpha \beta + {\lambda ^2})( - \lambda + \sqrt {{\lambda ^2} + 8\alpha \beta } ))}}{{16{{({\lambda ^2} + 9\alpha \beta )}^2}}} $, 则此时$ \Delta = - 18\alpha \beta - \frac{1}{2}\lambda (5\lambda + 3\sqrt {{\lambda ^2} + 8\alpha \beta } ) $, 有

① 当$ \Delta > 0 $时, 则$ R = \pm \frac{{\beta \gamma \sqrt \Delta }}{{2({\lambda ^2} + 9\alpha \beta )}} $, 结合(3.10)、(3.11)式, 代入(3.7)、(3.8)和(2.5)式, 可得

$ \begin{eqnarray} {a_1}(\xi ) & = & \frac{1}{{4\beta }}\left( { - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } } \right)(1 + f(\xi ) + {k_1}\xi + {k_2}), \end{eqnarray} $

(3.20)

$ \begin{eqnarray} {a_0}(\xi ) & = & - \frac{{( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{4\beta }}\left( {1 + f(\xi ) + {k_1}\xi + {k_2} - \tan \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}} \mp {C_1}} \right)} \right.{}\; \; \; \; \; \; \; \; \; \; \end{eqnarray} $

(3.21)

$ \begin{eqnarray} && {}\cdot \left. {\frac{{\beta \gamma \sqrt \Delta }}{{2({\lambda ^2} + 9\alpha \beta )}}} \right) + \frac{{\gamma ( - 48\alpha \beta - 5{\lambda ^2} + \sqrt {{\lambda ^2} + 8\alpha \beta } (\lambda \pm (3\sqrt {{\lambda ^2} + 8\alpha \beta } + \lambda ))}}{{16({\lambda ^2} + 9\alpha \beta )}}, \\ \left( {\frac{{G - G'}}{{G + G'}}} \right)& = & \frac{{ - 8\beta \gamma \sqrt \Delta \cos{C_1}}}{{2{C_2}\beta \gamma \sqrt \Delta \cos{{{C_1}}^2} + (72\alpha \beta + 8{\lambda ^2} \pm {C_2}\beta \gamma \sqrt \Delta \sin (2{C_1})) \tan\left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}}{} \\ && {} \cdot \frac{1}{{\cos\left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{2({\lambda ^2} + 9\alpha \beta )}} \mp {C_1}} \right) + \cos{C_1}}} \cdot \frac{1}{{1 + f(\xi ) + {k_1}\xi + {k_2}}} + 1, \end{eqnarray} $

(3.22)

其中$ {C_1}, {C_2} $为积分常数.再将(3.20)–(3.22)式代入(3.2)式, 则得到了方程(1.1)的解

$ \begin{equation*} \begin{split} u(\xi ) = & - \frac{{ - 4({\lambda ^2} + 9\alpha \beta )\cos \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right) + {C_2}\beta \gamma \sqrt \Delta \cos {C_1}\sin \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}} \mp {C_1}} \right)}}{{{C_2}\beta \gamma \sqrt \Delta \cos \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}} \mp {C_1}} \right)\cos {C_1} + 4({\lambda ^2} + 9\alpha \beta )\sin \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}}\\ & \cdot \frac{{2\gamma \sqrt \Delta (\lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{16({\lambda ^2} + 9\alpha \beta )}} + \frac{{\gamma ( - 48\alpha \beta - 5{\lambda ^2} + \sqrt {{\lambda ^2} + 8\alpha \beta } (\lambda \pm (3\sqrt {{\lambda ^2} + 8\alpha \beta } + \lambda ))}}{{16({\lambda ^2} + 9\alpha \beta )}}.\\ \end{split} \end{equation*} $

② 当$ \Delta < 0 $, 且$ {C_1} = 0 $时, 则$ R = \pm \frac{{\beta \gamma \sqrt { - \Delta } }}{{2({\lambda ^2} + 9\alpha \beta )}} $, 结合(3.12)、(3.13)式, 代入(3.7)、(3.8)和(2.5)式可得

$ \begin{align} {a_1}(\xi ) = &\frac{1}{{4\beta }}\left( { - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } } \right)(1 + f(\xi ) + {k_1}\xi + {k_2}), \end{align} $

(3.23)

$ \begin{align} {a_0}(\xi ) = & - \frac{{( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{4\beta }}\left( {1 + f(\xi ) + {k_1}\xi + {k_2} + \tanh \left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)} \right.{} \\ & {}\left. { \cdot \frac{{\beta \gamma \sqrt { - \Delta } }}{{2({\lambda ^2} + 9\alpha \beta )}}} \right) + \frac{{\gamma ( - 48\alpha \beta - 5{\lambda ^2} + \sqrt {{\lambda ^2} + 8\alpha \beta } (\lambda \pm (3\sqrt {{\lambda ^2} + 8\alpha \beta } + \lambda ))}}{{16({\lambda ^2} + 9\alpha \beta )}}, \end{align} $

(3.24)

$ \begin{align} \left( {\frac{{G - G'}}{{G + G'}}} \right) = & - \frac{{4\beta \gamma \sqrt { - \Delta } }}{{{C_2}\beta \gamma \sqrt { - \Delta } + 4({\lambda ^2} + 9\alpha \beta )\tanh \left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}} \cdot \frac{1}{{1 + \cosh\left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{2({\lambda ^2} + 9\alpha \beta )}}} \right)}}{} \\ & {} \cdot \frac{1}{{1 + f(\xi ) + {k_1}\xi + {k_2}}} + 1, \end{align} $

(3.25)

其中$ {C_2} $为积分常数.再将(3.23)–(3.25)式代入(3.2)式, 则得到了方程(1.1)的解

$ \begin{equation*} \begin{split} u(\xi ) = & \frac{{4({\lambda ^2} + 9\alpha \beta ) + {C_2}\beta \gamma \sqrt { - \Delta } \tanh \left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}}{{{C_2}\beta \gamma \sqrt { - \Delta } + 4({\lambda ^2} + 9\alpha \beta )\tanh \left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}} \cdot \frac{{2\gamma \sqrt { - \Delta } (\lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{16({\lambda ^2} + 9\alpha \beta )}}\\ &+ \frac{{\gamma ( - 48\alpha \beta - 5{\lambda ^2} + \sqrt {{\lambda ^2} + 8\alpha \beta } (\lambda \pm (3\sqrt {{\lambda ^2} + 8\alpha \beta } + \lambda ))}}{{16({\lambda ^2} + 9\alpha \beta )}}. \end{split} \end{equation*} $

(3)  $ c = \frac{\gamma }{4}( - \lambda - \sqrt {{\lambda ^2} + 8\alpha \beta } ) $, $ \delta = \frac{{{\gamma ^2}(8\alpha \beta (36\alpha \beta + 5{\lambda ^2}) - 2\lambda (6\alpha \beta + {\lambda ^2})( - \lambda - \sqrt {{\lambda ^2} + 8\alpha \beta } ))}}{{16{{({\lambda ^2} + 9\alpha \beta )}^2}}} $, 则此时$ \Delta = - 18\alpha \beta + \frac{1}{2}\lambda ( - 5\lambda + 3\sqrt {{\lambda ^2} + 8\alpha \beta } ) $, 有

① 当$ \Delta > 0 $时, 则$ R = \pm \frac{{\beta \gamma \sqrt \Delta }}{{2({\lambda ^2} + 9\alpha \beta )}} $, 结合(3.10)、(3.11)式, 代入(3.7)、(3.8)和(2.5)式, 可得

$ \begin{eqnarray} {a_1}(\xi ) & = & \frac{1}{{4\beta }}\left( { - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } } \right)(1 + f(\xi ) + {k_1}\xi + {k_2}), \end{eqnarray} $

(3.26)

$ \begin{eqnarray} {a_0}(\xi ) & = & - \frac{{( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{4\beta }}\left( {1 + f(\xi ) + {k_1}\xi + {k_2} - \tan \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}} \mp {C_1}} \right)} \right.{}\; \; \; \; \; \; \; \; \; \; \end{eqnarray} $

(3.27)

$ \begin{eqnarray} && {} \left. { \cdot \frac{{\beta \gamma \sqrt \Delta }}{{2({\lambda ^2} + 9\alpha \beta )}}} \right) + \frac{{\gamma ( - 48\alpha \beta - 5{\lambda ^2} + \sqrt {{\lambda ^2} + 8\alpha \beta } ( - \lambda \pm (3\sqrt {{\lambda ^2} + 8\alpha \beta } - \lambda ))}}{{16({\lambda ^2} + 9\alpha \beta )}}, \\ \left( {\frac{{G - G'}}{{G + G'}}} \right) & = &\frac{{ - 8\beta \gamma \sqrt \Delta \cos{C_1}}}{{2{C_2}\beta \gamma \sqrt \Delta \cos{{{C_1}}^2} + (72\alpha \beta + 8{\lambda ^2} \pm {C_2}\beta \gamma \sqrt \Delta \sin (2{C_1}))\tan\left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}}{} \\ && {} \cdot \frac{1}{{\cos\left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{2({\lambda ^2} + 9\alpha \beta )}} \mp {C_1}} \right) + \cos{C_1}}} \cdot \frac{1}{{1 + f(\xi ) + {k_1}\xi + {k_2}}} + 1, \end{eqnarray} $

(3.28)

其中$ {C_1}, {C_2} $为积分常数.再将(3.26)–(3.28)式代入(3.2)式, 则得到了方程(1.1)的解

$ \begin{equation*} \begin{split} u(\xi ) = & - \frac{{ - 4({\lambda ^2} + 9\alpha \beta )\cos \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right) + {C_2}\beta \gamma \sqrt \Delta \cos {C_1}\sin \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}} \mp {C_1}} \right)}}{{{C_2}\beta \gamma \sqrt \Delta \cos \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}} \mp {C_1}} \right)\cos {C_1} + 4({\lambda ^2} + 9\alpha \beta )\sin \left( {\frac{{\beta \gamma \sqrt \Delta \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}} \\ & \cdot \frac{{2\gamma \sqrt \Delta (\lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{16({\lambda ^2} + 9\alpha \beta )}} + \frac{{\gamma ( - 48\alpha \beta - 5{\lambda ^2} + \sqrt {{\lambda ^2} + 8\alpha \beta } ( - \lambda \pm (3\sqrt {{\lambda ^2} + 8\alpha \beta } - \lambda ))}}{{16({\lambda ^2} + 9\alpha \beta )}}. \end{split} \end{equation*} $

② 当$ \Delta < 0 $, 且$ {C_1} = 0 $时, 则$ R = \pm \frac{{\beta \gamma \sqrt { - \Delta } }}{{2({\lambda ^2} + 9\alpha \beta )}} $, 结合(3.12)、(3.13)式, 代入(3.7)、(3.8)和(2.5)式, 可得

$ \begin{eqnarray} {a_1}(\xi ) & = & \frac{1}{{4\beta }}\left( { - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } } \right)(1 + f(\xi ) + {k_1}\xi + {k_2}), \end{eqnarray} $

(3.29)

$ \begin{eqnarray} {a_0}(\xi ) & = & - \frac{{( - \lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{4\beta }}\left( {1 + f(\xi ) + {k_1}\xi + {k_2} + \tanh \left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)} \right.{} \end{eqnarray} $

(3.30)

$ \begin{eqnarray} && {} \left. { \cdot \frac{{\beta \gamma \sqrt { - \Delta } }}{{2({\lambda ^2} + 9\alpha \beta )}}} \right) + \frac{{\gamma ( - 48\alpha \beta - 5{\lambda ^2} + \sqrt {{\lambda ^2} + 8\alpha \beta } ( - \lambda \pm (3\sqrt {{\lambda ^2} + 8\alpha \beta } - \lambda ))}}{{16({\lambda ^2} + 9\alpha \beta )}}, \\ \left( {\frac{{G - G'}}{{G + G'}}} \right) & = & \frac{{ - 4\beta \gamma \sqrt { - \Delta } }}{{{C_2}\beta \gamma \sqrt { - \Delta } + 4({\lambda ^2} + 9\alpha \beta )\tanh \left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}} \cdot \frac{1}{{1 + \cosh\left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{2({\lambda ^2} + 9\alpha \beta )}}} \right)}}{} \\ && {} \cdot \frac{1}{{1 + f(\xi ) + {k_1}\xi + {k_2}}} + 1, \end{eqnarray} $

(3.31)

其中$ {C_2} $为积分常数.再将(3.29)–(3.31)式代入(3.2)式, 则得到了方程(1.1)的解

$ \begin{equation*} \begin{split} u(\xi ) = & \frac{{4({\lambda ^2} + 9\alpha \beta ) + {C_2}\beta \gamma \sqrt { - \Delta } \tanh \left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}}{{{C_2}\beta \gamma \sqrt { - \Delta } + 4({\lambda ^2} + 9\alpha \beta )\tanh \left( {\frac{{\beta \gamma \sqrt { - \Delta } \xi }}{{4({\lambda ^2} + 9\alpha \beta )}}} \right)}} \cdot \frac{{2\gamma \sqrt { - \Delta } (\lambda \pm \sqrt {{\lambda ^2} + 8\alpha \beta } )}}{{16({\lambda ^2} + 9\alpha \beta )}} \\ & + \frac{{\gamma ( - 48\alpha \beta - 5{\lambda ^2} + \sqrt {{\lambda ^2} + 8\alpha \beta } ( - \lambda \pm (3\sqrt {{\lambda ^2} + 8\alpha \beta } - \lambda ))}}{{16({\lambda ^2} + 9\alpha \beta )}}. \end{split} \end{equation*} $

本文构造了一种满足一类二阶非线性变系数常微分方程的$ \left( {\frac{{G - G'}}{{G + G'}}} \right) $展开法.通过此展开法, 并结合齐次平衡法和Mathematica符号计算软件, 求得了Burgers-KPP方程多个新的三角函数形式和双曲函数形式的显式行波解, 扩大了该类方程的解的范围, 同时也验证了该展开法对于求解非线性偏微分方程的精确解是简单有效的.