(3+1)维广义Kadomtsev-Petviashvili方程新的精确周期孤立波解

图 1 $k_3 = -2$ , ${\cal J}_3 = -5$ , $x = 2$ , ${\cal I}_1 = {\cal I}_2 = {\cal I}_3 = {\cal J}_1 = {\cal J}_2 = {\cal K}_2 = {\cal K}_3 = 1$ , $\epsilon_1 = 1$ , (a) $z = -5$ , (b) $z = 0$ , (c) $z = 5$

情形三

$\begin{eqnarray} k_1 & = & 0, {\cal J}_2 = -\frac{{\cal I} _1 {\cal J} _1}{{\cal I} _2}, {\cal J}_3 = -\frac{{\cal I} _1 {\cal J} _1}{{\cal I} _3}, {\cal L}_2 = \frac{{\cal I} _2 \left({\cal K} _2^2-4 {\cal I} _1 {\cal I} _2^2 {\cal J} _1\right)}{{\cal I} _2^2-{\cal I} _1 {\cal J} _1}, \\ {\cal L}_3& = & \frac{{\cal I} _3 \left({\cal K} _3^2-4 {\cal I} _1 {\cal I} _3^2 {\cal J} _1\right)}{{\cal I} _3^2-{\cal I} _1 {\cal J} _1}, {\cal L}_1 = \frac{{\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1}{{\cal I} _1+{\cal J} _1}, \\ {\cal K}_1 & = & [2 {\cal I} _2 \left({\cal I} _1+{\cal J} _1\right) \left({\cal I} _2^2-{\cal I} _1 {\cal J} _1\right) {\cal K} _2+\epsilon_2 [4 {\cal I} _2^2 \left({\cal I} _1+{\cal J} _1\right)^2 \left({\cal I} _2^2-{\cal I} _1 {\cal J} _1\right)^2 {\cal K} _2^2\\&&-4 \left({\cal I} _2^4-2 {\cal I} _1 {\cal J} _1 {\cal I} _2^2+{\cal I} _1^2 {\cal J} _1^2\right) [{\cal J} _1^2 {\cal I} _1^6-3 {\cal J} _1^3 {\cal I} _1^5-{\cal I} _2^2 {\cal J} _1 {\cal I} _1^5+{\cal I} _2^2 {\cal J} _1^2 {\cal I} _1^4\\ &&-6 {\cal I} _2^2 {\cal J} _1^3 {\cal I} _1^3-2 {\cal I} _2^4 {\cal J} _1 {\cal I} _1^3-{\cal I} _2^4 {\cal J} _1^2 {\cal I} _1^2-3 {\cal I} _2^4 {\cal J} _1^3 {\cal I} _1-{\cal I} _2^6 {\cal J} _1 {\cal I} _1-{\cal I} _2^6 {\cal J} _1^2\\ &&+{\cal I} _2^2 \left({\cal I} _1+{\cal J} _1\right)^2 {\cal K} _2^2]]^{1/2}]/ [2 \left({\cal I} _2^4-2 {\cal I} _1 {\cal J} _1 {\cal I} _2^2+{\cal I} _1^2 {\cal J} _1^2\right)], \\ {\cal K}_2 & = & [{\cal I} _2 {\cal I} _3 \left({\cal I} _2^2-{\cal I} _1 {\cal J} _1\right) \left({\cal I} _3^2-{\cal I} _1 {\cal J} _1\right) {\cal K} _3+\epsilon_3 [-{\cal I} _1 {\cal I} _2^2 \left({\cal I} _2-{\cal I} _3\right)^2 \left({\cal I} _2+{\cal I} _3\right)^2\\ &&\times {\cal J} _1 \left({\cal I} _3^2-{\cal I} _1 {\cal J} _1\right)^2 [{\cal I} _2^2 \left({\cal I} _3^2-{\cal I} _1 {\cal J} _1\right)-{\cal I} _1 {\cal J} _1 \left({\cal I} _3^2+3 {\cal I} _1 {\cal J} _1\right)]]^{1/2}]\\ &&/[{\cal I} _2^2 \left({\cal I} _3^2-{\cal I} _1 {\cal J} _1\right)^2], \end{eqnarray}$

(2.20)

其中 ${\cal I}_1$ , ${\cal I}_2$ , ${\cal I}_3$ , ${\cal K}_3$ , $k_2$ , $k_3$ 以及 ${\cal J}_1$ 是任意常数, $\epsilon _2 = \pm1$ , $\epsilon _3 = \pm1$ .将这些结果代入方程(2.3),可得

$\begin{eqnarray} \psi & = & \cos [x {\cal I} _2+\frac{t \left({\cal K} _2^2-4 {\cal I} _1 {\cal I} _2^2 {\cal J} _1\right) {\cal I} _2}{{\cal I} _2^2-{\cal I} _1 {\cal J} _1}+z {\cal K} _2-\frac{y {\cal I} _1 {\cal J} _1}{{\cal I} _2}] k_2+\sin [x {\cal I} _3+z {\cal K} _3\\&&+\frac{t \left({\cal K} _3^2-4 {\cal I} _1 {\cal I} _3^2 {\cal J} _1\right) {\cal I} _3}{{\cal I} _3^2-{\cal I} _1 {\cal J} _1}-\frac{y {\cal I} _1 {\cal J} _1}{{\cal I} _3}] k_3+e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}}. \end{eqnarray}$

(2.21)

因此,我们获得了方程(1.1)第二种新的周期孤立波解

$\begin{eqnarray} u_2 & = & [2 [-e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}} {\cal I} _1-\sin [x {\cal I} _2+\frac{t \left({\cal K} _2^2-4 {\cal I} _1 {\cal I} _2^2 {\cal J} _1\right) {\cal I} _2}{{\cal I} _2^2-{\cal I} _1 {\cal J} _1}\\&&+z {\cal K} _2-\frac{y {\cal I} _1 {\cal J} _1}{{\cal I} _2}] k_2 {\cal I} _2+\cos [x {\cal I} _3+\frac{t \left({\cal K} _3^2-4 {\cal I} _1 {\cal I} _3^2 {\cal J} _1\right) {\cal I} _3}{{\cal I} _3^2-{\cal I} _1 {\cal J} _1}+z {\cal K} _3-\frac{y {\cal I} _1 {\cal J} _1}{{\cal I} _3}]\\&&\times k_3 {\cal I} _3]]/[\cos [x {\cal I} _2+\frac{t \left({\cal K} _2^2-4 {\cal I} _1 {\cal I} _2^2 {\cal J} _1\right) {\cal I} _2}{{\cal I} _2^2-{\cal I} _1 {\cal J} _1}+z {\cal K} _2-\frac{y {\cal I} _1 {\cal J} _1}{{\cal I} _2}] k_2\\&&+\sin [x {\cal I} _3+\frac{t \left({\cal K} _3^2-4 {\cal I} _1 {\cal I} _3^2 {\cal J} _1\right) {\cal I} _3}{{\cal I} _3^2-{\cal I} _1 {\cal J} _1}+z {\cal K} _3-\frac{y {\cal I} _1 {\cal J} _1}{{\cal I} _3}] k_3 \\ &&+e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}}]. \end{eqnarray}$

(2.22)

方程(2.22)的物理结构被展示在图 2.

图 2

图 2 $k_2 = -2$ , $\epsilon_2 = -1$ , $t = 5$ , $\epsilon_3 = -1$ , $k_3 = {\cal J}_1 = 2$ , ${\cal I}_1 = {\cal I}_2 = {\cal I}_3 = {\cal K}_3 = 1$ , $\epsilon_1 = 1$ , (a) $z = -5$ , (b) $z = 0$ , (c) $z = 5$

情形四

$\begin{eqnarray} k_1 & = & 0, {\cal I} _3 = {\cal I} _2 \epsilon _4, {\cal L}_2 = \frac{4 {\cal J} _2 {\cal I} _2^3+{\cal K} _2^2}{{\cal I} _2+{\cal J} _2}, {\cal L}_3 = \frac{4 {\cal J} _3 \epsilon _4 {\cal I} _2^3+{\cal K} _3^2}{{\cal J} _3+{\cal I} _2 \epsilon _4}, \\ {\cal L}_1 & = & \frac{-{\cal J} _1 {\cal I} _1^3+3 {\cal I} _2 {\cal J} _3 \epsilon _4 {\cal I} _1^2+3 {\cal I} _2^2 {\cal J} _1 {\cal I} _1+{\cal K} _1^2+3 {\cal I} _2^3 {\cal J} _3 \epsilon _4}{{\cal I} _1+{\cal J} _1}, \\ {\cal K}_2& = &[2 [{\cal I} _2^4 [{\cal J} _3^4+{\cal I} _2^2 {\cal J} _3^2+{\cal J} _2^2 {\cal J} _3^2-4 {\cal I} _2 {\cal J} _2 {\cal J} _3^2+2 \left({\cal I} _2-{\cal J} _2\right) \left({\cal J} _3^2-{\cal I} _2 {\cal J} _2\right) \epsilon _4 {\cal J} _3\\ &&+{\cal I} _2^2 {\cal J} _2^2]]^{1/2} \epsilon _5+\left({\cal I} _2+{\cal J} _2\right) {\cal K} _3 \left({\cal J} _3+{\cal I} _2 \epsilon _4\right)]/[{\cal I} _2^2+2 {\cal J} _3 \epsilon _4 {\cal I} _2+{\cal J} _3^2], \end{eqnarray}$

(2.23)

其中 ${\cal I}_1$ , ${\cal I}_2$ , ${\cal J}_1$ , ${\cal J}_2$ , ${\cal J}_3$ , ${\cal K}_1$ , ${\cal K}_3$ , $k_2$ 以及 $k_3$ 是任意常数, $\epsilon _4 = \pm1$ , $\epsilon _5 = \pm1$ .将这些结果代入方程(2.3),可得

$\begin{eqnarray} \psi & = & \cos [x {\cal I} _2+y {\cal J} _2+z {\cal K} _2+\frac{t \left(4 {\cal J} _2 {\cal I} _2^3+{\cal K} _2^2\right)}{{\cal I} _2+{\cal J} _2}] k_2+e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-t {\cal L} _1}\\&&+\sin [y {\cal J} _3+z {\cal K} _3+x {\cal I} _2 \epsilon _4+\frac{t \left(4 {\cal I} _2^3 {\cal J} _3 \epsilon _4^3+{\cal K} _3^2\right)}{{\cal J} _3+{\cal I} _2 \epsilon _4}] k_3. \end{eqnarray}$

(2.24)

因此,我们获得了方程(1.1)第三种周期孤立波解

$\begin{eqnarray} u_3 & = & [2 [-e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-t {\cal L} _1} {\cal I} _1-\sin [x {\cal I} _2+y {\cal J} _2+z {\cal K} _2+\frac{t \left(4 {\cal J} _2 {\cal I} _2^3+{\cal K} _2^2\right)}{{\cal I} _2+{\cal J} _2}]\\ &&\times k_2 {\cal I} _2+\cos [y {\cal J} _3+z {\cal K} _3+x {\cal I} _2 \epsilon _4+\frac{t \left(4 {\cal I} _2^3 {\cal J} _3 \epsilon _4^3+{\cal K} _3^2\right)}{{\cal J} _3+{\cal I} _2 \epsilon _4}]k_3 {\cal I} _2 \epsilon _4]]/[\cos [x {\cal I} _2\\ &&+y {\cal J} _2+z {\cal K} _2+\frac{t \left(4 {\cal J} _2 {\cal I} _2^3+{\cal K} _2^2\right)}{{\cal I} _2+{\cal J} _2}] k_2+e^{-x {\cal I} _1 -y {\cal J} _1-z {\cal K} _1-t {\cal L} _1}\\ &&+\sin [y {\cal J} _3+z {\cal K} _3+x {\cal I} _2 \epsilon _4+\frac{t \left(4 {\cal I} _2^3 {\cal J} _3 \epsilon _4^3+{\cal K} _3^2\right)}{{\cal J} _3+{\cal I} _2 \epsilon _4}] k_3]. \end{eqnarray}$

(2.25)

方程(2.25)的物理结构被展示在图 3.

图 3

图 3 $\epsilon_4 = \epsilon_5 = 1$ , $k_3 = -2$ , ${\cal I}_1 = {\cal I}_2 = {\cal K}_1 = k_2 = {\cal J}_1 = {\cal J}_2 = {\cal J}_3 = {\cal K}_3 = 1$ , $z = 2$ , (a) $t = -2$ , (b) $t = 0$ , (c) $t = 2$

情形五

$\begin{eqnarray} k_2& = &0, {\cal J} _3 = -\frac{{\cal I} _1 {\cal J} _1}{{\cal I} _3}, {\cal L}_3 = \frac{{\cal K} _3^2-4 {\cal I} _1 {\cal I} _3^2 {\cal J} _1}{{\cal I} _3-\frac{{\cal I} _1 {\cal J} _1}{{\cal I} _3}}, {\cal L}_1 = \frac{{\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1}{{\cal I} _1+{\cal J} _1}, \\ {\cal K}_3& = &[{\cal I} _3 \left({\cal I} _1+{\cal J} _1\right) \left({\cal I} _3^2-{\cal I} _1 {\cal J} _1\right) {\cal K} _1+[{\cal I} _3^2 \left({\cal I} _1^2+{\cal I} _3^2\right)^2 {\cal J} _1 \left({\cal I} _1+{\cal J} _1\right)^2 [-{\cal J} _1 {\cal I} _1^2\\ &&+\left({\cal I} _3^2+3 {\cal J} _1^2\right) {\cal I} _1+{\cal I} _3^2 {\cal J} _1]]^{1/2} \epsilon _6]/[{\cal I} _3^2 \left({\cal I} _1+{\cal J} _1\right)^2], \end{eqnarray}$

(2.26)

其中 ${\cal I}_1$ , ${\cal I}_2$ , ${\cal I}_3$ , ${\cal J}_1$ , ${\cal J}_2$ , ${\cal L}_2$ , ${\cal K}_1$ , ${\cal K}_2$ , $k_1$ 和 $k_3$ 是任意常数, $\epsilon _6 = \pm1$ .将这些结果代入方程(2.3),可得

$\begin{eqnarray} \psi & = & e^{x {\cal I} _1+y {\cal J} _1+z {\cal K} _1+\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}} k_1+e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}}\\&&+\sin [x {\cal I} _3+z {\cal K} _3+\frac{t \left({\cal K} _3^2-4 {\cal I} _1 {\cal I} _3^2 {\cal J} _1\right)}{{\cal I} _3 -\frac{{\cal I} _1 {\cal J} _1}{{\cal I} _3}}-\frac{y {\cal I} _1 {\cal J} _1}{{\cal I} _3}] k_3. \end{eqnarray}$

(2.27)

因此,我们获得了方程(1.1)第四种新的周期孤立波解

$\begin{eqnarray} u_4 & = & [2 [-e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}} {\cal I} _1+e^{x {\cal I} _1+y {\cal J} _1+z {\cal K} _1+\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}} k_1 {\cal I} _1\\&&+\cos [x {\cal I} _3+z {\cal K} _3+\frac{t \left({\cal K} _3^2-4 {\cal I} _1 {\cal I} _3^2 {\cal J} _1\right)}{{\cal I} _3-\frac{{\cal I} _1 {\cal J} _1}{{\cal I} _3}}-\frac{y {\cal I} _1 {\cal J} _1}{{\cal I} _3}] k_3 {\cal I} _3]]\\ &&/[e^{x {\cal I} _1+y {\cal J} _1+z {\cal K} _1+\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}} k_1+e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}}\\ &&+\sin [x {\cal I} _3+z {\cal K} _3+\frac{t \left({\cal K} _3^2-4 {\cal I} _1 {\cal I} _3^2 {\cal J} _1\right)}{{\cal I} _3-\frac{{\cal I} _1 {\cal J} _1}{{\cal I} _3}}-\frac{y {\cal I} _1 {\cal J} _1}{{\cal I} _3}] k_3]. \end{eqnarray}$

(2.28)

方程(2.28)的物理结构被展示在图 4.

图 4

图 4 ${\cal I}_1 = {\cal K}_2 = z = 2$ , ${\cal I}_3 = -1$ , ${\cal I}_2 = 3$ , $k_1 = {\cal J}_1 = {\cal K}_1 = 1$ , $k_3 = -2$ , ${\cal J}_2 = 4$ , ${\cal L}_2 = 5$ , $\epsilon _6 = 1$ , (a) $t = -2$ , (b) $t = 0$ , (c) $t = 2$

情形六

$\begin{eqnarray} k_2 & = & 0, {\cal I} _3 = {\cal I} _1 \epsilon _7 {\rm i}, {\cal L}_3 = \frac{{\cal K} _3^2-4 {\rm i} {\cal I} _1^3 {\cal J} _3 \epsilon _7}{{\cal J} _3+{\rm i} {\cal I} _1 \epsilon _7}, \\ {\cal L}_1 & = & \frac{{\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1}{{\cal I} _1+{\cal J} _1}, {\cal K}_3 = \frac{2 {\rm i} {\cal J} _3 \epsilon _8 {\cal I} _1^2+2 {\cal J} _1 \epsilon _7 \epsilon _8 {\cal I} _1^2+{\rm i} {\cal K} _1 \epsilon _7 {\cal I} _1+{\cal J} _3 {\cal K} _1}{{\cal I} _1+{\cal J} _1}, \end{eqnarray}$

(2.29)

其中 ${\cal I}_1$ , ${\cal I}_2$ , ${\cal J}_1$ , ${\cal J}_2$ , ${\cal J}_3$ , ${\cal L}_2$ , ${\cal K}_1$ , ${\cal K}_2$ , $k_1$ 以及 $k_3$ 是任意常数, $\epsilon _7 = \pm1$ , $\epsilon _8 = \pm1$ .将这些结果代入方程(2.3),可得

$\begin{eqnarray} \psi & = & e^{x {\cal I} _1+y {\cal J} _1+z {\cal K} _1+\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}} k_1+e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}}\\ &&+\sin [y {\cal J} _3+z {\cal K} _3+{\rm i} x {\cal I} _1 \epsilon _7+\frac{t \left({\cal K} _3^2 -4 {\rm i} {\cal I} _1^3 {\cal J} _3 \epsilon _7\right)}{{\cal J} _3+{\rm i} {\cal I} _1 \epsilon _7}] k_3. \end{eqnarray}$

(2.30)

因此,我们获得了方程(1.1)第五种新的周期孤立波解

$\begin{eqnarray} u_5 & = & [2 [-e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}} {\cal I} _1+e^{x {\cal I} _1+y {\cal J} _1+z {\cal K} _1+\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}} k_1 {\cal I} _1\\ &&+{\rm i} \cos [y {\cal J} _3+z {\cal K} _3+{\rm i} x {\cal I} _1 \epsilon _7+\frac{t \left({\cal K} _3^2-4 {\rm i} {\cal I} _1^3 {\cal J} _3 \epsilon _7\right)}{{\cal J} _3+{\rm i} {\cal I} _1 \epsilon _7}] k_3 \epsilon _7 {\cal I} _1]]\\ &&/[e^{x {\cal I} _1+y {\cal J} _1+z {\cal K} _1+\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}} k_1+e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal K} _1^2-4 {\cal I} _1^3 {\cal J} _1\right)}{{\cal I} _1+{\cal J} _1}}\\ &&+\sin [y {\cal J} _3+z {\cal K} _3+{\rm i} x {\cal I} _1 \epsilon _7+\frac{t \left({\cal K} _3^2-4 {\rm i} {\cal I} _1^3 {\cal J} _3 \epsilon _7\right)}{{\cal J} _3+{\rm i} {\cal I} _1 \epsilon _7}] k_3]. \end{eqnarray}$

(2.31)

情形七

$\begin{eqnarray} {\cal I}_1 & = & \frac{{\cal I} _2 {\cal J} _3}{{\cal J} _1}, \quad {\cal I}_3 = -{\cal I} _2, \quad {\cal J} _2 = - {\cal J} _3, \\ {\cal L}_1 & = & \frac{{\cal J} _1^2 {\cal K} _1^2-4 {\cal I} _2^3 {\cal J} _3^3}{{\cal J} _1^3+{\cal I} _2 {\cal J} _3 {\cal J} _1}, \quad {\cal L}_2 = \frac{{\cal K} _2^2-4 {\cal I} _2^3 {\cal J} _3}{{\cal I} _2-{\cal J} _3}, \quad {\cal L}_3 = \frac{{\cal K} _3^2-4 {\cal I} _2^3 {\cal J} _3}{{\cal J} _3-{\cal I} _2}, \\ {\cal K}_2& = &[{\cal J} _1 [\left({\cal I} _2-{\cal J} _3\right) {\cal J} _3 {\cal K} _1 {\cal I} _2^3+{\cal J} _1^2 \left({\cal I} _2-{\cal J} _3\right) {\cal K} _1 {\cal I} _2^2\\ &&-{\cal J} _1 \sqrt{\frac{{\cal I} _2^7 \left({\cal J} _1^2+{\cal I} _2 {\cal J} _3\right)^2 \left({\cal J} _1^2+{\cal J} _3^2\right)^2 \left({\cal J} _3 {\cal I} _2^2+\left({\cal J} _1^2-{\cal J} _3^2\right) {\cal I} _2+3 {\cal J} _1^2 {\cal J} _3\right)}{{\cal J} _1^6}} \epsilon _{10}]]\\&&/[{\cal I} _2^2 \left({\cal J} _1^2+{\cal I} _2 {\cal J} _3\right)^2], {\cal K}_3 = [{\cal J} _1 [{\cal I} _2 \left({\cal J} _3-{\cal I} _2\right) {\cal K} _1 {\cal J} _3^5-{\cal J} _1^2 \left({\cal I} _2-{\cal J} _3\right) {\cal K} _1 {\cal J} _3^4\\ &&+{\cal J} _1 \sqrt{\frac{{\cal I} _2^3 {\cal J} _3^8 \left({\cal J} _1^2+{\cal I} _2 {\cal J} _3\right)^2 \left({\cal J} _1^2+{\cal J} _3^2\right)^2 \left({\cal J} _3 {\cal I} _2^2+\left({\cal J} _1^2- {\cal J} _3^2\right) {\cal I} _2+3 {\cal J} _1^2 {\cal J} _3\right)}{{\cal J} _1^6}} \epsilon _9]]\\ &&/[{\cal J} _3^4 \left({\cal J} _1^2+{\cal I} _2 {\cal J} _3\right)^2], \end{eqnarray}$

(2.32)

其中 ${\cal I}_2$ , ${\cal J}_1$ , ${\cal J}_3$ , ${\cal K}_1$ , $k_1$ , $k_2$ 以及 $k_3$ 是任意常数, $\epsilon _9 = \pm1$ , $\epsilon _{10} = \pm1$ .将这些结果代入方程(2.3),可得

$\begin{eqnarray} \psi & = & e^{y {\cal J} _1+z {\cal K} _1+\frac{t \left({\cal J} _1^2 {\cal K} _1^2-4 {\cal I} _2^3 {\cal J} _3^3\right)}{{\cal J} _1^3+{\cal I} _2 {\cal J} _3 {\cal J} _1}+\frac{x {\cal I} _2 {\cal J} _3}{{\cal J} _1}} k_1+\cos [x {\cal I} _2-y {\cal J} _3+z {\cal K} _2\\ &&+\frac{t \left({\cal K} _2^2-4 {\cal I} _2^3 {\cal J} _3\right)}{{\cal I} _2-{\cal J} _3}] k_2+e^{-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal J} _1^2 {\cal K} _1^2-4 {\cal I} _2^3 {\cal J} _3^3\right)}{{\cal J} _1^3+{\cal I} _2 {\cal J} _3 {\cal J} _1} -\frac{x {\cal I} _2 {\cal J} _3}{{\cal J} _1}}\\ &&-\sin [x {\cal I} _2-y {\cal J} _3-z {\cal K} _3-\frac{t \left({\cal K} _3^2 -4 {\cal I} _2^3 {\cal J} _3\right)}{{\cal J} _3-{\cal I} _2}] k_3. \end{eqnarray}$

(2.33)

因此,我们获得了方程(1.1)第六种新的周期孤立波解

$\begin{eqnarray} u_6 & = & [2 [-\sin [x {\cal I} _2-y {\cal J} _3+z {\cal K} _2+\frac{t \left({\cal K} _2^2-4 {\cal I} _2^3 {\cal J} _3\right)}{{\cal I} _2-{\cal J} _3}] k_2 {\cal I} _2-\cos [x {\cal I} _2-y {\cal J} _3-z {\cal K} _3\\ &&-\frac{t \left({\cal K} _3^2-4 {\cal I} _2^3 {\cal J} _3\right)}{{\cal J} _3-{\cal I} _2}] k_3 {\cal I} _2-\frac{e^{-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal J} _1^2 {\cal K} _1^2-4 {\cal I} _2^3 {\cal J} _3^3\right)}{{\cal J} _1^3+{\cal I} _2 {\cal J} _3 {\cal J} _1}-\frac{x {\cal I} _2 {\cal J} _3}{{\cal J} _1}} {\cal J} _3 {\cal I} _2}{{\cal J} _1}\\ &&+\frac{e^{y {\cal J} _1+z {\cal K} _1+\frac{t \left({\cal J} _1^2 {\cal K} _1^2-4 {\cal I} _2^3 {\cal J} _3^3\right)}{{\cal J} _1^3+{\cal I} _2 {\cal J} _3 {\cal J} _1}+\frac{x {\cal I} _2 {\cal J} _3}{{\cal J} _1}} k_1 {\cal J} _3 {\cal I} _2}{{\cal J} _1}]]/[e^{y {\cal J} _1+z {\cal K} _1+\frac{t \left({\cal J} _1^2 {\cal K} _1^2-4 {\cal I} _2^3 {\cal J} _3^3\right)}{{\cal J} _1^3+{\cal I} _2 {\cal J} _3 {\cal J} _1}+\frac{x {\cal I} _2 {\cal J} _3}{{\cal J} _1}} k_1\\ &&+e^{-y {\cal J} _1-z {\cal K} _1-\frac{t \left({\cal J} _1^2 {\cal K} _1^2-4 {\cal I} _2^3 {\cal J} _3^3\right)}{{\cal J} _1^3+{\cal I} _2 {\cal J} _3 {\cal J} _1}-\frac{x {\cal I} _2 {\cal J} _3}{{\cal J} _1}}+\cos [x {\cal I} _2+\frac{t \left({\cal K} _2^2-4 {\cal I} _2^3 {\cal J} _3\right)}{{\cal I} _2-{\cal J} _3}-y {\cal J} _3\\ &&+z {\cal K} _2] k_2-\sin [x {\cal I} _2-y {\cal J} _3-z {\cal K} _3-\frac{t \left({\cal K} _3^2-4 {\cal I} _2^3 {\cal J} _3\right)}{{\cal J} _3-{\cal I} _2}] k_3]. \end{eqnarray}$

(2.34)

方程(2.34)的物理结构被展示在图 5.

图 5

图 5 $k_3 = -2$ , $\epsilon _9 = \epsilon _{10} = 1$ , $x = 0.2$ , ${\cal I}_2 = k_1 = k_2 = {\cal J}_1 = {\cal K}_1 = 1$ , ${\cal J}_3 = 2$ , (a) $t = -1$ , (b) $t = 0$ , (c) $t = 1$

情形八

$\begin{eqnarray} {\cal I}_1 & = & {\rm i} {\cal I}_2 \epsilon_{11}, {\cal L}_3 = \frac{4 {\cal J} _3 {\cal I} _3^3+{\cal K} _3^2}{{\cal I} _3+{\cal J} _3}, {\cal L}_2 = \frac{4 {\cal J} _2 {\cal I} _2^3+{\cal K} _2^2}{{\cal I} _2+{\cal J} _2}, {\cal I}_3 = \frac{{\cal I} _2 {\cal J} _2}{{\cal J} _3}, \\ {\cal L}_1& = &\frac{4 {\rm i} {\cal J} _1 \epsilon _{11} {\cal I} _2^3+{\cal K} _1^2}{{\cal J} _1+{\rm i} {\cal I} _2 \epsilon _{11}}, {\cal J}_2 = -{\rm i} {\cal J} _1 \epsilon _{11}, {\cal K}_2 = -\frac{{\rm i} \left({\rm i} {\cal I} _2 {\cal K} _1+{\cal J} _1 \epsilon _{11} {\cal K} _1\right)}{{\cal J} _1+{\rm i} {\cal I} _2 \epsilon _{11}}, \\ {\cal K}_3& = &[{\cal J} _3^3 {\cal K} _1 [{\cal J} _1 [-{\cal I} _2^4+3 \left({\cal J} _1^2-{\cal J} _3^2\right) {\cal I} _2^2+{\cal J} _1^2 {\cal J} _3^2] \epsilon _{11}-{\rm i} {\cal I} _2 [{\cal J} _1^4-3 {\cal J} _3^2 {\cal J} _1^2\\&&+{\cal I} _2^2 \left({\cal J} _3^2-3 {\cal J} _1^2\right)]]-{\rm i} [{\cal I} _2^3 {\cal J} _3^4 \left({\cal J} _1^2+{\cal J} _3^2\right)^2 [{\cal I} _2 [-{\cal J} _1^8+21 {\cal I} _2^2 {\cal J} _1^6-35 {\cal I} _2^4 {\cal J} _1^4\\&&+7 {\cal I} _2^6 {\cal J} _1^2-\left({\cal I} _2^6+3 {\cal J} _1^2 {\cal I} _2^4-45 {\cal J} _1^4 {\cal I} _2^2+17 {\cal J} _1^6\right) {\cal J} _3^2]+{\rm i} {\cal J} _1 [{\cal I} _2^8-21 {\cal J} _1^2 {\cal I} _2^6\\&&+35 {\cal J} _1^4 {\cal I} _2^4-7 {\cal J} _1^6 {\cal I} _2^2+\left(3 {\cal I} _2^6+25 {\cal J} _1^2 {\cal I} _2^4-39 {\cal J} _1^4 {\cal I} _2^2+3 {\cal J} _1^6\right) {\cal J} _3^2] \epsilon _{11}]]^{1/2} \epsilon _{12}]\\&&/[{\cal J} _3^4 \left(\left({\cal I} _2^4-6 {\cal J} _1^2 {\cal I} _2^2+{\cal J} _1^4\right) \epsilon _{11}-4 {\rm i} {\cal I} _2 {\cal J} _1 \left({\cal I} _2^2-{\cal J} _1^2\right)\right)], \end{eqnarray}$

(2.35)

其中 ${\cal I}_2$ , ${\cal J}_1$ , ${\cal J}_3$ , ${\cal K}_1$ , $k_1$ , $k_2$ 以及 $k_3$ 是任意常数, $\epsilon _{11} = \pm1$ , $\epsilon _{12} = \pm1$ .将这些结果代入方程(2.3),可得

$\begin{eqnarray} \psi & = & e^{y {\cal J} _1+z {\cal K} _1+t {\cal L} _1+{\rm i} x {\cal I} _2 \epsilon _{11}} k_1+\sin [y {\cal J} _3+z {\cal K} _3+t {\cal L} _3-\frac{{\rm i} x {\cal I} _2 {\cal J} _1 \epsilon _{11}}{{\cal J} _3}] k_3\\ & &+e^{-y {\cal J} _1-z {\cal K} _1-t {\cal L} _1-{\rm i} x {\cal I} _2 \epsilon _{11}}+\cos [x {\cal I} _2+z {\cal K} _2+t {\cal L} _2-{\rm i} y {\cal J} _1 \epsilon _{11}] k_2. \end{eqnarray}$

(2.36)

因此,我们获得了方程(1.1)第七种新的周期孤立波解

$\begin{eqnarray} u_7 & = & [2 [-\sin [x {\cal I} _2+z {\cal K} _2+t {\cal L} _2-{\rm i} y {\cal J} _1 \epsilon _{11}] k_2 {\cal I} _2-{\rm i} e^{-y {\cal J} _1-z {\cal K} _1-t {\cal L} _1-{\rm i} x {\cal I} _2 \epsilon _{11}} \epsilon _{11} {\cal I} _2\\&&-\frac{{\rm i} \cos [y {\cal J} _3+z {\cal K} _3+t {\cal L} _3-\frac{{\rm i} x {\cal I} _2 {\cal J} _1 \epsilon _{11}}{{\cal J} _3}] k_3 {\cal J} _1 \epsilon _{11} {\cal I} _2}{{\cal J} _3}+{\rm i} e^{y {\cal J} _1+z {\cal K} _1+t {\cal L} _1+{\rm i} x {\cal I} _2 \epsilon _{11}} k_1\\ &&\times \epsilon _{11} {\cal I} _2]]/[e^{y {\cal J} _1+z {\cal K} _1+t {\cal L} _1+{\rm i} x {\cal I} _2 \epsilon _{11}} k_1+e^{-y {\cal J} _1-z {\cal K} _1-t {\cal L} _1-{\rm i} x {\cal I} _2 \epsilon _{11}}+\cos [x {\cal I} _2+z {\cal K} _2\\&&+t {\cal L} _2-{\rm i} y {\cal J} _1 \epsilon _{11}] k_2+\sin [y {\cal J} _3+z {\cal K} _3+t {\cal L} _3-\frac{{\rm i} x {\cal I} _2 {\cal J} _1 \epsilon _{11}}{{\cal J} _3}] k_3]. \end{eqnarray}$

(2.37)

情形九

$\begin{eqnarray} {\cal I}_1 & = & {\cal J}_2 = {\cal I}_3 = 0, {\cal L}_1 = \frac{{\cal K} _1^2}{{\cal J} _1}, {\cal L}_2 = \frac{{\cal K} _2^2}{{\cal I} _2}, {\cal L}_3 = \frac{{\cal K} _3^2}{{\cal J} _3}, \\ {\cal K}_3& = &\frac{{\cal J} _3 {\cal K} _1}{{\cal J} _1}, {\cal K}_2 = \frac{{\cal J} _1 \epsilon _{13} {\cal I} _2^2+{\cal K} _1 {\cal I} _2}{{\cal J} _1}, \end{eqnarray}$

(2.38)

其中 ${\cal I}_2$ , ${\cal J}_1$ , ${\cal J}_3$ , ${\cal K}_1$ , $k_1$ , $k_2$ 以及 $k_3$ 是任意常数, $\epsilon _{13} = \pm1$ .将这些参数的值代入方程(2.3),可得

$\begin{eqnarray} \psi & = & \cos [\frac{t \left({\cal J} _1 \epsilon _{13} {\cal I} _2^2+{\cal K} _1 {\cal I} _2\right)^2}{{\cal I} _2 {\cal J} _1^2}+\frac{z \left({\cal J} _1 \epsilon _{13} {\cal I} _2^2+{\cal K} _1 {\cal I} _2\right)}{{\cal J} _1}+x {\cal I} _2] k_2+e^{\frac{t {\cal K} _1^2}{{\cal J} _1}+z {\cal K} _1+y {\cal J} _1} k_1\\&&+e^{-\frac{t {\cal K} _1^2}{{\cal J} _1}-z {\cal K} _1-y {\cal J} _1}+\sin [\frac{t {\cal J} _3 {\cal K} _1^2}{{\cal J} _1^2}+\frac{z {\cal J} _3 {\cal K} _1}{{\cal J} _1}+y {\cal J} _3] k_3. \end{eqnarray}$

(2.39)

因此,我们获得了方程(1.1)第八种新的周期孤立波解

$\begin{eqnarray} u_8 & = & -[2 \sin [\frac{t \left({\cal J} _1 \epsilon _{13} {\cal I} _2^2+{\cal K} _1 {\cal I} _2\right)^2}{{\cal I} _2 {\cal J} _1^2}+\frac{z \left({\cal J} _1 \epsilon _{13} {\cal I} _2^2+{\cal K} _1 {\cal I} _2\right)}{{\cal J} _1}+x {\cal I} _2] k_2 {\cal I} _2]\\&&/[e^{\frac{t {\cal K} _1^2}{{\cal J} _1}+z {\cal K} _1+y {\cal J} _1} k_1+e^{-\frac{t {\cal K} _1^2}{{\cal J} _1}-z {\cal K} _1-y {\cal J} _1}+\sin [\frac{t {\cal J} _3 {\cal K} _1^2}{{\cal J} _1^2}+\frac{z {\cal J} _3 {\cal K} _1}{{\cal J} _1}+y {\cal J} _3] k_3\\&&+\cos [\frac{t \left({\cal J} _1 \epsilon _{13} {\cal I} _2^2+{\cal K} _1 {\cal I} _2\right)^2}{{\cal I} _2 {\cal J} _1^2}+\frac{z \left({\cal J} _1 \epsilon _{13} {\cal I} _2^2+{\cal K} _1 {\cal I} _2\right)}{{\cal J} _1}+x {\cal I} _2] k_2]. \end{eqnarray}$

(2.40)

方程(2.40)的物理结构被展示在图 6.

图 6

图 6 $k_3 = -2$ , $\epsilon _{13} = 1$ , $z = 2$ , ${\cal I}_2 = k_1 = k_2 = {\cal J}_1 = {\cal K}_1 = 1$ , ${\cal J}_3 = 2$ , (a) $y = -5$ , (b) $y = 0$ , (c) $y = 5$

情形十

$\begin{eqnarray} {\cal I}_2 & = & -\frac{{\cal I} _1 {\cal J} _1 \epsilon _{14}}{{\cal J} _3}, {\cal L}_3 = \frac{4 {\cal J} _3 {\cal I} _3^3+{\cal K} _3^2}{{\cal I} _3+{\cal J} _3}, {\cal L}_2 = \frac{4 {\cal J} _2 {\cal I} _2^3+{\cal K} _2^2}{{\cal I} _2+{\cal J} _2}, {\cal I}_3 = -\frac{{\cal I} _1 {\cal J} _1 \epsilon _{14}}{{\cal J} _3}, \\ {\cal J}_2& = &{\cal J} _3 \epsilon _{14}, {\cal L}_1 = \frac{-{\cal J} _1 {\cal I} _1^3+3 {\cal I} _2 {\cal J} _2 {\cal I} _1^2+3 {\cal I} _2^2 {\cal J} _1 {\cal I} _1+{\cal K} _1^2+3 {\cal I} _2^3 {\cal J} _2}{{\cal I} _1+{\cal J} _1}, {\cal K}_2 = {\cal K} _3 \epsilon _{14}, \\ {\cal K}_3& = &[\sqrt{{\cal I} _1^3 \left({\cal I} _1+{\cal J} _1\right)^2 {\cal J} _3^4 \left({\cal J} _1^2+{\cal J} _3^2\right)^2 \left({\cal I} _1 {\cal J} _1 \left({\cal I} _1+{\cal J} _1\right)-\left({\cal I} _1-3 {\cal J} _1\right) {\cal J} _3^2\right)} \epsilon _{15}\\&&+\left({\cal I} _1+{\cal J} _1\right) \left({\cal J} _3^2-{\cal I} _1 {\cal J} _1\right) {\cal K} _1 {\cal J} _3^3]/[\left({\cal I} _1+{\cal J} _1\right)^2 {\cal J} _3^4], \end{eqnarray}$

(2.41)

其中 ${\cal I}_1$ , ${\cal J}_1$ , ${\cal J}_3$ , ${\cal K}_1$ , $k_1$ , $k_2$ 和 $k_3$ 是任意常数, $\epsilon _{14} = \pm1$ , $\epsilon _{15} = \pm1$ .将这些结果代入方程(2.3),可得

$\begin{eqnarray} \psi & = & e^{x {\cal I} _1+y {\cal J} _1+z {\cal K} _1+t {\cal L} _1} k_1-\sin [\frac{x {\cal I} _1 {\cal J} _1}{{\cal J} _3}-y {\cal J} _3-z {\cal K} _3-t {\cal L} _3] k_3\\&&+e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-t {\cal L} _1}+\cos [t {\cal L} _2+y {\cal J} _3 \epsilon _{14}+z {\cal K} _3 \epsilon _{14}-\frac{x {\cal I} _1 {\cal J} _1 \epsilon _{14}}{{\cal J} _3}] k_2. \end{eqnarray}$

(2.42)

因此,我们获得了方程(1.1)第九种新的周期孤立波解

$\begin{eqnarray} u_{9} & = & [2 [-e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-t {\cal L} _1} {\cal I} _1+e^{x {\cal I} _1+y {\cal J} _1+z {\cal K} _1+t {\cal L} _1} k_1 {\cal I} _1\\&&+\frac{\sin [t {\cal L} _2+y {\cal J} _3 \epsilon _{14}+z {\cal K} _3 \epsilon _{14}-\frac{x {\cal I} _1 {\cal J} _1 \epsilon _{14}}{{\cal J} _3}] k_2 {\cal J} _1 \epsilon _{14} {\cal I} _1}{{\cal J} _3}\\&&-\frac{\cos [\frac{x {\cal I} _1 {\cal J} _1}{{\cal J} _3}-y {\cal J} _3-z {\cal K} _3-t {\cal L} _3] k_3 {\cal J} _1 {\cal I} _1}{{\cal J} _3}]]/[e^{x {\cal I} _1+y {\cal J} _1+z {\cal K} _1+t {\cal L} _1} k_1\\&&+e^{-x {\cal I} _1-y {\cal J} _1-z {\cal K} _1-t {\cal L} _1}+\cos [t {\cal L} _2+y {\cal J} _3 \epsilon _{14}+z {\cal K} _3 \epsilon _{14}-\frac{x {\cal I} _1 {\cal J} _1 \epsilon _{14}}{{\cal J} _3}] k_2\\&&-\sin [\frac{x {\cal I} _1 {\cal J} _1}{{\cal J} _3}-y {\cal J} _3-z {\cal K} _3-t {\cal L} _3] k_3]. \end{eqnarray}$

(2.43)

方程(2.43)的物理结构被展示在图 7.

图 7

DOI:10.3969/j.issn.1003-3998.2017.06.007

图 7 $k_3 = -2$ , $\epsilon _{14} = \epsilon _{15} = 1$ , $x = -5$ , ${\cal I}_1 = k_1 = k_2 = {\cal J}_1 = {\cal K}_1 = 1$ , ${\cal J}_3 = 2$ , (a) $z = -5$ , (b) $z = 0$ , (c) $z = 5$

情形十一

$\begin{eqnarray} {\cal I}_2 & = & {\cal I}_3 = {\cal J} _1 = 0, {\cal L}_1 = \frac{{\cal K} _1^2}{{\cal I} _1}, {\cal L}_2 = \frac{{\cal K} _2^2}{{\cal J} _2}, {\cal L}_3 = \frac{{\cal K} _3^2}{{\cal J} _3}, \\ {\cal K}_3& = &\frac{{\cal J} _3 {\cal K} _2}{{\cal J} _2}, {\cal K}_1 = {\cal I} _1 \left(\frac{{\cal K} _2}{{\cal J} _2}+{\rm i} {\cal I} _1 \epsilon _{16}\right), \end{eqnarray}$

(2.44)

其中 ${\cal I}_1$ , ${\cal J}_2$ , ${\cal J}_3$ , ${\cal K}_2$ , $k_1$ , $k_2$ 以及 $k_3$ 是任意常数, $\epsilon _{16} = \pm1$ .将这些结果代入方程(2.3),可得

$\begin{eqnarray} \psi & = & e^{\frac{t {\cal K} _1^2}{{\cal I} _1}+z {\cal K} _1+x {\cal I} _1} k_1+e^{-\frac{t {\cal K} _1^2}{{\cal I} _1}-z {\cal K} _1-x {\cal I} _1}+\cos [\frac{t {\cal K} _2^2}{{\cal J} _2}+z {\cal K} _2+y {\cal J} _2] k_2\\&&+\sin [\frac{t {\cal J} _3 {\cal K} _2^2}{{\cal J} _2^2}+\frac{z {\cal J} _3 {\cal K} _2}{{\cal J} _2}+y {\cal J} _3] k_3. \end{eqnarray}$

(2.45)

因此,我们获得了方程(1.1)第十种新的周期孤立波解

$\begin{eqnarray} u_{10} & = & [2 (e^{\frac{t {\cal K} _1^2}{{\cal I} _1}+z {\cal K} _1+x {\cal I} _1} k_1 {\cal I} _1-e^{-\frac{t {\cal K} _1^2}{{\cal I} _1}-z {\cal K} _1-x {\cal I} _1} {\cal I} _1)]/[e^{\frac{t {\cal K} _1^2}{{\cal I} _1}+z {\cal K} _1+x {\cal I} _1} k_1+e^{-\frac{t {\cal K} _1^2}{{\cal I} _1}-z {\cal K} _1-x {\cal I} _1}\\&&+ \cos [\frac{t {\cal K} _2^2}{{\cal J} _2}+z {\cal K} _2+y {\cal J} _2] k_2 +\sin [\frac{t {\cal J} _3 {\cal K} _2^2}{{\cal J} _2^2}+\frac{z {\cal J} _3 {\cal K} _2}{{\cal J} _2}+y {\cal J} _3] k_3]. \end{eqnarray}$

(2.46)

3 总结

本文通过使用Hirota的双线性形式和扩展的同宿测试方法,我们得到了(3+1)维广义KP方程新的精确周期孤立波解.此外,这些新的精确周期波解的物理性质和特性如图 1-图 7所示.

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