新 (3+1) 维 KP 方程的退化解与相互作用解

图1 (a) (d) 平行于 $ x $-轴的 1-呼吸解; (b) (e) 平行于 $ y $-轴的 1-呼吸解;(c) (f) 与斜率为 $ 5/7 $ 的直线平行的 1-呼吸解

平行于 $ y $ 轴的 1-呼吸解如图1(b)、1(e)所示; 若取

$\omega_{1,1}=0.5, \omega_{1,2}=a_{1,2}=b_{1,1}=b_{1,2}=1, a_{1,1}=0.6, $

与斜率为 $ 5/7 $ 的直线平行的 1-呼吸解的轨迹如图1(c)、1(f) 所示.

例Ⅱ $ P=2(N=4) $

此时, 辅助函数 $ H_4 $ 为

(2.12)$\begin{aligned}H_{4}= & 1+2 \exp \left(\xi_{1,1}\right) \cos \left(\xi_{1,2}\right)+\exp \left(K_{12}\right) \exp \left(2 \xi_{1,1}\right)+2 \exp \left(\xi_{3,1}\right) \cos \left(\xi_{3,2}\right)+\exp \left(K_{13}\right) \\& \times \exp \left(\xi_{1,1}+\xi_{3,1}\right)\left[\cos \left(\xi_{1,2}+\xi_{3,2}\right)+\mathrm{i} \sin \left(\xi_{1,2}+\xi_{3,2}\right)\right]+\exp \left(K_{14}\right) \exp \left(\xi_{1,1}+\xi_{3,1}\right) \\& \times\left[\cos \left(\xi_{1,2}-\xi_{3,2}\right)+\mathrm{i} \sin \left(\xi_{1,2}-\xi_{3,2}\right)\right]+\exp \left(K_{23}\right) \exp \left(\xi_{1,1}+\xi_{3,1}\right)\left[\cos \left(\xi_{3,2}-\xi_{1,2}\right)\right. \\& \left.+\mathrm{i} \sin \left(\xi_{3,2}-\xi_{1,2}\right)\right]+\exp \left(K_{24}\right) \exp \left(\xi_{1,1}+\xi_{3,1}\right)\left[\cos \left(\xi_{1,2}+\xi_{3,2}\right)-\mathrm{i} \sin \left(\xi_{1,2}+\xi_{3,2}\right)\right] \\& +\exp \left(K_{34}\right) \exp \left(2 \xi_{3,1}\right)+\exp \left(K_{12}+K_{13}+K_{23}\right) \exp \left(2 \xi_{1,1}+\xi_{3,1}\right)\left[\cos \left(\xi_{3,2}\right)+\mathrm{i} \sin \left(\xi_{3,2}\right)\right] \\& +\exp \left(K_{12}+K_{14}+K_{24}\right) \exp \left(2 \xi_{1,1}+\xi_{3,1}\right)\left[\cos \left(\xi_{3,2}\right)-\mathrm{i} \sin \left(\xi_{3,2}\right)\right]+\exp \left(K_{13}+K_{14}\right. \\& \left.+K_{34}\right) \exp \left(\xi_{1,1}+2 \xi_{3,1}\right)\left[\cos \left(\xi_{1,2}\right)+\mathrm{i} \sin \left(\xi_{1,2}\right)\right]+\exp \left(K_{23}+K_{24}+K_{34}\right) \exp \left(\xi_{1,1}\right. \\& \left.+2 \xi_{3,1}\right)\left[\cos \left(\xi_{1,2}\right)-\mathrm{i} \sin \left(\xi_{1,2}\right)\right]+\exp \left(K_{12}+K_{13}+K_{14}+K_{23}+K_{24}+K_{34}\right) \\& \times \exp \left[2\left(\xi_{1,1}+\xi_{3,1}\right)\right]\end{aligned}$

其中

$\begin{align*} \xi_{i,1}=&\omega_{i,1}x+(\omega_{i,1}a_{i,1}-\omega_{i,2}a_{i,2})y+(\omega_{i,1}b_{i,1}-\omega_{i,2}b_{i,2})z+[\omega_{i,1}(-\omega_{i,1}^2+3\omega_{i,2}^2+a_{i,1}^2\nonumber\\& -a_{i,2}^2-a_{i,1}-b_{i,1}-1)+\omega_{i,2}(-2a_{i,1}a_{i,2}+a_{i,2}+b_{i,2})]t,\nonumber\\ \xi_{i,2}=&\omega_{i,2}x+(\omega_{i,1}a_{i,2}-\omega_{i,2}a_{i,1})y+(\omega_{i,1}b_{i,2}-\omega_{i,2}b_{i,1})z+[\omega_{i,1}(-3\omega_{i,1}\omega_{i,2}+2a_{i,1}a_{i,2}\nonumber\\& -a_{i,2}-b_{i,2})+\omega_{i,2}(\omega_{i,2}^2+2a_{i,1}^2-a_{i,2}^2-a_{i,1}-b_{i,1}-1)]t,\nonumber\\ \exp(&K_{mn})=\frac{3(\omega_m-\omega_n)^2+(a_m+a_n)^2}{3(\omega_m+\omega_n)^2+(a_m+a_n)^2},\ i=1,3,\ m,n=1,2,3,4,\ m<n. \end{align*}$

将 (2.12) 式带入 (2.1) 式, 且取

$\begin{gather*} z=0, \omega_{1,1}=0.5, \omega_{1,2}=1, \omega_{3,1}=0.1, \omega_{3,2}=0.4, a_{1,1}=0.6, \\ a_{1,2}=1, a_{3,1}=0.3, a_{3,2}=1.6, b_{1,1}=b_{1,2}=1, b_{3,1}=0.25, b_{3,2}=0.5, \end{gather*}$

2-呼吸解随时间的演化过程如图2 所示.

图2

图2 2-呼吸解随时间的演化过程 (a) $ t=-10 $, (b) $ t=0 $, (c) $ t=10 $

3 Lump解

3.1 由同宿呼吸解退化为 Lump 解

根据文献 [18], 考虑将 $ H(x,y,z,t) $ 设为

(3.1)$\begin{aligned} H(x,y,z,t)=\exp(-\phi_1)+b_2\cos (\phi_2)+b_1\exp(\phi_1), \end{aligned}$

其中 $ \phi_1=k_1(x+p_1y+q_1z+r_1t), \phi_2=k_1(x+p_2y+q_2z+r_2t), k_i,b_i,p_i,q_i,r_i(i=1,2) $ 为待定参数. 将 (3.1) 式带入 (2.3) 式, 求解得到如下参数间的关系

$\begin{align*} \begin{cases} b_1=-\frac{b_2^2[12k_1^2+(p_1-p_2)^2]}{4[12k_1^2-(p_1+p_2)^2]},\\[3mm] q_1=2k_1^2+\frac{1}{2}(p_1^2-p_2^2+2p_1p_2)-p_1-r_1-1,\\[3mm] q_2=-2k_1^2-\frac{1}{2}(p_1^2-p_2^2-2p_1p_2)-p_2-r_2-1. \end{cases} \end{align*}$

同宿呼吸解为

(3.2)$\begin{aligned} u(x,y,z,t)=& \frac{-2k_1^2[2b_2\sin(\phi_2)(b_1\exp(\phi_1)+k_1\exp(-\phi_1))+\exp(-2\phi_1)(k_1^2-1)]}{[\exp(-\phi_1)+b_2\cos(\phi_2)+b_1 \exp(\phi_1)]^2}\nonumber\\[1mm] &+\frac{-2k_1^2(2b_1k_1+b_2^2-b_1)} {[\exp(-\phi_1)+b_2\cos(\phi_2)+b_1 \exp(\phi_1)]^2}, \end{aligned}$

其中

$\begin{align*} \begin{cases} \phi_1=k_1(x+p_1y+(2k_1^2+\frac{1}{2}(p_1^2-p_2^2+2p_1p_2) -p_1-r_1-1)z+r_1t),\\[3mm] \phi_2=k_1(x+p_2y+(-2k_1^2-\frac{1}{2}(p_1^2-p_2^2-2p_1p_2) -p_2-r_2-1)z+r_2t). \end{cases} \end{align*}$

取 $ k_1 = 0.3, p_1=r_1=2, p_2=0.5, b_2=1.5, r_2=1, z=t=0 $, 同宿呼吸解如图3(a) 所示. 为了从同宿呼吸解 $ u(x,y,z,t) $ 得到 Lump 解, 我们考虑令 $ k_1 \to 0, b_2=-2 $, 此时同宿呼吸解退化为

(3.3)$\begin{aligned} \hat{u}(x,y,z,t)=\frac{16(\varphi_1\varphi_2+\frac{12}{(p_1-p_2)^2})}{\left(\varphi_1^2+\varphi_2^2+\frac{24}{(p_1-p_2)^2}\right)^2}, \end{aligned}$

图3

图3 (a) 同宿呼吸解; (b) Lump解

其中

$\begin{gather*} \varphi_1=x+p_1y+\bigg(\frac{1}{2}(p_1^2-p_2^2+2p_1p_2)-p_1-r_1-1\bigg)z+r_1t,\\ \varphi_2=x+p_2y+\bigg(-\frac{1}{2}(p_1^2-p_2^2-2p_1p_2)-p_2-r_2-1\bigg)z+r_2t. \end{gather*}$

取 $ p_1=r_1=2, p_2=0.5, r_2=1, z=t=0 $, Lump 解如图3(a) 所示. 对比图形 3(a) 和3(b), Lump 波比呼吸波更为局域且展现出急速衰减的特点.

3.2 由 $ N $-孤子解退化为 $ Q $-Lump 解

若对 (2.3) 式应用参数极限法^[23], 则 $ N $-孤子解则可以退化为 $ Q $-Lump 解 ($ Q=N/2 $). 即令 (2.3) 式中的 $ \exp(d_i)=-1, \omega_i \to 0, \omega_i/\omega_j=O(1) $, (2.3) 式退化为

(3.4)$\begin{aligned} \hat{H}_{Q}=&\prod_{i=1}^N\Theta_i+\frac{1}{2}\sum_{i,j}^NM_{ij}\prod_{l\neq i,j}^N\Theta_{l}+\frac{1}{2!2^2}\sum_{i,j,r,s}^NM_{ij}M_{sr}\prod_{l\neq i,j,s,r}^N\Theta_{l}+\cdot\cdot\cdot\nonumber\\& +\frac{1}{Q!2^Q}\sum_{i,j,\cdots, m,n}^N\overbrace{M_{ij}M_{kl}\cdots M_{mn}}^M\prod_{p\neq i,j,k,l,\cdots, m,n}^N\Theta_{p}+\cdot\cdot\cdot, \end{aligned}$

其中 $ \Theta_i=x+a_iy+b_iz+(a_i^2-a_i-b_i-1)t, M_{ij}=-\frac{12}{(a_i-a_j)^2} $. 若参数同时满足 $ a_{i+M}^*=a_i, b_{i+M}^*=b_i $, 将 (3.4) 式带入 (2.1) 式即可得到方程的 $ Q $-Lump 解.

例 Ⅰ $ Q=1(N=2) $

此时, (3.4) 式为

(3.5)$\begin{aligned} \hat{H_1}=\Theta_1\Theta_2+M_{12}, \end{aligned}$

其中

$\Theta_i=x+a_iy+b_iz+(a_i^2-a_i-b_i-1)t, M_{12}=-\frac{12}{(a_1-a_2)^2}>0, i=1,2. $

将 (3.5) 式带入 (2.1), 且取 $ a_1=a_2^*=a_{1,1}+{\rm i}a_{1,2}, b_1=b_2^*=b_{1,1}+{\rm i}b_{1,2} $, 1-Lump 解为

(3.6)$\begin{aligned} u=\frac{4[-(\hat{x}+a_{1,1}\hat{y}+b_{1,1}\hat{z})^2+(a_{1,2}\hat{y}+b_{1,2}\hat{z})^2+\frac{3}{a_{1,2}^2}]}{[(\hat{x}+a_{1,1}\hat{y}+b_{1,1}\hat{z})^2+(a_{1,2}\hat{y}+b_{1,2}\hat{z})^2+\frac{3}{a_{1,2}^2}]^2}, \end{aligned}$

其中

(3.7)$\hat{x}=x+\left(-a_{1,1}^{2}-a_{1,2}^{2}-1\right) t, \quad \hat{y}=y+\left(2 a_{1,1}-1\right) t, \quad \hat{z}=z-t.$

由 (3.6) 式和 (3.7) 式易得 1-Lump 解得速度分量为

(3.8)$v_{x}=a_{1,1}^{2}+a_{1,2}^{2}+1, \quad v_{y}=-2 a_{1,1}+1, \quad v_{z}=1.$

在 $ (x,y) $ 平面上, 1-Lump 解在点

$\left(\frac{(a_{1,1}^2a_{1,2}+a_{1,2}^3-a_{1,1}b_{1,2}+a_{1,2}b_{1,1}+a_{1,2})t}{a_{1,2}},-\frac{(2a_{1,1}a_{1,2}-a_{1,2}-b_{1,2})t}{a_{1,2}}\right) $ 处达到最大值 $ \frac{4a_{12}^2}{3} $, 在点$\left(\frac{(a_{1,1}^2a_{1,2}+a_{1,2}^3-a_{1,1}b_{1,2}+a_{1,2}b_{1,1}+a_{1,2})t_0\pm 3}{a_{1,2}}, -\frac{(2a_{1,1}a_{1,2}-a_{1,2}-b_{1,2})t_0}{a_{1,2}}\right) $

处达到最小值 $-\frac{a_{12}^2}{6} $. 令$ a_{1,1}=a_{1,2}=b_{1,2}=1, b_{1,1}=2, z=0, t=0 $, 1-Lump 解的可视化图形如图4(a) 所示.

图4

图4 (a) 1-Lump 解; (b) 2-Lump 解; (c) 3-Lump 解

例Ⅱ $ Q=2(N=4) $

2-Lump 解的辅助函数 $ \hat{H_2} $ 为

(3.9)$\begin{aligned} \hat{H_2}=&\Theta_1\Theta_2\Theta_3\Theta_4+M_{12}\Theta_3\Theta_4+M_{13}\Theta_2\Theta_4+M_{14}\Theta_2\Theta_3+M_{23}\Theta_1\Theta_4+M_{24}\Theta_1\Theta_3\nonumber\\& +M_{34}\Theta_1\Theta_2+M_{12}M_{34}+M_{13}M_{24}+M_{14}M_{23}, \end{aligned}$

其中

$\Theta_i=x+a_iy+b_iz+(a_i^2-a_i-b_i-1)t, M_{ij}=-\frac{12}{(a_i-a_j)^2}>0, i,j=1,2,3,4,i<j. $

取

$a_1=a_3^*=1+{\rm i}, a_2=a_4^*=0.5+2{\rm i}, b_1=b_3^*=2+{\rm i}, b_2=b_4^*=1+1.5{\rm i}, z=0, t=0, $

2-Lump 解如图4(b) 所示.

例Ⅲ$ Q=3(N=6) $

当 $ N=6 $ 时, 由于 3-Lump 解的辅助函数的式子过长, 就不再具体展示. 令

$\begin{gather*} a_1=a_4^*=1.5+1.5{\rm i},\ a_2=a_5^*=1.1+2{\rm i},\ a_3=a_6^*=0.7+2.5{\rm i},\ b_1=b_4^*=2+{\rm i},\nonumber\\ b_2=b_5^*=1+1.5{\rm i},\ b_3=b_6^*=0.3+{\rm i}, \ z=0, \ t=0, \end{gather*}$

3-Lump 解如图4(c) 所示.

观察图4可以发现, 1-Lump 解沿着 “直线” 向四周传播, 其在传播时, 形状、振幅不发生改变. 但是在 2-Lump 解、3-Lump 解中, 我们观察到, 只有在 Lump 波的边缘地带是沿 “直线” 传播的, 在 Lump 波中间交会的地方, 其传播方式变得 “弯曲”, 形状发生了改变. 这是由于在 Lump 波的内侧, 能量进行了转换从而引发了它们形状的变化. 这种情况在现实中可能会引起类似海啸、地震等灾难现象.

4 相互作用解

4.1 由 $ P $-呼吸和 $ R $-孤子组成的相互作用解

如果 (2.3) 式中的参数满足

(4.1)$\begin{array}{c}N=2 P+R, \omega_{2 p-1}=\omega_{2 p}^{*}, a_{2 p-1}=a_{2 p}^{*}, b_{2 p-1}=b_{2 p}^{*}, d_{2 p-1}=d_{2 p}^{*}=0, \omega_{2 P+r}=\omega_{2 P+r} \\a_{2 P+r}=a_{2 P+r}, b_{2 P+r}=b_{2 P+r}, 1 \leq p \leq P, 1 \leq r \leq R,\end{array}$

其中 $ p,r,P,R $ 任意正整数, $ ^* $ 表示共轭复数, 就可以得到由 $ P $-呼吸与 $ R $-孤子组成的相互作用解.

例Ⅰ $ N=3(P=1,R=1) $

此时, $ H_3 $ 为

(4.2)$\begin{aligned} H_3=&\ 1+2\exp(\xi_{1,1})\cos(\xi_{1,2})+\exp(K_{12})\exp(2\xi_{1,1})+\exp(\xi_3) +\exp(K_{13})\exp(\xi_{1,1}\nonumber\\& +\xi_{3})[\cos(\xi_{1,2})+{\rm i}\sin(\xi_{1,2})]+\exp(K_{23})\exp(\xi_{1,1}+\xi_{3})[\cos(\xi_{1,2})-{\rm i}\sin(\xi_{1,2})] \nonumber\\&+\exp(K_{12}+K_{13}+K_{23})\exp(2\xi_{1,1}+\xi_{3}), \end{aligned}$

其中

$\begin{align*} \xi_{1,1}=&\ \omega_{1,1}x+(\omega_{1,1}a_{1,1}-\omega_{1,2}a_{1,2})y+(\omega_{1,1}b_{1,1}-\omega_{1,2}b_{1,2})z+[\omega_{1,1}(-\omega_{1,1}^2+3\omega_{1,2}^2+a_{1,1}^2\nonumber\\& -a_{1,2}^2-a_{1,1}-b_{1,1}-1)+\omega_{1,2}(-2a_{1,1}a_{1,2}+a_{1,2}+b_{1,2})]t,\\ \xi_{1,2}=&\ \omega_{1,2}x+(\omega_{1,1}a_{1,2}-\omega_{1,2}a_{1,1})y+(\omega_{1,1}b_{1,2}-\omega_{1,2}b_{1,1})z+[\omega_{1,1}(-3\omega_{1,1}\omega_{1,2}+2a_{1,1}a_{1,2}\nonumber\\& -a_{1,2}-b_{1,2})+\omega_{1,2}(\omega_{1,2}^2+2a_{1,1}^2-a_{1,2}^2-a_{1,1}-b_{1,1}-1)]t,\\ \xi_3=&\ \omega_3(x+a_3y+b_3z+(-\omega_3^2 + a_3^2-a_3-b_3-1)t)+d_3,\\ \exp(&K_{ij})=\frac{3(\omega_i-\omega_j)^2+(a_i-a_j)^2}{3(\omega_i+\omega_j)^2+(a_i-a_j)^2},\ i,j=1,2,3,i<j. \end{align*}$

将 (4.2) 式带入 (2.3) 式且取 $\omega_{1,1}=0.5, \omega_{1,2}=a_{1,2}=b_{1,1}=b_{1,2}=b_3=1, a_{1,1}=0.6, \omega_3=-1,a_3=2, z=0, $ 1-呼吸与 1- 孤子的相互作用情况如图5 所示.

图5

图5 1-呼吸与 1-孤子的相互作用过程 (a) $ t=-20 $ (b) $ t=0 $ (c) $ t=20 $

例Ⅱ $ N=4(P=1,R=2) $

当 $ N=4 $ 时, 将 (4.1) 式先后带入 (2.3) 式和 (2.1) 式, 且令

$\begin{gather*} \omega_{1,1}=0.5, \omega_{1,2}=a_{1,2}=b_{1,1}=b_{1,2}=1, a_{1,1}=0.6, \omega_3=-1, \\ \omega_4=0.5, a_3=2, a_4=0.5, b_3=1.5,b_4=3, z=0, \end{gather*}$

1-呼吸与 2-孤子组成的混合解如图6 所示.

图6

图6 1-呼吸与 2-孤子组成的混合解 (a) $ t=-5 $, (b) $ t=0 $, (c) $ t=5 $

例Ⅲ $ N=5(P=2,R=1) $

同样地, 令

$\begin{array}{c}\omega_{1,1}=0.5, \omega_{1,2}=a_{1,2}=b_{1,1}=b_{1,2}=b_{5}=1, \omega_{3,1}=0.1, \omega_{3,2}=0.4, a_{1,1}=0.6 \\a_{3,1}=0.3, a_{3,2}=1.6, b_{3,1}=0.25, b_{3,2}=0.5 \omega_{5}=0.7, a_{5}=1.5, z=0\end{array}$

2-呼吸与 1-孤子随时间的动力学行为如图7 所示.

图7

图7 2-呼吸与 1-孤子随时间的动力学行为 (a) $ t=-5 $, (b) $ t=0 $, (c) $ t=5 $

4.2 由 $ Q $-Lump 和 $ R $-孤子组成的相互作用解

为构造出新 (3+1) 维 KP 方程的 $ Q $-Lump 与 $ R $-孤子组成的相互作用解, 我们对 (2.3) 式添加如下的约束条件

(4.3)$\begin{array}{c}N=2 Q+R, \omega_{2 q-1}, \omega_{2 q} \rightarrow 0, a_{q}=a_{Q+q}^{*}, b_{q}=b_{Q+q}^{*}, \exp \left(d_{2 q-1}\right)=\exp \left(d_{2 q}^{*}\right)=-1, \\\omega_{2 Q+r}=\omega_{2 Q+r}, a_{2 Q+r}=a_{2 P+r}, b_{2 Q+r}=b_{2 Q+r}, 1 \leq q \leq Q, 1 \leq r \leq R.\end{array}$

例Ⅰ$ N=3(Q=1,R=1) $

1-Lump 与 1-孤子组成的相互作用解的辅助函数 $ H_3 $ 为

(4.4)$\begin{aligned} H_3=&\Theta_1\Theta_2+M_{12}+\exp(\xi_3)[M_{12}+(L_{13}+\Theta_1)(L_{23}+\Theta_2)], \end{aligned}$

其中

$\begin{gather*} \Theta_i=x+a_iy+b_iz+(a_i^2-a_i-b_i-1)t,\nonumber\\ \xi_3=\omega_3(x+a_3y+b_3z+(-\omega_3^2 + a_3^2-a_3-b_3-1)t)+d_3, \nonumber\\ \ M_{12}=-\frac{12}{(a_1-a_2)^2},\ L_{i3}=-\frac{12\omega_3}{(a_i-a_3)^2+3\omega_3^2},\ i=1,2. \end{gather*}$

令

$a_1=a_2^*=1+{\rm i}, b_1=b_2^*=2+{\rm i}, \omega_3=1, a_3=2, b_3=0.5, d_3=z=0, $

1-Lump 与 1-孤子的相互作用过程如图8 所示.

图8

图8 1-Lump 与 1-孤子组成的相互作用解随时间的演化过程 (a) $ t=-4 $, (b) $ t=0 $, (c) $ t=4 $

例Ⅱ $ N=4(Q=1,R=2) $

此时, 辅助函数 $ H_4 $ 的表达式为

(4.5)$\begin{aligned} H_4=&\Theta_1\Theta_2+M_{12}+\exp(\xi_3)[M_{12}+(L_{13}+\Theta_1)(L_{23}+\Theta_2)]+\exp(\xi_4)[M_{12}+(L_{14}+\Theta_1)\nonumber\\&\times(L_{24}+\Theta_2)]+\exp(\xi_3+\xi_4)\exp(K_{34})[M_{12}+(L_{13}+L_{14}+\Theta_1)(L_{23}+L_{24}+\Theta_2)], \end{aligned}$

其中

$\begin{gather*} \Theta_i=x+a_iy+b_iz+(a_i^2-a_i-b_i-1)t,\ \exp(K_{34})=\frac{3(\omega_3-\omega_4)^2+(a_3-a_4)^2}{3(\omega_3+\omega_4)^2+(a_3-a_4)^2},\nonumber\\ \xi_j=\omega_j(x+a_jy+b_jz+(-\omega_j^2 + a_j^2-a_j-b_j-1)t)+d_j,\nonumber\\ \ M_{12}=-\frac{12}{(a_1-a_2)^2},\ L_{ij}=-\frac{12\omega_j}{(a_i-a_j)^2+3\omega_j^2},\ i=1,2,j=3,4. \end{gather*}$

取

$\begin{gather*} a_1=a_2^*=1+{\rm i}, b_1=b_2^*=0.5+0.5{\rm i}, \omega_3=1,\omega_4=-0.7, a_3=2,\\ a_4=b_3=0.5, b_4=-1, d_3=d_4=z=0, \end{gather*}$

1-Lump 与 2-孤子的随时间的演化过程如图9 所示.

图9

图9 1-Lump 与 2-孤子组成的相互作用解随时间的演化过程 (a) $ t=-8 $, (b) $ t=0 $, (c) $ t=8 $

例Ⅲ $ N=5(Q=2,R=1) $

当 $ N=5(Q=2,R=1) $ 时, 2-Lump 与 1-孤子的组成的相互作用解的辅助函数 $ H_5 $ 为

(4.6)$\begin{aligned} H_5=\ &\Theta_1\Theta_2\Theta_3\Theta_4+\Xi+\exp(\xi_5)[\Xi+M_{12}L_{35}L_{45}+M_{13}L_{25}L_{45}+M_{14}L_{25}L_{35}+M_{23}L_{15}L_{45}\nonumber\\ & +M_{24}L_{15}L_{35}+M_{34}L_{15}L_{25}+(L_{15}+\Theta_1)(L_{25}+\Theta_2)(L_{35}+\Theta_3)(L_{45}+\Theta_4)+\Theta_1(M_{23}L_{45}\nonumber\\& +M_{24}L_{35}+M_{34}L_{25})+\Theta_2(M_{13}L_{45}+M_{14}L_{35}+M_{34}L_{15})+\Theta_3(M_{12}L_{45}+M_{14}L_{25}\nonumber\\& +M_{24}L_{15})+\Theta_1(M_{12}L_{35}+M_{13}L_{25}+M_{23}L_{15})], \end{aligned}$

其中

$\begin{gather*} \Xi=M_{12}\Theta_3\Theta_4+M_{13}\Theta_2\Theta_4+M_{14}\Theta_2\Theta_3+M_{23}\Theta_1\Theta_4+M_{24}\Theta_1\Theta_3+M_{34}\Theta_1\Theta_2\nonumber \\ +M_{12}M_{34}+M_{13}M_{24}+M_{14}M_{23},\ \Theta_i=x+a_iy+b_iz+(a_i^2-a_i-b_i-1)t,\\ \xi_5=\omega_5(x+a_5y+b_5z+(-\omega_5^2 + a_5^2-a_5-b_5-1)t)+d_5,\nonumber\\ M_{ij}=-\frac{12}{(a_i-a_j)^2},\ L_{i5}=-\frac{12\omega_5}{(a_i-a_5)^2+3\omega_5^2},\ i,j=1,2,3,4,i<j. \end{gather*}$

取

$\begin{gather*} a_1=a_3^*=1+{\rm i}, b_1=b_3^*=2+{\rm i},a_2=a_4^*=0.5+2{\rm i},\\ b_2=b_4^*=1+1.5{\rm i}, \omega_5=-1,a_5=2, b_5=0.5, d_5=z=0, \end{gather*}$

该相互作用过程如图10 所示.

图10

图10 2-Lump 与 1-孤子组成的相互作用解 (a) $ t=-3 $, (b) $ t=0 $, (c) $ t=3 $

4.3 由 $ Q $-Lump 和 $ P $-呼吸组成的相互作用解

若 (2.3) 式中的参数满足如下的约束条件

(4.7)$\begin{array}{c}N=2 Q+2 P, \omega_{2 q-1}, \omega_{2 q} \rightarrow 0, a_{q}=a_{Q+q}^{*}, b_{q}=b_{Q+q}^{*}, \exp \left(d_{2 q-1}\right)=\exp \left(d_{2 q}^{*}\right)=-1 \\\omega_{2 Q+2 p-1}=\omega_{2 Q+2 p}^{*}, a_{2 Q+2 p-1}=a_{2 Q+2 p}^{*}, b_{2 Q+2 p-1}=b_{2 Q+2 p}^{*}, d_{2 Q+2 p-1}=d_{2 Q+2 p}^{*}=0 \\1 \leq q \leq Q, 1 \leq p \leq P\end{array}$

可以得到目标方程的由 $ Q $-Lump 和 $ P $-呼吸组成的相互作用解.

例Ⅰ $ N=4(Q=1,P=1) $

此情形下, (2.3) 式变为

(4.8)$\begin{aligned} H_4=&\Theta_1\Theta_2+M_{12}+\exp(\xi_{3,1})[M_{12}+(L_{13}+\Theta_1)(L_{23}+\Theta_2)][\cos(\xi_{3,2})+{\rm i}\sin({\xi_{3,2}})]\nonumber\\&+\exp(\xi_{3,1})[M_{12}+(L_{14}+ \Theta_1)(L_{24}+\Theta_2)][\cos(\xi_{3,2})-{\rm i}\sin({\xi_{3,2}})] \nonumber\\& +\exp(2\xi_{3,1})\exp(K_{34})[M_{12}+(L_{13}+L_{14}+\Theta_1)(L_{23}+L_{24}+\Theta_2)], \end{aligned}$

其中

$\begin{align*} \Theta_i=&x+a_iy+b_iz+(a_i^2-a_i-b_i-1)t,\ M_{12}=-\frac{12}{(a_1-a_2)^2},\ L_{ij}=-\frac{12\omega_j}{(a_i-a_j)^2+3\omega_j^2},\nonumber\\ \xi_{3,1}=&\omega_{3,1}x+(\omega_{3,1}a_{3,1}-\omega_{3,2}a_{3,2})y+(\omega_{3,1}b_{3,1}-\omega_{3,2}b_{3,2})z+[\omega_{3,1}(-\omega_{3,1}^2+3\omega_{3,2}^2+a_{3,1}^2\nonumber\\& -a_{3,2}^2-a_{3,1}-b_{3,1}-1)+\omega_{3,2}(-2a_{3,1}a_{3,2}+a_{3,2}+b_{3,2})]t,\nonumber\\ \xi_{3,2}=&\omega_{3,2}x+(\omega_{3,1}a_{3,2}-\omega_{3,2}a_{3,1})y+(\omega_{3,1}b_{3,2}-\omega_{3,2}b_{3,1})z+[\omega_{3,1}(-3\omega_{3,1}\omega_{3,2}+2a_{3,1}a_{3,2}\nonumber\\& -a_{3,2}-b_{3,2})+\omega_{3,2}(\omega_{3,2}^2+2a_{3,1}^2-a_{3,2}^2-a_{3,1}-b_{3,1}-1)]t,\nonumber\\& \exp(K_{34})=\frac{3(\omega_3-\omega_4)^2+(a_3-a_4)^2}{3(\omega_3+\omega_4)^2+(a_3-a_4)^2},\ i=1,2,j=3,4,i<j. \end{align*}$

令

$\begin{gather*} a_1=a_2^*=1+{\rm i}, b_1=b_2^*=0.5+0.5{\rm i}, \omega_3=\omega_4^*=-0.5+{\rm i},\\ a_3=a_4^*=0.6+1.5{\rm i}, b_3=b_4^*=1+{\rm i}, z=0, \end{gather*}$

1-Lump 与 1-呼吸随时间的弹性碰撞过程如图11所示.

图11

图11 $ 1 $-Lump 与 $ 1 $-呼吸组成的相互作用解 (a) $ t=-10 $, (b) $ t=0 $, (c) $ t=10 $

例Ⅱ $ N=6(Q=2,P=1) $

当 $ N=6(Q=2,P=1) $ 时, 将 (4.7) 式先后带入 (2.3) 式和 (2.1) 式, 可得到 2-Lump 和 1- 呼吸组成的相互作用解. 取

$\begin{gather*} a_1=a_3^*=1.5+1.5{\rm i}, b_1=b_3^*=2+{\rm i},a_2=a_4^*=1.1+2{\rm i}, b_2=b_4^*=1+1.5{\rm i},\\ \omega_3=\omega_4^*=0.8+1.7{\rm i},a_3=a_4^*=0.6+2{\rm i}, b_3=b_4^*=1+{\rm i}, z=0,t=-2, \end{gather*}$

该相互作用解如图12(a) 所示.

图12

图12 (a) 2-Lump 与 1-呼吸组成的相互作用解; (b) 1-Lump 与 2-呼吸组成的相互作用解

例Ⅲ $ N=6(Q=1,P=2) $

同样地, 当 $ N=6(Q=1,P=2) $ 时, 取

$\begin{gather*} a_1=a_2^*=0.5+2{\rm i}, b_1=b_2^*=1+1.5{\rm i}, \omega_3=\omega_4^*=1+0.9{\rm i},a_3=a_4^*=2+1.5{\rm i}, b_3=b_4^*=0.6-{\rm i},\\ \omega_5=\omega_6^*=0.9+1.2{\rm i},a_5=a_5^*=1+1.8{\rm i}, b_3=b_4^*=1.3+0.8{\rm i},z=0,t=-2, \end{gather*}$

1-Lump与 2-呼吸组成的相互作用解如图12(b) 所示.

4.4 由 $ Q $-Lump, $ P $-呼吸和 $ R $-孤子组成的混合解

若对 (2.3) 式施加下列的约束条件

(4.9)$\begin{array}{c}N=2 Q+2 P+R, \omega_{2 q-1}, \omega_{2 q} \rightarrow 0, a_{q}=a_{Q+q}^{*}, b_{q}=b_{Q+q}^{*}, \exp \left(d_{2 q-1}\right)=\exp \left(d_{2 q}^{*}\right)=-1 \\\xi_{2 Q+2 p-1}=\xi_{2 Q+2 p}^{*}, d_{2 Q+2 p-1}=d_{2 Q+2 p}^{*}=0, \omega_{2 P+2 Q+r}=\omega_{2 P+2 Q+r}, a_{2 P+2 Q+r}=a_{2 P+2 Q+r}, \\b_{2 P+2 Q+r}=b_{2 P+2 Q+r}, 1 \leq q \leq Q, 1 \leq p \leq P, 1 \leq r \leq R.\end{array}$

则可以得到方程 (1.1) 的由 Q-Lump、P-呼吸与 R-孤子这三种基本解组成的混合解.

例Ⅰ $ N=5(Q=1,P=1,R=1) $

1-Lump、1-呼吸与 1-孤子组成的混合解的辅助函数 $ H_5 $ 为

(4.10)$\begin{aligned} H_5=&\Theta_1\Theta_2+M_{12}+\exp(\xi_{3,1})[M_{12}+(L_{13}+\Theta_1)(L_{23}+\Theta_2)][\cos(\xi_{3,2})+{\rm i}\sin({\xi_{3,2}})]\nonumber\\& +\exp(\xi_{3,1})[M_{12}+(L_{14}+\Theta_1)(L_{24}+\Theta_2)][\cos(\xi_{3,2})-{\rm i}\sin({\xi_{3,2}})]+\exp(\xi_5)[M_{12} +(L_{15}\nonumber\\& +\Theta_1)(L_{25}+\Theta_2)]+\exp(2\xi_{3,1})\exp(K_{34})[M_{12}+(L_{13}+L_{14}+\Theta_1)(L_{23}+L_{24}+\Theta_2)]\nonumber\\& +\exp(\xi_{3,1}+\xi_5)\exp(K_{35})[M_{12}+(L_{13}+L_{15}+\Theta_1)(L_{23}+L_{25}+\Theta_2)][\cos(\xi_{3,2})\nonumber\\& +{\rm i}\sin({\xi_{3,2}})]+\exp(\xi_{3,1}+\xi_5)\exp(K_{45})[M_{12} +(L_{14}+L_{15}+\Theta_1)(L_{24}+L_{25}+\Theta_2)]\nonumber\\&\times[\cos(\xi_{3,2})-{\rm i}\sin({\xi_{3,2}})]+\exp(2\xi_{3,1}+\xi_5)\exp(K_{34}+K_{35}+K_{45})[M_{12}+(L_{13}+L_{14}\nonumber\\& +L_{15}+\Theta_1)(L_{23}+L_{24}+L_{25}+\Theta_2)], \end{aligned}$

其中

$\begin{align*} \Theta_i=&x+a_iy+b_iz+(a_i^2-a_i-b_i-1)t,\ M_{12}=-\frac{12}{(a_1-a_2)^2},\ L_{ij}=-\frac{12\omega_j}{(a_i-a_j)^2+3\omega_j^2},\nonumber\\ \xi_{3,1}=&\omega_{3,1}x+(\omega_{3,1}a_{3,1}-\omega_{3,2}a_{3,2})y+(\omega_{3,1}b_{3,1}-\omega_{3,2}b_{3,2})z+[\omega_{3,1}(-\omega_{3,1}^2+3\omega_{3,2}^2+a_{3,1}^2\nonumber\\& -a_{3,2}^2-a_{3,1}-b_{3,1}-1)+\omega_{3,2}(-2a_{3,1}a_{3,2}+a_{3,2}+b_{3,2})]t,\nonumber\\ \xi_{3,2}=&\omega_{3,2}x+(\omega_{3,1}a_{3,2}-\omega_{3,2}a_{3,1})y+(\omega_{3,1}b_{3,2}-\omega_{3,2}b_{3,1})z+[\omega_{3,1}(-3\omega_{3,1}\omega_{3,2}+2a_{3,1}a_{3,2}\nonumber\\& -a_{3,2}-b_{3,2})+\omega_{3,2}(\omega_{3,2}^2+2a_{3,1}^2-a_{3,2}^2-a_{3,1}-b_{3,1}-1)]t,\nonumber\\ \xi_5=&\omega_5(x+a_5y+b_5z+(-\omega_5^2 + a_5^2-a_5-b_5-1)t)+d_5,\nonumber\\ \exp&(K_{mn})=\frac{3(\omega_m-\omega_n)^2+(a_m-a_n)^2}{3(\omega_m+\omega_n)^2+(a_m-a_n)^2},i=1,2,j=3,4,5,i<j,\ m,n=3,4,5,m<n. \end{align*}$

令

$\begin{gather*} a_1=a_2^*=1.5+2{\rm i}, b_1=b_2^*=0.5+1.5{\rm i}, \omega_{3,1}=-0.5,\omega_{3,2}=1, a_{3,1}=0.6, \\ a_{3,2}=1.5, b_{3,1}=b_{3,2}=\omega_5=1,a_5=2,b_5=0.5, d_5=z=0, \end{gather*}$

该相互作用过程如图13 所示.

图13

图13 1-Lump、1-呼吸与 1-孤子组成的混合解 (a)(d) $ t=-15 $; (b)(e) $ t=0 $; (c)(f) $ t=15 $

例Ⅱ $ N=6(Q=1,P=1,R=2) $

当 $ N=6(Q=1,P=1,R=2) $ 时, 令

$\begin{gather*} a_1=a_2^*=1.5+2{\rm i}, b_1=b_2^*=0.5+1.5{\rm i}, \omega_{3,1}=-0.5,\omega_{3,2}=1, a_{3,1}=0.6, a_{3,2}=1.5, \\ b_{3,1}=b_{3,2}=\omega_5=1,\omega_6=0.7, a_5=2, a_6=3, b_5=0.5, b_6=-1, d_5=d_6=z=0, \end{gather*}$

1-Lump、1-呼吸与 2-孤子的动力学行为如图14 所示.

图14

图14 1-Lump、1-呼吸与 2-孤子组成的混合解 (a)(d) $ t=-15 $; (b)(e) $ t=0 $; (c)(f) $ t=15 $

5 总结

本文得到了方程 (1.1) 的不同类型的非线性波解, 例如 $ N $-孤子解、 $ P $-呼吸解、 $ Q $-Lump 解和相互作用解. 首先, 基于 Hirota 双线性形式得到了 $ N $-孤子解, 这是下文构造 $ P $-呼吸解、 $ Q $-Lump 解和相互作用解的基础. 其次, 通过模共振技术, 由 $ N $ 孤子解转化得到 $ P $-呼吸解. 1-, 2-呼吸解的可视化图形如图1 和2 所示. 图1(a)-1(f) 也清楚地显示了参数对1-呼吸解轨迹的影响. 然后, 本文通过两种途径得到了 Lump 解. 第一种途径是由同宿测试法获得的同宿呼吸解进行退化得到的 Lump 解 (图3(b)). 第二种途径是由 $ N $-孤子解的完全退化得到了 $ Q $-Lump 解. 1-, 2-和 3-Lump 解如图4 所示. 同时, 我们还发现在高阶 Lump 解之间会发生非弹性碰撞 ($ Q\ge 2 $). 这种非弹性碰撞在现实中可能导致地震、海啸等自然灾害. 比较这两种方法, 第一种方法只能产生最简单的 Lump 解, 而第二种方法可以产生更复杂的 2 阶、3 阶乃至高阶 Lump 解.

最后, 通过 $ N $-孤子解的部分退化, 推导出了多种混合解. 例如, 若部分参数满足模共振条件, 则可以得到了由 $ P $-呼吸与 $ R $-孤子组成的相互作用解. 1-呼吸与 1-孤子、1-呼吸与 2-孤子、2-呼吸与 1-孤子的相互作用过程分别如图5-7 所示. 若部分参数取极限, 则可以得到由 $ Q $-Lump 与 $ R $-孤子组成的相互作用解. 图8-10 展示了 1-Lump 与 1-孤子、1-Lump 与 2-孤子和 2-Lump 与 1- 孤子的动力学行为. 如果部分参数取极限, 其余参数取复共轭, 则可以得到由 $ Q $-Lump 和 $ P $-呼吸组成的相互作用解. 图11-12 分别展示的是由 1-Lump 与 1-呼吸、2-Lump 与 1-呼吸、1-Lump 与 2-呼吸组成的相互作用解. 此外, 图13-14 展示的是由 1-Lump、1-呼吸与 1-孤子和 1-Lump、1-呼吸与 2-孤子组成的 ``三'' 混合解. 本文研究的方法和结果与文献 [24] 有所不同, 且得到了具有复杂性质的相互作用解. 这些解在海洋工程和流体力学等领域具有重要的应用价值.

基于 Hirota 双线性方法^[15⇓-17], 可通过同宿测试法^[18⇓-20]、三波法^[9,19] 和基于 $ N $-孤子解的模共振方法^[22,23] 等多种方法得到非线性可积系统的呼吸波解. Lump 解也可以通过多种方法获得, 例如正二次函数法^[27,28], 基于 $ N $-孤子解和同宿呼吸解的参数极限法^{[11⇓-13,21]}. 此外, 也可以通过不同的方式获得备受学者关注的混合解. 例如, 通过 $ N $-孤子解的部分退化^[9,11⇓-13]或将测试函数设置为正二次函数、指数函数和周期函数的组合^[8,29] 的方法, 得到由呼吸、Lump 和孤子组成的混合解. 除了 Hirota 双线性方法外, Darboux 变换^[30], Bäcklund 变换^[18] 和 Wronskian 行列式^[31]也常用于精确解的研究. 近年来, 无限多个李对称和守恒量的多分量可积方程^[32,33] 的孤子解引起了人们的广泛关注. 本文所采用的方法可能有助于解决这一问题, 但具体应用还需进一步研究.

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