# 求解薛定谔–泊松方程组的时间分裂紧致差分格式Time-Splitting Compact Difference Scheme for Solving Schrodinger-Poisson Equations

DOI: 10.12677/AAM.2019.81002, PDF, HTML, XML, 下载: 567  浏览: 1,872  国家自然科学基金支持

Abstract: In this paper, we have introduced fourth-order compact finite difference, the time splitting method and the Crank-Nicolson method to solve the nonlinear Schrödinger-Poisson equations. Based on fast Sine transform, we construct a fast solver for the fully discretized system. The presented numerical algorithm has been used to solve one-dimensional, two-dimensional and three-dimensional nonlinear Schrödinger-Poisson equations. We provide specific numerical examples. Through the MATLAB software, we write matlab programs based on the presented numerical algorithm, calcu-late approximated error and draw the approximated numerical solution. The numerical results prove that the presented algorithm has spectral accuracy in space direction. They also confirm its efficiency and stability.

1. 引言

1.1. 1.1. 问题背景

$i{\partial }_{t}\psi \left(x,t\right)=-\frac{1}{2}\Delta \psi +V\left(x\right)\psi +\varphi \left(x,t\right)\psi +\beta {|\psi |}^{\frac{4}{d}}\psi ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in {R}^{d},\text{\hspace{0.17em}}t>0,$ (1.1)

$\Delta \varphi \left(x,t\right)=-{|\psi |}^{2},\text{ }x\in {R}^{d},$ (1.2)

$\psi \left(x,t=0\right)={\varphi }_{0}\left(x\right),\text{ }x\in {R}^{d},$

$\underset{|x|\to \infty }{lim}|\psi \left(x,t\right)|=0,$ (1.3)

$V=V\left(x\right)$ 为外势函数，形式已知。ϕ为泊松位势，通常写成卷积形式。

$N\left(t\right)={\int }_{{ℝ}^{d}}{|\psi \left(x,t\right)|}^{2}\text{d}x,$ (1.4)

$P\left(t\right)=-i{\int }_{{ℝ}^{d}}\text{ }{\psi }^{*}\left(x,t\right)\nabla \psi \left(x,t\right)\text{d}x,$ (1.5)

$L\left(t\right)=-i{\int }_{{ℝ}^{d}}\text{ }{\psi }^{*}\left(x,t\right)x×\nabla \psi \left(x,t\right)\text{d}x,$ (1.6)

$E\left(t\right)={\int }_{{ℝ}^{d}}\left[\frac{1}{2}{|\nabla \psi |}^{2}+V{|\psi |}^{2}+\frac{{|\nabla \varphi |}^{2}}{2}+\frac{d}{2+d}{|\psi |}^{2+4/d}\right]\text{d}x.$ (1.7)

1.2. 紧致差分格式设计

${x}_{i}=a+ih\text{\hspace{0.17em}}\left(0 称为节点，记 ${u}_{i}:=u\left({x}_{i}\right)$${{u}^{\prime }}_{i}:=\left(\frac{\text{d}u}{\text{d}x}\right)\left({x}_{i}\right)$${{{u}^{\prime }}^{\prime }}_{i}:=\left(\frac{{\text{d}}^{2}u}{\text{d}{x}^{2}}\right)\left({x}_{i}\right)$${{{{u}^{\prime }}^{\prime }}^{\prime }}_{i}\text{\hspace{0.17em}}\text{\hspace{0.17em}}:=\left(\frac{{\text{d}}^{3}u}{\text{d}{x}^{3}}\right)\left({x}_{i}\right)$${{{{{u}^{\prime }}^{\prime }}^{\prime }}^{\prime }}_{i}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}:=\left(\frac{{\text{d}}^{4}u}{\text{d}{x}^{4}}\right)\left({x}_{i}\right)$ ，其中i与M是正整数，a与b是给定的区间端点。

$\begin{array}{l}\beta {{{u}^{\prime }}^{\prime }}_{i-2}+\alpha {{{u}^{\prime }}^{\prime }}_{i-1}+{{u}^{\prime }}_{i}+\alpha {{{u}^{\prime }}^{\prime }}_{i+1}+\beta {{{u}^{\prime }}^{\prime }}_{i+2}\\ =c\frac{{u}_{i+3}-2{u}_{i}+{u}_{i-3}}{9{h}^{2}}+b\frac{{u}_{i+2}-2{u}_{i}+{u}_{i-2}}{4{h}^{2}}+a\frac{{u}_{i+1}-2{u}_{i}+{u}_{i-1}}{{h}^{2}},\end{array}$ (1.8)

$\frac{1}{10}{{{u}^{\prime }}^{\prime }}_{i-1}+{{{u}^{\prime }}^{\prime }}_{i}+\frac{1}{10}{{{u}^{\prime }}^{\prime }}_{i+1}=\frac{6}{5}\cdot \frac{{u}_{i+1}-2{u}_{i}+{u}_{i-1}}{{h}^{2}}.$ (1.9)

$2{{{u}^{\prime }}^{\prime }}_{i-1}+11{{{u}^{\prime }}^{\prime }}_{i}+2{{{u}^{\prime }}^{\prime }}_{i+1}=\frac{3}{4}\cdot \frac{{u}_{i+2}-2{u}_{i}+{u}_{i-2}}{{h}^{2}}+12\cdot \frac{{u}_{i+1}-2{u}_{i}+{u}_{i-1}}{{h}^{2}}$ (1.10)

1.3. 一维的离散Sine变换与逆变换

${u}_{i}=\underset{k=1}{\overset{M-1}{\sum }}\text{ }{\stackrel{^}{u}}_{k}\mathrm{sin}\left(\frac{ik\text{π}}{M}\right),\text{ }i=1,2,\cdots ,M-1,$ (1.11)

${\stackrel{^}{u}}_{k}=\frac{2}{M}\underset{i=1}{\overset{M-1}{\sum }}{u}_{i}\mathrm{sin}\left(\frac{ik\text{π}}{M}\right),\text{ }k=1,2,\cdots ,M-1,$ (1.12)

${u}_{i+1}=\underset{k=1}{\overset{M-1}{\sum }}\text{ }{\stackrel{^}{u}}_{k}\mathrm{sin}\left(\frac{\left(i+1\right)k\text{π}}{M}\right),\text{ }{u}_{i-1}=\underset{k=1}{\overset{M-1}{\sum }}\text{ }{\stackrel{^}{u}}_{k}\mathrm{sin}\left(\frac{\left(i-1\right)k\text{π}}{M}\right),$

${{{u}^{\prime }}^{\prime }}_{i}=\underset{k=1}{\overset{M-1}{\sum }}{{{\stackrel{^}{u}}^{\prime }}^{\prime }}_{k}\mathrm{sin}\left(\frac{ik\text{π}}{M}\right),$ (1.13)

${{{u}^{\prime }}^{\prime }}_{i+1}=\underset{k=1}{\overset{M-1}{\sum }}{{{\stackrel{^}{u}}^{\prime }}^{\prime }}_{k}\mathrm{sin}\left(\frac{\left(i+1\right)k\text{π}}{M}\right),\text{ }{{{u}^{\prime }}^{\prime }}_{i-1}=\underset{k=1}{\overset{M-1}{\sum }}{{{\stackrel{^}{u}}^{\prime }}^{\prime }}_{k}\mathrm{sin}\left(\frac{\left(i-1\right)k\text{π}}{M}\right).$

$\begin{array}{l}\underset{k=1}{\overset{M-1}{\sum }}\text{ }{{{\stackrel{^}{u}}^{\prime }}^{\prime }}_{k}\left\{\frac{1}{10}\mathrm{sin}\left(\frac{\left(i-1\right)k\text{π}}{M}\right)+\mathrm{sin}\left(\frac{ik\text{π}}{M}\right)+\frac{1}{10}\mathrm{sin}\left(\frac{\left(i+1\right)k\text{π}}{M}\right)\right\}\\ =\frac{6}{5{h}^{2}}\underset{k=1}{\overset{M-1}{\sum }}\text{ }{\stackrel{^}{u}}_{k}\left\{\mathrm{sin}\left(\frac{\left(i-1\right)k\text{π}}{M}\right)+\mathrm{sin}\left(\frac{\left(i+1\right)k\text{π}}{M}\right)-2\mathrm{sin}\left(\frac{ik\text{π}}{M}\right)\right\}.\end{array}$

$\underset{k=1}{\overset{M-1}{\sum }}\text{ }{{{\stackrel{^}{u}}^{\prime }}^{\prime }}_{k}\left(\frac{1}{5}\mathrm{cos}\left(\frac{k\text{π}}{M}\right)+1\right)\mathrm{sin}\left(\frac{ik\text{π}}{M}\right)=\frac{6}{5{h}^{2}}\underset{k=1}{\overset{M-1}{\sum }}{\stackrel{^}{u}}_{k}\left(2\mathrm{cos}\left(\frac{k\text{π}}{M}\right)-2\right)\mathrm{sin}\left(\frac{ik\text{π}}{M}\right).$

${\stackrel{^}{u}}_{k}=-{{{\stackrel{^}{u}}^{\prime }}^{\prime }}_{k}{\left(\frac{24si{n}^{2}\left(\frac{k\text{π}}{2M}\right)}{{h}^{2}}\right)}^{-1}\left(cos\left(\frac{k\text{π}}{M}\right)+5\right),\text{\hspace{0.17em}}k=1,2,\cdots ,M-1.$ (1.14)

${\stackrel{^}{u}}_{k}=-{{{\stackrel{^}{u}}^{\prime }}^{\prime }}_{k}\frac{\frac{24}{{h}^{2}}\left(\mathrm{cos}\left(k\text{π}/M\right)-1\right)+\frac{6}{4{h}^{2}}\left(\mathrm{cos}\left(2k\text{π}/M\right)-1\right)}{4\mathrm{cos}\left(j\text{π}/M\right)+11},\text{\hspace{0.17em}}k=1,2,\cdots ,M-1.$ (1.15)

2. 快速高效的数值方法

2.1. 一维非线性薛定谔泊松方程组的求解方法

$i{\psi }_{t}=-\frac{1}{2}{\psi }_{xx}+V\left(x\right)\psi +\varphi \psi +\beta {|\psi |}^{2}\psi ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}a (2.16)

$-{\varphi }_{xx}={|\psi |}^{2},$

$\psi \left(a,t\right)=0,\psi \left(b,t\right)=0,$ (2.17)

$\psi \left(x,0\right)={\varphi }_{0}\left(x\right),$

$-i{\stackrel{¯}{\psi }}_{t}\psi =V\left(x\right){|\psi \left(x,t\right)|}^{2}+\beta {|\psi \left(x,t\right)|}^{4}+\varphi {|\psi \left(x,t\right)|}^{2},$ (2.18)

$i{\psi }_{t}\left(x,t\right)=V\left(x\right)\psi \left(x,t\right)+\beta {|\psi \left(x,{t}_{n}\right)|}^{2}\psi \left(x,t\right)+\varphi \left(x,{t}_{n}\right)\psi \left(x,t\right),$ (2.19)

$i\frac{{\psi }_{t}}{\psi }=V\left(x\right)+\beta {|\psi \left(x,{t}_{n}\right)|}^{2}+\varphi \left(x,{t}_{n}\right),$ (2.20)

$\psi \left(x,{t}_{n+1}\right)=\psi \left(x,{t}_{n}\right){\text{e}}^{-i\tau \left(V\left(x\right)+\beta {|\psi \left(x,{t}_{n}\right)|}^{2}+\varphi \left(x,{t}_{n}\right)\right)},$ (2.21)

$\psi \left(x,{t}_{*}\right)=\psi \left(x,{t}_{n+1}\right)$$\psi \left(x,{t}_{*}\right)$ 作为时间分裂法的中间变量，但在方程(2.21)中，仍有未知量 $\varphi \left(x,{t}_{n}\right)$ 。由已知条件 $-\Delta \varphi ={|\psi |}^{2}$ ，即 $-{\varphi }_{xx}\left({x}_{j},{t}_{n}\right)={f}_{j}$ ，其中记 ${f}_{j}={|\psi \left({x}_{j},{t}_{n}\right)|}^{2}$ 。可先利用四阶紧致差分格式离散一维泊松方程 $-\Delta \varphi ={|\psi |}^{2}$ ，然后对离散后的方程组做Sine变换得 $-{\stackrel{^}{\varphi }}_{xx}={\stackrel{^}{f}}_{j}$ (详细过程见 [25] )。该式子右边 ${\stackrel{^}{f}}_{j}$

${\stackrel{^}{\varphi }}_{j}^{xx}=-{\stackrel{^}{\varphi }}_{j}\left(\frac{24si{n}^{2}\left(\frac{j\text{π}}{2M}\right)}{{h}_{x}^{2}}\right){\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)}^{-1},$

$-{\stackrel{^}{\varphi }}_{j}^{xx}={\stackrel{^}{f}}_{j},$

${\stackrel{^}{\varphi }}_{j}={\stackrel{^}{f}}_{j}{\left(\frac{24si{n}^{2}\left(\frac{j\text{π}}{2M}\right)}{{h}_{x}^{2}}\right)}^{-1}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right),$ (2.22)

Crank-Nicolson公式离散：

$i\frac{{\psi }_{j}^{n+1}-{\psi }_{j}^{*}}{\tau }=-\frac{1}{4}\left({\left({\psi }_{j}^{*}\right)}_{xx}+{\left({\psi }_{j}^{n+1}\right)}_{xx}\right),$

$\left(1-\frac{i\tau }{4}\Delta \right){\psi }_{j}^{n+1}=\left(1+\frac{i\tau }{4}\Delta \right){\psi }_{j}^{*},$

${\psi }_{j}^{n+1}-\frac{i\tau }{4}{\left({\psi }_{j}^{n+1}\right)}_{xx}={\psi }_{j}^{*}+\frac{i\tau }{4}{\left({\psi }_{j}^{*}\right)}_{xx},$ (2.23)

${\left({\stackrel{^}{\psi }}_{j}^{*}\right)}_{xx}=-{\stackrel{^}{\psi }}_{j}^{*}\frac{24si{n}^{2}\left(\frac{j\text{π}}{2M}\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)},$ (2.24)

${\left({\stackrel{^}{\psi }}_{j}^{*}\right)}_{xx}$ 。再做Sine变换就得 ${\left({\psi }_{j}^{*}\right)}_{xx}$ 。这样方程(2.23)的右边函数值全部已知，记 ${g}_{j}={\psi }_{j}^{*}+\frac{i\tau }{4}{\left({\psi }_{j}^{*}\right)}_{xx}$ ，对(2.23)两边同时做Sine变换 ${\stackrel{^}{\psi }}_{j}^{n+1}-\frac{i\tau }{4}{\left({\stackrel{^}{\psi }}_{j}^{n+1}\right)}_{xx}={\stackrel{^}{g}}_{j}$ ，其中 ${\stackrel{^}{g}}_{j}$${g}_{j}$ 经过一维的Sine逆变换得到的。对方程

(2.23)左边，我们利用基于Sine变换的四阶紧致差分格式的关系式：

${\left({\stackrel{^}{\psi }}_{j}^{n+1}\right)}_{xx}=-{\stackrel{^}{\psi }}_{j}^{n+1}\frac{24si{n}^{2}\left(\frac{j\text{π}}{2M}\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)}$ (2.25)

${\stackrel{^}{\psi }}_{j}^{n+1}=\frac{{\stackrel{^}{g}}_{j}}{1+\frac{i\tau }{4}\frac{24si{n}^{2}\left(\frac{j\text{π}}{2M}\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)}},$ (2.26)

1) 输入a，b，T，M，L，β，以及 ${\varphi }_{0}\left(x\right)$

2) 计算 ${x}_{j}=a+j{h}_{x}$$j=0,1,\cdots ,M$${t}_{n}=n\tau$${\psi }_{0}^{n}={\psi }_{M}^{n}=0$$n=0,1,\cdots ,L$ 。从 $n=0$ 出发；

3) 计算 ${f}_{j}={|{\psi }_{j}^{n}|}^{2}$ ，通过Sine逆变换得 ${\stackrel{^}{f}}_{j}$ ，利用 ${\stackrel{^}{\varphi }}_{j}^{n}={\stackrel{^}{f}}_{j}{\left(\frac{24si{n}^{2}\left(\frac{j\text{π}}{2M}\right)}{{h}_{x}^{2}}\right)}^{-1}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)$ ，再对 ${\stackrel{^}{\varphi }}_{j}^{n}$ 做Sine 变换得 ${\varphi }_{j}^{n}$

4) 由式子 ${\psi }_{j}^{*}={\psi }_{j}^{n}{\text{e}}^{-i\tau \left(V\left({x}_{j}\right)+\beta {|\psi \left({x}_{j},{t}_{n}\right)|}^{2}+{\varphi }_{j}^{n}\right)}$ ，计算出 ${\psi }_{j}^{*}$ ( $j=0,1,\cdots ,M$ )；

5) 通过Sine逆变换求出 ${\stackrel{^}{\psi }}_{j}^{*}$ ，再利用关系式(2.24)，求出 ${\left({\stackrel{^}{\psi }}_{j}^{*}\right)}_{xx}$ 。利用Sine变换求出 ${\psi }_{j}^{*}{}_{xx}$ ，再从 ${g}_{j}$ 的Sine逆变换求出 ${\stackrel{^}{g}}_{j}$

6) 由 ${\stackrel{^}{\psi }}_{j}^{n+1}=\frac{{\stackrel{^}{g}}_{j}}{1+\frac{i\tau }{4}\frac{24si{n}^{2}\left(\frac{j\text{π}}{2M}\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)}}$ 得到 ${\stackrel{^}{\psi }}_{j}^{n+1}$ ，再对其做一次Sine变换就可得到 ${\psi }_{j}^{n+1}$ 。再令 $n=n+\text{1}$ ，重复3~6直到 $n=L$ 为止。

2.2. 二维非线性薛定谔泊松方程组的求解方法

$i{\psi }_{t}=-\frac{1}{2}\Delta \psi +V\psi +\beta {|\psi |}^{2}\psi +\varphi \left(x,y,t\right)\psi ,\text{\hspace{0.17em}}a0,$ (2.27)

$-\Delta \varphi \left(x,y,t\right)={|\psi |}^{2},$ (2.28)

$\psi \left(a,y,t\right)=0;\psi \left(b,y,t\right)=0,$

$\psi \left(x,c,t\right)=0;\psi \left(x,d,t\right)=0,$

$\psi \left(x,y,0\right)={\varphi }_{0}\left(x,y\right),$

${u}_{ij}=\underset{k=1}{\overset{M-1}{\sum }}\text{\hspace{0.17em}}\underset{l=1}{\overset{N-1}{\sum }}\text{\hspace{0.17em}}{\stackrel{^}{u}}_{kl}\mathrm{sin}\left(\frac{ik\text{π}}{M}\right)\mathrm{sin}\left(\frac{jl\text{π}}{N}\right),\text{\hspace{0.17em}}1\le i\le M-1,\text{\hspace{0.17em}}1\le j\le N-1,$ (2.29)

${\stackrel{^}{u}}_{kl}=\frac{2}{M}\frac{2}{N}\underset{i=1}{\overset{M-1}{\sum }}\text{\hspace{0.17em}}\underset{j=1}{\overset{N-1}{\sum }}\text{\hspace{0.17em}}{u}_{ij}\mathrm{sin}\left(\frac{ik\text{π}}{M}\right)\mathrm{sin}\left(\frac{jl\text{π}}{N}\right),\text{\hspace{0.17em}}1\le k\le M-1,\text{\hspace{0.17em}}1\le l\le N-1.$ (2.30)

$\psi \left(x,y,{t}_{*}\right)=\psi \left(x,y,{t}_{n}\right){\text{e}}^{-i\tau \left(V\left(x,y\right)+\beta {|\psi \left(x,y,{t}_{n}\right)|}^{2}+\varphi \left(x,y,{t}_{n}\right)\right)},$ (2.31)

$\psi \left(x,y,{t}_{*}\right)$ 作为时间分裂法的中间变量，即为第二部分的初始值。但是，等式右边仍然有未知量 $\varphi \left(x,y,{t}_{n}\right)$ 。为了求解它，我们要利用方程 $-{\left({\varphi }_{jk}^{n}\right)}_{xx}-{\left({\varphi }_{jk}^{n}\right)}_{yy}={|{\psi }_{jk}^{n}|}^{2}$ ，记 ${f}_{jk}={|{\psi }_{jk}^{n}|}^{2}$ 。对于上式做二维的Sine变换得： $-{\left({\stackrel{^}{\psi }}_{jk}^{n}\right)}_{xx}-{\left({\stackrel{^}{\psi }}_{jk}^{n}\right)}_{yy}={\stackrel{^}{f}}_{jk}$ ，观察发现：等式右边可由已知函数 ${f}_{jk}$ 做二维的Sine逆变换求得；等式

${\left({\stackrel{^}{\varphi }}_{jk}^{n}\right)}_{xx}=-{\stackrel{^}{\varphi }}_{jk}^{n}\frac{24si{n}^{2}\left(\frac{j\text{π}}{2M}\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)},1\le j\le M-1,$ (2.32)

${\left({\stackrel{^}{\varphi }}_{jk}^{n}\right)}_{yy}=-{\stackrel{^}{\varphi }}_{jk}^{n}\frac{24si{n}^{2}\left(\frac{k\text{π}}{2N}\right)}{{h}_{y}^{2}\left(cos\left(\frac{k\text{π}}{N}\right)+5\right)},1\le k\le N-1,$ (2.33)

${\stackrel{^}{\varphi }}_{jk}^{n}=\frac{{\stackrel{^}{f}}_{jk}}{\frac{24si{n}^{2}\left(\frac{j\text{π}}{2M}\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)}+\frac{24si{n}^{2}\left(\frac{k\text{π}}{2N}\right)}{{h}_{y}^{2}\left(cos\left(\frac{k\text{π}}{N}\right)+5\right)}}.$ (2.34)

$i\frac{{\psi }_{jk}^{n+1}-{\psi }_{jk}^{*}}{\tau }=-\frac{1}{4}\left(\Delta \left({\psi }_{jk}^{*}\right)+\Delta \left({\psi }_{jk}^{n+1}\right)\right)$

$\left(1-\frac{i\tau }{4}\Delta \right){\psi }_{jk}^{n+1}=\left(1+\frac{i\tau }{4}\Delta \right){\psi }_{jk}^{*}$

${\psi }_{jk}^{n+1}-\frac{i\tau }{4}\Delta \left({\psi }_{jk}^{n+1}\right)={\psi }_{jk}^{*}+\frac{i\tau }{4}\Delta \left({\psi }_{jk}^{*}\right)$

${\psi }_{jk}^{n+1}-\frac{i\tau }{4}\left({\left({\psi }_{jk}^{n+1}\right)}_{xx}+{\left({\psi }_{jk}^{n+1}\right)}_{yy}\right)={\psi }_{jk}^{*}+\frac{i\tau }{4}\left({\left({\psi }_{jk}^{*}\right)}_{xx}+{\left({\psi }_{jk}^{*}\right)}_{yy}\right),$ (2.35)

${\stackrel{^}{\psi }}_{jk}^{n+1}-\frac{i\tau }{4}\left({\left({\stackrel{^}{\psi }}_{jk}^{n+1}\right)}_{xx}+{\left({\stackrel{^}{\psi }}_{jk}^{n+1}\right)}_{yy}\right)={\stackrel{^}{\psi }}_{jk}^{*}+\frac{i\tau }{4}\left({\left({\stackrel{^}{\psi }}_{jk}^{*}\right)}_{xx}+{\left({\stackrel{^}{\psi }}_{jk}^{*}\right)}_{yy}\right),$ (2.36)

${\left({\stackrel{^}{\psi }}_{jk}^{*}\right)}_{xx}={\stackrel{^}{\psi }}_{jk}^{*}\frac{12}{{\left({h}_{x}\right)}^{2}}\frac{cos\left(\frac{j\text{π}}{M}\right)-1}{cos\left(\frac{j\text{π}}{M}\right)+5},$ (2.37)

${\left({\stackrel{^}{\psi }}_{jk}^{*}\right)}_{yy}={\stackrel{^}{\psi }}_{jk}^{*}\frac{12}{{\left({h}_{y}\right)}^{2}}\frac{cos\left(\frac{k\text{π}}{N}\right)-1}{cos\left(\frac{k\text{π}}{N}\right)+5},$ (2.38)

${\left({\stackrel{^}{\psi }}_{jk}^{n+1}\right)}_{xx}={\stackrel{^}{\psi }}_{jk}^{n+1}\frac{12}{{\left({h}_{x}\right)}^{2}}\frac{cos\left(\frac{j\text{π}}{M}\right)-1}{cos\left(\frac{j\text{π}}{M}\right)+5},$ (2.39)

${\left({\stackrel{^}{\psi }}_{jk}^{n+1}\right)}_{yy}={\stackrel{^}{\psi }}_{jk}^{n+1}\frac{12}{{\left({h}_{y}\right)}^{2}}\frac{cos\left(\frac{k\text{π}}{N}\right)-1}{cos\left(\frac{k\text{π}}{N}\right)+5}.$ (2.40)

${\stackrel{^}{\psi }}_{jk}^{n+1}=\frac{{\stackrel{^}{g}}_{jk}}{1-\frac{i\tau }{4}\left(\frac{12\left(cos\left(\frac{j\text{π}}{M}\right)-1\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)}+\frac{12\left(cos\left(\frac{k\text{π}}{N}\right)-1\right)}{{h}_{y}^{2}\left(cos\left(\frac{k\text{π}}{N}\right)+5\right)}\right)}.$ (2.41)

1) 输入a，b，c，d，T，M，N，L，β，以及 ${\varphi }_{0}\left(x,y\right)$${h}_{x}$${h}_{y}$$\tau$

2) 计算 ${x}_{j}=a+j{h}_{x}$$j=0,1,\cdots ,M$${y}_{k}=c+k{h}_{y}$$k=0,1,\cdots ,N$${t}_{n}=n\tau$${\psi }_{0k}^{n}={\psi }_{Mk}^{n}=0$${\psi }_{j0}^{n}={\psi }_{jN}^{n}=0$$n=0,1,\cdots ,L$ 。 从 $n=0$ 出发；

3) 计算 ${f}_{jk}={|{\psi }_{jk}^{n}|}^{2}$ ，通过二维的Sine逆变换得 ${\stackrel{^}{f}}_{jk}$ ，利用关系式：

${\stackrel{^}{\varphi }}_{jk}^{n}=\frac{{\stackrel{^}{f}}_{jk}}{-\frac{12\left(cos\left(\frac{j\text{π}}{M}\right)-1\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)}-\frac{12\left(cos\left(\frac{k\text{π}}{N}\right)-1\right)}{{h}_{y}^{2}\left(cos\left(\frac{k\text{π}}{N}\right)+5\right)}},$ (2.42)

4) 由 ${\psi }_{jk}^{*}={\psi }_{jk}^{n}{\text{e}}^{-i\tau \left(V\left({x}_{j},{y}_{k}\right)+\beta {|\psi \left({x}_{j},{y}_{k},{t}_{n}\right)|}^{2}+{\varphi }_{jk}^{n}\right)}$ ，计算出 ${\psi }_{jk}^{*}$

5) 通过二维的Sine逆变换求出 ${\stackrel{^}{\psi }}_{jk}^{*}$ ，再利用关系式(2.39)，(2.40)，求出 ${\left({\stackrel{^}{\psi }}_{jk}^{*}\right)}_{xx}$${\left({\stackrel{^}{\psi }}_{jk}\right)}_{yy}$

6) 最后由关系式(2.41)，求出 ${\stackrel{^}{\psi }}_{jk}^{n+1}$ ，对其作二维的Sine变换得 ${\psi }_{jk}^{n+1}$ 。再令 $n=n+\text{1}$ ，重复3~6直到 $n=L$ 为止。

2.3. 三维非线性薛定谔泊松方程组的求解方法

$-i{\psi }_{t}=-\frac{1}{2}\Delta \psi +V\psi +\beta {|\psi |}^{2}+\varphi \psi ,$ (2.43)

$-\Delta \varphi \left(x,y,z,t\right)={|\psi |}^{2},$ (2.44)

$\psi \left(a,y,z,t\right)=\psi \left(b,y,z,t\right)=0,$

$\psi \left(x,c,z,t\right)=\psi \left(x,d,z,t\right)=0,$

$\psi \left(x,y,e,t\right)=\psi \left(x,y,f,t\right)=0,$

$\psi \left(x,y,z,0\right)={\varphi }_{0}\left(x,y,z\right),$

${u}_{ijp}=\underset{k=1}{\overset{M-1}{\sum }}\text{\hspace{0.17em}}\underset{l=1}{\overset{N-1}{\sum }}\text{\hspace{0.17em}}\underset{q=1}{\overset{P-1}{\sum }}\text{ }{\stackrel{^}{u}}_{klq}sin\left(\frac{ik\text{π}}{M}\right)sin\left(\frac{jl\text{π}}{N}\right)sin\left(\frac{qp\text{π}}{P}\right),$ (2.45)

${\stackrel{^}{u}}_{klq}=\frac{2}{M}\frac{2}{N}\frac{2}{P}\underset{i=1}{\overset{M-1}{\sum }}\text{\hspace{0.17em}}\underset{j=1}{\overset{N-1}{\sum }}\text{\hspace{0.17em}}\underset{p=1}{\overset{P-1}{\sum }}\text{ }{u}_{ijp}\mathrm{sin}\left(\frac{ik\text{π}}{M}\right)\mathrm{sin}\left(\frac{jl\text{π}}{N}\right)\mathrm{sin}\left(\frac{qp\text{π}}{P}\right),$ (2.46)

1) 考虑非线性部分： $i{\psi }_{t}=V\psi +\beta {|\psi |}^{2}\psi +\varphi \psi$ ，利用共轭变换与积分性质得：

$\psi \left(x,y,z,{t}_{*}\right)=\psi \left(x,y,z,{t}_{n}\right){\text{e}}^{-i\tau \left(V\left(x,y,z\right)+\beta {|\psi \left(x,y,z,{t}_{n}\right)|}^{2}+\varphi \left(x,y,z,{t}_{n}\right)\right)}.$ (2.47)

$-{\varphi }_{xx}-{\varphi }_{yy}-{\varphi }_{zz}={|{\psi }_{jkl}^{n}|}^{2},$

$-{\stackrel{^}{\varphi }}_{xx}-{\stackrel{^}{\varphi }}_{yy}-{\stackrel{^}{\varphi }}_{zz}={|{\stackrel{^}{\psi }}_{jkl}^{n}|}^{2},$ (2.48)

${\stackrel{^}{f}}_{jlk}^{n}={|{\varphi }_{jkl}^{n}|}^{2}$ ，由基于Sine变换的四阶紧差分格式有如下关系式：

${\left({\stackrel{^}{\psi }}_{jkl}^{n}\right)}_{xx}={\stackrel{^}{\psi }}_{jlk}^{n}\frac{12\left(cos\left(\frac{j\text{π}}{M}\right)-1\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)},$

${\left({\stackrel{^}{\psi }}_{jkl}^{n}\right)}_{yy}={\stackrel{^}{\psi }}_{jlk}^{n}\frac{12\left(cos\left(\frac{k\text{π}}{N}\right)-1\right)}{{h}_{y}^{2}\left(cos\left(\frac{k\text{π}}{N}\right)+5\right)},$

${\left({\stackrel{^}{\psi }}_{jkl}^{n}\right)}_{zz}={\stackrel{^}{\psi }}_{jlk}^{n}\frac{12\left(cos\left(\frac{l\text{π}}{P}\right)-1\right)}{{h}_{z}^{2}\left(cos\left(\frac{l\text{π}}{P}\right)+5\right)}.$ (2.49)

${\stackrel{^}{\varphi }}_{jkl}^{n}=\frac{{\stackrel{^}{f}}_{jkl}^{n}}{-\left(\frac{12\left(cos\left(\frac{j\text{π}}{M}\right)-1\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)}+\frac{12\left(cos\left(\frac{k\text{π}}{N}\right)-1\right)}{{h}_{y}^{2}\left(cos\left(\frac{k\text{π}}{N}\right)+5\right)}+\frac{12\left(cos\left(\frac{l\text{π}}{P}\right)-1\right)}{{h}_{z}^{2}\left(cos\left(\frac{l\text{π}}{P}\right)+5\right)}\right)}.$ (2.50)

2) 考虑线性部分方程的求解： $i{\psi }_{t}=-\frac{1}{2}\Delta \psi$ ，其中 ${\psi }_{jlk}^{*}$ 作为起始值，在时间方向上，采用crank-nicolson

$i\frac{{\psi }_{jkl}^{n+1}-{\psi }_{jkl}^{*}}{\tau }=-\frac{1}{4}\left(\Delta {\psi }_{jkl}^{n+1}+\Delta {\psi }_{jkl}^{*}\right),$

${\psi }_{jkl}^{n+1}-\frac{i\tau }{4}\Delta \left({\psi }_{jkl}^{n+1}\right)={\psi }_{jkl}^{*}+\frac{i\tau }{4}\Delta \left({\psi }_{jkl}^{*}\right),$

${\stackrel{^}{\psi }}_{jkl}^{n+1}-\frac{i\tau }{4}\left[{\left({\stackrel{^}{\psi }}_{jkl}^{n+1}\right)}_{xx}+{\left({\stackrel{^}{\psi }}_{jkl}^{n+1}\right)}_{yy}+{\left({\stackrel{^}{\psi }}_{jkl}^{n+1}\right)}_{zz}\right]={\stackrel{^}{g}}_{jkl},$ (2.51)

${\stackrel{^}{g}}_{jkl}={\stackrel{^}{\psi }}_{jkl}^{*}+\frac{i\tau }{4}\left[{\left({\stackrel{^}{\psi }}_{jkl}^{*}\right)}_{xx}+{\left({\stackrel{^}{\psi }}_{jkl}^{*}\right)}_{yy}+{\left({\stackrel{^}{\psi }}_{jkl}^{*}\right)}_{zz}\right].$ (2.52)

${\left({\stackrel{^}{\psi }}_{jkl}^{*}\right)}_{xx}={\stackrel{^}{\psi }}_{jkl}^{*}\frac{12\left(cos\left(\frac{j\text{π}}{M}\right)-1\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)},$

${\left({\stackrel{^}{\psi }}_{jkl}^{*}\right)}_{yy}={\stackrel{^}{\psi }}_{jkl}^{*}\frac{12\left(cos\left(\frac{k\text{π}}{N}\right)-1\right)}{{h}_{y}^{2}\left(cos\left(\frac{k\text{π}}{N}\right)+5\right)},$

${\left({\stackrel{^}{\psi }}_{jkl}^{*}\right)}_{zz}={\stackrel{^}{\psi }}_{jkl}^{*}\frac{12\left(cos\left(\frac{l\text{π}}{P}\right)-1\right)}{{h}_{z}^{2}\left(cos\left(\frac{l\text{π}}{P}\right)+5\right)}.$

${\left({\stackrel{^}{\psi }}_{jkl}^{n+1}\right)}_{xx}={\stackrel{^}{\psi }}_{jkl}^{n+1}\frac{12\left(cos\left(\frac{j\text{π}}{M}\right)-1\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)},$

${\left({\stackrel{^}{\psi }}_{jkl}^{n+1}\right)}_{yy}={\stackrel{^}{\psi }}_{jkl}^{n+1}\frac{12\left(cos\left(\frac{k\text{π}}{N}\right)-1\right)}{{h}_{y}^{2}\left(cos\left(\frac{k\text{π}}{N}\right)+5\right)},$

${\left({\stackrel{^}{\psi }}_{jkl}^{n+1}\right)}_{zz}={\stackrel{^}{\psi }}_{jkl}^{n+1}\frac{12\left(cos\left(\frac{l\text{π}}{P}\right)-1\right)}{{h}_{z}^{2}\left(cos\left(\frac{l\text{π}}{P}\right)+5\right)}.$

${\stackrel{^}{\psi }}_{jkl}^{n+1}=\frac{{\stackrel{^}{g}}_{jkl}}{1-\frac{i\tau }{4}\left(\frac{12\left(\mathrm{cos}\left(\frac{j\text{π}}{M}\right)-1\right)}{{h}_{x}^{2}\left(cos\left(\frac{j\text{π}}{M}\right)+5\right)}+\frac{12\left(\mathrm{cos}\left(\frac{k\text{π}}{N}\right)-1\right)}{{h}_{y}^{2}\left(cos\left(\frac{k\text{π}}{N}\right)+5\right)}+\frac{12\left(\mathrm{cos}\left(\frac{l\text{π}}{P}\right)-1\right)}{{h}_{z}^{2}\left(cos\left(\frac{l\text{π}}{P}\right)+5\right)}\right)},$ (2.53)

1) 输入a，b，c，d，M，N，P，L，e，f，T，β，以及 ${\varphi }_{0}\left(x,y,z\right)$ ，hx，hy，hz，τ；

2) 计算 ${x}_{j}=a+j{h}_{x}$$j=0,1,\cdots ,M$${y}_{k}=c+k{h}_{y}$$k=0,1,\cdots ,N$${z}_{l}=e+l{h}_{z}$$l=0,1,\cdots ,P$${t}_{n}=n\tau$${\psi }_{0kl}^{n}={\psi }_{Mkl}^{n}=0$${\psi }_{j0l}^{n}={\psi }_{jNl}^{n}=0$${\psi }_{jk0}^{n}={\psi }_{jkP}^{n}=0$ 。从 $n=0$ 出发；

3) 计算 ${f}_{jkl}={|{\psi }_{jkl}^{n}|}^{2}$ ，通过三维Sine逆变换得 ${\stackrel{^}{f}}_{jkl}$ ，利用关系式(2.50)求出 ${\stackrel{^}{\varphi }}_{jkl}^{n}$ ，再做三维Sine变换得 ${\varphi }_{jkl}^{n}$

4) 由 ${\psi }_{jkl}^{*}={\psi }_{jkl}^{n}{\text{e}}^{-i\tau \left(V\left({x}_{j},{y}_{k},{z}_{l}\right)+\beta {|\psi \left({x}_{j},{y}_{k},{z}_{l},{t}_{n}\right)|}^{2}+{\varphi }_{jkl}^{n}\right)}$ ，计算出 ${\psi }_{jkl}^{*}$

5) 通过三维Sine逆变换求出 ${\stackrel{^}{\psi }}_{jkl}^{*}$ ，利用 ${\stackrel{^}{\psi }}_{jkl}^{*}$${\left({\stackrel{^}{\psi }}_{jkl}^{*}\right)}_{xx}$${\left({\stackrel{^}{\psi }}_{jkl}^{*}\right)}_{xx}$${\left({\stackrel{^}{\psi }}_{jkl}^{*}\right)}_{xx}$ 的关系，求出(2.52)中的 ${\stackrel{^}{g}}_{jkl}$

6) 利用方程(2.53)得 ${\stackrel{^}{\psi }}_{jkl}^{n+1}$ ，对其做三维Sine变换得 ${\psi }_{jkl}^{n+1}$ 。再令 $n=n+\text{1}$ ，重复3~6直到 $n=L$ 为止。

3. 数值计算例子

3.1. 一维情形

$V=V\left(x\right)=\frac{1}{2}{x}^{2}$${\varphi }_{0}\left(x\right)=\frac{1}{{\left(2\text{π}\right)}^{\frac{1}{4}}}{\text{e}}^{\frac{-{x}^{2}}{4}}$ 。通过改变空间步长h的大小做误差分析，在计算中我们取M = 512或空间步长 $h=\frac{1}{32}$ 所得的近似解看作是方程的准确解 ${u}_{exact}\left(x,t\right)$ 。我们通过第2.1节中所介绍的一维问题算法来求解，编写的代码在matlab软件中进行演算。误差定义 $error=\underset{0\le j\le M}{\mathrm{max}}|u\left({x}_{j},{t}_{n}=1\right)-{u}_{exact}\left({x}_{j},{t}_{n}=1\right)|$

3.2. 二维情形

$V=V\left(x,y\right)=\frac{1}{2}\left({x}^{2}+{y}^{2}\right)$ ，初值函数为 ${\varphi }_{0}\left(x,y\right)=\frac{1}{\sqrt{2\text{π}}}{\text{e}}^{\frac{-\left({x}^{2}+{y}^{2}\right)}{4}}$ 。通过改变空间步长hx,hy的大小做误差分析，在计算中我们取 $M=N=\text{512}$ 或空间步长 ${h}_{x}={h}_{y}=\frac{1}{32}$ 所得的近似解看作是方程的准确解 ${u}_{exact}\left(x,y,t\right)$ 。我

$error=\underset{0\le j\le M,0\le k\le P}{\mathrm{max}}|u\left({x}_{j},{y}_{k},{t}_{n}=1\right)-{u}_{exact}\left({x}_{j},{y}_{k},{t}_{n}=1\right)|$

Table 1. Error analysis of one-dimensional space calculation at time t = 1

Figure 1. The trend of the total number of particles N(t), impulse P(t), and total energy E(t) over time

Figure 2. Image of density function $|\psi \left(x,t\right)|$ at different times: (a) t = 2.5; (b) t = 5; (c) t = 7.5; (d) t = 10

Table 2. Error analysis of two-dimensional space calculation at time t = 1

Figure 3. The trend of the total number of particles N(t), the component of the impulse Px(t) Py(t), and the total energy E(t) over time

Figure 4. Image of density function $|\psi \left(x,y,t\right)|$ at different times: (a) t = 2.5; (b) t = 5; (c) t = 7.5; (d) t = 10

3.3. 3.3. 三维情形

$V=V\left(x,y,z\right)=\frac{1}{2}\left({x}^{2}+{y}^{2}+{z}^{2}\right)$ ，初值函数为 ${\varphi }_{0}\left(x,y,z\right)=\frac{1}{2{\text{π}}^{3/4}}{\text{e}}^{\frac{-\left({x}^{2}+{y}^{2}+{z}^{2}\right)}{4}}$ 。通过改变空间步长hx，hy，hz的大小做误差分析，在计算中我们取 $M=N=P=512$ 或空间步长 $h=\frac{1}{32}$ ，所得的近似解看作是方程的准确解 ${u}_{exact}\left(x,y,z,t\right)$ 。我们通过2.2节中 所介绍的二维问题算法来求解， 编写的代码在matlab软件中进行演算。误差定义 $error=\underset{0\le j\le M,0\le k\le N,0\le l\le P}{\mathrm{max}}|u\left({x}_{j},{y}_{k},{z}_{l},{t}_{n}=1\right)-{u}_{exact}\left({x}_{j},{y}_{k},{z}_{l},{t}_{n}=1\right)|$

Figure 5. The trend of the total number of particles N(t), the component of the impulse Px(t), Py(t), Pz(t), and the total energy E(t) over time

Figure 6. Density function $|\psi \left(0,y,z,t\right)|$ images at different times: (a) t = 5, (c) t = 10; Density function $|\psi \left(x,0,z,t\right)|$ images at different times: (b) t = 5, (d) t = 10

4. 结论

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