## General Propagation Lattice Boltzmann Model for a Variable-Coefficient Compound KdV-Burgers Equation

Zhang Zongning,, Li Chunguang,, Dong jianqiang

School of Mathematics and Information Science, North Minzu University, Yinchuan 750021

 基金资助: 国家自然科学基金.  11761005宁夏高等教育一流学科建设资助项目.  NXYLXK2017B09北方民族大学研究生创新项目.  YCX21156北方民族大学校级一般项目.  2020XYZSX02

 Fund supported: the NSFC.  11761005the First-Class Disciplines Foundation of Ningxia.  NXYLXK2017B09the Postgraduate Innovation Project of North Minzu University.  YCX21156the University-level Scientific Research Projects of North Minzu University.  2020XYZSX02

Abstract

This paper studies the numerical calculation method of a kind of general Kdv-Burgers equation with variable coefficients. Firstly, a lattice Boltzmann model of the generalized KdV-Burgers equation with variable coefficients is obtained by selecting the equilibrium distribution function and adding the correction function. The model could accurately recover the KdV-Burgers equation without any assumptions. Secondly, this paper studies the temporal and spatial change trend of the nonlinear high-order derivative term in the equation, and compares it with the analytical solution, and then gives an error analysis. Finally, This paper analyzes the precision of the space and time of the equation. According to the simulation experiment results, the model could reach 2nd order accuracy. Numerical results show that the current lattice Boltzmann model is a satisfactory and efficient algorithm.

Keywords： Lattice Boltzmann equation ; Chapman-Enskog analysis ; KdV-Burgers equation ; Variable coefficient

Zhang Zongning, Li Chunguang, Dong jianqiang. General Propagation Lattice Boltzmann Model for a Variable-Coefficient Compound KdV-Burgers Equation. Acta Mathematica Scientia[J], 2021, 41(5): 1283-1295 doi:

## 1 引言

$$$u_{t} -6f_{0}(t) uu_{x} +f_{2}(t) u_{xxx} -f_{3}(t) u_{x} +f_{4} (t)(Cu+xu_{x} ) = 0,$$$

$$$u_{t} +l(t)uu_{x} +m(t)u^{2p}u_{x} +n(t)u_{xx} -r(t)u_{xxx} +s(t)u_{x} = 0,$$$

$$$u(x, t) = \sum\limits_{i} f_{i}(x, t) = \sum\limits_{i} f_{i}^{(e q)}(x, t),$$$

$$$\sum\limits_{i} f_{i}^{(0)} = u, \quad \sum\limits_{i} f_{i}^{(n)} = 0, \quad n = 1, 2, \cdots.$$$

$$$\sum\limits_{i} h_{i} = \epsilon \sum\limits_{i} h_{i}^{(1)},$$$

$$$\sum\limits_{i} c_{i} h_{i} = \epsilon \sum\limits_{i} c_{i} h_{i}^{(1)},$$$

$$$\sum\limits_{i} c_{i}^{2} h_{i} = \epsilon \sum\limits_{i} c_{i}^{2}h_{i}^{(1)}.$$$

$$$\sum\limits_{i} c_{i} f_{i}^{(0)} = 0,$$$

$$$\sum\limits_{i} c_{i}^{2} f_{i}^{(0)} = \lambda_{1}n\left (t \right )u,$$$

$$$\sum\limits_{i} c_{i}^{3} f_{i}^{(0)} = \lambda_{2}r\left (t \right )u.$$$

$$$\sum\limits_{i} h_{i} = 0,$$$

$$$\sum\limits_{i} c_{i} h_{i} = \lambda _{3} l (t) u^{2}+\lambda _{4} m (t)u^{p+1},$$$

$$$\sum\limits_{i} c_{i}^{2} h_{i} = 0.$$$

$$$\frac{\partial u}{\partial t_{1}}+\frac{\partial}{\partial x_{1}}\left(\sum\limits_{i} c_{i} f_{i}^{(0)}\right) = \sum\limits_{i} h_{i}^{(1)} = 0, \ \mbox{ 即 } \quad \frac{\partial u}{\partial t_{1}} = 0.$$$

$$$\frac{\partial u}{\partial t_{2}}+\Delta t\left(\frac{1}{2}-\tau\right) \sum\limits_{i}\left(\frac{\partial}{\partial t_{1}}+c_{i} \frac{\partial}{\partial x_{1}}\right)^{2} f_{i}^{(0)}+\Delta t \tau \sum\limits_{i}\left(\frac{\partial}{\partial t_{1}}+c_{i} \frac{\partial}{\partial x_{1}}\right) h_{i}^{(1)} = 0,$$$

$\begin{eqnarray} \sum\limits_{i}\left(\frac{\partial}{\partial t_{1}}+c_{i} \frac{\partial}{\partial x_{1}}\right)^{2} f_{i}^{(0)} & = &\frac{\partial^{2}}{\partial t_{1}^{2}}\left(\sum\limits_{i} f_{i}^{(0)}\right)+2 \frac{\partial^{2}}{\partial t_{1} \partial x_{1}}\left(\sum\limits_{i} c_{i} f_{i}^{(0)}\right)+\frac{\partial^{2}}{\partial x_{1}^{2}}\left(\sum\limits_{i} c_{i}^{2} f_{i}^{(0)}\right) \\& = &\frac{\partial^{2}}{\partial x_{1}^{2}}\left(\sum\limits_{i} c_{i}^{2} f_{i}^{(0)}\right) = \lambda_{1} n(t)\frac{\partial ^{2} u}{\partial x_{1}^{2}}, \end{eqnarray}$

$\begin{eqnarray} \sum\limits_{i}\left(\frac{\partial}{\partial t_{1}}+c_{i} \frac{\partial}{\partial x_{1}}\right) h_{i}^{(1)} & = &\frac{\partial}{\partial t_{1}}\left(\sum\limits_{i} h_{i}^{(1)}\right)+\frac{\partial}{\partial x_{1}}\left(\sum\limits_{i} c_{i} h_{i}^{(1)}\right) \\ & = &\frac{\partial}{\partial x_{1}}\left[\lambda _{3} l (t) u^{2}+\lambda _{4}m (t)u^{p+1} \right] \\ & = &2 \lambda _{3} l (t)u\frac{\partial u}{\partial x_{1} } +(p+1)\lambda _{4}m(t) u^{p} \frac{\partial u}{\partial x_{1} }, \end{eqnarray}$

$$$\frac{\partial u}{\partial t_{2}}+\frac{\tau }{\epsilon } [2 \lambda _{3}\Delta t l(t) u \frac{\partial u}{\partial x_{1} }+(p+1)\lambda _{4}\Delta t n (t)u^{p}\frac{\partial u}{\partial x_{1} } ] +\Delta t(\frac{1}{2} -\tau )\lambda _{1} n(t)\frac{\partial^{2} u}{\partial x_{1}^{2} } = 0.$$$

$\begin{eqnarray} &&\frac{\partial u}{\partial t_{3}}+\Delta t^{2}\left(\tau^{2}-\tau+\frac{1}{6}\right) \sum\limits_{i}\left(\frac{\partial}{\partial t_{1}}+c_{i} \frac{\partial}{\partial x_{1}}\right)^{3} f_{i}^{(0)}\\&& +\Delta t^{2}\left(\frac{\tau}{2}-\tau^{2}\right) \sum\limits_{i}\left(\frac{\partial}{\partial t_{1}}+c_{i} \frac{\partial}{\partial x_{1}}\right)^{2} h_{i}^{(1)} = 0, \end{eqnarray}$

$\begin{eqnarray} \sum\limits_{i}\left(\frac{\partial}{\partial t_{1}}+c_{i} \frac{\partial}{\partial x_{1}}\right)^{3} f_{i}^{(0)} & = &\frac{\partial^{3}}{\partial t_{1}^{3}}\left(\sum\limits_{i} f_{i}^{(0)}\right)+3 \frac{\partial^{3}}{\partial x_{1} \partial t_{1}^{2}}\left(\sum\limits_{i} c_{i} f_{i}^{(0)}\right) \\&&+3 \frac{\partial^{3}}{\partial x_{1}^{2} \partial t_{1}}\left(\sum\limits_{i} c_{i}^{2} f_{i}^{(0)}\right)+\frac{\partial^{3}}{\partial x_{1}^{3}}\left(\sum\limits_{i} c_{i}^{3} f_{i}^{(0)}\right) \\& = &3\lambda _{1} \frac{\partial^{3} }{\partial x_{1}^{2} \partial t_{1} }[n(t) u] +\lambda _{2} r(t)\frac{\partial^{3} u}{\partial x_{1}^{3} }, \end{eqnarray}$

$\begin{eqnarray} \sum\limits_{i}\left(\frac{\partial}{\partial t_{1}}+c_{i} \frac{\partial}{\partial x_{1}}\right)^{2} h_{i}^{(1)}& = & \frac{\partial^{2}}{\partial t_{1}^{2}}\left(\sum\limits_{i} h_{i}^{(1)}\right)+2 \frac{\partial^{2}}{\partial x_{1} \partial t_{1}}\left(\sum\limits_{i} c_{i} h_{i}^{(2)}\right) +\frac{\partial^{2}}{\partial x_{1}^{2}}\left(\sum\limits_{i} c_{i}^{2} h_{i}^{(1)}\right) \\& = &\frac{2}{\epsilon }\frac{\partial^{2} }{\partial x_{1} \partial t_{1}}[\lambda _{3} l (t)u^{2} +\lambda _{4} m (t)u^{p+1} ], \end{eqnarray}$

$\begin{eqnarray} && \frac{\partial u}{\partial t_{3}}+\Delta t^{2}(\tau^{2}-\tau+\frac{1}{6})[3\lambda _{1} n(t)\frac{\partial^{3} u}{\partial x_{1}^{2} \partial t_{1} } +\lambda _{2} r(t)\frac{\partial^{3} u}{\partial x_{1}^{3} }]\\&&+\frac{1}{\epsilon }\Delta t^{2}(\tau-2\tau^{2} ) \frac{\partial^{2} }{\partial x_{1} \partial t_{1}}[\lambda _{3} l (t)u^{2} +\lambda _{4} m (t)u^{p+1}] = 0. \end{eqnarray}$

$\begin{eqnarray} &&u_{t}+2 \lambda _{3}\tau \Delta t l(t) u u_{x} +(p+1)\lambda _{4}\tau \Delta t m (t)u^{p}u_{x}+\Delta t(\frac{1}{2} -\tau )\lambda _{1} n(t)u_{xx}\\&&+\Delta t^{2}(\tau^{2}-\tau+\frac{1}{6})[3\lambda _{1}\epsilon u_{xx} \partial t_{1} n(t) +\lambda _{2} r(t)u_{xxx} ]\\& &+\epsilon \Delta t^{2}(\tau-2\tau^{2} ) [2\lambda _{3}uu_{x} \partial t_{1} l (t) +(p+1)\lambda _{4} u^{p}u_{x} \partial t_{1} m (t)]+O(\epsilon ^{4} ) = 0, \end{eqnarray}$

$$$\left\{\begin{array}{ll} { }\lambda _{1} = \frac{1}{\Delta t (\tau-0.5)}, \\ { }\lambda _{2} = \frac{1}{\Delta t^{2} (\tau^{2} -\tau+\frac{1}{6} )}, \\ { }\lambda _{3} = \frac{1}{2\Delta t \tau }, \\{ }\lambda _{4} = \frac{1}{(p+1)\Delta t \tau }. \end{array} \right.$$$

$$$u_{t} + l(t) u u_{x} + m (t)u^{p}u_{x}- n(t)u_{xx}+r(t)u_{xxx} = {\cal R},$$$

$$$f_{4}^{\left ( 0 \right ) } = \sigma u,$$$

$\begin{eqnarray} f_{i}^{(0)} = \left\{\begin{array}{lll}{ } {} [1+6\sigma -\frac{n(t)}{c^{2}\Delta t\left ( \tau -0.5 \right ) } +\frac{r(t)}{2c^{3}\Delta t^{2} \left (\tau ^{2} -\tau +\frac{1}{6} \right ) } ]u, \ \ &i = 0, \\{ } {} [-4\sigma +\frac{n(t)}{2 c^{2}\Delta t\left ( \tau -0.5 \right ) } -\frac{r(t)}{2 c^{3}\Delta t^{2} \left (\tau ^{2} -\tau +\frac{1}{6} \right )} ]u, \ \ &i = 1, \\ { } {}[-4\sigma +\frac{n(t)}{2 c^{2}\Delta t\left ( \tau -0.5 \right ) } -\frac{r(t)}{6 c^{3}\Delta t^{2} \left (\tau ^{2} -\tau +\frac{1}{6} \right )} ]u, \ \ &i = 2, \\{ }{} [\sigma +\frac{r(t)}{6 c^{3}\Delta t^{2} \left (\tau ^{2} -\tau +\frac{1}{6} \right )} ]u, \ \ &i = 3, \\ \sigma u, \ \ &i = 4.\end{array}\right. \end{eqnarray}$

$$$h_{i } = \left\{\begin{array}{ll} { }\frac{1}{2} {\cal H}, &i = 0, \\{ } -\frac{1}{3} {\cal H}, &i = 1, \\{ } -\frac{1}{3} {\cal H}, &i = 2, \\{ } \frac{1}{3} {\cal H}, &i = 3, \\{ }-\frac{1}{6} {\cal H}, &i = 4.\end{array}\right.$$$

$$${\cal H} = [\frac{l (t)}{2} u^{2} +\frac{m(t)}{p+1} u^{p+1}]/\left (\Delta t \tau c \right ).$$$

$$$\tau _{1} = \tau ^{2} -\tau +\frac{1}{6},$$$

$$$k = \frac{r(t)}{c^{3}\Delta t^{2} \tau _{1} }.$$$

$$$\tau (t) = \frac{1}{2} +\sqrt{\frac{1}{12} +\frac{\Delta t}{\Delta x^{3}k } r(t) },$$$

$$$\frac{1}{12} +\frac{\Delta t}{\Delta x^{3}k } r(t)\ge 0\Rightarrow k\ge \frac{-12\Delta t r(t)}{\Delta x^{3} }.$$$

(1) 参数$k$是单松弛时间$\tau$的调节因子, 一般取值为$0.0<k<1.5$ (参见文献[11]);

(2) 常数$\sigma$是平衡函数的一个调整因子, 它可以在演化过程中调整模拟结果, 一般$\sigma$是一个负的微小值, $-\frac{1}{12} <\sigma <-0.01$ (参见文献[12]).

## 3 数值模拟

$$$E_{2} = \frac{1}{N} \sqrt{\sum\limits_{j = 1}^{N}\left [ u\left ( x_{j} , t \right )-u^{*} \left ( x_{j}, t \right ) \right ]^{2} },$$$

$$$E_{\infty } = \max\limits_{j = 0, 1, 2\dots, N }\left | u\left ( x_{j} , t \right )-u^{*} \left ( x_{j}, t \right ) \right |,$$$

$$$GRE = \frac{ { \sum\limits_{j = 1}^{N}}\left | u\left ( x_{j} , t \right )-u^{*} \left ( x_{j}, t \right ) \right | }{ { \sum\limits_{j = 1}^{N} \left | u^{*} \left ( x_{j} , t \right ) \right | } },$$$

### 图 1

 Parameters t=0.5 t=1.0 t=1.5 t=2.0 E2 5.1836e-04 4.9568e-04 6.1178e-04 7.6003e-04 E∞ 0.0364 0.0349 0.0478 0.0585 GRE 0.0334 0.0303 0.0369 0.0429

### 图 3

$$$u_{t} +5 \alpha u^{p} u_{x} = \beta t^{\frac{2}{n}-1 } u_{xx}.$$$

### 图 6

 Parameters t=1.0 t=2.0 t=3.0 E2 1.0660e-04 7.9380e-05 7.0678e-05 E∞ 0.0230 0.0151 0.0125 GRE 0.0343 0.0267 0.0252

$$$u_{t} -15\cos\left ( \pi t/2 \right ) u_{x} -6uu_{x} +0.06u^{p} u_{x} = 0.01u_{xxx}.$$$

$p = 2$时, 方程的解析解可以表示为

 Parameters t=1.0 t=2.0 t=3.0 E2 2.2805e-04 2.8858e-04 2.2420e-04 E∞ 0.0043 0.0055 0.0043 GRE 0.0593 0.0750 0.0583

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