﻿ 一维Boussinesq方程反问题与正问题的联合求解 The United Solution for Positive and Inverse Problems of One-Dimensional Boussinesq Equation

Pure Mathematics
Vol. 08  No. 06 ( 2018 ), Article ID: 27523 , 6 pages
10.12677/PM.2018.86087

The United Solution for Positive and Inverse Problems of One-Dimensional Boussinesq Equation

Yuhong Jin, Leihao Zuo*

Department of Basic Courses, Naval University of Engineering, Wuhan Hubei

Received: Oct. 19th, 2018; accepted: Oct. 31st, 2018; published: Nov. 13th, 2018

ABSTRACT

In this paper, the inverse problem and the positive problem of the one-dimensional Boussinesq equation are solved jointly by using the finite difference method. In the solving process, the nonlinear least square method is used for optimization. The approximate solution is consistent with the truth value.

Keywords:Inverse Problems, Boussinesq Equation, Finite Difference Method, Nonlinear Least Square Method

1. 反问题的提出

$\left\{\begin{array}{l}\frac{\partial h}{\partial t}=\frac{\partial }{\partial x}\left[k\left(x\right)h\left(x,t\right)\frac{\partial h}{\partial x}\right]+f\left(x,t\right)\\ h\left(x,0\right)={\phi }_{1}\left(x\right)\\ h\left(0,t\right)={g}_{1}\left(t\right)\\ h\left(1,t\right)={g}_{2}\left(t\right)\\ 0\le x\le 1,0\le t\le 1\end{array}$

2. 有限差分法

2.1. 差分格式的建立

$\left\{\begin{array}{l}\frac{\partial h}{\partial t}=\frac{\partial }{\partial x}\left[k\left(x\right)h\left(x,t\right)\frac{\partial h}{\partial x}\right]+f\left(x,t\right)\\ h\left(x,0\right)={\phi }_{1}\left(x\right)\\ h\left(0,t\right)={g}_{1}\left(t\right)\\ h\left(1,t\right)={g}_{2}\left(t\right)\\ 0\le x\le 1,0\le t\le 1\end{array}$

$\frac{\partial }{\partial x}\left[k\left(x\right)h\left(x,t\right)\frac{\partial h}{\partial x}\right]=\frac{\partial k\left(x\right)}{\partial x}\ast h\left(x,t\right)\ast \frac{\partial h}{\partial x}+k\left(x\right)\ast \frac{\partial h\left(x,t\right)}{\partial x}\ast \frac{\partial h}{\partial x}+k\left(x\right)\ast h\left(x,t\right)\ast \frac{{\partial }^{2}h}{\partial {x}^{2}}$

$\left\{\begin{array}{l}{{h}^{\prime }}_{t}\left({x}_{i},{t}_{0}\right)=\frac{1}{2\tau }\left(-3h\left({x}_{i},{t}_{0}\right)+4h\left({x}_{i},{t}_{1}\right)-h\left({x}_{i},{t}_{2}\right)\right)\\ {{h}^{\prime }}_{t}\left({x}_{i},{t}_{j}\right)=\frac{1}{2\tau }\left(-h\left({x}_{i},{t}_{j-1}\right)+h\left({x}_{i},{t}_{j+1}\right)\right),\text{\hspace{0.17em}}0

$\frac{\partial k\left(x\right)}{\partial x}$ 离散化后的三点数值微分公式表示为：

$\frac{\partial h}{\partial x}$ 离散化后的三点数值微分公式：

$\left\{\begin{array}{l}{{h}^{\prime }}_{x}\left({x}_{0},{t}_{j}\right)=\frac{1}{2\omega }\left(-3h\left({x}_{0},{t}_{j}\right)+4h\left({x}_{1},{t}_{j}\right)-h\left({x}_{2},{t}_{j}\right)\right)\\ {{h}^{\prime }}_{x}\left({x}_{i},{t}_{j}\right)=\frac{1}{2\omega }\left(-h\left({x}_{j-1},{t}_{j}\right)+h\left({x}_{j+1},{t}_{j}\right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}0

$\frac{{\partial }^{2}h}{\partial {x}^{2}}$ 离散化后的三点数值微分公式：

$\left\{\begin{array}{l}{{h}^{″}}_{x}\left({x}_{0},{t}_{j}\right)=\frac{1}{{\omega }^{2}}\left(h\left({x}_{2},{t}_{j}\right)-2h\left({x}_{1},{t}_{j}\right)+h\left({x}_{0},{t}_{j}\right)\right)\\ {{h}^{″}}_{x}\left({x}_{i},{t}_{j}\right)=\frac{1}{{\omega }^{2}}\left(h\left({x}_{i+1},{t}_{j}\right)-2h\left({x}_{i},{t}_{j}\right)+h\left({x}_{i-1},{t}_{j}\right)\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}0

2.2. 方程组求解

3. 数值实验

Figure 1. $m=5,n=5$

Figure 2. $m=10,n=10$

Figure 3. $m=50,n=50$

Figure 4. Comparison

Table 1. Average residual sum of squares under different grid densities

The United Solution for Positive and Inverse Problems of One-Dimensional Boussinesq Equation[J]. 理论数学, 2018, 08(06): 650-655. https://doi.org/10.12677/PM.2018.86087

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