﻿ 一维Boussinesq方程反问题的不适定性实例构建 The Ill-Posed Example Construction of Inverse Problems of One-Dimensional Boussinesq Equation

Pure Mathematics
Vol. 09  No. 04 ( 2019 ), Article ID: 30832 , 5 pages
10.12677/PM.2019.94064

The Ill-Posed Example Construction of Inverse Problems of One-Dimensional Boussinesq Equation

Yuhong Jin, Leihao Zuo*

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

Received: May 27th, 2019; accepted: Jun. 6th, 2019; published: Jun. 19th, 2019

ABSTRACT

The ill-posed nature of the inverse problem includes both the ill-posed nature of the problem itself and the ill-posed nature of the numerical algorithm. In this paper, we consider the ill-posedness of the inverse problem of one-dimensional Boussinesq equation, namely, the uniqueness of the solution. This paper points out the correct way to deal with the additional conditions when solving the inverse problem, and constructs four relatively simple examples, which are used not only to illustrate the unfitness of the inverse problem, but also to carry out subsequent numerical simulation calculation with the help of these four examples.

Keywords:Inverse Problems, Boussinesq Equation, Ill-Posedness

Copyright © 2019 by author(s) and Hans Publishers Inc.

1. 问题描述

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

2. 一维Boussinesq方程反问题的不适定性实例构建

2.1. 实例一

$\left\{\begin{array}{l}{\phi }_{1}\left(x\right)=x{\text{e}}^{-x}\\ {g}_{1}\left(t\right)=t{\text{e}}^{-t}\\ {g}_{2}\left(t\right)=\left(1+t\right){\text{e}}^{\left(-1-t\right)}\\ {\phi }_{2}\left(x\right)=\left(x+1\right){\text{e}}^{\left(-x-1\right)}\end{array}$ (2)

$f\left(x,t\right)=-{\text{e}}^{-2u}\left(6{u}^{2}-12u+3\right){x}^{2}+{\text{e}}^{-2u}\left(2{u}^{2}+2u-2\right)x+{\text{e}}^{-2u}\left(2u-1\right)+{\text{e}}^{-u}\left(1-u\right)$

$\left(h\frac{\partial h}{\partial x}\right){k}^{\prime }\left(x\right)+\left(\frac{\partial }{\partial x}\left(h\frac{\partial h}{\partial x}\right)\right)k\left(x\right)=\frac{\partial h}{\partial t}-f\left(x,t\right)$ (3)

${k}^{\prime }\left(x\right)-2k\left(x\right)=2x-6{x}^{2}$ (4)

2.2. 实例二

$\left\{\begin{array}{l}{\phi }_{1}\left(x\right)=\left(x-{x}^{2}+1\right){\text{e}}^{-x}\\ {g}_{1}\left(t\right)={\text{e}}^{-2t}\\ {g}_{2}\left(t\right)={\text{e}}^{-1-2t}\\ {\phi }_{2}\left(x\right)=\left(x-{x}^{2}\right){\text{e}}^{-1-x}+{\text{e}}^{-2-x}\end{array}$ (5)

$h\left(x,t\right)=\left(x-{x}^{2}\right){\text{e}}^{-x-t}+{\text{e}}^{-x-2t}$

$\left\{\begin{array}{l}\frac{\partial h}{\partial t}=x\left(x-1\right){\text{e}}^{\left(-x-t\right)}-2{\text{e}}^{\left(-x-2t\right)}\\ h\frac{\partial h}{\partial x}=-{\text{e}}^{-2x-2t}\left({\text{e}}^{-t}-{x}^{2}+x\right)\left({\text{e}}^{-t}-{x}^{2}+3x-1\right)\\ \frac{\partial h}{\partial x}\left(h\frac{\partial h}{\partial x}\right)={\text{e}}^{-2x-2t}{\left({\text{e}}^{-t}-{x}^{2}+3x-1\right)}^{2}+{\text{e}}^{-2x-2t}\left({\text{e}}^{-t}+x-{x}^{2}\right)\left({\text{e}}^{-t}-{x}^{2}+5x-4\right)\end{array}$ (6)

$f\left(x,t\right)={\text{e}}^{-x-t}\left(\left({x}^{2}-x\right)+{\text{e}}^{-t}\left(-3+9x-16{x}^{2}+8{x}^{3}-{x}^{4}\right)+{\text{e}}^{-2t}\left(5-8x+2{x}^{2}\right)-{\text{e}}^{-3t}\right)$

2.3. 实例三

$\frac{\text{d}}{\text{d}x}\left(k\left(x\right)h\left(x\right)\frac{\text{d}h}{\text{d}x}\right)+f\left(x\right)=0$ (7)

$\left\{\begin{array}{l}{\phi }_{1}\left(x\right)={\text{e}}^{-x}\\ {g}_{1}\left(t\right)=1\\ {g}_{2}\left(t\right)={\text{e}}^{-1}\\ {\phi }_{2}\left(x\right)={\text{e}}^{-x}\\ f\left(x,t\right)={\text{e}}^{-2x}\left(-6{x}^{2}+2x\right)\end{array}$ (8)

$k\left(x\right)=1+2x+3{x}^{2}+C\text{e}{}^{2x}$ (9)

2.4. 实例四

${h}_{1}{{h}^{\prime }}_{2}=\left({h}_{1}{{h}^{\prime }}_{1}{h}_{2}^{2}\right){k}^{\prime }+{\left({h}_{1}{{h}^{\prime }}_{1}\right)}^{\prime }{h}_{2}^{2}k+f\left(x,t\right)$ (10)

${h}_{2}\left(t\right)=\int f\left(t\right)\text{d}t$ (11)

$k\left(x\right)=\frac{1}{x+1}\int K\left(x\right)\text{d}x$ (12)

3. 总结

The Ill-Posed Example Construction of Inverse Problems of One-Dimensional Boussinesq Equation[J]. 理论数学, 2019, 09(04): 481-485. https://doi.org/10.12677/PM.2019.94064

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6. NOTES

*通讯作者。