# 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

DOI: 10.12677/PM.2019.94064, PDF, HTML, XML, 下载: 342  浏览: 516

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.

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

NOTES

*通讯作者。

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