#### 期刊菜单

Reversing Inverse Problem of Source Term of Heat Conduction Equation
DOI: 10.12677/AAM.2019.81012, PDF, HTML, XML, 下载: 1,179  浏览: 1,475

Abstract: In the paper, terminal observations are applied to the inverse problem of heat-transfer equations that rely on heat sources. This problem has important applications in the field of applied science. Based on the optimal control framework, the existence and necessary conditions of minimization of control functions are established. The Landweber iterative algorithm is applied in this problem, and the numerical simulation results are obtained.

1. 引言

${u}_{t}-\Delta u=f\left(x\right)x,\left(x,t\right)\in Q.$

${C}_{t}+\frac{1}{2}{\sigma }^{2}{S}^{2}{C}_{SS}+\left(r-q\right)S{C}_{S}-rC=0,$

$\left\{\begin{array}{ll}{u}_{t}-a\left(x\right){u}_{x}{}_{x}+b\left(x\right)u=f\left(x\right),\hfill & \left(x,t\right)\in Q=\left(0,l\right)×\left(0,T\right],\hfill \\ {u|}_{x=0}={u|}_{x=l}=0,\hfill & t\in \left(0,T\right],\hfill \\ {u|}_{t=0}=\phi \left(x\right),\hfill & x\in \left(0,l\right)，\hfill \end{array}$ (1)

${u|}_{t=T}=g\left(x\right),x\in \left[0,l\right],$

2. 控制问题

$J\left(\stackrel{˜}{f}\right)=\underset{a\in Α}{\mathrm{min}}J\left(f\right),$ (2)

$J\left(f\right)=\frac{1}{2}{\int }_{0}^{l}{|u\left(x,T;f\right)-g\left(x\right)|}^{2}\text{d}x+\frac{N}{2}{\int }_{0}^{l}{|\nabla f|}^{2}\text{d}x,$ (3)

$Α=\left\{f\left(x\right)||f|\le M,f\in {H}^{1}\left(0,l\right)\right\},$ (4)

$u\left(x,t;f\right)$ 是方程(1)对应于给定系数 $f\left(x\right)\in Α$ 的解，N是正则化参数，M是一个给定的常数。

$g\left(x\right)\in {L}^{2}\left(0,l\right).$ (5)

$\underset{\stackrel{¯}{Q}}{\mathrm{max}}|u|{\int }_{0}^{l}{u}^{2}\text{d}x\text{d}t+{\int }_{0}^{T}{\int }_{0}^{l}\left({|{u}_{t}|}^{2}+{|{u}_{xx}|}^{2}\right)\text{d}x\text{d}t\le C$ ，其中C为一个常数。

$\underset{f\in Α}{\mathrm{inf}}J\left(f\right)\le J\left({f}_{n}\right)\le \underset{f\in Α}{\mathrm{inf}}J\left(f\right)+\frac{1}{n},$

${f}_{n}\left(x\right)\to \stackrel{˜}{f}\left(x\right)\in {C}^{\frac{1}{2}}\left(0,l\right)$ ，得到 ${C}^{\alpha }\left(0,l\right)\left(0\le \alpha \le \frac{1}{2}\right)$

${u}_{n}\left(x,t\right)\to \stackrel{˜}{u}\left(x,t\right)$ ，得到 ${C}^{\alpha ,\frac{\alpha }{2}}\left(\stackrel{¯}{Q}\right)\cap {C}_{loc}^{2+2,1+\frac{\alpha }{2}}\left(Q\right)$

$J\left(\stackrel{˜}{f}\right)\le \underset{n\to \infty }{\mathrm{lim}}\mathrm{inf}J\left({f}_{n}\right)=\underset{f\in Α}{\mathrm{min}}J\left(f\right),$

3. 必要条件

(6)

$\left\{\begin{array}{ll}{v}_{t}-a\left(x\right){v}_{x}{}_{x}+bv=h-f,\hfill & \left(x,t\right)\in Q,\hfill \\ v\left(0,t\right)=v\left(l,t\right)=0,\hfill & t\in \left(0,T\right],\hfill \\ v\left(x,0\right)=0,\hfill & x\in \left(0,l\right),\hfill \end{array}$ (7)

$N{\int }_{0}^{l}\nabla f\cdot \nabla \left(h-f\right)\text{d}x+{\int }_{0}^{l}v\left(h-f\right)\text{d}x\ge 0.$ (8)

${J}_{\delta }\equiv J\left({f}_{\delta }\right)=\frac{1}{2}{\int }_{0}^{l}{|u\left(x,T;{f}_{\delta }\right)-g\left(x\right)|}^{2}\text{d}x+\frac{N}{2}{\int }_{0}^{l}{|\nabla {f}_{\delta }|}^{2}\text{d}x.$ (9)

${u}_{\delta }$ 是方程(1)的解，其中 $f={f}_{\delta }$ ，我们有

${\frac{\text{d}{J}_{\delta }}{\text{d}\delta }|}_{\delta =0}={{\int }_{0}^{l}\left[u\left(x,T;f\right)-g\left(x\right)\right]\frac{\partial {u}_{\delta }}{\partial \delta }|}_{\delta =0}\text{d}x+N{\int }_{0}^{l}\nabla f\cdot \nabla \left(h-f\right)\text{d}x\ge 0.$ (10)

${\stackrel{˜}{u}}_{\delta }\equiv \frac{\partial {u}_{\delta }}{\partial \delta }$ ，直接计算得到以下方程

$\left\{\begin{array}{l}{\stackrel{˜}{u}}_{\delta ,t}-a{\stackrel{˜}{u}}_{\delta ,xx}+b{\stackrel{˜}{u}}_{\delta }=h-f,\\ {{\stackrel{˜}{u}}_{\delta }|}_{x=0}=0,{{\stackrel{˜}{u}}_{\delta }|}_{x=l}=0,\\ {\stackrel{˜}{u}}_{\delta }\left(x,0\right)=0.\end{array}$ (11)

$\xi ={{\stackrel{˜}{u}}_{\delta }|}_{\delta =0}$ ，则 $\xi$ 满足

$\left\{\begin{array}{l}{\xi }_{t}-a{\xi }_{x}{}_{x}+b\xi =h-f,\hfill \\ \xi \left(0,t\right)=\xi \left(l,t\right)=0,\hfill \\ \xi \left(x,0\right)=0.\hfill \end{array}$ (12)

${\int }_{0}^{l}\left[u\left(x,T;f\right)-g\left(x\right)\right]\xi \left(x,T\right)\text{d}x+N{\int }_{0}^{l}\nabla f\cdot \nabla \left(h-f\right)\text{d}x\ge 0.$ (13)

$L\xi ={\xi }_{t}-a{\xi }_{x}{}_{x}-b\xi$ ，令v是下列问题的解

$\left\{\begin{array}{l}{L}^{\ast }v=-{v}_{t}-a{v}_{x}{}_{x}+bv=0,\hfill \\ v\left(x,T\right)=u\left(x,T\right)-g\left(x\right),\hfill \\ v\left(0,t\right)=v\left(l,t\right)=0.\hfill \end{array}$ (14)

$\begin{array}{c}0={\int }_{0}^{T}{\int }_{0}^{l}\xi {L}^{\ast }v\text{d}x\text{d}t=-{\int }_{0}^{l}\xi \left(x,T\right)\left[u\left(x,T\right)-g\left(x\right)\right]\text{d}x+{\int }_{0}^{T}{\int }_{0}^{l}vL\xi \text{d}x\text{d}t\\ =-{\int }_{0}^{l}\xi \left(x,T\right)\left[u\left(x,T\right)-g\left(x\right)\right]\text{d}x+{\int }_{0}^{T}{\int }_{0}^{l}v\left(h-f\right)\text{d}x\text{d}t\end{array}$ (15)

$N{\int }_{0}^{l}\nabla f\cdot \nabla \left(h-f\right)\text{d}x+{\int }_{0}^{l}v\left(h-f\right)\text{d}x\ge 0.$

4. 迭代格式步骤

${f}_{0}\left(x\right)=x\left(l-x\right),x\in \left(0,l\right)$ ;

$‖{u}_{1}\left(x,T\right)-g\left(x\right)‖\le \epsilon$

$‖{u}_{1}\left(x,T\right)-g\left(x\right)‖\ge \epsilon$

5. 数值实验

$\phi \left(x\right)=\left(1-{\text{π}}^{2}\right)\mathrm{sin}\text{π}x,\text{ }x\in \left[0,1\right].$

$g\left(x\right)=\left(2-{\text{π}}^{2}-{\text{e}}^{-{\text{π}}^{2}}\right)\mathrm{sin}\text{π}x,\text{ }x\in \left[0,1\right].$

Figure 1. Source item numerical inversion

${g}^{\delta }={u}^{\delta }\left(x,1\right)=u\left(x,1\right)\left[1+\delta ×random\left(x\right)\right]$

Figure 2. Numerical value inversion of noisy source term

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