Recovery of a space-dependent vector source in thermoelastic systems

Figures & data

Table 1. The stopping iteration number $k=k(e,\mathit{\kappa })$ for Experiment 1 given by (1919 $$\begin{array}{c}\hfill E_{k,{\mathbf{u}}_T}=∥{\mathbf{v}}_k^e(·,T)-{\tilde{\mathit{\xi }}}_T^e∥.\end{array}$$19 ), with $e(\tilde{e})$ given by (2323 $$\begin{array}{c}\hfill ∥{\mathit{\xi }}_T-{\mathit{\xi }}_T^e∥\approx e(\tilde{e})=\left\{\begin{array}{cc}0.0047\hfill & \tilde{e}=1\%\hfill \\ 0.0148\hfill & \tilde{e}=3\%\hfill \\ 0.0222\hfill & \tilde{e}=5\%.\hfill \end{array}\right)\end{array}$$23 ).

$\mathit{\kappa }\backslash \tilde{e}$	1%	3%	5%
$1$	$151$	$108$	$107$
$10$	$14$	$10$	$10$
$50$	$3$	$2$	$2$

Figure 1. The exact solution (2424 $$\begin{array}{c}\hfill p(x)=\left\{\begin{array}{cc}0\hfill & \text{for}0⩽x<\frac{1}{3}\hfill \\ 1\hfill & \text{for}\frac{1}{3}⩽x<\frac{2}{3}\hfill \\ 0\hfill & \text{for}\frac{2}{3}<x⩽1\hfill \end{array}\right)\end{array}$$24 ) and the numerical solution for the source (2222 $$\begin{array}{cc}\hfill u(x,t)& ={(1+t)}^{2}x(x-1),\hfill \\ \hfill \mathit{\theta }(x,t)& =(1+t)x(1-x),\hfill \\ \hfill p(x)& =x(x-1).\hfill \end{array}$$22 ) for $\tilde{e}=1\%$ and $\tilde{e}=5\%$ for different values of $\mathit{\kappa }$.

Figure 2. The exact solution (2424 $$\begin{array}{c}\hfill p(x)=\left\{\begin{array}{cc}0\hfill & \text{for}0⩽x<\frac{1}{3}\hfill \\ 1\hfill & \text{for}\frac{1}{3}⩽x<\frac{2}{3}\hfill \\ 0\hfill & \text{for}\frac{2}{3}<x⩽1\hfill \end{array}\right)\end{array}$$24 ) and the numerical solution for the source (2424 $$\begin{array}{c}\hfill p(x)=\left\{\begin{array}{cc}0\hfill & \text{for}0⩽x<\frac{1}{3}\hfill \\ 1\hfill & \text{for}\frac{1}{3}⩽x<\frac{2}{3}\hfill \\ 0\hfill & \text{for}\frac{2}{3}<x⩽1\hfill \end{array}\right)\end{array}$$24 ) for $\tilde{e}=1\%$ and $\mathit{\kappa }=10$.