## 6 Examples
### Example 3: Inverse Cauchy Problem and Boundary Value Problem  
Consider the inverse Cauchy problem:  
\[\Delta u=u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta\theta}=0, \tag{103}\] \[u(\rho,\theta)=h(\theta),\ 0\leq\theta\leq\pi,\] (104) \[u_{n}(\rho,\theta)=g(\theta),\ 0\leq\theta\leq\pi. \tag{105}\]  
The method of fundamental solutions is taken as  
\[u(\mathbf{x})=\sum_{j=1}^{n}c_{j}U(\mathbf{x},\mathbf{s}_{j}),\ \ \mathbf{s}_{j}\in\Omega^{c}. \tag{106}\]  
where  
\[U(\mathbf{x},\mathbf{s}_{j})=\ln r_{j},\ \ r_{j}=\|\mathbf{x}-\mathbf{s}_{j}\|. \tag{107}\]  
We consider  
\[u(x,y)=\cos x\cosh y+\sin x\sinh y, \tag{108}\] \[\rho(\theta)=\exp(\sin\theta)\sin^{2}(2\theta)+\exp(\cos\theta) \cos^{2}(2\theta), \tag{109}\]  
where \(u(x,y)\) is an exact solution of the Laplace equation, and \(\rho(\theta),\ 0\leq\theta\leq 2\pi\) describes the boundary contour in the polar coordinates.  
By GMRES with \(n=60\), through \(60\) iterations the residual is shown in Figure 1(a), which does not satisfy the convergence criterion with \(\varepsilon=10^{-2}\). If we run it to \(70\) steps, GMRES will blow up. In contrast, ROGMRES converges with \(72\) steps. As compared in Figure 1(b), ME obtained by ROGMRES is smaller than \(0.27\), as shown in Figure 1(c). However, GMRES obtains ME = \(0.5\) with a worse result.  
\begin{table}
\begin{tabular}{l l l l l l l l} \hline  & **GMRES(m)** & **ROGMRES** & **ROGMRES(m)** & **GMRES(\(\eta\))** & **GMRES(m,\(\eta\))** & **MPA(m)** & **A-GMRES(m)** \\ \hline ME & \(7.39\times 10^{-14}\) & \(7.22\times 10^{-14}\) & \(6.50\times 10^{-14}\) & \(8.22\times 10^{-14}\) & \(9.04\times 10^{-14}\) & \(5.66\times 10^{-14}\) & \(5.88\times 10^{-14}\) \\ \hline NS & \(28\) & \(44\) & \(18\) & \(16\) & \(16\) & \(10\) & \(15\) \\ \hline \end{tabular}
\end{table}
Table 1: For example 1, ME and NS obtained by different methods.  
\begin{table}
\begin{tabular}{l l l l l l l l} \hline  & **GMRES(m)** & **ROGMRES** & **OMPA** & **A-GMRES** & **ROGMRES(m)** & **GMRES(\(\eta\))** & **GMRES(m,\(\eta\))** \\ \hline ME & \(3.51\times 10^{-2}\) & \(9.77\times 10^{-14}\) & \(1.88\times 10^{-13}\) & \(3.36\times 10^{-9}\) & \(3.51\times 10^{-2}\) & \(3.65\times 10^{-2}\) & \(3.25\times 10^{-2}\) \\ \hline NS & \(>\)550 & \(51\) & \(43\) & \(52\) & \(>\)550 & \(45\) & \(163\) \\ \hline \end{tabular}
\end{table}
Table 2: For example 2, ME and NS obtained by different methods.  
Because the inverse Cauchy problem is a highly ill-posed problem, GMRES does not converge within a loose convergence criterion \(\varepsilon=10^{-2}\), and after 70 steps it will blow up. It is apparent that GMRES cannot solve the Cauchy problem adequately. The ROGMRES method is stable for the Cauchy problem, and can improve the accuracy to 0.27; however, it leaves room to improve the methods based on the Krylov subspace for solving the Cauchy problem.  
Next, we consider the mixed boundary value problem. Under \(\varepsilon=10^{-3}\), for GMRES(m) with \(m_{0}=3\) and \(m_{1}=10\), as shown in Figure 3a, \(\eta_{k}\) after 12 steps is irregular; it causes the residuals to decrease non-monotonically, as shown in Figure 3b. The GMRES(m) does not converge within 200 steps, and obtains an incorrect solution with ME = 0.92. If \(\eta_{k}<1/2\) is used as a stopping criterion, through 14 steps, ME = \(6.64\times 10^{-3}\) is obtained. When the ROGMRES method is applied, we obtain ME = \(4.16\times 10^{-4}\) with 187 steps. As shown in Figure 3a by a dashed line, \(\eta_{k}=1\) is kept well, and the residuals monotonically decrease, as shown in Figure 3b by a dashed line for ROGMRES. ME = \(4.87\times 10^{-4}\) is obtained through 167 steps by OMPA with \(m_{0}=4\), which is convergent faster than ROGMRES. ME = \(4.09\times 10^{-4}\) is obtained through 109 steps by A-GMRES with \(m_{0}=4\), which is convergent faster than ROGMRES and OMPA.  
Figure 2: For example 3 of the inverse Cauchy problem: (**a**) the residuals, (**b**) comparing solutions, and (**c**) the errors obtained by GMRES and ROGMRES.