## 3 Results
### Determination of Absorption Inhomogeneity  
We perform a numerical experiment of diffuse optical tomography using the toy model. Let us consider when the iterative scheme fails. We see \(F^{\prime}(0)=\left.\frac{\partial}{\partial a}u(\mathbf{\rho}^{j},t^{k};\mathbf{\rho}^{ j};a)\right|_{a=0}=0\). For sufficiently small \(\epsilon>0\), we have \(F^{\prime}(\epsilon)>0\), \(F^{\prime}(-\epsilon)<0\), \(|F^{\prime}(\pm\epsilon)|\simeq 0\), and \(U-F(\pm\epsilon)<0\). Therefore, if we start the iteration from \(a_{0}=\epsilon\), we obtain  
\[|a_{1}|<\epsilon,\quad|a_{2}|<|a_{1}|,\quad\ldots. \tag{44}\]  
Thus, the sequence \(a_{k}\) approaches 0 and can never arrive at \(\tilde{a}\) (\(=1.5\)). Let us consider how the difference \(U-F(a)\) depends on \(a\). We introduce  
\[h(t)=\frac{1}{t}\exp\left(-\frac{y_{0}^{2}}{4 Dct}\right)\left[1-\frac{\sqrt{ \pi Dct}}{\ell}e^{\left(\frac{y_{0}}{2\sqrt{Dct}}+\frac{\sqrt{Dct}}{\ell} \right)^{2}}\text{erfc}\left(\frac{y_{0}}{2\sqrt{Dct}}+\frac{\sqrt{Dct}}{\ell }\right)\right]. \tag{45}\]  
The following form is obtained using Equation (39). By neglecting noise, we have  
\[ \begin{split}& u(\mathbf{\rho}^{j},t^{k};\mathbf{\rho}^{j};\tilde{a})-u( \mathbf{\rho}^{j},t^{k};\mathbf{\rho}^{i};a)\\ &=\frac{\eta\epsilon^{-\mu_{0}ct}}{(2\pi D)^{2}}\int_{0}^{t^{k}}h (t^{k}-s)h(s)\left[\int_{-\infty}^{\infty}d_{\tilde{a}}(a,x^{\prime})\left(1- \tanh x^{\prime 2}\right)e^{-\frac{(x^{\prime}-x^{\prime})^{2}}{4 Dc(\tilde{a}^{ \prime}-s)}}e^{-\frac{(x^{\prime}-x^{\prime})^{2}}{4 Dc^{\prime}}}\,dx^{\prime }\right]\,ds,\end{split}  \tag{46}\]  
where \(d_{\tilde{a}}(a,x^{\prime})=\xi(\tilde{a},x^{\prime})-\xi(a,x^{\prime})\) with  
\[\xi(a,x^{\prime})=a^{2}\left[a+3\left(1+\frac{\tanh{x^{\prime 2}}}{10} \right)\right]. \tag{47}\]  
For a given \(x^{\prime}\), the function \(|d_{\tilde{a}}(a;x^{\prime})|^{2}\) has one global minimum at \(a=\tilde{a}\), one local minimum at \(a=-2\left(1+\frac{\tanh{x^{\prime 2}}}{10}\right)\), and one local maximum at \(a=0\). Therefore, the above expression implies that Algorithm 1 (LM) fails when the initial value \(a_{0}\) is negative and the correct value \(\tilde{a}=1.5\) is reconstructed only for a positive initial guess. Indeed, the computation ends up with \(-2.05\) if we start the iteration from \(a_{0}=-0.1\) (see below) or \(-0.01\), and the value \(1.68\) is obtained when \(a_{0}=0.01\). Figure 7 shows \(|d_{\tilde{a}}(a,x)|^{2}\) for \(\tilde{a}=1.5\) and \(\tanh(x^{2})=0.5\). In Figure 8, we plot reconstructed values of \(a\) against iteration numbers. The initial value is set to \(a_{0}=-0.1\). The reconstruction by Algorithm 1 (LM) is shown in the top panel of Figure 8. As we can predict from Figure 7, Algorithm 1 (LM) converges to the local minimum and can never arrive at the global minimum. Monte Carlo simulation can jump over the local maximum and approach the global minimum, but keeps fluctuating as shown in the middle panel of Figure 8. We initially set \((\sigma_{\epsilon},\epsilon)=(10^{-6},0.5)\) for Algorithms 2 and 3. In Algorithm 2, we set \((\sigma_{\epsilon},\epsilon)=(10^{-7},0.005)\) after the _temperature_ decreases at the \(k_{\text{b}}\)th Monte Carlo step. In the bottom panel of Figure 8, Algorithm 3 (hybrid) successfully arrives at the global minimum. Using Matlab, the computation time was \(17\,\text{min}\) while we needed \(3\,\text{hr}\) for Algorithm 2 (SA) (\(1000\) steps). \begin{table}
\begin{tabular}{l c c} \hline \hline  & **Case 1 (\(\mu_{a}\), \(\mu_{s}^{\prime}\))** & **Case 2 (\(\mu_{a}\), \(\mu_{s}^{\prime}\))** \\ \hline initial values & (0.01, 1.0) & (0.5, 1.0) \\ Algorithm 1 (LM) & (0.016, 0.63) & (0.068, 1.75) \\ Algorithm 2 (SA) & (0.032, 1.02) & (0.028, 0.92) \\ Algorithm 3 (hybrid) & (0.016, 0.63) & (0.016, 0.63) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Reconstructed values of \(\mu_{a},\mu_{s}^{\prime}\) are shown for Case 1 and Case 2. The units of \(\mu_{a},\mu_{s}^{\prime}\) are both mm\({}^{-1}\). For Algorithm 2, values at \(120\)th Monte Carlo step are shown.