## 5 Application
\begin{array}{cccccc}2(\lambda+1)&-1&0&\cdots&\cdots&0\\ -1&2(\lambda+1)&-1&\ddots&&\vdots\\ 0&-1&2(\lambda+1)&-1&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&\ddots&0\\ \vdots&&&\ddots&\ddots&\ddots&-1\\ 0&\cdots&\cdots&0&-1&2(\lambda+1)\end{array} \right)_{N\times N},\]  
\(C=-\lambda I\), \(\lambda=\left(\frac{h}{2}\right)^{2}\), \(I\) is the identity matrix in \(\mathbb{R}^{N}\), \(\mathbf{x}=(x_{1},x_{2},\ldots,x_{m})^{t}\), \(q(\mathbf{x})=\left(x_{1}^{5/3},x_{2}^{5/3},\ldots,x_{m}^{5/3}\right)^{t}\) and \(\mathbf{v}\) is a vector formed from the boundary conditions (systems of this type are so-called mildly nonlinear systems). If we denote the previous system by \(F(\mathbf{x})=0\), where:  
\[F(\mathbf{x})=A\mathbf{x}+h^{2}q(\mathbf{x})-\mathbf{v}\qquad\text{and}\qquad F :\mathbb{R}^{m}\longrightarrow\mathbb{R}^{m},\quad m=NM, \tag{23}\]  
then \(F^{\prime}(\mathbf{x})\) is a linear operator, which is given by the matrix:  
\[F^{\prime}(\mathbf{x})=A+h^{2}Q(\mathbf{x}),\qquad Q(\mathbf{x})=\frac{5}{3} \text{diag}\left\{x_{1}^{2/3},x_{2}^{2/3},\ldots,x_{m}^{2/3}\right\}. \tag{24}\]  
We now choose \(N=M=4\) and the infinity norm. In addition, the number of equations is \(m=16\), \(h=k=\frac{1}{5}\) and \(\lambda=1\). Besides, \(q(\mathbf{x})=\left(x_{1}^{5/3},x_{2}^{5/3},\ldots,x_{16}^{5/3}\right)^{t}\) and:  
\[\mathbf{v}=\left(\frac{44}{25},\frac{23}{25},\frac{28}{25},\frac{87}{25}, \frac{23}{25},0,0,2,\frac{28}{25},0,0,2,\frac{87}{25},2,2,4\right)^{t}.\]  
In this case, we observe that a solution \(\mathbf{x}^{*}\) of the system \(F(x)=0\) with \(F\) defined in (23) satisfies:  
\[\|\mathbf{x}^{*}\|\leq\|A^{-1}\|\left(\|\mathbf{v}\|+h^{2}\|q(\mathbf{x})\| \right)\qquad\Rightarrow\qquad\|\mathbf{x}^{*}\|-\frac{5}{3}\left(4+\frac{1} {25}\|\mathbf{x}^{*}\|^{5/3}\right)\leq 0,\]  
where \(\|A^{-1}\|=\frac{5}{3}\) and \(\|\mathbf{v}\|=4\), so that \(\|\mathbf{x}^{*}\|\in[0,\varrho_{1}]\cup[\varrho_{2},+\infty]\), where \(\varrho_{1}=9.5149\ldots\) and \(\varrho_{2}=45.9125\ldots\) are the two positive real roots of the scalar equation \(t-\frac{5}{3}\left(4+\frac{\epsilon^{5/3}}{25}\right)=0\). Then, we can consider:  
\[F:\Omega\subset\mathbb{R}^{16}\longrightarrow\mathbb{R}^{16}\qquad\text{ with}\qquad\Omega=\left\{\mathbf{x}\in\mathbb{R}^{16}:\,\|\mathbf{x}\|<10\right\},\]since \(\varrho_{1}<10<\varrho_{2}\). Moreover, \(F^{\prime}(\mathbf{x})\) is the linear operator given by the matrix:  
\[A+\frac{1}{15}\,\text{diag}\left\{x_{1}^{2/3},x_{2}^{2/3},\ldots,x_{16}^{2/3}\right\}\]  
and:  
\[F^{\prime}(\mathbf{x})-F^{\prime}(\mathbf{y})=\frac{1}{15}\,\text{diag}\left\{ x_{1}^{2/3}-y_{1}^{2/3},x_{2}^{2/3}-y_{2}^{2/3},\ldots,x_{16}^{2/3}-y_{16}^{2/3 }\right\},\]  
where \(\mathbf{y}=(y_{1},y_{2},\ldots,y_{16})^{t}\). In addition,  
\[\|F^{\prime}(\mathbf{x})-F^{\prime}(\mathbf{y})\|\leq\frac{1}{15}\left\| \mathbf{x}^{1/3}+\mathbf{y}^{1/3}\right\|\left\|\mathbf{x}^{1/3}-\mathbf{y}^ {1/3}\right\|\leq\frac{2}{15}\sqrt[3]{10}\,\left\|\mathbf{x}-\mathbf{y}\right\| ^{1/3},\]  
Thus, \(K=0.2872\ldots\), \(K_{0}=0.2199\ldots\) and \(p=\frac{1}{3}\). If we choose the starting point \(\mathbf{x_{0}}=\left(\frac{3}{2},\frac{3}{2},\ldots,\frac{3}{2}\right)^{t}\), we obtain \(\beta=1.4862\ldots\) and \(\eta=0.5185\ldots\), so that the condition \(\mathpzc{h}=K\beta\eta^{p}\leq\xi(p)\) of Theorem 1 is not satisfied, since \(\mathpzc{h}=0.3429\ldots>\xi(p)=\xi\left(\frac{1}{3}\right)=0.3071\ldots\), where \(\xi\left(\frac{1}{3}\right)\) is the unique zero of the corresponding auxiliary function given by (2), \(\phi\left(x;\frac{1}{3}\right)=\frac{1}{3}\left(6^{2/3}(1-t)^{4/3}-3\sqrt[3]{t}\right)\), in the interval \(\left(0,\frac{1}{2}\right]\). As a consequence, we cannot use Theorem 1 to guarantee the convergence of [PERSON]'s method for approximating a solution of the system \(F(\mathbf{x})=0\), where \(F\) is defined in (23).