## 2 Analysis of Two-Step Method
### Analysis of Convergence Rate  
Let us analyze the convergence rate of method (6). At first, let us obtain the expression for the spectral radius of matrix \(T\). Let \(s=(h,\beta_{1},\beta_{2})\) and the spectral radius be presented as the function \(r(s,\lambda)\), where \(\lambda\in[l,L]\). The expression for \(r\) can be obtained with the use of an expression for the roots of (8):  
\[\zeta_{1,2}=\frac{1}{2}\Big{(}A_{1}\pm\sqrt{D}\Big{)},\]  
where \(D=A_{1}^{2}-4A_{2}\), \(A_{1}=1+\beta_{1}-h(1+\beta_{2})\lambda\), \(A_{2}=\beta_{1}-h\lambda\beta_{2}\) and is written as  
\[r(s,\lambda)=\frac{1}{2}\max(|\zeta_{1}|,|\zeta_{2}|). \tag{13}\]  
If \(r(s,\lambda)\) is considered a function of \(\lambda\), the following theorem on its extremal property can be formulated:  
**Theorem 2**. : _The maximum value of \(r(s,\lambda)\) as a function of \(\lambda\in[l,L]\) takes place for \(\lambda=l\) or \(\lambda=L\)._  
**Proof of Theorem 2. ** (1) Let us obtain the expression for \(r(s,\lambda)\). For \(D>0\) in the case of \(A_{1}>0\), the following inequality takes place: \(A_{1}+\sqrt{D}>0\) and \(|A_{1}-\sqrt{D}|=A_{1}-\sqrt{D}\) if \(A_{1}>\sqrt{D}\), and in this case \(A_{1}+\sqrt{D}>A_{1}-\sqrt{D}>0\). If \(A_{1}-\sqrt{D}<0\), we obtain that \(|A_{1}-\sqrt{D}|=\sqrt{D}-A_{1}\) and \(A_{1}+\sqrt{D}>\sqrt{D}-A_{1}\). So, if \(D>0\) and \(A_{1}>0\), we have that \(r(s,\lambda)=\frac{1}{2}\big{(}A_{1}+\sqrt{D}\big{)}\). If \(D>0\) and \(A_{1}<0\), we have that \(|A_{1}-\sqrt{D}|=\sqrt{D}-A_{1}\), and for \(|A_{1}+\sqrt{D}|\) we have that if \(A_{1}+\sqrt{D}>0\), then \(\sqrt{D}-A_{1}>A_{1}+\sqrt{D}\). If \(A_{1}+\sqrt{D}<0\), then \(|A_{1}+\sqrt{D}|=-A_{1}-\sqrt{D}\) and \(\sqrt{D}-A_{1}>-A_{1}-\sqrt{D}\). So if \(A_{1}<0\) and \(D>0\), we obtain that \(r(s,\lambda)=\frac{1}{2}(\sqrt{D}-A_{1})\). For \(D<0\), it is easy to see that \(r(s,\lambda)=\sqrt{A_{2}}\). The case of \(D=0\) and case \(A_{1}=0\) are trivial to analyze. Thus, it is demonstrated that  
\[r(s,\lambda)= \begin{cases}\frac{1}{2}(A_{1}+\sqrt{D}),&A_{1}\geq 0,D\geq 0,\\ \frac{1}{2}(\sqrt{D}-A_{1}),&A_{1}<0,D\geq 0,\\ \sqrt{A_{2}},&D<0.\end{cases} \]  
(2) Let us analyze the behavior of \(r(s,\lambda)\) for \(\lambda\in[l,L]\). The expression for \(D\) is written as  
\[D=(1+\beta_{2})^{2}\lambda^{2}h^{2}-2((1+\beta_{1})(1+\beta_{2})-2\beta_{2}) \lambda h+(1-\beta_{1})^{2}.\]  
So, the non-negative values of \(D\) are associated with the solutions of the following inequality:  
\[(1+\beta_{2})^{2}t^{2}-2((1+\beta_{1})(1+\beta_{2})-2\beta_{2})t+(1-\beta_{1} )^{2}\geq 0.\]  
The corresponding discriminant is equal to \(16(\beta_{1}+\beta_{2}(\beta_{1}-1))\). As can be seen, solutions to this inequality exist, when  
\[\beta_{2}\leq\frac{\beta_{1}}{1-\beta_{1}}. \tag{14}\]  
The opposite inequality guarantees that it is valid for all \(\lambda>0\). For analysis of the general situation of the sign of \(D\) this restriction is too strict, so we consider the case of condition (14). The case of \(D<0\) leads to the investigation of function \(\psi(\lambda)=\sqrt{A_{2}(\lambda)}=\sqrt{\beta_{1}-\lambda h\beta_{2}}\). Condition \(A_{2}(\lambda)>0\) leads to the restriction \(\lambda<\frac{\beta_{1}}{h\beta_{2}}\), which is valid for \(h<\frac{\beta_{1}}{h\beta_{2}}\). It  
so, when \(\lambda\in\left(0,\frac{(1+\beta_{1})}{h(1+\beta_{2})}\right]\) (corresponds to \(A_{1}\geq 0\)), \(r\) decreases when \(\lambda\in(0,\eta(\beta_{1},\beta_{2})]\) and increases when \(\lambda>\eta(\beta_{1},\beta_{2})\) and its maximum is realized in the boundary point.