## 2 Derivation of the Method  
Consider the initial value problem defined on \(t\in[0,T]\) by  
\[y^{\prime}(t)=f(t,y(t)),\ \ \ y(t_{0})=y_{0} \tag{1}\]  
where \(f(t,y)\) is a general, continuous nonlinear function. Let \(h\) be a prescribed step size on a mesh \(I\) defined by  
\[I=\{t_{n}:t_{0}<t_{1}<\ldots<t_{N}=T\}.\]  
We choose \(m+1\) collocation points, defined by  
\[t_{n+p_{i}}=t_{0}+hp_{i},\ \ i=0,1,2,\ldots,m\]  
in the first interval \([t_{0},t_{1}]\). For the intervals, \([t_{n},t_{n+1}]\), where \(1\leq n\leq N-1\), we choose \(m+2\) collocation points:  
\[\{t_{n-p_{1}}=t_{n}-hp_{1}\}\bigcup\{t_{n+p_{i}}=t_{n}+hp_{i}:i=0,1,2,\ldots,m\}.\]  
The parameters \(p_{i}\) are called collocation parameters, and they satisfy the condition  
\[0=p_{0}<p_{1}<p_{2}<\cdots<p_{m-2}<p_{m-1}<p_{m}=1.\]  
An illustrative representation of the grid is shown in Figure 1, for the special case when \(m=3\) and \(N=4\). It is worth noting that, for \(n>0\), the grid \(I_{n}\) overlaps with one grid point (\(t_{n-1+p_{m-1}}=t_{n-p_{1}}\)) from the previous grid \(I_{n-1}\). For this reason, the method will be referred to as the overlapping hybrid block method. In the non-overlapping block \(I_{1}\), we suppose that the solution can be approximated by the polynomial:  
\[y(t_{0}+sh),\ \ \ 0\leq s\leq 1, \tag{2}\]  
that satisfies the collocation conditions:  
\[y^{\prime}(t_{p_{i}})=f_{p_{i}}. \tag{3}\]  
By the Lagrange interpolation formula, the function \(y^{\prime}(t_{0}+sh)\) can be approximated as  
\[y^{\prime}(t_{0}+sh)=\sum_{j=0}^{m}y^{\prime}(t_{p_{j}})\ell_{j}(s) \tag{4}\]  
Figure 1: Illustration of the overlapping grid when \(N=4\) and \(m=3\). where  
\[\ell_{j}(s)=\prod_{i=0,j\
eq j}^{m}\frac{s-p_{i}}{p_{j}-p_{i}}.\]  
Using the collocation Equation (3), we obtain  
\[I=\{t_{n}:t_{0}<t_{1}<\ldots<t_{N}=T\}.\]0  
which is integrated from 0 to \(p_{i}\) to give  
\[I=\{t_{n}:t_{0}<t_{1}<\ldots<t_{N}=T\}.\]1  
where  
\[I=\{t_{n}:t_{0}<t_{1}<\ldots<t_{N}=T\}.\]2  
The matrix form of (6) is  
\[I=\{t_{n}:t_{0}<t_{1}<\ldots<t_{N}=T\}.\]3  
where  
\[A_{1}= \begin{bmatrix}1&0&0&\cdots&0\\ 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\\ 0&0&0&\cdots&1\end{bmatrix} ,\ \ A_{0}= \begin{bmatrix}1&0&0&\cdots&0\\ 1&0&0&\cdots&0\\ 1&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\\ 1&0&0&\cdots&0\end{bmatrix} \]  
\[B_{0}= \begin{bmatrix}\beta_{1,0}&0&0&\cdots&0\\ \beta_{2,0}&0&0&\cdots&0\\ \beta_{3,0}&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\\ \beta_{m,0}&0&0&\cdots&0\end{bmatrix} ,\ \ B_{1}= \begin{bmatrix}\beta_{1,1}&\beta_{1,2}& \beta_{1,3}&\cdots&\beta_{1,m}\\ \beta_{2,1}&\beta_{2,2}&\beta_{2,3}&\cdots&\beta_{2,m}\\ \beta_{3,1}&\beta_{3,2}&\beta_{3,3}&\cdots&\beta_{3,m}\\ \vdots&\vdots&\vdots&\ddots&\\ \beta_{m,1}&\beta_{m,2}&\beta_{m,3}&\cdots&\beta_{m,m}.\end{bmatrix} \]  
and  
\[ \begin{array}{l}F_{0}=\begin{bmatrix}f_{0},f_{-p_{1}},\ldots,f_{-p_{m-1}} \end{bmatrix},\ \ F_{1}=\begin{bmatrix}f_{p_{1}},f_{p_{2}},\ldots,f_{p_{m}}\end{bmatrix},\\ Y_{0}=\begin{bmatrix}y_{0},y_{-p_{1}},\ldots,y_{-p_{m-1}}\end{bmatrix},\ \ Y_{1}=\begin{bmatrix}y_{p_{1}},y_{p_{2}},\ldots,y_{p_{m}}\end{bmatrix}.\end{array} \]  
In the block \(I_{n}=[t_{n},t_{n+1}]\) (\(n>0\)), we consider the collocation parameters  
\[-p_{1}<0<p_{1}<p_{2}<\cdots<p_{m-1}<p_{m}\]  
and assume the following collocation conditions:  
\[y^{\prime}(t_{n+p_{i}})=f_{n+p_{i}},\ \ y^{\prime}(t_{n-p_{1}})=f_{n-p_{1}} \tag{8}\]  
for \(i=0,1,\ldots,m\). By the Lagrange interpolation formula, the function \(y^{\prime}(t_{n}+sh)\) can be approximated as  
\[y^{\prime}(t_{n}+sh)=\sum_{j=-1}^{m}y^{\prime}(t_{n+p_{j}})\ell_{j}(s). \tag{9}\]  
Integrating from 0 to \(p_{i}\) gives  
\[y_{n+p_{i}}=y_{n}+h\sum_{j=-1}^{m}\hat{\beta}_{i,j}f_{n+p_{j}} \tag{10}\]where  
\[\hat{\beta}_{i,j}=\int_{0}^{p_{i}}\ell_{j}(\tau)d\tau,\quad j=-1,0,1,\ldots,m.\]  
Equation (10) can be written in matrix form as  
\[A_{1}Y_{n+1}=A_{0}Y_{n}+h(\hat{B}_{0}F_{n}+\hat{B}_{1}F_{n+1}) \tag{11}\]  
where  
\[Y_{n+1}=