1 Introduction
The main focus of numerical methods for solving differential equations in recent times has been placed on presenting numerical methods with an improved level of accuracy, ranging from first order ordinary differential equations [1-3] to higher order ordinary differential equations [4-6]. Numerical methods adopted to solve such ODEs have been developed using two popular approaches, i.e. numerical integration and collocation. However, there is a need to introduce new approaches to develop these numerical methods so that they become easier and less burdensome compared to the conventional approaches. Referring to the work of [7], three basic methods can be highlighted for developing single linear multistep methods: numerical integration, interpolation and Taylor series expansion. Although the derivations originally involved just single multistep methods for first order ordinary differential equations, over time the numerical methods evolved into a family of block methods and a further improvement with the introduction of higher derivatives, sometimes referred to as Obrechkoff type [8,9].
The aim of this article is to present a novel algorithm that is convenient to adopt for developing new generalized algorithm k-step ሺ݉ 1ሻ௧ derivative block methods for solving any ݉௧ order ordinary differential equations. This work is
Received January 5th, 2017, 1st revision August 2nd, 2017, 2nd revision September 27th, 2017. Accepted for publication October 3rd, 2017.
Copyright © 2018 Published by ITB Journal Publisher, ISSN: 2337-5760, DOI: 10.5614/j.math.fund.sci.2018.50.1.4 an extension of [10], which proposed a generalized algorithm for developing block methods without the introduction of higher derivatives.
This article is organized as follows: Section 2 presents the generalized algorithm for developing k-step higher derivative block methods with sample second order and third order block methods developed in Sections 3 and 4 respectively. The basic properties of the block methods developed in Sections 3 and 4 are also highlighted. Certain numerical examples are considered in Section 5 and the article is concluded in Section 6.
2 Generalized Algorithm for k-Step Higher Derivative Block Methods
Considering the general form of \(m^{th}\) order ordinary differential equation
\[y^{(m)} = f(x, y, y', y'', \dots, y^{(m-1)}), x \in [a, b]\]
Extending the generalized algorithm presented by [10] for k-step block methods to accommodate the presence of higher derivatives, the following algorithm is presented for developing k-step \((m+1)^{th}\) derivative block methods for solving \(m^{th}\) order ODEs.
\[y_{n+\xi} = \sum_{i=0}^{m-1} \frac{\left(\xi h\right)^{i}}{i!} y_{n}^{(i)} + \sum_{i=0}^{k} \left(\phi_{\xi i} f_{n+i} + \tau_{\xi i} g_{n+i}\right), \ \xi = 1, \ 2, \ \cdots, \ k\] (1)
with corresponding derivatives obtained from
\[y_{n+\xi}^{(a)} = \sum_{i=0}^{m-(a+1)} \frac{(\xi h)^{i}}{i!} y_{n}^{(i+a)} + \sum_{i=0}^{k} (\omega_{\xi ia} f_{n+i} + \varphi_{\xi ia} g_{n+i}),\] \[a = 1_{(\xi)}, \ 2_{(\xi)}, \ \cdots, \ (m-1)_{(\xi)}\] (2)
where
\[f_{n+i} = f\left(x_{n+i}, y_{n+i}, y'_{n+i}, y''_{n+i}, \cdots, y_{n+i}^{(m-1)}\right),\] \[g_{n+i} = \frac{df_{n+i} = f\left(x_{n+i}, y_{n+i}, y'_{n+i}, y''_{n+i}, \cdots, y_{n+i}^{(m-1)}\right)}{dx},\] and the condition \(k \ge m\) is set to avoid an underdetermined system. The unknown \(\emptyset\), \(\tau\), \(\omega\) and \(\varphi\) coefficients are obtained from \((\phi_{\varepsilon_0}, \phi_{\varepsilon_1}, \dots, \phi_{\varepsilon_k}, \tau_{\varepsilon_0}, \tau_{\varepsilon_1}, \dots, \tau_{\varepsilon_k})^T = A^{-1}B\) and
\[B_{(2k+2)\times 1} = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 & 0 & 0 & 0 & \cdots & 0 \\ 0 & h & 2h & \cdots & kh & 1 & 1 & 1 & \cdots & 1 \\ 0 & \frac{(h)^2}{2!} & \frac{(2h)^2}{2!} & \cdots & \frac{(kh)^2}{2!} & 0 & h & 2h & \cdots & kh \\ 0 & \frac{(h)^3}{3!} & \frac{(2h)^3}{3!} & \cdots & \frac{(kh)^3}{3!} & 0 & \frac{(h)^2}{2!} & \frac{(2h)^2}{2!} & \cdots & \frac{(kh)^2}{2!} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots \\ 0 & \frac{(h)^{(2k+1)}}{3!} & \frac{(2h)^{(2k+1)}}{3!} & \cdots & \frac{(kh)^{(2k+1)}}{3!} & 0 & \frac{(h)^{(2k)}}{2!} & \frac{(2h)^{(2k)}}{2!} & \cdots & \frac{(kh)^{(2k)}}{2!} \\ \vdots & \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots \\ \vdots & \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots \\ 0 & \frac{(h)^{(2k+1)}}{(2k+1)!} & \frac{(2h)^{(2k+1)}}{(2k+1)!} & \cdots & \frac{(kh)^{(2k+1)}}{(2k+1)!} & 0 & \frac{(h)^{(2k)}}{(2k)!} & \frac{(2h)^{(2k)}}{(2k)!} & \cdots & \frac{(kh)^{(2k)}}{(2k)!} \end{pmatrix}\] \[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\]
To ascertain that there are no linearly dependent columns or rows \((\det(A) \neq 0)\), the rank of the matrix is investigated when developing the block methods. Note that the approach in [7] for developing multistep methods using Taylor series expansion is adopted to expand individual terms in Eq. (1) and Eq. (2) to obtain the unknown coefficients. The conventional Taylor expansions concept is defined for \(y_{n+a}^{(n)} = y^{(n)}(x_n + ah)\) about \(x_n\) as
\[y^{(n)}(x_n + ah) = y^{(n)}(x_n) + ahy^{(n+1)}(x_n) + \frac{(ah)^2}{2!}y^{(n+2)}(x_n) + \cdots\] (4)
A more detailed explanation is given in the following subsections, where the new generalized algorithm is used to develop k-step \((m + 1)^{th}\) derivative block methods for (m, k) = (2,2) and (m, k) = (3,3).
3 Development of Two-Step Third Derivative Block Methods for Second Order ODEs
Consider developing a two-step third derivative block method for second order ODEs, that is k = 2 and m = 2. Substituting k and m in Eq. (1) and Eq. (2) gives
\[y_{n+\xi} = \sum_{i=0}^{1} \frac{\left(\xi h\right)^{i}}{i!} y_{n}^{(i)} + \sum_{i=0}^{2} \left(\phi_{\xi i} f_{n+i} + \tau_{\xi i} g_{n+i}\right), \ \xi = 1, \ 2\] (5)
and
\[y_{n+\xi}^{(1)} = y_n^{(1)} + \sum_{i=0}^{2} \left(\omega_{\xi i1} f_{n+i} + \varphi_{\xi i1} g_{n+i}\right), \ \xi = 1, \ 2\] (6)
where a = 1.
Therefore,
\[y_{n+1} = y_n + hy_n^{(1)} + (\phi_{10}f_n + \phi_{11}f_{n+1} + \phi_{12}f_{n+2} + \tau_{10}g_n + \tau_{11}g_{n+1} + \tau_{12}g_{n+2}),\] \[y_{n+2} = y_n + 2hy_n^{(1)} + (\phi_{20}f_n + \phi_{21}f_{n+1} + \phi_{22}f_{n+2} + \tau_{20}g_n + \tau_{21}g_{n+1} + \tau_{22}g_{n+2}),\] \[y_{n+1}^{(1)} = y_n^{(1)} + (\omega_{101}f_n + \omega_{111}f_{n+1} + \omega_{121}f_{n+2} + \varphi_{101}g_n + \varphi_{111}g_{n+1} + \varphi_{121}g_{n+2}),\] \[y_{n+2}^{(1)} = y_n^{(1)} + (\omega_{201}f_n + \omega_{211}f_{n+1} + \omega_{221}f_{n+2} + \varphi_{201}g_n + \varphi_{211}g_{n+1} + \varphi_{221}g_{n+2})\] \[(7)\]
Expanding \(y_{n+1}\) using Taylor series yields:
\[y(x_{n}) + hy^{(1)}(x_{n}) + \frac{(h)^{2}}{2!}y^{(2)}(x_{n}) + \frac{(h)^{3}}{3!}y^{(3)}(x_{n}) + \frac{(h)^{4}}{4!}y^{(4)}(x_{n})\] \[+ \frac{(h)^{5}}{5!}y^{(5)}(x_{n}) + \frac{(h)^{6}}{6!}y^{(6)}(x_{n}) + \frac{(h)^{7}}{7!}y^{(7)}(x_{n}) + \cdots = y(x_{n})\] \[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\] \[+ \frac{(h)^{2}}{2!}y^{(4)}(x_{n}) + \frac{(h)^{3}}{3!}y^{(5)}(x_{n}) + \cdots) + \phi_{12}(y^{(2)}(x_{n}) + 2hy^{(3)}(x_{n})\] \[+ \frac{(2h)^{2}}{2!}y^{(4)}(x_{n}) + \frac{(2h)^{3}}{3!}y^{(5)}(x_{n}) + \cdots) + \tau_{10}(y^{(3)}(x_{n})) + \tau_{11}(y^{(3)}(x_{n})\] \[+ hy^{(4)}(x_{n}) + \frac{(h)^{2}}{2!}y^{(5)}(x_{n}) + \frac{(h)^{3}}{3!}y^{(6)}(x_{n}) + \cdots) + \tau_{12}(y^{(3)}(x_{n})\] \[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\]
Equating coefficients of \(y^{(n)}(x_n)\) on both the left- and right-hand sides of the equation gives the following system of equations:
\[\frac{(h)^2}{2!} = \phi_{10} + \phi_{11} + \phi_{12}\]
\[\frac{(h)^{3}}{3!} = h\phi_{11} + 2h\phi_{12} + \tau_{10} + \tau_{11} + \tau_{12}\] \[\frac{(h)^{4}}{4!} = \frac{(h)^{2}}{2!} \phi_{11} + \frac{(2h)^{2}}{2!} \phi_{12} + h\tau_{11} + 2h\tau_{12}\] \[\frac{(h)^{5}}{5!} = \frac{(h)^{3}}{3!} \phi_{11} + \frac{(2h)^{3}}{3!} \phi_{12} + \frac{(h)^{2}}{2!} \tau_{11} + \frac{(2h)^{2}}{2!} \tau_{12}\] \[\frac{(h)^{6}}{6!} = \frac{(h)^{4}}{4!} \phi_{11} + \frac{(2h)^{4}}{4!} \phi_{12} + \frac{(h)^{3}}{3!} \tau_{11} + \frac{(2h)^{3}}{3!} \tau_{12}\] \[\frac{(h)^{7}}{7!} = \frac{(h)^{5}}{5!} \phi_{11} + \frac{(2h)^{5}}{5!} \phi_{12} + \frac{(h)^{4}}{4!} \tau_{11} + \frac{(2h)^{4}}{4!} \tau_{12}\] (8)
where truncation is made such that the number of unknowns equals the number of equations. Rewriting Eq. (8) in matrix form Ax = B gives:
\[\begin{pmatrix} 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & h & 2h & 1 & 1 & 1 \\ 0 & \frac{(h)^{2}}{2!} & \frac{(2h)^{2}}{2!} & 0 & h & 2h \\ 0 & \frac{(h)^{3}}{3!} & \frac{(2h)^{3}}{3!} & 0 & \frac{(h)^{2}}{2!} & \frac{(2h)^{3}}{2!} \\ 0 & \frac{(h)^{4}}{4!} & \frac{(2h)^{4}}{4!} & 0 & \frac{(h)^{3}}{3!} & \frac{(2h)^{3}}{3!} \\ 0 & \frac{(h)^{5}}{5!} & \frac{(2h)^{5}}{5!} & 0 & \frac{(h)^{4}}{4!} & \frac{(2h)^{4}}{4!} \end{pmatrix} \begin{pmatrix} \phi_{10} \\ \phi_{11} \\ \phi_{12} \\ \tau_{10} \\ \tau_{11} \\ \tau_{12} \end{pmatrix} = \begin{pmatrix} \frac{(h)^{2}}{2!} \\ \frac{(h)^{3}}{3!} \\ \frac{(h)^{4}}{4!} \\ \frac{(h)^{5}}{5!} \\ \frac{(h)^{6}}{6!} \\ \frac{(h)^{7}}{7!} \end{pmatrix}\] (9)
where \(\begin{pmatrix} 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & h & 2h & 1 & 1 & 1 \\ 0 & \frac{(h)^2}{2!} & \frac{(2h)^2}{2!} & 0 & h & 2h \\ 0 & \frac{(h)^3}{3!} & \frac{(2h)^3}{3!} & 0 & \frac{(h)^2}{2!} & \frac{(2h)^2}{2!} \\ 0 & \frac{(h)^4}{4!} & \frac{(2h)^4}{4!} & 0 & \frac{(h)^3}{3!} & \frac{(2h)^3}{3!} \\ 0 & \frac{(h)^5}{5!} & \frac{(2h)^5}{5!} & 0 & \frac{(h)^4}{4!} & \frac{(2h)^4}{4!} \end{pmatrix}\) has rank = 6, which implies that there
are no linearly dependent columns or rows and the inverse exists. These matrices correspond to the definitions assigned in Eq. (3) above and to obtain the unknown coefficients, the matrix inverse approach is adopted to obtain
\[\left(\phi_{10},\phi_{11},\phi_{12},\tau_{10},\tau_{11},\tau_{12}\right)^{T} = \left(\frac{13h^{2}}{42},\frac{h^{2}}{6},\frac{h^{2}}{42},\frac{59h^{3}}{1680},-\frac{8h^{3}}{105},-\frac{11h^{3}}{1680}\right).\]
Following the same approach for \(y_{n+2}\), the unknown coefficients are obtained
\[\left(\phi_{20},\phi_{21},\phi_{22},\tau_{20},\tau_{21},\tau_{22}\right)^T = \left(\frac{79h^2}{105},\frac{16h^2}{15},\frac{19h^2}{105},\frac{2h^3}{21},-\frac{16h^3}{105},-\frac{4h^3}{105}\right).\]
Furthermore, in the case of obtaining the coefficients for \(y_{n+1}^{(1)}\), the same approach is followed as for \(y_{n+1}\) using Taylor series expansion to obtain:
\[y^{(1)}(x_n) + hy^{(2)}(x_n) + \frac{(h)^2}{2!}y^{(3)}(x_n) + \frac{(h)^3}{3!}y^{(4)}(x_n) + \frac{(h)^4}{4!}y^{(5)}(x_n)\] \[+ \frac{(h)^5}{5!}y^{(6)}(x_n) + \frac{(h)^6}{6!}y^{(7)}(x_n) + \cdots = y^{(1)}(x_n) + \{\omega_{101}(y^{(2)}(x_n))\}\] \[+ \omega_{111}(y^{(2)}(x_n) + hy^{(3)}(x_n) + \frac{(h)^2}{2!}y^{(4)}(x_n) + \frac{(h)^3}{3!}y^{(5)}(x_n) + \cdots)\] \[+ \omega_{121}(y^{(2)}(x_n) + 2hy^{(3)}(x_n) + \frac{(2h)^2}{2!}y^{(4)}(x_n) + \frac{(2h)^3}{3!}y^{(5)}(x_n) + \cdots)\] \[+ \varphi_{101}(y^{(3)}(x_n)) + \varphi_{111}(y^{(3)}(x_n) + hy^{(4)}(x_n) + \frac{(h)^2}{2!}y^{(5)}(x_n)\] \[+ \frac{(h)^3}{3!}y^{(6)}(x_n) + \cdots) + \varphi_{121}(y^{(3)}(x_n) + 2hy^{(4)}(x_n)\] \[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\]
Equating coefficients of \(y^{(n)}(x_n)\) on both the left- and right-hand side of the equation gives the following system of equations:
\[h = \omega_{101} + \omega_{111} + \omega_{121}\] \[\frac{(h)^{2}}{2!} = h\omega_{111} + 2h\omega_{121} + \varphi_{101} + \varphi_{111} + \varphi_{121}\] \[\frac{(h)^{3}}{3!} = \frac{(h)^{2}}{2!}\omega_{111} + \frac{(2h)^{2}}{2!}\omega_{121} + h\varphi_{111} + 2h\varphi_{121}\] \[\frac{(h)^{4}}{4!} = \frac{(h)^{3}}{3!}\omega_{111} + \frac{(2h)^{3}}{3!}\omega_{121} + \frac{(h)^{2}}{2!}\varphi_{111} + \frac{(2h)^{2}}{2!}\varphi_{121}\] \[\frac{(h)^{5}}{5!} = \frac{(h)^{4}}{4!}\omega_{111} + \frac{(2h)^{4}}{4!}\omega_{121} + \frac{(h)^{3}}{3!}\varphi_{111} + \frac{(2h)^{3}}{3!}\varphi_{121}\] \[\frac{(h)^{6}}{6!} = \frac{(h)^{5}}{5!}\omega_{111} + \frac{(2h)^{5}}{5!}\omega_{121} + \frac{(h)^{4}}{4!}\varphi_{111} + \frac{(2h)^{4}}{4!}\varphi_{121}\] \[\frac{(h)^{6}}{6!} = \frac{(h)^{5}}{5!}\omega_{111} + \frac{(2h)^{5}}{5!}\omega_{121} + \frac{(h)^{4}}{4!}\varphi_{111} + \frac{(2h)^{4}}{4!}\varphi_{121}\]
Rewriting Eq. (10) in matrix form Ax = B gives:
\[\begin{pmatrix} 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & h & 2h & 1 & 1 & 1 \\ 0 & \frac{(h)^{2}}{2!} & \frac{(2h)^{2}}{2!} & 0 & h & 2h \\ 0 & \frac{(h)^{3}}{3!} & \frac{(2h)^{3}}{3!} & 0 & \frac{(h)^{2}}{2!} & \frac{(2h)^{2}}{2!} \\ 0 & \frac{(h)^{4}}{4!} & \frac{(2h)^{4}}{4!} & 0 & \frac{(h)^{3}}{3!} & \frac{(2h)^{3}}{3!} \\ 0 & \frac{(h)^{5}}{5!} & \frac{(2h)^{5}}{5!} & 0 & \frac{(h)^{4}}{4!} & \frac{(2h)^{4}}{4!} \end{pmatrix} \begin{pmatrix} \omega_{101} \\ \omega_{121} \\ \varphi_{101} \\ \varphi_{101} \\ \varphi_{111} \\ \varphi_{121} \end{pmatrix} = \begin{pmatrix} h \\ \frac{(h)^{2}}{2!} \\ \frac{(h)^{3}}{3!} \\ \frac{(h)^{4}}{4!} \\ \frac{(h)^{5}}{5!} \\ \frac{(h)^{6}}{6!} \end{pmatrix}\] (11)
which also corresponds to the definitions assigned in Eq. (3) above. The matrix inverse approach is adopted to obtain the unknown coefficients as:
\[\left(\omega_{101},\omega_{111},\omega_{121},\varphi_{101},\varphi_{111},\varphi_{121}\right)^T = \left(\frac{101h}{240},\frac{8h}{15},\frac{11h}{240},\frac{13h^2}{240},-\frac{h^2}{6},-\frac{h^2}{80}\right)\]
Similarly, the unknown coefficients for \(y_{n+2}^{(1)}\) are obtained as:
\[\left(\omega_{201},\omega_{211},\omega_{221},\varphi_{201},\varphi_{211},\varphi_{221}\right)^T = \left(\frac{7h}{15},\frac{16h}{15},\frac{7h}{15},\frac{h^2}{15},0,-\frac{h^2}{15}\right)\]
Substituting all obtained coefficients back in Eq. (7) gives the two-step third derivative block method for solving second order ordinary differential equations as:
\[\begin{split} y_{n+1} &= y_n + h y_n^{(1)} + \frac{h^2}{42} \left( 13f_n + 7f_{n+1} + f_{n+2} \right) + \frac{h^3}{1680} \left( 59g_n - 128g_{n+1} - 11g_{n+2} \right), \\ y_{n+2} &= y_n + 2h y_n^{(1)} + \frac{h^2}{105} \left( 79f_n + 112f_{n+1} + 19f_{n+2} \right) + \frac{h^3}{105} \left( 10g_n - 16g_{n+1} - 4g_{n+2} \right) \\ y_{n+1}^{(1)} &= y_n^{(1)} + \frac{h}{240} \left( 101f_n + 128f_{n+1} + 11f_{n+2} \right) + \frac{h^2}{240} \left( 13g_n - 40g_{n+1} - 3g_{n+2} \right), \\ y_{n+2}^{(1)} &= y_n^{(1)} + \frac{h}{15} \left( 7f_n + 16f_{n+1} + 7f_{n+2} \right) + \frac{h^2}{15} \left( g_n - g_{n+2} \right) \end{split}\]
4 Properties of the Two-Step Third-Derivative Block Method
The following properties of the two-step third-derivative block method are discussed: order, zero-stability, consistency and convergence.
Following the definition of Equation (11), the individual terms of the two-step third derivative block method are expanded using Taylor series expansions about \(x = x_n\). The order of the two-step third derivative block method is obtained to be p = 6.
Secondly, to analyze the two-step third derivative block method for zero-stability, the modulus of the roots of its first characteristic polynomial is expected to be simple or less than one. Thus, the correctors of the two-step third derivative block method are normalized to give the first characteristic polynomial as \(\rho(r) = det \left[ rI_2 - \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \right]\) with roots satisfying \(|r_j| \le 1\).
The two-step third derivative block method is consistent if it has order \(p \ge 1\) as satisfied in the previous paragraphs. Therefore, the two-step third derivative block method is convergent [12].
5 Development of Three-Step Fourth Derivative Block Methods for Third Order ODEs
Consider developing a three-step higher derivative block method for third order ODEs, i.e. k = 3 and m = 3.
Substituting k and m in Eq. (1) and Eq. (2) gives:
\[y_{n+\xi} = \sum_{i=0}^{2} \frac{\left(\xi h\right)^{i}}{i!} y_{n}^{(i)} + \sum_{i=0}^{3} \left(\phi_{\xi i} f_{n+i} + \tau_{\xi i} g_{n+i}\right), \ \xi = 1, \ 2, \ 3\] (12)
and
\[y_{n+\xi}^{(1)} = \sum_{i=0}^{1} \frac{(\xi h)^{i}}{i!} y_{n}^{(i+1)} + \sum_{i=0}^{3} (\omega_{\xi i1} f_{n+i} + \varphi_{\xi i1} g_{n+i}), \quad \xi = 1, 2, 3\] \[y_{n+\xi}^{(2)} = \sum_{i=0}^{0} \frac{(\xi h)^{i}}{i!} y_{n}^{(i+2)} + \sum_{i=0}^{3} (\omega_{\xi i2} f_{n+i} + \varphi_{\xi i2} g_{n+i}), \quad \xi = 1, 2, 3\] (13)
where a = 1,2.
\[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\]
\[y_{n+2}^{(2)} = y_n^{(2)} + \begin{pmatrix} \omega_{202} f_n + \omega_{212} f_{n+1} + \omega_{222} f_{n+2} + \omega_{232} f_{n+3} \\ + \varphi_{202} g_n + \varphi_{212} g_{n+1} + \varphi_{222} g_{n+2} + \varphi_{232} g_{n+3} \end{pmatrix},\] \[y_{n+3}^{(2)} = y_n^{(2)} + \begin{pmatrix} \omega_{302} f_n + \omega_{312} f_{n+1} + \omega_{322} f_{n+2} + \omega_{332} f_{n+3} \\ + \varphi_{302} g_n + \varphi_{312} g_{n+1} + \varphi_{322} g_{n+2} + \varphi_{332} g_{n+3} \end{pmatrix}\] (14)
For \(y_{n+1}\):
\[y(x_{n}) + hy^{(1)}(x_{n}) + \frac{(h)^{2}}{2!}y^{(2)}(x_{n}) + \frac{(h)^{3}}{3!}y^{(3)}(x_{n}) + \frac{(h)^{4}}{4!}y^{(4)}(x_{n})\] \[+ \frac{(h)^{5}}{5!}y^{(5)}(x_{n}) + \frac{(h)^{6}}{6!}y^{(6)}(x_{n}) + \frac{(h)^{7}}{7!}y^{(7)}(x_{n}) + \frac{(h)^{8}}{8!}y^{(8)}(x_{n})\] \[+ \frac{(h)^{9}}{9!}y^{(9)}(x_{n}) + \cdots = y(x_{n}) + hy^{(1)}(x_{n}) + \frac{(h)^{2}}{2!}y^{(2)}(x_{n})\] \[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\] \[+ \frac{(h)^{3}}{3!}y^{(6)}(x_{n}) + \frac{(h)^{4}}{4!}y^{(7)}(x_{n}) + \cdots) + \phi_{12}(y^{(3)}(x_{n}) + 2hy^{(4)}(x_{n})\] \[+ \frac{(2h)^{2}}{2!}y^{(5)}(x_{n}) + \frac{(2h)^{3}}{3!}y^{(6)}(x_{n}) + \frac{(2h)^{4}}{4!}y^{(7)}(x_{n}) + \cdots) + \phi_{13}(y^{(3)}(x_{n})\] \[+ 3hy^{(4)}(x_{n}) + \frac{(3h)^{2}}{2!}y^{(5)}(x_{n}) + \frac{(3h)^{3}}{3!}y^{(6)}(x_{n}) + \frac{(3h)^{4}}{4!}y^{(7)}(x_{n}) + \cdots)\] \[+ \tau_{10}(y^{(4)}(x_{n})) + \tau_{11}(y^{(4)}(x_{n}) + hy^{(5)}(x_{n}) + \frac{(h)^{2}}{2!}y^{(6)}(x_{n})\] \[+ \frac{(h)^{3}}{3!}y^{(7)}(x_{n}) + \frac{(h)^{4}}{4!}y^{(8)}(x_{n}) + \cdots) + \tau_{12}(y^{(4)}(x_{n}) + 2hy^{(5)}(x_{n})\] \[+ \frac{(2h)^{2}}{2!}y^{(6)}(x_{n}) + \frac{(2h)^{3}}{3!}y^{(7)}(x_{n}) + \frac{(2h)^{4}}{4!}y^{(8)}(x_{n}) + \cdots) + \tau_{13}(y^{(4)}(x_{n})\] \[+ 3hy^{(5)}(x_{n}) + \frac{(3h)^{2}}{2!}y^{(6)}(x_{n}) + \frac{(3h)^{3}}{3!}y^{(7)}(x_{n}) + \frac{(3h)^{4}}{4!}y^{(8)}(x_{n}) + \cdots)\] \[+ \eta_{10}(y^{(5)}(x_{n}) + \frac{(3h)^{2}}{2!}y^{(6)}(x_{n}) + \frac{(3h)^{3}}{3!}y^{(7)}(x_{n}) + \frac{(3h)^{4}}{4!}y^{(8)}(x_{n}) + \cdots)\]
Equating coefficients of \(y^{(n)}(x_n)\) on both the left- and right-hand side of the equation gives the following system of equations:
\[\begin{split} \frac{(h)^3}{3!} &= \phi_{10} + \phi_{11} + \phi_{12} + \phi_{13} \\ \frac{(h)^4}{4!} &= h\phi_{11} + 2h\phi_{12} + 3h\phi_{13} + \tau_{10} + \tau_{11} + \tau_{12} + \tau_{13} \\ \frac{(h)^5}{5!} &= \frac{(h)^2}{2!} \phi_{11} + \frac{(2h)^2}{2!} \phi_{12} + \frac{(3h)^2}{2!} \phi_{13} + h\tau_{11} + 2h\tau_{12} + 3h\tau_{13} \\ \frac{(h)^6}{6!} &= \frac{(h)^3}{3!} \phi_{11} + \frac{(2h)^3}{3!} \phi_{12} + \frac{(3h)^3}{3!} \phi_{13} + \frac{(h)^2}{2!} \tau_{11} + \frac{(2h)^2}{2!} \tau_{12} + \frac{(3h)^2}{2!} \tau_{13} \\ \frac{(h)^7}{7!} &= \frac{(h)^4}{4!} \phi_{11} + \frac{(2h)^4}{4!} \phi_{12} + \frac{(3h)^4}{4!} \phi_{13} + \frac{(h)^3}{3!} \tau_{11} + \frac{(2h)^3}{3!} \tau_{12} + \frac{(3h)^3}{3!} \tau_{13} \\ \frac{(h)^8}{8!} &= \frac{(h)^5}{5!} \phi_{11} + \frac{(2h)^5}{5!} \phi_{12} + \frac{(3h)^5}{5!} \phi_{13} + \frac{(h)^4}{4!} \tau_{11} + \frac{(2h)^4}{4!} \tau_{12} + \frac{(3h)^4}{4!} \tau_{13} \end{split}\]
\[\frac{(h)^9}{9!} = \frac{(h)^6}{6!} \phi_{11} + \frac{(2h)^6}{6!} \phi_{12} + \frac{(3h)^6}{6!} \phi_{13} + \frac{(h)^5}{5!} \tau_{11} + \frac{(2h)^5}{5!} \tau_{12} + \frac{(3h)^5}{5!} \tau_{13} \frac{(h)^{10}}{10!} = \frac{(h)^7}{7!} \phi_{11} + \frac{(2h)^7}{7!} \phi_{12} + \frac{(3h)^7}{7!} \phi_{13} + \frac{(h)^6}{6!} \tau_{11} + \frac{(2h)^6}{6!} \tau_{12} + \frac{(3h)^6}{6!} \tau_{13}\] (15)
Rewriting Eq. (15) in matrix form Ax = B gives:
\[\begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & h & 2h & 3h & 1 & 1 & 1 & 1 \\ 0 & \frac{(h)^{2}}{2!} & \frac{(2h)^{2}}{2!} & \frac{(3h)^{2}}{2!} & 0 & h & 2h & 3h \\ 0 & \frac{(h)^{3}}{3!} & \frac{(2h)^{3}}{3!} & \frac{(3h)^{3}}{3!} & 0 & \frac{(h)^{2}}{2!} & \frac{(2h)^{2}}{2!} & \frac{(3h)^{2}}{2!} \\ 0 & \frac{(h)^{4}}{4!} & \frac{(2h)^{4}}{4!} & \frac{(3h)^{4}}{4!} & 0 & \frac{(h)^{3}}{3!} & \frac{(2h)^{3}}{3!} & \frac{(3h)^{3}}{3!} \\ 0 & \frac{(h)^{5}}{5!} & \frac{(2h)^{5}}{5!} & \frac{(3h)^{5}}{5!} & 0 & \frac{(h)^{4}}{4!} & \frac{(2h)^{4}}{4!} & \frac{(3h)^{4}}{4!} \\ 0 & \frac{(h)^{6}}{6!} & \frac{(2h)^{6}}{6!} & \frac{(3h)^{6}}{6!} & 0 & \frac{(h)^{5}}{5!} & \frac{(2h)^{5}}{5!} & \frac{(3h)^{5}}{5!} \\ 0 & \frac{(h)^{7}}{7!} & \frac{(2h)^{7}}{7!} & \frac{(3h)^{7}}{7!} & 0 & \frac{(h)^{6}}{6!} & \frac{(2h)^{6}}{6!} & \frac{(3h)^{6}}{6!} & \frac{(3h)^{6}}{6!} \\ \end{pmatrix} (16)\] where \[\begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & h & 2h & 3h & 1 & 1 & 1 & 1 \\ 0 & \frac{(h)^2}{2!} & \frac{(2h)^2}{2!} & \frac{(3h)^2}{2!} & 0 & h & 2h & 3h \\ 0 & \frac{(h)^3}{3!} & \frac{(2h)^3}{3!} & \frac{(3h)^3}{3!} & 0 & \frac{(h)^2}{2!} & \frac{(2h)^2}{2!} & \frac{(3h)^2}{2!} \\ 0 & \frac{(h)^4}{4!} & \frac{(2h)^4}{4!} & \frac{(3h)^4}{4!} & 0 & \frac{(h)^3}{3!} & \frac{(2h)^3}{3!} & \frac{(3h)^3}{3!} \\ 0 & \frac{(h)^5}{5!} & \frac{(2h)^5}{5!} & \frac{(3h)^5}{5!} & 0 & \frac{(h)^4}{4!} & \frac{(2h)^4}{4!} & \frac{(3h)^4}{4!} \\ 0 & \frac{(h)^6}{6!} & \frac{(2h)^6}{6!} & \frac{(3h)^6}{6!} & 0 & \frac{(h)^5}{5!} & \frac{(2h)^5}{5!} & \frac{(3h)^5}{5!} \\ 0 & \frac{(h)^7}{7!} & \frac{(2h)^7}{7!} & \frac{(3h)^7}{7!} & 0 & \frac{(h)^6}{6!} & \frac{(2h)^6}{6!} & \frac{(3h)^6}{6!} \end{pmatrix}\]
implies that there are also no linearly dependent columns or rows and the inverse exists. These matrices likewise correspond to the definitions in Eq. (3). The unknown coefficients are obtained below:
\[\begin{split} & \left(\phi_{10}, \phi_{11}, \phi_{12}, \phi_{13}, \tau_{10}, \tau_{11}, \tau_{12}, \tau_{13}\right)^T \\ & = \left(\frac{62387h^3}{544320}, \frac{89h^3}{3360}, \frac{439h^3}{20160}, \frac{1031h^3}{272160}, \frac{1879h^4}{181440}, -\frac{359h^4}{10080}, -\frac{13h^4}{960}, -\frac{17h^4}{18144}\right) \end{split}\]
Next the unknown coefficients for \(y_{n+2}\) and \(y_{n+2}\) are obtained as:
\[\begin{split} & \left(\phi_{20},\phi_{21},\phi_{22},\phi_{23},\tau_{20},\tau_{21},\tau_{22},\tau_{23}\right)^{T} \\ & = \left(\frac{5048h^{3}}{8505},\frac{164h^{3}}{315},\frac{4h^{3}}{21},\frac{244h^{3}}{8505},\frac{172h^{4}}{2835},-\frac{4h^{4}}{15},-\frac{34h^{4}}{315},-\frac{4h^{4}}{567}\right) \\ & \left(\phi_{30},\phi_{31},\phi_{32},\phi_{33},\tau_{30},\tau_{31},\tau_{32},\tau_{33}\right)^{T} \\ & = \left(\frac{657h^{3}}{448},\frac{2187h^{3}}{1120},\frac{2187h^{3}}{2240},\frac{117h^{3}}{1120},\frac{351h^{4}}{2240},-\frac{729h^{4}}{1120},-\frac{729h^{4}}{2240},-\frac{27h^{4}}{1120}\right) \end{split}\]
In obtaining the coefficients for \(y_{n+1}^{(1)}\), Taylor series expansion is used, which gives:
\[y^{(1)}(x_n) + hy^{(2)}(x_n) + \frac{(h)^2}{2!}y^{(3)}(x_n) + \frac{(h)^3}{3!}y^{(4)}(x_n) + \frac{(h)^4}{4!}y^{(5)}(x_n)\] \[+ \frac{(h)^5}{5!}y^{(6)}(x_n) + \frac{(h)^6}{6!}y^{(7)}(x_n) + \frac{(h)^7}{7!}y^{(8)}(x_n) + \frac{(h)^8}{8!}y^{(9)}(x_n) + \cdots\] \[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\] \[+ hy^{(4)}(x_n) + \frac{(h)^2}{2!}y^{(5)}(x_n) + \frac{(h)^3}{3!}y^{(6)}(x_n) + \frac{(h)^4}{4!}y^{(7)}(x_n) + \cdots\] \[+ \omega_{121}(y^{(3)}(x_n) + 2hy^{(4)}(x_n) + \frac{(2h)^2}{2!}y^{(5)}(x_n) + \frac{(2h)^3}{3!}y^{(6)}(x_n)\] \[+ \frac{(2h)^4}{4!}y^{(7)}(x_n) + \cdots) + \omega_{131}(y^{(3)}(x_n) + 3hy^{(4)}(x_n) + \frac{(3h)^2}{2!}y^{(5)}(x_n)\] \[+ \frac{(3h)^3}{3!}y^{(6)}(x_n) + \frac{(3h)^4}{4!}y^{(7)}(x_n) + \cdots) + \varphi_{101}(y^{(4)}(x_n)) + \varphi_{111}(y^{(4)}(x_n)\] \[+ hy^{(5)}(x_n) + \frac{(h)^2}{2!}y^{(6)}(x_n) + \frac{(h)^3}{3!}y^{(7)}(x_n) + \frac{(h)^4}{4!}y^{(8)}(x_n) + \cdots\] \[+ \varphi_{121}(y^{(4)}(x_n) + 2hy^{(5)}(x_n) + \frac{(2h)^2}{2!}y^{(6)}(x_n) + \frac{(2h)^3}{3!}y^{(7)}(x_n)\] \[+ \frac{(2h)^4}{4!}y^{(8)}(x_n) + \cdots) + \varphi_{131}(y^{(4)}(x_n) + 3hy^{(5)}(x_n) + \frac{(3h)^2}{2!}y^{(6)}(x_n)\] \[+ \frac{(3h)^3}{3!}y^{(7)}(x_n) + \frac{(3h)^4}{4!}y^{(8)}(x_n) + \cdots)\]
Equating coefficients of \(y^{(n)}(x_n)\) gives:
\[\begin{split} \frac{(h)^2}{2!} &= \omega_{101} + \omega_{111} + \omega_{121} + \omega_{131} \\ \frac{(h)^3}{3!} &= h\omega_{111} + 2h\omega_{121} + 3h\omega_{131} + \varphi_{101} + \varphi_{111} + \varphi_{121} + \varphi_{131} \\ \frac{(h)^4}{4!} &= \frac{(h)^2}{2!} \omega_{111} + \frac{(2h)^2}{2!} \omega_{121} + \frac{(3h)^2}{2!} \omega_{131} + h\varphi_{111} + 2h\varphi_{121} + 3h\varphi_{131} \\ \frac{(h)^5}{5!} &= \frac{(h)^3}{3!} \omega_{111} + \frac{(2h)^3}{3!} \omega_{121} + \frac{(3h)^3}{3!} \omega_{131} + \frac{(h)^2}{2!} \varphi_{111} + \frac{(2h)^2}{2!} \varphi_{121} + \frac{(3h)^2}{2!} \varphi_{131} \\ \frac{(h)^6}{6!} &= \frac{(h)^4}{4!} \omega_{111} + \frac{(2h)^4}{4!} \omega_{121} + \frac{(3h)^4}{4!} \omega_{131} + \frac{(h)^3}{3!} \varphi_{111} + \frac{(2h)^3}{3!} \varphi_{121} + \frac{(3h)^3}{3!} \varphi_{131} \end{split}\]
\[\frac{(h)^{7}}{7!} = \frac{(h)^{5}}{5!} \omega_{111} + \frac{(2h)^{5}}{5!} \omega_{121} + \frac{(3h)^{5}}{5!} \omega_{131} + \frac{(h)^{4}}{4!} \varphi_{111} + \frac{(2h)^{4}}{4!} \varphi_{121} + \frac{(3h)^{4}}{4!} \varphi_{131}\] \[\frac{(h)^{8}}{8!} = \frac{(h)^{6}}{6!} \omega_{111} + \frac{(2h)^{6}}{6!} \omega_{121} + \frac{(3h)^{6}}{6!} \omega_{131} + \frac{(h)^{5}}{5!} \varphi_{111} + \frac{(2h)^{5}}{5!} \varphi_{121} + \frac{(3h)^{5}}{5!} \varphi_{131}\] \[\frac{(h)^{9}}{9!} = \frac{(h)^{7}}{7!} \omega_{111} + \frac{(2h)^{7}}{7!} \omega_{121} + \frac{(3h)^{7}}{7!} \omega_{131} + \frac{(h)^{6}}{6!} \varphi_{111} + \frac{(2h)^{6}}{6!} \varphi_{121} + \frac{(3h)^{6}}{6!} \varphi_{131}\] \[(17)\]
Rewriting Eq. (17) in matrix form Ax = B gives:
\[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\]
corresponding to the definitions in Eq. (3). The values of the unknown coefficients are given below:
\[\begin{split} &\left(\boldsymbol{\omega}_{101}, \boldsymbol{\omega}_{111}, \boldsymbol{\omega}_{121}, \boldsymbol{\omega}_{131}, \boldsymbol{\varphi}_{101}, \boldsymbol{\varphi}_{111}, \boldsymbol{\varphi}_{121}, \boldsymbol{\varphi}_{131}\right)^T \\ &= \left(\frac{19519h^2}{68040}, \frac{1301h^2}{10080}, \frac{181h^2}{2520}, \frac{3329h^2}{272160}, \frac{371h^3}{12960}, -\frac{313h^3}{2520}, -\frac{89h^3}{2016}, -\frac{137h^3}{45360}\right) \end{split}\]
Similarly, the unknown coefficients for \(y_{n+2}^{(1)}\) and \(y_{n+3}^{(1)}\) are obtained as:
\[\begin{split} &\left(\varpi_{201},\varpi_{211},\varpi_{221},\varpi_{231},\varphi_{201},\varphi_{211},\varphi_{221},\varphi_{231}\right)^T\\ &=\left(\frac{5731h^2}{8505},\frac{296h^2}{315},\frac{109h^2}{315},\frac{344h^2}{8505},\frac{2606h^3}{2835},-\frac{20h^3}{63},-\frac{52h^3}{315},-\frac{4h^3}{405}\right)\\ &\left(\varpi_{301},\varpi_{311},\varpi_{321},\varpi_{331},\varphi_{301},\varphi_{311},\varphi_{321},\varphi_{331}\right)^T\\ &=\left(\frac{603h^2}{560},\frac{2187h^2}{1120},\frac{729h^2}{560},\frac{27h^2}{160},\frac{27h^2}{224},-\frac{243h^3}{560},-\frac{243h^3}{1120},-\frac{9h^3}{280}\right) \end{split}\] while the coefficients for \(y_{n+1}^{(2)},\,y_{n+2}^{(2)}\) and \(y_{n+3}^{(2)}\) are obtained as:
\[\begin{split} &\left(\boldsymbol{\omega}_{102}, \boldsymbol{\omega}_{112}, \boldsymbol{\omega}_{122}, \boldsymbol{\omega}_{132}, \boldsymbol{\varphi}_{102}, \boldsymbol{\varphi}_{112}, \boldsymbol{\varphi}_{122}, \boldsymbol{\varphi}_{132}\right)^T \\ &= \left(\frac{6893h}{18144}, \frac{313h}{672}, \frac{89h}{672}, \frac{397h}{18144}, \frac{1283h^2}{30240}, -\frac{851h^2}{3360}, -\frac{269h^2}{3360}, -\frac{163h^2}{30240}\right) \end{split}\]
\[\begin{split} &\left(\omega_{202},\omega_{212},\omega_{222},\omega_{232},\varphi_{202},\varphi_{212},\varphi_{222},\varphi_{232}\right)^T \\ &= \left(\frac{223h}{567},\frac{20h}{21},\frac{13h}{21},\frac{20h}{567},\frac{43h^2}{945},-\frac{16h^2}{105},-\frac{19h^2}{105},-\frac{8h^2}{945}\right) \\ &\left(\omega_{302},\omega_{312},\omega_{322},\omega_{332},\varphi_{302},\varphi_{312},\varphi_{322},\varphi_{332}\right)^T \\ &= \left(\frac{93h}{224},\frac{243h}{2224},\frac{243h}{2224},\frac{93h}{224},\frac{57h^2}{212},-\frac{81h^2}{1120},\frac{81h^2}{1120},-\frac{57h^2}{1120}\right) \end{split}\]
Substituting all obtained coefficients back in Eq. (14) gives the three-step fourth derivative block method for solving third order ordinary differential equations as:
\[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\]
6 Properties of the Three-Step Fourth-Derivative Block Method
The following properties of the three-step fourth derivative block method are discussed: order, zero-stability, consistency and convergence.
Following the same approach as the two-step third derivative block method, the correctors of the three-step fourth derivative block method are also expanded using Taylor series about \(= x_n\). The order of the three-step fourth derivative block method is obtained to be p = 8.
Secondly, to analyze the three-step fourth derivative block method for zerostability, the modulus of the roots of its first characteristic polynomial is expected to be simple or less than one. Thus, the correctors of the three-step fourth derivative block method are normalized to give the first characteristic
polynomial as \[\rho(r) = det \begin{bmatrix} rI_3 - \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \end{bmatrix}\] with roots satisfying \(|r_j| \le 1\).
The three-step fourth derivative block method is consistent if it has order p > 1as satisfied in the previous paragraphs. Therefore, the three-step fourth derivative block method is convergent [12].
7 Numerical Examples
This section considers certain linear and non-linear second and third order ordinary differential equations. The developed methods in Sections 3 and 4 above are adopted to solve these ODEs (encompassing both initial and boundary value problems) and a comparison is made with previous works.
Problem 1. Consider the non-linear second order initial value problem
\[y'' - x(y')^2 = 0, y(0) = 1, y'(0) = \frac{1}{2}, x \in [0,1]\] with exact solution: \(y = 1 + \frac{1}{2} \ln \left( \frac{2+x}{2-x} \right)\).
This initial value problem was compared with [9], where a hybrid order six block method was presented with \(h = \frac{1}{100}\) as presented in Table 1. The common ground for comparison is in terms of the order of the block methods since the two-step third-derivative block method and the hybrid block method presented by [9] are both of order six.
Problem 2. Consider the linear second order initial value problem
\[y'' + \frac{6}{x}y' + \frac{6}{x^2}y = 0, y(1) = 1, y'(1) = 1, x \in [0, hN]\] with exact solution: \(y = \frac{5}{3x} - \frac{2}{3x^4}\)
The number of steps (N) considered for this initial value problem is N=10 with respect to the selected step-size \(h=\frac{0.1}{32}\). The results as displayed in Table 2 were compared to the solution obtained by order six hybrid block method presented by [13]. The block method compared to [13] is the order six two-step third-derivative block method.
Problem 3. Consider the non-linear second order boundary value problem
\[y'' = \frac{1}{8} (32 + 2x^3 - yy'), y(1) = 17, y(3) = \frac{43}{3}, x \in [1,3]\] with exact solution: \(y = x^2 + \frac{16}{x}\)
The boundary value problem defined here was solved by [14] using an order six self-starting block method and h = 0.1. Thus, since the basis for comparison is the equal order of the methods, the suitable block method is the two-step third-derivative block method. The details of the results obtained can be seen in Table 3.
Problem 4. Consider the special third order initial value problem
\[y''' = -y, y(0) = 1, y'(0) = -1, y''(0) = 1, x \in [0,1]\] with exact solution: \(y = e^{-x}\).
This third order IVP was compared with [4] in which the authors present an order eight block method with h = 0.1. Table 4 shows the comparison of the method in [9] with the order eight three-step fourth-derivative block method.
Problem 5. Consider the non-linear third order boundary value problem
\[y''' = -2e^{-3y} + 4(1+x)^{-3}, y(0) = 0, y'(0) = 1, y(1) = \ln 2, x \in [0,1]\] with exact solution: \(y = \ln(1 - x)\).
The third order BVP defined here was solved by [15] using an order eight block method and compared with the three-step fourth-derivative method also of order eight. The step-size h for this problem varied with respect to the number of steps (N) as shown in Table 5. The maximum error (MAXE) at the boundary x = 1 was recorded for N = 7,14,28,56.
Table 1 Comparison of the Two-Step Third Derivative Block Method with [9] for Solving Problem 1.
| x | Exact Solution | Computed Solution | Error [9] | Error (TSTD) |
|---|---|---|---|---|
| 0.1 | 1.0500417292784914 | 1.0550555592082120 | 6.661338E-16 | 2.220446E-16 |
| 0.2 | 1.1003353477310756 | 1.1003353477310753 | 1.332268E-15 | 2.220446E-16 |
| 0.3 | 1.1511404359364668 | 1.1511404359364668 | 4.440892E-16 | 0.000000E+00 |
| 0.4 | 1.2027325540540823 | 1.2027325540540821 | 1.332268E-15 | 2.220446E-16 |
| 0.5 | 1.2554128118829952 | 1.2554128118829952 | 3.774758E-15 | 0.000000E+00 |
| 0.6 | 1.3095196042031119 | 1.3095196042031116 | 1.065814E-14 | 2.220446E-16 |
| 0.7 | 1.3654437542713964 | 1.3654437542713957 | 2.642331E-14 | 6.661338E-16 |
| 0.8 | 1.4236489301936019 | 1.4236489301936006 | 5.861978E-14 | 1.332268E-15 |
| 0.9 | 1.4847002785940520 | 1.4847002785940489 | 1.265654E-13 | 3.108624E-15 |
| 1.0 | 1.5493061443340550 | 1.5493061443340488 | 2.711165E-13 | 6.217249E-15 |
| CPU Time N/A ≈ 0.59 secs | ||||
Note: TSTD: Two-step third derivative block method.
Table 2 Comparison of the Two-Step Third Derivative Block Method with [13] for Solving Problem 2.
| x | Exact Solution | Computed Solution | Error [13] | Error (TSTD) |
|---|---|---|---|---|
| 1.003125 | 1.0030765258576961 | 1.0030765258576961 | N/A | 0.000000E+00 |
| 1.006250 | 1.0060575030835164 | 1.0060575030835164 | N/A | 0.000000E+00 |
| 1.009375 | 1.0089449950888376 | 1.0089449950888376 | 9.661126E-08 | 0.000000E+00 |
| 1.012500 | 1.0117410181679887 | 1.0117410181679887 | 9.425732E-08 | 0.000000E+00 |
| 1.015625 | 1.0144475426864139 | 1.0144475426864139 | 9.197108E-08 | 0.000000E+00 |
| 1.018750 | 1.0170664942356729 | 1.0170664942356726 | 8.975049E-08 | 2.220446E-16 |
| 1.021875 | 1.0195997547562881 | 1.0195997547562876 | 8.759359E-08 | 4.440892E-16 |
| 1.025000 | 1.0220491636294322 | 1.0220491636294318 | 8.549846E-08 | 4.440892E-16 |
| 1.028125 | 1.0244165187384029 | 1.0244165187384029 | 8.346327E-08 | 0.000000E+00 |
| 1.031250 | 1.0267035775008062 | 1.0267035775008062 | 8.148622E-08 | 0.000000E+00 |
| CPU Time | N/A | ≈ 0.06 secs |
Note: TSTD: Two-step third derivative block method.
Table 3 Comparison of the Two-Step Third Derivative Block Method with [14] for Solving Problem 3.
| x | Exact Solution | Computed Solution | Error [14] | Error (TSTD) |
|---|---|---|---|---|
| 1.1 | 15.755454545454546 | 15.755455083760621 | 1.50E-06 | 5.383061E-07 |
| 1.2 | 14.773333333333333 | 14.773334073676603 | 3.97E-07 | 7.403433E-07 |
| 1.3 | 13.997692307692306 | 13.997693050796402 | 8.12E-08 | 7.431041E-07 |
| 1.4 | 13.388571428571428 | 13.388572125044842 | 1.57E-06 | 6.964734E-07 |
| 1.5 | 12.916666666666666 | 12.916667287517622 | 1.83E-06 | 6.208510E-07 |
| 1.6 | 12.559999999999999 | 12.560000544356562 | 1.79E-06 | 5.443566E-07 |
| 1.7 | 12.301764705882352 | 12.301765175115765 | 1.58E-06 | 4.692334E-07 |
| 1.8 | 12.128888888888888 | 12.128889289986821 | 7.93E-07 | 4.010979E-07 |
| 1.9 | 12.031052631578946 | 12.031052970831665 | 1.07E-06 | 3.392527E-07 |
| 2.0 | 12.000000000000000 | 12.000000284503839 | 1.07E-06 | 2.845038E-07 |
| 2.1 | 12.029047619047619 | 12.029047854947255 | 6.42E-07 | 2.358996E-07 |
| 2.2 | 12.112727272727273 | 12.112727465928481 | 1.27E-06 | 1.932012E-07 |
| 2.3 | 12.246521739130435 | 12.246521894771508 | 2.18E-07 | 1.556411E-07 |
| 2.4 | 12.426666666666666 | 12.426666789477615 | 1.84E-06 | 1.228109E-07 |
| 2.5 | 12.650000000000000 | 12.650000094140001 | 3.82E-07 | 9.414000E-08 |
| \(\boldsymbol{x}\) | Exact Solution | Computed Solution | Error [14] | Error (TSTD) |
|---|---|---|---|---|
| 2.6 | 12.913846153846155 | 12.913846223094138 | 1.86E-06 | 6.924798E-08 |
| 2.7 | 13.215925925925927 | 13.215925973633468 | 1.35E-06 | 4.770754E-08 |
| 2.8 | 13.554285714285715 | 13.554285743482653 | 1.22E-06 | 2.919694E-08 |
| 2.9 | 13.927241379310345 | 13.927241392694905 | 2.75E-06 | 1.338456E-08 |
| 3.0 | 14.33333333333333 | 14.33333333333333 | 0.000000 | 0.000000E+00 |
| CPU Time | N/A | \(\approx 0.12 \text{ secs}\) | ||
Note: TSTD: Two-step third derivative block method.
Table 4 Comparison of the Three-Step Fourth Derivative Block Method with [4] for Solving Problem 4.
| \(\overline{x}\) | Exact Solution | Computed Solution | Error [4] | Error (TSFD) |
|---|---|---|---|---|
| 0.1 | 0.90483741803595952 | 0.90483741803595963 | 2.138401E-12 | 1.110223E-16 |
| 0.2 | 0.81873075307798182 | 0.81873075307798204 | 6.055156E-13 | 2.220446E-16 |
| 0.3 | 0.74081822068171777 | 0.74081822068171832 | 7.395751E-12 | 5.551115E-16 |
| 0.4 | 0.67032004603563933 | 0.67032004603564022 | 2.158163E-12 | 8.881784E-16 |
| 0.5 | 0.60653065971263342 | 0.60653065971263520 | 1.484579E-11 | 1.776357E-15 |
| 0.6 | 0.54881163609402639 | 0.54881163609402939 | 1.098521E-11 | 2.997602E-15 |
| 0.7 | 0.49658530379140947 | 0.49658530379141408 | 3.142886E-11 | 4.607426E-15 |
| 0.8 | 0.44932896411722156 | 0.44932896411722822 | 2.309530E-11 | 6.661338E-15 |
| 0.9 | 0.40656965974059905 | 0.40656965974060832 | 5.154149E-11 | 9.270362E-15 |
| 1.0 | 0.36787944117144222 | 0.36787944117145466 | 8.200535E-11 | 1.243450E-14 |
| CPU | Time | N/A | \(\approx 0.06 \text{ secs}\) |
TSFD: Three-step fourth derivative block method.
Table 5 Comparison of the MAXE for the Three-Step Fourth Derivative Block Method with [15] for Solving Problem 5.
| N | Exact Solution | Computed Solution (TSFD) | CPU Time | MAXE (TSFD) | MAXE [15] |
|---|---|---|---|---|---|
| 7 | 0.69314718055994529 | 0.69314718055994529 | \(\approx 0.10 \text{ secs}\) | 0.000000E+00 | 5.24E-09 |
| 14 | 0.69314718055994529 | 0.69314718055994529 | \(\approx 0.15 \text{ secs}\) | 0.000000E+00 | 2.39E-11 |
| 28 | 0.69314718055994529 | 0.69314718055994529 | \(\approx 0.30 \text{ secs}\) | 0.000000E+00 | 9.50E-14 |
| 56 | 0.69314718055994529 | 0.69314718055994529 | \(\approx 0.61 \text{ secs}\) | 0.000000E+00 | 3.62E-16 |
Note: TSTD: Two-step third derivative block method, TSFD: Three-step fourth derivative block method, N: Number of steps, MAXE: Maximum error, N/A: Not available.
8 Conclusion
This article presented a new approach to developing block methods for solving \(m^{th}\) order ordinary differential equations with the presence of \((m+1)^{th}\). The generalized approach is seen to be quite flexible as the algorithm can simultaneously produce block methods of step length k for solving any order m of ordinary differential equations. The sample methods developed using this new approach were seen to satisfy the basic properties to ensure convergence and their accuracy is also displayed (see Tables 1-5). Thus, this new generalized approach is quite suitable for developing block methods for solving higher order ODEs.