Direct integrator of block type methods with additional derivative for general third order initial value problems

Abstract

This paper presents the construction of the two-point and three-point block methods with additional derivatives for directly solving $y″\prime =f(t,y,y\prime y″)$ . The proposed block methods are formulated using Hermite Interpolating Polynomial and approximate the solution of the problem at two or three-point concurrently. The block methods obtain the numerical solutions directly without reducing the equation into the first order system of initial value problems (IVPs). The order and zero-stability of the proposed methods are also investigated. Numerical results are presented and comparisons with other existing block methods are made. The performance shows that the proposed methods are very efficient in solving the general third order IVPs.

Introduction

Olabode ¹ used block multistep method to solve special third order initial value problems (IVPs) of ordinary differential equations (ODEs). This new block method has desirable property of Runge-Kutta method, which is self-starting. Majid et al. ² derived a two-point implicit block direct integration method to solve third order IVPs. This method estimates the solutions of the IVPs at two points concurrently.

Mehrkanoon ³ used the Gauss Seidel approach to implement three-point implicit block multistep method for directly solving the third order IVPs. Majid et al. ⁴ applied the simple form of Adam’s Moulton method to present two-point 4-step implicit block method for solving the general third order ODEs directly. In the same vein, Bolaji et al. ⁵ developed a 1-step implicit block method using collocation technique to find the approximations of third order ODEs where three sub-step methods were considered. Jator et al. ⁶ constructed a new class of implicit continuous linear multistep methods (LMMs) which were used to solve general third order initial and boundary value problems. Jikantoro et al. ⁷ developed hybrid method for direct integration of special third order ODEs.

Omar et al. ⁸ solved the third order ODEs directly by developing a new predictor-corrector block method of order seven. Abdullahi et al. ⁹ derived predictor-corrector block method of order six to solve third order ODEs directly by using collocation and interpolation approach on power series. In the previously mentioned work, all of them did not use the additional derivative in the formulation of the methods. Most of the methods were also derived using interpolation and collocation technique, the points need to be collocated and interpolated after which a system of linear equations need to be solved simultaneously to obtain the coefficient of the block methods. Hence the process requires more computational effort compared to integration method used in this paper, whereby the function is replaced by the interpolating polynomial and integrating the polynomial at the specific points. This technique is simpler and straight forward.

In this paper, we focus on the derivation of two-point and three-point block methods for solving general third order IVPs, which approximate the solution to the IVP at two or three-point simultaneously. The block methods obtain the numerical solutions directly without reducing the IVP into a system of first order IVPs three times the dimension. In most of the previous work, the methods do not have the extra derivative of the functions. In the new proposed methods, the higher derivative of the function $f(t,y,y\prime ,y″)$ is incorporated into the methods. The purpose of introducing the additional derivative into the formula is to enhance the efficiency in terms of accuracy and computational time. Numerical experiments are carried out to show the superiority of these new methods compared to the existing methods. It is expected that the new methods will outperformed the existing methods in terms precision and they will require less CPU time to do all the computation. The order and zero-stability of the methods are also presented, which clearly show that the methods are convergence.

General third order IVP can be written as:

$$\begin{array}{l}{y}^{‴}=f\left(t,y,y^{\prime },y^{″}\right),y\left(a\right)=y_0,y^{\prime }\left(a\right)=y_0^{\prime }\mathrm{and}\\ y^{″}\left(a\right)=y_0^{\prime \prime }a\le t\le b\end{array}\}$$

(1)

The methods are two-point and three-point block methods. The second derivative of fwith respect to t can be written as

$$y^{\mathrm{iv}}=f\prime (t,y,y\prime ,y″)=f_t+y\prime f_y+y″f_{y\prime }+f{f}_{y″}=g(t,y,y\prime ,y″).$$

The derivation of the new methods are based on Hermite Interpolating Polynomial P, which can be defined by:

$$P\left(t\right)=\sum _{i=0}^n\sum _{k=0}^{m_{i-1}}{f}_i^{\left(k\right)}{L}_{i,k}\left(t\right),$$

(2)

where $f_i=f\left(t\right),t_i=a+\mathrm{ih},i=0,1,\dots ,n$ and $h=\frac{b-a}{n}$ , n is a positive integer. $L_{i,k}\left(t\right)$ can be defined by

$$L_{i,m_i}\left(t\right)={\ell }_{i,m_i}\left(t\right),i=0,1,\dots ,n,$$

$$\begin{array}{c}\multicolumn{1}{c}{{\ell }_{i,k}\left(t\right)=\frac{{(t-t_i)}^k}{k!}\underset{j=0,j\ne i}{\overset{n}{\Pi }}{\left(\frac{t-t_j}{t_i-t_j}\right)}^{m}j,i=0,1,\dots ,n,}\\ \multicolumn{1}{c}{k=0,1,\dots ,m_i.}\end{array}$$

And recursively for $k=m_i-2,m_i-3,\dots ,0.$

$$L_{i,k}\left(t\right)={\ell }_{i,k}\left(t\right)-\sum _{v=k+1}^{m_i-1}{\ell }_{i,k}^{\left(v\right)}\left(t_i\right)L_{i,v}\left(t\right).$$

Derivation of the new methods

In this section, we present the construction of two-point and three-point implicit block methods with additional derivatives.

Two-point additional derivative block method

In two-point block method, each block in the interval [a,b] contains two points. The values of $y_{n+1},$ ${y\prime }_{n+1}$ and ${y″}_{n+1}$ at the first point $t_{n+1}$ can be obtained by integrating (1) once, twice and thrice over the interval $[t_n,t_{n+1}]$ which gives

$$\underset{t_n}{\overset{t_{n+1}}{\int }}y″\prime \mathrm{dt}=\underset{t_n}{\overset{t_{n+1}}{\int }}f(t,y,y\prime ,y″)\mathrm{dt}.$$

(3)

and

$$\underset{t_n}{\overset{t_{n+1}}{\int }}\underset{t_n}{\overset{t}{\int }}y″\prime \mathrm{dt}\mathrm{dt}=\underset{t_n}{\overset{t_{n+1}}{\int }}\underset{t_n}{\overset{t}{\int }}f(t,y,y\prime ,y″)\mathrm{dt}\mathrm{dt}.$$

(4)

$$\underset{t_n}{\overset{t_{n+1}}{\int }}\underset{t_n}{\overset{t}{\int }}\underset{t_n}{\overset{t}{\int }}y″\prime \mathrm{dt}\mathrm{dt}\mathrm{dt}=\underset{t_n}{\overset{t_{n+1}}{\int }}\underset{t_n}{\overset{t}{\int }}\underset{t_n}{\overset{t}{\int }}f(t,y,y\prime ,y″)\mathrm{dt}\mathrm{dt}\mathrm{dt}.$$

(5)

Let $t_{n+1}=t_n+h$ and substituting into (3), (4) and (5), we have

$$y″\left(t_{n+1}\right)=y″\left(t\right)+\underset{t_n}{\overset{t_{n+1}}{\int }}f(t,y,y\prime ,y″)\mathrm{dt},$$

(6)

$$y\prime \left(t_{n+1}\right)=y\prime \left(t_n\right)+\mathrm{hy}″\left(t_n\right)+\underset{x_n}{\overset{x_{n+1}}{\int }}(t_{n+1}-t)f(t,y,y\prime ,y″)\mathrm{dt},$$

(7)

$$y\left(t_{n+1}\right)=y\left(t_n\right)+\mathrm{hy}\prime \left(t_n\right)+\frac{h^2}{2}y″\left(t_n\right)$$

$$+\underset{t_n}{\overset{t_{n+1}}{\int }}\frac{{(t_{n+1}-t)}^2}{2}f(t,y,y\prime ,y″)\mathrm{dt}.$$

(8)

Replace, $f(t,y,y\prime ,y″)$ in (6), (7) and (8) by Hermite interpolating polynomial in (2), which is defined as $P_2\left(t\right)$ , where

$$\begin{array}{l}{P}_2\left(t\right)=[{\left(\frac{t-t_{n+1}}{t_n-t_{n+1}}\right)}^2{\left(\frac{t-t_{n+2}}{t_n-t_{n+2}}\right)}^2+\left(\frac{2}{t_n-t_{n+1}}\right)\left(\frac{2}{t_n-t_{n+2}}\right)\left(t-t_n\right){\left(\frac{t-t_{n+1}}{t_n-t_{n+1}}\right)}^2\\ {\left(\frac{t-t_{n+2}}{t_n-t_{n+2}}\right)}^2\left]f_0+\right[{\left(\frac{t-t_{n+1}}{t_n-t_{n+1}}\right)}^2{\left(\frac{t-t_{n+2}}{t_n-t_{n+2}}\right)}^2\left]f_1+\right[{\left(\frac{t-t_n}{t_{n+2}-t_n}\right)}^2\\ {\left(\frac{t-t_{n+1}}{t_{n+2}-t_{n+1}}\right)}^2+\left(\frac{2}{t_{n+2}-t_{n+1}}\right)\left(\frac{2}{t_{n+2}-t_n}\right)\left(t-t_{n+2}\right){\left(\frac{t-t_n}{t_{n+2}-t_n}\right)}^2\\ \frac{t-t_{n+1}}{t_{n+2}-t_{n+1}}{)}^2\left]f_2+\right[\left(t-t_n\right){\left(\frac{t-t_{n+1}}{t_n-t_{n+1}}\right)}^2{\left(\frac{t-t_{n+2}}{t-t_{n+2}}\right)}^2\left]g_0+\right[\left(t-t_{n+1}\right)\\ {\left(\frac{t-t_n}{t_{n+1}-t_n}\right)}^2{\left(\frac{t-t_{n+2}}{t-t}\right)}^2\left]g_1+\right[\left(t-t_{n+2}\right)\\ {\left(\frac{t-t_n}{t_{n+2}-t_n}\right)}^2{\left(\frac{t-t_{n+1}}{t_{n+2}-t_{n+1}}\right)}^2]g_2\end{array}$$

(9)

Let t $=t_{n+2}+\mathrm{sh}$ and

$$s=\frac{t-t_{n+2}}{h}.$$

(10)

Substitute $\mathrm{dt}=\mathrm{hds}$ and change the integration limits from −2 to −1 in (6), (7) and (8) to obtain

$$y″\left(t_{n+1}\right)=y″\left(t_n\right)+\underset{-2}{\overset{-1}{\int }}[f_0{L}_{0,0}\left(s\right)+f_1{L}_{1,0}\left(s\right)+f_2{L}_{2,0}\left(s\right)$$

$$+g_0{L}_{0,1}\left(s\right)+g_2{L}_{2,1}\left(s\right)]\mathrm{hds}.$$

(11)

$$\begin{array}{c}\multicolumn{1}{c}{y\prime \left(t_{n+1}\right)=y\prime \left(t_n\right)+\mathrm{hy}″\left(t_n\right)}\\ +\underset{-2}{\overset{-1}{\int }}(-h-\mathrm{sh})[f_0{L}_{0,0}\left(s\right)+f_1{L}_{1,0}\left(s\right)+f_2{L}_{2,0}\left(s\right)\hfill \end{array}$$

$$+g_0{L}_{0,1}\left(s\right)+g_2{L}_{2,1}\left(s\right)]\mathrm{hds}.$$

(12)

$$\begin{array}{c}\multicolumn{1}{c}{y\left(t_{n+1}\right)=y\left(t_n\right)+\mathrm{hy}\prime \left(t_n\right)+\frac{h^2}{2}y″\left(t_n\right)}\\ +\underset{-2}{\overset{-1}{\int }}\frac{{(-h-\mathrm{sh})}^2}{2}[f_0{L}_{0,0}\left(s\right)+f_1{L}_{1,0}\left(s\right)+f_2{L}_{2,0}\left(s\right)\hfill \end{array}$$

$$+g_0{L}_{0,1}\left(s\right)+g_2{L}_{2,1}\left(s\right)]\mathrm{hds}.$$

(13)

where

$$L_{0,0}\left(s\right)=-\frac{1}{4}(s^3+s^2)-\frac{1}{2}(s^3+2s^2)(s+1),$$

$$L_{1,0}\left(s\right)=s^2(s+2{)}^2,$$

$$L_{2,0}\left(s\right)=\frac{1}{4}(s+2{)}^2(s+1)-\frac{1}{2}(s^2+s)(s+2{)}^2,$$

$$L_{0,1}\left(s\right)=-\frac{h}{4}{s}^2(s+2)(s+1),$$

$$L_{2,1}\left(s\right)=\frac{h}{4}s(s+2{)}^2(s+1).$$

Evaluating the integrals in (11), (12) and (13), results in the formula at the first point as follows,

$$\begin{array}{c}\multicolumn{1}{c}{y_{n+1}″=y_n″+\frac{h}{240}[131f_n+128f_{n+1}-19f_{n+2}]}\\ +\frac{h^2}{240}[23g_n+7g_{n+2}],\hfill \end{array}$$

(14)

$$\begin{array}{c}\multicolumn{1}{c}{y_{n+1}\prime =y_n\prime +{\mathrm{hy}}_n″+\frac{h^2}{30}[11f_n+5f_{n+1}-f_{n+2}]}\\ +\frac{h^3}{240}[13g_n+3g_{n+2}],\hfill \end{array}$$

(15)

$$\begin{array}{c}\multicolumn{1}{c}{y_{n+1}=y_n+{\mathrm{hy}}_n\prime +\frac{h^2}{2}{y}_n″+\frac{h^3}{3360}}\\ [461f_n+128f_{n+1}-29f_{n+2}]\hfill \\ +\frac{h^4}{3360}[59g_n+11g_{n+2}].\hfill \end{array}$$

(16)

Integrating (1) once, twice and thrice over the $[t_{n+1},t_{n+2}]$ to acquire the approximate solutions of $y_{n+2}$ , $y_{n+2}\prime$ and $y_{n+2}″,$ taking $t_{n+2}=t_{n+1}+h.$

$$y″\left(t_{n+2}\right)=y″\left(t_{n+1}\right)+\underset{t_{n+1}}{\overset{t_{n+2}}{\int }}f(t,y,y\prime ,y″)\mathrm{dt},$$

(17)

$$\begin{array}{c}\multicolumn{1}{c}{y\prime \left(t_{n+2}\right)=y\prime \left(t_{n+1}\right)+\mathrm{hy}″\left(t_{n+1}\right)}\\ +\underset{t_{n+1}}{\overset{t_{n+2}}{\int }}(t_{n+2}-t)f(t,y,y″,y\prime )\mathrm{dt},\hfill \end{array}$$

(18)

$$\begin{array}{c}\multicolumn{1}{c}{y\left(t_{n+2}\right)=y\left(t_{n+1}\right)+\mathrm{hy}\prime \left(t_{n+1}\right)+\frac{h^2}{2}y″\left(t_{n+1}\right)}\\ +\underset{t_{n+1}}{\overset{t_{n+2}}{\int }}\frac{{(t_{n+2}-t)}^2}{2}f(t,y,y\prime ,y″)\mathrm{dt}.\hfill \end{array}$$

(19)

Then, replace $f(t,y,y\prime ,y″)$ in (17), (18) and (19) by Hermite interpolating polynomial in (9). Letting $\mathrm{dt}=\mathrm{hds}$ and replacing the lower limit of the integration by −1 and the upper limit by 0, gives

$$\begin{array}{c}\multicolumn{1}{c}{y″\left(t_{n+2}\right)=y″\left(t_{n+1}\right)}\\ +\underset{-1}{\overset{0}{\int }}[f_0{L}_{0,0}\left(s\right)+f_1{L}_{1,0}\left(s\right)+f_2{L}_{2,0}\left(s\right)\hfill \end{array}$$

$$+g_0{L}_{0,1}\left(s\right)+g_2{L}_{2,1}\left(s\right)]\mathrm{hds}.$$

(20)

$$\begin{array}{c}\multicolumn{1}{c}{y\prime \left(t_{n+2}\right)=y\prime \left(t_{n+1}\right)+\mathrm{hy}″\left(t_{n+1}\right)}\\ +\underset{-1}{\overset{0}{\int }}(-\mathrm{sh})[f_0{L}_{0,0}\left(s\right)+f_1{L}_{1,0}\left(s\right)+f_2{L}_{2,0}\left(s\right)\hfill \end{array}$$

$$+g_0{L}_{0,1}\left(s\right)+g_2{L}_{2,1}\left(s\right)]\mathrm{hds}.$$

(21)

$$\begin{array}{c}\multicolumn{1}{c}{y\left(t_{n+2}\right)=y\left(t_{n+1}\right)+\mathrm{hy}\prime \left(t_{n+1}\right)+\frac{h^2}{2}y″\left(t_{n+1}\right)}\\ +\underset{-1}{\overset{0}{\int }}\frac{{(\mathrm{sh})}^2}{2}[f_0{L}_{0,0}\left(s\right)+f_1{L}_{1,0}\left(s\right)\hfill \end{array}$$

$$f_2{L}_{2,0}\left(s\right)+g_0{L}_{0,1}\left(s\right)+g_2{L}_{2,1}\left(s\right)]\mathrm{hds}.$$

(22)

Evaluating the integrals to obtain the formula at the second point as follows,

$$\begin{array}{c}\multicolumn{1}{c}{y_{n+2}″=y_{n+1}″+\frac{h}{240}[-19f_n+128f_{n+1}+131f_{n+2}]}\\ +\frac{h^2}{240}[-7g_n-23g_{n+2}].\hfill \end{array}$$

(23)

$$\begin{array}{c}{y}_{n+2}\prime =y_{n+1}\prime +{\mathrm{hy}}_{n+1}″+\frac{h^2}{240}\hfill \\ [-11f_n+88f_{n+1}+43f_{n+2}]+\frac{h^3}{120}[-2g_n-5g_{n+2}].\hfill \end{array}$$

(24)

$$\begin{array}{c}{y}_{n+2}=y_{n+1}+{\mathrm{hy}}_{n+1}\prime +\frac{h^2}{2}{y}_{n+1}″+\frac{h^3}{1680}\hfill \\ [-25f_n+232f_{n+1}+73f_{n+2}]+\frac{h^4}{1680}[-9g_n-19g_{n+2}].\hfill \end{array}$$

(25)

The set of formulae in equations (14), (15), (16), (39), (24) and (25) is denoted as 2PSDIBM(3) method.

Three-point additional derivative block method

In the three-point implicit block method, each block in the interval [a,b] contains three points. The values of $y_{n+1}$ , $y_{n+2}$ and $y_{n+3}$ at the point $t_{n+1}$ , $t_{n+2}$ and $t_{n+3}$ are concurrently evaluated in a block.

The derivation of the three-point block method is similar to the previous construction of the two-point block method. Equation (1) will be integrated once, twice and thrice on the interval $[t_n,t_{n+1}]$ , $[t_{n+1},t_{n+2}]$ and $[t_{n+2},t_{n+3}]$ to obtain the values of $y_{n+1},y_{n+1}\prime ,y_{n+1}″,$ $y_{n+2},y_{n+2}\prime ,y_{n+2}″,y_{n+3},$ $y_{n+3}\prime$ and $y_{n+3}″$ . Define $P_3\left(t\right)$ as follows,

$$\begin{array}{c}\multicolumn{1}{c}{P_3\left(t\right)=[\left(\frac{t-t_{n+1}}{t_n-t_{n+1}}\right)\left(\frac{t-t_{n+1}}{t_n-t_{n+1}}\right){\left(\frac{t-t_{n+3}}{t_n-t_{n+3}}\right)}^2}\\ +(\left(\frac{-1}{t_n-t_{n+1}}\right)+\left(\frac{-1}{t_n-t_{n+2}}\right)+\left(\frac{-2}{t_n-t_{n+3}}\right))\hfill \\ \left((t-t_n)\left(\frac{t-t_{n+1}}{t_n-t_{n+1}}\right)\left(\frac{t-t_{n+2}}{t_n-t_{n+2}}\right){\left(\frac{t-t_{n+3}}{t_n-t_{n+3}}\right)}^2\right)]f_0\hfill \\ +[{(\frac{t-t_n}{t_{n+1}-t_n})}^2\left(\frac{t-t_{n+2}}{t_{n+1}-t_{n+2}}\right){(\frac{t-t_{n+3}}{t_{n+1}-t_{n+3}})}^2]f_1\\ +[{(\frac{t-t_n}{t_{n+2}-t_n})}^2\left(\frac{t-t_{n+1}}{t_{n+2}-t_{n+1}}\right){(\frac{t-t_{n+3}}{t_{n+2}-t_{n+3}})}^2]f_2\\ +[{(\frac{t-t_n}{t_{n+3}-t_n})}^2\left(\frac{t-t_{n+1}}{t_{n+3}-t_{n+1}}\right)\left(\frac{t-t_{n+2}}{t_{n+3}-t_{n+2}}\right)\\ -(\left(\frac{2}{t_{n+3}-t_n}\right)+\left(\frac{1}{t_{n+3}-t_{n+1}}\right)+\left(\frac{1}{t_{n+3}-t_{n+2}}\right))\\ (\left(t-t_{n+3}\right){(\frac{t-t_n}{t_{n+3}-t_n})}^2\left(\frac{t-t_{n+1}}{t_{n+3}-t_{n+1}}\right)\left(\frac{t-t_{n+2}}{t_{n+3}-t_{n+2}}\right))]f_3\\ +(\left(t-t_n\right)\left(\frac{t-t_{n+1}}{t_n-t_{n+1}}\right)\left(\frac{t-t_{n+2}}{t_n-t_{n+2}}\right){(\frac{t-t_{n+3}}{t_n-t_{n+3}})}^2)g_0\\ +(\left(t-t_{n+3}\right)\left(\frac{t-t_n}{t_{n+3}-t_n}{)}^2\left(\frac{t-t_{n+1}}{t_{n+3}-t_{n+1}}\right)\left(\frac{t-t_{n+2}}{x_{n+3}-x_{n+2}}\right)\right)g_3.\end{array}$$

(26)

Replace $f(t,y,y\prime ,y″)$ by polynomial in (26), let t $=t_{n+3}+\mathrm{sh},$ hence, $s=\frac{t-t_{n+3}}{h}.$ The formula for the first, second and third point can be written as follows:

$$\begin{array}{c}{y}_{n+1}″=y_n″+\frac{h}{6480}\hfill \\ [3463f_n+3537f_{n+1}-783f_{n+2}+263f_{n+3}]\hfill \end{array}$$

$$+\frac{h^2}{1080}[97g_n-17g_{n+3}].$$

(27)

$$\begin{array}{c}{y}_{n+1}\prime =y_n\prime +{\mathrm{hy}}_n″+\frac{h^2}{45360}\hfill \\ [16384f_n+7857f_{n+1}-2376f_{n+2}+815f_{n+3}]\hfill \end{array}$$

$$+\frac{h^3}{7560}[388g_n-53g_{n+3}].$$

(28)

$$\begin{array}{c}{y}_{n+1}=y_n+{\mathrm{hy}}_n\prime +\frac{h^2}{2}{y}_n″+\frac{h^3}{20160}\hfill \\ [2735f_n+807f_{n+1}-279f_{n+2}+97f_{n+3}]\hfill \end{array}$$

$$+\frac{h^4}{10080}[169g_n-19g_{n+3}].$$

(29)

$$\begin{array}{c}{y}_{n+2}″=y_{n+1}″+\frac{h}{80}\hfill \\ [-7f_n+47f_{n+1}+47f_{n+2}-7f_{n+3}]\hfill \end{array}$$

$$+\frac{h^2}{360}[-11g_n+11g_{n+3}].$$

(30)

$$\begin{array}{c}{y}_{n+2}\prime =y_{n+1}\prime +{\mathrm{hy}}_{n+1}″+\frac{h^2}{15120}\hfill \\ [-715f_n+5832f_{n+1}+3051f_{n+2}-608f_{n+3}]\hfill \end{array}$$

$$+\frac{h^3}{2520}[-41g_n+36g_{n+3}].$$

(31)

$$\begin{array}{c}\multicolumn{1}{c}{y_{n+2}=y_{n+1}+{\mathrm{hy}}_{n+1}\prime +\frac{h^2}{2}{y}_{n+1}″+\frac{h^3}{60480}}\\ [-899f_n+8613f_{n+1}+3051f_{n+2}-685f_{n+3}]\hfill \end{array}$$

$$+\frac{h^4}{10080}[-51g_n+41g_{n+3}].$$

(32)

$$\begin{array}{c}{y}_{n+3}″=y_{n+2}″+\frac{h}{6480}\hfill \\ [263f_n-783f_{n+1}+3537f_{n+2}+3463f_{n+3}]\hfill \end{array}$$

$$+\frac{h^2}{1080}[17g_n-97g_{n+3}].$$

(33)

$$\begin{array}{c}{y}_{n+3}\prime =y_{n+2}\prime +{\mathrm{hy}}_{n+2}″+\frac{h^2}{1680}\hfill \\ [38f_n-115f_{n+1}+626f_{n+2}+291f_{n+3}]\hfill \end{array}$$

$$+\frac{h^3}{2520}[22g_n-97g_{n+3}].$$

(34)

$$\begin{array}{c}\multicolumn{1}{c}{y_{n+3}=y_{n+2}+{\mathrm{hy}}_{n+2}\prime +\frac{h^2}{2}{y}_{n+2}″}\\ +\frac{h^3}{181440}[1295f_n-3969f_{n+1}+25353f_{n+2}\hfill \end{array}$$

$$7561f_{n+3}]+\frac{h^4}{30240}[83g_n-313g_{n+3}].$$

(35)

This method is denoted as the 3PSDBI(3) method.

Order conditions and error constant of the methods

The local truncation error associated with the normalized form of the proposed method can be defined as the linear difference operator

$$\begin{array}{c}L[Z\left(t\right);h]=\hfill \\ \sum _{i=0}^k[{\alpha }_{i}Z(t+\mathrm{jh})-h\beta Z_i\prime (t+\mathrm{jh})-\cdots -h^d{\gamma }_i{Z}^{\left(d\right)}(t+\mathrm{jh})],\hfill \end{array}$$

(36)

where d is the order of the differential equation and $Z\left(t\right)$ is an arbitrary function, which satisfied the conditions for continuity and differentiability, see Fatunla ¹⁰ and Lambert ¹¹ for further details. Using Taylor series at point t to expand $Z(t+\mathrm{jh})$ and $Z^{\left(d\right)}(t+\mathrm{jh})$ in (36) gives

$L[Z\left(t\right);h]=C_{0}Z\left(t\right)+C_1\mathrm{hZ}\prime \left(t\right)+C_p{h}^p{Z}^p\left(t\right)+\dots +C_{p+1}{h}^{p+1}{Z}^{p+1}\left(t\right)+\dots ,$ the constant coefficients $\mathrm{Cp},p=0,1,\dots$ are given as follows:

$$\begin{array}{c}{C}_0=\sum _{i=0}^k{\alpha }_j,\hfill \\ C_1=\sum _{i=0}^{k}j{\alpha }_j-\sum _{i=0}^k{\beta }_j,\hfill \\ C_2=\frac{1}{2!}\sum _{i=0}^k{j}^2{\alpha }_j-\sum _{i=0}^{k}j{\beta }_j-\sum _{i=0}^k{\gamma }_j\hfill \\ \multicolumn{1}{c}{⋮}\\ C_p=\frac{1}{p!}\sum _{i=0}^k{j}^p{\alpha }_j-\frac{1}{(p-1)!}\sum _{i=0}^k{j}^{p-1}{\beta }_j\hfill \\ -\frac{1}{(p-2)!}\sum _{i=0}^k{j}^{p-2}{\gamma }_j-\frac{1}{(p-3)!}\sum _{i=0}^k{j}^{p-3}{\psi }_j-\frac{1}{(p-4)!}\hfill \\ \multicolumn{1}{c}{\sum _{i=0}^k{j}^{p-4}{\delta }_j,p=3,4,\dots }\end{array}$$

(37)

The proposed method has order p if $C_0=C_1=\dots C_p=C_{p+1}=C_{p+2}=0,C_{p+3}\ne 0.$ Therefore, $C_{p+3}$ is the error constant and $C_{p+3}{h}^{p+3}{Z}^{(p+3)}\left(t_n\right)$ is the principal local truncation error at the point $t_n$ . The two-point block method given by (14), (15), (16), (23), (24) and (25) can be written in a matrix as follows:

$$\alpha Y_m=h\beta Y_m\prime +h^2\gamma Y_m″+h^3\psi F_m+h^3\delta G_m$$

(38)

where $\alpha ,\beta ,\gamma$ , $\psi$ and $\delta$ are the coefficients with the m-vector $Y_m,Y_m\prime ,Y_m″,F_m$ and $G_m$ be defind as,

$$\begin{array}{c}\multicolumn{1}{c}{\alpha =\left[\begin{array}{cccccc}0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & -1\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & -1\hfill & 1\hfill \end{array}\right],}\\ \multicolumn{1}{c}{\beta =\left[\begin{array}{cccccc}0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 1\hfill & -1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & -1\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \end{array}\right],}\end{array}$$

$$\begin{array}{c}\multicolumn{1}{c}{\gamma =\left[\begin{array}{cccccc}0\hfill & 0\hfill & 0\hfill & 1\hfill & -1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{1}{2}\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & -1\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & \frac{1}{2}\hfill & 0\hfill \end{array}\right],}\\ \multicolumn{1}{c}{\psi =\left[\begin{array}{cccccc}0\hfill & 0\hfill & 0\hfill & \frac{131}{240}\hfill & \frac{128}{240}\hfill & \frac{-19}{240}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{11}{30}\hfill & \frac{5}{30}\hfill & \frac{-1}{30}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{461}{3360}\hfill & \frac{128}{3360}\hfill & \frac{-29}{3360}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{-19}{240}\hfill & \frac{128}{240}\hfill & \frac{131}{240}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{-11}{240}\hfill & \frac{88}{240}\hfill & \frac{43}{240}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{-25}{1680}\hfill & \frac{232}{1680}\hfill & \frac{73}{1680}\hfill \end{array}\right],}\end{array}$$

$$\delta =\left[\begin{array}{cccccc}0\hfill & 0\hfill & 0\hfill & \frac{23}{240}\hfill & 0\hfill & \frac{7}{240}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{13}{240}\hfill & 0\hfill & \frac{3}{240}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{59}{3360}\hfill & 0\hfill & \frac{11}{3360}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{-7}{240}\hfill & 0\hfill & \frac{-23}{240}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{-2}{120}\hfill & 0\hfill & \frac{-5}{120}\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & \frac{-9}{1680}\hfill & 0\hfill & \frac{-19}{1680}\hfill \end{array}\right],$$

$$\begin{array}{c}\multicolumn{1}{c}{Y_m=\left[\begin{array}{c}{y}_{n-3}\hfill \\ \multicolumn{1}{c}{y_{n-2}}\\ \multicolumn{1}{c}{y_{n-1}}\\ \multicolumn{1}{c}{y_n}\\ \multicolumn{1}{c}{y_{n+1}}\\ \multicolumn{1}{c}{y_{n+2}}\end{array}\right],Y_m\prime =\left[\begin{array}{c}{y\prime }_{n-3}\hfill \\ \multicolumn{1}{c}{{y\prime }_{n-2}}\\ \multicolumn{1}{c}{{y\prime }_{n-1}}\\ \multicolumn{1}{c}{{y\prime }_n}\\ \multicolumn{1}{c}{{y\prime }_{n+1}}\\ \multicolumn{1}{c}{y_{n+2}^{\mathrm{me}}}\end{array}\right],}\\ \multicolumn{1}{c}{Y_m″=\left[\begin{array}{c}{y″}_{n-3}\hfill \\ \multicolumn{1}{c}{{y″}_{n-2}}\\ \multicolumn{1}{c}{{y″}_{n-1}}\\ \multicolumn{1}{c}{{y″}_n}\\ \multicolumn{1}{c}{{y″}_{n+1}}\\ \multicolumn{1}{c}{y_{n+2}″}\end{array}\right],F_m=\left[\begin{array}{c}{f}_{n-3}\hfill \\ \multicolumn{1}{c}{f_{n-2}}\\ \multicolumn{1}{c}{f_{n-1}}\\ \multicolumn{1}{c}{f_n}\\ \multicolumn{1}{c}{f_{n+1}}\\ \multicolumn{1}{c}{f_{n+2}}\end{array}\right],G_m=\left[\begin{array}{c}{g}_{n-3}\hfill \\ \multicolumn{1}{c}{g_{n-2}}\\ \multicolumn{1}{c}{g_{n-1}}\\ \multicolumn{1}{c}{g_n}\\ \multicolumn{1}{c}{g_{n+1}}\\ \multicolumn{1}{c}{g_{n+2}}\end{array}\right],}\end{array}$$

By substituting these matrices into equation (37) we have

$$C_0=C_1=C_2=C_3=C_4=C_5=C_6=C_7=\overline{0},$$

where $\overline{0}$ is the zero vector which can be written as $\overline{0}=[0,0,0,0,0,0{]}^T.$

For $p=8,$ it is found that $C_8\ne \overline{0},$ where

$$\begin{array}{c}\multicolumn{1}{c}{C_8=\frac{1}{8!}\sum _{j=0}^5{j}^8{\alpha }_j-\frac{1}{\left(7\right)!}\sum _{j=0}^5{j}^7{\beta }_j-\frac{1}{\left(6\right)!}\sum _{j=0}^5{j}^6{\gamma }_j-\frac{1}{\left(5\right)!}}\\ \multicolumn{1}{c}{\sum _{j=0}^5{j}^5{\psi }_j-\frac{1}{\left(4\right)!}\sum _{j=0}^5{j}^4{\delta }_j\ne \overline{0}.}\end{array}$$

Then, the order of the two-point block method is $5$ with error constant, $C_8=[\frac{-1}{720},\frac{-1}{1575},\frac{-1}{5760},\frac{1}{720},\frac{19}{25200},\frac{47}{201600}{]}^T.$

The formulae of the three-point implicit block method given by equations (27) to (35) can be written as in (38) with $\alpha ,\beta ,\gamma$ and $\delta$ depend on the coefficients of the method. While

$$\begin{array}{c}\multicolumn{1}{c}{Y_m=\left[\begin{array}{c}{y}_{n-5}\hfill \\ \multicolumn{1}{c}{y_{n-4}}\\ \multicolumn{1}{c}{y_{n-3}}\\ \multicolumn{1}{c}{y_{n-2}}\\ \multicolumn{1}{c}{y_{n-1}}\\ \multicolumn{1}{c}{y_n}\\ \multicolumn{1}{c}{y_{n+1}}\\ \multicolumn{1}{c}{y_{n+2}}\\ \multicolumn{1}{c}{y_{n+3}}\end{array}\right],Y_m\prime =\left[\begin{array}{c}{y\prime }_{n-5}\hfill \\ \multicolumn{1}{c}{{y\prime }_{n-4}}\\ \multicolumn{1}{c}{{y\prime }_{n-3}}\\ \multicolumn{1}{c}{{y\prime }_{n-2}}\\ \multicolumn{1}{c}{{y\prime }_{n-1}}\\ \multicolumn{1}{c}{{y\prime }_n}\\ \multicolumn{1}{c}{{y\prime }_{n+1}}\\ \multicolumn{1}{c}{{y\prime }_{n+2}}\\ \multicolumn{1}{c}{{y\prime }_{n+3}}\end{array}\right],Y_m″=\left[\begin{array}{c}{y″}_{n-5}\hfill \\ \multicolumn{1}{c}{{y″}_{n-4}}\\ \multicolumn{1}{c}{{y″}_{n-3}}\\ \multicolumn{1}{c}{{y″}_{n-2}}\\ \multicolumn{1}{c}{{y″}_{n-1}}\\ \multicolumn{1}{c}{{y″}_n}\\ \multicolumn{1}{c}{{y″}_{n+1}}\\ \multicolumn{1}{c}{{y″}_{n+2}}\\ \multicolumn{1}{c}{{y″}_{n+3}}\end{array}\right],}\end{array}$$

$$F_m=\left[\begin{array}{c}{f}_{n-5}\hfill \\ \multicolumn{1}{c}{f_{n-4}}\\ \multicolumn{1}{c}{f_{n-3}}\\ \multicolumn{1}{c}{f_{n-2}}\\ \multicolumn{1}{c}{f_{n-1}}\\ \multicolumn{1}{c}{f_n}\\ \multicolumn{1}{c}{f_{n+1}}\\ \multicolumn{1}{c}{f_{n+2}}\\ \multicolumn{1}{c}{f_{n+3}}\end{array}\right],G_m=\left[\begin{array}{c}{g}_{n-5}\hfill \\ \multicolumn{1}{c}{g_{n-4}}\\ \multicolumn{1}{c}{g_{n-3}}\\ \multicolumn{1}{c}{g_{n-2}}\\ \multicolumn{1}{c}{g_{n-1}}\\ \multicolumn{1}{c}{g_n}\\ \multicolumn{1}{c}{g_{n+1}}\\ \multicolumn{1}{c}{g_{n+2}}\\ \multicolumn{1}{c}{g_{n+3}}\end{array}\right],$$

Substituting into equation (37), we obtained

$$C_0=C_1=C_2=C_3=C_4=C_5=C_6=C_7=C_8=\overline{0},$$

where $\overline{0}$ is the zero vector which can be written as $\overline{0}=[0,0,0,0,0,0,0,0,0{]}^T.$

For $p=9,$

$$\begin{array}{c}\multicolumn{1}{c}{C_9=\frac{1}{9!}\sum _{j=0}^8{j}^9{\alpha }_j-\frac{1}{\left(8\right)!}\sum _{j=0}^8{j}^8{\beta }_j-\frac{1}{\left(7\right)!}\sum _{j=0}^8{j}^7{\gamma }_j}\\ \multicolumn{1}{c}{-\frac{1}{\left(6\right)!}\sum _{j=0}^8{j}^6{\psi }_j-\frac{1}{\left(5\right)!}\sum _{j=0}^8{j}^5{\delta }_j}\end{array}$$

$$\ne \overline{0}.$$

Then, the order of the three-point block method is $6$ with error constant $C_9=[\frac{97}{100800},\frac{269}{604800},\frac{89}{725760},\frac{-113}{100800},\frac{-113}{201600},\frac{-121}{725760},\frac{97}{100800},\frac{313}{604800},\frac{577}{3628800}{]}^T.$ Hence, the method is said to be consistent since the order is higher than three.

Zero-stability of the methods

In this section, zero-stability of the proposed methods is discussed. For the two-point block method, substituting equation (14) into (23), gives

$$\begin{array}{c}{y}_{n+2}″=y_n″+\frac{h}{15}\hfill \\ [7f_n+16f_{n+1}+7f_{n+2}]+\frac{h^2}{15}[g_n-g_{n+2}].\hfill \end{array}$$

(39)

Substituting equations (14) and (15) into (24), gives

$$\begin{array}{c}{y}_{n+2}\prime =y_n\prime +2{\mathrm{hy}}_n″+\frac{h^2}{15}\hfill \\ [13f_n+16f_{n+1}+f_{n+2}]+\frac{h^3}{15}\left[2g_n\right],\hfill \end{array}$$

(40)

Also by substituting equations (14), (15) and (16) into (25), we have

$$\begin{array}{c}\multicolumn{1}{c}{y_{n+2}=y_n+2{\mathrm{hy}}_n\prime +2h^2{y}_n″}\\ \multicolumn{1}{c}{+h^3[\frac{16}{21}{f}_n+\frac{64}{105}{f}_{n+1}-\frac{4}{105}{f}_{n+2}]}\end{array}$$

$$+h^4[\frac{4}{35}{g}_n+\frac{2}{105}{g}_{n+2}].$$

(41)

Write the formulae in (14), (15), (16), (39), (40), and (41) matrix form and the zero-stability polynomial of the method can be obtained from

$$\begin{array}{c}\multicolumn{1}{c}{\left[\begin{array}{cccccc}1\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 1\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 1\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right]\left[\begin{array}{c}{y}_{n+1}″\hfill \\ \multicolumn{1}{c}{y_{n+1}\prime }\\ \multicolumn{1}{c}{y_{n+1}}\\ \multicolumn{1}{c}{y_{n+2}″}\\ \multicolumn{1}{c}{y_{n+2}\prime }\\ \multicolumn{1}{c}{y_{n+2}}\end{array}\right]=\left[\begin{array}{cccccc}0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right]}\\ \multicolumn{1}{c}{\left[\begin{array}{c}{y}_{n-1}″\hfill \\ \multicolumn{1}{c}{y_{n-1}\prime }\\ \multicolumn{1}{c}{y_{n-1}}\\ \multicolumn{1}{c}{y_n″}\\ \multicolumn{1}{c}{y_n\prime }\\ \multicolumn{1}{c}{y_n}\end{array}\right]}\end{array}$$

The zero-stability polynomial of the two-point block method is,

$$\rho (R){=}_{\det }[R{M}^{\left(0\right)}-M^{\left(1\right)}]=0,$$

where

$$\begin{array}{c}{M}^{\left(0\right)}=\left[\begin{array}{cccccc}1\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 1\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 1\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right]\mathrm{and}\hfill \\ M^{\left(1\right)}=\left[\begin{array}{cccccc}0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right].\hfill \end{array}$$

$$\begin{array}{c}\multicolumn{1}{c}{\rho \left(R\right)=R^3(R^3-3R^2+3R-1)}\\ \multicolumn{1}{c}{=0,R=0,0,0,1,1,1,\left|R\right|\le 1.}\end{array}$$

The two-point implicit block method is zero-stable, since, $\rho \left(R\right)$ has roots $,\left|R\right|\le 1$ , and that the multiplicity of the roots with modulus one be at most 3.

For the three-point implicit block method, substituting equation (27) into (30), we have

$$y_{n+2}″=y_n″+h[\frac{181}{405}{f}_n+\frac{17}{15}{f}_{n+1}+\frac{7}{15}{f}_{n+2}-\frac{19}{405}{f}_{n+3}]$$

$$+\frac{h^2}{135}[8g_n+2g_{n+3}].$$

(42)

Substituting equations (27) and (28) into (31), will give

$$\begin{array}{c}\multicolumn{1}{c}{y_{n+2}\prime =y_n\prime +2{\mathrm{hy}}_n″+h^2}\\ \multicolumn{1}{c}{[\frac{481}{567}{f}_n+\frac{116}{105}{f}_{n+1}+\frac{1}{35}{f}_{n+2}+\frac{52}{2835}{f}_{n+3}]}\end{array}$$

$$+\frac{h^3}{945}[118g_n-8g_{n+3}]].$$

(43)

Also by substituting equations (27), (28) and (29) into (32), we will have

$$\begin{array}{c}{y}_{n+2}=y_n+2{\mathrm{hy}}_n\prime +2h^2{y}_n″+h^3\hfill \\ [\frac{236}{315}{f}_n+\frac{22}{35}{f}_{n+1}-\frac{8}{105}{f}_{n+2}+\frac{2}{63}{f}_{n+3}]\hfill \end{array}$$

$$+h^4[\frac{34}{315}{g}_n-\frac{4}{315}{g}_{n+3}].$$

(44)

Substituting equation (42) into (33), we have

$$\begin{array}{c}{y}_{n+3}″=y_n″+\frac{h}{80}\hfill \\ [39f_n+81f_{n+1}+81f_{n+2}+39f_{n+3}]\hfill \\ +\frac{h^2}{40}[3g_n-3g_{n+3}].\hfill \end{array}$$

(45)

Substituting equations (42) and (43) into (34), we have

$$\begin{array}{c}{y}_{n+3}\prime =y_n\prime +3{\mathrm{hy}}_n″+h^2\hfill \\ [\frac{369}{280}{f}_n+\frac{243}{112}{f}_{n+1}+\frac{243}{280}{f}_{n+2}+\frac{81}{560}{f}_{n+3}]\hfill \end{array}$$

$$+h^3[\frac{27}{140}{g}_n-\frac{9}{280}{g}_{n+3}].$$

(46)

Also by substituting equations (42), (43) and (44) into (35), we have

$$\begin{array}{c}{y}_{n+3}=y_n+3{\mathrm{hy}}_n\prime +\frac{2}{9}{h}^2{y}_n″+h^3\hfill \\ [\frac{117}{64}{f}_n+\frac{729}{320}{f}_{n+1}+\frac{729}{2240}{f}_{n+2}+\frac{153}{2240}{f}_{n+3}]\hfill \end{array}$$

$$+\frac{h^4}{1120}[297g_n-27g_{n+3}].$$

(47)

Write the formulae in (27), (28), (29), (42), (43), (44), (45), (46), and (47) in matrix form and the zero-stability of the method can be obtained from

$$\begin{array}{c}\left[\begin{array}{ccccccccc}1\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 1\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 1\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right]\left[\begin{array}{c}{y}_{n+1}″\hfill \\ \multicolumn{1}{c}{y_{n+1}\prime }\\ \multicolumn{1}{c}{y_{n+1}}\\ \multicolumn{1}{c}{y_{n+2}″}\\ \multicolumn{1}{c}{y_{n+2}\prime }\\ \multicolumn{1}{c}{y_{n+2}}\\ \multicolumn{1}{c}{y_{n+3}″y_{n+3}\prime }\\ \multicolumn{1}{c}{y_{n+3}}\end{array}\right]=\hfill \\ \left[\begin{array}{ccccccccc}0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right]\left[\begin{array}{c}{y}_{n-2}″\hfill \\ \multicolumn{1}{c}{y_{n-2}\prime }\\ \multicolumn{1}{c}{y_{n-2}}\\ \multicolumn{1}{c}{y_{n-1}″}\\ \multicolumn{1}{c}{y_{n-1}\prime }\\ \multicolumn{1}{c}{y_{n-1}}\\ \multicolumn{1}{c}{y_n″}\\ \multicolumn{1}{c}{y_{n-2}}\\ \multicolumn{1}{c}{y_n}\end{array}\right]\hfill \end{array}$$

Let $M^{\left(0\right)}\mathrm{be}$ the matrix coefficient on the left and $M^{\left(1\right)}$ is the matrix coefficient on the right of the above equation.

Zero-stability polynomial of the method is $\rho \left(R\right)=$ det $[R{M}^{\left(0\right)}-M^{\left(1\right)}]=0$ ,

$$\begin{array}{c}\rho \left(R\right)=R^6(R^3-3R^2+3R-1)=0,\hfill \\ R=0,0,0,0,0,0,1,1,1,\left|R\right|\le 1.\hfill \end{array}$$

The three-point block method is zero-stable since $\left|R\right|\le 1$ and multiplicity of the roots $\left|R\right|=1$ is at most 3.

The zero stability and consistency (the order of the methods is greater than three) of the two-point and three-point methods implied that the methods are convergence, according to Ackleh et al. ¹²

Numerical Experiments

Based on the proposed methods, codes in C-programming are constructed for solving general third order IVPs, comparisons of the results are done with other existing methods. The comparisons are made with block methods of almost the same order or the same point number. The values of $y{\prime }_{n+1},y_{n+1},y{\prime }_{n+2},y_{n+2}$ in the two-point method and $y{\prime }_{n+1},y_{n+1},{y^{\prime }}_{n+2},y_{n+2},{y^{\prime }}_{n+3}$ and $y_{n+3}$ in the three-point method are approximated using the predictor-corrector equations. Where Taylor method is used as the predictor equation, this is the common practice in the implementation of other implicit block methods in the literature, see Majid et al. ¹³ for further details. We are also using Taylor method for the predictor equations in the implementation of the methods of comparison, hence, it is a very fair comparison. The predictor equations using Taylor method can be written as:

$$\begin{array}{c}{y}_{n+m}^{″p}=y_{n+(m-1)}^{″c}+{\mathrm{hf}}_{n+(m-1)}^c,\hfill \\ y_{n+m}^{\prime p}=y_{n+(m-1)}^{\prime c}+{\mathrm{hy}}_{n+(m-1)}^{″c}+\frac{h^2}{2!}{f}_{n+(m-1)}^c,\hfill \\ \begin{array}{c}{y}_{n+m}^p=y_{n+(m-1)}^c+{\mathrm{hy}}_{n+(m-1)}^{\prime c}\hfill \\ +\frac{h^2}{2!}{y}_{n+(m-1)}^{″c}+\frac{h^3}{3!}{f}_{n+(m-1)}^c,\hfill \end{array}\hfill \end{array}(48)$$

$$f_{n+m}^p=f(x_{n+m},y_{n+m}^p,y_{n+m}^{\prime p},y_{n+m}^{″p}),$$

$$g_{n+m}^p=f\prime (x_{n+m},y_{n+m}^p,y_{n+m}^{\prime p},y_{n+m}^{″p}).$$

$$f_{n+m}^c=f(x_{n+m},y_{n+m}^c,y_{n+m}^{\prime c},y_{n+m}^{″c}),$$

$$g_{n+m}^c=f\prime (x_{n+m},y_{n+m}^c,y_{n+m}^{″c},y_{n+m}^{″c}).$$

Where $m=1,2$ for the two-point method and $m=1,2,3$ for the three-point method.

Test problems:

Problem 1

$$\begin{array}{c}y″\prime =y″+\mathrm{co}{s}^2\left(t\right)+\sin \left(t\right)+1,\hfill \\ y\left(0\right)=0,y\prime \left(0\right)=1,y″\left(0\right)=0,[0,2].\hfill \end{array}$$

Exact Solution: $y\left(t\right)=\frac{1}{40}\cos \left(2t\right)-\frac{1}{20}\sin \left(2t\right)+\frac{1}{2}\cos \left(t\right)+\frac{1}{2}\sin \left(t\right)+\frac{21}{10}{e}^t-\frac{3}{4}{t}^2-\frac{3}{2}t-\frac{21}{8}.$

Problem 2

$$y″\prime =-\frac{3y″}{2y^2},y\left(0\right)=1,y\prime \left(0\right)=\frac{1}{2},y″\left(0\right)=-\frac{1}{4},[0,1].$$

Exact Solution: $y\left(t\right)=\sqrt{t+1}.$

Problem 3

$$\begin{array}{c}\multicolumn{1}{c}{y″\prime -2y″-3y\prime +10y=34t{e}^{-2t}}\\ -16e^{-2t}-10t^2+6t+34\hfill \end{array}$$

$$y\left(0\right)=3,y\prime \left(0\right)=0,y″\left(0\right)=0,[0,1].$$

Exact Solution: $y\left(t\right)=t^2{e}^{-2t}-t^2+3.$

Problem 4

$$y″\prime =\frac{3y\prime }{4y^4},y\left(0\right)=1,y\prime \left(0\right)=\frac{1}{2},y″\left(0\right)=-\frac{1}{4},[0,2].$$

Exact Solution: $y\left(x\right)=\sqrt{t+1}.$

Problem 5

$$y″\prime =x-4y\prime ,y\left(0\right)=0,y\prime \left(0\right)=0,y″\left(0\right)=1,[0,2].$$

Exact Solution: $y\left(t\right)=\frac{3}{16}(1-\cos \left(2t\right))+\frac{1}{8}{t}^2.$

Problem 6

$$y″\prime =y+3e^t,y\left(0\right)=0,y\prime \left(0\right)=1,y″\left(0\right)=2,[0,5].$$

Exact Solution: $y\left(x\right)=t{e}^t.$

Notations used:

Time	:	Seconds.
Maxerror	:	Maximum error $|y\left(t_i\right)-y_i|$ where $y\left(t_i\right)$ is the exact solution and $y_i$ the computed solution.
New 2PSDBI(3)	:	Order five, the newly proposed two-point additional derivative block method.
Abdullahi:	:	Order six, $k=5$ five-point block predictor-corrector method proposed by Abdullahi et al. 9
Olusola:	:	Order five, four-point block method proposed by Olusola. 14
New 3PSDBI(3)	:	Order six, the newly proposed three-point additional derivative block method.
Majid(3):	:	Order seven, two-point four step implicit block method proposed by Majid et al. 4
Omar	:	Order seven, predictor-corrector block method proposed in. 8

Numerical results for 2PSDBI(3) are given in Figures 1–6, whereas for 3PSDBI(3) are given in Figures 7–12 respectively.

Figure 1.

Comparison of the methods (New 2PSDBI(3)) for problem 1.

Figure 2.

Comparison of the methods (New 2PSDBI(3)) for problem 2.

Figure 3.

Comparison of the methods (New 2PSDBI(3)) for problem 3.

Figure 4.

Comparison of the methods (New 2PSDBI(3)) for problem 4.

Figure 5.

Comparison of the methods (New 2PSDBI(3)) for problem 5.

Figure 6.

Comparison of the methods (New 2PSDBI(3)) for problem 6.

Figure 7.

Comparison of the methods (New 3PSDBI(3)) for problem 1.

Figure 8.

Comparison of the methods (New 3PSDBI(3)) for problem 2.

Figure 9.

Comparison of the methods (New 3PSDBI(3)) for problem 3.

Figure 10.

Comparison of the methods (New 3PSDBI(3)) for problem 4.

Figure 11.

Comparison of the methods (New 3PSDBI(3)) for problem 5.

Figure 12.

Comparison of the methods (New 3PSDBI(3)) for problem 6.

Noted too that the methods can be used to solve physical problems such as the well-known thin film flow problem, ¹⁵ Blasius equation ¹⁶ and the nonlinear Genesio equation, ¹⁷ and we can treat it as further research.

Discussion and conclusion

Here, we presented the construction of the two and three-point additional derivative block methods which are zero-stable for directly solving $y″\prime =f(t,y,y\prime ,y″)$ . The methods are used to solve linear and nonlinear problems. Numerical results are presented in the form of efficiency curves, where the common logarithm of the maximum global errors, were plotted versus the computational time in seconds. Graphs in Figures 1–6 show the efficiency curves of the proposed 2PSDBI(3) method as compared to Abdullahi and Olusola methods. From the efficiency curves the graphs of the new 2PSDBI(3) lie well below the graphs of methods by Abdullahi and Olusola, this shows that 2PSDBI(3) method can achieved smaller maximum error compared to the other two methods for the same computational time. Hence 2PSDBI(3) method is the most efficient compared to Abdulahi and Olusola methods for solving the same test problems. Though all the three methods are almost of the same algebraic order, the 2PSDBI(3) method is order five, Abdullahi method is order six and Olusola method is order five and noted too that the new proposed method is a two-point block method, Abdullahi method is a five-point block method and Olusola method is a four-point block method.

Figures 7–12 show that the efficiency curves of the proposed 3PSDBI(3) order six method compared to Majid(3) two-point block method of order seven, and Omar methods which is order seven block method. From the efficiency curves, the graphs of the new 3PSDBI(3) method lie well below the graphs of methods by Majid(3) and Omar, this shows that 3PSDBI(3) method can achieved smaller maximum error compared to the other two methods for the same computational time. Based on the results, it can be concluded that the proposed 3PSDBI(3) method is the most efficient compared to the two existing methods, though the order of the existing methods is one order higher.

It can be concluded that introducing the additional derivatives into the proposed block methods do enhanced to efficiency of the methods. Thus they have a very high potential to be an efficient solver for general third order IVPs. The methods can also be adapted for solving any third order boundary value problems, third order delay differential equations as well as third order volterra-integro differential equations.