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A Modified Iterative Method for Solving Nonlinear Functional Equation

Accepted: 14 Apr 2020
Journal Details Format
Journal
First Published
01 Jan 2016
Publication timeframe
2 times per year
Languages
English

An iterative method (AIM) is one of the numerical method, which is easy to apply and very time convenient for solving nonlinear differential equations. However, if we want to work in a large interval, sometimes it may be difficult to apply AIM. Therefore, a multistage AIM named Multistage Modified Iterative Method (MMIM) is introduced in this article to work in a large computational interval. The applicability of MMIM for increasing the solution domain of the given problems is construed in this article. Some problems are solved numerically using MMIM, which provides a better result in the extended interval as compared to AIM. Comparison tables and some graphs are included to demonstrate the results.

MSC 2010

Introduction

Nonlinear differential and integral equations have been a key in the various fields of Engineering, Physics, Life Sciences, and Mathematics. There is no general method to solve the nonlinear differential equations. Also, it is not possible to convert a nonlinear differential equation into an appropriate form using variable transformations and sometimes not possible to get the close form analytical solution of the nonlinear differential equations. George Adomian developed a numerical method from 1970 to 1990. The important feature of the method is the deputation of ‘Adomian polynomials’, which acknowledge the approximation of nonlinear part without using perturbation and linearisation techniques. In articles [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], a number of renowned nonlinear problems have been solved by some other efficient numerical methods.

However, numerical and analytical methods are not limited in the fields of ODE(s) and PDE(s), but may have a crucial role in the field of nuclear magnetic resonance, astrophysics, cold plasma, porous media, and physics through fractional differential equations [12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]. In spite of these abovementioned methods, AIM  is getting a lot of attention in getting the solution for nonlinear differential equations. However, this method works for a small interval only. But to find a numerical solution for a large interval using AIM, we have to use large number of components. So, in most of the problems, after some iterations, either mathematical software declines to give the solution for a given problem or the iterations become so much complicated to move to successive steps of the solution. In Multistage Adomian Method , a comparatively large interval is split into a number of small subintervals and these small subintervals are used to find the solution.

The objective of this article is to solve a nonlinear problem for a large interval using AIM. In the present work, we introduce MMIM, developed by using AIM in each of subinterval separately. Therefore, MMIM is helpful to find the solution in a comparatively large interval in comparison to AIM.

The procedure given in this article can be extended to solve the various problems, like Riccati differential equation, Abel's differential equation, integral equations, partial differential equation and fractional differential equations.

In Section 2, the outline of AIM is provided. In Section 3, two-step modified iterative method is discussed. In Section 4, an analysis of the two-step modified iterative method has been discussed. MMIM is explained in Section 5. An absolute error formula is discussed in Section 6. Finally, some examples are solved to establish the competency of MMIM, in Section 7.

An Iterative Method 

Consider a general functional equation: $u=ψ+ξ(u),$ u = \psi + \xi \left(u \right), where ξ (u) is a nonlinear operator from a Banach space BB and called as known function or source term. To apply AIM, the solution of Eq. (1) has been assumed to be in convergent series form as $u=∑n=0∞un.$ u = \sum\limits_{n = 0}^\infty {u_n}.

The nonlinear terms can be decomposed as $ξ(u)=ξ(∑i=0∞ui), =ξ(u0)+∑i=1∞{ξ(∑j=0iuj)−ξ(∑j=0i−1uj)},$ \matrix{{\xi \left(u \right) = \xi \left({\sum\limits_{i = 0}^\infty {u_i}} \right),} \cr {{\kern 1pt} = \xi \left({{u_0}} \right) + \sum\limits_{i = 1}^\infty \left\{{\xi \left({\sum\limits_{j = 0}^i {u_j}} \right) - \xi \left({\sum\limits_{j = 0}^{i - 1} {u_j}} \right)} \right\},} \cr} and using Eqs (2) and (3), Eq. (1) can be written as $u=∑n=0∞un,=ψ+ξ(u0)+∑i=1∞{ξ(∑j=0iuj)−ξ(∑j=0i−1uj)},$ \matrix{{u = \sum\limits_{n = 0}^\infty {u_n},} \cr {= \psi + \xi \left({{u_0}} \right) + \sum\limits_{i = 1}^\infty \left\{{\xi \left({\sum\limits_{j = 0}^i {u_j}} \right) - \xi \left({\sum\limits_{j = 0}^{i - 1} {u_j}} \right)} \right\},} \cr} therefore a recurrence relation is defined as follows: ${u0=ψu1=ξ(u0)um+1=ξ(u0+⋯+um)−ξ(u0+⋯+um−1) m=1,2,...$ \left\{{\matrix{{{u_0} = \psi} \hfill \cr {{u_1} = \xi \left({{u_0}} \right)} \hfill \cr {{u_{m + 1}} = \xi \left({{u_0} + \cdots + {u_m}} \right) - \xi \left({{u_0} + \cdots + {u_{m - 1}}} \right)} \hfill \cr {\quad \quad m = 1,2,...} \hfill \cr}} \right.

This implies $u1+u2+⋯+um+1=ξ(u0+u1+⋯+um)$ {u_1} + {u_2} + \cdots + {u_{m + 1}} = \xi \left({{u_0} + {u_1} + \cdots + {u_m}} \right)

Therefore, using Eq. (6) in Eq. (1), $u=ψ+∑n=1∞un.$ u = \psi + \sum\limits_{n = 1}^\infty {u_n}.

If ξ is contraction, i.e., $‖ξ(x)−ξ(y)‖≤K‖x−y‖,0 \left\| {\xi \left(x \right) - \xi \left(y \right)} \right\| \le K\left\| {x - y} \right\|,0 < K < 1, then using Eq. (5), we have $‖um+1‖=‖ξ(u0+u1+⋯+um)−ξ(u0+u1+⋯+um−1)‖ ≤K‖um‖≤Km+1‖u0‖, where m=0,1,2,...$ \matrix{{\left\| {{u_{m + 1}}} \right\| = \left\| {\xi \left({{u_0} + {u_1} + \cdots + {u_m}} \right) - \xi \left({{u_0} + {u_1} + \cdots + {u_{m - 1}}} \right)} \right\|} \hfill \cr {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \le K\left\| {{u_m}} \right\| \le {K^{m + 1}}\left\| {{u_0}} \right\|,{\kern 1pt} {\kern 1pt} {\rm{where}}{\kern 1pt} {\kern 1pt} m = 0,1,2,...} \hfill \cr}

Therefore using Banach fixed point theorem  a solution for Eq. (1) is obtained.

Two-Step Modified Iterative Method (TSMIM)

Consider, the general functional Eq. (1). In TSMIM, source term ψ has been divided into two parts ψ1 and ψ2 (ψ1 and ψ2 may contain more than one term). Under this assumption, we set $ψ=ψ1+ψ2$ \psi = {\psi _1} + {\psi _2} and on applying a slight variation to the components u0 and u1, a modified recursive scheme defined as follows, ${u0=ψ1u1=ψ2+ξ(u0)um+1=ξ(u0+⋯+um)−ξ(u0+⋯+um−1) m=1,2,3,…$ \left\{{\matrix{{{u_0} = {\psi _1}} \hfill \cr {{u_1} = {\psi _2} + \xi \left({{u_0}} \right)} \hfill \cr {{u_{m + 1}} = \xi \left({{u_0} + \cdots + {u_m}} \right) - \xi \left({{u_0} + \cdots + {u_{m - 1}}} \right)} \hfill \cr {\quad m = 1,2,3, \ldots} \hfill \cr {} \hfill \cr}} \right.

When we analyse Eqs (5) and (9), it is observed that zeroth component u0 is defined by the function ψ1 as a part of ψ, the remaining part ψ2 is added to the definition of the component u1 in Eq. (9). This change in u0 may result as a lection in the computational work and accelerate the convergence of the solution. This slight variation in the definition of the components u0 and u1 may provide the solution in only two iterations.

It should be noted that the success of this method depends on proper choice of the part ψ1 or ψ2, but it cannot be said that which part of ψ, either ψ1 or ψ2, would be appropriate to take as a zeroth component.

Analysis of the Two-Step Modified Iterative Method

For the sake of convenience, let us consider Eq. (1)

In the study of TSMIM, the source term has been split into two parts ψ1 and ψ2. Now, modified Eq. (1) can be written as $u=ψ1+ψ2+ξ(u).$ u = {\psi _1} + {\psi _2} + \xi \left(u \right).

Using the recursive scheme Eq. (9) and choosing u0 such that u0 = ψ1 = u, we get $u1=ψ2+ξ(u0),$ {u_1} = {\psi _2} + \xi \left({{u_0}} \right), using Eq. (10), Eq. (11) implies $u1=u−ψ1=0$ {u_1} = u - {\psi _1} = 0 i.e., using the TSMIM, the solution can be obtained in two iterations. However, the drawback of this method is that, the zeroth component should satisfy the given problem.

If zeroth component does not satisfy the given problem, then the method discussed in article  can be used.

Multistage Modified Iterative Method

In most of the problems, the computational interval I = [a, J], JR cannot be always small. Now to accelerate the rate of convergence and to improve the accuracy of the solutions in a large interval using AIM, the entire interval I is divided into “n” subintervals. The main advantage of splitting the interval I into n subintervals (only a few components) gives a better approximation of the solution in the Jith subinterval, where $I=∑i=1nJi.$ I = \sum\limits_{i = 1}^n {J_i}. Let [a, J] is the interval, in which, the given problem to be solved. Here, the interval I = [a, J], JR is split into n subintervals, such that ti < ti+1 and for i = 0, 1, 2,..., n − 1.

Therefore, to solve the problem of type Eq. (1), initial guess of the solution should be known in the interval and the solution can be considered as $ϕi=∑n=0∞ϕin,$ {\phi _i} = \sum\limits_{n = 0}^\infty {\phi _{in}}, where ϕ1(t) is the approximate numerical solution for the first sub-interval [a, t1). This first subinterval will be used to obtain an initial condition at t = t1, i.e., y(t1) = a2 for the solution ϕ2(t) in the second subinterval [t1, t2). Similarly the initial condition for the third interval at t = t2 is y(t2) = a3 and get ϕ3(t). This procedure continues over n subintervals and we obtain a numeric solution ϕi(t) over the ith subinterval [ti−1, ti). Using this multistage method, the function ϕi(t) can be obtained throughout the entire interval for all i.

Note 1: Before applying this method, it should be noted that

The length of the subinterval should be small to get the higher accuracy.

It is not necessary that the length of each subinterval to be equal, it can also be changed to achieve better results.

Absolute Error Formula

Since MMIM is a modification of AIM, and in each subinterval AIM will be used separately, therefore the absolute error formula for the solution ϕi in ith subinterval [ti−1, ti] will be $Absolute Error=|u−ϕi|,=|u−∑n=0∞ϕin|, for i=1,2,3,…$ {\rm{Absolute}}\;{\rm{Error}} = |u - {\phi _i}|, = |u - \sum\limits_{n = 0}^\infty {\phi _{in}}|,\;\;{\rm{for}}\;i = 1,2,3, \ldots where n is the number of components taken for the numerical solution in that particular subinterval.

Illustrative Examples

In this section, some examples have been included for the verification of the MMIM. AIM and MMIM are applied separately to find the solution in a given interval. Further, solutions of the both methods are compared graphically and the comparative study is summarised in Tables 1 and 2, respectively.

Comparative study of AIM and MMIM

t Exact value Absolute error (AIM) Absolute error (MMIM)
0.4 0.16 3.49×10−9 3.496×10−9
0.8 0.64 5.833×10−5 5.833×10−5
1.2 1.44 1.761×10−2 1.671×10−5
1.6 2.56 0.989 5.644×10−4
2.0 4.0 5.71 1.009×10−3

Comparative study of AIM and MMIM

t Exact value Absolute error (AIM) Absolute error (MMIM)
0.4 1.4 2.878×10−3 4.475×10−6
0.8 1.8 0.147 1.260×10−5
1.2 2.2 1.203 2.865×10−5
1.6 2.6 3.393 5.664×10−5
2.0 3.0 5.401 1.012×10−4

Example 1. Consider the nonlinear integral equation $y=15x5+x2−∫0xy2(t)dt,$ y = {1 \over 5}{x^5} + {x^2} - \int_0^x {y^2}(t)dt, such that exact solution is y = x2 in [0, 2].

Solution: Here, it can be observed that the solution of the given problem is already inbuilt in the given problem.

Therefore, we can apply TSMIM.

Let the solution of Eq. (14) exists in a converging series form, $y=∑i=1∞yi,$ y = \sum\limits_{i = 1}^\infty {y_i}, using Eq. (9), we can define the following recurrence relation: $y0=x2y1=15x5−∫0xy02(t)dt=0y2=y3=…=yk+1=0,$ \matrix{{{y_0} = {x^2}} \cr {{y_1} = {1 \over 5}{x^5} - \int_0^x y_0^2(t)dt = 0} \cr {{y_2} = {y_3} = \ldots = {y_{k + 1}} = 0,} \cr} where k ≥ 0.

Hence, the required solution would be $y=y0+y1+y2+⋯=x2.$ y = {y_0} + {y_1} + {y_2} + \cdots = {x^2}.

It should be noted that, if we take the 0th component as the solution of the given problem, the next component becomes zero. But if we do not take 0th component as the solution of the given problem, we shall apply AIM.

Using AIM: Suppose the problem Eq. (14) has the series solution of the form Eq. (15). Therefore, the recurrence relation can be defined as, $y0=15x5,y1=x2−∫0xy02(t)dt=x2−x11275,y2=−∫0x(y0(t)+y1(t))2dt+∫0xy02dt=−x55−x820+⋯,y3=−∫0x(y0(t)+y1(t)+y2(t))2dt+∫0x(y0(t)+y1(t))2dt=x820+7x11550+⋯$ \matrix{{{y_0} = {1 \over 5}{x^5},} \cr {{y_1} = {x^2} - \int_0^x y_0^2(t)dt = {x^2} - {{{x^{11}}} \over {275}},} \cr {{y_2} = - \int_0^x {{({y_0}(t) + {y_1}(t))}^2}dt + \int_0^x y_0^2dt = - {{{x^5}} \over 5} - {{{x^8}} \over {20}} + \cdots ,} \cr {{y_3} = - \int_0^x {{({y_0}(t) + {y_1}(t) + {y_2}(t))}^2}dt + \int_0^x {{({y_0}(t) + {y_1}(t))}^2}dt = {{{x^8}} \over {20}} + {{7{x^{11}}} \over {550}} + \cdots} \cr}

The graph of the solution Eq. (15) at i = 1, 2, 3, 4 and the comparison with the exact solution y = x2 are shown in Figure 1.

Figure 1 shows a comparison between exact solution ‘_ _’ and computed solution ‘__’.

It can be illustrated from Figure 1 that the computed solution coincide with the exact solution in the interval [0, 1]. The behaviour of the exact solution changes rapidly for x ≥ 1.5, i.e., the approximated solution does not give a better result with respect to the exact solution in the interval [0, 2]. Therefore to increase the efficiency of the solution in the computational interval, one may use MMIM.

Using MMIM: As the given problem explain above, the solution of the problem Eq. (14) lies in the interval [0, 2], but to evaluate the solution using AIM in the interval [0, 2] may be comparatively difficult. Therefore, we can apply MMIM by dividing the given interval such that P = {0, 1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2}. For the success of MMIM, AIM will be applied in these intervals separately.

First, we find solution in the interval [0, 1) by taking sum of the five components, y1(x) = y10 + y11 + y12 + y13 + y14, where $y10=0,y12=15x5+x2−∫0xy102(t)dt=x2+x55,y11=−∫0x(y10(t)+y11(t))2dt+∫0xy102(t)dt=−x55−x820−x11275+⋯,y13=−∫0x(y10(t)+y11(t)+y12(t))2dt+∫0x(y10(t)+y11(t))2dt=x820+7x11550+x141925−x176800−x2055000−x231739375+⋯,$ \matrix{{{y_{10}} = 0,} \cr {{y_{12}} = {1 \over 5}{x^5} + {x^2} - \int_0^x y_{10}^2(t)dt = {x^2} + {{{x^5}} \over 5},} \cr {{y_{11}} = - \int_0^x {{({y_{10}}(t) + {y_{11}}(t))}^2}dt + \int_0^x y_{10}^2(t)dt = - {{{x^5}} \over 5} - {{{x^8}} \over {20}} - {{{x^{11}}} \over {275}} + \cdots ,} \cr {{y_{13}} = - \int_0^x {{({y_{10}}(t) + {y_{11}}(t) + {y_{12}}(t))}^2}dt + \int_0^x {{({y_{10}}(t) + {y_{11}}(t))}^2}dt} \cr {= {{{x^8}} \over {20}} + {{7{x^{11}}} \over {550}} + {{{x^{14}}} \over {1925}} - {{{x^{17}}} \over {6800}} - {{{x^{20}}} \over {55000}} - {{{x^{23}}} \over {1739375}} + \cdots ,} \cr} and $y14=−∫0x(y10(t)+y11(t)+y12(t)+y13(t))2dt+∫0x(y10(t)+y11(t)+y12(t))2dt=−x11110−x141925−⋯.$ {y_{14}} = - \int_0^x {({y_{10}}(t) + {y_{11}}(t) + {y_{12}}(t) + {y_{13}}(t))^2}dt + \int_0^x {({y_{10}}(t) + {y_{11}}(t) + {y_{12}}(t))^2}dt = - {{{x^{11}}} \over {110}} - {{{x^{14}}} \over {1925}} - \cdots .

The graph of the solution y1(x) = y10 + y11 + y12 + y13 + y14 in the interval [0, 1) is shown in Figure 2 The end point of the first interval [0, 1) will work as the initial point for the second interval [1, 1.1), such that $y1(1)=0.9986≈1=y20.$ {y_1}(1) = 0.9986 \approx 1 = {y_{20}}.

Therefore, solution in the interval [1, 1.1) is given by, $y20=1,y21=−x+x2,y22=−0.5435064935+0.35x+0.8x2+⋯,y23=−0.2295528059+0.5742110811+⋯,y24=−0.02507479237+0.06506311753x+⋯.$ \matrix{{{y_{20}} = 1,} \cr {{y_{21}} = - x + {x^2},} \cr {{y_{22}} = - 0.5435064935 + 0.35x + 0.8{x^2} + \cdots ,} \cr {{y_{23}} = - 0.2295528059 + 0.5742110811 + \cdots ,} \cr {{y_{24}} = - 0.02507479237 + 0.06506311753x + \cdots .} \cr}

The graph of the solution y2(x) = y20 + y21 + y22 + y23 + y24 for the interval is shown in Figure 3.

Take end point of the interval [1, 1.1) as the initial point for the interval [1.1, 1.2), such that $y2(1.1)=1.2099≈1.21=y30.$ {y_2}(1.1) = 1.2099 \approx 1.21 = {y_{30}}.

Since, solution for the interval [1.1, 1.2) will be given by, $y30=1.21,y31=0.40051−1.461x+⋯,y32=−0.6157294163+1.4641x+⋯,y33=−0.5432305405+1.207498697x+⋯,y34=−0.1195253165+0.4357396811x+⋯.$ \matrix{{{y_{30}} = 1.21,} \cr {{y_{31}} = 0.40051 - 1.461x + \cdots ,} \cr {{y_{32}} = - 0.6157294163 + 1.4641x + \cdots ,} \cr {{y_{33}} = - 0.5432305405 + 1.207498697x + \cdots ,} \cr {{y_{34}} = - 0.1195253165 + 0.4357396811x + \cdots .} \cr}

The graph of the solution y3 = y30 + y31 + y32 + y33 + y34 for the interval [1.1, 1.2) is shown in Figure 4.

Take end point of the interval [1.1, 1.2), as the initial point for the interval [1.2, 1.3), such that $y3(1.2)=1.4399≈1.44=y40.$ {y_3}(1.2) = 1.4399 \approx 1.44 = {y_{40}}.

Then, the solution in the interval [1.2, 1.3) will be given by, $y40=1.44,y41=1.04832+2.0736x+⋯,y42=−0.34573339+2.0736x+⋯,y43=−1.005231724+1.256973647x+⋯,y44=−0.5589816161+2.296545961+⋯.$ \matrix{{{y_{40}} = 1.44,} \cr {{y_{41}} = 1.04832 + 2.0736x + \cdots ,} \cr {{y_{42}} = - 0.34573339 + 2.0736x + \cdots ,} \cr {{y_{43}} = - 1.005231724 + 1.256973647x + \cdots ,} \cr {{y_{44}} = - 0.5589816161 + 2.296545961 + \cdots .} \cr}

Therefore, the graph of the solution y4 = y40 + y41 + y42 + y43 + y44 for the interval [1.2, 1.3) is shown in Figure 5.

Proceeding in similar way up to interval [1.9, 2], with length of the subinterval 0.1 and after combining or merging all subintervals, we get solution for the interval [0, 2] in Figure 6, which is quite similar to the graph of exact solution, therefore for more clarity we observe Table 1. It is clear from Table 1 that the solution using MMIM is more promising than AIM.

Example 2. Consider the differential equation $y′(x)=−2x−x2+y2(x), y(0)=1,$ y'(x) = - 2x - {x^2} + {y^2}(x),\;\;y(0) = 1, such that the exact solution is y(x) = 1 + x in [0, 2].

Solution: The solution of Eq. (16) can be written in an integral form as $y(x)=1−x2−x33+∫0xy2(t)dt$ y(x) = 1 - {x^2} - {{{x^3}} \over 3} + \int_0^x {y^2}(t)dt

Here, we see that the solution of the problem Eq. (17) is not inbuilt under the given problem. Therefore, we use AIM.

Using AIM: Let the solution of Eq. (17) is $y=∑i=1∞yi$ y = \sum\limits_{i = 1}^\infty {y_i} and the recurrence relation can be defined as, $y0=1,y1=x−x2−x33,y2=x2−x33−⋯,y3=2x33+x43−⋯,⋮$ \matrix{{{y_0} = 1,} \cr {{y_1} = x - {x^2} - {{{x^3}} \over 3},} \cr {{y_2} = {x^2} - {{{x^3}} \over 3} - \cdots ,} \cr {{y_3} = {{2{x^3}} \over 3} + {{{x^4}} \over 3} - \cdots ,} \cr \vdots \cr}

The graph of the computed solution using AIM, $y=∑i=04yi$ y = \sum\limits_{i = 0}^4 {y_i} and the exact solution are shown in Figure 7.

Using MMIM: Here, we can see that the solution of the problem Eq. (16) given by AIM works only for the interval [0, 0.3]. If we want to find out the solution of the given problem for the interval [0, 2], then we can apply MMIM to solve this problem. We shall apply this method by dividing the given interval [0, 2] such that P = {0, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2}.

First, we find solution for the interval [0, 0.3) by taking sum of the five components y1 = y10 + y11 + y12 + y13 + y14. Using Eq. (5), the graph of the solution in the interval [0, 0.3) is shown in Figure 8.

The end point of the interval [0, 0.3) will work as the initial point for second interval [0.3, 0.4) such that $y1(0.3)=1.299≈1.3=y20.$ {y_1}(0.3) = 1.299 \approx 1.3 = {y_{20}}.

Solution for the interval [0.3, 0.4) is given by, $y20=1.3,y21=−0.417+1.9x+⋯,y22=−0.3110044886−1.69x+⋯,y23=−0.02693861645+0.2508741836x−0.6935121566x2+⋯,y24=0.004469301741−0.05439096203x+0.2284531252x2⋯.$ \matrix{{{y_{20}} = 1.3,} \cr {{y_{21}} = - 0.417 + 1.9x + \cdots ,} \cr {{y_{22}} = - 0.3110044886 - 1.69x + \cdots ,} \cr {{y_{23}} = - 0.02693861645 + 0.2508741836x - 0.6935121566{x^2} + \cdots ,} \cr {{y_{24}} = 0.004469301741 - 0.05439096203x + 0.2284531252{x^2} \cdots .} \cr}

The graph of the solution y2 = y20 + y21 + y22 + y23 + y24 for the interval [0.3, 0.4) is shown in Figure 9.

In the similar manner, the solution of the given problem Eq. (16) can be calculated up to interval [1.9, 2] with length of the subinterval 0.1. Combining and merging all the solution graphs, the graph of the solution for the interval [0, 2] is shown in Figure 10. In Figure 10, which is quite similar to the graph of exact solution, therefore for more clarity we observe Table 2. It is clear from Table 1 that the solution using MMIM is more promising than AIM.

Conclusion

In our work, a multistage iterative method as a modification in AIM is introduced, which is known as MMIM. MMIM is the extension of AIM, although the solution given by AIM is limited for a very small interval. Therefore MMIM is useful to solve the functional equations for a comparatively large interval by integrating AIM. The numerical examples and associated graphs and tables have been defined and constructed in this article. Figures 25 are the graphs of the solution of Eq. (14) for the subintervals [0, 1) to [1.2, 1.3). These graphs show the behaviour of the numerical solution calculated using AIM separately in their respective subintervals and the procedure is continued up to the subinterval [1.9, 2]. Then all the solution graphs are merged and the final graph is drawn as in Figure 6 with the concept of MMIM. We can observe with the help of Figure 1 that for the interval [0, 2], MMIM converges to the exact solution more accurately than the AIM, and further the efficiency of the method is shown in Table 1.

In Table 1, a comparison between absolute error between AIM and MMIM in the computational interval [0, 2] at different points are shown when used them separately. Therefore with the help of solution graphs and table of Eq. (14), we can state that MMIM is more accurate than AIM.

Similar results are followed in all the figures and table in Example 2. Therefore, MMIM is simple and easy to apply, requires less computational complexity and also provides more reliable results.

Comparative study of AIM and MMIM

t Exact value Absolute error (AIM) Absolute error (MMIM)
0.4 1.4 2.878×10−3 4.475×10−6
0.8 1.8 0.147 1.260×10−5
1.2 2.2 1.203 2.865×10−5
1.6 2.6 3.393 5.664×10−5
2.0 3.0 5.401 1.012×10−4

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