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# The new modified Ishikawa iteration method for the approximate solution of different types of differential equations

## Abstract

In this article, the new Ishikawa iteration method is presented to find the approximate solution of an ordinary differential equation with an initial condition. Additionally, some numerical examples with initial conditions are given to show the properties of the iteration method. Furthermore, the results of absolute errors are compared with Euler, Runge-Kutta and Picard iteration methods. Finally, the present method, namely the new modified Ishikawa iteration method, is seen to be very effective and efficient in solving different type of the problem.

MSC:65K15, 65L07, 65L06, 65L70.

## Introduction

Various kinds of numerical methods, especially iterative methods , were used to solve different types of differential equations. In recent years, there has been a growing interest in the treatment of iterative approximation of fixed point theory on normed linear spaces , Banach spaces  and Hilbert spaces [19, 20], respectively.

A new modified Ishikawa iteration method has been developed to find an approximate solution for different types of differential equations with initial conditions [13, 1529] on metric spaces. The solutions are also obtained in terms of the Picard iteration. Also in Section 2, examples of these kinds of equations are solved using this new method which is called new modified Ishikawa iteration method and also the results are discussed and comparison using Euler [1, 2], Runge-Kutta [30, 31], and Picard iteration methods [1, 2, 27] is presented in tables and figures. Additionally, this approximated method can solve various different differential equations such as integral, difference, integro-differential and functional differential equations. Finally, numerical results shows that the new modified Ishikawa iteration method is more or less effective and also convenient for solving different types of differential equations.

Now, let us give some of the important theorems and definitions in order to solve linear and nonlinear differential equations using a contraction mapping.

## 1 Preliminaries

Theorem 1.1 (Banach contraction principle)

Let $(X,d)$ be a complete metric space and $T:X→X$ be a contraction with the Lipschitzian constant L. Then T has a unique fixed point $u∈X$. Furthermore, for any $x∈X$ we have

$lim n → ∞ T n (x)=u$

with

$d ( T n ( x ) , u ) ≤ L n 1 − L d ( x , F ( x ) )$

(see ).

Proof We first show uniqueness. Suppose there exist $x,y∈X$ with $T(x)=x$ and $T(y)=y$. Then

$d(x,y)=d ( T ( x ) , T ( y ) ) ≤Ld(x,y),$

therefore $d(x,y)=0$.

To show existence select, $x∈X$. We first show that ${ T n (x)}$ is a Cauchy sequence. Notice for $n∈{0,1,…}$ that

$d ( T n ( x ) , T n + 1 ( x ) ) ≤Ld ( T n − 1 ( x ) , T n x ) ≤⋯≤ L n d ( x , T ( x ) ) .$

Thus, for $m>n$ where $n∈{0,1,…}$,

$d ( T n ( x ) , T m ( x ) ) ≤ d ( T n ( x ) , T n + 1 ( x ) ) + d ( T n + 1 ( x ) , T n + 2 ( x ) ) + ⋯ + ≤ ⋯ ≤ d ( T m − 1 ( x ) , T m ( x ) ) ≤ L n d ( x , T ( x ) ) + ⋯ + L m − 1 d ( x , T ( x ) ) ≤ L n d ( x , T ( x ) ) [ 1 + L + L 2 + ⋯ ] = L n 1 − L d ( x , T ( x ) ) .$

That is, for $m>n$, $n∈{0,1,…}$,

$d ( T n ( x ) , T m ( x ) ) ≤ L n 1 − L d ( x , T ( x ) ) .$
(1.1)

This shows that ${ T n (x)}$ is a Cauchy sequence and since X is complete, there exists $u∈X$ with $lim n → ∞ T n (x)=u$. Moreover, the continuity of T yields

$u= lim n → ∞ T n + 1 (x)= lim n → ∞ T ( T n ( x ) ) =T(u),$

therefore u is a fixed point of T. Finally, letting $m→∞$ in (1.1) yields

$d ( T n ( x ) , u ) ≤ L n 1 − L d ( x , T ( x ) ) .$

□

Corollary 1.2 Let $(X,d)$ be a complete metric space and let $B( x 0 ,r)={x∈X:d(x, x 0 ), where $x 0 ∈X$ and $r>0$. Suppose $T:B( x 0 ,r)→X$ is a contraction (that is, $d(T(x),T(y))≤Ld(x,y)$ for all $x,y∈B( x 0 ,r)$ with $0≤L<1$) with $d(T( x 0 ), x 0 )<(1−L)r$. Then T has a unique fixed point in $B( x 0 ,r)$ (see ).

Definition 1.3 If the sequence ${ x n } n = 0 ∞$ provides the condition $x n + 1 =T x n$ for $n=0,1,2,…$ , then this is called the Picard iteration [1, 27].

Definition 1.4 Let $x 0 ∈X$ be arbitrary. If the ${ x n } n = 0 ∞$ sequence provides the condition

$x n + 1 = ( 1 − α n ) x n + { α n } T y n , y n = ( 1 − β n ) x n + { β n } T x n$

for $n=0,1,2,…$ , then this is called the Ishikawa iteration [16, 24] where $( α n )$ and $( β n )$ are sequences of positive numbers that satisfy the following conditions:

1. (i)

$0≤ α n ≤ β n <1$, for all positive integers n,

2. (ii)

$lim n → ∞ β n =0$,

3. (iii)

$∑ n ≥ 0 α n β n =∞$.

Definition 1.5 If $λ∈[0,1]$, $γ∈[0,1]$ and $y 0 ∈X$, T is defined as a contraction mapping with regard to Picard iteration and also the ${ y n } n = 0 ∞$ sequence provides the conditions

$y n + 1 = λ y n − 1 + ( 1 − λ ) T y n − 1 , y n = ( 1 − γ ) y n − 2 + γ T y n − 2 , n = 2 , 4 , … , y n + 1 = y 0 + ∫ x 0 x F ( t , y n ( t ) ) d t , n = 0 , T y n − 1 = y n , 0 < λ , μ < 1$

then this is called a new modified Ishikawa iteration where $T= ∫ x 0 x F(t, y n (t))dt$.

In order to illustrate the performance of the new modified Ishikawa iteration method in solving linear and nonlinear differential equations and justify the accuracy and efficiency of the method presented in this paper, we consider the following examples. In all examples, we used four types of iteration methods and the comparison is shown in figures and tables respectively.

## 2 Application of methods

Example 2.1 Let us consider the differential equation $y ′ = | y |$ subject to the initial condition $y(0)=1$.

Firstly, we obtained the exact solution of the equation as $|y|= 1 4 ( x + 2 ) 2 =1+x+ x 2 4$.

By Theorem 1.1 and Corollary 1.2, since $T= ∫ x 0 x F(t, y n (t))dt$, then

$| T ( x ) − T ( y ) | = | ∫ 0 x t d t − ∫ 0 y t d t | = | 2 3 t 3 | 0 x − 2 3 t 3 | 0 y | ≤ 2 3 | x 3 − y 3 | ≤ 2 3 | x − y | .$

So, $|T(x)−T(y)|≤ 2 3 |x−y|$ is found. Thus T has a unique fixed point, which is the unique solution of the integral equation $T= ∫ x 0 x F(t, y n (t))dt$ or the differential equation $y ′ = | y |$, $y(0)=1$.

Firstly, we approach the approximate solution using by the Picard iteration method. Thus

$y 1 = 1 + x , y 2 = 1 3 + 2 3 ( x + 1 ) 3 / 2$

are obtained. If we take the series expansion of the function $( x + 1 ) 3 / 2$ for the seven terms, then $y 2 =1+x+ x 2 4 + x 3 24 + x 4 64 + x 5 128 + 7 x 6 1536$ is found. Now, applying the new modified Ishikawa iteration method to the equation for $λ=0.5$, $γ=0.5$,

$y 1 = 1 + x , y 2 = 1 + 0.5 x , y 3 = 1 + 0.75 x , y 4 = 1 + 0.625 x , y 5 = 1 + 0.6875 x , y 6 = 1 + 0.65625 x , y 7 = 1 + 0.671875 x , y 8 = 1 + 0.6640625 x , y 9 = 1 + 0.66796875 x , y 10 = 1 + 0.666015625 x , y 11 = 1 + 0.666992187 x$

are found and also for $λ=0.5$, $γ=0.25$,

$y 1 = 1 + x , y 2 = 1 + 0.25 x , y 3 = 1 + 0.625 x , y 4 = 1 + 0.34375 x , y 5 = 1 + 0.484375 x , y 6 = 1 + 0.37890625 x , y 7 = 1 + 0.431640625 x , y 8 = 1 + 0.392089843 x , y 9 = 1 + 0.411865234 x , y 10 = 1 + 0.39703369 x , y 11 = 1 + 0.404449462 x$

are obtained. On the other hand, for $λ=0.25$, $γ=0.5$,

$y 1 = 1 + x , y 2 = 1 + 0.5 x , y 3 = 1 + 0.625 x , y 4 = 1 + 0.5625 x , y 5 = 1 + 0.578125 x , y 6 = 1 + 0.5703125 x , y 7 = 1 + 0.572265625 x , y 8 = 1 + 0.571289062 x , y 9 = 1 + 0.571533203 x , y 10 = 1 + 0.571411132 x , y 11 = 1 + 0.57144165 x$

are calculated. In the same way, for $λ=0.25$, $γ=0.25$,

$y 1 = 1 + x , y 2 = 1 + 0.25 x , y 3 = 1 + 0.4375 x , y 4 = 1 + 0.296875 x , y 5 = 1 + 0.33203125 x , y 6 = 1 + 0.305664062 x , y 7 = 1 + 0.312255859 x , y 8 = 1 + 0.307312011 x , y 9 = 1 + 0.308547973 x , y 10 = 1 + 0.307621001 x , y 11 = 1 + 0.307852744 x$

are found and also for $λ=0.75$, $γ=0.25$,

$y 1 = 1 + x , y 2 = 1 + 0.25 x , y 3 = 1 + 0.8125 x , y 4 = 1 + 0.390625 x , y 5 = 1 + 0.70703125 x , y 6 = 1 + 0.469726562 x , y 7 = 1 + 0.647705078 x , y 8 = 1 + 0.514221191 x , y 9 = 1 + 0.614334106 x , y 10 = 1 + 0.539249419 x , y 11 = 1 + 0.595562934 x$

are obtained. Similarly, for $λ=0.25$, $γ=0.75$,

$y 1 = 1 + x , y 2 = 1 + 0.75 x , y 3 = 1 + 0.8125 x , y 4 = 1 + 0.796875 x , y 5 = 1 + 0.80078125 x , y 6 = 1 + 0.799804687 x , y 7 = 1 + 0.800048828 x , y 8 = 1 + 0.799987792 x , y 9 = 1 + 0.800003051 x , y 10 = 1 + 0.799999236 x , y 11 = 1 + 0.80000019 x$

are calculated. Finally, for $λ=0.75$, $γ=0.75$,

$y 1 = 1 + x , y 2 = 1 + 0.75 x , y 3 = 1 + 0.9375 x , y 4 = 1 + 0.890625 x , y 5 = 1 + 0.92578125 x , y 6 = 1 + 0.916992187 x , y 7 = 1 + 0.923583984 x , y 8 = 1 + 0.921936035 x , y 9 = 1 + 0.923171996 x , y 10 = 1 + 0.922863006 x , y 11 = 1 + 0.922940253 x$

are obtained.

Now we calculate the approximate solution by the Euler method. At first we use the formula

$y n + 1 = y n +hF( x n , y n )$

with $F(x,y)= ∣ y ∣$, $h=0.2$ and $x 0 =0$, $y 0 =1$. From the initial condition $y(0)=1$, we have $F(0,1)=1$. We now proceed with the calculations as follows:

$F 0 = F ( 0 , 1 ) = 1 , y 1 = y 0 + h F 0 = 1 + 0.2 , y 1 = 1.2 , F 1 = F ( 0.2 , 1.2 ) = 1.095445115 , y 2 = 1.2 + 0.2 F 1 , y 2 = 1.419089023 , F 2 = F ( 0.4 , 1.419089023 ) = 1.19125523 , y 3 = 1.419089023 + 0.2 F 2 , y 3 = 1.657340069 .$

Finally, applying the Runge-Kutta method to the given initial value problem, we carry out the intermediate calculations in each step to give figures after the decimal point and round off the final results at each step to four such places. Here $F(x,y)= | y | x 0 =0$, $y 0 =1$ and we are to use $h=0.2$. Using these quantities, we calculated successively $k 1$, $k 2$, $k 3$, $k 4$ and $K 0$ defined by

$k 1 = h g ( y 0 , x 0 ) , k 2 = h g ( y 0 + h 2 , x 0 + k 1 2 ) , k 3 = h g ( y 0 + h 2 , x 0 + k 2 2 ) , k 4 = h g ( y 0 + h , x 0 + k 3 )$

and $K 0 = 1 6 ( k 1 +2 k 2 +2 k 3 + k 4 )$, $y n + 1 = y n + K 0$. Thus we find $k 1$, $k 2$, $k 3$, $k 4$ for $n=0$ as follows:

$k 1 = 0.2 F ( x 0 , y 0 ) = 0.2 F ( 0 , 1 ) = 0.2 , k 2 = 0.2 F ( x 0 + 0.1 , y 0 + 0.1 ) = 0.209761769 , k 3 = 0.2 F ( x 0 + 0.1 , y 0 + 0.104880884 ) = 0.210226628 , k 4 = 0.2 F ( x 0 + 0.2 , y 0 + 0.210226628 ) = 0.220020601 .$

So, $y 1 =1.209999565$ is obtained for $x 1 =0.2$. On the other hand, we calculated $k 1$, $k 2$, $k 3$, $k 4$ for $n=1$ as follows:

$k 1 = 0.2 F ( x 1 , y 1 ) = 0.2 F ( 0.2 , 1.209999565 ) = 0.21999996 , k 2 = 0.2 F ( x 1 + 0.1 , y 1 + 0.10999998 ) = 0.229782466 , k 3 = 0.2 F ( x 1 + 0.1 , y 1 + 0.114891233 ) = 0.230207801 , k 4 = 0.2 F ( x 1 + 0.2 , y 1 + 0.230207801 ) = 0.240017279 .$

Hence $y 2 =1.4399999194$ is calculated for $x 2 =0.4$. Finally we get $k 1$, $k 2$, $k 3$, $k 4$ for $n=2$ as follows:

$k 1 = h F ( x 2 , y 2 ) = 0.239999993 , k 2 = h F ( x 2 + h 2 , y 2 + k 1 2 ) = 0.249799913 , k 3 = h F ( x 2 + h 2 , y 2 + k 2 2 ) = 0.2250191916 , k 4 = h F ( x 2 + h , y 2 + k 3 ) = 0.260014756 .$

Thus, $y 3 =1.689999654$ is obtained for $x 3 =0.6$.

After the necessary calculation which is done above, the comparison is shown schematically in Figure 1.

On the other hand, we may give Table 1, Table 2, Table 3 and Table 4 using the modified Ishikawa iteration method for different values of λ and γ. Now we may give Table 5 which is expressed that absolute error of Example 2.1 for different values of λ and γ with $x=0.2$, $x=0.4$ and $x=0.6$ respectively.

Corollary 2.1 If we compare the approximate solution with the different values of λ and γ, then the conclusion may be indicated using Table 1, Table 2, Table 3 and Table 4 as follows.

The best approximation is obtained taking the different values of λ and γ and using the new modified Ishikawa iteration method for $x=0.2$, $x=0.4$ and $x=0.5$ getting ($λ=0.25$, $γ=0.25$; $λ=0.5$, $γ=0.25$; $λ=0.25$, $γ=0.5$; $λ=0.75$, $γ=0.25$; $λ=0.5$, $γ=0.5$; $λ=0.25$, $γ=0.75$; $λ=0.75$, $γ=0.75$) respectively.

Similarly, we calculated the solution for $x=0.6$, $x=1$ and $x=1.5$ then the approximation is found more sensitive taking ($λ=0.25$, $γ=0.25$; $λ=0.5$, $γ=0.25$; $λ=0.25$, $γ=0.5$; $λ=0.75$, $γ=0.25$; $λ=0.5$, $γ=0.5$; $λ=0.25$, $γ=0.75$; $λ=0.75$, $γ=0.75$) respectively.

Corollary 2.2 Absolute error of the modified Ishikawa iteration method is computed taking different values of λ and γ ($x=0.2$, $x=0.4$ and $x=0.6$), which is not more effective than Runge-Kutta, Picard and Euler iteration methods.

Example 2.2

Let us consider the differential equation

$y ′ =y+ x 2$

subject to the initial condition

$y(0)=0.$

Firstly, we obtained the exact solution of the equation as $y=2 e x − x 2 −2x−2$.

Using Theorem 1.1 and Corollary 1.2, since $T= ∫ x 0 x F(t, y n (t))dt$, then T has a unique fixed point, which is the unique solution of the differential equation $y ′ =y+ x 2$ with the initial condition $y(0)=0$.

At first, we approach the approximate solution by the Picard iteration method as follows:

$y 1 = x 3 3 , y 2 = x 3 3 + x 4 12 , y 3 = x 3 3 + x 4 12 x 5 60 , y 4 = x 3 3 + x 4 12 + x 5 60 + x 6 360 .$

Now applying the new modified Ishikawa iteration method to the equation for $λ=0.5$, $γ=0.5$, then

$y 1 = x 3 3 , y 2 = x 3 6 , y 3 = x 3 4 , y 4 = 0.208333 x 3 , y 5 = 0.2291666 x 3 , y 6 = 0.218749983 x 3 , y 7 = 0.223958321 x 3 , y 8 = 0.221354152 x 3 , y 9 = 0.222656236 x 3 , y 10 = 0.222005194 x 3 , y 11 = 0.222330715 x 3$

are found and also for $λ=0.5$, $γ=0.25$,

$y 1 = x 3 3 , y 2 = 0.08333333 x 3 , y 3 = 0.208333316 x 3 , y 4 = 0.114583304 x 3 , y 5 = 0.16145831 x 3 , y 6 = 0.126302055 x 3 , y 7 = 0.143880182 x 3 , y 8 = 0.130696586 x 3 , y 9 = 0.137288384 x 3 , y 10 = 0.132344535 x 3 , y 11 = 0.134816459 x 3$

are obtained. On the other hand, for $λ=0.25$, $γ=0.5$,

$y 1 = x 3 3 , y 2 = x 3 6 , y 3 = 0.208333333 x 3 , y 4 = 0.375 x 3 , y 5 = 0.192708333 x 3 , y 6 = 0.190104166 x 3 , y 7 = 0.190755207 x 3 , y 8 = 0.190429686 x 3 , y 9 = 0.190348306 x 3 , y 10 = 0.190388996 x 3 , y 11 = 0.190378823 x 3$

are calculated. In the same way, for $λ=0.25$, $γ=0.25$,

$y 1 = x 3 3 , y 2 = 0.083333333 x 3 , y 3 = 0.1458333333 x 3 , y 4 = 0.098958333 x 3 , y 5 = 0.110677083 x 3 , y 6 = 0.10188802 x 3 , y 7 = 0.104085285 x 3 , y 8 = 0.102437336 x 3 , y 9 = 0.102849323 x 3 , y 10 = 0.102540332 x 3 , y 11 = 0.10261758 x 3$

are found and also for $λ=0.75$, $γ=0.25$,

$y 1 = x 3 3 , y 2 = 0.083333333 x 3 , y 3 = 0.270833333 x 3 , y 4 = 0.130208333 x 3 , y 5 = 0.235677083 x 3 , y 6 = 0.15657552 x 3 , y 7 = 0.215901692 x 3 , y 8 = 0.171407063 x 3 , y 9 = 0.204778034 x 3 , y 10 = 0.179749805 x 3 , y 11 = 0.198520977 x 3$

are obtained. Similarly, for $λ=0.25$, $γ=0.75$,

$y 1 = x 3 3 , y 2 = 0.25 x 3 , y 3 = 0.270833333 x 3 , y 4 = 0.265625 x 3 , y 5 = 0.266927083 x 3 , y 6 = 0.266601562 x 3 , y 7 = 0.266682942 x 3 , y 8 = 0.266662597 x 3 , y 9 = 0.266667683 x 3 , y 10 = 0.266666411 x 3 , y 11 = 0.266666729 x 3$

are calculated. Finally, for $λ=0.75$, $γ=0.75$,

$y 1 = x 3 3 , y 2 = 0.25 x 3 , y 3 = 0.3125 x 3 , y 4 = 0.296875 x 3 , y 5 = 0.30859375 x 3 , y 6 = 0.305664062 x 3 , y 7 = 0.307861328 x 3 , y 8 = 0.307312011 x 3 , y 9 = 0.307723998 x 3 , y 10 = 0.307621001 x 3 , y 11 = 0.307698249 x 3$

are found. Now, we get the approximate solution using by the Euler method. Firstly, we use the formula

$y n + 1 = y n +hF( x n , y n )$

with $F(x,y)=y+ x 2$, $h=0.2$ and $x 0 =0$, $y 0 =0$. From the initial condition $y(0)=0$, we have $F(0,0)=0$. We now proceed with the calculations as follows:

$y 1 = y 0 + h F ( y 0 , x 0 ) = 0 + 0.0 = 0.0 , x 1 = x 0 + h = 0.0 + 0.2 = 0.2 , y 2 = y 1 + h F ( y 1 , x 1 ) = 0.0 + 0.2 ⋅ 0.04 = 0.008 , x 2 = x 1 + h = 0.2 + 0.2 = 0.4 , y 3 = y 2 + h F ( y 2 , x 2 ) = 0.008 + 0.0336 = 0.0416 , x 3 = x 2 + h = 0.4 + 0.2 = 0.6 .$

Finally, applying the Runge-Kutta method to the given initial value problem, we carry out the intermediate calculations in each step to give figures after decimal point and round off the final results each at step to four such places. $F(x,y)=y+ x 2$, $x 0 =0$, $y 0 =0$ and we are to use $h=0.2$. Using these quantities, we calculated successively $k 1$, $k 2$, $k 3$, $k 4$ and $K 0$ defined by

$k 1 = h g ( y 0 , x 0 ) , k 2 = h g ( y 0 + h 2 , x 0 + k 1 2 ) , k 3 = h g ( y 0 + h 2 , x 0 + k 2 2 ) , k 4 = h g ( y 0 + h , x 0 + k 3 )$

and $K 0 = 1 6 ( k 1 +2 k 2 +2 k 3 + k 4 )$, $y n + 1 = y n + K 0$. Thus we find $k 1$, $k 2$, $k 3$, $k 4$ for $n=0$ as follows:

$k 1 = 0.2 f ( x 0 , y 0 ) = 0 , k 2 = 0.2 f ( x 0 + 0.1 , y 0 + k 1 2 ) = 0.002 , k 3 = 0.2 f ( x 0 + 0.1 , y 0 + k 2 2 ) = 0.0022 , k 4 = 0.2 f ( x 0 + 0.2 , y 0 + k 3 ) = 0.00844 .$

So, $y 1 =0.0028066666666667$ is obtained for $x 1 =0.2$. On the other hand, we calculated $k 1$, $k 2$, $k 3$, $k 4$ for $n=1$ as follows:

$k 1 = 0.2 f ( x 1 , y 1 ) = 0.0085613333333333 , k 2 = 0.2 f ( x 1 + 0.1 , y 1 + k 1 2 ) = 0.0194174666666667 , k 3 = 0.2 f ( x 1 + 0.1 , y 1 + k 2 2 ) = 0.02050308 , k 4 = 0.2 f ( x 1 + 0.2 , y 1 + k 3 ) = 0.0366619493333333 .$

Hence $y 2 =0.0236507293333334$ is calculated for $x 2 =0.4$. Finally, we get $k 1$, $k 2$, $k 3$, $k 4$ for $n=2$ as follows:

$k 1 = 0.2 f ( x 2 , y 2 ) = 0.03673014586666668 , k 2 = 0.2 f ( x 2 + 0.1 , y 2 + k 1 2 ) = 0.058403160453333348 , k 3 = 0.2 f ( x 2 + 0.1 , y 2 + k 2 2 ) = 0.0605704619120000148 , k 4 = 0.2 f ( x 2 + 0.2 , y 2 + k 3 ) = 0.08884423704906680296 .$

Thus $y 3 =0.08423766727440010142666666666667$ is obtained for $x 3 =0.6$.

After the necessary calculation which is done above, the comparison is shown schematically in Figure 2.

On the other hand, we may give Table 6, Table 7, Table 8 and Table 9 by the new modified Ishikawa iteration method for different values of λ and γ. Now we may give Table 10 which is expressed that absolute error of Example 2.2 for different values of λ and γ with $x=0.2$, $x=0.4$ and $x=0.6$ respectively.

Corollary 2.3 If we compare the approximate solution with the different values of λ and γ, then the conclusion may be indicated using Table 6, Table 7, Table 8 and Table 9 as follows.

The best approximation is obtained taking the different values of λ and γ and using the modified Ishikawa iteration method for $x=0.2$, $x=0.4$ and $x=0.5$ getting ($λ=0.25$, $γ=0.25$; $λ=0.5$, $γ=0.25$; $λ=0.25$, $γ=0.5$; $λ=0.75$, $γ=0.25$; $λ=0.5$, $γ=0.5$; $λ=0.25$, $γ=0.75$; $λ=0.75$, $γ=0.75$) respectively.

Similarly, we calculated the solution for $x=0.6$, $x=1$ and $x=1.5$ then the approximation is found more sensitive taking ($λ=0.25$, $γ=0.25$; $λ=0.5$, $γ=0.25$; $λ=0.25$, $γ=0.5$; $λ=0.75$, $γ=0.25$; $λ=0.5$, $γ=0.5$; $λ=0.25$, $γ=0.75$; $λ=0.75$, $γ=0.75$) respectively.

Corollary 2.4 Absolute error of the modified Ishikawa iteration method is computed taking different values of λ and γ ($x=0.2$, $x=0.4$ and $x=0.6$), which is not more effective than Picard, Runge-Kutta and Euler iteration methods.

Example 2.3

Let us consider the differential equation

$y ′ =2x(y+1)$

subject to the initial condition

$y(0)=0.$

Using Theorem 1.1 and Corollary 1.2, since $T= ∫ x 0 x F(t, y n (t))dt$, then T has a unique fixed point, which is the unique solution of the differential equation $y ′ =2x(y+1)$ with the initial condition $y(0)=0$.

Firstly, we obtained the exact solution of the equation as $y= e x 2 −1$. Then we approach the approximate solution by the Picard iteration method as follows:

$y 1 = x 2 , y 2 = x 2 + x 4 2 , y 3 = x 2 + x 4 2 + x 6 6 , y 4 = x 2 + x 4 2 ! + x 6 3 ! + x 8 4 ! .$

Now, applying the new modified Ishikawa iteration method to the equation for $λ=0.5$, $γ=0.5$, then

$y 1 = x 2 , y 2 = 0.5 x 2 , y 3 = 0.75 x 2 , y 4 = 0.625 x 2 , y 5 = 0.6875 x 2 , y 6 = 0.65625 x 2 , y 7 = 0.671875 x 2 , y 8 = 0.6640625 x 2 , y 9 = 0.66796875 x 2 , y 10 = 0.666015625 x 2 , y 11 = 0.666992687 x 2$

are found and also for $λ=0.5$, $γ=0.25$,

$y 1 = x 2 , y 2 = 0.25 x 2 , y 3 = 0.625 x 2 , y 4 = 0.34375 x 2 , y 5 = 0.484375 x 2 , y 6 = 0.37890625 x 2 , y 7 = 0.431640625 x 2 , y 8 = 0.392089843 x 2 , y 9 = 0.411865234 x 2 , y 10 = 0.39703369 x 2 , y 11 = 0.404449462 x 2$

are obtained. On the other hand, for $λ=0.25$, $γ=0.5$,

$y 1 = x 2 , y 2 = 0.5 x 2 , y 3 = 0.625 x 2 , y 4 = 0.5625 x 2 , y 5 = 0.578125 x 2 , y 6 = 0.5703125 x 2 , y 7 = 0.572265625 x 2 , y 8 = 0.571289062 x 2 , y 9 = 0.571533203 x 2 , y 10 = 0.571411132 x 2 , y 11 = 0.57144165 x 2$

are calculated. In the same way, for $λ=0.25$, $γ=0.25$,

$y 1 = x 2 , y 2 = 0.25 x 2 , y 3 = 0.4375 x 2 , y 4 = 0.296875 x 2 , y 5 = 0.33203125 x 2 , y 6 = 0.305664062 x 2 , y 7 = 0.312255859 x 2 , y 8 = 0.307312011 x 2 , y 9 = 0.308547973 x 2 , y 10 = 0.307621001 x 2 , y 11 = 0.307852994 x 2$

are found and also for $λ=0.75$, $γ=0.25$,

$y 1 = x 2 , y 2 = 0.25 x 2 , y 3 = 0.8125 x 2 , y 4 = 0.390625 x 2 , y 5 = 0.70703125 x 2 , y 6 = 0.469726562 x 2 , y 7 = 0.647705078 x 2 , y 8 = 0.514221191 x 2 , y 9 = 0.614334106 x 2 , y 10 = 0.539249419 x 2 , y 11 = 0.595562934 x 2$

are obtained. Similarly, for $λ=0.25$, $γ=0.75$,

$y 1 = x 2 , y 2 = 0.75 x 2 , y 3 = 0.8125 x 2 , y 4 = 0.796875 x 2 , y 5 = 0.80078125 x 2 , y 6 = 0.799804687 x 2 , y 7 = 0.800048828 x 2 , y 8 = 0.799987792 x 2 , y 9 = 0.800003051 x 2 , y 10 = 0.799999236 x 2 , y 11 = 0.80000019 x 2$

are calculated. Finally, for $λ=0.75$, $γ=0.75$,

$y 1 = x 2 , y 2 = 0.75 x 2 , y 3 = 0.9375 x 2 , y 4 = 0.890625 x 2 , y 5 = 0.92578125 x 2 , y 6 = 0.916992187 x 2 , y 7 = 0.923583984 x 2 , y 8 = 0.921936035 x 2 , y 9 = 0.923171996 x 2 , y 10 = 0.922863006 x 2 , y 11 = 0.923094748 x 2$

are found. Now, we find the approximate solution using by the Euler method. Firstly, we use the formula

$y n + 1 = y n +hF( x n , y n )$

with $F(x,y)=2x(y+1)$, $h=0.2$ and $x 0 =0$, $y 0 =0$. From the initial condition $y(0)=0$, we have $F(0,0)=0$. We now proceed with the calculations as follows:

$y 1 = y 0 + h f ( x 0 , y 0 ) = 0 , x 1 = x 0 + h = 0.2 , y 2 = y 1 + h f ( x 1 , y 1 ) = 0.08 , x 2 = x 1 + h = 0.4 , y 3 = y 2 + h f ( x 2 , y 2 ) = 0.2528 , x 3 = x 2 + h = 0.6 .$

Finally, applying the Runge-Kutta method to the given initial value problem, we carry out the intermediate calculations in each step to give figures after the decimal point and round off the final results at each step to four such places.Here $F(x,y)=2x(y+1)$, $x 0 =0$, $y 0 =0$ and we are to use $h=0.2$. Using these quantities, we calculated successively $k 1$, $k 2$, $k 3$, $k 4$ and $K 0$ defined by

$k 1 = h g ( y 0 , x 0 ) , k 2 = h g ( y 0 + h 2 , x 0 + k 1 2 ) , k 3 = h g ( y 0 + h 2 , x 0 + k 2 2 ) , k 4 = h g ( y 0 + h , x 0 + k 3 )$

and $K 0 = 1 6 ( k 1 +2 k 2 +2 k 3 + k 4 )$, $y n + 1 = y n + K 0$. Thus we find $k 1$, $k 2$, $k 3$, $k 4$ for $n=0$ as follows:

$k 1 = 0.2 f ( x 0 , y 0 ) = 0 , k 2 = 0.2 f ( x 0 + 0.1 , y 0 + k 1 2 ) = 0.04 , k 3 = 0.2 f ( x 0 + 0.1 , y 0 + k 2 2 ) = 0.0408 , k 4 = 0.2 f ( x 0 + 0.2 , y 0 + k 3 ) = 0.083264 .$

So, $y 1 =0.040810666$ is obtained for $x 1 =0.2$. On the other hand, we calculated $k 1$, $k 2$, $k 3$, $k 4$ for $n=1$ as follows:

$k 1 = 0.2 f ( x 1 , y 1 ) = 0.083264853 , k 2 = 0.2 f ( x 1 + 0.1 , y 1 + k 1 2 ) = 0.129893171 , k 3 = 0.2 f ( x 1 + 0.1 , y 1 + k 2 2 ) = 0.13269087 , k 4 = 0.2 f ( x 1 + 0.2 , y 1 + k 3 ) = 0.187760245 .$

Hence $y 2 =0.173509529$ is calculated for $x 2 =0.4$. Finally we get $k 1$, $k 2$, $k 3$, $k 4$ for $n=2$ as follows:

$k 1 = 0.2 f ( x 2 , y 2 ) = 0.187761524 , k 2 = 0.2 f ( x 2 + 0.1 , y 2 + k 1 2 ) = 0.253478058 , k 3 = 0.2 f ( x 2 + 0.1 , y 2 + k 2 2 ) = 0.260049711 , k 4 = 0.2 f ( x 2 + 0.2 , y 2 + k 3 ) = 0.344054217 .$

Thus $y 3 =0.433321409$ is obtained for $x 3 =0.6$.

After the necessary calculation which is done above, the comparison is shown schematically in Figure 3.

On the other hand, we may give Table 11, Table 12, Table 13 and Table 14 by the new modified Ishikawa iteration method for different values of λ and γ. Now we may give Table 15 which is expressed that absolute error of Example 2.3 for different values of λ and γ with $x=0.2$, $x=0.4$ and $x=0.6$ respectively.

Corollary 2.5 If we compare the approximate solution with the different values of λ and γ, then the conclusion may be indicated using by Table 11, Table 12, Table 13 and Table 14 as follows.

The best approximation is obtained taking the different values of λ and γ and using the modified Ishikawa iteration method for $x=0.2$, $x=0.4$ and $x=0.5$ getting ($λ=0.25$, $γ=0.25$; $λ=0.5$, $γ=0.25$; $λ=0.25$, $γ=0.5$; $λ=0.75$, $γ=0.25$; $λ=0.5$, $γ=0.5$; $λ=0.25$, $γ=0.75$; $λ=0.75$, $γ=0.75$) respectively.

Similarly, we calculated the solution for $x=0.6$, $x=1$ and $x=1.5$, then the approximation is found more sensitive taking ($λ=0.25$, $γ=0.25$; $λ=0.5$, $γ=0.25$; $λ=0.25$, $γ=0.5$; $λ=0.75$, $γ=0.25$; $λ=0.5$, $γ=0.5$; $λ=0.25$, $γ=0.75$; $λ=0.75$, $γ=0.75$) respectively.

Corollary 2.6 Absolute error of the modified Ishikawa iteration method is computed taking different values of λ and γ ($x=0.2$, $x=0.4$ and $x=0.6$), which is not more effective than Picard, Runge-Kutta and Euler iteration methods.

## 3 Conclusion

A new technique, using the new modified Ishikawa iteration method, to numerically solve the different types of differential equations is presented. All the numerical results obtained using the new modified Ishikawa iteration method described earlier show a very good agreement with the exact solution. Comparing the new modified Ishikawa iteration method with several other methods that have been advanced for solving linear and nonlinear differential equations shows that the new technique is reliable, powerful and promising. We believe that the efficiency of the new modified Ishikawa iteration method gives it much wider applicability which should be explored further.

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Bildik, N., Bakır, Y. & Mutlu, A. The new modified Ishikawa iteration method for the approximate solution of different types of differential equations. Fixed Point Theory Appl 2013, 52 (2013). https://doi.org/10.1186/1687-1812-2013-52

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### Keywords

• ordinary differential equation
• Euler method
• fixed point
• numerical analysis
• modified Ishikawa iteration
• Picard successive iteration method 