<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">APM</journal-id><journal-title-group><journal-title>Advances in Pure Mathematics</journal-title></journal-title-group><issn pub-type="epub">2160-0368</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/apm.2019.910040</article-id><article-id pub-id-type="publisher-id">APM-95490</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  The Successive Approximation Method for Solving Nonlinear Fredholm Integral Equation of the Second Kind Using Maple
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Dalal</surname><given-names>Adnan Maturi</given-names></name><xref ref-type="aff" rid="aff1"><sub>1</sub></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib></contrib-group><aff id="aff1"><label>1</label><addr-line>Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah, KSA</addr-line></aff><pub-date pub-type="epub"><day>27</day><month>09</month><year>2019</year></pub-date><volume>09</volume><issue>10</issue><fpage>832</fpage><lpage>843</lpage><history><date date-type="received"><day>24,</day>	<month>August</month>	<year>2019</year></date><date date-type="rev-recd"><day>27,</day>	<month>September</month>	<year>2019</year>	</date><date date-type="accepted"><day>30,</day>	<month>September</month>	<year>2019</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  In this paper, we will use the successive approximation method for solving Fredholm integral equation of the second kind using Maple18. By means of this method, an algorithm is successfully established for solving the non-linear Fredholm integral equation of the second kind. Finally, several examples are presented to illustrate the application of the algorithm and results appear that this method is very effective and convenient to solve these equations.
 
</p></abstract><kwd-group><kwd>Nonlinear Fredholm Integral Equation of the Second Kind</kwd><kwd> Successive Approximation Method</kwd><kwd> Maple18</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>The current research intends to the successive approximation method for solving nonlinear Fredholm integral equation of the second kind using Maple18.</p><p>Homotopy perturbation technique in [<xref ref-type="bibr" rid="scirp.95490-ref1">1</xref>] . A coupling method of a homotopy technique and a perturbation technique [<xref ref-type="bibr" rid="scirp.95490-ref2">2</xref>] . Homotopy perturbation method: anew non-linear analytical technique [<xref ref-type="bibr" rid="scirp.95490-ref3">3</xref>] . Asymptotology by homotopy perturbation method [<xref ref-type="bibr" rid="scirp.95490-ref4">4</xref>] . Application of homotopy perturbation method to nonlinear wave equations [<xref ref-type="bibr" rid="scirp.95490-ref5">5</xref>] . Homotopy perturbation method for solving boundary value problems [<xref ref-type="bibr" rid="scirp.95490-ref6">6</xref>] . New interpretation of homotopy perturbation method [<xref ref-type="bibr" rid="scirp.95490-ref7">7</xref>] . Numerical Solution of Systems of Linear Volterra Integral Equations Using Block-Pulse Functions [<xref ref-type="bibr" rid="scirp.95490-ref8">8</xref>] . Numerical Analysis and Volterra Integral and Differential Equations [<xref ref-type="bibr" rid="scirp.95490-ref9">9</xref>] [<xref ref-type="bibr" rid="scirp.95490-ref10">10</xref>] . Biorthogonal Systems for Solving Volterra Integral equation Systems of the Second Kind [<xref ref-type="bibr" rid="scirp.95490-ref11">11</xref>] . Modified HPM for Solving Systems of Volterra Integral Equation of the Second Kind [<xref ref-type="bibr" rid="scirp.95490-ref12">12</xref>] . Analytical and Numerical Methods for Volterra Equations [<xref ref-type="bibr" rid="scirp.95490-ref13">13</xref>] . Numerical Computational Solution of the Linear Volterra Integral Equations System Via Rationalized Hear Functions [<xref ref-type="bibr" rid="scirp.95490-ref14">14</xref>] . Linear and Nonlinear Integral Equation: Methods and Application [<xref ref-type="bibr" rid="scirp.95490-ref15">15</xref>] . Optimal Control Approach for Solving Linear Volterra Integral Equations [<xref ref-type="bibr" rid="scirp.95490-ref16">16</xref>] . Numerical Solution of System of Two Nonlinear Volterra Integral Equations [<xref ref-type="bibr" rid="scirp.95490-ref17">17</xref>] [<xref ref-type="bibr" rid="scirp.95490-ref18">18</xref>] . Adomian Decomposition Method of Fredholm Integral Equation of the Second Kind Using Maple and applications in [<xref ref-type="bibr" rid="scirp.95490-ref19">19</xref>] [<xref ref-type="bibr" rid="scirp.95490-ref20">20</xref>] . Approximate solutions of linear Volterra integral equation systems with variable coeffi cients [<xref ref-type="bibr" rid="scirp.95490-ref21">21</xref>] . Numerical Solution of System of Three Nonlinear Volterra Integral Equation Using Implicit Trapezoidal [<xref ref-type="bibr" rid="scirp.95490-ref22">22</xref>] . Solving nth-Order Integro-Differential Equations Using the Combined Laplace Transfer-Successive Approximations Method [<xref ref-type="bibr" rid="scirp.95490-ref23">23</xref>] . Numerical approach based on Bernstein polynomials for solving mixed Volterra-Fredholm integral equations [<xref ref-type="bibr" rid="scirp.95490-ref24">24</xref>] . An inverse eigenproblem and an associated approximation problem for generalized reflexive and anti-reflexive matrices [<xref ref-type="bibr" rid="scirp.95490-ref25">25</xref>] . Discrete Adomian Decomposition solution of Nonlinear Fredholm Integral Equation [<xref ref-type="bibr" rid="scirp.95490-ref26">26</xref>] . He’s homotopy perturbation method: A strongly promising method for solving non-linear systems of the mixed Volterra-Fredholm integral equations [<xref ref-type="bibr" rid="scirp.95490-ref27">27</xref>] . On the convergence of Homotopy perturbation Method [<xref ref-type="bibr" rid="scirp.95490-ref28">28</xref>] . The homotopy perturbation method for solving neutral functional-differential equations with proportional delays [<xref ref-type="bibr" rid="scirp.95490-ref29">29</xref>] . Variational iteration method as a kernel constructive technique [<xref ref-type="bibr" rid="scirp.95490-ref30">30</xref>] . A modified homotopy perturbation method for solving the nonlinear mixed Volterra–Fredholm integral equation [<xref ref-type="bibr" rid="scirp.95490-ref31">31</xref>] . A note on the homotopy analysis method [<xref ref-type="bibr" rid="scirp.95490-ref32">32</xref>] . Application of homotopy analysis method to the solution of ninth order boundary value problems in AFTI-F16 fighters [<xref ref-type="bibr" rid="scirp.95490-ref33">33</xref>] . Modeling spectra of breaking waves propagating over beach [<xref ref-type="bibr" rid="scirp.95490-ref34">34</xref>] . Modified Adomian decomposition method for solving the problem of boundary layer convective heat transfer [<xref ref-type="bibr" rid="scirp.95490-ref35">35</xref>] . New version of Optimal Homotopy Asymptotic Method for the solution of nonlinear boundary value problems in finite and infinite intervals [<xref ref-type="bibr" rid="scirp.95490-ref36">36</xref>] .</p><p>Different types of analytical methods and numerical methods were used to solve the problem [<xref ref-type="bibr" rid="scirp.95490-ref1">1</xref>] - [<xref ref-type="bibr" rid="scirp.95490-ref36">36</xref>] . In this article we have applied the successive approximation method used by using the Maple algorithm by applying this algorithm to different examples, including finding the approximate solution and then comparing it to the exact solution and finding out the amount of error between the approximate solution and the exact solution.</p><p>The main objective of this work is to use the successive approximations method in solving the nonlinear Fredholm integral equation of the second kind using Maple18.</p><p>The paper is arranged as follows: In Section 2, the successive approximations method. In Section 3, numerical examples are also considered to show the ability of the proposed method, and the conclusion is drawn in Section 4.</p></sec><sec id="s2"><title>2. The Successive Approximation Method</title><p>The nonlinear Fredholm integral equation of the second kind</p><p>u ( x ) = f ( x ) + λ ∫ a b K ( x , t ) F ( u ( t ) ) d t (1)</p><p>where u ( x ) is the unknown function to be determined, K ( x , t ) is the kernel, F ( u ( t ) ) is a nonlinear function of u ( t ) , and λ is a parameter. u 0 ( x ) = any selective real valued function,</p><p>u n + 1 ( x ) = f ( x ) + λ ∫ a b K ( x , t ) u n ( t ) d t ,     n ≥ 0. (2)</p><p>The question of convergence of u n ( x ) is justified by noting the following theorem</p><p>Theorem 1 see [<xref ref-type="bibr" rid="scirp.95490-ref16">16</xref>] If f ( x ) in (1) is continuous for the interval a ≤ x ≤ b and the kernel K ( x , t ) is also continuous in the triangle a ≤ x ≤ b , a ≤ t ≤ b the sequence of successive approximations u n ( x ) , n ≥ 0 converges to the solution u ( x ) of the integral equation under discussion.</p></sec><sec id="s3"><title>3. Numerical Examples</title><p>In this section, we solve some examples, and we can compare the numerical results with the exact solution.</p><p>Example 1. Consider the nonlinear Fredholm integral equation of the second kind</p><p>u ( x ) = cos ( x ) − π 2 48 + 1 12 ∫ 0 π t u 2 ( t ) d t , (3)</p><p>with the exact solution u ( x ) = cos ( x ) .</p><p>Example 2. Consider the nonlinear Fredholm integral equation of the second kind</p><p>u ( x ) = ln x + 143 144 + 1 36 ∫ 0 1 t u 2 ( t ) d t , (4)</p><p>with the exact solution u ( x ) = x + ln x .</p><table-wrap id="table1" ><label><xref ref-type="table" rid="table1">Table 1</xref></label><caption><title> Numerical results and exact solution of Nonlinear Fredholm integral equation of the second kind for example 1</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x</th><th align="center" valign="middle" >E x a c t 1 = cos ( x )</th><th align="center" valign="middle" >u = cos ( x ) − π 2 48 + 0.2049572390</th><th align="center" valign="middle" >E r r o r = | E x a c t 1 − u |</th></tr></thead><tr><td align="center" valign="middle" >0.1</td><td align="center" valign="middle" >0.9950042</td><td align="center" valign="middle" >0.9943446</td><td align="center" valign="middle" >0.0006595</td></tr><tr><td align="center" valign="middle" >0.2</td><td align="center" valign="middle" >0.9800666</td><td align="center" valign="middle" >0.9794071</td><td align="center" valign="middle" >0.0006595</td></tr><tr><td align="center" valign="middle" >0.3</td><td align="center" valign="middle" >0.9553365</td><td align="center" valign="middle" >0.9546770</td><td align="center" valign="middle" >0.0006595</td></tr><tr><td align="center" valign="middle" >0.4</td><td align="center" valign="middle" >0.9210610</td><td align="center" valign="middle" >0.9204015</td><td align="center" valign="middle" >0.0006595</td></tr><tr><td align="center" valign="middle" >0.5</td><td align="center" valign="middle" >0.8775826</td><td align="center" valign="middle" >0.8769230</td><td align="center" valign="middle" >0.0006595</td></tr><tr><td align="center" valign="middle" >0.6</td><td align="center" valign="middle" >0.8253356</td><td align="center" valign="middle" >0.8246761</td><td align="center" valign="middle" >0.0006595</td></tr><tr><td align="center" valign="middle" >0.7</td><td align="center" valign="middle" >0.7648422</td><td align="center" valign="middle" >0.7641827</td><td align="center" valign="middle" >0.0006595</td></tr><tr><td align="center" valign="middle" >0.8</td><td align="center" valign="middle" >0.6967067</td><td align="center" valign="middle" >0.6960472</td><td align="center" valign="middle" >0.0006595</td></tr><tr><td align="center" valign="middle" >0.9</td><td align="center" valign="middle" >0.6216100</td><td align="center" valign="middle" >0.6209504</td><td align="center" valign="middle" >0.0006595</td></tr><tr><td align="center" valign="middle" >1.0</td><td align="center" valign="middle" >0.5403023</td><td align="center" valign="middle" >0.5396428</td><td align="center" valign="middle" >0.0006595</td></tr></tbody></table></table-wrap><table-wrap id="table2" ><label><xref ref-type="table" rid="table2">Table 2</xref></label><caption><title> Numerical results and exact solution of Nonlinear Fredholm integral equation of the second kind for example 2</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >E r r o r = | E x a c t 2 − u |</th><th align="center" valign="middle" >u = x + ln x + 1.34767 10000000</th><th align="center" valign="middle" >E x a c t 2 = x + ln x</th><th align="center" valign="middle" >x</th></tr></thead><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >−2.2025850</td><td align="center" valign="middle" >−2.2025851</td><td align="center" valign="middle" >0.1</td></tr><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >−1.4094378</td><td align="center" valign="middle" >−1.4094379</td><td align="center" valign="middle" >0.2</td></tr><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >−0.9039727</td><td align="center" valign="middle" >−0.9039728</td><td align="center" valign="middle" >0.3</td></tr><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >−0.5162906</td><td align="center" valign="middle" >−0.5162907</td><td align="center" valign="middle" >0.4</td></tr><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >−0.1931470</td><td align="center" valign="middle" >−0.1931472</td><td align="center" valign="middle" >0.5</td></tr><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >0.0891745</td><td align="center" valign="middle" >0.0891744</td><td align="center" valign="middle" >0.6</td></tr><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >0.3433252</td><td align="center" valign="middle" >0.3433251</td><td align="center" valign="middle" >0.7</td></tr><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >0.5768566</td><td align="center" valign="middle" >0.5768564</td><td align="center" valign="middle" >0.8</td></tr><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >0.7946396</td><td align="center" valign="middle" >0.7946395</td><td align="center" valign="middle" >0.9</td></tr><tr><td align="center" valign="middle" >0.0000001</td><td align="center" valign="middle" >1.0000001</td><td align="center" valign="middle" >1.0000000</td><td align="center" valign="middle" >1.0</td></tr></tbody></table></table-wrap><p>Example 3. Consider the nonlinear Fredholm integral equation of the second kind</p><p>u ( x ) = x e x − 1 288 ( 3 + e 2 ) x + 1 36 ∫ 0 1 x t u 2 ( t ) d t , (5)</p><p>with the exact solution u ( x ) = x e x .</p><p>Example 4. Consider the nonlinear Fredholm integral equation of the second kind</p><p>u ( x ) = e x + 1 144 ( 127 − e 2 ) + 1 36 ∫ 0 1 t ( u + u 2 ( t ) ) d t , (6)</p><p>with the exact solution u ( x ) = 1 + e x .</p><table-wrap id="table3" ><label><xref ref-type="table" rid="table3">Table 3</xref></label><caption><title> Numerical results and exact solution of Nonlinear Fredholm integral equation of the second kind for example 3</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >E r r o r = | E x a c t 3 − u |</th><th align="center" valign="middle" >u = x e x − ( 1 96 + 1 288 e x ) x + 0.03519196200 x</th><th align="center" valign="middle" >E x a c t 3 = x e x</th><th align="center" valign="middle" >x</th></tr></thead><tr><td align="center" valign="middle" >0.0000881</td><td align="center" valign="middle" >0.1104290</td><td align="center" valign="middle" >0.1105171</td><td align="center" valign="middle" >0.1</td></tr><tr><td align="center" valign="middle" >0.0001762</td><td align="center" valign="middle" >0.2441043</td><td align="center" valign="middle" >0.2442806</td><td align="center" valign="middle" >0.2</td></tr><tr><td align="center" valign="middle" >0.0002643</td><td align="center" valign="middle" >0.4046933</td><td align="center" valign="middle" >0.4049576</td><td align="center" valign="middle" >0.3</td></tr><tr><td align="center" valign="middle" >0.0003525</td><td align="center" valign="middle" >0.5963774</td><td align="center" valign="middle" >0.5967299</td><td align="center" valign="middle" >0.4</td></tr><tr><td align="center" valign="middle" >0.0004406</td><td align="center" valign="middle" >0.8239201</td><td align="center" valign="middle" >0.8243606</td><td align="center" valign="middle" >0.5</td></tr><tr><td align="center" valign="middle" >0.0005287</td><td align="center" valign="middle" >1.0927426</td><td align="center" valign="middle" >1.0932713</td><td align="center" valign="middle" >0.6</td></tr><tr><td align="center" valign="middle" >0.0006168</td><td align="center" valign="middle" >1.4090101</td><td align="center" valign="middle" >1.4096269</td><td align="center" valign="middle" >0.7</td></tr><tr><td align="center" valign="middle" >0.0007049</td><td align="center" valign="middle" >1.7797278</td><td align="center" valign="middle" >1.7804327</td><td align="center" valign="middle" >0.8</td></tr><tr><td align="center" valign="middle" >0.0007930</td><td align="center" valign="middle" >2.2128498</td><td align="center" valign="middle" >2.2136428</td><td align="center" valign="middle" >0.9</td></tr><tr><td align="center" valign="middle" >0.0008811</td><td align="center" valign="middle" >2.7174007</td><td align="center" valign="middle" >2.7182818</td><td align="center" valign="middle" >1.0</td></tr></tbody></table></table-wrap><table-wrap id="table4" ><label><xref ref-type="table" rid="table4">Table 4</xref></label><caption><title> Numerical results and exact solution of Nonlinear Fredholm integral equation of the second kind for example 4</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >E r r o r = | E x a c t 4 − u |</th><th align="center" valign="middle" >u = e x − 1 144 e 2 + 1.051185675</th><th align="center" valign="middle" >E x a c t 4 = 1 + e x</th><th align="center" valign="middle" >x</th></tr></thead><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >2.1050437</td><td align="center" valign="middle" >2.1051709</td><td align="center" valign="middle" >0.1</td></tr><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >2.2212755</td><td align="center" valign="middle" >2.2214028</td><td align="center" valign="middle" >0.2</td></tr><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >2.3497316</td><td align="center" valign="middle" >2.3498588</td><td align="center" valign="middle" >0.3</td></tr><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >2.4916975</td><td align="center" valign="middle" >2.4918247</td><td align="center" valign="middle" >0.4</td></tr><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >2.6485941</td><td align="center" valign="middle" >2.6487213</td><td align="center" valign="middle" >0.5</td></tr><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >2.8219916</td><td align="center" valign="middle" >2.8221188</td><td align="center" valign="middle" >0.6</td></tr><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >3.0136255</td><td align="center" valign="middle" >3.0137527</td><td align="center" valign="middle" >0.7</td></tr><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >3.2254137</td><td align="center" valign="middle" >3.2255409</td><td align="center" valign="middle" >0.8</td></tr><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >3.4594759</td><td align="center" valign="middle" >3.4596031</td><td align="center" valign="middle" >0.9</td></tr><tr><td align="center" valign="middle" >0.0001272</td><td align="center" valign="middle" >3.7181546</td><td align="center" valign="middle" >3.7182818</td><td align="center" valign="middle" >1.0</td></tr></tbody></table></table-wrap><p>Example 5. Consider the nonlinear Fredholm integral equation of the second kind</p><p>u ( x ) = sin ( x ) − π 2 64 + 1 48 ∫ 0 π t ( 1 + u 2 ( t ) ) d t , (7)</p><p>with the exact solution u ( x ) = sin ( x ) .</p><p>Example 6. Consider the nonlinear Fredholm integral equation of the second kind</p><p>u ( x ) = sin ( x ) + 1 − π 16 ( 1 + 5 π 12 ) + 1 48 ∫ 0 π t ( u + u 2 ( t ) ) d t (8)</p><table-wrap id="table5" ><label><xref ref-type="table" rid="table5">Table 5</xref></label><caption><title> Numerical results and exact solution of Nonlinear Fredholm integral equation of the second kind for example 5</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >E r r o r = | E x a c t 5 − u |</th><th align="center" valign="middle" >u = sin ( x ) − π 2 64 + 0.1543332664</th><th align="center" valign="middle" >E x a c t 5 = sin ( x )</th><th align="center" valign="middle" >x</th></tr></thead><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.0999541</td><td align="center" valign="middle" >0.0998334</td><td align="center" valign="middle" >0.1</td></tr><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.1987900</td><td align="center" valign="middle" >0.1986693</td><td align="center" valign="middle" >0.2</td></tr><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.2956409</td><td align="center" valign="middle" >0.2955202</td><td align="center" valign="middle" >0.3</td></tr><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.3895390</td><td align="center" valign="middle" >0.3894183</td><td align="center" valign="middle" >0.4</td></tr><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.4795462</td><td align="center" valign="middle" >0.4794255</td><td align="center" valign="middle" >0.5</td></tr><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.5647632</td><td align="center" valign="middle" >0.5646425</td><td align="center" valign="middle" >0.6</td></tr><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.6443384</td><td align="center" valign="middle" >0.6442177</td><td align="center" valign="middle" >0.7</td></tr><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.7174768</td><td align="center" valign="middle" >0.7173561</td><td align="center" valign="middle" >0.8</td></tr><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.7834476</td><td align="center" valign="middle" >0.7833269</td><td align="center" valign="middle" >0.9</td></tr><tr><td align="center" valign="middle" >0.0001207</td><td align="center" valign="middle" >0.8415917</td><td align="center" valign="middle" >0.8414710</td><td align="center" valign="middle" >1.0</td></tr></tbody></table></table-wrap><table-wrap id="table6" ><label><xref ref-type="table" rid="table6">Table 6</xref></label><caption><title> Numerical results and exact solution of Nonlinear Fredholm integral equation of the second kind for example 6</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >E r r o r = | E x a c t 6 − u |</th><th align="center" valign="middle" >u = sin ( x ) + 0.9808838911</th><th align="center" valign="middle" >E x a c t 6 = 1 + sin ( x )</th><th align="center" valign="middle" >x</th></tr></thead><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.0807173</td><td align="center" valign="middle" >1.0998334</td><td align="center" valign="middle" >0.1</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.1795532</td><td align="center" valign="middle" >1.1986693</td><td align="center" valign="middle" >0.2</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.2764041</td><td align="center" valign="middle" >1.2955202</td><td align="center" valign="middle" >0.3</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.3703022</td><td align="center" valign="middle" >1.3894183</td><td align="center" valign="middle" >0.4</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.4603094</td><td align="center" valign="middle" >1.4794255</td><td align="center" valign="middle" >0.5</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.5455264</td><td align="center" valign="middle" >1.5646425</td><td align="center" valign="middle" >0.6</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.6251016</td><td align="center" valign="middle" >1.6442177</td><td align="center" valign="middle" >0.7</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.6982400</td><td align="center" valign="middle" >1.7173561</td><td align="center" valign="middle" >0.8</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.7642108</td><td align="center" valign="middle" >1.7833269</td><td align="center" valign="middle" >0.9</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.8223549</td><td align="center" valign="middle" >1.8414710</td><td align="center" valign="middle" >1.0</td></tr></tbody></table></table-wrap><p>with the exact solution u ( x ) = 1 + sin ( x ) .</p><p>Example 7. Consider the nonlinear Fredholm integral equation of the second kind</p><p>u ( x ) = cos ( x ) + 7 6 − 5 π 2 144 + 1 36 ∫ 0 π t ( u + u 2 ( t ) ) d t , (9)</p><p>with the exact solution u ( x ) = 1 + cos (x)</p><table-wrap id="table7" ><label><xref ref-type="table" rid="table7">Table 7</xref></label><caption><title> Numerical results and exact solution of Nonlinear Fredholm integral equation of the second kind for example 7</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >E r r o r = | E x a c t 7 − u |</th><th align="center" valign="middle" >u = cos ( x ) + 1.345517155 − 5 π 2 144</th><th align="center" valign="middle" >E x a c t 7 = 1 + cos ( x )</th><th align="center" valign="middle" >x</th></tr></thead><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.0807173</td><td align="center" valign="middle" >1.0998334</td><td align="center" valign="middle" >0.1</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.1795532</td><td align="center" valign="middle" >1.1986693</td><td align="center" valign="middle" >0.2</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.2764041</td><td align="center" valign="middle" >1.2955202</td><td align="center" valign="middle" >0.3</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.3703022</td><td align="center" valign="middle" >1.3894183</td><td align="center" valign="middle" >0.4</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.4603094</td><td align="center" valign="middle" >1.4794255</td><td align="center" valign="middle" >0.5</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.5455264</td><td align="center" valign="middle" >1.5646425</td><td align="center" valign="middle" >0.6</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.6251016</td><td align="center" valign="middle" >1.6442177</td><td align="center" valign="middle" >0.7</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.6982400</td><td align="center" valign="middle" >1.7173561</td><td align="center" valign="middle" >0.8</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.7642108</td><td align="center" valign="middle" >1.7833269</td><td align="center" valign="middle" >0.9</td></tr><tr><td align="center" valign="middle" >0.0191161</td><td align="center" valign="middle" >1.8223549</td><td align="center" valign="middle" >1.8414710</td><td align="center" valign="middle" >1.0</td></tr></tbody></table></table-wrap></sec><sec id="s4"><title>4. Conclusion</title><p>In the paper, a successive approximations method is presented for solving the nonlinear Fredholm integral equation of the second kind using Maple18. The benefit of our method lies in the fact that, for some nonlinear problems, our method is still convergent as illustrated by figures and tables showing match the right accuracy, which shows the exact solution with the approximate solution is largely identical and noticeable Tables 1-7 represent the exact and numerical results of the examples in this article. Figures 1-7 readily show the comparison of exact solution and approximate solution. We can see from the figures that the approximate solution is very applicable to the exact solution and application is displayed through some examples. Numerical results show that the accuracy of the solutions obtained is good.</p></sec><sec id="s5"><title>Acknowledgements</title><p>This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University. The author, therefore, acknowledges with thanks to DSR technical and financial support.</p></sec><sec id="s6"><title>Conflicts of Interest</title><p>The author declares no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s7"><title>Cite this paper</title><p>Maturi, D.A. (2019) The Successive Approximation Method for Solving Nonlinear Fredholm Integral Equation of the Second Kind Using Maple. Advances in Pure Mathematics, 9, 832-843. https://doi.org/10.4236/apm.2019.910040</p></sec></body><back><ref-list><title>References</title><ref id="scirp.95490-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">He, J.H. (1999) Homotopy Perturbation Technique. Computer Methods in Applied Mechanics and Engineering, 178, 257-262.  
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