<?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">OALibJ</journal-id><journal-title-group><journal-title>Open Access Library Journal</journal-title></journal-title-group><issn pub-type="epub">2333-9705</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/oalib.1106601</article-id><article-id pub-id-type="publisher-id">OALibJ-102243</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Biomedical&amp;Life Sciences</subject><subject> Business&amp;Economics</subject><subject> Chemistry&amp;Materials Science</subject><subject> Computer Science&amp;Communications</subject><subject> Earth&amp;Environmental Sciences</subject><subject> Engineering</subject><subject> Medicine&amp;Healthcare</subject><subject> Physics&amp;Mathematics</subject><subject> Social Sciences&amp;Humanities</subject></subj-group></article-categories><title-group><article-title>
 
 
  Successive Approximation Method for Solving Wu-Zhang Systems of (1 + 1) Dimensional
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Abdulghafor</surname><given-names>M. Al-Rozbayani</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Abdulbaset</surname><given-names>H. Shammar</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib></contrib-group><aff id="aff1"><addr-line>Department of Mathematics, College of Computer Sciences and Mathematics, University of Mosul, Mosul, Iraq</addr-line></aff><pub-date pub-type="epub"><day>04</day><month>08</month><year>2020</year></pub-date><volume>07</volume><issue>08</issue><fpage>1</fpage><lpage>7</lpage><history><date date-type="received"><day>11,</day>	<month>July</month>	<year>2020</year></date><date date-type="rev-recd"><day>15,</day>	<month>August</month>	<year>2020</year>	</date><date date-type="accepted"><day>18,</day>	<month>August</month>	<year>2020</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 section the Successive approximate method (S.A.M) introduced for solving the Wu-Zhang systems, a (1 + 1)-dimensional nonlinear dispersive wave equation, this method shows us that the technique provided without disorder, in this model of convergence power series with a simple calculated ingredients and gives effective results. 
  
 
</p></abstract><kwd-group><kwd>Successive Approximate Method</kwd><kwd> Wu-Zhang Systems</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>The Nonlinear partial differential equations (NPDEs) have a big advantage in various fields of science like mathematical, physics, chemistry, biology, mechanics, and engineering. It is important to get a reliable solution and numerical solution for nonlinear systems. In this section we solved the Wu-Zhang Systems of one dimensional with initial condition, the Wu-Zhang system can be written as</p><p>{ u t + u u x + v x = 0 v t + v u x + u v x + 1 3 u x x x = 0 (1)</p><p>where (v) is height water,(u) is the surface speed of the water along x-direction (Mix, S. K. [<xref ref-type="bibr" rid="scirp.102243-ref1">1</xref>], Laing, Y. C.; Feng [<xref ref-type="bibr" rid="scirp.102243-ref2">2</xref>], Ja, Anwar; Jameel [<xref ref-type="bibr" rid="scirp.102243-ref3">3</xref>], Manaa, Saad A.; Easif [<xref ref-type="bibr" rid="scirp.102243-ref4">4</xref>], R. K., Saeed [<xref ref-type="bibr" rid="scirp.102243-ref5">5</xref>], Zheng, Xuedong; Chen [<xref ref-type="bibr" rid="scirp.102243-ref6">6</xref>], Manaa, Saad A.; Easif [<xref ref-type="bibr" rid="scirp.102243-ref4">4</xref>] [<xref ref-type="bibr" rid="scirp.102243-ref7">7</xref>], Hosseini, K.; Ansari [<xref ref-type="bibr" rid="scirp.102243-ref8">8</xref>], Khater, Mostafa M. A.; Attia [<xref ref-type="bibr" rid="scirp.102243-ref9">9</xref>], Jafari, M. A.; Aminataei, A. [<xref ref-type="bibr" rid="scirp.102243-ref10">10</xref>]).</p></sec><sec id="s2"><title>2. Successive Approximation Method (S.A.M) [<xref ref-type="bibr" rid="scirp.102243-ref5">5</xref>]</title><p>The successive approximation method from important and active methods to solve partial differential equations, and also is good method for solving any initial value problem</p><p>∂ u ∂ t = f ( u , t ) ,   u ( x , t 0 ) = u 0 (2)</p><p>It starts by observing that any solution to (2) must also be a solution to</p><p>u ( x , t ) = u 0 + ∫ t 0 t f ( u ( x , s ) , s ) d s , (3)</p><p>Thus iterative these steps of solutions, we obtain on solution closer to the accurate solutions of (3), the S.A.M depends on the integral equation (3), as in the following:</p><p>u 0 ( x , t ) = u 0 ,</p><p>u 1 ( x , t ) = u 0 + ∫ t 0 t f ( u 0 ( x , s ) , s ) d s ,</p><p>u 2 ( x , t ) = u 0 + ∫ t 0 t f ( u 1 ( x , s ) , s ) d s ,</p><p>u 3 ( x , t ) = u 0 + ∫ t 0 t f ( u 2 ( x , s ) , s ) d s ,</p><p>This process can be continued to obtain the n<sup>th</sup> approximate</p><p>u n ( x , t ) = u 0 + ∫ t 0 t f ( u n − 1 ( x , s ) , s ) d s ,     n = 1 , 2 , ⋯</p><p>Therefore determine if u n ( x , t ) closer the solution u ( x , t ) as n increases. This is done by proving the following:</p><p>The sequence { u n ( x , t ) } converges to a limit u ( x , t ) , that is:</p><p>lim n → ∞ u n ( x , t ) = u ( x , t ) ,     t 0 ≤ x ≤ t .</p><p>The limiting function u ( x , t ) is a solution of (3) on the interval t 0 ≤ x ≤ t . The solution u ( x , t ) of (3) is unique. A proof of these results can be depended along the lines of the corresponding proof for ordinary differential equations (Coddington, 1995).</p></sec><sec id="s3"><title>3. Application (S.A.M) to solve of Equation (1)</title><p>In this section, we solve the Wu-Zhang systems of one dimensional</p><p>∂ u ( x , s ) ∂ s = − u ( x , s ) ∂ u ( x , s ) ∂ x − ∂ v ( x , s ) ∂ x (4)</p><p>∂ v ( x , s ) ∂ s = − v ( x , s ) ∂ u ( x , s ) ∂ x − u ( x , s ) ∂ v ( x , s ) ∂ x − 1 3 ∂ 3 u ( x , s ) ∂ x 3 (5)</p><p>With initial condition:</p><p>u ( x , 0 ) = 1 3 k 1 [ b 0 + 18 k 1 3 + b 0 2 tanh ( 54 k 1 3 + 3 b 0 2 6 k 1 ( x + k 2 ) ) ] (6)</p><p>v ( x , 0 ) = k 1 + b 0 3 k 1 [ 1 3 k 1 [ b 0 + 18 k 1 2 + b 0 2 tanh ( 54 k 1 3 + 3 b 0 2 6 k 1 ( x + k 2 ) ) ] ]     − 1 2 [ 1 3 k 1 [ b 0 + 18 k 1 3 + b 0 2 tanh ( 54 k 1 3 + 3 b 0 2 6 k 1 ( x + k 2 ) ) ] ] 2 (7)</p><p>where b 0 , k 1 and k 2 are arbitrary constants.</p><p>By using S.A.M as follows:</p><p>Integrating both sides of Equations (4), (5) with respect to (s), from (0) to (t), we get</p><p>u ( x , s ) = u ( x , 0 ) − ∫ 0 t ( u ( x , s ) ∂ u ( x , s ) ∂ x + ∂ v ( x , s ) ∂ x ) d s (8)</p><p>v ( x , s ) = v ( x , 0 ) − ∫ 0 t ( v ( x , s ) ∂ u ( x , s ) ∂ x + u ( x , s ) ∂ v ( x , s ) ∂ x + 1 3 ∂ 3 u ( x , s ) ∂ x 3 ) d s (9)</p><p>Using the initial condition in Equations (8), (9) we obtain:</p><p>u ( x , s ) = u 0 − ∫ 0 t ( u ( x , s ) ∂ u ( x , s ) ∂ x + ∂ v ( x , s ) ∂ x ) d s (10)</p><p>v ( x , s ) = v 0 − ∫ 0 t ( v ( x , s ) ∂ u ( x , s ) ∂ x + u ( x , s ) ∂ v ( x , s ) ∂ x + 1 3 ∂ 3 u ( x , s ) ∂ x 3 ) d s (11)</p><p>We substituting initial conditions u ( x , 0 ) and v ( x , 0 ) in the integral Equations (10), (11) to get a first approximation u 1 ( x , t ) and v 1 ( x , t )</p><p>u 1 ( x , s ) = u 0 − ∫ 0 t ( u 0 ( x , s ) ∂ u 0 ( x , s ) ∂ x + ∂ v 0 ( x , s ) ∂ x ) d s (12)</p><p>v 1 ( x , s ) = v 0 − ∫ 0 t ( v 0 ( x , s ) ∂ u 0 ( x , s ) ∂ x + u 0 ( x , s ) ∂ v 0 ( x , s ) ∂ x + 1 3 ∂ 3 u 0 ( x , s ) ∂ x 3 ) d s (13)</p><p>Then this u 1 ( x , t ) and v 1 ( x , t ) is substituted again in the integral of (10), (11) after replacing (t) by (s) to get a second approximation u 2 ( x , t ) and v 2 ( x , t )</p><p>u 2 ( x , s ) = u 0 − ∫ 0 t ( u 1 ( x , s ) ∂ u 1 ( x , s ) ∂ x + ∂ v 1 ( x , s ) ∂ x ) d s (14)</p><p>v 2 ( x , s ) = v 0 − ∫ 0 t ( v 1 ( x , s ) ∂ u 1 ( x , s ) ∂ x + u 1 ( x , s ) ∂ v 1 ( x , s ) ∂ x + 1 3 ∂ 3 u 1 ( x , s ) ∂ x 3 ) d s (15)</p><p>This process can continued to get the n<sup>th</sup> approximation</p><p>u n ( x , s ) = u 0 − ∫ 0 t ( u n − 1 ( x , s ) ∂ u n − 1 ( x , s ) ∂ x + ∂ v n − 1 ( x , s ) ∂ x ) d s (16)</p><p>v n ( x , s ) = v 0 − ∫ 0 t ( v n − 1 ( x , s ) ∂ u n − 1 ( x , s ) ∂ x + u n − 1 ( x , s ) ∂ v n − 1 ( x , s ) ∂ x + 1 3 ∂ 3 u ( x , s ) ∂ x 3 ) d s (17)</p><p>For n = 1 , 2 , ⋯ to solve Equations (16), (17), we use the initial conditions u ( x , 0 ) and v ( x , 0 ) which are given in Equations (6) and (7) , respectively. by using iterative steps for Equations (16), (17) we can obtain u 1 ( x , t ) , u 2 ( x , t ) , u 3 ( x , t ) and v 1 ( x , t ) , v 2 ( x , t ) , v 3 ( x , t ) , after then substituting these values in the Equations (18) and (19)</p><p>u = ∑ n = 0 ∞ u n ( x , t ) (18)</p><p>v = ∑ n = 0 ∞ v n ( x , t ) (19)</p><p>We get the following series:</p><p>u ˜ 3 ( x , t ) = 1 3 k 1 b 0 + 18 k 1 3 + b 0 2 tanh ( 1 6 k 1 54 k 1 3 + 3 b 0 2 ( x + k 2 ) )     − 1 54 b 0 ( 18 k 1 3 + b 0 2 ) 3 t k 1 3 cosh ( 1 6 k 1 3 18 k 1 3 + b 0 2 ( x + k 2 ) ) 2 + ⋯</p><p>⋮</p><p>v ˜ 3 ( x , t ) = k 1 + 1 9 k 1 2 b 0 ( b 0 + 18 k 1 3 + b 0 2 tanh ( 1 6 k 1 54 k 1 3 + 3 b 0 2 ( x + k 2 ) ) )     − 1 18 k 1 2 ( b 0 + 18 k 1 3 + b 0 2 tanh ( 1 6 k 1 54 k 1 3 + 3 b 0 2 ( x + k 2 ) ) ) 2     + 1 162 sinh ( 1 6 k 1 54 k 1 3 + 3 b 0 2 ( x + k 2 ) ( 18 k 1 3 + b 0 2 ) 54 k 1 3 + 3 b 0 2 b 0 t ) cosh ( 1 6 k 1 54 k 1 3 + 3 b 0 2 ( x + k 2 ) 3 k 1 4 ) + ⋯</p><p>⋮</p><p>Therefore, the exact solution of Wu-Zhang equation is given by</p><p>u ( x , t ) = lim n → ∞ u ˜ n ( x , t ) ,</p><p>v ( x , t ) = lim n → ∞ v ˜ n ( x , t ) ,</p><p>This solution is convergent to the exact solution</p><p>u ( x , t ) = 1 3 k 1 [ b 0 + 18 k 1 3 + b 0 2 tanh ( 54 k 1 3 + 3 b 0 2 6 k 1 ( x − b 0 3 k 1 t + k 2 ) ) ]</p><p>v ( x , t ) = k 1 + b 0 3 k 1 [ 1 3 k 1 [ b 0 + 18 k 1 3 + b 0 2 tanh ( 54 k 1 3 + 3 b 0 2 6 k 1 ( x − b 0 3 k 1 t + k 2 ) ) ] ]     − 1 2 [ 1 3 k 1 [ b 0 + 18 k 1 3 + b 0 2 tanh ( 54 k 1 3 + 3 b 0 2 6 k 1 ( x − b 0 3 k 1 t + k 2 ) ) ] ] 2</p><p>The approximate results for u ( x , t ) and v ( x , t ) were compared with exact solution as showing in <xref ref-type="table" rid="table1">Table 1</xref> and <xref ref-type="table" rid="table2">Table 2</xref>. Also we compare figure approximate solution with figure exact solution as showing in <xref ref-type="fig" rid="fig1">Figure 1</xref> and <xref ref-type="fig" rid="fig2">Figure 2</xref>. <xref ref-type="fig" rid="fig3">Figure 3</xref>, error between approximate solution and exact solution.</p><p>In <xref ref-type="fig" rid="fig1">Figure 1</xref> and <xref ref-type="fig" rid="fig2">Figure 2</xref> the final results obtained from (S.A.M) were compared with the results of the exact solution when b 0 = 0.1 , k 1 = 0.05 , k 2 = 0.1 , t = 0.2 . The comparison shows a good agreement between the results.</p><table-wrap id="table1" ><label><xref ref-type="table" rid="table1">Table 1</xref></label><caption><title> Comparing exact solutions with approximate solutions (S.A.M) of the Wu-Zhang systems with initial conditions Equations (6), (7) where b 0 = 0.1 , k 1 = 0.05 , k 2 = 0.1 , t = 0.2 for u ( x , t ) </title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x</th><th align="center" valign="middle" >u SAM</th><th align="center" valign="middle" >u exact</th><th align="center" valign="middle" >| u SAM − u exact |</th></tr></thead><tr><td align="center" valign="middle" >−50</td><td align="center" valign="middle" >−0.07119812060</td><td align="center" valign="middle" >−0.07119812060</td><td align="center" valign="middle" >5.089992619 &#215; 10<sup>−29</sup></td></tr><tr><td align="center" valign="middle" >−40</td><td align="center" valign="middle" >−0.07119812060</td><td align="center" valign="middle" >−0.07119812060</td><td align="center" valign="middle" >1.807525608 &#215; 10<sup>−23</sup></td></tr><tr><td align="center" valign="middle" >−30</td><td align="center" valign="middle" >−0.07119812060</td><td align="center" valign="middle" >−0.07119812060</td><td align="center" valign="middle" >6.418769275 &#215; 10<sup>−18</sup></td></tr><tr><td align="center" valign="middle" >−20</td><td align="center" valign="middle" >−0.07119812060</td><td align="center" valign="middle" >−0.07119812060</td><td align="center" valign="middle" >2.279392226 &#215; 10<sup>−12</sup></td></tr><tr><td align="center" valign="middle" >−10</td><td align="center" valign="middle" >−0.07119413194</td><td align="center" valign="middle" >−0.07119413830</td><td align="center" valign="middle" >6.362404900 &#215; 10<sup>−9</sup></td></tr><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0.6478908251</td><td align="center" valign="middle" >0.6509522859</td><td align="center" valign="middle" >3061460790 &#215; 10<sup>−3</sup></td></tr><tr><td align="center" valign="middle" >10</td><td align="center" valign="middle" >1.404527229</td><td align="center" valign="middle" >1.404527118</td><td align="center" valign="middle" >1.113311969 &#215; 10<sup>−7</sup></td></tr><tr><td align="center" valign="middle" >20</td><td align="center" valign="middle" >1.404531454</td><td align="center" valign="middle" >1.404531454</td><td align="center" valign="middle" >1.765272343 &#215; 10<sup>−12</sup></td></tr><tr><td align="center" valign="middle" >30</td><td align="center" valign="middle" >1.404531454</td><td align="center" valign="middle" >1.404531454</td><td align="center" valign="middle" >4.971007506 &#215; 10<sup>−18</sup></td></tr><tr><td align="center" valign="middle" >40</td><td align="center" valign="middle" >1.404531454</td><td align="center" valign="middle" >1.404531454</td><td align="center" valign="middle" >1.399835854 &#215; 10<sup>−23</sup></td></tr><tr><td align="center" valign="middle" >50</td><td align="center" valign="middle" >1.404531454</td><td align="center" valign="middle" >1.404531454</td><td align="center" valign="middle" >3.941938236 &#215; 10<sup>−29</sup></td></tr></tbody></table></table-wrap><table-wrap id="table2" ><label><xref ref-type="table" rid="table2">Table 2</xref></label><caption><title> Comparing exact solutions with approximate solutions (S.A.M) of the Wu-Zhang systems with initial conditions Equations (6), (7) where b 0 = 0.1 , k 1 = 0.05 , k 1 = 0.05 , t = 0.2 for v ( x , t ) </title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x</th><th align="center" valign="middle" >v SAM</th><th align="center" valign="middle" >v exact</th><th align="center" valign="middle" >| v SAM − v exact |</th></tr></thead><tr><td align="center" valign="middle" >−50</td><td align="center" valign="middle" >7.000000000 &#215; 10<sup>−12</sup></td><td align="center" valign="middle" >7.000000000 &#215; 10<sup>−12</sup></td><td align="center" valign="middle" >7.000000000 &#215; 10<sup>−12</sup></td></tr><tr><td align="center" valign="middle" >−40</td><td align="center" valign="middle" >7.000000000 &#215; 10<sup>−12</sup></td><td align="center" valign="middle" >7.000000000 &#215; 10<sup>−12</sup></td><td align="center" valign="middle" >7.000000000 &#215; 10<sup>−12</sup></td></tr><tr><td align="center" valign="middle" >−30</td><td align="center" valign="middle" >6.999999972 &#215; 10<sup>−12</sup></td><td align="center" valign="middle" >7.000000000 &#215; 10<sup>−12</sup></td><td align="center" valign="middle" >7.000000028 &#215; 10<sup>−12</sup></td></tr><tr><td align="center" valign="middle" >−20</td><td align="center" valign="middle" >6.990087894 &#215; 10<sup>−12</sup></td><td align="center" valign="middle" >7.000000000 &#215; 10<sup>−12</sup></td><td align="center" valign="middle" >7.009912106 &#215; 10<sup>−12</sup></td></tr><tr><td align="center" valign="middle" >−10</td><td align="center" valign="middle" >2943074716 &#215; 10<sup>−6</sup></td><td align="center" valign="middle" >2938409000 &#215; 10<sup>−6</sup></td><td align="center" valign="middle" >4.671715510 &#215; 10<sup>−9</sup></td></tr><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0.2749015323</td><td align="center" valign="middle" >0.2720987513</td><td align="center" valign="middle" >2802780979 &#215; 10<sup>−3</sup></td></tr><tr><td align="center" valign="middle" >10</td><td align="center" valign="middle" >3117925519 &#215; 10<sup>−6</sup></td><td align="center" valign="middle" >3200100000 &#215; 10<sup>−6</sup></td><td align="center" valign="middle" >8.207448126 &#215; 10<sup>−8</sup></td></tr><tr><td align="center" valign="middle" >20</td><td align="center" valign="middle" >7.676417228 &#215; 10<sup>−5</sup></td><td align="center" valign="middle" >0.0000000000000</td><td align="center" valign="middle" >7.676417228 &#215; 10<sup>−15</sup></td></tr><tr><td align="center" valign="middle" >30</td><td align="center" valign="middle" >2.161679351 &#215; 10<sup>−20</sup></td><td align="center" valign="middle" >0.0000000000000</td><td align="center" valign="middle" >2.161679351 &#215; 10<sup>−20</sup></td></tr><tr><td align="center" valign="middle" >40</td><td align="center" valign="middle" >6.087289664 &#215; 10<sup>−26</sup></td><td align="center" valign="middle" >0.0000000000000</td><td align="center" valign="middle" >6.087289664 &#215; 10<sup>−26</sup></td></tr><tr><td align="center" valign="middle" >50</td><td align="center" valign="middle" >1.714180972 &#215; 10<sup>−31</sup></td><td align="center" valign="middle" >0.0000000000000</td><td align="center" valign="middle" >1.714180972 &#215; 10<sup>−31</sup></td></tr></tbody></table></table-wrap></sec><sec id="s4"><title>4. Conclusion</title><p>In this search, the Successive Approximate Method (S.A.M) was used to get the approximate solution of Wu-Zhang Systems. The results we got from this method are high efficient with big accurate and give a good convergence to the exact solution.</p></sec><sec id="s5"><title>Acknowledgements</title><p>The authors are very grateful to the University of Mosul/College of Computer Sciences and Mathematics for their provided facilities, which helped to improve the quality of this work.</p></sec><sec id="s6"><title>Conflicts of Interest</title><p>The authors declare no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s7"><title>Cite this paper</title><p>Al-Rozbayani, A.M. and Shammar, A.H. (2020) Successive Approximation Method for Solving Wu- Zhang Systems of (1 + 1) Dimensional. 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