<!DOCTYPE html>
<html lang="en">
<head>
<script>
  MathJax = { 
    tex: {
		    inlineMath: [['\\(','\\)']]
	} }
</script>
<script type="text/javascript" src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-chtml.js">
</script>
<meta name="generator" content="plasTeX" />
<meta charset="utf-8" />
<meta name="viewport" content="width=device-width, initial-scale=1" />
<title>Existence and approximation of solutions to boundary value problems for delay integro-differential equations: Existence and approximation of solutions to boundary value problems for delay integro-differential equations</title>
<link rel="stylesheet" href="/var/www/clients/client1/web1/web/files/jnaat-files/journals/1/articles/styles/theme-white.css" />
</head>

<body>

<div class="wrapper">

<div class="content">
<div class="content-wrapper">


<div class="main-text">


<div class="titlepage">
<h1>Existence and approximation of solutions to boundary value problems for delay integro-differential equations</h1>
<p class="authors">
<span class="author">Igor Cherevko\(^\ast \) Andrew Dorosh\(^\ast \)</span>
</p>
<p class="date">June 17, 2015.</p>
</div>
<p>\(^\ast \)Faculty of Mathematics and Informatics, Yuriy Fedkovych Chernivtsi National University, Ukraine, e-mail: <span class="tt">i.cherevko@chnu.edu.ua</span>, <span class="tt">uefa2012@ukr.net</span>. </p>

<div class="abstract"><p> The paper deals with existence, uniqueness and spline approximation of solutions to boundary value problems for delay integro-differential equations. An iterative approximation scheme based on the use of cubic splines with defect two is presented, and sufficient conditions for its convergence are obtained. </p>
<p><b class="bf">MSC.</b> 34A45, 34K10, 34K28, 65L03, 65L10 </p>
<p><b class="bf">Keywords.</b> Boundary value problems, integro-differential equations, delay, spline functions </p>
</div>
<h1 id="sec:1">1 Introduction</h1>

<p>Dynamic processes in many applied problems are described by delay differential and integral equations (Andreeva, Kolmanovsky and Shayhet 1992). An analytical solution of such equation exists only in the simplest cases, so the construction and study of approximate algorithms for solutions of these equations are important. </p>
<p>In the present note we study an approximate method of solving boundary value problems for delay integro-differential equations based on approximation of the solution by cubic splines with defect two. </p>
<p>Existence and uniqueness of a solution of delay boundary value problems in various function spaces were considered by Grim and Schmitt (1968), Kamensky and Myshkis (1972), Biga and Gaber (2007), Athanasiadou (2013). Applying spline functions for solving differential-difference equations was investigated by Nikolova and Bainov (1981), Cherevko and Yakimov (1989), Nastasyeva and Cherevko (1999). </p>
<h1 id="sec:2">2 Problem statement. Existence of a solution</h1>
<p>  Let us consider the following boundary value problem </p>
<div class="displaymath" id="eq:1">
  \begin{align}  {y}”\left( x \right)& =f\left( x,y\left( x \right),y\left( x-{{\tau }_{0}}\left( x \right) \right),{y}’\left( x \right),{y}’\left( x-{{\tau }_{1}}\left( x \right) \right) \right)+ \label{eq:1}\\ & \quad +\int \nolimits _a^b \, g\left( x,s,y\left( s \right),y\left( s-{{\tau }_{0}}\left( s \right) \right),{y}’\left( s \right),{y}’\left( s-{{\tau }_{1}}\left( s \right) \right) \right)ds,\nonumber \\ {{y}^{\left( i \right)}}\left( x \right)& ={{\varphi }^{\left( i \right)}}\left( x \right),~ ~ i=0,1,~ ~ x\in \left[ {{a}^{*}};a \right],~ ~ y\left( b \right)=\gamma ,\label{eq:2} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000002">
  \[  {{a}^{*}}=\min \Big\{  \underset {x\in \left[ a;b \right]}{\mathop{\inf }}\! \! \! \left( x-{{\tau }_{0}}\left( x \right) \right),\! \! ~ \underset {x\in \left[ a;b \right]}{\mathop{\inf }}\! \! \! \left( x-{{\tau }_{1}}\left( x \right) \right) \Big\} , \quad ~ \gamma \in R,~ {{\tau }_{0}}\left( x \right)\ge 0, ~ {{\tau }_{1}}\left( x \right)\ge 0.  \]
</div>
<p>Let \(f\left( x,{{u}_{0}},{{u}_{1}},{{v}_{0}},{{v}_{1}} \right)\), \(~ g\left( x,s,{{u}_{0}},{{u}_{1}},{{v}_{0}},{{v}_{1}} \right)\) be continuous functions in \(G=\left[ a,b \right]\times G_{1}^{2}\times G_{2}^{2}\) and \(Q=\left[ a,b \right]\times G\), where \({{G}_{1}}=\left\{  u~ \in R:~ \left| u \right|{\lt}{{P}_{1}} \right\} \), \(~ {{G}_{2}}=\left\{  v~ \in R:~ \left| v \right|\le {{P}_{2}} \right\} \), \({{P}_{1}}, {{P}_{2}}\) are positive constants, \(\varphi \left( t \right)\in {{C}^{1}}\left[ {{a}^{*}};a \right]\), delays \({{\tau }_{0}}\left( x \right)\) and \({{\tau }_{1}}\left( x \right)\) are continuous functions on \(\left[ a,b \right]\), and additionally, \({{\tau }_{1}}\left( x \right)\) is such that the set \(E=\{  {{x}_{i}}\in \left[ a,b \right]:~ {{x}_{i}}-{{\tau }_{1}}\left( {{x}_{i}} \right)=a,~ i=\overline{1,k} \} \) is finite. </p>
<p>We introduce the notations: </p>
<div class="displaymath" id="a0000000003">
  \begin{align*}  P=& \sup \biggl\{  \big| f\left( x,u,{{u}_{1}},v,{{v}_{1}} \right) \big|+\Big| \int \nolimits _a^b \, g( x,s,u,{{u}_{1}},v,{{v}_{1}} )ds \Big|:\\ & \qquad \qquad ~ \left| {{u}_{i}} \right|{\lt}{{P}_{1}}, \left| {{v}_{i}} \right|{\lt}{{P}_{2}}, ~ i=0,1,~ x,s\in \left[ a,b \right] \biggr\} ,\\ J=& \left[ {{a}^{*}};a \right], I=\left[ a,b \right],~ {{I}_{1}}=\left[ a,{{x}_{1}} \right],~ {{I}_{2}}=\left[ {{x}_{1}},{{x}_{2}} \right],~ \ldots , \\ {{I}_{k}}=& \left[ {{x}_{k-1}},{{x}_{k}} \right],\  {{I}_{k+1}}=\left[ {{x}_{k}},b \right],\\ B\left( J\mathop{\cup }^{}I \right)=& \Bigl\{  y\left( x \right):y\left( x \right)\in \Bigl( C( J\mathop{\cup }^{}I )\mathop{\cap }^{}\big( {{C}^{1}}( J )\mathop{\cup }^{}{{C}^{1}}( I ) \big)\mathop{\cap }^{}\Bigl( \textstyle \bigcup \limits _{j=1}^{k+1}{{{C}^{2}}\left( {{I}_{j}} \right)} \Bigr) \Bigr), \\ & \quad \left| y\left( x \right) \right|\le {{P}_{1}}, ~ | {y}’( x ) |\le {{P}_{2}} \Bigr\} . \end{align*}
</div>
<p>A function \(y=y\left( x \right)\) from the space \(B\left( J\mathop{\cup }^{}I \right)\) is called a solution of the problem (<a href="#eq:1">1</a>)-(<a href="#eq:2">2</a>) if it satisfies the equation (<a href="#eq:1">1</a>) on \(\left[ a;b \right]\) (with the possible exception of the set \(E\)) and boundary conditions (<a href="#eq:2">2</a>). </p>
<p>From the definition of the space \(B\left( J\mathop{\cup }^{}I \right)\) we conclude that the solution of the problem (<a href="#eq:1">1</a>)-(<a href="#eq:2">2</a>) is continuously differentiable for each \(x\in \left[ a,b \right]\) where \(y'\left( a \right)\) is the right derivative. </p>
<p>Let us introduce a norm in the space \(B\left( J\mathop{\cup }^{}I \right)\): </p>
<div class="displaymath" id="a0000000004">
  \begin{gather*}  {{\left\|  y \right\| }_{B}}=\max \bigg\{  \tfrac {8}{{{\left( b-a \right)}^{2}}}\underset {x\in J\mathop{\cup }^{}I}{\mathop{\max }}\, \left| y\left( x \right) \right|,\tfrac {2}{b-a}\max \Big\{  \underset {x\in J}{\mathop{\max }}\, \left| {y}’\left( x \right) \right|,\underset {x\in I}{\mathop{\max }}\, \left| {y}’\left( x \right) \right| \Big\}  \bigg\} . \end{gather*}
</div>
<p> The space \(B\left( J\mathop{\cup }^{}I \right)\) with this norm is a Banach space. </p>
<p>The boundary value problem (<a href="#eq:1">1</a>)-(<a href="#eq:2">2</a>) is equivalent to the following integral equation (Grim and Schmitt 1968; Kamensky and Myshkis 1972) </p>
<div class="displaymath" id="eq:3">
  \begin{align}  y\left( x \right)=& \int \nolimits _{{a}^{*}}^b \, \Biggl[ f\left( s,y\left( s \right),y\left( s-{{\tau }_{0}}\left( s \right) \right),{y}’\left( s \right),{y}’\left( s-{{\tau }_{1}}\left( s \right) \right) \right)+ \label{eq:3}\\ & \quad +\int \nolimits _a^b\, g\left( s,\xi ,y\left( \xi \right),y\left( \xi -{{\tau }_{0}}\left( \xi \right) \right),{y}’\left( \xi \right),{y}’\left( \xi -{{\tau }_{1}}\left( \xi \right) \right) \right)d\xi \Biggr]\bar{G}\left( x,s \right)ds\nonumber \\ & \quad +l\left( x \right),\qquad x\in J\mathop{\cup }^{}I, \nonumber \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  \bar{G}\left( x,s \right)=\begin{cases}  G\left( x,s \right),&  x,s\in I, \\ 0,& \text{otherwise}, \end{cases}\qquad l\left( x \right)=\begin{cases}  \varphi \left( x \right),& ~ x\in J , \\ \frac{\gamma -\varphi ( a )}{b-a}(x\! -\! a)+\varphi ( a ),& ~ x\in I, \end{cases}\end{equation*}
</div>
<p> and \(G\left( x,s \right)\) is the Green’s function of the following boundary value problem </p>
<div class="displaymath" id="a0000000006">
  \[ {y}''\left( x \right)=0,\quad x\in I,y\left( a \right)=y\left( b \right)=0. \]
</div>
<p>We define an operator \(T\) in the space \(B\left( J\mathop{\cup }^{}I \right)\) in the following way </p>
<div class="displaymath" id="a0000000007">
  \begin{align*}  \left( Ty \right)\left( x \right)=& \int \nolimits _{{a}^{*}}^b\, \Biggl[ f\left( s,y\left( s \right),y\left( s-{{\tau }_{0}}\left( s \right) \right),{y}’\left( s \right),{y}’\left( s-{{\tau }_{1}}\left( s \right) \right) \right)+ \\ & \quad +\int \nolimits _a^b\! \! g\left( s,\xi ,y\left( \xi \right),y\left( \xi \! -\! {{\tau }_{0}}\left( \xi \right) \right),{y}’\left( \xi \right),{y}’\left( \xi \! -\! {{\tau }_{1}}\left( \xi \right) \right) \right)d\xi \Biggr]\! \! \bar{G}\left( x,s \right)ds \\ & \quad +l\left( x \right),\quad x\in J\mathop{\cup }^{}I. \end{align*}
</div>
<p> Hence, </p>
<div class="displaymath" id="eq:4">
  \begin{align}  {{\left( Ty \right)}^{\prime }}\left( x \right)=& \int \nolimits _{{a}^{*}}^b\, \Biggl[ f\left( s,y\left( s \right),y\left( s-{{\tau }_{0}}\left( s \right) \right),{y}’\left( s \right),{y}’\left( s-{{\tau }_{1}}\left( s \right) \right) \right)+ \label{eq:4} \\ & \quad +\! \! \int \limits _a^b\! \! g\left( s,\xi ,y\left( \xi \right),y\left( \xi \! -\! {{\tau }_{0}}\left( \xi \right) \right),{y}’\left( \xi \right),{y}’\left( \xi \! -\! {{\tau }_{1}}\left( \xi \right) \right) \right)d\xi \Biggr]\! \! \bar{G}_{x}^{'}\left( x,s \right)ds\nonumber \\ & \quad +\tfrac {\gamma -\varphi \left( a \right)}{b-a},\quad x\in J\mathop{\cup }^{}I. \nonumber \end{align}
</div>
<p><div class="theorem_thmwrapper " id="th:1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let the following conditions hold: </p>
<ul class="itemize">
  <li><p>\(\max \left\{  \underset {x\in J}{\mathop{\max }}\, \left| \varphi \left( x \right) \right|,~ \frac{{{\left( b-a \right)}^{2}}}{8}P+\max \left( \left| \varphi \left( a \right) \right|,\left| \gamma \right| \right) \right\} \le {{P}_{1}},\) </p>
</li>
  <li><p>\(\max \left\{  \underset {x\in J}{\mathop{\max }}\, \left| {\varphi }'\left( x \right) \right|,~ \frac{b-a}{2}P+\left| \frac{\gamma -\varphi \left( a \right)}{b-a} \right| \right\} \le {{P}_{2}},\) </p>
</li>
  <li><p>the functions \(f\left( x,{{u}_{0}},{{u}_{1}},{{v}_{0}},{{v}_{1}} \right),~ g\left( x,s,{{u}_{0}},{{u}_{1}},{{v}_{0}},{{v}_{1}} \right)\) satisfy the Lipschitz condition for variables \({{u}_{i}},{{v}_{i}},i=\overline{0,1}\) with constants \(L_{j}^{1},L_{j}^{2},~ j=\overline{1,4}\) in \(G\) and \(Q\), </p>
</li>
  <li><p>\(\frac{{{\left( b-a \right)}^{2}}}{8}\underset {j=1}{\overset {2}{\mathop\sum }}\, \left( L_{j}^{1}+\left( b-a \right)L_{j}^{2} \right)+\frac{b-a}{2}\underset {j=3}{\overset {4}{\mathop\sum }}\, \left( L_{j}^{1}+\left( b-a \right)L_{j}^{2} \right){\lt}1\). </p>
</li>
</ul>
<p> Then there exists a unique solution of the problem <span class="rm">(<a href="#eq:1">1</a>)-(<a href="#eq:2">2</a>)</span> in \(B\left( J\mathop{\cup }^{}I \right)\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000008">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Based on Green’s function (Hartman 2002) </p>
<div class="displaymath" id="a0000000009">
  \[ G\left( t,s \right)=\left\{  \begin{matrix}  \frac{\left( s-a \right)\left( t-b \right)}{b-a},~ a\le s\le t\le b, \vspace{3mm}

\\ \frac{\left( t-a \right)\left( s-b \right)}{b-a},~ a\le t\le s\le b, 

\\ \end{matrix} \right. \]
</div>
<p> we obtain the following estimates </p>
<div class="equation" id="eq:5">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:5} \int \nolimits _a^b\, \left| G\left( t,s \right) \right|ds\le \tfrac {{{\left( b-a \right)}^{2}}}{8},\quad \int \nolimits _a^b\, G_{t}^{'}\left( t,s \right)ds\le \tfrac {b-a}{2}. \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p>When the conditions 1)-2) and the inequalities (<a href="#eq:5">9</a>) are true the operator \(T\) maps the space \(B\left( J\mathop{\cup }^{}I \right)\) on itself. </p>
<p>Let \({{y}_{1}},{{y}_{2}}\in B\left( J\mathop{\cup }^{}I \right)\). Considering the condition 3) and the estimates (<a href="#eq:5">9</a>), we get: </p>
<div class="displaymath" id="a0000000010">
  \begin{align*} & \left| \left( T{{y}_{1}} \right)\left( t \right)-\left( T{{y}_{2}} \right)\left( t \right) \right|\le \\ & \le \int \nolimits _{{a}^{*}}^b\, \biggl[ \left( L_{1}^{1}+L_{2}^{1} \right)\underset {t~ \in ~ J\mathop{\cup }^{}I}{\mathop{\max }}\, \left| {{y}_{1}}\left( t \right)-{{y}_{2}}\left( t \right) \right| \\ & \quad + \left( L_{3}^{1}+L_{4}^{1} \right)\max \biggl\{  \underset {t~ \in ~ I}{\mathop{\max }}\, \left| y_{1}^{'}\left( t \right)-y_{2}^{'}\left( t \right) \right|,~ \underset {t~ \in ~ J}{\mathop{\max }}\, \left| y_{1}^{'}\left( t \right)-y_{2}^{'}\left( t \right) \right| \biggr\}  \\ & \quad +\left( b-a \right)\left( L_{1}^{2}+L_{2}^{2} \right)\underset {t~ \in ~ J\mathop{\cup }^{}I}{\mathop{\max }}\, \left| {{y}_{1}}\left( t \right)-{{y}_{2}}\left( t \right) \right| \\ & \quad +\left( b\! -\! a \right)\! \left( L_{3}^{2}\! +\! L_{4}^{2} \right)\max \! \!  \left\{  \underset {t~ \in ~ I}{\mathop{\max }}\, \left| y_{1}^{'}\left( t \right)\! -\! y_{2}^{'}\left( t \right) \right|,\! \underset {t~ \in ~ J}{\mathop{\max }}\, \left| y_{1}^{'}\left( t \right)\! -\! y_{2}^{'}\left( t \right) \right| \right\}  \biggr]\! \bar{G}\left( t,s \right)ds \\ & \le \tfrac {{{\left( b-a \right)}^{2}}}{8}\biggl[ \tfrac {{{\left( b-a \right)}^{2}}}{8}\left( L_{1}^{1}+L_{2}^{1}+\left( b-a \right)\left( L_{1}^{2}+L_{2}^{2} \right) \right) \\ & \quad + \tfrac {b-a}{2}\left( L_{3}^{1}+L_{4}^{1}+\left( b-a \right)\left( L_{3}^{2}+L_{4}^{2} \right) \right) \biggr]{{\left\|  {{y}_{1}}-{{y}_{2}} \right\| }_{B}}, \\ & \left| {{\left( T{{y}_{1}} \right)}^{'}}\left( t \right)-{{\left( T{{y}_{2}} \right)}^{'}}\left( t \right) \right|\le \\ & \le \tfrac {b-a}{2}\biggl[ \tfrac {{{\left( b-a \right)}^{2}}}{8}\left( L_{1}^{1}+L_{2}^{1}+\left( b-a \right)\left( L_{1}^{2}+L_{2}^{2} \right) \right) \\ & \quad +\tfrac {b-a}{2}\left( L_{3}^{1}+L_{4}^{1}+\left( b-a \right)\left( L_{3}^{2}+L_{4}^{2} \right) \right) \biggr]{{\left\|  {{y}_{1}}-{{y}_{2}} \right\| }_{B}}. \end{align*}
</div>
<p>Based on the obtained estimates and on the definition of the norm in the space \(B\left( J\mathop{\cup }^{}I \right)\) we have: </p>
<div class="displaymath" id="eq:6">
  \begin{align} & {{\left\|  \left( T{{y}_{1}} \right)\left( t \right)-\left( T{{y}_{2}} \right)\left( t \right) \right\| }_{B}}\le \label{eq:6} \\ & \le \left[ \tfrac {{{\left( b-a \right)}^{2}}}{8}\underset {i=1}{\overset {2}{\mathop\sum }}\, \left( L_{i}^{1}+\left( b-a \right)L_{i}^{2} \right)+\tfrac {b-a}{2}\underset {i=3}{\overset {4}{\mathop\sum }}\, \left( L_{i}^{1}+\left( b-a \right)L_{i}^{2} \right) \right]{{\left\|  {{y}_{1}}-{{y}_{2}} \right\| }_{B}}. \nonumber \end{align}
</div>
<p>The inequality (<a href="#eq:6">10</a>) and the condition 4) imply that the operator \(T\) is a contraction in \(B\left( J\mathop{\cup }^{}I \right)\) and it has a single fixed point in this space, therefore the boundary value problem (<a href="#eq:1">1</a>)-(<a href="#eq:2">2</a>) has a unique solution \(y\left( t \right)\in B\left( J\mathop{\cup }^{}I \right)\). The proof is complete. </p>
<h1 id="sec:3">3 Cubic splines with defect two</h1>
<p>  Let us consider an irregular grid \(\Delta =\left\{  a={{x}_{0}}{\lt}{{x}_{1}}{\lt}\ldots {\lt}{{x}_{n}}=b \right\} \) on the segment \(\left[ a;b \right]\), \(E\subset \Delta \). We denote by \(S\left( y,x \right)\) an interpolating cubic spline with defect two on \(\Delta \) which belongs to the space \(B\left( J\mathop{\cup }^{}I \right)\). </p>
<p>We can obtain a formula of \(S\left( y,x \right)\) (Nikolova and Bainov 1981; Nastasyeva and Cherevko 1999; Dorosh and Cherevko 2014): </p>
<div class="displaymath" id="a0000000011">
  \begin{align}  S\left( y,x \right)=& M_{j-1}^{+}\tfrac {{{\left( {{x}_{j}}-x \right)}^{3}}}{6{{h}_{j}}}+M_{j}^{-}\tfrac {{{\left( x-{{x}_{j-1}} \right)}^{3}}}{6{{h}_{j}}}+\left( {{y}_{j-1}}-\tfrac {M_{j-1}^{+}h_{j}^{2}}{6} \right)\tfrac {{{x}_{j}}-x}{{{h}_{j}}} \nonumber \\ & \quad +\left( {{y}_{j}}-\tfrac {M_{j}^{-}h_{j}^{2}}{6} \right)\tfrac {x-{{x}_{j-1}}}{{{h}_{j}}},\, x\in \left[ {{x}_{j-1}};{{x}_{j}} \right],\, {{h}_{j}}={{x}_{j}}-{{x}_{j-1}},\, j=\overline{1,n}, \label{eq:7} \end{align}
</div>
<p> where \(M_{j}^{+}={S}''\left( y,{{x}_{j}}+0 \right),~ j=\overline{0,n-1},~ M_{j}^{-}={S}''\left( y,{{x}_{j}}-0 \right),~ j=\overline{1,n}\) satisfy the following system of equations </p>
<div class="displaymath" id="a0000000012">
  \begin{align} & {{h}_{j+1}}{{y}_{j-1}}-\left( {{h}_{j}}+{{h}_{j+1}} \right){{y}_{j}}+{{h}_{j}}{{y}_{j+1}} = \nonumber \\ & = \tfrac {{{h}_{j}}{{h}_{j+1}}}{6}\left( {{h}_{j}}M_{j-1}^{+}+2{{h}_{j}}M_{j}^{-}+2{{h}_{j+1}}M_{j}^{+}+{{h}_{j+1}}M_{j+1}^{-} \right),\, j=\overline{1,n-1}, \label{eq:8} \\ & {{y}_{0}}=\varphi \left( a \right),{{y}_{n}}=\gamma . \nonumber \end{align}
</div>
<p>We shall present the equations (<a href="#eq:8">12</a>) in a matrix form </p>
<div class="displaymath" id="eq:9">
  \begin{gather}  Ay=BM+d, \label{eq:9} \end{gather}
</div>
<p> where </p>
<div class="displaymath" id="a0000000013">
  \begin{equation*}  A=\left( \begin{matrix}  -\left( {{h}_{1}}+{{h}_{2}} \right) 

&  {{h}_{1}} 

&  0 

&  0 

&  \cdots 

&  0 

\\ {{h}_{3}} 

&  -\left( {{h}_{2}}+{{h}_{3}} \right) 

&  {{h}_{2}} 

&  0 

&  \cdots 

&  0 

\\ \cdots 

&  \cdots 

&  \cdots 

&  \cdots 

&  \cdots 

&  \cdots 

\\ 0 

&  0 

&  \cdots 

&  0 

&  {{h}_{n}} 

&  -\left( {{h}_{n-1}}+{{h}_{n}} \right) 

\\ \end{matrix} \right) \end{equation*}
</div>
<p> is an \(\left( n-1 \right)\times \left( n-1 \right)\) matrix, \(~ d={{\left( -{{h}_{2}}{{y}_{0}},~ 0,~ \ldots ,~ 0,~ -{{h}_{n-1}}{{y}_{n}} \right)}^{T}},\) \(B\) is a right side of the relations (<a href="#eq:8">12</a>) coefficient matrix with dimensions\(\left( n-1 \right)\times 2n\), </p>
<div class="displaymath" id="a0000000014">
  \[ M={{\left( M_{0}^{+},~ M_{1}^{-},M_{1}^{+},~ M_{2}^{-},~ M_{2}^{+},~ \ldots ,~ M_{n-1}^{-},~ M_{n-1}^{+},~ M_{n}^{-} \right)}^{T}}. \]
</div>
<p><div class="unlemma_thmwrapper " id="a0000000015">
  <div class="unlemma_thmheading">
    <span class="unlemma_thmcaption">
    Lemma
    </span>
  </div>
  <div class="unlemma_thmcontent">
  <p>The following correlations are true: </p>
<div class="displaymath" id="eq:10">
  \begin{flalign} &  1) \det \left( A \right)={{\left( -1 \right)}^{n-1}}{{h}_{2}}{{h}_{3}}\ldots {{h}_{n-1}}\left( b-a \right), &  \label{eq:10} \\ &  2) \| {{A}^{-1}}\|  \le \tfrac {{{K}^{2}}}{8{{h}^{3}}}\left( b-a \right), &  \label{eq:11} \\ &  3) \underset {1\le i{\lt}n-2}{\mathop{\max }}\, \underset {j=1}{\overset {n-1}{\mathop\sum }}\, \left( a_{i+1,j}^{-1}-a_{i,j}^{-1} \right)\le \tfrac {{{K}^{2}}\left( b-a \right)}{2{{h}^{2}}}, &  \label{eq:12} \\ &  4) \| B\|  \le {{H}^{3}}, &  \label{eq:13} \end{flalign}
</div>
<p> where \(h=\underset {i}{\mathop{\min }}\, {{h}_{i}},\, H=\underset {i}{\mathop{\max }}\, {{h}_{i}},\, K=\frac{H}{h}\) and \(a_{ij}^{-1}\) are elements of a matrix\({{A}^{-1}}\). </p>

  </div>
</div> </p>
<p>The proof of the lemma statements is easy to obtain by applying the principle of mathematical induction and using the structure of the matrices \(A,B\). </p>
<h1 id="sec:4">4 Computational scheme</h1>

<ul class="itemize">
  <li><p>Choose a cubic spline \(S\left( {{y}^{\left( 0 \right)}},x \right)\) randomly so that the boundary conditions (<a href="#eq:2">2</a>) are enforced, for instance, \(S\left( {{y}^{\left( 0 \right)}},x \right)=\frac{\gamma -\varphi \left( a \right)}{b-a}\left( x-a \right)+\varphi \left( a \right)\). </p>
</li>
  <li><p>Using the original equation (<a href="#eq:1">1</a>) and the spline \(S\left( {{y}^{\left( k \right)}},x \right)\), find </p>
<div class="displaymath" id="eq:14">
  \begin{align}  M_{j}^{+\left( k+1 \right)}=& f\biggl( {{x}_{j}},S\left( {{y}^{\left( k \right)}},{{x}_{j}}+0 \right),S\left( {{y}^{\left( k \right)}},{{x}_{j}}-{{\tau }_{0}}\left( {{x}_{j}} \right)+0 \right), \label{eq:14} \\ & {S}’\left( {{y}^{\left( k \right)}},{{x}_{j}}+0 \right),{S}’\left( {{y}^{\left( k \right)}},{{x}_{j}}-{{\tau }_{1}}\left( {{x}_{j}} \right)+0 \right) \biggr)+ \nonumber \\ & +\int \nolimits _a^b\, g\biggl( {{x}_{j}},s,S\left( {{y}^{\left( k \right)}},s \right),S\left( {{y}^{\left( k \right)}},s-{{\tau }_{0}}\left( s \right) \right),{S}’\left( {{y}^{\left( k \right)}},s \right), \nonumber \\ & {S}’\left( {{y}^{\left( k \right)}},s-{{\tau }_{1}}\left( s \right) \right) \biggr)ds,\, j=\overline{0,n-1}, \nonumber \end{align}
</div>
<div class="displaymath" id="eq:15">
  \begin{align}  M_{j}^{-\left( k+1 \right)}=& f\biggl( {{x}_{j}},S\left( {{y}^{\left( k \right)}},{{x}_{j}}-0 \right),S\left( {{y}^{\left( k \right)}},{{x}_{j}}-{{\tau }_{0}}\left( {{x}_{j}} \right)-0 \right), \label{eq:15} \\ & {S}’\left( {{y}^{\left( k \right)}},{{x}_{j}}-0 \right),{S}’\left( {{y}^{\left( k \right)}},{{x}_{j}}-{{\tau }_{1}}\left( {{x}_{j}} \right)-0 \right) \biggr)+ \nonumber \\ & +\int \nolimits _a^b\, g\biggl( {{x}_{j}},s,S\left( {{y}^{\left( k \right)}},s \right),S\left( {{y}^{\left( k \right)}},s-{{\tau }_{0}}\left( s \right) \right),{S}’\left( {{y}^{\left( k \right)}},s \right), \nonumber \\ & {S}’\left( {{y}^{\left( k \right)}},s-{{\tau }_{1}}\left( s \right) \right) \biggr)ds,\, j=\overline{1,n}. \nonumber \end{align}
</div>
<p> In (<a href="#eq:14">18</a>), (<a href="#eq:15">19</a>) put \({{S}^{\left( p \right)}}\left( {{y}^{\left( k \right)}},t \right)={{\varphi }^{\left( p \right)}}\left( t \right),\, p=0,1\) if \(t{\lt}a\). </p>
</li>
  <li><p>Compute \(\left\{  y_{j}^{k+1} \right\} ,\, j=\overline{0,n}\) from the equations (<a href="#eq:8">12</a>). </p>
</li>
  <li><p>Build a cubic spline \(S\left( {{y}^{\left( k+1 \right)}},x \right)\) according to (<a href="#eq:7">11</a>), using the values of \(\left\{  y_{j}^{k+1} \right\} ,\, M_{j}^{+\left( k+1 \right)},\, M_{j}^{-\left( k+1 \right)}.\) This spline will be the next approximation. </p>
</li>
</ul>
<p>Let us denote </p>
<div class="displaymath" id="a0000000016">
  \begin{gather}  {{\lambda }_{1}}=L_{1}^{1}+L_{2}^{1}+\left( b-a \right)\left( L_{1}^{2}+L_{2}^{2} \right),\, {{\lambda }_{2}}=L_{3}^{1}+L_{4}^{1}+\left( b-a \right)\left( L_{3}^{2}+L_{4}^{2} \right), \nonumber \\ u=\tfrac {{{K}^{5}}}{8}{{\left( b-a \right)}^{2}}+\tfrac {{{H}^{2}}}{8},\, v=\tfrac {{{K}^{5}}}{2}\left( b-a \right)+\tfrac {2H}{3}. \label{eq:16} \end{gather}
</div>
<p><div class="theorem_thmwrapper " id="th:2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Assume that the conditions of Theorem <span class="rm"><a href="#th:1">1</a></span> hold. If the following inequality is true </p>
<div class="equation" id="eq:17">
<p>
  <div class="equation_content">
    \begin{equation}  \theta =u{{\lambda }_{1}}+v{{\lambda }_{2}}<1, \label{eq:17} \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>
<p> then there exists \({{H}^{*}}{\gt}0\) such that for each \(0{\lt}H{\lt}{{H}^{*}}\) the sequence of splines \(\left\{  S\left( {{y}^{\left( k \right)}},x \right) \right\} ,\, k=0,1\ldots \) converges uniformly on \(\left[ a;b \right]\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000017">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>The equation (<a href="#eq:10">14</a>) implies that it is possible to construct an iterative spline sequence \(S\left( {{y}^{\left( k \right)}},x \right),~ k=0,1,\ldots \) using the scheme A)-D). We shall demonstrate that the series </p>
<div class="displaymath" id="a0000000018">
  \[ {{S}^{\left( p \right)}}\left( {{y}^{\left( 0 \right)}},x \right)+\underset {i=1}{\overset {\infty }{\mathop\sum }}\, \left[ {{S}^{\left( p \right)}}\left( {{y}^{\left( i \right)}},x \right)-{{S}^{\left( p \right)}}\left( {{y}^{\left( i-1 \right)}},x \right) \right],\, p=0,1 \]
</div>
<p> are uniformly convergent on \(\left[ a;b \right]\) and thus the sequences \({{S}^{\left( p \right)}}\left( {{y}^{\left( k \right)}},x \right),\, k=0,1,\ldots ,\, p=0,1\) are also uniformly convergent. </p>
<p>Let us define scalar functions \(y\left( x \right),\, M\left( x \right)\) on \(\left[ a;b \right]\) and denote the following vectors </p>
<div class="displaymath" id="a0000000019">
  \[  \bar{y}={{\Bigl( y\left( {{x}_{1}} \right),\ldots , y\left( {{x}_{n-1}} \right) \Bigr)}^{T}},  \]
</div>
<div class="displaymath" id="a0000000020">
  \[  \bar{M}=\Bigl( M\left( {{x}_{0}}+0 \right), M\left( {{x}_{1}}-0 \right), M\left( {{x}_{1}}+0 \right),~ \ldots , M\left( {{x}_{n-1}}-0 \right),  \]
</div>
<div class="displaymath" id="a0000000021">
  \[  M\left( {{x}_{n-1}}+0 \right), M\left( {{x}_{n}}-0 \right) \Bigr)^{T}. \]
</div>
<p> We shall write the iterative algorithm A)-D) in a matrix form </p>
<div class="displaymath" id="eq:18">
  \begin{gather}  {{\bar{y}}^{\left( k+1 \right)}}={{A}^{-1}}B{{\bar{M}}^{k+1}}+{{A}^{-1}}d, \label{eq:18} \end{gather}
</div>
<p> where the vector \(\bar{M}\) components are defined according to (<a href="#eq:14">18</a>)-(<a href="#eq:15">19</a>) and the constant vector \(d\) depends only on the boundary conditions (<a href="#eq:2">2</a>). </p>
<p>From (<a href="#eq:18">22</a>) we obtain the estimate </p>
<div class="displaymath" id="eq:19">
  \begin{align}  \left\|  {{y}^{\left( k+1 \right)}}-{{y}^{\left( k \right)}} \right\| & =\left\|  {{A}^{-1}}B{{M}^{k+1}}-{{A}^{-1}}B{{M}^{k}} \right\|  \label{eq:19} \\ & \le \left\|  {{A}^{-1}} \right\| \left\|  B \right\| \left\|  {{{\bar{M}}}^{k+1}}-{{{\bar{M}}}^{k}} \right\| . \nonumber \end{align}
</div>
<p>From (<a href="#eq:14">18</a>)-(<a href="#eq:15">19</a>) and the properties of the functions \(f\) and \(g\) we obtain the following inequalities </p>
<div class="displaymath" id="eq:20">
  \begin{align}  \left\|  M_{j}^{+\left( k+1 \right)}-M_{j}^{+\left( k \right)} \right\| & \le \label{eq:20} {{\lambda }_{1}}\underset {x\in \left[ a;b \right]}{\mathop{\max }}\, \left| S\left( {{y}^{\left( k \right)}},x \right)-S\left( {{y}^{\left( k-1 \right)}},x \right) \right| \\ & \quad + \! \! {{\lambda }_{2}}\! \underset {x\in \left[ a;b \right]}{\mathop{\max }}\! \! \left| {S}’\! \left( {{y}^{\left( k \right)}},x \right)\! -\! {S}’\! \left( {{y}^{\left( k-1 \right)}},x \right) \right|\! \! , j=0,1,\ldots ,n\! -\! 1, \nonumber \\ \left\|  M_{j}^{-\left( k+1 \right)}-M_{j}^{-\left( k \right)} \right\| & \le {{\lambda }_{1}}\underset {x\in \left[ a;b \right]}{\mathop{\max }}\, \left| S\left( {{y}^{\left( k \right)}},x \right)-S\left( {{y}^{\left( k-1 \right)}},x \right) \right| \nonumber \\ & \quad + {{\lambda }_{2}}\underset {x\in \left[ a;b \right]}{\mathop{\max }}\! \! \left| {S}’\left( {{y}^{\left( k \right)}},x \right)\! -\! {S}’\left( {{y}^{\left( k-1 \right)}},x \right) \right|, j=1,2,\ldots ,n. \nonumber \end{align}
</div>
<p> Therefore, taking into account the above mentioned lemma, (<a href="#eq:19">23</a>) can be written in the following way </p>
<div class="displaymath" id="eq:21">
  \begin{align}  \left\|  {{y}^{\left( k+1 \right)}}-{{y}^{\left( k \right)}} \right\| & \le \tfrac {{{K}^{5}}}{8}\left( b-a \right)\biggl[ {{\lambda }_{1}}\left\|  S\left( {{y}^{\left( k \right)}},x \right)-S\left( {{y}^{\left( k-1 \right)}},x \right) \right\|  \label{eq:21} \\ & \quad + {{\lambda }_{2}}\left\|  {S}’\left( {{y}^{\left( k \right)}},x \right)-{S}’\left( {{y}^{\left( k-1 \right)}},x \right) \right\|  \biggr]. \nonumber \end{align}
</div>
<p>Let \(x\in \left[ {{x}_{j-1}};{{x}_{j}} \right]\). Considering (<a href="#eq:7">11</a>), we have </p>
<div class="displaymath" id="a0000000022">
  \begin{align} & \left| S\left( {{y}^{\left( k+1 \right)}},x \right)-S\left( {{y}^{\left( k \right)}},x \right) \right|\le \left| \tfrac {{{x}_{j}}-x}{6{{h}_{j}}} \right|\left( {{\left( {{x}_{j}}-x \right)}^{2}}-h_{j}^{2} \right)+ \nonumber \\ & \qquad + \tfrac {x-{{x}_{j-1}}}{6{{h}_{j}}}\left( {{\left( x-{{x}_{j-1}} \right)}^{2}}-h_{j}^{2} \right)\left\|  {{{\bar{M}}}^{k+1}}-{{{\bar{M}}}^{k}} \right\| + \label{eq:22} \\ & \qquad +\left| y_{j-1}^{k+1}-y_{j}^{k} \right|\left| \tfrac {{{x}_{j}}-x}{{{h}_{j}}} \right|+\left| y_{j}^{k+1}-y_{j}^{k} \right|\left| \tfrac {x-{{x}_{j-1}}}{{{h}_{j}}} \right|. \nonumber \end{align}
</div>
<p> It is easy to show that </p>
<div class="displaymath" id="eq:23">
  \begin{gather}  \underset {x\in \left[ {{x}_{j-1}};{{x}_{j}} \right]}{\mathop{\max }}\, \left| \tfrac {{{x}_{j}}-x}{6{{h}_{j}}}\left( h_{j}^{2}-{{\left( {{x}_{j}}-x \right)}^{2}} \right)+\tfrac {x-{{x}_{j-1}}}{6{{h}_{j}}}\left( h_{j}^{2}-{{\left( x-{{x}_{j-1}} \right)}^{2}} \right) \right|\le \tfrac {{{H}^{2}}}{8}. \label{eq:23} \end{gather}
</div>
<p>Using (<a href="#eq:20">24</a>), (<a href="#eq:21">25</a>), (<a href="#eq:23">27</a>), from (<a href="#eq:22">26</a>) we obtain </p>
<div class="displaymath" id="eq:24">
  \begin{align} & \left\|  S\left( {{y}^{\left( k+1 \right)}},x \right)-S\left( {{y}^{\left( k \right)}},x \right) \right\|  \label{eq:24} \le \tfrac {{{H}^{2}}}{8}\left\|  {{{\bar{M}}}^{k+1}}-{{{\bar{M}}}^{k}} \right\| +\left\|  {{y}^{\left( k+1 \right)}}-{{y}^{\left( k \right)}} \right\|  \\ & \le \left( \tfrac {{{K}^{5}}}{8}{{\left( b-a \right)}^{2}}+\tfrac {{{H}^{2}}}{8} \right)\biggl( {{\lambda }_{1}}\left\|  S\left( {{y}^{\left( k \right)}},x \right)-S\left( {{y}^{\left( k-1 \right)}},x \right) \right\| + \nonumber \\ & \quad + {{\lambda }_{2}}\left\|  {S}’\left( {{y}^{\left( k \right)}},x \right)-{S}’\left( {{y}^{\left( k-1 \right)}},x \right) \right\|  \biggr). \nonumber \end{align}
</div>
<p> According to the spline (<a href="#eq:7">11</a>), we get </p>
<div class="displaymath" id="eq:25">
  \begin{align} & \left| {S}’\left( {{y}^{\left( k+1 \right)}},x \right)-{S}’\left( {{y}^{\left( k \right)}},x \right) \right|\le \label{eq:25} \\ & \le \left| \tfrac {{{h}_{j}}}{6}-\tfrac {{{\left( {{x}_{j}}-x \right)}^{2}}}{2{{h}_{j}}} \right|\left| M_{j-1}^{+\left( k+1 \right)}-M_{j-1}^{+\left( k \right)} \right| \nonumber \\ & \quad +\left| \tfrac {{{\left( x-{{x}_{j-1}} \right)}^{2}}}{2{{h}_{j}}}-\tfrac {{{h}_{j}}}{6} \right|\left| M_{j}^{-\left( k+1 \right)}-M_{j}^{-\left( k \right)} \right|+\tfrac {1}{{{h}_{j}}}\left| y_{j}^{k+1}-y_{j-1}^{k+1}-\left( y_{j}^{k}-y_{j-1}^{k} \right) \right|. \nonumber \end{align}
</div>
<p>One can show that </p>
<div class="displaymath" id="eq:26">
  \begin{gather}  \underset {x\in \left[ {{x}_{j-1}};{{x}_{j}} \right]}{\mathop{\max }}\, \left( \left| \tfrac {{{h}_{j}}}{6}-\tfrac {{{\left( {{x}_{j}}-x \right)}^{2}}}{2{{h}_{j}}} \right|+\left| \tfrac {{{\left( x-{{x}_{j-1}} \right)}^{2}}}{2{{h}_{j}}}-\tfrac {{{h}_{j}}}{6} \right| \right)\le \tfrac {2H}{3}, \label{eq:26} \\ \underset {1{\lt}j{\lt}n}{\mathop{\max }}\, \left| y_{j}^{k+1}-y_{j-1}^{k+1}-\left( y_{j}^{k}-y_{j-1}^{k} \right) \right|\le \tfrac {{{K}^{4}}}{2}\left( b-a \right)H\left\|  {{{\bar{M}}}^{k+1}}-{{{\bar{M}}}^{k}} \right\| . \label{eq:27} \end{gather}
</div>
<p> Due to (<a href="#eq:26">30</a>)-(<a href="#eq:27">31</a>), the inequality (<a href="#eq:25">29</a>) implies that </p>
<div class="displaymath" id="a0000000023">
  \begin{align} & \left\|  {S}’\left( {{y}^{\left( k+1 \right)}},x \right)-{S}’\left( {{y}^{\left( k \right)}},x \right) \right\| \le \nonumber \\ & \le \left( \tfrac {{{K}^{5}}}{2}\left( b-a \right)+\tfrac {2}{3}H \right)\biggl( {{\lambda }_{1}}\left\|  S\left( {{y}^{\left( k \right)}},x \right)-S\left( {{y}^{\left( k-1 \right)}},x \right) \right\|  \label{eq:28} \\ & \quad + {{\lambda }_{2}}\left\|  {S}’\left( {{y}^{\left( k \right)}},x \right)-{S}’\left( {{y}^{\left( k-1 \right)}},x \right) \right\|  \biggr). \nonumber \end{align}
</div>
<p> After iterating (<a href="#eq:24">28</a>), (<a href="#eq:28">32</a>) and considering the notations (<a href="#eq:16">20</a>)-(<a href="#eq:17">21</a>) we obtain </p>
<div class="displaymath" id="eq:29">
  \begin{align} & \left\|  S\left( {{y}^{\left( k+1 \right)}},x \right)-S\left( {{y}^{\left( k \right)}},x \right) \right\| \le \label{eq:29} \\ & \le u{{\theta }^{k-1}}\left( {{\lambda }_{1}}\left\|  S\left( {{y}^{\left( 1 \right)}},x \right)-S\left( {{y}^{\left( 0 \right)}},x \right) \right\| +{{\lambda }_{2}}\left\|  {S}’\left( {{y}^{\left( 1 \right)}},x \right)-{S}’\left( {{y}^{\left( 0 \right)}},x \right) \right\|  \right), \nonumber \\ & \left\|  {S}’\left( {{y}^{\left( k+1 \right)}},x \right)-{S}’\left( {{y}^{\left( k \right)}},x \right) \right\| \le \nonumber \\ & \le v{{\theta }^{k-1}}\left( {{\lambda }_{1}}\left\|  S\left( {{y}^{\left( 1 \right)}},x \right)-S\left( {{y}^{\left( 0 \right)}},x \right) \right\| +{{\lambda }_{2}}\left\|  {S}’\left( {{y}^{\left( 1 \right)}},x \right)-{S}’\left( {{y}^{\left( 0 \right)}},x \right) \right\|  \right). \nonumber \end{align}
</div>
<p>The correlations (<a href="#eq:29">33</a>) with the condition (<a href="#eq:17">21</a>) ensure the convergence of the sequences \(\left\{  {{S}^{\left( p \right)}}\left( {{y}^{\left( k \right)}},x \right) \right\} ,\, k=0,1,\ldots ,\, p=0,1\). Theorem <a href="#th:2">2</a> is proved. </p>
<p>Let us denote \(\underset {k\to \infty }{\mathop{\lim }}\, {{S}^{\left( p \right)}}\left( {{y}^{\left( k \right)}},x \right)={{S}^{\left( p \right)}}\left( \tilde{y},x \right),\, p=0,1,\ldots \) Note that the parameters \(\widetilde{M}_{j}^{+},\, \widetilde{M}_{j}^{-}\) of the spline \(S\left( \tilde{y},x \right)\) satisfy the system (<a href="#eq:8">12</a>) and equations (<a href="#eq:14">18</a>)-(<a href="#eq:15">19</a>). </p>
<p>Let \(S\left( y,x \right)\) be a cubic spline with defect 2 which interpolates the solution \(y\left( x \right)\) of the boundary value problem (<a href="#eq:1">1</a>)-(<a href="#eq:2">2</a>). Thus, </p>
<div class="displaymath" id="a0000000024">
  \begin{align}  \left\|  {{S}^{\left( p \right)}}\left( \tilde{y},x \right)-{{y}^{\left( p \right)}}\left( x \right) \right\| & \le \left\|  {{S}^{\left( p \right)}}\left( \tilde{y},x \right)-{{S}^{\left( p \right)}}\left( y,x \right) \right\| \nonumber \\ & \quad +\left\|  {{S}^{\left( p \right)}}\left( y,x \right)-{{y}^{\left( p \right)}}\left( x \right) \right\| ,\quad p=0,1. \label{eq:30} \end{align}
</div>
<p> For the second term on the right side of the inequality (<a href="#eq:30">34</a>) it is true (Alberg, Nilson and Walsh 1967) that </p>
<div class="displaymath" id="eq:31">
  \begin{align} & \left\|  {{S}^{\left( p \right)}}\left( y,x \right)-{{y}^{\left( p \right)}}\left( x \right) \right\| \le {{K}_{p}}{{H}^{2-p}}\omega \left( {y}”\left( x \right),H \right),\,  \label{eq:31} \\ & p=0,1,2,\, {{K}_{0}}=\tfrac {5}{2},\, {{K}_{1}}={{K}_{2}}=5, \nonumber \end{align}
</div>
<p> where \(\omega \left( {y}''\left( x \right),H \right)=\underset {1\le r\le k+1}{\mathop{\max }}\, {{\omega }_{r}}\left( {y}''\left( x \right),H \right)\), \({{\omega }_{r}}\left( {y}''\left( x \right),H \right)\) is a modulus of continuity for \({y}''\left( x \right)\) on \({{I}_{r}}=\left[ {{x}_{r-1}};{{x}_{r}} \right]\). </p>
<p>We shall denote </p>
<div class="displaymath" id="a0000000025">
  \[ \underset {x\in \left[ a;b \right]}{\mathop{\max }}\, \left| {{S}^{\left( p \right)}}\left( \tilde{y},x \right)-{{S}^{\left( p \right)}}\left( y,x \right) \right|={\alpha }_{p} ,\quad p=0,1. \]
</div>
<p> According to the properties of the functions \(f\), \(g\) and estimates (<a href="#eq:31">35</a>), we obtain </p>
<div class="displaymath" id="a0000000026">
  \begin{align} & \biggl| M_{j}^{+}-f\Bigl( {{x}_{j}},y\left( {{x}_{j}} \right),y\left( {{x}_{j}}-{{\tau }_{0}}\left( {{x}_{j}} \right) \right),{y}’\left( {{x}_{j}} \right),{y}’\left( {{x}_{j}}-{{\tau }_{1}}\left( {{x}_{j}} \right) \right) \Bigr) \nonumber \\ & \quad - \int \nolimits _a^b\, g\Bigl( {{x}_{j}},s,y\left( s \right),y\left( s-{{\tau }_{0}}\left( s \right) \right),{y}’\left( s \right),{y}’\left( s-{{\tau }_{1}}\left( s \right) \right) \Bigr)ds \biggr| \label{eq:32} \\ & \le 5\left( 1+\tfrac {1}{2}{{\lambda }_{1}}{{H}^{2}}+{{\lambda }_{2}}H \right)\omega \left( {y}”\left( x \right),H \right), \nonumber \\ & \biggl| M_{j}^{-}-f\Bigl( {{x}_{j}},y\left( {{x}_{j}} \right),y\left( {{x}_{j}}-{{\tau }_{0}}\left( {{x}_{j}} \right) \right),{y}’\left( {{x}_{j}} \right),{y}’\left( {{x}_{j}}-{{\tau }_{1}}\left( {{x}_{j}} \right) \right) \Bigr) \nonumber \\ & \quad - \int \nolimits _a^b\, g\Bigl( {{x}_{j}},s,y\left( s \right),y\left( s-{{\tau }_{0}}\left( s \right) \right),{y}’\left( s \right),{y}’\left( s-{{\tau }_{1}}\left( s \right) \right) \Bigr)ds \biggr| \label{eq:33} \\ & \le 5\left( 1+\tfrac {1}{2}{{\lambda }_{1}}{{H}^{2}}+{{\lambda }_{2}}H \right)\omega \left( {y}”\left( x \right),H \right). \nonumber \end{align}
</div>
<p>Using the formulas of \(S\left( \tilde{y},x \right),\, S\left( y,x \right)\) and the inequalities (<a href="#eq:32">36</a>)-(<a href="#eq:33">37</a>), one can get the following system of inequalities </p>
<div class="equation" id="eq:34">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:34} \begin{split}  {{\alpha }_{0}}\le u\left( {{\alpha }_{0}}{{\lambda }_{1}}+{{\alpha }_{1}}{{\lambda }_{2}}+5\left( 1+\tfrac {1}{2}{{\lambda }_{1}}{{H}^{2}}+{{\lambda }_{2}}H \right)\omega \left( {y}”\left( x \right),H \right) \right), \\ {{\alpha }_{1}}\le v\left( {{\alpha }_{0}}{{\lambda }_{1}}+{{\alpha }_{1}}{{\lambda }_{2}}+5\left( 1+\tfrac {1}{2}{{\lambda }_{1}}{{H}^{2}}+{{\lambda }_{2}}H \right)\omega \left( {y}”\left( x \right),H \right) \right). \end{split} \end{equation}
  </div>
  <span class="equation_label">38</span>
</p>
</div>
<p> Solving the system (<a href="#eq:34">38</a>), we find estimates for the first terms on the right side of (<a href="#eq:30">34</a>): </p>
<div class="displaymath" id="a0000000027">
  \begin{gather*}  {{\alpha }_{0}}\le \tfrac {5\left( 1+\frac{1}{2}{{\lambda }_{1}}{{H}^{2}}+{{\lambda }_{2}}H \right)u}{1-\theta }\omega \left( {y}”\left( x \right),H \right), \\ {{\alpha }_{1}}\le \tfrac {5\left( 1+\frac{1}{2}{{\lambda }_{1}}{{H}^{2}}+{{\lambda }_{2}}H \right)v}{1-\theta }\omega \left( {y}”\left( x \right),H \right). \end{gather*}
</div>
<p> Now the inequalities (<a href="#eq:30">34</a>) can be written in the following form </p>
<div class="displaymath" id="eq:35">
  \begin{gather}  \left\|  {{S}^{\left( p \right)}}\left( \tilde{y},x \right)-{{y}^{\left( p \right)}}\left( x \right) \right\| \le {{K}_{p}}\omega \left( {y}”\left( x \right),H \right),\quad p=0,1, \label{eq:35} \end{gather}
</div>
<p> where \({{K}_{0}}=\underset {H\le {{H}^{*}}}{\mathop{\sup }}\, \left( \frac{u\mu }{1-\theta }+\frac{5{{H}^{2}}}{2} \right),{{K}_{1}}=\underset {H\le {{H}^{*}}}{\mathop{\sup }}\, \left( \frac{v\mu }{1-\theta }+5H \right).\) </p>
<p>We can summarize the aforementioned arguments concerning accuracy of approximating the solution of the boundary value problem (<a href="#eq:1">1</a>)-(<a href="#eq:2">2</a>) based on the spline sequence as the following theorem. </p>
<p><div class="theorem_thmwrapper " id="th:3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let the solution of the boundary value problem <span class="rm">(<a href="#eq:1">1</a>)-(<a href="#eq:2">2</a>)</span> exist, be unique and belong to the space \(B\left[ {{a}^{*}};b \right]\). If the condition <span class="rm">(<a href="#eq:17">21</a>)</span> holds, then there exists \({{H}^{*}}{\gt}0\) such that for any \(H{\lt}{{H}^{*}}\) the spline sequence \(\left\{  S\left( {{y}^{\left( k \right)}},x \right) \right\} \) is approximating the solution of the boundary value problem <span class="rm">(<a href="#eq:1">1</a>)-(<a href="#eq:2">2</a>)</span> and the correlations <span class="rm">(<a href="#eq:35">41</a>)</span> are true. </p>

  </div>
</div> </p>
<h1 id="sec:5">5 Example</h1>
<p>  Let us consider the usage of this calculation scheme for finding an approximate solution of the following boundary value problem </p>
<div class="displaymath" id="a0000000028">
  \[ {y}''\left( x \right)=-\alpha {y}'\left( x-\tfrac {\pi }{2} \right)+\int \nolimits _{0}^{\tfrac {\pi }{2}}{y(t-\tfrac {\pi }{2})dt}+\cos x,\quad 0\le x\le \tfrac {\pi }{2}, \]
</div>
<div class="displaymath" id="a0000000029">
  \[ y\left( x \right)=\sin (x)+1,\quad -\tfrac {\pi }{2}\le x{\lt}0, \]
</div>
<div class="displaymath" id="a0000000030">
  \[ y\left( 0 \right)=1,\, y(\tfrac {\pi }{2})=2+\alpha . \]
</div>
<p>In this example \(L_{1}^{1}=L_{2}^{1}=L_{3}^{1}=0,\, L_{4}^{1}=\alpha ,\) \(L_{1}^{2}=L_{3}^{2}=L_{4}^{2}=0,\, L_{2}^{2}=1,\) so \({{\lambda }_{1}}=\frac{\pi }{2},\, {{\lambda }_{2}}=\alpha ,\) \(h=H=\frac{\pi }{40},\, K=1,\) \(u=\frac{{{\pi }^{2}}}{32}+\frac{{{H}^{2}}}{8},\, v=\frac{\pi }{4}+\frac{2}{3}H,\) \(\theta =\left( \frac{{{\pi }^{2}}}{32}+\frac{{{H}^{2}}}{8} \right)\frac{\pi }{2}+\left( \frac{\pi }{4}+\frac{2}{3}H \right)\alpha .\) If we put \(\alpha =\frac{1}{4}\), then \(\theta \approx 0.695{\lt}1\) and therefore the conditions of the Theorems 1 and 2 are satisfied. The precise solution \({{y}_{p}}\left( x \right)\) of this boundary value problem, which was found using the step method, is </p>
<div class="displaymath" id="a0000000031">
  \[ {{y}_{p}}\left( x \right)=\alpha \sin x-\cos x+\left( \tfrac {\pi }{2}-1 \right)\tfrac {{{x}^{2}}}{2}+\tfrac {\pi }{4}\left( 1-\tfrac {\pi }{2} \right)x+2. \]
</div>
<div class="table"  id="tab1">
   <div class="centered"><small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(x\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({{y}_{a}}(x)\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({{y}_{p}}(x)\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\Delta \) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\delta \) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 0 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 1 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 1 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 0 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 0% </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p>\(\frac{\pi }{8}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 1.03971 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 1.03976 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.00005 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.01% </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p>\(\frac{\pi }{4}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 1.2935 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 1.2936 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.00010 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.01% </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p>\(\frac{3\pi }{8}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 1.7162 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 1.7163 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.00010 </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.01% </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\frac{\pi }{2}\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 2.25 </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 2.25 </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0 </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0% </p>

    </td>
  </tr>
</table> </small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Precise and approximate solutions.</span> 
</figcaption>  </div>
</div>
<p>The results of the calculation are given in Table <a href="#tab1">1</a>, where \({{y}_{p}}(x)\) is the precise solution, \({{y}_{a}}(x)\) is the approximate solution obtained with \(h=\frac{\pi }{40}\) after 2 iterations, \(\Delta \) is the absolute error and \(\delta \) is the relative error. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="RefAlberg">1</a></dt>
  <dd><p><i class="sc">J. Alberg, E. Nilson, J. Walsh</i>, <i class="it">The theory of splines and their applications</i>, Academic, New York, 1967. </p>
</dd>
  <dt><a name="RefAndreeva">2</a></dt>
  <dd><p><i class="sc">E. Andreeva, V. Kolmanovsky, D. Shayhet</i>, <i class="it">Control of systems with aftereffect</i>, Nauka, Moscow, 1992. </p>
</dd>
  <dt><a name="RefAthanasiadou">3</a></dt>
  <dd><p><i class="sc">E.S. Athanasiadou</i>, <i class="it">On the existence and uniqueness of solutions of boundary problems for second order functional differential equations</i>, Mathematica Moravica, <b class="bf">17</b> (2013) 1, pp.&#160;51–57. </p>
</dd>
  <dt><a name="RefBiga">4</a></dt>
  <dd><p><i class="sc">A.M. Bica, R. Gabor</i>, <i class="it">Existence, uniqueness and approximation for the solution of a second order neutral differential equation with delay in Banach spaces</i>, Mathematica, <b class="bf">49</b> (2007) 2, pp.&#160;117–130. </p>
</dd>
  <dt><a name="RefDorosh">5</a></dt>
  <dd><p><i class="sc">A. Dorosh, I. Cherevko</i>, <i class="it">Application of spline functions for approximating solutions of linear boundary value problems with delay</i>, Mathematical and computer modelling. Series: Physical and mathematical sciences, <b class="bf">10</b> (2014), pp.&#160;80–88. </p>
</dd>
  <dt><a name="RefYakimov">6</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.1007/BF01060583"><i class="sc">I.M. Cherevko, I.V. Yakimov</i>, <i class="it">Numerical method of solving boundary value problems for integro-differential equations with deviating argument</i>, Ukrainian Mathematical Journal, <b class="bf">41</b> (1989) 6, pp.&#160;854–860. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="RefGrim">7</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.1090/S0002-9904-1968-12114-7"> <i class="sc">L.J. Grim, K. Schmitt</i>, <i class="it">Boundary value problems for delay differential equations</i>, Bull. Amer. Math. Soc., <b class="bf">74</b> (1968) 5, pp.&#160;997–1000. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="RefHartman">8</a></dt>
  <dd><p><i class="sc">F. Hartman</i>, <i class="it">Ordinary differential equations</i>, The Society for Industrial and Applied Mathematics, Philadelphia, 2002. </p>
</dd>
  <dt><a name="RefKamensky">9</a></dt>
  <dd><p><i class="sc">G. Kamensky, A. Myshkis</i>, <i class="it">Boundary value problems for nonlinear differential equations with deviating argument of neutral type</i>, Differential equations, <b class="bf">8</b> (1972) 12, pp.&#160;2171–2179. </p>
</dd>
  <dt><a name="RefNastasyeva">10</a></dt>
  <dd><p><i class="sc">N. Nastasyeva, I. Cherevko</i>, <i class="it">Cubic splines with defect two and their applications to boundary value problems</i>, Bulletin of Kyiv University. Physics and mathematics, <b class="bf">1</b> (1999), pp.&#160;69–73. </p>
</dd>
  <dt><a name="RefNikolova">11</a></dt>
  <dd><p><a href ="http://hdl.handle.net/10131/5437"> <i class="sc">T.S. Nikolova, D.D. Bainov</i>, <i class="it">Application of spline-functions for the construction of an approximate solution of boundary problems for a class of functional-differential equations</i>, Yokohama Math. J., <b class="bf">29</b> (1981) 1, pp.&#160;108–122. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
</dl>


</div>
</div> <!--main-text -->
</div> <!-- content-wrapper -->
</div> <!-- content -->
</div> <!-- wrapper -->

<nav class="prev_up_next">
</nav>

<script type="text/javascript" src="/var/www/clients/client1/web1/web/files/jnaat-files/journals/1/articles/js/jquery.min.js"></script>
<script type="text/javascript" src="/var/www/clients/client1/web1/web/files/jnaat-files/journals/1/articles/js/plastex.js"></script>
<script type="text/javascript" src="/var/www/clients/client1/web1/web/files/jnaat-files/journals/1/articles/js/svgxuse.js"></script>
</body>
</html>