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<h1>Fourier series approximation for the Cauchy singular integral equation</h1>
<p class="authors">
<span class="author">Hamid BOULARES\(^\ast \), Hamza GUEBBAI\(^\ast \) Amira ARBAOUI\(^{\ast }\)</span>
</p>
<p class="date">August 18, 2017. Accepted: December 19, 2017. Published online: August 6, 2018.</p>
</div>
<p>\(^\ast \)Department of Mathematics, University 8 May 1945 Guelma. B.P. 401 Guelma 24000 Algeria. e-mail: <span class="tt">boulareshamid@gmail.com</span>, <span class="tt">guebaihamza@yahoo.fr</span>, <span class="tt">arbaoui.amira.93@gmail.com</span>. </p>

<div class="abstract"><p> Using the Fourier series as a projection in the Galerkin method, we approach the solution of the Cauchy singular integral equation. This study is carried in \(L^2\). Numerical examples are developped to show the effectiveness of this method. </p>
<p><b class="bf">MSC.</b> 45B05, 65F10, 65J10, 42A16 </p>
<p><b class="bf">Keywords.</b> Singular integral equation, Cauchy operator, Galerkin method, Fourier series </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>Numerical solution methods of integral equations play a very important role in various scientific fields. With the advantage of numerical calculation machines, including computers, these methods have now become an essential tool for investigation in various fundamental problems of our assimilation of scientific phenomena that are difficult, i.e. impossible to solve in the past. </p>
<p>Integral equations with singular kernel represent a great numerical challenge. In addition, if the approximation is performed in a space with weak property as \(L^2\), it accentuates the difficulty. </p>
<p>In this article, we are interested in the following problem: given \(f\in L^{2}\left(0,1\right)\), find a function \(u\in L^{2}\left(0,1\right)\) such that </p>
<div class="displaymath" id="a0000000003">
  \begin{eqnarray}  u=Cu+f, \end{eqnarray}
</div>
<p> where, \(C\) is the Cauchy operator defined by: </p>
<div class="displaymath" id="a0000000004">
  \begin{eqnarray*}  \textrm{For\  all}\  s\in \left]0,1\right[,\  Cu(s)=\lim _{\varepsilon \rightarrow 0}\left( \int _{0}^{s-\varepsilon }\tfrac {u\left( t\right) }{s-t}dt+\int _{s+\varepsilon }^{1}\tfrac {u\left( t\right) }{s-t}dt\right). \end{eqnarray*}
</div>
<p>\(C\) is an integral operator with singular kernel. But, in <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span> the authors show that \(C\) is a bounded operator on \(L^{2}\left(0,1\right)\) to itself with \(\left\| C\right\| \leq \pi \). In addition, they show that it is a skew-hermitian operator, <i class="it">i.e.</i>, \(C^{*}=-C\). Which gives that \(\left(I-C\right)^{-1}\) exists and that \(\big\| \left(I-C\right)^{-1}\big\| \leq 1\) <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>. </p>
<p>Then, for all given \(f\in L^{2}\left(0,1\right)\), the solution \(u\) of \((1)\) exists and is unique. </p>
<p>Our goal is to construct an approximate solution for \((1)\), using Galerkin method. This method is well studied and very known <span class="cite">
	[
	<a href="#2" >2</a>
	, 
	<a href="#Kulkarni" >3</a>
	, 
	<a href="#Ahu" >4</a>
	]
</span>, but for compact operator only. Recently, this method is developed for the Cauchy operator using the piecewise constant projection<span class="cite">
	[
	<a href="#Men1" >5</a>
	, 
	<a href="#Men2" >6</a>
	]
</span>. Our aim is to use Fourier series as projection to make the Galerkin method more efficient. </p>
<h1 id="a0000000005">2 Numerical study</h1>
<p>Let \(\left\{ \pi _{n}\right\} _{n\geq 0}\) be the set of operators defined on \(L^{2}\left(0,1\right)\) to itself by: For all \(x\in L^{2}\left(0,1\right)\), \(n\geq 0\): </p>
<div class="displaymath" id="a0000000006">
  \begin{eqnarray*}  \pi _{n}x\left(t\right)=\sum _{k=0}^{n} a_{k}(x)\cos \left( k\pi t\right),\  t\in \left[0,1\right], \end{eqnarray*}
</div>
<p> where, </p>
<div class="displaymath" id="a0000000007">
  \begin{eqnarray*}  a_{0}(x)& =& \int _{0}^{1}x\left( t\right)dt,\\ a_{k}(x)& =& 2\int _{0}^{1}x\left( t\right) \cos \left( k\pi t\right) dt,\  \  k\geq 1. \end{eqnarray*}
</div>
<p> It is clear that for all \(n\geq 0\), \(\pi _{n}\) is a projection and \(\left\| \pi _{n}\right\| \leq 1\). \(\pi _{n}\) is the Fourier series, truncated at \(n\), of \(x\) extended over \(\mathop\mathbb {R}\nolimits \) in the following sense: </p>
<div class="displaymath" id="a0000000008">
  \begin{eqnarray*}  x(t)& :=& x(2-t),\  t\in \left[1,2\right],\\ x & \textrm{is}& \  2\textrm{-periodic\  over}\  \mathop\mathbb {R}\nolimits . \end{eqnarray*}
</div>
<p> We choose this approach to weaken the effect of the Gibbs phenomenon <span class="cite">
	[
	<a href="#3" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#4" >8</a>
	]
</span>. </p>
<p><div class="proposition_thmwrapper " id="a0000000009">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">1</span>
  </div>
  <div class="proposition_thmcontent">
  <p>For all \(n\geq 0\), \(\pi _{n}\) is a selfadjoint operator and </p>
<div class="displaymath" id="a0000000010">
  \begin{eqnarray*}  \lim _{n\rightarrow \infty }\left\Vert \left( I-\pi _{n}\right)x\right\Vert _{L^{2}\left(0,1\right)}=0. \end{eqnarray*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000011">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> For all \(n\geq 0\), for all \(x,y\in L^{2}\left(0,1\right)\), we have, </p>
<div class="displaymath" id="a0000000012">
  \begin{align*}  \langle \pi _{n}x,y \rangle =& \displaystyle \int ^{1}_{0}\pi _{n}x(t)y(t)dt \displaystyle \sum ^{n}_{k=0}a_{k}(x)\int ^{1}_{0}\cos (k\pi t)y(t)dt \\ =& a_{0}(x)a_{0}(y)+\displaystyle \tfrac {1}{2}\displaystyle \sum ^{n}_{k=0}a_{k}(x)a_{k}(y) =\displaystyle \sum ^{n}_{k=0}\left(\int ^{1}_{0}\cos (k\pi t)x(t)dt\right)a_{k}(y) \\ =& \displaystyle \int ^{1}_{0}x(t)\pi _{n}y(t)dt = \langle x,\pi _{n}y \rangle . \end{align*}
</div>
<p> For the pointwise convergence of \(\pi _{n}\) to the identity \(I\), notice that using the previous extension of \(x\), we obtain <span class="cite">
	[
	<a href="#3" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#4" >8</a>
	]
</span> </p>
<div class="displaymath" id="a0000000013">
  \begin{eqnarray*}  \left\Vert \left( I-\pi _{n}\right)x\right\Vert _{L^{2}\left(0,1\right)}=\tfrac {1}{2}\left\Vert \left( I-\pi _{n}\right)x\right\Vert _{L^{2}\left(-1,1\right)}\rightarrow _{n\rightarrow \infty }0. \end{eqnarray*}
</div>
<p> <div class="proof_wrapper" id="a0000000014">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>Galerkin’s method is to approximate \(C\) by \(C_{n}=\pi _{n}C\pi _{n}\), for \(n\geq 0\). <div class="theorem_thmwrapper " id="a0000000015">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>For all \(n\geq 1\), \(C_{n}^{\ast }\! =\! -C_{n}\), \(I-C_{n}\) is invertible and </p>
<div class="displaymath" id="a0000000016">
  \[  \| \left(I - C_{n}\right)^{-1} \|  \leq 1.  \]
</div>

  </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>We have, </p>
<div class="displaymath" id="a0000000018">
  \begin{eqnarray*}  C_{n}^{\ast } &  =& (\pi _{n}C\pi _{n})^{\ast } =\pi _{n}^{\ast }C^{\ast }\pi _{n}^{\ast }\\ & =& \pi _{n}\left(-C\right)\pi _{n}=-C_{n}. \end{eqnarray*}
</div>
<p> In the same way as in <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>, we define the operators \(A_{n}=iC_{n}\) for all \(n\geq 0\). Then, \(A_{n}\) is selfadjoint and its spectrum is real. Therefore, </p>
<div class="displaymath" id="a0000000019">
  \[ \textrm{sp}\left(C_{n}\right)\subset \left\{ i\alpha \; :\; \alpha \in \mathbb {R}\right\} ,\  n\geq 0. \]
</div>
<p> This means that \(\left(I-C_{n}\right)\) is invertible. But, for all \(x\in L^{2}\left(0,1\right)\), </p>
<div class="displaymath" id="a0000000020">
  \begin{eqnarray*}  \operatorname {Re}\left( \langle (I-C_{n})x,x \rangle \right)=\tfrac {1}{2}\left( \langle (I-C_{n})x,x \rangle +\overline{ \langle (I-C_{n})x,x \rangle }\right)= \langle x,x \rangle . \end{eqnarray*}
</div>
<p> Hence, </p>
<div class="displaymath" id="a0000000021">
  \begin{eqnarray*}  \left\| x\right\| _{L^{2}\left(0,1\right)}^{2}\leq \left| \langle (I-C_{n})x,x \rangle \right|\leq \left\| (I-C_{n})x\right\| _{L^{2}\left(0,1\right)}\left\| x\right\| _{L^{2}\left(0,1\right)}, \end{eqnarray*}
</div>
<p> and \(\| \left(I-C_{n}\right)^{-1}\| \leq 1\). <div class="proof_wrapper" id="a0000000022">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>The last theorem ensures, for all \(n\geq 0\) and all \(f\in L^{2}\left(0,1\right)\), the existence and the unicity of \(u_{n}\) the solution of the following approximate equation: </p>
<div class="displaymath" id="a0000000023">
  \[ u_{n}=C_{n}u_{n}+f. \]
</div>
<p> However, the following theorem ensures the convergence of \(u_{n}\) to \(u\). </p>
<p><div class="theorem_thmwrapper " id="a0000000024">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p>For all \(n\geq 0\), </p>
<div class="displaymath" id="a0000000025">
  \begin{eqnarray*}  \left\Vert u-u_{n}\right\Vert _{L^{2}\left(0,1\right)}\leq \pi \left\Vert \left( I-\pi _{n}\right)u\right\Vert _{L^{2}\left(0,1\right)}+\left\Vert \left( I-\pi _{n}\right)Cu\right\Vert _{L^{2}\left(0,1\right)}. \end{eqnarray*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000026">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We have, </p>
<div class="displaymath" id="a0000000027">
  \begin{eqnarray*}  u-u_{n} &  =& Cu-C_{n}u_{n}\\ &  =& Cu-C_{n}u+C_{n}\left( u-u_{n}\right), \\ \left( I-C_{n}\right) \left( u-u_{n}\right) &  =& Cu-C_{n}u. \end{eqnarray*}
</div>
<p> Then, </p>
<div class="displaymath" id="a0000000028">
  \begin{eqnarray*}  \Vert u-u_{n}\Vert _{L^{2}\left(0,1\right)} &  \leq & \Vert \left( I-C_{n}\right) ^{-1}\Vert \left\Vert \left( Cu-C_{n}u\right) \right\Vert _{L^{2}\left(0,1\right)}\\ & \leq & \left\Vert \left( Cu-C_{n}u\right) \right\Vert _{L^{2}\left(0,1\right)}. \end{eqnarray*}
</div>
<p> But, </p>
<div class="displaymath" id="a0000000029">
  \begin{align*}  \left\Vert Cu-C_{n}u\right\Vert _{L^{2}\left(0,1\right)} &  =\left\Vert Cu-\pi _{n}C\pi _{n}u\right\Vert _{L^{2}\left(0,1\right)}\nonumber \\ &  =\left\Vert Cu-\pi _{n}Cu+\pi _{n}Cu-\pi _{n}C\pi _{n}u\right\Vert _{L^{2}\left(0,1\right)}\nonumber \\ &  =\left\Vert \left( I-\pi _{n}\right) Cu+\pi _{n}C\left( I-\pi _{n}\right) u\right\Vert _{L^{2}\left(0,1\right)}\nonumber \\ &  \leq \left\Vert \left( I-\pi _{n}\right) Cu\right\Vert _{L^{2}\left(0,1\right)} +\left\Vert \pi _{n}C\left( I-\pi _{n}\right) u\right\Vert _{L^{2}\left(0,1\right)}. \end{align*}
</div>
<p> We use \(\left\| \pi _{n}C\right\| \leq \pi \) to conclude. <div class="proof_wrapper" id="a0000000030">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<h1 id="a0000000031">3 Numerical structure</h1>
<p>We have for: For all \( n\geq 0\), </p>
<div class="displaymath" id="a0000000032">
  \begin{eqnarray}  u_{n} &  =& \pi _{n}C\pi _{n}u_{n}+f ,\nonumber \\ &  =&  \pi _{n}C\left( \sum _{k=0}^{n}a_{k}\left(u_{n}\right)\cos \left( k\pi \cdot \right)\right)+f,\nonumber \\ &  =& \sum _{k=1}^{n}a_{k}\left(u_{n}\right)\pi _{n}\varphi _{k}+f \end{eqnarray}
</div>
<p> where </p>
<div class="displaymath" id="a0000000033">
  \begin{eqnarray*}  \varphi _{k}=C\left(\cos \left(k\pi \cdot \right)\right),\  k\geq 0. \end{eqnarray*}
</div>
<p>We have </p>
<div class="displaymath" id="a0000000034">
  \begin{eqnarray*}  \varphi _{0}(s)& =& \lim _{\varepsilon \rightarrow 0}\left( \int _{0}^{s-\varepsilon }\tfrac {1}{s-t}dt+\int _{s+\varepsilon }^{1}\tfrac {1}{s-t}dt\right) =\ln \left( \tfrac {s}{1-s}\right). \end{eqnarray*}
</div>
<p>For \(k\geq 1\), </p>
<div class="displaymath" id="a0000000035">
  \begin{eqnarray*}  \varphi _{k}(s)& =& \lim _{\varepsilon \rightarrow 0}\left( \int _{0}^{s-\varepsilon }\tfrac {\cos \left( k\pi t\right) }{s-t}dt+\int _{s+\varepsilon }^{1}\tfrac {\cos \left( k\pi t\right) }{s-t}dt\right)\\ &  =& \lim _{\varepsilon \rightarrow 0}\Big(\sin \left( k\pi s\right) [ \operatorname {Si}\left(k\pi \left( -\varepsilon \right) \right) - \operatorname {Si}\left( k\pi \left(\varepsilon \right) \right) ] \\ & \quad & +\sin \left( k\pi s\right) [ \operatorname {Si}\left(k\pi \left( 1-s\right) \right) - \operatorname {Si}\left( k\pi \left( -\varepsilon \right) \right) ] \\ & \quad &  -\cos \left( k\pi s\right) [ \operatorname {Ci}\left( k\pi \left( -\varepsilon \right)\right) - \operatorname {Ci}\left( k\pi \left( \varepsilon \right) \right) ] \\ & \quad & -\cos \left( k\pi s\right) [ \operatorname {Ci}\left( k\pi \left( 1-s\right) \right)- \operatorname {Ci}\left( k\pi \left( -\varepsilon \right) \right) ]\Big), \end{eqnarray*}
</div>
<p> where, \( \operatorname {Ci}\) is the cosine integral function and \( \operatorname {Si}\) is the sine integral function. Using the fact that, </p>
<div class="displaymath" id="a0000000036">
  \begin{eqnarray*}  \operatorname {Ci}\left( -x\right) & =&  \operatorname {Ci}\left( x\right)+i\pi ,\\ \operatorname {Si}\left( -x\right) & =& - \operatorname {Si}\left( x\right),\   \operatorname {Si}\left( 0\right) =0, \end{eqnarray*}
</div>
<p> we get, </p>
<div class="displaymath" id="a0000000037">
  \begin{eqnarray*}  \varphi _{k}(s)& =& \sin \left( k\pi s\right) \left[ \operatorname {Si}\left( k\pi \left( 1-s\right) \right) + \operatorname {Si}\left( k\pi s\right) \right] \\ & \quad & -\cos \left( k\pi s\right) \left[ \operatorname {Ci}\left( k\pi \left( 1-s\right) \right) + \operatorname {Ci}\left( k\pi s \right) \right],\  k\geq 1 \end{eqnarray*}
</div>
<p> Multiplying \((2)\) by \(\cos (i\pi s),\  0\leq i\leq n\) and integrating over \(\left(0,1\right)\), we get the following system: </p>
<div class="displaymath" id="a0000000038">
  \begin{eqnarray*} \begin{pmatrix}  a_{0}(u_{n})

\\ \vdots 

\\ a_{n}(u_{n}) 

\end{pmatrix}= M\begin{pmatrix}  a_{0}(u_{n})

\\ \vdots 

\\ a_{n}(u_{n}) 

\end{pmatrix}+ \begin{pmatrix}  a_{0}(f)

\\ \vdots 

\\ a_{n}(f) 

\end{pmatrix}, \end{eqnarray*}
</div>
<p> where, \(\left(a_{i}(u_{n})\right)_{0\leq i\leq n}\) and \(\left(a_{i}(f)\right)_{0\leq i\leq n}\) are respectively the Fourier series coefficients of \(u_n\) and \(f\) and, </p>
<div class="displaymath" id="a0000000039">
  \begin{eqnarray*}  M\left(i,k\right)=\left(2-\delta \left(i\right)\right)\int _{0}^{1}\varphi _{k}\left( s\right) \cos \left( i\pi s\right)ds, \end{eqnarray*}
</div>
<p> where, \(\delta \left(0\right)=1\) and \(\delta \left(k\right)=0\) if \(k\neq 0\). Once, the previous system solved, we construct \(u_{n}\) using the following formula: </p>
<div class="displaymath" id="a0000000040">
  \begin{eqnarray*}  \textrm{For\  all},\  s\in \left[0,1\right],\  u_{n}(s)=\sum _{k=0}^{n}a_{k}\left(u_{n}\right)\cos \left(k\pi s\right). \end{eqnarray*}
</div>
<p> The coefficients \(\left(a_{i}(f)\right)_{1\leq i\leq n}\) and those of the matrices \(M\) are approximated using the numerical integration method, mid-point, with the subdivision: \(M\geq 2,\  h=\frac{1}{M},\  t_{p+\frac{1}{2}}=\left(p+\frac{1}{2}\right)h\), for \(0\leq p\leq M-1\). </p>
<h1 id="a0000000041">4 Numerical example</h1>
<p>To show the effectiveness of our approximation method, we will compare it with the one developed in <span class="cite">
	[
	<a href="#Men1" >5</a>
	]
</span>, <span class="cite">
	[
	<a href="#Men2" >6</a>
	]
</span>. The latter consists in using the piecewise constant projection denoted \(\left\{ \tilde{\pi }_{n}\right\} _{n\geq 2}\). In what follows, \(u_{n,f}\) denotes the approximate solution calculated by our method and \(u_{n,pc}\) denotes the one calculated by the method developed in <span class="cite">
	[
	<a href="#Men1" >5</a>
	]
</span>, <span class="cite">
	[
	<a href="#Men2" >6</a>
	]
</span>. The theoretical error bounds obtained for the two methods are denoted by: </p>
<div class="displaymath" id="a0000000042">
  \begin{eqnarray*}  TEBF_{n}& :=& \pi \left\Vert \left( I-\pi _{n}\right)u\right\Vert _{L^{2}\left(0,1\right)}+\left\Vert \left( I-\pi _{n}\right)Cu\right\Vert _{L^{2}\left(0,1\right)},\\ TEBPC_{n}& :=& \pi \left\Vert \left( I-\tilde{\pi }_{n}\right)u\right\Vert _{L^{2}\left(0,1\right)}+\left\Vert \left( I-\tilde{\pi }_{n}\right)Cu\right\Vert _{L^{2}\left(0,1\right)}, \end{eqnarray*}
</div>
<p> where, \(\left\| \cdot \right\| _{L^{2}\left(0,1\right)}\) is approximated using the mid-point method described below with \(M=1000\). <div class="example_thmwrapper " id="ex 4">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">4</span>
  </div>
  <div class="example_thmcontent">
  <p> If we take </p>
<div class="displaymath" id="a0000000043">
  \begin{eqnarray*}  f\left(s\right)=s+1+s\ln \left(\tfrac {1-s}{s}\right), \end{eqnarray*}
</div>
<p> then the equation \((1)\) admits the following unique solution: </p>
<div class="displaymath" id="a0000000044">
  \begin{eqnarray*}  u\left( s\right)=s. \end{eqnarray*}
</div>
<p><small class="footnotesize"><div class="table"  id="a0000000045">
   <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(n\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(M\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(\displaystyle \left\|  u-u_{n,f}\right\|  _{L^{2}\left(0,1\right)}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(TEBF_{n}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(\displaystyle \left\|  u-u_{n,pc}\right\|  _{L^{2}\left(0,1\right)}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(TEBPC_{n}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>10 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 200 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>7.41E-2 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.30E-1 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.06E-1 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> 4.55E-1</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>50 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 300 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.85E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 8.89E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7.49E-2 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 1.65E-1</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>100 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 500 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1.79E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5.60E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4.82E-2 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 1.06E-1</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>200 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1000 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1.03E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3.21E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.99E-2 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 6.62E-2</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>300 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1000 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6.83E-3 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.13E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.16E-2 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 4.79E-2</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>400 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1000 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 4.71E-3 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1.46E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1.67E-2 </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3.71E-2</p>

    </td>
  </tr>
</table> </div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Fourier projection vs. piecewise constant projection for <span class="mbox" style="width: ">Ex. <a href="#ex 4">4</a>.</span></span> 
</figcaption>
</div> </small> </p>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="ex 5">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">5</span>
  </div>
  <div class="example_thmcontent">
  <p> If we take </p>
<div class="displaymath" id="a0000000046">
  \begin{eqnarray*}  f\left( s\right)=s^{-\frac{1}{4}}-s^{-\frac{1}{4}}\left(\ln \left(\tfrac {1+s^{\frac{1}{4}}}{1-s^{\frac{1}{4}}}\right)-2\tan ^{-1}\left(s^{-\frac{1}{4}}\right)\right), \end{eqnarray*}
</div>
<p> then the equation \((1)\) admits the following unique solution: </p>
<div class="displaymath" id="a0000000047">
  \begin{eqnarray*}  u\left( s\right)=s^{-\frac{1}{4}}. \end{eqnarray*}
</div>
<p><small class="footnotesize"><div class="table"  id="a0000000048">
   <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(n\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(M\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(\displaystyle \left\|  u-u_{n,f}\right\|  _{L^{2}\left(0,1\right)}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(TEBF_{n}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(\displaystyle \left\|  u-u_{n,pc}\right\|  _{L^{2}\left(0,1\right)}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(TEBPC_{n}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>10 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 200 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3.89E-1 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1.21E+0 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7.41E-1 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> 1.64E+0</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>50 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 300 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.09E-1 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6.52E-1 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4.23E-1 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 9.36E-1</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>100 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 500 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1.47E-1 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4.59E-1 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3.17E-1 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 7.01E-1</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>200 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1000 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 9.33E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.90E-1 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.23E-1 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 4.94E-1</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>300 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1000 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6.48E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.01E-1 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 1.70E-1 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 3.78E-1</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>400 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1000 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 4.60E-2 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1.43E-1 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1.52E-1 </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3.33E-1</p>

    </td>
  </tr>
</table> </div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Fourier projection vs. piecewise constant projection for <span class="mbox" style="width: ">Ex. <a href="#ex 5">5</a>.</span></span> 
</figcaption>
</div> </small> </p>

  </div>
</div> </p>
<h1 id="a0000000049">5 Conclusion</h1>
<p>We have constructed a Galerkin-type approximation method for the Cauchy singular integral equation, using the Fourier series as projection in \(L^{2}\left(0,1\right)\). </p>
<p>Based on the fact that the Cauchy operator is skew-hermitian and that the Fourier series is a selfadjoint projection, we show the convergence of our method. </p>
<p>The numerical examples developed show that, regarding the approximation error, our method based on Fourier series is more efficient than the procedure that uses piecewise constant functions. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="1">1</a></dt>
  <dd><p><a href ="https:doi.org/10.1017/cbo9781139172028"> <i class="sc">D. Porter, D.S.G. Stirling</i>, <i class="it">Integral Equations</i>, Cambridge Univ. Press, 1990. <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="2">2</a></dt>
  <dd><p><a href ="https:doi.org/10.1007/978-0-387-21526-6"> <i class="sc">K. Atkinson</i>, <i class="it">Theoretical Numerical Analysis: Functional Analysis Framework</i>, Springer, New York, 2001. <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="Kulkarni">3</a></dt>
  <dd><p><a href ="https:doi.org/10.1163/156939505774286139"> <i class="sc">Rekha P. Kulkarni</i>, <i class="it">On improvement of the iterated Galerkin solution of the second kind integral equations</i>, J. Numer. Math., <b class="bf">13</b> (2005) 3, pp. 205–218. <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="Ahu">4</a></dt>
  <dd><p><i class="sc">M. Ahues, F.D. d’Almeida, R.R. Fernandes</i>, <i class="it">Piecewise constant Galerkin approximations of wealkly singular integral equations</i>, Inter. J.P.A.M., <b class="bf">55</b> (2009) 4, 569–580. </p>
</dd>
  <dt><a name="Men1">5</a></dt>
  <dd><p><a href ="https:doi.org/10.1016/j.aml.2011.11.012"> <i class="sc">A. Mennouni</i>, <i class="it">A projection method for solving Cauchy singular integro-differential equations</i>, Appl. Math. Lett., <b class="bf">25</b> (2012) 6, 986–989. <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="Men2">6</a></dt>
  <dd><p><a href ="https:doi.org/10.1016/j.cam.2017.05.028"> <i class="sc">A. Mennouni</i>, <i class="it">Piecewise constant Galerkin method for a class of Cauchy singular integral equations of the second kind in \(L^2\)</i>, J. Comp. Appl. Math., <b class="bf">326</b> (2017), 268–272. <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="3">7</a></dt>
  <dd><p><i class="sc">A. Zygmund</i>, <i class="it">Trigonometric Series</i>, The University Press, Cambridge, 1959. </p>
</dd>
  <dt><a name="4">8</a></dt>
  <dd><p><a href ="https:doi.org/10.1007/978-0-387-98098-0"> <i class="sc">K.R. Davidson, A.P. Donsig</i>, <i class="it">Real Analysis and Applications</i>, Springer, New York, 2009. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
</dl>


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