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<h1>On the unique solvability and numerical study of absolute value equations</h1>
<p class="authors">
<span class="author">Mohamed Achache\(^\ast \)</span>
</p>
<p class="date">March 21, 2019; accepted: June 17, 2019; published online: January 29, 2020.</p>
</div>
<div class="abstract"><p> The aim of this paper is twofold. Firstly, we consider the unique solvability of absolute value equations (<i class="it"><i class="it">AVE</i></i>), \(Ax-B\vert x\vert =b,\) when the condition \(\Vert A^{-1}\Vert {\lt}\frac{1}{\left\Vert B\right\Vert }\) holds. This is a generalization of an earlier result by Mangasarian and Meyer for the special case where \(B=I.\) Secondly, a generalized Newton method for solving the <i class="it">AVE</i> is proposed. We show under the condition \(\Vert A^{-1}\Vert {\lt}\frac{1}{4\Vert B\Vert },\) that the algorithm converges linearly global to the unique solution of the <i class="it">AVE</i>. Numerical results are reported to show the efficiency of the proposed method and to compare with an available method. </p>
<p><b class="bf">MSC.</b> 90C33, 19K56 </p>
<p><b class="bf">Keywords.</b> Absolute value equations, Complementarity, Generalized Newton method, Global convergence. </p>
</div>
<p>\(^\ast \) Department of Mathematics, University of Abbas Ferhat Sétif 1, Sétif, Algeria, e-mail: <span class="tt">achache_m@univ-setif.dz</span>. </p>
<h1 id="a0000000002">1 Introduction</h1>
<p>We consider the following absolute value equations (<i class="it">AVE</i>) of the type: </p>
<div class="equation" id="f:1">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:1} Ax-B\left| x\right| =b, \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> where \(A,B\in \mathbb {R}^{n\times n}\) (\(B\neq 0\)), \(b\in \mathbb {R}^{n}\) and \(\left| x\right|\) denotes the vector with absolute values of each component of \(x\in \mathbb {R}^{n}\). When \(B=I\) where \(I\) is the identity matrix, the <i class="it">AVE</i> (<a href="#f:1">1</a>) reduces to the absolute value equations of the type: </p>
<div class="equation" id="f:2">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:2} Ax-\left| x\right| =b. \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> In <span class="cite">
	[
	<a href="#6" >6</a>
	]
</span>, <span class="cite">
	[
	<a href="#8" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#9" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#14" >14</a>
	]
</span>, some frequently optimization problems such as linear programming, convex quadratic programming, bimatrix games <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span> and the linear complementarity problem (<i class="it">LCP</i>) can be equivalently reformulated as the <i class="it">AVE</i> (<a href="#f:1">1</a>)–(<a href="#f:2">2</a>). Up to now several authors interested to study this topic, cf. <i class="it">e.g.</i> Mangasarian <span class="cite">
	[
	<a href="#6" >6</a>
	]
</span>, Mangasarian and Meyer <span class="cite">
	[
	<a href="#9" >9</a>
	]
</span>, Prokopyev <span class="cite">
	[
	<a href="#14" >14</a>
	]
</span>, Rohn <span class="cite">
	[
	<a href="#16" >16</a>
	]
</span>, <span class="cite">
	[
	<a href="#18" >18</a>
	]
</span> and Lotfi and Veiseh <span class="cite">
	[
	<a href="#12" >12</a>
	]
</span>. Besides, various numerical methods for solving the <i class="it">AVE</i> (<a href="#f:1">1</a>)–(<a href="#f:2">2</a>) have been proposed, cf. <i class="it">e.g.</i> Cacetta <span class="cite">
	[
	<a href="#3" >3</a>
	]
</span>, Haghani <i class="it">et al.</i> <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span>, Noor <i class="it">et al.</i> <span class="cite">
	[
	<a href="#10" >10</a>
	]
</span>, Jiang and Zhang <span class="cite">
	[
	<a href="#15" >15</a>
	]
</span>, Rhon <span class="cite">
	[
	<a href="#17" >17</a>
	]
</span>, Abdallah <i class="it">et al.</i> <span class="cite">
	[
	<a href="#2" >2</a>
	]
</span> and recently, Achache and Hazzam <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>. Generally, the <i class="it">AVE</i> problem is an <i class="it">NP</i>-hard in its general form since the <i class="it">LCP</i> can be formulated as the <i class="it">AVE</i> (see <span class="cite">
	[
	<a href="#8" >8</a>
	]
</span>). In <span class="cite">
	[
	<a href="#7" >7</a>
	]
</span>, a generalized Newton algorithm for the <i class="it">AVE</i> (<a href="#f:2">2</a>) was investigated, in which it was proved that this method converges globally linear from any starting point to the unique solution of the <i class="it">AVE</i> (<a href="#f:2">2</a>) under the condition that \(\|  A^{-1}\|  {\lt}\tfrac {1}{4}\). Later on, Li <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>, developed a modified generalized Newton for solving the <i class="it">AVE</i> (<a href="#f:2">2</a>). His approach is based on introducing an identity matrix such that a modified generalized Newton method for solving the <i class="it">AVE</i> (<a href="#f:2">2</a>) can be established. He showed under certain conditions that this method converges linearly global to a solution of <i class="it">AVE</i> (<a href="#f:2">2</a>). </p>
<p>In this paper we are interested in a first part of it in the unique solvability of the <i class="it">AVE</i> (<a href="#f:1">1</a>). We show that the condition \(\|  A^{-1}\|  {\lt}\tfrac {1}{\left\|  B\right\| }\) is sufficient to guarantee the unique solvability of the <i class="it">AVE</i> (<a href="#f:1">1</a>) for each \(b\in \mathbb {R}^{n}.\) This generalizes an earlier result by Mangasarian and Meyer who showed in <span class="cite">
	[
	<a href="#9" >9</a>
	, 
	Proposition 4
	]
</span>, that the absolute value equations of the type (<a href="#f:2">2</a>) is uniquely solvable for each \(b\in \mathbb {R}^{n}\) if \(\| A^{-1}\|  {\lt}1.\) To do so, we transform the <i class="it">AVE</i> (<a href="#f:1">1</a>) into a standard linear complementarity problem (<i class="it">LCP</i>) and show under the above mentioned condition that the <i class="it">LCP</i> has a unique solution, so is the <i class="it">AVE</i> (<a href="#f:1">1</a>) for every \(b\in \mathbb {R}^{n}\). </p>
<p>In the second part, inspired by the work of Mangasarian <span class="cite">
	[
	<a href="#7" >7</a>
	]
</span>, we propose a generalized Newton method to solve the <i class="it">AVE</i> (<a href="#f:1">1</a>). When \(\|  A^{-1}\|  {\lt}\tfrac {1}{4\left\|  B\right\|  },\) we show that the method is globally convergent to the unique solution of <i class="it">AVE</i> (<a href="#f:1">1</a>) for every \(b\in \mathbb {R}^{n}\). Finally, some numerical results are presented to show the efficiency of this approach. In addition, a numerical comparison is made with an available method. </p>
<p>The paper is built as follows. In section 2, the main result of the unique solvability of the <i class="it">AVE</i> (<a href="#f:1">1</a>) is stated. In section 3, a generalized Newton method is proposed for solving the <i class="it">AVE</i> (<a href="#f:1">1</a>). The linearly global convergence of the method is established. In section 4, numerical results are reported to validate the efficiency of our approach. In section 5, some concluding remarks are given. </p>
<p>The notation used throughout the paper is as follows. The scalar product of two vectors \(x\) and \(y\) in \(\mathbb {R}^{n}\) will be denoted by \(\langle x,y\rangle =x^{T}y.\) For \(x\in \mathbb {R}^{n},\) the norm \(\Vert x\Vert \) will denote the Euclidean norm \((x^{T}x)^{\frac{1}{2}}.\) For a matrix \(A\in \mathbb {R}^{n\times n},\) the transpose of \(A\) is denoted by \(A^{T}.\) The induced spectral norm of a matrix \(A\) is denoted by \(\Vert A\Vert =\max _{\left\Vert x\right\Vert =1}\Vert Ax\Vert .\) This definition implies that \(\Vert Ax\Vert \leq \Vert A\Vert \) \(\Vert x\Vert \) and \(\Vert AB\Vert \leq \Vert A\Vert \Vert B\Vert \) for any matrices \(A,B\in \mathbb {R}^{n\times n}.\) Let \(\left\{  \alpha _{k}\right\} \) be a non negative sequence of real numbers converging to zero, then the convergence of \(\left\{  \alpha _{k}\right\} \) is said to be linear if there exists \(\rho \in \left( 0,1\right)\) such that \(\alpha _{k+1}\leq \rho \alpha _{k}\) for \(k\geq k_{0}.\) For a matrix \(A\in {\mathbb R}^{n\times n}\), its eigenvalues are denoted by \(\lambda (A)\) with \(\lambda _{\min }(A)\) and \(\lambda _{\max }(A)\) are called the smallest and the greatest eigenvalue of \(A\), respectively. Finally, the notation \(x\geq 0,\) means that the components of \(x\) are greater or equal to zero. </p>
<h1 id="a0000000003">2 The unique solvability of the <i class="it">AVE</i></h1>
<p>Throughout the paper, we make the following assumption on the <i class="it">AVE</i> (<a href="#f:1">1</a>). </p>
<p><div class="assumption_thmwrapper " id="As:1">
  <div class="assumption_thmheading">
    <span class="assumption_thmcaption">
    Assumption
    </span>
    <span class="assumption_thmlabel">1</span>
  </div>
  <div class="assumption_thmcontent">
  <p> The matrix \(A\) is nonsingular and the pair of matrices \(\left[A,B\right]\), satisfies the condition </p>
<div class="displaymath" id="a0000000004">
  \[  \|  A^{-1}\|  {\lt}\tfrac {1}{\left\|  B\right\| }.  \]
</div>

  </div>
</div> </p>
<p>In the proof of the main result, we shall reformulate the <i class="it">AVE</i> (<a href="#f:1">1</a>) as an equivalent standard <i class="it">LCP</i>. Then, one may investigate the <i class="it">AVE</i> (<a href="#f:1">1</a>) across the <i class="it">LCP</i> by making use of the theory of <i class="it">LCP</i> where we invoke under our assumption the class of \(\mathcal{P}\)-matrix, to conclude the unique solvability of the <i class="it">AVE</i> (<a href="#f:1">1</a>) for each \(b\in \mathbb {R}^{n}.\) </p>
<p>Now, we recall some necessary definitions and auxiliary results that will be used later to prove the unique solvability of the <i class="it">AVE</i> (<a href="#f:1">1</a>). <div class="definition_thmwrapper " id="a0000000005">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">1</span>
  </div>
  <div class="definition_thmcontent">
  <p>A matrix \(M\in \mathbb {R}^{n\times n}\) is positive definite if \(\langle Mx,x\rangle {\gt}0\) for all non zero \(x\in \mathbb {R}^{n}.\) </p>

  </div>
</div> </p>
<p><div class="definition_thmwrapper " id="a0000000006">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">2</span>
  </div>
  <div class="definition_thmcontent">
  <p>A matrix \(M\in \mathbb {R}^{n\times n}\) is a \(\mathcal{P}\)-matrix if the determinants of all principal submatrices of \(M\) are positive. Consequently, any positive definite matrix is a \(\mathcal{P}\)-matrix. </p>

  </div>
</div> <div class="definition_thmwrapper " id="a0000000007">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">3</span>
  </div>
  <div class="definition_thmcontent">
  <p>The standard <i class="it">LCP</i> consists of finding \(x,y\in \mathbb {R}^{n}\) such that </p>
<div class="equation" id="f:3">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:3} x\geq 0,\, y\geq 0, y=Mx+q,\, \langle x,y\rangle =0, \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p> where \(M\in \mathbb {R}^{n\times n}\) and \(q\in \mathbb {R}^{n}\) are given. The standard <i class="it">LCP</i> with a \(\mathcal{P}\)-matrix is denoted by \(\mathcal{P}-\)<i class="it">LCP</i>. </p>

  </div>
</div> As a consequence an important result has been proved by Cottle <i class="it">et al.</i> <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>, where any \(\mathcal{P}\)-<i class="it">LCP</i> has a unique solution for every \(q\in \mathbb {R}^{n}.\) <div class="theorem_thmwrapper " id="a0000000008">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p><span class="cite">
	[
	<a href="#4" >4</a>
	, 
	Theorem 3.3.7
	]
</span> A matrix \(M\in \mathbb {R}^{n\times n}\) is a \(\mathcal{P}\)-matrix if and only if the <i class="it">LCP</i> has a unique solution for \(q\in \mathbb {R}^{n}.\) </p>

  </div>
</div> <div class="definition_thmwrapper " id="a0000000009">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">5</span>
  </div>
  <div class="definition_thmcontent">
  <p>For \(x\in \mathbb {R}^{n}\), we define the vectors \(x^{+}\) and \(x^{-}\) by \(x^{+}_{i}=\max (x_{i},0)\) and \(x^{-}_{i}=\max (-x_{i},0), 1\leq i\leq n\), respectively. Then </p>
<div class="equation" id="f:4">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:4} x^{+}\geq 0,\, x^{-}\geq 0,\, x=x^{+}-x^{-},\, \left\vert x\right\vert =x^{+}+x^{-},\, \langle x^{+}, x^{-}\rangle =0. \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>

  </div>
</div> <div class="lemma_thmwrapper " id="a0000000010">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">6</span>
  </div>
  <div class="lemma_thmcontent">
  <p>If Assumption (<a href="#f:1">1</a>) holds, then \((A-B)\) is nonsingular. </p>

  </div>
</div> <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>Indeed \((A-B)\) is nonsingular. For if not then for some \(x\neq 0,\) we have \((A-B)x=0,\) which will derive a contradiction. This implies that \(x=A^{-1}Bx.\) Hence by Assumption 1 it follows that \(\Vert x\Vert \leq \Vert A^{-1}\Vert \Vert B\Vert \Vert x\Vert {\lt}\Vert x\Vert \) which is a contradiction. Therefore \((A-B)\) is nonsingular. This achieves the proof. <span class="qed">â–¡</span></p>
<h2 id="a0000000012">2.1 Reformulation of the <i class="it">AVE</i> as a standard <i class="it">LCP</i></h2>
<p> <div class="proposition_thmwrapper " id="a0000000013">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">7</span>
  </div>
  <div class="proposition_thmcontent">
  <p>The <i class="it">AVE</i> is as a standard <i class="it">LCP</i>. </p>

  </div>
</div> <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>According to the decomposition of \(x\) and \(\vert x\vert \) in (<a href="#f:4">4</a>), the <i class="it">AVE</i> (<a href="#f:1">1</a>) can be reformulated to the following standard <i class="it">LCP</i>: find \(x^{+}\geq 0,x^{-}\geq 0,\) such that </p>
<div class="equation" id="f:5">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:5} x^{+}=Mx^{-}+q,\, \, \langle x^{+}, x^{-}\rangle =0, \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p>where \(M=(A-B)^{-1}(A+B)\) and \(q=(A-B)^{-1}b.\) This achieves the proof.<span class="qed">â–¡</span></p>
<p>We are now ready to state our main result concerning the unique solvability of the <i class="it">AVE</i>s. <div class="theorem_thmwrapper " id="a0000000015">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">8</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Under Assumption 1, the <i class="it">AVE</i> (<a href="#f:1">1</a>) is uniquely solvable for each \(b\in \mathbb {R}^{n}.\) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000016">
  <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 shall prove that the <i class="it">LCP</i> in (<a href="#f:5">5</a>) has a unique solution <i class="it">i.e.</i>, we prove \(M=(A-B)^{-1}(A+B)\) is positive definite. We have for all \(x\neq 0\), </p>
<div class="displaymath" id="a0000000017">
  \[  \langle (A-B)^{-1}(A+B)x,x\rangle =\langle (A+B)x,(A^{T}-B^{T})^{-1}x\rangle .  \]
</div>
<p> Letting \((A^{T}-B^{T})^{-1}x=z,\) hence we get,</p>
<div class="displaymath" id="a0000000018">
  \begin{eqnarray*}  \langle (A-B)^{-1}(A+B)x,x\rangle & =& \langle (A+B)(A^{T}-B^{T})z,z\rangle \\ & =& \langle (AA^{T}-BB^{T})z,z\rangle +\langle (BA^{T}-AB^{T})z,z\rangle \\ & =& \langle (AA^{T}-BB^{T})z,z\rangle , \end{eqnarray*}
</div>
<p>since \(\langle (BA^{T}-AB^{T})z,z\rangle =0,\) for all \(z\in \mathbb {R}^{n}.\) We have by Cauchy-Schwartz inequality and Assumption 1 that </p>
<div class="displaymath" id="a0000000019">
  \begin{eqnarray*}  \langle (AA^{T}-BB^{T})z,z\rangle & =& \langle AA^{T}z,z\rangle -\langle BB^{T}z,z\rangle \\ & =& \Vert Az\Vert ^{2}-\Vert Bz\Vert ^{2} \\ & =& \Vert y\Vert ^{2}-\Vert BA^{-1}y\Vert ^{2} \\ & \geq & (1-\Vert A^{-1}\Vert \Vert B\Vert )^{2}\left\Vert y\right\Vert ^{2}{\gt}0,\, \mbox{for all}\, y\neq 0, \end{eqnarray*}
</div>
<p>where \(y=Az.\) Therefore \((A-B)^{-1}(A+B)\) is positive definite and the <i class="it">LCP</i> in (<a href="#f:5">5</a>) is a \(\mathcal{P}-\)<i class="it">LCP</i>. By Theorem 4, the <i class="it">LCP</i> has a unique solution for each \(b\) and so the <i class="it">AVE</i> (<a href="#f:1">1</a>). This achieves the proof. <span class="qed">â–¡</span></p>
<p>In the next section, we propose a generalized Newton method for solving the <i class="it">AVE</i> (<a href="#f:1">1</a>). In the sequel of the analysis of the proposed method, we will guard the Assumption 1, where we prove under this latter that the generalized Newton method is well-defined <i class="it">i.e.</i>, the generalized Jacobian matrix is nonsingular. </p>
<h1 id="a0000000020">3 A generalized Newton method for the <i class="it">AVE</i></h1>
<p> By defining the function \(f:\mathbb {R}^{n}\rightarrow \mathbb {R}^{n}\) by </p>
<div class="equation" id="f:6">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:6} f(x)=Ax-B\vert x\vert -b, \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>it is clear that \(x\) is a solution of the <i class="it">AVE</i> (<a href="#f:1">1</a>) if and only if </p>
<div class="equation" id="f:7">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:7} f(x) = 0. \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p>Notice that \(|x|\) is not differentiable. A generalized Jacobian \(\partial \vert x\vert \) of \(\vert x\vert \) based on a subgradient <span class="cite">
	[
	<a href="#13" >13</a>
	]
</span> of its components is given by the diagonal matrix \(D(x)\): \(\partial \vert x\vert = D(x) = \operatorname {diag}(\operatorname {sign}(x))\), with </p>
<div class="displaymath" id="a0000000021">
  \[  \operatorname {sign}(x)=\left\{  \begin{smallmatrix} 1, &  \mbox{if } &  x_{i}{\gt}0, \\ 0, &  \mbox{if} &  x_{i}=0, \\ -1, &  \  \mbox{if } &  x_{i}{\lt}0.\end{smallmatrix}\right.  \]
</div>
<p> Noting that \(D(x)x=\vert x\vert \) and \(\left\Vert D(x)\right\Vert \leq 1\) for all \(x\). </p>
<p>To solve the equation (<a href="#f:7">7</a>), we use the Newton method with a generalized Jacobian \(\partial f(x)\) of \(f(x)\) given by: </p>
<div class="equation" id="f:8">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:8} \partial f(x) = A-BD(x). \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p>The generalized Newton method for finding a zero of the equation \(f(x) = 0\) consists of the following iteration: </p>
<div class="equation" id="f:9">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:9} f(x^{k})+\partial f(x^{k})(x^{k+1}-x^{k})=0. \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p>Based on the values of \(f(x)\) and \(\partial f(x)\) in (<a href="#f:6">6</a>) and (<a href="#f:8">8</a>), the generalized Newton iteration in (<a href="#f:9">9</a>), for solving the <i class="it">AVE</i> (<a href="#f:1">1</a>), simplifies to </p>
<div class="equation" id="f:10">
<p>
  <div class="equation_content">
    \begin{equation} \label{f:10} (A-BD(x^{k}))x^{k+1}=b. \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p>The next lemma establishes under Assumption 1 that the generalized Newton iteration in (<a href="#f:10">10</a>), is well-defined. <div class="lemma_thmwrapper " id="a0000000022">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">9</span>
  </div>
  <div class="lemma_thmcontent">
  <p>If Assumption 1 holds, then \((A-BD)\) is nonsingular for every diagonal matrix \(D\) whose diagonal elements are \(1\), \(-1\) or \(0\). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000023">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>If \((A-BD)\) is singular then for some \(x\neq 0,\) we have \((A-BD)x=0\). It follows that \(x=A^{-1}BDx.\) By Assumption 1, it follows that </p>
<div class="displaymath" id="a0000000024">
  \[  \Vert x\Vert \leq \Vert A^{-1}\Vert \Vert B\Vert \Vert D\Vert \Vert x\Vert {\lt}\Vert D\Vert \Vert x\Vert \leq \Vert x\Vert .  \]
</div>
<p> Then we have a contradiction. Hence \((A-BD)\) is nonsingular. This completes the proof.<span class="qed">â–¡</span></p>
<p>Now, we can formally describe the corresponding generalized Newton algorithm for solving the <i class="it">AVE</i> (<a href="#f:1">1</a>) as follows. </p>
<p><br /></p>
<h2 id="a0000000025">3.1 Algorithm</h2>
<div class="centered"> <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">Input: </b> </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>An accuracy \(\epsilon {\gt}0;\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>an initial starting point \(x^{0}\in \mathbb {R}^{n};\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>two matrices \(A\) and \(B\) and a vector \(b\);</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>set \(k:=0;\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p><b class="bfseries">while</b> \(\Vert f(x^{k})\Vert {\gt}\epsilon \, \) <b class="bfseries">do</b> </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p><b class="bfseries">begin</b> </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>compute \(x^{k}\) from the linear system \((A-BD(x^{k-1}))x^{k}=b;\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>update \(k:=k+1;\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p><b class="bfseries">end.</b> </p>

    </td>
  </tr>
</table><br /> A generalized Newton algorithm for the <i class="it">AVE</i>. </div>
<p> Following Mangasarian <span class="cite">
	[
	<a href="#7" >7</a>
	]
</span>, we establish the boundedness of the iterates of (<a href="#f:10">10</a>) and hence the existence of an accumulation point for them. <div class="proposition_thmwrapper " id="a0000000026">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">10</span>
  </div>
  <div class="proposition_thmcontent">
  <p>If Assumption 1 holds, then the iterates in (<a href="#f:10">10</a>) of the generalized Newton method are well-defined and bounded. Consequently, there exists an accumulation point \(\bar{x}\) such that \((A-BD)\bar{x}=b\) for some diagonal \(D\) whose diagonal elements equal \(\pm 1\) or \(0.\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000027">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>By Lemma 6, the matrix \((A-BD)^{-1}\) exists and therefore the sequence of the generalized Newton iterations in <a href="#f:10" class="eqref">10</a>, is well-defined. Suppose now on the contrary that the sequence \(\{  x^{k}\} \) is unbounded. Then there exists a subsequence \(\{  x^{k_{i}+1}\}  \rightarrow \infty \) with nonzero \(x^{k_{i}+1}\) such that \(D(\) \(x^{k_{i}})=\tilde{D}\) where \(\tilde{D}\) is a fixed diagonal matrix whose diagonal elements equal \(\pm 1\) or \(0\) extracted from the finite number of possible configurations for \(D(x^{k})\) in the sequence \(\{  D(x^{k})\} \) and such that the bounded subsequence \(\big\{  \tfrac {x^{k_{i}+1}}{\Vert x^{k_{i}+1}\Vert }\big\} \) converges to \(\tilde{x}\). Hence, </p>
<div class="displaymath" id="a0000000028">
  \[  (A-B\tilde{D})\tfrac {x^{k_{i}+1}}{\Vert x^{k_{i}+1}\Vert }=\tfrac {b}{\Vert x^{k_{i}+1}\Vert }.  \]
</div>
<p> Letting \(i\rightarrow \infty ,\) gives \((A-B\tilde{D})\tilde{x}=0,\Vert \tilde{x}\Vert =1.\) In summary, there exists a \(\tilde{x}\neq 0\) such that \((A-B\tilde{D})\tilde{x}=0,\) which is a contradiction to Lemma 9. Consequently, the sequence \(\{  x^{k}\} \) is bounded and there exists an accumulation point \(\bar{x},\) of \(\{  x^{k}\} \) such that \(f(\bar{x}) = 0\). This completes the proof. <span class="qed">â–¡</span></p>
<p>We are now ready to study the global convergence of the generalized Newton method. First we give the following lemma. <div class="lemma_thmwrapper " id="a0000000029">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">11</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Let \(x\) and \(y\) be points in \(\mathbb {R}^{n}\). Then, \(\big\Vert \left\vert x\right\vert -\left\vert y\right\vert \big\Vert \leq 2\left\Vert x-y\right\Vert .\) </p>

  </div>
</div> <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>For a detailed proof, see Lemma 5 in <span class="cite">
	[
	<a href="#7" >7</a>
	]
</span>. <span class="qed">â–¡</span><div class="lemma_thmwrapper " id="a0000000031">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">12</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Under the assumption \(\left\Vert (A-BD)^{-1}\right\Vert {\lt}\frac{1}{3\left\Vert B\right\Vert },\) the generalized Newton iteration converges linearly from any starting point to a solution \(x^{\ast }\) for any solvable <i class="it">AVE</i> (<a href="#f:1">1</a>). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000032">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Let \(x^{\ast }\) be a solution of the <i class="it">AVE</i> (<a href="#f:1">1</a>), then \((A-BD(x^{\ast }))x^{\ast }=b.\) Noting that \(\vert x^{\ast }\vert =D(x^{\ast })x^{\ast }\) and \(\vert x^{k}\vert =D(x^{k})x^{k}\). Now, subtracting \((A-BD(x^{\ast }))x^{\ast }=b\) from \((A-BD(x^{k}))x^{k+1}=b,\) we obtain </p>
<div class="displaymath" id="a0000000033">
  \begin{eqnarray*}  A(x^{k+1}-x^{\ast }) & =& BD(x^{k})x^{k+1}-BD(x^{\ast })x^{\ast } \\ & =& BD(x^{k})(x^{k+1}+x^{k}-x^{k})-BD(x^{\ast })x^{\ast } \\ & =& B(\vert x^{k}\vert -\left\vert x^{\ast }\right\vert )+BD(x^{k})(x^{k+1}-x^{\ast }+x^{\ast }-x^{k}) \\ & =& B(\vert x^{k}\vert -\left\vert x^{\ast }\right\vert )-BD(x^{k})(x^{k}-x^{\ast })+BD(x^{k})(x^{k+1}-x^{\ast }). \end{eqnarray*}
</div>
<p>Hence </p>
<div class="displaymath" id="a0000000034">
  \[  (A-BD(x^{k}))(x^{k+1}-x^{\ast })=B(\vert x^{k}\vert -\left\vert x^{\ast }\right\vert )-BD(x^{k})(x^{k}-x^{\ast }).  \]
</div>
<p> Consequently, </p>
<div class="displaymath" id="a0000000035">
  \[  (x^{k+1}-x^{\ast })=(A-BD(x^{k}))^{-1}B(\vert x^{k}\vert -\left\vert x^{\ast }\right\vert )-BD(x^{k})(x^{k}-x^{\ast })).  \]
</div>
<p> By Lemma 7, we have, </p>
<div class="displaymath" id="a0000000036">
  \[  \big\Vert x^{k+1}-x^{\ast }\big\Vert \leq \big\Vert (A-BD(x^{k}))^{-1}\big\Vert (2\left\Vert B\right\Vert \big\Vert x^{k}-x^{\ast }\big\Vert )+\left\Vert B\right\Vert \big\Vert x^{k}-x^{\ast }\big\Vert ).  \]
</div>
<p> Hence, \(\big\Vert x^{k+1}-x^{\ast }\big\Vert \leq 3\big\Vert (A-BD(x^{k}))^{-1}\big\Vert \left\Vert B\right\Vert (\big\Vert x^{k}-x^{\ast }\big\Vert )\). So by the condition \(\left\Vert (A-BD)^{-1}\right\Vert {\lt}\tfrac {1}{3\left\Vert B\right\Vert },\) it follows that \(\big\Vert x^{k+1}-x^{\ast }\big\Vert {\lt}\big\Vert x^{k}-x^{\ast }\big\Vert \). Hence the sequence \(\big\{  x^{k}\big\} \) converges linearly to \(x^{\ast }.\) This completes the proof. <span class="qed">â–¡</span></p>
<p>We are now ready to prove our main result of the global convergence. We quote first the following lemma. <div class="lemma_thmwrapper " id="a0000000037">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">13</span>
  </div>
  <div class="lemma_thmcontent">
  <p>If Assumption 1 holds, then \((A-BD)\) is nonsingular and </p>
<div class="displaymath" id="a0000000038">
  \[  \big\Vert (A-BD)^{-1}\big\Vert \leq \tfrac {\left\Vert A^{-1}\right\Vert }{1-\left\Vert A^{-1}B\right\Vert }.  \]
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000039">
  <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 first part follows directly from Lemma 9. For the proof of the second part, we have \((A-BD)^{-1}\) can be written in the form </p>
<div class="displaymath" id="a0000000040">
  \[  (A-BD)^{-1}=(I-A^{-1}BD)^{-1}A^{-1}.  \]
</div>
<p> But since \((I-A^{-1}BD)^{-1}(I-A^{-1}BD)=I\), it follows that </p>
<div class="displaymath" id="a0000000041">
  \[  (I-A^{-1}BD)^{-1}=I+(I-A^{-1}BD)^{-1}A^{-1}BD.  \]
</div>
<p> By introducing an induced matrix norm, we get </p>
<div class="displaymath" id="a0000000042">
  \[  \big\Vert (I-A^{-1}BD)^{-1}\big\Vert \leq \tfrac {1}{1-\left\Vert A^{-1}B\right\Vert \left\Vert D\right\Vert }\leq \tfrac {1}{1-\left\Vert A^{-1}B\right\Vert }.  \]
</div>
<p> Because, </p>
<div class="displaymath" id="a0000000043">
  \[  \big\Vert (A-BD)^{-1}\big\Vert =\big\Vert (I-A^{-1}BD)^{-1}A^{-1}\big\Vert ,  \]
</div>
<p> it implies that </p>
<div class="displaymath" id="a0000000044">
  \[  \big\Vert (A-BD)^{-1}\big\Vert \leq \tfrac {\left\Vert A^{-1}\right\Vert }{1-\left\Vert A^{-1}B\right\Vert }.  \]
</div>
<p> This completes the proof. <span class="qed">â–¡</span></p>
<p><div class="proposition_thmwrapper " id="a0000000045">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">14</span>
  </div>
  <div class="proposition_thmcontent">
  <p>Let \(\Vert A^{-1}\Vert {\lt}\frac{1}{4\left\Vert B\right\Vert }\) and \(D(x^{k})\neq 0\) for all \(k.\) Then the <i class="it">AVE</i> (<a href="#f:1">1</a>) is uniquely solvable for each \(b\in \mathbb {R}^{n}\) and the generalized Newton iteration is well-defined and converges globally from any starting point to a solution \(x^{\ast }\) for any solvable <i class="it">AVE</i>. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000046">
  <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 uniquely solvable of <i class="it">AVE</i> (<a href="#f:1">1</a>) for any \(b\) follows from Theorem 8 which requires \(\Vert A^{-1}\Vert {\lt}\frac{1}{\left\Vert B\right\Vert }.\) Now by Lemma 13, we have, </p>
<div class="displaymath" id="a0000000047">
  \[  \big\Vert (A-BD)^{-1}\big\Vert \leq \tfrac {\Vert A^{-1}\Vert }{1-\left\Vert A^{-1}B\right\Vert } {\lt}\tfrac {\frac{1}{4\left\Vert B\right\Vert }}{1-\frac{1}{4}}=\tfrac {1}{3\left\Vert B\right\Vert }.  \]
</div>
<p> Hence by Lemma 12, the sequence generated by the generalized Newton method converges to the unique solution of the <i class="it">AVE</i> (<a href="#f:1">1</a>) from any starting point \(x^{0}.\) This completes the proof. <span class="qed">â–¡</span></p>
<h1 id="a0000000048">4 Numerical results</h1>
<p> In this section, we report some numerical results of the algorithm on two examples of absolute value equations. The experiments are done by the software MATLAB on a PC. In the implementation, our tolerance is \(\epsilon =10^{-6}\) and the number of iterations and the time produced by the algorithm are denoted by <b class="bfseries">It</b> and <b class="bfseries">CPU</b>, respectively. To this end and in order to demonstrate the performance of the algorithm, we compare the obtained numerical results with those obtained by the modified generalized method in <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>. For these examples we will ensure that the sufficient condition of solvability of the <i class="it">AVE</i> problem is guaranteed <i class="it">i.e.</i>, the <i class="it">AVE</i> (<a href="#f:1">1</a>) has a unique solution for each \(b\). </p>
<p><div class="problem_thmwrapper " id="a0000000049">
  <div class="problem_thmheading">
    <span class="problem_thmcaption">
    Problem
    </span>
    <span class="problem_thmlabel">15</span>
  </div>
  <div class="problem_thmcontent">
  <p>The data \((A,B,b)\) of the <i class="it">AVE</i> problem is taken as </p>
<div class="displaymath" id="a0000000050">
  \[  A=(a_{ij})= \bigg\{ \begin{smallmatrix} 10 &  \  \  i=j, \\ -1 &  \  \  \left\vert i-j\right\vert =1, \\ 0 &  {\rm otherwise} \end{smallmatrix} \qquad B=(b_{ij})= \bigg\{ \begin{smallmatrix} 5, &  \  \   i=j, \\ -1, &  \  \   \left\vert i-j\right\vert =1, \\ 0, &  {\rm otherwise.} \end{smallmatrix}  \]
</div>
<p> Observe first that the two matrices \(A\) and \(B\) can be written as follows: </p>
<div class="displaymath" id="a0000000051">
  \[  A=10I-\bar{A},\; B=5I-\bar{A}  \]
</div>
<p>where </p>
<div class="displaymath" id="a0000000052">
  \[  \bar{A}=(\bar{a}_{ij})= \bigg\{ \begin{smallmatrix} 0, &  \  \   i=j, \\ -1, &  \  \   \left\vert i-j\right\vert =1, \\ 0, &  {\rm otherwise.} &  \end{smallmatrix} \]
</div>
<p> Since \(A\) and \(B\) are symmetric real matrices, then their eigenvalues are reals. Let us denote by \(\lambda (A)\) and \(\lambda (B)\) the eigenvalues of \(A\) and \(B\), respectively. It follows from the matrix calculus that their eigenvalues are given by </p>
<div class="displaymath" id="a0000000053">
  \[  \lambda (A)=10-\lambda (\bar{A}) \textrm{ and }\lambda (B)=5-\lambda (\bar{A}),  \]
</div>
<p> where \(\lambda (\bar{A})\) are the eigenvalues of \(\bar{A}\). Substituting \(\lambda (\bar{A})\) from the equation \(\lambda (B)=5-\lambda (\bar{A})\) into \(\lambda (A)=10-\lambda (\bar{A})\), it follows that \(\lambda (A)-\lambda (B)=5{\gt}0,\) <i class="it">i.e.</i>,\(\lambda (A){\gt}\lambda (B)\). Consequently, \(\lambda _{\min }(A){\gt}\lambda _{\max }(B)\) which is equivalent to say that \(\left\Vert A^{-1}\right\Vert {\lt}\frac{1}{\left\Vert B\right\Vert }\). Therefore, the Problem 15 is uniquely solvable for each \(b\in \mathbb {R}^{n}.\) For \(b=(A-B)e,\) the exact unique solution is given by \(x^{\ast }=e\). The obtained numerical results for different values of size \(n\) and with the initial point \(x^{0}=(1,\, 2,\, \ldots ,\, n)^{T}\) are summarized in Table <a href="#tab:1">1</a>. </p>

  </div>
</div> </p>
<div class="table"  id="tab:1">
   <div class="centered"><small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> Methods\(\  \rightarrow \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">Gen. Newt</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">M. Gen. Newt</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>size \(n\downarrow \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">It</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">CPU</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">It</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">CPU</b> </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(20\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(2\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(0.00058\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(12\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.00239\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(500\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(2\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(0.43522\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(15\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(2.75882\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(1000\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(2\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(2.73065\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(16\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(23.39024\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(1500\) </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(8.72754\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(16\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(69.01293\) </p>

    </td>
  </tr>
</table></small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Numerical results for Problem 15.</span> 
</figcaption>  </div>
</div>
<p><div class="problem_thmwrapper " id="a0000000054">
  <div class="problem_thmheading">
    <span class="problem_thmcaption">
    Problem
    </span>
    <span class="problem_thmlabel">16</span>
  </div>
  <div class="problem_thmcontent">
  <p>This example is an <i class="it">AVE</i> problem of type <a href="#f:2" class="eqref">2</a> and it is taken as: </p>
<div class="displaymath" id="a0000000055">
  \[  A=(a_{ij})= \bigg\{ \begin{smallmatrix} 100 &  \  \   i=j \\ 10 &  \  \  \left\vert i-j\right\vert =1, \\ 0 &  \  \  {\rm otherwise} \end{smallmatrix}\qquad \forall i,j=1,..,n,  \]
</div>
<p>and \(B=I\). Also with the same reasoning as in the first example we can checked that Assumption 1 in this example is satisfied and so the Problem 16 has a unique solution for each \(b\in \mathbb {R}^{n}\). For \(b=(A-I)e\in \mathbb {R}^{n}\), the exact solution is \(x^{*}=e\). The obtained numerical results for different values of size \(n\) and with the initial point \(x^{0}=(1,\, 3,\, \ldots ,\, 2n+1)^{T}\), are summarized in Table&#160;<a href="#tab:2">2</a>. </p>
<div class="table"  id="tab:2">
   <div class="centered"><small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> Methods\(\  \rightarrow \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">Gen. Newt</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">M. Gen. Newt</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>size \(n\downarrow \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">It</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">CPU</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">It</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> <b class="bfseries">CPU</b> </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(20\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(2\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(0.0006\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(6\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.00164\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(500\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(2\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(0.9280\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(7\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(1.38453\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(1000\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(2\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(7.4371\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(8\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(12.43491\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(1500\) </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(29.632463\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(8\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(37.71990\) </p>

    </td>
  </tr>
</table></small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Numerical results for Problem 16.</span> 
</figcaption>  </div>
</div>
<p>In these tables “Gen. Newt”  and “M. Gen. Newt”  are denoted for the Generalized Newton method and the Modified Generalized Newton method <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>, respectively. From Tables <a href="#tab:1">1</a> and <a href="#tab:2">2</a>, we see that the number of iterations and the CPU times of the generalized Newton method are always less than those obtained by the modified generalized Newton method which indicates the superiority of the generalized Newton method. </p>

  </div>
</div> </p>
<h1 id="a0000000056">5 Concluding remarks</h1>
<p> In this paper, we have studied the unique solvability of the <i class="it">AVE</i> under a suitable sufficient condition. Moreover, we have extended a generalized Newton method for solving the <i class="it">AVE</i> (<a href="#f:1">1</a>) which globally and finitely converges to the unique solution of the <i class="it">AVE</i> (<a href="#f:1">1</a>) when \(\left\Vert A^{-1}\right\Vert {\lt}\tfrac {1}{4\left\Vert B\right\Vert }\). We have also reported some numerical results which indicate the effectiveness of the proposed method compared with an available generalized modified Newton method. Possible future work may consist to consider Traub’s Newton method for solving the <i class="it">AVE</i> (<a href="#f:1">1</a>) which is known as the two-step Newton method. </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/10.23952/jnfa.2018.39"> <i class="sc">M. Achache, N. Hazzam</i>, <i class="it">Solving absolute value equations via complementarity and interior point methods</i>, J. Nonl. Funct. Anal., (2018), pp.&#160;1–10 <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/10.1016/j.cam.2017.06.019"> <i class="sc">L. Abdellah, M. Haddou, T. Migot,</i> <i class="it">Solving absolute value equation using complementarity and smoothing functions</i>, J. Comput. Appl. Math., <b class="bf">327</b> (2018), pp.&#160;196–207.<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">3</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s10589-009-9242-9"> <i class="sc">L. Caccetta, B. Qu, G. Zhou</i>, <i class="it">A globally and quadratically convergent method for absolute value equations</i>, Comput. Optim. Appl., <b class="bf">48</b> (2011), pp.&#160;45–58 <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="4">4</a></dt>
  <dd><p><i class="sc">R.W. Cottle, J.S. Pang, R.E. Stone</i>, <i class="it">The Linear Complementarity Problem</i>, Academic Press, New-York, 1992. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s10957-015-0712-1"> <i class="sc">F.K. Haghani</i>, <i class="it">On generalized Traub’s method for absolute value equations</i>, J. Optim. Theory Appl., 166 (2015), pp.&#160;619–625. <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="6">6</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s11590-009-0169-y"> <i class="sc">S.L. Hu, Z.H. Huang</i>, <i class="it">A note on absolute value equations</i>, Optimization Letters, <b class="bf">4</b> (2010), pp.&#160;417–424. <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="7">7</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s11590-008-0094-5"> <i class="sc">O.L. Mangasarian</i>, <i class="it">A generalized Newton method for absolute value equations</i>, Optimization Letters, <b class="bf">3</b> (2009), pp.&#160;101–108. <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="8">8</a></dt>
  <dd><p><a href ="https://https://doi.org/10.1007/s11590-013-0656-z"> <i class="sc">O.L. Mangasarian</i>, <i class="it">Linear complementarity as absolute value equation solution</i>, Optimization Letters, <b class="bf">8</b> (2014), pp.&#160;1529–1534. <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="9">9</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.laa.2006.05.004"> <i class="sc">O.L. Mangasarian, R.R. Meyer</i>, <i class="it">Absolute value equations</i>, Linear Algebra and its Applications, <b class="bf">419</b> (2006), pp.&#160;359–367. <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="10">10</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s11590-011-0332-0"> <i class="sc">M.A. Noor, J. Iqbal, K.I. Noor, A. Al-Said</i>, <i class="it">On a iterative method for solving absolute value equations</i>, Optimization Letters, <b class="bf">6</b> (2012), pp.&#160;1027–1033. <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="11">11</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s10957-016-0956-4"> <i class="sc">C.X. Li</i>, <i class="it">A modified generalized Newton method for absolute value equations</i>, J. Optim. Theory Appl., <b class="bf">170</b> (2016), pp.&#160;1055–1059. <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="12">12</a></dt>
  <dd><p><i class="sc">T. Lotfi, H. Veiseh</i>, <i class="it">A note on unique solvability of the absolute value equation</i>, Journal of Linear and Topological Algebra, <b class="bf">2</b> (2013), pp.&#160;77–81. </p>
</dd>
  <dt><a name="13">13</a></dt>
  <dd><p><i class="sc">B.T. Polyak</i>, <i class="it">Introduction to Optimization</i>, Optimization Software Inc., Publications Division, New York, 1987. </p>
</dd>
  <dt><a name="14">14</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s10589-007-9158-1"> <i class="sc">O. Prokopyev</i>, <i class="it">On equivalent reformulations for absolute value equations</i>, Computational Optimization and Applications, <b class="bf">44</b> (2009), pp.&#160;363–372. <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="15">15</a></dt>
  <dd><p><a href ="https://doi.org/10.3934/jimo.2013.9.789"> <i class="sc">X. Jiang, Y. Zhang</i>, <i class="it">A smoothing-type algorithm for absolute value equation</i>, Journal of Industrial and Management Optimization, <b class="bf">9</b> (2013), pp.&#160;789–798. <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="16">16</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s11590-009-0129-6"> <i class="sc">J. Rohn</i>, <i class="it">On unique solvability of the absolute value equations</i>, Optimization Letters, <b class="bf">3</b> (2009), pp.&#160;603–606. <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="17">17</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s11590-011-0305-3"> <i class="sc">J. Rohn</i>, <i class="it">An algorithm for computing all solutions of an absolute value equation</i>, Optimization Letters, <b class="bf">6</b> (2012), pp.&#160;851–856. <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="18">18</a></dt>
  <dd><p><a href ="https://doi.org/10.1080/0308108042000220686"> <i class="sc">J. Rohn</i>, <i class="it">A theorem of the alternatives for the equation \(Ax + B|x| = b\). Linear and Multilinear Algebra</i>, <b class="bf">52</b> (2004), pp.&#160;421–426. <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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