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<h1>Continuity of the quenching time in a semilinear heat equation with Neumann boundary condition</h1>
<p class="authors">
<span class="author">Firmin K. N’gohisse\(^\ast \) Théodore K. Boni\(^\S \)</span>
</p>
<p class="date">March 21, 2009.</p>
</div>
<p>\(^\ast \)Université d’Abobo-Adjamé, UFR-SFA, Département de Mathématiques et Informatiques, 02 BP 801 Abidjan 02, (Côte d’Ivoire), e-mail: <span class="tt">firmingoh@yahoo.fr.</span> </p>
<p>\(^\S \)Institut National Polytechnique Houphouet-Boigny de Yamoussoukro, BP&#160;1093 Yamoussoukro, (Côte d’Ivoire), e-mail: <span class="tt">theokboni@yahoo.fr</span>. </p>

<div class="abstract"><p> This paper concerns the study of a semilinear parabolic equation subject to Neumann boundary conditions and positive initial datum. Under some assumptions, we show that the solution of the above problem quenches in a finite time and estimate its quenching time. We also prove the continuity of the quenching time as a function of the initial datum. Finally, we give some numerical results to illustrate our analysis. </p>
<p><b class="bf">MSC.</b> 35B40, 35B50, 35K60, 65M06. </p>
<p><b class="bf">Keywords.</b> Quenching, semilinear parabolic equation, numerical quenching time. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>Let \(\Omega \) be a bounded domain in \(\mathbb {R}^N\) with smooth boundary \(\partial \Omega .\) Consider the following initial-boundary value problem </p>
<div class="displaymath" id="2.1.1">
  \begin{align}  u_t& =\Delta u-f(u)\quad \mbox{in}\quad \Omega \times (0,T),\label{2.1.1}\\ \tfrac {\partial u}{\partial \nu }& =0\quad \mbox{on}\quad \partial \Omega \times (0,T),\label{2.1.2}\\ u(x,0)& =u_0(x) \quad \mbox{in}\quad \overline{\Omega },\label{2.1.3} \end{align}
</div>
<p> where \(\Delta \) is the Laplacian, \(\nu \) is the exterior normal unit vector on \(\partial \Omega .\) For the nonlinear term \(f(u),\) our standing assumptions are the following<br />(A1)&#8195;\(f:(0,\infty )\rightarrow (0,\infty )\) is a \(C^1\) convex, nonincreasing function satisfying \(\lim _{s\rightarrow 0^+}f(s)=\infty ,\) \(\int _0^{\alpha }\tfrac {{\rm d}\sigma }{f(\sigma )}{\lt}\infty \) for any positive real \(\alpha ,\) and \(\int _0^{\infty }\tfrac {{\rm d}\sigma }{f(\sigma )}=\infty .\)<br />(A2)&#8195;There exists a positive constant \(C_0\) such that </p>
<div class="displaymath" id="2.1.4">
  \begin{eqnarray}  -sf’(H(s))\leq C_0\quad \mbox{for}\quad s\geq 0,\label{2.1.4} \end{eqnarray}
</div>
<p> where \(H(s)\) is the inverse of the function \(F(s)\) defined as follows </p>
<div class="displaymath" id="a0000000003">
  \begin{eqnarray*}  F(s)=\int _0^s\tfrac {{\rm d}\sigma }{f(\sigma )}. \end{eqnarray*}
</div>
<p> For the initial datum, we make the following hypotheses<br />(A3):&#8195;\(u_0\in C^2(\overline{\Omega }),\) \(u_0(x){\gt}0\) in \(\overline{\Omega },\) and there exists a positive constant \(B\) such that </p>
<div class="displaymath" id="2.1.5">
  \begin{eqnarray}  \Delta u_0(x)-f(u_0(x))\leq -Bf(u_0(x))\quad \mbox{in}\quad \Omega .\label{2.1.5} \end{eqnarray}
</div>
<p> It is worth noting that, if we choose \(f(s)=s^{-p}\) with \(p\) a positive real, then it is not hard to see that \(f\) satisfies the different assumptions listed in the introduction of the paper. Also, in this case, we note that \(F(s)=\tfrac {s^{p+1}}{p+1},\) \(H(s)=((p+1)s)^{1/(p+1)},\) and \(-sf'(H(s))=\tfrac {p}{p+1}.\)<br />Here \((0,T)\) is the maximal time interval of existence of the solution \(u,\) and by a solution we mean the following. <div class="definition_thmwrapper " id="def.2.1.1">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">1</span>
  </div>
  <div class="definition_thmcontent">
  <p> A solution of <span class="rm">(<a href="#2.1.1">1</a>)–(<a href="#2.1.3">3</a>)</span> is a function \(u(x,t)\) continuous verifying <span class="rm">(<a href="#2.1.1">1</a>)–(<a href="#2.1.3">3</a>)</span>, \(u(x,t){\gt}0\) in \(\overline{\Omega }\times [0,T),\) and twice continuously differentiable in \(x\) and once in \(t\) in \(\Omega \times (0,T).\) </p>

  </div>
</div> The time \(T\) may be finite or infinite. When \(T\) is infinite, then we say that the solution \(u\) exists globally. When \(T\) is finite, then the solution \(u\) develops a quenching in a finite time, namely, </p>
<div class="displaymath" id="a0000000004">
  \begin{eqnarray*}  \lim _{t\rightarrow T}u_{\min }(t)=0, \end{eqnarray*}
</div>
<p> where \(u_{\min }(t)=\min _{x\in \overline{\Omega }}u(x,t).\) In this last case, we say that the solution \(u\) quenches in a finite time, and the time \(T\) is called the quenching time of the solution \(u.\) Since the pioneering work of Kawarada in <span class="cite">
	[
	<a href="#K" >25</a>
	]
</span> regarding the phenomenon of quenching, solutions of semilinear parabolic equations which quench in a finite time have been the subject of investigation of many authors (see, <span class="cite">
	[
	<a href="#AW" >2</a>
	]
</span>–<span class="cite">
	[
	<a href="#BB" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#Bon" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#BonN" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#DX" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#FKL" >14</a>
	]
</span>, <span class="cite">
	[
	<a href="#FL" >15</a>
	]
</span>, <span class="cite">
	[
	<a href="#Gu" >22</a>
	]
</span>, <span class="cite">
	[
	<a href="#KR" >26</a>
	]
</span>, <span class="cite">
	[
	<a href="#L" >28</a>
	]
</span>–<span class="cite">
	[
	<a href="#Lev" >30</a>
	]
</span>, <span class="cite">
	[
	<a href="#P" >33</a>
	]
</span>, <span class="cite">
	[
	<a href="#SK" >37</a>
	]
</span>, and the references cited therein). In particular, in <span class="cite">
	[
	<a href="#Bon" >7</a>
	]
</span>, the problem (<a href="#2.1.1">1</a>)–(<a href="#2.1.3">3</a>) has been studied. By standard methods, it is well known that, making use of the assumptions made on the paper, one easily proves the existence and uniqueness of solutions (see, <span class="cite">
	[
	<a href="#Bon" >7</a>
	, 
	Theorems 2.1 and 2.2
	]
</span>, <span class="cite">
	[
	<a href="#Fr" >16</a>
	, 
	 Chapter 7, Theorem 13 (and its corollary)
	]
</span>, <span class="cite">
	[
	<a href="#LSU" >27</a>
	, 
	 Theorem 7.4
	]
</span>). It is also shown that the solution of (<a href="#2.1.1">1</a>)–(<a href="#2.1.3">3</a>) quenches in a finite time and its quenching time has been estimated (see, <span class="cite">
	[
	<a href="#Bon" >7</a>
	]
</span>). In this paper, we are interested in the continuity of the quenching time as a function of the initial datum. More precisely, we consider the following initial-boundary value problem </p>
<div class="displaymath" id="2.1.6">
  \begin{align}  v_t& =\Delta v-f(v)\quad \mbox{in}\quad \Omega \times (0,T_h),\label{2.1.6}\\ \tfrac {\partial v}{\partial \nu }& =0\quad \mbox{on}\quad \partial \Omega \times (0,T_h),\label{2.1.7}\\ v(x,0)& =u^h_0(x)\quad \mbox{in}\quad \overline{\Omega },\label{2.1.8} \end{align}
</div>
<p> where \(u_0^h\in C^2(\overline{\Omega }),\) \(u^h_0(x)\geq u_0(x)\) in \(\Omega ,\) and \(\lim _{h\rightarrow 0}\| u_0^h-u_0\| _{\infty }=0.\) Here \((0,T_h)\) is the maximal time interval on which the solution \(v\) of (<a href="#2.1.6">6</a>)–(<a href="#2.1.8">8</a>) exists. Definition <a href="#2.1.1">1</a> is valid for the solution \(v\) of (<a href="#2.1.6">6</a>)–(<a href="#2.1.8">8</a>) and we assume that this solution is sufficiently regular. It is worth noting that the regularity of solutions increases with respect to the regularity of initial data, and one may apply without difficulties the maximum principle (see, <span class="cite">
	[
	<a href="#Fr" >16</a>
	, 
	Chapter 7, Theorem 13 (and its corollary)
	]
</span>, <span class="cite">
	[
	<a href="#LSU" >27</a>
	, 
	Chapter 5, Theorem 53
	]
</span>, <span class="cite">
	[
	<a href="#Pa" >35</a>
	, 
	Chapter 4
	]
</span>). In the present paper, we prove that if \(h\) is small enough, then the solution \(v\) of (<a href="#2.1.6">6</a>)–(<a href="#2.1.8">8</a>) quenches in a finite time and its quenching time \(T_h\) goes to \(T\) as \(h\) goes to zero, where \(T\) is the quenching time of the solution \(u\) of (<a href="#2.1.1">1</a>)–(<a href="#2.1.3">3</a>). In addition we provide an upper bound of \(|T_h-T|\) in terms of \(\| u_0^h-u_0\| _\infty .\) This work is in parts the continuity of our earlier study in <span class="cite">
	[
	<a href="#BonN" >8</a>
	]
</span> where we considered a particular case of the problem (<a href="#2.1.1">1</a>)–(<a href="#2.1.3">3</a>) choosing \(f(s)=s^{-p}\) with \(p\) a positive constant. Under some hypotheses, we showed that the time \(T_h\) goes to \(T\) as \(h\) tends to zero. Let us point out that in <span class="cite">
	[
	<a href="#BonN" >8</a>
	]
</span>, we have not found an upper bound of \(|T_h-T|.\) Similar results have been obtained in <span class="cite">
	[
	<a href="#BC" >5</a>
	]
</span>, <span class="cite">
	[
	<a href="#CPEl" >10</a>
	]
</span>, <span class="cite">
	[
	<a href="#G" >18</a>
	]
</span>–<span class="cite">
	[
	<a href="#GVa" >20</a>
	]
</span>, <span class="cite">
	[
	<a href="#GR" >23</a>
	]
</span>, <span class="cite">
	[
	<a href="#GRZ" >24</a>
	]
</span>, <span class="cite">
	[
	<a href="#N" >32</a>
	]
</span>, <span class="cite">
	[
	<a href="#Q" >36</a>
	]
</span>, where the authors considered analogous problems within the framework of the phenomenon of blow-up (we say that a solution blows up in a finite time if it reaches the value infinity in a finite time). The remainder of the paper is written in the following manner. In the next section, we present some results concerning quenching solutions for our subsequent use. In the third section, we analyze the continuity of the quenching time as a function of the initial datum and finally, in the last section, we show some computational results to illustrate the theory given in the paper. </p>
<h1 id="a0000000005">2 Quenching time</h1>
<p>In this section, under some assumptions, we show that the solution \(v\) of (<a href="#2.1.6">6</a>)–(<a href="#2.1.8">8</a>) quenches in a finite time and estimate its quenching time. </p>
<p>We begin by proving the following result which is inspired of an idea of Friedman and McLeod in <span class="cite">
	[
	<a href="#FM" >17</a>
	]
</span>. <div class="theorem_thmwrapper " id="the.2.2.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Suppose that there exists a constant \(A\in (0,1]\) which is independent of \(h\) such that the initial datum at <span class="rm">(<a href="#2.1.8">8</a>)</span> satisfies </p>
<div class="displaymath" id="2.2.1">
  \begin{eqnarray}  \Delta u^h_0(x)-f(u^h_0(x))\leq -A f(u^h_0(x))\quad \mbox{in} \quad \Omega .\label{2.2.1} \end{eqnarray}
</div>
<p> Then, the solution \(v\) of <span class="rm">(<a href="#2.1.6">6</a>)–(<a href="#2.1.8">8</a>)</span> quenches in a finite time, and an upper bound of its quenching time is \(\tfrac {F(u^h_{0min})}{A}.\) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000006">
  <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 \(T_h\) be a time up to which \(v\) remains strictly positive everywhere. Our aim is to show that \(T_h\) is finite and \(\tfrac {F(u^h_{0min})}{A}\) is one of its upper bound. Introduce the function \(J(x,t)\) defined as follows </p>
<div class="displaymath" id="a0000000007">
  \[ J(x,t)=v_t(x,t)+A f(v(x,t))\quad \mbox{in} \quad \overline{\Omega } \times [0,T_h). \]
</div>
<p> A straightforward computation yields </p>
<div class="displaymath" id="2.2.2">
  \begin{eqnarray}  J_t-\Delta J =\left(v_t-\Delta v\right)_t+A f’(v)v_t- A\Delta f(v)\quad \mbox{in}\quad \Omega \times (0,T_h).\label{2.2.2} \end{eqnarray}
</div>
<p> Again by a direct calculation, we observe that </p>
<div class="displaymath" id="a0000000008">
  \begin{eqnarray*}  \Delta f(v)= f”(v)|\nabla v|^2+ f’(v)\Delta v \quad \mbox{in}\quad \Omega \times (0,T_h), \end{eqnarray*}
</div>
<p> which implies that \(\Delta f(v)\geq f'(v)\Delta v\) in \(\Omega \times (0,T_h).\) Using this estimate and (<a href="#2.2.2">10</a>), we arrive at </p>
<div class="displaymath" id="2.2.3">
  \begin{eqnarray}  J_t-\Delta J \leq \left(v_t-\Delta v\right)_t+A f’(v)(v_t- \Delta v)\quad \mbox{in} \quad \Omega \times (0,T_h). \label{2.2.3} \end{eqnarray}
</div>
<p> It follows from (<a href="#2.1.6">6</a>) and (<a href="#2.2.3">11</a>) that </p>
<div class="displaymath" id="a0000000009">
  \begin{eqnarray*}  J_t-\Delta J \leq - f’(v)v_t-A f’(v)f(v)\quad \mbox{in} \quad \Omega \times (0,T_h). \end{eqnarray*}
</div>
<p> Taking into account the expression of \(J,\) we find that </p>
<div class="displaymath" id="a0000000010">
  \begin{eqnarray*}  J_t- \Delta J \leq -f’(v)J \quad \mbox{in} \quad \Omega \times (0,T_h). \end{eqnarray*}
</div>
<p> Employing the condition (<a href="#2.1.7">7</a>), it is not hard to see that </p>
<div class="displaymath" id="a0000000011">
  \begin{eqnarray*}  \tfrac {\partial J}{\partial \nu }=\left(\tfrac {\partial v}{\partial \nu }\right)_t+A f’(v)\tfrac {\partial v}{\partial \nu }=0\quad \mbox{on} \quad \partial \Omega \times (0,T_h), \end{eqnarray*}
</div>
<p> and due to (<a href="#2.2.1">9</a>), we discover that </p>
<div class="displaymath" id="a0000000012">
  \begin{eqnarray*}  J(x,0)=\Delta u^h_0(x)-f(u^h_0(x))+A f(u^h_0(x))\leq 0\quad \mbox{in} \quad \Omega . \end{eqnarray*}
</div>
<p> We infer from the maximum principle that </p>
<div class="displaymath" id="a0000000013">
  \begin{eqnarray*}  J(x,t) \leq 0\quad \mbox{in} \quad \Omega \times (0,T_h), \end{eqnarray*}
</div>
<p> or equivalently </p>
<div class="displaymath" id="2.2.4">
  \begin{eqnarray}  v_t(x,t)+Af(v(x,t)) \leq 0\quad \mbox{in} \quad \Omega \times (0,T_h).\label{2.2.4} \end{eqnarray}
</div>
<p> This estimate may be rewritten in the following manner </p>
<div class="displaymath" id="2.2.5">
  \begin{eqnarray}  \tfrac {{\rm d}v}{f(v)} \leq -A{\rm d}t\quad \mbox{in} \quad \Omega \times (0,T_h). \label{2.2.5} \end{eqnarray}
</div>
<p> Integrate the above inequality over \((0,T_h)\) to obtain </p>
<div class="displaymath" id="2.2.6">
  \begin{eqnarray}  T_h\leq \tfrac {1}{A} \int _{v(x,T_h)}^{v(x,0)} \tfrac {{\rm d}\sigma }{f(\sigma )}\quad \mbox{for} \quad x\in \Omega . \label{2.2.6} \end{eqnarray}
</div>
<p> From (<a href="#2.2.4">12</a>), we observe that \(v\) is nonincreasing with respect to the second variable, which implies \(0\leq v(x,T_h)\leq v(x,0)\) in \(\Omega .\) It is easy to see that </p>
<div class="displaymath" id="a0000000014">
  \begin{eqnarray*}  \int _{v(x,T_h)}^{v(x,0)} \tfrac {{\rm d}\sigma }{f(\sigma )} \leq \int _{0}^{v(x,0)} \tfrac {{\rm d}\sigma }{f(\sigma )}\quad \mbox{for} \quad x\in \Omega . \end{eqnarray*}
</div>
<p> We deduce from (<a href="#2.2.6">14</a>) that \(T_h\leq \tfrac {1}{A} \int _{0}^{u^h_{0min}} \tfrac {{\rm d}\sigma }{f(\sigma )}\) or equivalently </p>
<div class="displaymath" id="a0000000015">
  \begin{eqnarray*}  T_h\leq \tfrac {F(u^h_{0min})}{A}. \end{eqnarray*}
</div>
<p> Consequently, \(v\) quenches in a finite time because the quantity on the right hand side of the above inequality is finite. This finishes the proof. </p>
<p><div class="remark_thmwrapper " id="2.2.7">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3</span>
  </div>
  <div class="remark_thmcontent">
  <p> Let \(t\in (0,T_h).\) Integrating the inequality in (<a href="#2.2.4">12</a>) from \(t\) to \(T_h,\) we get </p>
<div class="displaymath" id="a0000000016">
  \begin{eqnarray*}  T_{h}-t\leq \tfrac {1}{A} \int _0^{v(x,t)} \tfrac {{\rm d}\sigma }{f(\sigma )} \quad \mbox{for}\quad x \in \Omega . \end{eqnarray*}
</div>
<p> We deduce that </p>
<div class="displaymath" id="a0000000017">
  \begin{eqnarray*}  T_{h}-t\leq \tfrac {F(v_{\min }(t))}{A}\quad \mbox{for}\quad t\in (0,T_h).\label{2.2.7}\hfil \qed \end{eqnarray*}
</div>

  </div>
</div> <div class="remark_thmwrapper " id="rem.2.2.2">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">4</span>
  </div>
  <div class="remark_thmcontent">
  <p> In view of the condition (<a href="#2.1.5">5</a>) and reasoning as in the proof of Theorem <a href="#the.2.2.1">2</a>, it is not hard to see that there exists a positive constant \(C\) such that \(u_{\min }(t)\geq H(C(T-t))\) for \(t\in (0,T).\)<span class="qed">â–¡</span></p>

  </div>
</div> We also need the following result which shows an upper bound of \(u_{\min }(t)\) for \(t\in (0,T).\) <div class="theorem_thmwrapper " id="2.2.8">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(u\) be solution of <span class="rm">(<a href="#2.1.1">1</a>)–(<a href="#2.1.3">3</a>)</span>. Then, the following estimate holds </p>
<div class="displaymath" id="a0000000018">
  \begin{eqnarray*}  u_{\min }(t)\leq H(T-t) \quad \mbox{for}\quad t\in (0,T).\label{2.2.8} \end{eqnarray*}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000019">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>To prove the above estimate, we proceed as follows. Introduce the function \(w(t)\) defined as follows \(w(t)=u_{\min }(t)\) for \(t\in [0,T)\) and let \(t_1,\) \(t_2\in [0,T).\) Then, there exist \(x_1,\) \(x_2\in \Omega \) such that \(w(t_1)=u(x_1,t_1)\) and \(w(t_2)=u(x_2,t_2).\) Applying Taylor’s expansion, we observe that </p>
<div class="displaymath" id="a0000000020">
  \begin{eqnarray*}  w(t_2)-w(t_1)\geq u(x_2,t_2)-u(x_2,t_1)=(t_2-t_1)u_t(x_2,t_2)+o(t_2-t_1), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000021">
  \begin{eqnarray*}  w(t_2)-w(t_1)\leq u(x_1,t_2)-u(x_1,t_1)=(t_2-t_1)u_t(x_1,t_1)+o(t_2-t_1), \end{eqnarray*}
</div>
<p> which implies that \(w(t)\) is Lipschitz continuous. Further, if \(t_2{\gt}t_1,\) then </p>
<div class="displaymath" id="a0000000022">
  \begin{eqnarray*}  \tfrac {w(t_2)-w(t_1)}{t_2-t_1}\geq u_t(x_2,t_2)+o(1)= \Delta u(x_2,t_2)-f(u(x_2,t_2)) +o(1). \end{eqnarray*}
</div>
<p> Obviously, it is not hard to see that \(\Delta u(x_2,t_2)\geq 0.\) Letting \(t_1\rightarrow t_2,\) we obtain \(w'(t)\geq -f(w(t))\) for a.e. \(t\in (0,T)\) or equivalently \(\tfrac {{\rm d}w}{f(w)}\geq -{\rm d}t\) for a.e \(t\in (0,T).\) Integrate the above inequality over \((t,T)\) to obtain \(T-t\geq \int _0^{w(t)}\tfrac {{\rm d}\sigma }{f(\sigma )}\) for \(t\in (0,T).\) Since \(w(t)=u_{\min }(t),\) we arrive at \(u_{\min }(t)\leq H(T-t)\) for \(t\in (0,T)\) and the proof is complete. </p>
<p><div class="remark_thmwrapper " id="rem.2.2.3">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">6</span>
  </div>
  <div class="remark_thmcontent">
  <p> Regarding the last part of the proof of Theorem <a href="#2.2.8">5</a>, one sees that \(T\geq \int _0^{u_{0min}}\tfrac {{\rm d}\sigma }{f(\sigma )}.\) Thus, we have a lower bound of the quenching time of the solution \(u\) of (<a href="#2.1.1">1</a>)–(<a href="#2.1.3">3</a>). In the same way, it is not hard to see that \(\int _0^{u^h_{0min}}\tfrac {{\rm d}\sigma }{f(\sigma )}\) is a lower bound of the quenching time of the solution \(v\) of (<a href="#2.1.4">4</a>)–(<a href="#2.1.6">6</a>).<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000023">3 Continuity of the quenching time</h1>
<p>This section is dedicated to our main result. Our aim consists in proving that, if \(h\) is small enough, then the solution \(v\) of (<a href="#2.1.6">6</a>)–(<a href="#2.1.8">8</a>) quenches in a finite time and its quenching time \(T_h\) goes to \(T\) as \(h\) tends to zero. We also provide an upper bound of \(|T_h-T|\) in terms of \(\| u_0^h-u_0\| _\infty .\) Our result regarding the continuity of the quenching time is stated in the following theorem. <div class="theorem_thmwrapper " id="the.2.3.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Suppose that the problem <span class="rm">(<a href="#2.1.1">1</a>)–(<a href="#2.1.3">3</a>)</span> has a solution \(u\) which quenches at the time \(T.\) Then, under the assumption of Theorem <span class="rm"><a href="#the.2.2.1">2</a></span>, the solution \(v\) of <span class="rm">(<a href="#2.1.6">6</a>)–(<a href="#2.1.8">8</a>)</span> quenches in a finite time \(T_h,\) and there exist positive constants \(\alpha \) and \(\gamma \) such that for \(h\) small enough, the following estimate holds </p>
<div class="displaymath" id="a0000000024">
  \begin{eqnarray*}  |T_h-T| \leq \alpha \| u_0^h-u_0\| _{\infty }^{\gamma }. \end{eqnarray*}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000025">
  <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 know from Theorem <a href="#the.2.2.1">2</a> that the solution \(v\) quenches in a finite time \(T_h.\) Now, to achieve our objective, it remains to demonstrate the above estimate. We begin by proving that \(T_h\geq T.\) In order to obtain this result, we proceed as follows. Since \(u_0^h(x)\geq u_0(x)\) in \(\Omega ,\) we know from the maximum principle that \(v\geq u\) as long as all of them are defined. This implies that \(T_h\geq T,\) and consequently, we have \(T_h-T=|T_h-T|.\) In order to show the remaining part of the proof, we proceed by introducing the error function \(e(x,t)\) defined as follows </p>
<div class="displaymath" id="a0000000026">
  \begin{eqnarray*}  e(x,t)=v(x,t)-u(x,t)\quad \mbox{in}\quad \overline{\Omega }\times [0,T). \end{eqnarray*}
</div>
<p> Let \(t_0\) be any positive quantity satisfying \(t_0{\lt}T.\) A routine computation reveals that </p>
<div class="displaymath" id="a0000000027">
  \begin{eqnarray*}  e_t-\Delta e= -f’(\theta )e\quad \mbox{in}\quad \Omega \times (0,t_0), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000028">
  \begin{eqnarray*}  \tfrac {\partial e}{\partial \nu }=0\quad \mbox{on}\quad \partial \Omega \times (0,t_0), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000029">
  \begin{eqnarray*}  e(x,0)= u_0^h(x)-u_0(x) \quad \mbox{in}\quad \overline{\Omega }, \end{eqnarray*}
</div>
<p> where \(\theta \) is an intermediate value between \(u\) and \(v.\) Due to the fact that \(v(x,t)\geq u(x,t)\) in \(\Omega \times (0,t_0),\) then making use of Remark <a href="#rem.2.2.2">4</a>, it is easy to check that </p>
<div class="displaymath" id="2.3.1">
  \begin{eqnarray}  \theta (x,t)\geq u_{\min }(t) \geq H(C(T-t)) \quad \mbox{in}\quad \Omega \times (0,t_0), \label{2.3.1} \end{eqnarray}
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000030">
  \begin{eqnarray*}  e_t\leq \Delta e-f’(H(C(T-t)))e \quad \mbox{in}\quad \Omega \times (0,t_0). \end{eqnarray*}
</div>
<p> In view of the condition (<a href="#2.1.4">4</a>), we observe that there exists a positive constant \(C_1\) such that </p>
<div class="displaymath" id="a0000000031">
  \begin{eqnarray*}  e_t\leq \Delta e+\tfrac {C_1}{T-t}e \quad \mbox{in}\quad \Omega \times (0,t_0). \end{eqnarray*}
</div>
<p> Let \(Z(t)\) be the solution of the following ODE </p>
<div class="displaymath" id="a0000000032">
  \begin{eqnarray*}  Z’(t)=\tfrac {C_1Z(t)}{T-t} \quad \mbox{for}\quad t\in (0,t_0), \quad Z(0)= \| u_0^h-u_0\| _\infty . \end{eqnarray*}
</div>
<p> When we solve the above ODE, we observe that its solution \(Z(t)\) is given explicitly by </p>
<div class="displaymath" id="a0000000033">
  \begin{eqnarray*}  Z(t)= T^{C_1}\| u_0^h-u_0\| _\infty (T-t)^{-C_1} \quad \mbox{for}\quad t\in [0,t_0). \end{eqnarray*}
</div>
<p> On the other hand, an application of the maximum principle renders </p>
<div class="displaymath" id="a0000000034">
  \begin{eqnarray*}  e(x,t)\leq Z(t)= C_2\| u_0^h-u_0\| _\infty (T-t)^{-C_1}\quad \mbox{in}\quad \Omega \times [0,t_0), \end{eqnarray*}
</div>
<p> where \(C_2=T^{C_1}.\) Fix \(a\) a positive constant and let \(t_1\in (0,T)\) be a time such that \(\| e(\cdot ,t_1)\| _\infty \leq C_2\| u_0^h-u_0\| _\infty (T-t_1)^{-C_1}=a\) for \(h\) small enough. This implies that </p>
<div class="displaymath" id="2.3.2">
  \begin{eqnarray}  T-t_1=\left(\tfrac {C_2\| u_0^h-u_0\| _\infty }{a}\right)^{\tfrac {1}{C_1}}. \label{2.3.2} \end{eqnarray}
</div>
<p> Making use of Remark <a href="#2.2.7">3</a> and the triangle inequality, it is easy to see that </p>
<div class="displaymath" id="a0000000035">
  \begin{eqnarray*}  |T_h-t_1|\leq \tfrac {F(v_{\min }(t_1))}{A}\leq \tfrac {F(u_{\min }(t_1)+\| e(\cdot ,t_1)\| _\infty )}{A}. \end{eqnarray*}
</div>
<p> Since \(\| e(\cdot ,t_1)\| _\infty \leq a\) and due to the fact that the function \(F: [0,\infty )\rightarrow [0,\infty )\) is increasing, we infer from Theorem <a href="#2.2.8">5</a> that </p>
<div class="displaymath" id="2.3.3">
  \begin{eqnarray}  |T_h-t_1| \leq \tfrac {1}{A}F(H(T-t_1)+a).\label{2.3.3} \end{eqnarray}
</div>
<p> Having in mind that \(H\) is the inverse of \(F,\) we deduce that \(H: [0,\infty ) \rightarrow [0,\infty )\) is also increasing. We recall that \(\lim _{s\rightarrow \infty } F(s)=\infty ,\) which implies that \(\lim _{s\rightarrow \infty } H(s)=\infty .\) Introduce the function \(\varphi \) defined as follows </p>
<div class="displaymath" id="a0000000036">
  \begin{eqnarray*}  \varphi (x)=H(x(T-t_1)), \quad x\in [0,\infty ). \end{eqnarray*}
</div>
<p> It is clear that \(\varphi (x)\) is increasing for \(x\in [0,\infty ).\) In addition \(\lim _{x\rightarrow \infty } \varphi (x)=\infty .\) According to the fact that \(\varphi (1)+a\) belongs to \((0,\infty ),\) we conclude that there exists a positive constant \(C_3\) such that \( \varphi (1)+a\leq \varphi (C_3),\) which implies that \(H(T-t_1)+a\leq H(C_3(T-t_1)).\) Recalling that \(F(x)\) is increasing for \(x\in [0,\infty ),\) we deduce that \(\tfrac {1}{A}F(H(T-t_1)+a)\leq \tfrac {1}{A}F(H(C_3(T-t_1))).\) Exploiting the above inequality and (<a href="#2.3.3">17</a>), we find that </p>
<div class="displaymath" id="2.3.4">
  \begin{eqnarray}  |T_h-t_1| \leq \tfrac {1}{A}F(H(C_3(T-t_1)))=\tfrac {C_3}{A}|T-t_1|.\label{2.3.4} \end{eqnarray}
</div>
<p> We deduce from (<a href="#2.3.4">18</a>) and the triangle inequality that </p>
<div class="displaymath" id="a0000000037">
  \begin{eqnarray*}  |T-T_h|\leq |T-t_1|+ |T_h-t_1|\leq |T-t_1|+\tfrac {C_3}{A}|T-t_1|, \end{eqnarray*}
</div>
<p> which leads us to </p>
<div class="displaymath" id="a0000000038">
  \begin{eqnarray*}  |T-T_h| \leq (1+\tfrac {C_3}{A})|T-t_1|. \end{eqnarray*}
</div>
<p> Use the equality (<a href="#2.3.2">16</a>) to complete the rest of the proof. </p>
<h1 id="a0000000039">4 Numerical results</h1>
<p>In this section, we give some computational experiments to confirm the theory given in the previous section. We consider the radial symmetric solution of the following initial-boundary value problem </p>
<div class="displaymath" id="a0000000040">
  \begin{eqnarray*}  u_t= \Delta u-u^{-p}\quad \mbox{in}\quad B\times (0,T), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000041">
  \begin{eqnarray*}  \tfrac {\partial u}{\partial \nu }=0\quad \mbox{on}\quad S\times (0,T), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000042">
  \begin{eqnarray*}  u(x,0)=u_0(x) \quad \mbox{in}\quad \overline{B}, \end{eqnarray*}
</div>
<p> where \(p\) is a positive constant, \(u_0(x)=4+3\cos (\pi \| x\| )+\tfrac {\varepsilon }{2+\cos (\pi \| x\| )},\) with \(\varepsilon \) a nonnegative parameter, \(B=\{ x \in \mathbb {R}^{N}; \;  \| x\| {\lt}1\} ,\) \(S=\{ x \in \mathbb {R}^{N}; \| x\| =1\} ,\) \(1\leq N \leq 3.\) The above problem may be rewritten in the following form </p>
<div class="displaymath" id="2.4.1">
  \begin{eqnarray}  u_t=u_{rr}+\tfrac {N-1}{r}u_r-u^{-p},\quad r\in (0,1),\quad t\in (0,T), \label{2.4.1} \end{eqnarray}
</div>
<div class="displaymath" id="2.4.2">
  \begin{eqnarray}  u_r(0,t)=0,\quad u_r(1,t)=0,\quad t\in (0,T), \label{2.4.2} \end{eqnarray}
</div>
<div class="displaymath" id="2.4.3">
  \begin{eqnarray}  u(r,0)=\varphi (r),\quad r\in [0,1], \label{2.4.3} \end{eqnarray}
</div>
<p> where we take \(\varphi (r)=4+3\cos (\pi r)+\tfrac {\varepsilon }{2+cos(\pi r)}.\) We start by the construction of an adaptive scheme as follows. Let \(I\) be a positive integer and let \(h=1/I\). Define the grid \(x_i=ih,\) \(0\leq i \leq I,\) and approximate the solution \(u\) of (<a href="#2.4.1">19</a>)–(<a href="#2.4.3">21</a>) by the solution \(U_h^{(n)}=(U_0^{(n)}, ..., U_I^{(n)})^T\) of the following explicit scheme </p>
<div class="displaymath" id="a0000000043">
  \begin{align*}  \tfrac {U_0^{(n+1)}-U_0^{(n)}}{\Delta t_n}& = N\tfrac {2U_1^{(n)}-2U_0^{(n)}}{h^2}-(U_0^{(n)})^{-p},\\ \tfrac {U_i^{(n+1)}-U_i^{(n)}}{\Delta t_n}& = \tfrac {U_{i+1}^{(n)}-2U_i^{(n)}+U_{i-1}^{(n)}}{h^2}+ \tfrac {(N-1)}{ih} \tfrac {U_{i+1}^{(n)}-U_{i-1}^{(n)}}{2h}-(U_i^{(n)})^{-p},\  1\leq i \leq I-1,\\ \tfrac {U_{I}^{(n+1)}-U_{I}^{(n)}}{\Delta t_n}& = \tfrac {2U_{I-1}^{(n)}-2U_I^{(n)}}{h^2}+(N-1) \tfrac {U_{I}^{(n)}-U_{I-1}^{(n)}}{h}-(U_I^{(n)})^{-p},\\ U_i^{(0)}& =\varphi _i,\  0\leq i\leq I. \end{align*}
</div>
<p> After a little transformation, the above equations become </p>
<div class="displaymath" id="a0000000044">
  \begin{align*}  U_0^{(n+1)}& \geq (1-2\tfrac {N\Delta t_n}{h^2}-\Delta t_n(U_{hmin}^{(n)})^{-p-1})U_0^{(n)}+2\tfrac {N\Delta t_n}{h^2}U_1^{(n)},\\ U_i^{(n+1)}& \geq (\tfrac {\Delta t_n}{h^2}-\tfrac {(N-1)}{ih}\tfrac {\Delta t_n}{2h})U_{i-1}^{(n)}+(1-2\tfrac {\Delta t_n}{h^2}-\Delta t_n(U_{hmin}^{(n)})^{-p-1})U_i^{(n)}\\ & \quad +(\tfrac {\Delta t_n}{h^2}+\tfrac {(N-1)}{ih}\tfrac {\Delta t_n}{2h})U_{i+1}^{(n)},\quad 1\leq i \leq I-1,\\ U_I^{(n+1)}& \geq (2\tfrac {\Delta t_n}{h^2}-(N-1)\tfrac {\Delta t_n}{h})U_{I-1}^{(n)} \\ & \quad +(1-2\tfrac {\Delta t_n}{h^2}+(N-1)\tfrac {\Delta t_n}{h}-\Delta t_n(U_{hmin}^{(n)})^{-p-1})U_I^{(n)}, \end{align*}
</div>
<p> with \(U_{hmin}^{(n)}=\min _{0\leq i \leq I}U_i^{(n)}.\) Let us notice that, if \(\Delta t_n\leq \tfrac {h^2(U_{hmin}^{(n)})^{p+1}}{h^2+2N(U_{hmin}^{(n)})^{p+1}},\) then one easily sees by induction that \(U_{hmin}^{(n)}\) is positive for \(n\geq 0.\) Thus, the above condition is the CFL condition that ensures the stability of our scheme. It is important to note that if one chooses \(\Delta t_n=\min \{ \tfrac {(1-h^2)h^2}{2N}, h^2(U_{hmin}^{(n)})^{p+1}\} ,\) then the above CFL condition is fulfilled. Consequently, in the sequel, we shall pick the above time step for our explicit scheme. An important fact concerning the phenomenon of quenching is that, if the solution \(u\) quenches at the time \(T,\) then when the time \(t\) approaches the quenching time \(T,\) the solution \(u\) decreases to zero rapidly. Thus, in order to permit the discrete solution to reproduce the properties of the continuous one when the time \(t\) approaches the quenching time \(T,\) we need to adapt the size of the time step. This is the reason why we have chosen the above time step. For this time step, our explicit scheme becomes an adaptive scheme which is one of suitable schemes for problems whose solutions quench in a finite time. We also approximate the solution \(u\) of (<a href="#2.4.1">19</a>)-(<a href="#2.4.3">21</a>) by the solution \(U_h^{(n)}\) of the implicit scheme below </p>
<div class="displaymath" id="a0000000045">
  \begin{align*}  \tfrac {U_0^{(n+1)}-U_0^{(n)}}{\Delta t_n}& = N\tfrac {2U_1^{(n+1)}-2U_0^{(n+1)}}{h^2}-(U_0^{(n)})^{-p-1}U_0^{(n+1)}\\ \tfrac {U_i^{(n+1)}-U_i^{(n)}}{\Delta t_n}& = \tfrac {U_{i+1}^{(n+1)}-2U_i^{(n+1)}+U_{i-1}^{(n+1)}}{h^2} +\tfrac {(N-1)}{ih}\tfrac {U_{i+1}^{(n+1)}-U_{i-1}^{(n+1)}}{2h}\\ & \quad -(U_i^{(n)})^{-p-1}U_i^{(n+1)},\  1\leq i \leq I-1,\\ \tfrac {U_{I}^{(n+1)}-U_{I}^{(n)}}{\Delta t_n}& = \tfrac {2U_{I-1}^{(n+1)}-2U_I^{(n+1)}}{h^2}+(N-1) \tfrac {U_{I}^{(n+1)}-U_{I-1}^{(n+1)}}{h} -(U_I^{(n)})^{-p-1}U_I^{(n+1)},\\ U_i^{(0)}& =\varphi _i,\  0\leq i\leq I. \end{align*}
</div>
<p> As in the case of the explicit scheme, here again, we transform our scheme to an adaptive scheme by choosing \(\Delta t_n= h^2 (U_{hmin}^{(n)})^{p+1}.\) The implicit scheme gives the following equations </p>
<div class="equation" id="2.4.4">
<p>
  <div class="equation_content">
    \begin{equation}  (1+2\tfrac {N\Delta t_n}{h^2}+\Delta t_n(U_0^{(n)})^{-p-1})U_0^{(n+1)}-2\tfrac {N\Delta t_n}{h^2}U_1^{(n+1)}=U_0^{(n)}, \label{2.4.4} \end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<div class="displaymath" id="a0000000046">
  \begin{align} & (\tfrac {(N-1)}{ih}\tfrac {\Delta t_n}{2h}-\tfrac {\Delta t_n}{h^2})U_{i-1}^{(n+1)}+(1+2\tfrac {\Delta t_n}{h^2}+\Delta t_n(U_i^{(n)})^{-p-1})U_i^{(n+1)}-\nonumber \\ & \quad -(\tfrac {\Delta t_n}{h^2}+\tfrac {(N-1)}{ih}\tfrac {\Delta t_n}{2h})U_{i+1}^{(n+1)}=U_i^{(n)},\  1\leq i\leq I-1, \label{2.4.5} \end{align}
</div>
<div class="displaymath" id="a0000000047">
  \begin{align} & ((N-1)\tfrac {\Delta t_n}{h}-2\tfrac {\Delta t_n}{h^2})U_{I-1}^{(n+1)}+\nonumber \\ & \quad +(1+2\tfrac {\Delta t_n}{h^2}-(N-1)\tfrac {\Delta t_n}{h}+\Delta t_n(U_I^{(n)})^{-p-1})U_I^{(n+1)}=U_I^{(n)}. \label{2.4.6} \end{align}
</div>
<p> The above scheme leads us to the following tridiagonal linear system </p>
<div class="displaymath" id="a0000000048">
  \[ A_{h}^{(n)}U_{h}^{(n+1)}=U_{h}^{(n)}, \]
</div>
<p> where \(A_{h}^{(n)}\) is a \( (I+1)\times (I+1)\) tridiagonal matrix defined as follows </p>
<div class="displaymath" id="a0000000049">
  \begin{eqnarray*}  A_{h}^{(n)}=\left(\begin{array}{lllll} a_{0} &  b_{0} &  0 &  \cdots &  0 \\ c_{1} &  a_{1} &  b_{1} &  0 &  \cdots \\ 0 &  \ddots &  \ddots &  \ddots &  \ddots \\ \vdots &  \ddots &  \ddots &  \ddots &  b_{I-1} \\ 0 &  \cdots &  0 & c_{I} &  a_{I} \end{array}\right) \end{eqnarray*}
</div>
<p> with </p>
<div class="displaymath" id="a0000000050">
  \begin{align*}  a_{0}& =1+2\tfrac {N\Delta t_n}{h^2}+\Delta t_n(U_0^{(n)})^{-p-1},\\ a_{I}& =1+2\tfrac {\Delta t_n}{h^2}-(N-1)\tfrac {\Delta t_n}{h}+\Delta t_n(U_I^{(n)})^{-p-1},\\ a_{i}& =1+2\tfrac {\Delta t_n}{h^2}+\Delta t_n(U_i^{(n)})^{-p-1},\  i=1,...,I-1,\\ b_{0}& =-2\tfrac {N\Delta t_n}{h^2},\quad b_{i}=-(\tfrac {\Delta t_n}{h^2}+\tfrac {(N-1)}{ih}\tfrac {\Delta t_n}{2h}), i=1,...,I-1,\\ c_{i}& =(\tfrac {(N-1)}{ih}\tfrac {\Delta t_n}{2h}-\tfrac {\Delta t_n}{h^2}),\  i=1,...,I-1,\\ c_I& =-2\tfrac {\Delta t_n}{h^2}+(N-1)\tfrac {\Delta t_n}{h}. \end{align*}
</div>
<p> It is not hard to see that </p>
<div class="displaymath" id="a0000000051">
  \[ (A_h^{n})_{ii}{\gt}0,\quad (A_h^{n})_{ij}\leq 0,\quad i\neq j,\quad (A_h^{n})_{ii}{\gt}\sum _{i\neq j}|(A_h^{n})_{ij}|. \]
</div>
<p> These inequalities imply that the matrix \(A_h^{(n)}\) is invertible and its inverse \((A_h^{(n)})^{-1}\) is a positive matrix. Consequently, it is easy to see that, if \(U_h^{(n)}\) is positive, then \(U_h^{(n+1)}\) exists and is also positive. Thus, since \(U_h^{(0)}=\varphi _h\) is positive, we show by induction that \(U_h^{(n)}\) exists and is positive. It is not hard to see that \(u_{rr}(0,t)=\lim _{r\rightarrow 0} \tfrac {u_r(r,t)}{r}.\) Hence, if \(r=0,\) then we see that </p>
<div class="displaymath" id="a0000000052">
  \begin{eqnarray*}  u_t(0,t)=Nu_{rr}(0,t)-(u(0,t))^{-p}, \quad t\in (0,T), \end{eqnarray*}
</div>
<p> These observations have been taken into account in the construction of our schemes at the first node. We need the following definition. <div class="definition_thmwrapper " id="Tab.one">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">8</span>
  </div>
  <div class="definition_thmcontent">
  <p> We say that the discrete solution \(U_h^{(n)}\) of the explicit scheme or the implicit scheme quenches in a finite time if \(\lim _{n\rightarrow \infty }U_{hmin}^{(n)}=0,\) and the series \(\sum _{n=0}^{\infty } \Delta t_n\) converges. The quantity \(\sum _{n=0}^{\infty } \Delta t_n\) is called the numerical quenching time of the discrete solution \(U_h^{(n)}.\) </p>

  </div>
</div> In the following tables, in rows, we present the numerical quenching times, the numbers of iterations, the CPU times and the orders of the approximations corresponding to meshes of 16, 32, 64, 128. We take for the numerical quenching time \(t_{n}=\sum _{j=0}^{n-1}\Delta t_{j}\) which is computed at the first time when \(U_{hmin}^{(n)}\leq 10^{-10}.\) The order \((s)\) of the method is computed from </p>
<div class="displaymath" id="a0000000053">
  \begin{eqnarray*}  s=\tfrac {\log ((t_{2h}-t_{h})/(t_{4h}-t_{2h}))}{\log (2)}. \end{eqnarray*}
</div>
<p><b class="bfseries">Numerical experiments for \(p=1, N=2\)</b> </p>
<p><b class="bfseries">First case:</b> \(\varepsilon =0\) </p>
<div class="table"  id="Tab:two">
   <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:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> I </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> \(t_{n}\) </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> 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> CPU time </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> s</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; border-left: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:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3.604286 </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> 5415 </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> 12 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>32 </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> 3.731558 </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> 21476 </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> 71 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>64 </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> 3.796654 </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> 84141</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> 523 </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> 0.97</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; border-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3.828011 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 335561</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3782</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1.04 </p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations, CPU times (seconds) and orders of the approximations obtained with the explicit Euler method.</span> 
</figcaption></div>
</div>

<div class="table"  id="Tab:three">
   <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:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> I </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> \(t_{n}\) </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> 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> CPU time </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> s</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; border-left: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:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3.604107 </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> 5325 </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> 13 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>32 </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> 3.731511 </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> 21121 </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> 87 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>64 </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> 3.796641 </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> 84721 </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> 1106 </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> 0.97</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; border-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3.830302 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 331834</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 7718</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.95 </p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations, CPU times (seconds) and orders of the approximations obtained with the implicit Euler method.</span> 
</figcaption></div>
</div>
<p><br /><b class="bfseries">Second case:</b> \(\varepsilon =1/50\) </p>
<div class="table"  id="Tab:for">
   <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:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> I </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> \(t_{n}\) </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> 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> CPU time </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> s</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; border-left: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:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3.642767 </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> 5453 </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> 12 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>32 </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> 3.770560 </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> 21626 </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> 55 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>64 </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> 3.835916 </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> 84784 </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> 557 </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> 0.96</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; border-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3.870205 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 335873</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3684</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.93 </p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations, CPU times (seconds) and orders of the approximations obtained with the explicit Euler method.</span> 
</figcaption></div>
</div>

<div class="table"  id="Tab:five">
    <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:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> I </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> \(t_{n}\) </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> 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> CPU time </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> s</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; border-left: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:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3.642591 </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> 5453 </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> 15 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>32 </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> 3.770514 </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> 21636 </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> 89 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>64 </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> 3.835904 </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> 84784 </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> 972 </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> 0.97</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; border-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3.870181 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 335867</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 7821</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.93 </p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations, CPU times (seconds) and orders of the approximations obtained with the implicit Euler method.</span> 
</figcaption></div>
</div>
<p><b class="bfseries">Third case:</b> \(\varepsilon =1/100\) </p>
<div class="table"  id="Tab.six">
    <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:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> I </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> \(t_{n}\) </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> 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> CPU time </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> s</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; border-left: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:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3.623502 </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> 5433 </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> 12 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>32 </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> 3.751034 </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> 21556 </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> 60 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>64 </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> 3.816361 </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> 84462 </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> 492 </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> 0.97</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; border-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3.850163 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 333798</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3554</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.95 </p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">5</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations, CPU times (seconds) and orders of the approximations obtained with the explicit Euler method.</span> 
</figcaption></div>
</div>

<div class="table"  id="a0000000054">
    <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:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> I </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> \(t_{n}\) </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> 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> CPU time </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> s</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; border-left: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:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 3.623324 </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> 5433 </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> 12 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>32 </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> 3.750988 </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> 21556 </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> 69 </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> -</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; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>64 </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> 3.816249 </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> 84462 </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> 1034 </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> 0.97</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; border-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3.849163 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 332769</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 7763</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.959 </p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">6</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations, CPU times (seconds) and orders of the approximations obtained with the implicit Euler method.</span> 
</figcaption></div>
</div>
<p><div class="remark_thmwrapper " id="rem.2.4.1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
  </div>
  <div class="remark_thmcontent">
  <p> If we consider the problem (<a href="#2.4.1">19</a>)-(<a href="#2.4.3">21</a>) in the case where \(p=1\) and the initial datum \(\varphi (r)=4+3\cos (\pi r)+\tfrac {\varepsilon }{2+\cos (\pi r)}\) with \(\varepsilon =0,\) then we see that the numerical quenching time of the discrete solution for the explicit scheme or the implicit scheme is approximately equal to that in which the initial datum increases slightly, that is when \(\varepsilon \) is a small positive real (see, Tables 1-6 for an illustration). This result confirms the theory regarding the continuity of the quenching time as a function of the initial datum.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>In what follows, we give some plots to illustrate our analysis. In Figures <a href="#fig:1">1</a>, <a href="#fig:2">2</a> and <a href="#fig:3">3</a> we can appreciate that the discrete solution quenches in a finite time. We also remark that the representation of the discrete solution when \(\varepsilon =0\) is practically the same that the one when \(\varepsilon =1/50\) or \(\varepsilon =1/100.\) </p>
<figure id="fig:1">
  <div class="centered"> <img src="img-0001.png" alt="\includegraphics[ scale=0.32]{math0.png}" style="width:670.72px; height:496.0px" />
 </div>
<figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text"> Evolution of the discrete solution, \(\varepsilon =0\)</span> 
</figcaption>


</figure>

<figure >
  <div class="centered"> <div class="minipage" style="width: 125.19212598425196pt"> <img src="img-0002.png" alt="\includegraphics[ scale=0.32]{math100}" style="width:670.72px; height:496.0px" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text"> Evolution of the discrete solution, \(\varepsilon =1/50\)</span> 
</figcaption> </div> &#8195;&#8195;&#8195;&#8195;&#8195;&#8195;<div class="minipage" style="width: 125.19212598425196pt"> <img src="img-0003.png" alt="\includegraphics[ scale=0.32]{math50}" style="width:670.72px; height:496.0px" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text"> Evolution of the discrete solution, \(\varepsilon =1/100\)</span> 
</figcaption> </div>  </div>

</figure>
<p> <br /><div class="acknowledgement_thmwrapper " id="a0000000055">
  <div class="acknowledgement_thmheading">
    <span class="acknowledgement_thmcaption">
    Acknowledgement
    </span>
  </div>
  <div class="acknowledgement_thmcontent">
  <p>The authors want to thank the anonymous referees for the thorough reading of the manuscript and several suggestions that helped us improve the presentation of the paper. </p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
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</dd>
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</dl>


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