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<h1>Quenching for discretizations of a semilinear parabolic equation with nonlinear boundary outflux</h1>
<p class="authors">
<span class="author">Cyrille K. N’Dri,\(^1\) Ardjouma Ganon,\(^1\) Gozo Yoro,\(^2\) Kidjegbo A. Touré\(^1\) </span>
</p>
<p class="date">April 11, 2023; accepted: September 28, 2023; published online: December 22, 2023.</p>
<p class="thanks">\(^1\)UMRI Mathématiques et Nouvelles Technologies de l’Information, Institut National Polytechnique Félix HOUPHOUËT-BOIGNY, Yamoussoukro, Côte d’Ivoire, e-mail:  C. K. N’Dri: <span class="tt">ndri.pack@gmail.com</span><br />Ardjouma Ganon: <span class="tt">ardjganon@gmail.com</span><br />Kidjegbo A. Touré: <span class="tt">latoureci@gmail.com</span> <br />\(^2\) Département Mathématiques et Informatique, Université Nangui Abrogoua d’Abobo Adjamé, Abidjan, Côte d’Ivoire, e-mail: <span class="tt">yorocarol@yahoo.fr</span>.</p>
</div>



<div class="abstract"> In this paper, we study numerical approximations of a semilinear parabolic problem in one-dimension, of which the nonlinearity appears both in source term and in Neumann boundary condition. By a semidiscretization using finite difference method, we obtain a system of ordinary differential equations which is an approximation of the original problem. We obtain conditions under which the positive solution of our system quenches in a finite time and estimate its semidiscrete quenching time. Convergence of the numerical quenching time to the theoretical one is established. Next, we show that the quenching rate of the numerical scheme is different from the continuous one. Finally, we give numerical results to illustrate our analysis. </div>
<p><b class="bfseries">MSC. </b>35K61, 65M06, 35B51 </p>
<p><b class="bfseries">Keywords. </b>Nonlinear parabolic equations, nonlinear boundary outflux, semidiscretizations, numerical quenching, quenching rate. </p>
<h1 id="probleme">1 Introduction</h1>
<p>  Consider the following semilinear parabolic equation with nonlinear boundary outflux \( \left\{  \begin{array}{rl}& u_{t}(x,t)=u_{xx}(x,t)+f(x)(1-u(x,t))^{-p},\:  0{\lt}x{\lt}1,\:  0{\lt}t{\lt}\infty ,\\ & u_{x}(0,t)=u^{-q}(0,t),\quad 0{\lt}t{\lt}\infty ,\\ & u_{x}(1,t)=0 ,\quad 0{\lt}t{\lt}\infty ,\\ & u(x,0)=u_{0}(x),\quad 0\leq x\leq 1. \end{array} \right. \label{probleme} \) </p>
<p>Here \(p, q\) are positive constants, \(f\) is a non-negative function and initial datum \(u_{0}:[0,1]\rightarrow (0,1)\) is smooth enough and satisfies boundary conditions. We can regard this problem as a heat conduction model that incorporates the effects of nonlinear reaction (source) and nonlinear boundary outflux (emission). </p>
<p>Our problem derives from the general the problem \((1.1)\) of <span class="cite">
	[
	<a href="index.html#pao2007" >18</a>
	]
</span>, where the authors proved the existence and uniqueness of the classical solution.  </p>
<p>We say that the classical solution \(u(x,t)\) of <a href="index.html#probleme" class="eqref">1</a> quenches in a finite time if there exists a finite time \(T_{q}\) such that: </p>
<div class="displaymath" id="a0000000002">
  \begin{equation*}  \displaystyle \lim _{t\rightarrow T^{-}_{q}} \min _{0\leq x\leq 1}u(x,t)=0,\quad or \quad \displaystyle \lim _{t\rightarrow T^{-}_{q}} \max _{0\leq x\leq 1}u(x,t)=1. \end{equation*}
</div>
<p>From now, we denote the quenching time with \(T_{q}\). </p>
<p>The quenching problem has been the subject of investigations of many authors since several decades (see <span class="cite">
	[
	<a href="index.html#dz1" >7</a>
	, 
	<a href="index.html#hkawarada" >13</a>
	, 
	<a href="index.html#salin2003" >19</a>
	, 
	<a href="index.html#A135" >20</a>
	, 
	<a href="index.html#zhi2011" >22</a>
	, 
	<a href="index.html#zhi2007" >23</a>
	, 
	<a href="index.html#die-1997" >6</a>
	, 
	<a href="index.html#cyrille2" >17</a>
	]
</span> and the references cited therein). It was in \(1975\) that Kawarada <span class="cite">
	[
	<a href="index.html#hkawarada" >13</a>
	]
</span> introduced the first concept of quenching, he found sufficient conditions under which the solutions of the following problem quenches in finite time, </p>
<div class="displaymath" id="a0000000003">
  \begin{align*} & u_{t}=u_{xx}+\tfrac {1}{1-u},\quad t{\gt}0,\quad x\in (0,l),\\ & u(t,0)=u(t,l)=0,\quad t{\gt}0,\\ & u(0,x)=0,\quad x\in (0,l), \end{align*}
</div>
<p> where \(l\) is positive. </p>
<p>In our problem we have two singular heat sources, namely, a source term \(f(x)(1-u)^{-p}\) and the boundary outflux term \(u^{-q}\). Some authors studied such quenching problems with the nonlinearity both in source and in boundary conditions <span class="cite">
	[
	<a href="index.html#dz1" >7</a>
	, 
	<a href="index.html#A135" >20</a>
	, 
	<a href="index.html#zhi2011" >22</a>
	, 
	<a href="index.html#zhi2007" >23</a>
	, 
	<a href="index.html#cyrille2" >17</a>
	]
</span>. In <span class="cite">
	[
	<a href="index.html#A135" >20</a>
	]
</span>, Selcuk and Ozalp studied the quenching behavior of the solution of a semilinear heat equation with a singular boundary outflux. They showed that quenching occurs on the boundary under certain conditions and that the time derivative blows up at a quenching point.<br />Concerning problem <a href="index.html#probleme" class="eqref">1</a>, Zhi <span class="cite">
	[
	<a href="index.html#zhi2011" >22</a>
	]
</span> showed that if the initial datum satisfies </p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  u^{''}_{0}(x)+f(x)(1-u_{0}(x))^{-p}\leq 0\text{ and not equals} \  0 \text{ identically},\:  x\in (0,1), \end{equation*}
</div>
<p> then the classical solution \(u\) of <a href="index.html#probleme" class="eqref">1</a> quenches in a finite time \(T_{q}\) with the following estimate \(u(0,t)\sim (T_{q}-t)^{\frac{1}{2(q+1)}}\). This condition excludes the formation of a singularity for the source term, thus it is only sufficient to consider the case of quenching formation on the boundary for the solution of problem <a href="index.html#probleme" class="eqref">1</a>. He also asserted that the quenching can only occur on the point \(x=0\) if the given initial datum is monotone. A related problem was studied earlier in <span class="cite">
	[
	<a href="index.html#zhi2007" >23</a>
	]
</span> where the author prescribed \(f(x)=1\), similar results have been also obtained. </p>
<p>Here we are interested in the numerical study of the phenomenon of quenching (For other works on numerical approximations of quenching solutions we refer to <span class="cite">
	[
	<a href="index.html#numericalquenchin" >15</a>
	, 
	<a href="index.html#euclid" >5</a>
	, 
	<a href="index.html#Doc3" >14</a>
	, 
	<a href="index.html#cyrille2" >17</a>
	]
</span>). We give assumptions under which the solution of a semidiscrete form of <a href="index.html#probleme" class="eqref">1</a> quenches in a finite time and estimate its semidiscrete quenching time. We also prove that, under suitable assumptions on the initial datum, the semidiscrete quenching time converges to the theoretical one when the mesh size goes to zero. Our work was motived by the papers in <span class="cite">
	[
	<a href="index.html#edja1" >9</a>
	, 
	<a href="index.html#abia1996" >1</a>
	, 
	<a href="index.html#cyrille1" >16</a>
	, 
	<a href="index.html#adou2" >2</a>
	, 
	<a href="index.html#ganon1" >10</a>
	, 
	<a href="index.html#assale" >4</a>
	]
</span>, where the authors have used semidiscrete forms for some parabolic equations to study 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). </p>
<p>The paper is written in the following manner. In the next Section, we present a semidiscrete scheme of <a href="index.html#probleme" class="eqref">1</a>. In <a href="index.html#sec3">section 3</a>, we give properties concerning our semidiscrete scheme. In <a href="index.html#sec4">section 4</a>, under appropriate conditions, we prove that the solution of the semidiscrete form quenches in a finite time, estimate its semidiscrete quenching time and give results on the numerical quenching rate. In <a href="index.html#sec5">section 5</a>, we study the convergence of semidiscrete quenching time. Finally, in the last section, we give numerical experiments. </p>
<h1 id="sec2">2 The Semidiscrete Problem</h1>

<p>Let \(I\) be a positive integer, we set \(h=\frac{1}{I}\), and we define the grid \(x_{i}=ih\), \(i=0,\ldots ,I\). We approximate the solution \(u\) of the problem <a href="index.html#probleme" class="eqref">1</a> by the solution \(U_{h}=(U_{0},U_{1},\ldots ,U_{I})^{T}\) and the initial datum \(u_{0}\) by \(\varphi _{h}=(\varphi _{0},\varphi _{1},\ldots ,\varphi _{I})^{T}\) of the semidiscrete equations </p>
<div class="displaymath" id="semidiscret1">
  \begin{align} & \tfrac {dU_{i}(t)}{dt}=\delta ^{2}U_{i}(t)+ f_{i}(1-U_{i}(t))^{-p},\quad i=1,\ldots ,I-1, \:  t\in [0,T_{q}^{h}), \label{semidiscret1}\\ & \tfrac {dU_{0}(t)}{dt}=\delta ^{2}U_{0}(t)-\tfrac {2}{h}U^{-q}_{0}(t)+ f_{0}(1-U_{0}(t))^{-p},\quad [0,T_{q}^{h}), \label{semidiscret2}\\ & \tfrac {dU_{I}(t)}{dt}=\delta ^{2}U_{I}(t)+f_{I}(1-U_{I}(t))^{-p},\quad [0,T_{q}^{h}), \label{semidiscret3}\\ & U_{i}(0)=\varphi _{i}{\gt}0, \quad i=0,\ldots ,I, \label{semidiscret4} \intertext {where} & f_{i}\simeq f(x_{i}),\quad i=0,\ldots ,I,\notag \\ & \delta ^{2}U_{i}(t)=\displaystyle \tfrac {U_{i+1}(t)-2U_{i}(t)+U_{i-1}(t)}{h^{2}},\quad i=1,\ldots ,I-1,\notag \\ & \delta ^{2}U_{0}(t)=\displaystyle \tfrac {2U_{1}(t)-2U_{0}(t)}{h^{2}},\quad \delta ^{2}U_{I}(t)=\displaystyle \tfrac {2U_{I-1}(t)-2U_{I}(t)}{h^{2}}\notag . \end{align}
</div>
<p>Here \([0,T_{q}^{h})\) is the maximal time interval on which \(\| U_{h}(t)\| _{\inf }{\gt}0\), where \(\| U_{h}(t)\| _{\inf }=\displaystyle \min _{0\leq i \leq I}\lvert U_{i}(t)\rvert \). The time \(T^{h}_{q}\) can be finite or infinite. When \(T_{q}^{h}\) is finite, we say that \(U_{h}\) quenches in finite time and \(T^{h}_{q}\) is called the quenching time of the solution \(U_{h}\). Whereas when \(T_{q}^{h}\) is infinite we say that \(U_{h}\) quenches globally. </p>
<p>In the following, we give important results which will be used later. Here, \(V_{i}\) and \(W_{i}\) denote the respective \((i+1)\)-th components of \(V_{h}\) and \(W_{h}\). </p>
<p><div class="definition_thmwrapper " id="def sur solution">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">1</span>
  </div>
  <div class="definition_thmcontent">
  <p> A function \(V_{h}\in \mathcal{C}^{1}([0,T],\mathbb {R}^{I+1})\) is an upper solution of <a href="index.html#semidiscret1" class="eqref">1</a>–?? if </p>
<div class="displaymath" id="a0000000005">
  \begin{flalign*} & \displaystyle \tfrac {dV_{i}(t)}{dt}-\delta ^{2}V_{i}(t) \geq f_{i}(1-V_{i}(t))^{-p},\quad i=1,\ldots ,I, \:  t\in (0,T],\\ & \displaystyle \tfrac {dV_{0}(t)}{dt}-\delta ^{2}V_{0}(t)+\tfrac {2}{h}V^{-q}_{0}(t)\geq f_{0}(1-V_{0}(t))^{-p},\:  t\in (0,T],\\ & V_{i}(0)\geq \varphi _{i}, \quad i=0,\ldots ,I . \end{flalign*}
</div>
<p> On the other hand, we say that \(V_{h}\in \mathcal{C}^{1}([0,T],\mathbb {R}^{I+1})\) is a lower solution of <a href="index.html#semidiscret1" class="eqref">1</a>–?? if these inequalities are reversed. </p>

  </div>
</div> </p>
<p>The following results are semidiscrete forms of a Maximum Principle and will be an important tool in the study of the semidiscrete problem <a href="index.html#semidiscret1" class="eqref">1</a>–??. Their proofs are standard and will be omitted. </p>
<p><div class="lemma_thmwrapper " id="principe de maximum">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(a_{h}\in C^{0}([0,T],\mathbb {R}^{I+1})\) and \(V_{h}\in C^{1}([0,T],\mathbb {R}^{I+1})\) such that </p>
<div class="displaymath" id="lemma1-1">
  \begin{align} & \displaystyle \tfrac {d}{dt}V_{i}-\delta ^{2}V_{i}+a_{i}V_{i}\geq 0,\quad i=0,\ldots ,I,\:  t\in [0,T], \label{lemma1-1}\\ & V_{i}(0)\geq 0,\quad i=0,\ldots ,I, \label{lemma1-2} \end{align}
</div>
<p> then we have&#8195;\(V_{i}(t)\geq 0,\quad i=0,\ldots ,I,\:  t\in [0,T]\). </p>

  </div>
</div> </p>
<p>Another form of the Maximum Principle for semidiscrete equations are the following comparison lemma. </p>
<p><div class="lemma_thmwrapper " id="lemme de comparaison 2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(g\in C(\mathbb {R},\mathbb {R})\) and \(W_{h},V_{h}\in C^{1}([0,T],\mathbb {R}^{I+1})\) such that <small class="small"><div class="displaymath" id="C1">
  \begin{align} & \tfrac {d}{dt}V_{i}-\delta ^{2}V_{i}-g(V_{i})\leq \tfrac {d}{dt}W_{i}-\delta ^{2}W_{i}-g(W_{i}),\: i=1,\ldots ,I,\:  t\in (0,T],\label{C1}\\ & \tfrac {d}{dt}V_{0}+\tfrac {2}{h}V^{-q}_{0}-\delta ^{2}V_{0}-g(V_{0})\leq \tfrac {d}{dt}V_{0}+\frac{2}{h}W^{-q}_{0}-\delta ^{2}W_{0}-g(W_{0}),t\in (0,T],\label{C2}\\ & V_{i}(0)\leq W_{i}(0),\:  i=0,\ldots ,I,\label{C3} \end{align}
</div> </small> then&#8195;\(V_{i}(t)\leq W_{i}(t),\quad i=0,\ldots ,I,\: t\in [0,T]\). </p>

  </div>
</div> </p>
<h1 id="sec3">3 Quenching in the semidiscrete problem</h1>

<p>In this section, under appropriate assumptions, we show that solution \(U_{h}\) of the semidiscrete problem <a href="index.html#semidiscret1" class="eqref">1</a>–?? quenches in a finite time \(T^{h}_{q}\) and we estimate its semidiscrete quenching time. </p>
<p>The following result gives a property of the operator \(\delta ^{2}\). </p>
<p><div class="lemma_thmwrapper " id="egalite sur operateur">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">4</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(U_{h}\in \mathbb {R}^{I+1}\) be such that \(U_{h}{\gt}0\). Then, we have </p>
<div class="displaymath" id="a0000000006">
  \begin{align*} & \delta ^{2}(U^{-q}_{i})\geq -qU^{-q-1}_{i}\delta ^{2}U_{i},\:  i=0,\ldots ,I. \end{align*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000007">
  <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 us introduce function \(f(s)=s^{-q}\). Using Taylor’s expansion we get </p>
<div class="displaymath" id="a0000000008">
  \begin{align*} & \delta ^{2}f(U_{0})=f^{'}(U_{0})\delta ^{2}U_{0}+\tfrac {(U_{1}-U_{0})^{2}}{h^{2}}f^{''}(\zeta _{0}),\\ & \delta ^{2}f(U_{I})=f^{'}(U_{I})\delta ^{2}U_{I}+\tfrac {(U_{I-1}-U_{I})^{2}}{h^{2}}f^{''}(\zeta _{I}),\\ & \delta ^{2}f(U_{i})=f^{'}(U_{i})\delta ^{2}U_{i}+\tfrac {(U_{i+1}-U_{i})^{2}}{2h^{2}}f^{''}(\eta _{i})+\tfrac {(U_{i-1}-U_{i})^{2}}{2h^{2}}f^{''}(\zeta _{i}),\:  i=1,\ldots ,I-1. \end{align*}
</div>
<p> where \(\eta _{i}\) is an intermediate value between \(U_{i}\) and \(U_{i+1}\) and \(\zeta _{i}\) the one between \(U_{i}\) and \(U_{i-1}\).<br />The result follows taking into account the fact that \(U_{h}\) is nonnegative. <div class="proof_wrapper" id="a0000000009">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>In the rest of this paper, we assume that these conditions are satisfied </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000017247C97040>">
  \begin{align} & 0\leq f_{i}\leq f_{i+1},\quad i=0,\ldots ,I-1, \label{croissance_f} \\ & 0{\lt}\varphi _{i}{\lt}\varphi _{i+1},\quad i=0,\ldots ,I-1. \label{croissance_donnee_initiale} \end{align}
</div>
<p><div class="lemma_thmwrapper " id="propriete solution">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">5</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(U_{h}\) be the solution of <a href="index.html#semidiscret1" class="eqref">1</a>–??, then, we have </p>
<ol class="enumerate">
  <li><p>\(U_{i}(t)\geq \varphi _{i}\)&#8195;for&#8195;\(i=0,\ldots ,I\), \(t\in [0,T]\); </p>
</li>
  <li><p>\(U_{i}(t){\lt}U_{i+1}(t)\)&#8195;for&#8195;\(i=0,\ldots ,I-1\),  \(t\in [0,T]\). </p>
</li>
</ol>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000010">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<ol class="enumerate">
  <li><p>Using <a href="index.html#lemme de comparaison 2">lemma 3</a>, we obtain \(U_{i}(t)\geq \varphi _{i}{\gt}0,\quad i=0,\ldots ,I,\:  t\in [0,T]\). </p>
</li>
  <li><p>For \(i=0,\ldots ,I-1\), \(t\in [0,T]\), introduce \(Z_{i}\) such that </p>
<p>\(Z_{i}(t)=U_{i+1}(t)-U_{i}(t)\). Let \(t_{0}\) be the first \(t{\gt}0\) such that \(Z_{i}(t){\gt}0\) for \(t\in [0,t_{0})\), but \(Z_{i_{0}}(t_{0})=0\) for a certain \(i_{0}\in \{  0,\ldots ,I-1\} \). Without loss of generality, we suppose that \(i_{0}\) is the smallest integer checking the equality above. We observe that </p>
<div class="displaymath" id="a0000000011">
  \begin{align*} & \tfrac {d}{dt}Z_{i_{0}}(t_{0})=\displaystyle \lim _{k\rightarrow 0}\tfrac {Z_{i_{0}}(t_{0})-Z_{i_{0}}(t_{0}-k)}{k}\leq 0,\quad i_{0}=0,\ldots ,I-1,\\ & \delta ^{2}Z_{i_{0}}(t_{0})=\tfrac {Z_{i_{0}-1}(t_{0})-2Z_{i_{0}}(t_{0})+Z_{i_{0}+1}(t_{0})}{h^{2}}{\gt}0,\quad i_{0}=1,\ldots ,I-2,\\ & \delta ^{2}Z_{0}(t_{0})=\tfrac {Z_{1}(t_{0})-3Z_{0}(t_{0})}{h^{2}}{\gt}0,\\ & \delta ^{2}Z_{I-1}(t_{0})=\tfrac {Z_{I-2}(t_{0})-3Z_{I-1}(t_{0})}{h^{2}}{\gt}0. \end{align*}
</div>
<p>By a computation, we get </p>
<div class="displaymath" id="a0000000012">
  \begin{align*} & \tfrac {d}{dt}Z_{0}(t_{0})-\delta ^{2}Z_{0}(t_{0})-f_{1}(1-U_{1}(t_{0}))^{-p}+f_{0}(1-U_{0}(t_{0}))^{-p}-\tfrac {2}{h}U^{-q}_{0}(t_{0}){\lt}0,\\ & \tfrac {d}{dt}Z_{i_{0}}(t_{0})-\delta ^{2}Z_{i_{0}}(t_{0})-f_{i_{0}+1}(1-U_{i_{0}+1}(t_{0}))^{-p}+f_{i_{0}}(1-U_{i_{0}}(t_{0}))^{-p}{\lt}0,\:  i_{0}=1,\ldots ,I-1. \end{align*}
</div>
<p>But these inequalities contradict <a href="index.html#semidiscret1" class="eqref">1</a>–<a href="index.html#semidiscret3" class="eqref">3</a> and the proof is complete. </p>
</li>
</ol>
<p> <div class="proof_wrapper" id="a0000000013">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="quenches">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(U_{h}\) be the solution of <a href="index.html#semidiscret1" class="eqref">1</a>–??. Assume that the initial data at ?? verifies </p>
<div class="displaymath" id="condition theo quench1">
  \begin{align} & \delta ^{2}\varphi _{i}+f_{i}(1-\varphi _{i})^{-p}\leq -\epsilon \varphi _{i}^{-q},\quad i=1,\ldots ,I,\label{condition theo quench1}\\ & \delta ^{2}\varphi _{0}+f_{0}(1-\varphi _{0})^{-p}-\frac{2}{h}\varphi ^{-q}_{0}\leq - \epsilon \varphi ^{-q}_{0}\label{condition theo quench2}, \intertext {for a certain constant $\epsilon \in (0,1]$.} \intertext {Then, the solution $U_{h}$ quenches in a finite time $T_{q}^{h}$ and we have the following estimate} & T_{q}^{h}\leq \frac{\|  \varphi _{h}\| _{\inf }^{q+1}}{\epsilon (q+1)}.\notag \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000014">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Since \((0,T_{q}^{h})\) is the maximal time interval which on which \(\|  U_{h}\| _{\inf }{\gt}0\). We want to show that \(T_{q}^{h}\) is finite and satisfies the above inequality. Introduce the vector \(J_{h}(t)\) defined as follows </p>
<div class="displaymath" id="fonctionnel theo quench1">
  \begin{flalign} & J_{i}(t)=\tfrac {dU_{i}(t)}{dt}+\epsilon U^{-q}_{i}(t),\quad i=0,\ldots ,I,\label{fonctionnel theo quench1} \intertext {by a straightforward computation, we get} & \tfrac {d}{dt}J_{i}-\delta ^{2}J_{i}=\tfrac {d}{dt}(\tfrac {dU_{i}}{dt}-\delta ^{2}U_{i})-\epsilon (qU^{-q-1}_{i}\tfrac {dU_{i}}{dt}+\delta ^{2}U^{-q}_{i}),\quad i=0,\ldots ,I.\notag \end{flalign}
</div>
<p>Using <a href="index.html#egalite sur operateur">lemma 4</a> and equalities <a href="index.html#semidiscret1" class="eqref">1</a>–<a href="index.html#semidiscret3" class="eqref">3</a> and <a href="index.html#fonctionnel theo quench1" class="eqref">12</a>, the above equalities give </p>
<div class="displaymath" id="a0000000015">
  \begin{flalign} & \tfrac {d}{dt}J_{i}-\delta ^{2}J_{i}\leq pf_{i}(1-U_{i})^{-p-1}J_{i},\quad i=1,\ldots ,I,\\ & \tfrac {d}{dt}J_{0}-\delta ^{2}J_{0}\leq (\tfrac {2q}{h}U^{-q-1}_{0}+pf_{0}(1-U_{0})^{-p-1})J_{0}. \end{flalign}
</div>
<p>We observe from <a href="index.html#condition theo quench1" class="eqref">11</a>–?? that </p>
<div class="displaymath" id="a0000000016">
  \begin{flalign*} & J_{i}(0)=\delta ^{2}\varphi _{i}+f_{i}(1-\varphi _{i})^{-p}+\epsilon \varphi ^{-q}_{i}\leq 0,\quad i=1,\ldots ,I,\\ & J_{0}(0)= \delta ^{2}\varphi _{0}+f_{0}(1-\varphi _{0})^{-p}+(\epsilon -\tfrac {2}{h})\varphi ^{-q}_{0}\leq 0. \end{flalign*}
</div>
<p>We deduce from <a href="index.html#principe de maximum">lemma 2</a> that \(J_{h}(t)\leq 0,\:  \text{for}\:  t\in (0,T_{q}^{h})\), which implies that </p>
<div class="displaymath" id="a0000000017">
  \begin{flalign} & \tfrac {dU_{i}}{dt}+\epsilon U^{-q}_{i}\leq 0\quad \text{for}\quad t\in (0,T_{q}^{h}),\:  i=0,\ldots ,I. \end{flalign}
</div>
<p>By the above inequality we obtain the following form \(U^{q}_{i}dU_{i}\leq -\epsilon dt\: \text{for}\:  t\in (0,T_{q}^{h})\) and \(i=0,\ldots ,I\). Integrating this inequality over \([0,T_{q}^{h})\), we obtain </p>
<div class="displaymath" id="estimation temps1">
  \begin{flalign} & T_{q}^{h}-t\leq \tfrac {U^{q+1}_{i}(t)}{\epsilon (q+1)}.\label{estimation temps1} \end{flalign}
</div>
<p>From <a href="index.html#propriete solution">lemma 5</a>, we have \(U_{0}(0)=\|  \varphi _{h}\| _{\inf }\) and taking \(t=0\) in <a href="index.html#estimation temps1" class="eqref">15</a> we get the desired result. <div class="proof_wrapper" id="a0000000018">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>The following result concerns the lower bound for the quenching rate. </p>
<p><div class="remark_thmwrapper " id="sous solution rate">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">7</span>
  </div>
  <div class="remark_thmcontent">
  <p> Using the inequality <a href="index.html#estimation temps1" class="eqref">15</a> we obtain \(T_{q}^{h}-t_{0}\leq \tfrac {U^{q+1}_{0}(t_{0})}{\epsilon (q+1)}\quad \text{for}\quad t_{0}\in (0,T_{q}^{h})\), which implies that \(U_{0}(t)\geq (\epsilon (q+1))^{\frac{1}{q+1}}(T_{q}^{h}-t)^{\frac{1}{q+1}}\quad \text{for}\quad t\in (0,T_{q}^{h})\). </p>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="rate">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">8</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Assume that <a href="index.html#condition theo quench1" class="eqref">11</a>–?? remains true. Then, near the quenching time \(T_{q}^{h}\), the solution \(U_{h}\) to problem <a href="index.html#semidiscret1" class="eqref">1</a>–?? has following quenching rate estimate </p>
<p>\( U_{0}(t)\sim (T_{q}^{h}-t)^{\frac{1}{(q+1)}}\), in the sense that there exist two positive constants \(C_{1}\), \(C_{2}\) such that </p>
<div class="displaymath" id="a0000000019">
  \begin{flalign*}  \displaystyle C_{1}(T_{q}^{h}-t)^{\frac{1}{q+1}}\leq U_{0}(t)\leq C_{2}(T_{q}^{h}-t)^{\frac{1}{q+1}},\quad \text{for}\quad t\in (0,T_{q}^{h}). \end{flalign*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000020">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> <a href="index.html#sous solution rate">remark 7</a> ensures the term of left hand side. </p>
<p>Let \(i_{0}\) be such that \(U_{i_{0}}(t)=\displaystyle \min _{0\leq i\leq I} U_{i}(t)\), \(t\in (0,T_{q}^{h})\). From <a href="index.html#propriete solution">lemma 5</a> we obtain </p>
<div class="displaymath" id="a0000000021">
  \begin{align*} & \delta ^{2}U_{i_{0}}(t)=\displaystyle \tfrac {U_{i_{0}+1}(t)-2U_{i_{0}}(t)+U_{i_{0}-1}(t)}{h^{2}}\geq 0,\quad i_{0}=1,\ldots ,I-1,\\ & \delta ^{2}U_{0}(t)=\displaystyle \tfrac {2U_{1}(t)-2U_{0}(t)}{h^{2}}\geq 0,\\ & \delta ^{2}U_{I}(t)=\displaystyle \tfrac {2U_{I-1}(t)-2U_{I}(t)}{h^{2}}\geq 0. \end{align*}
</div>
<p> Which leads to \(\displaystyle \tfrac {dU_{0}}{dt}\geq -\tfrac {2}{h}U^{-q}_{0}\), integrating over \((t,T_{q}^{h})\) we have </p>
<div class="displaymath" id="a0000000022">
  \[  U_{0}\leq (\tfrac {2(q+1)}{h})^{\frac{1}{q+1}}(T_{q}^{h}-t)^{\frac{1}{q+1}}.  \]
</div>
<p>Hence, we have \(U_{0}\leq C_{2 }(T_{q}^{h}-t)^{\frac{1}{q+1}}\) and the proof is completed. <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> </p>
<p><div class="remark_thmwrapper " id="a0000000024">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
  </div>
  <div class="remark_thmcontent">
  <p>Let us point out that the quenching rate for the numerical scheme, </p>
<p>\((T_{q}^{h}-t)^{\frac{1}{q+1}}\), is different from the continuous one, \((T_{q}-t)^{\frac{1}{2(q+1)}}\) <span class="cite">
	[
	<a href="index.html#zhi2011" >22</a>
	]
</span>. </p>

  </div>
</div> </p>
<h1 id="sec4">4 Convergence of the semidiscrete Quenching time</h1>

<p>In this section, with suitable assumptions, we establish the convergence of the quenching time of the approximate semidiscrete solution to the quenching time of the theoretical solution.<br />The next theorem establishes that, for each fixed time interval \([0,T]\), \((T{\lt} T_{q})\) where \(u\) is defined, the solution of the semidiscrete problem <a href="index.html#semidiscret1" class="eqref">1</a>-?? approximates \(u\), as \(h\rightarrow 0\) . </p>
<p><div class="theorem_thmwrapper " id="convergence uh vers u">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">10</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Assume that the problem <a href="index.html#probleme" class="eqref">1</a> has a solution \(u\in C^{4,1}([0,1]\times [0,T])\) such that \(\displaystyle \inf _{t\in [0,T]}\| u(.,t)\| _{\infty }=\alpha {\gt}0\) and the initial condition \(\varphi _{h}\) at ?? verifies </p>
<div class="displaymath" id="uh moins u tend vers 0">
  \begin{align}  \| \varphi _{h}-u_{h}(0)\| _{\infty }=\circ (1), \quad as\quad h\rightarrow 0,\label{uh moins u tend vers 0} \end{align}
</div>
<p> where \(u_{h}(t)=(u(x_{0},t),\ldots ,u(x_{I},t))^{T}\), \(t\in [0,T]\). Then, for \(h\) small enough, the semidiscrete problem <a href="index.html#semidiscret1" class="eqref">1</a>–?? has a unique solution \(U_{h}\in C^{1}([0,T],\mathbb {R}^{I+1})\) such that </p>
<div class="displaymath" id="a0000000025">
  \begin{align*}  \displaystyle \max _{t\in [0,T]}(\| U_{h}(t)-u_{h}(t)\| _{\infty })=O( \| \varphi _{h}-u_{h}(0)\| _{\infty }+h^{2})\quad as\quad h\rightarrow 0. \end{align*}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000026">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> The semidiscrete problem (<a href="index.html#semidiscret1">1</a>)–(??) has for each \(h\), a unique solution \(U_{h}\in C^{1}([0,T_{q}^{h}),\mathbb {R}^{I+1})\). Let \(t(h)\) be the greatest value of \(t{\gt}0\) such that </p>
<div class="displaymath" id="majorationUh-uh">
  \begin{align}  \| U_{h}(t)-u_{h}(t)\| _{\infty }{\lt}\tfrac {\alpha }{2}\quad for\quad t\in (0,t(h)).\label{majorationUh-uh} \end{align}
</div>
<p> The relation (<a href="index.html#uh moins u tend vers 0">16</a>) implies \(t(h){\gt}0\) for \(h\) small enough. </p>
<p>Let \(t^{\star }(h)=\min \{ t(h),T\} \), using the triangle inequality we obtain </p>
<p>\(\| U_{h}(t)\| _{\inf }\geq \| u(.,t)\| _{\inf }-\| U_{h}(t)-u_{h}(t)\| _{\infty }\) for \(t\in (0,t^{\star }(h))\), which implies that </p>
<div class="displaymath" id="minoration Uh">
  \begin{align}  \| U_{h}(t)\| _{\inf }\geq \tfrac {\alpha }{2} \quad for \quad t\in (0,t^{\star }(h)). \label{minoration Uh} \end{align}
</div>
<p> Let \(e_{h}(t)=U_{h}(t)-u_{h}(t)\) be the error discretization and the vector \(z_{h}(t)\) defined by \(z_{i}(t)=e^{(K+1)(t-x^{2}+2x)}(\| \varphi _{h}-u_{h}(0)\| _{\infty }+Lh^{2}),\quad 0\leq i\leq I,:\  t\in (0,t^{\star }(h))\). Using the <a href="index.html#principe de maximum">lemma 2</a> we can prove that \(z_{i}(t){\gt}\lvert e_{i}(t)\rvert \) for \(t\in (0,t^{\star }(h)),\:  i=0,\ldots ,I\), which implies that </p>
<div class="displaymath" id="a0000000027">
  \begin{align*}  \| U_{h}(t)-u_{h}(t)\| _{\infty }\leq e^{(K+1)(t-x^{2}+2x)}(\| \varphi _{h}-u_{h}(0)\| _{\infty }+Lh^{2}),\quad t\in (0,t^{\star }(h)). \end{align*}
</div>
<p> Suppose that \(T{\gt}t(h)\), from (<a href="index.html#majorationUh-uh">17</a>) we have \(\frac{\alpha }{2}=\| U_{h}(t(h))-u_{h}(t(h))\| _{\infty }\leq e^{(K+1)(T-x^{2}+2x)}(\| \varphi _{h}-u_{h}(0)\| _{\infty }+Lh^{2})\). Since the term on the right hand side of the above inequality goes to zero as \(h\) goes to zero, we deduce that \(\tfrac {\alpha }{2}\leq 0\), which is impossible. <div class="proof_wrapper" id="a0000000028">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="a0000000029">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">11</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(T_{q}\) be the quenching time of the solution \(u\) of <a href="index.html#probleme" class="eqref">1</a> such that </p>
<p>\(u\in C^{4,1}([0,1]\times [0,T_{q}))\). We assume that the initial data at ?? satisfies </p>
<p>\(\| \varphi _{h}-u_{h}(0)\| _{\infty }=o(1),\:  \text{as} \quad h\rightarrow 0\). Then under the hypothesis of <a href="index.html#quenches">theorem 6</a>, the solution \(U_{h}\) of the problem <a href="index.html#semidiscret1" class="eqref">1</a>–?? quenches in finite time \(T^{h}_{q}\) and we have </p>
<div class="displaymath" id="a0000000030">
  \begin{align*}  \displaystyle \lim _{h\rightarrow 0}T^{h}_{q}=T_{q}. \end{align*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000031">
  <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 \(\eta {\gt}0\), there exists a positive constant \(\alpha \) such that </p>
<div class="displaymath" id="inegalie convergence thq">
  \begin{align}  \tfrac {1}{\epsilon }\frac{s^{q+1}}{q+1}\leq \tfrac {\eta }{2}\quad for\quad s\in [0,\alpha ].\label{inegalie convergence thq} \end{align}
</div>
<p>Since \(u\) quenches in a finite time \(T_{q}\), there exists \(T_{0}\in (T_{q}-\tfrac {\eta }{2},T_{q})\) such that<br />\(0{\lt}\| u(.,t)\| _{\inf }\leq \tfrac {\alpha }{2}\) for \(t\in [T_{0},T_{q})\). From <a href="index.html#convergence uh vers u">theorem 10</a>, the problem <a href="index.html#semidiscret1" class="eqref">1</a>-?? has a solution \(U_{h}(t)\) such that \(\| U_{h}(t)-u_{h}(t)\| _{\infty }\leq \tfrac {\alpha }{2}\) for \(t\in [0,T_{1}]\), where \(T_{1}=\tfrac {T_{0}+T_{q}}{2}\), which gives \(\| U_{h}(T_{1})-u_{h}(T_{1})\| _{\infty }\leq \tfrac {\alpha }{2}\). Using the triangle inequality, we get \(\| U_{h}(T_{1})\| _{\inf }\leq \| U_{h}(T_{1})-u_{h}(T_{1})\| _{\infty }+\| u_{h}(T_{1})\| _{\inf }\leq \alpha \). From <a href="index.html#quenches">theorem 6</a>, \(U_{h}\) quenches in a finite time \(T_{q}^{h}\). We deduce from <a href="index.html#sous solution rate">remark 7</a> and <a href="index.html#inegalie convergence thq" class="eqref">19</a> that </p>
<div class="displaymath" id="a0000000032">
  \begin{equation*}  \lvert T_{q}^{h}-T_{q}\rvert \leq \lvert T_{q}^{h}-T_{1}\rvert + \lvert T_{1}-T_{q}\rvert \leq \frac{1}{\epsilon }\tfrac {\| U_{h}(T_{1})\| ^{q+1}_{\inf }}{q+1}+\frac{\eta }{2}\leq \eta . \end{equation*}
</div>
<p> <div class="proof_wrapper" id="a0000000033">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<h1 id="sec5">5 Numerical experiments</h1>
<p>  In the section, we present numerical approximations to the quenching time of problem (<a href="index.html#probleme">1</a>) in the case where \(u_{0}(x) = -x^{\varepsilon }+\varepsilon x+\varepsilon ^{-1/q}\) with \(0{\lt}p\leq 1\), \(0{\lt}q\leq \frac{1}{2}\), \(\varepsilon =\tfrac {101}{100}\) and \(f(x)=\tfrac {1}{1000}\). We also discuss the quenching sets. To do this, we transform the semidiscrete scheme (<a href="index.html#semidiscret1">1</a>)–(??) into the following semidiscrete equations. Set us <small class="small">\(V_{i}(t)=\displaystyle \tfrac {1}{U_{i}(t)}\)</small>, we obtain <small class="small"><div class="displaymath" id="transformat1">
  \begin{align}  \tfrac {d}{dt}V_{i}(t)=\tfrac {1}{h^{2}}\Big(2V_{i}(t)-\tfrac {V^{2}_{i}(t)}{V_{i-1}(t)}-\tfrac {V^{2}_{i}(t)}{V_{i+1}(t)}\Big)-f_{i}V^{p+2}_{i}(t)(V_{i}(t)-1)^{-p},i=0,\ldots ,I-1,\label{transformat1}\\ \tfrac {d}{dt}V_{0}(t)=\tfrac {2}{h^{2}}\Big(V_{0}(t)-\tfrac {V^{2}_{0}(t)}{V_{1}(t)}\Big)+\tfrac {2}{h}V^{2+q}_{0}(t)-f_{0}V^{2+p}_{0}(t)(V_{0}(t)-1)^{-p},\label{transformat2}\\ \tfrac {d}{dt}V_{I}(t)=\tfrac {2}{h^{2}}\Big(V_{I}(t)-\tfrac {V^{2}_{I}(t)}{V_{I-1}(t)}\Big)-f_{I}V^{p+2}_{I}(t)(V_{I}(t)-1)^{-p}\label{transformat3},\\ V_{i}(0)=(\varphi _{i})^{-1},\quad i=0,\ldots ,I.\label{transformat4} \end{align}
</div> </small> Using the method presented by Hirota and Ozawa <span class="cite">
	[
	<a href="index.html#exploration1" >12</a>
	]
</span>, we transform the semidiscrete scheme (<a href="index.html#transformat1">20</a>)–(<a href="index.html#transformat4">23</a>) into a tractable form by the arc length transformation technique (see <span class="cite">
	[
	<a href="index.html#jlike" >21</a>
	]
</span>, [S. Moriguti, C. Okuno, R. Suekane, M. Iri, K. Takeuchi, Ikiteiru Suugaku - Suuri Kougaku no Hatten (in Japanese), Baifukan, Tokyo, 1979.]) like this: </p>
<div class="equation" id="arc">
<p>
  <div class="equation_content">
    \begin{equation} \label{arc} \begin{cases}  \dfrac {d}{d\eta }\begin{pmatrix}  t 

\\ V_0 

\\ \vdots 

\\ V_I 

\end{pmatrix} = \dfrac {1}{\sqrt{1+\sum _{i=0}^I f_i^2}}\begin{pmatrix}  1 

\\ f_0 

\\ \vdots 

\\ f_I 

\end{pmatrix}, \quad 0<\eta <\infty ,\\ t(0)=0, \quad V_i(0)= (\varphi _{i})^{-1}, \quad i=0,\ldots ,I, \end{cases} \end{equation}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p> where </p>
<div class="displaymath" id="a0000000034">
  \begin{align*} & f_{0}=\tfrac {2}{h^{2}}\Big(V_{0}(t)-\tfrac {V^{2}_{0}(t)}{V_{1}(t)}\Big)+\tfrac {2}{h}V^{2+q}_{0}(t)-f_{0}V^{2+p}_{0}(t)(V_{0}(t)-1)^{-p},\\ & f_{i}=\tfrac {1}{h^{2}}\Big(2V_{i}(t)-\tfrac {V^{2}_{i}(t)}{V_{i-1}(t)}-\tfrac {V^{2}_{i}(t)}{V_{i+1}(t)}\Big)-f_{i}V^{p+2}_{i}(t)(V_{i}(t)-1)^{-p},\quad i=1,\ldots ,I-1,\\ & f_{I}=\tfrac {2}{h^{2}}\Big(V_{I}(t)-\tfrac {V^{2}_{I}(t)}{V_{I-1}(t)}\Big)-f_{I}V^{p+2}_{I}(t)(V_{I}(t)-1)^{-p},\\ & V_{i}(0)=(\varphi _{i})^{-1},\quad i=0,\ldots ,I. \end{align*}
</div>
<p> “\(\eta \)” is the arc length and we have \(d\eta ^{2}=dt^{2}+\displaystyle \sum ^{I}_{i=0}dV^{2}_{i}\). Note that in the transformation below the variables \(t\) and \(V_{i}\) are functions of \(\eta \) such that \(\displaystyle \lim _{\eta \rightarrow \infty }t(\eta )=T^{h}\quad and \quad \displaystyle \lim _{\eta \rightarrow \infty }\| V_{h}(\eta )\| _{\infty }=\infty \). Now we introduce \(\{ \eta _{j}\} \) which is the sequence of the arc lengths and we apply an ODE solver to (<a href="index.html#arc">24</a>) for each value of \(\eta _{j}\) in order to generate a linearly convergent sequence to the blow-up time. The resulting sequence is accelerated by the Aitken \(\Delta ^{2}\) method <span class="cite">
	[
	<a href="index.html#aitken" >3</a>
	]
</span>. We use the DOP54 <span class="cite">
	[
	<a href="index.html#dop54" >11</a>
	]
</span> as the adaptive code for the integration of the ODEs. It has been written by Hairer <i class="it">et al.</i> <span class="cite">
	[
	<a href="index.html#dop54" >11</a>
	]
</span> based on explicit Runge-Kutta method of order \((4)5\) due to Dormand and Prince <span class="cite">
	[
	<a href="index.html#dorman" >8</a>
	]
</span>. Let us define the sequence \(\eta _{j}\) by \(\eta _{j}=2^{12}.2^{j}\) \((j=0,\ldots ,10)\), and the parameters in the DOP54 are \(InitialStep=0\) and \(AbsTol=RelTol=1.e-15\). The parameters \(AbsTol\) and \(RelTol\) specify the tolerances of the absolute and relative errors, respectively, and \(InitialStep\) is used to choose the manner in which the errors are controlled. In the following tables, in rows, we present the numerical quenching times \(T^{h}\) of problem (<a href="index.html#semidiscret1">1</a>)-(??), the Step and the orders of the approximations corresponding to meshes of \(16\), \(32\), \(64\), \(128\), \(256\) and \(512\). The order \((s)\) of the method is computed from </p>
<div class="displaymath" id="a0000000035">
  \begin{equation*}  \frac{\log ((T^{4h}-T^{2h})/(T^{2h}-T^{h}))}{\log (2)}. \end{equation*}
</div>
<div class="table"  id="a0000000036">
   <div class="minipage" style="width: 160.0pt"> <small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(T^{h}\) </p>

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>16</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.320236408</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>6304</p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>32</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.318644976</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>12122</p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>64</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.318150680</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>23261</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.69</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.318001392</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>44785</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.73</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>256</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.317957379</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>87402</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.76</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>512</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>0.317944663</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>178187</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>1.79 </p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Semidiscrete solution for \(p=1/4\), \(q=1/4\).</span> 
</figcaption> </div> <div class="minipage" style="width: 160.0pt"> <small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(T^{h}\) </p>

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>16</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.200847014</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>4490</p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>32</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.198905750</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>8498</p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>64</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.198304689</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>16154</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.69</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.198124110</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>30905</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.73</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>256</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.198071185</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>59982</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.77</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>512</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>0.198055981</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>120646</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>1.80</p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Semidiscrete solution for \(p=1/4\), \(q=1/2\).</span> 
</figcaption> </div> 
</div>
<div class="table"  id="a0000000037">
   <div class="minipage" style="width: 160.0pt"> <small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(T^{h}\) </p>

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>16</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.320475664</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>6306</p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>32</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.318883760</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>12127</p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>64</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.318389339</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>23270</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.69</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.318240017</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>44802</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.73</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>256</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.318195997</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>87437</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.76</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>512</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>0.318183278</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>178267</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>1.79</p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">Semidiscrete solution for \(p=1/2\), \(q=1/4\).</span> 
</figcaption> </div> <div class="minipage" style="width: 160.0pt"> <small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(T^{h}\) </p>

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>16</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.201026888</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>4492</p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>32</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.199084652</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 8502</p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>64</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.198483334</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>16161</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.69</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>128</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.198302688</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>30920</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.74</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>256</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p>0.198249747</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 60011</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>1.77</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>512</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>0.198234539</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>120710</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>1.80</p>

    </td>
  </tr>
</table> </small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4</span> 
  <span class="caption_text">Semidiscrete solution for \(p=1/2\), \(q=1/2\).</span> 
</figcaption> </div> 
</div>
<p><div class="remark_thmwrapper " id="a0000000038">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">12</span>
  </div>
  <div class="remark_thmcontent">
  <p>From the tables, we can see the convergence of \(T^{h}\) to the quenching time of the solution of <a href="index.html#probleme" class="eqref">1</a>, since the rate of convergence is near \(2\), which is just the accuracy of the difference approximation in space. The tables of our numerical results show that there is a relationship between the quenching time and the flow on the boundary on the one hand and the absorption in the interior of the domain on the other hand. Indeed, when the absorption in the interior of the domain is constant \((p=1/4)\) and that the flow on the boundary increases from \(1/4\) to \(1/2\), the quenching time decreases from \(0.318\) to \(0.198\) whereas when the flow on the boundary is constant \((q=1/4)\) and that the absorption in the interior of the domain increases from \(1/4\) to \(1/2\), the quenching time remains substantially the same at \(0.318\). The absorption in the interior of the domain has in fact no essential effect upon the quenching behavior of problem <a href="index.html#probleme" class="eqref">1</a>, whereas the flow on the boundary leads to the quenching, which is in agreement with the theoretical results <span class="cite">
	[
	<a href="index.html#zhi2007" >23</a>
	, 
	<a href="index.html#zhi2011" >22</a>
	]
</span>. </p>

  </div>
</div> </p>
<p>For other illustrations, in what follows, we present several graphs to illustrate our analysis. In Figures <a href="index.html#fig1:">1</a>–<a href="index.html#fig4:">4</a>, we have used the case where \(I=64\) and \(p=1/4\). We can appreciate in <a href="index.html#fig1:"> 1</a> and <a href="index.html#fig2:"> 2</a> that the discrete solution quenches in a finite time at the first node, which is well known in a theoretical point of view <span class="cite">
	[
	<a href="index.html#zhi2007" >23</a>
	, 
	<a href="index.html#zhi2011" >22</a>
	]
</span>. In <a href="index.html#fig3:"> 3</a> and <a href="index.html#fig4:"> 4</a> we see that the approximation of \(u(x,T)\) increases and gives the value zero at the first node. The time \(T\) represents the quenching time of the solution \(u\). </p>
<figure >
  <p> <div class="minipage" style="width: 160.0pt"> <div class="centered"><img src="images/img-0001.jpg" alt="\includegraphics[height=4in, width=5.0776in]{SOL2.jpg}" style="height:4in; width:5.0776in" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Evolution of the semidiscrete solution, \(q=1/4\).</span> 
</figcaption> </div></div><div class="minipage" style="width: 160.0pt"> <img src="images/img-0002.jpg" alt="\includegraphics[height=4in, width=5.0776in]{SOL1.jpg}" style="height:4in; width:5.0776in" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Evolution of the semidiscrete solution, \(q=1/2\).</span> 
</figcaption> <div class="centered"><div class="centered"></div></div></div> </p>

</figure>
<figure >
  <p> <div class="minipage" style="width: 160.0pt"> <div class="centered"><img src="images/img-0003.jpg" alt="\includegraphics[height=4in, width=5.0776in]{NODE1.jpg}" style="height:4in; width:5.0776in" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">Profile of the approximation of \(u(x,T)\), \(q=1/4\).</span> 
</figcaption> </div></div><div class="minipage" style="width: 160.0pt"> <img src="images/img-0004.jpg" alt="\includegraphics[height=4in, width=5.0776in]{NODE2.jpg}" style="height:4in; width:5.0776in" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">4</span> 
  <span class="caption_text">Profile of the approximation of \(u(x,T)\), \(q=1/2\).</span> 
</figcaption> <div class="centered"><div class="centered"></div></div></div> </p>

</figure>
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</dd>
</dl>

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