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<title>Quenching for semidiscretizations of a semilinear heat equation with potential and general nonlinearity: Quenching for semidiscretizations of a semilinear heat equation with potential and general nonlinearity</title>
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<h1>Quenching for semidiscretizations of a semilinear heat equation with potential and general nonlinearity</h1>
<p class="authors">
<span class="author">Halima Nachid\(^\ast \)</span>
</p>
<p class="date">December 23, 2010</p>
</div>
<p>\(^\ast \)Université d’Abobo-Adjamé, UFR-SFA, Département de Mathématiques et Informatiques, 02 BP 801 Abidjan 02, (Côte d’Ivoire), International University of Grand-Bassam Route de Bonoua Grand-Bassam BP 564 Grand-Bassam, (Côte d’Ivoire) and Laboratoire de Modélisation Mathématique et de Calcul Économique LM2CE settat, (Maroc), e-mail: <span class="ttfamily">nachidhalima@yahoo.fr</span>. </p>

<div class="abstract"><p>This paper concerns the study of the numerical approximation for the following boundary value problem </p>
<div class="displaymath" id="a0000000002">
  \begin{equation*} \begin{cases}  u_t(x,t)-u_{xx}(x,t)= -a(x)f(u(x,t)), &  0{\lt}x{\lt}1,\;  t{\gt}0, \\ u_x(0,t)= 0, u(1,t)=0, &  t{\gt}0, \\ u(x,0)= u_{0}(x){\gt}0, &  0\leq x \leq 1, \\ \end{cases}\end{equation*}
</div>
<p> where \(f:(0,\infty )\rightarrow (0,\infty )\) is a \(C^{1}\) convex, nondecreasing function, <br />\(\lim _{s\rightarrow 0^{+}}f(s)=\infty ,\) \(\int _0^{\alpha }\tfrac {{\rm d}s}{f(s)}{\lt}\infty \) for any positive real \(\alpha .\) The initial datum \(u_{0}\in C^{2}([0,1])\), \(u'_{0}(0)=0\) and \(u'_{0}(1)=0\). The potential \(a\in C^{1}((0,1))\), \(a(x){\gt}0,\) \(x \in (0,1),\) \(a'(0)=0,\) \(a'(1)=0.\) We find some conditions under which the solution of a semidiscrete form of the above problem quenches in a finite time and estimate its semidiscrete quenching time. We also prove that the semidiscrete quenching time converges to the real one when the mesh size goes to zero. A similar study has been also investigated taking a discrete form of the above problem. Finally, we give some numerical experiments to illustrate our analysis. </p>
<p><b class="bf">MSC.</b> 35B40, 35B50, 35K60, 65M06. </p>
<p><b class="bf">Keywords.</b> Semidiscretizations, semilinear parabolic equation, quenching, numerical quenching time, convergence, fulldiscretizations. </p>
</div>
<h1 id="a0000000003">1 Introduction</h1>
<p>Consider the following boundary value problem </p>
<div class="displaymath" id="a0000000004">
  \begin{align}  u_t(x,t)& = u_{xx}(x,t)-a(x)f(u(x,t)),\quad (x,t)\in (0,1)\times (0,T),\\ u_x(0,t)& =0,\quad u(1,t)=1,\quad t\in [0,T)\\ u(x,0)& =u_{0}(x){\gt}0, \quad x\in [0,1], \end{align}
</div>
<p> where \(f:(0,\infty )\rightarrow (0,\infty )\) is a \(C^{1}\) nondecreasing function, \(\int _0^{\alpha }\tfrac {{\rm d}s}{f(s)}{\lt}\infty ,\) for any positive real \(\alpha \). \(\lim _{s \rightarrow 0^{+}}f(0)=\infty ,\) \(a \in C^{1}([0,1]),\) \(a(x){\gt}0,\) \(x \in (0,1),\) \(a'(0)=0,\) \(a'(1)=0,\) \(u'_{0}(1)=0\) and \(u'_{0}(0)=0\). The initial data \(u_{0}\in C^{2}([0,1])\), \(u_{0}(x){\gt}0,\) \(x \in (0,1),\) \(u_{0}(x){\lt}1,\) \(x \in (0,1),\) </p>
<div class="displaymath" id="a0000000005">
  \begin{align} & u”_{0}(x)-a(x)f(u_{0}(x)){\lt}0,\  x\in (0,1), \\ & u’_{0}(x){\gt}0,\   x\in (0,1),\\ & u’_{0}(0)=0,\   u_{0}(1)=1. \end{align}
</div>
<p> Here \([0,T]\) is the maximal time interval on which the solution \(u\) of (1)–(3) exists. The time \(T \) may be finite or infinite. When \(T\) is infinite, then we say that the solution \(u\) of (1)–(3) exists globally. When \(T\) is finite, then the solution \(u\) of (1)–(3) develops a singularity in a finite time, namely, \(\lim _{t\rightarrow T}u_{min}(t)=0,\) where \(u_{min}(t)={\rm min}_{0\leq x \leq 1}u(x,t).\) In this last case,we say that the solution \(u\) of (1)–(3) quenches in a finite time and the time \(T\) is called the quenching time of the solution \(u.\) </p>
<p>The theoretical study of solutions for semilinear parabolic equations which quench in a finite time has been the subject of investigations of many authors (see [2], [4]–[7], [11], [12], [16] and the references cited therein). Local in time existence of a classical solution has been proved and this solution is unique. In addition, it is shown that if the initial data at (3) satisfies \(u''_{0}(x)-a(x)u^{-p}_{0}(x)\leq -Au^{-p}_{0}(x)\) in \([0,1]\) where \(A\in (0,1]\) and \(p{\gt}0\), then the classical solution \(u\) of (1)–(3) quenches in a finite time \(T\) and we have the following estimates </p>
<div class="displaymath" id="a0000000006">
  \begin{align*}  \tfrac {\min _{0\leq x\leq 1}(u_{0}(x))^{p+1}}{p+1}& \leq T\leq \tfrac {\min _{0\leq x\leq 1}(u_{0}(x))^{p+1}}{A(p+1)},\\ (A(p+1))^{\tfrac {1}{p+1}}(T-t)^{\tfrac {1}{p+1}}& \leq u_{\min }(t)\leq ((B(p+1))^{\tfrac {1}{p+1}}(T-t)^{\tfrac {1}{p+1}}, \end{align*}
</div>
<p> for \(t\in (0,T)\), (see, for instance [4]–[6]). </p>
<p>In this paper, we are interested in the numerical study of the phenomenon of quenching. Our aim is to build a semidiscrete scheme where solution obeys the property of the continuous one. In order to facilitate our discussion, let us notice that the first condition in (4) allows the solution \(u\) to attain its minimum at the point \(x=0,\) and the second one permits the solution \(u\) to decrease with respect to the second variable. The hypotheses (5) are compatibility condition which ensure the regularity of the solution \(u.\) </p>
<p>This paper is organized as follows. In the next section, we give some results about the discrete maximum principle. In the third section, under some conditions, we prove that the solution of a semidiscrete form of (1)–(3) quenches in a finite time and estimate its semidiscrete quenching time. In the fourth section, we prove the convergence of the semidiscrete quenching time. In the fifth section, we study the results of sections 3 and 4 taking a discrete form of (1)–(3). Finally, in the last section, we give some numerical results to illustrate our analysis. </p>
<h1 id="a0000000007">2 Properties of a semidiscrete problem</h1>
<p>We start our study by the construction of a semidiscrete scheme as follows. Let \(I\) be a positive integer, and let \(h=\tfrac {1}{I}\). Define the grid \(x_{i}=ih\), \(0\leq i \leq I,\) and approximate the solution \(u\) of the problem (1)–(3) by the solution \(U_{h}(t)=(U _{0}(t),U_{1}(t),\ldots ,U_{I}(t))^{T}\) of the following semidiscrete equations </p>
<div class="displaymath" id="a0000000008">
  \begin{align} & \tfrac {dU_i(t)}{dt}-\delta ^2U_{i}(t)=-\beta _{i}f(U_{i}(t)),\  0\leq i\leq I,\  t\in (0,T^{h}_{q}),\\ & U_I(t)=1,\  t\in (0,T^{h}_{q}),\\ & U_i(0)=\varphi _{i}{\gt}0,\  0\leq i\leq I, \end{align}
</div>
<p> where </p>
<div class="displaymath" id="a0000000009">
  \begin{eqnarray*}  \delta ^2U_i(t)=\tfrac {U_{i+1}(t)-2U_i(t)+U_{i-1}(t)}{h^2},& 1\leq i\leq I-1, \end{eqnarray*}
</div>
<p> where \(\beta _i {\gt}0,\) \(\varphi _{i}{\gt}0\) </p>
<div class="displaymath" id="a0000000010">
  \begin{eqnarray}  \delta ^2U_0(t)=\tfrac {2U_{1}(t)-2U_0(t)}{h^2},\quad \delta ^2U_I(t)=\tfrac {2U_{I-1}(t)-2U_{I}(t)}{h^2}. \end{eqnarray}
</div>
<p> \(\beta _i \) and \(\varphi _{i}\) are approximations of \(a(x_{i})\) and \(u_{0}(x_{i})\), respectively. There is another reason which has motivated our choice, namely we want to know the behavior of the quenching time when one perturbs slightly either the potential or the initial datum. </p>
<p>Here \((0,T^{h}_{q})\) is the maximal time interval on which \(\| U_{h}(t)\| _{\inf }{\gt}0\) where </p>
<div class="displaymath" id="a0000000011">
  \[ \| U_{h}(t)\| _{\inf }=\min _{0\leq i\leq I}U_{i}(t). \]
</div>
<p> When the time \(T^{h}_{q}\) is finite, we say that the solution \(U_{h}(t)\) of (7)–(9) quenches in a finite time and the time \(T^{h}_{q}\) is called the quenching time of the solution \(U_{h}(t)\). The following lemma is a semidiscrete form of the maximum principle. <div class="lem_thmwrapper " id="a0000000012">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">2.1</span>
  </div>
  <div class="lem_thmcontent">
  <p><i class="it">Let \(\alpha _{h}(t)\in C^{0}([0,T),\mathbb {R}^{I+1})\) and let \(V_{h}\in C^{1}([0,T),\mathbb {R}^{I+1})\) be such that <div class="displaymath" id="a0000000013">
  \begin{align} & \tfrac {dV_i(t)}{dt}-\delta ^2V_i(t)+\alpha _i(t)V_i(t)\geq 0,\  0\leq i\leq I,\  t\in (0,T),\\ & V_i(0)\geq 0, 0\leq i\leq I. \end{align}
</div> Then \(V_i(t)\geq 0\), \(0\leq i\leq I\), \(t\in (0,T)\).</i> </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000014">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Let \(T_{0}\) be any quantity satisfying the inequality \(T_{0}{\lt}T\) and define the vector \(Z_{h}(t)=e^{\lambda t}V_{h}(t)\) where \(\lambda \) is such that </p>
<div class="displaymath" id="a0000000015">
  \[ \alpha _{i}(t)-\lambda {\gt}0\quad \mbox{for}\quad 0\leq i\leq I,\quad t\in [0,T_{0}]. \]
</div>
<p> Set \(m=\min _{0\leq t\leq T_{0}}\| Z_{h}(t)\| _{\inf }\). Since \(Z_{h}(t)\) is a continuous vector on the compact \([0,T_0]\), there exist \(i_0\in \{ 0,...,I\} \) and \(t_0\in [0,T_0]\) such that \(m=Z_{i_0}(t_0)\). We observe that </p>
<div class="displaymath" id="a0000000016">
  \begin{align}  \tfrac {dZ_{i_0}(t_0)}{dt}=\lim _{k\rightarrow 0}\tfrac {Z_{i_0}(t_0)-Z_{i_0}(t_0-k)}{k}\leq 0, \end{align}
</div>
<div class="displaymath" id="a0000000017">
  \begin{eqnarray}  \delta ^2Z_{i_0}(t_0)\geq 0. \end{eqnarray}
</div>
<p> From (11), we obtain the following inequality </p>
<div class="displaymath" id="a0000000018">
  \begin{eqnarray}  \tfrac {dZ_{i_{0}}(t_0)}{dt}-\delta ^{2}Z_{i_{0}}(t_0)+(\alpha _{i_{0}}(t_0)-\lambda )Z_{i_{0}}(t_0)\geq 0. \end{eqnarray}
</div>
<p> We deduce from (13)–(15) that \((\alpha _{i_{0}}(t_{0})-\lambda )Z_{i_{0}}(t_{0})\geq 0\), which implies that \(Z_{i_{0}}(t_{0})\geq 0\). Therefore, \(V_{h}(t)\geq 0\) for \(t\in [0,T_{0}]\) and the proof is complete.</p>
<p>Another form of the maximum principle for semidiscrete equations is the following comparison lemma. </p>
<p><div class="lem_thmwrapper " id="a0000000019">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">2.2</span>
  </div>
  <div class="lem_thmcontent">
  <p><i class="it">Let \(f \in C^{0}(\mathbb {R}\times \mathbb {R},\mathbb {R}).\) If \(V_{h}\), \(W_{h} \in C^{1}([0,T),\mathbb {R}^{I+1})\) are such that <div class="displaymath" id="a0000000020">
  \begin{align*}  \tfrac {dV_{i}(t)}{dt}\! -\! \delta ^2V_{i}(t)\! +\! f(V_{i}(t),t)\! & {\lt}\! \tfrac {dW_{i}(t)}{dt}\! -\! \delta ^2W_i(t)\! +\! f(W_{i}(t),t)\! ,\  0\leq i\leq I,t\in (0,T),\\ V_{i}(0)& {\lt} W_{i}(0),\  0\leq i\leq I, \end{align*}
</div> then \(V_{i}(t){\lt}W_{i}(t),\) &#8195;\(0\leq i\leq I\),&#8195;t \(\in (0,T).\)</i> </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000021">
  <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 \(Z_{h}(t)=W_{h}(t)-V_{h}(t)\) and let \(t_{0}\) be the first \(t\in (0,T)\) such that \(Z_{h}(t){\gt}0\) for \(t \in [0,t_{0})\) but \(Z_{i_{0}}(t_{0})=0\) for a certain \(i_{0}\in \{ 0,...,I\} \). We see that </p>
<div class="displaymath" id="a0000000022">
  \begin{align*}  \tfrac {dZ_{i_{0}}(t_{0})}{dt}& =\lim _{k\rightarrow 0}\tfrac {Z_{i_0}(t_0)-Z_{i_0}(t_0-k)}{k}\leq 0,\\ \delta ^2Z_{i_0}(t_0)& \geq 0. \end{align*}
</div>
<p> Therefore, we have </p>
<div class="displaymath" id="a0000000023">
  \[ \tfrac {dZ_{i_0}(t_0)}{dt}-\delta ^2Z_{i_0}(t_0)+f(W_{i_{0}}(t_{0}),t_{0})-f(V_{i_{0}}(t_{0}),t_{0})\leq 0, \]
</div>
<p> which contradicts the first strict inequality of the lemma and this ends the proof. </p>
<h1 id="a0000000024">3 Quenching problem</h1>
<p>In this section, under some assumptions, we show that the solution \(U_{h}\) of (7)–(9) quenches in a finite time and estimate its semidiscrete quenching time. We need the following result about the operator \(\delta ^{2}\). </p>
<p><div class="lem_thmwrapper " id="a0000000025">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">3.1</span>
  </div>
  <div class="lem_thmcontent">
  <p><i class="it">Let \(U_{h}\in \mathbb {R}^{I+1}\) be such that \(U_{h}{\gt}0\). Then, we have <div class="displaymath" id="a0000000026">
  \[ \delta ^{2}(f(U))_i\geq f'(U_{i})\delta ^{2}U_{i}, \quad 0\leq i \leq I. \]
</div></i> </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000027">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Applying Taylor’s expansion, we find that </p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  \delta ^{2}f(U)_{i}=f’(U_{i})\delta ^{2} U_{i}+\tfrac {(U_{i+1}-U_i)^2}{h^{2}}f”(\theta _{i}) +\tfrac {(U_{i-1}-U_i)^2}{h^{2}}f”(\eta _{i}),\quad 0\leq i\leq I, \end{equation*}
</div>
<p> where \(\theta _{i}\) is an intermediate value between \(U_{i}\) and \(U_{i+1}\), \(\eta _{i}\) the one between \(U_{i-1}\) and \(U_{i}\), \(U_{-1}=U_{1}\), \(U_{I+1}=U_{I-1}\), \(\eta _{0}=\theta _{0}\), \(\eta _{I}=\theta _{I}\). Use the fact that \(U_{h}{\gt}0\) to complete the rest of the proof. </p>
<p>The statement of the result about solutions which quench in a finite time is the following. <div class="theo_thmwrapper " id="a0000000029">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">3.1</span>
  </div>
  <div class="theo_thmcontent">
  <p>Let \(U_h\) be the solution of <span class="rm">(7)–(9)</span> and assume that there exists a positive constant \(A\in (0,1]\) and the initial data at <span class="rm">(9)</span> satisfies </p>
<div class="displaymath" id="a0000000030">
  \begin{eqnarray}  \delta ^{2}\varphi _{i}-\beta _{i}f(\varphi _{i})\leq -A f(\varphi _{i}),&  0\leq i\leq I. \end{eqnarray}
</div>
<p> Then, the solution \(U_h\) quenches in a finite time \(T_{q}^{h}\) and we have the following estimate </p>
<div class="displaymath" id="a0000000031">
  \[ T_{q}^{h}\leq \tfrac {1}{A}\int _{0}^{\| \varphi _{h}\| _{inf}}\tfrac {{\rm d}\sigma }{f(\sigma )}. \]
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000032">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Since \((0,T_{q}^{h})\) is the maximal time interval on which \(\| U_{h}(t)\| _{\inf }{\gt}0\), our aim is 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="a0000000033">
  \begin{eqnarray*}  J_i(t)=\tfrac {{\rm d}U_i(t)}{{\rm d}t}+A f(U_i(t)),& 0\leq i\leq I. \end{eqnarray*}
</div>
<p> A straightforward calculation gives </p>
<div class="displaymath" id="a0000000034">
  \begin{eqnarray*}  \tfrac {{\rm d}J_i}{{\rm d}t}-\delta ^2J_i=\tfrac {{\rm d}}{{\rm d}t}(\tfrac {{\rm d}U_i}{{\rm d}t}-\delta ^2U_i)-A f’(U_i(t))\tfrac {{\rm d}U_{i}}{{\rm d}t}-A\delta ^{2}(f(U_i(t)))_{i},&  0\leq i\leq I. \end{eqnarray*}
</div>
<p> From Lemma 3.1, we have \(\delta ^{2}(f(U))_i\geq f'(U_{i})\delta ^2U_{i},\) \( 0\leq i\leq I,\) which implies that </p>
<div class="displaymath" id="a0000000035">
  \begin{eqnarray*}  \tfrac {{\rm d}J_i}{{\rm d}t}-\delta ^2J_i\leq \tfrac {{\rm d}}{{\rm d}t}(\tfrac {{\rm d}U_i}{{\rm d}t}-\delta ^2U_i)+A f’(U_{i})(\tfrac {{\rm d}U_i}{{\rm d}t}-\delta ^2U_i),& 0\leq i\leq I. \end{eqnarray*}
</div>
<p> Using (7), we arrive at </p>
<div class="displaymath" id="a0000000036">
  \begin{eqnarray*}  \tfrac {{\rm d}J_{i}}{{\rm d}t}-\delta ^2J_{i}\leq -\beta _{i}f’(U_{i})J_{i},&  0\leq i\leq I,\quad t\in (0,T_{q}^{h}). \end{eqnarray*}
</div>
<p> From (16), we observe that \(J_{h}(0)\leq 0\). We deduce from Lemma 2.1 that \(J_{h}(t)\leq 0\) for \(t\in (0,T_{q}^{h})\), which implies that </p>
<div class="displaymath" id="a0000000037">
  \begin{eqnarray}  \tfrac {{\rm d}U_i(t)}{{\rm d}t}\leq -A f(U_{i}(t)),\quad 0\leq i\leq I,\quad t\in (0,T_{q}^{h}). \end{eqnarray}
</div>
<p> These estimates may be rewritten in the following form \(\tfrac {{\rm d}U_{i}}{f(U_{i})}\leq -A{\rm d}t\), \(0\leq i\leq I\). Integrating the above inequalities over the interval \((t,T_{q}^{h})\), we get </p>
<div class="displaymath" id="a0000000038">
  \begin{eqnarray}  T_{q}^{h}-t\leq \tfrac {1}{A}\int _{0}^{U_i(t)}\tfrac {{\rm d}\sigma }{f(\sigma )}, & 0\leq i\leq I. \end{eqnarray}
</div>
<p> Using the fact that \(\| \varphi _{h}\| _{\inf }=U_{i_{0}}(0)\) for a certain \(i_{0}\in \{ 0,...,I\} \) and taking \(t=0\) in (18), we obtain the desired result. </p>
<p><div class="remark_thmwrapper " id="a0000000039">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.2</span>
  </div>
  <div class="remark_thmcontent">
  <p>The inequalities (18) imply that </p>
<div class="displaymath" id="a0000000040">
  \begin{eqnarray*}  T_{q}^{h}-t_{0}\leq \tfrac {1}{A}\int _{0}^{\| U_h(t_{0})\| _{\inf }}\tfrac {{\rm d}\sigma }{f(\sigma )}, \quad \text{for}\  t_{0}\in (0,T_{q}^{h}), \end{eqnarray*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000041">
  \begin{eqnarray*}  \| U_{h}(t)\| _{\inf }\geq H(A (T_{q}^{h}-t)),\quad \text{for} t\in (0,T_{q}^{h}), \end{eqnarray*}
</div>
<p> where \(H(s)\) is the inverse of the function \(F(s)=\int _{0}^{s}\tfrac {{\rm d}\sigma }{f(\sigma )}.\)<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000042">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.3</span>
  </div>
  <div class="remark_thmcontent">
  <p>Let \(U_{h}\) be the solution of (7)–(9). Then, we have </p>
<div class="displaymath" id="a0000000043">
  \[ T_{q}^{h}\geq \tfrac {1}{\| B_h\| _\infty }\int _{0}^{\| \varphi _{h}\| }\tfrac {{\rm d}\sigma }{f(\sigma )}, \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000044">
  \[ \| U_{h}(t)\| _{\inf }\leq \| B_h\| _\infty H(A (T_{q}^{h}-t)),\quad \text{for}\  t\in (0,T^{h}_{q}). \]
</div>
<p> To prove these estimates, we proceed as follows. Introduce the function \(v(t)\) defined as follows \(v(t)=\| U_{h}(t)\| _{\inf }\) for \(t\in [0,T^{h}_{q})\). Let \(t_{1},t_{2}\in [0,T^{h}_{q})\). Then, there exist \(i_{1},i_{2}\in \{ 0,...,I\} \) such that \(v(t_{1})=U_{i_{1}}(t_{1})\) and \(v(t_{2})=U_{i_{2}}(t_{2})\). We observe that </p>
<div class="displaymath" id="a0000000045">
  \[ v(t_{2})-v(t_{1})\geq U_{i_{2}}(t_{2})-U_{i_{2}}(t_{1})=(t_{2}-t_{1})\tfrac {{\rm d}U_{i_{2}}(t_{2})}{{\rm d}t}+o(t_{2}-t_{1}), \]
</div>
<div class="displaymath" id="a0000000046">
  \[ v(t_{2})-v(t_{1})\leq U_{i_{1}}(t_{2})-U_{i_{1}}(t_{1})=(t_{2}-t_{1})\tfrac {{\rm d}U_{i_{1}}(t_{1})}{{\rm d}t}+o(t_{2}-t_{1}), \]
</div>
<p> which implies that \(v(t)\) is Lipschitz continuous. Further, if \(t_{2}{\gt}t_{1}\), then </p>
<div class="displaymath" id="a0000000047">
  \[ \tfrac {v(t_{2})-v(t_{1})}{t_{2}-t_{1}}\geq \tfrac {{\rm d}U_{i_{2}}(t_{2})}{{\rm d}t}+o(1)=\delta ^{2}U_{i_{2}}(t_{2})-\beta _{i_{2}}f(U_{i_{2}}(t_{2}))+o(1). \]
</div>
<p> Obviously, \(\delta ^{2}U_{i_{2}}(t_{2})\geq 0\). Letting \(t_{1}\rightarrow t_{2}\), and using the fact that \(\beta _{i_{2}}\leq \| B_h\| _\infty ,\) we obtain \(\tfrac {{\rm d}v(t)}{{\rm d}t}\geq -\| B_h\| _\infty f(v(t))\) for \(t\in (0,T^{h}_{q})\) or equivalently \(\tfrac {{\rm d}v}{f(v(t))}\geq -\| B_h\| _\infty dt\) for \(t\in (0,T^{h}_{q})\). Integrate the above inequality over \((t,T^{h}_{q})\) to obtain \(T^{h}_{q}-t\geq \tfrac {1}{\| B_h\| _\infty }\int _{0}^{v(t)}\tfrac {{\rm d}\sigma }{f(\sigma )}\). Since \(v(t)=\| U_{h}(t)\| _{\inf }\), we arrive at \(T^{h}_{q}-t\geq \tfrac {1}{\| B_h\| _\infty }\int _{0}^{\| U_{h}(t)\| _{\inf }}\tfrac {{\rm d}\sigma }{f(\sigma )}\) and the second estimate follows. To obtain the first one, it suffices to replace t by \(0\) in the above inequality and use the fact that \(\| \varphi _{h}\| _{\inf }=\| U_{h}(0)\| _{\inf }\).<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000048">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.4</span>
  </div>
  <div class="remark_thmcontent">
  <p>If \(\varphi _{i}=\alpha \), \(0\leq i\leq I\), where \(\alpha \) is a positive constant, then one may take \(A=1\). It may imply that the potential equals to 1. In this case, </p>
<div class="displaymath" id="a0000000049">
  \[ T^{h}_{q}=\tfrac {\alpha ^{p+1}}{p+1}\quad {\rm and} \quad \| U_{h}(t)\| _{\inf }=(p+1)^{\tfrac {1}{p+1}}(T^{h}_{q}-t)^{\tfrac {1}{p+1}}\quad {\rm for} \quad t\in (0,T^{h}_{q}). \]
</div>
<p><span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000050">4 Convergence of the semidiscrete quenching time</h1>
<p>In this section, under some assumptions, we show that the solution of the semidiscrete problem quenches in a finite time and its semidiscrete quenching time converges to the real one when the mesh size goes to zero. We denote </p>
<div class="displaymath" id="a0000000051">
  \[ u_{h}(t)=(u(x_{0},t),...,u(x_{I},t))^{T}\quad \mbox{and}\quad \| U_{h}(t)\| _{\infty }=\max _{0 \leq i\leq I}|U_{i}(t)|. \]
</div>
<p> In order to obtain the convergence of the semidiscrete quenching time, we firstly prove the following theorem about the convergence of the semidiscrete scheme. </p>
<p><div class="theo_thmwrapper " id="a0000000052">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">4.1</span>
  </div>
  <div class="theo_thmcontent">
  <p>Assume that the problem <span class="rm">(1)–(3)</span> has a solution <br />\(u\in C^{4,1}([0,1]\times [0,T])\) such that \(\min _{t\in [0,T]}u_{\min }(t)=\varrho {\gt}0.\) Suppose that the potential at <span class="rm">(7)</span> and the initial data at <span class="rm">(9)</span> satisfy </p>
<div class="displaymath" id="a0000000053">
  \begin{eqnarray}  \| \varphi _{h}-u_{h}(0)\| _{\infty }=o(1)\quad as\quad h\rightarrow 0, \end{eqnarray}
</div>
<div class="displaymath" id="a0000000054">
  \begin{eqnarray}  \| \beta _{h}-a_{h}\| _{\infty }=o(1)\quad as\quad h\rightarrow 0. \end{eqnarray}
</div>
<p> Then, for h sufficiently small, the problem <span class="rm">(7)–(9)</span> has a unique solution \( U_{h}\in C^{1}([0,T],\mathbb {R}^{I+1})\) such that the following relation holds </p>
<div class="displaymath" id="a0000000055">
  \begin{eqnarray*}  \max _{0\leq t \leq T}\| \beta _{h}-a_{h}\| _{\infty }= 0(\| \varphi _{h}-u_{h}(0)\| _{\infty }+h^{2})\quad as\quad h\rightarrow 0. \end{eqnarray*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000056">
  <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 \(K{\gt}0\) and \(L{\gt}0\) be such that </p>
<div class="displaymath" id="a0000000057">
  \begin{eqnarray}  \tfrac {\| u_{xxxx}\| _{\infty }}{12}\leq K, f(\tfrac {\rho }{2})\leq K \quad \mbox{and}\quad -(\| a_{h}\| _\infty +1)f’(\tfrac {\rho }{2})\leq L. \end{eqnarray}
</div>
<p> The problem (7)–(9) has for each \(h\), a unique solution \(U_{h}\in C^{1}([0,T_{q}^{h}),\mathbb {R}^{I+1})\). Let \(t(h)\leq \min \{ T,T_{q}^{h}\} \) be the greatest value of \(t{\gt}0\) such that </p>
<div class="displaymath" id="a0000000058">
  \begin{eqnarray}  \| U_{h}(t)-u_{h}(t)\| _{\infty }{\lt}\tfrac {\varrho }{2}&  \mbox{for} & t\in (0,t(h)). \end{eqnarray}
</div>
<p> The relation (19) implies that \(t(h){\gt}0\) for \(h\) sufficiently small. By the triangle inequality, we obtain </p>
<div class="displaymath" id="a0000000059">
  \begin{eqnarray*}  \| U_{h}(t)\| _{\inf }\geq \| u_{h}(t)\| _{\inf }-\| U_{h}(t)-u_{h}(t)\| _{\infty }&  \mbox{for} & t\in (0,t(h)), \end{eqnarray*}
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000060">
  \begin{eqnarray}  \| U_{h}(t)\| _{\inf }\geq \varrho -\tfrac {\varrho }{2}=\tfrac {\varrho }{2} &  \mbox{for} & t\in (0,t(h)). \end{eqnarray}
</div>
<p> Since \(u\in C^{4,1}\), taking the derivative in \(x\) on both sides of (1) and due to the fact that \(u_{x}\), \(u_{xt}\) vanish at \(x=0\) and \(x=1\), we observe that \(u_{xxx}\) also vanishes at \(x=0\) and \(x=1\). Applying Taylor’s expansion, we discover that </p>
<div class="displaymath" id="a0000000061">
  \[ u_{xx}(x_{i},t)=\delta ^{2}u(x_{i},t)-\tfrac {h^{2}}{12}u_{xxxx}(\widetilde{x}_{i},t),\quad 0\leq i\leq I,\  t\in (0,t(h)). \]
</div>
<p> To establish the above equalities for \(i=0\) and \(i=I\), we have used the fact that \(u_{x}\) and \(u_{xxx}\) vanish at \(x=0\) and \(x=1\). A direct calculation yields </p>
<div class="displaymath" id="a0000000062">
  \begin{eqnarray*}  u(x_{i},t)-\delta ^{2}u(x_{i},t)=-\beta _{i}f(u(x_{i},t))-\tfrac {h^{2}}{12}u_{xxxx}({\widetilde{x}}_{i},t)+(\beta _{i}-a(x_{i}))f(u(x_{i},t)), \end{eqnarray*}
</div>
<p> for \(1\leq i\leq I-1\). Let \(e_{h}(t)=U_{h}(t)-u_{h}(t)\) be the error of discretization. From the mean value theorem, we have </p>
<div class="displaymath" id="a0000000063">
  \begin{eqnarray*}  \tfrac {{\rm d}e_{i}(t)}{{\rm d}t}-\delta ^{2}e_{i}(t)=-\beta _{i}f’(\theta _{i}e_{i}+\tfrac {h^{2}}{12}u_{xxxx}({\widetilde{x}}_{i},t)- (\beta _{i}-a(x_{i}))f(u(x_{i},t)), \end{eqnarray*}
</div>
<p> for \(0\leq i\leq I,\  t\in (0,t(h))\), where \(\theta _{i}\) is an intermediate value between \(U_{i}(t)\) and \(u(x_{i},t)\). Using (21), (22), we arrive at </p>
<div class="displaymath" id="a0000000064">
  \begin{eqnarray*}  \tfrac {{\rm d}e_{i}(t)}{{\rm d}t}-\delta ^{2}e_{i}(t)\leq L|e_{i}(t)|+Kh^{2}+ K\| \beta _{h}-a_{h}\| _{\infty }, & 0\leq i \leq I, t\in (0,t(h)). \end{eqnarray*}
</div>
<p> Introduce the vector \(z_{h}(t)\) defined as follows </p>
<div class="displaymath" id="a0000000065">
  \begin{eqnarray*}  z_{i}(t)={\rm e}^{(L+1)t}(\| \varphi _{h}-u_{h}(0)\| _{\infty }+Kh^{2}+K \| \beta _{h}-a_{h}\| _{\infty }),\  0\leq i \leq I,\; t\in (0,t(h)). \end{eqnarray*}
</div>
<p> A straightforward computation reveals that </p>
<div class="displaymath" id="a0000000066">
  \begin{eqnarray*}  \tfrac {{\rm d}z_{i}}{{\rm d}t}-\delta ^{2}z_{i}{\gt}L|z_{i}|+Kh^{2}+ K \| \beta _{h}-a_{h}\| _{\infty },& 0\leq i \leq I,\quad t\in (0,t(h)), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000067">
  \begin{eqnarray*}  z_{i}(0){\gt}e_{i}(0),& 0\leq i \leq I. \end{eqnarray*}
</div>
<p> It follows from Comparison Lemma 2.2 that </p>
<div class="displaymath" id="a0000000068">
  \begin{eqnarray*}  z_{i}(t){\gt}e_{i}(t)&  \mbox{for} & t\in (0,t(h)), \  0\leq i\leq I. \end{eqnarray*}
</div>
<p> In the same way, we also prove that </p>
<div class="displaymath" id="a0000000069">
  \begin{eqnarray*}  z_{i}(t){\gt}-e_{i}(t)&  \mbox{for} &  t\in (0,t(h)), \  0\leq i\leq I, \end{eqnarray*}
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000070">
  \begin{eqnarray*}  \| U_{h}(t)-u_{h}(t)\| _{\infty }\leq {\rm e}^{(L+1)t}( \| \varphi _{h}-u_{h}(0)\| _{\infty }+Kh^{2}+ K \| \beta _{h}-a_{h}\| _{\infty }), \end{eqnarray*}
</div>
<p> for \(t\in (0,t(h))\). Let us show that \(t(h)=\min \{ T,T^{h}_{q}\} \). Suppose that \(t(h){\lt}\min \{ T,T^{h}_{q}\} \). From (22), we obtain </p>
<div class="displaymath" id="a0000000071">
  \begin{eqnarray*}  \tfrac {\varrho }{2}\! \leq \! \| U_{h}(t(h))-u_{h}(t(h))\| _{\infty }\! \leq \!  {\rm e}^{(L+1)T}(\| \varphi _{h}-u_{h}(0)\| _{\infty }+Kh^{2}+ K \| \beta _{h}-a_{h}\| _{\infty }). \end{eqnarray*}
</div>
<p> Let us notice that both last formulas for \(t(h)\) are valid for sufficiently small \(h\). Since the term on the right hand side of the above inequality goes to zero as \(h\) goes to zero, we deduce that \(\tfrac {\varrho }{2}\leq 0\), which is impossible. Consequently \(t(h)=\min \{ T,T^{h}_{q}\} \). </p>
<p>Now, let us show that \(t(h)=T\). Suppose that \(t(h)=T^{h}_{q}{\lt}T\). Reasoning as above, we prove that we have a contradiction and the proof is complete. </p>
<p>Now, we are in a position to prove the main theorem of this section. </p>
<p><div class="theo_thmwrapper " id="a0000000072">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">4.2</span>
  </div>
  <div class="theo_thmcontent">
  <p>Suppose that the problem <span class="rm">(1)–(3)</span> has a solution \(u\) which quenches in a finite time \(T_{q}\) such that \(u\in C^{4,1}([0,1]\times [0,T_{q})).\) Assume that the potential at <span class="rm">(7)</span> and the initial data at <span class="rm">(9)</span> satisfy the conditions <span class="rm">(19)</span> and <span class="rm">(20)</span>, respectively. Under the hypothesis of Theorem <span class="rm">3.1</span>, the problem <span class="rm">(7)–(9)</span> has a solution \(U_{h}\) which quenches in a finite time \(T_{q}^{h}\) and we have </p>
<div class="displaymath" id="a0000000073">
  \[ \lim _{h\rightarrow 0}T_{q}^{h}=T_{q}. \]
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000074">
  <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 \(0{\lt}\varepsilon {\lt}T_{q}/2\). There exists \(\varrho \in (0,1)\) such that </p>
<div class="displaymath" id="a0000000075">
  \begin{eqnarray}  \tfrac {1}{A}\int _{0}^{\varrho }\tfrac {{\rm d}\sigma }{f(\sigma )}\leq \tfrac {\varepsilon }{2}. \end{eqnarray}
</div>
<p> Since \(u\) quenches in a finite time \(T_{q}\), there exist \(h_{0}(\varepsilon ){\gt}0\) and a time \(T_{0}\in (T_{q}-\tfrac {\varepsilon }{2},T_{q})\) such that \(0{\lt}u_{\min }(t){\lt}\tfrac {\varrho }{2}\) \(\quad \) for \(t\in [T_{0},T_{q})\), \(h\leq h_{0}(\varepsilon )\). It is not hard to see that \(u_{\min }(t){\gt}0\) for \(t\in [0,T_{0}]\), \(h\leq h_{0}(\varepsilon )\). From Theorem 4.1, the problem (7)–(9) has a solution \(U_{h}(t)\) and we get \(\| U_{h}(t)-u_{h}(t)\| _{\infty }\leq \tfrac {\varrho }{2}\) for \(t\in [0,T_{0}]\), \(h\leq h_{0}(\varepsilon )\), which implies that \(\| U_{h}(T_{0})-u_{h}(T_{0})\| _{\infty }\leq \tfrac {\varrho }{2}\) for \(h\leq h_{0}(\varepsilon )\). Applying the triangle inequality, we find that </p>
<div class="displaymath" id="a0000000076">
  \begin{eqnarray*}  \| U_{h}(T_{0})\| _{\inf }\leq \| U_{h}(T_{0})-u_{h}(T_{0})\| _{\infty }+\| u_{h}(T_{0})\| _{\inf }\leq \tfrac {\varrho }{2}+\tfrac {\varrho }{2}=\varrho \quad \mbox{for}\quad h\leq h_{0}(\varepsilon ). \end{eqnarray*}
</div>
<p> From Theorem 3.1, \(U_{h}(t)\) quenches at the time \(T_{q}^{h}\). We deduce from Remark 3.1 and (22) that for \(h\leq h_{0}(\varepsilon )\), </p>
<div class="displaymath" id="a0000000077">
  \begin{eqnarray*}  |T_{q}^{h}-T_{q}|\leq |T_{q}^{h}-T_{0}|+|T_{0}-T_{q}|\leq \tfrac {1}{A}\int _{0}^{\| U_{h}(T_{0})\| _{\inf }}\tfrac {{\rm d}\sigma }{f(\sigma )}+\tfrac {\varepsilon }{2}\leq \varepsilon , \end{eqnarray*}
</div>
<p> which leads us to the desired result. </p>
<h1 id="a0000000078">5 Full discretizations</h1>
<p>In this section, we study the phenomenon of quenching using a full discrete explicit scheme of (1)–(3). Approximate the solution \(u(x,t)\) of the problem (1)–(3) by the solution \(U^{{(n)}}_{h}=(U_{0}^{(n)},U_{1}^{(n)},\ldots ,U_{I}^{(n)})^{T}\) of the following explicit scheme </p>
<div class="displaymath" id="a0000000079">
  \begin{eqnarray}  \delta _tU_i^{(n)}=\delta ^2U_i^{(n)}-\beta _{i}f(U_i^{(n)}),&  0\leq i\leq I, \end{eqnarray}
</div>
<div class="displaymath" id="a0000000080">
  \begin{eqnarray}  U_i^{(0)}=\varphi _{i}{\gt}0,&  0\leq i\leq I, \end{eqnarray}
</div>
<p> where \(n\geq 0\), </p>
<div class="displaymath" id="a0000000081">
  \begin{align*}  \delta _{t}U_{i}^{(n)}& =\tfrac {U_{i}^{(n+1)}-U_{i}^{(n)}}{\Delta t_{n}}.\\ (\tfrac {f(s)}{s})’& =\tfrac {f’(s)s-f(s)}{s^2}\leq 0,\  \mbox{for} s{\gt}0. \end{align*}
</div>
<p> If \(U_h^{(n)}{\gt}0\), then \(-\tfrac {f(U_i^{(n)})}{U_i^{(n)}}\geq -\tfrac {f(\| U_h^{(n)}\| _{\inf })}{\| U_h^{(n)}\| _{\inf }}\), \(0\leq i\leq I\), and a straightforward computation reveals that </p>
<div class="displaymath" id="a0000000082">
  \begin{align*}  U_{0}^{(n+1)}& \geq \tfrac {2\Delta t_n}{h^2}U_{1}^{(n)}+(1-2\tfrac {\Delta t_n}{h^2}-\| \beta _{h}\| _\infty \Delta t_n\tfrac {f(\| U_h^{(n)}\| _{\inf })}{\| U_h^{(n)}\| _{\inf }})U_{0}^{(n)},\\ U_{i}^{(n+1)}& \geq \! \tfrac {\Delta t_n}{h^2}U_{i\! +\! 1}^{(n)}\! \! +\! \! \bigg(\! 1\! \! -\! \! 2\tfrac {\Delta t_n}{h^2}\! \! -\! \! \| \beta _{h}\| _\infty \Delta t_n\tfrac {f(\| U_h^{(n)}\| _{\inf })}{\| U_h^{(n)}\| _{\inf }}\! \bigg)U_i^{(n)}\! \! +\! \! \tfrac {\Delta t_n}{h^2}U_{i\! -\! 1}^{(n)}\! ,\;  1\! \! \leq i\! \! \leq I\! -\! 1,\\ U_{I}^{(n+1)}& \geq \tfrac {2\Delta t_n}{h^2}U_{I-1}^{(n)}+(1-2\tfrac {\Delta t_n}{h^2}-\| \beta _{h}\| _\infty \Delta t_n\tfrac {f(\| U_h^{(n)}\| _{\inf })}{\| U_h^{(n)}\| _{\inf }})U_{I}^{(n)}. \end{align*}
</div>
<p> In order to permit the discrete solution to reproduce the properties of the continuous one when the time \(t\) approaches the quenching time \(T_{q}\), we need to adapt the size of the time step so that we choose <br />\(\Delta t_n= \min \{ \tfrac {(1-\tau )h^2}{2},\tau \tfrac {f(\| U_h^{(n)}\| _{\inf })}{\| U_h^{(n)}\| _{\inf }}\} \) with \(0{\lt}\tau {\lt}1\). We observe that \(1-2\tfrac {\Delta t_n}{h^2}-\| \beta _{h}\| _\infty \Delta t_n \tfrac {f(\| U_h^{(n)}\| _{\inf })}{\| U_h^{(n)}\| _{\inf }}\geq 0\), which implies that \(U_{h}^{(n+1)}{\gt}0\). Thus, since by hypothesis \(U_{h}^{(0)}=\varphi _{h}{\gt}0\), if we take \(\Delta t_n\) as defined above, then using a recursion argument, we see that the positivity of the discrete solution is guaranteed. Here, \(\tau \) is a parameter which will be chosen later to allow the discrete solution \(U_{h}^{(n)}\) to satisfy certain properties useful to get the convergence of the numerical quenching time defined below. </p>
<p>If necessary, we may take \(\Delta t_n= \min \{ \tfrac {(1-\tau )h^2}{K},\tau \tfrac {f(\| U_h^{(n)}\| _{\inf })}{\| U_h^{(n)}\| _{\inf }}\} \) with \(K{\gt}2\) because in this case, the positivity of the discrete solution is also guaranteed.<br />The following lemma is a discrete form of the maximum principle. <div class="lem_thmwrapper " id="a0000000083">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">5.1</span>
  </div>
  <div class="lem_thmcontent">
  <p><i class="it">Let \(a_{h}^{(n)}\) and \(V_h^{(n)}\) be two sequences such that \(a_{h}^{(n)}\) is bounded and <div class="displaymath" id="a0000000084">
  \begin{eqnarray}  \delta _tV_i^{(n)}-\delta ^2V_i^{(n)}+a_{i}^{(n)}V_i^{(n)}\geq 0,&  0\leq i\leq I,&  n\geq 0, \end{eqnarray}
</div> <div class="displaymath" id="a0000000085">
  \begin{eqnarray}  V_i^{(0)}\geq 0,&  0\leq i\leq I. \end{eqnarray}
</div> Then \(V_i^{(n)}\geq 0\) for \(n\geq 0\),  \(0\leq i\leq I\) if \(\Delta t_n\leq \tfrac {h^2}{2+\| a_{h}^{(n)}\| _{\infty }h^2}\).</i> </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000086">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>If \(V_{h}^{(n)}\geq 0\), then a routine computation yields </p>
<div class="displaymath" id="a0000000087">
  \begin{eqnarray*}  V_{0}^{(n+1)}\geq \tfrac {2\Delta t_n}{h^2}V_{1}^{(n)}+(1-2\tfrac {\Delta t_n}{h^2}-\Delta t_n\| a_{h}^{(n)}\| _{\infty })V_{0}^{(n)}, \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000088">
  \begin{eqnarray*}  V_{i}^{(n+1)}\geq \tfrac {\Delta t_n}{h^2}V_{i+1}^{(n)}+(1-2\tfrac {\Delta t_n}{h^2}-\Delta t_n\| a_{h}^{(n)}\| _{\infty })V_i^{(n)}+\tfrac {\Delta t_n}{h^2}V_{i-1}^{(n)},\;  1\leq i\leq I-1, \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000089">
  \begin{eqnarray*}  V_{I}^{(n+1)}\geq \tfrac {2\Delta t_n}{h^2}V_{I-1}^{(n)}+(1-2\tfrac {\Delta t_n}{h^2}-\Delta t_n\| a_{h}^{(n)}\| _{\infty })V_{I}^{(n)}. \end{eqnarray*}
</div>
<p> Since \(\Delta t_n\leq \tfrac {h^2}{2+\| a_{h}^{(n)}\| _{\infty }h^2}\), we see that \(1-2\tfrac {\Delta t_n}{h^2}-\Delta t_n\| a_{h}^{(n)}\| _{\infty }\) is nonnegative. From (27), we deduce by induction that \(V_h^{(n)}\geq 0\) which ends the proof. </p>
<p>A direct consequence of the above result is the following comparison lemma. Its proof is straightforward. <div class="lem_thmwrapper " id="a0000000090">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">5.2</span>
  </div>
  <div class="lem_thmcontent">
  <p><i class="it">Let \(V_h^{(n)}\), \(W_h^{(n)}\) and \(a_{h}^{(n)}\) be three sequences such that \(a_{h}^{(n)}\) is bounded and <div class="displaymath" id="a0000000091">
  \begin{eqnarray*}  \delta _tV_i^{(n)}-\delta ^2V_i^{(n)}+a_{i}^{(n)}V_i^{(n)}\leq \delta _tW_i^{(n)}-\delta ^2W_i^{(n)}+a_i^{(n)}W_i^{(n)}, \end{eqnarray*}
</div> <div class="displaymath" id="a0000000092">
  \begin{eqnarray*}  0\leq i\leq I,\quad n\geq 0, \end{eqnarray*}
</div> <div class="displaymath" id="a0000000093">
  \begin{eqnarray*}  V_i^{(0)}\leq W_i^{(0)},&  0\leq i\leq I. \end{eqnarray*}
</div> Then \(V_i^{(n)}\leq W_i^{(n)}\) for \(n\geq 0\), \(0\leq i\leq I\) if \(\Delta t_n\leq \tfrac {h^2}{2+\| a_{h}^{(n)}\| _{\infty }h^2}\).</i> </p>

  </div>
</div> Now, let us give a property of the operator \(\delta _{t}\) stated in the following lemma. Its proof is quite similar to that of Lemma 3.1, so we omit it here. <div class="lem_thmwrapper " id="a0000000094">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">5.3</span>
  </div>
  <div class="lem_thmcontent">
  <p><i class="it">Let \(U^{(n)}\in \mathbb {R}\) be such that \(U^{(n)}{\gt}0\) for \(n\geq 0\). Then, we have <div class="displaymath" id="a0000000095">
  \[ \delta _{t}f(U^{(n)})\geq f'(U^{(n)})\delta _{t}U^{(n)},\quad n\geq 0. \]
</div></i> </p>

  </div>
</div> <div class="lem_thmwrapper " id="a0000000096">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">5.4</span>
  </div>
  <div class="lem_thmcontent">
  <p><i class="it">Let \(a\), \(b\) be two positive numbers such that \(b{\lt}1.\) Then the following estimate holds <div class="displaymath" id="a0000000097">
  \begin{eqnarray*}  \sum _{n=0}^{\infty }\tfrac {ab^{n}}{f(ab^{n})}\leq \tfrac {a}{f(a)}-\tfrac {1}{\ln (b)}\int _{0}^{a}\tfrac {{\rm d}\sigma }{f(\sigma )}. \end{eqnarray*}
</div></i> </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000098">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>We have \(\int _{0}^{\infty }\tfrac {ab^{x}}{f(ab^{x})}= \sum _{n=0}^{\infty }\int _{n}^{n+1}\tfrac {ab^{x}{\rm d}x}{f(ab^{x})}.\) We observe that \(ab^{x}\geq ab^{n+1}\) for \(n\leq x\leq n+1,\) which that \( \int _{n}^{n+1}\tfrac {ab^{x}{\rm d}x}{f(ab^{x})}\geq \tfrac {ab^{n+1}}{f(ab^{n+1})}.\) Consequently, we get </p>
<div class="equation" id="a0000000099">
<p>
  <div class="equation_content">
    \begin{equation}  \int _{n}^{n+1}\tfrac {ab^{x}{\rm d}x}{f(ab^{x})}\geq \sum _{n=0}^{\infty }\int _{n}^{n+1}\tfrac {ab^{x}{\rm d}x}{f(ab^{x})}=-\tfrac {a}{f(a)}+\\ \sum _{n=0}^{\infty }\tfrac {ab^{n}}{f(ab^{n})}. \end{equation}
  </div>
  <span class="equation_label">33</span>
</p>
</div>
<p> Use the fact that \(\int _{0}^{\infty }\tfrac {ab^{x}}{f(ab^{x})}=-\tfrac {1}{\ln (b)}\int _{0}^{a}\tfrac {{\rm d}\sigma }{f(\sigma )}\) to complete the rest of the proof. </p>
<p>The theorem below is the discrete version of Theorem 4.1. <div class="theo_thmwrapper " id="a0000000100">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">5.1</span>
  </div>
  <div class="theo_thmcontent">
  <p>Suppose that the problem <span class="rm">(1)–(3)</span> has a solution <br />\(u\in C^{4,2}([0,1]\times [0,T])\) such that \(\min _{t\in [0,T]}u_{\min }(t)=\rho {\gt}0\). Assume that the initial data at <span class="rm">(26)</span> satisfies the condition <span class="rm">(16)</span>. Then, the problem <span class="rm">(25)–(26)</span> has a solution \(U_{h}^{(n)}\) for h sufficiently small, \(0\leq n\leq J\) and the following relation holds </p>
<div class="displaymath" id="a0000000101">
  \begin{eqnarray*}  \max _{0\leq n\leq J}\| U_h^{(n)}-u_h(t_n)\| _{\infty }=O(\| \varphi _{h}-u_h(0)\| _{\infty }+\| a_h-\beta _h\| _{\infty }+h^2)&  as &  h\rightarrow 0, \end{eqnarray*}
</div>
<p> where \(J\) is any quantity satisfying the inequality \(\sum _{n=0}^{J-1}\Delta t_n\leq T\) and \(t_n=\sum _{j=0}^{n-1}\Delta t_j\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000102">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>For each \(h\), the problem (25)–(26) has a solution \(U_{h}^{(n)}\). Let \(N\leq J\) be the greatest value of \(n\) such that </p>
<div class="displaymath" id="a0000000103">
  \begin{eqnarray}  \| U_h^{(n)}-u_h(t_n)\| _{\infty }{\lt}\tfrac {\rho }{2}\quad \mbox{for} \quad n{\lt}N. \end{eqnarray}
</div>
<p> We know that \(N\geq 1\) because of (16). Applying the triangle inequality, we have </p>
<div class="displaymath" id="a0000000104">
  \begin{eqnarray}  \| U_h^{(n)}\| _{\inf }\geq \| u_h(t_n)\| _{\inf }-\| U_h^{(n)}-u_h(t_n)\| _{\infty }\geq \tfrac {\rho }{2} \mbox{for}\  n{\lt}N. \end{eqnarray}
</div>
<p> As in the proof of Theorem 4.1, using Taylor’s expansion, we find that for \(n{\lt}N\), \(0\leq i\leq I\), </p>
<div class="displaymath" id="a0000000105">
  \begin{align*} & \delta _tu(x_i,t_n)-\delta ^2u(x_i,t_n)+ \beta _{i}f(u(x_{i},t_n))+(a(x_{i}-\beta _{i}))f(u(x_{i},t_n))\\ & =-\tfrac {h^2}{12}u_{xxxx}(\widetilde{x}_{i},t_n)+\tfrac {\Delta t_n}{2}u_{tt}(x_i,\widetilde{t}_n). \end{align*}
</div>
<p> Let \(e_h^{(n)}=U_h^{(n)}-u_h(t_n)\) be the error of discretization. From the mean value theorem, we get for \(n{\lt}N\), \(0\leq i\leq I\), </p>
<div class="displaymath" id="a0000000106">
  \begin{align*}  \delta _{t}e_{i}^{(n)}-\delta ^{2}e_{i}^{(n)}& =-\beta _{i}f’(\xi _{i}^{(n)})e_{i}^{(n)}+\tfrac {h^2}{12}u_{xxxx}(\widetilde{x}_{i},t_n)-\tfrac {\Delta t_{n}}{2}u_{tt}(x_i,\widetilde{t}_n)\\ & \quad +(a(x_{i}-\beta _{i}))f(u(x_{i},t_n)), \end{align*}
</div>
<p> where \(\xi _{i}^{(n)}\) is an intermediate value between \(u(x_i,t_n)\) and \(U_{i}^{(n)}\). Since <br />\(u_{xxxx}(x,t)\), \(u_{tt}(x,t)\) are bounded,\(u(x,t)\geq \rho \) and \(\Delta t_{n}=O(h^{2})\), then there exists a positive constant \(M\) such that </p>
<div class="displaymath" id="a0000000107">
  \begin{eqnarray*}  \delta _te_i^{(n)}-\delta ^2e_i^{(n)}\leq -\beta _if’(\xi _i^{(n)})e_i^{(n)}+M\| a_h-\beta _h\| _{\infty }+Mh^2,\; 0 \leq i\leq I,\;  n{\lt}N. \end{eqnarray*}
</div>
<p> Set \(L=-(\| a_h\| \infty +1)f'(\tfrac {\rho }{2})\) and introduce the vector \(V_h^{(n)}\) defined as follows </p>
<div class="displaymath" id="a0000000108">
  \[ V_i^{(n)}=e^{(L+1)t_n}(\| \varphi _{h}-u_h(0)\| _{\infty }+Mh^2+M\| a_{h}-\beta _{h}\| _{\infty }),\quad 0\leq i\leq I,\;  n{\lt}N. \]
</div>
<p> A straightforward computation gives </p>
<div class="displaymath" id="a0000000109">
  \begin{align*}  \delta _tV_i^{(n)}-\delta ^2V_i^{(n)}& {\gt}-\beta _{i}f’(\xi _{i}^{(n)})V_i^{(n)}\! +\! Mh^2\! +\! M\| a_{h}-\beta _{h}\| _{\infty },\; 0 \leq i\leq I,\; n{\lt}N,\\ V_i^{(0)}& {\gt}e_{i}^{(0)},\  0\leq i \leq I. \end{align*}
</div>
<p> We observe from (29) that \(-\beta _{i}f'(\xi _{i}^{(n)})\) is bounded from above by \(L\). It follows from Comparison Lemma 5.2 that \(V_{h}^{(n)}\geq e_{h}^{(n)}\). By the same way, we also prove that \(V_{h}^{(n)}\geq -e_{h}^{(n)}\), which implies that </p>
<div class="displaymath" id="a0000000110">
  \begin{eqnarray*}  \| U_{h}^{(n)}-u_h(t_n)\| _{\infty }\leq {\rm e}^{(L+1)t_n}(\| \varphi _{h}-u_{h}(0)\| _{\infty }+Mh^2+M\| a_{h}-\beta _{h}\| _{\infty }),\; n{\lt}N. \end{eqnarray*}
</div>
<p> Let us show that \(N=J\). Suppose that \(N{\lt}J\). If we replace \(n\) by \(N\) in (29) and use (30), we find that </p>
<div class="displaymath" id="a0000000111">
  \[ \tfrac {\rho }{2}\leq \| U_h^{(N)}-u_h(t_N)\| _{\infty }\leq e^{(L+1)T}(\| \varphi _{h}-u_h(0)\| _{\infty }+Mh^2+M\| a_{h}-\beta _{h}\| _{\infty }). \]
</div>
<p> Since the term on the right hand side of the second inequality goes to zero as \(h\) goes to zero, we deduce that \(\tfrac {\rho }{2}\leq 0\), which is a contradiction and the proof is complete. </p>
<p>To handle the phenomenon of quenching for discrete equations, we need the following definition. <div class="defi_thmwrapper " id="a0000000112">
  <div class="defi_thmheading">
    <span class="defi_thmcaption">
    Definition
    </span>
    <span class="defi_thmlabel">5.1</span>
  </div>
  <div class="defi_thmcontent">
  <p>We say that the solution \(U_h^{(n)}\) of <span class="rm">(25)–(26)</span> quenches in a finite time if  \(\| U_h^{(n)}\| _{\inf }{\gt}0\) for \(n\geq 0\), but </p>
<div class="displaymath" id="a0000000113">
  \begin{eqnarray*}  \lim _{n\rightarrow +\infty }\| U_h^{(n)}\| _{\inf }=0\quad and\quad T_{h}^{\Delta t}=\lim _{n\rightarrow \infty }\sum _{i=0}^{n-1}\Delta t_i{\lt}\infty . \end{eqnarray*}
</div>
<p> The number \(T_{h}^{\Delta t}\) is called the numerical quenching time of  \(U_h^{(n)}\). </p>

  </div>
</div> The following theorem reveals that the discrete solution \(U_h^{(n)}\) of (25)-(26) quenches in a finite time under some hypotheses. <div class="theo_thmwrapper " id="a0000000114">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">5.2</span>
  </div>
  <div class="theo_thmcontent">
  <p>Let \(U_h^{(n)}\) be the solution of <span class="rm">(25)–(26)</span>. Suppose that there exists a constant \(A\in (0,1]\) such that the initial data at <span class="rm">(26)</span> satisfies </p>
<div class="displaymath" id="a0000000115">
  \begin{eqnarray}  \delta ^2\varphi _{i}-\beta _{i}f(\varphi _{i})\leq -A f(\varphi _{i}),\quad 0\leq i \leq I. \end{eqnarray}
</div>
<p> Then \(U_h^{(n)}\) is nonincreasing and quenches in a finite time \(T_{h}^{\Delta t}\) which satisfies the following estimate </p>
<div class="displaymath" id="a0000000116">
  \begin{eqnarray*}  T_{h}^{\Delta t}\leq \tfrac {\tau \| \varphi _h\| _{\inf }}{f(\| \varphi _h\| _{\inf })}-\tfrac {\tau }{\ln (1-\tau ’)}\int _{0}^{\| \varphi _h\| _{\inf }}\tfrac {{\rm d}\sigma }{f(\sigma )}, \end{eqnarray*}
</div>
<p> where \(\tau '=A\min \{ \tfrac {(1-\tau )h^{2}f(\| \varphi _{h}\| _{\inf })}{2\| \varphi _{h}\| _{\inf }} ,\tau \} \). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000117">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Introduce the vector \(J_h^{(n)}\) defined as follows </p>
<div class="displaymath" id="a0000000118">
  \[ J_i^{(n)}=\delta _tU_i^{(n)}+Af(U_i^{(n)}),\quad 0\leq i\leq I,\  n\geq 0. \]
</div>
<p> A straightforward computation yields for \(0\leq i\leq I\), \(n\geq 0\), </p>
<div class="displaymath" id="a0000000119">
  \begin{eqnarray*}  \delta _tJ_i^{(n)}-\delta ^2J_i^{(n)}=\delta _t\left(\delta _tU_i^{(n)}-\delta ^2U_i^{(n)}\right)+A\delta _tf(U_i^{(n)})-A\delta ^2f(U_i^{(n)}). \end{eqnarray*}
</div>
<p> Using (25), we arrive at </p>
<div class="displaymath" id="a0000000120">
  \begin{eqnarray*}  \delta _tJ_i^{(n)}-\delta ^2J_i^{(n)}=-(\beta _{i}-A)\delta _tf(U_i^{(n)})-A\delta ^2f(U_i^{(n)}),\quad 0\leq i\leq I,\  n\geq 0. \end{eqnarray*}
</div>
<p> It follows from Lemmas 5.3 and 3.1 that for \(0\leq i\leq I\), \(n\geq 0\), </p>
<div class="displaymath" id="a0000000121">
  \begin{eqnarray*}  \delta _tJ_i^{(n)}-\delta ^2J_i^{(n)}& \leq & -(\beta _{i}-A)f’(U_i^{(n)})\delta _tU_i^{(n)}-A f’(U_i^{(n)})\delta ^{2}U_i^{(n)}. \end{eqnarray*}
</div>
<p> We deduce from (25) that </p>
<div class="displaymath" id="a0000000122">
  \begin{eqnarray*}  \delta _tJ_i^{(n)}-\delta ^2J_i^{(n)}\leq -\beta _{i}f’(U_i^{(n)})J_i^{(n)},&  0\leq i\leq I,\  n\geq 0. \end{eqnarray*}
</div>
<p> Obviously, the inequalities (31) ensure that \(J_h^{(0)}\leq 0\). Applying Lemma 5.1, we get \(J_h^{(n)}\leq 0\) for \(n\geq 0\), which implies that </p>
<div class="displaymath" id="a0000000123">
  \begin{eqnarray}  U_i^{(n+1)}\leq U_i^{(n)}(1-A\Delta t_n\tfrac {f((U_i^{(n)}))}{U_i^{(n)}}, & \quad 0\leq i\leq I,\  n\geq 0. \end{eqnarray}
</div>
<p> These estimates reveal that the sequence \(U_h^{(n)}\) is nonincreasing. By induction, we obtain \(U_{h}^{(n)}\leq U_{h}^{(0)}=\varphi _{h}\). Thus, the following holds </p>
<div class="displaymath" id="a0000000124">
  \begin{eqnarray}  A\Delta t_n \tfrac {f(\| U_{h}^{(n)}\| _{\inf })}{\| U_{h}^{(n)}\| _{\inf }}\geq A\min \{ \tfrac {(1-\tau )h^{2}f(\| \varphi _{h}\| _{\inf })}{2\| \varphi _{h}\| _{\inf }},\tau \} =\tau ’. \end{eqnarray}
</div>
<p> Let \(i_{0}\) be such that \(\| U^{(n)}_{h}\| _{\inf }=U^{(n)}_{i_{0}}\). Replacing \(i\) by \(i_{0}\) in (38), we obtain </p>
<div class="displaymath" id="a0000000125">
  \begin{eqnarray}  \| U_h^{(n+1)}\| _{\inf }\leq \| U_h^{(n)}\| _{\inf }(1-\tau ’),\quad n\geq 0, \end{eqnarray}
</div>
<p> and by iteration, we arrive at </p>
<div class="displaymath" id="a0000000126">
  \begin{eqnarray}  \| U_h^{(n)}\| _{\inf }\leq \| U_h^{(0)}\| _{\inf }(1-\tau ’)^{n}=\| \varphi _{h}\| _{\inf }(1-\tau ’)^{n},\quad n\geq 0. \end{eqnarray}
</div>
<p> Since the term on the right hand side of the above equality goes to zero as \(n\) approaches infinity, we conclude that \(\| U_h^{(n)}\| _{\inf }\) tends to zero as \(n\) approaches infinity. Now, let us estimate the numerical quenching time. Due to (33) and the restriction \(\Delta t_n\leq \tfrac {\tau \| U_h^{(n)}\| _{\inf }}{f(\| U_h^{(n)}\| _{\inf })}\), it is not hard to see that </p>
<div class="displaymath" id="a0000000127">
  \begin{eqnarray*}  \Sigma _{n=0}^{+\infty }\Delta t_n\leq \tau \Sigma _{n=0}^{+\infty }\tfrac {\| \varphi _{h}\| _{\inf }(1-\tau ’)^{n}}{f(\| \varphi _{h}\| _{\inf }(1-\tau ’)^{n})}, \end{eqnarray*}
</div>
<p> because \(\tfrac {s}{f(s)}\) is nondecreasing for \(s{\gt}0.\) It follows from Lemma 5.4 that </p>
<div class="displaymath" id="a0000000128">
  \begin{eqnarray*}  \Sigma _{n=0}^{+\infty }\Delta t_n\leq \tfrac {\tau \| \varphi _h\| _{\inf }}{f(\| \varphi _h\| _{\inf })}-\tfrac {\tau }{\ln (1-\tau ’)}\int _{0}^{\| \varphi _h\| _{\inf }}\tfrac {{\rm d}\sigma }{f(\sigma )}. \end{eqnarray*}
</div>
<p> Use the fact that the quantity on the right hand side of the above inequality converges towards is finite to complete the rest of the proof. </p>
<p><div class="remark_thmwrapper " id="a0000000129">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">5.3</span>
  </div>
  <div class="remark_thmcontent">
  <p>From (35), we deduce by induction that </p>
<div class="displaymath" id="a0000000130">
  \[ \| U_h^{(n)}\| _{\inf }\leq \| U_{h}^{(q)}\| _{\inf }(1-\tau ')^{n-q}\  \mbox{for}\  n\geq q, \]
</div>
<p> and we see that </p>
<div class="displaymath" id="a0000000131">
  \begin{eqnarray*}  T_{h}^{\Delta t}-t_{q}=\Sigma _{n=q}^{+\infty }\Delta t_n\leq \tau \Sigma _{n=q}^{\infty }\tfrac {\| U_h^{(q)}\| _{\inf }(1-\tau ’)^{n-q}}{f(\| U_h^{(q)}\| _{\inf }(1-\tau ’)^{n-q})}, \end{eqnarray*}
</div>
<p> because \(\tfrac {s}{f(s)}\) is nondecreasing for \(s{\gt}0.\) It follows from Lemma 5.4 that </p>
<div class="displaymath" id="a0000000132">
  \begin{eqnarray*}  T_{h}^{\Delta t}-t_q\leq \tfrac {\tau \| U_h^{(q)}\| _{\inf }}{f(\| U_h^{(q)}\| _{\inf })}-\tfrac {\tau }{\ln (1-\tau ’)}\int _{0}^{\| U_h^{(q)}\| _{\inf }}\tfrac {{\rm d}\sigma }{f(\sigma )}. \end{eqnarray*}
</div>
<p> Since \(\tau ^{'}=A\min \{ \tfrac {(1-\tau )h^{2}f(\| \varphi _{h}\| _{\inf })}{2\| \varphi _{h}\| _{\inf }},\tau \} \), if we take \(\tau =h^{2}\), we get </p>
<div class="displaymath" id="a0000000133">
  \[ \tfrac {\tau ^{'}}{\tau }=A\min \{ \tfrac {(1-h^{2})h^{2}f(\| \varphi _{h}\| _{\inf })}{2\| \varphi _{h}\| _{\inf }},1\} \geq A\min \{ \tfrac {f(\| \varphi _{h}\| _{\inf })}{4\| \varphi _{h}\| _{\inf }},1\} . \]
</div>
<p> Therefore, there exist constants \(c_{0}\), \(c_{1}\) such that \(0\leq c_{0}\leq \tau /\tau ^{'}\leq c_{1}\) and \(\tfrac {-\tau }{\ln (1-\tau ^{'})}=O(1)\), for the choice \(\tau =h^{2}\).<span class="qed">â–¡</span></p>

  </div>
</div> In the sequel, we take \(\tau =h^{2}\). </p>
<p>Now, we are in a position to state the main theorem of this section. <div class="theo_thmwrapper " id="a0000000134">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">5.4</span>
  </div>
  <div class="theo_thmcontent">
  <p>Suppose that the problem <span class="rm">(1)–(3)</span> has a solution \(u\) which quenches in a finite time \(T_q\) and \(u\in C^{4,2}([0,1]\times [0,T_{q}))\). Assume that the initial data at <span class="rm">(25)</span> satisfies the condition <span class="rm">(16)</span>. Under the assumption of Theorem <span class="rm">5.2</span>, the problem <span class="rm">(25)–(26)</span> has a solution \(U_h^{(n)}\) which quenches in a finite time \(T_{h}^{\Delta t}\) and the following relation holds </p>
<div class="displaymath" id="a0000000135">
  \[ \lim _{h\rightarrow 0}T^{\Delta t}_{h}=T_q. \]
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000136">
  <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 Remark 5.1 that \(\tfrac {\tau }{\ln (1-\tau ^{'})}\) is bounded. Letting \(0{\lt}\varepsilon {\lt}T_{q}/2\), there exists a constant \(R\in (0,1)\) such that </p>
<div class="displaymath" id="a0000000137">
  \begin{eqnarray}  \tfrac {\tau R}{f(R)}-\tfrac {\tau }{\ln (1-\tau ’)}\int _{0}^{R}\tfrac {{\rm d}\sigma }{f(\sigma )}{\lt} \tfrac {\varepsilon }{2}. \end{eqnarray}
</div>
<p> Since \(u\) quenches at the time \(T_{q}\), there exist \(T_{1}\in (T_{q}-\tfrac {\varepsilon }{2},T_{q})\) and \(h_{0}(\varepsilon ){\gt}0\) such that \(0{\lt}u_{\min }(t){\lt}\tfrac {R}{2}\) for \(t\in [T_1,T_q)\), \(h\leq h_{0}(\varepsilon )\). Let \(q\) be a positive integer such that \(t_q=\sum _{n=0}^{q-1}\Delta t_n\in [T_1,T_q)\) for \(h\leq h_{0}(\varepsilon )\). It follows from Theorem 5.1 that the problem (25)–(26) has a solution \(U_h^{(n)}\) which obeys \(\| U_h^{(n)}-u_h(t_n)\| _{\infty }{\lt}\tfrac {R}{2}\) for \(n\leq q\), \(h\leq h_{0}(\varepsilon )\), which implies that </p>
<div class="displaymath" id="a0000000138">
  \begin{eqnarray*}  \| U_h^{(q)}\| _{\inf }\leq \| U_h^{(q)}-u_h(t_q)\| _{\infty }+\| u_h(t_q)\| _{\inf } {\lt}\tfrac {R}{2}+\tfrac {R}{2}=R,\quad h\leq h_{0}(\varepsilon ). \end{eqnarray*}
</div>
<p> From Theorem 5.2, \(U_h^{(n)}\) quenches at the time \(T_{h}^{\Delta t}\). It follows from Remark 5.1 and (36) that \(|T_{h}^{\Delta t}-t_q|\leq \tfrac {\tau \| U_h^{(q)}\| _{\inf }}{f(\| U_h^{(q)}\| _{\inf })}-\tfrac {\tau }{\ln (1-\tau ’)}\int _{0}^{\| U_h^{(q)}\| _{\inf }}\tfrac {{\rm d}\sigma }{f(\sigma )}{\lt}\tfrac {\varepsilon }{2}\) because \(\| U_h^{(q)}\| _{\inf }{\lt}R\) for \(h\leq h_{0}(\varepsilon )\). We deduce that for \(h\leq h_{0}(\varepsilon )\), </p>
<div class="displaymath" id="a0000000139">
  \[ |T_q-T_{h}^{\Delta t}|\leq |T_q-t_q|+|t_q-T_{h}^{\Delta t}|\leq \tfrac {\varepsilon }{2}+\tfrac {\varepsilon }{2}\leq \varepsilon , \]
</div>
<p> which leads us to the result. </p>
<h1 id="a0000000140">6 Numerical results</h1>
<p>In this section, we present some numerical approximations to the quenching time for the solution of the problem (1)–(3) in the case where \(p=1\) and \(u_{0}(x)=\tfrac {2+\varepsilon \cos ({\pi }x)}{4}\) with \(0{\lt}\varepsilon \leq 1\). Firstly, we take the explicit scheme in (25)–(26). Secondly, we use the following implicit scheme </p>
<div class="displaymath" id="a0000000141">
  \begin{eqnarray*}  \tfrac {U_i^{(n+1)}-U_i^{(n)}}{\Delta t_n}=\delta ^{2}U_i^{(n+1)}-a_{i}(U_i^{(n)})^{-p-1}U_i^{(n+1)},& 0\leq i\leq I, \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000142">
  \begin{eqnarray*}  U_i^{(0)}=\varphi _{i}{\gt}0,&  0\leq i\leq I, \end{eqnarray*}
</div>
<p> where \(n\geq 0\), \(\Delta t_n=K\| U_h^{(n)}\| _{\inf }^{p+1}\) with \(K=10^{-3}\). </p>
<p>In both cases, \(\varphi _{i}=\tfrac {2+\varepsilon \cos ({\pi }ih)}{4}\), \(0\leq i\leq I\). For the above implicit scheme, the existence and positivity of the discrete solution \(U_{h}^{(n)}\) is guaranteed using standard methods (see [3]). In the tables 1–8, in rows, we present the numerical quenching times, the numbers of iterations and the CPU times 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 </p>
<div class="displaymath" id="a0000000143">
  \[ \Delta t_{n}=|t_{n+1}-t_{n}|\leq 10^{-16}. \]
</div>
<div class="table"  id="a0000000144">
  <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

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

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.062132 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4102 </p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.062253 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 15883 </p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.062312 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 61257 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 60 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.062322 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 235525 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1245 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations and CPU times (seconds) obtained with the explicit Euler method for \(\varepsilon =1\).</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000145">
  <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

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

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.062302 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4017 </p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.062317 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 15499 </p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.062323 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 59679 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 138 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.062324 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 229179 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 4260 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations and CPU times (seconds) obtained with the implicit Euler method for \(\varepsilon =1\).</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000146">
  <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

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

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.121368 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2389 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.121210 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 8882 </p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.121170 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 32769 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 222 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.121157 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 119887 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3887</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations and CPU times (seconds) obtained with the explicit Euler method for \(\varepsilon =1/10\).</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000147">
  <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

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

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.121316 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 14047 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 25 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.121326 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 14071 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 45 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.121328 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 14091 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 168 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.121329 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 14098 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 795 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations and CPU times (seconds) obtained with the implicit Euler method for \(\varepsilon =1/10\).</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000148">
  <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

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

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.124875 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2356 </p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.124694 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 8728 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 17 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.124649 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 32091 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 236 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.124638 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 112964 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3974</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">5</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations and CPU times (seconds) obtained with the explicit Euler method for \(\varepsilon =1/100\).</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000149">
  <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

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

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.124822 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 13915 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 24 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.1248195 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 13920 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 44 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.1248193 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 13923 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 168</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.1248191 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 13925 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 793</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">6</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations and CPU times (seconds) obtained with the implicit Euler method for \(\varepsilon =1/100\).</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000150">
  <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

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

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.125208 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2351 </p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.125024 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 8708 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 17 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.124979 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 32006 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 191</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.124957 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 112873 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 3852</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">7</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations and CPU times (seconds) obtained with the explicit Euler method for \(\varepsilon =1/1000\).</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000151">
  <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(I\) </p>

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

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.125155 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 13914 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 26 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.12515090 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 13917 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 52 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.12515091 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 13918 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 154</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-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:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.12515093 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 13919 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 781</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">8</span> 
  <span class="caption_text">Numerical quenching times, numbers of iterations and CPU times (seconds) obtained with the implicit Euler method for \(\varepsilon =1/1000\).</span> 
</figcaption> 
</div>
<p><div class="remark_thmwrapper " id="a0000000152">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">6.1</span>
  </div>
  <div class="remark_thmcontent">
  <p>When \(\varepsilon =0\) and \(p=1\), we know that the quenching time of the continuous solution of (1)–(3) is equal \(0.125\). We have also seen in Remark 3.3 that the quenching time of the semidiscrete solution is equal \(0.125\). We observe from Tables 1–8 that when \(\varepsilon \) decays to zero, then the numerical quenching time of the discrete solution goes to \(0.125\). When one examines tables 1,2,3 and 4 one sees that an important perturbation on the potential and the initial datum has a meaningful impact on the numerical quenching time.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p><div class="acknowledgement_thmwrapper " id="a0000000153">
  <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 throughout reading of the manuscript and several suggestions that help us improve the presentation of the paper. </p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
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</dd>
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  <dd><p><i class="sc">H. A. Levine</i>, <i class="it">Quenching, nonquenching and beyond quenching for solutions of some parabolic equations</i>, Annali Math. Pura Appl., <b class="bfseries">155</b>, pp.&#160;243–260, 1990. </p>
</dd>
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</dd>
  <dt><a name="14">14</a></dt>
  <dd><p><i class="sc">K. W. Liang, P. Lin, M. T. Ong</i> and <i class="sc">R. C. E. Tan</i>, <i class="it">A splitting moving mesh method for reaction-diffusion equations of quenching type</i>, J. Comput. Phys., to appear. </p>
</dd>
  <dt><a name="15">15</a></dt>
  <dd><p><i class="sc">T. Nakagawa</i>, <i class="it">Blowing up on the finite difference solution to \(u_{t}=u_{xx}+u^{2}\)</i>, Appl. Math. Optim., <b class="bfseries">2</b>, pp.&#160;337–350, 1976. </p>
</dd>
  <dt><a name="16">16</a></dt>
  <dd><p><i class="sc">D. Phillips</i>, <i class="it">Existence of solutions of quenching problems</i>, Appl. Anal., <b class="bfseries">24</b>, pp.&#160;253–264, 1987. </p>
</dd>
  <dt><a name="17">17</a></dt>
  <dd><p><i class="sc">M. H. Protter</i> and <i class="sc">H. F. Weinberger</i>, <i class="it">Maximum principles in differential equations</i>, Prentice Hall, Englewood Cliffs, NJ, 1967. </p>
</dd>
  <dt><a name="18">18</a></dt>
  <dd><p><i class="sc">Q. Sheng</i> and <i class="sc">A. Q. M. Khaliq</i>, <i class="it">Adaptive algorithms for convection-diffusion-reaction equations of quenching type</i>, Dyn. Contin. Discrete Impuls. Syst. Ser. A, Math. Anal., <b class="bfseries">8</b>, pp.&#160;129–148, 2001. </p>
</dd>
  <dt><a name="19">19</a></dt>
  <dd><p><i class="sc">Q. Sheng</i> and <i class="sc">A. Q. M. Khaliq</i>, <i class="it">A compound adaptive approach to degenerate nonlinear quenching problems</i>, Numer. Methods PDE, <b class="bfseries">15</b>, pp.&#160;29–47, 1999. </p>
</dd>
  <dt><a name="20">20</a></dt>
  <dd><p><i class="sc">W. Walter</i>, <i class="it">Differential-und Integral-Ungleichungen</i>, Springer, Berlin, 1964. </p>
</dd>
</dl>


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