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<h1>Remarks on the quenching estimate for a nonlocal diffusion problem with a reaction term</h1>
<p class="authors">
<span class="author">Halima Nachid\(^\ast \)</span>
</p>
<p class="date">May 1st, 2012</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="tt">nachidhalima@yahoo.fr</span>. </p>

<div class="abstract"><p> In this paper, we address the following initial value problem <br /></p>
<div class="displaymath" id="a0000000002">
  \[  \left.\begin{array}{ll} \hbox{$u_t=\int _{\Omega }J(x-y)(u(y, t)-u(x, t)){\rm d}y+f(u(x, t))\quad \mbox{in}\quad \overline{\Omega }\times (0,T)$,} \\ \hbox{$u(x,0)=u_{0}(x)\geq 0\quad \mbox{in}\quad \overline{\Omega }$,} \\ \end{array}\right.  \]
</div>
<p> <br />where \(\Omega \) is a bounded domain in \(\mathbb {R}^N\) with smooth boundary \(\partial \Omega \), \(f: (-\infty , b)\rightarrow (0, \infty )\) is a \(C^1\) convex nondecreasing function, \(\lim _{s\rightarrow b^{-}}f(s)=\infty \), \(\int ^{\infty }\tfrac {{\rm d}\sigma }{f(\sigma )}{\lt}\infty \), with \(b\) a positive constant, \(J: \mathbb {R}^N\rightarrow \mathbb {R}\) is a kernel which is measurable, nonnegative and bounded in \(\mathbb {R}^N\). Under some conditions, we show that the solution of a perturbed form of the above problem quenches in a finite time and estimate its quenching time. We also prove the continuity of the quenching time as a function of the initial datum. Finally, we give some numerical results to illustrate our analysis. </p>
<p><b class="bf">MSC.</b> 35B40, 45A07, 45G10, 65M06. </p>
<p><b class="bf">Keywords.</b> Nonlocal diffusion, quenching, continuity, numerical quenching time, reaction-diffusion equation. </p>
</div>
<h1 id="a0000000003">1 Introduction</h1>
<p>Let \(\Omega \) be a bounded domain in \(\mathbb {R}^N\) with smooth boundary \(\partial \Omega .\) Consider the following initial value problem </p>
<div class="displaymath" id="a0000000004">
  \begin{eqnarray}  u_t=\int _{\Omega }J(x-y)(u(y, t)-u(x, t)) {\rm d}y+f(u(x, t))\quad \mbox{ in }\quad \overline{\Omega }\times (0, T), \end{eqnarray}
</div>
<div class="displaymath" id="a0000000005">
  \begin{eqnarray}  u(x,0)=u_{0}(x)\geq 0&  \mbox{in}&  \overline{\Omega }, \end{eqnarray}
</div>
<p> where \(f: (-\infty , b)\rightarrow (0, \infty )\) is a \(C^1\) convex nondecreasing function, \(\int ^{\infty }\tfrac {{\rm d}\sigma }{f(\sigma )}{\lt}\infty \), \(\lim _{s\rightarrow b^{-}}f(s)=\infty \), with \(b\) a positive constant, \(J: \mathbb {R}^N\rightarrow \mathbb {R}\) is a kernel which is measurable, nonnegative and bounded in \(\mathbb {R}^N\). In addition, \(J\) is symmetric (\(J(z)=J(-z)\)) and \(\int _{\mathbb {R}^N}J(z) {\rm d}z=1\). The initial datum \(u_0\in C^0(\overline{\Omega })\), \(0\leq u_0(x){\lt}b\) in \(\overline{\Omega }\). Let us notice that, if \(f(s)=(b-s)^{-p}\) with \(p\) a positive constant, then \(f\) satisfies the above conditions. Here, \((0,T)\) is the maximal time interval on which the solution \(u\) exists. The time \(T\) may be finite or infinite. When \(T\) is infinite, then we say that the solution \(u\) exists globally. When \(T\) is finite, then the solution \(u\) develops a singularity in a finite time, namely, </p>
<div class="displaymath" id="a0000000006">
  \[  \lim _{t\rightarrow T}\| u(\cdot , t)\| _{\infty }= b,  \]
</div>
<p> where \(\| u(\cdot , t)\| _{\infty }=\sup _{x\in \Omega }|u(x, t)|\). In this last case, we say that the solution \(u\) quenches in a finite time, and the time \(T\) is called the quenching time of the solution \(u\). Recently, nonlocal diffusion has been the subject of investigation of many authors (see, [2], [8], [11], [13], [15], [17], [19], [20], [25], [29], [32], and the references cited therein). Nonlocal evolution equations of the form </p>
<div class="displaymath" id="a0000000007">
  \[  u_t=\int _{\mathbb {R}^N}J(x-y)(u(y, t)-u(x, t)) {\rm d}y,  \]
</div>
<p> and variations of it, have been used by several authors to model diffusion processes (see, [4], [5], [11], [19], [20]). The solution \(u(x, t)\) can be interpreted as the density of a single population at the point \(x\), at the time \(t\), and \(J(x-y)\) as the probability distribution of jumping from location \(y\) to location \(x\). Then the convolution \((J*u)(x, t)=\int _{\mathbb {R}^N}J(x-y)u(y, t) {\rm d}y\) is the rate at which individuals are arriving to position \(x\) from all other places, and \(-u(x, t)=-\int _{\mathbb {R}^N}J(x-y)u(y, t) {\rm d}y\) is the rate at which they are leaving location \(x\) to travel to any other site (see, [19]). For the problem described in (1)–(2), the integral is taken over \(\Omega \). Consequently, there is no individuals that arrive or leave the domain \(\Omega \). It is the reason why in the title of the paper, we have added Neumann boundary condition. On the other hand, the term of the source \(f(u)\) can be rewritten as follows </p>
<div class="displaymath" id="a0000000008">
  \[  f(u(x, t))=\int _{\mathbb {R}^N}J(x-y)f(u(x, t)) {\rm d}y.  \]
</div>
<p> Therefore, in view of the above equality, the term \(f(u)\) can be interpreted as a force that decreases the rate of individuals which are leaving location \(x\) to travel to any other site, provoking as we shall see later, the phenomenon of quenching of the solution \(u\). For local diffusions, solutions which quench in a finite time has been the subject of investigation of many authors (see, for instance [10], [18], [23], [24], [26], [28], [30], and the reference cited therein). </p>
<p>Similar results have been obtained in [32], where the authors considered analogous problems within the framework of the phenomenon of blow-up (we say that a solution blows up in a finite time if it reaches the value infinity in a finite time). </p>
<p>The first paper which deals with blow-up of (1)–(2) that we are aware as that of Perez-LLanos and Rossi in [32], where they considered the problem (1)-(2) in the case where \(f(u)=u^{p}\) with \(p=\) const \({\gt}1\). They proved that the solution \(u\) of (1)–(2) blows up in a finite time and localized the blow-up set. Some results about blow-up rate are also given. In the same way, in [29], Nabongo and Boni examined the initial value problem </p>
<div class="displaymath" id="a0000000009">
  \begin{eqnarray*}  u_t=\varepsilon \int _{\Omega }J(x-y)(u(y, t)-u(x, t)) {\rm d}y+f(u)\quad \mbox{ in }\quad \overline{\Omega }\times (0, T), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000010">
  \begin{eqnarray*}  u=0 &  \mbox{ on }&  (\mathbb {R}-\Omega )\times (0, T), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000011">
  \begin{eqnarray*}  u(x,0)=u_{0}(x)\geq 0 &  \mbox{ in }&  \overline{\Omega }, \end{eqnarray*}
</div>
<p> where \(\varepsilon \) is a positive parameter. They showed that, if \(\varepsilon \) is small enough, then the solution \(u\) of the above problem blows up in a finite time, and its blow-up time goes to that of the solution of the following ODE </p>
<div class="displaymath" id="a0000000012">
  \[  \left\{  \begin{array}{ll} \hbox{$\alpha '(t)=f(\alpha (t)),\quad t{\gt}0$,} \\ \hbox{$\alpha (0)=\max _{x\in \overline{\Omega }}u_{0}(x)$,} \\ \end{array}\right.  \]
</div>
<p> as \(\varepsilon \) goes to zero. In this paper, we are interested in the the phenomenon of quenching of the solution \(u\), and the continuity of the quenching time for the problem describe in (1)-(2). More precisely, consider the following initial value problem </p>
<div class="displaymath" id="a0000000013">
  \begin{eqnarray}  v_t=\int _{\Omega }J(x-y)(v(y, t)-v(x, t)) {\rm d}y+f(v)\quad \mbox{ in } \quad \overline{\Omega }\times (0, T_h), \end{eqnarray}
</div>
<div class="displaymath" id="a0000000014">
  \begin{eqnarray}  v(x,0)=u^h_{0}(x)\quad \mbox{ in }\quad \overline{\Omega }, \end{eqnarray}
</div>
<p> where \(u^h_{0}\in C^0(\overline{\Omega })\), \(0\leq u^h_{0}(x)\leq u_{0}(x)\) in \(\overline{\Omega }\), \(\lim _{h\rightarrow 0}u^h_{0}=u_{0}\). Here \((0, T_h)\) is the maximal time interval of existence of the solution \(v\). In the current paper, under some hypotheses, we show that the solution \(v\) of (3)–(4) quenches in a finite time and estimate its quenching time. We demonstrate in passing that, when the norm of the initial datum is large enough, then the solution \(v\) of (3)–(4) quenches in a finite time and its quenching time goes to that of the solution of a certain differential equation </p>
<div class="displaymath" id="a0000000015">
  \[ \alpha '(t)=f(\alpha (t)),\quad t{\gt}0,\quad \alpha (0)=\| u^h_{0}\| _{\infty }, \]
</div>
<p> as \(\| u^h_{0}\| _{\infty }\) goes to \(b\). Finally, under some hypotheses, we prove that the solution \(v\) of (3)–(4) quenches in a finite time and its quenching time goes to that of the solution \(u\) of (1)–(2) when \(h\) goes to zero. The remainder of the paper is organized in the following manner. In the next section, we prove the local existence and uniqueness of solutions. In the third section, under some conditions, we show that the solution \(v\) of (3)–(4) quenches in a finite time and estimate its quenching time. We also show that its quenching time goes to that of the solution \(u\) of (1)–(2) when \(h\) goes to zero, in the last section, we give some computational results to illustrate our analysis. </p>
<h1 id="a0000000016">2 Local existence</h1>
<p>In this section, we shall establish the existence and uniqueness of nonnegative solutions of (1)–(2) in \(\Omega \times (0, T)\) for all small \(T\). We shall also prove some results concerning the maximum principle within the framework of nonlocal diffusion problems for our subsequent use. Let us notice that results on local existence and uniqueness are known for our problem if one modifies slightly the proof given by Perez-LLanos and Rossi in [32]. However, for the sake of completeness, we outline them. Let \(t_0\) be fixed, and define the function space \(Y_{t_0}=\{ u; u\in C([0, t_0], C(\overline{\Omega }))\} \) equipped with the norm defined by \(\| u\| _{Y_{t_0}}=\max _{0\leq t\leq t_0}\| u(\cdot , t)\| _{\infty }\) for \(u\in Y_{t_0}\). It is easy to see that \(Y_{t_0}\) is a Banach space. Introduce the set </p>
<div class="displaymath" id="a0000000017">
  \[  X_{t_0}=\{ u; u\in Y_{t_0}, \| u\| _{Y_{t_0}}\leq b_0\} ,  \]
</div>
<p> where \(b_0=\tfrac {\| u_0\| _{\infty }+b}{2}\). We observe that \(X_{t_0}\) is a nonempty bounded closed convex subset of \(Y_{t_0}\). Define the map \(R\) as follows </p>
<div class="displaymath" id="a0000000018">
  \[  R: X_{t_0}\rightarrow X_{t_0}  \]
</div>
<div class="displaymath" id="a0000000019">
  \[  R(v)(x, t)=u_0(x)+\int _0^t\int _{\Omega }J(x-y)(v(y, s)-v(x, s)) {\rm d}y{\rm d}s+\int _0^tf(v(x, s)) {\rm d}s.  \]
</div>
<p> <div class="theo_thmwrapper " id="a0000000020">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">1</span>
  </div>
  <div class="theo_thmcontent">
  <p>Assume that \(u_0\in C^0(\overline{\Omega })\), and \(0\leq u_0(x) {\lt} b\) in \(\overline{\Omega }.\) Then \(R\) maps \(X_{t_0}\) into \(X_{t_0}\), and \(R\) is strictly contractive if \(t_0\) is approximately small relative to \(\| u_0\| _{\infty }\). </p>

  </div>
</div> </p>
<p><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>Due to the fact that \(\int _{\Omega }J(x-y) {\rm d}y\leq \int _{\mathbb {R}^N}J(x-y) {\rm d}y=1,\) a straightforward computation reveals that </p>
<div class="displaymath" id="a0000000022">
  \[  |R(v)(x, t)-u_0(x)|\leq 2\| v\| _{Y_{t_0}}t+f(\| v\| _{Y_{t_0}})t,  \]
</div>
<p> which implies that \( \| R(v)\| _{Y_{t_0}}\leq \| u_0\| _{\infty }+2b_0t_0+f(b_0)t_0.\) If </p>
<div class="displaymath" id="a0000000023">
  \begin{eqnarray}  \qquad \qquad \qquad \qquad \qquad t_0\leq \tfrac {b_0-\| u_0\| _{\infty }}{2b_0+f(b_0)}, \end{eqnarray}
</div>
<p> then </p>
<div class="displaymath" id="a0000000024">
  \[  \| R(v)\| _{Y_{t_0}}\leq b_0.  \]
</div>
<p> Therefore, if (5) holds, then \(R\) maps \(X_{t_0}\) into \(X_{t_0}\). Now, we are going to prove that the map \(R\) is strictly contractive. Let \(v, z\in X_{t_0}\). Setting \(\alpha =v-z\), we discover that </p>
<div class="displaymath" id="a0000000025">
  \begin{eqnarray*}  |(R(v)-R(z))(x, t)|\leq \left|\int _0^t\int _{\Omega }J(x-y)(\alpha (y, s)-\alpha (x, s)) {\rm d}y{\rm d}s\right| \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000026">
  \begin{eqnarray*}  \qquad \qquad \qquad \qquad \qquad +\left|\int _0^t(f(v(x, s))-f(z(x, s))) {\rm d}s\right|. \end{eqnarray*}
</div>
<p> Use Taylor’s expansion to obtain </p>
<div class="displaymath" id="a0000000027">
  \begin{eqnarray*}  |(R(v)-R(z))(x, t)|\leq 2\| \alpha \| _{Y_{t_0}}t+t\| v-z\| _{Y_{t_0}}f’(\| \beta \| _{Y_{t_0}}), \end{eqnarray*}
</div>
<p> where \(\beta \) is an intermediate function between \(v\) and \(z\). We deduce that </p>
<div class="displaymath" id="a0000000028">
  \begin{eqnarray*}  \| R(v)-R(z)\| _{Y_{t_0}}\leq 2\| \alpha \| _{Y_{t_0}}t_0+t_0\| v-z\| _{Y_{t_0}}f’(\| \beta \| _{Y_{t_0}}), \end{eqnarray*}
</div>
<p> which implies that \(\| R(v)-R(z)\| _{Y_{t_0}}\leq (2t_0+t_0f'(b_0))\| v-z\| _{Y_{t_0}}\). If </p>
<div class="displaymath" id="a0000000029">
  \begin{eqnarray}  \qquad \qquad \qquad \qquad \qquad t_0\leq \tfrac {1}{4+2f’(b_0)}, \end{eqnarray}
</div>
<p> then \(\| R(v)-R(z)\| _{Y_{t_0}}\leq \tfrac {1}{2}\| v-z\| _{Y_{t_0}}\). Hence, we see that \(R(v)\) is a strict contraction in \(Y_{t_0}\) and the proof is complete. </p>
<p>It follows from the contraction mapping principle that for appropriately chosen \(t_0{\gt}0\), \(R\) has a unique fixed point \(u(x, t)\in Y_{t_0}\) which is a solution of (1)-(2). If \(\| u\| _{Y_{t_0}}{\lt}b\), then taking as initial datum \(u(\cdot , t_0)\in C^{0}(\overline{\Omega })\) and arguing as before, it is possible to extend the solution up to some interval \([0, t_1)\) for certain \(t_1{\gt}t_0\). Now, to end this section, we shall provide some results about the maximum principle tailored to our study. The following lemma is a version of the maximum principle for nonlocal problems. <div class="lem_thmwrapper " id="a0000000030">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">2</span>
  </div>
  <div class="lem_thmcontent">
  <p>Let \(a\in C^0(\overline{\Omega }\times [0, T))\), and let \(u\in C^{0, 1}(\overline{\Omega }\times [0, T))\) satisfying the following inequalities </p>
<div class="equation" id="a0000000031">
<p>
  <div class="equation_content">
    \begin{equation}  u_t-\int _{\Omega }J(x-y)(u(y, t)-u(x, t)) {\rm d}y+a(x, t)u(x, t)\geq 0 \; \;  \mbox{ in } \; \; \overline{\Omega }\times (0, T), \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<div class="displaymath" id="a0000000032">
  \begin{eqnarray}  u(x, 0)\geq 0 \quad \mbox{ in } \quad \overline{\Omega }. \end{eqnarray}
</div>
<p> Then, we have \(u(x, t)\geq 0\)   in  \(\overline{\Omega }\times (0, T)\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000033">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Let \(T_0\) be any positive quantity satisfying \(T_0{\lt}T\). Since \(a(x, t)\) is bounded in \(\overline{\Omega }\times [0, T_0]\), then there exists \(\lambda \) such that \(a(x, t)-\lambda {\gt}0\) in \(\overline{\Omega }\times [0, T]\). Define \(z(x, t)=e^{\lambda t}u(x, t)\) and let \(m=\min _{x\in \overline{\Omega }, t\in [0, T_0]}z(x, t)\). Due to the fact that \(z\) is continuous in \(\overline{\Omega }\times [0, T_0]\), then it achieves its minimum in \(\overline{\Omega }\times [0, T_0]\). Consequently, there exists \((x_0, t_0)\in \overline{\Omega }\times [0, T_0]\) such that \(m=z(x_0, t_0)\). We get \(z(x_0, t_0)\leq z(x_0, t)\) for \(t\leq t_0\) and \(z(x_0, t_0)\leq z(y, t_0)\) for \(y\in \Omega \). This implies that </p>
<div class="displaymath" id="a0000000034">
  \begin{eqnarray}  z_t(x_0, t_0)\leq 0, \quad \int _{\Omega }J(x_0-y)(z(y, t_0)-z(x_0, t_0)) {\rm d}y\geq 0. \end{eqnarray}
</div>
<p> With the aid of the first inequality of the lemma, it is not hard to see that </p>
<div class="displaymath" id="a0000000035">
  \[  z_t(x_0, t_0)-\int _{\Omega }J(x_0-y)(z(y, t_0)-z(x_0, t_0)) {\rm d}y+ a(x_0, t_0)-\lambda )z(x_0, t_0)\geq 0.  \]
</div>
<p> We deduce from (9) that \((a(x_0, t_0)-\lambda )z(x_0, t_0\geq 0\). Since \(a(x_0, t_0)-\lambda {\gt}0\), we get \(z(x_0, t_0)\geq 0\). This implies that \(u(x, t)\geq 0\) in \(\overline{\Omega }\times [0, T_0]\), and the proof is complete. </p>
<p>An immediate consequence of the above lemma is the following comparison lemma. Its proof is straightforward. <div class="lem_thmwrapper " id="a0000000036">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">3</span>
  </div>
  <div class="lem_thmcontent">
  <p>Let \(a\in C^0(\overline{\Omega }\times [0, T))\) and let \(u\), \(v\in C^{0, 1}(\overline{\Omega }\times [0, T))\) satisfying the following inequalities </p>
<div class="displaymath" id="a0000000037">
  \begin{eqnarray*}  u_t-\int _{\Omega }J(x-y)(u(y, t)-u(x, t)) {\rm d}y+a(x, t)u(x, t) \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000038">
  \begin{eqnarray*}  \geq v_t-\int _{\Omega }J(x-y)(v(y, t)-v(x, t)) {\rm d}y+a(x, t)v(x, t) \; \;  \mbox{ in } \; \; \overline{\Omega }\times (0, T), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000039">
  \begin{eqnarray*}  u(x, 0)\geq v(x, 0) \quad \mbox{ in } \quad \overline{\Omega }. \end{eqnarray*}
</div>
<p> Then, we have \(u(x, t)\geq v(x, t)\)   in  \(\overline{\Omega }\times (0, T)\). </p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000040">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">4</span>
  </div>
  <div class="remark_thmcontent">
  <p>Invoking the mean value theorem and Lemma 2.2, it is not hard to see that \(v(x, t)\leq u(x, t)\) as long as all of them are defined. We infer that \(T_h\geq T\).<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000041">3 The quenching time</h1>
<p>In this section, under some conditions, we show that the solution \(v\) of (3)–(4) quenches in a finite time and estimate its quenching time. We demonstrate in passing that, if the \(L^{\infty }\) norm of the initial datum is large enough, then the solution \(v\) of (3)–(4) quenches in a finite time and its quenching time goes to that of the solution of a differential equation as \(\| u_0^h\| _{\infty }\) goes to \(b\). Finally, we gather some results that we deem useful to prove the continuity of the quenching time time. Our first result says that the solution \(v\) of (3)–(4) always quenches in a finite time if the initial datum is nonnegative. It is stated in the following theorem. </p>
<p><div class="theo_thmwrapper " id="a0000000042">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">5</span>
  </div>
  <div class="theo_thmcontent">
  <p>Let \(v\) be the solution of <span class="rm">(3)–(4)</span>. Then \(v\) quenches in a finite time, and its quenching time \(T_h\) obeys the following estimate </p>
<div class="displaymath" id="a0000000043">
  \[  T_h\leq \int _A^{b}\tfrac {{\rm d}s}{f(s)},  \]
</div>
<p> where \(A=\tfrac {1}{|\Omega |}\int _{\Omega }u_0^h(x){\rm d}x\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000044">
  <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 \((0, T_h)\) is the maximal time interval on which <br />\(\| v(.,t)\| _{\infty }{\lt}b \), our aim is to show that \(T_h\) is finite and satisfies the above inequality. Due to the fact that the initial datum \(u_0^h\) is nonnegative in \(\overline{\Omega }\), we know from Lemma 2.1 that the solution \(v\) is also nonnegative in \(\overline{\Omega }\times (0, T_h)\). Integrating both sides of (3) over \((0, t)\), we find that </p>
<div class="displaymath" id="a0000000045">
  \begin{eqnarray*}  v(x, t)-u_0^h(x)=\int _0^t\int _{\Omega }J(x-y)(v(y, s)-v(x, s)){\rm d}y{\rm d}s \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000046">
  \begin{eqnarray*}  \qquad \qquad \qquad \qquad +\int _0^tf(v(x, s)) {\rm d}s\quad \mbox{ for }\quad t\in (0, T_h). \end{eqnarray*}
</div>
<p> Integrate again in the \(x\) variable and apply Fubini’s theorem to obtain </p>
<div class="displaymath" id="a0000000047">
  \begin{eqnarray}  \int _{\Omega }v(x, t) {\rm d}x-\int _{\Omega }u_0^h(x) {\rm d}x=\int _0^t\Big(\int _{\Omega }f(v(x, s)\Big) {\rm d}s\quad \mbox{ for }\quad t\in (0, T_h). \end{eqnarray}
</div>
<p> Set </p>
<div class="displaymath" id="a0000000048">
  \[  w(t)=\tfrac {1}{|\Omega |}\int _{\Omega }v(x, t) {\rm d}x\quad \mbox{ for }\quad t\in [0, T_h).  \]
</div>
<p> Taking the derivative of \(w\) in \(t\) and using (10), we arrive at </p>
<div class="displaymath" id="a0000000049">
  \[  w'(t)=\int _{\Omega }\tfrac {1}{|\Omega |}f(v(x, t)) {\rm d}x,\quad \mbox{ for }\quad t\in (0, T_h)  \]
</div>
<p> It follows from Jensen’s inequality that \(w'(t)\geq f(w(t))\) for \(t\in (0, T_h)\), or equivalently </p>
<div class="displaymath" id="a0000000050">
  \begin{eqnarray*}  \qquad \qquad \qquad \quad \quad \tfrac {{\rm d}w}{f(w)}\geq {\rm d}t\quad \mbox{ for }\quad t\in (0, T_h). \end{eqnarray*}
</div>
<p> Integrate the above inequality over \((0, T_h)\) to obtain </p>
<div class="displaymath" id="a0000000051">
  \[  T_h\leq \int _{w(0)}^{b}\tfrac {{\rm d}s}{f(s)}.  \]
</div>
<p> Since the quantity on the right hand side of the above inequality is finite, we deduce that \(v\) quenches in a finite time at the time \(T_h\) which obeys the above inequality. Use the fact that \(w(0)=A\) to complete the rest of the proof. </p>
<p>The above theorem allows us to obtain an estimate which depends on the \(L^{1}\) norm of the initial datum. This kinds of estimation is not interesting in order to obtain the continuity of the quenching time as a function of the initial datum. Therefore, we shall give another result which reveals an estimate of the of the quenching time that depends on the \(L^{\infty }\) norm of the initial datum. This result is stated in the theorem below. </p>
<p><div class="theo_thmwrapper " id="a0000000052">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">6</span>
  </div>
  <div class="theo_thmcontent">
  <p>Let \(A=\int _0^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}\). If \(A{\lt}1\), then the solution \(v\) of <span class="rm">(3)–(4)</span> quenches in a finite time, and its quenching time \(T_h\) obeys the following estimate </p>
<div class="displaymath" id="a0000000053">
  \[  T_h\leq \tfrac {1}{1-A}\int _{\| u_0^h\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}.  \]
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000054">
  <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 \((0, T_h)\) is the maximal time interval of existence of the solution \(v\), our aim is to show that \(T_h\) is finite and satisfies the above inequality. As in the proof of Theorem 3.1, an application of Lemma 2.1 reveals that the solution \(v\) is nonnegative in \(\overline{\Omega }\times (0, T_h)\). Due to the fact that \(J(z)\) is nonnegative for \(z\in \mathbb {R}^N\), and </p>
<div class="displaymath" id="a0000000055">
  \[  \int _{\Omega }J(x-y) {\rm d}y\leq \int _{\mathbb {R}^N}J(x-y) {\rm d}y=1 \quad \mbox{ for }\quad x\in \overline{\Omega },  \]
</div>
<p> we note that </p>
<div class="displaymath" id="a0000000056">
  \[  v_t(x, t)\geq -v(x, t)+f(v(x, t))\quad \mbox{ in }\quad \overline{\Omega }\times (0, T_h),  \]
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000057">
  \begin{eqnarray*}  v_t(x, t)\geq f(v(x, t))\left(1-\tfrac {v(x, t)}{f(v(x, t))}\right)& \mbox{ in }&  \overline{\Omega }\times (0, T_h). \end{eqnarray*}
</div>
<p> It is not hard to see that </p>
<div class="displaymath" id="a0000000058">
  \[  \int _0^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}\geq \sup _{0\leq t{\lt}b}\int _0^t\tfrac {{\rm d}\sigma }{f(\sigma )}\geq \sup _{0\leq t{\lt}b}\tfrac {t}{f(t)},  \]
</div>
<p> because \(f(s)\) is nondecreasing for \(s\in [0, b)\). We infer that </p>
<div class="displaymath" id="a0000000059">
  \begin{eqnarray*}  v_t(x, t)\geq (1-A)f(v(x, t)) &  \mbox{ in }&  \overline{\Omega }\times (0, T_h), \end{eqnarray*}
</div>
<p> or equivalently </p>
<div class="displaymath" id="a0000000060">
  \begin{eqnarray}  \tfrac {{\rm d}v}{f(v)}\geq (1-A){\rm d}t &  \mbox{ in }&  \overline{\Omega }\times (0, T_h). \end{eqnarray}
</div>
<p> Integrate the above inequality over \((0, T_h)\) to obtain </p>
<div class="displaymath" id="a0000000061">
  \begin{eqnarray*}  (1-A)T_h\leq \int _{u_0^h(x)}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}\quad \mbox{ in }\quad \overline{\Omega }. \end{eqnarray*}
</div>
<p> It follows that </p>
<div class="displaymath" id="a0000000062">
  \begin{eqnarray*}  T_h\leq \tfrac {1}{1-A}\int _{\| u_0^h\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}. \end{eqnarray*}
</div>
<p> We conclude that the solution \(v\) of (3)–(4) quenches in a finite time, because the quantity on the right hand side of the above inequality is finite. This finishes the proof. </p>
<p><div class="remark_thmwrapper " id="a0000000063">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">7</span>
  </div>
  <div class="remark_thmcontent">
  <p>Let \(t_0\in (0, T_h)\). Integrating the inequality (11) over \((t_0, T_h)\), we find that </p>
<div class="displaymath" id="a0000000064">
  \[  (1-A)(T_h-t_0)\leq \int _{v(x, t_0)}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )} \quad \mbox{ for } \quad x\in \overline{\Omega },  \]
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000065">
  \[  T_h-t_0\leq \tfrac {1}{1-A}\int _{\| v(\cdot , t_0)\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}.\hfil \qed  \]
</div>

  </div>
</div> </p>
<p>It is worth noting that the above estimate is crucial to obtain the continuity of the quenching time as a function of the initial datum. Let us notice that the condition \(A{\lt}1\) of the above theorem is very restrictive in certain situations. By the following theorem, we avoid this condition in the case where the \(L^{\infty }\) norm of the initial datum is large enough. </p>
<p><div class="theo_thmwrapper " id="a0000000066">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">8</span>
  </div>
  <div class="theo_thmcontent">
  <p>Let \(v\) be the solution of <span class="rm">(3)–(4)</span>, and suppose that the initial datum at <span class="rm">(4)</span> obeys the following condition \(f(\| u_0^h\| _{\infty }){\gt}b.\) Then, the solution \(v\) quenches in a finite time, and its quenching time \(T_h\) is estimated as follows </p>
<div class="displaymath" id="a0000000067">
  \[  T_h\leq \tfrac {f(\| u_0^h\| _{\infty })}{f(\| u_0^h\| _{\infty })- b}\int _{\| u_0^h\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}.  \]
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000068">
  <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 \((0, T_h)\) is the maximal time interval of existence of the solution \(v\), our aim is to show that \(T_h\) is finite and satisfies the above inequality. Owing to Lemma 2.1, we know that the solution \(v\) is nonnegative in \(\overline{\Omega }\times (0, T_h)\) because the initial datum \(u_0^h\) is nonnegative in \(\overline{\Omega }\). We note that </p>
<div class="displaymath" id="a0000000069">
  \begin{eqnarray*}  \int _{\Omega }J(x-y) {\rm d}y\leq \int _{\mathbb {R}^N}J(x-y) {\rm d}y=1 \quad \mbox{ for }\quad x\in \overline{\Omega }, \end{eqnarray*}
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000070">
  \begin{eqnarray}  v_t(x, t)\geq -v(x, t)+f(v(x, t))\quad \mbox{ in }\quad \overline{\Omega }\times (0, T_h). \end{eqnarray}
</div>
<p> Let \(x_0(t)\in \overline{\Omega }\) be such that </p>
<div class="displaymath" id="a0000000071">
  \[  U(t)=\max _{x\in \overline{\Omega }}v(x, t)=v(x_0(t), t) \quad \mbox{ for }\quad t\in (0, T_h).  \]
</div>
<p> It is easy to see that </p>
<div class="displaymath" id="a0000000072">
  \[  U'(t)=\max _{x\in \overline{\Omega }}v_t(x, t) \quad \mbox{ for }\quad t\in (0, T_h).  \]
</div>
<p> Consequently, replacing \(x\) by \(x_0(t)\) in (12), we have </p>
<div class="displaymath" id="a0000000073">
  \begin{eqnarray*}  U’(t) \geq v_t(x, t)\geq -U(t)+f(U(t)) \quad \mbox{ for }\quad t\in (0, T_h). \end{eqnarray*}
</div>
<p> This estimate may be rewritten as follows </p>
<div class="displaymath" id="a0000000074">
  \begin{eqnarray*}  U’(t) \geq f(U(t))\left(1-\tfrac {U(t)}{f(U(t))}\right) \quad \mbox{ for }\quad t\in (0, T_h). \end{eqnarray*}
</div>
<p> According to the fact that \(U(t)\leq b\), the above estimate becomes </p>
<div class="displaymath" id="a0000000075">
  \begin{eqnarray}  U’(t) \geq f(U(t))\left(1-\tfrac {b}{f(U(t))}\right) \quad \mbox{ for }\quad t\in (0, T_h). \end{eqnarray}
</div>
<p> We note that \(U'(0){\gt}0\), and we claim that \(U'(t){\gt}0\) for \(t\in (0, T_h)\). To prove the claim, we argue by contradiction. Let \(t_0\) be the first \(t\in (0, T_h)\) such that \(U'(t){\gt}0\) for \(t\in (0, t_0)\), but \(U'(t_0)=0\). This implies that \(U(t_0)\geq U(0)=\| u_0^h\| _{\infty }\). Therefore, we get </p>
<div class="displaymath" id="a0000000076">
  \[  0=U'(t_0)\geq f(\| u_0^h\| _{\infty })\left(1-\tfrac {b}{f(\| u_0^h\| _{\infty })}\right){\gt}0,  \]
</div>
<p> which is a contradiction, and the claim is proved. In view of the claim, we find that \(U(t)\geq \| u_0^h\| _{\infty }\) for \(t\in (0, T_h)\), and making use of (13), we arrive at </p>
<div class="displaymath" id="a0000000077">
  \begin{eqnarray*}  U’(t)\geq \left(1-\tfrac {b}{f(\| u_0^h\| _{\infty })}\right)f(U(t))\quad \mbox{ for }\quad t\in (0, T_h), \end{eqnarray*}
</div>
<p> or equivalently </p>
<div class="displaymath" id="a0000000078">
  \begin{eqnarray}  \tfrac {dU}{f(U)}\geq \left(1-\tfrac {b}{f(\| u_0^h\| _{\infty })}\right) dt\quad \mbox{ for }\quad t\in (0, T_h). \end{eqnarray}
</div>
<p> Integrating the above estimate over \((0, T_h)\) we obtain </p>
<div class="displaymath" id="a0000000079">
  \[  \left(1-\tfrac {b}{f(\| u_0^h\| _{\infty })}\right)T_h\leq \int _{\| u_0^h\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )},  \]
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000080">
  \[  T_h\leq \tfrac {f(\| u_0^h\| _{\infty })}{f(\| u_0^h\| _{\infty })-b}\int _{\| u_0^h\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )} .  \]
</div>
<p> We use the fact that the quantity on the right hand side of the above inequality is finite to complete the rest of the proof. </p>
<p><div class="remark_thmwrapper " id="a0000000081">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
  </div>
  <div class="remark_thmcontent">
  <p>Let \(t_{0}\in (0,T_h)\). Integrating the estimate (14) over \((t_{0} ,T_h)\), we discover that </p>
<div class="displaymath" id="a0000000082">
  \[  T_h-t_0\leq \tfrac {f(\| u_{0}^{h}\| _{\infty })}{f(\| u_{0}^{h}\| _{\infty })-b}\int _{\| v(\cdot , t_0)\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}.\hfil \qed  \]
</div>

  </div>
</div> </p>
<p>Up to now, the results obtained allow us to see some upper bounds of the quenching time. In the theorem below, we derive a lower bound of the quenching time when quenching occurs. </p>
<p><div class="theo_thmwrapper " id="a0000000083">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">10</span>
  </div>
  <div class="theo_thmcontent">
  <p>Suppose that the solution \(v\) of <span class="rm">(3)–(4)</span> quenches in a finite time \(T_h\). Then, we have </p>
<div class="displaymath" id="a0000000084">
  \[  T_h\geq \int _{\| u_0^h\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}.  \]
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000085">
  <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 \(\alpha (t)\) be the solution of the following ordinary differential equation </p>
<div class="displaymath" id="a0000000086">
  \[  \alpha '(t)=f(\alpha (t)), \quad t\in (0, T_e),\quad \alpha (0)=\| u_0^h\| _{\infty },  \]
</div>
<p> where \((0, T_e)\) is the maximal time interval of existence of the solution \(\alpha (t)\). By a routine computation, one easily sees that \(T_e=\int _{\| u_0^h\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}\). Now, let us introduce the function \(z\) defined as follows </p>
<div class="displaymath" id="a0000000087">
  \[  z(x, t)=\alpha (t) \quad \mbox{ in }\quad \overline{\Omega }\times [0, T_e).  \]
</div>
<p> A straightforward calculation yields </p>
<div class="displaymath" id="a0000000088">
  \[  z_t(x, t)=\int _{\Omega }J(x-y)(z(y, t)-z(x, t)) {\rm d}y+f(z(x, t))\quad \mbox{ in }\quad \overline{\Omega }\times (0, T_e),  \]
</div>
<div class="displaymath" id="a0000000089">
  \[  z(x, 0)\geq v(x, 0)\quad \mbox{ in }\quad \overline{\Omega }.  \]
</div>
<p> Set </p>
<div class="displaymath" id="a0000000090">
  \[  w(x, t)=z(x, t)-v(x, t)\quad \mbox{ in }\quad \overline{\Omega }\times [0, T_{*}),  \]
</div>
<p> where \(T_{*}=\min \{ T_h, T_e\} \). Making use of the mean value theorem, we find that </p>
<div class="displaymath" id="a0000000091">
  \begin{align*}  w_t(x, t)& \geq \int _{\Omega }J(x-y)(w(y, t)-w(x, t)) {\rm d}y\\ & \quad +f’(\xi (x, t))w(x, t)\quad \mbox{ in }\quad \overline{\Omega }\times (0, T_{*}),\\ w(x, 0)& \geq 0 \quad \mbox{ in }\quad \overline{\Omega }, \end{align*}
</div>
<p> where \(\xi (x, t)\) is an intermediate value between \(v(x, t)\) and \(z(x, t)\). It follows from Lemma 2.1 that </p>
<div class="displaymath" id="a0000000092">
  \[  w(x, t)\geq 0 \quad \mbox{ in }\quad \overline{\Omega }\times (0, T_{*}),  \]
</div>
<p> or equivalently </p>
<div class="displaymath" id="a0000000093">
  \begin{eqnarray}  v(x, t)\leq \alpha (t)\quad \mbox{ in }\quad \overline{\Omega }\times (0, T_{*}). \end{eqnarray}
</div>
<p> We claim that \(T_h\geq T_e\). To prove the claim, we argue by contradiction. Suppose that \(T_h{\lt}T_e\). In view of (15), we see that \(\| v(\cdot , T_h)\| _{\infty }\leq \alpha (T_h){\lt}b\), which contradicts the fact that \((0, T_h)\) is the maximum time interval of existence of the solution \(v\). This demonstrates the claim, and the proof is complete. </p>
<p><div class="remark_thmwrapper " id="a0000000094">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">11</span>
  </div>
  <div class="remark_thmcontent">
  <p>Combining Theorems 3.1 and 3.4, we note that, if the initial datum at (4) satisfies \(u_{0}^{h}=\beta =\) const \(\geq 0,\) then the solution \(v\) of (3)–(4) quenches in a finite time \(T_{h}=\int _{\beta }^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}.\)<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>With the aid of Theorems 3.3 and 3.4, we can derive the following interesting result. </p>
<p><div class="theo_thmwrapper " id="a0000000095">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">12</span>
  </div>
  <div class="theo_thmcontent">
  <p>Let \(v\) be the solution of <span class="rm">(3)–(4)</span>, and suppose that the initial datum <span class="rm">(4)</span> obeys the following condition \(f(\| u_0^h\| _{\infty }){\gt}b.\) Then, the solution \(v\) quenches in a finite time, and its quenching time \(T_h\) obeys the following estimates </p>
<div class="displaymath" id="a0000000096">
  \[  0\leq T_h-T_e\leq \tfrac {bT_e}{f(\| u_0^h\| _{\infty })}+o\left(\tfrac {T_e}{f(\| u_0^h\| _{\infty })}\right) \quad \mbox{ as } \quad \| u_0^h\| _{\infty }\rightarrow b,  \]
</div>
<p> where \(T_e=\int _{\| u_0^h\| _{\infty }}^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000097">
  <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 \((0, T_h)\) is the maximal time interval of existence of the solution \(u\), our aim is to show that \(T_h\) is finite and satisfies the above estimates. Making use of Theorems 3.3 and 3.4, we find that \(T_e\) is finite and obeys the following estimates </p>
<div class="displaymath" id="a0000000098">
  \begin{eqnarray}  T_e\leq T_h\leq \tfrac {T_e}{1-\tfrac {b}{f(\| u_0^h\| _{\infty })}}. \end{eqnarray}
</div>
<p> Apply Taylor’s expansion to obtain </p>
<div class="displaymath" id="a0000000099">
  \[  \tfrac {1}{1-\tfrac {b}{f(\| u_0^h\| _{\infty })}}=1+\tfrac {b}{f(\| u_0^h\| _{\infty })}+o\left(\tfrac {1}{f(\| u_0^h\| _{\infty })}\right)\quad \mbox{ as }\quad \| u_0^h\| _{\infty }\rightarrow b.  \]
</div>
<p> Use (16) and the above relation to complete the rest of the proof. </p>
<p><div class="remark_thmwrapper " id="a0000000100">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">13</span>
  </div>
  <div class="remark_thmcontent">
  <p>The estimates of Theorem 3.5 can be rewritten as follows </p>
<div class="displaymath" id="a0000000101">
  \[  0\leq \tfrac {T_h}{T_e}-1\leq \tfrac {b}{f(\| u_0^h\| _{\infty })}+o\left(\tfrac {b}{f(\| u_0^h\| _{\infty })}\right) \quad \mbox{ as }\quad \| u_0^h\| _{\infty }\rightarrow b.  \]
</div>
<p> We infer that </p>
<div class="displaymath" id="a0000000102">
  \[  \lim _{\| u_0^h\| _{\infty }\rightarrow b}\tfrac {T_h}{T_e}=1.\hfil \qed  \]
</div>

  </div>
</div> </p>
<h1 id="a0000000103">4 Continuity of the quenching time</h1>
<p>In this section, under some assumptions, we show that the solution \(v\) of (3)–(4) quenches in a finite time, and its quenching time goes to that of the solution \(u\) of (1)–(2) when the parameter \(h\) goes to zero. In order to obtain the above result, we firstly reveal that the solution \(v\) approaches the solution \(u\) in any interval \(\overline{\Omega }\times [0, T-\tau ]\) where \(\tau \  in (0, T)\). This result is stated in the following theorem. </p>
<p><div class="theo_thmwrapper " id="a0000000104">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">14</span>
  </div>
  <div class="theo_thmcontent">
  <p>Assume that the problem <span class="rm">(1)–(2)</span> has a solution \(u\in C^{0, 1}(\overline{\Omega }\times [0, T))\) such that \(\sup _{t\in [0, T-\tau ]}\| u(\cdot , t)\| _{\infty }\leq b-\alpha ,\) where \(\alpha \in (0,b)\) and \( \tau \in (0, T)\). Suppose that the initial datum at \(u_0^h\) satisfies the following condition </p>
<div class="displaymath" id="a0000000105">
  \begin{eqnarray}  \| u_0^h-u_0\| _{\infty }=o(1) &  \mbox{ as }&  h\rightarrow 0. \end{eqnarray}
</div>
<p> Then, the problem <span class="rm">(3)–(4)</span> admits a unique solution \(v\in C^{0, 1}(\overline{\Omega }\times [0, T_h))\), and the following relation holds </p>
<div class="displaymath" id="a0000000106">
  \begin{eqnarray*}  \sup _{t\in [0, T-\tau ]}\| v(\cdot , t)-u(\cdot , t)\| _{\infty }=O(\| u_0^h-u_0\| _{\infty }) &  \mbox{ as }&  h\rightarrow 0, \end{eqnarray*}
</div>
<p> where \(\tau \in (0, T)\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000107">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>The problem (3)–(4) admits a unique solution \(v\in C^{0, 1}(\overline{\Omega }\times [0, T_h))\). In Remark 2.1, we have mentioned that \(T_h\geq T\). Let \(t(h)\leq T-\tau \) be the first \(t\) such that </p>
<div class="displaymath" id="a0000000108">
  \begin{eqnarray}  \| v(\cdot , t)-u(\cdot , t)\| _{\infty }{\lt} \tfrac {\alpha }{2}&  \mbox{ for }&  t\in (0, t(h)). \end{eqnarray}
</div>
<p> We know from (17) that \(t(h){\gt}0\) for \(h\) small enough. An application of the triangle inequality yields </p>
<div class="displaymath" id="a0000000109">
  \begin{eqnarray*}  \| v(\cdot , t)\| _{\infty }\leq \| u(\cdot , t)\| _{\infty }+\| v(\cdot , t)-u(\cdot , t)\| _{\infty }&  \mbox{ for }&  t\in (0, t(h)), \end{eqnarray*}
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000110">
  \begin{eqnarray}  \| v(\cdot , t)\| _{\infty }\leq b-\alpha + \tfrac {\alpha }{2}\leq b-\tfrac {\alpha }{2}&  \mbox{ for }&  t\in (0, t(h)). \end{eqnarray}
</div>
<p> Introduce the error \(e\) defined as follows </p>
<div class="displaymath" id="a0000000111">
  \[  e(x, t)=v(x, t)-u(x, t)\quad \mbox{ in }\quad \overline{\Omega }\times [0, t(h)).  \]
</div>
<p> Making use of the mean value theorem, we find that </p>
<div class="displaymath" id="a0000000112">
  \[  e_t(x, t)=\int _{\Omega }J(x-y)(e(y, t)-e(x, t)) {\rm d}y+f'(\xi (x, t))e(x, t) \quad \mbox{ in }\quad \overline{\Omega }\times (0, t(h)),  \]
</div>
<div class="displaymath" id="a0000000113">
  \[  e(x, 0)=u_0^h(x)-u_0(x)\quad \mbox{ in }\quad \overline{\Omega },  \]
</div>
<p> where \(\xi (x, t)\) is an intermediate value between \(v(x, t)\) and \(u(x, t)\). Set </p>
<div class="displaymath" id="a0000000114">
  \[  z(x, t)=e^{(L+1)t}\| u_0^h-u_0\| _{\infty }\quad \mbox{ in }\quad \overline{\Omega }\times [0, T],  \]
</div>
<p> where \(L=f'( b-\tfrac {\alpha }{2})\). Due the fact that \(\| u(\cdot , t)\| _{\infty }\leq b-\alpha \) for \(t \in (0, t(h)),\) having in mind (19), it is not hard to see that \(L\geq f'(\xi (x, t)) \in \overline{\Omega }\times (0, t(h)).\) A straightforward computation reveals that </p>
<div class="displaymath" id="a0000000115">
  \[  z_t(x, t)\geq \int _{\Omega }J(x-y)(z(y, t)-z(x, t)) {\rm d}y+f'(\xi (x, t))z(x, t) \quad \mbox{ in }\quad \overline{\Omega }\times (0, t(h)),  \]
</div>
<div class="displaymath" id="a0000000116">
  \[  z(x, 0)\geq e(x, 0)\quad \mbox{ in }\quad \overline{\Omega }.  \]
</div>
<p> Invoking Lemma 2.2, we obtain </p>
<div class="displaymath" id="a0000000117">
  \[  z(x, t)\geq e(x, t)\quad \mbox{ in }\quad \overline{\Omega }\times (0, t(h)).  \]
</div>
<p> In the same way, we also prove that </p>
<div class="displaymath" id="a0000000118">
  \[  z(x, t)\geq -e(x, t)\quad \mbox{ in }\quad \overline{\Omega }\times (0, t(h)),  \]
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000119">
  \begin{eqnarray}  \| v(\cdot , t)-u(\cdot , t)\| _{\infty }\leq {\rm e}^{(L+1)t}\| u_0^h-u_0\| _{\infty }\quad \mbox{ for }\quad t\in (0, t(h)). \end{eqnarray}
</div>
<p> Now, we claim that \(t(h)=T-\tau \). To prove the claim, we argue by contradiction. Suppose that \(t(h){\lt}T-\tau \). In view of (18) and (20), it is easy to check that </p>
<div class="displaymath" id="a0000000120">
  \[  \tfrac {\alpha }{2}\leq \| v(\cdot , t(h))-u(\cdot , t(h))\| _{\infty }\leq {\rm e}^{(L+1)T}\| u_0^h-u_0\| _{\infty }.  \]
</div>
<p> Since the term on the right hand side of the above inequality goes to zero as \(h\) goes to zero, infer that \(\tfrac {\alpha }{2}\leq 0\), which is a contradiction. This demonstrates the claim, and the proof is complete. </p>
<p>At the moment, we are in a position to prove the main result of this section. </p>
<p><div class="theo_thmwrapper " id="a0000000121">
  <div class="theo_thmheading">
    <span class="theo_thmcaption">
    Theorem
    </span>
    <span class="theo_thmlabel">15</span>
  </div>
  <div class="theo_thmcontent">
  <p>Assume that the problem <span class="rm">(1)–(2)</span> has a solution \(u\) which quenches in a finite time \(T\) such that \(u\in C^{0, 1}(\overline{\Omega }\times [0, T))\). Suppose that the initial datum at \(u_0^h\) satisfies the condition <span class="rm">(17)</span>. Then, under the assumption of Theorem <span class="rm">3.2</span>, the problem <span class="rm">(3)–(4)</span> admits a unique solution \(v\) which quenches in a finite time, and the following relation holds </p>
<div class="displaymath" id="a0000000122">
  \[  \lim _{h\rightarrow 0}T_h=T.  \]
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000123">
  <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 \(0{\lt}\varepsilon {\lt}T/2\). There exists a positive constant \(\alpha \in (0,b)\) such that </p>
<div class="displaymath" id="a0000000124">
  \begin{eqnarray}  \tfrac {1}{1-A}\int _{b-\alpha }^{b}\tfrac {{\rm d}\sigma }{f(\sigma )}{\lt}\tfrac {\varepsilon }{2}. \end{eqnarray}
</div>
<p> Since \(u\) quenches at the time \(T\), then there exists a time \(T_0\in (T-\varepsilon /2, T)\) such that \(\| u(\cdot , t)\| _{\infty }\geq b-\tfrac {\alpha }{2}\) for \(t\in [T_0, T)\). Invoking Theorem 4.1, we note that the problem (3)–(4) admits a unique solution \(v\), and the following estimate holds \(\| v(\cdot , T_0)-u(\cdot , T_0)\| _{\infty }\leq \tfrac {\alpha }{2}\). Making use of the triangle inequality, we find that </p>
<div class="displaymath" id="a0000000125">
  \[  \| v(\cdot , T_0)\| _{\infty }\geq \| u(\cdot , T_0)\| _{\infty }-\| v(\cdot , T_0)-u(\cdot , T_0)\| _{\infty },  \]
</div>
<p> which implies that </p>
<div class="displaymath" id="a0000000126">
  \[  \| v(\cdot , T_0)\| _{\infty }\geq b-\tfrac {\alpha }{2}-\tfrac {\alpha }{2}=b-\alpha .  \]
</div>
<p> In Remark 2.1 of the paper, we have revealed that \(T_h\geq T\). We infer from (21) and Remark 3.1 that </p>
<div class="displaymath" id="a0000000127">
  \[  0\leq T_h-T\leq T_h-T_0+T_0-T\leq \tfrac {\varepsilon }{2}+\tfrac {\varepsilon }{2}=\varepsilon ,  \]
</div>
<p> and the proof is complete. </p>
<p><div class="remark_thmwrapper " id="a0000000128">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">16</span>
  </div>
  <div class="remark_thmcontent">
  <p>If in Theorem 4.2 we replace the assumption of Theorem 3.2 by that of Theorem 3.3, then the result of Theorem 4.2 remains valid.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000129">5 Numerical results</h1>
<p>In this section, we give some computational experiments to confirm the theory given in the previous section. We consider the problem (1)-(2) in the case where \(\Omega =(-1, 1)\), \(f(u)= (1-u)^{p}\) with \(p{\gt}1\), </p>
<div class="displaymath" id="a0000000130">
  \begin{align*}  J(x)= \begin{cases}  \tfrac {3}{2}x^2, &  {\rm if}\  |x|{\lt}1, \\ 0, &  {\rm if}\  |x|\geq 1, \end{cases}\end{align*}
</div>
<p> \(u_0(x)=\gamma (\tfrac {2-\varepsilon (\sin (\pi x))^2}{4})\) with \(\gamma {\gt}0\), \(\varepsilon \in (0, 1]\). We start by the construction of some adaptive schemes as follows. Let \(I\) be a positive integer, and let \(h=2/I\). Define the grid \(x_i=-1+ih,\) \(0\leq i \leq I\), and approximate the solution \(u\) of (1)-(2) by the solution \(U_h^{(n)}=(U_0^{(n)}, \cdots , U_I^{(n)})^T\) of the following explicit scheme </p>
<div class="displaymath" id="a0000000131">
  \begin{align*}  \tfrac {U_i^{(n+1)}-U_i^{(n)}}{\Delta t_n}= &  \sum _{j=0}^{I-1}hJ(x_i-x_j)(U_j^{(n)}-U_i^{(n)})+(1-U_i^{(n)})^{-p}, \quad 0\leq i \leq I, \\ U_i^{(0)}=&  \varphi _i, \quad 0\leq i\leq I, \end{align*}
</div>
<p> where \(\varphi _i=\gamma (\tfrac {2-\varepsilon (\sin (\pi x_i))^2}{4})\). In order to permit the discrete solution to reproduce the properties of the continuous one when the time \(t\) approaches the quenching time \(T\), we need to adapt the size of the time step so that we take </p>
<div class="displaymath" id="a0000000132">
  \[  \Delta t_n=h^2(1- \| U_h^{(n)}\| _{\infty })^{p+1}  \]
</div>
<p> with \(\| U_h^{(n)}\| _{\infty }=\max _{0\leq i \leq I}|U_i^{(n)}|.\) Let us notice that the restriction on the time step ensures the nonnegativity of the discrete solution. We also approximate the solution \(u\) of (1)–(2) by the solution \(U_h^{(n)}\) of the implicit scheme below </p>
<div class="displaymath" id="a0000000133">
  \begin{align*}  \tfrac {U_i^{(n+1)}-U_i^{(n)}}{\Delta t_n}=&  \sum _{j=0}^{I-1}hJ(x_i-x_j)(U_j^{(n+1)}-U_i^{(n+1)})+ (1-U_i^{(n)})^{-p}, \quad 0\leq i \leq I, \\ U_i^{(0)}=& \varphi _i, \quad 0\leq i\leq I. \end{align*}
</div>
<p> As in the case of the explicit scheme, here, we also choose </p>
<div class="displaymath" id="a0000000134">
  \[  \Delta t_n=h^2(1- \| U_h^{(n)}\| _{\infty })^{p+1}.  \]
</div>
<p> Let us again remark that for the above implicit scheme, existence and nonnegativity of the discrete solution are also guaranteed using standard methods (see, for instance [9]). We need the following definition. </p>
<p><div class="defi_thmwrapper " id="a0000000135">
  <div class="defi_thmheading">
    <span class="defi_thmcaption">
    Definition
    </span>
    <span class="defi_thmlabel">17</span>
  </div>
  <div class="defi_thmcontent">
  <p>We say that the discrete solution \(U_h^{(n)}\) of the explicit scheme or the implicit scheme quenches in a finite time if \(\lim _{n\rightarrow \infty }\| U_h^{(n)}\| _{\infty }=1\), and the series \(\sum _{n=0}^{\infty } \Delta t_n\) converges. The quantity \(\sum _{n=0}^{\infty } \Delta t_n\) is called the numerical quenching time of the discrete solution \(U_h^{(n)}.\) </p>

  </div>
</div> </p>
<p>In the following tables, in rows, we present the numerical quenching times, the numbers of iterations, the CPU times and the orders of the approximations corresponding to meshes of 16, 32, 64, 128. We take for the numerical blow-up time \(t_n=\sum _{j=0}^{n-1}\Delta t_{j}\) which is computed at the first time when </p>
<div class="displaymath" id="a0000000136">
  \[  \Delta t_{n}=|t_{n+1}-t_{n}|\leq 10^{-16}.  \]
</div>
<p> The order \((s)\) of the method is computed from </p>
<div class="displaymath" id="a0000000137">
  \[  s=\tfrac {\log ((T_{4h}-T_{2h})/(T_{2h}-T_{h}))}{\log (2)}.  \]
</div>
<p>Tables 1–8 show the numerical quenching times, numbers of iterations, CPU times (seconds) and orders of the approximations obtained with the explicit and implicit Euler method. </p>
<div class="table"  id="a0000000138">
   <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>
    <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> \(s\)</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> 0.1267303</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> 74 </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> 1.1 </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> -</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> 0.1254584</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> 295 </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> 1.8 </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> -</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> 0.1251387</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> 1179 </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> 6.26 </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>1.992 </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> 0.1250586</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> 4715 </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> 78 </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> 1.996 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Explicit Euler method, \(p=1\), \(\gamma =1\), \(\varepsilon =1/100\)</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000139">
   <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>
    <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> \(s\)</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> 0.1267353</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> 75 </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> 1.8 </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> -</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> 0.1254588</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> 295 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.1 </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> -</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> 0.1251389</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> 1179 </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> 6.26 </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>1.995 </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> 0.1250587</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> 4715 </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> 78 </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> 1.996 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">The implicit Euler method, \(p=1\), \(\gamma =1\), \(\varepsilon =1/100\)</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000140">
   <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>
    <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> \(s\)</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> 0.0326408</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> 35 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.2 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -</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> 0.0319626</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> 139 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 15.6 </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> -</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> 0.0317912</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> 554 </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> 74 </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> 1.98 </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> 0.0317480</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> 2215 </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> 79 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 2.01 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">The explicit Euler method, \(p=1\), \(\gamma =1.5\), \(\varepsilon =1\)</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000141">
   <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>
    <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> \(s\)</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> 0.0326418</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> 35 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.4 </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> -</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> 0.0319629</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> 139 </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> 16.1 </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> -</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> 0.0317915</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> 554 </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> 75.4 </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> 1.981 </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> 0.0317481</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> 2215 </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> 79 </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> 1.985 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4</span> 
  <span class="caption_text">The implicit Euler method, \(p=1\), \(\gamma =1.5\), \(\varepsilon =1\)</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000142">
   <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>
    <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> \(s\)</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> 0.0321300</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> 35 </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> 1.4 </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> -</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> 0.0314844</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> 137 </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> 6.1 </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> -</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> 0.0313305</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> 548 </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> 62 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.89 </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> 0.0312703</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> 2189 </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> 96 </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> 1.35 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">5</span> 
  <span class="caption_text">The explicit Euler method, \(p=1\), \(\gamma =1.5\), \(\varepsilon =1/100\)</span> 
</figcaption> 
</div>
<div class="table"  id="a0000000143">
   <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>
    <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> \(s\)</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> 0.0321305</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> 35 </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> 1.6 </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> -</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> 0.0314846</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> 137 </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> 6.1 </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> -</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> 0.0313325</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> 548 </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> 63 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.08 </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> 0.0312713</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> 2189 </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> 98 </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> 1.31 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">6</span> 
  <span class="caption_text">The implicit Euler method, \(p=1\), \(\gamma =1.5\), \(\varepsilon =1/100\)</span> 
</figcaption> 
</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>
    <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> \(s\)</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> 3.684595 e-4</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>
    <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> 1.5 </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> -</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> 3.371334 e-4</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> 13 </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> 9.6 </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> -</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> 3.190056 e-4</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>
    <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> 78 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.02 </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> 3.145623 e-4</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> 207 </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> 499 </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> 0.78 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">7</span> 
  <span class="caption_text">The explicit Euler method, \(p=1\), \(\gamma =1.95\), \(\varepsilon =1\)</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>
    <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> \(s\)</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> 3.684596 e-4</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>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 2.5 </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> -</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> 3.371394 e-4</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> 13 </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> 10.6 </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> -</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> 3.190086 e-4</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>
    <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> 79 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>2.02 </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> 3.145624 e-4</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> 207 </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> 499 </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> 0.78 </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">8</span> 
  <span class="caption_text">The implicit Euler method, \(p=1\), \(\gamma =1.95\), \(\varepsilon =1\)</span> 
</figcaption> 
</div>
<p><div class="remark_thmwrapper " id="a0000000146">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">18</span>
  </div>
  <div class="remark_thmcontent">
  <p>If we consider the problem (1)–(2) in the case where \(f(u)=(1-u)^{-1}\) and \(u_0(x)=1/2\), then we know from Remark 3.3 that the quenching time of the solution \(u\) equals \(0.125\). We observe from Tables 1 to 2 that when \(\varepsilon \) is small enough, then the numerical quenching time is approximately equal to \(0.125\). This fact confirms the result established within the framework of the continuity. On the other hand, the quenching time \(T_e\) of the solution of the following differential equation \(\alpha '(t)=(1-\alpha (t))^{-1}\), \(t{\gt}0\), \(\alpha (0)=\tfrac {\gamma }{2}\) is given explicitly by \(T_e=(2-\gamma )^{2}/8,\) and </p>
<div class="displaymath" id="a0000000147">
  \[  T_e= \left\{ \begin{array}{ll} \hbox{3.145 e-4 \quad \mbox{when} \quad $\gamma $=1.95 } \\ \end{array}\right\} .  \]
</div>
<p> We note from Tables 3 to 4 that, when \(\gamma =1.95\), then numerical quenching time of the discrete solution is approximately equal that to \(T_e\). These results illustrate the idea of Theorem 3.5.</p>

  </div>
</div> </p>
<p>For other illustrations, in what follows, we shall give some plots. In the following figures, we can appreciate that the discrete solution quenches in a finite time. </p>

<figure >
  <div class="centered"><div class="minipage" style="width: 156.49015748031493pt"> <img src="img-0001.png" alt="\includegraphics[ scale=0.4]{fig11-eps-converted-to.png}" style="width:700.0px; height:527.2px" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Evolution of the discrete solution, \(\gamma =1\), \(\varepsilon =\frac1{100}\) (explicit scheme).</span> 
</figcaption> </div> &#8195;&#8195;&#8195;<div class="minipage" style="width: 156.49015748031493pt"> <img src="img-0002.png" alt="\includegraphics[ scale=0.4]{fig12-eps-converted-to.png}" style="width:700.0px; height:527.2px" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Evolution of the discrete solution, \(\gamma =1\), \(\varepsilon =\frac1{100}\) (implicit scheme).</span> 
</figcaption>  </div> </div>

</figure>

<figure >
  <div class="centered"><div class="minipage" style="width: 156.49015748031493pt"> <img src="img-0003.png" alt="\includegraphics[ scale=0.4]{fig05-eps-converted-to.png}" style="width:700.0px; height:527.2px" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">Evolution of the discrete solution, \(\gamma =1.5\), \(\varepsilon =1\) (explicit scheme).</span> 
</figcaption>  </div> &#8195;&#8195;&#8195;<div class="minipage" style="width: 156.49015748031493pt"> <img src="img-0004.png" alt="\includegraphics[ scale=0.4]{fig06-eps-converted-to.png}" style="width:700.0px; height:527.2px" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">4</span> 
  <span class="caption_text">Evolution of the discrete solution, \(\gamma =1.5\), \(\varepsilon =1\) (implicit scheme).</span> 
</figcaption>  </div> </div>

</figure>
<p><br /></p>
<figure >
  <div class="centered"><div class="minipage" style="width: 156.49015748031493pt"> <img src="img-0005.png" alt="\includegraphics[scale=0.4]{fig3exp-eps-converted-to.png}" style="width:700.0px; height:527.2px" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">5</span> 
  <span class="caption_text">Evolution of the discrete solution, \(\gamma =1.95\), \(\varepsilon =1\) (explicit scheme).</span> 
</figcaption>  </div> &#8195;&#8195;&#8195;<div class="minipage" style="width: 156.49015748031493pt"> <img src="img-0006.png" alt="\includegraphics[ scale=0.4]{fig4imp-eps-converted-to.png}" style="width:700.0px; height:527.2px" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">6</span> 
  <span class="caption_text">Evolution of the discrete solution, \(\gamma =1.95\), \(\varepsilon =1\) (implicit scheme).</span> 
</figcaption>  </div> </div>

</figure>
<p><div class="acknowledgement_thmwrapper " id="a0000000148">
  <div class="acknowledgement_thmheading">
    <span class="acknowledgement_thmcaption">
    Acknowledgement
    </span>
  </div>
  <div class="acknowledgement_thmcontent">
  <p>The authors would like to thank the referees for the throughout reading of the manuscript and several suggestions that helped us improve the presentation of the paper. </p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="1">1</a></dt>
  <dd><p><i class="sc">A. Acker</i> and <i class="sc">B. Kawohl</i>, <i class="it">Remarks on quenching</i>, Nonl. Anal. TMA, <b class="bfseries">13</b> (1989), pp. 53–61. </p>
</dd>
  <dt><a name="2">2</a></dt>
  <dd><p><i class="sc">F. Andreu, J. M. Mazon, J. D. Rossi</i> and <i class="sc">J. Toledo</i>, <i class="it">The Neumann problem for nonlocal nonlinear diffusion equations</i>, J. Evol. Equations, <b class="bfseries">8</b>(1) (2008), pp. 189–215. </p>
</dd>
  <dt><a name="3">3</a></dt>
  <dd><p><i class="sc">F. Andreu, J. M. Mazon, J. D. Rossi</i> and <i class="sc">J. Toledo</i>, <i class="it">A nonlocal p-Laplacian evolution equation with Neumann boundary conditions</i>, Preprint. </p>
</dd>
  <dt><a name="4">4</a></dt>
  <dd><p><i class="sc">P. Bates</i> and <i class="sc">A. Chmaj</i>, <i class="it">An intergrodifferential model for phase transitions: stationary solutions in higher dimensions</i>, J. Statistical Phys., <b class="bfseries">95</b> (1999), pp. 1119–1139. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><i class="sc">P. Bates</i> and <i class="sc">A. Chmaj</i>, <i class="it">A discrete convolution model for phase transitions</i>, Arch. Rat. Mech. Anal., <b class="bfseries">150</b> (1999), pp. 281–305. </p>
</dd>
  <dt><a name="6">6</a></dt>
  <dd><p><i class="sc">P. Bates</i> and <i class="sc">J. Han</i>, <i class="it">The Dirichlet boundary problem for a nonlocal Cahn-Hilliard equation</i>, J. Math. Anal. Appl., <b class="bfseries">311</b>(1) (2005), pp.&#160;289–312. </p>
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</dl>


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