ABSTRACT

This paper studies quenching properties of solutions of a semilinear parabolic system with localized reaction sources in a square domain. The system has the homogeneous Dirichlet boundary condition and null initial condition. We prove that solutions quench simultaneously, and compute approximated critical values of the system using a numerical method.

Keywords:

Simultaneous Quenching, Parabolic System, Conformal Mapping

1. Introduction

Let $\Gamma \in \left(0,\infty \right]$, a and b be positive constants such that $a\le b$,$D=\left(-1,1\right)\times -\left(1,1\right)$,$\bar{D}=\left[-1,1\right]\times \left[-1,1\right]$,$\partial D=\left\{\left\{-1\right\}\times \left[-1,1\right]\right\}\cup \left\{\left\{1\right\}\times \left[-1,1\right]\right\}\cup \left\{\left[-1,1\right]\times \left\{-1\right\}\right\}\cup \left\{\left[-1,1\right]\times \left\{1\right\}\right\}$. We also let L be the operator such that $Lu=u_t-\Delta u$. In this paper, we study the following semilinear parabolic system:

$\begin{array}{l}Lu=\frac{a}{1-v\left(0,0,t\right)}\text{in}D\times \left(0,\Gamma \right),\\ Lv=\frac{b}{1-u\left(0,0,t\right)}\text{in}D\times \left(0,\Gamma \right),\end{array}\}$ (1.1)

$\begin{array}{l}u\left(x,y,0\right)=0\text{and}v\left(x,y,0\right)=0\text{on}\bar{D},\\ u\left(x,y,t\right)=0\text{and}v\left(x,y,t\right)=0\text{on}\partial D\times \left(0,\Gamma \right).\end{array}\}$ (1.2)

Problem (1.1)-(1.2) illustrates the instabilities in some dynamical systems in which certain reactions are localized to electrodes, catalytic membranes, or other surfaces and local sites are immersed in a bulk medium happening at the origin (0, 0), see [1] [2] . Additionally, (1.1) describes a thermal ignition driven by the temperature at a single point, see [3] . Chadam, Peirce, and Yin [4] examined the blow-up set of solutions when the initial data are nontrivial and nonnegative bounded functions.

The quenching problem was initiated by Kawarada [5] . The model describes polarization phenomena in ionic conductors. Quenching also illustrates the phase transition between liquids and solids [6] . Chang, Hsu, and Liu [7] discussed the quenching rate of the problem (1.1)-(1.2) in an n-dimensional ball.

The solutions u and v are said to quench if there exists a finite time $\Gamma$ such that

$\begin{array}{l}\max \left\{u\left(x,y,t\right):\left(x,y\right)\in \bar{D}\right\}\to 1^{-}\\ \text{and}\text{max}\left\{v\left(x,y,t\right):\left(x,y\right)\in \bar{D}\right\}\to 1^{-}\text{as}t\to {\Gamma }^{-}.\end{array}$

To problem (1.1)-(1.2), there are critical values a* and b* (both are positive) such that the maximum of solutions u and v reaches 1 in a finite time if a > a* and b > b* while u and v exist globally and are bounded above by 1 if a < a* and b < b*, see [8] .

The purposes of this paper are to prove solutions u and v to quench simultaneously at (0,0), and use a numerical method to determine approximated values of a* and b* of the problem (1.1)-(1.2).

This paper is organized as follows. In Section 2, we prove that there are unique solutions u and v of the problem (1.1)-(1.2). In Section 3, we prove that either u or v quenches in a finite time. Then, we show that solutions u and v quench simultaneously at (0, 0). In Section 4, we calculate approximated values of a* and b*. This a* and b* associate with the existence of solutions of their steady state problem of the problem (1.1)-(1.2). Our numerical method is to evaluate an approximation of the steady solutions expressed in an integral representation form. For illustration, some examples are provided.

2. Properties of u and v

Let ${\beta }_1\left(x,y,t\right)$ and ${\beta }_2\left(x,y,t\right)$ be nontrivial, nonnegative, and bounded functions on $\bar{D}\times \left(0,\infty \right)$. Here is a comparison theorem

Lemma 2.1. Suppose that $u\left(x,y,t\right)$ and $v\left(x,y,t\right)$ are solutions of the following system:

$\begin{array}{l}Lu\ge {\beta }_1\left(x,y,t\right)v\left(0,0,t\right)\text{in}D\times \left(0,\Gamma \right),\\ Lv\ge {\beta }_2\left(x,y,t\right)u\left(0,0,t\right)\text{in}D\times \left(0,\Gamma \right),\\ u\left(x,y,0\right)\ge 0\text{and}v\left(x,y,0\right)\ge 0\text{on}\bar{D},\\ u\left(x,y,t\right)\ge 0\text{and}v\left(x,y,t\right)\ge 0\text{on}\partial D\times \left(0,\Gamma \right).\end{array}$

Then, $u\left(x,y,t\right)\ge 0$ and $v\left(x,y,t\right)\ge 0$ on $\bar{D}\times \left[0,\Gamma \right)$.

Proof. Let $\epsilon$ be a positive real number, and

$\begin{array}{l}\Phi \left(x,y,t\right)=u\left(x,y,t\right)+\epsilon \left[\sin \left(\frac{\text{π}}{2}\left(x+1\right)\right)+\frac{1}{2}\right]{\text{e}}^{\gamma t},\\ \Psi \left(x,y,t\right)=v\left(x,y,t\right)+\epsilon \left[\sin \left(\frac{\text{π}}{2}\left(x+1\right)\right)+\frac{1}{2}\right]{\text{e}}^{\gamma t},\end{array}$

where $\gamma$ is a positive real number to be determined. From the construction, $\Phi \left(x,y,0\right)>0$ and $\Psi \left(x,y,0\right)>0$ on $\bar{D}$. By assumptions and a direction computation,

$\begin{array}{l}{\Phi }_t-\Delta \Phi -{\beta }_1\Psi \left(0,0,t\right)\\ =u_t-\Delta u+\epsilon {\text{e}}^{\gamma t}\left[\left(\gamma +\frac{{\text{π}}^2}{4}\right)\sin \left(\frac{\text{π}}{2}\left(x+1\right)\right)+\frac{\gamma }{2}\right]-{\beta }_1\left(v\left(0,0,t\right)+\frac{3\epsilon }{2}{\text{e}}^{\gamma t}\right)\\ \ge \frac{\epsilon {\text{e}}^{\gamma t}}{2}\left(\gamma -3{\beta }_1\right).\end{array}$

We choose $\gamma$ such that $\gamma >3{\beta }_1$. Thus,

${\Phi }_t-\Delta \Phi -{\beta }_1\Psi \left(0,0,t\right)>0\text{in}D\times \left(0,\Gamma \right).$

Suppose $\Phi \left(x,y,t\right)\le 0$ somewhere in $D\times \left(0,\Gamma \right)$. Then, the set

$\left\{t:\Phi \left(x,y,t\right)\le 0\text{for}\text{some}\left(x,y\right)\in D\right\}$

is non-empty. Let $\tilde{t}$ denote its infimum. Then, $0<\tilde{t}<\Gamma$ because $\Phi \left(x,y,0\right)>0$ on $\bar{D}$. Thus, there exists some $\left(x_1,y_1\right)\in D$ such that $\Phi \left(x_1,y_1,\tilde{t}\right)=0$ and ${\Phi }_t\left(x_1,y_1,\tilde{t}\right)\le 0$. On the other hand, $\Phi$ attains its local minimum at $\left(x_1,y_1,\tilde{t}\right)$. Therefore, $\Delta \Phi \left(x_1,y_1,\tilde{t}\right)>0$. Then, at $t=\tilde{t}$,

${\Phi }_t\left(x_1,y_1,\tilde{t}\right)-{\beta }_1\Psi \left(0,0,\tilde{t}\right)>L\Phi \left(x_1,y_1,\tilde{t}\right)-{\beta }_1\Psi \left(0,0,\tilde{t}\right)>0.$ (2.1)

Follow a similar argument, we assume that $\Psi \left(x,y,t\right)\le 0$ somewhere in $D\times \left(0,\Gamma \right)$. Then, there exist some $\hat{t}\in \left(0,\Gamma \right)$ and $\left(x_2,y_2\right)\in D$ such that $\Psi \left(x_2,y_2,\hat{t}\right)=0$,${\Psi }_t\left(x_2,y_2,\hat{t}\right)\le 0$, and $\Psi$ attains its local minimum at $\left(x_2,y_2,\hat{t}\right)$. Then at $t=\tilde{t}$

${\Psi }_t\left(x_2,y_2,\hat{t}\right)-{\beta }_2\Phi \left(0,0,\hat{t}\right)>L\Psi \left(x_2,y_2,\hat{t}\right)-{\beta }_2\Phi \left(0,0,\hat{t}\right)>0.$ (2.2)

Let us assume that $\hat{t}<\tilde{t}$. Since $\Phi$ attains its local minimum at $\left(x_1,y_1,\tilde{t}\right)$,$\Phi \left(0,0,\hat{t}\right)>0$. By inequality (2.2),

$0\ge {\Psi }_t\left(x_2,y_2,\hat{t}\right)\ge {\Psi }_t\left(x_2,y_2,\hat{t}\right)-{\beta }_2\Phi \left(0,0,\hat{t}\right)>0.$

This gives a contradiction. Hence, $\Psi \left(x,y,t\right)>0$ in $D\times \left(0,\Gamma \right)$. Then, by (2.1), we show that $\Phi \left(x,y,t\right)>0$ in $D\times \left(0,\Gamma \right)$. Through a similar calculation, we obtain the same result when $\hat{t}\ge \tilde{t}$. Let $\epsilon \to 0$, we have $u\left(x,y,t\right)\ge 0$ and $v\left(x,y,t\right)\ge 0$ in $D\times \left(0,\Gamma \right)$. Follow $u\ge 0$ and $v\ge 0$ on $\bar{D}$ and $\partial D\times \left(0,\Gamma \right)$, u and v are non-negative on $D\times \left[0,\Gamma \right)$. The proof is complete. ,

By Lemma 2.1, 0 is a lower solution of the problem (1.1)-(1.2). On the other hand, if u and v do not quench, then ${\max }_{\bar{D}}u<1$ and ${\max }_{\bar{D}}v<1$, and solutions u and v are bounded above by 1. Further, u and v cease to exist when $u\ge 1$ and $v\ge 1$. Therefore, $0\le u<1$ and $0\le v<1$ on $\bar{D}\times \left[0,\Gamma \right)$. The existence of classical solutions of the problem (1.1)-(1.2) is able to obtain by the Schauder fixed point theorem of [ [9] , pp. 502-504], and by Lemma 2.1 u and v are unique.

Theorem 2.2. Problem (1.1)-(1.2) has unique classical solutions u and $v\in C^{2+\alpha ,1+\alpha /2}\left(\bar{D}\times \left[0,\Gamma \right)\right)$ for some $\alpha \in \left(0,1\right)$ such that $0\le u,v<1$ on $\bar{D}\times \left[0,\Gamma \right)$.

Lemma 2.3. $u_t\ge 0$ and $v_t\ge 0$ on $\bar{D}\times \left[0,\Gamma \right)$. Further, $u_t>0$ and $v_t>0$ in $D\times \left(0,\Gamma \right)$.

Proof. From Theorem 2.2, $u\left(x,y,t\right)$ and $v\left(x,y,t\right)$ are nonnegative on $\bar{D}\times \left[0,\Gamma \right)$. Let h be a real number in $\left(0,\Gamma \right)$. Then, $u\left(x,y,h\right)\ge 0$ and $v\left(x,y,h\right)\ge 0$ on $\bar{D}$, and $u\left(x,y,t+h\right)=0$ and $v\left(x,y,t+h\right)=0$ on $\partial D\times \left[0,\Gamma -h\right)$. By the mean value theorem, there exist u₁ (between $u\left(0,0,t+h\right)$ and $u\left(0,0,t\right)$ ) and v₁ (between $v\left(0,0,t+h\right)$ and $v\left(0,0,t\right)$ ) such that

$\begin{array}{l}Lu\left(x,y,t+h\right)-Lu\left(x,y,t\right)=\frac{a}{{\left(1-v_1\right)}^2}\left(v\left(0,0,t+h\right)-v\left(0,0,t\right)\right),\\ Lv\left(x,y,t+h\right)-Lv\left(x,y,t\right)=\frac{b}{{\left(1-u_1\right)}^2}\left(u\left(0,0,t+h\right)-u\left(0,0,t\right)\right).\end{array}$

By Lemma 2.1, $u\left(x,y,t+h\right)\ge u\left(x,y,t\right)$ and $v\left(x,y,t+h\right)\ge v\left(x,y,t\right)$ on $\bar{D}\times \left[0,\Gamma -h\right)$. This gives $\frac{u\left(x,y,t+h\right)-u\left(x,y,t\right)}{h}\ge 0$ and $\frac{v\left(x,y,t+h\right)-v\left(x,y,t\right)}{h}\ge 0$ on $\bar{D}\times \left[0,\Gamma -h\right)$.

Taking $h\to 0$, u_t and v_t are nonnegative on $\bar{D}\times \left[0,\Gamma \right)$, respectively. To show that u_t and v_t are positive, let us differentiate (1.1) with respect to t. Then, u_t and v_t satisfy

$\begin{array}{l}L{u}_t=\frac{a}{{\left(1-v\left(0,0,t\right)\right)}^2}{v}_t\left(0,0,t\right)\ge 0\text{in}D\times \left(0,\Gamma \right),\\ L{v}_t=\frac{b}{{\left(1-u\left(0,0,t\right)\right)}^2}{u}_t\left(0,0,t\right)\ge 0\text{in}D\times \left(0,\Gamma \right),\end{array}\}$

$\begin{array}{l}{u}_t\left(x,y,0\right)\ge 0\text{and}{v}_t\left(x,y,0\right)\ge 0\text{on}\bar{D},\\ u_t\left(x,y,t\right)=0\text{and}{v}_t\left(x,y,t\right)=0\text{on}\partial D\times \left(0,\Gamma \right).\end{array}\}$

By the maximum principle [ [10] , p. 54], $u_t>0$ and $v_t>0$ in $D\times \left(0,\Gamma \right)$. ,

3. Simultaneous Quenching and Global Existence

In this section, we show that either u or v quenches in a finite time first. Then, we prove that u and v quench in the same time at (0, 0). Afterward, we prove that the problem (1.1)-(1.2) has a global solution when a and b are sufficiently small.

Lemma 3.1. u and v both attain their maximum at (0, 0) for all $t\in \left(0,\Gamma \right)$.

Proof. It suffices to prove that u attains its maximum along the x and y axes. Let us consider the first equation of (1.1) along the x-axis, we have

$Lu\left(x,0,t\right)=u_t\left(x,0,t\right)-u_{xx}\left(x,0,t\right)=\frac{a}{1-v\left(0,0,t\right)}.$

Differentiate the above equation with respect to x to yield

$L{u}_x\left(x,0,t\right)=0.$

By the symmetry of D with respect to the x and y axes, $u_x\left(0,0,t\right)=0$ for all $t\in \left(0,\Gamma \right)$. By the Hopf’s lemma [11] , p. 170], $u_x\left(-1,0,t\right)>0$ for all $t\in \left(0,\Gamma \right)$. At t = 0, $u_x\left(x,0,0\right)=0$ for all $x\in \left[-1,0\right]$. By the maximum principle, $u_x\left(x,0,t\right)>0$ for all $x\in \left[-1,0\right)$ when $t\in \left(0,\Gamma \right)$. Similarly, for all $t\in \left(0,\Gamma \right)$, we obtain $u_x\left(x,0,t\right)<0$ when $x\in \left(0,1\right]$. Therefore, $u\left(0,0,t\right)\ge u\left(x,0,t\right)$ for all $x\in \left[-1,1\right]$ when $t\in \left(0,\Gamma \right)$. Likewise, $u\left(0,0,t\right)\ge u\left(0,y,t\right)$ for all $y\in \left[-1,1\right]$ when $t\in \left(0,\Gamma \right)$. Thus, $u\left(0,0,t\right)\ge u\left(x,y,t\right)$ on $\bar{D}\times \left(0,\Gamma \right)$. Similarly, $v\left(0,0,t\right)\ge v\left(x,y,t\right)$ on $\bar{D}\times \left(0,\Gamma \right)$. Hence, u and v both attain their maximum at (0, 0) for all $t\in \left(0,\Gamma \right)$. ,

Let ${\varphi }_1$ be the eigenfunction corresponding to the first eigenvalue ${\lambda }_1\left(>0\right)$ of the Sturm-Liuoville problem below,

$\Delta \varphi +\lambda \varphi =0\text{in}D,\varphi =0\text{on}\partial D.$

This eigenfunction has the properties: ${\varphi }_1>0$ in D and ${\displaystyle \underset{D}{\iint }{\varphi }_1\text{d}x\text{d}y}=1$ [ [12] , p. 10]. Let c be a positive real number such that $c\le ab/\left(a+b\right)$. We show that either u or v quenches in a finite time.

Lemma 3.2. If $2\sqrt{2}c>{\lambda }_1$, where $c\le ab/\left(a+b\right)$, then either u or v quenches on $\bar{D}$ in a finite time $\tilde{\Gamma }$.

Proof. By Lemma 3.1, $u\left(0,0,t\right)\ge u\left(x,y,t\right)$ and $v\left(0,0,t\right)\ge v\left(x,y,t\right)$ on $\bar{D}\times \left(0,\Gamma \right)$. Let $\hat{u}\left(x,y,t\right)$ and $\hat{v}\left(x,y,t\right)$ be the solutions to the following parabolic system:

$\begin{array}{l}L\hat{u}=\frac{a}{1-\hat{v}\left(x,y,t\right)}\text{in}D\times \left(0,\Gamma \right),\\ L\hat{v}=\frac{b}{1-\hat{u}\left(x,y,t\right)}\text{in}D\times \left(0,\Gamma \right),\end{array}\}$ (3.1)

$\begin{array}{l}\hat{u}\left(x,y,0\right)=0\text{and}\hat{v}\left(x,y,0\right)=0\text{on}\bar{D},\\ \hat{u}\left(x,y,t\right)=0\text{and}\hat{v}\left(x,y,t\right)=0\text{on}\partial D\times \left(0,\Gamma \right).\end{array}\}$ (3.2)

By the maximum principle, $\hat{u}\left(x,y,t\right)\ge 0$ and $\hat{v}\left(x,y,t\right)\ge 0$ on $\bar{D}\times \left[0,\Gamma \right)$. Further, $u-\hat{u}$ and $v-\hat{v}$ satisfy the system below:

$\begin{array}{l}L\left(u-\hat{u}\right)=\frac{a}{1-v\left(0,0,t\right)}-\frac{a}{1-\hat{v}\left(x,y,t\right)}\ge \frac{a}{1-v\left(x,y,t\right)}-\frac{a}{1-\hat{v}\left(x,y,t\right)},\\ L\left(v-\hat{v}\right)=\frac{b}{1-u\left(0,0,t\right)}-\frac{b}{1-\hat{u}\left(x,y,t\right)}\ge \frac{b}{1-u\left(x,y,t\right)}-\frac{b}{1-\hat{u}\left(x,y,t\right)}.\end{array}$

By $u-\hat{u}=0$ and $v-\hat{v}=0$ on $\bar{D}$ and $\partial D\times \left(0,\Gamma \right)$, and the maximum principle, we have $u\ge \hat{u}$ and $v\ge \hat{v}$ on $\bar{D}\times \left[0,\Gamma \right)$. It suffices to prove either $\hat{u}$ or $\hat{v}$ to quench over $\bar{D}$ in a finite time.

Multiplying ${\varphi }_1$ on both sides of (3.1) and integrating expressions over the domain D, we obtain

$\begin{array}{l}{\displaystyle \underset{D}{\iint }{\hat{u}}_t{\varphi }_1\text{d}x\text{d}y}-{\displaystyle \underset{D}{\iint }\Delta \hat{u}{\varphi }_1\text{d}x\text{d}y}=a{\displaystyle \underset{D}{\iint }\frac{{\varphi }_1}{1-\hat{v}}\text{d}x\text{d}y},\\ {\displaystyle \underset{D}{\iint }{\hat{v}}_t{\varphi }_1\text{d}x\text{d}y}-{\displaystyle \underset{D}{\iint }\Delta \hat{v}{\varphi }_1\text{d}x\text{d}y}=b{\displaystyle \underset{D}{\iint }\frac{{\varphi }_1}{1-\hat{u}}\text{d}x\text{d}y}.\end{array}$

Use the Green’s second identity [ [10] , p. 96] and (3.2), it yields

$\begin{array}{l}{\left({\displaystyle \underset{D}{\iint }\hat{u}{\varphi }_1\text{d}x\text{d}y}\right)}_t=-{\lambda }_1{\displaystyle \underset{D}{\iint }\hat{u}{\varphi }_1\text{d}x\text{d}y}+a{\displaystyle \underset{D}{\iint }\frac{{\varphi }_1}{1-\hat{v}}\text{d}x\text{d}y},\\ {\left({\displaystyle \underset{D}{\iint }\hat{v}{\varphi }_1\text{d}x\text{d}y}\right)}_t=-{\lambda }_1{\displaystyle \underset{D}{\iint }\hat{v}{\varphi }_1\text{d}x\text{d}y}+b{\displaystyle \underset{D}{\iint }\frac{{\varphi }_1}{1-\hat{u}}\text{d}x\text{d}y}.\end{array}$

By the Maclaurin’s series, we have

$\begin{array}{l}{\left({\displaystyle \underset{D}{\iint }\hat{u}{\varphi }_1\text{d}x\text{d}y}\right)}_t\ge -{\lambda }_1{\displaystyle \underset{D}{\iint }\hat{u}{\varphi }_1\text{d}x\text{d}y}+a{\displaystyle \underset{D}{\iint }{\hat{v}}^2{\varphi }_1\text{d}x\text{d}y}+a{\displaystyle \underset{D}{\iint }{\varphi }_1\text{d}x\text{d}y},\\ {\left({\displaystyle \underset{D}{\iint }\hat{v}{\varphi }_1\text{d}x\text{d}y}\right)}_t\ge -{\lambda }_1{\displaystyle \underset{D}{\iint }\hat{v}{\varphi }_1\text{d}x\text{d}y}+b{\displaystyle \underset{D}{\iint }{\hat{u}}^2{\varphi }_1\text{d}x\text{d}y}+b{\displaystyle \underset{D}{\iint }{\varphi }_1\text{d}x\text{d}y}.\end{array}$

Let $R\left(t\right)={\displaystyle \underset{D}{\iint }\hat{u}{\varphi }_1\text{d}x\text{d}y}$ and $P\left(t\right)={\displaystyle \underset{D}{\iint }\hat{v}{\varphi }_1\text{d}x\text{d}y}$. Adding above inequalities together and using the Jensen’s inequality [ [12] , p. 11], we obtain

$\frac{\text{d}}{\text{d}t}\left(R+P\right)\ge -{\lambda }_1\left(R+P\right)+a{P}^2+b{R}^2+a+b.$ (3.3)

As $c\le ab/\left(a+b\right)\le a\le b$, we have

$\frac{\left(a-c\right)P^2+\left(b-c\right)R^2}{2}\ge \sqrt{\left(a+b\right)\left(\frac{ab}{a+b}-c\right)+c^2}PR\ge cPR.$

Hence,

$a{P}^2+b{R}^2\ge c{\left(R+P\right)}^2.$

Then, differential inequality (3.3) becomes

$\frac{\text{d}}{\text{d}t}\left(R+P\right)\ge -{\lambda }_1\left(R+P\right)+c{\left(R+P\right)}^2+2c.$

Let $E\left(t\right)=R\left(t\right)+P\left(t\right)$. Then, $E\left(t\right)\ge 0$ in $\left[0,T\right)$ and

$\frac{\text{d}}{\text{d}t}E\ge -{\lambda }_{1}E+c{E}^2+2c.$

Using separation of variables and integrating both sides over (0, t), we obtain

$t\le \frac{2}{\sqrt{8c^2-{\lambda }_1^2}}\left[{\tan }^{-1}\left(\frac{2cE\left(t\right)-{\lambda }_1}{\sqrt{8c^2-{\lambda }_1^2}}\right)+{\tan }^{-1}\left(\frac{{\lambda }_1}{\sqrt{8c^2-{\lambda }_1^2}}\right)\right].$

Suppose that $E\left(t\right)$ exists for all $t>0$. By the assumption $2\sqrt{2}c>{\lambda }_1$, we have

${\tan }^{-1}\left[\left(2cE\left(t\right)-{\lambda }_1\right)/\sqrt{8c^2-{\lambda }_1^2}\right]\to \infty \text{if}t\to \infty .$

But, ${\tan }^{-1}\left[\left(2cE\left(t\right)-{\lambda }_1\right)/\sqrt{8c^2-{\lambda }_1^2}\right]$ is bounded above by $\text{π}/2$. This is a contradiction. It implies that $E\left(t\right)$ ceases to exist in a finite time $\hat{\Gamma }$. This shows that either $R\left(t\right)$ or $P\left(t\right)$ does not exist when t approaches $\hat{\Gamma }$. Thus, either $\hat{u}$ or $\hat{v}$ quenches on $\bar{D}$ at $\hat{\Gamma }$. Since $u\ge \hat{u}$ and $v\ge \hat{v}$, either u or v quenches on $\bar{D}$ in a finite time $\tilde{\Gamma }$ where $\tilde{\Gamma }\le \hat{\Gamma }$. ,

From the result of Lemma 3.1, we know that (0, 0) is a quenching point of u and v if they quench. Let ${\Gamma }^{*}$ be the supremum of the time $\Gamma$ for which the problem (1.1)-(1.2) has unique solutions u and v.

Theorem 3.3. If ${\Gamma }^{*}<\infty$, then either $u\left(0,0,t\right)$ or $v\left(0,0,t\right)$ quenches at ${\Gamma }^{*}$.

Proof. Suppose that both u and v do not quench at (0,0) when $t={\Gamma }^{*}$. Then, there exist positive constants $k_1$ and $k_2$ such that $u\left(0,0,t\right)\le k_1<1$ and $v\left(0,0,t\right)\le k_2<1$ for all $t\in \left[0,{\Gamma }^{*}\right]$. This shows that $a/\left(1-v\left(0,0,t\right)\right)<Q_1$ and $b/\left(1-u\left(0,0,t\right)\right)<Q_2$ for some positive constants $Q_1$ and $Q_2$ when $t\in \left[0,{\Gamma }^{*}\right]$. Then, by Theorem 4.2.1 of [ [13] , p. 139], u and $v\in C^{2+\alpha ,1+\alpha /2}\left(\bar{D}\times \left[0,{\Gamma }^{*}\right]\right)$. This implies that there exist positive constants $k_3$ and $k_4$ such that $u\left(x,y,t\right)\le k_3<1$ and $v\left(x,y,t\right)\le k_4<1$ for all $\left(x,y,t\right)\in \bar{D}\times \left[0,{\Gamma }^{*}\right]$. In order to arrive a contradiction, we need to show that u and v can continue to exist in a larger time interval $\left[0,{\Gamma }^{*}+t_1\right)$ for some positive $t_1$. This can be achieved by extending the upper bound. Let us construct upper solutions $\psi \left(x,y,t\right)=k_{3}f\left(t\right)$ and $\sigma \left(x,y,t\right)=k_{4}g\left(t\right)$, where f(t) and g(t) are solutions of the following differential system:

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{k}_{3}f\left(t\right)=\frac{a}{1-k_{4}g\left(t\right)}\text{for}t>{\Gamma }^{*},f\left({\Gamma }^{*}\right)=1,\\ \frac{\text{d}}{\text{d}t}{k}_{4}g\left(t\right)=\frac{b}{1-k_{3}f\left(t\right)}\text{for}t>{\Gamma }^{*},g\left({\Gamma }^{*}\right)=1.\end{array}$

By $k_{3}f\left({\Gamma }^{*}\right)<1,k_{4}g\left({\Gamma }^{*}\right)<1$, and the Picard iteration, f(t) and g(t) are positive functions, and $a/\left[1-k_{4}g\left(t\right)\right]>0$ and $b/\left[1-k_{3}f\left(t\right)\right]>0$. This implies that f(t) and g(t) are increasing functions of t. Let $t_1$ be a positive real number determined by $k_{3}f\left({\Gamma }^{*}+t_1\right)=k_5<1$ and $k_{4}g\left({\Gamma }^{*}+t_1\right)=k_6<1$ for some positive constants $k_5$ and $k_6$ greater than $k_3$ and $k_4$ respectively. By our construction, $\psi \left(x,y,t\right)=\psi \left(0,0,t\right)$ and $\sigma \left(x,y,t\right)=\sigma \left(0,0,t\right)$ satisfy,

$\begin{array}{l}L\psi \left(x,y,t\right)=\frac{a}{1-\sigma \left(0,0,t\right)}\text{in}D\times \left({\Gamma }^{*},{\Gamma }^{*}+t_1\right),\\ L\sigma \left(x,y,t\right)=\frac{b}{1-\psi \left(0,0,t\right)}\text{in}D\times \left({\Gamma }^{*},{\Gamma }^{*}+t_1\right),\end{array}$

$\begin{array}{l}\psi \left(x,y,{\Gamma }^{*}\right)=k_3\ge u\left(x,y,{\Gamma }^{*}\right)\text{and}\sigma \left(x,y,{\Gamma }^{*}\right)=k_4\ge v\left(x,y,{\Gamma }^{*}\right)\text{on}\bar{D},\\ \psi \left(x,y,t\right)=k_{3}f\left(t\right)>0\text{and}\sigma \left(x,y,t\right)=k_{4}g\left(t\right)>0\text{on}\partial D\times \left[{\Gamma }^{*},{\Gamma }^{*}+t_1\right).\end{array}$

By Lemma 2.1, $\psi \left(x,y,t\right)\ge u\left(x,y,t\right)$ and $\sigma \left(x,y,t\right)\ge v\left(x,y,t\right)$ on $\bar{D}\times \left[{\Gamma }^{*},{\Gamma }^{*}+t_1\right)$. Therefore, we find solutions u and v to the problems (1.1)-(1.2) on $\bar{D}\times \left[{\Gamma }^{*},{\Gamma }^{*}+t_1\right)$. This contradicts the definition of ${\Gamma }^{*}$. Hence, either $u\left(0,0,t\right)$ or $v\left(0,0,t\right)$ quenches at ${\Gamma }^{*}$. ,

Let ${\phi }_0\left(x,y\right)\in C\left(\bar{D}\right)\cap C^2\left(D\right)$ such that $\Delta {\phi }_0\left(x,y\right)<0$,${\phi }_0\left(x,y\right)>0$ in D, and ${\phi }_0\left(x,y\right)=0$ on $\partial D$ and ${\max }_{x\in \bar{D}}{\phi }_0\left(x,y\right)\le 1$. Let $\phi \left(x,y,t\right)$ be the solution to the following first initial-boundary value problem:

$\begin{array}{l}Lw=0\text{in}D\times \left(0,\infty \right),\\ w\left(x,y,0\right)={\phi }_0\left(x,y\right)\text{on}\bar{D},\\ w\left(x,y,t\right)=0\text{on}\partial D\times \left(0,\infty \right).\end{array}$

By the maximum principle, $\phi \left(x,y,t\right)>0$ in $D\times \left[0,\infty \right)$ and is bounded above by ${\phi }_0\left(x,y\right)$, and it satisfies

$\underset{\left(x,y,t\right)\in \bar{D}\times \left[0,\infty \right)}{\max }\phi \left(x,y,t\right)\le 1.$

Let $t_2\in \left(0,\Gamma \right)$ such that $v\left(0,0,t_2\right)\le k_7<1$ for some positive constant $k_7$. Then,

$\frac{a\phi \left(x,y,t_2\right)}{1-k_7}\ge \frac{a\phi \left(x,y,t_2\right)}{1-v\left(0,0,t_2\right)}.$ (3.4)

As $u_t\left(x,y,t_2\right)>0$ and $\phi \left(x,y,t_2\right)>0$ in D, and $u_t\left(x,y,t_2\right)=\phi \left(x,y,t_2\right)=0$ on $\partial D$, we choose a positive real number ${\mu }_1$ less than 1 such that

$u_t\left(x,y,t_2\right)\ge \frac{{\mu }_{1}a\phi \left(x,y,t_2\right)}{1-k_7}\text{on}\bar{D}.$ (3.5)

Also, $u_t\left(x,y,t\right)={\mu }_{1}a\phi \left(x,y,t\right)/\left[1-v\left(0,0,t\right)\right]$ for all $\left(x,y,t\right)\in \partial D\times \left[0,\Gamma \right)$. Let us define $I\left(x,y,t\right)=u_t\left(x,y,t\right)-{\mu }_{1}a\phi \left(x,y,t\right)/\left[1-v\left(0,0,t\right)\right]$. By inequalities (3.4) and (3.5), $I\left(x,y,t_2\right)\ge 0$ on $\bar{D}$. Let $H\left(x,y,t\right)=v_t\left(x,y,t\right)-{\mu }_{2}b\phi \left(x,y,t\right)/\left[1-u\left(0,0,t\right)\right]$ where ${\mu }_2$ is a positive real number less than 1. Similar to the previous argument, we choose ${\mu }_2$ such that $H\left(x,y,t_2\right)\ge 0$ on $\bar{D}$. We modify the proof of Lemma 3.4 of [7] to obtain the result below.

Lemma 3.4. $I\left(x,y,t\right)\ge 0$ and $H\left(x,y,t\right)\ge 0$ on $\bar{D}\times \left[t_2,\Gamma \right)$.

Proof. By a direct computation,

$\begin{array}{l}{I}_t=u_{tt}-\frac{{\mu }_{1}a\phi }{{\left[1-v\left(0,0,t\right)\right]}^2}{v}_t\left(0,0,t\right)-\frac{{\mu }_{1}a{\phi }_t}{1-v\left(0,0,t\right)},\\ \Delta I=\Delta u_t-\frac{{\mu }_{1}a}{1-v\left(0,0,t\right)}\Delta \phi .\end{array}$

From the above expression, we have

$LI=\frac{a}{{\left[1-v\left(0,0,t\right)\right]}^2}{v}_t\left(0,0,t\right)\left(1-{\mu }_1\phi \right)\text{in}D\times \left(0,\Gamma \right).$

By $\phi \le 1$ on $\bar{D}\times \left[0,\infty \right)$,${\mu }_1<1$, and $v_t\left(0,0,t\right)>0$ for all $t\in \left(0,\Gamma \right)$, it gives $LI\ge 0$ in $D\times \left(0,\Gamma \right)$. In addition, $I\left(x,y,t_2\right)\ge 0$ on $\bar{D}$, and $I\left(x,y,t\right)=0$ on $\partial D\times \left(t_2,\Gamma \right)$. By the maximum principle, $I\left(x,y,t\right)\ge 0$ on $\bar{D}\times \left[t_2,\Gamma \right)$. Similarly, $H\left(x,y,t\right)\ge 0$ on $\bar{D}\times \left[t_2,\Gamma \right)$. ,

Here is the result of simultaneous quenching.

Theorem 3.5. If either u or v quenches at (0, 0) when $t=\Gamma$, then u and v quench simultaneously at (0, 0) when $t=\Gamma$.

Proof. If not, let us assume that v quenches at (0, 0) when $t=\Gamma$ but u continues to exist beyond $\Gamma$. That is, there exists a positive constant $k_8$ such that $0\le u\le k_8<1$ for all $\left(x,y,t\right)\in \bar{D}\times \left[0,\Gamma +t_3\right)$ for some $t_3>0$. By Lemma 3.4, we have

$u_t\left(0,0,t\right)\ge \frac{{\mu }_{1}a\phi \left(0,0,t\right)}{1-v\left(0,0,t\right)}\text{and}{v}_t\left(0,0,t\right)\ge \frac{{\mu }_{2}b\phi \left(0,0,t\right)}{1-u\left(0,0,t\right)}\text{forall}t\in \left[t_2,\Gamma \right).$

By Lemma 3.1, u and v both attain their maximum at (0,0) for all $t\in \left(0,\Gamma \right)$. Then, $\Delta u\left(0,0,t\right)<0$ and $\Delta v\left(0,0,t\right)<0$ on $\left[t_2,\Gamma \right)$. Combine (1.1) with above inequalities to give

$\begin{array}{l}\frac{{\mu }_{1}a\phi \left(0,0,t\right)}{1-v\left(0,0,t\right)}\le u_t\left(0,0,t\right)<\frac{a}{1-v\left(0,0,t\right)},\\ \frac{{\mu }_{2}b\phi \left(0,0,t\right)}{1-u\left(0,0,t\right)}\le v_t\left(0,0,t\right)<\frac{b}{1-u\left(0,0,t\right)}.\end{array}$

From them, we get a compound inequality

$\frac{{\mu }_{1}a\phi \left(0,0,t\right)}{1-v\left(0,0,t\right)}\frac{1-u\left(0,0,t\right)}{b}\le \frac{\text{d}u\left(0,0,t\right)}{\text{d}v\left(0,0,t\right)}\le \frac{a}{1-v\left(0,0,t\right)}\frac{1-u\left(0,0,t\right)}{{\mu }_{2}b\phi \left(0,0,t\right)}.$ (3.5)

From the left-side inequality, we have

$\frac{{\mu }_{1}a\phi \left(0,0,t\right)}{b\left[1-v\left(0,0,t\right)\right]}\text{d}v\left(0,0,t\right)\le \frac{\text{d}u\left(0,0,t\right)}{1-u\left(0,0,t\right)}.$

Since $0<\phi \left(0,0,t\right)\le 1$ for all $t\in \left(0,\infty \right)$, there exists a positive constant $\delta >0$ such that $\phi \left(0,0,t\right)>\delta$ for all $t\in \left[t_2,\Gamma \right)$. Integrating both sides over the interval $\left[t_2,s\right]$ where $s\in \left(t_2,\Gamma \right)$, we obtain

$\begin{array}{l}\frac{{\mu }_{1}a\delta }{b}\left\{\ln \left[1-v\left(0,0,s\right)\right]-\ln \left[1-v\left(0,0,t_2\right)\right]\right\}\\ \ge \ln \left[1-u\left(0,0,s\right)\right]-\ln \left[1-u\left(0,0,t_2\right)\right].\end{array}$

By assumption, v quenches at (0, 0) when $t=\Gamma$, we have $\ln \left[1-v\left(0,0,s\right)\right]\to -\infty$ as $s\to {\Gamma }^{-}$. Since $\ln \left[1-v\left(0,0,t_2\right)\right]$ and $\ln \left[1-u\left(0,0,t_2\right)\right]$ are both bounded, the above inequality implies $\ln \left[1-u\left(0,0,s\right)\right]\to -\infty$ as $s\to {\Gamma }^{-}$. Therefore, u quenches at (0, 0) when $t=\Gamma$. It contradicts that u exists on $\bar{D}\times \left[0,\Gamma +t_3\right)$. Follow the second half of inequality (3.5), we can prove that v quenches at (0, 0) when $t=\Gamma$ if u quenches at (0, 0) when $t=\Gamma$. The proof is complete. ,

Now, we prove that u and v exist globally when a and b are sufficiently small. Our method is to construct global-exist upper solutions of the problem (1.1)-(1.2).

Lemma 3.6. If a and b are sufficiently small, then there is a global solution to the problem (1.1)-(1.2).

Proof. It suffices to construct upper solutions which exist all time. Let $q\left(x,y\right)=A\left(2-x^2-y^2\right)$ and $m\left(x,y\right)=B\left(2-x^2-y^2\right)$ where A and B are positive real numbers such that $A<1/2$ and $B<1/2$. Clearly, $0\le q,m<1$ for all $\left(x,y\right)\in \bar{D}$. In addition,

$Lq-\frac{a}{1-m\left(0,0\right)}=4A-\frac{a}{1-2B}.$

If a is sufficiently small, then we have the inequality: $4A\ge a/\left(1-2B\right)$. This leads to