Simultaneous and non-simultaneous quenching for a coupled semilinear parabolic system in a $n$-dimensional ball with singular localized sources

1.   Introduction

Let $a$ and $b$ be positive real numbers, $c$ be a positive constant, $x_{0}$ be a fixed point in a $n$-dimensional space $\mathbb{R} ^{n}$ with $n = 1, \ 2, ...$, and $B_{1}\left(x_{0}\right)$ be a $n$ -dimensional open ball with the center $x_{0}$ and radius $1$ such that $B_{1}\left(x_{0}\right) = \left\{ x\in \mathbb{R} ^{n}:\left\vert \left\vert x-x_{0}\right\vert \right\vert < 1\right\}$ where $\left\vert \left\vert x-x_{0}\right\vert \right\vert$ represents the Euclidean distance between $x$ and $x_{0}$. We also let $\overline{ B_{1}\left(x_{0}\right) }$ and $\partial B_{1}\left(x_{0}\right)$ denote the closure and boundary of $B_{1}\left(x_{0}\right)$, respectively. Let $L$ be the parabolic operator such that $Lu = u_{t}-\Delta u$. In this paper, we deal with the quenching problem of a coupled semilinear parabolic system with nonlinear singular localized sources at $x_{0}$. This problem is described below:

In the problem (1.1)-(1.2), we assume that the source functions $f$ and $g$ are differentiable over the interval $\left[0, c\right)$ and satisfy the following hypotheses:

(H$_{1}$) $f > 0$, $f^{\prime } > 0$, $f^{\prime \prime } > 0$, $g > 0$, $g^{\prime } > 0$, $g^{\prime \prime } > 0$;

(H$_{2}$) both $f$ and $g$ being unbounded when $u$ and $v$ tend to $c$, that is, $f\left(v\right) \rightarrow \infty$ when $v\rightarrow c^{-}$ (that is, $v$ approaches $c$ from the left) and $g\left(u\right) \rightarrow \infty$ when $u\rightarrow c^{-}$.

The problem (1.1)-(1.2) describes the instabilities in some dynamic systems of certain reactions that have localized electrodes immersed in a bulk medium at the point $x_{0}$, see ^[1,12]. Li and Wang ^[10] used the equation (1.1) to explore a thermal ignition driven by the temperature at a single point. Chadam et al. ^[2] examined the blow-up set of solutions.

The quenching problem is able to illustrate the polarization phenomena in ionic conductors and the phase transition between liquids and solids, see ^[11]. We say that the solution $\left(u, v\right)$ quenches at a point in $\overline{B_{1}\left(x_{0}\right) }$ if there exists a finite time $T\ \left(>0\right)$ such that

where $t\rightarrow T^{-}$ represents $t$ approaching $T$ from the left. $T$ is called the quenching time. Quenching and blow-up problems are related. Under some transformations, quenching problems are able to change to blow-up problems, see ^[5,6].

Ji et al. ^[7] studied simultaneous and non-simultaneous quenching of one-dimensional coupled system with the singular nonlinear reaction sources on the boundary. They used this model to describe heat propagations between two different materials. The multi-dimensional quenching problem of coupled semilinear parabolic systems describes non-Newtonian filtration systems incorporated with the effect of singular nonlinear reaction sources inside the domain, see Jia et al. ^[8]. Their model is

where $p_{1}$, $p_{2}$, $q_{1}$, and $q_{2}$ are positive real numbers, and $\Omega$ is a bounded domain in $\mathbb{R} ^{n}$. When $\Omega = B_{R}\left(x_{0}\right)$, they proved that the solution $\left(u, v\right)$ quenches simultaneously if $p_{2}\geq p_{1}+1$ and $q_{1}\geq q_{2}+1$. Depending on the initial data $u_{0}$ and $v_{0}$, they also showed that both simultaneous and non-simultaneous quenching may occur when $p_{2} < p_{1}+1$ and $q_{1} < q_{2}+1$. Zheng and Wang ^[16] studied simultaneous and non-simultaneous quenching for the coupled system: $Lu = v^{-p}$, $Lv = u^{-q}$ in $B_{R}\left(x_{0}\right) \times \left(0, T\right)$ subject to the Dirichlet boundary condition. When $\Omega$ is a square domain in $\mathbb{R} ^{2}$, Chan ^[3] studied the simultaneous quenching for the coupled system: $Lu = a/\left(1-v\left(0, 0, t\right) \right)$, $Lv = b/\left(1-u\left(0, 0, t\right) \right)$ in $\Omega \times \left(0, T\right)$ with the homogeneous first boundary condition. He also computed an approximated critical value of $a$ and $b$ by a numerical method.

The main goals of this paper are to study (a) simultaneous quenching and (b) non-simultaneous quenching of the solution $\left(u, v\right)$ under some conditions on $\int_{0}^{c}f\left(\omega \right) d\omega$ and $\int_{0}^{c}g\left(\omega \right) d\omega$. In this article, simultaneous quenching means that the maximum of $u$ and $v$\tends to $c$ in the same finite time. Non-simultaneous quenching means that either the maximum of $u$ or $v$ tends to $c$\ in a finite time, but the other remains bounded by $c$. We are going to study cases (a) and (b) of the problem (1.1)-(1.2) when these two integrals are either infinite or finite. Without loss of generality, let us assume $x_{0}$ being the origin $0$. The problem (1.1)-(1.2) becomes

Similar consideration is also available in ^[4,8,16]. In section 2, we provide some properties of the solution $\left(u, v\right)$. The results of simultaneous and non-simultaneous quenching are going to illustrate in section 3.

2.   Properties of the solution

In this section, we are going to show some properties of the solution $\left(u, v\right)$. One of the main results is to prove that $u$ and $v$ attain their maximum at $x = 0$, and they both quench only at $x = 0$. In the sequel, we assume that $k_{j}$ are positive constants for $j = 1, \ 2, ..., \ 19.$ We also let $Y\left(x, t\right)$ and $Z\left(x, t\right)$ be nontrivial and nonnegative bounded functions on $\overline{B_{1}\left(0\right) }\times \left[0, \infty \right)$. Here is the comparison theorem.

Lemma 2.1. Assume that $\left(u, v\right)$ is the solution to the problem below:

then $u\left(x, t\right) \geq 0$ and $v\left(x, t\right) \geq 0$ on $\overline{B_{1}\left(0\right) }\times \left[0, T\right)$.

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

where $\gamma$ is a positive real number to be determined and $\hat{\phi} _{1}$ is the first eigenfunction of the following eigenvalue problem:

where $\partial /\partial \nu$ is the outward normal derivative on $\partial B_{1}\left(0\right)$. Let $\hat{\lambda}_{1}$ be the corresponding eigenvalue. By Theorem 3.1.2 of ^[13], $\hat{\phi}_{1}$ exists and $\hat{\phi}_{1} > 0$ on $\overline{B_{1}\left(0\right) }$ and $\hat{\lambda}_{1} > 0$. Based on the construction, we know that $\Phi \left(x, 0\right) > 0$ and $\Psi \left(x, 0\right) > 0$ on $\overline{B_{1}\left(0\right) }$. By a direct calculation, we obtain the inequality below

Since $\hat{\phi}_{1} > 0$ on $\overline{B_{1}\left(0\right) }$, $Y$ is nonnegative and bounded, and $\hat{\lambda}_{1} > 0$, we are able to choose $\gamma$ such that $\gamma > Y\hat{\phi}_{1}\left(0\right) /\hat{\phi}_{1}- \hat{\lambda}_{1}$ in $B_{1}\left(0\right)$. Thus,

Suppose $\Phi \left(x, t\right) \leq 0$ somewhere in $B_{1}\left(0\right) \times \left(0, T\right)$. Then, the set $\left\{ t:\Phi \left(x, t\right) \leq 0\text{ for some }x\in B_{1}\left(0\right) \right\}$ is non-empty. Let $\tilde{t}$ denote the infimum of this set. Then, $0 < \tilde{t} < T$ because $\Phi \left(x, 0\right) > 0$ on $\overline{B_{1}\left(0\right) }$. Thus, there exists some point $x_{1}\in B_{1}\left(0\right)$ such that $\Phi \left(x_{1}, \tilde{t}\right) = 0$ and $\Phi _{t}\left(x_{1}, \tilde{t} \right) \leq 0$. On the other hand, $\Phi$ attains its local minimum at $\left(x_{1}, \tilde{t}\right)$. Then, $\Delta \Phi \left(x_{1}, \tilde{t} \right) \geq 0$. Let us consider $t = \tilde{t}$, we get

Follow a similar argument, if we assume that $\Psi \left(x, t\right) \leq 0$ somewhere in $B_{1}\left(0\right) \times \left(0, T\right)$, then there exist some $\hat{t}\in \left(0, T\right)$ and $x_{2}\in B_{1}\left(0\right)$ such that $\Psi \left(x_{2}, \hat{t}\right) = 0$, $\Psi _{t}\left(x_{2}, \hat{t}\right) \leq 0$, and $\Psi$ attains its local minimum at $\left(x_{2}, \hat{t}\right)$. Then, at $t = \hat{t}$

Let us assume that $\hat{t} < \tilde{t}$. As $\Phi$ attains its local minimum at $\left(x_{1}, \tilde{t}\right)$, we have $\Phi \left(0, \hat{t}\right) > 0$. From the expression (2.2) and $Z$ is nonnegative and bounded, we have the inequality below:

This is a contradiction. Hence, $\Psi \left(x, t\right) > 0$ in $B_{1}\left(0\right) \times \left(0, T\right)$. Then by (2.1), we show that $\Phi \left(x, t\right) > 0$ in $B_{1}\left(0\right) \times \left(0, T\right)$. Through a similar calculation, we obtain the same result when $\hat{t}\geq \tilde{t}$. Let $\varepsilon \rightarrow 0$, we have $u\left(x, t\right) \geq 0$ and $v\left(x, t\right) \geq 0$ in $B_{1}\left(0\right) \times \left(0, T\right)$. Following the homogeneous initial-boundary conditions, we conclude that $u$ and $v$ are non-negative on $\overline{B_{1}\left(0\right) }\times \left[0, T\right)$. The proof is complete.

By Lemma 2.1, $\left(0, 0\right)$ is a lower solution of the problem (1.3)-(1.4). On the other side, $u < c$ and $v < c$ on $\overline{B_{1}\left(0\right) }\times \left[0, T\right)$. Since $u$ and $v$ stop to exist for $u\geq c$ and $v\geq c$, it follows from Theorem 2.1 of ^[2] that the problem (1.3)-(1.4) has the unique classical solution $\left(u, v\right) \in C\left(\overline{B_{1}\left(0\right) }\times \left[0, T\right) \right) \cap C^{2+\alpha, 1+\alpha /2}\left(B_{1}\left(0\right) \times \left[0, T\right) \right)$ for some $\alpha \in \left(0, 1\right)$ such that $0\leq u < c$ and $0\leq v < c$ on $\overline{B_{1}\left(0\right) }\times \left[0, T\right)$. As $f$ and $g$ are differentiable, it follows from Theorem 8.9.2 of Pao ^[13] that the solution $\left(u, v\right)$ exists either in a finite time or globally.

Based on the result of Lemma 2.1, we prove $u_{t}$ and $v_{t}$ being positive over the domain.

Lemma 2.2. The solution $\left(u, v\right)$ has the properties: (i) $u_{t}\geq 0$ and $v_{t}\geq 0$ on $\overline{B_{1}\left(0\right) }\times \left[0, T\right)$, and (ii) $u_{t} > 0$ and $v_{t} > 0$ in $B_{1}\left(0\right) \times \left(0, T\right)$.

Proof. (i) For $\theta _{1} > 0$, let us consider the first equation of the problem (1.3) at $t+\theta _{1}$. We have $Lu\left(x, t+\theta _{1}\right) = af\left(v\left(0, t+\theta _{1}\right) \right)$ in $B_{1}\left(0\right) \times \left(0, T-\theta _{1}\right)$. Subtract the first equation of the problem (1.3) from this equation, and based on the mean value theorem, there exists some $\zeta _{1}$ where $\zeta _{1}$ is between $v\left(0, t+\theta _{1}\right)$ and $v\left(0, t\right)$ such that

Since $u\geq 0$ on $\overline{B_{1}\left(0\right) }\times \left[0, T\right)$, we have $u\left(x, \theta _{1}\right) -u\left(x, 0\right) \geq 0$ for $x\in \overline{B_{1}\left(0\right) }$. From the boundary condition, $u\left(x, t+\theta _{1}\right) -u\left(x, t\right) = 0$ for $x\in \partial B_{1}\left(0\right)$ and $t > 0$. By Lemma 2.1, $\left(u(x, t+\theta _{1})-u(x, t)\right) /\theta _{1}\geq 0$ on $\overline{B_{1}\left(0\right) } \times \left[0, T-\theta _{1}\right)$. As $\theta _{1}\rightarrow 0^{+}$, $u_{t}\geq 0$ on $\overline{B_{1}\left(0\right) }\times \left[0, T\right)$. Similarly, we obtain $v_{t}\geq 0$ on $\overline{B_{1}\left(0\right) } \times \left[0, T\right)$.

(ii) To show that $u_{t}$ is positive, we differentiate the first equation of the problem (1.3) with respect to $t$ to get

From (i), we know $v_{t}\geq 0$ on $\overline{B_{1}\left(0\right) }\times \left[0, T\right)$. By (H$_{1}$) (see section 1) and the strong maximum principle, we have $u_{t} > 0$ in $B_{1}\left(0\right) \times \left(0, T\right)$. We follow the similar procedure to conclude $v_{t} > 0$ in $B_{1}\left(0\right) \times \left(0, T\right)$.

By the symmetry of $B_{1}\left(0\right)$, we represent the problem (1.3)-(1.4) in the polar coordinate system

Lemma 2.3. The solution $\left(u, v\right)$ to the problem (2.4) attains its maximum at $r = 0$ for $t\in \left(0, T\right)$.

Proof. It is noticed that the solution to the problem (2.4) is radial symmetric with respect to $r = 0$. To show $u$ and $v$ attaining their maximum at $r = 0$, we are going to prove $u_{r} < 0$ and $v_{r} < 0$ for $r\in \left(0, 1\right]$. We let $H\left(r, t\right) = u_{r}\left(r, t\right)$. Differentiating the first equation of the problem (2.4) with respect to $r$, we have

At $t = 0$, $H\left(r, 0\right) = 0$ for $r\in \lbrack 0, 1]$. By Lemma 2.2(ii), $u_{t} > 0$ in $B_{1}\left(0\right) \times \left(0, T\right)$. By Hopf's Lemma, $H\left(1, t\right) < 0$ for $t\in \left(0, T\right)$. Also, $H\left(0, t\right) = u_{r}\left(0, t\right) = 0$ for $t\in \left[0, T\right)$. By the maximum principle ^[13], $H < 0$ for $\left(r, t\right) \in \left(0, 1\right] \times \left(0, T\right)$. Therefore, $u\left(0, t\right) \geq u\left(r, t\right)$ for $\left(r, t\right) \in \left[0, 1\right] \times \left(0, T\right)$. Similarly, we prove that $v_{r} < 0$ for $\left(r, t\right) \in \left(0, 1\right] \times \left(0, T\right)$. Hence, $u$ and $v$ achieve their maximum at $r = 0$ for $t\in \left(0, T\right).$

Let $\phi _{1}$ be the eigenfunction corresponding to the first eigenvalue $\lambda _{1}\left(>0\right)$ of the eigenvalue problem below:

This eigenfunction has the properties: $0 < \phi _{1}\leq 1$ in $B_{1}\left(0\right)$ and $\int_{B_{1}\left(0\right) }\phi _{1}dx = 1$ ^[15]. Let $k_{1} = abf^{\prime \prime }\left(0\right) g^{\prime \prime }\left(0\right) / \left[2\left(af^{\prime \prime }\left(0\right) +bg^{\prime \prime }\left(0\right) \right) \right]$ and $k_{2} = af\left(0\right) +bg\left(0\right)$. By (H$_{1}$), $k_{1}$ and $k_{2}$ are positive. We show that either $u$ or $v$ quenches in a finite time.

Lemma 2.4. If $2\sqrt{k_{1}k_{2}} > \lambda _{1}$, then either $u$ or $v$ quenches on $\overline{ B_{1}\left(0\right) }$ in a finite time $\tilde{T}$.

Proof. By Lemma 2.3, $u\left(0, t\right) \geq u\left(x, t\right)$ and $v\left(0, t\right) \geq v\left(x, t\right)$ on $\overline{ B_{1}\left(0\right) }\times \left(0, T\right)$. Let $\hat{u}\left(x, t\right)$ and $\hat{v}\left(x, t\right)$ be the solutions to the following auxiliary parabolic system:

By the comparison theorem ^[13], $\hat{u}\left(x, t\right) \geq 0$ and $\hat{v}\left(x, t\right) \geq 0$ on $\overline{B_{1}\left(0\right) }\times \left(0, T\right)$. Further, $u-\hat{u}$ and $v-\hat{v}$ satisfy the expression below:

By $u-\hat{u} = 0$ and $v-\hat{v} = 0$ on $\overline{B_{1}\left(0\right) }$ and $\partial B_{1}\left(0\right) \times \left(0, T\right)$, and the comparison theorem, we have $u\geq \hat{u}$ and $v\geq \hat{v}$ on $\overline{B_{1}\left(0\right) }\times \left(0, T\right)$. It suffices to prove either $\hat{u}$ or $\hat{v}$ to quench over $\overline{B_{1}\left(0\right) }$ in a finite time. Multiplying $\phi _{1}$ on both sides of (2.5) and integrating expressions over the domain $B_{1}\left(0\right)$, we obtain

Using the Green's second identity and (2.6), it gives

Applying the Maclaurin's series on the functions $f$ and $g$, we have

By $0 < \phi _{1}\leq 1$ in $B_{1}\left(0\right)$ and the Jensen's inequality ^[15], we have

Let $R\left(t\right) = \int_{B_{1}\left(0\right) }\hat{u}\phi _{1}dx$ and $P\left(t\right) = \int_{B_{1}\left(0\right) }\hat{v}\phi _{1}dx$. From these two inequalities above, we have the following inequality:

Then, by the inequality below:

we obtain this expression

Then, the differential inequality (2.7) becomes

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

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

Suppose that $E\left(t\right)$ exists for all $t > 0$. By the assumption $2 \sqrt{k_{1}k_{2}} > \lambda _{1}$, we have

But, $\tan ^{-1}\left[\left(2k_{1}E\left(t\right) -\lambda _{1}\right) / \sqrt{4k_{1}k_{2}-\lambda _{1}^{2}}\right]$ is bounded above by $\pi /2$. This is a contradiction. It implies that $E(t)$ ceases to exist in a finite time $\hat{T}$. This shows that either $P\left(t\right)$ or $R\left(t\right)$ does not exist when $t$ tends to $\hat{T}$. Thus, either $\hat{u}$ or $\hat{v}$ quenches on $\overline{B_{1}\left(0\right) }$ at $\hat{T}$. Since $u\geq \hat{u}$ and $v\geq \hat{v}$, we then have either $u$ or $v$ quenches on $\overline{B_{1}\left(0\right) }$ in a finite time $\tilde{T}$ where $\tilde{T}\leq \hat{T}$.

Let $M_{1}$ and $M_{2}$ be positive constants such that $M_{1}/\left(2n\right) < c$ and $M_{2}/\left(2n\right) < c$. We are going to prove the global existence of solutions when $a$ and $b$ are sufficiently small. Our method is to construct a global-existed upper solution of the problem (1.1)-(1.2).

Lemma 2.5. If $a$ and $b$ are sufficiently small, then the solution $\left(u, v\right)$ exists globally.

Proof. It suffices to construct an upper solution which exists all time. Let $\bar{u}\left(x\right) = M_{1}\left(1-\left\Vert x\right\Vert ^{2}\right) /\left(2n\right)$ and $\bar{v}\left(x\right) = M_{2}\left(1-\left\Vert x\right\Vert ^{2}\right) /\left(2n\right)$. Clearly, $0\leq \bar{u}$, $\bar{v} < c$ for all $x\in \overline{B_{1}\left(0\right) }$. Let us consider the following problem:

If $a$ and $b$ are sufficiently small, then

Then, we subtract Eq (1.3) from above inequalities and by the mean value theorem to obtain

where $\chi _{1}$ is between $\bar{v}\left(0\right)$ and $v\left(0, t\right)$ and $\chi _{2}$ is between $\bar{u}\left(0\right)$ and $u\left(0, t\right)$. On $\partial B_{1}\left(0\right)$, $\bar{u}-u = 0$ and $\bar{v}-v = 0$. By Lemma 2.1, $u\left(x, t\right) \leq \bar{u}\left(x\right)$ and $v\left(x, t\right) \leq \bar{v}\left(x\right)$ on $\overline{B_{1}\left(0\right) }\times \left[0, \infty \right)$. Thus, the solution $\left(u, v\right)$ exists globally. The proof is complete.

From the result of Lemma 2.3, we know that $x = 0$ is a quenching point of $u$ and $v$ if they quench. Let $T^{\ast }$ be the supremum of the time $T$ for which the problem (1.3)-(1.4) has the unique solution $\left(u, v\right)$.

Theorem 2.6. If $T^{\ast } < \infty$, then either $u\left(0, t\right)$ or $v\left(0, t\right)$ quenches at $T^{\ast }$.

Proof. Suppose that both $u$ and $v$ do not quench at $x = 0$ when $t = T^{\ast }$. Then, there exist $k_{3}$ and $k_{4}$ such that $u\left(0, t\right) \leq k_{3} < c$ and $v\left(0, t\right) \leq k_{4} < c$ for $t\in \left[0, T^{\ast }\right]$. This shows that $af\left(v\left(0, t\right) \right) < k_{5}$ and $bg\left(u\left(0, t\right) \right) < k_{6}$ for $t\in \left[0, T^{\ast }\right]$. Then, by Theorem 4.2.1 of ^[9], $u$ and $v\in C^{2+\alpha, 1+\alpha /2}\left(\overline{ B_{1}\left(0\right) }\times \lbrack 0, T^{\ast }]\right)$. This implies that there exist $k_{7}$ and $k_{8}$ such that $u\left(x, t\right) \leq k_{7} < c$ and $v\left(x, t\right) \leq k_{8} < c$ for $\left(x, t\right) \in \overline{B_{1}\left(0\right) }\times \left[0, T^{\ast }\right]$. In order to arrive at a contradiction, we need to show that $u$ and $v$ can continue to exist in a longer time interval $\left[0, T^{\ast }+t_{1}\right)$ for some positive $t_{1}$. This can be accomplished by extending the upper bound of $u$ and $v$. Let us construct upper solutions $\psi \left(x, t\right) = k_{7}h\left(t\right)$ and $\sigma \left(x, t\right) = k_{8}i\left(t\right)$, where $h\left(t\right)$ and $i\left(t\right)$ are solutions to the following system:

From $af\left(k_{8}i\left(t\right) \right) > 0$ and $bg\left(k_{7}h\left(t\right) \right) > 0$, this implies that $h\left(t\right)$ and $i\left(t\right)$ are increasing functions of $t$. Let $t_{1}$ be a positive real number determined by $k_{7}h\left(T^{\ast }+t_{1}\right) = k_{9} < c$ and $k_{8}i\left(T^{\ast }+t_{1}\right) = k_{10} < c$ for some $k_{9}\left(>k_{7}\right)$ and $k_{10}\left(>k_{8}\right)$. By our construction, $\psi \left(x, t\right) = \psi \left(0, t\right)$ and $\sigma \left(x, t\right) = \sigma \left(0, t\right)$ satisfy

By Lemma 2.1, $\psi \left(x, t\right) \geq u\left(x, t\right)$ and $\sigma \left(x, t\right) \geq v\left(x, t\right)$ on $\overline{B_{1}\left(0\right) }\times \left[T^{\ast }, T^{\ast }+t_{1}\right)$. Therefore, we find the solution $\left(u, v\right)$ to the problem (1.3)-(1.4) on $\overline{B_{1}\left(0\right) }\times \left[T^{\ast }, T^{\ast }+t_{1}\right)$. This contradicts the definition of $T^{\ast }$. Hence, either $u\left(0, t\right)$ or $v\left(0, t\right)$ quenches at $T^{\ast }$.

Let $y = u_{t}$ and $z = v_{t}$. We differentiate the problem (2.4) with respect to $t$ to obtain the following system

The result below shows that $u_{t}\left(r, t\right)$ and $v_{t}\left(r, t\right)$ are decreasing functions in $r$.

Lemma 2.7. $u_{t}\left(r_{2}, t\right) < u_{t}\left(r_{1}, t\right)$ and $v_{t}\left(r_{2}, t\right) < v_{t}\left(r_{1}, t\right)$ for $0 < r_{1} < r_{2} < 1$ and $t\in \left(0, T\right)$.

Proof. We differentiate the first equation of problem (2.8) with respect to $r$ to obtain the following differential equation

For $r\in \left[0, 1\right)$, $u_{rt}\left(r, 0\right)$ is given by

Using $u_{r}\left(r, 0\right) = 0$ and Lemma 2.3, we have $u_{rt}\left(r, 0\right) \leq 0$. Thus, $y_{r}\left(r, 0\right) \leq 0$ for $r\in \left[0, 1\right)$. By Lemma 2.2(i),

By the Hopf's lemma, $\partial y\left(1, t\right) /\partial r < 0$ for $t > 0$. By the symmetry of $B_{1}\left(0\right)$ with respect to $0$, $\partial y\left(0, t\right) /\partial r = 0$ for $t\geq 0$. Let $U = y_{r}\left( = u_{rt}\right)$. $U$ satisfies the following initial-boundary value problem:

By the maximum principle, $U\left(r, t\right) < 0$ for $\left(0, 1\right] \times \left(0, T\right)$. We integrate $U\left(r, t\right) < 0$ with respect to $r$ over $\left(r_{1}, r_{2}\right)$ to yield $y\left(r_{2}, t\right) < y\left(r_{1}, t\right)$. That is, $u_{t}\left(r_{2}, t\right) < u_{t}\left(r_{1}, t\right)$ for $0 < r_{1} < r_{2} < 1$ and $t\in \left(0, T\right)$. We follow a similar procedure to obtain $v_{t}\left(r_{2}, t\right) < v_{t}\left(r_{1}, t\right)$ for $0 < r_{1} < r_{2} < 1$ and $t\in \left(0, T\right)$.

Here is the corollary of above lemma. It illustrates that $u_{t}$ and $v_{t}$ attain their maximum value at $r = 0$ for $t\in \left(0, T\right)$.

Corollary 2.8. $u_{t}\left(r, t\right) < u_{t}\left(0, t\right)$ and $v_{t}\left(r, t\right) < v_{t}\left(0, t\right)$ for $\left(r, t\right) \in \left(0, 1\right) \times \left(0, T\right)$.

Now, we are going to prove that the solution $\left(u, v\right)$ quenches at $x = 0$ only.

Theorem 2.9. The solution $\left(u, v\right)$ quenches only at $x = 0$.

Proof. To establish this result, we let $V = v_{rt}\left( = z_{r}\right)$ and $t_{2}\in \left(0, T\right)$. $V$ satisfies the problem (2.9) with $U$ substituting by $V$. By Lemma 2.7, $U\left(r_{2}, t\right) < 0$ and $V\left(r_{2}, t\right) < 0$ for $r_{2}\in \left(0, 1\right)$ and $t\in \left[t_{2}, s\right)$ where $s\leq T$. Also, $U\left(r, t_{2}\right) < 0$ and $V\left(r, t_{2}\right) < 0$ for $r\in \left(0, r_{2}\right]$. Let $J$ be the parabolic operator such that $JW = W_{t}-W_{rr}-\left(n-1\right) W_{r}/r+\left(n-1\right) W/r^{2}$. Let us consider the following auxiliary problem below:

By the maximum principle, $W\left(r, t\right) < 0$ for $\left(0, 1\right) \times \left(t_{2}, T\right)$. For $\left(r, t\right) \in \left[0, 1\right] \times \left[t_{2}, T\right)$, the integral representation form of $W$ is given by

where $K$ is the Green's function of the parabolic operator $J$. $K$ is able to determine using the method of separation of variables and it would be represented in the form of infinite series, see ^[14]. Since $W$ is negative in $\left(0, 1\right) \times \left(t_{2}, T\right)$ and $K$ is positive in the set $\{\left(r, \xi, t\right) :r$ and $\xi$ are in $\left(0, 1\right)$, and $t > t_{2}\}$, there exists a positive constant $\rho$ such that $W\left(r, t\right) < -\rho$ for $\left(r, t\right) \in \left(0, 1\right) \times \left(t_{2}, T\right)$. By $U\left(1, t\right) < 0$ for $t\in \left(0, T\right)$ and the comparison theorem, $U\left(r, t\right) \leq W\left(r, t\right)$ for $\left(r, t\right) \in \left[0, 1\right] \times \left[t_{2}, T\right)$. Thus, $U\left(r, t\right) \leq W\left(r, t\right) < -\rho$ for $\left(r, t\right) \in \left(0, 1\right) \times \left(t_{2}, T\right)$. Now, we integrate $U\left(r, t\right) \left( = u_{rt}\left(r, t\right) \right) < -\rho$ with respect to $r$ over $\left(r_{3}, r_{4}\right)$ and then with respect to $t$ over $\left(t, t_{3}\right)$ where $r_{3}, r_{4}\in \left(0, r_{2}\right]$ to obtain

Since $u_{r} < 0$ in $\left(0, 1\right] \times \left(0, T\right)$, $u$ has no maximum except $r = 0$. Suppose that $u$ quenches for $r\in \left(0, 1-r_{2}\right)$. Let us assume that $u\left(r_{3}, t\right)$ and $u\left(r_{4}, t\right)$ both quench at $T$. Therefore, $u\left(r_{3}, t_{3}\right) \rightarrow c^{-}$ and $u\left(r_{4}, t_{3}\right) \rightarrow c^{-}$ as $t_{3}\rightarrow T^{-}$. From the above inequality, we have

Equivalently,

This contradicts $u_{r}\left(r, t\right) < 0$ for $\left(r, t\right) \in \left(0, 1\right] \times \left(0, T\right)$. Hence, $u$ quenches only at $x = 0$. Similarly, $v$ quenches only at $x = 0$ also.

3.   Simultaneous and non-simultaneous quenching

In this section, we prove the solution $\left(u, v\right)$ to quench either (i) simultaneously or (ii) non-simultaneously under some conditions. Let $\varphi _{0}\left(x\right) \in C\left(\overline{B_{1}\left(0\right) } \right) \cap C^{2}\left(B_{1}\left(0\right) \right)$ such that $\Delta \varphi _{0}\left(x\right) < 0$, $\varphi _{0}\left(x\right) > 0$ in $B_{1}\left(0\right)$, and $\varphi _{0}\left(x\right) = 0$ on $\partial B_{1}\left(0\right)$ and $\max_{x\in \overline{B_{1}\left(0\right) } }\varphi _{0}\left(x\right) \leq 1$. Let $\varphi \left(x, t\right)$ be the solution to the following first initial-boundary value problem:

By the maximum principle, $\varphi \left(x, t\right) > 0$ in $B_{1}\left(0\right) \times \left[0, \infty \right)$ and is bounded above by $\varphi _{0}\left(x\right)$, and $\varphi \left(x, t\right)$ satisfies

Let $t_{4}\in \left(0, T\right)$ such that $v\left(0, t_{4}\right) \leq k_{11} < c$. Then,

By Lemma 2.2(ii), $u_{t}\left(x, t\right) > 0$ in $B_{1}\left(0\right) \times \left(0, T\right)$. Since $u_{t}\left(x, t_{4}\right) > 0$\ and $\varphi \left(x, t_{4}\right) > 0$ in $B_{1}\left(0\right)$, and $u_{t}\left(x, t_{4}\right) = \varphi \left(x, t_{4}\right) = 0$ on $\partial B_{1}\left(0\right)$, we choose a positive real number $\eta _{1}\left(<1\right)$ such that

Clearly, $u_{t}\left(x, t\right) = a\eta _{1}\varphi \left(x, t\right) f\left(v\left(0, t\right) \right)$ for $\left(x, t\right) \in \partial B_{1}\left(0\right) \times \left[0, T\right)$. Let $I\left(x, t\right) = u_{t}\left(x, t\right) -a\eta _{1}\varphi \left(x, t\right) f\left(v\left(0, t\right) \right)$. By inequalities (3.1) and (3.2), $I\left(x, t_{4}\right) \geq 0$ on $\overline{B_{1}\left(0\right) }$. Let $Q\left(x, t\right) = v_{t}\left(x, t\right) -b\eta _{2}\varphi \left(x, t\right) g\left(u\left(0, t\right) \right)$ for some positive $\eta _{2}$ less than $1$. We follow a similar computation to get $Q\left(x, t_{4}\right) \geq 0$ on $\overline{B_{1}\left(0\right) }$. We modify the proof of Lemma 3.4 of ^[4] to obtain the result below.

Lemma 3.1. $I\left(x, t\right) \geq 0$ and $Q\left(x, t\right) \geq 0$ on $\overline{B_{1}\left(0\right) } \times \left[t_{4}, T\right)$.

Proof. By a direct computation,

Then, we have

By $\varphi \leq 1$ on $\overline{B_{1}\left(0\right) }\times \left[0, \infty \right)$, $\eta _{1} < 1$, and $v_{t}\left(0, t\right) > 0$ for $t\in \left(0, T\right)$, it gives $LI\geq 0$ in $B_{1}\left(0\right) \times \left(0, T\right)$. In addition, $I\left(x, t_{4}\right) \geq 0$ on $\overline{B_{1}\left(0\right) }$, and $I\left(x, t\right) = 0$ on $\partial B_{1}\left(0\right) \times \left(t_{4}, T\right)$. By the maximum principle, $I\left(x, t\right) \geq 0$ on $\overline{B_{1}\left(0\right) } \times \left[t_{4}, T\right)$. Similarly, we have $Q\left(x, t\right) \geq 0$ on $\overline{B_{1}\left(0\right) }\times \left[t_{4}, T\right)$.

Now, we provide the result of simultaneous quenching of the solution $\left(u, v\right)$ when $\int_{0}^{c}f\left(\omega \right) d\omega = \infty$ and $\int_{0}^{c}g\left(\omega \right) d\omega = \infty$. With these two integrals and (H$_{2}$) (see section 1), we know that $\int_{m}^{c}f\left(\omega \right) d\omega = \infty$ and $\int_{m}^{c}g\left(\omega \right) d\omega = \infty$, and $\int_{0}^{m}f\left(\omega \right) d\omega < \infty$ and $\int_{0}^{m}g\left(\omega \right) d\omega < \infty$ for $m\in \left[0, c\right)$.

Theorem 3.2. If $\int_{0}^{c}f\left(\omega \right) d\omega = \infty$ and $\int_{0}^{c}g\left(\omega \right) d\omega = \infty$, and either $u$ or $v$ quenches at $x = 0$ in $T$, then $u$ and $v$ both quench at $x = 0$ in the same time $T$.

Proof. Suppose not, let us assume that $v\left(0, t\right)$ quenches at $T$ but $u\left(0, t\right)$ remains bounded on $\left[0, T\right]$. Then, $0\leq u\left(0, t\right) \leq k_{12} < c$ for $t\in \left[0, T\right]$. From Lemma 3.1, we have

By Lemma 2.3, $u$ and $v$ both attain the maximum at $x = 0$ for $t\in \left(0, T\right)$. Then, $\Delta u\left(0, t\right) < 0$ and $\Delta v\left(0, t\right) < 0$ over $\left(0, T\right)$. From the equation (1.3), we obtain the following inequalities:

By $g > 0$ and $\varphi \left(0, t\right) > 0$ for $t\in \left[0, \infty \right)$, we divide the first inequality by the second one to achieve

From the first-half inequality, it yields the expression below:

Let $\delta$ be a positive real number such that $\delta = \min_{\left[0, T \right] }\varphi \left(0, t\right)$. Then, we integrate both sides over $\left[t_{4}, s\right)$ for $s\in \left(t_{4}, T\right]$ to attain

When $s\rightarrow T^{-}$, $v\left(0, s\right) \rightarrow c^{-}$. By assumption $\int_{0}^{c}f\left(\omega \right) d\omega = \infty$, $\lim_{s\rightarrow T^{-}}\int_{v\left(0, t_{4}\right) }^{v\left(0, s\right) }f\left(v\left(0, t\right) \right) dv\left(0, t\right) = \infty$. If $u\left(0, s\right) \leq k_{12} < c$ as $s\rightarrow T^{-}$, then there exists $k_{13}$ such that

Therefore,

It leads to a contradiction. Hence, $u\left(0, t\right)$ quenches at $T$. From the second-half of inequality (3.4) and $\int_{0}^{c}g\left(\omega \right) d\omega = \infty$, we prove that $v\left(0, t\right)$ quenches at $t = T$ if $u\left(0, t\right)$ quenches. This completes the proof.

Theorem 3.3. Suppose that $\int_{0}^{c}f\left(\omega \right) d\omega < \infty$ and $\int_{0}^{c}g\left(\omega \right) d\omega < \infty$, and depending on $a$ and $b$ , then the following three cases could happen: (i) $u$ and $v$ both quench in $T$ at $x = 0$, (ii) either $u$ or $v$ quenches in $T$ at $x = 0$, or (iii) both $u$ and $v$ do not quench.

Proof. From (3.3), we have the inequality below:

Thus,

We integrate both sides with respect to $t$ over $\left[t_{4}, s\right)$ for $s\in \left(t_{4}, T\right]$ to obtain

(i) In this case, we prove simultaneous quenching of $u$ and $v$ in $T$ at $x = 0$.

Let us assume that $v\left(0, t\right)$ quenches at $t = T$ but $u\left(0, t\right)$ remains bounded on $\left[0, T\right]$. We integrate the inequality (3.5) with respect to $t$ over $\left[t_{4}, s\right)$ to obtain

By the mean value theorem for definite integrals, there exists $t_{5}\in \left(t_{4}, s\right)$ such that $\int_{t_{4}}^{s}u_{t}\left(0, t\right) v_{t}\left(0, t\right) dt = v_{t}\left(0, t_{5}\right) \int_{t_{4}}^{s}u_{t}\left(0, t\right) dt$. This gives

We evaluate the integral of middle expression to yield

As $v_{t}\left(0, t_{5}\right) > 0$, it is equivalent to

By $v_{t}\left(0, t\right) \leq bg\left(u\left(0, t\right) \right)$ and $u\left(0, t\right)$ remains bounded on $\left[0, T\right]$, then there exists $k_{14}$\ such that $v_{t}\left(0, t_{5}\right) \leq k_{14}$ for $t_{5}\in \left[t_{4}, s\right]$ for $s\in \left(t_{4}, T\right]$. This implies

If we choose $b$ being sufficiently large such that $b\eta _{2}\delta \lim_{s\rightarrow T^{-}}\int_{u\left(0, t_{4}\right) }^{u\left(0, s\right) }g\left(u\left(0, t\right) \right) du\left(0, t\right) /k_{14}\geq$ $c$, then we have

This leads to a contradiction. Therefore, $u$ quenches in $T$ at $x = 0$ also when $b$ is sufficient large. Hence, $u$ and $v$ quench simultaneously in $T$ at $x = 0$.

(ii) We prove non-simultaneous quenching.

Let us assume that both $v\left(0, t\right)$ and $u\left(0, t\right)$ do not quench in any finite time. From the inequality (3.6),

Then, there exists $k_{15}$ such that

Since $\lim_{s\rightarrow T^{-}}\int_{u\left(0, t_{4}\right) }^{u\left(0, s\right) }g\left(u\left(0, t\right) \right) du\left(0, t\right) < \infty$, we choose a sufficiently large $b$ such that

This leads to a contradiction. Therefore, either $u$ or $v$ quenches in $T$ at $x = 0$, or $u$ and $v$ quench simultaneously at $x = 0$.

Now, let us assume that the solution $\left(u, v\right)$ quenches simultaneously at $x = 0$. By Lemma 2.2(ii), $v_{t}\left(0, t\right) > 0$ for $t > 0$. Then, there exists $k_{16}$ such that $v_{t}\left(0, t\right) > k_{16}$ for $t\in \left[t_{4}, s\right)$ where $s\in \left(t_{4}, T\right]$. From the inequality (3.5), we have

We integrate this expression with respect to $t$ over $\left(t_{4}, s\right)$ to achieve

We take the limit $s$ to $T$ on both sides and by $v_{t}\left(0, t\right) > k_{16}$ to get

Evaluating the integration on the left side of the above expression, we have

Let us assume that $u\left(0, t_{4}\right) = k_{17}\left(<c\right)$ and $u\left(0, T\right) = c$. We choose $a$ being small enough so that $\left(a\int_{0}^{c}f\left(\omega \right) d\omega \right) /k_{16} < c-k_{17}$. Then,

It leads to a contradiction. Hence, $u$ and $v$ quench non-simultaneously at $x = 0$.

(iii) By Lemma 2.5, the solution $\left(u, v\right)$ exists globally if $a$ and $b$ are sufficiently small. Thus, both $u$ and $v$ do not quench.

Theorem 3.4. Suppose that $\int_{0}^{c}f\left(\omega \right) d\omega < \infty$ and $\int_{0}^{c}g\left(\omega \right) d\omega = \infty$, then any quenching in the problem (1.3)-(1.4) is non-simultaneous with $\lim_{s\rightarrow T^{-}}u\left(0, s\right) \leq k_{18} < c$. That is, $u$ does not quench in $T$ at $x = 0$.

Proof. From the expression (3.4)

we have

Then, we integrate the expression over the time interval $\left[0, s\right)$ for $s\in \left(0, T\right]$ and by the mean value theorem for definite integrals to give

for some $t_{6}\in \left(0, s\right)$ with $\varphi \left(0, t_{6}\right) > 0$. Suppose that $v\left(0, s\right) \rightarrow c^{-}$ as $s\rightarrow T^{-}$. By assumption $\int_{0}^{c}f\left(\omega \right) d\omega < \infty$, it implies that $\lim_{s\rightarrow T^{-}}\int_{0}^{u\left(0, s\right) }g\left(u\left(0, t\right) \right) du\left(0, t\right) < \infty$. Thus, $\lim_{s\rightarrow T^{-}}u\left(0, s\right) \leq k_{18} < c$. Hence, $u$ does not quench in $T$ at $x = 0$.

Based on a similar proof of Theorem 3.4, we also prove that any quenching in the problem (1.3)-(1.4) is non-simultaneous with $\lim_{s\rightarrow T^{-}}v\left(0, s\right) \leq k_{19} < c$ when $\int_{0}^{c}g\left(\omega \right) d\omega < \infty$ and $\int_{0}^{c}f\left(\omega \right) d\omega = \infty$.

4.   Conclusions

In this article, we prove that the solution $\left(u, v\right)$ to the problem (1.3)-(1.4) attains its maximum value at the center $x = 0$ over the domain $B_{1}\left(0\right)$. Further, we obtain the main result that $x = 0$ is the only quenching point. Then, we show that the solution $\left(u, v\right)$ quenches simultaneously at $x = 0$ when $\int_{0}^{c}f\left(\omega \right) d\omega = \infty$ and $\int_{0}^{c}g\left(\omega \right) d\omega = \infty$. When the integrals $\int_{0}^{c}f\left(\omega \right) d\omega$ and $\int_{0}^{c}g\left(\omega \right) d\omega$ are both finite, the solution $\left(u, v\right)$ could quench simultaneously or non-simultaneously, or $\left(u, v\right)$ exists globally. When one of the integrals is finite and the other is unbounded, we show that $\left(u, v\right)$ quenches non-simultaneously.

Acknowledgments

The author thanks the anonymous referee for careful reading. This research did not receive any specific grant funding agencies in the public, commercial, or not-for-profit sectors.

Conflict of interest

The author declares that there are no conflicts of interest in this paper.