Mass spectrometry imaging for early discovery and development of cancer drugs

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.