In this section, we investigate the quenching phenomenon for the problem (1.1) under the conditions (A_{2}) and (A_{4}).

### 3.1 Quenching on the boundary and blow-up of \(u_{t}\)

In this section, we prove the solution quenches in finite time and blowing up of \(u_{t}\) at the only quenching point \(x = 1\). First of all, we have the following:

### Lemma 3.1

*Assume that* (A_{2}) *and* (A_{4}) *hold and the solution*
*u*
*of the problem* (1.1) *exists in*
\((0,\tilde{T_{0}})\)
*for some*
\(\tilde{T_{0}}>0\). *Then*
\(u_{x}(x,t)<0\)
*and*
\(u_{t}(x,t)<0\)
*in*
\((0,1]\times(0,\tilde{T_{0}})\).

The proof is similar to Lemma 2.1, so we omit it.

### Theorem 3.1

*Assume that* (A_{2}) *and* (A_{4}) *hold*. *Then there exists a finite time*
*T*, *such that every solution of* (1.1) *quenches in this time*, *and the only quenching point is*
\(x=1\).

### Proof

By the maximum principle, we can obtain \(0< u(\cdot, t)< 1\) for all *t* in the existence interval. Together with (A_{4}), we get

$$\beta=-u^{-q(p-1)}(1,0)+ \int_{0}^{1}\bigl(1-u(x,0)\bigr)^{-h}\,dx< 0. $$

Denote \(I(t)=\int_{0}^{1} u(x,t)\,dx\). By Lemma 3.1, it is easy to see that

$$\begin{aligned} I'(t)= \int_{0}^{1}u_{t}(x,t)\,dx =-u^{-q(p-1)}(1,t)-+ \int_{0}^{1}\bigl(1-u(x,t)\bigr)^{-h}\,dx\le \beta. \end{aligned}$$

Thus \(I(t)\le I(0)+\beta t\), which means that \(I(\tilde{t_{0}})=0\) for some \(\tilde{t_{0}}>0\). In addition, notice that \(u_{x}<0\) for \(0< x\le1\), we can see that there exists *T* (\(0< T<\tilde{t_{0}}\)) such that \(\lim_{t\rightarrow T^{-}}u(1,t)=0\). Combining with the singular nonlinearity of the boundary flux, *u* must occur quenching on the boundary \(x=1\). As in Theorem 2.1, in the following, we only need to show that the solutions *u* cannot take place quenching in \((1/2,1)\times(\gamma, T)\) for some *γ* (\(0<\gamma<T\)).

Define

$$H(x,t)=u_{x}+\varepsilon \biggl(x-\frac{1}{4} \biggr), \quad(x,t) \in\biggl(\frac{1}{4},1\biggr)\times (\gamma,T), $$

where *ε* is sufficiently small. Since \(u_{x}(x,t)<0\) in \((0,1]\times[0,T)\), \(H(x,t)\) satisfies

$$\begin{aligned} &H_{t}-(p-1)|u_{x}|^{p-2}H_{xx} =-(p-1) (p-2)|u_{x}|^{p-3}(H_{x}- \varepsilon)^{2}+h(1-u)^{-h-1}u_{x}< 0, \\ &\quad\mbox{for } (x,t)\in\biggl(\frac{1}{4},1\biggr)\times(\gamma,T). \end{aligned}$$

Further, on the parabolic boundary, since \(u_{x}(x, t) < 0\) in \((0,1]\times[0,T)\) and choosing *ε* small enough, we have

$$\begin{aligned} &H\biggl(\frac{1}{4},t\biggr)=u_{x}\biggl(\frac{1}{4},t \biggr)< 0, \quad\mbox{for } t\in[\gamma,T), \\ &H(1,t)=-u^{-q}(1,t)+\frac{3}{4}\varepsilon\le-1+ \frac{3}{4}\varepsilon< 0, \quad\mbox{for } t\in[\gamma,T), \\ &H(x,\tilde{\eta})\le u_{x}\biggl(\frac{1}{4},\gamma\biggr)+ \frac{3}{4}\varepsilon< 0, \quad\mbox{for } x\in\biggl[\frac{1}{4},1 \biggr]. \end{aligned}$$

Making use of the maximum principle, we obtain \(H(x,t)\le0\) in \((1/4,1)\times(\gamma,T)\), which yields

$$\begin{aligned} -u_{x}\geq\varepsilon\biggl(x-\frac{1}{4} \biggr),\quad (x,t)\in\biggl(\frac{1}{4},1\biggr)\times(\tilde{\eta},T). \end{aligned}$$

(3.1)

Integrating (3.1) with respect to *x* from *x* to 1 gives

$$u(x,t)\geq u(1,t)+ \int_{x}^{1}\varepsilon\biggl(x-\frac{1}{4} \biggr)\,dx \geq \int_{x}^{1}\varepsilon\biggl(x-\frac{1}{4} \biggr)\,dx>0, $$

which implies that \(u(x,t)\) if \(x<1\). This completes the proof of Theorem 3.1. □

### Theorem 3.2

\(u_{t}\)
*blows up at the quenching point*
\(x=1\).

### Proof

We prove the theorem by contradiction. Assume that \(u_{t}\) is bounded on \([0,1]\times[0,T)\). Then there exists a positive constant *L* such that \(u_{t}>-L\). Thus, we have

$$\bigl(|u_{x}|^{p-2}u_{x}\bigr)_{x}+(1-u)^{-h}>-L. $$

Integrating with respect to *x* from *x* to 1 yields

$$-u^{-q(p-1)}(1,t)>-u^{-q(p-1)}(x,t)-L-\bigl(1-u(0,t) \bigr)^{-h}. $$

Therefore, it is found that the left-hand side tends to negative infinity as \(t\rightarrow T^{-}\), while the right-hand side is finite. This completes the proof of Theorem 3.2. □

### 3.2 Quenching rate

Now, we are in a position to investigate the bounds on the quenching rate. First of all, we will show the lower bound of the quenching rate.

### Theorem 3.3

*Assume that* (A_{2}) *and* (A_{4}) *hold*. *Then there exists a positive constant*
\(C_{2}\)
*such that*

$$u(1,t)\geq C_{2}(T-t)^{\frac{1}{pq+2}} $$

*for*
*t*
*sufficiently close to*
*T*.

### Proof

Let \(k(u)=-qu^{-q(p-1)(\delta-1)-q-1}\), where \(1-\frac {q+1}{q(p-1)}<\delta<1-\frac{1}{q(p-1)}\). It is easy to see that \(k(u)<0\), \(k'(u)>0\), and \(k''(u)<0\). Letting *τ* be close to *T*, we introduce the function

$$Q(x,t)=u_{t}-\varepsilon k(u) (-u_{x})^{(p-1)(2-\delta)}, \quad\mbox{in } (1-T+\tau,1)\times(\tau,T), $$

where *ε* is a positive constant. Through a fairly complicated calculation, one has

$$\begin{aligned} Q_{t}=(p-1)|u_{x}|^{p-2}Q_{xx}+(p-1) (p-2) (-u_{x})^{p-3}(-u_{xx})Q_{x} +J(x,t)Q+W(x,t). \end{aligned}$$

Here

$$\begin{aligned} J(x,t)={}&h(1-u)^{-h-1} +\varepsilon(2-\delta)\bigl[(p-1) (2-\delta)-1 \bigr](-u_{x})^{(p-1)(1-\delta)-1}k(u)u_{t} \\ &{}+\varepsilon\bigl[(p-1) (5-2\delta)-1\bigr]k'(u) (-u_{x})^{(p-1)(2-\delta)} \\ &{}+\varepsilon^{2}(2-\delta)\bigl[(p-1) (2-\delta )-1 \bigr]k^{2}(u) (-u_{x})^{(p-1)(3-2\delta)-1} \end{aligned}$$

and

$$\begin{aligned} W(x,t)={}&\varepsilon(p-1)k''(u) (-u_{x})^{(p-1)(3-\delta)+1} +\varepsilon^{2}\bigl[(p-1) (5-2\delta)-1\bigr]k'(u)k(u) (-u_{x})^{2(p-1)(2-\delta)} \\ &{}+\varepsilon^{3}(2-\delta)\bigl[(p-1) (2-\delta )-1 \bigr]k^{3}(u) (-u_{x})^{(p-1)(5-3\delta)-1} \\ &{}-2\varepsilon(2-\delta)\bigl[(p-1) (2-\delta)-1\bigr]k(u) (-u_{x})^{(p-1)(1-\delta )-1}u_{t}(1-u)^{-h} \\ &{}+\varepsilon(2-\delta)\bigl[(p-1) (2-\delta)-1\bigr]k(u) (-u_{x})^{(p-1)(1-\delta )}(1-u)^{-2h} \\ &{}-\varepsilon(p-1)\bigl[(5-2\delta)k'(u) (1-u)+(2-\delta )k(u)h \bigr](-u_{x})^{(p-1)(2-\delta)}(1-u)^{-h-1}. \end{aligned}$$

Notice that \(k(u)<0\), \(k'(u)>0\), \(k''(u)<0\), and *τ* is sufficiently close to *T*, then \(J(x,t)>0\) and \(W(x,t)<0\). Therefore,

$$\begin{aligned} &Q_{t}< (p-1)|u_{x}|^{p-2}Q_{xx}+(p-1) (p-2) (-u_{x})^{p-3}(-u_{xx})Q_{x} +J(x,t)Q,\\ &\quad (x,t)\in(1-T+\tau, 1)\times(\tau, T). \end{aligned}$$

Further, on the parabolic boundary, in view of the only quenching point \(x=1\) and provided *ε* sufficient small, both \(Q(1-T+\tau, t)\) and \(Q(x,\tau)\) are negative. On the right boundary \(x=1\), we get

$$\begin{aligned} Q_{x}(1,t)={}&q\bigl[1-\varepsilon(2-\delta)\bigr]u^{-q-1}(1,t)Q(1,t) +\varepsilon q\bigl\{ \bigl[\varepsilon q(2-\delta)+q(p-1) (\delta-1)+1\bigr] \\ &{}\times u^{-qp-1}(1,t)+(2-\delta) \bigl(1-u(1,t)\bigr)^{-h}\bigr\} u^{-q-1}(1,t)\\ \le{}&q\bigl[1-\varepsilon(2-\delta)\bigr]u^{-q-1}(1,t)Q(1,t), \end{aligned}$$

provided *ε* is sufficiently small and *τ* is sufficiently close to *T*. Thus, take advantage of the maximum principle, \(Q(x,t)\le0\) on \([1-T+\tau,1]\times[\tau,T)\). Then we have \(Q(1,t)\le0\), that is,

$$\begin{aligned} u_{t}(1,t)\le\varepsilon k\bigl(u(1,t)\bigr) \bigl(-u_{x}(1,t)\bigr)^{(p-1)(2-\delta)} =-\varepsilon q u^{-qp-1}(1,t). \end{aligned}$$

(3.2)

Integrating (3.2) with respect to *t* from *t* to *T*, it gives

$$\begin{aligned} u(1,t)\geq\bigl[\varepsilon q(qp+2)\bigr]^{\frac{1}{qp+2}}(T-t)^{\frac{1}{qp+2}} =C_{2}(T-t)^{\frac{1}{qp+2}}, \end{aligned}$$

where \(C_{2}=[\varepsilon q(qp+2)]^{\frac{1}{qp+2}}\). This completes the proof of Theorem 3.3. □

To end this section, we present the upper bound on the quenching rate.

### Theorem 3.4

*Assume that* (A_{2}) *and* (A_{4}) *hold*. *Then there exists a positive constant*
\(C_{3}\)
*such that*

$$u(1,t)\le C_{3}(T-t)^{\frac{1}{pq+2}} $$

*for*
*t*
*sufficiently close to*
*T*.

### Proof

Denote \(E(x,t)=|u_{x}(x,t)|^{p-2}u_{x}(x,t)+\varrho^{p-1}(x)u^{-q(p-1)}(x,t)\) in \((0,1)\times(0,T)\), where

$$\begin{aligned} \varrho(x)= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} 0,& x\in[0,x_{0}],\\ \frac{(x-x_{0})^{r}}{(1-x_{0})^{r}}, & x\in(x_{0},1], \end{array}\displaystyle \right . \end{aligned}$$

with some \(x_{0}<1\) and choosing \(r\geq3\) large enough so that \(\varrho(x)\le-u_{0}'(x)u^{q}_{0}(x)\) for \(x_{0}< x\le1\). We can easily obtain \(E(0,t)=E(1,t)=0\), and \(E(x,0)\le0\). In addition, in \((0,1)\times(0,T)\), *E* satisfies

$$\begin{aligned} E_{t}={}&(p-1)|u_{x}|^{p-2}E_{xx} \\ &{}-(p-1)^{2}\bigl[pq^{2}+q\bigr]\varrho^{p-1}(x)|u_{x}|^{p}u^{-q(p-1)-2} \\ &{}-2(p-1)^{3}q\varrho^{p-2}(x)\varrho'(x)|u_{x}|^{p-1}u^{-q(p-1)-1} \\ &{}-(p-1)^{2}|u_{x}|^{p-2}\varrho^{p-3}(x) \bigl[(p-2)\varrho^{\prime2}(x) +\varrho(x) \varrho''(x)\bigr]u^{-q(p-1)} \\ &{}-q(p-1)\varrho^{p-1}(x)u^{-q(p-1)-1}(1-u)^{-h} -(p-1)h|u_{x}|^{p-1}(1-u)^{-h-1}. \end{aligned}$$

According to the definition of \(\varphi(x)\), it is easy to see that \(\varrho(x)\geq0\), \(\varrho'(x)\geq0\), and \(\varrho''(x)\geq0\). Then we have

$$E_{t}\le(p-1)|u_{x}|^{p-2}E_{xx}. $$

Making use of the maximum principle, we get \(E(x,t)\le0\), that is,

$$\varrho(x)u^{-q}(x,t)\le-u_{x}(x,t), \quad\mbox{for } (x,t) \in[0,1]\times[0,T). $$

Furthermore, because of \(E(x,t)\le0\), we have \(E_{x}(1,t)\geq0\). In fact,

$$\begin{aligned} E_{x}(1,t)=\lim_{x\rightarrow1^{-}}\frac{E(x,t)-E(1,t)}{x-1}\geq0, \end{aligned}$$

which implies

$$\begin{aligned} u_{t}(1,t)&\geq -(p-1)\bigl[ \varrho'(1)+qu^{-q-1}(1,t)\bigr]u^{-q(p-1)}(1,t)+ \bigl(1-u(1,t)\bigr)^{-h} \\ &\geq -\tilde{C}q(p-1)u^{-pq-1}(1,t). \end{aligned}$$

(3.3)

Integrating (3.3) with respect to *t* from *t* to *T* gives

$$u(1,t)\le\bigl[\tilde{C}q(p-1) (pq+2)\bigr]^{\frac{1}{pq+2}}(T-t)^{\frac{1}{pq+2}} =C_{3}(T-t)^{\frac{1}{pq+2}}, $$

where \(C_{3}=[\tilde{C}q(p-1)(pq+2)]^{\frac{1}{pq+2}}\), which produces the asserted result. This completes the proof of Theorem 3.4. □

From Theorem 3.3 and Theorem 3.4, we have the following exact quenching rate.

### Corollary 3.1

*Assume that* (A_{2}) *and* (A_{4}) *hold*. *Then the solution of the problem* (1.1) *satisfies*

$$C_{2}(T-t)^{\frac{1}{pq+2}}\le u(1,t)\le C_{3}(T-t)^{\frac{1}{pq+2}}, $$

*that is*,

$$u(1,t)\sim(T-t)^{\frac{1}{pq+2}} $$

*for*
*t*
*sufficiently close to*
*T*. *Here*
\(C_{2}\), \(C_{3}\)
*are positive constants which are given in Theorem *
3.3
*and Theorem *
3.4.