The following theorem shows that the solution *v* of the integral equation (5) quenches at the point \(x_{0}\).

### Theorem 4.1

*Let*
\([0,t_{q})\)
*be the maximum time interval such that the integral equation* (5) *has a continuously differentiable solution*
\(v(x,t)\)
*for any*
\((x,t)\in[0,l]\times[0,t_{q})\). *If*
\(v(x_{0},t)\)
*converges to* 1^{−}
*as*
*t*
*converges to*
\(t_{q}\), *then*
\(v_{t}(x_{0},t)\)
*goes to* ∞ *as*
\(t\rightarrow{t_{q}}\).

### Proof

Let us consider the derivative of (5): for any \((x,t) \in (0,l)\times(0,t_{q})\),

$$\begin{aligned}[b] v_{t}(x,t)={}&G_{\alpha}(x,t;x_{0} ,0)f(1) \\ &+ \int_{0}^{t}G_{\alpha }(x,t- \tau;x_{0},0)f_{v} \bigl(1-v(x_{0},\tau) \bigr)v_{\tau}(x,\tau)\,d\tau. \end{aligned}$$

(7)

From the positivity of \(G_{\alpha}\) and the properties of *f*, we have that

$$ v_{t}(x,t)>0\quad\text{for } (x,t) \in(0,l)\times(0,t_{q}). $$

(8)

Equation (7):

$$\begin{aligned} v_{t}(x_{0},t)& \geq G_{\alpha}(x_{0},t;x_{0} ,0) \biggl[f(1)+ \int_{0}^{t}f_{v} \bigl(1-v(x_{0}, \tau) \bigr)v_{\tau}(x_{0},\tau)\,d\tau \biggr] \\ & = G_{\alpha}(x_{0},t;x_{0} ,0)f \bigl(1-v(x_{0},t) \bigr). \end{aligned}$$

Since \(\lim_{v\rightarrow1^{-}}f(1-v)=\infty\), this shows that \(v_{t}(x_{0},t)\rightarrow{\infty}\) as \(t\rightarrow{t_{q}}\). Therefore, we can conclude that *v* quenches at the point \(x_{0}\) when \(t\rightarrow t_{q}\). □

We next give the condition that guarantees the occurrence of quenching in a finite time.

### Theorem 4.2

*There exists a finite time*
\(t^{*}\)
*such that the integral equation* (5) *has no continuous nonnegative solutions*
*v*
*with*
\(v(x,t)<1\)
*for every*
\(x \in[0,l]\)
*and for*
\(t>t^{*}\).

### Proof

Suppose that the integral equation (5) has a continuous solution *v* with \(v(x,t)<1\) for any \((x,t)\in[0,l] \times[0,t_{2}]\), where \(t_{2}\) is determined later. Without loss of generality, let us consider integral (5) at the point \(x=x_{0}\)

$$ v(x_{0},t)= \int_{0}^{t}G_{\alpha}(x_{0},t- \tau ;x_{0},0)f \bigl(1-v(x_{0},\tau) \bigr)\,d\tau $$

for \(t \in[0,t_{2}]\). We then choose \(t_{2}\) where \(t_{2}\) satisfies

$$ \int_{0}^{t_{2}}G_{\alpha}(x_{0},t_{2}- \tau;x_{0},0)\,d\tau\geq \int _{0}^{1}\frac{1}{f(1-r)}\,dr. $$

Since \(G_{\alpha}\) is decreasing with respect to *t*, we obtain that

$$ v(x_{0},t)= \int_{0}^{t}G_{\alpha}(x_{0},t- \tau ;x_{0},0)f \bigl(1-v(x_{0},\tau) \bigr)\,d\tau\geq{J(t)} $$

with \(J(t) = \int_{0}^{t}G_{\alpha}(x_{0},t_{2}-\tau ;x_{0},0)f(1-v(x_{0},\tau))\,d\tau\). We then have that

$$\begin{aligned} \frac{dJ(t)}{dt}&=G_{\alpha}(x_{0},t_{2}-t;x_{0},0)f \bigl(1-v(x_{0},t) \bigr) \\ &\geq{G_{\alpha}(x_{0},t_{2}-t;x_{0},0)f \bigl(1-J(t) \bigr)} \end{aligned}$$

for any \(t \in[0,t_{2}]\). Then, since \(J(t_{2})=v(x_{0},t)<1\), we have

$$ \int_{0}^{1}\frac{1}{f(1-r)}\,dr> \int_{0}^{J({t}_{2})}\frac {1}{f(1-J(t))}\,dJ\geq{ \int_{0}^{t_{2}}G_{\alpha}(x_{0},t_{2}-t;x_{0},0)\,dt}, $$

which contradicts the definition of \(t_{2}\). Therefore, this theorem is proved completely. □