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. □