In the following, we derive the lower and upper bounds of \(T(\varepsilon)\) and give the proof of Theorem 1.2. First, we prove an upper estimate.
Proposition 4.1
If
φ
satisfies (1.4), \(\varphi(x_{0})=\max_{x\in\overline{\Omega}}\varphi(x)\), and
\(\varepsilon>0\)
is small enough, then we have
$$ T(\varepsilon) \leq e^{-\varphi(x_{0})} +e^{-2\varphi(x_{0})}\bigl\vert \Delta\varphi (x_{0})\bigr\vert \varepsilon+\circ(\varepsilon). $$
Proof
Let \(w(x,t)=-\ln(V(x,t)-t)\), where \(V(x,t)=U(x,\varepsilon t)\), and \(U(x,\tau)\) satisfies \(U_{\tau}=\Delta U\). Then
$$\begin{aligned} w_{t}-\varepsilon\Delta w-e^{w} &=-\frac{V_{t} -1}{V-t}+ \varepsilon\frac{\Delta V}{V-t}-\varepsilon \frac{|\nabla V|^{2}}{(V-t)^{2}} -\frac{1}{V-t} \\ &\leq -\frac{V_{t} -\varepsilon\Delta V}{V-t}=0. \end{aligned}$$
So we see that w is a subsolution.
Next, let \(U(x,\tau)\) satisfy
$$\begin{aligned}& U(x,0)= e^{-\varphi(x)}, \quad x\in\Omega, \\& U(x,\tau)= \frac{\tau}{\varepsilon}+1,\quad (x,t)\in\partial \Omega\times(0, C \varepsilon), \end{aligned}$$
where \(C>0\) is a constant. Then
$$\begin{aligned}& w(x,0)= -\ln\bigl(U(x,0)\bigr)=\varphi(x)=u(x,0),\quad x\in\Omega, \\& w(x,t)= -\ln\bigl(U(x,\varepsilon t)-t\bigr)=0, \quad (x,t)\in\partial\Omega \times(0, C). \end{aligned}$$
Thus, by the comparison principle
$$ u(x,t)\geq w(x,t), \quad (x,t)\in\Omega\times(0, C). $$
Notice that the boundary values of \(U(x,\tau)\) are bounded, that is, \(0\leq U(x,\tau) \leq C+1\) for \((x,\tau)\in\partial\Omega\times (0, C\varepsilon)\), so we can write
$$ U(x_{0},\varepsilon t)=U(x_{0},0)+ \varepsilon t U_{\tau}(x_{0},0)+\circ (\varepsilon) $$
(4.1)
for \(0< t< C\).
We take \(\tilde{t}=U(x_{0},\varepsilon\tilde{t})\). By the definition of \(w(x,t)\), \(w(x,t)\rightarrow\infty\) as \(t\rightarrow\tilde{t}\), so that the function \(w(x,t)\) blows up at time ≤t̃.
Choosing \(C>U(x_{0},0)+1\), by (4.1) we have
$$\begin{aligned} \tilde{t}&=U(x_{0},\varepsilon\tilde{t}) \\ &=U(x_{0},0)+\varepsilon\widetilde{t}U_{\tau}(x_{0},0)+o( \varepsilon) \\ &=U(x_{0},0)+\varepsilon U(x_{0},\varepsilon \tilde{t})U_{\tau}(x_{0},0)+o(\varepsilon) \\ &=U(x_{0},0)+\varepsilon\bigl[U(x_{0},0)+\varepsilon \tilde{t} U_{\tau}(x_{0},0)+o(\varepsilon) \bigr]U_{\tau}(x_{0},0)+o(\varepsilon) \\ &=U(x_{0},0)+\varepsilon U(x_{0},0)\Delta U+o( \varepsilon). \end{aligned}$$
Since \(U(x_{0},0)=e^{-\varphi(x_{0})}\) and \(\nabla\varphi(x_{0})=0\), we have \(\Delta U(x_{0},0)=-e^{-\varphi(x_{0})} \Delta\varphi(x_{0})\). Thus, as \(\varepsilon\rightarrow0\),
$$ \tilde{t}=e^{-\varphi(x_{0})} +e^{-2\varphi(x_{0})}\bigl\vert \Delta\varphi (x_{0})\bigr\vert \varepsilon+\circ(\varepsilon). $$
Since \(T(\varepsilon)\leq\tilde{t}\), Proposition 4.1 is proved. □
Now, we show a lower bound of \(T(\varepsilon)\).
Proposition 4.2
If
φ
satisfies (1.4), \(\varphi(x_{0})=\max_{x\in\overline{\Omega}}\varphi(x)\), \(\Delta\varphi(x_{0})<0\), and
\(\varepsilon>0\)
is small enough, then we have
$$ T(\varepsilon)\geq e^{-\varphi(x_{0})} +\frac{1}{4}e^{-2\varphi (x_{0})}\bigl\vert \Delta\varphi(x_{0})\bigr\vert \varepsilon+\circ( \varepsilon). $$
Proof
Without loss of generality, let \(x_{0}=0\). We construct a supersolution of the form
$$ w(x,t)=-\ln\bigl(e^{-\varphi(0)}-t\bigr)+W(x,t),\quad (x,t)\in\Omega\times (0, C), $$
where \(C\in(0,e^{-\varphi(0)})\) is a constant, \(W(x,t)=Z(x,\varepsilon t)\), and \(Z(x,\tau)\) satisfies
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} Z_{\tau}=\Delta Z,& x\in\Omega, 0< \tau< C\varepsilon, \\ Z(x,0)=\varphi(x)-\varphi(0),& x\in\Omega, \\ Z(x,\tau)=-\varphi(0),& x\in\partial\Omega, 0< \tau< C\varepsilon. \end{array}\displaystyle \right . $$
(4.2)
Clearly, by the maximum principle, \(Z\leq0\), and hence \(W(x,t)=Z(x,\varepsilon t)\leq0\) for \((x,t)\in\Omega\times (0, C)\). We get
$$\begin{aligned} w_{t}-\varepsilon\Delta w-e^{w} &=\frac{1}{e^{\varphi(0)}-t}+W_{t} -\varepsilon\Delta W-\frac {e^{W}}{e^{-\varphi(0)}-t} \\ &=\frac{1-e^{W}}{e^{-\varphi(0)}-t}\geq0 \end{aligned}$$
and also
$$ w(x,0)=\varphi(x)=u(x,0), \quad x\in\Omega, $$
and
$$ w(x,t)=-\ln\bigl(e^{\varphi(0)}-t\bigr)-\varphi(0)>0,\quad x\in\partial \Omega, 0< t< C. $$
It follows that
$$ w(x,t)\geq u(x,t),\quad x\in\partial\Omega, 0< t< C. $$
Choose \(\mu>0\) small enough and \(\eta>0\) sufficiently small depending on μ such that
$$ -\Delta\varphi(x)>-\Delta\varphi(0)-\mu \quad \mbox{for } |x|< \eta. $$
Then, when \(|x|<\eta\),
$$\begin{aligned} Z(x,\tau)&=Z(x,0)+Z_{\tau}(x,0)\tau+o(\tau) \\ &=\varphi(x)-\varphi(0)+\Delta\varphi(x)\tau+o(\tau) \\ &\leq\Delta\varphi(x)\tau+o(\tau) \\ &\leq\bigl(\Delta\varphi(0)+\mu\bigr)\tau+o(\tau) \end{aligned}$$
as \(\tau\rightarrow0\). When \(|x|\geq\eta\) and \(x\in\Omega\), \(Z(x,\tau)\leq0\). It follows that, when \(|x|<\eta\),
$$\begin{aligned} u(x,C)&\leq w(x,C)=-\ln\bigl(e^{-\varphi(0)}-C\bigr)+W(x,C) \\ &=-\ln\bigl(e^{-\varphi(0)}-C\bigr)+\bigl(\Delta\varphi(0)+\mu\bigr)\varepsilon C+o(\varepsilon) \end{aligned}$$
as \(\varepsilon\rightarrow0\), and, when \(|x|\geq\eta\),
$$\begin{aligned} u(x,C)&\leq w(x,C)=-\ln\bigl(e^{-\varphi(0)}-C\bigr)+W(x,C) \\ &\leq-\ln\bigl(e^{-\varphi(0)}-C\bigr). \end{aligned}$$
Denote by \(\tilde{u} (x,t)\) the solution of
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \tilde{u}_{t}=\varepsilon\Delta\tilde{u}+e^{\tilde{u}},&(x,t)\in\Omega\times(C,+\infty), \\ \tilde{u}(x,t)=0,&(x,t)\in\partial\Omega\times(C,+\infty), \\ \tilde{u}(x,C)=w(x,C), &x\in\Omega, \end{array}\displaystyle \right . $$
(4.3)
and by T̃ the blowup time for ũ. By the comparison principle,
$$ T(\varepsilon)\geq\widetilde{T}\geq C+e^{-\max_{\Omega} u(\cdot,C)}. $$
On the one hand, when \(\max_{\Omega} u(\cdot,C)\) arrives at \(\{x| |x|<\eta\}\), we have
$$\begin{aligned} T(\varepsilon)&\geq C+e^{-\max_{\Omega} u(\cdot,C)} \\ &\geq C+e^{[\ln(e^{-\varphi(0)}-C)-(\Delta\varphi(0)+\mu )\varepsilon C+\circ(\varepsilon)]} \\ &\geq C+\bigl(e^{-\varphi(0)}-C\bigr)\bigl[1-\bigl(\Delta\varphi(0)+\mu\bigr) \varepsilon C+\circ(\varepsilon)\bigr] \\ &\geq e^{-\varphi(0)}-\varepsilon C \bigl(e^{-\varphi(0)}-C\bigr)\bigl[\Delta \varphi(0)+\mu\bigr]+\circ(\varepsilon). \end{aligned}$$
Since \(\mu>0\) is arbitrary and \(\Delta\varphi(0)<0\), we get
$$ T(\varepsilon)\geq e^{-\varphi(0)}+\varepsilon C \bigl(e^{-\varphi (0)}-C\bigr)\bigl\vert \Delta\varphi(0)\bigr\vert +\circ( \varepsilon). $$
(4.4)
In order to get the optimal estimate, we choose \(C=\frac{1}{2}e^{-\varphi(0)}\), which maximizes the coefficient of ε, that is,
$$ T(\varepsilon)\geq e^{-\varphi(x_{0})} +\frac{1}{4}e^{-2\varphi (x_{0})}\bigl\vert \Delta\varphi(x_{0})\bigr\vert \varepsilon+\circ( \varepsilon). $$
Since \(\max_{\Omega} u(\cdot,C)\) arrives at \(\{x| |x|\geq \eta\}\), we have
$$\begin{aligned} T(\varepsilon)&\geq C+e^{-\max_{\Omega} u(\cdot,C)} \\ &\geq C+e^{\ln(e^{-\varphi(0)}-C)} \\ &=e^{-\varphi(0)}. \end{aligned}$$
In conclusion,
$$ T(\varepsilon)\geq e^{-\varphi(0)} +\frac{1}{4}e^{-2\varphi (0)}\bigl\vert \Delta\varphi(0)\bigr\vert \varepsilon+\circ(\varepsilon). $$
Proposition 4.2 is proved. □
Combining Propositions 4.1 and 4.2, we get the proof of Theorem 1.2.