In this section, we prove Theorem 1.1 by using the following lemma (see [17]).
Lemma 2.1
Suppose that a nonnegative, twice-differentiable function
\(\theta(t)\)
satisfies the inequality
$$\theta''(t)\theta(t)-(1+r) \bigl( \theta'(t)\bigr)^{2} \geq0, \quad t>0, $$
where
\(r>0\)
is some constant. If
\(\theta(0) > 0\)
and
\(\theta'(0)>0\), then there exists
\(0 < t_{1}\leq\frac{\theta(0)}{r\theta'(0)}\)
such that
\(\theta(t)\rightarrow+\infty\)
as
\(t\rightarrow t_{1}^{-}\).
Proof of Theorem 1.1
We give the proof in the following two steps.
Step 1: Blow-up. Let \(u(t)\) be the solution of problem (1.1) with the initial data satisfying (1.7). We may assume \(J(u(t))\geq0\); otherwise, there exists some \(t_{0}\geq0\) such that \(J(u(t_{0}))<0\), then \(u(t)\) will blow up in finite time by (RES1), the proof of this step is complete. So, in the following, we give our proof by contradiction and assume that \(u(t)\) exists globally and \(J(u(t))\geq0\) for all \(t\geq0\).
Differentiating (1.3) and making use of (1.1) and (1.4), we have the following equalities:
$$\begin{aligned}& \frac{d}{dt}J\bigl(u(t)\bigr)=-\|u_{t} \|_{2}^{2}-\|\nabla u_{t}\|_{2}^{2}=- \|u_{t}\|^{2}_{H_{0}^{1}}, \end{aligned}$$
(2.1)
$$\begin{aligned}& \begin{aligned}[b] \frac{d}{dt} \biggl(\frac{1}{2} \bigl\Vert u(t) \bigr\Vert _{H^{1}_{0}}^{2} \biggr)&=-\| \nabla u \|_{2}^{2}-\|\nabla u\|_{2q+2}^{2q+2}+\|u \|_{p+1}^{p+1} \\ &=-I\bigl(u(t)\bigr). \end{aligned} \end{aligned}$$
(2.2)
Since
$$ \int_{0}^{t} \bigl\Vert u_{s}(s) \bigr\Vert _{H_{0}^{1}}\,ds\geq \biggl\Vert \int_{0}^{t}u_{s}(s)\,ds \biggr\Vert _{H_{0}^{1}}= \bigl\Vert u(t)-u_{0} \bigr\Vert _{H_{0}^{1}} \geq \bigl\Vert u(t) \bigr\Vert _{H_{0}^{1}}- \Vert u_{0} \Vert _{H_{0}^{1}}, \quad t\geq0, $$
by Hölder’s inequality, (2.1), and \(J(u_{0})\geq J(u(t))\geq0\), we obtain that
$$ \begin{aligned}[b] \bigl\Vert u(t) \bigr\Vert _{H_{0}^{1}}&\leq \|u_{0}\|_{H_{0}^{1}}+t^{\frac{1}{2}}\biggl[ \int_{0}^{t} \bigl\Vert u_{s}(s) \bigr\Vert _{H_{0}^{1}}\,ds\biggr]^{\frac{1}{2}} \\ &= \|u_{0}\|_{H_{0}^{1}}+t^{\frac{1}{2}}\bigl[J(u_{0})-J \bigl(u(t)\bigr)\bigr]^{\frac{1}{2}} \\ &\leq \|u_{0}\|_{H_{0}^{1}}+t^{\frac{1}{2}}\bigl(J(u_{0}) \bigr)^{\frac{1}{2}},\quad t\geq0. \end{aligned} $$
(2.3)
Combining (1.5) and Hölder’s inequality, we deduce that
$$ \bigl\Vert \nabla u(t) \bigr\Vert _{2q+2}^{2q+2}\geq| \Omega|^{-q} \biggl(\frac{\lambda _{1}}{1+\lambda_{1}} \biggr)^{q+1} \bigl\Vert u(t) \bigr\Vert _{H_{0}^{1}}^{2q+2}. $$
On the other hand, by (1.3), (1.4), (2.2), and \(0\leq 2q< p-1\), we obtain
$$\begin{aligned} \frac{d}{dt} \biggl(\frac{1}{2} \bigl\Vert u(t) \bigr\Vert _{H^{1}_{0}}^{2} \biggr)&=\frac{p-1}{2} \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}+\frac{p-2q-1}{2q+2} \bigl\Vert \nabla u(t) \bigr\Vert _{2q+2}^{2q+2}-(p+1)J\bigl(u(t)\bigr) \\ &\geq\frac{(p-1)\lambda_{1}}{2(1+\lambda_{1})} \bigl\Vert u(t) \bigr\Vert ^{2}_{H_{0}^{1}}+ \frac {p-2q-1}{2q+2} \biggl(\frac{\lambda_{1}}{1+\lambda_{1}} \biggr)^{q+1}|\Omega |^{-q} \bigl\Vert u(t) \bigr\Vert _{H_{0}^{1}}^{2q+2} \\ &\quad {}-(p+1)J \bigl(u(t)\bigr) \\ &\geq\frac{(p-1)\lambda_{1}}{1+\lambda_{1}} \biggl[\frac{1}{2} \bigl\Vert u(t) \bigr\Vert _{H^{1}_{0}}^{2}-\frac{(p+1)(1+\lambda_{1})}{(p-1)\lambda_{1}}J\bigl(u(t)\bigr) \biggr] . \end{aligned}$$
Since \(\frac{d}{dt}(J(u(t)))\leq0\), it follows from the above inequality that
$$\begin{aligned}& \frac{d}{dt} \biggl[\frac{1}{2} \bigl\Vert u(t) \bigr\Vert _{H^{1}_{0}}^{2}-\frac{(p+1)(1+\lambda _{1})}{(p-1)\lambda_{1}}J\bigl(u(t)\bigr) \biggr] \\& \quad \geq \frac{(p-1)\lambda_{1}}{1+\lambda _{1}} \biggl[\frac{1}{2} \bigl\Vert u(t) \bigr\Vert _{H^{1}_{0}}^{2}-\frac{(p+1)(1+\lambda _{1})}{(p-1)\lambda_{1}}J\bigl(u(t)\bigr) \biggr]. \end{aligned}$$
Let
$$H(t)=\frac{1}{2} \bigl\Vert u(t) \bigr\Vert _{H^{1}_{0}}^{2}- \frac{(p+1)(1+\lambda _{1})}{(p-1)\lambda_{1}}J\bigl(u(t)\bigr), $$
then
$$\frac{d}{dt}H(t)\geq\frac{(p-1)\lambda_{1}}{1+\lambda_{1}}H(t) $$
for all \(t\geq0\). By using Gronwall’s inequality, we get
$$H(t)\geq e^{\frac{(p-1)\lambda_{1}}{1+\lambda_{1}}t}H(0). $$
Noticing that \(H(0)>0\) via (1.7) and the assumption \(J(u(t))\geq 0\) for \(t\geq0\), we deduce
$$ \bigl\| u(t)\bigr\| _{H_{0}^{1}}\geq\sqrt{2H(0)}e^{\frac{(p-1)(\lambda_{1}}{2(1+\lambda _{1})}t},\quad t\geq0, $$
which is a contradiction with (2.3) for t sufficiently large. Hence, \(u(t)\) blows up at some finite time, i.e., \(T<\infty\).
Step 2: Lifespan. We will find an upper bound for T. Firstly, we claim that
$$ I\bigl(u(t)\bigr)= \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}+ \bigl\Vert \nabla u(t) \bigr\Vert _{2q+2}^{2q+2}- \bigl\Vert u(t) \bigr\Vert _{p+1}^{p+1}< 0,\quad t\in[0, T). $$
(2.4)
Indeed, combining (1.3) and (1.4), after a simple calculation, we get
$$\begin{aligned} J\bigl(u(t)\bigr) =&\frac{p-1}{2(p+1)} \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}+\frac {p-2q-1}{2(q+1)(p+1)} \bigl\Vert \nabla u(t) \bigr\Vert _{2q+2}^{2q+2} \\ &{}+\frac {1}{p+1}I\bigl(u(t) \bigr),\quad t\in[0,T). \end{aligned}$$
(2.5)
It follows from (1.5), (1.7), and (2.5) that
$$ \frac{(p-1)\lambda_{1}}{2(p+1)(1+\lambda_{1})}\|u_{0}\|^{2}_{H_{0}^{1}}>J(u_{0}) \geq \frac{p-1}{2(p+1)}\frac{\lambda_{1}}{1+\lambda_{1}}\|u_{0}\|_{H_{0}^{1}}^{2}+ \frac {1}{p+1}I(u_{0}), $$
where we also use \(0\leq2q< p-1\), which implies \(I(u_{0})<0\). Hence, if (2.4) does not hold, there must exist \(t_{0}\in(0,T)\) such that \(I(u(t_{0}))=0\), \(I(u(t))<0\) for \(t\in[0,t_{0})\). Then, by (2.2), we obtain that \(\|u(t)\|_{H_{0}^{1}}^{2}\) is strictly increasing on \([0, t_{0})\). Then it follows from (1.7) that
$$ \begin{aligned}[b] J(u_{0})&< \frac{(p-1)\lambda_{1}}{2(p+1)(1+\lambda_{1})} \Vert u_{0} \Vert ^{2}_{H_{0}^{1}} \\ &< \frac{(p-1)\lambda_{1}}{2(p+1)(1+\lambda_{1})} \bigl\Vert u(t_{0}) \bigr\Vert ^{2}_{H_{0}^{1}}. \end{aligned} $$
(2.6)
On the other hand, combining (2.1) and (2.5), we get
$$\begin{aligned} J(u_{0})&\geq J\bigl(u(t_{0})\bigr)=\frac{p-1}{2(p+1)} \bigl\Vert \nabla u(t_{0}) \bigr\Vert _{2}^{2}+ \frac {p-2q-1}{2(q+1)(p+1)} \bigl\Vert \nabla u(t_{0}) \bigr\Vert _{2q+2}^{2q+2}+\frac {1}{p+1}I\bigl(u(t_{0})\bigr) \\ &\geq\frac{(p-1)\lambda_{1}}{2(p+1)(1+\lambda_{1})} \bigl\Vert u(t_{0}) \bigr\Vert ^{2}_{H_{0}^{1}}, \end{aligned}$$
which is a contradiction with (2.6). Hence, \(I(u(t))<0\) and \(\| u(t)\|_{H_{0}^{1}}^{2}\) is strictly increasing on \([0, T)\).
We define the functional
$$ F(t)= \int_{0}^{t} \bigl\Vert u(s) \bigr\Vert _{H_{0}^{1}}^{2}\,ds+(T-t) \Vert u_{0} \Vert _{H_{0}^{1}}^{2}+\beta(t+\gamma )^{2}, \quad t\in[0,T), $$
with two positive constants β, γ to be chosen later. Since \(\|u(t)\|_{H_{0}^{1}}^{2}\) is strictly increasing, we get
$$ \begin{aligned}[b] F'(t)&= \bigl\Vert u(t) \bigr\Vert _{H_{0}^{1}}^{2}- \Vert u_{0} \Vert _{H_{0}^{1}}^{2}+2\beta (t+\gamma) \\ &= \int_{0}^{t}\frac{d}{ds} \bigl\Vert u(s) \bigr\Vert _{H_{0}^{1}}^{2}\,ds+2\beta(t+\gamma)\geq2\beta (t+ \gamma)>0 \end{aligned} $$
(2.7)
and
$$ \begin{aligned}[b] F''(t)&= \frac{d}{dt} \bigl\Vert u(t) \bigr\Vert _{H_{0}^{1}}^{2}+2 \beta \\ &=(p-1) \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}+ \frac{p-2q-1}{q+1} \bigl\Vert \nabla u(t) \bigr\Vert _{2q+2}^{2q+2}-2(p+1)J \bigl(u(t)\bigr)+2\beta \\ &\geq\frac{(p-1)\lambda_{1}}{1+\lambda_{1}} \bigl\Vert u(t) \bigr\Vert _{H_{0}^{1}}^{2}+2(p+1) \int _{0}^{t} \Vert u_{s} \Vert _{H_{0}^{1}}^{2}\,ds-2(p+1)J(u_{0}). \end{aligned} $$
(2.8)
Noticing that
$$F(0)=T\|u_{0}\|^{2}_{H_{0}^{1}}+\beta \gamma^{2}>0 $$
and
$$F'(0)=2\beta\gamma>0, $$
by using Young’s inequality, Hölder’s inequality, and the element algebraic inequality
$$ab+cd\leq\sqrt{a^{2}+c^{2}}\sqrt{b^{2}+d^{2}}, $$
we can deduce
$$ \begin{aligned} \xi(t)&:= \biggl( \int_{0}^{t} \bigl\Vert u(s) \bigr\Vert _{H_{0}^{1}}^{2}\,ds+\beta(t+\gamma)^{2} \biggr) \biggl( \int_{0}^{t} \Vert u_{s} \Vert _{H_{0}^{1}}^{2}\,ds+\beta \biggr) \\ &\quad {}- \biggl( \int_{0}^{t}\frac{1}{2}\frac {d}{ds} \bigl\Vert u(s) \bigr\Vert _{H_{0}^{1}}^{2}\,ds+\beta(t+\gamma) \biggr)^{2} \\ &\geq0. \end{aligned} $$
Hence, it follows from the above inequality and (2.7) that
$$ \begin{aligned} -\bigl(F'(t)\bigr)^{2}&=-4 \biggl[\frac{1}{2} \int_{0}^{t}\frac{d}{ds} \bigl\Vert u(s) \bigr\Vert _{H_{0}^{1}}^{2}\,ds+2\beta(t+\gamma) \biggr]^{2} \\ &=4 \biggl(\xi(t)- \bigl(F(t)-(T-t) \Vert u_{0} \Vert _{H_{0}^{1}}^{2} \bigr) \biggl( \int_{0}^{2}\frac {d}{ds} \bigl\Vert u(s) \bigr\Vert _{H_{0}^{1}}^{2}\,ds+\beta \biggr) \biggr) \\ &\geq-4F(t) \biggl( \int_{0}^{t}\frac{d}{ds} \bigl\Vert u(s) \bigr\Vert _{H_{0}^{1}}^{2}\,ds+\beta \biggr). \end{aligned} $$
By the above equality, (2.8), and the fact that \(\|u(t)\| _{H_{0}^{1}}^{2}\) is strictly increasing, we have
$$ \begin{aligned} F(t)F''(t)- \frac{p+1}{2}\bigl(F'(t)\bigr)^{2}&\geq F(t) \biggl[F''(t)-2(p+1) \biggl( \int _{0}^{t}\frac{d}{ds} \bigl\Vert u(s) \bigr\Vert _{H_{0}^{1}}^{2}\,ds+\beta \biggr) \biggr] \\ &\geq2(p+1)F(t) \biggl[\frac{(p-1)\lambda_{1}}{2(p+1)(1+\lambda_{1})} \Vert u_{0} \Vert _{H_{0}^{1}}^{2}-J(u_{0})-\beta \biggr]. \end{aligned} $$
From (1.7), we can choose β sufficiently small such that
$$ 0< \beta\leq\beta_{0}:=\frac{(p-1)\lambda_{1}}{2(p+1)(1+\lambda_{1})} \|u_{0}\| _{H_{0}^{1}}^{2}-J(u_{0}). $$
(2.9)
Then the conditions of Lemma 2.1 are satisfied with \(r=\frac{p-1}{2}\), so we have
$$ T\leq\frac{2F(0)}{(p-1)F'(0)}=\frac{\|u_{0}\|_{H_{0}^{1}}^{2}}{(p-1)\beta\gamma }T+\frac{\gamma}{p-1}. $$
(2.10)
Fixing arbitrary β satisfying (2.9), then let γ be sufficiently large such that
$$ \frac{\|u_{0}\|_{H_{0}^{1}}^{2}}{(p-1)\beta}< \gamma< +\infty, $$
then it follows from (2.10) that
$$ T\leq\frac{\beta\gamma^{2}}{(p-1)\beta\gamma-\|u_{0}\|_{H_{0}^{1}}^{2}}. $$
(2.11)
Define a function \(T_{\beta}(\gamma)\) by
$$ T_{\beta}(\gamma)=\frac{\beta\gamma^{2}}{(p-1)\beta\gamma-\|u_{0}\| _{H_{0}^{1}}^{2}},\quad \gamma\in \biggl( \frac{\|u_{0}\|_{H_{0}^{1}}^{2}}{(p-1)\beta}, +\infty \biggr). $$
It is easy to prove that the function \(T_{\beta}(\gamma)\) has a unique minimum at
$$\gamma_{\beta}:=\frac{2\|u_{0}\|_{H_{0}^{1}}^{2}}{(p-1)\beta}\in\biggl(\frac{\|u_{0}\| _{H_{0}^{1}}^{2}}{(p-1)\beta}, +\infty \biggr). $$
Then it follows from (2.11) that
$$ T\leq\inf_{\gamma\in (\frac{\|u_{0}\|_{H_{0}^{1}}^{2}}{(p-1)\beta}, +\infty )}T_{\beta}(\gamma)=T_{\beta}( \gamma_{\beta})=\frac{4\|u_{0}\| _{H_{0}^{1}}^{2}}{(p-1)^{2}\beta} $$
for any β satisfying (2.9). Finally, we obtain
$$ T\leq\inf_{\beta\in(0,\beta_{0}]}\frac{4\|u_{0}\|_{H_{0}^{1}}^{2}}{(p-1)^{2}\beta }=\frac{4\|u_{0}\|_{H_{0}^{1}}^{2}}{(p-1)^{2}\beta_{0}} = \frac{8(p+1)(1+\lambda_{1})\|u_{0}\|^{2}_{H_{0}^{1}}}{(p-1)^{2}[(p-1)\lambda_{1}\| u_{0}\|^{2}_{H_{0}^{1}}-2(p+1)(1+\lambda_{1})J(u_{0})]}. $$
This completes the proof of Theorem 1.1. □
Corollary 2.1
For all
\(0\leq2q< p-1\)
and any
\(M>0\), there exists initial
\(u_{0M}\in W_{0}^{1, 2q+2}(\Omega)\)
such that the weak solution for corresponding problem (1.1) will blow up in finite time.
Proof
Let \(M>0\), and \(\Omega_{1}\) and \(\Omega_{2}\) be two arbitrary disjoint open subdomains of Ω. We assume that \(v\in W_{0}^{1, 2q+2}(\Omega _{1})\subset W_{0}^{1, 2q+2}(\Omega)\subset H_{0}^{1}(\Omega)\) is an arbitrary nonzero function, then we can take \(\alpha_{1}>0\) sufficiently large such that
$$\begin{aligned} \|\alpha_{1}v\|_{H_{0}^{1}}^{2}&=\alpha_{1}^{2} \int_{\Omega}|v^{2}|\,dx+\alpha_{1}^{2} \int _{\Omega}|\nabla v|^{2}\,dx=\alpha_{1}^{2} \int_{\Omega_{1}}|v^{2}|\,dx+\alpha_{1}^{2} \int _{\Omega_{1}}|\nabla v|^{2}\,dx \\ &>\frac{2(p+1)(1+\lambda_{1})}{(p-1)\lambda_{1}}M. \end{aligned}$$
We claim that there exist \(w\in W_{0}^{1, 2q+2}(\Omega_{2})\subset W_{0}^{1, 2q+2}(\Omega) \) and \(\alpha>\alpha_{1}\) such that \(J(w)=M-J(\alpha v)\).
In fact, we choose a function \(w_{k}\in C_{0}^{1}(\Omega_{2})\) such that \(\| \nabla w_{k}\|_{2}\geq k\) and \(\|w_{k}\|_{\infty}\leq c_{0}\). Hence,
$$\begin{aligned} &\frac{1}{2} \int_{\Omega_{2}}|\nabla w_{k}|^{2}\,dx+ \frac{1}{2q+2} \int_{\Omega _{2}}|\nabla w_{k}|^{2q+2}\,dx- \frac{1}{p+1} \int_{\Omega_{2}}|w_{k}|^{p+1}\,dx \\ &\quad \geq\frac{1}{2} \int_{\Omega_{2}}|\nabla w_{k}|^{2}\,dx+ \frac{1}{2q+2}|\Omega _{2}|^{-q} \biggl( \int_{\Omega_{2}}|\nabla w_{k}|^{2}\,dx \biggr)^{q+1}-\frac {1}{p+1}c_{0}^{p+1}| \Omega_{2}|. \end{aligned}$$
On the other hand, since \(0\leq2q< p-1\), it holds that
$$\begin{aligned} M-J(\alpha v)&=M- \frac{\alpha^{2}}{2} \int_{\Omega_{1}}|\nabla v|^{2}\,dx-\frac {\alpha^{2q+2}}{2q+2} \int_{\Omega_{1}}|\nabla|^{2q+2}\,dx \\ &\quad {}+\frac{\alpha^{p+1}}{p+1} \int_{\Omega_{1}}|v|^{p+1}\,dx\rightarrow+\infty , \quad \mbox{as }\alpha\rightarrow+\infty. \end{aligned}$$
Hence, there exist \(k>0\) and \(\alpha>\alpha_{1}\) both sufficiently large such that
$$ M-J(\alpha v)=\frac{1}{2} \int_{\Omega_{2}}|\nabla w_{k}|^{2}\,dx+ \frac {1}{2q+2} \int_{\Omega_{2}}|\nabla w_{k}|^{2q+2}\,dx- \frac{1}{p+1} \int_{\Omega _{2}}|w_{k}|^{p+1}\,dx. $$
Then we choose \(w=w_{k}\) and denote \(u_{0M}:=\alpha v+w\). Hence, we have
$$\begin{aligned} \|u_{0M}\|_{H_{0}^{1}}^{2}&= \int_{\Omega} \bigl\vert u_{0M}^{2} \bigr\vert \,dx+ \int_{\Omega} \vert \nabla u_{0M} \vert ^{2}\,dx\geq\alpha^{2} \int_{\Omega_{1}} \bigl\vert v^{2} \bigr\vert \,dx+ \alpha^{2} \int_{\Omega _{1}} \vert \nabla v \vert ^{2}\,dx \\ &>\frac{2(p+1)(1+\lambda_{1})}{(p-1)\lambda_{1}}M \end{aligned}$$
and
$$M=J(\alpha v)+J(w)=J(u_{0M})< \frac{(p-1)\lambda_{1}}{2(p+1)(1+\lambda _{1})}\|u_{0M} \|^{2}_{H_{0}^{1}}. $$
The proof is complete. □
Remark 2.1
In this remark, we establish the blow-up rate for \(J(u_{0})<0\). We define the functionals \(\varphi(t)=\|u(t)\|_{H_{0}^{1}}^{2}\) and \(\psi(t)=-2(p+1)J(u(t))\) as these in [12]. It was shown in (4.8) of [12] that
$$ \frac{\varphi'(t)}{[\varphi(t)]^{\frac{p+1}{2}}}\geq\frac{\psi (0)}{[\varphi(0)]^{\frac{p+1}{2}}}. $$
Now, we integrate the inequality from t to T, noticing \(\lim_{t\rightarrow T^{-}}\varphi(t)=+\infty\) (by (RES1)), we obtain
$$ \varphi(t)\leq \biggl[\frac{(p-1)\psi(0)}{2[\varphi(0)]^{\frac {p+1}{2}}} \biggr]^{\frac{2}{1-p}}. $$
Then it follows from the definitions of \(\varphi(t)\) and \(\psi(t)\) that
$$ \bigl\| u(t)\bigr\| _{H_{0}^{1}}\leq \biggl[\frac{(1-p^{2})J(u_{0})}{\|u_{0}\| _{H_{0}^{1}}^{p+1}} \biggr]^{\frac{1}{1-p}}(T-t)^{-\frac{1}{p-1}}. $$