Now, we give our main results.
Theorem 3.1
Assuming that (A) holds and
\(u_{0},u_{1}\in H^{2} \cap H^{1}_{0}\), then any solution
u
of the problem (1.1)-(1.3) with initial data satisfying
\(E(0)< E_{1}\)
and
\(\Vert \nabla u_{0} \Vert ^{2}+\Vert \nabla u_{0}\Vert ^{2(r+1)}\ge \lambda_{1}^{2(r+1)}\)
will blow up in finite time.
Proof
We set
$$ H(t)=E_{2}-E(t),\quad \textit{for }t\ge 0, $$
(3.1)
where \(E_{2} \in (E(0),E_{1})\). From (2.3) and (3.1), we get
$$ H'(t)=-E'(t)=\Vert u_{t}\Vert ^{m}_{m}+\Vert \nabla u_{t}\Vert ^{2}> 0; $$
(3.2)
then \(H(t)\) is an increasing function and
$$ H(t)\ge H(0)=E_{2}-E(0)>0. $$
(3.3)
On the other hand, by Lemma 2.3, we have
$$\begin{aligned} H(t) < & E_{2}-\frac{1}{2}\biggl[\Vert u_{t} \Vert ^{2}+ \Vert \nabla u \Vert ^{2}+ \frac{1}{r+1}\Vert \nabla u \Vert ^{2(r+1)}\biggr]+\frac{1}{p}\Vert u\Vert ^{p}_{p} \\ < & E_{1}-\frac{1}{2(r+1)}\bigl[ \Vert \nabla u \Vert ^{2}+\Vert \nabla u \Vert ^{2(r+1)}\bigr]+ \frac{1}{p} \Vert u\Vert ^{p}_{p} \\ \le& E_{1}-\frac{1}{2(r+1)}\lambda_{1}^{2(r+1)}+ \frac{1}{p}\Vert u\Vert ^{p} _{p} \\ =&- \frac{1}{p}\lambda_{1}^{2(r+1)}+\frac{1}{p}\Vert u \Vert ^{p}_{p}\leq \frac{1}{p}\Vert u\Vert ^{p}_{p}. \end{aligned}$$
(3.4)
Hence, combining (3.3) and (3.4) with the embedding \(H_{0}^{1}\hookrightarrow L^{p}\), we have
$$ 0< H(0)\le H(t)\le \frac{1}{p}\Vert u\Vert ^{p}_{p} \le \frac{B^{p}}{p}\Vert \nabla u \Vert ^{p}. $$
(3.5)
We set
$$G(t)=(u,u_{t})+\frac{1}{2}\Vert \nabla u\Vert ^{2}, $$
and then we define
$$ L(t)=H^{k(1-\alpha)}(t)+\epsilon G(t), $$
(3.6)
where \(\alpha, k,\epsilon >0\) are small enough to be chosen later. By the definition of the solution, we have
$$\begin{aligned}& G'(t)=\Vert u_{t}\Vert ^{2}-\Vert \nabla u\Vert ^{2} -\Vert \nabla u\Vert ^{2(r+1)}- \int_{\Omega }\vert u_{t}\vert ^{m-2}u_{t} u\,dx+\Vert u\Vert _{p}^{p}. \end{aligned}$$
(3.7)
Adding the term \(p(H(t)-E_{2}+E(t))\) and using the definition of \(E(t)\) in (2.2), then (3.7) becomes
$$\begin{aligned} G'(t) \ge& \Vert u_{t}\Vert ^{2}-\Vert \nabla u\Vert ^{2} -\Vert \nabla u\Vert ^{2(r+1)} - \int_{\Omega }\vert u_{t}\vert ^{m-2}u_{t} u\,dx \\ &{}+\frac{p}{2}\biggl[\Vert u_{t}\Vert ^{2}+\Vert \nabla u\Vert ^{2} +\frac{1}{r+1}\Vert \nabla u\Vert ^{2(r+1)}\biggr]+pH(t)-pE _{2} \\ =&\frac{p+2}{2}\Vert u_{t}\Vert ^{2} + \frac{p-2}{2}\bigl\Vert \nabla u(t)\bigr\Vert ^{2} + \frac{p-2(r+1)}{2(r+1)}\Vert \nabla u\Vert ^{2(r+1)} \\ &{}- \int_{\Omega }\vert u_{t}\vert ^{m-2}u_{t} u\,dx +pH(t)-pE_{2}. \end{aligned}$$
(3.8)
By \(r>0\) and Lemma 2.3 again, we have
$$\begin{aligned}& \frac{p-2}{2}\bigl\Vert \nabla u(t)\bigr\Vert ^{2} + \frac{p-2(r+1)}{2(r+1)}\Vert \nabla u\Vert ^{2(r+1)}-pE _{2} \\& \quad \ge \frac{p-2(r+1)}{2(r+1)}\bigl[\bigl\Vert \nabla u(t)\bigr\Vert ^{2} +\Vert \nabla u\Vert ^{2(r+1)}\bigr]-pE _{2} \\& \quad \ge \frac{p-2(r+1)}{2(r+1)} \frac{\lambda_{2}^{2(r+1)}-\lambda_{1} ^{2(r+1)}}{\lambda_{2}^{2(r+1)}}\bigl[\bigl\Vert \nabla u(t)\bigr\Vert ^{2} +\Vert \nabla u\Vert ^{2(r+1)}\bigr] \\& \qquad {}+\frac{p-2(r+1)}{2(r+1)}\frac{\lambda_{1}^{2(r+1)}[\Vert \nabla u(t)\Vert ^{2} +\Vert \nabla u\Vert ^{2(r+1)}]}{\lambda_{2}^{2(r+1)}}-pE_{2} \\& \quad \ge \frac{p-2(r+1)}{2(r+1)} \frac{\lambda_{2}^{2(r+1)}-\lambda_{1} ^{2(r+1)}}{\lambda_{2}^{2(r+1)}}\bigl[\bigl\Vert \nabla u(t)\bigr\Vert ^{2} \\& \qquad {} +\Vert \nabla u\Vert ^{2(r+1)}\bigr]+ \frac{p-2(r+1)}{2(r+1)}\lambda_{1}^{2(r+1)}-pE_{2}. \end{aligned}$$
(3.9)
From the fact that \(p>2(r+1)\), Lemma 2.3 and \(E_{2}< E_{1}\), we see that
$$\begin{aligned}& \begin{aligned}&\frac{p-2(r+1)}{2(r+1)} \frac{\lambda_{2}^{2(r+1)}-\lambda_{1}^{2(r+1)}}{ \lambda_{2}^{2(r+1)}}>0, \\ &\frac{p-2(r+1)}{2(r+1)}\lambda_{1}^{2(r+1)}-pE _{2}>\frac{p-2(r+1)}{2(r+1)}\lambda_{1}^{2(r+1)}-pE_{1} =0. \end{aligned} \end{aligned}$$
(3.10)
It follows from (3.8), (3.9) and (3.10) that
$$\begin{aligned} G'(t) \ge& \frac{p+2}{2}\Vert u_{t}\Vert ^{2}+\frac{p-2(r+1)}{2(r+1)} \frac{ \lambda_{2}^{2(r+1)}-\lambda_{1}^{2(r+1)}}{\lambda_{2}^{2(r+1)}}\bigl[\bigl\Vert \nabla u(t)\bigr\Vert ^{2} +\Vert \nabla u\Vert ^{2(r+1)}\bigr] \\ &{}- \int_{\Omega }\vert u_{t}\vert ^{m-2}u_{t} u\,dt+pH(t). \end{aligned}$$
(3.11)
From the Hölder inequality, \(p>m\) and (3.5), we have
$$\begin{aligned} \biggl\vert \int_{\Omega }\vert u_{t}\vert ^{m-2}u_{t} u\,dt\biggr\vert \leq& \int_{\Omega }\vert u_{t}\vert ^{m-1}\vert u\vert \,dx \\ \leq& \Vert u\Vert _{m}\Vert u_{t}\Vert ^{m-1} _{m} \\ \leq& C\Vert u\Vert ^{1-\frac{p}{m}}_{p} \Vert u\Vert ^{\frac{p}{m}}_{p}\Vert u_{t}\Vert ^{m-1} _{m} \\ \leq& C\Vert u\Vert ^{\frac{p}{m}}_{p} H^{\frac{1}{p}-\frac{1}{m}}(t) \Vert u_{t}\Vert ^{m-1}_{m}. \end{aligned}$$
(3.12)
From (3.5), Young’s inequality and the fact that \(\Vert u_{t}\Vert ^{m} _{m}\le H'(t)\), we get
$$\begin{aligned}& \biggl\vert \int_{\Omega }\vert u_{t}\vert ^{m-2}u_{t} u\,dt\biggr\vert \leq C\bigl[\epsilon_{1}^{m} \Vert u \Vert ^{p} _{p} + \epsilon_{1}^{-\frac{m}{m-1}}H'(t) \bigr]H^{-\alpha_{1}}(t), \end{aligned}$$
(3.13)
where \(\alpha_{1}=\frac{1}{m}-\frac{1}{p}>0\), \(\epsilon_{1}>0\). Now, we take α and k satisfying
$$\begin{aligned}& \begin{aligned}&0< \alpha < \min \biggl\{ \alpha_{1},\frac{1}{2}- \frac{1}{p}, \frac{p-m}{p(m-1)}\biggr\} , \\ & \max\biggl\{ \frac{1}{2},1- \alpha_{1}, \frac{1}{r+1}, \frac{p+2}{2p}\biggr\} < k(1-\alpha)< 1, \end{aligned} \end{aligned}$$
(3.14)
and then we have
$$ 1-k(1-\alpha)-\alpha_{1}< 0,\qquad 1< \frac{1}{k(1-\alpha)}< r+1,\qquad k(1-\alpha)> \frac{p+2}{2p}. $$
(3.15)
Furthermore, from (3.13) and (3.5), we have
$$\begin{aligned}& \biggl\vert \int_{\Omega }\vert u_{t}\vert ^{m-2}u_{t} u\,dt\biggr\vert \\& \quad \leq C\bigl[\epsilon_{1}^{m} H^{- \alpha_{1}}(0)\Vert u\Vert ^{p}_{p} + \epsilon_{1}^{-\frac{m}{m-1}}H^{1-k(1- \alpha)-\alpha_{1}}(0)H^{k(1-\alpha)-1}(t)H'(t) \bigr]. \end{aligned}$$
(3.16)
By differentiating (3.6), we see from (3.11) and (3.16) that
$$\begin{aligned} L'(t) \ge& \bigl[k(1-\alpha)- \epsilon C \epsilon_{1}^{-\frac{m}{m-1}}H ^{1-k(1-\alpha)-\alpha_{1}}(0)\bigr]H^{k(1-\alpha)-1}(t)H'(t) \\ &{}+\epsilon \frac{p+2}{2}\Vert u_{t}\Vert ^{2}+ \epsilon pH(t)- \epsilon C \epsilon_{1}^{m} H^{-\alpha_{1}}(0)\Vert u\Vert ^{p}_{p} \\ &{}+\epsilon \frac{p-2(r+1)}{2(r+1)} \frac{\lambda_{2}^{2(r+1)}-\lambda _{1}^{2(r+1)}}{\lambda_{2}^{2(r+1)}}\bigl[\bigl\Vert \nabla u(t) \bigr\Vert ^{2} +\Vert \nabla u\Vert ^{2(r+1)}\bigr] . \end{aligned}$$
(3.17)
Letting \(\delta =\frac{1}{2}\min \{\frac{p+2}{2}, \frac{p}{2}, \frac{p-2(r+1)}{2(r+1)} \frac{\lambda_{2}^{2(r+1)}-\lambda_{1}^{2(r+1)}}{ \lambda_{2}^{2(r+1)}} \}\) and decomposing \(\epsilon pH(t)\) in (3.17) by \(\epsilon pH(t)=2\delta \epsilon H(t)+(p-2\delta) \epsilon H(t)\), we find from (3.3) and (3.17) that
$$\begin{aligned} L'(t) \ge& \bigl[k(1-\alpha)- \epsilon C \epsilon_{1}^{-\frac{m}{m-1}}H ^{1-k(1-\alpha)-\alpha_{1}}(0)\bigr]H^{k(1-\alpha)-1}(t)H'(t) \\ &{}+\epsilon \biggl[\frac{2\delta }{p}-C\epsilon_{1}^{m} H^{-\alpha_{1}}(0)\biggr]\Vert u\Vert ^{p} _{p}+\epsilon \biggl[\frac{p+2}{2}-\delta \biggr]\Vert u_{t}\Vert ^{2} \\ &{}+\epsilon \biggl[ \frac{p-2(r+1)}{2(r+1)} \frac{\lambda_{2}^{2(r+1)}- \lambda_{1}^{2(r+1)}}{\lambda_{2}^{2(r+1)}}-\delta \biggr] \bigl[ \bigl\Vert \nabla u(t)\bigr\Vert ^{2} +\Vert \nabla u\Vert ^{2(r+1)}\bigr] \\ &{}+(p-2\delta) \epsilon H(t). \end{aligned}$$
(3.18)
Choosing \(\epsilon_{1} >0\) small enough so that \(\epsilon_{1}^{m} < \frac{ \delta }{pC}H^{\alpha_{1}}(0)\) and \(0<\epsilon < \frac{k(1-\alpha)}{2C}H^{-(1-k(1-\alpha)-\alpha_{1})}(0)\epsilon _{1}^{\frac{m}{m-1}}\), we have from (3.18)
$$\begin{aligned}& L'(t)\ge C\epsilon \bigl( \Vert u\Vert ^{p}_{p}+ \Vert u_{t}\Vert ^{2}+H(t)+ \bigl\Vert \nabla u(t)\bigr\Vert ^{2} +\Vert \nabla u \Vert ^{2(r+1)} \bigr), \end{aligned}$$
(3.19)
for a positive constant C. Therefore, \(L(t)\) is a nondecreasing function. Letting ϵ in (3.6) be small enough, we get \(L(0) > 0\). Consequently, we obtain \(L(t) \ge L(0) > 0\) for \(t\ge 0\).
We claim the inequality
$$\begin{aligned}& L'(t)\ge C L(t)^{\frac{1}{k(1-\alpha)}}. \end{aligned}$$
(3.20)
For the proof of (3.20), we consider two alternatives:
(i) If there exists a \(t > 0\) so that \(G(t) < 0\), then
$$\begin{aligned}& L(t)^{\frac{1}{k(1-\alpha)}}=\bigl[H^{k(1-\alpha)}(t)+\epsilon G(t) \bigr]^{\frac{1}{k(1- \alpha)}}\le H(t). \end{aligned}$$
(3.21)
Thus (3.20) follows from (3.21).
(ii) If there exists a \(t > 0\) so that \(G(t) \ge 0\), since \(1<\frac{1}{k(1- \alpha)}<1+r\) by (3.15), then we deduce from (3.6), the Young inequality, the Hölder inequality and the embedding \(L^{p}\hookrightarrow L^{2}\) that
$$\begin{aligned} L(t)^{\frac{1}{k(1-\alpha)}} \le& \biggl[H^{k(1-\alpha)}(t)+\Vert u\Vert ^{\tau }+ \Vert u_{t}\Vert ^{s}+\frac{1}{2} \Vert \nabla u\Vert ^{2}\biggr]^{\frac{1}{k(1-\alpha)}} \\ \le& C\bigl[H(t)+\Vert u\Vert _{p}^{\frac{\tau }{k(1-\alpha)}}+ \Vert u_{t}\Vert ^{\frac{s}{k(1- \alpha)}}+\Vert \nabla u\Vert ^{\frac{2}{k(1-\alpha)}} \bigr], \end{aligned}$$
(3.22)
for \(\frac{1}{\tau }+\frac{1}{s}=1\), \(\tau >0\), \(s>0\). If wee take \(s=2k(1-\alpha)\), then \(s>1\) by (3.14), and \(\frac{s}{k(1- \alpha)}=2\). By (3.15), we have \(\frac{\tau }{k(1-\alpha)}=\frac{2}{2k(1- \alpha)-1}< p\), \(\frac{2}{k(1-\alpha)}<2(r+1)\). Furthermore, we get
$$ \Vert u_{t}\Vert ^{\frac{s}{k(1-\alpha)}}=\Vert u_{t}\Vert ^{2},\qquad \Vert u\Vert _{p}^{\frac{ \tau }{k(1-\alpha)}}=\Vert u\Vert _{p}^{\frac{2}{2k(1-\alpha)-1}}. $$
(3.23)
Thus from (3.22), (3.23) and (3.5), we have
$$\begin{aligned} L(t)^{\frac{1}{k(1-\alpha)}} \le& C\bigl[H(t)+ \Vert u_{t}\Vert ^{2}+\Vert u\Vert _{p}^{\frac{2}{2k(1- \alpha)-1}-p}\Vert u\Vert _{p}^{p}+\Vert \nabla u\Vert ^{\frac{2}{k(1-\alpha)}-2(r+1)}\Vert \nabla u\Vert ^{2(r+1)}\bigr] \\ \le& C\biggl[H(t)+ \Vert u_{t}\Vert ^{2}+\bigl(pH(0) \bigr)^{\frac{1}{p}(\frac{2}{2k(1-\alpha)-1}-p)}\Vert u\Vert _{p}^{p} \\ &{}+\biggl( \frac{p}{B^{p}}H(0)\biggr)^{\frac{1}{p}(\frac{2}{k(1- \alpha)}-2(r+1))}\Vert \nabla u\Vert ^{2(r+1)}\biggr] \\ \le& C\bigl[H(t)+ \Vert u_{t}\Vert ^{2}+\Vert u\Vert _{p}^{p}+\Vert \nabla u\Vert ^{2(r+1)}\bigr]. \end{aligned}$$
(3.24)
This inequality together with (3.19) implies (3.20).
Then, by integrating both sides of (3.20) over \([0,t]\), it follows that there exists a \(T_{0}>0\) so that
$$ \lim_{t \rightarrow T_{0}^{-}} L(t)= \lim_{t \rightarrow T_{0}^{-}} \bigl(H^{k(1-\alpha)}(t)+\epsilon G(t)\bigr)= \infty. $$
(3.25)
This combined with (3.24), (3.21) and (3.5) gives
$$\lim_{t \rightarrow T_{0}^{-}} \bigl(\Vert u\Vert _{p}^{p}+ \Vert \nabla u\Vert ^{2}+\Vert \nabla u\Vert ^{2(r+1)}+ \Vert u_{t}\Vert ^{2}\bigr)=\infty. $$
This theorem is proved. □
Theorem 3.2
Assuming that
\(u_{0}\in H^{2}\cap H^{1}_{0}\), \(u_{1} \in H^{1}_{0}\), and
\(p >\max\{2(r+1), m\}\), \(E(0) < 0\), then the local solution of the problem (1.1)-(1.3) blows up in finite time.
Proof
Setting \(H(t) = -E(t)\) instead of \(H(t)\) in (3.1) and then applying the same arguments as in Theorem 3.1, we get the desired result. □
Remark 3.3
We point out that the method can also be extended to equation (1.1) with the general function \(M(s)\), \(h(s)\) and \(g(s)\) as in [26], and it can also be extended to equation (1.5) as in [27].