In this section we state and prove the blow-up result.
Theorem 3.1
Assume that (G1) and (G2) hold and
$$ \int_{0}^{\infty}g(s)\,ds< \frac{p/2-1}{p/2-1+1/2p}. $$
(3.1)
Assume further that
\(p>\rho+2\), and
\((u_{0}, u_{1})\in H^{1}_{0}(\Omega )\times L^{2}(\Omega)\)
is given. Then the solution
\(u(t)\)
of problem (1.1) blows up in finite time, i.e. there exists
\(T_{0}<+\infty\)
such that
$$ {\lim_{t\rightarrow T_{0}^{-}}} \bigl(\Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+\Vert \nabla u\Vert ^{2}_{2}+ \Vert u\Vert ^{p}_{p} \bigr)=\infty, $$
(3.2)
if
$$ E(0)< \biggl(1-\frac{1}{p(p-2)}\frac{1-l}{l} \biggr) \biggl( \frac{1}{2}-\frac{1}{p} \biggr)B^{-{\frac{2p}{p-2}}}_{1}, $$
(3.3)
and
$$ \Vert \nabla u_{0}\Vert _{2}>B^{-{\frac{p}{p-2}}}_{1}. $$
(3.4)
Proof
Assume that there exists some positive constant C such that for \(t>0\) the solution \(u(t)\) of (1.1) satisfies
$$ \Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+ \Vert \nabla u\Vert ^{2}_{2}+\Vert u\Vert ^{p}_{p}\leq C. $$
(3.5)
We set
where the constant \(E_{2}\in(E(0), E_{1})\) shall be chosen later. By Lemma 2.2,
$$ H'(t)=-E'(t)\geq0. $$
(3.6)
Then, for \(0\leq s\leq t\), we have
$$ 0< H(0)\leq H(s)\leq H(t)=E_{2}-E(t). $$
(3.7)
From (2.6), we have
$$\begin{aligned} H(t) =& E_{2}-E(t) \\ =& E_{2}-\frac{1}{\rho+2}\Vert u_{t}\Vert ^{\rho+2}_{\rho+2}-\frac{1}{2}\biggl(1- \int_{0}^{t} g(s)\,ds\biggr)\bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}-\frac{1}{2}(g\circ\nabla u) (t)+ \frac{1}{p}\bigl\Vert u(t)\bigr\Vert ^{p}_{p} \\ \leq& E_{2}-\frac{1}{2}\biggl(1- \int_{0}^{t} g(s)\,ds\biggr)\bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+\frac{1}{p}\bigl\Vert u(t)\bigr\Vert ^{p}_{p} \\ \leq& E_{1}-\frac{1}{2}\biggl(1- \int_{0}^{t} g(s)\,ds\biggr)\bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}+\frac{1}{p}\bigl\Vert u(t)\bigr\Vert ^{p}_{p} \\ \leq& E_{1}-\frac{1}{2}B^{-{\frac{2p}{p-2}}}_{1}+ \frac{1}{p}\bigl\Vert u(t)\bigr\Vert ^{p}_{p} \\ \leq& \biggl(\frac{1}{2}-\frac{1}{p}\biggr)B^{-{\frac{2p}{p-2}}}_{1}- \frac{1}{2}B^{-{\frac{2p}{p-2}}}_{1}+\frac{1}{p}\bigl\Vert u(t)\bigr\Vert ^{p}_{p} \\ \leq& \frac{1}{p}\bigl\Vert u(t)\bigr\Vert ^{p}_{p}. \end{aligned}$$
(3.8)
Define
$$ L(t)=H^{1-\sigma}(t)+\frac{\varepsilon}{\rho+1} \int_{\Omega} \vert u_{t}\vert ^{\rho}u_{t}u \,dx+\frac{\varepsilon}{2} \int_{\Omega} \vert \nabla u\vert ^{2}\,dx, $$
(3.9)
where the constant \(\varepsilon>0\) shall be chosen later and the constant σ satisfies
$$ 0< \sigma< \frac{1}{\rho+2}-\frac{1}{p}. $$
(3.10)
Taking a derivative of (3.9) and using Lemma 2.2, we have
$$\begin{aligned} L'(t) =& (1-\sigma)H^{-\sigma}(t) \biggl(\bigl\Vert \nabla u_{t}(t)\bigr\Vert ^{2}_{2}- \frac{1}{2}\bigl(g'\circ\nabla u\bigr) (t)+ \frac{1}{2}g(t)\bigl\Vert \nabla u(t)\bigr\Vert ^{2}_{2} \biggr) \\ & {}+\frac{\varepsilon}{\rho+1}\Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+ \varepsilon \int_{\Omega} \vert u_{t}\vert ^{\rho}u_{tt}u \,dx+\varepsilon \int_{\Omega}\nabla u_{t}\nabla u \,dx \\ \geq& (1-\sigma) H^{-\sigma}(t)\bigl\Vert \nabla u_{t}(t) \bigr\Vert ^{2}_{2}+\frac{\varepsilon}{\rho+1}\Vert u\Vert ^{\rho+2}_{\rho+2}-\varepsilon \Vert \nabla u\Vert ^{2}_{2}+\varepsilon \Vert u\Vert ^{p}_{p} \\ &{} -\varepsilon \int_{\Omega} \int_{0}^{t} g(t-\tau)\triangle u(\tau)\,d\tau u(t)\,dx. \end{aligned}$$
(3.11)
For the last term on the right side of (3.11), using the Green formula, we get
$$\begin{aligned}& - \int_{\Omega} \int_{0}^{t} g(t-\tau)\triangle u(\tau)\,d\tau u(t)\,dx \\& \quad= \int_{0}^{t} g(t-\tau) \int_{\Omega}\nabla u(\tau)\nabla u(t)\,dx \,d\tau \\& \quad= \int_{0}^{t} g(t-\tau) \int_{\Omega}\nabla u(t)\nabla\bigl(u(\tau)-u(t)\bigr)\,dx \,d\tau+ \int_{0}^{t} g(t-\tau)\bigl\Vert \nabla u(t)\bigr\Vert ^{2}_{2}\,d\tau \\& \quad= \int_{0}^{t} g(t-\tau) \int_{\Omega}\nabla u(t) \bigl(\nabla u(\tau)-\nabla u(t)\bigr)\,dx\, d \tau+ \int_{0}^{t} g(\tau)\,d\tau\bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}. \end{aligned}$$
(3.12)
Substituting (3.12) into (3.11), we obtain
$$\begin{aligned} L'(t) \geq& (1-\sigma) H^{-\sigma}(t)\bigl\Vert \nabla u_{t}(t)\bigr\Vert ^{2}_{2}+\frac{\varepsilon}{\rho+1} \Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+\varepsilon \Vert u\Vert ^{p}_{p} \\ &{} +\varepsilon \int_{0}^{t} g(t-\tau) \int_{\Omega}\nabla u(t) \bigl(\nabla u(\tau)-\nabla u(t)\bigr)\,dx\, d \tau \\ &{} -\varepsilon\biggl(1- \int_{0}^{t} g(\tau)\,d\tau\biggr)\bigl\Vert \nabla u(t)\bigr\Vert ^{2}_{2}. \end{aligned}$$
(3.13)
Using the Cauchy inequality, for \(0<\varepsilon_{1}<1\) we have
$$\begin{aligned}& \int_{0}^{t} g(t-\tau) \int_{\Omega}\nabla u(t) \bigl(\nabla u(\tau)-\nabla u(t)\bigr)\,dx\, d \tau \\& \quad\geq-\frac{p(1-\varepsilon_{1})}{2} \int_{0}^{t} g(t-\tau)\bigl\Vert \nabla u(\tau)- \nabla u(t)\bigr\Vert ^{2}_{2}\,d\tau \\& \qquad{} -\frac{1}{(1-\varepsilon_{1})2p} \int_{0}^{t} g(\tau)\,d\tau\bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}. \end{aligned}$$
(3.14)
By (3.14), we know
$$\begin{aligned} L'(t) \geq& (1-\sigma) H^{-\sigma}(t)\bigl\Vert \nabla u_{t}(t)\bigr\Vert ^{2}_{2}+ \frac{\varepsilon}{\rho+1}\Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+ \varepsilon \Vert u\Vert ^{p}_{p} \\ &{}-\varepsilon\biggl(1- \int_{0}^{t} g(\tau)\,d\tau\biggr)\bigl\Vert \nabla u(t)\bigr\Vert ^{2}_{2} \\ & {}-\varepsilon\biggl(\frac{p(1-\varepsilon_{1})}{2} \int_{0}^{t} g(t-\tau)\bigl\Vert \nabla u(\tau)- \nabla u(t)\bigr\Vert ^{2}_{2}\,d\tau \\ &{}+\frac{1}{(1-\varepsilon_{1})2p} \int_{0}^{t} g(\tau)\,d\tau\bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2} \biggr) \\ \geq& (1-\sigma) H^{-\sigma}(t)\bigl\Vert \nabla u_{t}(t) \bigr\Vert ^{2}_{2}+\varepsilon\biggl(\frac{1}{\rho+1}+ \frac{p(1-\varepsilon_{1})}{\rho+2} \biggr)\Vert u_{t}\Vert ^{\rho +2}_{\rho+2}+ \varepsilon(1-\varepsilon_{1})p\bigl(E_{2}-E(t)\bigr) \\ & {}+\varepsilon\biggl( \biggl(\frac{p(1-\varepsilon_{1})}{2}-1 \biggr) \biggl(1- \int_{0}^{t} g(\tau)\,d\tau\biggr)\bigl\Vert \nabla u(t)\bigr\Vert ^{2}_{2} \\ &{}-\frac{1}{(1-\varepsilon_{1})2p} \int_{0}^{t} g(\tau)\,d\tau\bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2} \biggr) \\ &{} -\varepsilon(1-\varepsilon_{1})pE_{2}+\varepsilon \varepsilon_{1}\Vert u\Vert ^{p}_{p} \\ =& (1-\sigma) H^{-\sigma}(t)\bigl\Vert \nabla u_{t}(t)\bigr\Vert ^{2}_{2}+\varepsilon\biggl(\frac{1}{\rho+1}+ \frac{p(1-\varepsilon_{1})}{\rho+2} \biggr)\Vert u_{t}\Vert ^{\rho +2}_{\rho+2}+ \varepsilon(1-\varepsilon_{1})pH(t) \\ & {}+\varepsilon\biggl( \biggl(\frac{p(1-\varepsilon_{1})}{2}-1 \biggr) \biggl(1- \int_{0}^{t} g(\tau)\,d\tau\biggr)\bigl\Vert \nabla u(t)\bigr\Vert ^{2}_{2} \\ &{}-\frac{1}{(1-\varepsilon_{1})2p} \int_{0}^{t} g(\tau)\,d\tau\bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2} \biggr) \\ &{} -\varepsilon(1-\varepsilon_{1})pE_{2}+\varepsilon \varepsilon_{1}\Vert u\Vert ^{p}_{p}. \end{aligned}$$
(3.15)
For the fourth term on the right side of (3.15), by (2.2) and Lemma 2.3, we obtain
$$\begin{aligned}& \biggl(\frac{p(1-\varepsilon_{1})}{2}-1 \biggr) \biggl(1- \int_{0}^{t} g(\tau)\,d\tau\biggr)\bigl\Vert \nabla u(t)\bigr\Vert ^{2}_{2}-\frac{1}{(1-\varepsilon_{1})2p} \int_{0}^{t} g(\tau)\,d\tau\bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2} \\& \quad= \frac{ (\frac{p(1-\varepsilon_{1})}{2}-1) (1-\int_{0}^{t} g(\tau)\, d\tau)-\frac{1}{(1-\varepsilon_{1})2p}\int_{0}^{t} g(\tau)\,d\tau}{1-\int _{0}^{t} g(\tau)\,d\tau} \biggl(1- \int_{0}^{t} g(\tau)\,d\tau\biggr)\bigl\Vert \nabla u(t)\bigr\Vert ^{2}_{2} \\& \quad\geq\frac{ (\frac{p(1-\varepsilon_{1})}{2}-1)l -\frac{1}{(1-\varepsilon_{1})2p}(1-l)}{1-\int_{0}^{t} g(\tau)\,d\tau}\beta^{2}. \end{aligned}$$
(3.16)
Then, by (3.15) and (3.16), we have
$$\begin{aligned} L'(t) \geq& (1-\sigma) H^{-\sigma}(t)\bigl\Vert \nabla u_{t}(t)\bigr\Vert ^{2}_{2}+ \varepsilon\biggl(\frac{1}{\rho+1}+\frac{p(1-\varepsilon_{1})}{\rho+2} \biggr)\Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+ \varepsilon(1- \varepsilon_{1})pH(t) \\ & {}+\frac{ (\frac{p(1-\varepsilon_{1})}{2}-1)l -\frac{1}{(1-\varepsilon_{1})2p}(1-l)}{1-\int_{0}^{t} g(\tau)\,d\tau }\varepsilon\beta^{2}-\varepsilon(1- \varepsilon_{1})pE_{2}+\varepsilon\varepsilon_{1} \Vert u\Vert ^{p}_{p}. \end{aligned}$$
(3.17)
Since
$$\begin{aligned} \int_{0}^{\infty}g(s)\,ds< \frac{p/2-1}{p/2-1+1/2p}, \end{aligned}$$
we have
$$\begin{aligned} \biggl(\frac{p}{2}-1 \biggr) \biggl(1- \int_{0}^{\infty}g(\tau)\,d\tau\biggr)- \frac{1}{2p} \int_{0}^{\infty}g(\tau)\,d\tau>0. \end{aligned}$$
It is easy to see that there exists \(\varepsilon^{*}_{1}>0\), such that, for \(0<\varepsilon_{1}<\varepsilon^{*}_{1}\),
$$\begin{aligned} \frac{ (\frac{p(1-\varepsilon_{1})}{2}-1)l-\frac{1}{(1-\varepsilon _{1})2p}(1-l)}{1-\int_{0}^{t} g(\tau)\,d\tau}\beta^{2}>\frac{ (\frac {p(1-\varepsilon_{1})}{2}-1)l-\frac{1}{(1-\varepsilon _{1})2p}(1-l)}{1-\int_{0}^{t} g(\tau)\,d\tau}B^{-\frac{2p}{p-2}}_{1}. \end{aligned}$$
(3.18)
Since
$$\begin{aligned} E(0)< \biggl(\frac{1}{2}-\frac{1}{p} \biggr) \biggl(1- \frac{1}{p(p-2)}\frac{1-l}{l} \biggr)B^{-\frac{2p}{p-2}}_{1}= \frac{(\frac{p}{2}-1)l- \frac{1}{2p}(1-l)}{pl}B^{-\frac{2p}{p-2}}_{1}, \end{aligned}$$
we may choose \(0<\varepsilon_{1}<1\) sufficiently small, and \(E_{2}\in (E(0),E_{1})\) sufficiently near E(0), such that
$$\begin{aligned} \frac{(\frac{p(1-\varepsilon_{1})}{2}-1)l-\frac {1}{(1-\varepsilon_{1})2p}(1-l)}{1-\int_{0}^{t} g(\tau)\,d\tau}B^{-\frac {2p}{p-2}}_{1}-p(1- \varepsilon_{1})E_{2}\geq0. \end{aligned}$$
(3.19)
Then, for \(t>0\), by (3.17) and (3.19), we obtain
$$\begin{aligned} L'(t) \geq& (1-\sigma) H^{-\sigma}(t)\bigl\Vert \nabla u_{t}(t)\bigr\Vert ^{2}_{2}+\varepsilon\biggl( \frac{1}{\rho+1}+\frac{p(1-\varepsilon_{1})}{\rho+2} \biggr)\Vert u_{t}\Vert ^{\rho+2}_{\rho+2} \\ & {}+\varepsilon(1-\varepsilon_{1})pH(t)+\varepsilon \varepsilon_{1}\Vert u\Vert ^{p}_{p}. \end{aligned}$$
(3.20)
From the above estimate we know there exists a constant \(\gamma>0\) such that
$$\begin{aligned} L'(t)\geq\varepsilon\gamma\bigl(\Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+H(t)+\Vert u\Vert ^{p}_{p} \bigr)\geq0,\quad t>0, \end{aligned}$$
(3.21)
where
$$\begin{aligned} \gamma=\mbox{min} \biggl\{ \biggl(\frac{1}{\rho+1}+\frac{p(1-\varepsilon _{1})}{\rho+2} \biggr), (1-\varepsilon_{1})p, \varepsilon_{1} \biggr\} . \end{aligned}$$
Since
$$\begin{aligned} L(0)=H^{1-\sigma}(0)+\frac{\varepsilon}{\rho+1} \int_{\Omega} \vert u_{t}\vert ^{\rho}u_{1}u_{0}\,dx+\frac {\varepsilon}{2} \int_{\Omega} \vert \nabla u_{0}\vert ^{2}\,dx>0, \end{aligned}$$
combining (3.21), we have
$$\begin{aligned} L(t)\geq L(0)>0,\quad t>0. \end{aligned}$$
We now estimate the term \(\int_{\Omega} \vert u_{t}\vert ^{\rho }u_{t}u \,dx\) as follows:
$$\begin{aligned} \biggl\vert \int_{\Omega} \vert u_{t}\vert ^{\rho}u_{t}u \,dx\biggr\vert \leq \Vert u_{t}\Vert ^{\rho+1}_{\rho+2} \Vert u\Vert _{\rho+2}\leq C \Vert u_{t}\Vert ^{\rho+1}_{\rho+2}\Vert u\Vert _{p}. \end{aligned}$$
Using Young’s inequality, we have
$$\begin{aligned} \biggl(\biggl\vert \int_{\Omega} \vert u_{t}\vert ^{\rho}u_{t}u \,dx\biggr\vert \biggr)^{\frac{1}{1-\sigma}}\leq C\Vert u_{t}\Vert ^{\frac{\rho+1}{1-\sigma}}_{\rho+2}\Vert u\Vert ^{\frac{1}{1-\sigma }}_{p}\leq C \bigl(\Vert u_{t}\Vert ^{\frac{\rho+1}{1-\sigma}\mu}_{\rho +2}+\Vert u \Vert ^{\frac{1}{1-\sigma}\theta}_{p} \bigr), \end{aligned}$$
where \(\frac{1}{\mu}+\frac{1}{\theta}=1\). By choosing
$$\begin{aligned} \mu=\frac{(1-\sigma)(\rho+2)}{\rho+1}>1, \end{aligned}$$
we have
$$\begin{aligned} \frac{\theta}{1-\sigma}=\frac{\rho+2}{(1-\sigma)(\rho+2)-(\rho+1)}. \end{aligned}$$
By (3.10), we know
$$\begin{aligned} \frac{\theta}{1-\sigma}< p. \end{aligned}$$
(3.22)
Then, by (3.8), we have
$$\begin{aligned} \Vert u\Vert ^{\frac{\theta}{1-\sigma}}_{p}=\Vert u\Vert ^{p-(p-\frac{\theta}{1-\sigma})}_{p}=\Vert u\Vert ^{p}_{p} \Vert u\Vert ^{-k}_{p}\leq \Vert u\Vert ^{p}_{p}CH^{-\frac{k}{p}}(t)\leq C\Vert u\Vert ^{p}_{p}H^{-\frac{k}{p}}(0), \end{aligned}$$
(3.23)
where \(k=p-\frac{\theta}{1-\sigma}\) is a positive constant. Now, from (3.23), we have
$$\begin{aligned} \biggl(\biggl\vert \int_{\Omega} \vert u_{t}\vert ^{\rho}u_{t}u \,dx\biggr\vert \biggr)^{\frac{1}{1-\sigma}}\leq C\bigl(\Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+\Vert u\Vert ^{p}_{p}H^{-\frac{p}{k}}(0) \bigr). \end{aligned}$$
Therefore it follows
$$\begin{aligned} L^{\frac{1}{1-\sigma}}(t) =& \biggl(H^{1-\sigma}(t)+ \frac{\varepsilon}{\rho+1} \int_{\Omega} \vert u_{t}\vert ^{\rho}u_{t}u \,dx+\frac{\varepsilon}{2} \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{2}\,dx \biggr)^{\frac{1}{1-\sigma}} \\ \leq& C \biggl( H(t)+\biggl\vert \int_{\Omega} \vert u_{t}\vert ^{\rho}u_{t}u \,dx\biggr\vert ^{\frac{1}{1-\sigma}}+\biggl\vert \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{2}\,dx \biggr\vert ^{\frac{1}{1-\sigma}} \biggr) \\ \leq& C \bigl(H(t)+\Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+ \Vert u\Vert ^{p}_{p}+\bigl\Vert \nabla u(t)\bigr\Vert ^{\frac{2}{1-\sigma}}_{2} \bigr). \end{aligned}$$
(3.24)
From (3.5) and (3.7), we have
$$\begin{aligned} \bigl\Vert \nabla u(t)\bigr\Vert ^{\frac{2}{1-\sigma}}_{2} \leq C^{\frac{1}{1-\sigma}}\leq\frac{C^{\frac{1}{1-\sigma}}}{H(0)}H(t). \end{aligned}$$
(3.25)
It follows from (3.24) and (3.25) that
$$\begin{aligned} L^{\frac{1}{1-\sigma}}(t)\leq C \bigl(H(t)+\Vert u_{t} \Vert ^{\rho+2}_{\rho+2}+\Vert u\Vert ^{p}_{p} \bigr). \end{aligned}$$
(3.26)
Combining (3.21) and (3.26), we arrive at
$$\begin{aligned} L'(t)>\frac{\varepsilon\gamma}{C}L^{\frac{1}{1-\sigma }}(t),\quad t>0. \end{aligned}$$
(3.27)
By a simple integration of (3.27) over \((0, t)\), we obtain
$$\begin{aligned} L^{\sigma/(1-\sigma)}(t)\geq\frac{1}{L^{-\sigma/(1-\sigma )}(0)-\varepsilon\gamma t\sigma/[C(1-\sigma)]},\quad t>0. \end{aligned}$$
This shows that \(L(t)\) blows up in finite time \(T_{0}\), and
$$\begin{aligned} T_{0}\leq\frac{C(1-\sigma)}{\varepsilon\gamma\sigma L^{\sigma/(1-\sigma)}(0)}. \end{aligned}$$
Furthermore, we have
$$\begin{aligned} {\lim_{t\rightarrow T_{0}^{-}}} \bigl(\Vert u_{t}\Vert ^{\rho+2}_{\rho+2}+\Vert \nabla u\Vert ^{2}_{2}+ \Vert u\Vert ^{p}_{p} \bigr)=\infty. \end{aligned}$$
(3.28)
This leads to a contradiction with (3.5). Thus, the solution of problem (1.1) blows up in finite time. □