In the current section, we follow the same approach given in [11] to prove the blow-up of solution of problem (2.20).
Remark 1
By integration of (3.2) over \((0,t)\), we have
$$ \begin{aligned} E(t)& =E(0)-a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds \\ &\quad{}+\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}+\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ &\quad{}-b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds. \end{aligned} $$
(5.1)
Now, let us define \(F(t)\):
$$\begin{aligned} F(t) =& \Vert u \Vert _{2}^{2}+a \int _{0}^{t} \Vert u \Vert _{2}^{2} \,ds \\ &{}-\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) +b_{1}H(t), \end{aligned}$$
(5.2)
where
$$ H(t)= \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl(\xi ^{2}+ \eta \bigr) \biggl( \int _{0}^{s}\phi (\xi ,z)\,dz \biggr) ^{2}\,d\xi \,d\rho \,ds. $$
(5.3)
Lemma 10
Assume that \(\| \nabla u\| _{2}^{2}\) is bounded on \([0,T)\), then
$$ H(t)\leq C< +\infty . $$
(5.4)
More precisely
$$ H(t)\leq \frac{1}{2}C_{1}B_{q}e^{-\eta C_{2}} \bigl[ C_{2}^{2\alpha -1} \alpha +C_{2}^{3-2\alpha }\eta \bigr] \Gamma (\alpha )T^{4}, $$
where
$$ C_{1}=\sup_{t\in {[} 0,T)} \bigl\{ \Vert \nabla u \Vert _{2}^{2},1 \bigr\} . $$
Proof
Using (2.18), we obtain
$$ \phi (\xi ,t)= \int _{0}^{t}\mu (\xi )e^{-(\xi ^{2}+\eta )(t-s)}u(x,s)\,ds, \quad \forall x\in \Gamma _{0}. $$
(5.5)
Hölder’s inequality yields
$$ \phi (\xi ,t)\leq \biggl( \int _{0}^{t}\mu ^{2}(\xi )e^{-2(\xi ^{2}+ \eta )(t-s)}\,ds \biggr) ^{\frac{1}{2}} \biggl( \int _{0}^{t} \bigl\vert u(x,s) \bigr\vert ^{2}\,ds \biggr) ^{\frac{1}{2}},\quad \forall x\in \Gamma _{0}. $$
(5.6)
On the other hand,
$$ \biggl( \int _{0}^{t}\phi (\xi ,s)\,ds \biggr) ^{2} \leq T \int _{0}^{t} \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,ds. $$
(5.7)
From (5.6) in (5.7), we obtain
$$ \biggl( \int _{0}^{t}\phi (\xi ,s)\,ds \biggr) ^{2} \leq T \int _{0}^{t} \biggl[ \int _{0}^{s}\mu ^{2}(\xi )e^{-2(\xi ^{2}+\eta )(s-z)}\,dz \int _{0}^{s} \bigl\vert u(x,z) \bigr\vert ^{2}\,dz \biggr] \,ds. $$
(5.8)
Applying Lemma 2 leads to
$$ \int _{\Gamma _{0}} \biggl( \int _{0}^{t}\phi (\xi ,s)\,ds \biggr) ^{2}\,d \rho \leq B_{q}C_{1}T \int _{0}^{t} \biggl[ \int _{0}^{s}\mu ^{2}(\xi )e^{-2( \xi ^{2}+\eta )(s-z)}\,dz \biggr] \,ds. $$
(5.9)
Since \(z\in (0,s)\), we choose \(\exists C_{2}\geq 0\) such that \(s-z\geq \frac{C_{2}}{2}\) to term (5.9) into
$$ \int _{\Gamma _{0}} \biggl( \int _{0}^{t}\phi (\xi ,s)\,ds \biggr) ^{2}\,d \rho \leq \frac{1}{2}B_{q}C_{1}T^{3} \mu ^{2}(\xi )e^{-C_{2}(\xi ^{2}+ \eta )}. $$
(5.10)
Multiplication by \(\xi ^{2}+\eta \) followed by integration over \((0,t)\times (-\infty ,+\infty )\) yields
$$ \begin{aligned} H(t)& \leq C_{1}B_{q}e^{-\eta C_{2}}T^{3} \int _{0}^{t} \biggl[ \int _{0}^{+\infty }\xi ^{2\alpha +1}e^{-C_{2}\xi ^{2}}\,d \xi \biggr] \,ds \\ &\quad{}+C_{1}B_{q}e^{-\eta C_{2}}\eta T^{3} \int _{0}^{t} \biggl[ \int _{0}^{+ \infty }\xi ^{2\alpha -1}e^{-C_{2}\xi ^{2}}\,d \xi \biggr] \,ds. \end{aligned} $$
(5.11)
Then
$$ \begin{aligned} H(t)& \leq \frac{1}{2}C_{1}B_{q}e^{-\eta C_{2}}C_{2}^{2 \alpha -1}T^{3} \int _{0}^{t} \biggl[ \int _{0}^{+\infty }y^{\alpha }e^{-y}\,dy \biggr] \,ds \\ &\quad{}+\frac{1}{2}C_{1}B_{q}e^{-\eta C_{2}}C_{2}^{3-2\alpha } \eta T^{3} \int _{0}^{t} \biggl[ \int _{0}^{+\infty }y^{\alpha -1}e^{-y}\,dy \biggr] \,ds. \end{aligned} $$
(5.12)
Applying a special integral (Euler gamma function), we obtain
$$ H(t)\leq \frac{1}{2}C_{1}B_{q}e^{-\eta C_{2}} \bigl[ C_{2}^{2\alpha -1} \alpha +C_{2}^{3-2\alpha }\eta \bigr] \Gamma (\alpha )T^{4}. $$
(5.13)
□
Lemma 11
Suppose \(p>2\), then
$$ \begin{aligned} F^{\prime \prime }(t)&\geq (p+2) \Vert u_{t} \Vert _{2}^{2} \\ &\quad{}+2p \biggl\{ -E(0)+a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds- \frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ &\quad {} +b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \biggr\} . \end{aligned} $$
(5.14)
Proof
We differentiate with respect to t in (5.2), then we get
$$ \begin{aligned} F^{\prime }(t)& =2 \int _{\Omega }uu_{t}\,dx+a \Vert u \Vert _{2}^{2} \\ &\quad{}+\frac{1}{2}g ( t ) \Vert \nabla u \Vert _{2}^{2}- \frac{1}{2} \bigl( g^{\prime }\circ \nabla u \bigr) ( t ) \\ &\quad{}+2b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \phi (\xi ,s) \int _{0}^{s}\phi (\xi ,z)\,dz \,d\xi \,d\rho \,ds. \end{aligned} $$
(5.15)
Using divergence theorem and (2.20), we obtain
$$ \begin{aligned} F^{\prime \prime }(t)& =2 \Vert u_{t} \Vert _{2}^{2}-2 \int _{\Omega }\nabla u \int _{0}^{t}g ( t-s ) \nabla u ( s ) \,ds \,dx \\ &\quad{}+2 \Vert u \Vert _{p}^{p}+2b_{1} \int _{\Gamma _{0}}u(x,t) \int _{-\infty }^{+\infty }\mu (\xi )\phi (\xi ,t)\,d\xi \,d\rho \\ &\quad{}+2b_{1} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl(\xi ^{2}+ \eta \bigr) \phi (\xi ,t) \int _{0}^{t}\phi (\xi ,s)\,ds \,d\xi \,d\rho . \end{aligned} $$
(5.16)
By definition of energy functional in (3.1) and relation (5.1), we give the following evaluation of the third term of (5.16):
$$ \begin{aligned} 2 \Vert u \Vert _{p}^{p}& =p \Vert u_{t} \Vert _{2}^{2}+p \Vert \nabla u \Vert _{2}^{2}+pb_{1} \int _{\Gamma _{0}} \int _{- \infty }^{+\infty } \bigl\vert \phi (\xi ,t) \bigr\vert ^{2}\,d\xi \,d\rho -2pE(0) \\ &\quad{}+2p \biggl[ a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds- \frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ & \quad {} +b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \biggr] . \end{aligned} $$
(5.17)
We can also estimate the last term of (5.16) using Lemma 8:
$$ \begin{aligned} & \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl(\xi ^{2}+ \eta \bigr) \phi (\xi ,t) \int _{0}^{t}\phi (\xi ,s)\,ds \,d\xi \,d\rho \\ & \quad = \int _{\Gamma _{0}}u(x,t) \int _{-\infty }^{+\infty }\phi (\xi ,t) \mu (\xi )\,d\xi \,d\rho - \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl\vert \phi (\xi ,t) \bigr\vert ^{2}\,d\xi \,d\rho . \end{aligned} $$
(5.18)
From (5.17), (5.18), and (5.16), we get
$$ \begin{aligned} F^{\prime \prime }(t)&\geq (p+2) \Vert u_{t} \Vert _{2}^{2}+(p-2) \Vert \nabla u \Vert _{2}^{2}+b_{1}(p-2) \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl\vert \phi (\xi ,t) \bigr\vert ^{2}\,d\xi \,d\rho \\ &\quad{}+2p \biggl[ -E(0)+a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds- \frac{1}{2} \biggl(1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}- \frac{1}{2} ( g\circ \nabla u ) ( t ) \\ & \quad {} +b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \biggr] . \end{aligned} $$
(5.19)
Taking \(p>2\), we obtain the needed estimation
$$ \begin{aligned} F^{\prime \prime }(t)&\geq (p+2) \Vert u_{t} \Vert _{2}^{2} \\ &\quad{}+2p \biggl\{ -E(0)+a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds- \frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ &\quad {} +b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \biggr\} . \end{aligned}$$
□
Lemma 12
Suppose that \(p>2\) and that either one of the next assumptions is verified:
(i) \(E(0)<0\);
(ii) \(E(0)=0\), and
$$ F^{\prime }(0)>a \Vert u_{0} \Vert _{2}^{2}; $$
(5.20)
(iii) \(E(0)>0\), and
$$ F^{\prime }(0)> \bigl[ F(0)+l_{0} \bigr] +a \Vert u_{0} \Vert _{2}^{2}, $$
(5.21)
where
$$ r=p-2\sqrt{p^{2}-p} $$
and
$$ l_{0}=a \Vert u_{0} \Vert _{2}^{2}-2E(0). $$
(5.22)
Then \(F^{\prime }(t)>a\| u_{0}\| _{2}^{2}\) for \(t>t_{0}\), where
$$ t^{\ast }>\max \biggl\{ 0, \frac{F^{\prime }(0)-a \Vert u_{0} \Vert _{2}^{2}]}{2pE(0)} \biggr\} , $$
(5.23)
where \(t_{0}=t^{\ast }\) in case (i), and \(t_{0}=0\) in cases (ii) and (iii).
Proof
(i) Case of \(E(0)<0\).
From (5.14), we have
$$ F^{{\prime \prime }}(t)\geq -2pE(0), $$
which clearly leads to
$$ F^{{\prime }}(t)\geq F^{{\prime }}(0)-2pE(0)t. $$
Then
$$ F^{{\prime }}(t)>a \Vert u_{0} \Vert _{2}^{2},\quad \forall t \geq t^{\ast }, $$
where \(t^{\ast }\) as given in (5.23).
(ii) Case \(E(0)=0\).
Using (5.14) we get
$$ F^{\prime \prime }(t)\geq 0,\quad \forall t\geq 0. $$
Thus
$$ F^{\prime }(t)\geq F^{\prime }(0),\quad \forall t\geq 0. $$
Then, by (5.20),
$$ F^{{\prime }}(t)>a \Vert u_{0} \Vert _{2}^{2},\quad \forall t \geq 0. $$
(iii) Case \(E(0)>0\).
The proof of this case consists of getting to a differential inequality: \(B^{\prime \prime }(t)-pB^{\prime }(t)+pB(t)\geq 0\) pursued by a use of Lemma 3. Indeed, from (5.15) we have
$$ \begin{aligned} F^{\prime }(t)&=2 \int _{\Omega }uu_{t}\,dx+a \Vert u \Vert _{2}^{2} \\ &\quad{}+\frac{1}{2}g ( t ) \Vert \nabla u \Vert _{2}^{2}- \frac{1}{2} \bigl( g^{\prime }\circ \nabla u \bigr) ( t ) \\ &\quad{}+2b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \phi (\xi ,s) \int _{0}^{s}\phi (\xi ,z)\,dz \,d\xi \,d\rho \,ds. \end{aligned} $$
(5.24)
Or, the last term in (5.24) can be estimated using Young’s inequality
$$ \begin{aligned} & \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \phi (\xi ,s) \int _{0}^{s}\phi (\xi ,z)\,dz\,d \xi \,d\rho \,ds \\ &\quad \leq \frac{1}{2} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \\ &\qquad {} +\frac{1}{2} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \biggl( \int _{0}^{s}\phi (\xi ,z)\,dz \biggr) ^{2}\,d \xi \,d\rho \,ds. \end{aligned} $$
(5.25)
On the other hand,
$$ 2 \int _{0}^{t} \int _{\Omega }u_{s}u\,dx \,ds= \int _{0}^{t}\frac{d}{ds} \Vert u_{s} \Vert _{2}^{2}\,ds= \Vert u \Vert _{2}^{2}- \Vert u_{0} \Vert _{2}^{2}. $$
(5.26)
By Young’s inequality, we get
$$ \Vert u \Vert _{2}^{2}\leq \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds+ \int _{0}^{t} \Vert u \Vert _{2}^{2} \,ds+ \Vert u_{0} \Vert _{2}^{2}. $$
(5.27)
Now, we remake (5.24) using (5.25) and (5.27):
$$ \begin{aligned} F^{\prime }(t)& \leq \Vert u \Vert _{2}^{2}+ \Vert u_{t} \Vert _{2}^{2}+a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds+a \int _{0}^{t} \Vert u \Vert _{2}^{2} \,ds+a \Vert u_{0} \Vert _{2}^{2} \\ &\quad{}-\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ &\quad {}+b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl(\xi ^{2}+ \eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \\ &\quad{}+b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \biggl( \int _{0}^{s}\phi (\xi ,z)\,dz \biggr) ^{2}\,d\xi \,d \rho \,ds. \end{aligned} $$
(5.28)
From the definition of F in (5.2), inequality (5.28) also becomes
$$ \begin{aligned} F^{\prime }(t)&\leq F(t)+ \Vert u_{t} \Vert _{2}^{2}+b_{1}\int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl(\xi ^{2}+ \eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \\ &\quad{}+a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds+a \Vert u_{0} \Vert _{2}^{2}. \end{aligned} $$
(5.29)
Thus, by (5.14), we get
$$ \begin{aligned} F^{\prime \prime }(t)-p \bigl\{ F^{\prime }(t)-F(t) \bigr\} & \geq 2 \Vert u_{t} \Vert _{2}^{2}+ap \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds-pa \Vert u_{0} \Vert _{2}^{2}-2pE(0) \\ &\quad{}+pb_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds. \end{aligned} $$
(5.30)
Hence
$$ F^{\prime \prime }(t)-pF^{\prime }(t)+pF(t)+pl_{0}\geq 0, $$
(5.31)
where
$$ l_{0}=a \Vert u_{0} \Vert _{2}^{2}-2E(0). $$
Posing
leads to
$$ B^{\prime \prime }(t)-pB^{\prime }(t)+pB(t)\geq 0. $$
(5.32)
By Lemma 3 and for \(p=\delta +1\), we conclude that if
$$ B^{\prime }(t)> \bigl(p-2\sqrt{p^{2}-p} \bigr)B(0)+a \Vert u_{0} \Vert _{2}^{2}, $$
(5.33)
then
$$ F^{\prime }(t)=B^{\prime }(t)>a \Vert u_{0} \Vert _{2}^{2} \quad \forall t\geq 0. $$
□
Theorem 5
Suppose that \(p>2\) and that either one of the next assumptions is verified:
(i) \(E(0)<0\);
(ii) \(E(0)=0\) and (5.20) holds;
(iii) \(0< E(0)< \frac{(2p-4) ( F^{\prime }(t_{0})-a\| u_{0}\| _{2}^{2} ) ^{2}J(t_{0})^{\frac{1}{\gamma _{1}}}}{16p}\) and (5.21) holds.
Then, in the sense of Definition 1, the solution \((u,\phi )\) blows up at finite time \(T^{\ast }\).
For case (i):
$$ T^{\ast }\leq t_{0}-\frac{J(t_{0})}{J^{\prime }(t_{0})}. $$
(5.34)
Moreover, if \(J(t_{0})<\min \{ 1,\sqrt{\frac{\sigma }{-b}} \} \), we get
$$ T^{\ast }\leq t_{0}+\frac{1}{\sqrt{-b}}\ln \frac{\sqrt{\frac{\sigma }{-b}}}{\sqrt{\frac{\sigma }{-b}}-J(t_{0})}. $$
(5.35)
For case (ii), we get either
$$ T^{\ast }\leq t_{0}-\frac{J(t_{0})}{J^{\prime }(t_{0})} $$
(5.36)
or
$$ T^{\ast }\leq t_{0}+\frac{J(t_{0})}{J^{\prime }(t_{0})}. $$
(5.37)
For case (iii):
$$ T^{\ast }\leq \frac{J(t_{0})}{\sqrt{\sigma }}, $$
(5.38)
or else
$$ T^{\ast }\leq t_{0}+2^{\frac{3\gamma _{1}+1}{2\gamma _{1}}} \frac{\gamma _{1}c}{\sqrt{\sigma }} \bigl\{ 1- \bigl[1-cJ(t_{0}) \bigr]^{\frac{1}{2\gamma _{1}}} \bigr\} , $$
(5.39)
where \(\gamma _{1}=\frac{p-4}{4}\), \(c=(\frac{b}{\sigma })^{\frac{\gamma _{1}}{2+\gamma _{1}}}\), \(J(t)\), b and σ are as in (5.40) and (5.54) respectively.
Note that \(t_{0} =0\) in cases (ii) and (iii). For case (i), we have as in (5.23): \(t_{0}=t^{*}\).
Proof
Consider
$$ J(t)= \bigl[ F(t)+a(T-t) \Vert u_{0} \Vert _{2}^{2} \bigr] ^{- \gamma _{1}},\quad t\in {[} t_{0},T]. $$
(5.40)
We differentiate on \(J(t)\) to get
$$ J^{{\prime }}(t)=-\gamma _{1}J(t)^{1+\frac{1}{\gamma _{1}}} \bigl[ F^{ \prime }(t)-a \Vert u_{0} \Vert _{2}^{2} \bigr] $$
(5.41)
and again
$$ J^{{\prime \prime }}(t)=-\gamma _{1}J(t)^{1+\frac{2}{\gamma _{1}}}G(t), $$
(5.42)
where
$$ G(t)=F^{{\prime \prime }}(t) \bigl[ F(t)+a(T-t) \Vert u_{0} \Vert _{2}^{2} \bigr] -(1+\gamma _{1}) \bigl\{ F^{{\prime }}(t)-a \Vert u_{0} \Vert _{2}^{2} \bigr\} ^{2}. $$
(5.43)
Using (5.14), we obtain
$$ \begin{aligned} F^{\prime \prime }(t)&\geq (p+2) \Vert u_{t} \Vert _{2}^{2} \\ &\quad{}+2p \biggl\{ -E(0)+a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds- \frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ &\quad {} +b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \biggr\} . \end{aligned}$$
Consequently,
$$ \begin{aligned} F^{\prime \prime }(t)&\geq -2pE(0) \\ &\quad {}\times p \biggl\{ \Vert u_{t} \Vert _{2}^{2}+a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds-\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ &\quad {} +b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \biggr\} . \end{aligned} $$
(5.44)
Or, from (5.15) and the fact that \(\| u\| _{2}^{2}-\| u_{0}\| _{2}^{2}=2\int _{0}^{t}\int _{\Omega }u_{s}u\,dx \,ds\), we attain
$$ \begin{aligned} F^{\prime }(t)-a \Vert u_{0} \Vert _{2}^{2}& =2 \int _{\Omega }uu_{t}\,dx+2a \int _{0}^{t} \int _{\Omega }u_{s}u\,dx \,ds \\ &\quad{}+2b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \phi (\xi ,s) \int _{0}^{s}\phi (\xi ,z)\,dz \,d\xi \,d\rho \,ds. \end{aligned} $$
(5.45)
Going back to (5.43) with (5.44) and (5.45) in hand, we get
$$ \begin{aligned} G(t) &\geq -2pE(0)J(t)^{\frac{-1}{\gamma _{1}}} \\ &\quad{}+p \biggl\{ \Vert u_{t} \Vert _{2}^{2}+a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds-\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ & \quad {} +b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds \biggr\} \\ &\quad{}\times \biggl[ \Vert u \Vert _{2}^{2}+a \int _{0}^{t} \Vert u \Vert _{2}^{2}\,ds-\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ & \quad {} +b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \biggl( \int _{0}^{s}\phi (\xi ,z)\,dz \biggr) ^{2}\,d \xi \,d\rho \,ds \biggr] \\ &\quad{}-4(1+\gamma _{1}) \biggl\{ \int _{\Omega }uu_{t}\,dx+a \int _{0}^{t} \int _{\Omega }u_{s}u\,dx \,ds+\frac{1}{2}g ( t ) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} \bigl( g^{\prime }\circ \nabla u \bigr) ( t ) \\ & \quad {} +b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+ \infty } \bigl(\xi ^{2}+\eta \bigr) \phi (\xi ,s) \int _{0}^{s}\phi (\xi ,z)\,dz\,d \xi \,d\rho \,ds \biggr\} ^{2}. \end{aligned} $$
(5.46)
For the sake of simplicity, we introduce the following notations:
$$\begin{aligned}& \begin{aligned} \mathbf{A}& = \Vert u \Vert _{2}^{2}+a \int _{0}^{t} \Vert u \Vert _{2}^{2}\,ds-\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ &\quad{}+b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \biggl( \int _{0}^{s}\phi (\xi ,z)\,dz \biggr) ^{2}\,d\xi \,d \rho \,ds, \end{aligned} \\& \begin{aligned} \mathbf{B}& = \int _{\Omega }uu_{t}\,dx+a \int _{0}^{t} \int _{\Omega }u_{s}u\,dx \,ds+\frac{1}{2}g ( t ) \Vert \nabla u \Vert _{2}^{2}- \frac{1}{2} \bigl( g^{\prime }\circ \nabla u \bigr) ( t ) \\ &\quad{}+b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \phi (\xi ,s) \int _{0}^{s}\phi (\xi ,z)\,dz \,d\xi \,d\rho \,ds, \end{aligned} \\& \begin{aligned} \mathbf{C}& = \Vert u_{t} \Vert _{2}^{2}+a \int _{0}^{t} \Vert u_{s} \Vert _{2}^{2}\,ds-\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \\ &\quad{}+b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \bigl\vert \phi (\xi ,s) \bigr\vert ^{2}\,d\xi \,d\rho \,ds. \end{aligned} \end{aligned}$$
Therefore
$$ Q(t)\geq -2pE(0)J(t)^{\frac{-1}{\gamma _{1}}}+p \bigl\{ \mathbf{A} \mathbf{C}-\mathbf{B}^{2} \bigr\} . $$
(5.47)
Note that, \(\forall w\in R\) and \(\forall t>0\),
$$ \begin{aligned} \mathbf{A}w^{2}+2\mathbf{B}w+\mathbf{C}& = \biggl[ w^{2} \Vert u \Vert _{2}^{2}+2w \int _{\Omega }uu_{t}\,dx+ \Vert u_{t} \Vert _{2}^{2} \biggr] \\ &\quad{}+a \int _{0}^{t} \biggl[ w^{2} \Vert u \Vert _{2}^{2}+2w \int _{ \Omega }uu_{s}\,dx+ \Vert u_{s} \Vert _{2}^{2} \biggr] \,ds \\ &\quad{}+ \bigl( w^{2}+1 \bigr) \biggl( -\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \biggr) \\ &\quad{}+w \biggl( \frac{1}{2}g ( t ) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} \bigl( g^{\prime }\circ \nabla u \bigr) ( t ) \biggr) \\ &\quad{}+b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \biggl[ w^{2} \biggl( \int _{0}^{s}\phi (\xi ,z)\,dz \biggr) ^{2} \\ & \quad {} +2w\phi (\xi ,s) \int _{0}^{s}\phi (\xi ,z)\,dz+ \bigl\vert \phi ( \xi ,s) \bigr\vert ^{2} \biggr] \,d\xi \,d\rho \,ds. \end{aligned} $$
(5.48)
Hence
$$ \begin{aligned} &\mathbf{A}w^{2} +2\mathbf{B}w+\mathbf{C}\\ &\quad = \bigl[ w \Vert u \Vert _{2}+ \Vert u_{t} \Vert _{2} \bigr] ^{2}+a \int _{0}^{t} \bigl[ w \Vert u \Vert _{2}+ \Vert u_{s} \Vert _{2} \bigr] ^{2}\,ds \\ &\qquad{}+ \bigl( w^{2}+1 \bigr) \biggl( -\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} ( g\circ \nabla u ) ( t ) \biggr) \\ &\qquad {}+w \biggl( \frac{1}{2}g ( t ) \Vert \nabla u \Vert _{2}^{2}-\frac{1}{2} \bigl( g^{\prime } \circ \nabla u \bigr) ( t ) \biggr) \\ &\qquad{}+b_{1} \int _{0}^{t} \int _{\Gamma _{0}} \int _{-\infty }^{+\infty } \bigl( \xi ^{2}+\eta \bigr) \biggl[ w \int _{0}^{s}\phi (\xi ,z)\,dz+ \bigl\vert \phi ( \xi ,s) \bigr\vert \biggr] ^{2}\,d\xi \,d\rho \,ds. \end{aligned} $$
(5.49)
It is clear that
$$ \mathbf{A}w^{2}+2\mathbf{B}+\mathbf{C}\geq 0 $$
and
$$ \mathbf{B}^{2}-\mathbf{A}\mathbf{C}\leq 0. $$
(5.50)
Then, from (5.47) and (5.50), we obtain
$$ G(t)\geq -2pE(0)J(t)^{\frac{-1}{\gamma _{1}}},\quad t\geq t_{0}. $$
(5.51)
Hence, by (5.42) and (5.51),
$$ J^{\prime \prime }(t)\leq \frac{p^{2}-4p}{2}E(0)J(t)^{1+ \frac{1}{\gamma _{1}}},\quad t\geq t_{0}. $$
(5.52)
Or, by Lemma [6], \(J^{{\prime }}(t)<0\), where \(t\geq t_{0}\).
Multiplication by \(J^{{\prime }}(t)\) in (5.52), followed by integration from \(t_{0}\) to t, leads to
$$ J^{{\prime }}(t)^{2}\geq \sigma +bJ(t)^{2+\frac{1}{\gamma _{1}}}, $$
(5.53)
where
$$ \textstyle\begin{cases} \sigma = [ \frac{(p-4)^{2}}{16} ( F^{{ \prime }}(t_{0})- \Vert u_{0} \Vert _{2}^{2} ) ^{2}-\frac{p(p-4)^{2}}{2p-4}E(0)J(t_{0})^{\frac{-1}{\gamma _{1}}} ] J(t_{0})^{2+\frac{2}{\gamma _{1}}}, \\ b =\frac{p(p-4)^{2}}{2p-4}E(0). \end{cases} $$
(5.54)
Note that \(\sigma >0\) is equivalent to \(E(0)< \frac{(2p-4) ( F^{\prime }(t_{0})-a\| u_{0}\| _{2}^{2} ) ^{2}J(t_{0})^{\frac{1}{\gamma _{1}}}}{16p}\), which by Lemma 4 ensures the existence of a finite time \(T^{\ast }>0\) such that
$$ \lim_{t\rightarrow T^{\ast -}}J ( t ) =0. $$
That involves
$$ \begin{aligned} &\lim_{t\rightarrow T^{\ast -}} \biggl[ \Vert u \Vert _{2}^{2}+a \int _{0}^{t} \Vert u \Vert _{2}^{2} \,ds-\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}\\ &\quad {}- \frac{1}{2} ( g\circ \nabla u ) ( t ) +b_{1}H(t) \biggr] ^{-1}=0, \end{aligned} $$
(5.55)
i.e.,
$$ \begin{aligned} &\lim_{t\rightarrow T^{\ast -}} \biggl[ \Vert u \Vert _{2}^{2}+a \int _{0}^{t} \Vert u \Vert _{2}^{2} \,ds-\frac{1}{2} \biggl( 1- \int _{0}^{t}g(s)\,ds \biggr) \Vert \nabla u \Vert _{2}^{2}\\ &\quad {}- \frac{1}{2} ( g\circ \nabla u ) ( t ) +b_{1}H(t) \biggr] =+ \infty . \end{aligned} $$
(5.56)
So, there exists T such that \(t_{0}< T\leq T^{*}\) and \(\| \nabla u\| _{2}^{2}\longrightarrow +\infty \) as \(t\longrightarrow T^{-}\).
Indeed, if it is not the case, then \(\| \nabla u\| _{2}^{2}\) remained bounded on \([t_{0},T^{\ast })\), which by Lemma 10 leads to
$$ \lim_{t\rightarrow T^{\ast -}} \bigl[ \Vert u \Vert _{2}^{2}+b_{1}H(t) \bigr] =C< +\infty , $$
contradicting (5.56). □