In this section, we deal with the blow-up solutions of equation (1.1). Before we prove our blow-up result, we need the following lemmas.
Lemma 3.1
(see [16], Lemma 2.1)
Assume that
\(g(t)\)
satisfies the assumptions (A3) and (A2), and
\(\Lambda(t)\)
is a function that is twice continuously differentiable, satisfying
$$ \textstyle\begin{cases} \Lambda''(t)+\Lambda'(t)>\int_{0}^{t}g(t-\tau)\int_{\Omega}\nabla u(\tau,x)\nabla u(t,x)\, dx\, d\tau, \\ \Lambda(0)>0,\qquad \Lambda'(0)>0 \end{cases} $$
for every
\(t\in[0,T_{0})\), where
\(u(t)\)
is the corresponding solution of (1.1) with
\(u_{0}\)
and
\(u_{1}\). Then the function
\(\Lambda(t)\)
is strictly increasing on
\([0,T_{0})\).
Lemma 3.2
Assume
\(u_{0}\in H_{0}^{1}(\Omega)\cap H^{2}(\Omega)\), \(u_{1}\in L^{2}(\Omega)\), satisfy
$$\int_{\Omega}u_{0}(x)u_{1}(x)\, dx>0. $$
If the local solution
\(u(t)\)
of (1.1) satisfies
then
\(\|u(t)\|_{2}^{2}\)
is strictly increasing on
\([0,T)\).
Proof
Since \(u(t)\) is the local solution of (1.1), by a simple computation we have
$$\begin{aligned} \frac{1}{2}\frac{d^{2}}{dt^{2}}\int_{\Omega}\bigl\vert u(t,x) \bigr\vert ^{2}\, dx =&\int_{\Omega}\bigl( \bigl\vert u_{t}(t) \bigr\vert ^{2}+uu_{tt} \bigr)\, dx \\ =&\int_{\Omega}\bigl\vert u_{t}(t,x) \bigr\vert ^{2}\, dx-\int_{\Omega}uu_{t}\, dx-M \bigl( \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2} \bigr) \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2} \\ &{}+\int _{\Omega}f(u)u\, dx +\int_{0}^{t}g(t-\tau)\int _{\Omega}\nabla u(t)\cdot\nabla u(\tau)\, dx\, d\tau \\ >&-\int_{\Omega}uu_{t}\, dx+\int _{0}^{t}g(t-\tau)\int_{\Omega}\nabla u(t)\cdot \nabla u(\tau)\, dx\, d\tau, \end{aligned}$$
where the last inequality uses \(I(u)<0\), which implies
$$\frac{d^{2}}{dt^{2}}\int_{\Omega}\bigl\vert u(t,x) \bigr\vert ^{2}\, dx+\frac{d}{dt}\int_{\Omega}\bigl\vert u(t,x) \bigr\vert ^{2}\, dx>\int_{0}^{t}g(t- \tau)\int_{\Omega}\nabla u(t)\cdot\nabla u(\tau)\, dx\, d\tau. $$
Therefore, this lemma comes from Lemma 3.1. □
Proof of Theorem 2.2
We next prove Theorem 2.2 in two steps. First, by a contradiction argument we claim that
$$ I \bigl(u(t) \bigr)< 0 $$
(3.1)
and
$$ \bigl\Vert u(t) \bigr\Vert ^{2}_{2}> \frac{(4\delta+2)E(0)}{ (2\delta m_{0}-(2\delta+1)l )\min\{\lambda,1\}} $$
(3.2)
for every \(t\in[0,T)\). If this was not the case, then there would exist a time \(t_{1}\) such that
$$ t_{1}=\min \bigl\{ t\in(0,T): I \bigl(u(t) \bigr)=0 \bigr\} >0. $$
(3.3)
By the continuity of the solution \(u(t)\) as a function of t, we see that \(I (u(t) )<0\) when \(t\in(0,t_{1})\) and \(I (u(t_{1}) )=0\). Thus by Lemma 3.1 we have
$$\bigl\Vert u(t) \bigr\Vert ^{2}_{2}>\|u_{0} \|^{2}_{2}>\frac{(4\delta+2)E(0)}{ (2\delta m_{0}-(2\delta+1)l )\min\{\lambda,1\}} $$
for every \(t\in[0,t_{1})\). In addition, it is obvious that \(\|u(t)\|^{2}_{2}\) is continuous on \([0, t_{1}]\). Thus the following inequality is obtained:
$$ \bigl\Vert u(t_{1}) \bigr\Vert ^{2}_{2}> \frac{(4\delta+2)E(0)}{ (2\delta m_{0}-(2\delta+1)l )\min\{\lambda,1\}}. $$
(3.4)
On the other hand, it follows from the definition of \(E(t)\) and (2.2) that
$$\begin{aligned}& \frac{1}{2}\overline{M} \bigl( \bigl\Vert \nabla u(t_{1}) \bigr\Vert _{2}^{2} \bigr)- \frac{1}{2}\int_{0}^{t_{1}} g(\tau)\, d\tau \bigl\Vert \nabla u(t_{1}) \bigr\Vert ^{2}_{2} \\& \quad {}+ \frac{1}{2}(g\circ\nabla u) (t_{1})-\int_{\Omega}F \bigl(u(t_{1}) \bigr)\, dx\leq E(0). \end{aligned}$$
(3.5)
Using the assumptions (A1) and (A2), we have
$$\begin{aligned}& M \bigl( \bigl\Vert \nabla u(t_{1}) \bigr\Vert _{2}^{2} \bigr) \bigl\Vert \nabla u(t_{1}) \bigr\Vert _{2}^{2}+2 \delta m_{0} \bigl\Vert \nabla u(t_{1}) \bigr\Vert _{2}^{2}-(2\delta+1)l \bigl\Vert \nabla u(t_{1}) \bigr\Vert _{2}^{2} \\& \quad {}-\int _{\Omega}f \bigl(u(t_{1}) \bigr)u(t_{1})\, dx \leq(4\delta+2)E(0). \end{aligned}$$
Noting the fact that \(I (u(t_{1}) )=0\), we then have
$$\bigl(2\delta m_{0}-(2\delta+1)l \bigr) \bigl\Vert \nabla u(t_{1}) \bigr\Vert ^{2}_{2}\leq(4\delta+2)E(0). $$
Thus, by the Poincaré inequality, we have
$$ \bigl\Vert u(t_{1}) \bigr\Vert ^{2}_{2} \leq\frac{(4\delta+2)E(0)}{ (2\delta m_{0}-(2\delta+1)l )\min\{\lambda,1\}}. $$
(3.6)
Obviously, there is a contradiction between (3.4) and (3.6). Thus, we have proved that (3.1) is true for every \(t\in[0,T)\). Furthermore, by Lemma 3.2 we see that (3.2) is also valid on \(t\in[0,T)\).
Secondly, we prove that the solution of problem (1.1) blows up in a finite time. Assume by contradiction that the solution u is global. Then, for sufficiently large \(T>0\), we consider \(H(t):[0,T]\rightarrow\mathbb{R}_{+}\) defined by
$$H(t)=\bigl\Vert u(t)\bigr\Vert _{2}^{2}+\int_{0}^{t} \bigl\Vert u(\tau) \bigr\Vert ^{2}_{2}\, d\tau +(T-t) \|u_{0}\|_{2}^{2}+\alpha(t_{0}+t)^{2}, $$
where \(t_{0}\) and α are positive constants, which will be determined in the sequel. A direct computation yields
$$\begin{aligned} H'(t)&=2\int_{\Omega}u(t)u_{t}(t)\, dx+ \bigl\Vert u(t) \bigr\Vert ^{2}_{2}-\|u_{0} \|_{2}^{2}+2\alpha (t_{0}+t) \\ &=2\int_{\Omega}u(t)u_{t}(t)\, dx+2\int _{0}^{t} \bigl(u(\tau),u_{t}(\tau) \bigr) \, d\tau+2\alpha(t_{0}+t) \end{aligned}$$
and
$$\begin{aligned} H''(t) =&2 \bigl( u_{tt},u(t) \bigr)+2 \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}+2 \bigl(u(t),u_{t}(t) \bigr)+2\alpha \\ =&2 \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}-2M \bigl(\Vert \nabla u\Vert _{2}^{2} \bigr)\Vert \nabla u \Vert _{2}^{2}+2\int_{0}^{t}g(t- \tau)\int_{\Omega}\nabla u(\tau)\nabla u(t)\, dx\, d\tau \\ &{}+2\int _{\Omega}f(u)u\, dx+2\alpha. \end{aligned}$$
Therefore, we have
$$\begin{aligned}& H(t)H''(t)-(\delta+1)H'(t)^{2} \\& \quad =2H(t) \biggl( \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}-M \bigl(\Vert \nabla u\Vert _{2}^{2} \bigr)\Vert \nabla u\Vert _{2}^{2}+\int _{0}^{t}g(t-\tau)\int_{\Omega}\nabla u(\tau)\nabla u(t)\, dx\, d\tau \\& \qquad {}+\int_{\Omega}f(u)u\, dx+ \alpha \biggr)-4(\delta+1) \biggl(\int_{\Omega}u(t)u_{t}(t) \, dx+\int_{0}^{t} \bigl(u(\tau),u_{t}( \tau) \bigr)\, d\tau+\alpha (t_{0}+t) \biggr)^{2} \\& \quad =2H(t) \biggl( \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}-M \bigl(\Vert \nabla u\Vert _{2}^{2} \bigr)\Vert \nabla u\Vert _{2}^{2}+\int _{0}^{t}g(t-\tau)\int_{\Omega}\nabla u(\tau)\nabla u(t)\, dx\, d\tau \\& \qquad {}+\int_{\Omega}f(u)u\, dx+ \alpha \biggr)+4(\delta+1) \bigl(G(t)- \bigl(H(t)-(T-t)\Vert u_{0} \Vert _{2}^{2} \bigr)\Psi (t) \bigr), \end{aligned}$$
(3.7)
where \(\Psi(t),G(t):[0,T]\rightarrow\mathbb{R}_{+}\) are the functions defined by
$$\Psi(t)= \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}+ \int_{0}^{t} \bigl\Vert u_{t}(\tau) \bigr\Vert ^{2}_{2}\, d\tau+\alpha $$
and
$$\begin{aligned} G(t) =&\Psi(t) \biggl( \bigl\Vert u(t) \bigr\Vert _{2}^{2}+ \int_{0}^{t} \bigl\Vert u(\tau) \bigr\Vert ^{2}_{2}\, d\tau+\alpha (t_{0}+t)^{2} \biggr) \\ &{}- \biggl(\int_{\Omega}u(t)u_{t}(t)\, dx+\int _{0}^{t}(u,u_{t})\, d\tau+ \alpha(t_{0}+t) \biggr)^{2}. \end{aligned}$$
Using the Schwarz inequality, we have
$$\begin{aligned}& \biggl(\int_{\Omega}uu_{t}\, dx \biggr)^{2} \leq \bigl\Vert u(t) \bigr\Vert _{2}^{2} \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}, \\& \biggl(\int _{0}^{t}(u,u_{t})\, d\tau \biggr)^{2}\leq \int_{0}^{t} \bigl\Vert u(\tau) \bigr\Vert ^{2}_{2}\, d\tau\int _{0}^{t} \bigl\Vert u_{t}(\tau) \bigr\Vert ^{2}_{2}\, d \tau \end{aligned}$$
and
$$\begin{aligned}& \int_{\Omega}u(t)u_{t}(t)\, dx\int _{0}^{t} \bigl(u(\tau),u_{t}(\tau) \bigr) \, d\tau \\& \quad \leq \bigl\Vert u(t) \bigr\Vert _{2} \biggl(\int _{0}^{t} \bigl\Vert u_{t}(\tau) \bigr\Vert ^{2}_{2}\, d\tau \biggr)^{\frac {1}{2}} \bigl\Vert u_{t}(t) \bigr\Vert _{2} \biggl(\int_{0}^{t} \bigl\Vert u(\tau) \bigr\Vert ^{2}_{2}\, d\tau \biggr)^{\frac {1}{2}} \\& \quad \leq \frac{1}{2} \bigl\Vert u(t) \bigr\Vert _{2}^{2} \int_{0}^{t} \bigl\Vert u_{t}(\tau) \bigr\Vert ^{2}_{2}\, d\tau+\frac {1}{2} \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}\int _{0}^{t} \bigl\Vert u(\tau) \bigr\Vert ^{2}_{2}\, d\tau \end{aligned}$$
and
$$\begin{aligned}& \alpha(t+t_{0})\int_{\Omega}u(t)u_{t}(t)\, dx \\& \quad \leq \sqrt{\alpha}\sqrt{\alpha}(t+s_{0}) \bigl\Vert u(t) \bigr\Vert _{2} \bigl\Vert u_{t}(t) \bigr\Vert _{2} \\& \quad \leq \frac{1}{2}\alpha \bigl\Vert u(t) \bigr\Vert _{2}^{2}+ \frac{1}{2}\alpha(t+t_{0})^{2} \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}. \end{aligned}$$
Similarly, we have
$$ \alpha(t+t_{0})\int_{0}^{t} \bigl( u( \tau), u_{t}(\tau) \bigr)\, d\tau\leq \frac{1}{2}\alpha\int _{0}^{t} \bigl\Vert u(\tau) \bigr\Vert ^{2}_{2}\, d\tau+\frac{1}{2}\alpha(t+t_{0})^{2} \int_{0}^{t}\bigl\Vert u_{t}(\tau)\bigr\Vert ^{2}_{2}\, d\tau. $$
The previous inequalities entail \(G(t)\geq0\) for every \([0,T]\). Using (3.7), we get
$$ H(t)H''(t)-(\delta+1)H'(t)^{2} \geq H(t)L(t), $$
(3.8)
where
$$\begin{aligned} L(t) =&-(4\delta+2) \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}-2M \bigl(\Vert \nabla u\Vert _{2}^{2} \bigr)\Vert \nabla u\Vert _{2}^{2} \\ &{}+2 \int_{0}^{t}g(t-\tau)\int_{\Omega}\nabla u(\tau)\nabla u(t)\, dx\, d\tau \\ &{}+2\int_{\Omega}f(u)u\, dx -4(\delta+1)\int _{0}^{t} \bigl\Vert u_{t}(\tau) \bigr\Vert ^{2}_{2}\, d\tau-(4\delta+2)\alpha \\ =&-(4\delta+2) \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}-2 \biggl(M \bigl(\Vert \nabla u\Vert _{2}^{2} \bigr)-\int_{0}^{t} g(\tau)\, d\tau \biggr)\Vert \nabla u \Vert _{2}^{2} \\ &{}+2 \int_{\Omega}f(u)u\, dx -(4 \delta+2)\alpha-4(\delta+1)\int_{0}^{t} \bigl\Vert u_{t}(\tau) \bigr\Vert ^{2}_{2}\, d\tau \\ &{}+2\int _{0}^{t}g(t-\tau)\int_{\Omega}\nabla u(t) \bigl(\nabla u(\tau)-\nabla u(t) \bigr)\, dx\, d\tau. \end{aligned}$$
(3.9)
Using Young’s inequality, we have
$$\begin{aligned}& \int_{0}^{t}g(t-\tau)\int _{\Omega}\nabla u(t)\nabla \bigl(u(\tau)-u(t) \bigr)\, dx\, d\tau \\& \quad \geq-(2\delta+1) (g\circ\nabla u) (t)-\frac{1}{(8\delta+4)}\int_{0}^{t}g( \tau)\, d\tau \bigl\Vert \nabla u(t) \bigr\Vert ^{2}_{2}. \end{aligned}$$
(3.10)
Inserting (3.10) into (3.9), we have
$$\begin{aligned} L(t) \geq&-(4\delta+2) \bigl\Vert u_{t}(t) \bigr\Vert ^{2}_{2}-(4\delta+2) (g\circ\nabla u) (t) \\ &{}-2 \biggl(M \bigl( \Vert \nabla u\Vert _{2}^{2} \bigr)-\int _{0}^{t} g(\tau)\, d\tau \biggr)\Vert \nabla u \Vert _{2}^{2} \\ &{} +2\int_{\Omega}f(u)u\,dx -4(\delta+1)\int _{0}^{t} \bigl\Vert u_{t}(\tau) \bigr\Vert ^{2}_{2}\, d\tau \\ &{}-\frac{1}{4\delta+2}\int _{0}^{t} g(\tau)\, d\tau \Vert \nabla u\Vert _{2}^{2}-(4\delta+2)\alpha \\ \geq&-(8\delta+4)E(t)+2 \bigl((2\delta+1)\overline{M} \bigl(\Vert \nabla u \Vert _{2}^{2} \bigr)-M \bigl(\Vert \nabla u\Vert _{2}^{2} \bigr)\Vert \nabla u\Vert _{2}^{2} \bigr) \\ &{}-4(\delta+1)\int_{0}^{t} \bigl\Vert u_{t}(\tau) \bigr\Vert ^{2}_{2}\, d\tau -2\int_{\Omega}\bigl[(4\delta+2)F(u)-f(u)u \bigr]\, dx \\ &{}- \biggl(4\delta+\frac{1}{4\delta+2} \biggr) \int_{0}^{t} g(\tau)\, d\tau \Vert \nabla u\Vert _{2}^{2}-(4\delta+2) \alpha \\ \geq&-(8\delta+4)E(0)+2 \bigl((2\delta+1)\overline{M} \bigl(\Vert \nabla u \Vert _{2}^{2} \bigr)-M \bigl(\Vert \nabla u\Vert _{2}^{2} \bigr)\Vert \nabla u\Vert _{2}^{2} \bigr) \\ &{}+4\delta\int_{0}^{t} \bigl\Vert u_{t}(\tau) \bigr\Vert ^{2}_{2}\, d\tau -2\int_{\Omega}\bigl[(4\delta+2)F(u)-f(u)u \bigr]\, dx \\ &{}- \biggl(4\delta+\frac{1}{4\delta+2} \biggr) \int_{0}^{t} g(\tau)\, d\tau \Vert \nabla u\Vert _{2}^{2}-(4\delta+2) \alpha. \end{aligned}$$
Using the assumptions (A1) and (A2), we have
$$\begin{aligned} L(t) \geq&-(8\delta+4)E(0)+ \biggl(4\delta m_{0}- \biggl(4\delta+ \frac{1}{4\delta+2} \biggr)\int_{0}^{t} g(\tau)\, d \tau \biggr)\|\nabla u\|_{2}^{2} -(4\delta+2)\alpha \\ \geq& -(8\delta+4)E(0)+ \bigl(4\delta m_{0}- (4\delta+1)l \bigr) \lambda \bigl\Vert u(t) \bigr\Vert _{2}^{2} -(4\delta+2) \alpha \\ \geq&-(8\delta+4)E(0)+ \bigl(4\delta m_{0}- (4\delta+1)l \bigr) \lambda\|u_{0}\|_{2}^{2} -(4\delta+2)\alpha, \end{aligned}$$
where the last inequality follows from Lemma 3.2 and the Poincaré inequality. From (2.4), we have
$$(8\delta+4)E(0)< \bigl(4\delta m_{0}-(4\delta+2)l \bigr)\lambda \|u_{0}\|_{2}^{2}< \bigl(4\delta m_{0}-(4\delta+1)l \bigr)\lambda\|u_{0} \|_{2}^{2}. $$
Thus, we can let α satisfy
$$(4\delta+2)\alpha< \bigl(4\delta m_{0}-(4\delta+2)l \bigr)\lambda \|u_{0}\|_{2}^{2}-(8\delta+4)E(0), $$
which implies that there exists \(\theta>0\) (independent of T) such that
$$ L(t)\geq\theta \quad \mbox{for } t\in[0,T]. $$
(3.11)
By (3.8) and (3.11), it follows that
$$H(t)H''(t)-(\delta+1)H'(t)^{2}>0. $$
Moreover, we let \(t_{0}\) satisfy
$$\alpha t_{0}+\int_{\Omega}u_{0}u_{1} \, dx>0, $$
which means \(H'(0)>0\). Thus by \(H''(t)>0\) we see that \(H(t)\) and \(H'(t)\) are strictly increasing on \([0,T]\).
Setting \(y(t)=H(t)^{-\delta}\), then we have
$$y'(t)=-\delta H(t)^{-(\delta+1)}H'(t)< 0 $$
and
$$y''(t)=-\delta H^{-\delta-2} \bigl(H(t)H''(t)-( \delta+1)H'(t)^{2} \bigr)< 0 $$
for all \(t\in[0,T]\), which implies that \(y(t)\) reaches 0 in finite time, say as \(t\rightarrow T^{*}\). Since \(T^{*}\) is independent of the initial choice of T, we may assume that \(T^{*} < T\). This tells us that
$$\lim_{t\rightarrow T^{*}}H(t)=\infty. $$
□