Now, we will state the blow-up result of the solutions to the problem (1.1).
Theorem 3.1
Assume that
\(p>0\)
and
\(u_{0}\in{H}_{0}^{2}(\Omega)\). If
\(u(x,t)\)
is a solution of the problem
\((1,1)\)
and the initial data
\(u_{0}(x)\)
satisfies
$$\begin{aligned} \int_{\Omega} \bigl(|u_{0}|^{2}+| \nabla{u}_{0}|^{2} \bigr)\,dx>\eta{E}({0}), \end{aligned}$$
(3.1)
then the solution of problem (1.1) blows up in finite time; that is, the maximum existence time
\(T_{\max}\)
of
\(u(x,t)\)
is finite and
$$\begin{aligned} \lim_{t\rightarrow{T_{\max}^{-}}}\int_{0}^{t}\int _{\Omega} \bigl(|u|^{2}+|\nabla{u}|^{2} \bigr) \,dx\,d\tau=+\infty, \end{aligned}$$
where
\(\eta=\frac{\alpha}{m}\); \(m=(\frac{\alpha}{2}-1)\lambda_{1}\); \(2\leq\alpha\leq2p+2\); \(\lambda_{1}\)
is the first eigenvalue of operator −△ under homogeneous Dirichlet boundary conditions.
Proof
The proof makes use of the so-called ‘concavity method’. Multiplying (1.1) by u and integrating over Ω, we have
$$\begin{aligned} &\frac{1}{2}\frac{d}{dt}\int_{\Omega}|u|^{2} \,dx+ \frac {1}{2}\frac {d}{dt}\int_{\Omega}| \nabla{u}|^{2}\,dx+\int_{\Omega}|\nabla {u}|^{2}\,dx+\int_{\Omega}|\triangle{u}|^{2} \,dx \\ &\quad=\int_{\Omega}u^{p}(x,t) \biggl[\int _{\Omega }K(x,y)u^{p+1}(y,t)\,dy \biggr]u(x,t)\,dx. \end{aligned}$$
Hence
$$\begin{aligned} &\frac{1}{2}\frac{d}{dt}\int_{\Omega}|u|^{2} \,dx+ \frac {1}{2}\frac {d}{dt}\int_{\Omega}| \nabla{u}|^{2}\,dx+\int_{\Omega}|\nabla {u}|^{2}\,dx+\int_{\Omega}|\triangle{u}|^{2} \,dx \\ &\qquad{} -\int_{\Omega}\int_{\Omega}K(x,y)u^{p+1}(x,t)u^{p+1}(y,t) \,dx \,dy+\alpha{E}(u)-\frac{\alpha}{2}\int_{\Omega}| \nabla{u}|^{2}\,dx \\ &\qquad{} -\frac{\alpha}{2}\int_{\Omega}|\triangle{u}|^{2} \,dx+ \frac {\alpha }{2p+2}\int_{\Omega}\int_{\Omega}K(x,y)u^{p+1}(x,t)u^{p+1}(y,t) \,dx \,dy \\ &\quad=\frac{1}{2}\frac{d}{dt} \biggl[\int_{\Omega}|u|^{2} \,dx+ \int_{\Omega }|\nabla{u}|^{2}\,dx \biggr]+ \alpha{E}(u) \\ &\qquad{} + \biggl(\frac{\alpha}{2p+2}-1 \biggr)\int_{\Omega}\int _{\Omega }K(x,y)u^{p+1}(x,t)u^{p+1}(y,t)\,dx\,dy \\ &\qquad{} + \biggl(1-\frac{\alpha}{2} \biggr)\int_{\Omega}| \nabla{u}|^{2}\,dx+ \biggl(1-\frac {\alpha }{2} \biggr)\int _{\Omega}|\triangle{u}|^{2}\,dx=0. \end{aligned}$$
(3.2)
We consider the following function:
$$\begin{aligned} H(t)=\int_{\Omega} \bigl(|u|^{2}\,dx+| \nabla{u}|^{2} \bigr)\,dx-\eta{E}({0}). \end{aligned}$$
(3.3)
From (3.2), (3.3), Lemma 2.1, and Poincaré’s inequality, we have
$$\begin{aligned} &\frac{1}{2}\frac{d}{dt}H(t) \\ &\quad=\frac{1}{2}\frac{d}{dt}\int_{\Omega } \bigl(|u|^{2}+|\nabla{u}|^{2} \bigr)\,dx \\ &\quad= \biggl(\frac{\alpha}{2}-1 \biggr)\int_{\Omega} \bigl(| \nabla {u}|^{2}+|\triangle {u}|^{2} \bigr)\,dx-\alpha{E}(u) \\ &\qquad{} + \biggl(1-\frac{\alpha}{2p+2} \biggr)\int_{\Omega}\int _{\Omega }K(x,y)u^{p+1}(x,t)u^{p+1}(y,t)\,dx \,dy \\ &\quad= \biggl(\frac{\alpha}{2}-1 \biggr)\int_{\Omega} \bigl(| \nabla {u}|^{2}+|\triangle {u}|^{2} \bigr)\,dx- \alpha{E}(u_{0})+\alpha\int_{0}^{t}\int _{\Omega } \bigl(|u_{t}|^{2}+| \nabla{u}_{t}|^{2} \bigr)\,dx\,d\tau \\ &\qquad{} + \biggl(1-\frac{\alpha}{2p+2} \biggr)\int_{\Omega}\int _{\Omega }K(x,y)u^{p+1}(x,t)u^{p+1}(y,t)\,dx \,dy \\ &\quad\geq \biggl(\frac{\alpha}{2}-1 \biggr)\int_{\Omega} \bigl(|\nabla {u}|^{2}+|\triangle {u}|^{2} \bigr)\,dx- \alpha{E}(u_{0}) \\ &\quad\geq \biggl(\frac{\alpha}{2}-1 \biggr)\lambda_{1}\int _{\Omega} \bigl(|\nabla {u}|^{2}+|{u}|^{2} \bigr)\,dx-\alpha{E}(u_{0}) \\ &\quad=m \biggl[\int_{\Omega} \bigl(|\nabla{u}|^{2}+|{u}|^{2} \bigr)\,dx-\eta {E}(u_{0}) \biggr]=mH(t), \end{aligned}$$
(3.4)
where \(\eta=\frac{\alpha}{m}\); \(m=(\frac{\alpha}{2}-1)\lambda_{1}\); \(2\leq\alpha\leq2p+2\); \(\lambda_{1}\) is the first eigenvalue of operator −△ under homogeneous Dirichlet boundary conditions.
Due to the conditions (3.1), it follows that
$$\begin{aligned} H(0)=\int_{\Omega} \bigl(|u_{0}|^{2}+| \nabla{u}_{0}|^{2} \bigr)\,dx-\eta {E}(u_{0})>0. \end{aligned}$$
(3.5)
Multiplying (3.4) by \(e^{-2mt}\), we have
$$\begin{aligned} e^{-2mt}\frac{d}{dt}H(t)-2me^{-2mt}H(t)=\frac {d}{dt} \bigl[e^{-2mt}H(t) \bigr]\geq0. \end{aligned}$$
From the last inequality above and (3.5), we obtain
$$\begin{aligned} H(t)\geq{H}(0)e^{2mt}>0. \end{aligned}$$
(3.6)
From what has been discussed above, we find
$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\int_{\Omega} \bigl(|u|^{2}+|\nabla {u}|^{2} \bigr)\,dx>\alpha\int _{0}^{t}\int_{\Omega} \bigl(|u_{t}|^{2}+|\nabla {u}_{t}|^{2} \bigr)\,dx\,d\tau. \end{aligned}$$
(3.7)
Now we define
$$\begin{aligned} G(t)=\int_{0}^{t}\int _{\Omega} \bigl(|u|^{2}+|\nabla{u}|^{2} \bigr) \,dx\,d\tau. \end{aligned}$$
(3.8)
Differentiating the identity (3.8) with respect to t, we deduce that
$$\begin{aligned}& G'(t)=\int_{\Omega} \bigl(|u|^{2}+| \nabla{u}|^{2} \bigr)\,dx, \\& G''(t)=\frac{d}{dt}\int_{\Omega} \bigl(|u|^{2}+|\nabla {u}|^{2} \bigr)\,dx\geq2\alpha\int _{0}^{t}\int_{\Omega} \bigl(|u_{t}|^{2}+|\nabla {u}_{t}|^{2} \bigr)\,dx\,d\tau, \end{aligned}$$
so we have
$$\begin{aligned} G''(t)G(t) \geq&2\alpha\int _{0}^{t}\int_{\Omega} \bigl(|u_{t}|^{2}+|\nabla{u}_{t}|^{2} \bigr)\,dx\,d\tau\cdot\int_{0}^{t}\int _{\Omega} \bigl(|u|^{2}+|\nabla{u}|^{2} \bigr) \,dx\,d\tau \\ \geq& 2\alpha\int_{0}^{t}\int _{\Omega}|u_{t}|^{2}\,dx\,d\tau\cdot \int _{0}^{t}\int_{\Omega}|u|^{2} \,dx\,d\tau \\ &{} +2\alpha\int_{0}^{t}\int _{\Omega}|u_{t}|^{2}\,dx\,d\tau\cdot\int _{0}^{t}\int_{\Omega}| \nabla{u}|^{2}\,dx\,d\tau \\ &{} +2\alpha\int_{0}^{t}\int _{\Omega}|\nabla{u}_{t}|^{2}\,dx\,d\tau \cdot \int_{0}^{t}\int_{\Omega}|u|^{2} \,dx\,d\tau \\ &{} +2\alpha\int_{0}^{t}\int _{\Omega}|\nabla{u}_{t}|^{2}\,dx\,d\tau \cdot \int_{0}^{t}\int_{\Omega}| \nabla{u}|^{2}\,dx\,d\tau. \end{aligned}$$
(3.9)
Using Schwarz’s inequality, we get
$$\begin{aligned}& \biggl(\int_{0}^{t}\int_{\Omega}uu_{t} \,dx\,d\tau \biggr)^{2}\leq\int_{0}^{t} \int_{\Omega}|u_{t}|^{2}\,dx\,d\tau\cdot\int _{0}^{t}\int_{\Omega }|u|^{2} \,dx\,d\tau, \end{aligned}$$
(3.10)
$$\begin{aligned}& \biggl(\int_{0}^{t}\int_{\Omega} \nabla{u}\nabla{u}_{t}\,dx\,d\tau \biggr)^{2}\leq\int _{0}^{t}\int_{\Omega}| \nabla{u}_{t}|^{2}\,dx\,d\tau\cdot \int_{0}^{t} \int_{\Omega}|\nabla{u}|^{2}\,dx\,d\tau, \end{aligned}$$
(3.11)
and
$$\begin{aligned} &2\int_{0}^{t}\int _{\Omega}uu_{t}\,dx\,d\tau\cdot\int_{0}^{t} \int_{\Omega}\nabla{u}\nabla{u}_{t}\,dx\,d\tau \\ &\quad\leq2 \biggl(\int_{0}^{t}\int _{\Omega}|u_{t}|^{2}\,dx\,d\tau \biggr)^{\frac {1}{2}}\cdot \biggl(\int_{0}^{t}\int _{\Omega}|u|^{2}\,dx\,d\tau \biggr)^{\frac{1}{2}} \\ &\qquad{} \cdot \biggl(\int_{0}^{t}\int _{\Omega}|\nabla {u}_{t}|^{2}\,dx\,d\tau \biggr)^{\frac{1}{2}}\cdot \biggl(\int_{0}^{t}\int _{\Omega}|\nabla {u}|^{2}\,dx\,d\tau \biggr)^{\frac{1}{2}} \\ &\quad\leq\int_{0}^{t}\int_{\Omega}| \nabla{u}_{t}|^{2}\,dx\,d\tau\cdot \int_{0}^{t} \int_{\Omega}|{u}|^{2}\,dx\,d\tau \\ &\qquad{} +\int_{0}^{t}\int_{\Omega}|u_{t}|^{2} \,dx\,d \tau\cdot\int_{0}^{t}\int _{\Omega}| \nabla{u}|^{2}\,dx\,d\tau. \end{aligned}$$
(3.12)
Inserting (3.10)-(3.12) into (3.9), we find
$$\begin{aligned} G''(t)G(t) \geq&2\alpha \biggl(\int _{0}^{t}\int_{\Omega }uu_{t} \,dx\,d \tau \biggr)^{2}+2\alpha \biggl(\int_{0}^{t} \int_{\Omega }\nabla {u}\nabla{u}_{t}\,dx\,d\tau \biggr)^{2} \\ &{}+4\alpha\int_{0}^{t}\int_{\Omega}uu_{t} \,dx\,d \tau\cdot\int_{0}^{t}\int _{\Omega} \nabla{u}\nabla{u}_{t}\,dx\,d\tau \\ =&2\alpha \biggl[\int_{0}^{t}\int _{\Omega}(uu_{t}+\nabla{u}\nabla {u}_{t})\,dx \,d\tau \biggr]^{2} \\ =&\frac{\alpha}{2} \biggl(\int_{0}^{t}G''( \tau)\,d\tau \biggr)=\frac{\alpha }{2} \bigl(G'(t)-G'(0) \bigr)^{2}. \end{aligned}$$
(3.13)
Thus, we obtain
$$\begin{aligned} G''(t)G(t)-\frac{\alpha}{2} \bigl(G'(t)-G'(0) \bigr)^{2}\geq0. \end{aligned}$$
(3.14)
On the other hand, from (3.6), we know
$$\begin{aligned} \lim_{t\rightarrow\infty}H(t)=+\infty. \end{aligned}$$
This implies
$$\begin{aligned} G'(t)=\int_{\Omega} \bigl[|u|^{2}+|\nabla{u}|^{2} \bigr]\,dx\rightarrow +\infty, \quad t\rightarrow\infty. \end{aligned}$$
(3.15)
Hence, for \(2<\beta<\alpha\) there exists a \(T_{\beta}\), such that for all \(t\geq{T}_{\beta}\)
$$\begin{aligned} \alpha \bigl(G'(t)-G'(0) \bigr)^{2}\geq\beta{G}'(t)^{2}. \end{aligned}$$
(3.16)
By (3.14) and (3.16), we have
$$\begin{aligned} G''(t)G(t)-\frac{\beta}{2}{G}'(t)^{2} \geq0,\quad t\geq{T}_{\beta}. \end{aligned}$$
(3.17)
We consider the function \(G(t)^{-q}\) for \(0< q<\frac{\beta}{2}\), we see that
$$\begin{aligned} \bigl(G(t)^{-q} \bigr)'' =&qG(t)^{-q-2} \bigl[(q+1){G}'(t)^{2}-G''(t)G(t) \bigr] \\ \leq&{q}G(t)^{-q-2} \biggl[\frac{2(q+1)}{\beta}-1 \biggr]G''(t)G(t)< 0, \quad t\geq{T}_{\beta}. \end{aligned}$$
(3.18)
Since a concave function must always lie below any tangent line, we see that \(G(t)^{-q}\) reaches 0 in finite time as \(t\rightarrow{T}^{-}\), where \(T>T_{\beta}\). This means
$$\begin{aligned} \lim_{t\rightarrow{T}^{-}}G(t)=+\infty, \end{aligned}$$
or
$$\begin{aligned} \lim_{t\rightarrow{T}^{-}}\int_{0}^{t} \int_{\Omega} \bigl(|u|^{2}+|\nabla {u}|^{2} \bigr)\,dx\,d\tau=+\infty. \end{aligned}$$
(3.19)
Then the desired assertion immediately follows. □