In order to state and prove our result, we introduce
$$\begin{aligned} Z =& L^{\infty}\bigl([0,T); W^{1,\alpha} (\Omega) \bigr) \cap W^{1,\infty} \bigl( [0,T) ; L^{2} (\Omega) \bigr) \\ & {}\cap W^{1,\beta} \bigl( [0,T); W^{1,\beta} (\Omega)\bigr)\cap W^{1,m} \bigl( [0,T) ; L^{m} (\Omega) \bigr) \end{aligned}$$
for \(T>0\) and the energy functional
$$\begin{aligned}& E(t)= {1 \over 2} \int_{\Omega} u_{t}^{2} \,dx + {1\over \alpha} \int_{\Omega} \vert \nabla u\vert ^{\alpha} \,dx - {b \over p} \int_{\Omega} \vert u\vert ^{p} \,dx + {1 \over 2} \int_{\Gamma_{1}}h(x)q(x)y^{2} (t) \,d\Gamma. \end{aligned}$$
(2.1)
Theorem 2.1
Assume that
\(\alpha, \beta, m, p \geq2\)
such that
\(\beta<\alpha\), and
\(max\{m,\alpha\} < p< r_{\alpha}\), where
\(r_{\alpha}\)
is the Sobolev critical exponent of
\(W^{1, \alpha}(\Omega)\). Assume further that
Then the solution
\((u,y) \in Z \times L^{2}(R^{+}; L^{2}(\Gamma_{1}))\)
of (1.1)-(1.6) can not exist for all time.
Remark 2.2
If the solution u of (1.1)-(1.6) is smooth enough, then it blows up in finite time.
Proof
We suppose that the solution exists for all time, and we reach a contradiction. For this purpose, we multiply Eq. (1.1) by \(u_{t}\) and, using (1.2)-(1.4), we obtain
$$\begin{aligned} E^{\prime}(t) =& - \int_{\Omega} \bigl\vert \nabla u_{t}(t)\bigr\vert ^{2} \,dx - \int_{\Omega} \bigl\vert \nabla u_{t}(t)\bigr\vert ^{\beta} \,dx \\ &{}-a \int_{\Omega} \bigl\vert u_{t}(t)\bigr\vert ^{m} \,dx - \int _{\Gamma_{1}}h(x)f(x)y_{t}^{2} (t) \,d\Gamma \leq0 \end{aligned}$$
(2.3)
for any regular solution. Hence we get \(E(t) \leq E(0)\)
\(\forall t \geq0\).
By setting \(H(t)=-E(t)\), we deduce
$$ 0 < H(0) \leq H(t) \leq{b \over p} \int_{\Omega} \bigl\vert u(t)\bigr\vert ^{p} \,dx,\quad \forall \geq0. $$
(2.4)
Now, we define
$$\begin{aligned}& L(t)= H^{1-\sigma}(t) +\varepsilon \int_{\Omega} u(t)u_{t}(t) \,dx -{\varepsilon\over 2} \int_{\Gamma_{1}} h(x)f(x)y^{2} (t) \,d\Gamma -\varepsilon \int_{\Gamma_{1}}h(x)u(t)y(t) \,d\Gamma \end{aligned}$$
(2.5)
for ε small to be chosen later and
$$ 0 < \sigma\leq \min\biggl\{ \frac{\alpha-2}{p} , \frac{\alpha-\beta}{ p(\beta-1)}, \frac{p -m}{p(m -1)} , \frac{\alpha-2}{2\alpha} \biggr\} . $$
(2.6)
Our goal is to show that \(L(t)\) satisfies a differential inequality of the form
$$ L^{\prime}(t) \geq\xi L^{q}(t),\quad q>1. $$
(2.7)
This, of course, will lead to a blow-up in finite time.
By taking a derivative of (2.5), we get
$$\begin{aligned} L^{\prime}(t) =& (1-\sigma)H^{-\sigma}(t)H^{\prime}(t)+ \varepsilon \int _{\Omega} u_{t}^{2}(t) \,dx +\varepsilon \int_{\Omega} u(t)u_{tt}(t) \,dx \\ & {}-{\varepsilon} \int_{\Gamma_{1}} h(x)f(x)y(t)y_{t}(t) \,d\Gamma -\varepsilon \int_{\Gamma_{1}}h(x)u_{t}(t)y(t) \,d\Gamma \\ &{}-\varepsilon \int _{\Gamma_{1}}h(x)u(t)y_{t}(t) \,d\Gamma. \end{aligned}$$
(2.8)
By using Eqs. (1.1)-(1.4), estimate (2.8) becomes
$$\begin{aligned} L^{\prime}(t) = & (1-\sigma)H^{-\sigma}(t)H^{\prime}(t)+ \varepsilon \int _{\Omega} u_{t}^{2}(t) \,dx \\ & {}+\varepsilon \int_{\Omega} u(t) \bigl[ \Delta u_{t}(t) + \operatorname{div} \bigl( \bigl\vert \nabla u(t)\bigr\vert ^{\alpha-2} \nabla u(t)\bigr) +\operatorname{div} \bigl( \bigl\vert \nabla u_{t}(t)\bigr\vert ^{\beta-2} \nabla u_{t}(t)\bigr) \\ & {}-a \bigl\vert u_{t}(t)\bigr\vert ^{m -2} u_{t}(t) +b\bigl\vert u(t)\bigr\vert ^{p -2} u(t)\bigr] \,dx -{\varepsilon} \int_{\Gamma_{1}} h(x)f(x)y(t)y_{t}(t) \,d\Gamma \\ & {}-\varepsilon \int_{\Gamma_{1}}h(x)u_{t}(t)y(t) \,d\Gamma-\varepsilon \int _{\Gamma_{1}}h(x)u(t)y_{t}(t) \,d\Gamma \\ =& (1-\sigma)H^{-\sigma}(t)H^{\prime}(t)+\varepsilon \int_{\Omega} u_{t}^{2}(t) \,dx -\varepsilon \int_{\Omega} \nabla u_{t}(t) \nabla u(t) \,dx \\ & {}-\varepsilon \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx -\varepsilon \int_{\Omega}\bigl(\bigl\vert \nabla u_{t}(t)\bigr\vert ^{\beta-2} \nabla u_{t}(t)\bigr) \nabla u(t) \,dx \\ & {}-a\varepsilon \int_{\Omega} \bigl\vert u_{t}(t)\bigr\vert ^{m -2} u_{t}(t)u(t) \,dx +b\varepsilon \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx \\ & {}+ \varepsilon \int_{\Gamma_{1}}\biggl( \frac{\partial u_{t}(t)}{\partial\nu} + \bigl\vert \nabla u(t) \bigr\vert ^{\alpha-2} \frac{\partial u(t)}{\partial\nu} + \bigl\lvert \nabla u_{t}(t)\bigr\rvert ^{\beta-2} \frac{\partial u_{t}(t)}{\partial\nu}\biggr)u(t) \,d \Gamma \\ & {}-\varepsilon \int_{\Gamma_{1}} h(x)f(x)y(t)y_{t}(t) \,d\Gamma -\varepsilon \int_{\Gamma_{1}}h(x)u_{t}(t)y(t) \,d\Gamma-\varepsilon \int _{\Gamma_{1}}h(x)u(t)y_{t}(t) \,d\Gamma \\ = & (1-\sigma)H^{-\sigma}(t)H^{\prime}(t)+\varepsilon \int_{\Omega} u_{t}^{2}(t) \,dx -\varepsilon \int_{\Omega} \nabla u_{t}(t) \nabla u(t) \,dx \\ & {}-\varepsilon \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx -\varepsilon \int_{\Omega}\bigl(\bigl\vert \nabla u_{t}(t)\bigr\vert ^{\beta-2} \nabla u_{t}(t)\bigr) \nabla u(t) \,dx \\ & {}-a\varepsilon \int_{\Omega} \bigl\vert u_{t}(t)\bigr\vert ^{m -2} u_{t}(t)u(t) \,dx +b\varepsilon \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx + \varepsilon \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma. \end{aligned}$$
(2.9)
Exploiting Hölder’s and Young’s inequalities, for any \(\eta,\mu ,\delta>0\), we obtain
$$\begin{aligned}& \int_{\Omega} \bigl\vert u_{t}(t)\bigr\vert ^{m -2} u_{t}(t)u(t) \,dx \leq\frac{\eta^{m}}{m} \int _{\Omega} \bigl\vert u(t)\bigr\vert ^{m } \,dx + \frac{m-1}{m} \eta^{-\frac{m}{m-1}} \int _{\Omega} \bigl\vert u_{t}(t)\bigr\vert ^{m} \,dx, \end{aligned}$$
(2.10)
$$\begin{aligned}& \int_{\Omega} \nabla u_{t}(t) \nabla u(t) \,dx \leq \frac{1}{4\mu} \int _{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{2} \,dx + \mu \int_{\Omega} \bigl\vert \nabla u_{t}(t)\bigr\vert ^{2} \,dx, \end{aligned}$$
(2.11)
$$\begin{aligned}& \int_{\Omega}\bigl\vert \nabla u_{t}(t)\bigr\vert ^{\beta-2} \nabla u_{t}(t) \nabla u(t) \,dx \leq\frac{\delta^{\beta}}{\beta} \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\beta} \,dx + \frac{\beta-1}{\beta} \delta^{-\frac{\beta}{\beta-1}} \int_{\Omega }\bigl\vert \nabla u_{t}(t)\bigr\vert ^{\beta} \,dx. \end{aligned}$$
(2.12)
A substitution of (2.10)-(2.12) in (2.9) yields
$$\begin{aligned} L^{\prime}(t) \geq & (1-\sigma)H^{-\sigma}(t)H^{\prime }(t)+ \varepsilon \int_{\Omega} u_{t}^{2}(t) \,dx - \frac{\varepsilon}{4\mu} \int _{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{2} \,dx \\ & {}- \varepsilon\mu \int_{\Omega} \bigl\vert \nabla u_{t}(t)\bigr\vert ^{2} \,dx -\varepsilon \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx - \frac {\varepsilon\delta^{\beta}}{\beta} \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\beta} \,dx \\ & {}- \frac{\varepsilon(\beta-1)}{\beta} \delta^{-\frac{\beta}{\beta -1}} \int_{\Omega}\bigl\vert \nabla u_{t}(t)\bigr\vert ^{\beta} \,dx -\frac{a\varepsilon\eta ^{m}}{m} \int_{\Omega} \bigl\vert u(t)\bigr\vert ^{m } \,dx \\ & {}- \frac{a\varepsilon(m-1)}{m} \eta^{-\frac{m}{m-1}} \int_{\Omega} \bigl\vert u_{t}(t)\bigr\vert ^{m} \,dx +b\varepsilon \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx \\ &{}+ \varepsilon \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma. \end{aligned}$$
(2.13)
Therefore, by choosing \(\eta,\mu,\delta\) so that
$$\begin{aligned}& \eta^{-\frac{m}{m-1}} = M_{1} H^{-\sigma}(t), \\ & \mu= M_{2} H^{-\sigma}(t), \\ & \delta^{-\frac{\beta}{\beta-1}}= M_{3} H^{-\sigma}(t) \end{aligned}$$
for \(M_{1}, M_{2}, M_{3}\) to be specified later, and using (2.13), we arrive at
$$\begin{aligned} L^{\prime}(t) \geq & (1-\sigma)H^{-\sigma}(t)H^{\prime }(t)+ \varepsilon \int_{\Omega} u_{t}^{2}(t) \,dx - \frac{\varepsilon }{4M_{2}}H^{\sigma}(t) \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{2} \,dx \\ &{}-\varepsilon \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx -\frac {\varepsilon M_{3}^{-(\beta-1)}}{\beta}H^{\sigma(\beta-1)}(t) \int _{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\beta} \,dx \\ & {}-\frac{a\varepsilon}{m} M_{1} ^{-(m-1)}H^{\sigma(m-1)}(t) \int_{\Omega } \bigl\vert u(t)\bigr\vert ^{m } \,dx +b \varepsilon \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx \\ & {}-\varepsilon \biggl[ M_{2} \int_{\Omega} \bigl\vert \nabla u_{t}(t)\bigr\vert ^{2} \,dx + \frac{\beta-1}{\beta}M_{3} \int_{\Omega}\bigl\vert \nabla u_{t}(t)\bigr\vert ^{\beta} \,dx \\ &{}+\frac{a(m-1)}{m} M_{1} \int_{\Omega} \bigl\vert u_{t}(t)\bigr\vert ^{m} \,dx \biggr] H^{-\sigma}(t) +\varepsilon \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma. \end{aligned}$$
(2.14)
If \(M = M_{2}+ \frac{(\beta-1)M_{3}}{\beta}+\frac{(m-1)M_{1}}{m}\), then (2.14) takes the form
$$\begin{aligned} L^{\prime}(t) \geq & (1-\sigma-\varepsilon M)H^{-\sigma}(t)H^{\prime }(t)+\varepsilon \int_{\Omega} u_{t}^{2}(t) \,dx - \frac{\varepsilon }{4M_{2}}H^{\sigma}(t) \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{2} \,dx \\ &{}-\varepsilon \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx -\frac {\varepsilon M_{3}^{-(\beta-1)}}{\beta}H^{\sigma(\beta-1)}(t) \int _{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\beta} \,dx \\ & {}-\frac{a\varepsilon}{m} M_{1} ^{-(m-1)}H^{\sigma(m-1)}(t) \int_{\Omega } \bigl\vert u(t)\bigr\vert ^{m } \,dx +b \varepsilon \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx \\ & {}+\varepsilon M H^{-\sigma}(t) \int_{\Gamma_{1}}h(x)f(x)y_{t}^{2}(t) \,d\Gamma + \varepsilon \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma. \end{aligned}$$
(2.15)
Then we use the embedding \(L^{p}(\Omega) \hookrightarrow L^{m}(\Omega)\) and (2.4) to get
$$ H^{\sigma(m-1)}(t) \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{m} \,dx \leq \biggl(\frac{b}{p}\biggr)^{\sigma (m-1)} \biggl( \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx \biggr)^{\frac{m+\sigma p (m-1)}{p}}. $$
(2.16)
We also exploit the inequality
$$\int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{2} \,dx \leq c \biggl( \int_{\Omega}\bigl\lvert \nabla u(t)\bigr\rvert ^{\alpha}\,dx \biggr)^{\frac{2}{\alpha}}, $$
the embedding \(W^{1,\alpha}(\Omega) \hookrightarrow H^{1} (\Omega)\) and (2.4) to obtain
$$\begin{aligned} & H^{\sigma}(t) \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{2} \,dx \leq c\biggl(\frac {b}{p}\biggr)^{\sigma} \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{\frac {p\sigma+2}{\alpha}}. \end{aligned}$$
(2.17)
Since \(\alpha> \beta\), we obtain
$$\begin{aligned} & \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\beta}\,dx \leq c \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{\frac{\beta}{\alpha}}, \end{aligned}$$
we derive
$$\begin{aligned} & H^{\sigma(\beta-1)} (t) \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\beta}\,dx \leq c\biggl(\frac{b}{p}\biggr)^{\sigma(\beta-1)} \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{\frac{p\sigma(\beta-1)+\beta}{\alpha}}, \end{aligned}$$
(2.18)
where c is a constant depending on Ω only. By using (2.6) and the inequality
$$\begin{aligned}& z^{\nu} \leq z+1 \leq\biggl(1+ \frac{1}{a}\biggr) (z+a),\quad \forall z \geq0, 0 < \nu< 1, a \geq0, \end{aligned}$$
(2.19)
we get the following inequalities:
$$\begin{aligned}& \biggl( \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx \biggr)^{\frac{m+\sigma p (m-1)}{p}} \leq c \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{\frac{m+\sigma p (m-1)}{\alpha}} \\& \phantom{\biggl( \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx \biggr)^{\frac{m+\sigma p (m-1)}{p}}} \leq d \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx+ H(0) \biggr) \\& \phantom{\biggl( \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx \biggr)^{\frac{m+\sigma p (m-1)}{p}}} \leq d \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx+ H(t) \biggr),\quad \forall t \geq0, \end{aligned}$$
(2.20)
$$\begin{aligned}& \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{\frac{p\sigma +2}{\alpha}} \leq d \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx + H(t) \biggr),\quad \forall t \geq0, \end{aligned}$$
(2.21)
and
$$\begin{aligned}& \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{\frac{p\sigma (\beta-1) +\beta}{\alpha}} \leq d \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx + H(t) \biggr),\quad \forall t \geq0, \end{aligned}$$
(2.22)
where \(d=1+1/H(0), a=H(0)\). Inserting (2.16)-(2.18) and (2.20)-(2.22) into (2.15), we deduce
$$\begin{aligned} L^{\prime}(t) \geq & (1-\sigma-\varepsilon M )H^{-\sigma }(t)H^{\prime}(t) \\ &{}+ kH(t) +\biggl(\varepsilon+\frac{k}{2}\biggr) \int_{\Omega} u_{t}^{2}(t) \,dx \\ &{}-\frac{\varepsilon c_{2}}{ M_{2}} \biggl( \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx +H(t) \biggr) -\varepsilon \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx \\ &{}-\frac{\varepsilon c_{3}}{ M_{3}^{\beta-1}} \biggl( \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx +H(t) \biggr) + \frac{k}{\alpha} \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx \\ &{}-\frac{\varepsilon c_{1}}{ M_{1}^{m -1}} \biggl( \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx +H(t) \biggr) +b\biggl(\varepsilon-\frac{k}{p}\biggr) \int _{\Omega}\bigl\vert u(t)\bigr\vert ^{p} \,dx \\ & {}+\varepsilon M H^{-\sigma}(t) \int_{\Gamma_{1}}h(x)f(x)y_{t}^{2}(t) \,d\Gamma + \biggl(\varepsilon+\frac{k}{2}\biggr) \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma \end{aligned}$$
for some constant k and \(c_{1}= \frac{acd}{m} (\frac{b}{p})^{\sigma (m-1)}\), \(c_{2}= \frac{cd}{4} (\frac{b}{p})^{\sigma}\), \(c_{3}= \frac {cd}{\beta} (\frac{b}{p})^{\sigma(\beta-1)}\).
Using \(k=\varepsilon p\), we arrive at
$$\begin{aligned} L^{\prime}(t) \geq & (1-\sigma-\varepsilon M )H^{-\sigma }(t)H^{\prime}(t) + \varepsilon\biggl(\frac{p+2}{2} \biggr) \int_{\Omega} u_{t}^{2}(t) \,dx \\ & {}+\varepsilon \biggl( p - \frac{ c_{2}}{ M_{2}} -\frac{c_{3}}{ M_{3}^{\beta -1}} - \frac{c_{1}}{ M_{1}^{m -1}} \biggr) H(t) \\ & {}+ \varepsilon \biggl( \frac{p}{\alpha} - \frac{ c_{2}}{ M_{2}} - \frac {c_{3}}{ M_{3}^{\beta-1}} -\frac{c_{1}}{ M_{1}^{m -1}} -1 \biggr) \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx \\ & {}+\varepsilon M H^{-\sigma}(t) \int_{\Gamma_{1}}h(x)f(x)y_{t}^{2}(t) \,d\Gamma + \varepsilon\biggl(\frac{p+2}{2}\biggr) \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma. \end{aligned}$$
At this point, by choosing \(M_{1}, M_{2}, M_{3}\) large enough and using
$$\varepsilon M H^{-\sigma}(t) \int_{\Gamma_{1}}h(x)f(x)y_{t}^{2}(t) \,d\Gamma >0, $$
we have
$$\begin{aligned} L^{\prime}(t) \geq & (1-\sigma-\varepsilon M )H^{-\sigma }(t)H^{\prime}(t) \\ & {}+r\varepsilon \biggl( H(t) + \int_{\Omega} u_{t}^{2}(t) \,dx + \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx + \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma \biggr), \end{aligned}$$
(2.23)
where r is a positive constant (this is possible since \(p>\alpha\)).
We choose \(0<\varepsilon< \frac{1-\sigma}{M}\) so that
$$\begin{aligned} L(0) = H^{1-\sigma}(0)+\varepsilon \int_{\Omega} u_{0}u_{1} \,dx - \frac {\varepsilon}{2} \int_{\Gamma_{1}}h(x)f(x)y_{0}^{2} \,d\Gamma - \varepsilon \int _{\Gamma_{1}}h(x)u_{0}y_{0} \,d\Gamma>0. \end{aligned}$$
Then from (2.23) we get
$$\begin{aligned} L(t) \geq L(0) >0, \quad\forall t \geq0, \end{aligned}$$
and
$$\begin{aligned} & L^{\prime}(t) \geq r\varepsilon \biggl( H(t) + \int_{\Omega} u_{t}^{2}(t) \,dx + \int_{\Omega} \bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx + \int_{\Gamma _{1}}h(x)q(x)y^{2}(t) \,d\Gamma \biggr). \end{aligned}$$
(2.24)
On the other hand, from (2.5) and \(f, h >0\), we have
$$\begin{aligned} L(t)\leq H^{1-\sigma}(t) +\varepsilon \int_{\Omega} u(t)u_{t}(t) \,dx -\varepsilon \int_{\Gamma_{1}}h(x)u(t)y(t) \,d\Gamma. \end{aligned}$$
Consequently, the above estimate leads to
$$\begin{aligned} L^{\frac{1}{1-\sigma}}(t)\leq C(\varepsilon,\sigma) \biggl[ H(t) + \biggl( \int_{\Omega} u(t)u_{t}(t) \,dx \biggr)^{\frac{1}{1-\sigma}} + \biggl( \int _{\Gamma_{1}}h(x)u(t)y(t) \,d\Gamma \biggr)^{\frac{1}{1-\sigma}} \biggr]. \end{aligned}$$
(2.25)
From Hölder’s inequality, we obtain
$$\begin{aligned} \int_{\Omega} u(t)u_{t}(t) \,dx \leq & \biggl( \int_{\Omega}u_{t}^{2}(t) \,dx \biggr)^{\frac{1}{2}} \biggl( \int_{\Omega}u^{2}(t) \,dx \biggr)^{\frac{1}{2}} \\ \leq & c \biggl( \int_{\Omega}u_{t}^{2}(t) \,dx \biggr)^{\frac{1}{2}} \biggl( \int_{\Omega }\bigl\vert u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{\frac{1}{\alpha}}, \end{aligned}$$
where c is the positive constant which comes from the embedding \(L^{\alpha}(\Omega) \hookrightarrow L^{2}(\Omega)\). This inequality implies that there exists a positive constant \(c_{4}>0\) such that
$$\begin{aligned} \biggl( \int_{\Omega} u(t)u_{t}(t) \,dx \biggr)^{\frac{1}{1-\sigma}} \leq c_{4} \biggl( \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{\alpha} \,dx \biggr)^{\frac{1}{(1-\sigma)\alpha }} \biggl( \int_{\Omega}u_{t}^{2}(t) \,dx \biggr)^{\frac{1}{2(1-\sigma)}}. \end{aligned}$$
Applying Young’s inequality to the right-hand side of the preceding inequality, we have a positive constant, also denoted by \(c>0\), such that
$$\begin{aligned} \biggl( \int_{\Omega} u(t)u_{t}(t) \,dx \biggr)^{\frac{1}{1-\sigma}} \leq c \biggl[ \biggl( \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{\alpha} \,dx \biggr)^{\frac{\mu}{(1-\sigma )\alpha}} + \biggl( \int_{\Omega}u_{t}^{2}(t) \,dx \biggr)^{\frac{\theta }{2(1-\sigma)}} \biggr] \end{aligned}$$
for \(\frac{1}{\mu} + \frac{1}{\theta}=1\). We take \(\theta=2(1-\sigma)\), hence \(\mu=2(1-\sigma)/(1-2\sigma)\), to get
$$\begin{aligned} \biggl( \int_{\Omega} u(t)u_{t}(t) \,dx \biggr)^{\frac{1}{1-\sigma}} \leq c \biggl[ \biggl( \int_{\Omega}\bigl\vert u(t)\bigr\vert ^{\alpha} \,dx \biggr)^{\frac{2}{(1-2\sigma )\alpha}} + \int_{\Omega}u_{t}^{2}(t) \,dx \biggr]. \end{aligned}$$
By Poincare’s inequality, we obtain
$$\begin{aligned} \biggl( \int_{\Omega} u(t)u_{t}(t) \,dx \biggr)^{\frac{1}{1-\sigma}} \leq c \biggl[ \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha} \,dx \biggr)^{\frac {2}{(1-2\sigma)\alpha}} + \int_{\Omega}u_{t}^{2}(t) \,dx \biggr]. \end{aligned}$$
We use (2.6) and the algebraic inequality (2.19) with \(z=\Vert \nabla u(t) \Vert _{\alpha}^{\alpha}\), \(d= 1+ 1/H(0)\), \(a=H(0)\), \(\nu =2/{\alpha(1-2\sigma)}\), condition (2.6) on σ ensures that \(0<\nu<1\), and it follows that
$$\begin{aligned} z^{\nu}\leq d\bigl(z+H(0)\bigr) \leq d\bigl(z+H(t)\bigr). \end{aligned}$$
Therefore, from (2.20), there exists a positive constant, denoted by \(c_{4}\), such that for all \(t \geq0\),
$$\begin{aligned} \biggl( \int_{\Omega} u(t)u_{t}(t) \,dx \biggr)^{\frac{1}{1-\sigma}} \leq c_{4} \bigl[H(t) + \bigl\Vert \nabla u(t)\bigr\Vert _{\alpha}^{\alpha} + \bigl\Vert u_{t}(t)\bigr\Vert _{2}^{2} \bigr]. \end{aligned}$$
(2.26)
Furthermore, by the same method, we have
$$\begin{aligned} \int_{\Gamma_{1}}h(x)u(t)y(t) \,d\Gamma =& \biggl\vert \int_{\Gamma_{1}}\frac {h(x)q(x)}{q(x)}u(t)y(t) \,d\Gamma\biggr\vert \\ \leq & \frac{\Vert h\Vert _{\infty}^{{1}\over {2}} \Vert q\Vert _{\infty}^{{1}\over {2}}}{q_{0}} \biggl( \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma \biggr)^{{1}\over {2}} \biggl( \int_{\Gamma_{1}}u^{2}(t) \,d\Gamma \biggr)^{{1}\over {2}}. \end{aligned}$$
Using the embedding \(W_{0}^{1,\alpha}(\Omega) \hookrightarrow L^{2} (\Gamma _{1})\) and Hölder’s inequality, we get
$$\begin{aligned} \int_{\Gamma_{1}}h(x)u(t)y(t) \,d\Gamma \leq c_{5} \frac{\Vert h\Vert _{\infty}^{{1}\over {2}} \Vert q\Vert _{\infty}^{{1}\over {2}}}{q_{0}} \biggl( \int_{\Gamma _{1}}h(x)q(x)y^{2}(t) \,d\Gamma \biggr)^{{1}\over {2}} \biggl( \int_{\Omega }\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{{1}\over {\alpha}}. \end{aligned}$$
Consequently, there exists a positive constant \(c_{5}=c_{5}( \Vert h\Vert _{\infty}, \Vert q\Vert _{\infty}, q_{0}, \sigma, \alpha)\) such that
$$\begin{aligned} \biggl( \int_{\Gamma_{1}}h(x)u(t)y(t) \,d\Gamma \biggr)^{{1} \over {1-\sigma}} \leq c_{5} \biggl( \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma \biggr)^{{1}\over {2(1-\sigma)}} \biggl( \int_{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{{1}\over {\alpha(1-\sigma)}}. \end{aligned}$$
Using Young’s inequality exactly as in (2.26), we write
$$\begin{aligned} \biggl( \int_{\Gamma_{1}}h(x)u(t)y(t) \,d\Gamma \biggr)^{{1} \over {1-\sigma}} \leq c_{6} \biggl[ \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma+ \biggl( \int _{\Omega}\bigl\vert \nabla u(t)\bigr\vert ^{\alpha}\,dx \biggr)^{{2}\over {\alpha(1-2\sigma)}} \biggr], \end{aligned}$$
where \(c_{6}\) is a positive constant depending on \(c_{5}\) and α. Consequently, applying once again the algebraic inequality (2.19) with \(z=\Vert \nabla u(t)\Vert _{\alpha}^{\alpha}\), \(\nu=2/{\alpha (1-2\sigma)}\) and making use of (2.6), we obtain by the same method as above
$$\begin{aligned} \biggl( \int_{\Gamma_{1}}h(x)u(t)y(t) \,d\Gamma \biggr)^{{1} \over {1-\sigma}} \leq c_{7} \biggl[ H(t) + \bigl\Vert \nabla u(t)\bigr\Vert _{\alpha}^{\alpha}+ \int_{\Gamma _{1}}h(x)q(x)y^{2}(t) \,d\Gamma \biggr], \end{aligned}$$
(2.27)
where \(c_{7}\) is a positive constant. From (2.25), (2.26) and (2.27), we arrive at
$$\begin{aligned} L^{{1} \over {1-\sigma}}(t) \leq c \biggl[ H(t) + \bigl\Vert \nabla u(t) \bigr\Vert _{\alpha}^{\alpha}+\bigl\Vert u_{t}(t)\bigr\Vert _{2}^{2} + \int_{\Gamma_{1}}h(x)q(x)y^{2}(t) \,d\Gamma \biggr], \end{aligned}$$
(2.28)
where c is a positive constant. Consequently, a combination of (2.24) and (2.28), for some \(\xi>0\), yields
$$\begin{aligned} L^{\prime}(t) \geq\xi L^{{1} \over {1-\sigma}}(t), \quad\forall t \geq0. \end{aligned}$$
(2.29)
Integration of (2.29) over \((0,t)\) gives
$$\begin{aligned} L^{{\sigma} \over {1-\sigma}}(t) \geq\frac{1}{ L^{-{\sigma} \over {1-\sigma}}(0)-{{\xi\sigma} \over {1-\sigma}}t},\quad \forall t \geq0. \end{aligned}$$
Hence \(L(t)\) blows up in finite time
$$\begin{aligned} T^{*} \leq \frac{1-\sigma}{\xi\sigma L^{{\sigma} \over {1-\sigma}}(0)}. \end{aligned}$$
Thus the proof of Theorem 2.1 is complete. □