Open Access

Global nonexistence of solutions for a quasilinear wave equation with acoustic boundary conditions

Boundary Value Problems20172017:42

https://doi.org/10.1186/s13661-017-0773-1

Received: 7 December 2016

Accepted: 13 February 2017

Published: 28 March 2017

Abstract

We consider the quasilinear wave equation
$$u_{tt} -\triangle u_{t} -\operatorname{div}\bigl(\vert \nabla u\vert ^{\alpha-2} \nabla u\bigr) - \operatorname{div}\bigl(\vert \nabla u_{t}\vert ^{\beta-2} \nabla u_{t} \bigr) +a \vert u_{t}\vert ^{m-2} u_{t} =b|u|^{p-2} u $$
\(a,b>0\), associated with initial and Dirichlet boundary conditions at one part and acoustic boundary conditions at another part, respectively. We prove, under suitable conditions on α, β, m, p and for negative initial energy, a global nonexistence of solutions.

Keywords

quasilinear wave equationblow-upacoustic boundary

MSC

35B4035B4435L72

1 Introduction

In this paper, we consider the following quasilinear wave equation with acoustic boundary conditions:
$$\begin{aligned}& u_{tt} -\triangle u_{t} -\operatorname{div}\bigl(\vert \nabla u\vert ^{\alpha-2} \nabla u\bigr) - \operatorname{div}\bigl(\vert \nabla u_{t}\vert ^{\beta-2} \nabla u_{t} \bigr) \\& \quad{}+a\vert u_{t}\vert ^{m-2} u_{t} =b \vert u\vert ^{p-2} u\quad \text{in } \Omega\times (0,\infty), \end{aligned}$$
(1.1)
$$\begin{aligned}& u=0\quad \text{on } \Gamma_{0} \times(0,\infty), \end{aligned}$$
(1.2)
$$\begin{aligned}& \frac{\partial u_{t}}{\partial\nu} + \vert \nabla u\vert ^{\alpha-2} \frac {\partial u}{\partial\nu}+ \vert \nabla u_{t}\vert ^{\beta-2} \frac{\partial u_{t}}{\partial\nu} = h(x)y_{t}\quad \text{on } \Gamma_{1} \times(0,\infty), \end{aligned}$$
(1.3)
$$\begin{aligned}& u_{t} +f(x)y_{t} +q(x)y = 0 \quad\text{on } \Gamma_{1} \times(0,\infty), \end{aligned}$$
(1.4)
$$\begin{aligned}& u(x,0)=u_{0}(x), \qquad u_{t}(x,0)=u_{1}(x) \quad\text{in } \Omega, \end{aligned}$$
(1.5)
$$\begin{aligned}& y(x,0) = y_{0}(x)\quad \mbox{on } \Gamma_{1} \times(0,\infty), \end{aligned}$$
(1.6)
where \(a,b>0, \alpha,\beta,m,p >2\), Ω is a regular and bounded domain of \(R^{n}(n \geq1)\) and \(\partial\Omega(=\Gamma):= \Gamma_{0} \cup \Gamma_{1}\). Here \(\Gamma_{0}, \Gamma_{1}\) are closed and disjoint, and \(\frac {\partial}{\partial\nu}\) denotes the unit outer normal derivative. The functions \(f,q,h : \Gamma _{1} \longrightarrow R^{+}\) are essentially bounded and \(0< q_{0} \leq q(x)\) on \(\Gamma_{1}\).

The system (1.1)-(1.6) is a model of a quasilinear wave equation with acoustic boundary conditions. The acoustic boundary conditions were introduced by Morse and Ingard [1] in 1968 and developed by Beale and Rosencrans in [2], where the authors proved the global existence and regularity of the linear problem. Furthermore, Boukhatem and Benabderrahmane [3, 4] studied the existence, blow-up and decay of solutions for viscoelastic wave equations with acoustic boundary conditions. Graber and Said-Houari [5] studied the blow-up solutions for the wave equation with semilinear porous acoustic boundary conditions. Moreover, Wu [6] also considered blow-up solutions for a nonlinear wave equation with porous acoustic boundary conditions. The global nonexistence of solutions for a class of wave equations with nonlinear damping and source terms was proved by Messaoudi and Said-Houari [79] (see [1013] for more details). Recently, Piskin [14] investigated the energy decay and blow-up of solutions for quasilinear hyperbolic equations with nonlinear damping and source terms (see [1518] for more details).

Motivated by the previous works, in this paper, we study the global nonexistence of solutions for quasilinear wave equations with acoustic boundary conditions. To the best of our knowledge, there are no results of a quasilinear wave equation with acoustic boundary conditions. This work is meaningful. The outline of the paper is the following. In Section 2, we prove the main result.

2 Blow-up results

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
$$ E(0) < 0. $$
(2.2)
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. □

Declarations

Acknowledgements

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2016R1A1B03930361).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Division of Mathematical Sciences, Pukyong National University
(2)
Department of Mathematics, Pusan National University
(3)
Institute of Liberal Education, Catholic University of Daegu

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