On the decay and blow up of solutions for a quasilinear hyperbolic equations with nonlinear damping and source terms

Abstract

In this work we investigate the global existence, decay, and blow up of solutions for a quasilinear hyperbolic equation. We prove the decay estimates of the energy function by using Nakao’s inequality. Also, we obtain the blow up of solutions and lifespan estimates in three different ranges of the initial energy.

1 Introduction

In this work we study the following quasilinear hyperbolic equations:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} u_{tt}-\operatorname{div} ( \vert \nabla u\vert ^{m}\nabla u ) -\bigtriangleup u_{t}+\vert u_{t}\vert ^{q-1}u_{t}=\vert u\vert ^{p-1}u,& x\in\Omega, t>0, \\ u ( x,0 ) =u_{0} ( x ) ,\qquad u_{t} ( x,0 ) =u_{1} ( x ) , & x\in\Omega, \\ u ( x,t ) =0,& x\in\partial\Omega, t>0 , \end{array}\displaystyle \right . $$

(1)

where Ω is a bounded domain with smooth boundary ∂Ω in $R^{n}$ ($n\geq1$); $m>0$, $p,q\geq1$.

Problems of this type arise in physics. For example, this problem represents the longitudinal motion of a viscoelastic configuration which obeys a nonlinear Voight model [1, 2].

When $m=0$, (1) becomes the following wave equation with nonlinear and strong damping terms:

$$ u_{tt}-\bigtriangleup u-\bigtriangleup u_{t}+\vert u_{t}\vert ^{q-1}u_{t}=\vert u\vert ^{p-1}u. $$

(2)

Gerbi and Houari [3] studied the exponential decay, Chen and Liu [4] studied the global existence, decay, and exponential growth of solutions of the problem (2). Also, Gazzola and Squassina [5] studied the global existence and blow up of solutions of the problem (2), for $q=1$.

In the absence of the strong damping term $\bigtriangleup u_{t}$ and $m=0$, the problem (1) can be reduced to the following wave equation with nonlinear damping and source terms:

$$ u_{tt}-\bigtriangleup u+\vert u_{t}\vert ^{q-1}u_{t}= \vert u\vert ^{p-1}u. $$

(3)

Many authors have investigated the local existence, blow up, and asymptotic behavior of solutions of (3); see [6–11]. The interaction between the damping $(\vert u_{t}\vert ^{q-1}u_{t})$ and the source term $(\vert u\vert ^{p-1}u)$ makes the problem more interesting. Levine [7, 8] first studied the interaction between the linear damping ($q=1$) and source term by using a concavity method. But this method cannot be applied in the case of a nonlinear damping term. Georgiev and Todorova [6] extended Levine’s result to the nonlinear case ($q>1$). They showed that solutions with a negative initial energy blow up in finite time. Later, Vitillaro [11] extended these results to the case of a nonlinear damping and a positive initial energy.

In [12], Messaoudi studied decay of solutions of the problem (1), using the techniques combination of the perturbed energy and potential well methods. Recently, the problem (1) was studied by Wu and Xue [13]. They proved the uniform energy decay rates of the solutions, by utilizing the multiplier method.

In this work, we established the polynomial and exponential decay of solutions of the problem (1) by using Nakao’s inequality. After that, we show the blow up of solutions with negative and nonnegative initial energy, using the same techniques as in [14].

This work is organized as follows: In the next section, we present some lemmas, notations, and a local existence theorem. In Section 3, the global existence and decay of solutions are given. In Section 4, we show the blow up of solutions, for $q=1$.

2 Preliminaries

In this section, we shall give some assumptions and lemmas which will be used throughout this paper. Let $\Vert \cdot \Vert$ and $\Vert \cdot \Vert _{p}$ denote the usual $L^{2} ( \Omega )$ norm and $L^{p} ( \Omega )$ norm, respectively.

Lemma 1

(Sobolev-Poincaré inequality) [15]

Let p be a number with $2\leq p<\infty$ ($n=1,2$) or $2\leq p\leq\frac{2n}{n-2}$ ($n\geq3$), then there is a constant $C_{\ast}=C_{\ast} ( \Omega , p )$ such that

$$ \Vert u\Vert _{p}\leq C_{\ast} \Vert \nabla u\Vert \quad \textit{for }u\in H_{0}^{1} ( \Omega ) . $$

Lemma 2

[16]

Let $\phi ( t )$ be a nonincreasing and nonnegative function defined on $[ 0,T ]$, $T>1$, satisfying

$$ \phi^{1+\alpha} ( t ) \leq w_{0} \bigl( \phi ( t ) -\phi ( t+1 ) \bigr) , \quad t\in [ 0,T ] $$

for $w_{0}$ a positive constant and α a nonnegative constant. Then we have, for each $t\in [ 0,T ]$,

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \phi ( t ) \leq\phi ( 0 ) e^{-w_{1} [ t-1 ] ^{+}},& \alpha=0, \\ \phi ( t ) \leq ( \phi ( 0 ) ^{-\alpha }+w_{0}^{-1}\alpha [ t-1 ] ^{+} ) ^{-\frac{1}{\alpha}},& \alpha>0,\end{array}\displaystyle \right . $$

where $[ t-1 ] ^{+}=\max \{ t-1,0 \}$ and $w_{1}=\ln ( \frac{w_{0}}{w_{0}-1} )$.

Next, we state the local existence theorem that can be established by combining the arguments of [6, 17, 18].

Theorem 3

(Local existence)

Suppose that $m+2< p+1<\frac{n ( m+2 ) }{n- ( m+2 ) }$, $m+2< n$, and further $u_{0}\in W_{0}^{1,m+2} ( \Omega )$ and $u_{1}\in L^{2} ( \Omega )$ such that problem (1) has a unique local solution,

$$ u\in C \bigl( [ 0,T ) ;W_{0}^{1,m+2} ( \Omega ) \bigr) \quad \textit{and}\quad u_{t}\in C \bigl( [ 0,T ) ;L^{2} ( \Omega ) \bigr) \cap L^{q+1} \bigl( \Omega\times [ 0,T ) \bigr) . $$

Moreover, at least one of the following statements holds true:

• (i) $T=\infty$,

• (ii) $\Vert u_{t}\Vert ^{2}+\Vert \nabla u\Vert _{m+2}^{m+2}\rightarrow\infty$ as $t\rightarrow T^{-}$.

3 Global existence and decay of solutions

In this section, we discuss the global existence and decay of the solution for problem (1).

We define

$$ J ( t ) =\frac{1}{m+2}\Vert \nabla u\Vert _{m+2}^{m+2}-\frac{1}{p+1}\Vert u\Vert _{p+1}^{p+1} $$

(4)

and

$$ I ( t ) =\Vert \nabla u\Vert _{m+2}^{m+2}-\Vert u\Vert _{p+1}^{p+1}. $$

(5)

We also define the energy function as follows:

$$ E ( t ) =\frac{1}{2}\Vert u_{t}\Vert ^{2}+ \frac {1}{m+2}\Vert \nabla u\Vert _{m+2}^{m+2}- \frac{1}{p+1}\Vert u\Vert _{p+1}^{p+1}. $$

(6)

Finally, we define

$$ W= \bigl\{ u:u\in W_{0}^{1,m+2} ( \Omega ) , I ( u ) >0 \bigr\} \cup \{ 0 \} . $$

(7)

The next lemma shows that our energy functional (6) is a nonincreasing function along the solution of (1).

Lemma 4

$E ( t )$ is a nonincreasing function for $t\geq0$ and

$$ E^{\prime} ( t ) =- \bigl( \Vert u_{t}\Vert _{q+1}^{q+1}+\Vert \nabla u_{t}\Vert ^{2} \bigr) \leq0. $$

(8)

Proof

Multiplying the equation of (1) by $u_{t}$ and integrating over Ω, using integrating by parts, we get

$$ E ( t ) -E ( 0 ) =-\int_{0}^{t} \bigl( \Vert u_{\tau }\Vert _{q+1}^{q+1}+\Vert \nabla u_{\tau} \Vert ^{2} \bigr) \, d\tau\quad \text{for }t\geq0. $$

(9)

 □

Lemma 5

Let $u_{0}\in W$ and $u_{1}\in L^{2} ( \Omega )$. Suppose that $p>m+1$ and

$$ \beta=C_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{p-m-1}{m+2}}< 1, $$

(10)

then $u\in W$ for each $t\geq0$.

Proof

Since $I ( 0 ) >0$, it follows by the continuity of $u ( t )$ that

$$ I ( t ) >0 $$

for some interval near $t=0$. Let $T_{m}>0$ be a maximal time, when (5) holds on $[ 0,T_{m} ]$.

From (4) and (5), we have

$$\begin{aligned} J ( t ) =&\frac{1}{p+1}I ( t ) +\frac{p-m-1}{ ( p+1 ) ( m+2 ) }\Vert \nabla u\Vert _{m+2}^{m+2} \\ \geq&\frac{p-m-1}{ ( p+1 ) ( m+2 ) }\Vert \nabla u\Vert _{m+2}^{m+2}. \end{aligned}$$

(11)

Thus, from (6) and $E ( t )$ being nonincreasing by (8), we have

$$\begin{aligned} \Vert \nabla u\Vert _{m+2}^{m+2} \leq&\frac{ ( p+1 ) ( m+2 ) }{p-m-1}J ( t ) \\ \leq&\frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( t ) \\ \leq&\frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) . \end{aligned}$$

(12)

And so, exploiting Lemma 1, (10), and (12), we obtain

$$\begin{aligned} \Vert u\Vert _{p+1}^{p+1} \leq&C_{\ast} \Vert \nabla u\Vert ^{p+1} \\ \leq&C_{\ast} \Vert \nabla u\Vert _{m+2}^{p+1} \\ =&C_{\ast} \Vert \nabla u\Vert _{m+2}^{p-m-1}\Vert \nabla u\Vert _{m+2}^{m+2} \\ \leq&C_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{p-m-1}{m+2}}\Vert \nabla u \Vert _{m+2}^{m+2} \\ =&\beta \Vert \nabla u\Vert _{m+2}^{m+2} \\ < &\Vert \nabla u\Vert _{m+2}^{m+2} \quad \text{on }t\in [ 0,T_{m} ] . \end{aligned}$$

(13)

Therefore, by using (5), we conclude that $I ( t ) >0$ for all $t\in [ 0,T_{m} ]$. By repeating the procedure, $T_{m}$ is extended to T. The proof of Lemma 5 is completed. □

Lemma 6

Let the assumptions of Lemma  5 hold. Then there exists $\eta _{1}=1-\beta$ such that

$$ \Vert u\Vert _{p+1}^{p+1}\leq ( 1-\eta_{1} ) \Vert \nabla u\Vert _{m+2}^{m+2}. $$

Proof

From (13), we get

$$ \Vert u\Vert _{p+1}^{p+1}\leq\beta \Vert \nabla u\Vert _{m+2}^{m+2}. $$

 □

Let $\eta_{1}=1-\beta$, then we have the following result.

Remark 7

From Lemma 6, we can deduce that

$$ \Vert \nabla u\Vert _{m+2}^{m+2}\leq\frac{1}{\eta _{1}}I ( t ) . $$

(14)

Theorem 8

Suppose that $m+2< p+1<\frac{n ( m+2 ) }{n- ( m+2 ) }$, $m+2< n$ holds. Let $u_{0}\in W$ satisfying (10). Then the solution of problem (1) is global.

Proof

It is sufficient to show that $\Vert u_{t}\Vert ^{2}+ \Vert \nabla u\Vert _{m+2}^{m+2}$ is bounded independently of t. To achieve this we use (5) and (6) to obtain

$$\begin{aligned} E ( 0 ) \geq&E ( t ) =\frac{1}{2}\Vert u_{t}\Vert ^{2}+\frac{1}{m+2}\Vert \nabla u\Vert _{m+2}^{m+2}- \frac{1}{p+1}\Vert u\Vert _{p+1}^{p+1} \\ =&\frac{1}{2}\Vert u_{t}\Vert ^{2}+ \frac{p-m-1}{ ( p+1 ) ( m+2 ) }\Vert \nabla u\Vert _{m+2}^{m+2}+\frac{1}{p+1}I ( t ) \\ \geq&\frac{1}{2}\Vert u_{t}\Vert ^{2}+ \frac{p-m-1}{ ( p+1 ) ( m+2 ) }\Vert \nabla u\Vert _{m+2}^{m+2} \end{aligned}$$

since $I ( t ) \geq0$. Therefore,

$$ \Vert u_{t}\Vert ^{2}+\Vert \nabla u\Vert _{m+2}^{m+2}\leq CE ( 0 ) , $$

where $C=\max \{ 2,\frac{ ( p+1 ) ( m+2 ) }{p-m-1} \}$. Then by Theorem 3, we have the global existence result. □

Theorem 9

Suppose that $m+2< p+1<\frac{n ( m+2 ) }{n- ( m+2 ) }$, $m+2< n$, and (10) hold, and further $u_{0}\in W$. Thus, we have the following decay estimates:

$$ E ( t ) \leq \left \{ \textstyle\begin{array}{l@{\quad}l} E ( 0 ) e^{-w_{1} [ t-1 ] ^{+}}, &\textit{if }q=1, m=0, \\ ( E ( 0 ) ^{-\alpha}+C_{7}^{-1}\alpha [ t-1 ] ^{+} ) ^{-\frac{1}{\alpha}},&\textit{if }q>\frac{1}{m+1},\end{array}\displaystyle \right . $$

where $w_{1}$, α, and $C_{7}$ are positive constants which will be defined later.

Proof

By integrating (8) over $[ t,t+1 ]$, $t>0$, we have

$$ E ( t ) -E ( t+1 ) =D^{q+1} ( t ) , $$

(15)

where

$$ D^{q+1} ( t ) =\int_{t}^{t+1} \bigl( \Vert u_{\tau} \Vert _{q+1}^{q+1}+\Vert \nabla u_{\tau} \Vert ^{2} \bigr)\, d\tau. $$

(16)

By virtue of (16) and Hölder’s inequality, we observe that

$$ \int_{t}^{t+1}\int_{\Omega} \vert u_{t}\vert ^{2}\, dx\, dt\leq \vert \Omega \vert ^{\frac{q+1}{q+2}}D^{2} ( t ) =CD^{2} ( t ) . $$

(17)

Hence, from (17), there exist $t_{1}\in [ t,t+\frac {1}{4} ]$ and $t_{2}\in [ t+\frac{3}{4},t+1 ]$ such that

$$ \bigl\Vert u_{t} ( t_{i} ) \bigr\Vert \leq CD ( t ) , \quad i=1,2 . $$

(18)

Multiplying (1) by u and integrating it over $\Omega\times [ t_{1},t_{2} ]$, using integration by parts, we get

$$\begin{aligned} \int_{t_{1}}^{t_{2}}I ( t ) \, dt =&-\int _{t_{1}}^{t_{2}}\int_{\Omega }uu_{tt} \,dx\,dt-\int_{t_{1}}^{t_{2}}\int_{\Omega} \nabla u_{t}\nabla \,dx\,dt \\ &{}-\int_{t_{1}}^{t_{2}}\int_{\Omega} \vert u_{t}\vert ^{q-1}u_{t}u\, dx\, dt. \end{aligned}$$

(19)

By using (1) and integrating by parts and applying the Cauchy-Schwarz inequality in the first term and the Hölder inequality in the second term of the right-hand side of (19), we obtain

$$\begin{aligned} \int_{t_{1}}^{t_{2}}I ( t )\, dt \leq&\bigl\Vert u_{t} ( t_{1} ) \bigr\Vert \bigl\Vert u ( t_{1} ) \bigr\Vert +\bigl\Vert u_{t} ( t_{2} ) \bigr\Vert \bigl\Vert u ( t_{2} ) \bigr\Vert \\ &{}+\int_{t_{1}}^{t_{2}}\bigl\Vert u_{t} ( t ) \bigr\Vert ^{2}\, dt+\int_{t_{1}}^{t_{2}} \Vert \nabla u_{t}\Vert \Vert \nabla u\Vert \, dt \\ &{}-\int_{t_{1}}^{t_{2}}\int_{\Omega} \vert u_{t}\vert ^{q-1}u_{t}u\, dx\, dt. \end{aligned}$$

(20)

Now, our goal is to estimate the last term in the right-hand side of inequality (20). By using Hölder inequality, we obtain

$$ \int_{t_{1}}^{t_{2}}\int_{\Omega} \vert u_{t}\vert ^{q-1}u_{t}u\, dx\, dt\leq\int _{t_{1}}^{t_{2}}\bigl\Vert u_{t} ( t ) \bigr\Vert _{q+1}^{q}\bigl\Vert u ( t ) \bigr\Vert _{q+1}\, dt. $$

(21)

By applying the Sobolev-Poincaré inequality and (12), we find

$$\begin{aligned}& \int_{t_{1}}^{t_{2}}\bigl\Vert u_{t} ( t ) \bigr\Vert _{q+1}^{q}\bigl\Vert u ( t ) \bigr\Vert _{q+1}\,dt \\& \quad \leq C_{\ast}\int_{t_{1}}^{t_{2}}\bigl\Vert u_{t} ( t ) \bigr\Vert _{q+1}^{q}\Vert \nabla u\Vert \,dt \\& \quad \leq C_{\ast}\int_{t_{1}}^{t_{2}}\bigl\Vert u_{t} ( t ) \bigr\Vert _{q+1}^{q}\Vert \nabla u\Vert _{m+2}\,dt \\& \quad \leq C_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}}\int_{t_{1}}^{t_{2}}\bigl\Vert u_{t} ( t ) \bigr\Vert _{q+1}^{q}E^{\frac{1}{m+2}} ( s ) \,dt \\& \quad \leq C_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) \int _{t_{1}}^{t_{2}}\bigl\Vert u_{t} ( t ) \bigr\Vert _{q+1}^{q}\,dt \\& \quad \leq C_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) D^{q} ( t ) . \end{aligned}$$

(22)

Now, we estimate the fourth term of the right-hand side of inequality (20). By using the embedding $L^{m+2} ( \Omega ) \hookrightarrow L^{2} ( \Omega )$, we have

$$\begin{aligned}& \int_{t_{1}}^{t_{2}}\Vert \nabla u_{t} \Vert \Vert \nabla u\Vert \,dt \\& \quad \leq C_{\ast}\int_{t_{1}}^{t_{2}}\Vert \nabla u_{t}\Vert \bigl\Vert \nabla u ( t ) \bigr\Vert _{m+2}\,dt \\& \quad \leq C_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}}\int_{t_{1}}^{t_{2}}\Vert \nabla u_{t}\Vert E^{\frac{1}{m+2}} ( s ) \,dt \\& \quad \leq C_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) \int _{t_{1}}^{t_{2}}\Vert \nabla u_{t}\Vert \,dt, \end{aligned}$$

which implies

$$\begin{aligned} \int_{t_{1}}^{t_{2}}\Vert \nabla u_{t} \Vert \,dt \leq& \biggl( \int_{t_{1}}^{t_{2}}1\,dt \biggr) ^{\frac{1}{2}} \biggl( \int_{t_{1}}^{t_{2}} \Vert \nabla u_{t}\Vert ^{2}\,dt \biggr) ^{\frac{1}{2}} \\ \leq&CD ( t ). \end{aligned}$$

Then

$$ \int_{t_{1}}^{t_{2}}\Vert \nabla u_{t} \Vert \Vert \nabla u\Vert \,dt\leq CC_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}}\sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) D ( t ). $$

(23)

From (12), (18), and the Sobolev-Poincaré inequality, we have

$$ \bigl\Vert u_{t} ( t_{i} ) \bigr\Vert \bigl\Vert u ( t_{i} ) \bigr\Vert \leq C_{1}D ( t ) \sup _{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) , $$

(24)

where $C_{1}=2C_{\ast} ( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) ) ^{\frac{1}{m+2}}$. Then by (20)-(24) we have

$$\begin{aligned} \int_{t_{1}}^{t_{2}}I ( t ) \,dt \leq&C_{1}D ( t ) \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) +D^{2} ( t ) \\ &{}+CC_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) D ( t ) \\ &{}+C_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) D^{q} ( t ) . \end{aligned}$$

(25)

On the other hand, from (5), (6), and Remark 7, we obtain

$$ E ( t ) \leq\frac{1}{2}\Vert u_{t}\Vert ^{2}+C_{3}I ( t ) , $$

(26)

where $C_{3}=\frac{1}{\eta_{1}}\frac{p-m-1}{ ( p+1 ) ( m+2 ) }+\frac{1}{p+1}$.

By integrating (26) over $[ t_{1},t_{2} ]$, we have

$$ \int_{t_{1}}^{t_{2}}E ( t ) \,dt\leq\frac{1}{2} \int_{t_{1}}^{t_{2}}\Vert u_{t} \Vert ^{2}\,dt+C_{3}\int_{t_{1}}^{t_{2}}I ( t ) \,dt. $$

(27)

Then by (18), (25), and (27), we get

$$\begin{aligned} \int_{t_{1}}^{t_{2}}E ( t ) \,dt \leq& \frac{1}{2}CD^{2} ( t ) +C_{3}\biggl[ C_{1}D ( t ) \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) +D^{2} ( t ) \\ &{}+CC_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) D ( t ) \\ &{} +C_{\ast} \biggl( \frac{ ( p+1 ) ( m+2 ) }{p-m-1}E ( 0 ) \biggr) ^{\frac{1}{m+2}} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{m+2}} ( s ) D^{q} ( t ) \biggr] . \end{aligned}$$

(28)

By integrating (8) over $[ t,t_{2} ]$, we obtain

$$ E ( t ) =E ( t_{2} ) +\int_{t}^{t_{2}} \bigl( \Vert u_{\tau} \Vert _{q+1}^{q+1}+\Vert \nabla u_{\tau} \Vert ^{2} \bigr)\, d\tau. $$

(29)

Therefore, since $t_{2}-t_{1}\geq\frac{1}{2}$, we conclude that

$$ \int_{t_{1}}^{t_{2}}E ( t ) \,dt\geq ( t_{2}-t_{1} ) E ( t_{2} ) \geq\frac{1}{2}E ( t_{2} ) . $$

That is,

$$ E ( t_{2} ) \leq2\int_{t_{1}}^{t_{2}}E ( t ) \,dt. $$

(30)

Consequently, exploiting (15), (28)-(30), and since $t_{1},t_{2}\in [ t,t+1 ]$, we get

$$\begin{aligned} E ( t ) \leq&2\int_{t_{1}}^{t_{2}}E ( t ) \,dt+\int _{t}^{t+1} \bigl( \Vert u_{\tau} \Vert _{q+1}^{q+1}+\Vert \nabla u_{\tau} \Vert ^{2} \bigr) \, d\tau \\ =&2\int_{t_{1}}^{t_{2}}E ( t ) \,dt+D^{q+1} ( t ) . \end{aligned}$$

(31)

Then, by (28), we have

$$\begin{aligned} E ( t ) \leq& \biggl( \frac{1}{2}C+C_{3} \biggr) D^{2} ( t ) +D^{q+1} ( t ) \\ &{}+C_{4} \bigl[ D ( t ) +D^{q} ( t ) \bigr] E^{\frac {1}{m+2}} ( t ) . \end{aligned}$$

Hence, we obtain

$$ E ( t ) \leq C_{5} \bigl[ D^{2} ( t ) +D^{q+1} ( t ) +D^{\frac{m+2}{m+1}} ( t ) +D^{\frac{m+2}{m+1}q} ( t ) \bigr] . $$

(32)

Note that, since $E ( t )$ is nonincreasing and $E ( t ) \geq0$ on $[ 0,\infty )$,

$$\begin{aligned} D^{q+1} ( t ) =&E ( t ) -E ( t+1 ) \\ \leq&E ( 0 ) . \end{aligned}$$

Thus, we have

$$ D ( t ) \leq E^{\frac{1}{q+1}} ( 0 ) . $$

(33)

It follows from (32) and (33) that

$$\begin{aligned} E ( t ) \leq&C_{5} \bigl[ D^{\frac{m}{m+1}} ( t ) +D^{q-\frac{1}{m+1}} ( t ) +1+D^{\frac{ ( m+2 ) ( q-1 ) }{m+1}} ( t ) \bigr] D^{\frac{m+2}{m+1}} ( t ) \\ \leq&C_{5} \bigl[ E^{\frac{m}{ ( m+1 ) ( q+1 ) }} ( 0 ) +E^{ ( q-\frac{1}{m+1} ) \frac{1}{q+1}} ( 0 ) +1+E^{\frac{ ( m+2 ) ( q-1 ) }{ ( m+1 ) ( q+1 ) }} \bigr] D^{\frac{m+2}{m+1}} ( t ) \\ =&C_{6}D^{\frac{m+2}{m+1}} ( t ) . \end{aligned}$$

Thus, we get

$$ E^{1+\frac{ ( m+1 ) q-1}{m+2}} ( t ) \leq C_{7}D^{q+1} ( t ). $$

(34)

Case 1: When $q=1$ and $m=0$ from (34), we obtain

$$ E ( t ) \leq C_{7}D^{2} ( t ) =C_{7} \bigl[ E ( t ) -E ( t+1 ) \bigr] . $$

By Lemma 2, we get

$$ E ( t ) \leq E ( 0 ) e^{-w_{1} [ t-1 ] ^{+}}, $$

where $w_{1}=\ln\frac{C_{7}}{C_{7}-1}$.

Case 2: When $( m+1 ) q>1$, applying Lemma 2 to (34) yield

$$ E ( t ) \leq \bigl( E ( 0 ) ^{-\alpha }+C_{7}^{-1} \alpha [ t-1 ] ^{+} \bigr) ^{-\frac{1}{\alpha}}, $$

where $\alpha=\frac{ ( m+1 ) q-1}{m+2}$. The proof of Theorem 9 is completed. □

4 Blow up of solutions

In this section, we deal with the blow up of the solution for the problem (1), when $q=1$. Let us begin by stating the following two lemmas, which will be used later.

Lemma 10

[14]

Let us have $\delta>0$ and let $B ( t ) \in C^{2} ( 0,\infty )$ be a nonnegative function satisfying

$$ B^{\prime\prime} ( t ) -4 ( \delta+1 ) B^{\prime } ( t ) +4 ( \delta+1 ) B ( t ) \geq0. $$

(35)

If

$$ B^{\prime} ( 0 ) >r_{2}B ( 0 ) +K_{0}, $$

(36)

with $r_{2}=2 ( \delta+1 ) -2\sqrt{ ( \delta+1 ) \delta}$, then $B^{\prime} ( t ) >K_{0}$ for $t>0$, where $K_{0}$ is a constant.

Lemma 11

[14]

If $H ( t )$ is a nonincreasing function on $[ t_{0},\infty )$ and satisfies the differential inequality

$$ \bigl[ H^{\prime} ( t ) \bigr] ^{2}\geq a+b \bigl[ H ( t ) \bigr] ^{2+\frac{1}{\delta}}\quad \textit{for }t\geq t_{0}, $$

(37)

where $a>0$, $b\in R$, then there exists a finite time $T^{\ast}$ such that

$$ \lim_{t\rightarrow T^{\ast-}}H ( t ) =0. $$

Upper bounds for $T^{\ast}$ are estimated as follows:

• (i) If $b<0$ and $H ( t_{0} ) <\min \{ 1,\sqrt{-\frac {a}{b}} \}$ then

  $$ T^{\ast}\leq t_{0}+\frac{1}{\sqrt{-b}}\ln\frac{\sqrt{-\frac {a}{b}}}{\sqrt{-\frac{a}{b}}-H ( t_{0} ) }. $$

• (ii) If $b=0$, then

  $$ T^{\ast}\leq t_{0}+\frac{H ( t_{0} ) }{H^{\prime} ( t_{0} ) }. $$

• (iii) If $b>0$, then

  $$ T^{\ast}\leq\frac{H ( t_{0} ) }{\sqrt{a}}\quad \textit{or}\quad T^{\ast }\leq t_{0}+2^{\frac{3\delta+1}{2\delta}}\frac{\delta c}{\sqrt{a}} \bigl[ 1- \bigl( 1+cH ( t_{0} ) \bigr) ^{-\frac{1}{2\delta}} \bigr] , $$

  where $c= ( \frac{a}{b} ) ^{2+\frac{1}{\delta}}$.

Definition 12

A solution u of (1) with $q=1$ is called blow up if there exists a finite time $T^{\ast}$ such that

$$ \lim_{t\rightarrow T^{\ast-}} \biggl[ \int_{\Omega }u^{2} \,dx+\int_{0}^{t}\int_{\Omega} \bigl( u^{2}+\vert \nabla u\vert ^{2} \bigr) \, dx\, d\tau \biggr] =\infty. $$

(38)

Let

$$ a ( t ) =\int_{\Omega}u^{2}\,dx+\int _{0}^{t}\int_{\Omega} \bigl( u^{2}+\vert \nabla u\vert ^{2} \bigr) \, dx\, d\tau\quad \text{for }t\geq 0. $$

(39)

Lemma 13

Assume $m+2< p+1<\frac{n ( m+2 ) }{n- ( m+2 ) }$, $m+2< n$, and that $m\leq4\delta\leq p-1$, then we have

$$ a^{\prime\prime} ( t ) \geq4 ( \delta+1 ) \int_{\Omega }u_{t}^{2} \,dx-4 ( 2\delta+1 ) E ( 0 ) +4 ( 2\delta +1 ) \int_{0}^{t} \bigl( \Vert u_{t}\Vert ^{2}+\Vert \nabla u_{t}\Vert ^{2} \bigr)\, d\tau. $$

(40)

Proof

From (39), we have

$$\begin{aligned}& a^{\prime} ( t ) =2\int_{\Omega}uu_{t}\,dx+ \Vert u \Vert ^{2}+\Vert \nabla u\Vert ^{2}, \end{aligned}$$

(41)

$$\begin{aligned}& a^{\prime\prime} ( t ) =2\int_{\Omega }u_{t}^{2} \,dx+2\int_{\Omega}uu_{tt}\,dx+2\int _{\Omega} ( uu_{t}+\nabla u\nabla u_{t} ) \,dx \\& \hphantom{a^{\prime\prime} ( t )}=2\Vert u_{t}\Vert ^{2}-2\Vert \nabla u\Vert _{m+2}^{m+2}+2\Vert u\Vert _{p+1}^{p+1}. \end{aligned}$$

(42)

Then from (6) and (42), we have

$$\begin{aligned} a^{\prime\prime} ( t ) =&4 ( \delta+1 ) \int_{\Omega }u_{t}^{2} \,dx-4 ( 2\delta+1 ) E ( 0 ) +4 ( 2\delta +1 ) \int_{0}^{t} \bigl( \Vert u_{t}\Vert ^{2}+\Vert \nabla u_{t}\Vert ^{2} \bigr)\, d\tau \\ &{}+ \biggl( \frac{8\delta+4}{m+2}-2 \biggr) \Vert \nabla u\Vert _{m+2}^{m+2}+ \biggl( 2-\frac{8\delta+4}{p+1} \biggr) \Vert u \Vert _{p+1}^{p+1}. \end{aligned}$$

Since $m\leq4\delta\leq p-1$, we obtain (40). □

Lemma 14

Assume $m+2< p+1<\frac{n ( m+2 ) }{n- ( m+2 ) }$, $m+2< n$ and one of the following statements are satisfied:

• (i) $E ( 0 ) <0$,

• (ii) $E ( 0 ) =0$ and $\int_{\Omega}u_{0}u_{1}\,dx>0$,

• (iii) $E ( 0 ) >0$ and

  $$ a^{\prime} ( 0 ) >r_{2} \biggl[ a ( 0 ) +\frac{K_{1}}{4 ( \delta+1 ) } \biggr] +\Vert u_{0}\Vert ^{2} $$

  (43)

  holds.

Then $a^{\prime} ( t ) >\Vert u_{0}\Vert ^{2}$ for $t>t^{\ast}$, where $t_{0}=t^{\ast}$ is given by (44) in case (i) and $t_{0}=0$ in cases (ii) and (iii).

Here $K_{1}$ and $t^{\ast}$ are defined in (48) and (44), respectively.

Proof

(i) If $E ( 0 ) <0$, then from (40), we have

$$ a^{\prime} ( t ) \geq a^{\prime} ( 0 ) -4 ( 2\delta +1 ) E ( 0 ) t, \quad t\geq0. $$

Thus we get $a^{\prime} ( t ) >\Vert u_{0}\Vert ^{2}$ for $t>t^{\ast}$, where

$$ t^{\ast}=\max \biggl\{ \frac{a^{\prime} ( t ) -\Vert u_{0}\Vert ^{2}}{4 ( 2\delta+1 ) E ( 0 ) }, 0 \biggr\} . $$

(44)

(ii) If $E ( 0 ) =0$ and $\int_{\Omega}u_{0}u_{1}\,dx>0$, then $a^{\prime\prime} ( t ) \geq0$ for $t\geq0$. We have $a^{\prime} ( t ) >\Vert u_{0}\Vert ^{2}$, $t\geq0$.

(iii) If $E ( 0 ) >0$, we first note that

$$ 2\int_{0}^{t}\int_{\Omega}uu_{t} \, dx\, d\tau=\Vert u\Vert ^{2}-\Vert u_{0}\Vert ^{2}. $$

(45)

By the Hölder inequality and the Young inequality, we have

$$ \Vert u\Vert ^{2}\leq \Vert u_{0}\Vert ^{2}+\int_{0}^{t}\Vert u\Vert ^{2}\, d\tau+\int_{0}^{t} \Vert u_{t}\Vert ^{2}\, d\tau. $$

(46)

By the Hölder inequality, the Young inequality, and (46), we have

$$ a^{\prime} ( t ) \leq a ( t ) +\Vert u_{0}\Vert ^{2}+\int_{\Omega}u_{t}^{2}\,dx+ \int_{0}^{t}\Vert u_{t}\Vert ^{2}\, d\tau. $$

(47)

Hence, by (40) and (47), we obtain

$$ a^{\prime\prime} ( t ) -4 ( \delta+1 ) a^{\prime } ( t ) +\Vert u_{0}\Vert ^{2}a ( t ) +K_{1}\geq0, $$

where

$$ K_{1}=4 ( 2\delta+1 ) E ( 0 ) +4 ( \delta +1 ) \Vert u_{0}\Vert ^{2}. $$

(48)

Let

$$ b ( t ) =a ( t ) +\frac{K_{1}}{4 ( \delta+1 ) }, \quad t>0. $$

Then $b ( t )$ satisfies Lemma 10. Consequently, we get from (43) $a^{\prime} ( t ) >\Vert u_{0}\Vert ^{2}$, $t>0$, where $r_{2}$ is given in Lemma 10. □

Theorem 15

Assume $m+2< p+1<\frac{n ( m+2 ) }{n- ( m+2 ) }$, $m+2< n$ and one of the following statements are satisfied:

• (i) $E ( 0 ) <0$,

• (ii) $E ( 0 ) =0$ and $\int_{\Omega}u_{0}u_{1}\,dx>0$,

• (iii) $0< E ( 0 ) <\frac{ ( a^{\prime} ( t_{0} ) -\Vert u_{0}\Vert ^{2} ) ^{2}}{8 [ a ( t_{0} ) + ( T_{1}-t_{0} ) \Vert u_{0}\Vert ^{2} ] }$ and (43) holds.

Then the solution u blow up in finite time $T^{\ast}$ in the case of (38). In case (i),

$$ T^{\ast}\leq t_{0}-\frac{H ( t_{0} ) }{H^{\prime} ( t_{0} ) }. $$

(49)

Furthermore, if $H ( t_{0} ) <\min \{ 1,\sqrt{-\frac {a}{b}} \}$, we have

$$ T^{\ast}\leq t_{0}+\frac{1}{\sqrt{-b}}\ln\frac{\sqrt{-\frac {a}{b}}}{\sqrt{-\frac{a}{b}}-H ( t_{0} ) }, $$

(50)

where

$$\begin{aligned}& a=\delta^{2}H^{2+\frac{2}{\delta}} ( t_{0} ) \bigl[ \bigl( a^{\prime} ( t_{0} ) -\Vert u_{0}\Vert ^{2} \bigr) ^{2}-8E ( 0 ) H^{-\frac{1}{\delta}} ( t_{0} ) \bigr] >0, \end{aligned}$$

(51)

$$\begin{aligned}& b=8\delta^{2}E ( 0 ) . \end{aligned}$$

(52)

In case (ii),

$$ T^{\ast}\leq t_{0}-\frac{H ( t_{0} ) }{H^{\prime} ( t_{0} ) }. $$

(53)

In case (iii),

$$ T^{\ast}\leq\frac{H ( t_{0} ) }{\sqrt{a}}\quad \textit{or}\quad T^{\ast }\leq t_{0}+2^{\frac{3\delta+1}{2\delta}} \biggl( \frac{a}{b} \biggr) ^{2+\frac {1}{\delta}}\frac{\delta}{\sqrt{a}} \biggl\{ 1- \biggl[ 1+ \biggl( \frac{a}{b} \biggr) ^{2+\frac{1}{\delta}}H ( t_{0} ) \biggr] ^{-\frac{1}{2\delta}} \biggr\} , $$

(54)

where a and b are given; see (51), (52).

Proof

Let

$$ H ( t ) = \bigl[ a ( t ) + ( T_{1}-t ) \Vert u_{0}\Vert ^{2} \bigr] ^{-\delta}\quad \text{for }t\in [ 0,T_{1} ] , $$

(55)

where $T_{1}>0$ is a certain constant which will be specified later. Then we get

$$\begin{aligned}& H^{\prime} ( t ) = -\delta \bigl[ a ( t ) + ( T_{1}-t ) \Vert u_{0}\Vert ^{2} \bigr] ^{-\delta -1} \bigl[ a^{\prime} ( t ) -\Vert u_{0}\Vert ^{2} \bigr] \\& \hphantom{H^{\prime} ( t )} = -\delta H^{1+\frac{1}{\delta}} ( t ) \bigl[ a^{\prime } ( t ) - \Vert u_{0}\Vert ^{2} \bigr] , \end{aligned}$$

(56)

$$\begin{aligned}& H^{\prime\prime} ( t ) = -\delta H^{1+\frac{2}{\delta }} ( t ) a^{\prime\prime} ( t ) \bigl[ a ( t ) + ( T_{1}-t ) \Vert u_{0}\Vert ^{2} \bigr] \\& \hphantom{H^{\prime\prime} ( t ) =}{} +\delta H^{1+\frac{2}{\delta}} ( t ) ( 1+\delta ) \bigl[ a^{\prime} ( t ) -\Vert u_{0}\Vert ^{2} \bigr] ^{2} \end{aligned}$$

(57)

and

$$ H^{\prime\prime} ( t ) =-\delta H^{1+\frac{2}{\delta}} ( t ) V ( t ) , $$

(58)

where

$$ V ( t ) =a^{\prime\prime} ( t ) \bigl[ a ( t ) + ( T_{1}-t ) \Vert u_{0}\Vert ^{2} \bigr] - ( 1+\delta ) \bigl[ a^{\prime} ( t ) -\Vert u_{0}\Vert ^{2} \bigr] ^{2}. $$

(59)

For simplicity of the calculation, we define

$$\begin{aligned}& P_{u}=\int_{\Omega}u^{2}\,dx, \qquad R_{u}=\int_{\Omega}u_{t}^{2}\,dx, \\& Q_{u}=\int_{0}^{t}\Vert u\Vert ^{2}\,dt, \qquad S_{u}=\int_{0}^{t} \Vert u_{t}\Vert ^{2}\,dt. \end{aligned}$$

From (41), (45), and the Hölder inequality, we get

$$\begin{aligned} a^{\prime} ( t ) =&2\int_{\Omega}uu_{t}\,dx+ \Vert u_{0}\Vert ^{2}+2\int_{0}^{t} \int_{\Omega}uu_{t}\,dx\,dt \\ \leq&2 ( \sqrt{R_{u}P_{u}}+\sqrt{Q_{u}S_{u}} ) +\Vert u_{0}\Vert ^{2}. \end{aligned}$$

(60)

If case (i) or (ii) holds, by (40) we have

$$ a^{\prime\prime} ( t ) \geq ( -4-8\delta ) E ( 0 ) +4 ( 1+\delta ) ( R_{u}+S_{u} ) . $$

(61)

Thus, from (59)-(61) and (55), we obtain

$$\begin{aligned} V ( t ) \geq& \bigl[ ( -4-8\delta ) E ( 0 ) +4 ( 1+\delta ) ( R_{u}+S_{u} ) \bigr] H^{-\frac {1}{\delta}} ( t ) \\ &{}-4 ( 1+\delta ) ( \sqrt{R_{u}P_{u}}+ \sqrt{Q_{u}S_{u}} ) ^{2}. \end{aligned}$$

From (39),

$$ a ( t ) =\int_{\Omega}u^{2}\,dx+\int _{0}^{t}\int_{\Omega }u^{2} \, dx\, ds=P_{u}, $$

and (55), we get

$$ V ( t ) \geq ( -4-8\delta ) E ( 0 ) H^{-\frac{1}{\delta}} ( t ) +4 ( 1+\delta ) \bigl[ ( R_{u}+S_{u} ) ( T_{1}-t ) \Vert u_{0}\Vert ^{2}+\Theta ( t ) \bigr] , $$

where

$$ \Theta ( t ) = ( R_{u}+S_{u} ) ( P_{u}+Q_{u} ) - ( \sqrt{R_{u}P_{u}}+\sqrt{Q_{u}S_{u}} ) ^{2}. $$

By the Schwarz inequality, and $\Theta ( t )$ being nonnegative, we have

$$ V ( t ) \geq ( -4-8\delta ) E ( 0 ) H^{-\frac{1}{\delta}} ( t ) ,\quad t\geq t_{0}. $$

(62)

Therefore, by (58) and (62), we get

$$ H^{\prime\prime} ( t ) \leq4\delta ( 1+2\delta ) E ( 0 ) H^{1+\frac{1}{\delta}} ( t ) , \quad t\geq t_{0}. $$

(63)

By Lemma 13, we know that $H^{\prime} ( t ) <0$ for $t\geq t_{0}$. Multiplying (63) by $H^{\prime} ( t )$ and integrating it from $t_{0}$ to t, we get

$$ H^{\prime2} ( t ) \geq a+bH^{2+\frac{1}{\delta}} ( t ) $$

for $t\geq t_{0}$, where a, b are defined in (51) and (52) respectively.

If case (iii) holds, by the steps of case (i), we get $a>0$ if and only if

$$ E ( 0 ) < \frac{ ( a^{\prime} ( t_{0} ) - \Vert u_{0}\Vert ^{2} ) ^{2}}{8 [ a ( t_{0} ) + ( T_{1}-t_{0} ) \Vert u_{0}\Vert ^{2} ] }. $$

Then by Lemma 11, there exists a finite time $T^{\ast}$ such that $\lim_{t\rightarrow T^{\ast-}}H ( t ) =0$ and the upper bound of $T^{\ast}$ is estimated according to the sign of $E ( 0 )$. This means that (38) holds. □