1. Introduction
Let \(\Omega\) be a bounded domain in \(\mathbb{R}^{N}\) \((N\geq1)\) with smooth boundary \(\partial\Omega\). We
consider the initial-boundary value problem:
\begin{equation}
\label{1.1}
\left\{
\begin{array}{ll}
u_{tt}-\Delta u-\Delta u_{tt}+\Delta^{2}u-g\ast\Delta^{2}u-\Delta u_{t}=|u|^{p-2}u,& x\in\Omega,\ t>0,\\
u=0,\quad \frac{\partial u}{\partial\nu}=0,& x\in\partial\Omega,\ t>0,\\
u(x,0)=u_{0}(x),\quad u_{t}(x,0)=u_{1}(x),& x\in\Omega,
\end{array}
\right.
\end{equation}
(1)
where \(p>2\) and \(\nu\) represents the unit outward normal to \(\partial\Omega\). Here, \(g(t)\) is a positive function that represents the kernel of the memory term,
which will be specified in Section 2 and
\begin{equation*}
g\ast\Delta^{2}u(t)=\int_{0}^{t}g(t-\tau)\Delta^{2}u(\tau)d\tau.
\end{equation*}
The motivation of our work is due to the initial boundary problem of the double
dispersive-dissipative wave equation with nonlinear damping and source terms
\begin{equation}
\label{1.2}
\left\{
\begin{array}{ll}
u_{tt}-\Delta u-\Delta u_{tt}+\Delta^{2}u-\Delta u_{t}+a|u_{t}|^{m-2}u_{t}=b|u|^{p-2}u,& x\in\Omega,\ t>0,\\
u=0,\quad \frac{\partial u}{\partial\nu}=0,& x\in\partial\Omega,\ t>0,\\
u(x,0)=u_{0}(x),\quad u_{t}(x,0)=u_{1}(x),& x\in\Omega,\\
a,b>0, &
\end{array}
\right.
\end{equation}
(2)
which has been discussed by Di and Shang [
1] by considering the existence of global solutions
and the asymptotic behavior of global solutions with \(m\geq p\).
In the absence of the dispersive term and the nonlinear damping term, model \(2\) reduces
to the following wave equation
\begin{equation}
\label{1.3}
u_{tt}-\Delta u-\Delta u_{tt}-\Delta u_{t}=f(u).
\end{equation}
(3)
Shang [
2] studied the well-posedness, asymptotic behavior, and the finite time
blow-up of the solutions under some suitable conditions on \(f\) and for \(N=1,2,3\). Zhang and Hu [
3] showed the existence and the stability of global weak solutions. Xie and Zhong [
4] obtained the existence of global attractors in \(H^{1}_{0}(\Omega)\times H^{1}_{0}(\Omega)\), where the
nonlinear term \(f\) satisfies a critical exponential growth assumption. Xu
et al., [
5]
used the multiplier method to investigate the asymptotic behavior of solutions for (3).
Mellah [6] considered the following initial-boundary value problem
\begin{equation*}\label{W}
\left\{
\begin{array}{ll}
u_{tt}-\Delta u+\Delta^{2}u-g\ast\Delta^{2}u+u_{t}=|u|^{p-1}u,& x\in\Omega,\ t>0,\\
u=0,\quad \frac{\partial u}{\partial\nu}=0,& x\in\partial\Omega,\ t>0,\\
u(x,0)=u_{0}(x),\quad u_{t}(x,0)=u_{1}(x),& x\in\Omega,
\end{array}
\right.
\end{equation*}
in a bounded domain and \(p>1\). He investigated the small data global weak solutions and general decay of solutions, respectively.
Motivated by previous works, it is interesting to prove that problem (1) has a global weak solution assuming small initial
data. In addition, we show the general decay of solutions. The global solutions are constructed
by means of the Galerkin approximations and the general decay is
obtained by employing the technique used in [7].
2. Preliminaries
In this section, we present some materials needed in the proof of our main result. We use the following abbreviations; \(\|\cdot\|_{p}=\|\cdot\|_{L^{p}(\Omega)}\) \((1\leq p\leq+\infty)\) denotes
usual \(L^{p}\) norm, \((\cdot,\cdot)\) denotes the \(L^{2}\)-inner product, and consider the Sobolev spaces \(H^{1}_{0}(\Omega)\) and \(H^{2}_{0}(\Omega)\) with their
usual scalar products and norms. We also use the embedding \(H^{1}_{0}(\Omega)\hookrightarrow L^{p}(\Omega)\) for \(2< p\leq\frac{2N}{N-2}\) if \(N\geq3\) or \(2< p< \infty\) if \(N=1,2\). In this case, the
embedding constant is denoted by \(C_{*}\), that is
\(
\|u\|_{p}\leq C_{*}\|\nabla u\|_{2}.
\)
We define
\begin{equation*}
Q(z)=\frac{1}{2}z^{2}-\frac{C_{*}^{p}}{p}z^{p}.
\end{equation*}
By the direct computation, we deduce that \(Q\) is increasing in \([0,z_{0}]\), where
\(
z_{0}=C_{*}^{\frac{p}{2-p}}
\)
is its unique local maximum.
Next, we give the assumptions for problem (1).
- (G1) The relaxation function \(g:\mathbb{R}_{+}\rightarrow \mathbb{R}_{+}\) is a bounded \(C^{1}\) function such that
\begin{equation*}
g(0)>0,\quad 0< \eta=1-\int_{0}^{\infty}g(\tau)d\tau\leq1-\int_{0}^{t}g(\tau)d\tau=\eta(t).
\end{equation*}
- (G2) There exist positive constants \(\xi_{1}\) and \(\xi_{2}\) such that
\begin{equation*}
-\xi_{1}g(t)\leq g'(t)\leq-\xi_{2}g(t)\quad \forall t\geq0.
\end{equation*}
- (G3) We also assume that
\begin{equation*}
2< p\leq \frac{2N}{N-2}\ \ \mbox {if} \ \ N\geq3 \ \ \mbox {and} \ \ \ p>2 \ \ \ \mbox {if} \ \ N=1,2,
\end{equation*}
where \(\lambda_{1}\) is the first eigenvalue of the following problem
\begin{equation}
\label{2.4}
\Delta^{2}u=\lambda_{1}u \quad \text{in} \ \Omega,\quad u=\frac{\partial u}{\partial\nu}=0 \quad \text{in} \ \partial\Omega.
\end{equation}
(4)
Remark 1. [8] Assuming \(\lambda_{1}\) is the first eigenvalue of the problem (4), we have
\begin{equation}
\label{2.5}
\|\Delta u\|_{2}^{2}\geq\lambda_{1}\|\nabla u\|_{2}^{2}.
\end{equation}
(5)
The energy associated with problem (1) is given by
\begin{eqnarray}
\label{2.6}
E(t)&=&\frac{1}{2}\|u_{t}\|^{2}_{2}+\frac{1}{2}\|\nabla u_{t}\|^{2}_{2}+\frac{1}{2}\left(1-\int_{0}^{t}g(\tau)d\tau\right)\|\Delta u\|^{2}_{2}+\frac{1}{2}\|\nabla u\|^{2}_{2}
+\frac{1}{2}(g\circ \Delta u)(t)-\frac{1}{p}\|u\|_{p}^{p},
\end{eqnarray}
(6)
for \(u\in H^{2}_{0}(\Omega)\), where
\begin{equation*}
(g\circ\Delta u)(t)=\int_{0}^{t}g(t-\tau)\|\Delta u(\tau)-\Delta u(t)\|_{2}^{2}d\tau.
\end{equation*}
Now, we are in a position to state our main results.
3. Main results
In this section, we are going to obtain the existence of global weak solutions for problem (1) with the initial conditions \(\|\nabla u_{0}\|_{2}< z_{0}\) and \(E(0)< Q(z_{0})\).
Theorem 1.
Assume that \((G1)-(G3)\) hold, and that \(\left\{u_{0},u_{1}\right\}\) belong to \(H^{2}_{0}(\Omega)\times H^{1}_{0}(\Omega)\). Further assume that \(\|\nabla u_{0}\|_{2}< z_{0}\) and \(E(0)< Q(z_{0})\). Then, problem (1) admits a global weak solution, which satisfies
\[u\in L^{\infty}(0,\infty;H^{2}_{0}(\Omega)),\quad u_{t}\in L^{\infty}(0,\infty;H^{1}_{0}(\Omega)).\]
Moreover, the identity
\begin{eqnarray}
\label{3.7}
E(t)+\int_{0}^{t}\|\nabla u_{t}(\tau)\|_{2}^{2}d\tau-\frac{1}{2}\int_{0}^{t}(g'\circ \Delta u)(\tau)d\tau+\frac{1}{2}\int_{0}^{t}g(\tau)\|\Delta u(\tau)\|_{2}^{2}d\tau=E(0),
\end{eqnarray}
(7)
holds for \(0\leq t< \infty\). Also, for an increasing \(C^{2}\) function \(\zeta:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}\) satisfying
\begin{equation}
\label{3.8}
\zeta(0)=0,\quad \zeta_{t}(0)>0,\quad \lim_{t\rightarrow+\infty}\zeta(t)=+\infty,\quad \zeta_{tt}(t)< 0\quad \forall t\geq0,
\end{equation}
(8)
and, if \(\|g\|_{L^{1}(0,\infty)}\) is sufficiently small, we have for \(\kappa>0\)
\begin{equation*}
E(t)\leq E(0)e^{-\kappa \zeta(t)},\quad \forall t\geq0.
\end{equation*}
Remark 2.
From (8) and \((G2)\), we obtain
\begin{eqnarray}
\label{3.9}
\frac{d}{dt}E(t)&=&-\|\nabla u_{t}(t)\|_{2}^{2}+\frac{1}{2}(g'\circ\Delta u)(t)
-\frac{1}{2}g(t)\|\Delta u(t)\|_{2}^{2}\nonumber\\
&\leq&-\|\nabla u_{t}(t)\|_{2}^{2}-\frac{1}{2}\xi_{2}(g\circ\Delta u)(t)
-\frac{1}{2}g(t)\|\Delta u(t)\|_{2}^{2}\leq0.
\end{eqnarray}
(9)
Proof of Theorem 1 (Main result)
We divide the proof into two steps. In step 1, we prove the small data global
existence of weak solutions by using the Faedo-Galerkin approximation and in step 2, we establish the general decay of energy employing the method used in [
7].
Step 1: Global existence of weak solutions
Let \(\left\{\omega_{j}\right\}_{j=1}^{\infty}\)
be an orthogonal basis of \(H^{2}_{0}(\Omega)\) with \(\omega_{j}\) being the eigenfunction of the
following problem:
\begin{equation*}
-\Delta \omega_{j}=\lambda_{j}\omega_{j},\quad x\in\Omega,\quad \omega_{j}=0,\quad x\in\partial\Omega.
\end{equation*}
Let \(V^{n}=\text{Span}\left\{\omega_{1},\omega_{2},\cdot\cdot\cdot,\omega_{n}\right\}\). By the standard method of ODE, we know that there
exists only one local solution
\begin{equation*}
u^{n}(t)=\sum_{j=1}^{n}b^{n}_{j}(t)\omega_{j}
\end{equation*}
of the Cauchy problem as follows:
\begin{eqnarray}
\label{4.10}
&&\int_{\Omega}u^{n}_{tt}\omega dx+\int_{\Omega}\nabla u^{n}\cdot\nabla \omega dx+\int_{\Omega}\nabla u^{n}_{tt}\cdot\nabla \omega dx+\int_{\Omega}\Delta u^{n}\cdot\Delta \omega dx\nonumber\\
&&-\int_{0}^{t}g(t-\tau)\int_{\Omega}\Delta u^{n}(\tau)\cdot\Delta \omega dxd\tau+\int_{\Omega}\nabla u^{n}_{t}\cdot\nabla \omega dx-\int_{\Omega}|u^{n}|^{p-2}u^{n}\omega dx=0,
\end{eqnarray}
(10)
\begin{equation}
\label{4.11}
u^{n}(0)=u^{n}_{0}\rightarrow u_{0},\ \ \mbox {in} \ \ H^{2}_{0}(\Omega),\quad u^{n}_{t}(0)=u^{n}_{1}\rightarrow u_{1}\ \ \mbox {in} \ \ \ H^{1}_{0}(\Omega).
\end{equation}
(11)
By the standard theory of ODE
system, we prove the existence of solutions of problem (10)-(11) on some interval
\([0, t_{n})\), \(0< t_{n}< T\) for arbitrary \(T>0\), then, this solution can be extended to the whole
interval \([0,T]\) using the first estimate given below.
A Priori Estimates
Setting \(\omega=u^{n}_{t}(t)\) in (10), we have
\begin{eqnarray}
\label{4.12}
&&\frac{1}{2}\frac{d}{dt}\|u^{n}_{t}\|^{2}_{2}+\frac{1}{2}\frac{d}{dt}\|\nabla u^{n}_{t}\|^{2}_{2}+\frac{1}{2}\frac{d}{dt}\|\nabla u^{n}\|^{2}_{2}+\frac{1}{2}\frac{d}{dt}\|\Delta u^{n}\|^{2}_{2}-\frac{1}{p}\frac{d}{dt}\|u^{n}\|_{p}^{p}+\|\nabla u^{n}_{t}\|^{2}_{2}\nonumber\\
&&-\int_{0}^{t}g(t-\tau)\int_{\Omega}\Delta u^{n}(\tau)\cdot\Delta u^{n}_{t}(t)dxd\tau=0.
\end{eqnarray}
(12)
A direct computation shows that
\begin{eqnarray}
\label{4.13}
&&-\int_{0}^{t}g(t-\tau)\int_{\Omega}\Delta u^{n}(\tau)\cdot\Delta u^{n}_{t}(t)dxd\tau \nonumber\\ &&=\frac{1}{2}\frac{d}{dt}(g\circ\Delta u^{n})(t)-\frac{1}{2}\frac{d}{dt}\left(\int_{0}^{t}g(\tau)d\tau\right)\|\Delta u^{n}(t)\|^{2}_{2}-\frac{1}{2}(g'\circ\Delta u^{n})(t)+\frac{1}{2}g(t)\|\Delta u^{n}(t)\|^{2}_{2}.
\end{eqnarray}
(13)
Inserting (13) into (12) and integrating over \([0,t]\subset[0, T]\), we obtain
\begin{eqnarray}
\label{4.14}
&&\frac{1}{2}\|u^{n}_{t}\|^{2}_{2}+\frac{1}{2}\|\nabla u^{n}_{t}\|^{2}_{2}+\frac{\eta(t)}{2}\|\Delta u^{n}(t)\|^{2}_{2}+\frac{1}{2}\|\nabla u^{n}\|^{2}_{2}-\frac{1}{p}\|u^{n}\|_{p}^{p}
+\int_{0}^{t}\|\nabla u^{n}_{t}(\tau)\|^{2}_{2}d\tau+\frac{1}{2}(g\circ\Delta u^{n})(t)\nonumber\\
&&-\frac{1}{2}\int_{0}^{t}(g'\circ\Delta u^{n})(\tau)d\tau+\frac{1}{2}\int_{0}^{t}g(\tau)\|\Delta u^{n}(\tau)\|^{2}_{2}d\tau=E^{n}(0).
\end{eqnarray}
(14)
From assumption \((G3)\) and the Sobolev embedding, we have
\begin{equation*}
\|u^{n}\|^{p}_{p}\leq C_{*}^{p}\|\nabla u^{n}\|^{p}_{2},
\end{equation*}
and then we have
\begin{eqnarray}
\label{4.15}
&&\frac{1}{2}\|u^{n}_{t}\|^{2}_{2}+\frac{1}{2}\|\nabla u^{n}_{t}\|^{2}_{2}+\frac{\eta(t)}{2}\|\Delta u^{n}(t)\|^{2}_{2}+\mathcal{Q}(\|\nabla u^{n}\|^{2}_{2})+\int_{0}^{t}\|\triangle u^{n}_{t}(\tau)\|^{2}_{2}d\tau+\frac{1}{2}(g\circ\Delta u^{n})(t)\nonumber\\
&&-\frac{1}{2}\int_{0}^{t}(g'\circ\Delta u^{n})(\tau)d\tau+\frac{1}{2}\int_{0}^{t}g(\tau)\|\Delta u^{n}(\tau)\|^{2}_{2}d\tau\leq E^{n}(0).
\end{eqnarray}
(15)
By using the fact that
\begin{equation*}
-\int_{0}^{t}(g'\circ\Delta u^{n})(\tau)d\tau+\int_{0}^{t}g(\tau)\|\Delta u^{n}(\tau)\|^{2}_{2}d\tau\geq0,
\end{equation*}
estimate (15) yields
\begin{equation}
\label{4.16}
\frac{1}{2}\|u^{n}_{t}\|^{2}_{2}+\frac{1}{2}\|\nabla u^{n}_{t}\|^{2}_{2}+\frac{\eta(t)}{2}\|\Delta u^{n}(t)\|^{2}_{2}+\frac{1}{2}(g\circ\Delta u^{n})(t)+\mathcal{Q}(\|\nabla u^{n}\|^{2}_{2})+\int_{0}^{t}\|\nabla u^{n}_{t}(\tau)\|^{2}_{2}d\tau\leq E^{n}(0).
\end{equation}
(16)
From \(E(0)< \mathcal{Q}(z_{0})\) and (11), it follows that
\begin{equation}
\label{4.17}
E^{n}(0)< \mathcal{Q}(z_{0})
\end{equation}
(17)
for sufficiently large \(n\). We claim that there exists an integer \(N\) such that
\begin{equation}
\label{4.18}
\|\nabla u^{n}(t)\|^{2}_{2}< z_{0}\quad \forall t\in[0,t_{n})\quad n>N.
\end{equation}
(18)
Suppose the claim is proved, then \(\mathcal{Q}(\|\nabla u^{n}\|^{2}_{2})\geq0\) and from (16) and (17),
\begin{equation}
\label{4.19}
\frac{1}{2}\|u^{n}_{t}\|^{2}_{2}+\frac{1}{2}\|\nabla u^{n}_{t}\|^{2}_{2}+\frac{\eta(t)}{2}\|\Delta u^{n}(t)\|^{2}_{2}+\frac{1}{2}(g\circ\Delta u^{n})(t)+\int_{0}^{t}\|\nabla u^{n}_{t}(\tau)\|^{2}_{2}d\tau\leq E^{n}(0)< \mathcal{Q}(z_{0}),
\end{equation}
(19)
for sufficiently large \(n\) and \(0\leq t< \infty\).
Proof of the claim
Suppose that (18) false, then for each \(n>N\), there exists \(t\in[0,t_{n})\) such that \(\|\nabla u^{n}(t)\|_{2}\geq z_{0}\). Note that from \(\|\nabla u_{0}\|_{2}< z_{0}\) and (11) there exists \(N_{0}\) such that
\begin{equation*}
\|\nabla u^{n}(0)\|_{2}< z_{0}\quad \forall n>N_{0}.
\end{equation*}
Then by continuity there exits a first \(\widetilde{t_{n}}\in[0,t_{n})\) such that
\begin{equation}
\label{4.20}
\|\nabla u^{n}(\widetilde{t_{n}})\|_{2}=z_{0},
\end{equation}
(20)
from where
\begin{equation*}
\mathcal{Q}(\|\nabla u^{n}(t)\|_{2})\geq0 \quad \forall t\in[0,\widetilde{t_{n}}].
\end{equation*}
From \(E(0)< \mathcal{Q}(z_{0})\) and (19), there exists \(N>N_{0}\) and \(\gamma\in(0,z_{0})\) such that
\begin{eqnarray*}
0&\leq&\frac{1}{2}\|u^{n}_{t}(t)\|^{2}_{2}+\frac{1}{2}\|\nabla u^{n}_{t}(t)\|^{2}_{2}+\frac{\eta(t)}{2}\|\Delta u^{n}(t)\|^{2}_{2}+\frac{1}{2}(g\circ\Delta u^{n})(t)+\mathcal{Q}(\|\nabla u^{n}(t)\|^{2}_{2})\nonumber\\
&\leq&\mathcal{Q}(\gamma)\quad \forall\;\;\; t\in[0,\widetilde{t_{n}}]\quad \forall n>N.
\end{eqnarray*}
The monotonicity of \(\mathcal{Q}\) in \([0,z_{0}]\) implies that
\begin{equation*}
0\leq\|\nabla u^{n}(t)\|^{2}_{2}\leq\gamma< z_{0}\quad \forall t\in[0,\widetilde{t_{n}}],
\end{equation*}
in particular, \(\|\nabla u^{n}(t)\|^{2}_{2}< z_{0}\), which is a contradiction to (20).
From (19), we have
\begin{align}
\label{4.21}
\|\Delta u^{n}\|^{2}_{2}&< \frac{2\mathcal{Q}(z_{0})}{\eta},& 0\leq t< \infty,\\
\end{align}
(21)
\begin{align}
\label{4.22}
\| u^{n}_{t}\|^{2}_{2}&< 2\mathcal{Q}(z_{0}),& 0\leq t< \infty,\\
\end{align}
(22)
\begin{align}
\label{4.23}
\|\nabla u^{n}_{t}\|^{2}_{2}&< 2\mathcal{Q}(z_{0}),& 0\leq t< \infty,\\
\end{align}
(23)
\begin{align}
\label{4.24}
\int_{0}^{t}\|\nabla u^{n}_{t}(\tau)\|^{2}_{2}d\tau&<\mathcal{Q}(z_{0}),& 0\leq t< \infty.
\end{align}
(24)
Using Sobolev inequality, (5) and (21), it follows that
\begin{eqnarray}
\label{4.25}
\|u^{n}\|^{2}_{p}&\leq& C_{*}^{2}\|\nabla u^{n}\|^{2}_{2}\leq C_{*}^{2}\lambda_{1}^{-1}\|\Delta u^{n}\|^{2}_{2}< \frac{2C_{*}^{2}\lambda_{1}^{-1}\mathcal{Q}(z_{0})}{\eta},\quad 0\leq t< \infty.
\end{eqnarray}
(25)
Moreover, by (25), we get
\begin{eqnarray}
|(|u^{n}|^{p-2}u^{n},u^{n})|&\leq& \|u^{n}\|^{p}_{p}< C_{*}^{p}\left(\frac{2C_{*}^{2}\lambda_{1}^{-1}\mathcal{Q}(z_{0})}{\eta}\right)^{\frac{p}{2}},\quad 0\leq t< \infty.
\end{eqnarray}
(26)
Therefore, there exist \(u\), \(\chi\) and a subsequence still denotes \(\left\{u_{n}\right\}\) such that
\begin{equation}
u_{n}\rightarrow u \ \ \mbox {weak star in} \ \ \
L^{\infty}(0,\infty;H_{0}^{2}(\Omega)),\quad n\rightarrow+\infty,
\end{equation}
(27)
\begin{equation}
u^{n}_{t}\rightarrow u_{t} \ \ \mbox {weak star in} \ \ \
L^{\infty}(0,\infty;H_{0}^{1}(\Omega)),\quad n\rightarrow+\infty,
\end{equation}
(28)
\begin{equation}
|u^{n}|^{p-2}u^{n}\rightarrow \chi \ \ \mbox {weak star in} \ \ \
L^{\infty}(0,\infty;L^{\frac{p}{p-1}}(\Omega)),\quad n\rightarrow+\infty,
\end{equation}
(29)
Besides, from Lions-Aubin Lemma we also have
\begin{equation}
u^{n}\rightarrow u \ \ \mbox {strongly in} \ \ \
L^{2}(0,\infty;L^{2}(\Omega)),\quad n\rightarrow+\infty,
\end{equation}
(30)
and consequently, making use of the Lemma 1.3 in [
9], we deduce
\begin{equation}
|u^{n}|^{p-2}u^{n}\rightarrow \chi=|u|^{p-2}u \ \ \mbox {weak star in} \ \ \
L^{\infty}(0,\infty;L^{\frac{p}{p-1}}(\Omega)),\quad n\rightarrow+\infty.
\end{equation}
(31)
Thus, we obtain that \(u\) is a global weak of problem (1). In order to prove (7), we use the mean value theorem, we see that there exists \(0< \theta_{n}< 1\) such that
\begin{eqnarray*}
\|u^{n}\|_{p}^{p}-\|u\|_{p}^{p}
&\leq& p\left|\int_{\Omega}|u+\theta_{n}u^{n}|^{p-2}(u+\theta_{n}u^{n})(u^{n}-u)dx\right|\nonumber\\
&\leq& p\|u+\theta_{n}u^{n}\|_{p}^{p-1}\|u^{n}-u\|_{p}\nonumber\\
&\leq&c\|u^{n}-u\|_{p}\rightarrow 0 \ \ \mbox {as} \ \ \ n\rightarrow +\infty,
\end{eqnarray*}
and for each fixed \(t>0\), we obtain
\begin{eqnarray*}
|(g\circ\Delta u)(t)-(g\circ\Delta u^{n})(t)|&=&\left|\int_{0}^{t}g(t-\tau)\|\Delta u(\tau)-\Delta u(t)\|^{2}_{2}d\tau-\int_{0}^{t}g(t-\tau)\|\Delta u^{n}(\tau)-\Delta u^{n}(t)\|^{2}_{2}d\tau\right|\nonumber\\
&\leq&
\int_{0}^{t}g(t-\tau)\|\Delta u(\tau)-\Delta u^{n}(\tau)\|_{2}\|\Delta u(\tau)+\Delta u^{n}(\tau)\|_{2}d\tau\nonumber\\
&&+\int_{0}^{t}g(t-\tau)\|\Delta u(\tau)-\Delta u^{n}(\tau)\|_{2}d\tau\|\Delta u(t)+\Delta u^{n}(t)\|_{2}\nonumber\\
&&+\int_{0}^{t}g(t-\tau)\|\Delta u(\tau)+\Delta u^{n}(\tau)\|_{2}d\tau\|\Delta u(t)-\Delta u^{n}(t)\|_{2}\nonumber\\
&&+\int_{0}^{t}g(\tau)d\tau\|\Delta u(t)+\Delta u^{n}(t)\|_{2}\|\Delta u(t)-\Delta u^{n}(t)\|_{2}\nonumber\\
&\leq&c\int_{0}^{t}g(t-\tau)\|\Delta u(\tau)-\Delta u^{n}(\tau)\|_{2}d\tau\\
&&+c\int_{0}^{t}g(\tau)d\tau\|\Delta u(t)-\Delta u^{n}(t)\|_{2}\rightarrow 0 \ \ \mbox {as} \ \ \ n\rightarrow +\infty.
\end{eqnarray*}
Thus, we have
\begin{equation*}
\lim_{n\rightarrow +\infty}\|u^{n}\|_{p}^{p}=\|u\|_{p}^{p},\quad \lim_{n\rightarrow +\infty}(g\circ\Delta u^{n})(t)=(g\circ\Delta u)(t).
\end{equation*}
From (11), it follows that \(E^{n}(0)\rightarrow E(0)\) as \(n\rightarrow+\infty\). Finally, taking \(n\rightarrow +\infty\) in (14), we deduce that the energy identity (7) holds for \(0\leq t< \infty \).
Step 2: General decay of the energy
Here, we prove the energy decay estimate of the global solutions obtained
in the previous section. To obtain the decay result, we use the following lemmas which are of crucial importance
in the proof.
Lemma 1. Let \(u\in L^{\infty}(0,\infty;H^{2}_{0}(\Omega))\) with \(u_{t}\in L^{\infty}(0,\infty;H^{1}_{0}(\Omega))\) be the solution of (1) and \(E(0)< \mathcal{Q}(z_{0})\), \(\|\nabla u_{0}\|_{2}< z_{0}\),
then we have
\begin{eqnarray}
\label{4.32}
0\leq E(t)\leq C_{1}\|\nabla u_{t}\|_{2}^{2}+C_{2}\|\Delta u\|^{2}_{2}+\frac{1}{2}(g\circ \Delta u)(t),
\end{eqnarray}
(32)
where \(C_{1}=\frac{1}{2}(1+B^{2})\), \(C_{2}=\frac{1}{2}(1+\lambda_{1}^{-1})\) and \(B\) is the optimal constant satisfying the Poincare inequality \(\|u_{t}\|_{2}\leq B\|\nabla u_{t}\|_{2}\).
Proof.
From \(E(0)< \mathcal{Q}(z_{0})\) and \(\|\nabla u_{0}\|_{2}< z_{0}\), we can obtain \(\mathcal{Q}(\|\nabla u(t)\|_{2})\geq0\) for
\(0\leq t< \infty\). Thus we have
\begin{eqnarray*}
E(t)&=&\frac{1}{2}\|u_{t}\|^{2}_{2}+\frac{1}{2}\|\nabla u_{t}\|^{2}_{2}+\frac{1}{2}\|\nabla u\|^{2}_{2}+\frac{1}{2}(g\circ \Delta u)(t)+\frac{1}{2}\left(1-\int_{0}^{t}g(\tau)d\tau\right)\|\Delta u\|^{2}_{2}-\frac{1}{p}\|u\|_{p}^{p}\nonumber\\
&\geq&\frac{1}{2}\|u_{t}\|^{2}_{2}+\frac{1}{2}\|\nabla u_{t}\|^{2}_{2}+\frac{\eta}{2}\|\Delta u\|^{2}_{2}+\frac{1}{2}(g\circ \Delta u)(t)+\mathcal{Q}(\|\nabla u(t)\|_{2})\\&\geq&0,
\end{eqnarray*}
and
\begin{eqnarray*}
E(t)&\leq&\frac{1}{2}\|u_{t}\|^{2}_{2}+\frac{1}{2}\|\nabla u_{t}\|^{2}_{2}+\frac{1}{2}\|\Delta u\|^{2}_{2}+\frac{1}{2}(g\circ \Delta u)(t)+\frac{1}{2}\|\nabla u\|^{2}_{2}\nonumber\\
&\leq& \frac{1}{2}B^{2}\|\nabla u_{t}\|_{2}^{2}+\frac{1}{2}\|\nabla u_{t}\|_{2}^{2}+\frac{1}{2}\lambda_{1}^{-1}\|\Delta u\|^{2}_{2}+\frac{1}{2}\|\Delta u\|^{2}_{2}+\frac{1}{2}(g\circ \Delta u)(t).
\end{eqnarray*}
Let \(C_{1}=\frac{1}{2}(1+B^{2})\) and \(C_{2}=\frac{1}{2}(1+\lambda_{1}^{-1})\), then we have (32).
Lemma 2.
The energy \(E(t)\) satisfies
\begin{eqnarray}
\label{4.33}
\frac{d E(t)}{dt}&\leq&-\|\nabla u_{t}(t)\|^{2}_{2}-\frac{1}{2}\xi_{2}(g\circ\Delta u)(t)-\frac{1}{2}\left[g(0)-\xi_{1}\|g\|_{L^{1}(0,\infty)}\right]\|\Delta u(t)\|^{2}_{2}\quad \forall \;\;\; t\geq0.
\end{eqnarray}
(33)
Proof.
From (9), we have
\begin{eqnarray}
\label{4.34}
\frac{d E(t)}{dt}&\leq&-\|\nabla u_{t}(t)\|^{2}_{2}-\frac{\xi_{2}}{2}(g\circ \Delta u)(t)-\frac{1}{2}g(t)\|\Delta u(t)\|^{2}_{2}.
\end{eqnarray}
(34)
From assumptions \((G2)\) and since \(\int_{0}^{t}g'(\tau)d\tau=g(t)-g(0)\), we obtain
\begin{eqnarray}
\label{4.35}
-\frac{1}{2}g(t)\|\Delta u(t)\|^{2}_{2}&=&-\frac{1}{2}g(0)\|\Delta u(t)\|^{2}_{2}-\frac{1}{2}\left(\int_{0}^{t}g'(\tau)d\tau\right)\|\Delta u(t)\|^{2}_{2}\nonumber\\
&\leq&-\frac{1}{2}g(0)\|\Delta u(t)\|^{2}_{2}+\frac{\xi_{1}}{2}\|g\|_{L^{1}(0,\infty)}\|\Delta u(t)\|^{2}_{2}\nonumber\\
&=&-\frac{1}{2}\left[g(0)-\xi_{1}\|g\|_{L^{1}(0,\infty)}\right]\|\Delta u(t)\|^{2}_{2}.
\end{eqnarray}
(35)
Then, Combining (34) and (35) our conclusion holds. Multiplying (33) by \(e^{\kappa\zeta(t)}\) \((\kappa>0)\) and using (32), we have
\begin{eqnarray}
\label{4.36}
\frac{d}{dt}\left(e^{\kappa\zeta(t)}E(t)\right)&\leq&-\|\nabla u_{t}(t)\|^{2}_{2}e^{\kappa\zeta(t)}E(t)
-\frac{1}{2}\xi_{2}(g\circ\Delta u)(t)e^{\kappa\zeta(t)}E(t)\nonumber\\
&&-\frac{1}{2}\left[g(0)-\xi_{1}\|g\|_{L^{1}(0,\infty)}\right]\|\Delta u(t)\|^{2}_{2}e^{\kappa\zeta(t)}E(t)+\kappa\zeta_{t}(t)e^{\kappa\zeta(t)}E(t)\nonumber\\
&\leq&-\left[1-\kappa C_{1}\zeta_{t}(t)\right]\|\nabla u_{t}(t)\|^{2}_{2}e^{\kappa\zeta(t)}E(t)-\frac{1}{2}\left[\xi_{2}-\kappa\zeta_{t}(t)\right](g\circ\Delta u)(t)e^{\kappa\zeta(t)}E(t)\nonumber\\
&&-\frac{1}{2}\left[g(0)-\xi_{1}\|g\|_{L^{1}}-2C_{2}\kappa\zeta_{t}(t)\right]\|\Delta u(t)\|^{2}_{2}e^{\kappa\zeta(t)}E(t).
\end{eqnarray}
(36)
Using the fact that \(\zeta_{t}\) is decreasing by (8), we conclude that
\begin{eqnarray}
\label{4.37}
\frac{d}{dt}\left(e^{\kappa\zeta(t)}E(t)\right)&\leq&-\left[1-\kappa C_{1}\zeta_{t}(0)\right]
\|\nabla u_{t}(t)\|^{2}_{2}e^{\kappa\zeta(t)}E(t)-\frac{1}{2}\left[\xi_{2}-\kappa\zeta_{t}(0)\right](g\circ\Delta u)(t)e^{\kappa\zeta(t)}E(t)\nonumber\\
&&-\frac{1}{2}\left[g(0)-\xi_{1}\|g\|_{L^{1}(0,\infty)}-2C_{2}\kappa\zeta_{t}(0)\right]\|\Delta u(t)\|^{2}_{2}e^{\kappa\zeta(t)}E(t).
\end{eqnarray}
(37)
Choosing \(\|g\|_{L^{1}(0,\infty)}\) sufficiently small so that \[g(0)-\xi_{1}\|g\|_{L^{1}(0,\infty)}=K>0\]
and defining \[\kappa_{0}=\min\left\{\frac{1}{C_{1}\zeta_{t}(0)},\frac{\xi_{2}}{\zeta_{t}(0)},\frac{K}{2C_{2}\zeta_{t}(0)}\right\},\]
we conclude by taking \(\kappa\in(0, \kappa_{0}]\) in (37) that
\begin{equation}
\label{4.38}
\frac{d}{dt}\left(e^{\kappa\zeta(t)}E(t)\right)\leq0,\quad t>0.
\end{equation}
(38)
Integrating (38) over \((0,t)\), it follows that
\begin{equation}
E(t)\leq E(0)e^{-\kappa\zeta(t)},\quad t>0.
\end{equation}
(39)
Example 1.
For \(\zeta(t)=t+\frac{t}{t+1}\), we can get the exponential decay rate
\(
E(t)\leq E(0)e^{-\kappa t},\quad \forall t\geq0.
\)
For \(\zeta(t)=ln(1+t)\), we can get polynomial decay rate
\(
E(t)\leq E(0)(1+t)^{-\kappa },\quad \forall t\geq0.
\)
Conflicts of Interests
The author declares no conflict of interest.