1. Introduction
This paper deals with the initial boundary value problem of the dispersive
wave equation with memory and source terms
\begin{equation}\label{eq1.1}
u_{tt}-\Delta u+\alpha\Delta^{2} u-\int_{0}^{t}g(t-\tau)\Delta^{2} u(\tau)d\tau+u_{t}=|u|^{p-1}u,\quad x\in\Omega,\ t>0,
\end{equation}
(1)
where \(\Omega\) is a bounded domain in \(\mathbb{R}^{d}\) \((d\geq1)\) with a smooth boundary \(\partial\Omega\), \(\alpha\) is a positive constant and \(g(t)\) is a positive function that represents the kernel of the memory term, which will be specified in Section 2. Here, we understand \(\Delta^{2}u\) to be the dispersive term. In the absence of the viscoelastic term and the dispersive term (that is, if \(g=\alpha= 0\)), the model (1) reduces to the weakly damped wave equation
\begin{equation}
u_{tt}-\Delta u+u_{t}=|u|^{p-1}u,\quad x\in\Omega,\ t>0.
\end{equation}
(2)
The interaction between the weak damping term and the source term
are considered by many authors. We refer the reader to, Haraux and Zuazua [
1], Ikehata [
2]
and Levine [
3,
4]. If \(\alpha=0\) and \(g\) is not trivial on \(\mathbb{R}\), but replacing the fourth
order memory term in (1) by a weaker memory of the form \(\int_{0}^{t}g(t-\tau)\Delta u(\tau)d\tau\), then (1) can be rewritten
as follows
\begin{equation}\label{eq1.3}
u_{tt}-\Delta u+\int_{0}^{t}g(t-\tau)\Delta u(\tau)d\tau+u_{t}=|u|^{p-1}u,\quad x\in\Omega,\ t>0,
\end{equation}
(3)
The Equation (3) has been considered by Wang
et al [
5]. Under some appropriate assumptions on \(g\), by introducing potential wells they obtained the
existence of global solution and the explicit exponential energy decay estimates. Our main goal in the present paper is to discuss the global solutions and general decay to the following weakly damped wave equation with dispersive term, the fourth order memory term and the nonlinear source term
\begin{equation}\label{eq1.4}
u_{tt}-\Delta u+\Delta^{2} u-\int_{0}^{t}g(t-\tau)\Delta^{2} u(\tau)d\tau+u_{t}=|u|^{p-1}u \quad \text{in}\ \Omega\times\mathbb{R}^{+},
\end{equation}
(4)
with simply supported boundary condition
\begin{equation}\label{eq1.5}
u=0,\quad \frac{\partial u}{\partial\nu}=0 \quad \text{on}\ \partial\Omega\times\mathbb{R}^{+},
\end{equation}
(5)
and initial conditions
\begin{equation}\label{eq1.6}
u(\cdot,0)=u_{0} \quad \text{and} \quad u_{t}(\cdot,0)=u_{1} \quad \text{in}\ \Omega,
\end{equation}
(6)
where \(\Omega\) is a bounded domain of \(\mathbb{R}^{d}\) with a smooth boundary \(\partial\Omega\) and \(p>1\). Here, \(\nu\) is the unit
outward normal to \(\partial\Omega\), and \(g(t)\) is a positive function that represents the kernel of the memory term,
which will be specified in Section 2. We prove that Problem (4)-(6) 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 [
6].
2. Preliminaries
Before proceeding to our analysis, we use the following abbreviations \(\|\cdot\|_{q}=\|\cdot\|_{L^{q}(\Omega)}\) \((1\leq q\leq+\infty)\) denotes
usual \(L^{q}\) 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 \(H^{1}_{0}(\Omega)\hookrightarrow L^{q}(\Omega)\) for \(2< q< \frac{2d}{d-2}\) if \(d\geq3\) or \(2< q< \infty\) if \(d=1,2\). In this case, the
embedding constant is denoted by \(C_{*}\), that is
\(
\|u\|_{q}\leq C_{*}\|\nabla u\|_{2}.
\)
We define the polynomial \(Q\) by
\(
Q(z)=\frac{1}{2}z^{2}-\frac{C_{*}^{p+1}}{p+1}z^{p+1},
\)
which is increasing in \([0,z_{0}]\), where
\(
z_{0}=C_{*}^{\frac{p+1}{1-p}}
\)
is its unique local maximum. Next, we give the assumptions for Problem (4)-(6).
(G1) The relaxation function \(g:\mathbb{R}_{+}\rightarrow \mathbb{R}_{+}\) is a bounded \(C^{1}\) function such that
\(
g(0)>0,\quad 0< \eta=1-\int_{0}^{\infty}g(\tau)d\tau\leq1-\int_{0}^{t}g(\tau)d\tau=\eta(t).
\)
(G2) There exist positive constants \(\xi_{1}\) and \(\xi_{2}\) such that
\(
-\xi_{1}g(t)\leq g'(t)\leq-\xi_{2}g(t)\quad \forall t\geq0.
\)
(G3) We also assume that
\(
1
1 \ \ \ \mbox {if} \ \ d=1,2.
\)
where \(\lambda_{1}\) is the first eigenvalue of the following problem
\begin{equation}\label{eq2.7}
\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}
(7)
Remark 1. [7] Assuming \(\lambda_{1}\) is the first eigenvalue of the problem (7), we have
\begin{equation}\label{eq2.8}
\|\Delta u\|_{2}^{2}\geq\lambda_{1}\|\nabla u\|_{2}^{2}.
\end{equation}
(8)
Now, we define the following energy function associated with a solution \(u\) of the Problem (4)-(6)
\begin{equation}
E(t)=\frac{1}{2}\|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+1}\|u\|_{p+1}^{p+1}
\end{equation}
(9)
for \(u\in H^{2}_{0}(\Omega)\), and
\begin{equation}
E(0)=\frac{1}{2}\|u_{1}\|^{2}_{2}+\frac{1}{2}\|\Delta u_{0}\|^{2}_{2}+\frac{1}{2}\|\nabla u_{0}\|^{2}_{2}-\frac{1}{p+1}\|u_{0}\|_{p+1}^{p+1}
\end{equation}
(10)
is the initial total energy. To facilitate further on our analysis, we use the following notation
\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
Theorem 1.
Assume that \((G1)-(G3)\) hold, \(u_{0}\in H^{2}_{0}(\Omega)\), \(u_{1}\in L^{2}(\Omega)\). Further assume that \(\|\nabla u_{0}\|_{2}< z_{0}\) and \(E(0)< Q(z_{0})\), then the Problem
(4)-(6) possesses a global weak solution satisfying;
\(u\in L^{\infty}(0,\infty;H^{2}_{0}(\Omega)),\quad u_{t}\in L^{\infty}(0,\infty;L^{2}(\Omega))\)
for \(0\leq t< \infty\), and the energy identity
\begin{equation}\label{eq3.11}
E(t)+\int_{0}^{t}\|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{equation}
(11)
holds for \(0\leq t< \infty\). Moreover, for \(\zeta:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}\) a increasing \(C^{2}\) function satisfying
\begin{equation}\label{eq3.12}
\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}
(12)
and, if \(\|g\|_{L^{1}(0,\infty)}\) is sufficiently small, we have for \(\kappa>0\);
\(
E(t)\leq E(0)e^{-\kappa \zeta(t)},\quad \forall t\geq0.
\)
Remark 2.
From (11) and \((G2)\), we can easily obtain
\begin{equation}\label{eq3.13}
\frac{d}{dt}E(t)=-\|u_{t}(t)\|_{2}^{2}+\frac{1}{2}(g’\circ\Delta u)(t)-\frac{1}{2}g(t)\|\Delta u(t)\|_{2}^{2}\leq-\|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{equation}
(13)
Remark 3.
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
\(
E(t)\leq E(0)(1+t)^{-\kappa },\quad \forall t\geq0.
\)
4. Proof of main results
In this section, we shall divide the proof into two steps. In Step 1, we prove the global existence of weak solutions by using Galerkin’s approximations. In Step 2, we establish the general decay of energy employing the method used in [
6].
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 problem
\(
-\Delta \omega_{j}=\lambda_{j}\omega_{j},\quad x\in\Omega,\quad \omega_{j}=0,\quad x\in\partial\Omega.
\)
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
\(
u^{n}(t)=\sum_{j=1}^{n}b^{n}_{j}(t)\omega_{j}(x)
\)
of the Cauchy problem as follows
\begin{eqnarray}\label{eq4.14}
&&\int_{\Omega}u^{n}_{tt}\omega dx+\int_{\Omega}\nabla u^{n}\cdot\nabla \omega dx+\int_{\Omega}\Delta u^{n}\cdot\Delta \omega dx-\int_{0}^{t}g(t-\tau)\int_{\Omega}\Delta u^{n}(\tau)\cdot\Delta \omega dxd\tau\nonumber\\
&&+\int_{\Omega}u^{n}_{t}\omega dx-\int_{\Omega}|u^{n}|^{p-1}u^{n}\omega dx=0,
\end{eqnarray}
(14)
\begin{equation}\label{eq4.15}
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} \ \ \ L^{2}(\Omega).
\end{equation}
(15)
By the standard theory of ODE
system, we prove the existence of solutions of Problem (14)-(15) on some interval
\([0, t_{n})\), \(0< t_{n}0\), then, this solution can be extended to the whole
interval \([0,T]\) using the first estimate given below. Taking \(\omega=u^{n}_{t}(t)\) in (14), we obtain
\begin{eqnarray}\label{eq4.16}
&&\frac{1}{2}\frac{d}{dt}\|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+1}\frac{d}{dt}\|u^{n}\|_{p+1}^{p+1}
+\|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}
(16)
For the last term on the left hand side of (16) we have
\begin{eqnarray}\label{eq4.17}
&&-\int_{0}^{t}g(t-\tau)\int_{\Omega}\Delta u^{n}(\tau)\cdot\Delta u^{n}_{t}(t)dxd\tau =\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}\nonumber\\
&&-\frac{1}{2}(g’\circ\Delta u^{n})(t)+\frac{1}{2}g(t)\|\Delta u^{n}(t)\|^{2}_{2}.
\end{eqnarray}
(17)
Inserting (17) into (16) and integrating over \([0,t]\subset[0, T]\), we obtain
\begin{eqnarray}\label{eq4.18}
&&\frac{1}{2}\|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+1}\|u^{n}\|_{p+1}^{p+1}+\int_{0}^{t}\|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}
(18)
Now from assumption \((G3)\) and the Sobolev embedding, we have that
\begin{equation}\label{eq4.19}
\|u^{n}\|^{p+1}_{p+1}\leq C_{*}^{p+1}\|\nabla u^{n}\|^{p+1}_{2},
\end{equation}
(19)
and then we have
\begin{eqnarray}\label{eq4.20}
&&\frac{1}{2}\|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}\|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}
(20)
By using the fact that
\(
-\int_{0}^{t}(g’\circ\Delta u^{n})(\tau)d\tau+\int_{0}^{t}g(\tau)\|\Delta u^{n}(\tau)\|^{2}_{2}d\tau\geq0,
\)
estimate (20) yields
\begin{eqnarray}\label{eq4.21}
\frac{1}{2}\|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}\|u^{n}_{t}(\tau)\|^{2}_{2}d\tau\leq E^{n}(0).
\end{eqnarray}
(21)
From \(E(0)< \mathcal{Q}(z_{0})\) and (15), it follows that
\begin{equation}\label{eq4.22}
E^{n}(0)< \mathcal{Q}(z_{0}),
\end{equation}
(22)
for sufficiently large \(n\). We claim that there exists an integer \(N\) such that
\begin{equation}\label{eq4.23}
\|\nabla u^{n}(t)\|^{2}_{2}N.
\end{equation}
(23)
Suppose the claim is proved. Then \(\mathcal{Q}(\|\nabla u^{n}\|^{2}_{2})\geq0\) and from (21) and (22),
\begin{equation}\label{eq4.24}
\frac{1}{2}\|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}\|u^{n}_{t}(\tau)\|^{2}_{2}d\tau\leq E^{n}(0)< \mathcal{Q}(z_{0})
\end{equation}
(24)
for sufficiently large \(n\) and \(0\leq t< \infty\).
Proof. [Proof of the claim] Suppose that (23) false. Then for each \(n>N\), there exists \(t\in[0,t_{n})\) such that \(\|\nabla u^{n}(t)\|_{2}\geq z_{0}\). We note that from \(\|\nabla u_{0}\|_{2}< z_{0}\) and (15) there exists \(N_{0}\) such that
\(
\|\nabla u^{n}(0)\|_{2}N_{0}.
\)
Then by continuity there exits a first \(t_{n}^{*}\in[0,t_{n})\) such that
\begin{equation}\label{eq4.25}
\|\nabla u^{n}(t_{n}^{*})\|_{2}=z_{0},
\end{equation}
(25)
from where
\(
\mathcal{Q}(\|\nabla u^{n}(t)\|_{2})\geq0 \quad \forall t\in[0,t_{n}^{*}].
\)
Now from \(E(0)N_{0}\) and \(\gamma\in(0,z_{0})\) such that
\(
0\leq\frac{1}{2}\|u^{n}_{t}(t)\|^{2}_{2}+\frac{\eta(t)}{2}\|\Delta u^{n}(t)\|^{2}_{2}+\frac{1}{2}(g\circ\Delta u^{n})(t)+\mathcal{Q}(\|\nabla u^{n}(t)\|^{2}_{2})\leq\mathcal{Q}(\gamma)\quad \forall t\in[0,t_{n}^{*}]\quad \forall n>N.
\)
Then the monotonicity of \(\mathcal{Q}\) in \([0,z_{0}]\) implies that
\(
0\leq\|\nabla u^{n}(t)\|^{2}_{2}\leq\gamma< z_{0}\quad \forall t\in[0,t_{n}^{*}],
\)
and in particular, \(\|\nabla u^{n}(t)\|^{2}_{2}< z_{0}\), which is a contradiction to (24).
From (24), we have
\begin{equation}\label{eq4.26}
\|\Delta u^{n}\|^{2}_{2}< \frac{2\mathcal{Q}(z_{0})}{\eta},\quad 0\leq t< \infty,
\end{equation}
(26)
\begin{equation}\label{eq4.27}
\| u^{n}_{t}\|^{2}_{2}< 2\mathcal{Q}(z_{0}),\quad 0\leq t< \infty,
\end{equation}
(27)
\begin{equation}\label{eq4.28}
\int_{0}^{t}\|u^{n}_{t}(\tau)\|^{2}_{2}d\tau< \mathcal{Q}(z_{0}),\quad 0\leq t< \infty.
\end{equation}
(28)
Using Sobolev inequality, (8) and (26), it follows that
\begin{eqnarray}\label{eq4.29}
\|u^{n}\|^{2}_{p+1}\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}
(29)
Furthermore, by (29), we get
\begin{eqnarray}\label{eq4.30}
|(|u^{n}|^{p-1}u^{n},u^{n})|\leq \|u^{n}\|^{p+1}_{p+1}< C_{*}^{p+1}\left(\frac{2C_{*}^{2}\lambda_{1}^{-1}\mathcal{Q}(z_{0})}{\eta}\right)^{\frac{p+1}{2}},\quad 0\leq t< \infty.
\end{eqnarray}
(30)
The estimates (26)-(30) permit us to obtain a subsequences of \(\left\{u_{n}\right\}\) which from now
on will be also denoted by \(\left\{u_{n}\right\}\) and functions \(u\), \(\chi\) such that
\begin{equation}\label{eq4.31}
u_{n}\rightarrow u \ \ \mbox {weak star in} \ \ \
L^{\infty}(0,\infty;H_{0}^{2}(\Omega)),\quad n\rightarrow+\infty,
\end{equation}
(31)
\begin{equation}\label{eq4.32}
u^{n}_{t}\rightarrow u_{t} \ \ \mbox {weak star in} \ \ \
L^{\infty}(0,\infty;L^{2}(\Omega)),\quad n\rightarrow+\infty,
\end{equation}
(32)
\begin{equation}\label{eq4.33}
|u^{n}|^{p-1}u^{n}\rightarrow \chi \ \ \mbox {weak star in} \ \ \
L^{\infty}(0,\infty;L^{\frac{p+1}{p}}(\Omega)),\quad n\rightarrow+\infty.
\end{equation}
(33)
Besides, from Lions-Aubin Lemma we also have
\begin{equation}\label{eq4.34}
u^{n}\rightarrow u \ \ \mbox {strongly in} \ \ \
L^{2}(0,\infty;L^{2}(\Omega)),\quad n\rightarrow+\infty,
\end{equation}
(34)
and consequently, making use of the Lemma 1.3 in [
8], we deduce
\begin{equation}\label{eq4.35}
|u^{n}|^{p-1}u^{n}\rightarrow \chi=|u|^{p-1}u \ \ \mbox {weak star in} \ \ \
L^{\infty}(0,\infty;L^{\frac{p+1}{p}}(\Omega)),\quad n\rightarrow+\infty.
\end{equation}
(35)
Thus, we obtain that \(u\) is a global weak of problem (4)-(6).
Next, we shall prove that \(u\) satisfies (11). From the discussion above, we obtain for
each fixed \(t>0\) that
\begin{equation}
\lim_{n\rightarrow +\infty}(g\circ\Delta u^{n})(t)=(g\circ\Delta u)(t),\quad \lim_{n\rightarrow +\infty}\|u^{n}\|_{p+1}^{p+1}=\|u\|_{p+1}^{p+1}.
\end{equation}
(36)
We obtain for
each fixed \(t>0\) that
\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,\nonumber\\&&
\end{eqnarray}
(37)
as \(n\rightarrow +\infty\), and
\begin{eqnarray}
\|u^{n}\|_{p+1}^{p+1}-\|u\|_{p+1}^{p+1}
&\leq&(p+1)\left|\int_{\Omega}|u+\theta_{n}u^{n}|^{p-1}(u+\theta_{n}u^{n})(u^{n}-u)dx\right|\nonumber\\
&\leq&(p+1)\|u+\theta_{n}u^{n}\|_{p+1}^{p}\|u^{n}-u\|_{p+1}\leq C\|u^{n}-u\|_{p+1}\rightarrow 0,
\end{eqnarray}
(38)
as \(n\rightarrow +\infty\), where \(0< \theta_{n}< 1\). Hence, we have
\begin{equation}
\lim_{n\rightarrow +\infty}(g\circ\Delta u^{n})(t)=(g\circ\Delta u)(t),\quad \lim_{n\rightarrow +\infty}\|u^{n}\|_{p+1}^{p+1}=\|u\|_{p+1}^{p+1}.
\end{equation}
(39)
From (15), it follows that \(E^{n}(0)\rightarrow E(0)\) as \(n\rightarrow+\infty\). Finally, taking \(n\rightarrow +\infty\) in (18), we deduce that the energy identity (11) holds for \(0\leq t< \infty\).
Step 2 General decay of the energy
Firstly, we state several Lemmas to prove the decay rate estimate of the energy.
Lemma 1.
Let \(u\in L^{\infty}(0,\infty;H^{2}_{0}(\Omega))\) be the solution of (4)-(6) and \(E(0)< \mathcal{Q}(z_{0})\), \(\|\nabla u_{0}\|_{2}< z_{0}\),
then we have
\begin{eqnarray}\label{eq4.40}
0\leq E(t)\leq\frac{1}{2}\|u_{t}\|_{2}^{2}+C_{1}\|\Delta u\|^{2}_{2}+\frac{1}{2}(g\circ \Delta u)(t),
\end{eqnarray}
(40)
where \(C_{1}=\frac{1}{2}+(2\lambda_{1})^{-1}\).
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}\label{eq4.41}
E(t)&=&\frac{1}{2}\|u_{t}\|^{2}_{2}+\frac{1}{2}\left(1-\int_{0}^{t}g(\tau)d\tau\right)\|\Delta u\|^{2}_{2}+\frac{1}{2}(g\circ \Delta u)(t)+\frac{1}{2}\|\nabla u\|^{2}_{2}-\frac{1}{p+1}\|u\|_{p+1}^{p+1}\nonumber\\
&\geq&\frac{1}{2}\|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})\geq0,
\end{eqnarray}
(41)
and
\begin{eqnarray}\label{eq4.42}
E(t)\leq\frac{1}{2}\|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}
\leq\frac{1}{2}\|u_{t}\|_{2}^{2}+C_{1}\|\Delta u\|^{2}_{2}+\frac{1}{2}(g\circ \Delta u)(t).
\end{eqnarray}
(42)
Lemma 2.
The energy \(E(t)\) satisfies
\begin{eqnarray}\label{eq4.43}
\frac{d E(t)}{dt}&\leq&-\|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}
(43)
Proof.
From (13), we have
\begin{eqnarray}\label{eq4.44}
\frac{d E(t)}{dt}&\leq&-\|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}
(44)
From assumptions \((G2)\) and since \(\int_{0}^{t}g'(\tau)d\tau=g(t)-g(0)\), we obtain
\begin{eqnarray}\label{eq4.45}
-\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}
(45)
Then, Combining (45) and (44) our conclusion holds. Multiplying (43) by \(e^{\kappa\zeta(t)}\) \((\kappa>0)\) and using (40), we have
\begin{eqnarray}\label{eq4.46}
\frac{d}{dt}\left(e^{\kappa\zeta(t)}E(t)\right)&\leq&-\|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&-\frac{1}{2}\left[2-\kappa\zeta_{t}(t)\right]\|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_{1}\kappa\zeta_{t}(t)\right]\|\Delta u(t)\|^{2}_{2}e^{\kappa\zeta(t)}E(t).
\end{eqnarray}
(46)
Using the fact that \(\zeta_{t}\) is decreasing by (12), we arrive at
\begin{eqnarray}\label{eq4.47}
\frac{d}{dt}\left(e^{\kappa\zeta(t)}E(t)\right)&\leq&-\frac{1}{2}\left[2-\kappa\zeta_{t}(0)\right]
\|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_{1}\kappa\zeta_{t}(0)\right]\|\Delta u(t)\|^{2}_{2}e^{\kappa\zeta(t)}E(t).
\end{eqnarray}
(47)
Choosing \(\|g\|_{L^{1}(0,\infty)}\) sufficiently small so that \(g(0)-\xi_{1}\|g\|_{L^{1}(0,\infty)}=B>0\)
and defining \(\kappa_{0}=\min\left\{\frac{2}{\zeta_{t}(0)},\frac{\xi_{2}}{\zeta_{t}(0)},\frac{B}{2C_{1}\zeta_{t}(0)}\right\},\)
we conclude by taking \(\kappa\in(0, \kappa_{0}]\) in (47) that
\(
\frac{d}{dt}\left(e^{\kappa\zeta(t)}E(t)\right)\leq0,\quad t>0.
\)
Integrating the above inequality over \((0,t)\), it follows that
\(
E(t)\leq E(0)e^{-\kappa\zeta(t)},\quad t>0.
\)
Competing Interests
The author declares no conflict of interest.
