In order to derive the desired result of Theorem 2.2 by the integral method, we establish the following weighted integral inequality.
Lemma 3.1
Let
u
be the solution of (1.1) under Assumption 2.1, then
$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant C E(S) \end{aligned}$$
for some constant
\(C>0\).
To prove the above inequality, we need the following two lemmas.
Lemma 3.2
Let
u
be the solution of (1.1) under Assumption 2.1, then
$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant & C \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{2}\,dt+ C \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt+CE(S). \end{aligned}$$
Proof
Multiplying by \(\xi (t)u(t)\) both sides of equation (1.1), integrating the resulting equation over \(\varOmega \times [S,T]\) (\(0\leqslant S\leqslant T\)), then using the boundary conditions and Lemma 2.2, we have
$$\begin{aligned} 0={}& \int_{S}^{T}\xi (t) \biggl( u, \vert u_{t} \vert ^{\rho }u_{tt}-\Delta u-\Delta u_{tt}+ \int^{t}_{0}g(t-s)\Delta u(s)\,ds \biggr) \,dt \\ ={}& \int_{S}^{T}\xi (t) \bigl( u, \vert u_{t} \vert ^{\rho }u_{tt} \bigr) \,dt+ \int _{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt+ \int_{S}^{T}\xi (t) ( \nabla u, \nabla u_{tt} ) \,dt \\ & {} - \int_{S}^{T}\xi (t) \biggl( \nabla u, \int^{t}_{0}g(t-s)\nabla u(s)\,ds \biggr) \,dt \\ ={}& \int_{S}^{T}\xi (t) \bigl( u, \vert u_{t} \vert ^{\rho }u_{tt} \bigr) \,dt + \int _{S}^{T}\xi (t) \bigl( 1-G(t) \bigr) \Vert \nabla u \Vert ^{2}\,dt+ \int_{S}^{T} \xi (t) ( \nabla u,\nabla u_{tt} ) \,dt \\ & {} - \int_{S}^{T}\xi (t) \biggl( \nabla u, \int^{t}_{0}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) \,ds \biggr) \,dt. \end{aligned}$$
(3.1)
According to the definition of the energy functional \(E(t)\), we get
$$\begin{aligned} \bigl( 1-G(t) \bigr) \Vert \nabla u \Vert ^{2}=2E(t)-\frac{2}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}- \Vert \nabla u_{t} \Vert ^{2}-g\circ \nabla u(t). \end{aligned}$$
(3.2)
Combining (3.1) with (3.2), we deduce that
$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt ={}& \frac{1}{\rho +2} \int_{S}^{T}\xi (t) \Vert u _{t} \Vert ^{\rho +2}_{\rho +2}\,dt+\frac{1}{2} \int_{S}^{T}\xi (t) \Vert \nabla u _{t} \Vert ^{2}\,dt \\ & {}+\frac{1}{2} \int_{S}^{T}\xi (t) (g\circ \nabla u) (t)\,dt \\ & {} -\frac{1}{2} \int_{S}^{T}\xi (t) \bigl( u, \vert u_{t} \vert ^{\rho }u_{tt} \bigr) \,dt - \frac{1}{2} \int_{S}^{T}\xi (t) ( \nabla u,\nabla u_{tt} ) \,dt \\ & {} +\frac{1}{2} \int_{S}^{T}\xi (t) \biggl( \nabla u, \int^{t}_{0}g(t-s) \bigl( \nabla u(s)-\nabla (t) \bigr) \,ds \biggr) \,dt. \end{aligned}$$
(3.3)
From Lemma 2.3, we see that
$$\begin{aligned} \frac{d}{dt}E(t) =\frac{1}{2}g'\circ \nabla u(t)- \frac{1}{2}g(t) \Vert \nabla u \Vert ^{2} \leqslant \frac{1}{2}g'\circ \nabla u(t), \end{aligned}$$
that is,
$$\begin{aligned} -g'\circ \nabla u(t)\leqslant -2E'(t), \end{aligned}$$
which, together with Assumption 2.1, implies
$$\begin{aligned} \int_{S}^{T}\xi (t) ( g\circ \nabla u ) (t)\,dt \leqslant - \int _{S}^{T} \bigl( g'\circ \nabla u \bigr) (t)\,dt\leqslant -2 \int_{S}^{T}E'(t)\,dt. \end{aligned}$$
(3.4)
For the fourth term on the right-hand side of (3.3), integrating by parts and using Lemma 2.2, we have
$$\begin{aligned} - \int_{S}^{T}\xi (t) \bigl( u, \vert u_{t} \vert ^{\rho }u_{tt} \bigr) \,dt ={} & - \frac{1}{\rho +1} \bigl( \xi (t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr)\big| _{S}^{T}+ \frac{1}{\rho +1} \int_{S}^{T} \bigl( \bigl(\xi (t)u \bigr)_{t}, \vert u _{t} \vert ^{\rho }u_{t} \bigr) \,dt \\ ={}& -\frac{1}{\rho +1} \bigl( \xi (t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr)\big| _{S}^{T}+ \frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi '(t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr) \,dt \\ & {} +\frac{1}{\rho +1} \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt. \end{aligned}$$
(3.5)
By Young inequality, Lemma 2.1 and the definition of \(E(t)\), we have
$$\begin{aligned} \biggl\vert -\frac{1}{\rho +1} \bigl( \xi (t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr) \biggr\vert \leqslant{} & \frac{1}{\rho +1} \biggl[ \xi (t) \biggl( \frac{1}{ \rho +2} \Vert u \Vert _{\rho +2}^{\rho +2}+\frac{\rho +1}{\rho +2} \Vert u_{t} \Vert _{\rho +2}^{\rho +2} \biggr) \biggr] \\ \leqslant {}& \xi (t) \biggl( \frac{1}{(\rho +1)(\rho +2)} \biggl( \frac{2}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}+1}B^{\rho +2}E^{\frac{\rho }{2}}(0)+1 \biggr) E(t) \\ \leqslant {}& k_{1}\xi (t)E(t), \end{aligned}$$
with some positive constant \(k_{1}\). Hence,
$$\begin{aligned} \biggl\vert -\frac{1}{\rho +1} \bigl( \xi (t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr)\big| _{S}^{T} \biggr\vert \leqslant 2k_{1}\xi (0)E(S). \end{aligned}$$
(3.6)
Similarly,
$$\begin{aligned} \biggl\vert \frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi '(t) u, \vert u_{t} \vert ^{\rho }u _{t} \bigr) \,dt \biggr\vert \leqslant {}& k_{1} \int_{S}^{T} \bigl\vert \xi '(t) \bigr\vert E(t)\,dt=-k _{1} \int_{S}^{T}\xi '(t)E(t)\,dt. \end{aligned}$$
(3.7)
For the fifth term on the right-hand side of (3.3), integrating by parts, we have
$$\begin{aligned} &- \int_{S}^{T}\xi (t) ( \nabla u,\nabla u_{tt} ) \,dt \\ &\quad =- \bigl( \xi (t)\nabla u,\nabla u_{t} \bigr)\big| _{S}^{T}+ \int_{S} ^{T} \bigl( \bigl( \xi (t)\nabla u \bigr) _{t},\nabla u_{t} \bigr) \,dt \\ & \quad \leqslant \xi (t) \biggl( \frac{1}{2} \Vert \nabla u \Vert ^{2}+ \frac{1}{2} \Vert \nabla u_{t} \Vert ^{2} \biggr)\bigg| _{S}^{T}+ \int_{S}^{T} \bigl( \xi '(t)\nabla u, \nabla u_{t} \bigr) \,dt + \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ &\quad \leqslant \biggl( \frac{1}{1-G(\infty )}+1 \biggr) \biggl[ 2\xi (0)E(S)+ \int_{S}^{T} \bigl\vert \xi '(t) \bigr\vert E(t)\,dt \biggr] + \int_{S}^{T}\xi (t) \Vert \nabla u _{t} \Vert ^{2}\,dt \\ &\quad \leqslant 2 k_{2}\xi (0)E(S)-k_{2} \int_{S}^{T}\xi '(t)E(t)\,dt+ \int _{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt, \end{aligned}$$
(3.8)
with some positive constant \(k_{2}\).
For the sixth term on the right-hand side of (3.3), we have
$$\begin{aligned} & \biggl( \nabla u, \int_{0}^{t}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr) \\ &\quad \leqslant \varepsilon \Vert \nabla u \Vert ^{2}+ \frac{1}{4\varepsilon } \int_{\varOmega } \biggl( \int_{0}^{t} g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr) ^{2}\,dx \\ &\quad \leqslant \varepsilon \Vert \nabla u \Vert ^{2}+ \frac{1}{4\varepsilon } \int_{0}^{t}g(s)\,ds \int_{\varOmega } \int_{0}^{t}g(t-s) \bigl\vert \nabla u(s)- \nabla u(t) \bigr\vert ^{2}\,ds\,dx \\ &\quad \leqslant \frac{2\varepsilon }{1-G(\infty )} E(t)+\frac{G( \infty )}{2\varepsilon }(g\circ \nabla u) (t). \end{aligned}$$
Combining with (3.4), we obtain
$$\begin{aligned} & \int_{S}^{T}\xi (t) \biggl( \nabla u(t), \int_{0}^{t}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \frac{2\varepsilon }{1-G(\infty )} \int_{S}^{T} \xi (t)E(t)\,dt -\frac{G(\infty )}{\varepsilon } \int_{S}^{T}E'(t)\,dt. \end{aligned}$$
(3.9)
By (3.3)–(3.9), we get
$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant {}& \biggl( \frac{1}{\rho +2}+\frac{1}{2( \rho +1)} \biggr) \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt+ \frac{3}{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ & {} +\frac{\varepsilon }{1-G(\infty )} \int_{S}^{T}\xi (t)E(t)\,dt- \frac{1}{2} ( k_{1}+k_{2} ) \int_{S}^{T}\xi '(t)E(t)\,dt \\ & {} - \biggl( 1+\frac{G(\infty )}{2\varepsilon } \biggr) \int_{S}^{T}E'(t)\,dt+ ( k_{1}+k_{2} ) \xi (0)E(S). \end{aligned}$$
(3.10)
Integrating by parts (and noting \(E'(t)\leqslant 0\)), we have
$$\begin{aligned} - \int_{S}^{T}\xi '(t)E(t)\,dt &= -\xi (t)E(t)| _{S}^{T}+ \int _{S}^{T}\xi (t)E'(t)\,dt \\ &= -\xi (T)E(T)+\xi (S)E(S)+ \int_{S}^{T}\xi (t)E'(t)\,dt \\ &\leqslant \xi (0)E(S), \end{aligned}$$
(3.11)
and
$$\begin{aligned} - \int_{S}^{T}E'(t)\,dt=E(S)-E(T) \leqslant E(S). \end{aligned}$$
(3.12)
Owing to (3.10)–(3.12), we get
$$\begin{aligned} & \int_{S}^{T}\xi (t)E(t)\,dt \\ &\quad \leqslant \biggl( \frac{1}{\rho +2}+\frac{1}{2( \rho +1)} \biggr) \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt+ \frac{3}{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ & \quad\quad {} +\frac{\varepsilon }{1-G(\infty )} \int_{S}^{T}\xi (t)E(t)\,dt + \biggl( 1+ \frac{G(\infty )}{2\varepsilon }+\frac{3}{2}(k_{1}+k_{2})\xi (0) \biggr) E(S). \end{aligned}$$
(3.13)
Choosing ε small enough, we obtain from (3.13) that
$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant & C \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{2}\,dt+ C \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt+CE(S). \end{aligned}$$
The proof of Lemma 3.2 is completed. □
Lemma 3.3
Let
u
be the solution of (1.1) under Assumption 2.1, then
$$\begin{aligned} \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt+ \int_{S}^{T} \xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt\leqslant \varepsilon C \int_{S}^{T} \xi (t)E(t)\,dt+ C(\varepsilon )E(S). \end{aligned}$$
Proof
Multiplying by \(\xi (t)\int_{0}^{t}g(t-s) ( u(s)-u(t) ) \,ds\) both sides of equation (1.1) and then integrating the resulting equation over \(\varOmega \times [S,T]\) (\(0\leqslant S\leqslant T\)) gives
$$\begin{aligned} \begin{aligned}[b] & \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds, \\ &\quad \vert u _{t} \vert ^{\rho }u_{tt}-\Delta u-\Delta u_{tt}+ \int^{t}_{0}g(t-s)\Delta u(s)\,ds \biggr) \,dt=0.\end{aligned} \end{aligned}$$
(3.14)
Integrating by parts and using Lemma 2.2, we obtain
$$\begin{aligned} &\int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{tt},\xi (t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad =\frac{1}{\rho +1} \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \xi (t) \int _{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr)\bigg| _{S}^{T} \\ &\quad \quad {} -\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho } u_{t}, \xi '(t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \quad {} -\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \xi (t) \int_{0}^{t}g'(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \quad {} +\frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi (t)G(t) \bigr) \bigl\Vert u _{t}(t) \bigr\Vert _{\rho +2}^{\rho +2}\,dt. \end{aligned}$$
Moreover, we have
$$\begin{aligned} &\int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds,- \Delta u_{tt} \biggr) \,dt \\ &\quad = \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds,\nabla u_{tt} \biggr) \,dt \\ &\quad = \biggl( \nabla u_{t},\xi (t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr)\,ds \biggr)\bigg| _{S}^{T} \\ & \quad\quad {} - \int_{S}^{T} \biggl( \nabla u_{t},\xi '(t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr) \,dt \\ &\quad\quad {} - \int_{S}^{T} \biggl( \nabla u_{t},\xi (t) \int^{t}_{0}g'(t-s) \bigl(\nabla u(s)- \nabla u(t)\bigr)\,ds \biggr) \,dt + \int_{S}^{T} \xi (t)G(t) \Vert \nabla u_{t} \Vert ^{2}\,dt \end{aligned}$$
and
$$\begin{aligned} & \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds,- \Delta u+ \int_{0}^{t} g(t-s)\Delta u(s)\,ds \biggr) \,dt \\ &\quad = \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds,\nabla u(t)\,dx \biggr) \,dt \\ &\quad \quad {} - \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds, \int_{0}^{t} g(t-s)\nabla u(s)\,ds \biggr) \,dt \\ &\quad =- \int_{S}^{T}\xi (t) \biggl\Vert \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds \biggr\Vert ^{2}\,dt \\ &\quad \quad {} + \int_{S}^{T}\xi (t) \bigl( 1-G(t) \bigr) \biggl( \int_{0}^{t}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) \,ds,\nabla u(t) \biggr) \,dt. \end{aligned}$$
Therefore, plugging the above three identities into (3.14), we get
$$\begin{aligned} &\frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi (t)G(t) \bigr) \bigl\Vert u_{t}(t) \bigr\Vert _{\rho +2}^{\rho +2}\,dt+ \int_{S}^{T} \xi (t)G(t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ &\quad = -\frac{1}{\rho +1} \biggl( \vert u_{t} \vert ^{\rho }u_{t},\xi (t) \int _{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr)\bigg| _{S}^{T} \\ & \quad\quad {} +\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{t},\xi '(t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ & \quad\quad {} +\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho } u_{t},\xi (t) \int_{0}^{t}g'(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ & \quad\quad {} - \biggl( \nabla u_{t},\xi (t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr)\,ds \biggr)\bigg| _{S}^{T} \\ & \quad\quad {} + \int_{S}^{T} \biggl( \nabla u_{t},\xi '(t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr) \\ & \quad\quad {} + \int_{S}^{T} \biggl( \nabla u_{t},\xi (t) \int^{t}_{0}g'(t-s) \bigl(\nabla u(s)- \nabla u(t)\bigr)\,ds \biggr) \\ & \quad\quad {} + \int_{S}^{T}\xi (t) \biggl\Vert \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds \biggr\Vert ^{2}\,dt \\ &\quad\quad {} - \int_{S}^{T}\xi (t) \bigl( 1-G(t) \bigr) \biggl( \int_{0}^{t}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) ,\nabla u(t)\,dx \biggr) \,dt. \end{aligned}$$
(3.15)
Using Cauchy and Hölder inequalities as well as Lemma 2.1, we have
$$\begin{aligned} & \biggl\vert \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \int_{0}^{t} g(t-s) \bigl(u(s)-u(t)\bigr) \,ds \biggr) \biggr\vert \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2} \int_{\varOmega } \biggl( \int_{0}^{t} g(t-s) \bigl\vert u(s)-u(t) \bigr\vert \,ds \biggr) ^{\rho +2}\,dx \\ &\quad =\frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+ \frac{1}{ \rho +2} \int_{\varOmega } \biggl( \int_{0}^{t} g^{\frac{\rho +1}{\rho +2}}(t-s)g ^{\frac{1}{\rho +2}}(t-s) \bigl\vert u(s)-u(t) \bigr\vert \,ds \biggr) ^{\rho +2}\,dx \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2} \biggl( \int_{0}^{t} g(t-s)\,ds \biggr) ^{\rho +1} \int_{\varOmega } \biggl( \int_{0}^{t} g(t-s) \bigl\vert u(s)-u(t) \bigr\vert ^{\rho +2} \,ds \biggr) \,dx \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2}G^{\rho +1}(t) \int_{0}^{t} g(t-s) \bigl\Vert u(s)- u(t) \bigr\Vert ^{\rho +2}_{ \rho +2} \,ds \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2}G^{\rho +1}(t)B^{\rho +2} \int_{0}^{t} g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{\rho +2} \,ds \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2}G^{\rho +1}(t) \\ &\quad \quad{}\times B^{\rho +2} \int_{0}^{t} g(t-s) \bigl( \bigl\Vert \nabla u(s) \bigr\Vert + \bigl\Vert \nabla u(t) \bigr\Vert \bigr) ^{\rho } \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2}G^{\rho +1}(t) \\ &\quad \quad{}\times B^{\rho +2} \int_{0}^{t} g(t-s) \biggl[ 2 \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{1}{2}} \biggr] ^{\rho } \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \\ &\quad =\frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+ \frac{2^{ \rho }}{\rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}} \int_{0}^{t} g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{2} \,ds \\ &\quad \leqslant (\rho +1)E(t)+\frac{2^{\rho }}{\rho +2}G^{\rho +1}(t)B ^{\rho +2} \biggl( \frac{2E(0)}{1-G(\infty )} \biggr) ^{\frac{\rho }{2}} g\circ \nabla u \\ &\quad \leqslant \biggl( (\rho +1)+\frac{{2^{\rho +1}}}{\rho +2}G^{ \rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G(\infty )} \biggr) ^{\frac{ \rho }{2}} \biggr) E(t), \end{aligned}$$
(3.16)
which implies that
$$\begin{aligned} & {-} \frac{1}{\rho +1} \biggl( \vert u_{t} \vert ^{\rho }u_{t},\xi (t) \int _{0}^{t} g(t-s) \bigl(u(s)-u(t)\bigr) \,ds \biggr)\bigg| _{S}^{T} \\ &\quad \leqslant \frac{2\xi (0)}{\rho +1} \biggl( (\rho +1)+\frac{ {2^{\rho +1}}}{\rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}} \biggr) E(S) \\ &\quad \leqslant \frac{2\xi (0)}{\rho +1} \biggl( (\rho +1)+\frac{ {2^{\rho +1}}}{\rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}} \biggr) E(S). \end{aligned}$$
(3.17)
In order to estimate the second term on the right-hand side of (3.15), we apply (3.16) and (3.11) to get
$$\begin{aligned} &\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho } u_{t},\xi '(t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \frac{1}{\rho +1} \biggl( (\rho +1)+\frac{{2^{\rho +1}}}{ \rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G(\infty )} \biggr) ^{\frac{\rho }{2}} \biggr) \int_{S} ^{T} \bigl\vert \xi '(t) \bigr\vert E(t)\,dt \\ &\quad = -\frac{1}{\rho +1} \biggl( (\rho +1)+\frac{{2^{\rho +1}}}{ \rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G(\infty )} \biggr) ^{\frac{\rho }{2}} \biggr) \int_{S} ^{T}\xi '(t)E(t)\,dt \\ &\quad \leqslant \frac{\xi (0)}{\rho +1} \biggl( (\rho +1)+\frac{{2^{ \rho +1}}}{\rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}} \biggr) E(S). \end{aligned}$$
(3.18)
In addition, for any \(\delta >0\), using Young inequality, we have
$$\begin{aligned} & \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \xi (t) \int_{0}^{t} g'(t-s) \bigl(u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad = \int_{S}^{T} \biggl( \xi^{\frac{\rho +1}{\rho +2}}(t) \vert u_{t} \vert ^{ \rho }u_{t}, \xi^{\frac{1}{\rho +2}}(t) \int_{0}^{t} g'(t-s) \bigl(u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \delta \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{\rho +2}_{ \rho +2} \,dt+C(\delta ) \int_{S}^{T}\xi (t) \biggl[ \int_{\varOmega } \biggl( \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\vert u(s)-u(t) \bigr\vert \,ds \biggr) ^{\rho +2}\,dx \biggr] \,dt. \end{aligned}$$
Using Hölder inequality and Lemmas 2.1 and 2.3, we have
$$\begin{aligned} & \int_{S}^{T}\xi (t) \biggl[ \int_{\varOmega } \biggl( \int_{0}^{t} \bigl\vert g' \bigr\vert ^{\frac{ \rho +1}{\rho +2}}(t-s) \bigl\vert g' \bigr\vert ^{\frac{1}{\rho +2}}(t-s) \bigl\vert u(s)-u(t) \bigr\vert \,ds \biggr) ^{\rho +2}\,dx \biggr] \,dt \\ &\quad \leqslant \int_{S}^{T} \xi (t) \biggl( \int_{0}^{t} -g'(t-s)\,dt \biggr) ^{\rho +1} \biggl[ \int_{\varOmega } \biggl( \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\vert u(s)-u(t) \bigr\vert ^{\rho +2}\,ds \biggr) \,dx \biggr] \,dt \\ &\quad \leqslant g^{\rho +1}(0)\xi (0) \int_{S}^{T} \biggl( \int_{0} ^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert u(s)-u(t) \bigr\Vert ^{\rho +2}_{\rho +2} \,ds \biggr) \,dt \\ &\quad \leqslant g^{\rho +1}(0)\xi (0)B^{\rho +2} \int_{S}^{T} \biggl( \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{\rho +2}_{\rho +2} \,ds \biggr) \,dt \\ &\quad \leqslant 2^{\rho }g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) \int_{S}^{T} \biggl( \int _{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad =-2^{\rho }g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) \int_{S}^{T} \biggl( \int _{0}^{t} g'(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds \biggr) \,dt \\ &\quad \leqslant -2^{\rho +1}g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) \int_{S}^{T}E'(t)\,dt \\ &\quad \leqslant 2^{\rho +1}g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) E(S). \end{aligned}$$
Therefore,
$$\begin{aligned} \begin{aligned}[b] & \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \xi (t) \int_{0}^{t} g'(t-s) \bigl(u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \delta \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{\rho +2}_{ \rho +2} \,dt+2^{\rho +1}C(\delta )g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) E(S). \end{aligned} \end{aligned}$$
(3.19)
Since
$$\begin{aligned} & \biggl\vert \biggl( \nabla u_{t}, \xi (t) \int_{0}^{t} g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr) \,ds \biggr) \biggr\vert \\ &\quad \leqslant \frac{1}{2}\xi (t) \Vert \nabla u_{t} \Vert ^{2}+\frac{1}{2} \xi (t) \int_{\varOmega } \biggl( \int_{0}^{t} g(t-s) \bigl\vert \nabla u(s)-\nabla u(t) \bigr\vert \,ds \biggr) ^{2}\,dx \\ &\quad =\frac{1}{2}\xi (t) \Vert \nabla u_{t} \Vert ^{2}+\frac{1}{2}\xi (t)G(t) \int_{0}^{t} g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds\,dx \\ &\quad \leqslant \xi (0)E(t)+\frac{1}{2}\xi (t)G(t)g\circ \nabla u \leqslant 2\xi (0)E(t), \end{aligned}$$
(3.20)
one obtains
$$\begin{aligned} - \biggl( \nabla u_{t},\xi (t) \int_{0}^{t} g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr) \,ds \biggr)\bigg| _{S}^{T}\leqslant 4\xi (0)E(S). \end{aligned}$$
(3.21)
Using (3.20) and (3.11), we have
$$\begin{aligned} \int_{S}^{T} \biggl( \nabla u_{t},\xi '(t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr)\,ds \biggr) \leqslant 2 \int_{S}^{T} \bigl\vert \xi '(t) \bigr\vert E(t)\leqslant 2\xi (0)E(S). \end{aligned}$$
(3.22)
Similarly,
$$\begin{aligned} & \int_{S}^{T} \biggl( \nabla u_{t}, \xi (t) \int_{0}^{t} g'(t-s) \bigl(\nabla u(s)- \nabla u(t)\bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt \\ & \quad \quad {} +\frac{1}{2\delta } \int_{S}^{T} \xi (t) \biggl( - \int_{0}^{t}g'(t-s)\,ds \biggr) \cdot \biggl( \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds \biggr) \,dt \\ &\quad =\frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt-\frac{1}{2 \delta } \int_{S}^{T}\xi (t) \int_{0}^{t} g'(s) \,ds \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds\,dt \\ &\quad \leqslant \frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt+\frac{g(0)}{2\delta } \int_{S}^{T}\xi (t) \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds\,dt \\ &\quad \leqslant \frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt-\frac{g(0)}{\delta } \int_{S}^{T}\xi (t)E'(t)\,dt \\ &\quad \leqslant \frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt+\frac{g(0)\xi (0)}{\delta }E(S). \end{aligned}$$
(3.23)
Now, since \(g(t)\geqslant 0\) and \(G(\infty )<1\), we get
$$\begin{aligned} & \int_{S}^{T}\xi (t) \biggl\Vert \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds \biggr\Vert ^{2}\,dt \\ &\quad \leqslant \int_{S}^{T}\xi (t) \biggl( \int_{0}^{t}g(s)\,ds \biggr) \biggl( \int_{0}^{t}g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant G(\infty ) \int_{S}^{T}\xi (t) \biggl( \int_{0}^{t}g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant G(\infty ) \int_{S}^{T} \biggl( \int_{0}^{t}\frac{ \xi (t)}{\xi (t-s)}\xi (t-s)g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant G(\infty ) \int_{S}^{T} \biggl( \int_{0}^{t}\xi (t-s)g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant -G(\infty ) \int_{S}^{T} \biggl( \int_{0}^{t}g'(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant - 2G(\infty ) \int_{S}^{T}E'(t)\,dt\leqslant 2G( \infty )E(S)\leqslant 2E(S) \end{aligned}$$
(3.24)
and
$$\begin{aligned} & {-} \int_{S}^{T}\xi (t) \bigl( 1-G(t) \bigr) \biggl( \int_{0}^{t}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) ,\nabla u(t) \biggr) \,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt+\frac{G( \infty )}{4\varepsilon } \int_{S}^{T}\xi (t) \biggl( \int_{0}^{t}g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad =\varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt \\ &\quad\quad{} +\frac{G( \infty )}{4\varepsilon } \int_{S}^{T} \biggl( \int_{0}^{t}\frac{\xi (t)}{ \xi (t-s)}\xi (t-s)g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt+\frac{G( \infty )}{4\varepsilon } \int_{S}^{T} \biggl( \int_{0}^{t}\xi (t-s)g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt-\frac{G( \infty )}{4\varepsilon } \int_{S}^{T} \biggl( \int_{0}^{t}g'(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt-\frac{G( \infty )}{2\varepsilon } \int_{S}^{T}E'(t)\,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt+\frac{1}{2 \varepsilon }E(S)\leqslant \varepsilon \int_{S}^{T}\xi (t)E(t)\,dt+\frac{1}{2 \varepsilon }E(S). \end{aligned}$$
(3.25)
Combining (3.15)–(3.25), we obtain
$$\begin{aligned} &\frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi (t)G(t) \bigr) \Vert u_{t} \Vert _{ \rho +2}^{\rho +2}\,dt+ \int_{S}^{T} \xi (t)G(t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ &\quad \leqslant \delta \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{\rho +2}_{\rho +2} \,dt+\frac{ \delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt+\varepsilon \int _{S}^{T}\xi (t)E(t)\,dt \\ &\quad \quad {} + \biggl[ C_{1}+\frac{g(0)\xi (0)}{\delta }+\frac{1}{2\varepsilon } \biggr] E(S). \end{aligned}$$
(3.26)
Since g is continuous and \(g(0)>0\), for any \(t_{0}>0\), we have
$$\begin{aligned} G(t)= \int_{0}^{t}g(s)\,ds\geqslant \int_{0}^{t_{0}}g(s)\,dx>0,\quad \forall t\geqslant t_{0}. \end{aligned}$$
Now, if we fix \(\delta >0\) small enough such that
$$\begin{aligned} \delta < \int_{0}^{t_{0}}g(s)\,ds, \end{aligned}$$
then by (3.26), for \(T>S\geqslant t_{0}\), we have
$$\begin{aligned} \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt+ \int_{S}^{T} \xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt\leqslant \varepsilon C \int_{S}^{T} \xi (t)E(t)\,dt+ C(\varepsilon )E(S). \end{aligned}$$
□
Proof of Lemma 3.1
Plugging the estimate of Lemma 3.3 into the inequality of Lemma 3.2, and fixing ε small enough, we obtain
$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant C E(S) \end{aligned}$$
for some constant \(C>0\). Letting \(T\rightarrow +\infty \), we have
$$\begin{aligned} \int_{t}^{+\infty }\psi '(s)E(s)\,ds\leqslant {cE(t)},\quad \forall t\geqslant t_{0}, \end{aligned}$$
(3.27)
with
$$\begin{aligned} \psi (t):= \int_{t_{0}}^{t}\xi (s)\,ds. \end{aligned}$$
□
Proof of Theorem 2.2
From Assumption 2.1 and Lemma 2.3, we know that \(E(t)\) is a non-increasing function and \(\psi :[t_{0},+ \infty )\rightarrow \mathbb{R}^{+}\) is a strictly increasing \(C^{2}\) function such that \(\psi (t_{0})=0\) and \(\lim_{t\rightarrow +\infty } \psi (t)=+\infty \). Firstly, we define a new function \(f:[t_{0},\infty )\rightarrow \mathbb{R}^{+}\) as follows:
$$\begin{aligned} f(\tau )=E\bigl(\psi^{-1}(\tau )\bigr), \end{aligned}$$
then f is a non-increasing function such that
$$\begin{aligned} \int_{\psi (S)}^{\psi (T)}f(\tau )\,d\tau &= \int_{\psi (S)}^{\psi (T)}E\bigl( \psi^{-1}(\tau ) \bigr)\,d\tau = \int_{S}^{T}E(t)\psi '(t)\,dt \\ &\leqslant \int_{S}^{+\infty }E(t)\psi '(t)\,dt\leqslant cE(S)=cf\bigl( \psi (S)\bigr),\quad \forall t_{0}\leqslant S< T< \infty . \end{aligned}$$
Set \(t=\psi (S)\). Since \(\lim_{T\rightarrow +\infty }\psi (T)=+ \infty \), we get
$$\begin{aligned} \int_{t}^{+\infty }f(\tau )\,d\tau \leqslant cf(t),\quad \forall t \geqslant t_{0}. \end{aligned}$$
(3.28)
Next, we define the following function:
$$\begin{aligned} h(x)=e^{\frac{1}{c}x} \int_{x}^{+\infty }f(s)\,ds, \end{aligned}$$
(3.29)
where \(c>0\) is a constant. Noting (3.28), we obtain
$$\begin{aligned} h'(x)=\frac{1}{c}e^{\frac{1}{c}x} \biggl( \int_{x}^{+\infty }f(s)\,ds-cf(x) \biggr) \leqslant 0. \end{aligned}$$
(3.30)
Integrating (3.30) over \([t_{0},t]\) and noting (3.28), we have
$$\begin{aligned} h(t)\leqslant h(t_{0})=e^{\frac{1}{c}t_{0}} \int_{t_{0}}^{+\infty }f(s)\,ds \leqslant c_{1}f(t_{0}). \end{aligned}$$
(3.31)
Furthermore, using (3.29) and (3.31), we obtain
$$\begin{aligned} \int_{t}^{+\infty }f(s)\,ds\leqslant c_{1}f(t_{0})e^{-\frac{1}{c}t}. \end{aligned}$$
(3.32)
On the other hand, noting that f is a positive non-increasing function, it is easy to see that
$$\begin{aligned} \int_{t}^{+\infty }f(s)\,ds\geqslant \int_{t}^{t+c}f(s)\,ds\geqslant cf(t+c). \end{aligned}$$
(3.33)
Combining (3.32) with (3.33), we get
$$\begin{aligned} f(t+c)\leqslant c_{2}f(t_{0})e^{-\frac{1}{c}t}. \end{aligned}$$
(3.34)
Letting \(s=t+c\) in (3.34), we have
$$\begin{aligned} f(s)\leqslant c_{2}f(t_{0})e^{1-\frac{1}{c}s}, \end{aligned}$$
that is,
$$\begin{aligned} E\bigl(\psi^{-1}(s)\bigr)\leqslant c_{2}f(t_{0})e^{1-\frac{1}{c}s}. \end{aligned}$$
(3.35)
Moreover, letting \(t=\psi^{-1}(s)\) in (3.35), we get
$$\begin{aligned} E(t)\leqslant c_{2}f(t_{0})e^{1-\frac{1}{c}\psi (t)}, \end{aligned}$$
that is,
$$\begin{aligned} E(t)\leqslant C_{0}e^{-\omega \psi (t)}=C_{0}e^{-\omega \int_{t_{0}} ^{t}\xi (s)\,ds}, \end{aligned}$$
for some constants \(C_{0}\) and \(\omega =\frac{1}{c}>0\).
The proof of Theorem 2.2 is completed. □
Remark 3.1
From Theorem 2.2, if we choose different \(\xi (t)\), we can get different decay results. Choosing \(\xi (t) \equiv {a}\), we get the exponential decay result
$$\begin{aligned} E(t)\leqslant Ce^{-\omega t},\quad \forall t\geqslant t_{0}. \end{aligned}$$
Now consider \(\xi (t)=\frac{1}{(1+t)^{\gamma }}\) (\(0<\gamma \leqslant 1\)). If \(\gamma =1\), we get the polynomial decay result
$$\begin{aligned} E(t)\leqslant C (1+t)^{-\omega },\quad \omega >0, \forall t\geqslant t_{0}. \end{aligned}$$
If \(0<\gamma <1\), we get a decay result of the form
$$\begin{aligned} E(t)\leqslant Ce^{-\frac{\omega }{1-\gamma }(1+t)^{1-\gamma }},\quad \omega >0, \forall t\geqslant t_{0}. \end{aligned}$$