In this section, we discuss the decay rates of the energy functional associated with problem (1.6). To achieve this, we state and prove several lemmas that will be fundamental in establishing the main result.
4.1 Technical lemmas
The energy functional associated with problem (1.6) is given by
$$ {E(t)= \frac{1}{2} \int_{\Omega}|u_{t}|^{2} + \frac{1}{2} \|u\| _{H_{*}^{2}(\Omega)}^{2} + \int_{\Omega}H(u)}. $$
(4.1)
Lemma 4.1
The energy functional defined in (4.1) satisfies
$$ {\frac{dE(t)}{dt}= -\beta(t) \int_{\Omega }u_{t}g(u_{t}) \leq0.} $$
(4.2)
Proof
Multiplying (1.6) by \(u_{t}\) and integrating over Ω, we obtain
$$ {\frac{d}{dt} \biggl( \frac{1}{2} \int_{\Omega }|u_{t}|^{2} + \frac{1}{2} \|u\|_{H_{*}^{2}(\Omega)}^{2} + \int_{\Omega }H(u) \biggr) + \beta(t) \int_{\Omega}u_{t}g(u_{t}) = 0.} $$
From (A3) we get that \({sg(s)>0}\) for all \({s\neq0}\). Thus, by using (A1), we obtain
$$ {\frac{dE(t)}{dt}= -\beta(t) \int_{\Omega}u_{t}g(u_{t})\leq0.} $$
(4.3)
We note here that the calculations are justified for regular solutions. However, the result in (4.3) remains true for a weak solution by a density argument. □
Define the functional
$$ {F(t) = mE(t)+ \int_{\Omega}uu_{t},} $$
(4.4)
where m is a positive constant to be specified later.
Lemma 4.2
Assume that (A1)-(A3) hold. Then the functional
F
satisfies, along the solution of (1.6), the estimates
$${F'(t)\leq- E(t)+ C \int_{\Omega} \bigl( u_{t}^{2} + \bigl\vert ug(u_{t}) \bigr\vert \bigr)} $$
and
where
C
is a positive constant.
Proof
By using (1.6), definition (3.2), Lemma 4.1 and exploiting assumptions (A1) and (A2), direct differentiation gives
$$\begin{aligned} { F'(t)} =&{ mE'(t)+ \int_{\Omega}u_{t}^{2} + \int_{\Omega }uu_{tt}} \\ =& {-m\beta(t) \int_{\Omega}u_{t}g(u_{t})+ \int_{\Omega}u_{t}^{2} - \| u \|_{H_{*}^{2}(\Omega)}^{2}- \beta(t) \int_{\Omega}ug(u_{t}) - \int_{\Omega }uh(u)} \\ \leq& { \int_{\Omega}u_{t}^{2} - \frac{1}{2}\|u \|_{H_{*}^{2}(\Omega )}^{2} - \int_{\Omega}H(u)- \beta(t) \int_{\Omega}ug(u_{t})+ \int _{\Omega} \bigl( H(u)-uh(u) \bigr)} \\ \leq&{-E(t) + \frac{3}{2} \int_{\Omega}u_{t}^{2} + \beta(t) \int _{\Omega} \bigl\vert ug(u_{t}) \bigr\vert } \\ \leq& {-E(t)+ C \int_{\Omega} \bigl( u_{t}^{2} + \bigl\vert ug(u_{t}) \bigr\vert \bigr).} \end{aligned}$$
(4.5)
Next, we show that \({F\sim E}\). Using Young’s inequality and Lemma 2.2, we have
$$\begin{aligned} { F(t)} \leq&{ mE(t) + \frac{1}{2} \int_{\Omega}u_{t}^{2} + \frac{1}{2}\|u \|_{L^{2}(\Omega)}^{2}} \\ \leq&{ mE(t)+\frac{1}{2} \int_{\Omega}u_{t}^{2} + \frac{C_{*}}{2}\| u \|_{H_{*}^{2}(\Omega)}^{2} \leq\lambda_{2} E(t).} \end{aligned}$$
(4.6)
Also,
$$\begin{aligned} \begin{aligned} { F(t)} &\geq { mE(t) - \frac{1}{2} \int_{\Omega} u_{t}^{2} - \frac{1}{2} \|u\|_{L^{2}(\Omega)}^{2}} \\ &\geq { mE(t)- \frac{1}{2} \int_{\Omega}u_{t}^{2} - \frac {C_{*}}{2} \|u\|_{H_{*}^{2}(\Omega)}^{2}} \\ &= {\frac{(m-1)}{2} \int_{\Omega}u_{t}^{2} + \frac{(m-C_{*})}{2}\| u \|_{H_{*}^{2}(\Omega)}^{2} + m \int_{\Omega}H(u).} \end{aligned} \end{aligned}$$
We choose \({m>0}\) large enough so that \({(m-1), (m-C_{*}) >0 }\) and arrive at
$$ {F(t) \geq\lambda_{1} E(t).} $$
(4.7)
Thus, we get from (4.6) and (4.7) that
$${\lambda_{1} E(t)\leq F(t)\leq\lambda_{2} E(t).} $$
This completes the proof. □
Next, we choose \({0<\epsilon_{1}\leq\epsilon}\) so that
$$ {sg(s)\leq\min\bigl\lbrace \epsilon,M(\epsilon)\bigr\rbrace ,\quad \forall |s|\leq \epsilon_{1}.} $$
(4.8)
Then, for \({|s|\geq\epsilon_{1}}\), the function \({s\longmapsto \frac{|g(s)|}{|s|}}\) is continuous on compact intervals and thus attains its extrema. Thus, it follows from assumption (A3) that
$$ \textstyle\begin{cases} c'_{1}|s|\leq|g(s)|\leq c'_{2}|s|, & \mbox{if } |s|\geq\epsilon_{1}, \\ s^{2} + g^{2}(s)\leq M^{-1}(sg(s)), & \mbox{if } |s|\leq\epsilon_{1}. \end{cases} $$
(4.9)
As in [23], let us partition Ω as follows:
$${\Omega_{1} = \bigl\lbrace (x,y)\in\Omega: |u_{t}|\leq \epsilon_{1} \bigr\rbrace \quad \mbox{and} \quad \Omega_{2} = \bigl\lbrace (x,y)\in\Omega: |u_{t}| > \epsilon_{1} \bigr\rbrace .} $$
Lemma 4.3
The following inequalities hold for any
\(\epsilon>0\)
along the solution of (1.6):
$$ { \int_{\Omega_{1}} \bigl( u_{t}^{2} + \bigl\vert ug(u_{t}) \bigr\vert \bigr)\leq \int _{\Omega_{1}} u_{t}^{2} + C_{*} \epsilon E(t) + C_{\epsilon} \int_{\Omega _{1}} \bigl\vert g(u_{t}) \bigr\vert ^{2}} $$
(4.10)
and
$$ { \int_{\Omega_{2}} \bigl( u_{t}^{2} + \bigl\vert ug(u_{t}) \bigr\vert \bigr) \leq C\epsilon E(t)-C_{\epsilon}E'(t),} $$
(4.11)
where
\({C_{*}}\)
is the embedding constant defined in Lemma 2.2
and
\({C_{\epsilon}}\)
is a generic positive constant depending on
ϵ.
Proof
For the first inequality (4.10), we use Young’s inequality and Lemma 2.2 to get
$$\begin{aligned} { \int_{\Omega_{1}} \bigl( u_{t}^{2} + \bigl\vert ug(u_{t}) \bigr\vert \bigr)} \leq&{ \int_{\Omega_{1}} u_{t}^{2} + \epsilon \int_{\Omega_{1}} \vert u \vert ^{2} + C_{\epsilon} \int_{\Omega_{1}} \bigl\vert g(u_{t}) \bigr\vert ^{2}} \\ \leq&{ \int_{\Omega_{1}} u_{t}^{2} + C_{*} \epsilon\|u\|_{H_{*}^{2}(\Omega )}^{2} + C_{\epsilon} \int_{\Omega_{1}} \bigl\vert g(u_{t}) \bigr\vert ^{2}} \\ \leq& { \int_{\Omega_{1}} u_{t}^{2} + C_{*} \epsilon E(t) + C_{\epsilon } \int_{\Omega_{1}} \bigl\vert g(u_{t}) \bigr\vert ^{2}.} \end{aligned}$$
(4.12)
For the second inequality (4.11), we use Lemma 2.2 and Hölder’s inequality to obtain
$$\begin{aligned} \begin{aligned}[b] { \int_{\Omega_{2}} \bigl\vert ug(u_{t}) \bigr\vert } & \leq { \biggl( \int_{\Omega _{2}} \vert u \vert ^{2} \biggr)^{\frac{1}{2}} \biggl( \int_{\Omega_{2}} \bigl\vert g(u_{t}) \bigr\vert ^{2} \biggr)^{\frac{1}{2}} } \\ &\leq{\|u\|_{L^{2}(\Omega)} \biggl( \int_{\Omega_{2}} \bigl\vert g(u_{t}) \bigr\vert ^{2} \biggr)^{\frac{1}{2}}} \\ &\leq{ C_{*}\|u\|_{H_{*}^{2}(\Omega)} \biggl( \int_{\Omega_{2}} \bigl\vert g(u_{t}) \bigr\vert ^{2} \biggr)^{\frac{1}{2}}.} \end{aligned} \end{aligned}$$
(4.13)
Now, from (4.9)1 we observe that
$$ {|s|^{2}\leq c_{1}''sg(s) \quad \mbox{and} \quad \bigl\vert g(s) \bigr\vert ^{2}\leq c_{2}''sg(s) \quad \mbox{for some positive constants } c_{1}'', c_{2}''.} $$
Thus, with this in mind and Young’s inequality, we obtain
$$\begin{aligned} { \int_{\Omega_{2}} \bigl( u_{t}^{2} + \bigl\vert ug(u_{t}) \bigr\vert \bigr)} \leq&{ C \int_{\Omega_{2}} u_{t}g(u_{t}) + C \bigl( \|u \|_{H_{*}^{2}(\Omega)}^{2} \bigr) ^{\frac{1}{2}} \biggl( \int_{\Omega_{2}} u_{t}g(u_{t}) \biggr)^{\frac {1}{2}}} \\ \leq&{-C E'(t) + C \bigl( E(t) \bigr) ^{\frac{1}{2}} \bigl(-E'(t) \bigr)^{\frac{1}{2}}} \\ \leq& {-C E'(t) + C \bigl( \epsilon \bigl( E(t) \bigr) -C_{\epsilon}E'(t) \bigr)} \\ =& { C\epsilon E(t) - C_{\epsilon} E'(t).} \end{aligned}$$
(4.14)
□
Lemma 4.4
For
ϵ
small enough and two positive constants
d, C, the functional defined by
$${L(t)= F_{1}(t)+ C_{\epsilon}E(t), \quad \textit{where } F_{1}(t)= F(t)+ C_{\epsilon}E(t)} $$
satisfies, along the solution of (1.6), the estimate
$$ {L'(t)\leq-dE(t)+ C \int_{\Omega_{1}} \bigl( u_{t}^{2} + \bigl\vert g(u_{t}) \bigr\vert ^{2} \bigr)} $$
(4.15)
and
Proof
Using Lemmas 4.2 and 4.3, direct computations give
$$\begin{aligned} { F_{1}'(t)} =& { F'(t)+ C_{\epsilon}E'(t)} \\ \leq& {-E(t)+ C \int_{\Omega_{1}} \bigl( u_{t}^{2} + \bigl\vert ug(u_{t}) \bigr\vert \bigr) + C \int_{\Omega_{2}} \bigl( u_{t}^{2} + \bigl\vert ug(u_{t}) \bigr\vert \bigr)} \\ \leq& { -E(t)+ C \int_{\Omega_{1}} u_{t}^{2} + CC_{e} \epsilon E(t) + C_{\epsilon} \int_{\Omega_{1}} \bigl\vert g(u_{t}) \bigr\vert ^{2} + C\epsilon E(t)-C_{\epsilon }E'(t)} \\ \leq& {-(1-C\epsilon)E(t)+ C_{\epsilon} \int_{\Omega_{1}} \bigl( u_{t}^{2} + \bigl\vert g(u_{t}) \bigr\vert ^{2} \bigr)-C_{\epsilon}E'(t).} \end{aligned}$$
That is,
$$ { \bigl(F_{1}(t)+ C_{\epsilon}E(t) \bigr)'\leq -(1-C \epsilon )E(t)+ C_{\epsilon} \int_{\Omega_{1}} \bigl( u_{t}^{2} + \bigl\vert g(u_{t}) \bigr\vert ^{2} \bigr).} $$
(4.16)
This implies
$$ {L'(t)\leq -(1-C\epsilon)E(t)+ C_{\epsilon} \int_{\Omega _{1}} \bigl( u_{t}^{2} + \bigl\vert g(u_{t}) \bigr\vert ^{2} \bigr).} $$
(4.17)
We then choose ϵ small enough so that \({ (1-C\epsilon)>0}\) and obtain the result. It is easy to see that \({ L\sim E}\) since \({F\sim E}\). This completes the proof. □
4.2 Main decay result
Now, we state and prove our main decay result.
Theorem 4.1
Assume that (A1)-(A3) hold. Then there exist positive constants
\(k_{1}\), \(k_{2}\), \(k_{3}\), \(\epsilon_{0}\)
such that the solution of (1.6) satisfies
$$ {E(t)\leq k_{3}M_{1}^{-1} \biggl(k_{1} \int_{0}^{t}\beta(s)\, ds + k_{2} \biggr), \quad \forall t\geq0,} $$
(4.18)
where
$$ {M_{1}(t)= \int_{t}^{1}\frac{1}{M_{2}(s)}\, ds,\qquad M_{2}(t)=tM'(\epsilon_{0} t)} $$
(4.19)
and
\({M_{1}}\)
is strictly decreasing on
\((0,1]\)
and
\({\lim_{t\rightarrow0}M_{1}(t)=+\infty}\).
Proof
We have two cases as follows.
Case I. M is linear on \({(0,\epsilon]}\): Multiplying (4.15) by \(\beta(t)\) and using (4.9)2, we deduce that
$$\begin{aligned} {\beta(t)L'(t)} \leq& {-d\beta(t)E(t)+ C\beta(t) \int _{\Omega_{1}} \bigl( u_{t}^{2} + \bigl|g(u_{t})\bigr|^{2} \bigr)} \\ \leq&{-d\beta(t)E(t)+ C\beta(t) \int_{\Omega _{1}}M^{-1}\bigl(u_{t}g(u_{t}) \bigr)} \\ =&{ -d\beta(t)E(t)+ C\beta(t) \int_{\Omega_{1}}u_{t}g(u_{t})} \\ \leq& {-d\beta(t)E(t)+ C\beta(t) \int_{\Omega}u_{t}g(u_{t})} \\ =& {-d\beta(t)E(t)- CE'(t).} \end{aligned}$$
By using (A1), we obtain
$$ { \bigl(\beta(t)L(t) + CE(t) \bigr)'\leq-d \beta(t)E(t).} $$
(4.20)
Let \({J_{1} = \beta L+ C E} \). Then \({J_{1} \sim E} \) since \({L \sim E} \), and we get from (4.20)
$$ {J_{1}'(t)\leq-k_{1} \beta(t)J_{1}(t).} $$
(4.21)
Simple integration of (4.21) over \({(0,t)}\) and using the fact that \({J_{1}\sim E}\) give
$$ {E(t)\leq k_{2}e^{-k_{1} \int_{0}^{t}\beta(s)\, ds}=c M_{1}^{-1} \biggl(c \int_{0}^{t}\beta(s)\, ds \biggr) .} $$
Case II. M is nonlinear on \({(0,\epsilon]}\). In this case, we consider the functional \({I(t)}\) defined by
$$ {I(t)= \frac{1}{|\Omega_{1}|} \int_{\Omega_{1}} u_{t}g(u_{t}).} $$
We know that M is convex, so \(M^{-1}\) is concave. Thus, Jensen’s inequality yields
$$ {M^{-1}\bigl(I(t)\bigr)\geq\frac{1}{|\Omega_{1}|} \int_{\Omega_{1}}M^{-1}\bigl(u_{t}g(u_{t}) \bigr).} $$
(4.22)
By using (4.9)2, we obtain
$$ {\beta(t) \int_{\Omega_{1}} \bigl( u_{t}^{2} + \bigl\vert g(u_{t}) \bigr\vert ^{2} \bigr)\leq \beta(t) \int_{\Omega_{1}}M^{-1}\bigl(u_{t}g(u_{t}) \bigr)\leq C\beta(t)M^{-1}\bigl(I(t)\bigr).} $$
(4.23)
We multiply (4.15) by \(\beta(t)\) and use (4.23) to arrive at
$$\begin{aligned} {\beta(t)L'(t)} \leq& {-d\beta(t)E(t)+ C\beta(t) \int _{\Omega_{1}} \bigl( u_{t}^{2} + \bigl\vert g(u_{t}) \bigr\vert ^{2} \bigr)} \\ \leq& {-d\beta(t)E(t)+ C\beta(t)M^{-1}\bigl(I(t) \bigr).} \end{aligned}$$
(4.24)
This implies
$$ {\beta(t)L'(t) + E'(t)\leq-d\beta(t)E(t)+ C \beta(t)M^{-1}\bigl(I(t)\bigr)} $$
since \({E'\leq0}\). Using (A1), we obtain
$$ {R'_{0}(t)\leq-d\beta(t)E(t)+ C\beta(t)M^{-1} \bigl(I(t)\bigr),} $$
where
$$ {R_{0} = \beta L + E\sim E.} $$
(4.25)
Let \({\epsilon_{0}<\epsilon}\), \({C_{0}>0}\) and define the functional
$$ {R_{1}(t)=M' \biggl(\epsilon_{0} \frac{E(t)}{E(0)} \biggr)R_{0}(t)+ C_{0}E(t).} $$
(4.26)
Let us note here that \({E(0)>0}\), otherwise \(E(t)=0\), \(\forall t\in\mathbb{R}^{+}\), and thus the theorem is verified since \({E'(t)\leq0}\). Now, since \({R_{0}\sim E}\) and \({E'\leq0}\), \({M'>0}\) (M is increasing), \({M''>0}\) (M is convex) on \((0,\epsilon]\), then \({R_{1}}\) satisfies the following:
$$ \alpha_{1} R_{1}(t)\leq E(t)\leq \alpha_{2} R_{1}(t) \quad \mbox{for some }\alpha _{1}, \alpha_{2}>0 $$
(4.27)
and it follows from (4.26) that
$$\begin{aligned} { R'_{1}(t)} =& {\epsilon_{0} \biggl(\frac{E'(t)}{E(0)} \biggr) M'' \biggl( \epsilon_{0}\frac{E(t)}{E(0)} \biggr)R_{0}(t)+ M' \biggl(\epsilon_{0}\frac{E(t)}{E(0)} \biggr)R'_{0}(t) + C_{0}E'(t)} \\ \leq& { M' \biggl(\epsilon_{0} \frac{E(t)}{E(0)} \biggr) \bigl[-d\beta(t)E(t)+ C\beta(t)M^{-1} \bigl(I(t)\bigr) \bigr] + C_{0}E'(t)} \\ =& {-d\beta(t)E(t)M' \biggl(\epsilon_{0} \frac{E(t)}{E(0)} \biggr)} \\ &{}+ { C\beta(t)M' \biggl(\epsilon_{0} \frac{E(t)}{E(0)} \biggr)M^{-1}\bigl(I(t)\bigr)+ C_{0}E'(t).} \end{aligned}$$
(4.28)
Now, let \({M^{\ast}}\) be the convex conjugate of M in the sense of Young. Then
$$ {M^{\ast}(s) = s\bigl(M' \bigr)^{-1}(s) - M\bigl(\bigl(M'\bigr)^{-1}(s) \bigr), \quad \mbox{if } s\in\bigl(0, M'(\epsilon)\bigr)} $$
(4.29)
and \(M^{\ast}\) satisfies the generalised Young’s inequality
$$ {XY\leq M^{\ast}(X) + M(Y), \quad \mbox{if } X\in \bigl(0,M'(\epsilon)\bigr), Y\in (0,\epsilon).} $$
(4.30)
Next, we set \({X = M'(\epsilon_{0}\frac{E(t)}{E(0)})}\) and \({Y =M^{-1}(I(t))}\). By using Lemma 4.1, the fact that \(sg(s)\leq\min\lbrace\epsilon,G(\epsilon)\rbrace\), if \({|s|\leq\epsilon_{1}}\) and (4.28)-(4.30), we obtain
$$\begin{aligned} { R'_{1}(t)} \leq& {-d\beta(t)E(t)M' \biggl(\epsilon_{0}\frac {E(t)}{E(0)} \biggr) + C_{0}E'(t)} \\ &{}+ { C\beta(t) \biggl[M^{\ast} \biggl( M' \biggl( \epsilon_{0}\frac {E(t)}{E(0)} \biggr) \biggr) + M\bigl(M^{-1} \bigl(I(t)\bigr)\bigr) \biggr]} \\ =&{ -d\beta(t)E(t)M' \biggl( \epsilon_{0} \frac{E(t)}{E(0)} \biggr)+ C_{0}E'(t)} \\ &{}+{ C\beta(t)M^{\ast} \biggl( M' \biggl( \epsilon_{0}\frac {E(t)}{E(0)} \biggr) \biggr) + C\beta(t)I(t)} \\ =& {-d\beta(t)E(t)M' \biggl( \epsilon_{0} \frac{E(t)}{E(0)} \biggr)+ C_{0}E'(t)} \\ &{}+ { C\epsilon_{0}\beta(t) \biggl(\frac{E(t)}{E(0)} \biggr) M' \biggl(\epsilon_{0}\frac{E(t)}{E(0)} \biggr) -C \beta(t)M \biggl( \epsilon _{0}\frac{E(t)}{E(0)} \biggr) + C \beta(t)I(t)} \\ \leq& {-E(0)d\beta(t) \biggl(\frac{E(t)}{E(0)} \biggr) M' \biggl(\epsilon_{0}\frac{E(t)}{E(0)} \biggr) + C\epsilon_{0} \beta(t) \biggl(\frac{E(t)}{E(0)} \biggr) M' \biggl( \epsilon_{0}\frac{E(t)}{E(0)} \biggr)} \\ &{}+ { C\beta(t) \int_{\Omega} u_{t}g(u_{t})+ C_{0}E'(t)} \\ \leq& {-E(0)d\beta(t) \biggl(\frac{E(t)}{E(0)} \biggr) M' \biggl(\epsilon_{0}\frac{E(t)}{E(0)} \biggr) + C\epsilon_{0} \beta(t) \biggl( \frac{E(t)}{E(0)} \biggr) M' \biggl( \epsilon_{0}\frac{E(t)}{E(0)} \biggr)} \\ &{}- { CE'(t)+ C_{0}E'(t).} \end{aligned}$$
We choose \({C_{0}}\) large enough and \(\epsilon_{0}\) small enough such that
$${C-C_{0} < 0, \qquad E(0)d - C\epsilon_{0}>0,} $$
and arrive at
$$ {R'_{1}(t)\leq- k\beta(t) \biggl( \frac{E(t)}{E(0)} \biggr) M' \biggl(\epsilon_{0} \frac{E(t)}{E(0)} \biggr)= -k\beta(t)M_{2} \biggl( \epsilon _{0}\frac{E(t)}{E(0)} \biggr),} $$
(4.31)
where \({M_{2}(t)=tM'(\epsilon_{0}t).}\) We have that
$$ {M'_{2}(t)= M'(\epsilon_{0}t) + \epsilon_{0} tM''(\epsilon_{0}t).} $$
Thus, using the strict convexity of M on \({(0,\epsilon]}\), we get that \({M_{2}, M'_{2}>0}\) on \((0,1]\). It follows from (4.27) and (4.31) that the functional
$${R_{2}(t)=\alpha_{1}\frac{R_{1}(t)}{E(0)}} $$
satisfies
$$ {R_{2}\sim E} $$
(4.32)
and
$$ {R_{2}'(t)\leq-k_{1} \beta(t)M_{2}\bigl(R_{2}(t)\bigr) \quad \mbox{for some } k_{1}>0.} $$
(4.33)
Inequality (4.33) implies that
$${\bigl(M_{1}\bigl(R_{2}(t)\bigr)\bigr)'\geq k_{1}\beta(t),} $$
where
$$ {M_{1}(\tau)= \int_{\tau}^{1}\frac{1}{M_{2}(s)}\, ds,\quad \tau \in(0,1].} $$
Thus, integrating (4.33) over \((0,t)\) and noting that \({ M_{1}}\) is strictly decreasing on \((0,1]\) give
$$ {R_{2}(t)\leq M_{1}^{-1} \biggl(k_{1} \int_{0}^{t} \beta(s)\, ds + k_{2} \biggr) \quad \mbox{for some } k_{2}>0.} $$
(4.34)
Combining (4.32) and (4.34), we get the result. This completes the proof. □