In this section, we prove some important lemmas which will be essential in establishing the main result. Let μ̄ be a positive constant satisfying
$$ \frac{\gamma _{2}(1-\alpha _{1})}{\alpha _{1}(1-d)}< \bar{\mu}< \frac{\gamma _{1}-\gamma _{2}\alpha _{2}}{\alpha _{2}} $$
(3.1)
and set
$$ \mu (t)=\bar{\mu}a(t). $$
The energy functional of system (2.14)–(2.16) is defined by
$$ \begin{aligned} E(t)={}& \frac{1}{2} \int _{0}^{1} \bigl[\rho u_{t}^{2} + \rho _{1} \varphi _{t}^{2} + \rho _{2} \psi _{t}^{2} + \alpha u_{x}^{2} + k( \varphi _{x}+\psi )^{2}+b\psi _{x}^{2}+ \lambda (\varphi -u)^{2} \bigr]\,dx \\ &{}+ \frac{1}{2} \int _{0}^{1}\rho _{3}\theta ^{2}\,dx + \mu (t)\tau (t) \int _{0}^{1} \int _{0}^{1} G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx. \end{aligned} $$
(3.2)
Lemma 3.1
Let \((u,\varphi , \psi , \theta ,z)\) be the solution of system (2.14)–(2.16). Then, the energy functional (3.2) satisfies
$$ \begin{aligned} \frac{dE(t)}{dt}\leq {}&-a(t) [ \gamma _{1}-\bar{\mu}\alpha _{2}- \gamma _{2}\alpha _{2} ] \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx \\ &{}-a(t) \bigl[\bar{\mu}\bigl(1-\tau '(t)\bigr)\alpha _{1}- \gamma _{2}(1-\alpha _{1}) \bigr] \int _{0}^{1}z(x,1,t)g_{2}\bigl(z(x,1,t) \bigr)\,dx \\ & {}-\gamma _{3} \int _{0}^{1}\varphi _{t}^{2}\,dx- \beta \int _{0}^{1} \theta _{x}^{2}\,dx \\ \leq{} &0, \quad \forall t\geq 0. \end{aligned} $$
(3.3)
Proof
Multiplying (2.14)1 by \(u_{t}\), (2.14)2 by \(\varphi _{t}\), (2.14)3 by \(\psi _{t}\), and (2.14)4 by θ, integrating the outcome over \((0,1)\), and applying integration by parts and the boundary conditions, we get
$$\begin{aligned} \begin{aligned} &\frac {1}{2}\frac {d}{dt} \int _{0}^{1} \bigl[ \rho u^{2}_{t} + \alpha u_{x}^{2}+\lambda (\varphi -u)^{2} \bigr]\,dx \\ & \quad =\lambda \int _{0}^{1}\varphi _{t}(\varphi -u)\,dx- \gamma _{1}a(t) \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx- \gamma _{2}a(t) \int _{0}^{1}u_{t} g_{2} \bigl(z(x,1,t)\bigr)\,dx, \end{aligned} \end{aligned}$$
(3.4)
$$\begin{aligned} \begin{aligned} &\frac {1}{2}\frac {d}{dt} \int _{0}^{1} \bigl[ \rho _{1}\varphi _{t}^{2} +k(\varphi _{x}+\psi )^{2} \bigr]\,dx \\ & \quad =-\gamma _{3} \int _{0}^{1}\varphi _{t}^{2}\,dx- \lambda \int _{0}^{1} \varphi _{t}(\varphi -u)\,dx +k \int _{0}^{1} \psi _{t}(\varphi _{x}+ \psi )\,dx, \end{aligned} \end{aligned}$$
(3.5)
$$\begin{aligned} &\frac {1}{2}\frac {d}{dt} \int _{0}^{1} \bigl[ \rho _{2}\psi ^{2}_{t} + b\psi _{x}^{2} \bigr]\,dx= m \int _{0}^{1} \psi _{t}\theta _{x}\,dx-k \int _{0}^{1} \psi _{t}(\varphi _{x}+\psi )\,dx, \end{aligned}$$
(3.6)
$$\begin{aligned} &\frac {1}{2} \int _{0}^{1} \rho _{3}\theta ^{2}\,dx=-\beta \int _{0}^{1} \theta _{x}^{2}\,dx-m \int _{0}^{1}\psi _{t}\theta _{x}\,dx. \end{aligned}$$
(3.7)
Adding (3.4)–(3.7), we arrive at
$$\begin{aligned}& \begin{aligned} &\frac{1}{2} \int _{0}^{1} \bigl( \rho u^{2}_{t} +\alpha u_{x}^{2}+ \lambda (\varphi -u)^{2}+\rho _{1}\varphi _{t}^{2} +k(\varphi _{x}+ \psi )^{2}+\rho _{2}\psi ^{2}_{t} + b\psi _{x}^{2}+\rho _{3} \theta ^{2} \bigr)\,dx \\ &\quad = -\gamma _{1}a(t) \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx- \gamma _{2}a(t) \int _{0}^{1}u_{t} g_{2} \bigl(z(x,1,t)\bigr)\,dx \\ &\qquad {}-\gamma _{3} \int _{0}^{1}\varphi _{t}^{2}\,dx- \beta \int _{0}^{1} \theta _{x}^{2}\,dx. \end{aligned} \end{aligned}$$
(3.8)
Now, multiplying equation (2.14)5 by \(\mu (t)g_{2}(z(x,\sigma ,t))\) and integrating over \((0,1)\times (0,1)\), we obtain
$$ \begin{aligned} &\mu (t)\tau (t) \int _{0}^{1} \int _{0}^{1} z_{t}(x,\sigma ,t) g_{2}\bigl(z(x, \sigma ,t)\bigr)\,d\sigma \,dx \\ &\quad {} +\mu (t) \int _{0}^{1} \int _{0}^{1}\bigl(1-\tau '(t)\sigma \bigr)z_{\sigma}(x, \sigma , t) g_{2}\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx=0. \end{aligned} $$
(3.9)
On account of (2.6), we can write
$$ \frac{\partial }{\partial \sigma} \bigl[ G\bigl(z(x,\sigma ,t)\bigr) \bigr]=z_{ \sigma}(x, \sigma , t) g_{2}\bigl(z(x,\sigma ,t)\bigr). $$
(3.10)
Therefore, (3.9) becomes
$$ \begin{aligned} &\mu (t)\tau (t) \int _{0}^{1} \int _{0}^{1}z_{t}(x,\sigma ,t) g_{2}\bigl(z(x, \sigma ,t)\bigr)\,d\sigma \,dx \\ &\quad =-\mu (t) \int _{0}^{1} \int _{0}^{1}\bigl(1-\tau '(t)\sigma \bigr) \frac{\partial }{\partial \sigma} \bigl[ G\bigl(z(x,\sigma ,t)\bigr) \bigr]\,d\sigma \,dx. \end{aligned} $$
(3.11)
It follows that
$$ \begin{aligned} &\frac{d}{dt} \biggl(\mu (t)\tau (t) \int _{0}^{1} \int _{0}^{1}G\bigl(z(x, \sigma ,t)\bigr)\,d\sigma \,dx \biggr) \\ &\quad =-\mu (t) \int _{0}^{1} \int _{0}^{1}\frac{\partial}{\partial \sigma} \bigl[\bigl(1-\tau '(t)\sigma \bigr)G\bigl(z(x,\sigma ,t)\bigr) \bigr]\,d\sigma \,dx \\ &\qquad {} + \mu '(t)\tau (t) \int _{0}^{l} \int _{0}^{1}G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx \\ &\quad =\mu (t) \int _{0}^{1} \bigl( G\bigl(z(x,0,t)\bigr)-G \bigl(z(x,1,t)\bigr) \bigr)\,dx +\mu (t) \tau '(t) \int _{0}^{1} G\bigl(z(x,1,t)\bigr)\,dx \\ &\qquad {}+\mu '(t)\tau (t) \int _{0}^{1} \int _{0}^{1}G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx \\ &\quad =\mu (t) \int _{0}^{1}G\bigl(u_{t}(x,t)\bigr)\,dx-\mu (t) \bigl(1-\tau '(t)\bigr) \int _{0}^{1}G(z(x,1,t)\,dx \\ &\qquad {}+\mu '(t)\tau (t) \int _{0}^{1} \int _{0}^{1}G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx. \end{aligned} $$
(3.12)
Recalling the definition of the energy functional (3.2), and adding (3.8) and (3.12), we obtain
$$ \begin{aligned} \frac{dE(t)}{dt}={}& -\gamma _{1}a(t) \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx- \gamma _{2}a(t) \int _{0}^{1}u_{t} g_{2} \bigl(z(x,1,t)\bigr)\,dx \\ &{} +\mu (t) \int _{0}^{1}G\bigl(u_{t}(x,t)\bigr)\,dx-\mu (t) \bigl(1-\tau '(t)\bigr) \int _{0}^{1}G(z(x,1,t)\,dx \\ & {}-\gamma _{3} \int _{0}^{1}\varphi _{t}^{2}\,dx- \beta \int _{0}^{1} \theta _{x}^{2}\,dx +\mu '(t)\tau (t) \int _{0}^{1} \int _{0}^{1}G\bigl(z(x, \sigma ,t)\bigr)\,d\sigma \,dx. \end{aligned} $$
(3.13)
On the account of \((A_{1})\) and (2.5), we get
$$ \begin{aligned} \frac{dE(t)}{dt}\leq{} &- \bigl( \gamma _{1}a(t)-\mu (t)\alpha _{2} \bigr) \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx- \gamma _{2}a(t) \int _{0}^{1}u_{t} g_{2} \bigl(z(x,1,t)\bigr)\,dx \\ &{} -\mu (t) \bigl(1-\tau '(t)\bigr) \int _{0}^{1}G(z(x,1,t)\,dx -\gamma _{3} \int _{0}^{1}\varphi _{t}^{2}\,dx- \beta \int _{0}^{1}\theta _{x}^{2}\,dx . \end{aligned} $$
(3.14)
Now, we consider the convex conjugate of G defined by
$$ G^{*}(s)=s\bigl(G'\bigr)^{-1}(s)-G \bigl(\bigl(G'\bigr)^{-1}(s)\bigr), \quad \forall s\geq 0, $$
(3.15)
which satisfies the generalized Young inequality (see [2])
$$ AB\leq G^{*}(A)+G(B), \quad \forall A,B >0. $$
(3.16)
Using (2.5) and the definition of G, we get
$$ G^{*}(s)=sg_{2}^{-1}(s)-G \bigl(g_{2}^{-1}(s)\bigr), \quad \forall s\geq 0. $$
(3.17)
Therefore, on account of (2.5) and (3.17), we have
$$ \begin{aligned} G^{*}\bigl(g_{2} \bigl(z(x,1,t)\bigr)\bigr)&=z(x,1,t)g_{2}\bigl(z(x,1,t)\bigr)-G \bigl(z(x,1,t)\bigr) \\ &\leq (1-\alpha _{1})z(x,1,t)g_{2}\bigl(z(x,1,t)\bigr). \end{aligned} $$
(3.18)
A combination of (3.14), (3.16), and (3.18) leads to
$$\begin{aligned} \frac{dE(t)}{dt} \leq &- \bigl( \gamma _{1}a(t)-\mu (t)\alpha _{2} \bigr) \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx \\ &{} +\gamma _{2}a(t) \int _{0}^{1}(G(u_{t})+G^{*} \bigl( g_{2}\bigl(z(x,1,t)\bigr)\bigr)\,dx \\ &{} -\mu (t) \bigl(1-\tau '(t)\bigr) \int _{0}^{1}G(z(x,1,t)\,dx -\gamma _{3} \int _{0}^{1}\varphi _{t}^{2}\,dx- \beta \int _{0}^{1}\theta _{x}^{2}\,dx \\ \leq& - \bigl( \gamma _{1}a(t)-\mu (t)\alpha _{2} \bigr) \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx +\gamma _{2}a(t)\alpha _{2} \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx \\ &{} +\gamma _{2}a(t) (1-\alpha _{1}) \int _{0}^{1}z(x,1,t)g_{2}\bigl(z(x,1,t) \bigr)\,dx \\ &{} -\mu (t) \bigl(1-\tau '(t)\bigr) \int _{0}^{1}G(z(x,1,t)\,dx-\gamma _{3} \int _{0}^{1}\varphi _{t}^{2}\,dx- \beta \int _{0}^{1}\theta _{x}^{2}\,dx \\ \leq& - \bigl( \gamma _{1}a(t)-\mu (t)\alpha _{2}-\gamma _{2}a(t) \alpha _{2} \bigr) \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx \\ &{} - \bigl(\mu (t) \bigl(1-\tau '(t)\bigr)\alpha _{1}- \gamma _{2}a(t) (1- \alpha _{1}) \bigr) \int _{0}^{1}z(x,1,t)g_{2}\bigl(z(x,1,t) \bigr)\,dx \\ &{} -\gamma _{3} \int _{0}^{1}\varphi _{t}^{2}\,dx- \beta \int _{0}^{1} \theta _{x}^{2}\,dx. \end{aligned}$$
(3.19)
Recalling that \(\mu (t)=\bar{\mu}a(t)\), it follows from (3.19) that
$$ \begin{aligned} \frac{dE(t)}{dt}\leq{}& -a(t) [ \gamma _{1}-\bar{\mu}\alpha _{2}- \gamma _{2}\alpha _{2} ] \int _{0}^{1}u_{t} g_{1}(u_{t})\,dx \\ & {}-a(t) \bigl[\bar{\mu}\bigl(1-\tau '(t)\bigr)\alpha _{1}-\gamma _{2}(1- \alpha _{1}) \bigr] \int _{0}^{1}z(x,1,t)g_{2}\bigl(z(x,1,t) \bigr)\,dx \\ & {}-\gamma _{3} \int _{0}^{1}\varphi _{t}^{2}\,dx- \beta \int _{0}^{1} \theta _{x}^{2}\,dx. \end{aligned} $$
(3.20)
Using (2.9) and (3.1), we obtain the desired result. This finishes the proof. □
Lemma 3.2
The functional \(F_{1}\), defined by
$$ F_{1}(t):=-\rho _{2}\rho _{3} \int _{0}^{1}\psi _{t} \int _{0}^{x} \theta (y,t)\,dy\,dx, $$
satisfies, along the solution of system (2.14)–(2.16) and for any \(\epsilon _{1}, \epsilon _{2}>0\), the estimate
$$ \begin{aligned} F'_{1}(t)\le{}& - \frac {m\rho _{2}}{2} \int _{0}^{1}\psi _{t}^{2}\,dx + \epsilon _{1} \int _{0}^{1}\psi _{x}^{2}\,dx+ \epsilon _{2} \int _{0}^{1}( \varphi _{x}+\psi )^{2}\,dx \\ &{}+c \biggl(1+\frac {1}{\epsilon _{1}} + \frac {1}{\epsilon _{2}} \biggr) \int _{0}^{1}\theta _{x}^{2}\,dx. \end{aligned} $$
(3.21)
Proof
Differentiating \(F_{1}\), using (2.14)3 and (2.14)4, then integrating by parts and exploiting the boundary conditions lead to
$$ \begin{aligned} F_{1}'(t)={}& b \rho _{3} \int _{0}^{1}\psi _{x}\theta \,dx +k\rho _{3} \int _{0}^{1}(\varphi _{x}+\psi ) \int _{0}^{x}\theta (y,t)\,dy\,dx \\ & {}+ m\rho _{3} \int _{0}^{1}\theta ^{2}\,dx -\rho _{2}\beta \int _{0}^{1} \psi _{t}\theta _{x}\,dx-\rho _{2}m \int _{0}^{1}\psi _{t}^{2}\,dx. \end{aligned} $$
(3.22)
Making use of Cauchy–Schwarz, Young’s, and Poincaré’s inequalities, we get (3.21). □
Lemma 3.3
The functional \(F_{2}\), defined by
$$ F_{2}(t):= \int _{0}^{1} \biggl(\rho uu_{t} +\rho _{1}\varphi \varphi _{t} + \rho _{2}\psi \psi _{t}+\frac{\gamma _{3}}{2}\varphi ^{2} \biggr)\,dx, $$
satisfies, along the solution of system (2.14)–(2.16), the estimate
$$ \begin{aligned} F'_{2}(t)\le{}& - \int _{0}^{1} \biggl(\frac{\alpha}{2} u_{x}^{2}+ \lambda (\varphi -u)^{2}+k(\varphi _{x} +\psi )^{2}+\frac {b}{2}\psi _{x}^{2} \biggr)\,dx \\ & {}+ \int _{0}^{1} \bigl(\rho u_{t}^{2}+ \rho _{1}\varphi ^{2}_{t} + \rho _{2} \psi _{t}^{2} \bigr)\,dx +c \int _{0}^{1}\theta _{x}^{2}\,dx \\ &{} +c \int _{0}^{1} \bigl\vert g_{1}(u_{t}) \bigr\vert ^{2}\,dx+ c \int _{0}^{1} \bigl\vert g_{2} \bigl(z(x,1,t)\bigr) \bigr\vert ^{2}\,dx, \quad \forall t\geq 0. \end{aligned} $$
(3.23)
Proof
Directly differentiating \(F_{2}\), using (2.14)1, (2.14)2, and (2.14)3, then applying integration by parts and boundary conditions, we obtain
$$ \begin{aligned} F'_{2}(t)=&{}- \int _{0}^{1} \bigl( \alpha u_{x}^{2}+ \lambda (\varphi -u)^{2}+k( \varphi _{x} +\psi )^{2}+b\psi _{x}^{2} \bigr)\,dx \\ &{}+ \int _{0}^{1} \bigl(\rho u_{t}^{2} + \rho _{1}\varphi ^{2}_{t} + \rho _{2} \psi _{t}^{2} \bigr)\,dx +m \int _{0}^{1} \psi \theta _{x}\,dx \\ &{}-\gamma _{1}a(t) \int _{0}^{1} ug_{1}(u_{t})\,dx- \gamma _{2}a(t) \int _{0}^{1}ug_{2}\bigl(z(x,1,t)\bigr)\,dx. \end{aligned} $$
(3.24)
Using \((A_{1})\), Young’s and Poincaré’s inequalities, we obtain (3.23). □
Lemma 3.4
The functional
$$ F_{3}(t):=\bar{\mu}\tau (t) \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t) \sigma}G\bigl(z(x,\sigma ,t) \bigr)\,d\sigma \,dx, $$
satisfies, along the solution of system (2.14)–(2.16), the estimate
$$ \begin{aligned} F'_{3}(t) \le &-2F_{3}(t)+ \frac{\bar{\mu}\alpha _{2}}{2} \int _{0}^{1} \bigl( u_{t}^{2}+ \bigl\vert g_{1}(u_{t}) \bigr\vert ^{2} \bigr)\,dx, \quad \forall t \geq 0. \end{aligned} $$
(3.25)
Proof
Differentiating \(F_{3}\), we get
$$ \begin{aligned} F'_{3}(t) ={}& \bar{\mu}\tau '(t) \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t) \sigma}G\bigl(z(x,\sigma ,t) \bigr)\,d\sigma \,dx \\ &{} -2\bar{\mu}\tau (t)\tau '(t) \int _{0}^{1} \int _{0}^{1}\sigma e^{-2 \tau (t)\sigma}G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx \\ & {}+\bar{\mu}\tau (t) \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t) \sigma}z_{t}(x, \sigma ,t)g_{2}\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx. \end{aligned} $$
(3.26)
Using the last equation in (2.14), we can express the last term on the right hand-side of (3.26) as
$$\begin{aligned}& \tau (t) \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t)\sigma}z_{t}(x, \sigma ,t)g_{2}\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx \\& \quad = \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t)\sigma} \bigl(\tau '(t) \sigma -1 \bigr) z_{\sigma}(x,\sigma ,t)g_{2} \bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx \\& \quad = \int _{0}^{1} \int _{0}^{1}\frac{\partial}{\partial \sigma} \bigl[ e^{-2 \tau (t)\sigma} \bigl(\tau '(t)\sigma -1 \bigr)G\bigl(z(x,\sigma ,t)\bigr) \bigr]\,d\sigma \,dx \\& \begin{aligned}&\qquad {}+2\tau (t) \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t)\sigma} \bigl(\tau '(t)\sigma -1 \bigr)G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx \\ &\qquad {}-\tau '(t) \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t)\sigma}G\bigl(z(x, \sigma ,t) \bigr)\,d\sigma \,dx \end{aligned}\\& \quad =-\bigl(1-\tau '(t)\bigr)e^{-2\tau (t)} \int _{0}^{1}G\bigl(z(x,1,t)\bigr)\,dx + \int _{0}^{l}G(u_{t})\,dx \\& \qquad {}+2\tau (t) \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t)\sigma} \bigl(\tau '(t)\sigma -1 \bigr)G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx \\& \qquad {}-\tau '(t) \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t)\sigma}G\bigl(z(x, \sigma ,t) \bigr)\,d\sigma \,dx. \end{aligned}$$
(3.27)
Substituting (3.27) into (3.26), we arrive at
$$ \begin{aligned} F'_{3}(t) ={}&-2 \bar{\mu}\tau (t) \int _{0}^{1} \int _{0}^{1} e^{-2\tau (t) \sigma}G\bigl(z(x,\sigma ,t) \bigr)\,d\sigma \,dx +\bar{\mu} \int _{0}^{1}G(u_{t})\,dx \\ &{} -\bigl(1-\tau '(t)\bigr)e^{-2\tau (t)} \int _{0}^{1}G\bigl(z(x,1,t)\bigr)\,dx. \end{aligned} $$
(3.28)
Using condition (2.5) and Young’s inequality, we obtain (3.26). □
Lemma 3.5
Let \((u,\varphi , \psi ,\theta , z)\) be the solution of system (2.14)–(2.16). Then, for \(N, N_{1}, N_{2}>0\) sufficiently large, the Lyapunov functional L, defined by
$$\begin{aligned} L(t):=NE(t)+N_{1}F_{1}(t)+ N_{2}F_{2}(t)+F_{3}(t), \end{aligned}$$
(3.29)
satisfies, for some positive constants \(c_{1}\), \(c_{2}\), η,
$$ c_{1}E(t) \le L(t) \le c_{2}E(t), \quad \forall t \ge 0, $$
(3.30)
and
$$ \begin{aligned} L'(t)\le -\eta E(t)+c \int _{0}^{1}\bigl(u^{2}_{t}+ \bigl\vert g_{1}(u_{t}) \bigr\vert ^{2} \bigr)\,dx+ c \int _{0}^{1} \bigl\vert g_{2} \bigl(z(x,1,t)\bigr) \bigr\vert ^{2}\,dx, \quad \forall t \ge 0. \end{aligned} $$
(3.31)
Proof
Applying Cauchy–Schwarz, Young’s, and Poincaré’s inequalities, we have
$$\begin{aligned} \bigl\vert L(t)-NE(t) \bigr\vert \leq& N_{1} \biggl\vert -\rho _{2} \int _{0}^{1}\psi _{t} \int _{0}^{x} \theta (y,t)\,dy\,dx \biggr\vert \\ &{} + N_{2} \biggl\vert \int _{0}^{1} \biggl(\rho uu_{t} +\rho _{1} \varphi \varphi _{t} + \rho _{2}\psi \psi _{t}+\frac{\gamma _{3}}{2} \varphi ^{2} \biggr)\,dx \biggr\vert \\ &{} + \biggl\vert \bar{\mu}\tau (t) \int _{0}^{1} \int _{0}^{1} e^{-2 \tau (t)\sigma}G\bigl(z(x,\sigma ,t) \bigr)\,d\sigma \,dx \biggr\vert \\ \leq& \frac{N_{2}\rho}{2} \int _{0}^{1}u_{t}^{2}\,dx + \frac{N_{2}\rho _{1}}{2} \int _{0}^{1}\varphi _{t}^{2}\,dx+ \frac{(N_{1}+N_{2})\rho _{2}}{2} \int _{0}^{1}\psi _{t}^{2}\,dx \\ &{} +\frac{N_{2}\gamma _{3}}{2} \int _{0}^{1}\varphi ^{2}\,dx+ \frac{N_{2}\rho}{2} \int _{0}^{1}u_{x}^{2}\,dx + \frac{N_{2}\rho _{1}}{2} \int _{0}^{1}\varphi _{x}^{2}\,dx \\ &{} + \frac{N_{2}\rho _{2}}{2} \int _{0}^{1}\psi _{x}^{2}\,dx + \frac{ N_{1}\rho _{2}}{2} \int _{0}^{1} \biggl( \int _{0}^{x}\theta (y,t)\,dy \biggr)^{2}\,dx \\ &{} +\bar{\mu}\tau (t) \int _{0}^{1} \int _{0}^{1} G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx. \end{aligned}$$
(3.32)
Using the relations
$$ \begin{aligned} & \int _{0}^{1}\varphi ^{2}\,dx\leq 2 \int _{0}^{l}(\varphi -u)^{2}\,dx + 2 \int _{0}^{1} u_{x}^{2}\,dx, \\ & \int _{0}^{1}\varphi _{x}^{2}\,dx \leq 2 \int _{0}^{1}(\varphi _{x}+ \psi )^{2}\,dx +2 \int _{0}^{1}\psi _{x}^{2}\,dx, \end{aligned} $$
we arrive at
$$ \begin{aligned} \bigl\vert L(t)-NE(t) \bigr\vert \leq{} &\frac{N_{2}\rho}{2} \int _{0}^{1}u_{t}^{2}\,dx + \frac{N_{2}\rho _{1}}{2} \int _{0}^{1}\varphi _{t}^{2}\,dx+ \frac{(N_{1}+N_{2})\rho _{2}}{2} \int _{0}^{1}\psi _{t}^{2}\,dx \\ &{} +N_{2}\gamma _{3} \int _{0}^{l}(\varphi -u)^{2}\,dx+ \biggl( N_{2} \gamma _{3}+\frac{N_{2}\rho}{2} \biggr) \int _{0}^{1}u_{x}^{2}\,dx \\ & {}+N_{2}\rho _{1} \int _{0}^{1}(\varphi _{x}+\psi )^{2}\,dx + \biggl( \frac{N_{2}(\rho _{1}+\rho _{2})}{2} \biggr) \int _{0}^{1}\psi _{x}^{2}\,dx \\ &{} +\frac{ N_{1}\rho _{2}}{2} \int _{0}^{1}\theta ^{2}\,dx+\bar{\mu} \tau (t) \int _{0}^{1} \int _{0}^{1} G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx. \end{aligned} $$
(3.33)
From (3.33), we obtain
$$ \bigl\vert L(t)-NE(t) \bigr\vert \leq \bar{c}E(t). $$
(3.34)
By choosing N large enough such that
$$ c_{1}=N- \bar{c}>0,\qquad c_{2}=N+ \bar{c}>0, $$
(3.35)
estimate (4.14) follows. Next, we establish (3.31). Using Lemmas 3.1–3.4, we get
$$ \begin{aligned} L'(t)\le{}& -\rho \int _{0}^{1}u_{t}^{2}\,dx- [N \gamma _{3}-N_{2} \rho _{1} ] \int _{0}^{1}\varphi _{t}^{2}\,dx- \biggl[N_{1} \frac{m\rho _{2}}{2}-N_{2}\rho _{2} \biggr] \int _{0}^{1}\psi _{t}^{2}\,dx \\ & {}-\frac{N_{2}\alpha}{2} \int _{0}^{1}u_{x}^{2}\,dx-N_{2} \lambda \int _{0}^{1}(\varphi -u)^{2}\,dx- [N_{2}k-N_{1}\epsilon _{2} ] \int _{0}^{1}(\varphi _{x}+\psi )^{2}\,dx \\ &{} - \biggl[N_{2}\frac{b}{2}-N_{1}\epsilon _{1} \biggr] \int _{0}^{1} \psi _{x}^{2}\,dx- \biggl[N\beta -N_{1}c \biggl(1+\frac{1}{\epsilon _{1}} + \frac{1}{\epsilon _{2}} \biggr)-N_{2}c \biggr] \int _{0}^{1}\theta _{x}^{2}\,dx \\ &{} -\frac{2e^{-2\tau _{1}}}{a(0)}\mu (t)\tau (t) \int _{0}^{1} \int _{0}^{1}G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx + \biggl[\rho +N_{2}\rho + \frac{\bar{\mu }\alpha _{2}}{2} \biggr] \int _{0}^{1}u_{t}^{2}\,dx \\ &{} + \biggl[cN_{2}+ \frac{\bar{\mu}\alpha _{2}}{2} \biggr] \int _{0}^{1} \bigl\vert g_{1}(u_{t}) \bigr\vert ^{2}\,dx +cN_{2} \int _{0}^{1} \bigl\vert g_{2} \bigl(z(x,1,t)\bigr) \bigr\vert ^{2}\,dx. \end{aligned} $$
Choosing
$$ N_{2}=1, \qquad \epsilon _{1}=\frac {N_{2}b}{4N_{1}}, \qquad \epsilon _{2}= \frac {N_{2}k}{2N_{1}}, $$
we arrive at
$$ \begin{aligned} L'(t)\le{}& -\rho \int _{0}^{1}u_{t}^{2}\,dx- [N \gamma _{3}-\rho _{1} ] \int _{0}^{1}\varphi _{t}^{2}\,dx- \biggl[N_{1}\frac{m\rho _{2}}{2}- \rho _{2} \biggr] \int _{0}^{1}\psi _{t}^{2}\,dx \\ & {}-\frac{\alpha}{2} \int _{0}^{1}u_{x}^{2}\,dx-\lambda \int _{0}^{1}( \varphi -u)^{2}\,dx- \frac {k}{2} \int _{0}^{1}(\varphi _{x}+\psi )^{2}\,dx \\ & {}-\frac{b}{4} \int _{0}^{1}\psi _{x}^{2}\,dx- \biggl[N\beta -N_{1}c \biggl(1+\frac{4N_{1}}{b} +\frac{2N_{1}}{k} \biggr)-c \biggr] \int _{0}^{1} \theta _{x}^{2}\,dx \\ & {}-\frac{2e^{-2\tau _{1}}}{a(0)}\mu (t)\tau (t) \int _{0}^{1} \int _{0}^{1}G\bigl(z(x,\sigma ,t)\bigr)\,d\sigma \,dx + \biggl[2\rho + \frac{\bar{\mu }\alpha _{2}}{2} \biggr] \int _{0}^{1}u_{t}^{2}\,dx \\ &{} + \biggl[c+ \frac{\bar{\mu }\alpha _{2}}{2} \biggr] \int _{0}^{1} \bigl\vert g_{1}(u_{t}) \bigr\vert ^{2}\,dx +c \int _{0}^{1} \bigl\vert g_{2} \bigl(z(x,1,t)\bigr) \bigr\vert ^{2}\,dx. \end{aligned} $$
(3.36)
Now, we choose \(N_{1}\) large such that
$$ N_{1}\frac{m\rho _{2}}{2}-\rho _{2}>0. $$
Next, we select N very large so that (4.14) remains true and
$$ N\gamma _{3}-\rho _{1}>0,\qquad N\beta -N_{1}c \biggl(1+ \frac{4N_{1}}{b} +\frac{2N_{1}}{k} \biggr)-c>0. $$
Therefore, using the energy functional defined by (3.2), we obtain (3.31). □