In this section, we shall prove Theorem 2.2, which will be divided into the following lemmas.
Lemma 4.1
Let
\((\varphi,\varphi_{t},\psi,\psi_{t},z)\)
be the solution of problem (2.5)-(2.6). The functional
\(I_{1}\)
defined by
$$\begin{aligned} I_{1}=- \int^{1}_{0}(\rho_{1}\varphi \varphi_{t}+\rho_{2}\psi\psi_{t})\,dx- \frac {\mu_{1}}{2} \int^{1}_{0}\psi^{2}\,dx \end{aligned}$$
(4.1)
satisfies that for any
\(\varepsilon>0\),
$$\begin{aligned} \frac{d}{dt}I_{1}(t) \leq&- \int^{1}_{0} \bigl(\rho_{1} \varphi_{t}^{2}+\rho_{2}\psi _{t}^{2} \bigr)\,dx+k \int^{1}_{0}|\varphi_{x}+ \psi|^{2}\,dx+ (b+C_{1}+\varepsilon) \int^{1}_{0}\psi_{x}^{2}\,dx \\ &{}+\frac{\mu_{2}^{2}}{4\lambda_{1}\varepsilon} \int^{1}_{0}z^{2}(x,1,t)\,dx, \end{aligned}$$
(4.2)
hereafter
\(\lambda_{1}>0\)
is the first eigenvalue of −Δ in
\(H^{1}_{0}(0,1)\).
Proof
A straightforward calculation gives
$$\frac{dI_{1}}{dt}=- \int^{1}_{0}(\rho_{1}\varphi_{tt} \varphi+\rho_{2}\psi _{tt}\psi)\,dx- \int^{1}_{0} \bigl(\rho_{1} \varphi_{t}^{2}+\rho_{2}\psi_{t}^{2} \bigr)\,dx-\mu _{1} \int^{1}_{0}\psi\psi_{t}\,dx. $$
Using (2.5) and integrating by parts, we see that
$$\begin{aligned} \frac{dI_{1}}{dt} =&-\rho_{1} \int^{1}_{0}\varphi_{t}^{2}\,dx- \rho_{2} \int^{1}_{0}\psi _{t}^{2}\,dx+k \int^{1}_{0}|\varphi_{x}+ \psi|^{2}\,dx+b \int^{1}_{0}\psi_{x}^{2}\,dx \\ &{}+\mu_{2} \int^{1}_{0}z(x,1,t)\psi \,dx+ \int^{1}_{0}f(\psi)\psi \,dx. \end{aligned}$$
(4.3)
It follows from Young’s inequality and Poincaré’s inequality that for any \(\varepsilon>0\),
$$\begin{aligned}& \begin{aligned}[b] \int^{1}_{0}\bigl|z(x,1,t)\psi\bigr|\,dx&\leq\varepsilon \lambda_{1} \int^{1}_{0}\psi^{2}\,dx+\frac {1}{4\varepsilon\lambda_{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx \\ &\leq\varepsilon \int^{1}_{0}\psi_{x}^{2}\,dx+ \frac{1}{4\varepsilon\lambda _{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx, \end{aligned} \end{aligned}$$
(4.4)
$$\begin{aligned}& \int^{1}_{0}\bigl|f(\psi)\psi\bigr|\,dx\leq \int^{1}_{0}|\psi|^{\theta}|\psi||\psi|\,dx \leq\|\psi\|^{\theta}_{2(\theta+1)}\|\psi\|_{2(\theta+1)}\|\psi\| \leq C_{1} \int^{1}_{0}\psi^{2}_{x}\,dx, \end{aligned}$$
(4.5)
which, together with (4.3)-(4.4), gives us (4.2). The proof is now complete. □
Lemma 4.2
Let
\((\varphi,\varphi_{t},\psi,\psi_{t},z)\)
be the solution of problem (2.5)-(2.6). We define the functional
\(I_{2}\)
by
$$\begin{aligned} I_{2}(t)= \int^{1}_{0}(\rho_{2}\psi_{t} \psi+\rho_{1}\varphi_{t}g)\,dx+\frac{\mu _{1}}{2} \int^{1}_{0}\psi^{2}\,dx, \end{aligned}$$
(4.6)
where
g
is the solution of
$$\begin{aligned} -g_{xx}=\psi_{x}, \qquad g|_{x=0,1}=0. \end{aligned}$$
(4.7)
Then the functional
\(I_{2}\)
satisfies, for any
\(\eta,\tilde{\eta}>0\),
$$\begin{aligned} \frac{d}{dt}I_{2}(t) \leq& (\mu_{2} \eta-b ) \int^{1}_{0}\psi _{x}^{2}\,dx+ \biggl(\rho_{2}+\frac{\rho_{2}}{4\tilde{\eta}} \biggr) \int ^{1}_{0}\psi_{t}^{2}\,dx + \frac{\rho_{1}}{\lambda_{1}}\tilde{\eta} \int^{1}_{0}\varphi _{t}^{2}\,dx \\ &{}+\frac{\mu_{2}}{4\eta\lambda_{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx- \int^{1}_{0}\hat {f}(\psi)\,dx. \end{aligned}$$
(4.8)
Proof
We know from (2.5) that
$$\begin{aligned} \frac{d}{dt}I_{2}(t) =&-b \int^{1}_{0}\psi_{x}^{2}\,dx+ \rho_{2} \int^{1}_{0}\psi _{t}^{2}\,dx-k \int^{1}_{0}\psi^{2}\,dx+k \int^{1}_{0}g_{x}^{2}\,dx \\ &{}+\rho_{1} \int^{1}_{0}\varphi_{t}g_{t}\,dx- \mu_{2} \int^{1}_{0}\psi z(x,1,t)\,dx- \int^{1}_{0}f(\psi)\psi \,dx. \end{aligned}$$
(4.9)
By (4.7), we can get
$$ \left \{ \textstyle\begin{array}{@{}l} \int^{1}_{0}g_{x}^{2}\,dx\leq \int^{1}_{0}\psi^{2}\,dx\leq\int^{1}_{0}\psi_{x}^{2}\,dx, \\ \int^{1}_{0}g_{t}^{2}\,dx\leq \int^{1}_{0}g_{xt}^{2}\,dx\leq\int^{1}_{0}\psi_{t}^{2}\,dx. \end{array}\displaystyle \right . $$
(4.10)
Using Young’s inequality and Poincaré’s inequality, we have
$$\begin{aligned}& \begin{aligned}[b] \mu_{2} \int^{1}_{0}\bigl|\psi z(x,1,t)\bigr|\,dx&\leq\mu_{2} \eta\lambda_{1} \int^{1}_{0}\psi^{2}\,dx+\frac{\mu _{2}}{4\eta\lambda_{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx \\ &\leq\mu_{2}\eta \int^{1}_{0}\psi_{x}^{2}\,dx+ \frac{\mu_{2}}{4\eta\lambda _{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx, \end{aligned} \end{aligned}$$
(4.11)
$$\begin{aligned}& \begin{aligned}[b] \rho_{1} \int^{1}_{0}|\varphi_{t}g_{t}| \,dx& \leq\frac{\rho _{1}}{\lambda_{1}}\tilde{\eta} \int^{1}_{0}\varphi^{2}_{t}\,dx+ \frac{\rho _{1}\lambda_{1}}{4\tilde{\eta}} \int^{1}_{0}g_{t}^{2}\,dx \\ &\leq\frac{\rho_{1}}{\lambda_{1}}\tilde{\eta} \int^{1}_{0}\varphi ^{2}_{t}\,dx+ \frac{\rho_{1}}{4\tilde{\eta}} \int^{1}_{0}\psi_{t}^{2} \,dx. \end{aligned} \end{aligned}$$
(4.12)
Combining (2.2) and (4.11)-(4.12) with (4.9) and (2.2), we can complete the proof. □
Now we define the following functional:
$$\begin{aligned} J(t):=\rho_{2} \int^{1}_{0}\psi_{t}(\varphi_{x}+ \psi)\,dx+\rho_{2} \int^{1}_{0}\psi _{x}\varphi_{t} \,dx. \end{aligned}$$
(4.13)
Then we may get the following lemma.
Lemma 4.3
Let
\((\varphi,\varphi_{t},\psi,\psi_{t},z)\)
be the solution of problem (2.5)-(2.6), and assume that (1.4) holds. Then the functional
\(J(t)\)
satisfies, for any
\(\varepsilon>0\),
$$\begin{aligned} \frac{d}{dt}J(t) \leq&b[\psi_{x} \varphi_{x}]^{x=1}_{x=0}-\frac {k}{2} \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+ \biggl(\frac{\varepsilon }{b^{2}\lambda_{1}} +\frac{b^{2}}{2\varepsilon\lambda_{1}} \biggr) \int^{1}_{0}\psi _{x}^{2}\,dx \\ &{} + \biggl(\rho_{2}+\frac{\mu_{1}^{2}}{k} \biggr) \int^{1}_{0}\psi_{t}^{2}\,dx + \frac{\mu_{2}^{2}}{k} \int^{1}_{0}z^{2}(x,1,t)\,dx \\ &{}- \int^{1}_{0}\hat{f}(\psi)\,dx. \end{aligned}$$
(4.14)
Proof
By taking a derivative of (4.13), we arrive at
$$\begin{aligned} \frac{d}{dt}J(t) =&\rho_{2} \int^{1}_{0}\psi_{tt}(\varphi_{x}+ \phi)\,dx+\rho _{2} \int^{1}_{0}\psi_{t}(\varphi_{x}+ \psi)_{t}\,dx \\ &{}+\rho_{2} \int^{1}_{0}\psi_{xt}\varphi_{t} \,dx +\rho_{2} \int^{1}_{0}\psi_{x}\varphi_{tt} \,dx. \end{aligned}$$
Using (2.5), (1.4) and integration by parts, we get
$$\begin{aligned} \frac{d}{dt}J(t) =&b[\psi_{x}\varphi_{x}]^{x=1}_{x=0}+ \rho_{2} \int ^{1}_{0}\psi_{t}^{2}\,dx-k \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx \\ &{}-\mu_{1} \int^{1}_{0}\psi_{t}(\varphi_{x}+ \psi)\,dx-\mu_{2} \int^{1}_{0}(\varphi _{x}+\psi)z(x,1,t) \,dx \\ &{}- \int^{1}_{0}\varphi_{x}f(\psi)\,dx- \int^{1}_{0}f(\psi)\psi \,dx. \end{aligned}$$
By using Young’s inequality and Poincaré’s inequality, we know that for any \(\varepsilon>0\),
$$\begin{aligned}& {-}\mu_{1} \int^{1}_{0}\psi_{t}(\varphi_{x}+ \psi)\,dx \leq \frac{k}{4} \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+\frac{\mu_{1}^{2}}{k} \int ^{1}_{0}\psi_{t}^{2} \,dx, \end{aligned}$$
(4.15)
$$\begin{aligned}& {-}\mu_{2} \int^{1}_{0}(\varphi_{x}+\psi)z(x,1,t)\,dx \leq \frac{k}{4} \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+\frac{\mu_{2}^{2}}{k} \int ^{1}_{0}z^{2}(x,1,t) \,dx, \end{aligned}$$
(4.16)
and
$$\begin{aligned} \int^{1}_{0}\bigl|\varphi_{x}f(\psi)\bigr|\,dx \leq& \| \varphi_{x}\|\|\psi\|^{\theta}_{2(\theta+1)}\|\psi \|_{2(\theta+1)} \\ \leq&\frac{\varepsilon}{2b^{2}} \int^{1}_{0}\varphi_{x}^{2}\,dx+ \frac {b^{2}}{2\varepsilon\lambda_{1}} \int^{1}_{0}\psi_{x}^{2}\,dx \\ \leq&\frac{\varepsilon}{b^{2}} \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+\frac {\varepsilon}{b^{2}} \int^{1}_{0}\psi^{2}\,dx+\frac{b^{2}}{2\varepsilon\lambda _{1}} \int^{1}_{0}\psi_{x}^{2}\,dx \\ \leq&\frac{\varepsilon}{b^{2}} \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+ \biggl(\frac{\varepsilon}{b^{2}\lambda_{1}}+\frac{b^{2}}{2\varepsilon\lambda _{1}} \biggr) \int^{1}_{0}\psi_{x}^{2}\,dx, \end{aligned}$$
(4.17)
which, together with (4.15)-(4.16), gives us (4.14). The proof is now complete. □
Next we deal with the boundary term in (4.14). As in [13], we define the function
$$\begin{aligned} q(x)=-4x+2,\quad x\in(0,1). \end{aligned}$$
(4.18)
Lemma 4.4
Let
\((\varphi,\varphi_{t},\psi,\psi_{t},z)\)
be the solution of problem (2.5)-(2.6), then the following estimate holds for any
\(\varepsilon>0\):
$$\begin{aligned} b[\psi_{x}\varphi_{x}]^{x=1}_{x=0} \leq&-\frac{\varepsilon\rho _{1}}{k}\frac{d}{dt} \int^{1}_{0}q\varphi_{t} \varphi_{x}\,dx-\frac{\rho _{2}b}{4\varepsilon} \frac{d}{dt} \int^{1}_{0}q\psi_{t}\psi_{x} \,dx+ \frac{2\rho_{1}\varepsilon}{k} \int^{1}_{0}\varphi_{t}^{2}\,dx \\ &{}+ \biggl(\varepsilon+\frac{b^{2}}{2}+\frac{b^{2}}{2\varepsilon}+ \frac {b^{2}}{4\varepsilon^{3}}+\frac{3b^{2}}{4}+\frac{\varepsilon}{b^{2}\lambda_{1}} +\frac{b^{2}}{2\varepsilon\lambda_{1}} \biggr) \int^{1}_{0}\psi _{x}^{2}\,dx \\ &{}+ \biggl(\frac{\rho_{2}b}{2\varepsilon}+\frac{\mu_{1}^{2}}{4\varepsilon ^{2}} \biggr) \int^{1}_{0}\psi_{t}^{2}\,dx + \frac{k^{2}\varepsilon}{4} \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx \\ &{}+\frac{\mu_{2}^{2}}{4\varepsilon^{2}} \int^{1}_{0}z^{2}(x,1,t)\,dx. \end{aligned}$$
(4.19)
Proof
The same argument as in [13], we know that for any \(\varepsilon>0\),
$$\begin{aligned} b[\psi_{x}\varphi_{x}]^{x=1}_{x=0} \leq \varepsilon \bigl[\varphi_{x}^{2}(1)+ \varphi_{x}^{2}(0) \bigr]+\frac {b^{2}}{4\varepsilon} \bigl[ \psi_{x}^{2}(1)+\psi_{x}^{2}(0) \bigr]. \end{aligned}$$
(4.20)
By using (2.5), Young’s inequality, integration by parts and the following fact
$$\frac{d}{dt} \int^{1}_{0}b\rho_{2}q\psi_{t} \psi_{x}\,dx= \int^{1}_{0}b\rho_{2}\psi _{tt} \psi_{x}\,dx+ \int^{1}_{0}b\rho_{2}q\psi_{t} \psi_{xt}\,dx, $$
we see that
$$\begin{aligned} \frac{d}{dt} \int^{1}_{0}b\rho_{2}q\psi_{t} \psi_{x}\,dx \leq& -b^{2} \bigl[\psi_{x}^{2}(1)+ \psi_{x}^{2}(0) \bigr]+2b^{2} \int^{1}_{0}\psi _{x}^{2}\,dx+2 \rho_{2}b \int^{1}_{0}\psi_{t}^{2}\,dx \\ &{}+ \biggl(\frac{\varepsilon}{b^{2}}+\varepsilon^{2}k^{2} \biggr) \int ^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+2\varepsilon b^{2} \int^{1}_{0}\psi_{x}^{2}\,dx \\ &{}+ \biggl(b+\frac{b^{2}}{\varepsilon^{2}} +\frac{\varepsilon}{b^{2}\lambda_{1}} +\frac{b^{2}}{2\varepsilon\lambda_{1}} \biggr) \int^{1}_{0}\psi _{x}^{2}\,dx \\ &{}+\frac{\mu_{1}^{2}}{\varepsilon} \int^{1}_{0}\psi_{t}^{2}\,dx+ \frac{\mu _{2}^{2}}{\varepsilon} \int^{1}_{0}z^{2}(x,1,t)\,dx. \end{aligned}$$
(4.21)
Similarly,
$$\begin{aligned} \frac{d}{dt} \int^{1}_{0}\rho_{1}q\varphi_{t} \varphi_{x}\,dx\leq-k \bigl[\varphi_{x}^{2}(1)+ \varphi_{x}^{2}(0) \bigr]+3k \int^{1}_{0}\varphi_{x}^{2} \,dx+k \int ^{1}_{0}\psi_{x}^{2}\,dx +2\rho_{1} \int^{1}_{0}\varphi_{t}^{2}\,dx, \end{aligned}$$
which, along with (4.20)-(4.21), gives us (4.19). The proof is now complete. □
In order to handle the term \(z(x,\rho,t)\), we introduce the functional
$$\begin{aligned} I_{3}(t):= \int^{1}_{0} \int^{1}_{0}e^{-2\tau\rho}z^{2}(x, \rho,t)\,d\rho \,dx. \end{aligned}$$
(4.22)
Then we can find the following result in [13].
Lemma 4.5
Let
\((\varphi,\varphi_{t},\psi,\psi_{t},z)\)
be the solution of problem (2.5)-(2.6), then the following estimate holds:
$$\begin{aligned} \frac{d}{dt}I_{3}(t)\leq-I_{3}(t)- \frac{c}{2\tau} \int ^{1}_{0}z^{2}(x,1,t)\,dx+ \frac{1}{2\tau} \int^{1}_{0}\psi_{t}^{2}\,dx, \end{aligned}$$
(4.23)
where
c
is a positive constant.
Now we define the following Lyapunov functional \(\mathscr{L}(t)\) by
$$\begin{aligned} \mathscr{L}(t) :=&ME(t)+\frac{1}{8}I_{1}(t)+NI_{2}(t)+J(t)+ \frac {\varepsilon}{k} \int^{1}_{0}\rho_{1}q\varphi_{t} \varphi_{x}\,dx \\ &{}+\frac{\rho_{2}b}{4\varepsilon} \int^{1}_{0}q\psi_{t}\psi _{x} \,dx+I_{3}(t). \end{aligned}$$
(4.24)
Then we may obtain the following lemma.
Lemma 4.6
Let
\((\varphi,\varphi_{t},\psi,\psi_{t},z)\)
be the solution of problem (2.5)-(2.6). For
M
large enough, there exist two positives
\(\gamma_{1}\)
and
\(\gamma_{2}\)
depending on
M, N
and
ε
such that for any
\(t\geq0\),
$$\begin{aligned} \gamma_{1}E(t) \leq\mathscr{L}(t)\leq \gamma_{2}E(t). \end{aligned}$$
(4.25)
Proof
The same argument as in [13], we can deduce
$$\begin{aligned} \bigl|\mathscr{L}(t)-ME(t)\bigr| \leq&\alpha_{1} \int^{1}_{0}\varphi_{t}^{2}\,dx+ \alpha _{2} \int^{1}_{0}\psi_{t}^{2}\,dx+ \alpha_{3} \int^{1}_{0}(\varphi_{x}+\psi )^{2}\,dx \\ &{}+\alpha_{4} \int^{1}_{0}\psi_{x}^{2}\,dx+ \int^{1}_{0} \int^{1}_{0}z^{2}(x,\rho,t)\,d\rho \,dt \\ &{}+ \int^{1}_{0}\hat{f}(\psi)\,dx, \end{aligned}$$
(4.26)
where the positive constants \(\alpha_{i}\) (\(i=1,2,3,4\)) are determined as in [13].
Performing Young’s inequality and using the fact
$$\int^{1}_{0}\varphi_{x}^{2}\,dx \leq2 \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+2 \int ^{1}_{0}\psi^{2}\,dx, $$
we easily get
$$\begin{aligned} E(t) \geq& \frac{1}{4}\min\{1,\xi\} \biggl( \int^{1}_{0}\varphi_{t}^{2}\,dx+ \int^{1}_{0}\psi _{t}^{2}\,dx+ \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx \\ &{}+ \int^{1}_{0}\psi_{x}^{2}\,dx+ \int^{1}_{0} \int^{1}_{0}z^{2}(x,\rho,t)\,d\rho \,dt+ \int^{1}_{0}\hat{f}(\psi)\,dx \biggr), \end{aligned}$$
(4.27)
It follows from (4.26)-(4.27) that there exists a positive constant C̃ such that
$$\begin{aligned} \bigl|\mathscr{L}(t)-ME(t)\bigr| \leq&\tilde{C}E(t). \end{aligned}$$
(4.28)
Then choosing M so large that \(\gamma_{1}:=M-\tilde{C}>0\) and \(\gamma_{2}=M+\tilde{C}>0\), we complete the proof. □
Proof of Theorem 2.2
It follows from (3.1), (4.2), (4.8), (4.14), (4.19), (4.23) and (4.24) that
$$\begin{aligned} \frac{d}{dt}\mathscr{L}(t) \leq{}& \biggl[-MC-\frac{\rho_{2}}{8}+N \biggl(\rho_{2}+ \frac{\rho_{1}}{4\tilde {\eta}} \biggr)+\frac{\rho_{2}b}{2\varepsilon}+\frac{\mu _{1}^{2}}{4\varepsilon^{2}} + \rho_{2}+\frac{\mu_{1}^{2}}{k}+\frac{1}{2\tau} \biggr] \int^{1}_{0}\psi _{t}^{2}\,dx \\ &{} + \biggl[-MC+\frac{\mu_{2}^{2}}{32\lambda_{1}\varepsilon}+\frac {N\mu_{2}}{4\eta\lambda_{1}}+\frac{\mu_{2}^{2}}{k}+ \frac{\mu _{2}^{2}}{4\varepsilon^{2}}-\frac{c}{2\tau} \biggr] \int^{1}_{0}z^{2}(x,1,t)\,dx \\ &{} + \biggl[-\frac{\rho_{1}}{8}+N\tilde{\eta}\frac{\rho _{1}}{\lambda_{1}}+ \frac{2\rho_{1}\varepsilon}{k} \biggr] \int^{1}_{0}\varphi_{t}^{2}\,dx + \biggl[-\frac{3k}{8}+\frac{k^{2}\varepsilon}{4} \biggr] \int ^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx \\ &{} + \biggl[\frac{b+C_{1}}{8}+\varepsilon-N (b-\mu_{2}\eta )+ \frac{b^{2}}{2\varepsilon} +\frac{b^{2}}{4\varepsilon^{3}}+\frac{b^{2}}{2}+\frac{2\varepsilon }{b^{2}\lambda_{1}} + \frac{b^{2}}{\varepsilon\lambda_{1}} \biggr] \int^{1}_{0}\psi _{x}^{2}\,dx \\ &{} +(-N-1) \int^{1}_{0}\hat{f}(\psi)\,dx. \end{aligned}$$
(4.29)
First we choose η so small that
$$\eta\leq\frac{b}{2C\mu_{2}}, $$
and then we choose ε small enough so that
$$\varepsilon\leq \frac{5k}{128+4k^{2}}. $$
Then we take N so large that
$$\frac{Nb}{4}\geq \frac{b+C_{1}}{8}+\varepsilon+\frac{b^{2}}{2\varepsilon} + \frac{b^{2}}{4\varepsilon^{3}}+\frac{b^{2}}{2}+\frac{2\varepsilon }{b^{2}\lambda_{1}} +\frac{b^{2}}{\varepsilon\lambda_{1}}. $$
After that, we select η̃ small enough so that
$$\tilde{\eta}\leq \frac{1}{32N}. $$
Then we take M so large that there exists a positive constant δ such that
$$\begin{aligned} \frac{d}{dt}\mathscr{L}(t) \leq& -\delta \int^{1}_{0} \bigl(\varphi_{t}^{2}+ \psi_{t}^{2}+(\varphi_{x}+\psi)^{2}+\psi _{x}^{2}+z^{2}(x,1,t) \bigr)\,dx \\ &{}-\delta \int^{1}_{0} \int^{1}_{0}z^{2}(x,\rho,t)\,d\rho \,dx- \delta \int^{1}_{0}\hat{f}(\psi)\,dx. \end{aligned}$$
(4.30)
Noting that (2.13), we know that there exists a positive constant β such that
$$\begin{aligned} \frac{d}{dt}\mathscr{L}(t) \leq&-\beta E(t), \end{aligned}$$
(4.31)
which, together with (4.25), yields
$$\frac{d}{dt}\mathscr{L}(t)\leq -\frac{\beta}{\gamma_{2}} \mathscr{L}(t). $$
Then we can get
$$\begin{aligned} \mathscr{L}(t)\leq \mathscr{L}(0)e^{-\frac{\beta}{\gamma_{2}}t}. \end{aligned}$$
(4.32)
Using again (4.25), we find that
$$E(t)\leq \frac{\gamma_{2}}{\gamma_{1}}E(0)e^{-\frac{\beta}{\gamma _{2}}t}, $$
which gives us that the exponential stability holds for any \(U_{0}\in D(\mathcal{A})\). Noting that \(D(\mathcal{A})\) is dense in \(\mathscr{H}\), we can extend the energy inequalities to phase space \(\mathscr{H}\). Thus we complete the proof of Theorem 2.2. □