In this section, we shall state and prove our decay result.
Theorem 4.1
Suppose that Assumption
2.1
holds and
\(\mu_{1}>\mu_{2}\). For any initial value
\(\mathcal{U}_{0}\in\mathcal{H}\), there exist positive constants
C
and
α
such that the energy of problem (2.4) satisfies
$$E(t)\leq CE(0)e^{-\alpha t} \quad \textit{if } \frac{\rho_{1}}{K}=\frac {\rho_{2}}{b}. $$
Moreover, if the initial value
\(\mathcal{U}_{0}\in D(\mathcal{A})\), we have that, for some constants
\(C>0\)
and
\(M_{1}>0\), the energy of problem (2.4) satisfies
$$E(t)\leq C\bigl(E(0)+E_{2}(0) \bigr)t^{-1} \quad \textit{if } 0< \bigg|\frac{\rho _{1}}{K}- \frac{\rho_{2}}{b}\bigg|< \frac{M_{1}\gamma K}{4(K+b)}. $$
In order to prove this result, we introduce various functionals and establish several lemmas. The construction of the auxiliary function \(I_{1}(t)-I_{3}(t)\), \(I_{5}(t)\) comes from [16].
Lemma 4.2
Let
\((\phi,\psi,\theta,z)\)
be the solution of (2.4). The functional
\(I_{1}\)
defined by
$$ I_{1}(t):=- \int_{0}^{1}(\rho_{1}\phi_{t} \phi+\rho_{2}\psi_{t} \psi) \, dx-\frac{\mu_{1}}{2} \int_{0}^{1}\psi^{2}\,dx $$
(4.1)
satisfies
$$ \begin{aligned} I'_{1}(t)\leq{}&{-} \int_{0}^{1}\bigl(\rho_{1} \phi_{t}^{2}+\rho_{2}\psi_{t}^{2} \bigr)\,dx+ \int_{0}^{1} K(\phi_{x}+ \psi)^{2}\,dx+(b+C_{1}+2) \int_{0}^{1}\psi_{x}^{2}\,dx \\ &+\frac{\beta^{2}}{4} \int_{0}^{1}\theta_{tx}^{2}\,dx+ \frac{\mu_{2}^{2}}{4} \int_{0}^{1} z^{2}(x,1,t)\,dx. \end{aligned} $$
Proof
By differentiating \(I_{1}\) and using (2.4), we conclude that
$$ \begin{aligned}[b] I'_{1}(t)= {}&{-} \int_{0}^{1}\bigl(\rho_{1} \phi_{t}^{2}+\rho_{2}\psi_{t}^{2} \bigr)\,dx+ \int_{0}^{1} K(\phi_{x}+ \psi)^{2}\,dx+b \int_{0}^{1}\psi_{x}^{2}\,dx+ \int_{0}^{1}f(\psi)\psi\,dx \\ &+\beta \int_{0}^{1}\theta_{tx}\psi\,dx+ \mu_{2} \int_{0}^{1} z(x,1,t)\psi\,dx. \end{aligned} $$
(4.2)
By using Young’s inequality and the fact \(\int_{0}^{1}\psi^{2}\,dx\leq\int _{0}^{1}\psi_{x}^{2}\,dx\), we have
$$\begin{aligned}& \beta \int_{0}^{1}\theta_{tx}\psi\,dx \leq \frac{\beta^{2}}{4} \int_{0}^{1}\theta _{tx}^{2} \,dx+ \int_{0}^{1}\psi_{x}^{2}\,dx, \end{aligned}$$
(4.3)
$$\begin{aligned}& \mu_{2} \int_{0}^{1} z(x,1,t)\psi\,dx \leq\frac{\mu_{2}^{2}}{4} \int_{0}^{1} z^{2}(x,1,t)\,dx+ \int_{0}^{1}\psi_{x}^{2}\,dx. \end{aligned}$$
(4.4)
For the fourth term in (4.2), using (2.2) and the generalized Hölder inequality, we obtain
$$\begin{aligned} \int_{0}^{1}\big|f(\psi)\psi\big| \,dx \leq k_{0} \int_{0}^{1}|\psi|^{\varsigma}|\psi||\psi |\,dx \leq k_{0}\|\psi\|_{2(\varsigma+1)}^{\varsigma}\|\psi\|_{2(\varsigma+1)} \| \psi\|. \end{aligned} $$
By Sobolev–Poincaré inequality and \(\dot{E}(t)\leq0\), we get
$$ \|\psi\|_{2(\varsigma+1)}\leq C \|\psi_{x}\|\leq C \biggl( \frac{2}{b\gamma }E(t) \biggr)^{\frac{1}{2}}\leq C \biggl(\frac{2}{b\gamma} \biggr)^{\frac {1}{2}}E(0)^{\frac{1}{2}}, $$
(4.5)
in which \(C>0\) is a constant. Thus, together with the above two inequalities, Young’s inequality and the Sobolev embedding theorem for ψ, we obtain
$$ \int_{0}^{1}\big|f(\psi)\psi\big| \,dx \leq C_{1} \int_{0}^{1}\psi_{x}^{2}\,dx. $$
(4.6)
Insert (4.3), (4.4), and (4.6) into (4.3), then Lemma 4.2 follows. □
Lemma 4.3
Let
\((\phi,\psi,\theta,z)\)
be the solution of (2.4). The functional
\(I_{2}\)
defined by
$$ I_{2}(t):= \int_{0}^{1}(\rho_{2}\psi_{t} \psi+\rho_{1}\phi_{t}g)\, dx+\frac{\mu_{1}}{2} \int_{0}^{1}\psi^{2}\,dx, $$
(4.7)
where
g
is the solution of
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} -g_{xx}=\psi_{x},& 0< x< 1,\\ g(0)=g(1)=0,& \end{array}\displaystyle \right . $$
(4.8)
satisfies that, for any
\(\varepsilon_{2} > 0\),
$$ \begin{aligned}[b] I'_{2}(t)\leq{}&(-b+2 \varepsilon_{2}) \int_{0}^{1}\psi_{x}^{2}\,dx+ \rho _{1}\varepsilon_{2} \int_{0}^{1}\phi_{t}^{2}\,dx+ \biggl(\rho_{2}+\frac{\rho_{1}}{4\varepsilon _{2}}\biggr) \int_{0}^{1}\psi_{t}^{2}\,dx \\ &+\frac{\beta^{2}}{4\varepsilon_{2}} \int_{0}^{1}\theta_{tx}^{2}\,dx+ \frac{\mu _{2}^{2}}{4\varepsilon_{2}} \int_{0}^{1} z^{2}(x,1,t)\,dx- \int_{0}^{1} \hat{f}\bigl(\psi(t)\bigr)\,dx. \end{aligned} $$
(4.9)
Proof
Using (2.4) and integration by parts, we conclude that
$$ \begin{aligned}[b] I'_{2}(t)={}&{-}b \int_{0}^{1}\psi_{x}^{2}\,dx+ \rho_{2} \int_{0}^{1}\psi_{t}^{2}\,dx- \beta \int _{0}^{1}\theta_{tx}\psi\,dx- \mu_{2} \int_{0}^{1}z(x,1,t)\psi\,dx \\ &- \int_{0}^{1} f(\psi)\psi\,dx+\rho_{1} \int_{0}^{1}\varphi_{t}g_{t} \,dx-K \int_{0}^{1}\psi ^{2}\,dx+K \int_{0}^{1}g_{x}^{2}\,dx. \end{aligned} $$
(4.10)
By the fact following from (4.8) that
$$\begin{aligned}& \int_{0}^{1}g_{x}^{2}\,dx\leq \int_{0}^{1}\psi^{2}\,dx\leq \int_{0}^{1}\psi_{x}^{2}\,dx, \end{aligned}$$
(4.11)
$$\begin{aligned}& \int_{0}^{1}g_{t}^{2}\,dx\leq \int_{0}^{1}g_{xt}^{2}\,dx\leq \int_{0}^{1}\psi_{t}^{2}\,dx, \end{aligned}$$
(4.12)
we obtain that
$$ \begin{aligned}[b] I'_{2}(t)\leq{}&{-}b \int_{0}^{1}\psi_{x}^{2}\,dx+ \rho_{2} \int_{0}^{1}\psi_{t}^{2}\,dx- \beta \int_{0}^{1}\theta_{tx}\psi\,dx- \mu_{2} \int_{0}^{1}z(x,1,t)\psi\,dx \\ &- \int_{0}^{1} \hat{f}\bigl(\psi(t)\bigr)\,dx+ \rho_{1} \int_{0}^{1}\varphi_{t}g_{t} \,dx. \end{aligned} $$
(4.13)
By using Young’s inequality and (4.11)–(4.12), for any \(\varepsilon_{2} > 0 \), we have
$$\begin{aligned}& \beta \int_{0}^{1}\theta_{tx}\psi\,dx \leq \frac{\beta^{2}}{4\varepsilon _{2}} \int_{0}^{1}\theta_{tx}^{2}\,dx+ \varepsilon_{2} \int_{0}^{1}\psi_{x}^{2} \,dx, \end{aligned}$$
(4.14)
$$\begin{aligned}& \mu_{2} \int_{0}^{1} z(x,1,t)\psi\,dx \leq\frac{\mu_{2}^{2}}{4\varepsilon_{2}} \int _{0}^{1} z^{2}(x,1,t)\,dx+ \varepsilon_{2} \int_{0}^{1}\psi_{x}^{2} \,dx, \end{aligned}$$
(4.15)
$$\begin{aligned}& \begin{aligned}[b]\rho_{1} \int_{0}^{1}\varphi_{t}g_{t}\,dx &\leq\rho_{1}\varepsilon_{2} \int_{0}^{1}\phi_{t}^{2}\,dx+ \frac{\rho_{1}}{4\varepsilon _{2}} \int_{0}^{1}g_{t}^{2}\,dx\\ &\leq \rho_{1}\varepsilon_{2} \int_{0}^{1}\phi_{t}^{2}\,dx+ \frac{\rho_{1}}{4\varepsilon _{2}} \int_{0}^{1}\psi_{t}^{2} \,dx.\end{aligned} \end{aligned}$$
(4.16)
Combining (4.13)–(4.16), we have (4.9). □
Lemma 4.4
Let
\((\phi,\psi,\theta,z)\)
be the solution of (2.4). The functional
\(I_{3}\)
defined by
$$ I_{3}(t):=\rho_{2} \int_{0}^{1}\psi_{t}(\phi_{x}+ \psi)\,dx+\rho_{2} \int _{0}^{1}\psi_{x}\phi_{t} \,dx $$
(4.17)
satisfies for any
\(\varepsilon_{3}>0\)
that
$$ \begin{aligned}[b] I'_{3}(t)\leq{}& b \phi_{x}\psi_{x}|_{x=0}^{x=1}+\biggl( \frac{\rho_{2}K}{\rho _{1}}-b\biggr) \int_{0}^{1}(\phi_{x}+ \psi)_{x}\psi_{x}\,dx-\biggl(\frac{K}{4}- \frac{\varepsilon _{3} C_{2}}{b^{2}}\biggr) \int_{0}^{1}(\phi_{x}+\psi)^{2} \,dx \\ &+\biggl(\rho_{2}+\frac{\mu_{1}^{2}}{K}\biggr) \int_{0}^{1}\psi_{t}^{2}\,dx+ \frac{\beta^{2}}{K} \int _{0}^{1}\theta_{tx}^{2} \,dx+\frac{\mu_{2}^{2}}{K} \int_{0}^{1}z^{2}(x,1,t)\,dx \\ &- \int_{0}^{1} \hat{f}(\psi)\,dx +\biggl( \frac{\varepsilon_{3} C_{2}}{b^{2}}+\frac {b^{2}}{2\varepsilon_{3}}\biggr) \int_{0}^{1}\psi_{x}^{2}\,dx. \end{aligned} $$
(4.18)
Proof
By differentiating \(I_{3}\) and using (2.4), we conclude that
$$ \begin{aligned}[b] I'_{3}(t)={}&b \phi_{x}\psi_{x}|_{x=0}^{x=1}+\biggl( \frac{\rho_{2}K}{\rho_{1}}-b\biggr) \int _{0}^{1}(\phi_{x}+ \psi)_{x}\psi_{x}\,dx\\ &-K \int_{0}^{1}(\phi_{x}+\psi)^{2} \,dx+\rho_{2} \int _{0}^{1}\psi_{t}^{2}\,dx \\ &-\beta \int_{0}^{1}\theta_{tx}( \phi_{x}+\psi)\,dx-\mu_{1} \int_{0}^{1}\psi_{t}(\phi _{x}+\psi)\,dx-\mu_{2} \int_{0}^{1}z(x,1,t) (\phi_{x}+\psi)\,dx \\ &- \int_{0}^{1} f(\psi)\psi\,dx- \int_{0}^{1} f(\psi)\phi_{x}\,dx. \end{aligned} $$
(4.19)
By using Young’s inequality and Poincaré’s inequality, we have
$$\begin{aligned}& -\beta \int_{0}^{1}\theta_{tx}( \phi_{x}+\psi)\,dx\leq\frac{K}{4} \int_{0}^{1}(\phi _{x}+ \psi)^{2}\,dx+\frac{\beta^{2}}{K} \int_{0}^{1}\theta_{tx}^{2} \,dx, \end{aligned}$$
(4.20)
$$\begin{aligned}& -\mu_{1} \int_{0}^{1}\psi_{t}(\phi_{x}+ \psi)\,dx \leq\frac{K}{4} \int_{0}^{1}(\phi _{x}+ \psi)^{2}\,dx+\frac{\mu_{1}^{2}}{K} \int_{0}^{1}\psi_{t}^{2} \,dx, \end{aligned}$$
(4.21)
$$\begin{aligned}& -\mu_{2} \int_{0}^{1}z(x,1,t) (\phi_{x}+\psi)\,dx \leq\frac{K}{4} \int_{0}^{1}(\phi _{x}+ \psi)^{2}\,dx+\frac{\mu_{2}^{2}}{K} \int_{0}^{1}z^{2}(x,1,t) \,dx. \end{aligned}$$
(4.22)
By using the fact that
$$ \int_{0}^{1}\phi_{x}^{2}\,dx \leq2 \int_{0}^{1}(\phi_{x}+\psi)^{2} \, dx+2 \int_{0}^{1}\psi_{x}^{2} \,dx, $$
(4.23)
we arrive at, for any \(\varepsilon_{3} > 0 \),
$$ \begin{aligned}[b] \int_{0}^{1}\big| f(\psi)\phi_{x}\big|\,dx&\leq k_{0}\|\psi\|_{2(\varsigma +1)}^{\varsigma}\|\psi\|_{2(\varsigma+1)}\| \phi_{x}\| \\ &\leq\frac{\varepsilon_{3}C_{2}}{2b^{2}} \int_{0}^{1}\phi_{x}^{2}\,dx+ \frac {b^{2}}{2\varepsilon_{3}} \int_{0}^{1}\psi_{x}^{2}\,dx \\ &\leq\frac{\varepsilon_{3}C_{2}}{b^{2}} \int_{0}^{1}(\phi_{x}+\psi)^{2} \,dx+\frac {\varepsilon_{3}C_{2}}{b^{2}} \int_{0}^{1}\psi^{2}\,dx+ \frac{b^{2}}{2\varepsilon_{3}} \int _{0}^{1}\psi_{x}^{2}\,dx \\ &\leq\frac{\varepsilon_{3}C_{2}}{b^{2}} \int_{0}^{1}(\phi_{x}+\psi)^{2} \,dx +\biggl(\frac {\varepsilon_{3}C_{2}}{b^{2}}+\frac{b^{2}}{2\varepsilon_{3}}\biggr) \int_{0}^{1}\psi_{x}^{2}\,dx, \end{aligned} $$
(4.24)
in which \(C_{2}\) is a positive constant. Combining (4.20)–(4.24) yields the conclusion. □
Next we deal with the boundary term in (4.18). We introduce the function
$$ q(x)=-4x+2,\quad x\in(0,1). $$
(4.25)
Lemma 4.5
Let
\((\phi,\psi,\theta,z)\)
be the solution of (2.4), then for
\(\varepsilon_{3}>0\)
the following estimate holds:
$$ \begin{aligned}[b] b\phi_{x}\psi_{x}|_{x=0}^{x=1} \leq{} &{-}\frac{b\rho_{2}}{4\varepsilon_{3}}\frac{d}{dt} \int_{0}^{1}q\psi_{t}\psi_{x} \,dx -\frac{\varepsilon_{3}\rho_{1}}{K}\frac{d}{dt} \int_{0}^{1}q\phi_{t}\phi_{x} \,dx \\ &+\biggl(7\varepsilon_{3}+\frac{b^{2}}{2\varepsilon_{3}}+\frac{b^{2}}{4\varepsilon _{3}^{3}}+ \frac{3b^{2}}{4}+\frac{C_{3}}{4\varepsilon_{3}}+\frac{b}{4\varepsilon _{3}^{2}}\biggr) \int_{0}^{1}\psi_{x}^{2}\,dx \\ &+\biggl(\frac{\mu_{1}^{2}}{4\varepsilon_{3}^{2}}+\frac{\rho_{2}b}{2\varepsilon_{3}}\biggr) \int _{0}^{1}\psi_{t}^{2}\,dx + \biggl(\frac{1}{4}K^{2}\varepsilon_{3}+6 \varepsilon_{3}\biggr) \int_{0}^{1}(\phi_{x}+\psi)^{2} \, dx \\ &+\frac{2\rho_{1}\varepsilon_{3}}{K} \int_{0}^{1}\phi_{t}^{2}\,dx + \frac{\beta^{2}}{4\varepsilon_{3}^{2}} \int_{0}^{1}\theta_{tx}^{2}\,dx+ \frac{\mu _{2}^{2}}{4\varepsilon_{3}^{2}} \int_{0}^{1}z^{2}(x,1,t)\,dx. \end{aligned} $$
(4.26)
Proof
By using Young’s inequality, for \(\varepsilon_{3} > 0 \), we have
$$ b\phi_{x}\psi_{x}|_{x=0}^{x=1}\leq \frac{b^{2}}{4\varepsilon_{3}}\bigl[\psi _{x}^{2}(1)+ \psi_{x}^{2}(0)\bigr]+\varepsilon_{3}\bigl[ \phi_{x}^{2}(1)+\phi_{x}^{2}(0) \bigr]. $$
(4.27)
Also, we have
$$ \begin{aligned} \frac{d}{dt} \int_{0}^{1}b\rho_{2}q\psi_{t} \psi_{x}\,dx={}& \frac{1}{2}b^{2}q\psi_{x}^{2}|_{x=0}^{x=1}- \frac{1}{2} \int_{0}^{1}b^{2}q_{x} \psi_{x}^{2}\,dx -\frac{1}{2}b\rho_{2} \int_{0}^{1}q_{x}\psi_{t}^{2} \,dx \\ &-bK \int_{0}^{1}q(\phi_{x}+\psi) \psi_{x}\,dx-b\beta \int_{0}^{1}q\theta_{tx} \psi_{x}\, dx \\ &-\mu_{1}b \int_{0}^{1}q\psi_{t}\psi_{x} \,dx-\mu_{2}b \int_{0}^{1}qz(x,1,t)\psi_{x}\,dx\\&-b \int_{0}^{1}qf(\psi)\psi_{x}\,dx. \end{aligned} $$
By using Young’s inequality and Poincaré’s inequality, for \(\varepsilon_{3} > 0 \), we have
$$ \begin{aligned}[b] \frac{d}{dt} \int_{0}^{1}b\rho_{2}q\psi_{t} \psi_{x}\,dx\leq {}& {-}b^{2}\bigl[\psi_{x}^{2}(1)+ \psi_{x}^{2}(0)\bigr]\\&+\biggl(2b^{2}+ \frac{b^{2}}{\varepsilon _{3}^{2}}+3\varepsilon_{3}b^{2}+\frac{b}{\varepsilon_{3}}+C_{3} \biggr) \int_{0}^{1}\psi_{x}^{2}\, dx \\ &+\biggl(2\rho_{2}b+\frac{\mu_{1}^{2}}{\varepsilon_{3}}\biggr) \int_{0}^{1}\psi_{t}^{2}\, dx+K^{2}\varepsilon_{3}^{2} \int_{0}^{1}(\phi_{x}+\psi)^{2} \,dx \\ &+\frac{\beta ^{2}}{\varepsilon_{3}} \int_{0}^{1}\theta_{tx}^{2}\,dx+\frac{\mu_{2}^{2}}{\varepsilon_{3}} \int_{0}^{1}z^{2}(x,1,t)\,dx. \end{aligned} $$
(4.28)
Similarly,
$$\begin{aligned}[b] \frac{d}{dt} \int_{0}^{1}\rho_{1}q\phi_{t} \phi_{x}\,dx\leq{}&{-}K\bigl[\phi_{x}^{2}(1)+\phi _{x}^{2}(0)\bigr]\\&+3K \int_{0}^{1}\phi_{x}^{2}\,dx+K \int_{0}^{1}\psi_{x}^{2}\,dx+2 \rho_{1} \int _{0}^{1}\phi_{t}^{2} \,dx.\end{aligned} $$
(4.29)
Together with (4.27)–(4.29), using (4.23) gives us (4.26). □
Lemma 4.6
Let
\((\phi,\psi,\theta,z)\)
be the solution of (2.4). The functional
\(I_{4}\)
defined by
$$ I_{4}(t):= \int_{0}^{1}\biggl(\rho_{3} \theta_{t}\theta+\frac{k}{2}\theta _{x}^{2}+ \gamma\psi_{x}\theta\biggr)\,dx, $$
(4.30)
and its time derivative
\(I'_{4}(t)\)
satisfies
$$ I'_{4}(t)\leq-\delta \int_{0}^{1}\theta_{x}^{2}\,dx+ \int_{0}^{1}\psi_{x}^{2}\, dx+C_{4}\biggl(\frac{\gamma^{2}}{4}+\rho_{3}\biggr) \int_{0}^{1}\theta_{tx}^{2}\,dx, $$
where
\(C_{4}>0\)
is the Sobolev embedding constant.
Proof
By differentiating \(I_{4}\) and using (2.4), we conclude that
$$ I'_{4}(t)= -\delta \int_{0}^{1}\theta_{x}^{2}\,dx+ \rho_{3} \int_{0}^{1}\theta_{t}^{2}\, dx+\gamma \int_{0}^{1}\psi_{x}\theta_{t} \,dx. $$
Using Young’s inequality and Poincaré’s inequality clearly implies the conclusion (4.30). □
Lemma 4.7
Let
\((\phi,\psi,\theta,z)\)
be the solution of (2.4). The functional
\(I_{5}\)
defined by
$$I_{5}(t):= \int_{0}^{1} \int_{0}^{1}e^{-2\tau\rho}z^{2}(x, \rho,t) \,d\rho\,dx $$
for some constant
\(m>0\)
satisfies
$$ I'_{5}(t)\leq-m \int_{0}^{1}z^{2}(x,1,t)\,dx-m \int_{0}^{1} \int_{0}^{1}z^{2}(x,\rho ,t)\,d\rho\,dx+ \frac{1}{\tau} \int_{0}^{1}\psi_{t}^{2} \,dx. $$
(4.31)
Proof
By differentiating \(I_{5}\) and using (2.4), we conclude that
$$ \begin{aligned} I'_{5}(t)&= - \frac{2}{\tau} \int_{0}^{1} \int_{0}^{1}e^{-2\tau\rho}zz_{\rho }(x, \rho,t)\,d\rho\,dx = -\frac{1}{\tau} \int_{0}^{1} \int_{0}^{1}e^{-2\tau\rho}\frac{\partial}{\partial \rho}z^{2}(x, \rho,t) \,d\rho\,dx \\ &=-2 \int_{0}^{1} \int_{0}^{1}e^{-2\tau\rho}z^{2}(x, \rho,t) \,d\rho\,dx-\frac{1}{\tau } \int_{0}^{1} \int_{0}^{1}\frac{\partial}{\partial\rho}\bigl(e^{-2\tau\rho}z^{2}(x, \rho ,t)\bigr) \,d\rho\,dx \\ &\leq-m \int_{0}^{1}z^{2}(x,1,t)\,dx-m \int_{0}^{1} \int_{0}^{1}z^{2}(x,\rho,t)\,d\rho\, dx+ \frac{1}{\tau} \int_{0}^{1}\psi_{t}^{2}\,dx. \end{aligned} $$
This gives (4.31). □
Now we define the Lyapunov functional \(\mathscr{L}(t)\) as follows:
$$ \begin{aligned}[b] \mathscr{L}(t):={}&NE(t)+\frac {1}{8}I_{1}(t)+N_{2}I_{2}(t)+I_{3}(t)+I_{4}(t)+I_{5}(t) \\ &+\frac{b\rho_{2}}{4\varepsilon_{3}} \int_{0}^{1}q\psi_{t}\psi_{x} \,dx +\frac{\varepsilon_{3}\rho_{1}}{K} \int_{0}^{1}q\phi_{t}\phi_{x} \,dx, \end{aligned} $$
(4.32)
where N, \(N_{2}\) are positive constants to be chosen properly later. For N large enough, it is not difficult to prove that there exist two positive constants \(\gamma_{1}\) and \(\gamma_{2}\) such that, for any \(t>0\),
$$ \gamma_{1}E(t)\leq\mathscr{L}(t)\leq\gamma_{2}E(t). $$
(4.33)
Proof of Theorem 4.1
Combining Lemmas 4.2–4.7, we have
$$ \begin{aligned}[b] \mathscr{L}'(t)\leq{}& \biggl(-CN+ \frac{\beta^{2}}{32}+N_{2}\frac{\beta ^{2}}{4\varepsilon_{2}} +\frac{\beta^{2}}{K}+ \frac{\beta^{2}}{4\varepsilon_{3}^{2}}+C_{4}\biggl(\frac{\gamma ^{2}}{4}+\rho_{3} \biggr)\biggr) \int_{0}^{1}\theta_{tx}^{2}\,dx \\ &+\biggl(-CN-\frac{1}{8}\rho_{2}+N_{2}\biggl( \frac{\rho_{1}}{4\varepsilon_{2}}+\rho_{2}\biggr)+\rho _{2}+ \frac{\mu_{1}^{2}}{K}+\frac{\mu_{1}^{2}}{4\varepsilon_{3}^{2}}+\frac{\rho _{2}b}{2\varepsilon_{3}} +\frac{1}{\tau} \biggr) \int_{0}^{1}\psi_{t}^{2}\,dx \\ &+\biggl(-CN-m+\frac{\mu_{2}^{2}}{32}+N_{2}\frac{\mu_{2}^{2}}{4\varepsilon_{2}}+ \frac{\mu _{2}^{2}}{K}+\frac{\mu_{2}^{2}}{4\varepsilon_{3}^{2}}\biggr) \int_{0}^{1}z^{2}(x,1,t)\,dx \\ &+\biggl(-\frac{1}{8}\rho_{1}+N_{2}\rho_{1} \varepsilon_{2}+\frac{2\rho_{1}\varepsilon _{3}}{K}\biggr) \int_{0}^{1}\phi_{t}^{2}\,dx + \biggl[N_{2}(-b+2\varepsilon_{2})+\frac{1}{8}(b+C_{1}+2) \\ &+7\varepsilon_{3}+\frac {b^{2}}{2\varepsilon_{3}}+\frac{b^{2}}{4\varepsilon_{3}^{3}}+ \frac{3b^{2}}{4}+\frac {C_{3}}{4\varepsilon_{3}}+\frac{b}{4\varepsilon_{3}^{2}} +\frac{\varepsilon_{3}C_{2}}{b^{2}}+ \frac{b^{2}}{2\varepsilon_{3}}+1\biggr] \int_{0}^{1}\psi _{x}^{2}\,dx \\ &+\biggl(-\frac{1}{8}K+\frac{1}{4}K^{2} \varepsilon_{3}+6\varepsilon_{3}+\frac {\varepsilon_{3}C_{2}}{b^{2}}\biggr) \int_{0}^{1}(\phi_{x}+\psi)^{2} \,dx \\ &+(-\delta) \int_{0}^{1}\theta_{x}^{2} \,dx+(-m) \int_{0}^{1} \int_{0}^{1}z^{2}(x,\rho ,t)\,d\rho \,dx+(-N_{2}-1) \int_{0}^{1} \hat{f}(\psi)\,dx \\ &+\biggl(\frac{\rho_{2}K}{\rho_{1}}-b\biggr) \int_{0}^{1}(\phi_{x}+ \psi)_{x}\psi_{x}\,dx. \end{aligned} $$
(4.34)
Firstly, we take \(\varepsilon_{3}\) small enough such that
$$\left\{ \textstyle\begin{array}{l} -\frac{1}{8}K+\frac{1}{4}K^{2} \varepsilon_{3}+6\varepsilon_{3}+\frac {\varepsilon_{3}C_{2}}{b^{2}}< 0, \\ -\frac{1}{8}+\frac{2\varepsilon_{3}}{K}< 0. \end{array}\displaystyle \right . $$
Then we choose \(N_{2}\) so large that
$$N_{2}b>2\biggl[\frac{1}{8}(b+C_{1}+2)+7 \varepsilon_{3}+\frac{b^{2}}{2\varepsilon _{3}}+\frac{b^{2}}{4\varepsilon_{3}^{3}} + \frac{3b^{2}}{4}+\frac{C_{3}}{4\varepsilon_{3}}+\frac{b}{4\varepsilon _{3}^{2}}+\frac{\varepsilon_{3}C_{2}}{b^{2}}+ \frac{b^{2}}{2\varepsilon_{3}}+1\biggr]=:\Xi, $$
thus we have
$$-N_{2}b+\frac{1}{2}\Xi< -\frac{1}{2}\Xi. $$
After that, we select \(\varepsilon_{2}\) small enough such that
$$-\frac{1}{2}\Xi+N_{2}\varepsilon_{2}< 0 $$
and
$$-\frac{1}{8}+\frac{2\varepsilon_{3}}{K}+N_{2}\varepsilon_{2}< 0. $$
Finally, we choose N so large that
$$\begin{aligned}& -CN+\frac{\beta^{2}}{32}+N_{2}\frac{\beta^{2}}{4\varepsilon_{2}} +\frac{\beta^{2}}{K}+ \frac{\beta^{2}}{4\varepsilon_{3}^{2}}+C_{4}\biggl(\frac{\gamma ^{2}}{4}+\rho_{3} \biggr)< 0, \\& -CN-\frac{1}{8}\rho_{2}+N_{2}\biggl( \frac{\rho_{1}}{4\varepsilon_{2}}+\rho_{2}\biggr)+\rho _{2}+ \frac{\mu_{1}^{2}}{K}+\frac{\mu_{1}^{2}}{4\varepsilon_{3}^{2}}+\frac{\rho _{2}b}{2\varepsilon_{3}} +\frac{1}{\tau}< 0, \\& -CN-m+\frac{\mu_{2}^{2}}{32}+N_{2}\frac{\mu_{2}^{2}}{4\varepsilon_{2}}+\frac{\mu _{2}^{2}}{K}+ \frac{\mu_{2}^{2}}{4\varepsilon_{3}^{2}}< 0. \end{aligned}$$
Therefore (4.34) changes to
$$ \begin{aligned}[b] \mathscr{L}'(t)&\leq-M \int_{0}^{1}\bigl(\theta_{tx}^{2}+ \psi_{t}^{2}+z^{2}(x,1,t)+\phi _{t}^{2}+ \psi_{x}^{2}+(\phi_{x}+\psi)^{2}+ \theta_{x}^{2}\bigr)\,dx \\ &\quad-M \int_{0}^{1} \int_{0}^{1}z^{2}(x,\rho,t)\,d\rho\,dx+ \biggl(\frac{\rho_{2}K}{\rho _{1}}-b\biggr) \int_{0}^{1}(\phi_{x}+ \psi)_{x}\psi_{x}\,dx \\ &\leq-M_{1}E(t)+\biggl(\frac{\rho_{2}K}{\rho_{1}}-b\biggr) \int_{0}^{1}(\phi_{x}+ \psi)_{x}\psi_{x}\,dx, \end{aligned} $$
(4.35)
where M, \(M_{1}\) are positive constants.
Case 1: \({\frac{\rho_{1}}{K}=\frac{\rho_{2}}{b}}\).
In this case, (4.35) takes the form
$$\mathscr{L}'(t) \leq-M_{1}E(t). $$
Using (4.33), we get, for \(\alpha=\frac{M_{1}}{\gamma_{2}}\),
$$ \mathscr{L}'(t) \leq-\alpha\mathscr{L}(t). $$
(4.36)
A simple integration of (4.36) over \((0,t)\) leads to
$$\mathscr{L}(t)\leq\mathscr{L}(0)e^{-\alpha t}. $$
Recalling (4.33), we obtain
$$E(t)\leq CE(0)e^{-\alpha t}. $$
Case 2: \(0<|\frac{\rho_{1}}{K}-\frac{\rho_{2}}{b}|<\frac{M_{1}\gamma K}{4(K+b)}\).
Let
represent the first order energy defined in (3.6). By computation we have the estimate of the derivative of the second order energy (3.7) as
$$ E_{2}'(t)\leq-C \int_{0}^{1}\bigl(\theta_{ttx}^{2}+ \psi _{tt}^{2}+z_{t}^{2}(x,1,t)\bigr) \,dx. $$
(4.37)
Let us estimate the last term in (4.35). By setting \(\Lambda =\frac{1}{K}(\frac{\rho_{2}K}{\rho_{1}}-b)\rho_{1}\neq0\) and using (2.4), (4.23), we have
$$ \begin{aligned}[b] \biggl(\frac{\rho_{2}K}{\rho_{1}}-b\biggr) \int_{0}^{1}(\phi_{x}+ \psi)_{x}\psi_{x}\,dx&=\frac {1}{K}\biggl( \frac{\rho_{2}K}{\rho_{1}}-b\biggr) \int_{0}^{1}\rho_{1}\phi_{tt} \psi_{x}\, dx\\&=\Lambda \int_{0}^{1}\phi_{tt}\psi_{x} \,dx \\ &=-\Lambda\biggl(\frac{d}{dt} \int_{0}^{1}\phi_{xt}\psi\,dx- \frac{d}{dt} \int _{0}^{1}\phi_{x}\psi_{t} \,dx\biggr)\\ &\quad-\Lambda \int_{0}^{1}\phi_{x}\psi_{tt} \,dx \\ &\leq-\Lambda\biggl(\frac{d}{dt} \int_{0}^{1}\phi_{xt}\psi\,dx- \frac{d}{dt} \int _{0}^{1}\phi_{x}\psi_{t} \,dx\biggr)\\&\quad+\frac{|\Lambda|}{4} \int_{0}^{1}\psi_{tt}^{2}\,dx \\ &\quad+2|\Lambda| \int_{0}^{1} (\phi_{x}+ \psi)^{2}\,dx+ 2|\Lambda| \int_{0}^{1}\psi _{x}^{2}\,dx. \end{aligned} $$
(4.38)
Let
$$\mathcal{N}(t):= \int_{0}^{1}\phi_{xt}\psi\,dx- \int_{0}^{1}\phi_{x}\psi_{t} \,dx. $$
Then (4.35) becomes
$$ \mathscr{L}'(t)+\Lambda\mathcal{N}'(t) \leq-M_{2}E_{1}(t)+\frac{|\Lambda |}{4} \int_{0}^{1}\psi_{tt}^{2} \,dx $$
(4.39)
for \(M_{2}=M_{1}-\frac{4|\Lambda|}{\gamma}(\frac{1}{K}+\frac{1}{b})>0\). Let
$$ F(t)=\mathscr{L}(t)+\Lambda\mathcal {N}(t)+N_{3}\bigl(E_{1}(t)+E_{2}(t) \bigr)\geq0 $$
(4.40)
if \(N_{3}> \max\{C_{0}|\Lambda|-\gamma_{1},|\Lambda|,\frac{|\Lambda|}{4 C}\} \). Indeed, by using (4.11), (4.23), and \(ab\leq\frac {1}{2}a^{2}+\frac{1}{2}b^{2}\), we obtain
$$\begin{aligned} \big|\mathcal{N}(t)\big|&\leq\bigg| \int_{0}^{1}\phi_{xt}\psi\,dx\bigg|+\bigg| \int_{0}^{1}\phi_{x}\psi _{t} \,dx\bigg| \\ &\leq\frac{1}{2} \int_{0}^{1}\phi_{xt}^{2}\,dx+ \frac{1}{2} \int_{0}^{1}\psi^{2}\,dx + \frac{1}{2} \int_{0}^{1}\phi_{x}^{2}\,dx+ \frac{1}{2} \int_{0}^{1}\psi_{t}^{2}\,dx \\ &\leq\frac{1}{2} \int_{0}^{1}\phi_{xt}^{2}\,dx+ \frac{1}{2} \int_{0}^{1}\psi_{x}^{2}\,dx + \int_{0}^{1}(\phi_{x}+\psi)^{2} \,dx+ \int_{0}^{1}\psi_{x}^{2}\,dx+ \frac{1}{2} \int _{0}^{1}\psi_{t}^{2}\,dx \\ &\leq E_{2}(t)+\max\biggl\{ \frac{3}{b\gamma},\frac{2}{K\gamma}, \frac{1}{\rho _{2}\gamma}\biggr\} E_{1}(t):=E_{2}(t)+C_{0}E_{1}(t), \end{aligned} $$
where \(C_{0}=\max\{\frac{3}{b\gamma},\frac{2}{K\gamma},\frac{1}{\rho _{2}\gamma}\}\). With the help of (4.33), we obtain
$$\begin{aligned} F(t)&\geq\gamma_{1}E_{1}(t)-| \Lambda |\bigl(E_{2}(t)+C_{0}E_{1}(t) \bigr)+N_{3}\bigl(E_{1}(t)+E_{2}(t)\bigr) \\ &\geq\bigl(N_{3}+\gamma_{1}-C_{0}| \Lambda|\bigr)E_{1}(t)+\bigl(N_{3}-|\Lambda|\bigr)E_{2}(t) \geq0. \end{aligned} $$
It is easy to prove that
$$ c_{1}\bigl(E_{1}(t)+E_{2}(t)\bigr)\leq F(t)\leq c_{2}\bigl(E_{1}(t)+E_{2}(t)\bigr) $$
(4.41)
for some positive constants \(c_{1}\) and \(c_{2}\). By using (4.39) and (4.40), we obtain
$$ \begin{aligned} F'(t)&=\mathscr{L}'(t)+ \Lambda\mathcal{N}'(t)+N_{3}\bigl(E_{1}'(t)+E_{2}'(t) \bigr) \\ &\leq-M_{2}E_{1}(t)+\biggl(-CN_{3}+ \frac{|\Lambda|}{4}\biggr) \int_{0}^{1}\psi_{tt}^{2}\,dx. \end{aligned} $$
(4.42)
Thanks to the choice of \(N_{3}\), we have
$$ F'(t)\leq-M_{2}E_{1}(t). $$
(4.43)
Integrating (4.43) over \((0,t)\) yields
$$\int_{0}^{t}E_{1}(r)\,dr\leq \frac{1}{M_{2}}\bigl(F(0)-F(t)\bigr)\leq\frac{1}{M_{2}}F(0)\leq \frac{c_{2}}{M_{2}}\bigl(E_{1}(0)+E_{2}(0)\bigr). $$
Using the fact that
$$\bigl(tE_{1}(t)\bigr)'=tE_{1}'(t)+E_{1}(t) \leq E_{1}(t), $$
we get that
$$tE_{1}(t)\leq\frac{c_{2}}{M_{2}}\bigl(E_{1}(0)+E_{2}(0) \bigr). $$
This completes the proof of Theorem 4.1. □