- Research
- Open Access
Global existence and exponential stability for a nonlinear Timoshenko system with delay
- Baowei Feng1Email author and
- Maurício L Pelicer2
- Received: 18 June 2015
- Accepted: 29 October 2015
- Published: 10 November 2015
Abstract
This paper is concerned with a nonlinear Timoshenko system modeling clamped thin elastic beams with time delay. The delay is defined on a feedback term associated to the equation for rotation angle. Under suitable assumptions on the data, we establish the well-posedness of the problem with respect to weak solutions. We also establish the exponential stability of the system under the usual equal wave speeds assumption.
Keywords
- Timoshenko system
- time delay
- global existence
- exponential stability
MSC
- 35B40
1 Introduction
In this paper, we are concerned with a Timoshenko system with time delay, where \((x,t)\in(0,1)\times\mathbb{R}^{+}\). When \(\mu_{1}=\mu_{2}=f=0\), this system was proposed by Timoshenko [1] as a model for vibrations of a thin elastic beam of length 1. Here, \(\varphi=\varphi(x,t)\) denotes the transverse displacement of the beam, \(\psi=\psi(x,t)\) denotes the rotation angle of the beam’s filament and \(\rho_{1}\), \(\rho_{2}\), k, b are positive constants related to physical properties of the beam. In the system, \(\mu_{1}\psi_{t}\) represents a frictional damping and \(f(\psi)\) is a forcing term. The time delay is given by \(\mu_{2} \psi_{t}(x,t-\tau)\), where \(\mu_{1}\), \(\mu_{2}\), τ are positive constants.
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} \rho_{1} \varphi_{tt} -k(\varphi_{x}+\psi)_{x} = 0, \\ \rho_{2} \psi_{tt} - b \psi_{xx} + k(\varphi_{x}+\psi) + \mu_{1}\psi_{t} + \mu_{2} \psi_{t}(x,t-\tau) + f(\psi) = 0, \end{array}\displaystyle \right . \end{aligned}$$
(1.1)
To the system we add the initial conditions where \(f_{0}\) is prescribed, and the Dirichlet boundary conditions
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} \varphi(x,0)=\varphi_{0}, \qquad \varphi_{t}(x,0)=\varphi_{1},\qquad \psi(x,0)=\psi_{0}, \qquad \psi_{t}(x,0)= \psi_{1}, \\ \psi_{t}(x,t-\tau) = f_{0}(x,t-\tau), \quad t \in(0,\tau), \end{array}\displaystyle \right . \end{aligned}$$
(1.2)
$$\begin{aligned} \varphi(0,t)= \varphi(1,t) = \psi(0,t)= \psi(1,t)=0,\quad \forall t \ge0. \end{aligned}$$
(1.3)
We observe that our problem is set in a context where: (a) the damping is defined only on the equation for rotation angle; (b) the presence of a time delay; (c) exponential stability under a nonlinear forcing. Under this scenario we briefly comment some of related early works.
For partially damped Timoshenko systems, an important result was presented by Soufyane [2, 3]. He showed that the linear system is exponentially stable if and only if This assumption, which means that both waves on the system have equal propagation speed, was later extended to several other problems based on Timoshenko systems. We refer the reader to the references [4–13] among others.
$$\begin{aligned}& \rho_{1} \varphi_{tt} - k(\varphi_{x}+ \psi)_{x} = 0, \\& \rho_{2} \psi_{tt} - b \psi_{xx} + k( \varphi_{x}+\psi) + \psi_{t} = 0 \end{aligned}$$
$$ \frac{\rho_{1}}{\rho_{2}} = \frac{k}{b}. $$
(1.4)
On the other hand, dynamics of delay systems have been a major research subject in differential equations (see, e.g., [14, 15]). It is known that a time delay on the feedback term (internal or at the boundary) in a wave equation can destabilize the system, depending on the weight of each term, as discussed in Datko et al. [16] and Nicaise and Pignotti [17, 18]. Following that context, Said-Houari and Laskri [13] studied the stability of system (1.1) with \(f(\psi)=0\). They proved that, under condition (1.4) and \(\mu_{2} < \mu_{1}\), the system is exponentially stable.
In the present paper our objective is to extend the result of Said-Houari and Laskri [13] to a nonlinear framework by adding a forcing term \(f(\psi)\). The rest of the paper is organized as follows. In Section 2, we present some preliminary remarks and the main results. In Section 3, we prove the well-posedness of system (1.1)-(1.3) by using semigroup theory. In Section 4, we prove the exponential stability of system (1.1)-(1.3) by using energy methods.
2 Preliminaries and main results
In this paper we use standard Lebesgue and Sobolev spaces In the case \(q=2\) we write \(\Vert u \Vert\) instead of \(\Vert u \Vert_{2}\).
$$L^{q}(0,1), \quad 1\le q \le\infty, \mbox{and } H^{1}_{0}(0,1). $$
Now we give some hypotheses on the forcing term \(f(\psi(x,t))\). We assume \(f: \mathbb{R}\rightarrow\mathbb{R}\) satisfying where \(k_{0}>0\), \(\theta>0\). In addition we assume that with \(\hat{f}(z)=\int^{z}_{0}f(s)\,ds\).
$$\begin{aligned} \bigl|f \bigl(\psi^{2} \bigr)-f \bigl(\psi^{1} \bigr)\bigr| \leq k_{0} \bigl(\bigl|\psi^{1}\bigr|^{\theta}+\bigl| \psi^{2}\bigr|^{\theta} \bigr)\bigl|\psi^{1}-\psi^{2}\bigr| \quad\mbox{for all } \psi^{1},\psi^{2}\in\mathbb{R}, \end{aligned}$$
(2.1)
$$\begin{aligned} 0 \leq\hat{f}(\psi)\leq f(\psi)\psi \quad \mbox{for all } \psi\in \mathbb{R}, \end{aligned}$$
(2.2)
In order to deal with the delay feedback term, motivated by [13, 17, 18], we define the following new dependent variable: Then it is easy to verify Thus, equations (1.1) are transformed to with \(x\in(0,1)\), \(\rho\in(0,1)\) and \(t>0\), and the initial and boundary conditions are First of all, we shall show the well-posedness of problem (2.5)-(2.6).
$$\begin{aligned} z(x,\rho,t)=\psi_{t}(x,t-\tau\rho), \quad x\in(0,1),\rho \in(0,1), t>0. \end{aligned}$$
(2.3)
$$\begin{aligned} \tau z_{t}(x,\rho,t)+z_{\rho}(x,\rho,t)=0, \quad \mbox{in } (0,1)\times(0,1)\times(0,\infty). \end{aligned}$$
(2.4)
$$ \left \{ \textstyle\begin{array}{@{}l} \rho_{1}\varphi_{tt}(x,t)-k(\varphi_{x}+\psi)_{x}(x,t)=0,\\ \rho_{2}\psi_{tt}(x,t)-b\psi_{xx}(x,t)+k(\varphi_{x}+\psi)(x,t)\\ \quad{} +\mu_{1}\psi_{t}(x,t)+\mu _{2}z(x,1,t)+f(\psi(x,t))=0,\\ \tau z_{t}(x,\rho,t)+z_{\rho}(x,\rho,t)=0, \end{array}\displaystyle \right . $$
(2.5)
$$ \left \{ \textstyle\begin{array}{@{}ll} \varphi(x,0)=\varphi_{0},\quad \varphi_{t}(x,0)=\varphi_{1},\quad \psi(x,0)=\psi_{0},\quad \psi_{t}(x,0)=\psi_{1},\quad x\in(0,1)\\ z(x,\rho,0)=f_{0}(x,-\rho\tau), \quad (x,t)\in(0,1)\times(0,\tau),\\ \varphi(0,t)=\varphi(1,t)=\psi(0,t)=\psi(1,t)=0,\quad t>0,\\ z(x,0,t)=\psi_{t}(x,t),\quad x\in(0,1), t>0. \end{array}\displaystyle \right . $$
(2.6)
Before using the semigroup theory, we introduce two new dependent variables \(u=\varphi_{t}\) and \(v=\psi_{t}\), then problem (2.5)-(2.6) is reduced to the following problem for an abstract first-order evolutionary equation: where \(U=(\varphi,u,\psi,v,z)^{T}\), and with the domain where We define the energy space \(\mathscr{H}\) by For \(U=(\varphi,u,\psi,v,z)^{T}\), \(\overline{U}=(\overline{\varphi},\overline{u},\overline{\psi },\overline{v},\overline{z})^{T}\) and for ξ a positive constant satisfying we equip \(\mathscr{H}\) with the inner product
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} \frac{dU}{dt}(t)=\mathcal{A}U+F, \quad t>0, \\ U(0)=U_{0}=(\varphi_{0},\varphi_{1},\psi_{0},\psi_{1},f_{0}(\cdot,-\tau))^{T}, \end{array}\displaystyle \right . \end{aligned}$$
(2.7)
$$\mathcal{A}U= \begin{pmatrix} u \\ \frac{k}{\rho_{1}}(\varphi_{xx}+\psi_{x}) \\ v \\ \frac{b}{\rho_{2}}\psi_{xx}-\frac{k}{\rho_{2}}(\varphi_{x}+\psi)-\frac {\mu_{1}}{\rho_{2}}v-\frac{\mu_{2}}{\rho_{2}}z(\cdot,1) \\ -\frac{1}{\tau}z_{\rho} \end{pmatrix}, \qquad F= \begin{pmatrix} 0 \\ 0 \\ 0 \\ -\frac{1}{\rho_{2}}f(\psi) \\ 0 \end{pmatrix}, $$
$$\begin{aligned} D(\mathcal{A})= \bigl\{ (\varphi,u,\psi,v,z)^{T}\in H: v=z(\cdot,0), \mbox{ in } (0,1) \bigr\} , \end{aligned}$$
(2.8)
$$\begin{aligned}[b] H={}& \bigl(H^{2}(0,1)\cap H^{1}_{0}(0,1) \bigr) \times H^{1}_{0}(0,1)\times \bigl(H^{2}(0,1)\cap H^{1}_{0}(0,1) \bigr) \\ &{}\times H^{1}_{0}(0,1)\times L^{2} \bigl(0,1;H^{1}_{0}(0,1) \bigr). \end{aligned} $$
$$\begin{aligned} \mathscr{H}:=H^{1}_{0}(0,1)\times L^{2}(0,1)\times H^{1}_{0}(0,1)\times L^{2}(0,1)\times L^{2} \bigl((0,1)\times(0,1) \bigr). \end{aligned}$$
(2.9)
$$\begin{aligned} \tau\mu_{2}\leq\xi\leq\tau(2\mu_{1}- \mu_{2}), \end{aligned}$$
(2.10)
$$\begin{aligned} \langle U,\overline{U}\rangle_{\mathscr{H}} =& \int^{1}_{0} \bigl[\rho _{1}u\overline{u}+ \rho_{2}v\overline{v}+k(\varphi_{x}+\psi) (\overline { \varphi}_{x}+\overline{\psi})+b\psi_{x} \overline{ \psi}_{x} \bigr]\,dx \\ &{}+\xi \int^{1}_{0} \int^{1}_{0}z(x,\rho)\overline{z}(x,\rho)\,d\rho \,dx. \end{aligned}$$
(2.11)
Now we give the result of the well-posedness of solutions to problem (2.7).
Theorem 2.1
Assume that (2.1)-(2.2) and
\(\mu_{2}\leq\mu_{1}\)
hold, then we have the following results.
- (i)
If \(U_{0}\in\mathscr{H}\), then problem (2.7) has a unique mild solution \(U\in C([0,\infty),\mathscr{H})\) with \(U(0)=U_{0}\).
- (ii)If \(U_{1}\) and \(U_{2}\) are two mild solutions of problem (2.7), then there exists a positive constant \(C_{0}=C(U_{1}(0),U_{2}(0))\) such that$$\begin{aligned} \bigl\| U_{1}(t)-U_{2}(t)\bigr\| _{\mathscr{H}}\leq e^{C_{0}T}\bigl\| U_{1}(0)-U_{2}(0)\bigr\| _{\mathscr{H}} \quad \textit{for any }0\leq t\leq T. \end{aligned}$$(2.12)
- (iii)
If \(U_{0}\in D(\mathcal{A})\), then the above mild solution is a strong solution.
The functional energy of solutions of problem (2.5)-(2.6) is defined by Below we shall give the stability result.
$$\begin{aligned} E(t) =&\frac{1}{2} \int^{1}_{0} \bigl[\rho_{1} \varphi_{t}^{2}+\rho_{2}\psi _{t}^{2}+k| \varphi_{x}+\psi|^{2}+b\psi_{x}^{2} \bigr]\,dx \\ &{}+\frac{\xi}{2} \int^{1}_{0} \int^{1}_{0}z^{2}(x,\rho,t)\,d\rho \,dx+ \int^{1}_{0}\hat{f} \bigl(\psi(t) \bigr)\,dx. \end{aligned}$$
(2.13)
3 The well-posedness
In this section, we shall study the well-posedness of solutions to problem (2.5)-(2.6) to complete the proof of Theorem 2.1.
Lemma 3.1
The energy
\(E(t)\)
defined by (2.13) is a nonincreasing function along the solution trajectories, i.e., there exists a positive constant
C
such that for any
\(t\geq0\),
and there exist two positive constants
\(\delta_{0}\)
and
\(C_{1}\), independent of initial data in
\(\mathscr{H}\), such that for any
\(t\geq0\),
$$\begin{aligned} E'(t)\leq-C \int^{1}_{0}\psi_{t}^{2}(x,t) \,dx-C \int^{1}_{0}z^{2}(x,1,t)\,dx\leq0, \end{aligned}$$
(3.1)
$$\begin{aligned} E(t) \geq&\delta_{0} \biggl( \int^{1}_{0}\varphi_{t}^{2}\,dx+ \int^{1}_{0}\psi _{t}^{2}\,dx+ \int^{1}_{0}|\varphi_{x}+ \psi|^{2}\,dx+ \int^{1}_{0}\psi_{x}^{2}\,dx \\ &{} + \int^{1}_{0} \int^{1}_{0}z^{2}(x,\rho)\,d\rho \,dx \biggr)-C_{1}. \end{aligned}$$
(3.2)
Proof
Multiplying the first equation in (2.5) by \(\varphi_{t}\), the second equation by \(\phi_{t}\), integrating the result over \((0,1)\) with respect to x and using Young’s inequality, we obtain We multiply the third equation in (2.5) by \(\frac{\xi}{\tau}z\) and integrate the result over \((0, 1)\times(0,1)\) with respect to ρ and x, respectively, to get which, together with (3.3), (2.10) and the fact \(\frac{d}{dt}\hat{f}(\psi)=f(\psi)\psi_{t}\), gives us (3.1).
$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \biggl( \int^{1}_{0} \bigl[\rho_{1}\varphi _{t}^{2}+\rho_{2}\psi_{t}^{2}+k| \varphi_{x}+\psi|^{2}+b\psi_{x}^{2} \bigr]\,dx \biggr) \\ &\quad=-\mu_{1} \int^{1}_{0}\psi_{t}\,dx-\mu_{2} \int^{1}_{0}\psi_{t}z(x,1,t)\,dx \\ &\quad\leq \biggl(-\mu_{1}+\frac{\mu_{2}}{2} \biggr) \int^{1}_{0}\psi_{t}\,dx+\frac {\mu_{2}}{2} \int^{1}_{0}z^{2}(x,1,t)\,dx. \end{aligned}$$
(3.3)
$$\begin{aligned} \frac{\xi}{2}\frac{d}{dt} \int^{1}_{0} \int^{1}_{0}z(x,\rho,t)\,d\rho \,dx =&- \frac{\xi}{2\tau} \int^{1}_{0} \int^{1}_{0}\frac{\partial}{\partial \rho}z^{2}(x, \rho,t)\,d\rho \,dx \\ =&\frac{\xi}{2\tau} \int^{1}_{0} \bigl(z^{2}(x,0,t)-z^{2}(x,1,t) \bigr)\,dx, \end{aligned}$$
It is easy to get (3.2) by using (2.2) with \(\delta_{0}=\min\{\frac{1}{2},\frac{\xi}{2}\}\). The proof is therefore complete. □
Lemma 3.2
The operator \(\mathcal{A}\) defined in (2.7) is the infinitesimal generator of a \(C^{0}\)-semigroup in \(\mathscr{H}\).
Proof
It follows from (3.1) that for all \(U(t)\in D(\mathcal{A})\), which implies that the operator \(\mathcal{A}\) is a dissipative operator.
$$\langle\mathcal {A}U,U\rangle_{\mathscr{H}}\leq-C \int^{1}_{0}\psi_{t}^{2}(x,t) \,dx-C \int ^{1}_{0}z^{2}(x,1,t)\,dx\leq0, $$
Next we will prove that the operator \(I-\mathcal{A}\): \(D(\mathcal {A})\rightarrow\mathscr{H}\) is onto, that is, given \(U^{*}=(f_{1},f_{2},f_{3},f_{4},f_{5})^{T}\in\mathscr{H}\), we seek \(U=(\varphi ,u,\psi,v,z)^{T}\in D(\mathcal {A})\) is a solution of \((I-\mathcal{A})U=U^{*}\). We have These equations can be solved following Said-Houari and Laskri [13] or Nicaise and Pignotti [17].
$$\left \{ \textstyle\begin{array}{@{}ll} \varphi-u=f_{1}, \\ u-\frac{k}{\rho_{1}}(\varphi_{x}+\psi)_{x}=f_{2}, \\ \psi-v=f_{3}, \\ v-\frac{b}{\rho_{2}}\psi_{xx}+\frac{b}{\rho_{2}}(\varphi_{x}+\psi )+\frac{\mu_{1}}{\rho_{2}}v+\frac{\mu_{2}}{\rho_{2}}z(\cdot,1)=f_{4}, \\ z+\frac{1}{\tau}z_{\rho}=f_{5}. \end{array}\displaystyle \right . $$
Then we can infer that the operator \(\mathcal{A}\) is m-dissipative in \(\mathscr{H}\). Since \(D(\mathcal{A})\) is dense in \(\mathscr{H}\), thus we can conclude that the operator \(\mathcal{A}\) is the infinitesimal generator of a \(C^{0}\)-semigroup in \(\mathscr{H}\) by the Lumer-Phillips theorem (see, for example, Pazy [19]). The proof is now complete. □
Lemma 3.3
The operator F defined in (2.7) is locally Lipschitz in \(\mathscr{H}\).
Proof
Let \(U_{1}=(\varphi^{1},u^{1},\psi^{1},v^{1},z^{1})\) and \(U_{1}=(\varphi^{2},u^{2},\psi^{2},v^{2},z^{2})\), then we have By using (2.1), Hölder’s and Poincaré’s inequalities, we can obtain which gives us Then the operator F is locally Lipschitz in \(\mathscr{H}\). The proof is hence complete. □
$$\bigl\| F(U_{1})-F(U_{2})\bigr\| _{\mathscr{H}}\leq \bigl\| f \bigl( \psi^{1} \bigr)-f \bigl(\psi^{2} \bigr)\bigr\| _{L^{2}}. $$
$$\begin{aligned} \bigl\| f \bigl(\psi^{1} \bigr)-f \bigl(\psi^{2} \bigr) \bigr\| _{L^{2}} \leq& \bigl(\bigl\| \psi^{1}\bigr\| ^{\theta}_{2\theta }+ \bigl\| \psi^{2}\bigr\| ^{\theta}_{2\theta} \bigr)\bigl\| \psi^{1}- \psi^{2}\bigr\| \\ \leq&C_{1}\bigl\| \psi^{1}_{x}-\psi^{2}_{x} \bigr\| , \end{aligned}$$
$$\bigl\| F(U_{1})-F(U_{2})\bigr\| _{\mathscr{H}}\leq C_{1} \|U_{1}-U_{2}\|_{\mathscr{H}}. $$
Proof of Theorem 2.1
It follows from Lemmas 3.2-3.3 that the Cauchy problem has a unique local mild solution defined in a maximal interval \((0,t_{\max})\).
$$\begin{aligned} U(t)=e^{\mathcal {A}t}U_{0}+ \int^{t}_{0}e^{\mathcal{A}(t-s)}F \bigl(U(s) \bigr)\,ds \end{aligned}$$
(3.4)
If \(t_{\max}<\infty\), then Let \(U(t)\) be a mild solution with \(U_{0}\in D(\mathcal{A})\). By using Theorem 6.1.5 in Pazy [19], we conclude that it is a strong solution. It follows from (3.2) that for all \(t\geq0\), which, by density, holds for mild solutions. Then it is a contradiction with (3.5) and therefore \(t_{\max}=\infty\), that is, the solution is global. The proof of (i) of Theorem 2.1 is complete.
$$\begin{aligned} \lim_{t\rightarrow\infty}\bigl\| U(t)\bigr\| _{\mathscr{H}}=+\infty. \end{aligned}$$
(3.5)
$$\bigl\| U(t)\bigr\| ^{2}_{\mathscr{H}}\leq\frac{1}{\delta_{0}} \bigl(E(0)+C_{1} \bigr), $$
It is easy to get inequality (2.12) by using (3.4), the local Lipschitz behavior of F and Gronwall’s inequality. Then we can obtain the continuous dependence on the initial data for mild solutions. This proves the item (ii) of Theorem 2.1.
By using Theorem 6.1.5 in Pazy [19] (see also [20]), we know that any mild solutions with initial data in \(D(\mathcal {A})\) are strong. Then the proof of Theorem 2.1 is therefore complete. □
4 Exponential stability
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
satisfies that for any
\(\varepsilon>0\),
hereafter
\(\lambda_{1}>0\)
is the first eigenvalue of −Δ in
\(H^{1}_{0}(0,1)\).
$$\begin{aligned} I_{1}=- \int^{1}_{0}(\rho_{1}\varphi \varphi_{t}+\rho_{2}\psi\psi_{t})\,dx- \frac {\mu_{1}}{2} \int^{1}_{0}\psi^{2}\,dx \end{aligned}$$
(4.1)
$$\begin{aligned} \frac{d}{dt}I_{1}(t) \leq&- \int^{1}_{0} \bigl(\rho_{1} \varphi_{t}^{2}+\rho_{2}\psi _{t}^{2} \bigr)\,dx+k \int^{1}_{0}|\varphi_{x}+ \psi|^{2}\,dx+ (b+C_{1}+\varepsilon) \int^{1}_{0}\psi_{x}^{2}\,dx \\ &{}+\frac{\mu_{2}^{2}}{4\lambda_{1}\varepsilon} \int^{1}_{0}z^{2}(x,1,t)\,dx, \end{aligned}$$
(4.2)
Proof
A straightforward calculation gives Using (2.5) and integrating by parts, we see that It follows from Young’s inequality and Poincaré’s inequality that for any \(\varepsilon>0\),
which, together with (4.3)-(4.4), gives us (4.2). The proof is now complete. □
$$\frac{dI_{1}}{dt}=- \int^{1}_{0}(\rho_{1}\varphi_{tt} \varphi+\rho_{2}\psi _{tt}\psi)\,dx- \int^{1}_{0} \bigl(\rho_{1} \varphi_{t}^{2}+\rho_{2}\psi_{t}^{2} \bigr)\,dx-\mu _{1} \int^{1}_{0}\psi\psi_{t}\,dx. $$
$$\begin{aligned} \frac{dI_{1}}{dt} =&-\rho_{1} \int^{1}_{0}\varphi_{t}^{2}\,dx- \rho_{2} \int^{1}_{0}\psi _{t}^{2}\,dx+k \int^{1}_{0}|\varphi_{x}+ \psi|^{2}\,dx+b \int^{1}_{0}\psi_{x}^{2}\,dx \\ &{}+\mu_{2} \int^{1}_{0}z(x,1,t)\psi \,dx+ \int^{1}_{0}f(\psi)\psi \,dx. \end{aligned}$$
(4.3)
$$\begin{aligned}& \begin{aligned}[b] \int^{1}_{0}\bigl|z(x,1,t)\psi\bigr|\,dx&\leq\varepsilon \lambda_{1} \int^{1}_{0}\psi^{2}\,dx+\frac {1}{4\varepsilon\lambda_{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx \\ &\leq\varepsilon \int^{1}_{0}\psi_{x}^{2}\,dx+ \frac{1}{4\varepsilon\lambda _{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx, \end{aligned} \end{aligned}$$
(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)
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
where
g
is the solution of
Then the functional
\(I_{2}\)
satisfies, for any
\(\eta,\tilde{\eta}>0\),
$$\begin{aligned} I_{2}(t)= \int^{1}_{0}(\rho_{2}\psi_{t} \psi+\rho_{1}\varphi_{t}g)\,dx+\frac{\mu _{1}}{2} \int^{1}_{0}\psi^{2}\,dx, \end{aligned}$$
(4.6)
$$\begin{aligned} -g_{xx}=\psi_{x}, \qquad g|_{x=0,1}=0. \end{aligned}$$
(4.7)
$$\begin{aligned} \frac{d}{dt}I_{2}(t) \leq& (\mu_{2} \eta-b ) \int^{1}_{0}\psi _{x}^{2}\,dx+ \biggl(\rho_{2}+\frac{\rho_{2}}{4\tilde{\eta}} \biggr) \int ^{1}_{0}\psi_{t}^{2}\,dx + \frac{\rho_{1}}{\lambda_{1}}\tilde{\eta} \int^{1}_{0}\varphi _{t}^{2}\,dx \\ &{}+\frac{\mu_{2}}{4\eta\lambda_{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx- \int^{1}_{0}\hat {f}(\psi)\,dx. \end{aligned}$$
(4.8)
Proof
We know from (2.5) that By (4.7), we can get Using Young’s inequality and Poincaré’s inequality, we have
$$\begin{aligned} \frac{d}{dt}I_{2}(t) =&-b \int^{1}_{0}\psi_{x}^{2}\,dx+ \rho_{2} \int^{1}_{0}\psi _{t}^{2}\,dx-k \int^{1}_{0}\psi^{2}\,dx+k \int^{1}_{0}g_{x}^{2}\,dx \\ &{}+\rho_{1} \int^{1}_{0}\varphi_{t}g_{t}\,dx- \mu_{2} \int^{1}_{0}\psi z(x,1,t)\,dx- \int^{1}_{0}f(\psi)\psi \,dx. \end{aligned}$$
(4.9)
$$ \left \{ \textstyle\begin{array}{@{}l} \int^{1}_{0}g_{x}^{2}\,dx\leq \int^{1}_{0}\psi^{2}\,dx\leq\int^{1}_{0}\psi_{x}^{2}\,dx, \\ \int^{1}_{0}g_{t}^{2}\,dx\leq \int^{1}_{0}g_{xt}^{2}\,dx\leq\int^{1}_{0}\psi_{t}^{2}\,dx. \end{array}\displaystyle \right . $$
(4.10)
$$\begin{aligned}& \begin{aligned}[b] \mu_{2} \int^{1}_{0}\bigl|\psi z(x,1,t)\bigr|\,dx&\leq\mu_{2} \eta\lambda_{1} \int^{1}_{0}\psi^{2}\,dx+\frac{\mu _{2}}{4\eta\lambda_{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx \\ &\leq\mu_{2}\eta \int^{1}_{0}\psi_{x}^{2}\,dx+ \frac{\mu_{2}}{4\eta\lambda _{1}} \int^{1}_{0}z^{2}(x,1,t)\,dx, \end{aligned} \end{aligned}$$
(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: Then we may get the following lemma.
$$\begin{aligned} J(t):=\rho_{2} \int^{1}_{0}\psi_{t}(\varphi_{x}+ \psi)\,dx+\rho_{2} \int^{1}_{0}\psi _{x}\varphi_{t} \,dx. \end{aligned}$$
(4.13)
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 Using (2.5), (1.4) and integration by parts, we get By using Young’s inequality and Poincaré’s inequality, we know that for any \(\varepsilon>0\),
and which, together with (4.15)-(4.16), gives us (4.14). The proof is now complete. □
$$\begin{aligned} \frac{d}{dt}J(t) =&\rho_{2} \int^{1}_{0}\psi_{tt}(\varphi_{x}+ \phi)\,dx+\rho _{2} \int^{1}_{0}\psi_{t}(\varphi_{x}+ \psi)_{t}\,dx \\ &{}+\rho_{2} \int^{1}_{0}\psi_{xt}\varphi_{t} \,dx +\rho_{2} \int^{1}_{0}\psi_{x}\varphi_{tt} \,dx. \end{aligned}$$
$$\begin{aligned} \frac{d}{dt}J(t) =&b[\psi_{x}\varphi_{x}]^{x=1}_{x=0}+ \rho_{2} \int ^{1}_{0}\psi_{t}^{2}\,dx-k \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx \\ &{}-\mu_{1} \int^{1}_{0}\psi_{t}(\varphi_{x}+ \psi)\,dx-\mu_{2} \int^{1}_{0}(\varphi _{x}+\psi)z(x,1,t) \,dx \\ &{}- \int^{1}_{0}\varphi_{x}f(\psi)\,dx- \int^{1}_{0}f(\psi)\psi \,dx. \end{aligned}$$
$$\begin{aligned}& {-}\mu_{1} \int^{1}_{0}\psi_{t}(\varphi_{x}+ \psi)\,dx \leq \frac{k}{4} \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+\frac{\mu_{1}^{2}}{k} \int ^{1}_{0}\psi_{t}^{2} \,dx, \end{aligned}$$
(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)
$$\begin{aligned} \int^{1}_{0}\bigl|\varphi_{x}f(\psi)\bigr|\,dx \leq& \| \varphi_{x}\|\|\psi\|^{\theta}_{2(\theta+1)}\|\psi \|_{2(\theta+1)} \\ \leq&\frac{\varepsilon}{2b^{2}} \int^{1}_{0}\varphi_{x}^{2}\,dx+ \frac {b^{2}}{2\varepsilon\lambda_{1}} \int^{1}_{0}\psi_{x}^{2}\,dx \\ \leq&\frac{\varepsilon}{b^{2}} \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+\frac {\varepsilon}{b^{2}} \int^{1}_{0}\psi^{2}\,dx+\frac{b^{2}}{2\varepsilon\lambda _{1}} \int^{1}_{0}\psi_{x}^{2}\,dx \\ \leq&\frac{\varepsilon}{b^{2}} \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+ \biggl(\frac{\varepsilon}{b^{2}\lambda_{1}}+\frac{b^{2}}{2\varepsilon\lambda _{1}} \biggr) \int^{1}_{0}\psi_{x}^{2}\,dx, \end{aligned}$$
(4.17)
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 \(\varepsilon>0\), By using (2.5), Young’s inequality, integration by parts and the following fact we see that Similarly, which, along with (4.20)-(4.21), gives us (4.19). The proof is now complete. □
$$\begin{aligned} b[\psi_{x}\varphi_{x}]^{x=1}_{x=0} \leq \varepsilon \bigl[\varphi_{x}^{2}(1)+ \varphi_{x}^{2}(0) \bigr]+\frac {b^{2}}{4\varepsilon} \bigl[ \psi_{x}^{2}(1)+\psi_{x}^{2}(0) \bigr]. \end{aligned}$$
(4.20)
$$\frac{d}{dt} \int^{1}_{0}b\rho_{2}q\psi_{t} \psi_{x}\,dx= \int^{1}_{0}b\rho_{2}\psi _{tt} \psi_{x}\,dx+ \int^{1}_{0}b\rho_{2}q\psi_{t} \psi_{xt}\,dx, $$
$$\begin{aligned} \frac{d}{dt} \int^{1}_{0}b\rho_{2}q\psi_{t} \psi_{x}\,dx \leq& -b^{2} \bigl[\psi_{x}^{2}(1)+ \psi_{x}^{2}(0) \bigr]+2b^{2} \int^{1}_{0}\psi _{x}^{2}\,dx+2 \rho_{2}b \int^{1}_{0}\psi_{t}^{2}\,dx \\ &{}+ \biggl(\frac{\varepsilon}{b^{2}}+\varepsilon^{2}k^{2} \biggr) \int ^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+2\varepsilon b^{2} \int^{1}_{0}\psi_{x}^{2}\,dx \\ &{}+ \biggl(b+\frac{b^{2}}{\varepsilon^{2}} +\frac{\varepsilon}{b^{2}\lambda_{1}} +\frac{b^{2}}{2\varepsilon\lambda_{1}} \biggr) \int^{1}_{0}\psi _{x}^{2}\,dx \\ &{}+\frac{\mu_{1}^{2}}{\varepsilon} \int^{1}_{0}\psi_{t}^{2}\,dx+ \frac{\mu _{2}^{2}}{\varepsilon} \int^{1}_{0}z^{2}(x,1,t)\,dx. \end{aligned}$$
(4.21)
$$\begin{aligned} \frac{d}{dt} \int^{1}_{0}\rho_{1}q\varphi_{t} \varphi_{x}\,dx\leq-k \bigl[\varphi_{x}^{2}(1)+ \varphi_{x}^{2}(0) \bigr]+3k \int^{1}_{0}\varphi_{x}^{2} \,dx+k \int ^{1}_{0}\psi_{x}^{2}\,dx +2\rho_{1} \int^{1}_{0}\varphi_{t}^{2}\,dx, \end{aligned}$$
In order to handle the term \(z(x,\rho,t)\), we introduce the functional Then we can find the following result in [13].
$$\begin{aligned} I_{3}(t):= \int^{1}_{0} \int^{1}_{0}e^{-2\tau\rho}z^{2}(x, \rho,t)\,d\rho \,dx. \end{aligned}$$
(4.22)
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:
where
c
is a positive constant.
$$\begin{aligned} \frac{d}{dt}I_{3}(t)\leq-I_{3}(t)- \frac{c}{2\tau} \int ^{1}_{0}z^{2}(x,1,t)\,dx+ \frac{1}{2\tau} \int^{1}_{0}\psi_{t}^{2}\,dx, \end{aligned}$$
(4.23)
Now we define the following Lyapunov functional \(\mathscr{L}(t)\) by Then we may obtain the following lemma.
$$\begin{aligned} \mathscr{L}(t) :=&ME(t)+\frac{1}{8}I_{1}(t)+NI_{2}(t)+J(t)+ \frac {\varepsilon}{k} \int^{1}_{0}\rho_{1}q\varphi_{t} \varphi_{x}\,dx \\ &{}+\frac{\rho_{2}b}{4\varepsilon} \int^{1}_{0}q\psi_{t}\psi _{x} \,dx+I_{3}(t). \end{aligned}$$
(4.24)
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 where the positive constants \(\alpha_{i}\) (\(i=1,2,3,4\)) are determined as in [13].
$$\begin{aligned} \bigl|\mathscr{L}(t)-ME(t)\bigr| \leq&\alpha_{1} \int^{1}_{0}\varphi_{t}^{2}\,dx+ \alpha _{2} \int^{1}_{0}\psi_{t}^{2}\,dx+ \alpha_{3} \int^{1}_{0}(\varphi_{x}+\psi )^{2}\,dx \\ &{}+\alpha_{4} \int^{1}_{0}\psi_{x}^{2}\,dx+ \int^{1}_{0} \int^{1}_{0}z^{2}(x,\rho,t)\,d\rho \,dt \\ &{}+ \int^{1}_{0}\hat{f}(\psi)\,dx, \end{aligned}$$
(4.26)
Performing Young’s inequality and using the fact we easily get It follows from (4.26)-(4.27) that there exists a positive constant C̃ such that Then choosing M so large that \(\gamma_{1}:=M-\tilde{C}>0\) and \(\gamma_{2}=M+\tilde{C}>0\), we complete the proof. □
$$\int^{1}_{0}\varphi_{x}^{2}\,dx \leq2 \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx+2 \int ^{1}_{0}\psi^{2}\,dx, $$
$$\begin{aligned} E(t) \geq& \frac{1}{4}\min\{1,\xi\} \biggl( \int^{1}_{0}\varphi_{t}^{2}\,dx+ \int^{1}_{0}\psi _{t}^{2}\,dx+ \int^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx \\ &{}+ \int^{1}_{0}\psi_{x}^{2}\,dx+ \int^{1}_{0} \int^{1}_{0}z^{2}(x,\rho,t)\,d\rho \,dt+ \int^{1}_{0}\hat{f}(\psi)\,dx \biggr), \end{aligned}$$
(4.27)
$$\begin{aligned} \bigl|\mathscr{L}(t)-ME(t)\bigr| \leq&\tilde{C}E(t). \end{aligned}$$
(4.28)
Proof of Theorem 2.2
It follows from (3.1), (4.2), (4.8), (4.14), (4.19), (4.23) and (4.24) that First we choose η so small that and then we choose ε small enough so that Then we take N so large that After that, we select η̃ small enough so that Then we take M so large that there exists a positive constant δ such that Noting that (2.13), we know that there exists a positive constant β such that which, together with (4.25), yields Then we can get Using again (4.25), we find that 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. □
$$\begin{aligned} \frac{d}{dt}\mathscr{L}(t) \leq{}& \biggl[-MC-\frac{\rho_{2}}{8}+N \biggl(\rho_{2}+ \frac{\rho_{1}}{4\tilde {\eta}} \biggr)+\frac{\rho_{2}b}{2\varepsilon}+\frac{\mu _{1}^{2}}{4\varepsilon^{2}} + \rho_{2}+\frac{\mu_{1}^{2}}{k}+\frac{1}{2\tau} \biggr] \int^{1}_{0}\psi _{t}^{2}\,dx \\ &{} + \biggl[-MC+\frac{\mu_{2}^{2}}{32\lambda_{1}\varepsilon}+\frac {N\mu_{2}}{4\eta\lambda_{1}}+\frac{\mu_{2}^{2}}{k}+ \frac{\mu _{2}^{2}}{4\varepsilon^{2}}-\frac{c}{2\tau} \biggr] \int^{1}_{0}z^{2}(x,1,t)\,dx \\ &{} + \biggl[-\frac{\rho_{1}}{8}+N\tilde{\eta}\frac{\rho _{1}}{\lambda_{1}}+ \frac{2\rho_{1}\varepsilon}{k} \biggr] \int^{1}_{0}\varphi_{t}^{2}\,dx + \biggl[-\frac{3k}{8}+\frac{k^{2}\varepsilon}{4} \biggr] \int ^{1}_{0}(\varphi_{x}+ \psi)^{2}\,dx \\ &{} + \biggl[\frac{b+C_{1}}{8}+\varepsilon-N (b-\mu_{2}\eta )+ \frac{b^{2}}{2\varepsilon} +\frac{b^{2}}{4\varepsilon^{3}}+\frac{b^{2}}{2}+\frac{2\varepsilon }{b^{2}\lambda_{1}} + \frac{b^{2}}{\varepsilon\lambda_{1}} \biggr] \int^{1}_{0}\psi _{x}^{2}\,dx \\ &{} +(-N-1) \int^{1}_{0}\hat{f}(\psi)\,dx. \end{aligned}$$
(4.29)
$$\eta\leq\frac{b}{2C\mu_{2}}, $$
$$\varepsilon\leq \frac{5k}{128+4k^{2}}. $$
$$\frac{Nb}{4}\geq \frac{b+C_{1}}{8}+\varepsilon+\frac{b^{2}}{2\varepsilon} + \frac{b^{2}}{4\varepsilon^{3}}+\frac{b^{2}}{2}+\frac{2\varepsilon }{b^{2}\lambda_{1}} +\frac{b^{2}}{\varepsilon\lambda_{1}}. $$
$$\tilde{\eta}\leq \frac{1}{32N}. $$
$$\begin{aligned} \frac{d}{dt}\mathscr{L}(t) \leq& -\delta \int^{1}_{0} \bigl(\varphi_{t}^{2}+ \psi_{t}^{2}+(\varphi_{x}+\psi)^{2}+\psi _{x}^{2}+z^{2}(x,1,t) \bigr)\,dx \\ &{}-\delta \int^{1}_{0} \int^{1}_{0}z^{2}(x,\rho,t)\,d\rho \,dx- \delta \int^{1}_{0}\hat{f}(\psi)\,dx. \end{aligned}$$
(4.30)
$$\begin{aligned} \frac{d}{dt}\mathscr{L}(t) \leq&-\beta E(t), \end{aligned}$$
(4.31)
$$\frac{d}{dt}\mathscr{L}(t)\leq -\frac{\beta}{\gamma_{2}} \mathscr{L}(t). $$
$$\begin{aligned} \mathscr{L}(t)\leq \mathscr{L}(0)e^{-\frac{\beta}{\gamma_{2}}t}. \end{aligned}$$
(4.32)
$$E(t)\leq \frac{\gamma_{2}}{\gamma_{1}}E(0)e^{-\frac{\beta}{\gamma _{2}}t}, $$
Declarations
Acknowledgements
This work was supported by the Fundamental Research Funds for the Central Universities with contract No. JBK150128.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
(1)
Faculty of Economic Mathematics, Southwestern University of Finance and Economics, Chengdu, 611130, P.R. China
(2)
Departamento de Ciências, Campus Regional de Goioerê, Universidade Estadual de Maringá, Goioerê, PR, 87360-000, Brazil
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