Exponential Energy Decay of Solutions for a Transmission Problem With Viscoelastic Term and Delay

Abstract

In our previous work (Journal of Nonlinear Science and Applications 9: 1202–1215, 2016), we studied the well-posedness and general decay rate for a transmission problem in a bounded domain with a viscoelastic term and a delay term. In this paper, we continue to study the similar problem but without the frictional damping term. The main difficulty arises since we have no frictional damping term to control the delay term in the estimate of the energy decay. By introducing suitable energy and Lyapunov functionals, we establish an exponential decay result for the energy.

1. Introduction

In our previous work [1], we considered the following transmission system with a viscoelastic term and a delay term:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle u_{tt}(x,t)-a{u}_{xx}(x,t)+{\int }_0^{t}g(t-s)u_{xx}(x,s)\mathrm{d}s}\hfill \\ {\displaystyle +{\mu }_1{u}_t(x,t)+{\mu }_2{u}_t(x,t-\tau )=0,}\hfill & (x,t)\in \Omega \times (0,+\infty )\hfill \\ {\displaystyle v_{tt}(x,t)-b{v}_{xx}(x,t)=0,}\hfill & (x,t)\in (L_1,L_2)\times (0,+\infty )\hfill \end{array}\right.\end{array}$$

(1)

where $0<L_1<L_2<L_3$, $\Omega =(0,L_1)\cup (L_2,L_3)$, a, b, ${\mu }_1$, ${\mu }_2$ are positive constants, and $\tau >0$ is the delay. In that work, we first proved the well-posedness by using the Faedo–Galerkin approximations together with some energy estimates when ${\mu }_2\le {\mu }_1$. Then, a general decay rate result was established under the hypothesis that ${\mu }_2<{\mu }_1$. As for the previous results and developments of transmission problems, and the research of wave equations with viscoelastic damping or time delay effects, we have stated and summarized in great detail in our previous work [1], thus we just omit it here. The readers, for a better understanding of present work, are strongly recommended to [1] and the reference therein (see [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33]).

It is worth pointing out that, in our previous work, the assumption “${\mu }_2<{\mu }_1$" plays an important role in the proof of the above-mentioned general decay result. In this paper, we intend to investigate system (1) with ${\mu }_1=0$. That is, we study the exponential decay rate of the solutions for the following transmission system with a viscoelastic term and a delay term but without the frictional damping:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle u_{tt}(x,t)-a{u}_{xx}(x,t)+{\int }_0^{t}g(t-s)u_{xx}(x,s)\mathrm{d}s}\hfill \\ {\displaystyle +{\mu }_2{u}_t(x,t-\tau )=0,}\hfill & (x,t)\in \Omega \times (0,+\infty )\hfill \\ {\displaystyle v_{tt}(x,t)-b{v}_{xx}(x,t)=0,}\hfill & (x,t)\in (L_1,L_2)\times (0,+\infty )\hfill \end{array}\right.\end{array}$$

(2)

under the boundary and transmission conditions

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle u(0,t)=u(L_3,t)=0,}\hfill \\ {\displaystyle u(L_i,t)=v(L_i,t),}\hfill & i=1,2\hfill \\ {\displaystyle \left(a-{\int }_0^{t}g\left(s\right)\mathrm{d}s\right)u_x(L_i,t)=b{v}_x(L_i,t),}\hfill & i=1,2\hfill \end{array}\right.\end{array}$$

(3)

and the initial conditions

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{\displaystyle u(x,0)=u_0\left(x\right),u_t(x,0)=u_1\left(x\right),}\hfill & x\in \Omega \hfill \\ {\displaystyle u_t(x,t-\tau )=f_0(x,t-\tau ),}\hfill & x\in \Omega ,t\in [0,\tau ]\hfill \\ {\displaystyle v(x,0)=v_0\left(x\right),v_t(x,0)=v_1\left(x\right),}\hfill & x\in (L_1,L_2)\hfill \end{array}\right.\end{array}$$

(4)

where ${\mu }_2$ is a real number, a, b are positive constants and $u_1\left(x\right)=f_0(x,0)$.

The main difficulty in dealing with this problem is that in the first equation of system (2), we have no frictional damping term to control the delay term in the estimate of the energy decay. To overcome this difficulty, our basic idea is to control the delay term by making use of the viscoelastic term. In order to achieve this goal, a restriction of the size between the parameter ${\mu }_2$ and the relaxation function g and a suitable energy is needed. This is motivated by Dai and Yang’s work [34], in which the viscoelastic wave equation with delay term but without a frictional damping term was studied and an exponential decay result was established. In the work here, we will establish an exponential decay rate result for the energy.

The remaining part of this paper is organized as follows. In Section 2, we give some notations and hypotheses needed for our work and state the main results. In Section 3, under some restrictions of ${\mu }_2$ (see (35) below), we prove the exponential decay of the solutions for the relaxation function satisfying assumption $\left(H_1\right)$ and $\left(H_2\right)$.

2. Preliminaries and Main Results

In this section, we present some materials that shall be used in order to prove our main result. Let us first introduce the following notations:

$$\begin{array}{cc}& (g*h)\left(t\right):={\int }_0^{t}g(t-s)h\left(s\right)\mathrm{d}s\hfill \\ & (g\diamond h)\left(t\right):={\int }_0^{t}g(t-s)(h\left(t\right)-h\left(s\right))\mathrm{d}s\hfill \\ & (g\square h)\left(t\right):={\int }_0^{t}g(t-s){|h\left(t\right)-h\left(s\right)|}^2\mathrm{d}s\hfill \end{array}$$

We easily see that the above operators satisfy

$$\begin{array}{cc}& (g*h)\left(t\right)=\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right)h\left(t\right)-(g\diamond h)\left(t\right)\hfill \\ & {|(g\diamond h)\left(t\right)|}^2\le \left({\int }_0^t|g\left(s\right)|\mathrm{d}s\right)(|g|\square h)\left(t\right)\hfill \end{array}$$

Lemma 1. 

For any $g,h\in C^1\left(\mathbb{R}\right)$, the following equation holds

$$2[g*h]h^{\prime }=g^{\prime }\square h-g\left(t\right){|h|}^2-\frac{d}{dt}\left\{g\square h-\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right){|h|}^2\right\}$$

For the relaxation function g, we assume

(H1) g: ${\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ is a $C^1$ function satisfying

$$g\left(0\right)>0,{\beta }_0:=a-{\int }_0^{\infty }g\left(s\right)\mathrm{d}s=a-\overline{g}>0$$

(H2) There exists a positive constant ξ satisfying $\xi >{\xi }_0>0$ (${\xi }_0$ defined by (39) below) and

$$g^{\prime }\left(t\right)\le -\xi g\left(t\right),\forall t\ge 0$$

According to previous results in the literature (see [1]), we state the following well-posedness result, which can be proved by using the Faedo–Galerkin method.

Theorem 2. 

Assume that $\left(H1\right)$ and $\left(H2\right)$ hold. Then, given $(u_0,v_0)\in \mathcal{V}$, $(u_1,v_1)\in {\mathcal{L}}^2$, and $f_0\in L^2((0,1),H^1(\Omega ))$, Equations (2)–(4) have a unique weak solution in the following class:

$$\begin{array}{c}\hfill (u,v)\in C(0,\infty ;\mathcal{V})\cap C^1(0,\infty ;{\mathcal{L}}^2)\end{array}$$

where

$$\begin{array}{cc}\hfill \mathcal{V}=\{& (u,v)\in H^1(\Omega )\cap H^1(L_1,L_2):u(0,t)=u(L_3,0)=0,u(L_i,t)=v(L_i,t)\hfill \\ & \left(a-{\int }_0^{t}g\left(s\right)\mathrm{d}s\right)u_x(L_i,t)=b{v}_x(L_i,t),i=1,2\}\hfill \end{array}$$

and

$${\mathcal{L}}^2=L^2(\Omega )\times L^2(L_1,L_2)$$

To state our decay result, we introduce the following energy functional:

$$\begin{array}{cc}\hfill E\left(t\right)=& \frac{1}{2}{\int }_{\Omega }{u}_t^2(x,t)\mathrm{d}x+\frac{1}{2}\left(a-{\int }_0^{t}g\left(s\right)\mathrm{d}s\right){\int }_{\Omega }{u}_x^2(x,t)\mathrm{d}x+\frac{1}{2}{\int }_{\Omega }(g\square u_x)\mathrm{d}x\hfill \\ & +\frac{1}{2}{\int }_{L_1}^{L_2}\left[v_t^2(x,t)+b{v}_x^2(x,t)\right]\mathrm{d}x+\frac{\zeta }{2}{\int }_{t-\tau }^t{\int }_{\Omega }{e}^{\sigma (s-t)}{u}_s^2((x,s)\mathrm{d}x\mathrm{d}s\hfill \end{array}$$

(5)

where σ and ζ are positive constants to be determined later.

Remark 1. 

We note that the energy functional defined here is different from that of [1] in the construction of the last term. This is motivated by the idea of [28], in which wave equations with time dependent delay was studied.

Our decay results read as follows:

Theorem 3. 

Let $(u,v)$ be the solution of Equations (2)–(4). Assume that $\left(H1\right)$, $\left(H2\right)$ and

$$a>\frac{8(L_2-L_1)}{L_1+L_3-L_2}{\beta }_0,b>\frac{8(L_2-L_1)}{L_1+L_3-L_2}{\beta }_0$$

(6)

hold. Let $a_0$ be the constants defined by (35) below. If $|{\mu }_2|<a_0$, then there exists constants ${\gamma }_1,{\gamma }_2>0$ such that

$$E\left(t\right)\le {\gamma }_2{e}^{-{\gamma }_1\xi (t-t_0)},t\ge t_0$$

(7)

3. Proof of Theorem 3

For the proof of Theorem 3, we use the following lemmas.

Lemma 4. 

Let $(u,v)$ be the solution of Equations (2)–(4). Then, we have the inequality

$$\begin{array}{cc}\hfill \frac{d}{dt}E\left(t\right)\le & \frac{1}{2}{\int }_{\Omega }(g^{\prime }\square u_x)\left(t\right)\mathrm{d}x+\left(\frac{|{\mu }_2|}{2}+\frac{\zeta }{2}\right){\int }_{\Omega }{u}_t^2(x,t)\mathrm{d}x-\left(\frac{\zeta }{2}{e}^{-\sigma \tau }-\frac{|{\mu }_2|}{2}\right){\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x\hfill \\ & -\frac{1}{2}g\left(t\right){\int }_{\Omega }{u}_x^2(x,t)\mathrm{d}x-\frac{\sigma \zeta }{2}{\int }_{t-\tau }^t{\int }_{\Omega }{e}^{-\sigma (t-s)}{u}_s^2((x,s)\mathrm{d}x\mathrm{d}s\hfill \end{array}$$

(8)

Proof. 

Differentiating (5) and using (2), we have

$$\begin{array}{cc}\hfill \frac{d}{dt}E\left(t\right)=& {\int }_{\Omega }\left[u_t{u}_{tt}+\left(a-{\int }_0^{t}g\left(s\right)\mathrm{d}s\right)u_x{u}_{xt}-\frac{1}{2}g\left(t\right)u_x^2\right]\mathrm{d}x+{\int }_{L_1}^{L_2}\left[v_t{v}_{tt}+b{v}_x{v}_{xt}\right]\mathrm{d}x\hfill \\ & +{\int }_0^{t}g(t-s){\int }_{\Omega }{u}_{xt}(u_x\left(t\right)-u_x\left(s\right))\mathrm{d}x\mathrm{d}s+\frac{1}{2}{\int }_{\Omega }{g}^{\prime }\square u_x\mathrm{d}x+\frac{\zeta }{2}{\int }_{\Omega }{u}_t^2(x,t)\mathrm{d}x\hfill \\ & -\frac{\zeta }{2}{\int }_{\Omega }{e}^{-\sigma \tau }{u}_t^2(x,t-\tau )\mathrm{d}x-\frac{\sigma \zeta }{2}{\int }_{t-\tau }^t{\int }_{\Omega }{e}^{-\sigma (t-s)}{u}_s^2(x,s)\mathrm{d}x\mathrm{d}s\hfill \\ \hfill =& \frac{1}{2}{\int }_{\Omega }{g}^{\prime }\square u_x\mathrm{d}x-\frac{1}{2}g\left(t\right){\int }_{\Omega }{u}_x^2\mathrm{d}x-|{\mu }_2|{\int }_{\Omega }{u}_t\left(t\right)u_t(t-\tau )\mathrm{d}x+\frac{\zeta }{2}{\int }_{\Omega }{u}_t^2(x,t)\mathrm{d}x\hfill \\ & -\frac{\zeta }{2}{\int }_{\Omega }{e}^{-\sigma \tau }{u}_t^2(x,t-\tau )\mathrm{d}x-\frac{\sigma \zeta }{2}{\int }_{t-\tau }^t{\int }_{\Omega }{e}^{-\sigma (t-s)}{u}_s^2(x,s)\mathrm{d}x\mathrm{d}s\hfill \end{array}$$

(9)

By Cauchy inequalities, we get

$$\begin{array}{cc}\hfill \frac{d}{dt}E\left(t\right)\le & \frac{1}{2}{\int }_{\Omega }(g^{\prime }\square u_x)\left(t\right)\mathrm{d}x+\left(\frac{|{\mu }_2|}{2}+\frac{\zeta }{2}\right){\int }_{\Omega }{u}_t^2(x,t)\mathrm{d}x+\left(\frac{|{\mu }_2|}{2}-\frac{\zeta }{2}{e}^{-\sigma \tau }\right){\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x\hfill \\ & -\frac{1}{2}g\left(t\right){\int }_{\Omega }{u}_x^2(x,t)\mathrm{d}x-\frac{\sigma \zeta }{2}{\int }_{t-\tau }^t{\int }_{\Omega }{e}^{-\sigma (t-s)}{u}_s^2((x,s)\mathrm{d}x\mathrm{d}s\hfill \end{array}$$

The proof is complete. ☐

Remark 2. 

In ([1] Lemma 4.1), we proved that the energy functional defined in [1] is non-increasing. However, since ${\displaystyle \left(\frac{|{\mu }_2|}{2}+\frac{\zeta }{2}\right){\int }_{\Omega }{u}_t^2\mathrm{d}x\ge 0}$, $E\left(t\right)$ may not be non-increasing here.

Now, we define the functional $\mathcal{D}\left(t\right)$ as follows:

$$\mathcal{D}\left(t\right)={\int }_{\Omega }u{u}_t\mathrm{d}x+{\int }_{L_1}^{L_2}v{v}_t\mathrm{d}x$$

Then, we have the following estimate:

Lemma 5. 

The functional $\mathcal{D}\left(t\right)$ satisfies

$$\begin{array}{cc}\hfill \frac{d}{dt}\mathcal{D}\left(t\right)\le & {\int }_{\Omega }{u}_t^2\mathrm{d}x+{\int }_{L_1}^{L_2}{v}_t^2\mathrm{d}x+\left({\delta }_1|{\mu }_2|L^2+{\delta }_1-\left(a-{\int }_0^{t}g\left(s\right)\mathrm{d}s\right)\right){\int }_{\Omega }{u}_x^2\mathrm{d}x\hfill \\ & +\frac{1}{4{\delta }_1}{\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }(g\square u_x)\mathrm{d}x+\frac{|{\mu }_2|}{4{\delta }_1}{\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x-{\int }_{L_1}^{L_2}b{v}_x^2\mathrm{d}x\hfill \end{array}$$

(10)

Proof. 

Taking the derivative of $\mathcal{D}\left(t\right)$ with respect to t and using (2), we have

$$\begin{array}{cc}\hfill \frac{d}{dt}\mathcal{D}\left(t\right)=& {\int }_{\Omega }{u}_t^2\mathrm{d}x-{\int }_{\Omega }(a{u}_x-g*u_x)u_x\mathrm{d}x-|{\mu }_2|{\int }_{\Omega }{u}_t(x,t-\tau )u\mathrm{d}x+{\int }_{L_1}^{L_2}{v}_t^2\mathrm{d}x-{\int }_{L_1}^{L_2}b{v}_x^2\mathrm{d}x\hfill \\ \hfill =& {\int }_{\Omega }{u}_t^2\mathrm{d}x-\left(a-{\int }_0^{t}g\left(s\right)\mathrm{d}s\right){\int }_{\Omega }{u}_x^2\mathrm{d}x-{\int }_{\Omega }(g\diamond u_x)u_x\mathrm{d}x-|{\mu }_2|{\int }_{\Omega }{u}_t(x,t-\tau )u\mathrm{d}x\hfill \\ & +{\int }_{L_1}^{L_2}{v}_t^2\mathrm{d}x-{\int }_{L_1}^{L_2}b{v}_x^2\mathrm{d}x\hfill \end{array}$$

(11)

By the boundary condition (3), we have

$$u^2(x,t)={\left({\int }_0^x{u}_x(x,t)\mathrm{d}x\right)}^2\le L_1{\int }_0^{L_1}{u}_x^2(x,t)\mathrm{d}x,x\in [0,L_1]$$

$$u^2(x,t)\le (L_3-L_2){\int }_{L_2}^{L_3}{u}_x^2(x,t)\mathrm{d}x,x\in [L_2,L_3]$$

which implies

$${\int }_{\Omega }{u}^2(x,t)\mathrm{d}x\le L^2{\int }_{\Omega }{u}_x^2\mathrm{d}x,x\in \Omega$$

(12)

where $L=max\{L_1,L_3-L_2\}$. By exploiting Young’s inequality and (12), we get for any ${\delta }_1>0$

$$-|{\mu }_2|{\int }_{\Omega }{u}_t(x,t-\tau )u\mathrm{d}x\le \frac{|{\mu }_2|}{4{\delta }_1}{\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x+{\delta }_1|{\mu }_2|L^2{\int }_{\Omega }{u}_x\mathrm{d}x$$

(13)

Young’s inequality implies that

$${\int }_{\Omega }(g\diamond u_x)u_x\mathrm{d}x\le {\delta }_1{\int }_{\Omega }{u}_x^2\mathrm{d}x+\frac{1}{4{\delta }_1}{\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }(g\square u_x)\mathrm{d}x$$

(14)

Inserting the estimates (13) and (14) into (11), then (10) is fulfilled. The proof is complete. ☐

Now, as in Lemma 4.5 of [24], we introduce the function

$$\begin{array}{c}\hfill q\left(x\right)=\left\{\begin{array}{cccc}& x-\frac{L_1}{2},\hfill & & x\in [0,L_1]\hfill \\ & \frac{L_1}{2}-\frac{L_1+L_3-L_2}{2(L_2-L_1)}(x-L_1),\hfill & & x\in (L_1,L_2)\hfill \\ & x-\frac{L_2+L_3}{2},\hfill & & x\in [L_2,L_3]\hfill \end{array}\right.\end{array}$$

(15)

It is easy to see that $q\left(x\right)$ is bounded, that is, $|q(x)|\le M$, where ${\displaystyle M=max\left\{\frac{L_1}{2},\frac{L_3-L_2}{2}\right\}}$ is a positive constant. In addition, we define the functionals:

$${\mathcal{F}}_1\left(t\right)=-{\int }_{\Omega }q\left(x\right)u_t(a{u}_x-g*u_x)\mathrm{d}x,{\mathcal{F}}_2\left(t\right)=-{\int }_{L_1}^{L_2}q\left(x\right)v_x{v}_t\mathrm{d}x$$

(16)

Then, we have the following estimates.

Lemma 6. 

The functionals ${\mathcal{F}}_1\left(t\right)$ and ${\mathcal{F}}_2\left(t\right)$ satisfy

$$\begin{array}{cc}\hfill \frac{d}{dt}{\mathcal{F}}_1\left(t\right)\le & {\left[-\frac{q\left(x\right)}{2}{(a{u}_x-g*u_x)}^2\right]}_{\partial \Omega }-{\left[\frac{a}{2}q\left(x\right)u_t^2\right]}_{\partial \Omega }+\left[\frac{a}{2}+\frac{M^2}{4{\delta }_2}\right]{\int }_{\Omega }{u}_t^2\mathrm{d}x\hfill \\ & +\left[2a^2+{\delta }_2{M}^2{a}^2|{\mu }_2|+g^2\left(0\right){\delta }_2+(4+{\delta }_2|{\mu }_2|){\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right)}^2\right]{\int }_{\Omega }{u}_x^2\mathrm{d}x\hfill \\ & +\left[{\delta }_2|{\mu }_2|M^2+\frac{|{\mu }_2|}{4{\delta }_2}+\frac{|{\mu }_2|M^2}{4{\delta }_2}\right]{\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x\hfill \\ & +\left(4+\frac{|{\mu }_2|}{4{\delta }_2}\right)\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right){\int }_{\Omega }(g\square u_x)\mathrm{d}x-g\left(0\right){\delta }_2{\int }_{\Omega }(g^{\prime }\square u_x)\mathrm{d}x\hfill \end{array}$$

(17)

and

$$\begin{array}{cc}\hfill \frac{d}{dt}{\mathcal{F}}_2\left(t\right)\le & -\frac{L_1+L_3-L_2}{4(L_2-L_1)}\left({\int }_{L_1}^{L_2}{v}_t^2\mathrm{d}x+{\int }_{L_1}^{L_2}b{v}_x^2\mathrm{d}x\right)+\frac{L_1}{4}{v}_t^2\left(L_1\right)+\frac{L_3-L_2}{4}{v}_t^2\left(L_2\right)\hfill \\ & +\frac{b}{4}\left((L_3-L_2)v_x^2(L_2,t)+L_1{v}_x^2(L_1,t)\right)\hfill \end{array}$$

(18)

Proof. 

Taking the derivative of ${\mathcal{F}}_1\left(t\right)$ with respect to t and using (2), we get

$$\begin{array}{cc}\hfill \frac{d}{dt}{\mathcal{F}}_1\left(t\right)=& -{\int }_{\Omega }q\left(x\right)u_{tt}(a{u}_x-g*u_x)\mathrm{d}x-{\int }_{\Omega }q\left(x\right)u_t\left(a{u}_{xt}-g\left(t\right)u_x\left(t\right)+(g^{\prime }\diamond u_x)\left(t\right)\right)\mathrm{d}x\hfill \\ \hfill =& {\left[-\frac{q\left(x\right)}{2}{(a{u}_x-g*u_x)}^2\right]}_{\partial \Omega }+\frac{1}{2}{\int }_{\Omega }{q}^{\prime }\left(x\right){(a{u}_x-g*u_x)}^2\mathrm{d}x-{\left[\frac{a}{2}q\left(x\right)u_t^2\right]}_{\partial \Omega }\hfill \\ & +\frac{a}{2}{\int }_{\Omega }{q}^{\prime }\left(x\right)u_t^2\mathrm{d}x-{\int }_{\Omega }q\left(x\right)|{\mu }_2|u_t(x,t-\tau )(g*u_x)\mathrm{d}x\hfill \\ & +{\int }_{\Omega }q\left(x\right)a{u}_x|{\mu }_2|u_t(x,t-\tau )\mathrm{d}x-{\int }_{\Omega }q\left(x\right)u_t[(g^{\prime }\diamond u_x)\left(t\right)-g\left(t\right)u_x]\mathrm{d}x\hfill \end{array}$$

(19)

We note that

$$\begin{array}{cc}\hfill & \frac{1}{2}{\int }_{\Omega }{q}^{\prime }\left(x\right){(a{u}_x-g*u_x)}^2\mathrm{d}x\hfill \\ \hfill \le & 2{\int }_{\Omega }{a}^2{u}_x^2\mathrm{d}x+2{\int }_{\Omega }{(g*u_x)}^2\mathrm{d}x\hfill \\ \hfill \le & 2{\int }_{\Omega }{a}^2{u}_x^2\mathrm{d}x+2{\int }_{\Omega }{\left({\int }_0^{t}g(t-s)(u_x\left(s\right)-u_x\left(t\right)+u_x\left(t\right))\mathrm{d}s\right)}^2\mathrm{d}x\hfill \\ \hfill \le & 2a^2{\int }_{\Omega }{u}_x^2\mathrm{d}x+4{\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right)}^2{\int }_{\Omega }{u}_x^2\mathrm{d}x+4\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right){\int }_{\Omega }(g\square u_x)\mathrm{d}x\hfill \end{array}$$

(20)

Young’s inequality gives us for any ${\delta }_2>0$,

$$\begin{array}{c}\hfill {\int }_{\Omega }q\left(x\right)a{u}_x|{\mu }_2|u_t(x,t-\tau )\mathrm{d}x\le {\delta }_2{M}^2{a}^2|{\mu }_2|{\int }_{\Omega }{u}_x^2\mathrm{d}x+\frac{|{\mu }_2|}{4{\delta }_2}{\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x\end{array}$$

(21)

$$\begin{array}{cc}\hfill & {\int }_{\Omega }q\left(x\right)|{\mu }_2|u_t(x,t-\tau )(g*u_x)\mathrm{d}x\hfill \\ \hfill =& |{\mu }_2|{\int }_{\Omega }(g\diamond u_x)q\left(x\right)u_t(x,t-\tau )\mathrm{d}x+|{\mu }_2|{\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }q\left(x\right)u_t(x,t-\tau )u_x\mathrm{d}x\hfill \\ \hfill \le & {\delta }_2{M}^2|{\mu }_2|{\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x+\frac{|{\mu }_2|}{4{\delta }_2}{\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }(g\square u_x)\mathrm{d}x+{\delta }_2|{\mu }_2|{\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right)}^2{\int }_{\Omega }{u}_x^2\mathrm{d}x\hfill \\ & +\frac{|{\mu }_2|M^2}{4{\delta }_2}{\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x\hfill \end{array}$$

(22)

and

$$\begin{array}{cc}\hfill & -{\int }_{\Omega }q\left(x\right)u_t[(g^{\prime }\diamond u_x)\left(t\right)-g\left(t\right)u_x]\mathrm{d}x\hfill \\ \hfill \le & \frac{M^2}{4{\delta }_2}{\int }_{\Omega }{u}_t^2\mathrm{d}x+g^2\left(0\right){\delta }_2{\int }_{\Omega }{u}_x^2\mathrm{d}x-g\left(0\right){\delta }_2{\int }_{\Omega }(g^{\prime }\square u_x)\mathrm{d}x\hfill \end{array}$$

(23)

Inserting (20)–(23) into (19), we get (17).

By the same method, taking the derivative of ${\mathcal{F}}_1\left(t\right)$ with respect to t, we obtain

$$\begin{array}{cc}\hfill \frac{d}{dt}{\mathcal{F}}_2\left(t\right)=& -{\int }_{L_1}^{L_2}q\left(x\right)v_{xt}{v}_t\mathrm{d}x-{\int }_{L_1}^{L_2}q\left(x\right)v_x{v}_{tt}\mathrm{d}x\hfill \\ \hfill =& {\left[-\frac{1}{2}q\left(x\right)v_t^2\right]}_{L_1}^{L_2}+\frac{1}{2}{\int }_{L_1}^{L_2}{q}^{\prime }\left(x\right)v_t^2\mathrm{d}x+\frac{1}{2}{\int }_{L_1}^{L_2}b{q}^{\prime }\left(x\right)v_x^2\mathrm{d}x+{\left[-\frac{b}{2}q\left(x\right)v_x^2\right]}_{L_1}^{L_2}\hfill \\ \hfill \le & -\frac{L_1+L_3-L_2}{4(L_2-L_1)}\left({\int }_{L_1}^{L_2}{v}_t^2\mathrm{d}x+{\int }_{L_1}^{L_2}b{v}_x^2\mathrm{d}x\right)+\frac{L_1}{4}{v}_t^2\left(L_1\right)+\frac{L_3-L_2}{4}{v}_t^2\left(L_2\right)\hfill \\ & +\frac{b}{4}\left((L_3-L_2)v_x^2(L_2,t)+L_1{v}_x^2(L_1,t)\right)\hfill \end{array}$$

Thus, the proof of Lemma 6 is finished. ☐

In [1], the authors pointed out that if ${\mu }_2<{\mu }_1$, then the energy is non-increasing. Thus, the negative term ${\displaystyle -{\int }_{\Omega }{u}_t^2\mathrm{d}x}$ appeared in the derivative energy can be used to stabilize the system. However, in this paper, the energy is not non-increasing. In this case, we need some additional negative term ${\displaystyle -{\int }_{\Omega }{u}_t^2\mathrm{d}x.}$ For this purpose, let us introduce the functional

$${\mathcal{F}}_3\left(t\right)=-{\int }_{\Omega }{u}_t(g\diamond u)\mathrm{d}x$$

Then, we have the following estimate.

Lemma 7. 

The functionals ${\mathcal{F}}_3\left(t\right)$ satisfies

$$\begin{array}{cc}\hfill \frac{d}{dt}{\mathcal{F}}_3\left(t\right)\le & -\left({\int }_0^{t}g\left(s\right)\mathrm{d}s-\frac{{\alpha }_4}{2}\right){\int }_{\Omega }{u}_t^2(x,t)\mathrm{d}x+\left[{\delta }_4+{\delta }_4{\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right)}^2\right]{\int }_{\Omega }{u}_x^2(x,t)\mathrm{d}x\hfill \\ & +{\delta }_4|{\mu }_2|{\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x+\left[\left({\delta }_4+\frac{1}{2{\delta }_4}+\frac{a^2}{4{\delta }_4}+\frac{|{\mu }_2|L^2}{4{\delta }_4}\right){\int }_0^{t}g\left(s\right)\mathrm{d}s\right]{\int }_{\Omega }(g\square u_x)\mathrm{d}x\hfill \\ & -\frac{g\left(0\right)L^2}{2{\alpha }_4}{\int }_{\Omega }(g^{\prime }\square u_x)\mathrm{d}x\hfill \end{array}$$

(24)

Proof. 

Taking the derivative of ${\mathcal{F}}_3\left(t\right)$ with respect to t and using (2), we get

$$\begin{array}{cc}\hfill \frac{d}{dt}{\mathcal{F}}_3\left(t\right)=& -{\int }_{\Omega }{u}_{tt}(g\diamond u)\mathrm{d}x-{\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }{u}_t^2(x,t)\mathrm{d}x-{\int }_{\Omega }{u}_t(g^{\prime }\diamond u)\mathrm{d}x\hfill \\ \hfill =& -{\int }_{\Omega }\left(a{u}_{xx}(x,t)-{\int }_0^{t}g(t-s)u_{xx}(x,s)\mathrm{d}s-|{\mu }_2|u_t(x,t-\tau )\right)(g\diamond u)\mathrm{d}x\hfill \\ & -{\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }{u}_t^2(x,t)\mathrm{d}x-{\int }_{\Omega }{u}_t(g^{\prime }\diamond u)\mathrm{d}x\hfill \\ \hfill =& -{\int }_{\Omega }\left({\int }_0^{t}g(t-s)u_x(x,s)\mathrm{d}s\right)(g\diamond u_x)\mathrm{d}x+a{\int }_{\Omega }{u}_x(g\diamond u_x)\mathrm{d}x\hfill \\ & +|{\mu }_2|{\int }_{\Omega }{u}_t(x,t-\tau )(g\diamond u)\mathrm{d}x-{\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }{u}_t^2(x,t)\mathrm{d}x-{\int }_{\Omega }{u}_t(g^{\prime }\diamond u)\mathrm{d}x\hfill \end{array}$$

(25)

Young’s inequality implies that for any ${\delta }_4>0$:

$$\begin{array}{cc}\hfill & -{\int }_{\Omega }\left({\int }_0^{t}g(t-s)u_x(x,s)\mathrm{d}s\right)(g\diamond u_x)\mathrm{d}x\hfill \\ \hfill \le & \frac{{\delta }_4}{2}{\int }_{\Omega }{\left({\int }_0^{t}g(t-s)(u_x\left(t\right)-u_x\left(s\right)-u_x\left(t\right))\mathrm{d}s\right)}^2\mathrm{d}x+\frac{1}{2{\delta }_4}{\int }_{\Omega }{(g\diamond u_x)}^2\mathrm{d}x\hfill \\ \hfill \le & {\delta }_4{\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right)}^2{\int }_{\Omega }{u}_x^2(x,t)\mathrm{d}x+\left({\delta }_4+\frac{1}{2{\delta }_4}\right){\int }_{\Omega }{(g\diamond u_x)}^2\mathrm{d}x\hfill \\ \hfill \le & {\delta }_4{\left({\int }_0^{t}g\left(s\right)\mathrm{d}s\right)}^2{\int }_{\Omega }{u}_x^2(x,t)\mathrm{d}x+\left({\delta }_4+\frac{1}{2{\delta }_4}\right){\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }(g\square u_x)\mathrm{d}x\hfill \end{array}$$

(26)

and

$$a{\int }_{\Omega }{u}_x(g\diamond u_x)\mathrm{d}x\le {\delta }_4{\int }_{\Omega }{u}_x^2(x,t)\mathrm{d}x+\frac{a^2}{4{\delta }_4}{\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }(g\square u_x)\mathrm{d}x$$

(27)

By Young’s inequality and (12), we get for any ${\delta }_4>0$, ${\alpha }_1>0$

$$|{\mu }_2|{\int }_{\Omega }{u}_t(x,t-\tau )(g\diamond u)\mathrm{d}x\le {\delta }_4|{\mu }_2|{\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x+\frac{|{\mu }_2|L^2}{4{\delta }_4}{\int }_0^{t}g\left(s\right)\mathrm{d}s{\int }_{\Omega }(g\square u_x)\mathrm{d}x$$

(28)

and

$$\begin{array}{cc}\hfill -{\int }_{\Omega }{u}_t(g^{\prime }\diamond u)\mathrm{d}x& \le \frac{{\alpha }_4}{2}{\int }_{\Omega }{u}_x^2(x,t)\mathrm{d}x+\frac{1}{2{\alpha }_4}{\int }_0^t(-g^{\prime }\left(s\right))\mathrm{d}s{\int }_{\Omega }(-g^{\prime }\square u)\mathrm{d}x\hfill \\ & \le \frac{{\alpha }_4}{2}{\int }_{\Omega }{u}_x^2(x,t)\mathrm{d}x-\frac{g\left(0\right)L^2}{2{\alpha }_4}{\int }_{\Omega }(g^{\prime }\square u_x)\mathrm{d}x\hfill \end{array}$$

(29)

Inserting (26)–(29) into (25), we get (24). ☐

Now, we are ready to prove Theorem 3.

Proof. 

We define the Lyapunov functional:

$$L\left(t\right)=N_{1}E\left(t\right)+N_2\mathcal{D}\left(t\right)+{\mathcal{F}}_1\left(t\right)+N_4{\mathcal{F}}_2\left(t\right)+N_5{\mathcal{F}}_3\left(t\right)$$

(30)

where $N_1,N_2,N_4$ and $N_5$ are positive constants that will be fixed later.

Since g is continuous and $g\left(0\right)>0$, then for any $t\ge t_0>0$, we obtain

$${\int }_0^{t}g\left(s\right)\mathrm{d}s\ge {\int }_0^{t_0}g\left(s\right)\mathrm{d}s=g_0$$

(31)

Taking the derivative of (30) with respect to t and making the use of the above lemmas, we have

$$\begin{array}{cc}\hfill \frac{d}{dt}L\left(t\right)\le & -\left\{N_5\left(g_0-\frac{{\alpha }_4}{2}\right)-N_1\left(\frac{|{\mu }_2|}{2}+\frac{\zeta }{2}\right)-N_2-\left(\frac{a}{2}+\frac{M^2}{4{\delta }_2}\right)\right\}{\int }_{\Omega }{u}_t^2\mathrm{d}x\hfill \\ & -\{N_2{\beta }_0-N_2({\delta }_1+L^2{\delta }_1|{\mu }_2|)-(2a^2+4{\overline{g}}^2+{\delta }_2|{\mu }_2|{\overline{g}}^2+{\delta }_2{M}^2{a}^2|{\mu }_2|+g^2\left(0\right){\delta }_2)\hfill \\ & -N_5({\delta }_4+{\delta }_4{\overline{g}}^2)\}{\int }_{\Omega }{u}_x^2\mathrm{d}x\hfill \\ & +\left\{N_1\left(\frac{|{\mu }_2|}{2}-\frac{\zeta }{2}{e}^{-\sigma \tau }\right)+\frac{N_2|{\mu }_2|}{4{\delta }_1}+\left({\delta }_2|{\mu }_2|M^2+\frac{|{\mu }_2|(1+M^2)}{4{\delta }_2}\right)+N_5{\delta }_4|{\mu }_2|\right\}{\int }_{\Omega }{u}_t^2(x,t-\tau )\mathrm{d}x\hfill \\ & -\left\{\frac{b(L_1+L_3-L_2)}{4(L_2-L_1)}{N}_4+N_{2}b\right\}{\int }_{L_1}^{L_2}{v}_x^2\mathrm{d}x-\left\{\frac{L_1+L_3-L_2}{4(L_2-L_1)}{N}_4-N_2\right\}{\int }_{L_1}^{L_2}{v}_t^2\mathrm{d}x\hfill \\ & -(b-N_4)\frac{b}{4}\left((L_3-L_2)v_x^2(L_2,t)+L_1{v}_x^2(L_1,t)\right)-(a-N_4)\left[\frac{L_1}{4}{v}_t^2(L_1,t)+\frac{L_3-L_2}{4}{v}_t^2(L_2,t)\right]\hfill \\ & +\left[\frac{N_2\overline{g}}{4{\delta }_1}+\left(4\overline{g}+\frac{|{\mu }_2|\overline{g}}{4{\delta }_2}\right)+N_5\left({\delta }_4+\frac{1}{2{\delta }_4}+\frac{(a^2+|{\mu }_2|L^2)}{4{\delta }_4}\right)\overline{g}\right]{\int }_{\Omega }(g\square u_x)\mathrm{d}x\hfill \\ & +\left[\frac{N_1}{2}-g\left(0\right){\delta }_2-\frac{N_{5}g\left(0\right)L^2}{2{\alpha }_4}\right]{\int }_{\Omega }(g^{\prime }\square u_x)\mathrm{d}x\hfill \end{array}$$

(32)

At this moment, we wish all coefficients except the last two in (32) will be negative. We want to choose $N_2$ and $N_4$ to ensure that

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle a-N_4\ge 0}\hfill \\ b-N_4\ge 0\hfill \\ {\displaystyle \frac{L_1+L_3-L_2}{4(L_2-L_1)}{N}_4-N_2>0}\hfill \end{array}\right.\end{array}$$

(33)

For this purpose, since ${\displaystyle \frac{8l(L_2-L_1)}{L_1+L_3-L_2}<min\{a,b\}}$, we first choose $N_4$ satisfying

$$\begin{array}{c}\hfill \frac{8l(L_2-L_1)}{L_1+L_3-L_2}<N_4\le min\{a,b\}\end{array}$$

Then, we pick

$${\displaystyle {\alpha }_4=g_0,{\delta }_1<\frac{{\beta }_0}{8}\mathrm{and}{\delta }_2<\frac{1}{g^2\left(0\right)}}$$

such that

$$N_5\left(g_0-\frac{{\alpha }_4}{2}\right)=\frac{N_5{g}_0}{2},N_2{\beta }_0-N_2{\delta }_1>\frac{7N_2{\beta }_0}{8},\mathrm{and}{g}^2\left(0\right){\delta }_2<1$$

Once ${\delta }_2$ is fixed, we take $N_2$ satisfying

$$N_2>\frac{8(2a^2+4\overline{g}+g^2\left(0\right){\delta }_2)}{{\beta }_0}$$

such that

$$(2a^2+4\overline{g}+g^2\left(0\right){\delta }_2)<\frac{N_2{\beta }_0}{8}$$

Furthermore, we choose $N_5$ satisfying

$$\frac{N_5{g}_0}{8}>N_2+\frac{a}{2}+\frac{M^2}{4{\delta }_2}$$

such that

$$N_2<\frac{N_5{\beta }_0}{8},\frac{a}{2}<\frac{N_5{\beta }_0}{8},\frac{M^2}{4{\delta }_2}<\frac{N_5{\beta }_0}{8}\mathrm{and}\frac{N_5{g}_0}{8}-N_2-\left(\frac{a}{2}+\frac{M^2}{4{\delta }_2}\right)>0$$

Then, we pick ${\delta }_4$ satisfying

$${\delta }_4<\frac{{\beta }_0{N}_2}{8N_5(1+{\overline{g}}^2)}$$

such that

$$N_5({\delta }_4+{\delta }_4{\overline{g}}^2)<\frac{N_2{\beta }_0}{8}$$

Once the above constants are fixed, we choose $N_1$ satisfying

$$\frac{N_1}{2}>g\left(0\right){\delta }_2+\frac{N_{5}g\left(0\right)L^2}{2{\alpha }_4}$$

Now, we need to choose suitable $|{\mu }_2|$ and ζ such that

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle K_1-N_1\left(\frac{|{\mu }_2|}{2}+\frac{\zeta }{2}\right)>0}\hfill \\ {\displaystyle \frac{5{\beta }_0{N}_2}{8}-K_2|{\mu }_2|>0}\hfill \\ {\displaystyle K_3\frac{|{\mu }_2|}{2}-N_1\frac{\zeta }{2}{e}^{-\sigma \tau }<0}\hfill \end{array}\right.\end{array}$$

(34)

where

$$\begin{array}{cc}& K_1=\frac{N_5{g}_0}{8}-N_2-\left(\frac{a}{2}+\frac{M^2}{4{\delta }_2}\right),K_2=N_2{\delta }_1{L}^2+{\delta }_2{\overline{g}}^2+{\delta }_2{M}^2{a}^2\hfill \\ & K_3=N_1+\frac{N_2}{2{\delta }_1}+2{\delta }_2{M}^2+\frac{1}{2{\delta }_2}+\frac{M^2}{2{\delta }_2}+2N_5{\delta }_4\hfill \end{array}$$

We first choose ζ satisfying

$$\frac{2K_1}{N_1}-\zeta >0$$

Then, we pick $|{\mu }_2|$ satisfying

$$|{\mu }_2|<min\left\{\frac{5{\beta }_0{N}_2}{8K_2},\frac{N_1\zeta }{K_3{e}^{\sigma \tau }},\frac{2K_1}{N_1}-\zeta \right\}:=a_0$$

(35)

From the above, we deduce that there exist two positive constants ${\alpha }_5$ and ${\alpha }_6$ such that (32) becomes

$$\frac{d}{dt}L\left(t\right)\le -{\alpha }_{5}E\left(t\right)+{\alpha }_6{\int }_{\Omega }(g\square u_x)\mathrm{d}x$$

(36)

Multiplying (36) by ξ, we have

$$\xi \frac{d}{dt}L\left(t\right)\le -{\alpha }_5\xi E\left(t\right)+{\alpha }_6\xi {\int }_{\Omega }(g\square u_x)\mathrm{d}x$$

On the other hand, by the definition of the functionals $\mathcal{D}\left(t\right)$, ${\mathcal{F}}_1\left(t\right)$, ${\mathcal{F}}_2\left(t\right)$, ${\mathcal{F}}_3\left(t\right)$ and $E\left(t\right)$, for $N_1$ large enough, there exists a positive constant ${\alpha }_3$ satisfying

$$|N_2\mathcal{D}\left(t\right)+N_3{\mathcal{F}}_1\left(t\right)+N_4{\mathcal{F}}_2\left(t\right)+{\mathcal{F}}_3\left(t\right)|\le {\eta }_{1}E\left(t\right)$$

which implies that

$$(N_1-{\eta }_1)E\left(t\right)\le L\left(t\right)\le (N_1+{\eta }_1)E\left(t\right)$$

Exploiting (H2) and (8), we have

$$\begin{array}{c}\hfill \xi {\int }_{\Omega }(g\square u_x)\mathrm{d}x\le -{\int }_{\Omega }(g^{\prime }\square u_x)\mathrm{d}x\le -2\frac{d}{dt}E\left(t\right)+\frac{2K_1}{N_1}{\int }_{\Omega }{u}_t^2\mathrm{d}x\end{array}$$

(37)

Thus, (36) becomes

$$\xi \frac{d}{dt}L\left(t\right)\le -{\alpha }_5\xi E\left(t\right)-2{\alpha }_6\frac{d}{dt}E\left(t\right)+\frac{4K_1{\alpha }_6}{N_1}E\left(t\right)$$

(38)

We add a restriction condition on ξ, that is, we suppose that

$$\xi >\frac{4K_1{\alpha }_6}{N_1}:={\xi }_0$$

(39)

Then, (38) becomes, for some positive constants

$$\xi \frac{d}{dt}L\left(t\right)\le -{\alpha }_7\xi E\left(t\right)-{\alpha }_8\frac{d}{dt}E\left(t\right)$$

Now, we define functionals $\mathcal{L}\left(t\right)$ as

$$\mathcal{L}\left(t\right)=\xi L\left(t\right)+{\alpha }_{8}E\left(t\right)$$

It is clear that

$$\mathcal{L}\left(t\right)\sim E\left(t\right)$$

(40)

Then, we have

$$\frac{d}{dt}\mathcal{L}\left(t\right)\le -{\alpha }_7\xi E\left(t\right)$$

(41)

A simple integration of (41) over $(t_0,t)$ leads to

$$\mathcal{L}\left(t\right)\le \mathcal{L}\left(t_0\right)e^{-c\xi (t-t_0)},\forall t\ge t_0$$

(42)

Recalling (40), Equation (42) yields the desired result (7). This completes the proof of Theorem 3. ☐

4. Conclusions

The main purpose of present work is to investigate decay rate for a transmission problem with a viscoelastic term and a delay term but without the frictional damping term. It is based upon our previous work ([1]), in which we studied the well-posedness and general decay rate for a transmission problem in a bounded domain with a viscoelastic term and a delay term. The main difficulty in dealing with the problem here is that in the first equation of system (2), we have no frictional damping term to control the delay term in the estimate of the energy decay. To overcome this difficulty, our basic idea is to control the delay term by making use of the viscoelastic term. In order to achieve this target, a restriction of the size between the parameter ${\mu }_2$ and the relaxation function g and a suitable energy is needed. This is motivated by Dai and Yang’s work [34], in which the viscoelastic wave equation with delay term but without a frictional damping term was considered and an exponential decay result was established. In Section 2, we give some notations and hypotheses needed for our work and state the main results. In Section 3, because the energy is not non-increasing, we introduce the additional functional to produce negative term ${\displaystyle -{\int }_{\Omega }{u}_t^2\mathrm{d}x}$. Then by introducing suitable Lyaponov functionals, we prove the exponential decay of the solutions for the relaxation function satisfying assumption $\left(H_1\right)$ and $\left(H_2\right)$.