Global solutions and general decay for the dispersive wave equation with memory and source terms

1. Introduction

This paper deals with the initial boundary value problem of the dispersive wave equation with memory and source terms where \(\Omega\) is a bounded domain in \(\mathbb{R}^{d}\) \((d\geq1)\) with a smooth boundary \(\partial\Omega\), \(\alpha\) is a positive constant and \(g(t)\) is a positive function that represents the kernel of the memory term, which will be specified in Section 2. Here, we understand \(\Delta^{2}u\) to be the dispersive term. In the absence of the viscoelastic term and the dispersive term (that is, if \(g=\alpha= 0\)), the model (1) reduces to the weakly damped wave equation The interaction between the weak damping term and the source term are considered by many authors. We refer the reader to, Haraux and Zuazua [1], Ikehata [2] and Levine [3, 4]. If \(\alpha=0\) and \(g\) is not trivial on \(\mathbb{R}\), but replacing the fourth order memory term in (1) by a weaker memory of the form \(\int_{0}^{t}g(t-\tau)\Delta u(\tau)d\tau\), then (1) can be rewritten as follows The Equation (3) has been considered by Wang et al [5]. Under some appropriate assumptions on \(g\), by introducing potential wells they obtained the existence of global solution and the explicit exponential energy decay estimates. Our main goal in the present paper is to discuss the global solutions and general decay to the following weakly damped wave equation with dispersive term, the fourth order memory term and the nonlinear source term with simply supported boundary condition and initial conditions where \(\Omega\) is a bounded domain of \(\mathbb{R}^{d}\) with a smooth boundary \(\partial\Omega\) and \(p>1\). Here, \(\nu\) is the unit outward normal to \(\partial\Omega\), and \(g(t)\) is a positive function that represents the kernel of the memory term, which will be specified in Section 2. We prove that Problem (4)-(6) has a global weak solution assuming small initial data. In addition, we show the general decay of solutions. The global solutions are constructed by means of the Galerkin approximations and the general decay is obtained by employing the technique used in [6].

2. Preliminaries

Before proceeding to our analysis, we use the following abbreviations \(\|\cdot\|_{q}=\|\cdot\|_{L^{q}(\Omega)}\) \((1\leq q\leq+\infty)\) denotes usual \(L^{q}\) norm, \((\cdot,\cdot)\) denotes the \(L^{2}\)-inner product, and consider the Sobolev spaces \(H^{1}_{0}(\Omega)\) and \(H^{2}_{0}(\Omega)\) with their usual scalar products and norms. We also use the embedding \(H^{1}_{0}(\Omega)\hookrightarrow L^{q}(\Omega)\) for \(2< q< \frac{2d}{d-2}\) if \(d\geq3\) or \(2< q< \infty\) if \(d=1,2\). In this case, the embedding constant is denoted by \(C_{*}\), that is \( \|u\|_{q}\leq C_{*}\|\nabla u\|_{2}. \) We define the polynomial \(Q\) by \( Q(z)=\frac{1}{2}z^{2}-\frac{C_{*}^{p+1}}{p+1}z^{p+1}, \) which is increasing in \([0,z_{0}]\), where \( z_{0}=C_{*}^{\frac{p+1}{1-p}} \) is its unique local maximum. Next, we give the assumptions for Problem (4)-(6).
(G1) The relaxation function \(g:\mathbb{R}_{+}\rightarrow \mathbb{R}_{+}\) is a bounded \(C^{1}\) function such that \( g(0)>0,\quad 0< \eta=1-\int_{0}^{\infty}g(\tau)d\tau\leq1-\int_{0}^{t}g(\tau)d\tau=\eta(t). \)
(G2) There exist positive constants \(\xi_{1}\) and \(\xi_{2}\) such that \( -\xi_{1}g(t)\leq g'(t)\leq-\xi_{2}g(t)\quad \forall t\geq0. \)
(G3) We also assume that \( 1< p\leq \frac{d+2}{d-2}\ \ \mbox {if} \ \ d\geq3 \ \ \mbox {and} \ \ \ p>1 \ \ \ \mbox {if} \ \ d=1,2. \) where \(\lambda_{1}\) is the first eigenvalue of the following problem

Remark 1. [7] Assuming \(\lambda_{1}\) is the first eigenvalue of the problem (7), we have

Now, we define the following energy function associated with a solution \(u\) of the Problem (4)-(6) for \(u\in H^{2}_{0}(\Omega)\), and is the initial total energy. To facilitate further on our analysis, we use the following notation \begin{equation*} (g\circ\Delta u)(t)=\int_{0}^{t}g(t-\tau)\|\Delta u(\tau)-\Delta u(t)\|_{2}^{2}d\tau. \end{equation*} Now, we are in a position to state our main results.

3. Main results

Theorem 1. Assume that \((G1)-(G3)\) hold, \(u_{0}\in H^{2}_{0}(\Omega)\), \(u_{1}\in L^{2}(\Omega)\). Further assume that \(\|\nabla u_{0}\|_{2}< z_{0}\) and \(E(0)< Q(z_{0})\), then the Problem (4)-(6) possesses a global weak solution satisfying; \(u\in L^{\infty}(0,\infty;H^{2}_{0}(\Omega)),\quad u_{t}\in L^{\infty}(0,\infty;L^{2}(\Omega))\) for \(0\leq t< \infty\), and the energy identity

holds for \(0\leq t< \infty\). Moreover, for \(\zeta:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}\) a increasing \(C^{2}\) function satisfying and, if \(\|g\|_{L^{1}(0,\infty)}\) is sufficiently small, we have for \(\kappa>0\); \( E(t)\leq E(0)e^{-\kappa \zeta(t)},\quad \forall t\geq0. \)

Remark 2. From (11) and \((G2)\), we can easily obtain

Remark 3. For \(\zeta(t)=t+\frac{t}{t+1}\), we can get the exponential decay rate \( E(t)\leq E(0)e^{-\kappa t},\quad \forall t\geq0\). For \(\zeta(t)=ln(1+t)\), we can get polynomial decay rate \( E(t)\leq E(0)(1+t)^{-\kappa },\quad \forall t\geq0. \)

4. Proof of main results

In this section, we shall divide the proof into two steps. In Step 1, we prove the global existence of weak solutions by using Galerkin's approximations. In Step 2, we establish the general decay of energy employing the method used in [6].

Step 1 Global existence of weak solutions

Let \(\left\{\omega_{j}\right\}_{j=1}^{\infty}\) be an orthogonal basis of \(H^{2}_{0}(\Omega)\) with \(\omega_{j}\) being the eigenfunction of the problem \( -\Delta \omega_{j}=\lambda_{j}\omega_{j},\quad x\in\Omega,\quad \omega_{j}=0,\quad x\in\partial\Omega. \) Let \(V^{n}=\text{Span}\left\{\omega_{1},\omega_{2},\cdot\cdot\cdot,\omega_{n}\right\}\). By the standard method of ODE, we know that \( u^{n}(t)=\sum_{j=1}^{n}b^{n}_{j}(t)\omega_{j}(x) \) of the Cauchy problem as follows By the standard theory of ODE system, we prove the existence of solutions of Problem (14)-(15) on some interval \([0, t_{n})\), \(0< t_{n}< T\) for arbitrary \(T>0\), then, this solution can be extended to the whole interval \([0,T]\) using the first estimate given below. Taking \(\omega=u^{n}_{t}(t)\) in (14), we obtain For the last term on the left hand side of (16) we have Inserting (17) into (16) and integrating over \([0,t]\subset[0, T]\), we obtain Now from assumption \((G3)\) and the Sobolev embedding, we have that and then we have By using the fact that \( -\int_{0}^{t}(g'\circ\Delta u^{n})(\tau)d\tau+\int_{0}^{t}g(\tau)\|\Delta u^{n}(\tau)\|^{2}_{2}d\tau\geq0, \) estimate (20) yields From \(E(0)< \mathcal{Q}(z_{0})\) and (15), it follows that for sufficiently large \(n\). We claim that there exists an integer \(N\) such that Suppose the claim is proved. Then \(\mathcal{Q}(\|\nabla u^{n}\|^{2}_{2})\geq0\) and from (21) and (22), for sufficiently large \(n\) and \(0\leq t< \infty\).

Proof. [Proof of the claim] Suppose that (23) false. Then for each \(n>N\), there exists \(t\in[0,t_{n})\) such that \(\|\nabla u^{n}(t)\|_{2}\geq z_{0}\). We note that from \(\|\nabla u_{0}\|_{2}< z_{0}\) and (15) there exists \(N_{0}\) such that \( \|\nabla u^{n}(0)\|_{2}< z_{0}\quad \forall n>N_{0}. \) Then by continuity there exits a first \(t_{n}^{*}\in[0,t_{n})\) such that

from where \( \mathcal{Q}(\|\nabla u^{n}(t)\|_{2})\geq0 \quad \forall t\in[0,t_{n}^{*}]. \) Now from \(E(0)< \mathcal{Q}(z_{0})\) and (24), there exists \(N>N_{0}\) and \(\gamma\in(0,z_{0})\) such that \( 0\leq\frac{1}{2}\|u^{n}_{t}(t)\|^{2}_{2}+\frac{\eta(t)}{2}\|\Delta u^{n}(t)\|^{2}_{2}+\frac{1}{2}(g\circ\Delta u^{n})(t)+\mathcal{Q}(\|\nabla u^{n}(t)\|^{2}_{2})\leq\mathcal{Q}(\gamma)\quad \forall t\in[0,t_{n}^{*}]\quad \forall n>N. \) Then the monotonicity of \(\mathcal{Q}\) in \([0,z_{0}]\) implies that \( 0\leq\|\nabla u^{n}(t)\|^{2}_{2}\leq\gamma< z_{0}\quad \forall t\in[0,t_{n}^{*}], \) and in particular, \(\|\nabla u^{n}(t)\|^{2}_{2}< z_{0}\), which is a contradiction to (24). From (24), we have Using Sobolev inequality, (8) and (26), it follows that Furthermore, by (29), we get The estimates (26)-(30) permit us to obtain a subsequences of \(\left\{u_{n}\right\}\) which from now on will be also denoted by \(\left\{u_{n}\right\}\) and functions \(u\), \(\chi\) such that Besides, from Lions-Aubin Lemma we also have and consequently, making use of the Lemma 1.3 in [8], we deduce Thus, we obtain that \(u\) is a global weak of problem (4)-(6). Next, we shall prove that \(u\) satisfies (11). From the discussion above, we obtain for each fixed \(t>0\) that We obtain for each fixed \(t>0\) that as \(n\rightarrow +\infty\), and as \(n\rightarrow +\infty\), where \(0< \theta_{n}< 1\). Hence, we have From (15), it follows that \(E^{n}(0)\rightarrow E(0)\) as \(n\rightarrow+\infty\). Finally, taking \(n\rightarrow +\infty\) in (18), we deduce that the energy identity (11) holds for \(0\leq t< \infty\).

Step 2 General decay of the energy

Firstly, we state several Lemmas to prove the decay rate estimate of the energy.

Lemma 1. Let \(u\in L^{\infty}(0,\infty;H^{2}_{0}(\Omega))\) be the solution of (4)-(6) and \(E(0)< \mathcal{Q}(z_{0})\), \(\|\nabla u_{0}\|_{2}< z_{0}\), then we have

where \(C_{1}=\frac{1}{2}+(2\lambda_{1})^{-1}\).

Proof. From \(E(0)< \mathcal{Q}(z_{0})\) and \(\|\nabla u_{0}\|_{2}< z_{0}\), we can obtain \(\mathcal{Q}(\|\nabla u(t)\|_{2})\geq0\) for \(0\leq t< \infty\). Thus we have

and

Lemma 2. The energy \(E(t)\) satisfies

Proof. From (13), we have

From assumptions \((G2)\) and since \(\int_{0}^{t}g'(\tau)d\tau=g(t)-g(0)\), we obtain Then, Combining (45) and (44) our conclusion holds. Multiplying (43) by \(e^{\kappa\zeta(t)}\) \((\kappa>0)\) and using (40), we have Using the fact that \(\zeta_{t}\) is decreasing by (12), we arrive at Choosing \(\|g\|_{L^{1}(0,\infty)}\) sufficiently small so that \(g(0)-\xi_{1}\|g\|_{L^{1}(0,\infty)}=B>0\) and defining \(\kappa_{0}=\min\left\{\frac{2}{\zeta_{t}(0)},\frac{\xi_{2}}{\zeta_{t}(0)},\frac{B}{2C_{1}\zeta_{t}(0)}\right\},\) we conclude by taking \(\kappa\in(0, \kappa_{0}]\) in (47) that \( \frac{d}{dt}\left(e^{\kappa\zeta(t)}E(t)\right)\leq0,\quad t>0. \) Integrating the above inequality over \((0,t)\), it follows that \( E(t)\leq E(0)e^{-\kappa\zeta(t)},\quad t>0. \)

Competing Interests

The author declares no conflict of interest.