General and Optimal Decay Result for a Viscoelastic Problem with Nonlinear Boundary Feedback

Abstract

In this paper, we consider a viscoleastic equation with a nonlinear feedback localized on a part of the boundary and a relaxation function satisfying g^′(t) ≤−ξ(t)G(g(t)). We establish an explicit and general decay rate results, using the multiplier method and some properties of the convex functions. Our results are obtained without imposing any restrictive growth assumption on the damping term. This work generalizes and improves earlier results in the literature, in particular those of Messaoudi (Topological Methods in Nonlinear Analysis 51(2):413–427, 2018), Messaoudi and Mustafa (Nonlinear Analysis: Theory Methods & Applications 72(9–10):3602–3611, 2010), Mustafa (Mathematical Methods in the Applied Sciences 41(1): 192–204, 2018) and Wu (Zeitschrift für angewandte Mathematik und Physik 63(1):65–106, 2012).

1 Introduction

In this paper, we consider the following viscoelastic problem:

$$ \left\{\begin{array}{llll} u_{tt}(t)- {\Delta} u(t)+{{\int}_{0}^{t}}g(t-s){\Delta} u(s)ds= 0, \qquad \text{ in } {\Omega} \times \mathbb{R}^{+}\\ \frac{\partial u}{\partial \nu}(t)-{{\int}_{0}^{t}}g(t-s)\frac{\partial u}{\partial \nu}(s)ds+h(u_{t}(t))= 0,\quad \text{on } {\Gamma}_{1} \times \mathbb{R}^{+}\\ u(t)= 0,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \text{ on } {\Gamma}_{0} \times \mathbb{R}^{+}\\ u(x,0)=u_{0}(x), u_{t}(x,0)=u_{1}(x),\qquad\qquad\quad \text{in } {\Omega} \times \mathbb{R}^{+} \end{array}\right. $$

(1.1)

where u denotes the transverse displacement of waves, Ω is a bounded domain of $\mathbb {R}^{N} (N\ge 1)$ with a smooth boundary ∂Ω = Γ₀ ∪Γ₁ such that Γ₀ and Γ₁ are closed and disjoint, with meas. (Γ₀) > 0, ν is the unit outer normal to ∂Ω, and g and h are specific functions.

During the last half century, a great attention has been devoted to the study of viscoelastic problems and many existence and long-time behavior results have been established. We start with the pioneer work of Dafermos [5, 6], where he considered a one-dimensional viscoelastic problem of the form

$$\rho u_{tt}=c u_{xx}-{\int}_{-\infty}^{t}g(t-s)u_{xx}(s)ds, $$

and established various existence results and then proved, for smooth monotone decreasing relaxation functions, that the solutions go to zero as t goes to infinity. However, no rate of decay has been specified. Hrusa [7] considered a one-dimensional nonlinear viscoelastic equation of the form

$$u_{tt}-c u_{xx}+{{\int}_{0}^{t}} m(t-s)(\psi(u_{x}(s)))_{x}ds=f(x,t), $$

and proved several global existence results for large data. He also proved an exponential decay result for strong solutions when m(s) = e^−s and ψ satisfies certain conditions. In [8] Dassios and Zafiropoulos considered a viscoelatic problem in $\mathbb {R}^{3}$ and proved a polynomial deacy result for exponentially decaying kernels. In their book, Fabrizio and Morro [9] established a uniform stability of some problems in linear viscoelasticity. In all the above mentioned works, the rates of decay in relaxation functions were either of exponential or polynomial type. In 2008, Messaoudi [10, 11] generalized the decay rates allowing an extended class of relaxation functions and gave general decay rates from which the exponential and the polynomial decay rates are only special cases. However, the optimality in the polynomial decay case was not obtained. Precisely, he considered relaxation functions that satisfy

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

(1.2)

where $\xi :\mathbb {R}^{+} \to \mathbb {R}^{+}$ is a nonincreasing differentiable function and showed that the rate of the decay of the energy is the same rate of decay of g, which is not necessarily of exponential or polynomial decay type. After that a series of papers using Eq. 1.2 has appeared see, for instance, [12,13,14,15,16,17,18] and [19].

Inspired by the experience with frictional damping initiated in the work of Lasiecka and Tataru [20], another step forward was done by considering relaxation functions satisfying

$$ g^{\prime}(t)\le -\chi(g(t)). $$

(1.3)

This condition, where χ is a positive function, χ(0) = χ^′(0) = 0, and χ is strictly increasing and strictly convex near the origin, with some additional constraints imposed on χ, was used by several authors with different approaches. We refer to previous studies [21,22,23,24,25,26,27] and [28], where general decay results in terms of χ were obtained. Here, it should be mentioned that, in [26], it was the first time where Lasiecka and Wang established not only general but also optimal results in which the decay rates are characterized by an ODE of the same type as the one generated by the inequality (1.3) satisfied by g. Mustafa and Messaoudi [29] established an explicit and general decay rate for relaxation function satisfying

$$ g^{\prime}(t)\le -H(g(t)), $$

(1.4)

where $H\in C^{1}(\mathbb {R})$, with H(0) = 0 and H is linear or strictly increasing and strictly convex function C² near the origin. In [30], Cavalcanti et al. considered the following problem

$$ \left\{\begin{array}{llll} {\vert u_{t}\vert}^{\rho}u_{tt}-{\Delta} u-{\Delta} u_{tt}+{{\int}_{0}^{t}}g(t-s){\Delta} u(s)ds= 0,\qquad\qquad\text{in } {\Omega}\times\mathbb{R}^{+}, \\ u(x,t)= 0, \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \text{on } {\Gamma}\times\mathbb{R}^{+}, \\ u(x,0)=u_{0}(x),\quad u_{t}(x,0)=u_{1}(x),\qquad\qquad\qquad\qquad\quad \text{in}\hspace{0.05in}{\Omega}\times\mathbb{R}^{+}, \end{array}\right. $$

(1.5)

with a relaxation function satisfying Eq. 1.4 and the additional requirement:

$$\lim\inf_{x\to 0^{+}}{x^{2} H^{\prime\prime} -xH^{\prime}+H(x)}\ge 0, $$

and that $y^{1-\alpha _{0}}\in L^{1}(1,\infty )$, for some α₀ ∈ [0, 1), where y(t) is the solution of the problem

$$y^{\prime}(t)+H(y(t))= 0,\text{ } y(0)=g(0)>0. $$

They characterized the decay of the energy by the solution of a corresponding ODE as in [20]. Recently, Messaoudi and Al-Khulaifi [31] treated (1.5) with a relaxation function satisfying

$$ g^{\prime}(t)\le -\xi(t) g^{p}(t), \text{ } \forall t \ge0,\text{ } 1 \le p<\frac{3}{2}. $$

(1.6)

They obtained a more general stability result for which the results of [10, 11] are only special cases. Moreover, the optimal decay rate for the polynomial case is achieved without any extra work and conditions as in [25] and [20]. For stabilization by mean of boundary feedback, Cavalcanti et al. [32] studied (1.1) and proved a global existence result for weak and strong solutions. Moreover, they gave some uniform decay rate results under some restrictive assumptions on both the kernel g and the damping function h. These restrictions had been relaxed by Cavalcanti et al. [33] and further they established a uniform stability depending on the behavior of h near the origin and on the behavior of g at infinity. In the absence of the viscoelastic term (g = 0), problem (1.1) has been investigated by many authors and several stability results were established. We refer the reader to the work of Lasiecka and Tataru [20], Alabau-Boussouira [34], Cavalcanti et al. [35], Guesmia [36, 37], Cavalcanti [38] and the references therein.

2 Preliminaries

In this section, we present some materials needed in the proof of our results. We use the standard Lebesgue space L²(Ω) and the Sobolev space ${H_{0}^{1}}({\Omega })$ with their usual scalar products and norms and denote by V the following spac

$$V=\{v\in H^{1}({\Omega}):v = 0\hspace{0.05in}\text{on}\hspace{0.05in}{\Gamma}_{0}\}.$$

Throughout this paper, c is used to denote a generic positive constant.

We consider the following hypotheses:

(A1):

$g: \mathbb {R}^{+}\to \mathbb {R}^{+}$ is a C¹ nonincreasing function satisfying

$$ g(0) > 0, \qquad 1-{\int}_{0}^{+\infty}g(s)ds={\ell} > 0, $$

(2.1)

and there exists a C¹ function G : (0, ∞) → (0, ∞) which is linear or it is strictly increasing and strictly convex C² function on (0, r₁], r₁ ≤ g(0), with G(0) = G^′(0) = 0, such that

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

(2.2)

where ξ(t) is a positive nonincreasing differentiable function.

(A2):

$h: \mathbb {R}\to \mathbb {R}$ is a nondecreasing C⁰ function such that there exists a strictly increasing function $h_{0}\in C^{1}(\mathbb {R}^{+})$, with h₀(0) = 0, and positive constants c₁, c₂, ε such that

$$\begin{array}{@{}rcl@{}} &&h_{0}(\vert s \vert)\le \vert h(s) \vert \le h_{0}^{-1}(\vert s \vert)\qquad \text{for all}\qquad \vert s \vert \le \varepsilon,\\ &&\qquad c_{1}\vert s \vert \le \vert h(s) \vert \le c_{2}\vert s \vert \qquad \text{ for all}\qquad \vert s \vert \ge \varepsilon. \end{array} $$

(2.3)

In addition, we assume that the function H, defined by $H(s)=\sqrt {s}h_{0} (\sqrt {s})$, is a strictly convex C² function on (0, r₂], for some r₂ > 0, when h₀ is nonlinear.

Remark 2.1

It is worth noting that condition (2.3) was considered first in [20].

Remark 2.2

Hypothesis (A2) implies that sh(s) > 0, for all s≠ 0.

Remark 2.3

If G is a strictly increasing and strictly convex C² function on (0, r₁], with G(0) = G^′(0) = 0, then it has an extension $\overline {G}$, which is strictly increasing and strictly convex C² function on (0, ∞). For instance, if G(r₁) = a, G^′(r₁) = b, G^″(r₁) = c, we can define $\overline {G}$, for t > r₁, by

$$ \overline{G}(t)=\frac{c}{2} t^{2}+ (b-cr_{1})t+ \left( a+\frac{c}{2} {r_{1}}^{2}-b r_{1} \right). $$

(2.4)

The same remark can be established for $\overline {H}$.

For completeness we state, without proof, the existence result of [32].

Proposition 2.4

Let (u₀, u₁) ∈ V × L²(Ω) be given.Assume that (A1) and (A2) are satisfied, then the problem (1.1) has a unique global (weak) solution

$$u\in C(\mathbb{R}^{+};V)\cap C^{1}(\mathbb{R}^{+};L^{2}({\Omega}).$$

Moreover, if

$$(u_{0},u_{1})\in (H^{2}({\Omega})\cap V)\times V,$$

and satisfies the compatibility condition

$$\frac{\partial u_{0}}{\partial \nu}+h(u_{1})= 0\hspace{0.05in}\text{on}\hspace{0.05in}{\Gamma}_{1},$$

then the solution

$$u\in L^{\infty}(\mathbb{R}^{+};H^{2}({\Omega})\cap V)\cap W^{1,\infty}(\mathbb{R}^{+};V)\cap W^{2,\infty}(\mathbb{R}^{+};L^{2}({\Omega})).$$

We introduce the “modified” energy associated to problem (1.1):

$$ E(t)=\frac{1}{2} {\vert\vert{u_{t}(t)}\vert\vert}_{2}^{2} + \frac{1}{2}\left( 1-{{\int}_{0}^{t}}g(s)ds\right){\vert \vert\ {\nabla u(t)} \vert\vert}_{2}^{2}+\frac{1}{2}(g \circ \nabla u)(t), $$

(2.5)

where

$$(g \circ \nabla u)(t)={{\int}_{0}^{t}} g(t-s) {\vert\vert \nabla u(t)- \nabla u(s) \vert\vert}_{2}^{2}ds. $$

Direct differentiation, using Eq. 1.1, leads to

$$ E^{\prime}(t)=\frac{1}{2}(g^{\prime} \circ \nabla u)(t)-\frac{1}{2}g(t){\int}_{\Omega}\vert \nabla u\vert^{2} dx-{\int}_{{\Gamma}_{1}}u_{t}(t) h(u_{t}(t))d{\Gamma} \le0. $$

(2.6)

3 Technical Lemmas

In this section, we establish several lemmas needed for the proof of our main result.

Lemma 3.1

Under the assumptions (A1) and (A2), the functional

$$\psi_{1}(t):={\int}_{\Omega}u(t) u_{t}(t)dx$$

satisfies, along the solution of Eq. 1.1, the estimate

$$ \psi_{1}^{\prime}(t)\le -\frac{\ell}{2}{\vert \vert\ {\nabla u(t)} \vert\vert}_{2}^{2}+{\vert\vert{u_{t}(t)}\vert\vert}_{2}^{2}+ \frac{cC_{\alpha}}{2\ell}(k \circ \nabla u)(t)+c{\int}_{{\Gamma}_{1}} h^{2}(u_{t}(t))d{\Gamma}, \hspace{0.15in} \forall t \in \mathbb{R}^{+}, $$

(3.1)

where, for any 0 < α < 1,

$$ C_{\alpha}={\int}_{0}^{\infty}\frac{g^{2}(s)}{\alpha g(s)-g^{\prime}(s)}ds \hspace{0.15in}\text{ and } \hspace{0.15in}k(t)=\alpha g(t)-g^{\prime}(t). $$

(3.2)

Proof

Direct computations, using Eq. 1.1, yield

$$\begin{array}{@{}rcl@{}} &&\psi^{\prime}(t)={\int}_{\Omega}{u^{2}_{t}}dx+{\int}_{\Omega}u{\Delta} udx-{\int}_{\Omega}u{{\int}_{0}^{t}}g(t-s){\Delta} u(s)dsdx\\ &&\hspace{0.3in}={\int}_{\Omega}{u^{2}_{t}}dx-\left( 1-{{\int}_{0}^{t}}g(s)ds\right){\int}_{\Omega}{\vert \nabla u\vert}^{2}dx-{\int}_{{\Gamma}_{1}}uh(u_{t})d{\Gamma}\\ &&\hspace{0.4in}+{\int}_{\Omega}\nabla u.{{\int}_{0}^{t}}g(t-s)(\nabla u(s)-\nabla u(t))dsdx. \end{array} $$

(3.3)

Using Young’s and Cauchy Schwarz’ inequalities, we obtain

$$\begin{array}{@{}rcl@{}} &&{\int}_{\Omega}\nabla u.{{\int}_{0}^{t}}g(t-s)(\nabla u(s)-\nabla u(t))dsdx\\ &&\hspace{0.15in}\le \delta {\int}_{\Omega}{\vert \nabla u\vert}^{2}dx+\frac{1}{4\delta}{\int}_{\Omega}\left( {{\int}_{0}^{t}}g(t-s)\vert \nabla u(s)-\nabla u(t)\vert ds\right)^{2}dx\\ &&\hspace{0.15in}\le \delta {\int}_{\Omega}{\vert \nabla u\vert}^{2}dx+\\ &&\hspace{0.15in}\frac{1}{4 \delta}{\int}_{\Omega}\left( {{\int}_{0}^{t}}\frac{g(t-s)}{\sqrt{\alpha g(t-s)-g^{\prime}(t-s)}}\sqrt{\alpha g(t-s)-g^{\prime}(t-s)}\vert\nabla u(s)-\nabla u(t)\vert ds\right)^{2}dx\\ &&\hspace{0.15in}\le \delta {\int}_{\Omega}{\vert \nabla u\vert}^{2}dx+\\ &&\hspace{0.15in}\frac{1}{4 \delta}\left( {{\int}_{0}^{t}}\frac{g^{2}(s)}{\alpha g(s)-g^{\prime}(s)}ds\right){\int}_{\Omega}{{\int}_{0}^{t}}\left[\alpha g(t-s)-g^{\prime}(t-s)\right] \vert \nabla u(s)-\nabla u(t)\vert^{2} dsdx\\ &&\hspace{0.15in}\le\delta {\int}_{\Omega}{\vert \nabla u\vert}^{2}dx+\frac{1}{4\delta}C_{\alpha}(k \circ\nabla u)(t). \end{array} $$

(3.4)

Also, use of Young’s and Poincaré’s inequalities and the trace theorem gives

$$\begin{array}{@{}rcl@{}} -{\int}_{{\Gamma}_{1}}uh(u_{t})d{\Gamma} &\le& \delta {\int}_{{\Gamma}_{1}}u^{2}d{\Gamma}+\frac{1}{4\delta}{\int}_{{\Gamma}_{1}}h^{2}(u_{t})d{\Gamma}\\ &\le& c_{p}\delta {\int}_{\Omega}{\vert \nabla u\vert}^{2}d{\Gamma}+\frac{1}{4\delta}{\int}_{{\Gamma}_{1}}h^{2}(u_{t})d{\Gamma}. \end{array} $$

(3.5)

From Eqs. 3.4 and 3.5, we have

$$\begin{array}{@{}rcl@{}} &&{\int}_{\Omega}\nabla u.{{\int}_{0}^{t}}g(t-s)(\nabla u(s)-\nabla u(t))dsdx-{\int}_{{\Gamma}_{1}}uh(u_{t})d{\Gamma}\\ &&\hspace{0.5 in} \leq (1+c_{p})\delta {\int}_{\Omega}{\vert \nabla u\vert}^{2}dx+\frac{1}{4\delta}C_{\alpha}(k \circ \nabla u)(t)+\frac{1}{4\delta}{\int}_{{\Gamma}_{1}}h^{2}(u_{t})d{\Gamma}\\ &&\hspace{0.5in} \leq c\delta {\int}_{\Omega}{\vert \nabla u\vert}^{2}dx+\frac{1}{4\delta}C_{\alpha}(k \circ \nabla u)(t)+\frac{1}{4\delta}{\int}_{{\Gamma}_{1}}h^{2}(u_{t})d{\Gamma} \end{array} $$

(3.6)

Combining Eqs. 3.3 and 3.6 and choosing $\delta =\frac {\ell }{2c}$ leads to Eq. 3.1. □

Lemma 3.2

Under the assumptions (A1) and (A2), thefunctional

$$\psi_{2}(t):=-{\int}_{\Omega} u_{t}(t){{\int}_{0}^{t}}g(t-s)(u(t)-u(s))dsdx $$

satisfies, along the solution of Eq. 1.1, the estimate

$$\begin{array}{@{}rcl@{}} &\psi_{2}^{\prime}(t) \le \delta{\vert \vert\ {\nabla u(t)} \vert\vert}_{2}^{2}-\left( {{\int}_{0}^{t}}g(s)ds-\delta \right){\vert\vert{u_{t}(t)}\vert\vert}_{2}^{2}+\left( (\frac{3c}{\delta}+ 1)C_{\alpha}+\frac{c}{\delta}\right)(ko \nabla u)(t)\\ &\hspace{0.4in}+c{\int}_{{\Gamma}_{1}} h^{2}(u_{t}(t))d{\Gamma}, \hspace{0.2in} \forall t \in \mathbb{R}^{+}\hspace{0.05in}\text{ and }\hspace{0.08in}\forall \delta>0. \end{array} $$

(3.7)

Proof

By exploiting Eq. 1.1 and performing integration by parts, we arrive at

$$\begin{array}{@{}rcl@{}} \psi_{2}^{\prime}(t)&=&{\int}_{\Omega}\nabla u.{{\int}_{0}^{t}}g(t-s)(\nabla u(t)-\nabla u(s))dsdx\\ &&\hspace{0.3in}-{\int}_{\Omega}\left( {{\int}_{0}^{t}}g(t-s)\nabla u(s)ds\right).\left( {{\int}_{0}^{t}}g(t-s)(\nabla u(t)-\nabla u(s))ds\right)dx\\ &&\hspace{0.3in}+{\int}_{{\Gamma}_{1}}\left( {{\int}_{0}^{t}}g(t-s)(u(t)-u(s))ds\right)h(u_{t})d{\Gamma}\\ &&\hspace{0.3in}-{\int}_{\Omega}u_{t}{{\int}_{0}^{t}}g^{\prime}(t-s)(u(t)-u(s))dsdx-\left( {{\int}_{0}^{t}}g(s)ds\right){\int}_{\Omega}{u_{t}}^{2}dx\\ \hspace{0.25in}&=&\left( 1-{{\int}_{0}^{t}}g(s)ds\right){\int}_{\Omega}\nabla u.{{\int}_{0}^{t}}g(t-s)(\nabla u(t)-\nabla u(s))dsdx\\ &&\hspace{0.2in}+{\int}_{\Omega}{\left| {{\int}_{0}^{t}}g(t-s)(\nabla u(t)-\nabla u(s))ds\right|}^{2}dx\\ &&\hspace{0.3in}+{\int}_{{\Gamma}_{1}}\left( {{\int}_{0}^{t}}g(t-s)(u(t)-u(s))ds\right)h(u_{t})d{\Gamma}\\ &&\hspace{0.3in}-{\int}_{\Omega}u_{t}{{\int}_{0}^{t}}g^{\prime}(t-s)(u(t)-u(s))dsdx-\left( {{\int}_{0}^{t}}g(s)ds\right){\int}_{\Omega}{{u^{2}_{t}}}dx. \end{array} $$

Using Young’s inequality and performing similar calculations as in Eq. 3.4, we obtain

$$\left( 1 - {{\int}_{0}^{t}}g(s)ds\right) {\int}_{\Omega}\nabla u.{{\int}_{0}^{t}}g(t-s)(\nabla u(s)-\nabla u(t))dsdx\!\le \delta {\int}_{\Omega}{\vert \nabla u\vert}^{2}dx+\frac{c C_{\alpha}}{\delta}(k \circ \nabla u)(t)$$

and

$${\int}_{{\Gamma}_{1}}\left( {{\int}_{0}^{t}}g(t-s)(u(t)-u(s))ds\right)h(u_{t})d{\Gamma}\le \frac{c C_{\alpha}}{\delta} (k \circ \nabla u)(t)+ \delta {\int}_{{\Gamma}_{1}}h^{2}(u_{t})d{\Gamma}.$$

Also,

$$\begin{array}{@{}rcl@{}} &&-{\int}_{\Omega}u_{t}{{\int}_{0}^{t}}g^{\prime}(t-s)(u(t)-u(s))dsdx={\int}_{\Omega}u_{t} {{\int}_{0}^{t}}k(t-s)(u(t)-u(s))dsdx\\ &&-{\int}_{\Omega}u_{t} {{\int}_{0}^{t}}\alpha g(t-s)(u(t)-u(s))dsdx\le \delta {\int}_{\Omega}{u_{t}^{2}}dx\\ &&+\frac{1}{2\delta}{\int}_{\Omega}\left( {{\int}_{0}^{t}}\sqrt{k(t-s)} \sqrt{k(t-s)} \vert u(t)-u(s)\vert ds\right)^{2} dx\\ &&+\frac{\alpha^{2}}{2\delta}{\int}_{\Omega}\left( {{\int}_{0}^{t}}g(t-s)\vert u(t)-u(s)\vert ds\right)^{2} dx\le \delta {\int}_{\Omega}{u_{t}^{2}} dx\\ &&+\frac{\left( {{\int}_{0}^{t}}k(s)ds\right)}{2\delta} (k \circ u)(t)+\frac{\alpha^{2} C_{\alpha}}{2\delta}(k \circ \nabla u)(t)\le \delta {\int}_{\Omega}{u_{t}^{2}} dx\\ &&+\frac{c}{\delta}(k \circ \nabla u)(t)+\frac{c C_{\alpha}}{\delta} (k \circ \nabla u)(t). \end{array} $$

Combining all the above estimates, Eq. 3.7 is established. □

Lemma 3.3

Under the assumptions (A1) and (A2), the functional

$$ \psi_{3}(t)={\int}_{\Omega}{{\int}_{0}^{t}}r(t-s)\vert \nabla u(s)\vert^{2}dsdx, $$

(3.8)

satisfies, along the solution of Eq. 1.1, the estimate

$$ \psi_{3}^{\prime}(t)\le -\frac{1}{2}(g \circ \nabla u)(t)+ 3(1-\ell){\int}_{\Omega}\vert \nabla u(t)\vert^{2}dx. $$

(3.9)

where$r(t)={\int }_{t}^{+\infty }g(s)ds$.

Proof

By Young’s inequality and the fact that r^′(t) = −g(t), we see that

$$\begin{array}{@{}rcl@{}} &&\psi_{3}^{\prime}(t)=r(0){\int}_{\Omega}\vert \nabla u(t)\vert^{2}dx-{\int}_{\Omega}{{\int}_{0}^{t}}g(t-s)\vert \nabla u(s)\vert^{2} dx\\ &&\hspace{0.3in}=-{\int}_{\Omega}{{\int}_{0}^{t}}g(t-s)\vert \nabla u(s)-\nabla u(t)\vert^{2} ds dx\\ &&\hspace{0.3in}-2{\int}_{\Omega}\nabla u(t).{{\int}_{0}^{t}}g(t-s)(\nabla u(s)-\nabla u(t))dsdx+r(t){\int}_{\Omega}\vert \nabla u(t)\vert^{2} dx. \end{array} $$

Now,

$$\begin{array}{@{}rcl@{}} &&-2{\int}_{\Omega}\nabla u(t).{{\int}_{0}^{t}}g(t-s)(\nabla u(s)-\nabla u(t))dsdx\\ &&\hspace{0.3in}\le 2(1-\ell){\int}_{\Omega}\vert \nabla u(t)\vert^{2} dx+\frac{{{\int}_{0}^{t}}g(s)ds}{2(1-\ell)}{\int}_{\Omega}{{\int}_{0}^{t}}g(t-s)\vert \nabla u(s)-\nabla u(t)\vert^{2} dsdx. \end{array} $$

Using the facts that r(t) ≤ r(0) = 1 − ℓ and ${{\int }_{0}^{t}}g(s)ds \le 1-\ell$, Eq. 3.9 is established. □

Lemma 3.4

There exist positive constants d andt₁such that

$$ g^{\prime}(t)\le -d g(t),\hspace{0.1in}\forall t\in [0,t_{1}]. $$

(3.10)

Proof

By (A1), we easily deduce that $\lim _{t\to +\infty }g(t)= 0$. Hence, there is t₁ ≥ 0 large enough such that

$$g(t_{1})=r$$

and

$$g(t) \le r,\hspace{0.2in}\forall t\ge t_{1}.$$

As g and ξ are positive nonincreasing continuous and H is a positive continuous function, then, for all t ∈ [0, t₁],

$$\left\{\begin{array}{llll} 0<g(t_{1})\le g(t) \le g(0)\\ 0<\xi(t_{1})\le \xi(t) \le \xi(0), \end{array}\right. $$

which implies that there are two positive constants a and b such that

$$a \le \xi(t) H(g(t))\le b.$$

Consequently, for all t ∈ [0, t₁],

$$ g^{\prime}(t)\le -\xi(t) H(g(t))\le -\frac{a}{g(0)}g(0)\le -\frac{a}{g(0)}g(t). $$

(3.11)

□

Remark 3.5

Using the fact that $\frac {\alpha g^{2}(s)}{\alpha g(s)-g^{\prime }(s)} <g(s)$ and recalling the Lebesgue dominated convergence theorem, we can easily deduce that

$$ \alpha C_{\alpha}={\int}_{0}^{\infty}\frac{\alpha g^{2}(s)}{\alpha g(s)-g^{\prime}(s)}ds \to 0 \text{ as } \alpha \to 0. $$

(3.12)

Lemma 3.6

Assume that (A1) and (A2) hold. Then there exist constantsN, N₁, N₂, m, m₀, c > 0 such thatthe functional

$$L(t)=NE(t)+N_{1}{\psi_{1}} (t)+N_{2} {\psi_{2}}(t)+m_{0}E(t)$$

satisfies, for allt ≥ t₁,

$$ L^{\prime}(t) \le -mE(t)+c{\int}_{t_{1}}^{t}g(t-s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(s)\vert}^{2}dxds +c {\int}_{{\Gamma}_{1}} h^{2}(u_{t}(t))d {\Gamma} $$

(3.13)

Proof

By using Eqs. 2.6, 3.1 and 3.7, recalling that g^′ = (αg − k) and taking $\delta =\frac {\ell }{4N_{2}}$, we easily see that

$$\begin{array}{@{}rcl@{}} &&L^{\prime}(t) \le -\left( \frac{\ell}{2}N_{1}-\frac{\ell}{4}\right){\vert \vert\ {\nabla u} \vert\vert}_{2}^{2}-\left( N_{2}g_{1}-\frac{\ell}{4}-N_{1}\right){\vert\vert{u_{t}}\vert\vert}_{2}^{2}+\frac{\alpha}{2}N(g \circ \nabla u)(t)\\ &&\hspace{0.35in}-\left( \frac{1}{2}N-\frac{4c}{\ell}{N_{2}^{2}}-C_{\alpha}\left( \frac{c}{2 \ell}N_{1}+\frac{12c}{\ell}{N_{2}^{2}}+N_{2}\right)\right)(k \circ \nabla u)(t)]\\ &&\hspace{0.35in}+ c(N_{1}+N_{2}) {\int}_{{\Gamma}_{1}} h^{2}(u_{t})d {\Gamma} +m_{0}E^{\prime}(t). \end{array} $$

(3.14)

At this point, we choose N₁ large enough so that

$$\frac{\ell}{2}N_{1}-\frac{\ell}{4}>4(1-\ell)$$

and then N₂ large enough so that

$$N_{2} g_{1}-\frac{\ell}{4}-N_{1}-1 >0.$$

Now, using Remark 3.5, there is 0 < α₀ < 1 such that if α < α₀, then

$$ \alpha C_{\alpha}<\frac{1}{8\left( \frac{cN_{1}}{2\ell}+\frac{12c {N_{2}^{2}}}{\ell}+N_{2}\right)}. $$

(3.15)

Now, we choose N large enough and α so that

$$\frac{1}{4}N-\frac{4c}{{N^{2}_{2}}}>0\text{ and } \alpha=\frac{1}{2N}<\alpha_{0},$$

which gives

$$\frac{1}{2}N-\frac{4c}{\ell}{N_{2}^{2}}-C_{\alpha}\left( \frac{c}{2\ell}N_{1}+\frac{12c}{\ell}{N_{2}^{2}}+N_{2}\right)>0.$$

Therefore, we arrive at

$$ L^{\prime}(t) \le -4(1-\ell){\vert \vert\ {\nabla u} \vert\vert}_{2}^{2}- {\vert\vert{u_{t}}\vert\vert}_{2}^{2}+\frac{1}{4}(g \circ \nabla u)(t)+ c {\int}_{{\Gamma}_{1}} h^{2}(u_{t})d {\Gamma}+m_{0}E^{\prime}(t). $$

(3.16)

Using Eqs. 2.6 and 3.10 we conclude that, for any t ≥ t₁,

$$\begin{array}{@{}rcl@{}} &&{\int}_{0}^{t_{1}}g(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\le \frac{-1}{d}{\int}_{0}^{t_{1}}g^{\prime}(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\\ &&\hspace{2.65in}\le -cE^{\prime}(t) \end{array} $$

(3.17)

Combining Eqs. 3.16 and 3.17 and selecting a suitable choice of m₀, Eq. 3.13 is established. On the other hand (see [39]), we can choose N even larger (if needed) so that

$$ L \sim E. $$

(3.18)

□

4 Stability

In this section, we state and prove the main result of our work. For this purpose, we have the following lemmas and remarks.

Lemma 4.1

Under the assumptions (A1) and (A2), the solution of Eq. 1.1satisfies the estimates

$$ {\int}_{{\Gamma}_{1}}h^{2}(u_{t})d{\Gamma} \le c {\int}_{{\Gamma}_{1}}u_{t} h(u_{t})d{\Gamma},\hspace{0.3in}\text{ if } h_{0} \text{ is linear} $$

(4.1)

$$ {\int}_{{\Gamma}_{1}}h^{2}(u_{t})d{\Gamma} \le cH^{-1}(J(t))-cE^{\prime}(t),\hspace{0.3in}\text{ if } h_{0} \text{ is nonlinear} $$

(4.2)

where

$$ J(t):=\frac{1}{\vert{\Gamma}_{12}\vert}{\int}_{{\Gamma}_{12}}u_{t}(t)h(u_{t}(t))d{\Gamma} \le -cE^{\prime}(t) $$

(4.3)

and

$${\Gamma}_{12}=\{x\in {\Gamma}_{1}:\vert u_{t}(t) \vert \le \varepsilon_{1} \}.$$

Proof

Case 1::

h₀ is linear. Then, using (A2) we have

$$c^{\prime}_{1} \vert u_{t} \vert \le \vert h(u_{t}) \vert \le c_{2}^{\prime} \vert u_{t} \vert,$$

and hence

$$ h^{2}(u_{t}) \le c_{2}^{\prime} u_{t}h(u_{t}), $$

(4.4)

So, Eq. 4.1 is established.

Case 2::

h₀ is nonlinear on [0, ε].

We establish this case, borrowing some ideas from [20]. So, we first assume that max{r₂, h₀(r₂)} < ε; otherwise we take r₂ smaller. Let ε₁ = min{r₂, h₀(r₂)}. Using (A2), we have, for ε₁ ≤|s|≤ ε,

$$\vert h(s) \vert \le \frac{h_{0}^{-1}(\vert s \vert)}{\vert s \vert}\vert s \vert \le \frac{h_{0}^{-1}(\vert \varepsilon \vert)}{\vert \varepsilon_{1} \vert}\vert s \vert $$

and

$$\vert h(s) \vert \ge \frac{h_{0}(\vert s \vert)}{\vert s \vert}\vert s \vert \ge \frac{h_{0}(\vert \varepsilon_{1} \vert)}{\vert \varepsilon \vert}\vert s \vert $$

So, we deduce that

$$ \left\{\begin{array}{llll} h_{0}(\vert s \vert) \le \vert h(s)\vert \le h^{-1}_{0}(\vert s \vert)\hspace{0.1in}\text{for all}\hspace{0.08in} \vert s \vert < \varepsilon_{1}\\ c^{\prime}_{1} \vert s \vert \le \vert h(s) \vert \le c^{\prime}_{2} \vert s \vert \hspace{0.1in}\text{for all}\hspace{0.08in} \vert s \vert \ge \varepsilon_{1} \end{array}\right. $$

(4.5)

Then Eq. 4.5, yields, for all |s|≤ ε₁,

$$H(h^{2}(s))=\vert h(s) \vert h_{0}(\vert h(s) \vert) \le sh(s)$$

which gives

$$ h^{2}(s) \le H^{-1}(sh(s))\hspace{0.1in}\text{for all}\hspace{0.08in} \vert s \vert \le\varepsilon_{1}. $$

(4.6)

Now, we define the following partition which was first introduced by Komornik [40]:

$${\Gamma}_{11}=\{x\in {\Gamma}_{1}:\vert u_{t}(t) \vert > \varepsilon_{1} \},\hspace{0.2in}{\Gamma}_{12}=\{x\in {\Gamma}_{1}:\vert u_{t}(t) \vert \le \varepsilon_{1} \}$$

Using Eq. 4.5, we get on Γ₁₂

$$ u_{t}h(u_{t}(t)) \le\varepsilon_{1} h_{0}^{-1}(\varepsilon_{1}) \le h_{0}(r_{2})r_{2}=H({r_{2}^{2}}). $$

(4.7)

Then, Jensen’s inequality gives (note that H^− 1 is concave)

$$ H^{-1}\left( J(t)\right)\ge c {\int}_{{\Gamma}_{12}}H^{-1}(u_{t}(t)h(u_{t}(t)))d{\Gamma}. $$

(4.8)

Thus, combining Eqs. 4.6 and 4.8, we arrive at

$$\begin{array}{@{}rcl@{}} &&{\int}_{{\Gamma}_{1}}h^{2}(u_{t}(t))d{\Gamma}={\int}_{{\Gamma}_{12}}h^{2}(u_{t}(t))d{\Gamma}+{\int}_{{\Gamma}_{11}}h^{2}(u_{t}(t))d{\Gamma}\\ &&\hspace{1.1in}\le {\int}_{{\Gamma}_{12}}H^{-1}\left( u_{t}h(u_{t}(t))\right)d{\Gamma}+{\int}_{{\Gamma}_{11}}h^{2}(u_{t}(t))d{\Gamma}\\ &&\hspace{1.1in}\le cH^{-1}(J(t))-cE^{\prime}(t) \end{array} $$

(4.9)

□

Lemma 4.2

Assume that (A1) and(A2) hold andh₀is linear. Then, the energy functional satisfies the following estimate

$$ {\int}_{0}^{+\infty}E(s)ds < \infty $$

(4.10)

Proof

Let F(t) = L(t) + ψ₃(t), then using Eqs. 3.9 and 3.16, we obtain

$$ F^{\prime}(t)\le -(1-\ell){\int}_{\Omega}\vert \nabla u\vert dx-{\int}_{\Omega}{u_{t}^{2}} dx-\frac{1}{4}(go\nabla u)(t)+ c {\int}_{{\Gamma}_{1}} h^{2}(u_{t})d {\Gamma} $$

(4.11)

Using Eqs. 2.6, 4.1 and 4.11, we obtain

$$\begin{array}{@{}rcl@{}} &&F^{\prime}(t)\le -b E(t)+c{\int}_{\Omega}u_{t} h(u_{t})dx\\ &&\hspace{0.3in}\le -bE(t)-cE^{\prime}(t), \end{array} $$

where b is some positive constant. Therefore,

$$ b{\int}_{t_{1}}^{t}E(s)ds\le F_{1}(t_{1})-F_{1}(t)\le F_{1}(t_{1})<\infty, $$

(4.12)

where F₁(t) = F(t) + cE(t) ∼ E. □

Let’s define

$$ I(t):=-{\int}_{t_{1}}^{t}g^{\prime}(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\le -cE^{\prime}(t), $$

(4.13)

Lemma 4.3

Under the assumptions (A1) and (A2), we have the following estimates

$$ {\int}_{t_{1}}^{t}g(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\le \frac{1}{q} \overline{G}^{-1}\left( \frac{qI(t)}{\xi(t)}\right),\hspace{0.1in}\text{if } h_{0} \text{ is linear} $$

(4.14)

$$ {\int}_{t_{1}}^{t}g(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\le \frac{(t-t_{1})}{q} \overline{G}^{-1}\left( \frac{qI(t)}{(t-t_{1})\xi(t)}\right),\hspace{0.1in}\text{if } h_{0} \text{ is nonlinear} $$

(4.15)

whereq ∈ (0, 1) and$\overline {G}$is an extension ofGsuch that$\overline {G}$is strictly increasing and strictly convexC²function on (0, ∞); see Remark 2.3.

Proof

First we establish (4.14). For this, we define the following quantity

$$\lambda (t):=q{\int}_{t_{1}}^{t}{\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds,$$

where, by Eq. 4.10, q is chosen so small that, for all t ≥ t₁,

$$ \lambda (t)<1. $$

(4.16)

Since G is strictly convex on (0, r₁] and G(0) = 0, then

$$ G(\theta z)\le \theta G(z),\text{ } 0\le \theta \le 1\text{ and } z\in (0,r_{1}]. $$

(4.17)

The use of Eqs. 2.2, 4.16, and 4.17 and Jensen’s inequality leads to

$$\begin{array}{@{}rcl@{}} &&I(t)=\frac{1}{q\lambda(t)}{\int}_{t_{1}}^{t}\lambda (t)(-g^{\prime}(s)){\int}_{\Omega}{q\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\\ &&\hspace{0.3in}\ge \frac{1}{q\lambda(t)}{\int}_{t_{1}}^{t}\lambda (t)\xi(s) G(g(s)){\int}_{\Omega}{q\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\\ &&\hspace{0.3in}\ge \frac{\xi(t)}{q\lambda(t)}{\int}_{t_{1}}^{t}G(\lambda (t)g(s)){\int}_{\Omega}{q\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\\ &&\hspace{0.3in}\ge \frac{\xi(t)}{q}G\left( q{\int}_{t_{1}}^{t}g(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\right)\\ &&\hspace{0.3in}=\frac{\xi(t)}{q}\overline{G}\left( q{\int}_{t_{1}}^{t}g(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(s)\vert}^{2}dxds\right) \end{array} $$

(4.18)

This gives (4.14).

For the proof of (4.15), we define the following

$$\lambda_{1} (t):=\frac{q}{(t-t_{1})}{\int}_{t_{1}}^{t}{\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds,$$

then using (2.5) and (2.6), we easily see that

$$\lambda_{1}(t)\le \frac{8q E(0)}{\ell},$$

then choosing q ∈ (0, 1) small enough so that, for all t ≥ t₁,

$$ \lambda_{1} (t)<1. $$

(4.19)

The use of Eqs. 2.2, 4.17 and 4.19 and Jensen’s inequality leads to

$$\begin{array}{@{}rcl@{}} &&I(t)=\frac{1}{q\lambda_{1}(t)}{\int}_{t_{1}}^{t}\lambda_{1} (t)(-g^{\prime}(s)){\int}_{\Omega}{q\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\\ &&\hspace{0.3in}\ge \frac{1}{q\lambda_{1}(t)}{\int}_{t_{1}}^{t}\lambda_{1} (t)\xi(s) G(g(s)){\int}_{\Omega}{q\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\\ &&\hspace{0.3in}\ge \frac{\xi(t)}{q\lambda_{1}(t)}{\int}_{t_{1}}^{t}G(\lambda_{1} (t)g(s)){\int}_{\Omega}{q\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\\ &&\hspace{0.3in}\ge \frac{(t-t_{1})\xi(t)}{q} G\left( \frac{q}{(t-t_{1})}{\int}_{t_{1}}^{t}g(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\right)\\ &&\hspace{0.3in}=\frac{(t-t_{1})\xi(t)}{q}\overline{G}\left( \frac{q}{(t-t_{1})} {\int}_{t_{1}}^{t}g(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\right). \end{array} $$

(4.20)

This implies that

$${\int}_{t_{1}}^{t}g(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(t-s)\vert}^{2}dxds\le \frac{(t-t_{1})}{q} \overline{G}^{-1}\left( \frac{qI(t)}{(t-t_{1})\xi(t)}\right)$$

□

Theorem 4.4

Let (u₀, u₁) ∈ V × L²(Ω) be given. Assume that (A1) and (A2) are satisfied andh₀is linear. Then there exist strictly positive constantsc₁, c₂, k₁andk₂such that the solution of Eq. 1.1satisfies, for allt ≥ t₁,

$$ E (t)\le c_{1} e^{-c_{2}{\int}_{t_{1}}^{t}\xi(s)ds},\text{ if \textit{G} is linear} $$

(4.21)

$$ E (t)\le k_{2} G_{1}^{-1}\left( k_{1} {\int}_{t_{1}}^{t}\xi(s)ds\right),\text{ if } G \text{ is nonlinear,} $$

(4.22)

where$G_{1}(t)={\int }_{t}^{r_{1}}\frac {1}{sG^{\prime }(s)}ds$.

Proof

Case 1::

G is linear

Multiplying (3.13) by ξ(t) and using Eqs. 2.2, 2.6, 4.1, 4.3 and 4.13, we get

$$\begin{array}{@{}rcl@{}} &&\xi(t) L^{\prime}(t) \le\\ &&-m \xi(t) E(t)+c \xi(t) {\int}_{t_{1}}^{t}g(t-s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(s)\vert}^{2}dxds+c \xi(t) {\int}_{{\Gamma}_{1}} h^{2}(u_{t}(t))d {\Gamma}\\ & &\le -m \xi(t) E(t)+c {\int}_{t_{1}}^{t}\xi(s) g(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(s)\vert}^{2}dxds +c \xi(t) {\int}_{{\Gamma}_{1}} h^{2}(u_{t}(t))d {\Gamma}\\ & &\le -m \xi(t) E(t)-c {\int}_{t_{1}}^{t} g^{\prime}(s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(s)\vert}^{2}dxds +c \xi(t) {\int}_{{\Gamma}_{1}} u_{t}h(u_{t}(t))d {\Gamma}\\ &&\le -m \xi(t) E(t)-2c E^{\prime}(t) \end{array} $$

which gives, as ξ(t) is non-increasing,

$$ (\xi L + 2c E)^{\prime} \leq -m \xi(t) E(t), \forall t \geq t_{1}. $$

(4.23)

Hence, using the fact that ξL + 2cE ∼ E, we easily obtain

$$ E (t)\le c^{\prime} e^{-\bar{c}{\int}_{t_{1}}^{t} \xi (s) ds}. $$

(4.24)

Case 2::

G is non-linear.

Using Eqs. 3.13, 4.1 and 4.14, we obtain

$$ L^{\prime}(t)\le -m E(t)+c\left( \overline{G}\right)^{-1}\left( \frac{qI(t)}{\xi(t)}\right)-cE^{\prime}(t),\hspace{0.1in} $$

(4.25)

Let $\mathcal {F}_{1}(t)=L(t)+cE(t)\sim E$, then Eq. 4.25 becomes

$$ \mathcal{F}_{1}^{\prime}(t)\le -m E(t)+c\left( \overline{G}\right)^{-1}\left( \frac{qI(t)}{\xi(t)}\right),\hspace{0.1in} $$

(4.26)

we find that the functional $\mathcal {F}_{2}$, defined by

$$\mathcal{F}_{2}(t):=\overline{G}^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)\mathcal{F}_{1}(t)$$

satisfies, for some α₁, α₂ > 0.

$$ \alpha_{1}\mathcal{F}_{2}(t)\le E(t)\le \alpha_{2}\mathcal{F}_{2}(t) $$

(4.27)

and

$$\begin{array}{@{}rcl@{}} \mathcal{F}_{2}^{\prime}(t)&=&\varepsilon_{0}\frac{E^{\prime}(t)}{E(0)}\overline{G}^{\prime\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)\mathcal{F}_{1}(t)+\overline{G}^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right){\mathcal{F}_{1}}^{\prime}(t)\\ \hspace{0.3in}&\le& -m E(t)\overline{G}^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)+c \overline{G}^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)\overline{G}^{-1}\left( \frac{qI(t)}{\xi(t)}\right). \end{array} $$

(4.28)

Let $\overline {G}^{*}$ be the convex conjugate of $\overline {G}$ in the sense of Young (see [41]), then

$$ \overline{G}^{*}(s)=s(\overline{G}^{\prime})^{-1}(s)-\overline{G}\left[(\overline{G}^{\prime})^{-1}(s)\right],\hspace{0.1in}\text{if}\hspace{0.05in}s\in (0,\overline{G}^{\prime}(r_{1})] $$

(4.29)

and $\overline {G}^{*}$ satisfies the following generalized Young inequality

$$ A B\le \overline{G}^{*}(A)+\overline{G}(B),\hspace{0.15in}\text{if}\hspace{0.05in}A\in (0,\overline{G}^{\prime}(r_{1})],\hspace{0.05in}B\in(0,r_{1}]. $$

(4.30)

So, with $A=\overline {G}^{\prime }\left (\varepsilon _{0}\frac {E^{\prime }(t)}{E(0)}\right )$ and $B=\overline {G}^{-1}\left (\frac {qI(t)}{\xi (t)}\right )$ and using Eqs. 2.6 and 4.28–4.30, we arrive at

$$\begin{array}{@{}rcl@{}} \mathcal{F}_{2}^{\prime}(t)&\le& -m E(t)\overline{G}^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)+c \overline{G}^{*}\left( \overline{G}^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)\right)+c \left( \frac{qI(t)}{\xi(t)}\right)\\ \hspace{0.3in}&\le& -m E(t)\overline{G}^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)+c\varepsilon_{0}\frac{E(t)}{E(0)}\overline{G}^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)+c\left( \frac{qI(t)}{\xi(t)}\right). \end{array} $$

(4.31)

So, multiplying Eq. 4.31 by ξ(t) and using the fact that $\varepsilon _{0}\frac {E(t)}{E(0)}<r_{1}$, $\overline {G}^{\prime }\left (\varepsilon _{0}\frac {E(t)}{E(0)}\right )=G^{\prime }\left (\varepsilon _{0}\frac {E(t)}{E(0)}\right )$, gives

$$\begin{array}{@{}rcl@{}} &&\xi(t) \mathcal{F}_{2}^{\prime}(t)\le -m \xi(t) E(t)G^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)+c \xi(t)\varepsilon_{0}\frac{E(t)}{E(0)}G^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)+c q I(t)\\ &&\hspace{0.3in}\le-m \xi(t) E(t)G^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)+c \xi(t)\varepsilon_{0}\frac{E(t)}{E(0)}G^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)-c E^{\prime}(t) \end{array} $$

Consequently, with a suitable choice of ε₀, we obtain, for all t ≥ t₁,

$$ \mathcal{F}_{3}^{\prime}(t)\le -k \xi(t)\left( \frac{E(t)}{E(0)}\right)G^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)=-k\xi(t) G_{2}\left( \frac{E(t)}{E(0)}\right), $$

(4.32)

where $\mathcal {F}_{3}=\xi \mathcal {F}_{2}+c E \sim E$ and G₂(t) = tG^′(ε₀t). Since $G^{\prime }_{2}(t)=G^{\prime }(\varepsilon _{0}t)+\varepsilon _{0}t G^{\prime \prime }(\varepsilon _{0}t)$, then, using the strict convexity of G on (0, r₁], we find that $G_{2}^{\prime }(t), G_{2}(t)>0$ on (0, 1]. Thus, with

$$R(t)=\varepsilon \frac{\alpha_{1}\mathcal{F}_{3}(t)}{E(0)},\hspace{0.1in}0<\varepsilon<1, $$

taking in account (4.27) and (4.32), we have

$$ R(t)\sim E(t) $$

(4.33)

and, for some k₁ > 0.

$$R^{\prime}(t)\le -k_{1}\xi(t) G_{2}(R(t)),\hspace{0.1in}\forall t\ge t_{1}. $$

Then, the integration over (t₁, t) yields

$${\int}_{t_{1}}^{t}\frac{-R^{\prime}(s)}{G_{2}(R(s))}ds \ge k_{1}{\int}_{t_{1}}^{t}\xi(s)ds. $$

Hence, by an approprite change of variable, we get

$${\int}_{\varepsilon_{0} R(t)}^{\varepsilon_{0} R(t_{1})}\frac{1}{\tau G^{\prime}(\tau)}d\tau \ge k_{1} {\int}_{t_{1}}^{t} \xi(s)ds $$

Thus, we have

$$ R(t)\le \frac{1}{\varepsilon_{0}}G_{1}^{-1}\left( k_{1}{\int}_{t_{1}}^{t}\xi(s)ds\right), $$

(4.34)

where $G_{1}(t)={\int }_{t}^{r_{1}}\frac {1}{sG^{\prime }(s)}ds$. Here, we have used the fact that G₁ is strictly decreasing on (0, r₁]. Therefore Eq. 4.22 is established by virtue of Eq. 4.33. □

Remark 4.5

The decay rate of E(t) given by Eq. 2.2 is optimal because it is consistent with the decay rate of g(t) given by Eq. 4.22. In fact,

$$g(t)\le G_{0}^{-1}\left( {\int}_{g^{-1}(r_{1})}^{t}\xi(s)ds\right),\hspace{0.1in}\forall t \ge g^{-1}(r_{1}),$$

where $G_{0}(t)={{\int }_{t}^{r}}\frac {1}{G(s)}$. Using the properties of G, G₀ and G₁, we can see that

$$G_{1}(t)={\int}_{t}^{r_{1}}\frac{1}{s G^{\prime}(s)}ds\le {\int}_{t}^{r_{1}}\frac{1}{G(s)}ds=G_{0}(t). $$

This implies

$$G_{1}^{-1}(t)\le G_{0}^{-1}(t).$$

This shows that Eq. 4.22 provides the best decay rates expected under the very general assumption (2.2).

Theorem 4.6

Let (u₀, u₁) ∈ V × L²(Ω) be given. Assume that (A1) and (A2) are satisfied andh₀is nonlinear. Then there exist strictly positive constantsc₃, c₄, k₂, k₃andε₂such that the solution of Eq. 1.1satisfies, for allt ≥ t₁,

$$ E(t)\le H_{1}^{-1}\left( c_{3}{\int}_{t_{1}}^{t}\xi(s)ds+c_{4}\right),\text{ if \textit{G} is linear,} $$

(4.35)

where$H_{1}(t)={{\int }_{t}^{1}}\frac {1}{H_{2}(s)}ds$.

$$ E(t) \leq k_{3} (t-t_{1}) {W_{2}}^{-1} \left( \frac{k_{2}}{(t-t_{1}){\int}_{t_{1}}^{t}\xi(s) ds} \right), \text{ if \textit{G} is non-linear,} $$

(4.36)

whereW₂(t) = tW^′(ε₂t) and$W=\left (\overline {G}^{-1}+\overline {H}^{-1}\right )^{-1}$.

Proof

Case 1::

G is linear

Multiplying Eq. 3.13 by ξ(t) and using Eq. 4.2, we get

$$\begin{array}{@{}rcl@{}} &&\xi(t) L^{\prime}(t) \le -m \xi(t) E(t)+c \xi(t) {\int}_{t_{1}}^{t}g(t-s){\int}_{\Omega}{\vert \nabla u(t)-\nabla u(s)\vert}^{2}dxds \\ &&\hspace{0.95in}+c \xi(t) {\int}_{{\Gamma}_{1}} h^{2}(u_{t}(t))d {\Gamma}\\ &&\hspace{0.3in} \le -m \xi(t) E(t)-cE^{\prime}(t)+c \xi(t) {\int}_{{\Gamma}_{1}} h^{2}(u_{t}(t))d {\Gamma}\\ &&\hspace{0.3in} \le -m \xi(t) E(t)-c E^{\prime}(t) + c \xi(t) H^{-1}(J(t)) -c \xi(t) E^{\prime}(t)\\ &&\hspace{0.3in} \le -m \xi(t) E(t)-c E^{\prime}(t) + c \xi(t) H^{-1}(J(t)) -c \xi(0) E^{\prime}(t)\\ &&\hspace{0.3in} \le -m \xi(t) E(t)-c E^{\prime}(t) + c \xi(t) H^{-1}(J(t)) \end{array} $$

which gives, as ξ(t) is non-increasing,

$$ (\xi L +c E)^{\prime} \leq -m \xi(t) E(t)+ c \xi(t) H^{-1}(J(t)), \forall t \geq t_{1}. $$

(4.37)

Therefore, Eq. 4.37 becomes

$$ {\mathcal{L}}^{\prime}(t) \leq -m \xi(t) E(t)+ c \xi(t) H^{-1}(J(t)), \forall t \geq t_{1}, $$

(4.38)

where $\mathcal {L}:=\xi L + 2c E$, which is clearly equivalent to E. Now, for ε₁ < r₂ and c₀ > 0, using Eq. 4.38 and the fact that E^′≤ 0, H^′ > 0, H^″ > 0 on (0, r₂], we find that the functional $\mathcal {L}_{1}$, defined by

$$\mathcal{L}_{1}(t):=H^{\prime} \left( \varepsilon_{1}\frac{E(t)}{E(0)}\right)\mathcal{L}(t)+c_{0}E(t)$$

satisfies, for some α₃, α₄ > 0.

$$ \alpha_{3} \mathcal{L}_{1}(t)\le E(t)\le \alpha_{4}\mathcal{L}_{1}(t) $$

(4.39)

and

$$\begin{array}{@{}rcl@{}} \mathcal{L}_{1}^{\prime}(t)&=&\varepsilon_{0}\frac{E^{\prime}(t)}{E(0)}H^{\prime\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right) \mathcal{L}(t)+ H^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right){\mathcal{L}}^{\prime}(t)+c_{0}E^{\prime}(t)\\ \hspace{0.3in}&\le& \!-m E(t)H^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)\!+c \xi(t) H^{\prime}\left( \varepsilon_{0}\frac{E(t)}{E(0)}\right)H^{-1}(J(t))+c_{0}E^{\prime}(t) \end{array} $$

(4.40)

Let H^∗ be the convex conjugate of H in the sense of Young (see [41]), then, as in Eqs. 4.29 and 4.30, with $A=H^{\prime }\left (\varepsilon _{1}\frac {E(t)}{E(0)}\right )$ and B = H^− 1(J(t)), using Eqs. 2.6 and 4.7, we arrive at

$$\begin{array}{@{}rcl@{}} \mathcal{L}_{1}^{\prime}(t)&\le& -m E(t)H^{\prime}\left( \varepsilon_{1}\frac{E(t)}{E(0)}\right)+c \xi(t) H^{*}\left( H^{\prime}\left( \varepsilon_{1}\frac{E(t)}{E(0)}\right)\right)+c \xi(t) J(t)+c_{0}E^{\prime}(t)\\ &\le& -m E(t)H^{\prime}\left( \varepsilon_{1}\frac{E(t)}{E(0)}\right)+c\varepsilon_{1} \xi(t) \frac{E(t)}{E(0)}H^{\prime}\left( \varepsilon_{1}\frac{E(t)}{E(0)}\right)-c E^{\prime}(t)+c_{0} E^{\prime}(t) \end{array} $$

Consequently, with a suitable choice of ε₁ and c₀, we obtain, for all t ≥ t₁,

$$ \mathcal{L}_{1}^{\prime}(t)\le -c \xi(t) \frac{E^{\prime}(t)}{E(0)}H^{\prime}\left( \varepsilon_{1}\frac{E(t)}{E(0)}\right)=-c \xi(t) H_{2}\left( \varepsilon_{1}\frac{E(t)}{E(0)}\right), $$

(4.41)

where H₂(t) = tH^′(ε₁t). Since $H^{\prime }_{2}(t)=H^{\prime }(\varepsilon _{1}t)+\varepsilon _{1}t H^{\prime \prime }(\varepsilon _{1}t)$, then, using the strict convexity of H on (0, r₂], we find that $H_{2}^{\prime }(t), H_{2}(t)>0$ on (0, 1]. Thus, with

$$R_{1}(t)=\varepsilon \frac{\alpha_{3}{\mathcal{L}_{1}}(t)}{E(0)},\hspace{0.1in}0<\varepsilon<1, $$

taking in account (4.39) and (4.41), we have

$$ R_{1}(t)\sim E(t) $$

(4.42)

and, for some c₃ > 0.

$$R_{1}^{\prime}(t)\le -c_{3} \xi(t) H_{2}(R_{1}(t)),\hspace{0.1in}\forall t\ge t_{1}.$$

Then, a simple integration gives, for some c₄ > 0,

$$ R_{1}(t)\le H_{1}^{-1}(c_{3}{\int}_{t_{1}}^{t} \xi(s) ds +c_{4}),\hspace{0.1in}\forall t\ge t_{1}, $$

(4.43)

where $H_{1}(t)={{\int }_{t}^{1}}\frac {1}{H_{2}(s)}ds$.

Case 2. :

G is non-linear.

Using Eqs. 3.13, 4.2 and 4.15, we obtain

$$ L^{\prime}(t)\le -m E(t)+c(t-t_{1})\left( \overline{G}\right)^{-1}\left( \frac{qI_{1}(t)}{(t-t_{1})\xi(t)}\right)+c H^{-1}(J(t))-cE^{\prime}(t). $$

(4.44)

Since $\lim _{t\to +\infty } \frac {1}{t-t_{1}}= 0$, there exists t₂ > t₁ such that $\frac {1}{t-t_{1}} < 1$ whenever t > t₂. Combining this with the strictly increasing and strictly convex properties of $\overline {H}$, setting $\theta =\frac {1}{t-t_{1}} < 1$ and using Eq. 4.17, we obtain

$$\overline{H}^{-1}(J(t)) \leq (t-t_{1})\overline{H}^{-1}\left( \frac{J(t)}{(t-t_{1})}\right), \forall t \ge t_{2}$$

and, then, Eq. 4.44 becomes

$$\begin{array}{@{}rcl@{}} &&L^{\prime}(t)\le -m E(t)+c(t-t_{1})\left( \overline{G}\right)^{-1}\left( \frac{qI_{1}(t)}{(t-t_{1})\xi(t)}\right)+c(t-t_{1})\overline{H}^{-1}\left( \frac{J(t)}{(t-t_{1})}\right)\\ &&\hspace{1.5in}-cE^{\prime}(t),\hspace{0.5in}\forall t\ge t_{2}. \end{array} $$

(4.45)

Let L₁(t) = L(t) + cE(t) ∼ E, then Eq. 4.45 takes the form

$$ L_{1}^{\prime}(t)\le -m E(t)+c(t-t_{1})\left( \overline{G}\right)^{-1}\left( \frac{qI_{1}(t)}{(t-t_{1})\xi(t)}\right)+c(t-t_{1})\overline{H}^{-1}\left( \frac{J(t)}{(t-t_{1})})\right),\hspace{0.1in} $$

(4.46)

$$\text{Let } r_{0}=\min{\{r_{1},r_{2}\}},\chi(t)=\max{\{\frac{qI_{1}(t)}{(t-t_{1})\xi(t)},\frac{J(t)}{(t-t_{1})}}\}\text{ and } W=\left( \left( \overline{G}\right)^{-1}+\overline{H}^{-1}\right)^{-1}. $$

So, Eq. 4.46 reduces to

$$ L_{1}^{\prime}(t)\le -m E(t)+c(t-t_{1}) W^{-1}(\chi(t)),\forall t\ge t_{2}\hspace{0.1in} $$

(4.47)

Now, for ε₂ < r₀ and using Eq. 4.44 and the fact that E^′≤ 0, W^′ > 0, W^″ > 0 on (0, r₀], we find that the functional L₂, defined by

$$L_{2}(t):=W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right)L_{1}(t), \hspace{0.1in}\forall t\ge t_{2}, $$

satisfies, for some α₅, α₆ > 0.

$$ \alpha_{5} L_{2}(t)\le E(t)\le \alpha_{6}L_{2}(t) $$

(4.48)

and, for all t ≥ t₂,

$$\begin{array}{@{}rcl@{}} L_{2}^{\prime}(t)&=&\left( \frac{-\varepsilon_{2}}{(t-t_{1})^{2}} \frac{E(t)}{E(0)}+\frac{\varepsilon_{2}}{(t-t_{1})} \frac{E^{\prime}(t)}{E(0)}\right)W^{\prime\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot\frac{E(t)}{E(0)}\right){L_{1}}(t)\\ &&+W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot\frac{E(t)}{E(0)}\right){L}^{\prime}_{1}(t)\le -m E(t)W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot\frac{E(t)}{E(0)}\right)\\ &&+c(t-t_{1})W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot\frac{E(t)}{E(0)}\right)W^{-1}(\chi(t)). \end{array} $$

(4.49)

Let W^∗ be the convex conjugate of W in the sense of Young (see [41]), then, as in Eqs. 4.29 and 4.30, and with $A=W^{\prime }\left (\frac {\varepsilon _{2}}{t-t_{1}} \cdot \frac {E(t)}{E(0)}\right )$ and B = W^− 1(χ(t)), using (2.6), we arrive at

$$\begin{array}{@{}rcl@{}} &&L_{2}^{\prime}(t)\le -m E(t)W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot\frac{E(t)}{E(0)}\right)+c (t-t_{1})W^{*}\left( W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot\frac{E(t)}{E(0)}\right)\right)\\ &&\hspace{0.9in}+c (t-t_{1})\chi(t)\\ &&\hspace{0.3in}\le -m E(t)W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right)+c(t-t_{1})\frac{\varepsilon_{2}}{t-t_{1}} \cdot\frac{E(t)}{E(0)}W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot\frac{E(t)}{E(0)}\right)\\ &&\hspace{0.9in}+c(t-t_{1})\chi(t). \end{array} $$

(4.50)

Using Eqs. 4.3 and 4.13, we observe that

$$\begin{array}{@{}rcl@{}} (t-t_{1})\xi(t) \chi(t)&\le& qI(t)+\xi(t)J(t)\\ &\le& qI(t)+\xi(0)J(t)\\ &\le& -cE^{\prime}(t)-cE^{\prime}(t)\\ &\le& -cE^{\prime}(t) \end{array} $$

So, multiplying Eq. 4.50 by ξ(t) and using the fact that, $\varepsilon _{2}\frac {E(t)}{E(0)}<r_{0}$, give

$$\begin{array}{@{}rcl@{}} \xi(t)L_{2}^{\prime}(t)&\le& -m \xi(t) E(t)W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right)+c \varepsilon_{2} \xi(t) \cdot \frac{E(t)}{E(0)}W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right)\\ &&-cE^{\prime}(t), \forall t\ge t_{2}. \end{array} $$

Using the non-increasing property of ξ, we obtain, for all t ≥ t₂,

$$\begin{array}{@{}rcl@{}} (\xi L_{2}+cE)^{\prime}(t)&\le& -m \xi(t) E(t)W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right)\\ &&+c \varepsilon_{2}\xi(t) \frac{E(t)}{E(0)}W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right) \end{array} $$

Therefore, by setting L₃ := ξL₂ + cE ∼ E, we get

$$\begin{array}{@{}rcl@{}} &&L_{3}^{\prime}(t)\le -m \xi(t) E(t)W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right)+c \varepsilon_{2} \xi(t) \cdot \frac{E(t)}{E(0)}W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right)\\ &&\hspace{0.3in} \end{array} $$

This gives, for a suitable choice of ε₂,

$$L_{3}^{\prime}(t)\le -k \xi(t) \left( \frac{E(t)}{E(0)}\right)W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right), \hspace{0.3in}\forall t\ge t_{2} $$

or

$$ k\left( \frac{E(t)}{E(0)}\right)W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right)\xi(t)\leq - L_{3}^{\prime}(t), \hspace{0.3in}\forall t\ge t_{2} $$

(4.51)

An integration of Eq. 4.51 yields

$$ {\int}_{t_{2}}^{t} k \left( \frac{E(s)}{E(0)} \right)W^{\prime}\left( \frac{\varepsilon_{2}}{s-t_{1}} \cdot \frac{E(s)}{E(0)}\right)\xi(s) ds \leq - {\int}_{t_{2}}^{t} L_{3}^{\prime}(s)ds\le L_{3}(t_{2}). $$

(4.52)

Using the facts that W^′, W^″ > 0 and the non-increasing property of E, we deduce that the map $t \mapsto E(t)W^{\prime }\left (\frac {\varepsilon _{2}}{t-t_{1}} \cdot \frac {E(t)}{E(0)}\right )$ is non-increasing and consequently, we have

$$\begin{array}{@{}rcl@{}} && k \left( \frac{E(t)}{E(0)} \right)W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right){\int}_{t_{2}}^{t} \xi(s) ds\\ &&\hspace{0.3in}\leq {\int}_{t_{2}}^{t} k \left( \frac{E(s)}{E(0)} \right)W^{\prime}\left( \frac{\varepsilon_{2}}{s-t_{1}} \cdot \frac{E(s)}{E(0)}\right)\xi(s) ds\le L_{3}(t_{2}),\hspace{0.2in}\forall t\ge t_{2} \end{array} $$

(4.53)

Multiplying each side of Eq. 4.53 by $\frac {1}{t-t_{1}}$, we have

$$ k \left( \frac{1}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right) W^{\prime}\left( \frac{\varepsilon_{2}}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right){\int}_{t_{2}}^{t} \xi(s) ds \leq \frac{k_{2}}{t-t_{1}},\hspace{0.3in}\forall t\ge t_{2} $$

(4.54)

Next, we set W₂(s) = sW^′(ε₂s) which is strictly increasing, then we obtain,

$$ k W_{2} \left( \frac{1}{t-t_{1}} \cdot \frac{E(t)}{E(0)}\right) {\int}_{t_{2}}^{t} \xi(s) ds \leq \frac{k_{2}}{t-t_{1}},\hspace{0.3in}\forall t\ge t_{2} $$

(4.55)

Finally, for two positive constants k₂ and k₃, we obtain

$$ E(t) \leq k_{3} (t-t_{1}) {W_{2}}^{-1} \left( \frac{k_{2}}{(t-t_{1}){\int}_{t_{2}}^{t}\xi(s) ds} \right). $$

(4.56)

This finishes the proof. □

Example 4.7

The following examples illustrate our results:

• 1. h₀ and G are linear

  Let g(t) = ae^{−b(1 + t)}, where b > 0 and a > 0 is small enough so that Eq. 2.1 is satisfied, then g^′(t) = −ξ(t)G(g(t)) where G(t) = t and ξ(t) = b. For the frictional nonlinearity, assume that h₀(t) = ct and $H(t)=\sqrt {t} h_{0}(\sqrt {t})=ct$. Therefore,, we can use Eq. 4.21 to deduce

  $$ E(t) \leq c_{1} e^{-c_{2}t} $$

  (4.57)

  which is the exponential decay.

• 2. h₀ is linear and G is non-linear

  Let $g(t)=a e^{-t^{q}}$, where 0 < q < 1 and a > 0 is small enough so that g satisfies (2.1), then g^′(t) = −ξ(t)G (g(t)) where ξ(t) = 1 and $G(t)=\frac {q^{t}}{\left (ln (a/t)\right )^{\frac {1}{q}-1}}$. For, the boundary feedback, let h₀(t) = ct, and $H(t)=\sqrt {t} h_{0}(\sqrt {t})=ct$. Since

  $$\begin{array}{@{}rcl@{}} &&G^{\prime}(t)=\frac{(1-q)+q ln (a/t)}{\left( ln (a/t)\right)^{1/q}}\\ &&\hspace{0.45in}\text{and}\\ &&G^{\prime\prime}(t)=\frac{(1-q) \left( ln (a/t)+ 1/q \right)}{\left( ln (a/t)\right)^{\frac{1}{q}+ 1}}. \end{array} $$

  then the function G satisfies the condition (A1) on (0, r₁] for any 0 < r₁ < a.

  $$\begin{array}{@{}rcl@{}} G_{1}(t)&=&{\int}_{t}^{r_{1}}\frac{1}{sG^{\prime}(s)}ds={\int}_{t}^{r_{1}}\frac{\left[\ln{\frac{a}{s}}\right]^{\frac{1}{q}}}{s\left[1-q+q\ln{\frac{a}{s}}\right]}ds\\ &=&{\int}_{\ln{\frac{a}{r_{1}}}}^{\ln{\frac{a}{t}}}\frac{u^{\frac{1}{q}}}{1-q+qu}du\\ &=&\frac{1}{q}{\int}_{\ln{\frac{a}{r_{1}}}}^{\ln{\frac{a}{t}}} u^{\frac{1}{q}-1}\left[\frac{u}{\frac{1-q}{q}+u}\right]du\\ &\le& \frac{1}{q}{\int}_{\ln{\frac{a}{r_{1}}}}^{\ln{\frac{a}{t}}} u^{\frac{1}{q}-1}du \le \left( \ln{\frac{a}{t}}\right)^{\frac{1}{q}}. \end{array} $$

  Then, Eq. 4.22 gives

  $$ E(t)\leq k e^{-k t^{q}} $$

  (4.58)

• 3. h₀ is non-linear and G is linear

  Let g(t) = ae^{−b(1 + t)}, where b > 0 and a > 0 is small enough so that Eq. 2.1 is satisfied, then g^′(t) = −ξ(t)G(g(t)) where G(t) = t and ξ(t) = b. Also, assume that h₀(t) = ct^q, where q > 1 and $H(t)=\sqrt {t} h_{0}(\sqrt {t})=ct^{\frac {q + 1}{2}}$. Then,

  $${H_{1}}^{-1}(t)= \left( ct+ 1\right)^{\frac{-2}{q-1}}.$$

  Therefore, applying Eq. 4.35, we obtain

  $$ E(t)\leq \left( c_{1} t +c_{2} \right)^{\frac{-2}{q-1}} $$

  (4.59)

• 4. h₀ is non-linear and G is non-linear

  Let $g(t)=\frac {a}{(1+t)^{2}}$, where a is chosen so that hypothesis (2.1) remains valid. Then

  $$g^{\prime}(t)=-b G(g(t)),\hspace{0.2in}\text{with}\hspace{0.2in}G(s)=s^{\frac{3}{2}},$$

  where b is a fixed constant. For the boundary feedback, let h₀(t) = ct⁵ and H(t) = ct³. Then,

  $$W(s)=(G^{-1}+H^{-1})^{-1}=\left( \frac{-1+\sqrt{1 + 4s}}{2}\right)^{3}$$

  and

  $$\begin{array}{@{}rcl@{}} &&W_{2}(s)=\frac{3s}{\sqrt{1 + 4s}}\left( \frac{-1+\sqrt{1 + 4s}}{2}\right)^{2}\\ &&\hspace{0.3in}=\frac{3s}{2\sqrt{1 + 4s}}+\frac{3s^{2}}{\sqrt{1 + 4s}}-\frac{3s}{2}\\ &&\hspace{0.3in}\le \frac{3s}{2}+\frac{3 s^{2}}{2\sqrt{s}}-\frac{3s}{2}=c s^{\frac{3}{2}} \end{array} $$

  Therefore, applying Eq. 4.36, we obtain

  $$E(t)\le \frac{c}{(t-t_{1})^{\frac{1}{3}}}$$