Stabilization of a Non-linear Fractional Problem by a Non-Integer Frictional Damping and a Viscoelastic Term

Abstract

In this paper, we consider a fractional equation of order between one and two which may be looked at as an interpolation between the heat and wave equations. The problem is non-linear as it involves a power-type non-linearity. We investigate the possibilities of stabilizing the system by a lower-order fractional term and/or a memory term involving the Laplacian. We prove a global Mittag–Leffler stability result in case a fractional frictional damping is active and a local Mittag–Leffler stability result when the material is viscoelastic in case of small relaxation functions. Unlike the integer-order problems, additional serious difficulties arise in the present case. These difficulties are highlighted clearly in the introduction. They are mainly due to the memory dependence of the fractional derivatives which is the cause of the invalidity of the product rule in particular. We utilize several properties in fractional calculus. Moreover, we introduce new Lyapunov-type functionals in the context of the multiplier technique.

1. Introduction

For many phenomena in engineering and science, fractional calculus offers an appropriate platform for their studies [1,2,3,4,5,6,7]. In fact, for anomalous processes and disordered systems, where the mean square displacement MSD (measure of the deflection of the actual position of a particle with regards to a given reference position) grows in a non-linear fashion, one is forced to consider fractional derivatives. Complex diffusion phenomena are therefore better described by non-integer-order fractional derivatives.

The problem to be discussed here is the following non-linear fractional initial boundary value problem of order between 1 and 2

$$\left\{\begin{array}{c}{}^C{D}^{2\alpha }v(t,x)+a^C{D}^{\alpha }v(t,x)+{\left|v\right|}^{\gamma }v=\Delta v(t,x)-b{\int }_0^{t}k(t-s)\Delta v(s,x)ds,\hfill \\ t>0,x\in \Omega ,1/2<\alpha <1,\hfill \\ v(t,x)=0,t>0,x\in \partial \Omega ,\hfill \\ v(0,x)=v_0\left(x\right),v^{\prime }(0,x)=v_1\left(x\right),x\in \Omega \hfill \end{array}\right.$$

(1)

where ${}^C{D}^{\alpha }$ is the Caputo fractional derivative of order $\alpha$ defined below, k is a non-negative function, $v_0\left(x\right)$ and $v_1\left(x\right)$ are the initial data of the state, $a,b\ge 0$ will be equal to 1 or 0 depending on whether they are effective or not in the equation, $\gamma >0$ is to be determined later and $\Omega$ is a bounded domain in $R^n$ with smooth boundary $\partial \Omega$.

It is well known that the relationship between the heat flux vector and the temperature gradient is given by

$$\mathbf{q}=-k\nabla T$$

where k is the thermal conductivity. Then, having in mind the conservation of energy

$$\gamma \frac{\partial T}{\partial t}=-\mathrm{div}\mathbf{q}$$

where $\gamma$ is the heat capacity, the basic heat equation

$$\frac{\partial T}{\partial t}=\frac{k}{\gamma }\Delta T$$

is obtained. Taking into account the history dependence between the gradient and the flux

$$\mathbf{q}=-\frac{k}{\Gamma \left(\mu \right)}\frac{\partial }{\partial t}{\int }_0^t{(t-s)}^{\mu -1}\nabla T\left(s\right)ds=-k{}^{RL}{D}^{\mu }\nabla T,0<\mu <1,$$

we are lead to the fractional heat conduction equation

$${}^C{D}^{\alpha }T=\delta \Delta T$$

for $0<\alpha <1$ and $\delta >0$ (see [5]).

A sub-diffusive problem has been derived in [8] from the Rayleigh–Stokes problem

$$\frac{\partial w}{\partial t}=(1+\gamma {}^{RL}{D}^{\beta })\Delta w.$$

The authors proved that the solutions of Stoke’s first problem follow as a special case.

In case of a material with memory, the relationship between the heat flux vector and the temperature gradient will involve a term of the form

$${\int }_0^{t}g(t-s)\nabla T\left(s\right)ds.$$

The fractional equation

$${}^C{D}^{2\alpha }v(t,x)=\Delta v(t,x)$$

(2)

is a fractional version interpolating the heat and wave equations. It describes anomalous processes and corresponds to the (fast) super-diffusion (particles move faster than in the normal case) (see [9,10]) when the mean square displacement grows non-linearly in time ($<v\left(t\right)>{}^{~}{t}^{2\alpha },$ $1<2\alpha <2$). The importance of super-diffusion is explained and highlighted in many works, see [3,6,7,11] to cite but a few.

A linear growth of the MSD ($<v\left(t\right)>{}^{~}t$) corresponds to the ordinary case (normal diffusion), when ($<v\left(t\right)>{}^{~}{t}^{\alpha },$$0<\alpha <1$), we are in presence of sub-diffusion (particles move slower than in the ordinary case), whilst $\alpha =2$ corresponds to the ballistic diffusion. Apart from the normal diffusion and the ballistic diffusion, the other cases cannot be described adequately by the standard heat or wave equations. Therefore, accordingly, the equation

$${}^C{D}^{2\alpha }v(t,x)=\Delta v(t,x)-b{\int }_0^{t}k(t-s)\Delta v(s,x)ds,1/2<\alpha <1,$$

(3)

interpolates the heat conduction equation [12]

$$u^{\prime }\left(t\right)=\Delta u\left(t\right)-{\int }_0^{t}k(t-s)\Delta u\left(s\right)ds$$

(4)

and the viscoelastic equation [13]

$$w^{″}\left(t\right)=\Delta w\left(t\right)-{\int }_0^{t}k(t-s)\Delta w\left(s\right)ds.$$

(5)

The stability of these problems has been investigated by many researchers: as parabolic problems with memory [14,15] and as viscoelastic problems [16,17,18,19,20,21]. For the viscoelastic problem, it has been proved, roughly, that the memory term produces a weak damping which is able by itself to drive the system to rest. It is the kernel which determines the rate of stability. Roughly, the stability rate is similar to the decay rate of the kernel. Various kinds of stability rates (exponential, polynomial and arbitrary) may be found in the literature [12,13,14,15,16,17,18,19,20,21,22]. It is clear now that these models are not only interesting from the mathematical point of view but also from the point of view of applications. They may describe the vibrations of viscoelastic structures in anomalous media.

Moreover, having in mind the telegraph equation [1,2,22]

$$\frac{{\partial }^{2}v}{\partial t^2}+c\frac{\partial v}{\partial t}=\Delta v,$$

(6)

Orsingher and Beghin [4] showed that the law of iterated Brownian motion and the telegraph process with Brownian time are expressed by the fractional telegraph equation

$${}^C{D}^{2\alpha }v(t,x)+a^C{D}^{\alpha }v(t,x)=\Delta v(t,x),0<\alpha <1.$$

(7)

Therefore, our present model (1) is a kind of non-linear fractional telegraph problem with memory.

The telegraph equation in (6) is dissipative. This is due to the presence of a first-order derivative which is often called ’frictional’ damping. In fact, it is easy to see that the energy of the system tends to zero in an exponential manner as time tends to infinity. Hence, it is interesting to know whether the lower-order fractional derivative is a dissipative term for the fractional problem in (1). In the affirmative, what would be the rate of convergence?

While we can find a good number of results on the well-posedness of similar problems to ours for orders between 0 and 1 and for problems without the Laplacian memory term, there are very few papers dealing with the exact form in the problem in (1). We identified three papers: one by El-Sayed and Herzallah [23], one by Ponce [24] (for an infinite history) and one by Agarwal et al. [25], dealing with abstract operators. We mention here the result in [25].

Let $(X,∥.∥)$ be a Banach space and $M,$${\left(N\left(t\right)\right)}_{t\ge 0}$ be closed linear operators defined on domains $D\left(M\right)$ and $D\left(N\right(t\left)\right)\supseteq D\left(M\right)$ dense in X. We denote by ${∥.∥}_1$ the graph norm in $D\left(M\right),$$R(\nu ,M):={(\nu I-M)}^{-1}$ and $\rho \left(M\right)$ the resolvent of $M.$ The problem

$${}^C{D}^{\beta }v(t,x)=Mv(t,x)-{\int }_0^{t}N(t-s)v(s,x)ds,1<\beta <2$$

with initial data ($v\left(0\right)=v_0,$$v^{\prime }\left(0\right)=0$) in $D\left(M\right)$ (see [24,25]), admits a classical solution $v\in C((0,\infty );D\left(M\right))\cap C^1((0,\infty );X)$ such that $\frac{t^{1-\beta }}{\Gamma (2-\beta )}\ast (v-v_0)\in C^2((0,\infty );X)$ under the following assumptions

(A1) For some $0<{\theta }_0\le \pi /2$ and every $\theta <{\theta }_0,$ there is a constant $C>0$ such that

$${\Sigma }_{0,\beta \omega }:=\{\nu \in \mathbf{C}:\nu \ne 0,\left|\arg \left(\nu \right)\right|<\beta \omega \}\subset \rho \left(M\right)$$

and $∥R(\nu ,M)∥<C/\left|\nu \right|,$ $\nu \in {\Sigma }_{0,\beta \omega }$ with $\omega =\theta +\pi /2.$

(A2) For each $v\in D\left(M\right),$ $N(.)v$ is strongly measurable on $(0,\infty ).$ There exists a locally integrable function $f\left(t\right)$ with Laplace transform $\widehat{f}\left(\nu \right),$$\mathrm{Re}\left(\nu \right)>0$ and $∥N\left(t\right)v∥\le f\left(t\right){∥v∥}_1,$ $t>0,$ $v\in D\left(M\right).$ In addition, $\widehat{N}:{\Sigma }_{0,\pi /2}\to \mathcal{L}(D\left(M\right);X)$ (the space of bounded linear operators from $D\left(M\right)$ into X) has an analytic extension $\tilde{N}$ to ${\Sigma }_{0,\omega }$ verifying $∥\tilde{N}\left(\nu \right)v∥\le ∥\tilde{N}\left(\nu \right)∥{∥v∥}_1,$ $v\in D\left(M\right)$ and $∥\tilde{N}\left(\nu \right)∥=O\left(\frac{1}{\left|\nu \right|}\right)$ as $\left|\nu \right|\to \infty .$

(A3) There exists a subspace S dense in $(D\left(M\right),{∥.∥}_1)$ and $\tilde{C}>0$ such that $M\left(S\right)\subseteq D\left(M\right),$ $\widehat{N}\left(\nu \right)\left(S\right)\subseteq D\left(M\right),$ $∥M\widehat{N}\left(\nu \right)v∥\le \tilde{C}∥v∥,$ $v\in S$ and $\nu \in {\Sigma }_{0,\omega }.$

In the present situation our operator is the well-known Laplacian which fortunately satisfies the above conditions in an appropriate space. Therefore, we can profit from this result to ensure the well-posedness of our problem in (1).

In our case, because of the non-linearity present in the equation, we only obtain a local existence result (see [26]). Assuming that the initial data are in $D\left(A\right)$ (A is the operator in our problem in (1)), there exists a unique local solution in the space

$$C^{*}:=\left\{v\in C\left([0,T),D\left(A\right)\right)\cap C^1\left([0,T),H_0^1(\Omega )\right):t^{1-2\alpha }\ast (w-w_0)\in C^2\left([0,T),L^2(\Omega )\right)\right\},$$

for $T>0,$ where

$$D\left(A\right):=H_0^1(\Omega )\cap H^2(\Omega ).$$

For simplicity and convenience, we shall rather consider the problem

$$\left\{\begin{array}{c}{}^C{D}^{\alpha }\left({}^C{D}^{\alpha }v(t,x)\right)+a^C{D}^{\alpha }v(t,x)+{\left|v\right|}^{\gamma }v=\Delta v(t,x)-b{\int }_0^{t}k(t-s)\Delta v(s,x)ds,\hfill \\ t>0,x\in \Omega ,1/2<\alpha <1,\hfill \\ v(t,x)=0,t>0,x\in \partial \Omega ,\hfill \\ v(0,x)=v_0\left(x\right),x\in \Omega .\hfill \end{array}\right.$$

(8)

Because the fractional composition ${}^C{D}^{\alpha +\beta }v(t,x){=}^C{D}^{\alpha }\left({}^C{D}^{\beta }v(t,x)\right)$ is not correct in general, the two problems are not exactly the same. The case $\alpha =1/2$ corresponds to the parabolic case [12,15]. In the case where $\alpha =1$, we recover the viscoelastic problem studied extensively by many authors [7,14,18,19,20,21,22].

We shall shed some light on the effects of the non-linear term, the fractional lower-order term and the memory term on the stability of the system. The lower-order term and the memory term will be considered separately. It is well known, in the integer-order case, that having both at the same time enhances the dissipation. Thus, the challenging question is to consider only one at a time. We have purposefully written the lower-order term as a half-order fractional derivative of the leading fractional derivative as it corresponds to the one in (7). It would be interesting, at least from a mathematical point of view, to treat other orders as well. We will show below that it is a dissipative term. More precisely, we will prove that this lower-order term is by itself (without the memory term) sufficient to guarantee the stability of the system in the Mittag–Leffler manner (solutions of the system will decay as the Mittag–Leffler functions under certain norms). However, without the lower-order fractional term, the Mittag–Leffler stability is only shown for ’small’ kernels. We are unable to decide on the dissipativity of the system in the early stages. This is justified by the fact that the memory term alone is not capable of producing a considerable amount of damping. The product rule (Proposition 5 below) gives rise to a complicated term. It is defined by three singular integrals involving functions of undefined signs. Moreover, the stability is only of local character due to the difficulty to control the non-linear terms. In fact, this local character is not suitable to fraction, it is also the case in integer-order problems. The objective of this paper is to suggest a way to remedy to these difficulties. Unfortunately, we could not compare our results with existing ones as, to the best of our knowledge, this is the first paper treating this as a non-linear issue.

In case both the lower-order fractional term and memory term are present, the system remains Mittag–Leffler stable for reasonable classes of kernels. In fact, the memory term in this situation does not contribute significantly. On the contrary, it only adds some complexity to the estimations. To keep the size of the paper reasonable, we refrain from treating this case. Instead a simple combination of the first two cases is pursued (see Remark 1 below).

The next section contains some preliminary definitions and results. It is followed by Section 3 on the existence and uniqueness issue. In Section 4 we treat the case where only the lower-order term is present. In Section 5 we determine an appropriate function to work with. Section 6 contains the Mittag–Leffler stability in the presence of the memory term. We end the paper by Section 7.

2. Preliminaries

In this section, we give some definitions, formulas and inequalities involving fractional derivatives [27,28]. We also provide the fractional product rule.

Definition 1.

The Riemann–Liouville fractional integral of order $\alpha >0$

$$I^{\alpha }\phi \left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^t{(t-s)}^{\alpha -1}\phi \left(s\right)ds,\alpha >0$$

for any measurable function φ provided that the right-hand side exists. Here, $\Gamma \left(\alpha \right)$ is the usual Gamma function.

Definition 2

([28]). The fractional derivative of order α in the Caputo is defined by

$${}^C{D}^{\alpha }\phi \left(t\right)=\frac{1}{\Gamma (1-\alpha )}{\int }_0^t{(t-s)}^{-\alpha }{\phi }^{\prime }\left(s\right)ds,0<\alpha <1,$$

$${}^C{D}^{\alpha }\phi \left(t\right)=\frac{1}{\Gamma (2-\alpha )}{\int }_0^t{(t-s)}^{1-\alpha }{\phi }^{{}^{″}}\left(s\right)ds,1<\alpha <2,$$

whereas the fractional derivative of order α in the Riemann–Liouville is defined by

$${}^{RL}{D}^{\alpha }\phi \left(t\right)=\frac{1}{\Gamma (1-\alpha )}\frac{d}{dt}{\int }_0^t{(t-s)}^{-\alpha }\phi \left(s\right)ds,0<\alpha <1$$

provided that the integrals exist. Here, the primes denote time differentiation.

The relationship between both derivatives is

$${}^{RL}{D}^{\alpha }\phi \left(t\right)=\frac{\phi \left(0\right)t^{-\alpha }}{\Gamma (1-\alpha )}{+}^C{D}^{\alpha }\phi \left(t\right),0<\alpha <1,t>0.$$

(9)

The one-parametric and two-parametric Mittag-Leffler functions are defined by [25]

$$E_{\alpha }\left(z\right):={\sum }_{n=0}^{\infty }\frac{z^n}{\Gamma (\alpha n+1)},\mathrm{Re}\left(\alpha \right)>0,$$

and

$$E_{\alpha ,\beta }\left(z\right):={\sum }_{n=0}^{\infty }\frac{z^n}{\Gamma (\alpha n+\beta )},\mathrm{Re}\left(\alpha \right)>0,\mathrm{Re}\left(\beta \right)>0,$$

respectively. Notice that $E_{\alpha ,1}\left(z\right)\equiv E_{\alpha }\left(z\right)$.

The power $\gamma$ is assumed to satisfy

$$\gamma >0,n=1,2,0<\gamma <\frac{2}{n-2},n>2.$$

These conditions are imposed so as to ensure an embedding later in our proofs.

Proposition 1

([29]). If $\phi \left(t\right)$ is a differentiable function satisfying

$${}^C{D}^{\alpha }\phi \left(t\right)\le -\gamma \phi \left(t\right),0<\alpha <1$$

for $\gamma >0,$ then $\phi \left(t\right)\le \phi \left(0\right)E_{\alpha }(-\gamma t^{\alpha }),$$t\ge 0.$ In case the derivative is of Riemann–Liouville-type, then the decay is of the form $t^{\alpha -1}{E}_{\alpha ,\alpha }(-\gamma t^{\alpha }).$

Proposition 2

([27]). For $\alpha ,\beta >0,$ we have

$$\lambda t^{\alpha }{E}_{\alpha ,\alpha +\beta }\left(\lambda t^{\alpha }\right)=E_{\alpha ,\beta }\left(\lambda t^{\alpha }\right)-\frac{1}{\Gamma \left(\beta \right)}.$$

Proposition 3

([27], p. 61). For $\mu ,\alpha ,\beta >0,$ it holds that

$$\frac{1}{\Gamma \left(\mu \right)}{\int }_0^t{(t-s)}^{\mu -1}{E}_{\alpha ,\beta }\left(\lambda s^{\alpha }\right)s^{\beta -1}ds=t^{\mu +\beta -1}{E}_{\alpha ,\mu +\beta }\left(\lambda t^{\alpha }\right),t>0.$$

Proposition 4

([28], p. 99). If $I^{1-\alpha }\phi \left(t\right)\in C^1\left([0,\infty )\right),$ $0<\alpha <1$ and $\psi \left(t\right)$ is a continuous function, then

$${}^{RL}{D}^{\alpha }{\int }_0^t\phi (t-s)\psi \left(s\right)ds={\int }_0^t\psi {\left(s\right)}^{RL}{D}^{\alpha }\phi (t-s)ds+\psi \left(t\right)\underset{t\to 0^{+}}{lim}{I}^{1-\alpha }\phi \left(t\right),t>0.$$

A fractional version of the well-known product rule in the integer-order case is

Proposition 5

([30]). Let $\phi \left(t\right)$ and $\psi \left(t\right)$ be absolutely continuous functions. Then, for $t\ge 0$ and $0<\alpha <1,$ we have

$$\begin{array}{c}\phi {\left(t\right)}^C{D}^{\alpha }\psi \left(t\right)+\psi {\left(t\right)}^C{D}^{\alpha }\phi \left(t\right)\\ {=}^C{D}^{\alpha }\left(\phi \psi \right)\left(t\right)+\frac{\alpha }{\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{{\phi }^{\prime }\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}{\int }_0^{\xi }\frac{{\psi }^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}.\end{array}$$

From now on, we shall drop the letter “C” referring to “Caputo"” in ${}^C{D}^{\alpha },$ for notation convenience.

Proposition 6.

Let $v\left(t\right)$ be a real-valued differentiable function, then for $0<\alpha <1$

$$D^{\alpha }\left(\frac{{\left|v\right|}^{\rho +2}}{\rho +2}\right)\le {\left|v\right|}^{\rho }v{D}^{\alpha }v.$$

Proof. 

This is an immediate consequence of a combination of a result in [31]

$$D^{\alpha }\left(\frac{{\left|v\right|}^{\rho +2}}{\rho +2}\right)\le {\left|v\right|}^{\rho +1}{D}^{\alpha }\left|v\right|$$

and one in [32]

$$D^{\alpha }\left|v\right|\le sgn\left(v\right)D^{\alpha }v.$$

□

3. Existence and Uniqueness

Clearly, we can rewrite the above equation in (8) as a system of two equations of order $\alpha$ as follows

$$\left\{\begin{array}{c}{}^C{D}^{\alpha }v(t,x)=w(t,x),\hfill \\ {}^C{D}^{\alpha }w(t,x)=\Delta v(t,x)-av(t,x)-b{\int }_0^{t}k(t-s)\Delta v(s,x)ds-{\left|v\right|}^{\gamma }v\hfill \end{array}\right.$$

or in matrix form

$$\begin{array}{c}{}^C{D}^{\alpha }\left(\begin{array}{c}v(t,x)\\ w(t,x)\end{array}\right)=\left(\begin{array}{cc}0& I\\ \Delta & -aI\end{array}\right)\left(\begin{array}{c}v(t,x)\\ w(t,x)\end{array}\right)\hfill \\ -{\int }_0^t\left(\begin{array}{cc}0& 0\\ bk(t-s)\Delta & 0\end{array}\right)\left(\begin{array}{c}v(s,x)\\ w(s,x)\end{array}\right)ds+\left(\begin{array}{c}0\\ -{\left|v\right|}^{\gamma }(t,x)v(t,x)\end{array}\right).\hfill \end{array}$$

Notice that, if $v^{\prime }\left(t\right)$ is continuous near zero then $w\left(0\right){=}^C{D}^{\alpha }v\left(0\right)=0.$

It is proven in [13] that the abstract problem

$$\left\{\begin{array}{c}{}^C{D}^{\gamma }z(t,x)=\mathcal{P}z(t,x)+{\int }_0^t\mathcal{Q}(t-s)z(s,x)ds+f(t,z\left(t\right)),0<\gamma <1\hfill \\ z(0,x)=z_0\left(x\right)\in X\hfill \end{array}\right.$$

admits a unique solution under the above assumptions (A1)–(A3) and $f(t,z(t\left)\right)$ is locally Lipschitz with respect to its second argument. For the fractional case, instead of semigroups, we rather use the notion of resolvents defined next (see also [33,34]).

Definition 3.

The family of bounded linear operators ${\left(R_{\gamma }\left(t\right)\right)}_{t\ge 0}$ determines a $\gamma$-resolvent for this problem if

(a) The mapping $R_{\gamma }\left(t\right):[0,\infty )\to L\left(X\right):=L(X;X)$ is strongly continuous and $R_{\gamma }\left(0\right)=I.$

(b) $\forall z\in D\left(P\right),$$R_{\gamma }(.)z\in C\left([0,\infty );D\left(\mathcal{P}\right)\right)\cap C^{\gamma }\left((0,\infty );X\right)$ ($C^{\gamma }\left((0,\infty );X\right)$ designates the space of continuous functions z such that ${}^C{D}^{\gamma }z$ exists and is continuous), and

$$\begin{array}{c}{}^C{D}^{\gamma }{R}_{\gamma }\left(t\right)z=\mathcal{P}{R}_{\gamma }\left(t\right)z+{\int }_0^t\mathcal{Q}(t-s)R_{\gamma }\left(s\right)zds\\ =R_{\gamma }\left(t\right)\mathcal{P}u+{\int }_0^t{R}_{\gamma }(t-s)\mathcal{Q}\left(s\right)zds,t\ge 0.\end{array}$$

The following family

$$R_{\gamma }\left(t\right):=\frac{1}{2\pi i}{\int }_{\Gamma }{\mu }^{\gamma -1}{e}^{\mu t}{\left({\mu }^{\gamma }I-\mathcal{P}-{\mathcal{Q}}^{^}\left(\mu \right)\right)}^{-1}d\mu ,t\ge 0$$

is shown to be a $\gamma$-resolvent for a suitable path $\Gamma :=\{t{e}^{i\theta }:t\ge r\}\cup \{r{e}^{i\zeta }:-\theta \le \zeta \le \theta \}\cup \{t{e}^{-i\theta }:t\ge r\}$ oriented counterclockwise where $\pi /2<\theta <\varphi$ and r is determined in the proof. An associated family to ${\left(R_{\gamma }\left(t\right)\right)}_{t\ge 0}$ is defined by

$$S_{\gamma }\left(t\right):=\frac{t^{1-\gamma }}{2\pi i}{\int }_{\Gamma }{e}^{\mu t}{\left({\mu }^{\gamma }I-\mathcal{P}-{\mathcal{Q}}^{^}\left(\mu \right)\right)}^{-1}d\mu ,t\ge 0.$$

Theorem 1

([35]). For $z_0\in D\left(\mathcal{P}\right),$ the function

$$z\left(t\right):=R_{\gamma }\left(t\right)z_0+{\int }_0^t{(t-s)}^{\gamma -1}{S}_{\gamma }(t-s)f(s,z\left(s\right))ds$$

belongs to $C\left([0,T);D\left(\mathcal{P}\right)\right)\cap C^{\gamma }\left((0,T);X\right)$ and is a solution to the problem above for $T>0.$

As in our case the Laplacian is a sectorial operator (with open sector spectrum and a resolvent which is uniformly bounded from above outside any large sector) and taking into account the assumptions below on the kernel, the conditions (A1)–(A3) are satisfied. Consequently, there exists a unique local solution, to our problem, in the space

$$C^{*}:=\left\{w\in C\left([0,T),D\left(A\right)\right)\cap C^1\left([0,T),H_0^1(\Omega )\right):t^{-\alpha }\ast w_t\in C^1\left([0,T),L^2(\Omega )\right)\right\},$$

for some $T>0,$ where

$$D\left(A\right):=H_0^1(\Omega )\cap H^2(\Omega ).$$

4. The Case $\mathit{a}\ne \mathbf{0}$ and $\mathit{b}=\mathbf{0}$

Here, we present and prove our stability result. Without loss of generality, we assume $a=1$ and consider the functional

$${\mathcal{E}}_0\left(t\right):=\frac{1}{2}\left({∥D^{\alpha }v∥}^2+{∥\nabla v∥}^2\right)+\frac{1}{\gamma +2}{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2},t\ge 0.$$

From Proposition 5, it appears that

$$\begin{array}{c}{D}^{\alpha }{\mathcal{E}}_0\left(t\right)=-{∥D^{\alpha }v∥}^2+\frac{1}{\gamma +2}{D}^{\alpha }{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}-{\int }_{\Omega }{\left|v\right|}^{\gamma }v{D}^{\alpha }vdx\\ -\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left[{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^2+{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right|}^2\right]dx,t\ge 0.\end{array}$$

We shall assume that our initial data satisfy $(v_0,v_1)\in {\left(H_0^1(\Omega )\right)}^2$ and start the process of looking for a suitable functional to work with. In the next lemma, we introduce the first functional.

Lemma 1.

For the functional

$${\Psi }_0\left(t\right):={\int }_{\Omega }v{D}^{\alpha }vdx,t\ge 0,$$

it holds that

$$\begin{array}{c}{D}^{\alpha }{\Psi }_0\left(t\right)\le \left(1+\frac{C_p}{2}\right){∥D^{\alpha }v∥}^2-\frac{1}{2}{∥\nabla v∥}^2-{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}\\ +\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left[C_p{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right|}^2+{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right)}^2\right]dx,t\ge 0\end{array}$$

where $C_p$ is the Poincaré constant.

Proof. 

As $D^{\alpha }v$ is absolutely continuous, we may apply Proposition 5 to find, for $t\ge 0$

$$\begin{array}{c}{D}^{\alpha }{\Psi }_0\left(t\right)={∥D^{\alpha }v∥}^2+{\int }_{\Omega }v{D}^{\alpha }\left(D^{\alpha }v\right)dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{v^{\prime }\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right)dx\\ ={∥D^{\alpha }v∥}^2-{∥\nabla v∥}^2-{\int }_{\Omega }v{D}^{\alpha }vdx-{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{v^{\prime }\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right)dx.\end{array}$$

This expression is evaluated as follows

$$\begin{array}{c}{D}^{\alpha }{\Psi }_0\left(t\right)\le \left(1+\frac{C_p}{2}\right){∥D^{\alpha }v∥}^2-\frac{1}{2}{∥\nabla v∥}^2-{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}\\ +\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left[C_p{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right|}^2+{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right)}^2\right]dx,t\ge 0.\end{array}$$

The proof is complete. □

Theorem 2.

The problem in (8) with $a=1$ and $b=0,$ is Mittag–Leffler stable. This means that there exist positive constants $B_0$ (depending on ${\mathcal{E}}_0\left(0\right)$) and ${\nu }_0$ such that

$${\mathcal{E}}_0\left(t\right)\le B_0{E}_{\alpha }(-{\nu }_0{t}^{\alpha }),t\ge 0.$$

Proof. 

Let

$${\mathcal{L}}_0\left(t\right):=N{\mathcal{E}}_0\left(t\right)+M{\Psi }_0\left(t\right),t\ge 0$$

for some positive constants M and N, to be determined. Then, in view of Proposition 6, we may write

$$\begin{array}{c}{D}^{\alpha }{\mathcal{L}}_0\left(t\right)\le -N{∥D^{\alpha }v∥}^2+M\left(1+\frac{C_p}{2}\right){∥D^{\alpha }v∥}^2-\frac{M}{2}{∥\nabla v∥}^2-M{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}\\ -\frac{\alpha N}{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left[{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^2+{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right|}^2\right]dx\\ +\frac{\alpha M}{2\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left[C_p{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right|}^2+{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right)}^2\right]dx,t\ge 0\end{array}$$

or

$$\begin{array}{c}{D}^{\alpha }{\mathcal{L}}_0\left(t\right)\le \left[M\left(1+\frac{C_p}{2}\right)-N\right]{∥D^{\alpha }v∥}^2-\frac{M}{2}{∥\nabla v∥}^2\\ -M{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}+\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\\ \times \left[\left(C_{p}M-N\right){\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right|}^2+\left(M-N\right){\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^2\right]dx\end{array}$$

(10)

for $t\ge 0.$ To obtain $-C_1{\mathcal{E}}_0\left(t\right)$ for some positive constant $C_1$ in the right hand side of (10), we need

$$\left\{\begin{array}{c}M\left(1+\frac{C_p}{2}\right)<N,\hfill \\ C_{p}M\le N,\hfill \\ M\le N.\hfill \end{array}\right.$$

This requires

$$N>max\left\{C_p,1+\frac{C_p}{2}\right\}M.$$

Consequently

$$D^{\alpha }{\mathcal{L}}_0\left(t\right)\le -C_1{\mathcal{E}}_0\left(t\right),t\ge 0$$

for some positive constant $C_1$ and as ${\mathcal{L}}_0\left(t\right)$ and ${\mathcal{E}}_0\left(t\right)$ are equivalent, we obtain

$$D^{\alpha }{\mathcal{L}}_0\left(t\right)\le -C_2{\mathcal{L}}_0\left(t\right),t\ge 0$$

for some positive constant $C_2.$ The conclusion follows from Proposition 1 and the equivalence of ${\mathcal{L}}_0\left(t\right)$ and ${\mathcal{E}}_0\left(t\right)$ again. □

5. A Useful Functional

In this section, we determine an appropriate functional before we tackle the case of the memory term. We shall use the notation

$$(k\square \nabla v)\left(t\right):={\int }_{\Omega }{\int }_0^{t}k(t-s){\left|\nabla v\left(t\right)-\nabla v\left(s\right)\right|}^{2}dsdx,t\ge 0.$$

Proposition 7.

If $I^{1-\alpha }k\left(t\right)\in C^1\left([0,\infty )\right),$$0<\alpha <1$, then the following identity holds

$$\begin{array}{c}{\int }_{\Omega }{D}^{\alpha }\nabla v.{\int }_0^{t}k(t-s)\nabla v\left(s\right)dsdx=\frac{1}{2}\left\{{(}^{RL}{D}^{\alpha }k\square \nabla v)+\left({\int }_0^{t}k\left(s\right)ds\right)D^{\alpha }{∥\nabla v∥}^2\right\}\\ -\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}{\int }_0^{\xi }\frac{{\left[{∥\nabla v\left(s\right)∥}^2\right]}^{\prime }ds}{{(t-s)}^{\alpha }}-\frac{1}{2}{D}^{\alpha }(k\square \nabla v)\left(t\right)+\frac{\alpha }{\Gamma (1-\alpha )}\\ \times {\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{{\left[\left(\nabla v\right)\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}.\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }ds\right)dx,t\ge 0.\end{array}$$

Proof. 

By virtue of Proposition 5 and the identity

$$\begin{array}{c}{\int }_{\Omega }{\int }_0^{t}k(t-s){\left|\nabla v\left(t\right)-\nabla v\left(s\right)\right|}^{2}dsdx\\ ={∥\nabla v∥}^2{\int }_0^{t}k(t-s)ds+{\int }_0^{t}k(t-s){∥\nabla v\left(s\right)∥}^{2}ds\\ -2{\int }_{\Omega }\nabla v.{\int }_0^{t}k(t-s)\nabla v\left(s\right)dsdx,t\ge 0\end{array}$$

we see that

$$\begin{array}{c}{D}^{\alpha }(k\square \nabla v)\left(t\right)=\left(D^{\alpha }{\int }_0^{t}k(t-s)ds\right){∥\nabla v∥}^2+\left({\int }_0^{t}k(t-s)ds\right)D^{\alpha }{∥\nabla v∥}^2\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}{\int }_0^{\xi }\frac{{\left[{∥\nabla v\left(s\right)∥}^2\right]}^{\prime }ds}{{(t-s)}^{\alpha }}+D^{\alpha }{\int }_0^{t}k(t-s){∥\nabla v\left(s\right)∥}^{2}ds\\ -2{\int }_{\Omega }{D}^{\alpha }\nabla v.{\int }_0^{t}k(t-s)\nabla v\left(s\right)dsdx-2{\int }_{\Omega }\nabla v.D^{\alpha }{\int }_0^{t}k(t-s)\nabla v\left(s\right)dsdx+\frac{2\alpha }{\Gamma (1-\alpha )}\\ \times {\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{{\left(\nabla v\right)}^{\prime }\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}.\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }ds\right)dx,t\ge 0.\end{array}$$

(11)

Notice that

$$\begin{array}{c}({}^{RL}{D}^{\alpha }k\square \nabla v)\left(t\right)={∥\nabla v∥}^2{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)ds+{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s){∥\nabla v\left(s\right)∥}^{2}ds\\ -2{\int }_{\Omega }\nabla v.{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)\nabla v\left(s\right)dsdx,t\ge 0\end{array}$$

(12)

and by Proposition 4

$$D^{\alpha }{\int }_0^{t}k(t-s)ds={\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)ds+\underset{t\to 0^{+}}{lim}{I}^{1-\alpha }k\left(t\right),t\ge 0.$$

(13)

Again, by Proposition 4 and the summability of k, we deduce that

$$\begin{array}{c}{D}^{\alpha }{\int }_0^{t}k(t-s){∥\nabla v\left(s\right)∥}^{2}ds\\ ={}^{RL}{D}^{\alpha }{\int }_0^{t}k(t-s){∥\nabla v\left(s\right)∥}^{2}ds-\frac{t^{-\alpha }}{\Gamma (1-\alpha )}{\left.\left({\int }_0^{t}k(t-s){∥\nabla v\left(s\right)∥}^{2}ds\right)\right|}_{t=0}\\ ={\int }_0^t{}^{RL}{D}^{\alpha }k(t-s){∥\nabla v\left(s\right)∥}^{2}ds+{∥\nabla v\left(t\right)∥}^2\underset{t\to 0^{+}}{lim}{I}^{1-\alpha }k\left(t\right),t>0\end{array}$$

(14)

and

$$\begin{array}{c}{\int }_{\Omega }\nabla v.D^{\alpha }{\int }_0^{t}k(t-s)\nabla v\left(s\right)dsdx\\ ={\int }_{\Omega }\nabla v.{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)\nabla v\left(s\right)dsdx+{∥\nabla v\left(t\right)∥}^2\underset{t\to 0^{+}}{lim}{I}^{1-\alpha }k\left(t\right),t>0.\end{array}$$

(15)

Gathering the previous relations (13)–(15) in (11), we find

$$\begin{array}{c}{D}^{\alpha }(k\square \nabla v)\left(t\right)={∥\nabla v∥}^2{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)ds+\left({\int }_0^{t}k(t-s)ds\right)D^{\alpha }{∥\nabla v∥}^2\\ +{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s){∥\nabla v\left(s\right)∥}^{2}ds-\frac{\alpha }{\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}{\int }_0^{\xi }\frac{{\left[{∥\nabla v\left(s\right)∥}^2\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\\ -2{\int }_{\Omega }{D}^{\alpha }\nabla v.{\int }_0^{t}k(t-s)\nabla v\left(s\right)dsdx-2{\int }_{\Omega }\nabla v.{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)\nabla v\left(s\right)dsdx+\frac{2\alpha }{\Gamma (1-\alpha )}\\ \times {\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{{\left[\left(\nabla v\right)\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}.\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }ds\right)dx,t\ge 0.\end{array}$$

(16)

Therefore, in view of the relation (12) and (16), we obtain for $t\ge 0$

$$\begin{array}{c}{D}^{\alpha }(k\square \nabla v)\left(t\right)=({}^{RL}{D}^{\alpha }k\square \nabla v)+\left({\int }_0^{t}k\left(s\right)ds\right)D^{\alpha }{∥\nabla v∥}^2\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}{\int }_0^{\xi }\frac{{\left[{∥\nabla v\left(s\right)∥}^2\right]}^{\prime }ds}{{(t-s)}^{\alpha }}-2{\int }_{\Omega }{D}^{\alpha }\nabla v.{\int }_0^{t}k(t-s)\nabla v\left(s\right)dsdx\\ +\frac{2\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{{\left[\left(\nabla v\right)\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}.\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }ds\right)dx.\end{array}$$

(17)

This finishes the proof. □

A multiplication of the equation in (8) (with $a=0$ and $b=1$) by $D^{\alpha }v$ gives

$$\begin{array}{c}{\int }_{\Omega }{D}^{\alpha }v{D}^{\alpha }\left(D^{\alpha }v\right)dx+{\int }_{\Omega }{\left|v\right|}^{\gamma }v{D}^{\alpha }vdx\\ =-{\int }_{\Omega }{D}^{\alpha }\nabla v.\nabla vdx+{\int }_{\Omega }{D}^{\alpha }\nabla v.{\int }_0^{t}k(t-s)\nabla v(s,x)dsdx,t\ge 0\end{array}$$

and by the above Proposition 6, we see that

$$\begin{array}{c}{\int }_{\Omega }{D}^{\alpha }v{D}^{\alpha }\left(D^{\alpha }v\right)dx+{\int }_{\Omega }{\left|v\right|}^{\gamma }v{D}^{\alpha }vdx+{\int }_{\Omega }{D}^{\alpha }\nabla v.\nabla vdx\\ -\frac{1}{2}{D}^{\alpha }{∥\nabla v∥}^2{\int }_0^{t}k\left(s\right)ds+\frac{1}{2}{D}^{\alpha }(k\square \nabla v)\left(t\right)=\frac{1}{2}{(}^{RL}{D}^{\alpha }k\square \nabla v)\\ -\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}{\int }_0^{\xi }\frac{{\left[{∥\nabla v\left(s\right)∥}^2\right]}^{\prime }ds}{{(t-s)}^{\alpha }}+\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\\ \times {\int }_0^{\xi }\frac{{\left[\left(\nabla v\right)\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}.\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }ds\right)dx,t\ge 0.\end{array}$$

This identity is the main reason behind choosing the functional

$${\mathcal{E}}_1\left(t\right):=\frac{1}{2}\left\{{∥D^{\alpha }v∥}^2+\left(1-{\int }_0^{t}k\left(s\right)ds\right){∥\nabla v∥}^2+\left(k\square \nabla v\right)\left(t\right)\right\}+\frac{1}{\gamma +2}{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}.$$

6. The Case $\mathbf{a}=\mathbf{0}$ and $\mathit{b}\ne \mathit{0}$

Here, in the absence of the fractional damping, there remains only the viscoelastic term. Without loss of generality, we pick $b=1$. Our assumptions on the kernel k are:

(K) $k\in L_{loc}^1\left(R^{+}\right)$ is a non-negative continuous function satisfying $\widehat{k}\left(\lambda \right)=O\left({\left|\lambda \right|}^{-1}\right)$ and $I^{1-\alpha }k\left(t\right)\in C^1\left([0,\infty )\right),$

$${}^{RL}{D}^{\alpha }k\left(t\right)\le -C_{k}k\left(t\right),t>0$$

and

$$\overline{k}:={\int }_0^{\infty }k\left(s\right)ds<1$$

for some positive constant $C_k.$

To deal with the non-linearity we need to assume that

(Γ) $0<\gamma <\frac{n}{n-2}-1=\frac{2}{n-2}$ if $n\ge 3$ and $\gamma >0$ if $n=1,2.$

This assumption will allow us to use the embedding $H_0^1(\Omega )\subset L^q(\Omega )$ when $2\le q\le 2n/(n-2)$ if $n\ge 3$ and $q\ge 2$ if $n=1,2.$

It is easy to see from Proposition 1 that $k\left(t\right)\le K{t}^{\alpha -1}{E}_{\alpha ,\alpha }(-C_k{t}^{\alpha }),$ $t\ge 0$ for some $K>0$ and from Propositions 2 and 3 that such kernels are summable and

$${\int }_0^{t}k\left(s\right)ds\le K{\int }_0^t{s}^{\alpha -1}{E}_{\alpha ,\alpha }(-C_k{s}^{\alpha })ds=K{t}^{\alpha }{E}_{\alpha ,\alpha +1}(-C_k{t}^{\alpha })\le K/C_k,t>0.$$

In fact, this family of functions $t^{\alpha -1}{E}_{\alpha ,\alpha }(-C_k{t}^{\alpha })$ satisfies (K). Moreover, our proofs below are valid for initial data in ${\left(H_0^1(\Omega )\right)}^2.$

Lemma 2.

The functional ${\mathcal{E}}_1\left(t\right)$ satisfies along solutions of (8), for $t\ge 0$

$$\begin{array}{c}{D}^{\alpha }{\mathcal{E}}_1\left(t\right)\le \frac{1}{2}{(}^{RL}{D}^{\alpha }k\square \nabla v)-\frac{1}{2}{I}^{1-\alpha }k\left(t\right){∥\nabla v∥}^2\\ -\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^{2}dx\\ -\frac{\alpha }{4\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right|}^{2}dx\\ +\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }ds\right|}^{2}dx.\end{array}$$

Proof. 

Clearly, a direct fractional differentiation of ${\mathcal{E}}_1\left(t\right)$, using Proposition 5, yields for $t>0$

$$\begin{array}{c}{D}^{\alpha }{\mathcal{E}}_1\left(t\right)=\frac{1}{2}{D}^{\alpha }\left\{{∥D^{\alpha }v∥}^2+\left(1-{\int }_0^{t}k\left(s\right)ds\right){∥\nabla v∥}^2+\left(k\square \nabla v\right)\left(t\right)\right\}\\ +\frac{1}{\gamma +2}{D}^{\alpha }{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}={\int }_{\Omega }{D}^{\alpha }v{D}^{\alpha }\left(D^{\alpha }v\right)dx\\ -\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^{2}dx+\left(1-{\int }_0^{t}k\left(s\right)ds\right)\\ \times \left[{\int }_{\Omega }\nabla v.D^{\alpha }\left(\nabla v\right)dx-\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right|}^{2}dx\right]\\ -\frac{1}{2}\left(D^{\alpha }{\int }_0^{t}k\left(s\right)ds\right){∥\nabla v∥}^2+\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{{\left[{∥\nabla v\left(s\right)∥}^2\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right){\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\\ +\frac{1}{2}{D}^{\alpha }\left(k\square \nabla v\right)\left(t\right)+\frac{1}{\gamma +2}{D}^{\alpha }{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}.\end{array}$$

Along solutions of (8) (see (17)) and in view of Proposition 4, we find

$$\begin{array}{c}{D}^{\alpha }{\mathcal{E}}_1\left(t\right)=\frac{1}{2}{(}^{RL}{D}^{\alpha }k\square \nabla v)-\frac{1}{2}{I}^{1-\alpha }k\left(t\right){∥\nabla v∥}^2+\frac{1}{\gamma +2}{D}^{\alpha }{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}\\ -{\int }_{\Omega }{\left|v\right|}^{\gamma }v{D}^{\alpha }vdx-\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^{2}dx\\ -\frac{\alpha }{2\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right|}^{2}dx+\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\\ \times {\int }_0^{\xi }\frac{{\left[\left(\nabla v\right)\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}.\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }ds\right)dx,t\ge 0.\end{array}$$

We conclude by using Proposition 6 and the evaluation

$$\begin{array}{c}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{{\left[\nabla v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right).\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }ds\right)dx\\ \le \frac{1}{4}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right|}^{2}dx\\ +{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }ds\right|}^{2}dx.\end{array}$$

This finishes the proof. □

The last term in the estimation of $D^{\alpha }{\mathcal{E}}_1\left(t\right)$ needs to be controlled somehow. This term does not appear in the integer-order case (second-order) and, whatever evaluation we derive, its weight is going to be considerable. We suggest a new functional in the next lemma.

Lemma 3.

The fractional derivative of the functional

$${\Psi }_1\left(t\right):={\int }_{\Omega }{\left|{\int }_0^{t}k(t-s)\nabla v\left(s\right)ds\right|}^{2}dx,t\ge 0$$

satisfies

$$\begin{array}{c}{D}^{\alpha }{\Psi }_1\left(t\right)\le 2{\lambda }_1\overline{k}{\int }_{\Omega }\left({\int }_0^{t}k(t-s){\left|\nabla v\left(s\right)\right|}^{2}ds\right)dx\\ +\frac{K}{{\lambda }_1}(\left|{}^{RL}{D}^{\alpha }k\right|\square \nabla v)\left(t\right)+\frac{{\left[K{E}_{\alpha }(-C_k{t}^{\alpha })\right]}^2}{{\lambda }_1}{∥\nabla v∥}^2\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }{(t-\eta )}^{-\alpha }{\left({\int }_0^{\eta }k(\eta -\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }d\eta \right|}^{2}dx,\end{array}$$

for ${\lambda }_1>0,t\ge 0.$

Proof. 

With the help of Proposition 5, we can write

$$\begin{array}{c}{D}^{\alpha }{\Psi }_1\left(t\right)=2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\nabla v\left(s\right)ds\right).D^{\alpha }\left({\int }_0^{t}k(t-s)\nabla v\left(s\right)ds\right)dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }{(t-\eta )}^{-\alpha }{\left({\int }_0^{\eta }k(\eta -\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }d\eta \right|}^{2}dx,t\ge 0\end{array}$$

and the relation (9) gives

$$\begin{array}{c}{D}^{\alpha }{\Psi }_1\left(t\right)=2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\nabla v\left(s\right)ds\right)\\ \times .\left[{}^{RL}{D}^{\alpha }\left({\int }_0^{t}k(t-s)\nabla v\left(s\right)ds\right)dx-\frac{t^{-\alpha }}{\Gamma (1-\alpha )}{\left.\left({\int }_0^{t}k(t-s)\nabla v\left(s\right)ds\right)\right|}_{t=0}\right]dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }{(t-\eta )}^{-\alpha }{\left({\int }_0^{\eta }k(\eta -\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }d\eta \right|}^{2}dx,t\ge 0.\end{array}$$

Next, the summability of k and Proposition 4 imply

$$\begin{array}{c}{D}^{\alpha }{\Psi }_1\left(t\right)=2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\nabla v\left(s\right)ds\right).\left[{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)\left[\nabla v\left(s\right)-\nabla v\left(t\right)\right]ds\right.\\ \left.+\nabla v\left(t\right){\int }_0^t{}^{RL}{D}^{\alpha }k\left(s\right)ds+\nabla v\left(t\right)\underset{t\to 0^{+}}{lim}{I}^{1-\alpha }k\left(t\right)\right]dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }{(t-\eta )}^{-\alpha }{\left({\int }_0^{\eta }k(\eta -\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }d\eta \right|}^{2}dx\\ =2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\nabla v\left(s\right)ds\right).{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)\left[\nabla v\left(s\right)-\nabla v\left(t\right)\right]dsdx\\ +2I^{1-\alpha }k\left(t\right){\int }_{\Omega }\left({\int }_0^{t}k(t-s)\nabla v\left(s\right)ds\right).\nabla v\left(t\right)dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }{(t-\eta )}^{-\alpha }{\left({\int }_0^{\eta }k(\eta -\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }d\eta \right|}^{2}dx,t\ge 0\end{array}$$

or, for ${\lambda }_1>0$

$$\begin{array}{c}{D}^{\alpha }{\Psi }_1\left(t\right)\le 2{\lambda }_1\overline{k}{\int }_{\Omega }\left({\int }_0^{t}k(t-s){\left|\nabla v\left(s\right)\right|}^{2}ds\right)dx\\ +\frac{1}{{\lambda }_1}\left({\int }_0^t\left|{}^{RL}{D}^{\alpha }k\right|\left(s\right)ds\right)(\left|{}^{RL}{D}^{\alpha }k\right|\square \nabla v)\left(t\right)+\frac{{\left[I^{1-\alpha }k\left(t\right)\right]}^2}{{\lambda }_1}{∥\nabla v∥}^2\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }{(t-\eta )}^{-\alpha }{\left({\int }_0^{\eta }k(\eta -\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }d\eta \right|}^{2}dx,t\ge 0.\end{array}$$

Observing that (see Proposition 3)

$$I^{1-\alpha }k\left(t\right)\le \frac{K}{\Gamma (1-\alpha )}\underset{0}{\overset{t}{\int }}{(t-s)}^{\alpha -1}{E}_{\alpha ,\alpha }(-C_k{(t-s)}^{\alpha })s^{-\alpha }ds=K{E}_{\alpha }(-C_k{t}^{\alpha })$$

and

$${\int }_0^t\left|{}^{RL}{D}^{\alpha }k\left(s\right)\right|ds=-{\int }_0^{t}D{I}^{1-\alpha }k\left(s\right)ds=-{\left[I^{1-\alpha }k\left(s\right)\right]}_0^t=I^{1-\alpha }k\left(0\right)-I^{1-\alpha }k\left(t\right)\le K$$

it appears that

$$\begin{array}{c}{D}^{\alpha }{\Psi }_1\left(t\right)\le 2{\lambda }_1\overline{k}{\int }_{\Omega }\left({\int }_0^{t}k(t-s){\left|\nabla v\left(s\right)\right|}^{2}ds\right)dx\\ +\frac{K}{{\lambda }_1}(\left|{}^{RL}{D}^{\alpha }k\right|\square \nabla v)\left(t\right)+\frac{{\left[K{E}_{\alpha }(-C_k{t}^{\alpha })\right]}^2}{{\lambda }_1}{∥\nabla v∥}^2\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }{(t-\eta )}^{-\alpha }{\left({\int }_0^{\eta }k(\eta -\tau )\nabla v\left(\tau \right)d\tau \right)}^{\prime }d\eta \right|}^{2}dx,t\ge 0\end{array}$$

for ${\lambda }_1>0.$□

The second functional we introduce is in the next lemma.

Lemma 4.

For the functional

$${\Psi }_2\left(t\right):={\int }_{\Omega }v{D}^{\alpha }vdx,t\ge 0,$$

it holds that

$$\begin{array}{c}{D}^{\alpha }{\Psi }_2\left(t\right)\le {∥D^{\alpha }v∥}^2-\left(1-{\lambda }_2\right){∥\nabla v∥}^2-{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}\\ +\frac{\overline{k}}{4{\lambda }_2}{\int }_{\Omega }\left({\int }_0^{t}k(t-s){\left|\nabla v\left(s\right)\right|}^{2}ds\right)dx+\frac{\alpha {\lambda }_3{C}_p}{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }\frac{\nabla v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right|}^{2}dx\\ +\frac{\alpha }{4{\lambda }_3\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right)}^{2}dx,t\ge 0\end{array}$$

for${\lambda }_2,{\lambda }_3>0,$where$C_p$is the Poincaré constant.

Proof. 

By Proposition 5 we have

$$\begin{array}{c}{D}^{\alpha }{\Psi }_2\left(t\right)={∥D^{\alpha }v∥}^2+{\int }_{\Omega }v{D}^{\alpha }\left(D^{\alpha }v\right)dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{v^{\prime }\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right)dx\\ ={∥D^{\alpha }v∥}^2-{∥\nabla v∥}^2+{\int }_{\Omega }\nabla v.{\int }_0^{t}k(t-s)\nabla v\left(s\right)dsdx-{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{v^{\prime }\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right)dx,t\ge 0.\end{array}$$

Therefore, for ${\lambda }_2,{\lambda }_3>0$ and $t\ge 0$

$$\begin{array}{c}{D}^{\alpha }{\Psi }_2\left(t\right)\le {∥D^{\alpha }v∥}^2-\left(1-{\lambda }_2\right){∥\nabla v∥}^2+\frac{\overline{k}}{4{\lambda }_2}{\int }_{\Omega }\left({\int }_0^{t}k(t-s){\left|\nabla v\left(s\right)\right|}^{2}ds\right)dx-{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}\\ +\frac{\alpha {\lambda }_3{C}_p}{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }\frac{\nabla v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right|}^{2}dx+\frac{\alpha }{4{\lambda }_3\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(s\right)\right]}^{\prime }ds}{{(t-s)}^{\alpha }}\right)}^{2}dx.\end{array}$$

The proof is complete. □

Lemma 5.

The functional

$${\Psi }_3\left(t\right):={\int }_0^{t}k(t-s){∥\nabla v\left(s\right)∥}^{2}ds,t\ge 0$$

satisfies

$$D^{\alpha }{\Psi }_3\left(t\right)\le -C_k{\Psi }_3\left(t\right)+\left(\underset{t\to 0^{+}}{lim}{I}^{1-\alpha }k\left(t\right)\right){∥\nabla v∥}^2,t\ge 0.$$

Proof. 

This follows immediately from Proposition 4 and (K). □

As we have removed the lower-order fractional term, which represented the damping and was responsible for the appearance of the nice term $-{∥D^{\alpha }v∥}^2,$ we need to find a way to come up with this missing and essential term. The next suggested functional will take care of this matter.

Lemma 6.

If the exponent$\gamma$satisfies (Γ), then for the functional

$${\Psi }_4\left(t\right):=-{\int }_{\Omega }{D}^{\alpha }v{\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx,t\ge 0,$$

we have, along solutions of problem (8)

$$\begin{array}{c}{D}^{\alpha }{\Psi }_4\left(t\right)={\lambda }_5{∥\nabla v∥}^2+\overline{k}\left(\frac{1}{4{\lambda }_5}+\frac{C_p}{4{\lambda }_6}+1\right)(k\square \nabla v)\left(t\right)+\left({\lambda }_4-\frac{k_0}{2}\right){∥D^{\alpha }v∥}^2\\ +{\lambda }_6{C}_1{\mathcal{E}}_1^{\gamma +1}\left(t\right)-\frac{K{C}_p}{2k_0}{(}^{RL}{D}^{\alpha }k\square \nabla v)\left(t\right)+\frac{C_4{E}_{\alpha }(-C_k{t}^{\alpha })}{{\lambda }_4}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)}^{2}dx\\ +\frac{\alpha {\lambda }_7}{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^{2}dx\\ +\frac{\alpha }{4{\lambda }_7\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx,\end{array}$$

for ${\lambda }_i>0,i=4,5,6,7,t\ge t_0>0$ where $k_0={\int }_0^{t_0}k\left(s\right)ds$ for some $t_0>0$.

Proof. 

By Proposition 5, we see that

$$\begin{array}{c}{D}^{\alpha }{\Psi }_4\left(t\right)=-{\int }_{\Omega }{D}^{\alpha }\left(D^{\alpha }v\right){\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx\\ -{\int }_{\Omega }{D}^{\alpha }v{D}^{\alpha }{\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx+\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)\\ \times \left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)dx,t\ge 0\end{array}$$

and along solutions of (8), we find for $t\ge 0$

$$\begin{array}{c}{D}^{\alpha }{\Psi }_4\left(t\right)=-{\int }_{\Omega }\left[\Delta v-{\int }_0^{t}k(t-s)\Delta v\left(s\right)ds-{\left|v\right|}^{\gamma }v\right]{\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx\\ -{\int }_{\Omega }{D}^{\alpha }v\left[D^{\alpha }\left(v\left(t\right){\int }_0^{t}k(t-s)ds\right)-D^{\alpha }{\int }_0^{t}k(t-s)v\left(s\right)ds\right]dx+\frac{\alpha }{\Gamma (1-\alpha )}\\ \times {\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)dx.\end{array}$$

Next, we apply Proposition 4

$$\begin{array}{c}{D}^{\alpha }{\Psi }_4\left(t\right)={\int }_{\Omega }\left[\left(1-{\int }_0^{t}k\left(s\right)ds\right)\nabla v+{\int }_0^{t}k(t-s)\left[\nabla v\left(t\right)-\nabla v\left(s\right)\right]ds\right]\\ .{\int }_0^{t}k(t-s)\left[\nabla v\left(t\right)-\nabla v\left(s\right)\right]dsdx\\ -{\int }_{\Omega }{D}^{\alpha }v\left[D^{\alpha }\left(v\left(t\right){\int }_0^{t}k(t-s)ds\right)-{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)v\left(s\right)ds-v\left(t\right)I^{1-\alpha }k\left(0\right)\right]dx\\ +{\int }_{\Omega }{\left|v\right|}^{\gamma }v{\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx+\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\\ \times \left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)dx\end{array}$$

or, for $t\ge 0$

$$\begin{array}{c}{D}^{\alpha }{\Psi }_4\left(t\right)=\left(1-{\int }_0^{t}k\left(s\right)ds\right){\int }_{\Omega }\nabla v.{\int }_0^{t}k(t-s)\left[\nabla v\left(t\right)-\nabla v\left(s\right)\right]dsdx\\ +{\int }_{\Omega }{\left|{\int }_0^{t}k(t-s)\left[\nabla v\left(t\right)-\nabla v\left(s\right)\right]ds\right|}^{2}dx+{\int }_{\Omega }{\left|v\right|}^{\gamma }v{\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx\\ -{\int }_{\Omega }{D}^{\alpha }v\left[D^{\alpha }\left(v\left(t\right){\int }_0^{t}k(t-s)ds\right)-{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)v\left(s\right)ds\right]dx\\ +I^{1-\alpha }k\left(0\right){\int }_{\Omega }v{D}^{\alpha }vdx+\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\\ \left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)dx.\end{array}$$

(18)

Another application of Proposition 5 gives

$$\begin{array}{c}{D}^{\alpha }\left(v\left(t\right){\int }_0^{t}k(t-s)ds\right)=\left({\int }_0^{t}k\left(s\right)ds\right)D^{\alpha }v\left(t\right)+v\left(t\right)\left[{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)ds+I^{1-\alpha }k\left(0\right)\right]\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right),t\ge 0.\end{array}$$

Therefore, for ${\lambda }_4>0,t\ge 0$

$$\begin{array}{c}-{\int }_{\Omega }{D}^{\alpha }v{D}^{\alpha }\left(v\left(t\right){\int }_0^{t}k\left(s\right)ds\right)dx=-{\int }_{\Omega }v{D}^{\alpha }v\left[{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)ds+I^{1-\alpha }k\left(0\right)\right]dx\\ -\left({\int }_0^{t}k\left(s\right)ds\right){∥D^{\alpha }v∥}^2+\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{D}^{\alpha }v{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)dx\\ \le -{\int }_{\Omega }v{D}^{\alpha }v{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)dsdx-I^{1-\alpha }k\left(0\right){\int }_{\Omega }v{D}^{\alpha }vdx-\left({\int }_0^{t}k\left(s\right)ds\right){∥D^{\alpha }v∥}^2\\ +{\lambda }_4{∥D^{\alpha }v∥}^2+\frac{{\alpha }^2}{4{\lambda }_4{\Gamma }^2(1-\alpha )}{\int }_{\Omega }{\left[{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)\right]}^{2}dx.\end{array}$$

(19)

We can apply the Cauchy-Schwarz inequality to the last term in (19) to get

$$\begin{array}{c}{\int }_{\Omega }{\left[{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)\right]}^{2}dx\\ \le {\int }_{\Omega }\left[{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)}^2{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)}^2\right]dx\\ \le {\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)}^2{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)}^{2}dx,t\ge 0.\end{array}$$

In view of assumption (K), Proposition 3 and the fact that $E_{\alpha }(-C_k{t}^{\alpha })\le C_2{t}^{-\alpha }$ (for some positive constant $C_2$) away from zero

$${\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)}^2\le {\left[K\Gamma (1-\alpha )E_{\alpha }(-C_k{t}^{\alpha })\right]}^2{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\le C_3{E}_{\alpha }(-C_k{t}^{\alpha }),$$

for $t\ge t_0>0$ where $C_3:=K^2{\Gamma }^2(1-\alpha )C_2.$ Therefore

$$\begin{array}{c}-{\int }_{\Omega }{D}^{\alpha }v{D}^{\alpha }\left(v\left(t\right){\int }_0^{t}k\left(s\right)ds\right)dx\le -\left({\int }_0^{t}k\left(s\right)ds\right){∥D^{\alpha }v∥}^2\\ -{\int }_{\Omega }v{D}^{\alpha }v{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)dsdx+{\lambda }_4{∥D^{\alpha }v∥}^2\\ -I^{1-\alpha }k\left(0\right){\int }_{\Omega }v{D}^{\alpha }vdx+\frac{C_4{E}_{\alpha }(-C_k{t}^{\alpha })}{{\lambda }_4}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)}^{2}dx,\end{array}$$

(20)

for ${\lambda }_4>0,$$t\ge t_0>0$ where $C_4={\alpha }^2{C}_3/4{\Gamma }^2(1-\alpha ).$ Moreover, it is clear that

$${\int }_{\Omega }\nabla v.{\int }_0^{t}k(t-s)\left[\nabla v\left(t\right)-\nabla v\left(s\right)\right]dsdx\le {\lambda }_5{∥\nabla v∥}^2+\frac{\overline{k}}{4{\lambda }_5}(k\square \nabla v)\left(t\right),{\lambda }_5>0,t\ge 0$$

(21)

and the embedding mentioned above ($H_0^1(\Omega )\subset L^q(\Omega )$ when $2\le q\le 2n/(n-2)$ if $n\ge 3$ and $q\ge 2$ if $n=1,2$) gives

$$\begin{array}{c}{\int }_{\Omega }{\left|v\right|}^{\gamma }v{\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx\le {\lambda }_6{\int }_{\Omega }{\left|v\right|}^{2(\gamma +1)}dx+\frac{\overline{k}{C}_p}{4{\lambda }_6}(k\square \nabla v)\left(t\right)\\ \le {\lambda }_6{C}_1{\mathcal{E}}_1^{\gamma +1}\left(t\right)+\frac{\overline{k}{C}_p}{4{\lambda }_6}(k\square \nabla v)\left(t\right),{\lambda }_6>0\end{array}$$

(22)

where $C_1$ is the embedding constant. Observe also that

$$0\le {\int }_0^t\left|{}^{RL}{D}^{\alpha }k\left(s\right)\right|ds=-{\int }_0^{t}D{I}^{1-\alpha }k\left(s\right)ds=-{\left[I^{1-\alpha }k\left(s\right)\right]}_0^t=I^{1-\alpha }k\left(0\right)-I^{1-\alpha }k\left(t\right)$$

implies

$$\begin{array}{c}{\int }_{\Omega }{D}^{\alpha }v{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)v\left(s\right)dsdx-{\int }_{\Omega }v{D}^{\alpha }v{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)dsdx\\ =-{\int }_{\Omega }{D}^{\alpha }v{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)\left(v\left(t\right)-v\left(s\right)\right)dsdx\\ \le \frac{k_0}{2}{∥D^{\alpha }v∥}^2+\frac{C_p}{2k_0}\left({\int }_0^t\left|{}^{RL}{D}^{\alpha }k\right|\left(s\right)ds\right)(\left|{}^{RL}{D}^{\alpha }k\right|\square \nabla v)\left(t\right)\\ \le \frac{k_0}{2}{∥D^{\alpha }v∥}^2-\frac{C_p}{2k_0}{I}^{1-\alpha }k\left(0\right){(}^{RL}{D}^{\alpha }k\square \nabla v)\left(t\right)\\ \le \frac{k_0}{2}{∥D^{\alpha }v∥}^2-\frac{K{C}_p}{2k_0}{(}^{RL}{D}^{\alpha }k\square \nabla v)\left(t\right),t\ge 0.\end{array}$$

(23)

Finally, for ${\lambda }_7>0$

$$\begin{array}{c}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)dx\\ \le {\lambda }_7{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^{2}dx\\ +\frac{1}{4{\lambda }_7}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx.\end{array}$$

(24)

Taking into account the relations (20)–(24) in (18), we obtain

$$\begin{array}{c}{D}^{\alpha }{\Psi }_4\left(t\right)\le {\lambda }_5{∥\nabla v∥}^2+\overline{k}\left(\frac{1}{4{\lambda }_5}+\frac{C_p}{4{\lambda }_6}+1\right)(k\square \nabla v)\left(t\right)\\ +\left({\lambda }_4-\frac{k_0}{2}\right){∥D^{\alpha }v∥}^2-\frac{K{C}_p}{2k_0}{(}^{RL}{D}^{\alpha }k\square \nabla v)\left(t\right)+{\lambda }_6{C}_1{\mathcal{E}}_1^{\gamma +1}\left(t\right)\\ +\frac{C_4{E}_{\alpha }(-C_k{t}^{\alpha })}{{\lambda }_4}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)}^{2}dx+\frac{\alpha {\lambda }_7}{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\\ \times {\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^{2}dx+\frac{\alpha }{4{\lambda }_7\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\\ \times {\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx,t\ge t_0>0.\end{array}$$

The proof is complete. □

One more term needs to be controlled. To this end we introduce the functional

$${\Psi }_5\left(t\right):={\int }_{\Omega }{\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)}^{2}dx,t\ge 0.$$

Lemma 7.

The above functional ${\Psi }_5\left(t\right)$ fulfills

$$\begin{array}{c}{D}^{\alpha }{\Psi }_5\left(t\right)\le {\lambda }_8{∥D^{\alpha }v∥}^2+C_p\left({\lambda }_9\overline{k}+\frac{{\overline{k}}^3}{{\lambda }_8}+\frac{{\lambda }_{10}\alpha \overline{k}}{\Gamma (1-\alpha )}\right)\left(k\square \nabla v\right)\left(t\right)\\ +\frac{C_{p}K}{{\lambda }_9}\left(\left|{}^{RL}{D}^{\alpha }k\right|\square \nabla v\right)\left(t\right)+\frac{C_5{E}_{\alpha }(-C_k{t}^{\alpha })}{{\lambda }_{10}}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)}^{2}dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx,t\ge 0.\end{array}$$

Proof. 

Clearly,

$$\begin{array}{c}{D}^{\alpha }{\Psi }_5\left(t\right)=2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)D^{\alpha }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx\end{array}$$

and

$$\begin{array}{c}{\int }_{\Omega }{D}^{\alpha }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)dx={\int }_{\Omega }{D}^{\alpha }\left(v\left(t\right){\int }_0^{t}k(t-s)ds\right)dx\\ -{\int }_{\Omega }{D}^{\alpha }\left({\int }_0^{t}k(t-s)v\left(s\right)ds\right)dx\\ =\left({\int }_0^{t}k\left(s\right)ds\right)D^{\alpha }v\left(t\right)+v\left(t\right)\left[{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)ds+I^{1-\alpha }k\left(0\right)\right]\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)-{\int }_{\Omega }{D}^{\alpha }\left({\int }_0^{t}k(t-s)v\left(s\right)ds\right)dx.\end{array}$$

Therefore,

$$\begin{array}{c}{D}^{\alpha }{\Psi }_5\left(t\right)=2\left({\int }_0^{t}k\left(s\right)ds\right){\int }_{\Omega }{D}^{\alpha }v\left(t\right)\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)dx\\ +2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)v\left(t\right)\left[{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)ds+I^{1-\alpha }k\left(0\right)\right]dx\\ -\frac{2\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right){\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)dx\\ -2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)D^{\alpha }\left({\int }_0^{t}k(t-s)v\left(s\right)ds\right)dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx,t\ge 0\end{array}$$

or, using (14), we find

$$\begin{array}{c}{D}^{\alpha }{\Psi }_5\left(t\right)=2\left({\int }_0^{t}k\left(s\right)ds\right){\int }_{\Omega }{D}^{\alpha }v\left(t\right)\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)dx\\ +2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)v\left(t\right)\left[{\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)ds+I^{1-\alpha }k\left(0\right)\right]dx\\ -\frac{2\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right){\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)dx\\ -2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)\left({}^{RL}{D}^{\alpha }{\int }_0^{t}k(t-s)v\left(s\right)ds+v\left(t\right)I^{1-\alpha }k\left(0\right)\right)dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right)}^{\prime }ds\right)}^{2}dx,t\ge 0\end{array}$$

that is

$$\begin{array}{c}{D}^{\alpha }{\Psi }_5\left(t\right)\le {\lambda }_8{∥D^{\alpha }v∥}^2+\frac{{\overline{k}}^3{C}_p}{{\lambda }_8}\left(k\square \nabla v\right)\left(t\right)\\ +2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right){\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx\\ -\frac{2\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right){\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx,t\ge 0.\end{array}$$

Furthermore, we may evaluate

$$\begin{array}{c}2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right){\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx\\ \le {\lambda }_9\overline{k}{\int }_{\Omega }{\int }_0^{t}k(t-s){\left[v\left(t\right)-v\left(s\right)\right]}^{2}dsdx\\ +\frac{1}{{\lambda }_9}\left({\int }_0^t\left|{}^{RL}{D}^{\alpha }k\left(s\right)\right|ds\right){\int }_{\Omega }{\int }_0^t\left|{}^{RL}{D}^{\alpha }k(t-s)\right|{\left[v\left(t\right)-v\left(s\right)\right]}^{2}dsdx\\ \le {\lambda }_9\overline{k}{C}_p\left(k\square \nabla v\right)\left(t\right)+\frac{C_p}{{\lambda }_9}\left({\int }_0^t\left|{}^{RL}{D}^{\alpha }k\left(s\right)\right|ds\right)\left(\left|{}^{RL}{D}^{\alpha }k\right|\square \nabla v\right)\left(t\right),{\lambda }_9>0\end{array}$$

and as

$$0\le {\int }_0^t\left|{}^{RL}{D}^{\alpha }k\left(s\right)\right|ds=-{\int }_0^{t}D{I}^{1-\alpha }k\left(s\right)ds=I^{1-\alpha }k\left(0\right)-I^{1-\alpha }k\left(t\right)\le K$$

we deduce that

$$\begin{array}{c}2{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right){\int }_0^t{}^{RL}{D}^{\alpha }k(t-s)\left[v\left(t\right)-v\left(s\right)\right]dsdx\\ \le {\lambda }_9\overline{k}{C}_p\left(k\square \nabla v\right)\left(t\right)+\frac{C_{p}K}{{\lambda }_9}\left(\left|{}^{RL}{D}^{\alpha }k\right|\square \nabla v\right)\left(t\right).\end{array}$$

Therefore,

$$\begin{array}{c}{D}^{\alpha }{\Psi }_5\left(t\right)\le {\lambda }_8{∥D^{\alpha }v∥}^2+\frac{{\overline{k}}^3{C}_p}{{\lambda }_8}\left(k\square \nabla v\right)\left(t\right)+{\lambda }_9\overline{k}{C}_p\left(k\square \nabla v\right)\left(t\right)+\frac{C_{p}K}{{\lambda }_9}\left(\left|{}^{RL}{D}^{\alpha }k\right|\square \nabla v\right)\left(t\right)\\ -\frac{2\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }\left({\int }_0^{t}k(t-s)\left[v\left(t\right)-v\left(s\right)\right]ds\right){\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}\left({\int }_0^{\xi }\frac{k\left(\eta \right)d\eta }{{(t-\eta )}^{\alpha }}\right)\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx\end{array}$$

for ${\lambda }_8,{\lambda }_9>0,t\ge 0.$ Further, using a similar argument to the one used to derive (20), for $C_5=\frac{\alpha C_3}{\Gamma (1-\alpha )}$ and ${\lambda }_8,{\lambda }_9,{\lambda }_{10}>0,$ we obtain

$$\begin{array}{c}{D}^{\alpha }{\Psi }_5\left(t\right)\le {\lambda }_8{∥D^{\alpha }v∥}^2+C_p\left({\lambda }_9\overline{k}+\frac{{\overline{k}}^3}{{\lambda }_8}+\frac{{\lambda }_{10}\alpha \overline{k}}{\Gamma (1-\alpha )}\right)\left(k\square \nabla v\right)\left(t\right)\\ +\frac{C_{p}K}{{\lambda }_9}\left(\left|{}^{RL}{D}^{\alpha }k\right|\square \nabla v\right)\left(t\right)+\frac{C_5{E}_{\alpha }(-C_k{t}^{\alpha })}{{\lambda }_{10}}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{v^{\prime }\left(s\right)ds}{{(t-s)}^{\alpha }}\right)}^{2}dx\\ -\frac{\alpha }{\Gamma (1-\alpha )}{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx,t\ge 0.\end{array}$$

This ends the proof. □

Let

$${\mathcal{L}}_1\left(t\right):=N{\mathcal{E}}_1\left(t\right)+{\sum }_{i=1}^5{N}_i{\Psi }_i\left(t\right),t>0$$

for some positive constants $N,$$N_i,$$i=1,\dots ,5$ to be determined later.

Theorem 3.

Assume that k satisfies (K) and γ satisfies (Γ). Then, the solution to the problem in (8) (with $a=0$) is locally Mittag–Leffler stable, that is there exist positive constants $B_1$ and ${\nu }_1$ such that

$${\mathcal{E}}_1\left(t\right)\le B_1{E}_{\alpha }(-{\nu }_1{t}^{\alpha }),t\ge 0$$

for small K (or, alternatively, a large $C_k).$

Proof. 

The evaluations obtained in the previous lemmas, with $N_1=N,$ imply

$$\begin{array}{c}{D}^{\alpha }{\mathcal{L}}_1\left(t\right)\le \left[N\frac{{\left[K{E}_{\alpha }(-C_k{t}^{\alpha })\right]}^2}{{\lambda }_1}+N_{3}K+N_4{\lambda }_5-N_2\left(1-{\lambda }_2\right)\right]{∥\nabla v∥}^2-N_2{∥v\left(t\right)∥}_{\gamma +2}^{\gamma +2}\\ +\left[N_2+{\lambda }_8{N}_5+N_4\left({\lambda }_4-\frac{k_0}{2}\right)\right]{∥D^{\alpha }v∥}^2+\left(2N{\lambda }_1\overline{k}+\frac{\overline{k}{N}_2}{4{\lambda }_2}-N_3{C}_k\right){\Psi }_3\left(t\right)\\ +{\lambda }_6{N}_4{C}_1{\mathcal{E}}_1^{\gamma +1}\left(t\right)+\left[\frac{N}{2}-\frac{K{N}_4{C}_p}{2k_0}-\frac{KN}{{\lambda }_1}-\frac{C_p{N}_{5}K}{{\lambda }_9}\right]{(}^{RL}{D}^{\alpha }k\square \nabla v)\left(t\right)\\ +\left[N_4\overline{k}\left(\frac{1}{4{\lambda }_5}+\frac{C_p}{4{\lambda }_6}+1\right)+\overline{k}{N}_5{C}_p\left({\lambda }_9+\frac{{\overline{k}}^2}{{\lambda }_8}+\frac{{\lambda }_{10}\alpha }{\Gamma (1-\alpha )}\right)\right]\left(k\square \nabla v\right)\left(t\right)\\ +\frac{\alpha }{4\Gamma (1-\alpha )}\left(\frac{N_2}{{\lambda }_3}+4{\lambda }_7{N}_4-2N\right){\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }\frac{{\left[D^{\alpha }v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right)}^{2}dx\\ +\left[\left(\frac{N_4{C}_4}{{\lambda }_4}+\frac{N_5{C}_5}{{\lambda }_{10}}\right)C_p{E}_{\alpha }(-C_k{t}^{\alpha })+\frac{\alpha \left(4{\lambda }_3{C}_p{N}_2-N\right)}{4\Gamma (1-\alpha )}\right]{\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left|{\int }_0^{\xi }\frac{{\left[\nabla v\left(\eta \right)\right]}^{\prime }d\eta }{{(t-\eta )}^{\alpha }}\right|}^{2}dx\\ +\frac{\alpha }{4\Gamma (1-\alpha )}\left(\frac{N_4}{{\lambda }_7}-4N_5\right){\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-s)}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left[v\left(s\right)-v\left(\tau \right)\right]d\tau \right)}^{\prime }ds\right)}^{2}dx.\end{array}$$

(25)

The strategy consists of choosing the parameters so as to give rise to a term of the form $-C{\mathcal{E}}_1\left(t\right),$$C>0$, in addition to ${\lambda }_6{N}_4{C}_1{\mathcal{E}}_1^{\gamma +1}\left(t\right)$, on the right-hand side of (25). Observe that, as $E_{\alpha }(-C_k{t}^{\alpha })$ decreases towards zero, choosing $t_0$ large enough allows us to ignore this term. Incidentally, we may also ignore ${\lambda }_4$ and ${\lambda }_{10}$. Let ${\lambda }_1=1,$${\lambda }_2=1/2$ and recall that $\overline{k}\le K/C_k.$ Therefore, for small values of K (or large values of $C_k$) and consequently small ${\lambda }_5$,${\lambda }_6,{\lambda }_8,$ and ${\lambda }_9$, it suffices that

$$\left\{\begin{array}{c}\frac{N_2}{{\lambda }_3}+4{\lambda }_7{N}_4\le 2N,\hfill \\ 4{\lambda }_3{C}_p{N}_2\le N,\hfill \\ N_2<N_4\frac{k_0}{2},\hfill \\ \frac{N_4}{{\lambda }_7}\le 4N_5.\hfill \end{array}\right.$$

To this end, we may pick ${\lambda }_3={\lambda }_7=1/4,$$N_5=N_4,$$N_2=\frac{k_0{N}_4}{4}$ and $N\ge N_{4}max$$\left\{C_p\frac{k_0}{4},k_0+1\right\}$. Finally, we go back and select the remaining parameters to ensure the fractional inequality

$$D^{\alpha }{\mathcal{L}}_1\left(t\right)\le -C_6{\mathcal{E}}_1\left(t\right)+M_4{C}_1{\mathcal{E}}_1^{\gamma +1}\left(t\right),t>t_0$$

for some $C_6>0$ and by the equivalence of ${\mathcal{L}}_1\left(t\right)$ and ${\mathcal{E}}_1\left(t\right)$

$$D^{\alpha }{\mathcal{L}}_1\left(t\right)\le -C_7{\mathcal{L}}_1\left(t\right)+C_8{\mathcal{L}}_1^{\gamma +1}\left(t\right),C_7,C_8,t>t_0.$$

Let $\overline{t},\omega >0$ be real numbers (we can assume that $t_0<\overline{t}$) such that ${\mathcal{L}}_1\left(t_0\right)<\omega$ and $C_8{\mathcal{L}}_1^{\gamma }\left(t\right)\le \omega <C_7,$$0\le t\le \overline{t}$ so that

$$D^{\alpha }{\mathcal{L}}_1\left(t\right)\le -\left(C_7-\omega \right){\mathcal{L}}_1\left(t\right),t_0\le t\le \overline{t}.$$

Consequently,

$${\mathcal{L}}_1\left(t\right)\le {\mathcal{L}}_1\left(t_0\right)E_{\alpha }(-\left(C_7-\omega \right){\left(t-t_0\right)}^{\alpha }),t_0\le t\le \overline{t}.$$

Assume that there exists $t_0\le t_1\le \overline{t}$ such that ${\mathcal{L}}_1\left(t_1\right)=\omega ,$ then as

$$E_{\alpha }(-\left(C_7-\omega \right){\left(t-t_0\right)}^{\alpha })\le 1,$$

we have

$$\nu ={\mathcal{L}}_1\left(t_1\right)\le {\mathcal{L}}_1\left(t_0\right)E_{\alpha }(-\left(\chi -\omega \right){\left(t_1-t_0\right)}^{\alpha })<\omega .$$

This is a contradiction. Starting from $\overline{t},$ we can continue the process forever; thus, the proof is complete. □

Remark 1.

The case $a\ne 0$ and $b\ne 0$, results from the above arguments. The functionals in the first case are good enough to obtain an explicit upper bound for the kernels ensuring Mittag–Leffler stability.

7. Conclusions

A half-order fractional derivative of the leading term is found to be capable of driving the system to rest. The obtained and expected rate is of Mittag–Leffler-type which is characteristic of the fractional derivatives. The stability is global in time. Therefore, this term is a damping term just as its integer-order counterpart. In the case of a memory term (viscoelastic), a serious problem arises from the evaluation of the expression

$${\int }_{\Omega }{\int }_0^t\frac{d\xi }{{(t-\xi )}^{1-\alpha }}{\left({\int }_0^{\xi }{(t-\eta )}^{-\alpha }{\left({\int }_0^{s}k(s-\tau )\left|\nabla v\left(\tau \right)\right|d\tau \right)}^{\prime }d\eta \right)}^{2}dx$$

appearing in the energy derivative. To cope with this situation, we introduced a new functional ${\Psi }_1\left(t\right)$. Regarding non- linearity, we assumed some addition assumptions on the exponent of non-linearity so as to profit from an embedding result. Unfortunately, the obtained fractional inequality differs from the one corresponding to the lower-order fractional term. To extract information about the concerned functional we had to consider small kernels and use an argument to deal with the arising non-linearity. The price to pay is a stability of the local character only.