**4. Dissipativity**

In this section, we derive some uniform estimates of solutions to problem (*P*) by using Proposition 2. Besides, in this section, we assume that *g*(*<sup>t</sup>*, *ut*) = *g*(*u*(*<sup>t</sup>* − *<sup>τ</sup>*(*t*))).

**Definition 7.** *The system* (*P*) *is said to be dissipative in CH if there exists a bounded set B* ⊂ *CH, such that for any given bounded set A* ⊂ *CH, there is a time t*∗ = *<sup>t</sup>*<sup>∗</sup>(*A*)*, such that for any given initial function φ* ∈ *A, for all t* ∈ [−*h*, 0]*, the values of the corresponding solution u*(*t*) *of the problem* (*P*) *are contained in B for all t* ≥ *t*<sup>∗</sup>*. The set B is called an absorbing set of the system* (*P*)*.*

We assume that

$$
\lambda\_1 \nu > \sqrt{2}L\_{\mathbb{X}}.\tag{32}
$$

**Theorem 3.** *(Existence of absorbing sets in CH) Assume that* (*g*<sup>1</sup>) − (*g*<sup>3</sup>)*,* (17) *and* (32) *hold. Then there exists T* > 0*, such that for all t* ≥ *T, the solution of problem* (*P*) *satisfies*

$$\|\mu\_t\|\_{C\_H}^2 \le \frac{\lambda\_1 \nu f\_0}{(\lambda\_1 \nu)^2 - 2L\_\mathcal{S}^2} + 1, \forall \ t \ge T\_\nu$$

*where f*0 = *ν*<sup>−</sup><sup>1</sup> sup *t*≥0 *f*(*t*)<sup>2</sup>∗.

**Proof.** Multiplying (1) by *u*, integrating over Ω, we have

$$\begin{split} \|D\_t^u |u(t)|^2 + \nu \|u(t)\|^2 &\le \|f(t)\|\_\ast \|u(t)\| + |\mathcal{g}(u(t-\tau(t)))| |u(t)| \\ &\le \frac{\|f(t)\|\_\ast^2}{\nu} + \frac{2L\_\varkappa^2}{\lambda\_1 \nu} \sup\_{t-\tau(t)\le s \le t} |u(s)|^2. \end{split} \tag{33}$$

Then we obtain

$$\left|D\_t^{\alpha}|u(t)|^2 \le f\_0 - \lambda\_1 \nu |u(t)|^2 + \frac{2L\_{\mathcal{S}}^2}{\lambda\_1 \nu} \sup\_{t-\tau(t) \le s \le t} |u(s)|^2, \ t \in (0, T],$$

$$\left|u(t)\right|^2 = \left|\phi(t)\right|^2, \ t \in [-h, 0],$$

where *f*0 = *ν*<sup>−</sup><sup>1</sup> sup *t*≥0 *f*(*t*)<sup>2</sup>∗. Using Proposition 2, we find that

$$\left|\mu(t)\right|^2 \le \frac{\lambda\_1 \nu f\_0}{(\lambda\_1 \nu)^2 - 2L\_\%^2} + \mathcal{M}E\_a(\lambda^\* t^a),$$

for all *t* ≥ *<sup>τ</sup>*(*t*), where *M* = *φ*<sup>2</sup>*CH* = sup −*h*≤*t*≤0 |*φ*(*t*)|2, and the parameter *λ*∗ is defined by

$$\lambda^\* = \sup\_{t-\tau(t)\geq 1} \{\lambda : \lambda - (-\lambda\_1 \nu) - \frac{2L\_\mathcal{S}^2}{\lambda\_1 \nu} \frac{E\_a(\lambda(t-\tau(t))^a)}{E\_a(\lambda t^a)} = 0\},$$

is strictly negative, namely, there exists some positive constants 0 satisfying −*λ*1*<sup>ν</sup>* + <sup>2</sup>*L*2*g λ*1*<sup>ν</sup>* < −0 such that *λ*∗ ∈ [−*λ*1*<sup>ν</sup>* + <sup>2</sup>*L*2*g λ*1*<sup>ν</sup>* , <sup>−</sup>0], and the estimate in (9) holds for all *t* such that *t* ≥ *τ*(*t*) + 1. In other words, for *λ*∗ ∈ [−*λ*1*<sup>ν</sup>* + <sup>2</sup>*L*2*g λ*1*<sup>ν</sup>* , <sup>−</sup>0], we have

$$|u(t)|^2 \le \frac{\lambda\_1 \nu f\_0}{(\lambda\_1 \nu)^2 - 2L\_\mathcal{S}^2} + M E\_a(\lambda^\* t^a), \quad \forall \ t \ge \tau(t) + 1.$$

For the case of *t* < *τ*(*t*) + 1, in order to analyze the dissipativity of problem (*P*) in phase space *CH* by Proposition 3, we first need to consider the following fractional differential equation,

$$\begin{aligned} D\_t^\kappa w(t) + \lambda\_1 \nu w(t) &= f\_0 + \frac{2L\_\chi^2}{\lambda\_1 \nu} w(t - \tau(t)), \ 0 < t \le h + 1, \\ w(t) &= \left| \phi(t) \right|^2, \ t \in [-h, 0], \end{aligned} \tag{34}$$

Then, by using the method of steps [28] (Theorem 1), we have that the initial value problem (34) has, on the interval [0, *kh*], a unique solution that can be represented by *w*(*t*) = *wih*(*t*), if (*i* − 1)*h* ≤ *t* ≤ *ih*,

$$w\_{ih}(t) = \int\_0^t \mathbb{E}\_{a,a}(-\lambda\_1 \nu(t-s)^a) f\_{ih}(s)ds + c\_{ih} \mathbb{E}\_a(-\lambda\_1 \nu t^a), \ t \in [(i-1)h, ih].$$

where *cih* is a constant, *i* = 1, 2, ··· , *k*.

$$f\_{kl}(t) := \begin{cases} \frac{2L\_k^2}{\lambda\_1 \nu} w\_{0l}(t - h) + f\_{0\prime} & 0 < t \le h\_{\prime} \\ \frac{2L\_k^2}{\lambda\_1 \nu} w\_{1h}(t - h) + f\_{0\prime} & h < t \le 2h\_{\prime} \\ \dots & \\ \frac{2L\_k^2}{\lambda\_1 \nu} w\_{(k-1)h}(t - h) + f\_{0\prime} & (k - 1)h < t \le kh\_{\prime} \end{cases}$$

is continuous and *<sup>w</sup>*0*h*(*t*) = |*φ*(*t*)|2. *k* is a smallest integer such that *kh* ≥ *h* + 1. Therefore, we obtain that

$$|w(t)| \le \sum\_{0 \le i \le k} |w\_{ih}(t)| \le \mathcal{C}E\_a(-\lambda\_1 \nu t^a), \forall 0 \le t \le \tau(t) + 1. \tag{35}$$

Now, we estimate the solution of (1), for *t* < *τ*(*t*) + 1. By (33), we have

$$\begin{split} |D\_t^s|u(t)|^2 + \lambda\_1 \nu |u(t)|^2 \le \frac{\|f(t)\|\_\ast^2}{\nu} + \frac{2L\_\xi^2}{\lambda\_1 \nu} |u(t - \tau(t))|, \; 0 \le t < \tau(t) + 1, \\ |u(t)|^2 = |\phi(t)|^2, \; t \in [-h, 0]. \end{split} \tag{36}$$

Then, by Proposition 3 and (34)–(36), we have

$$|u(t)|^2 \le \mathcal{C}E\_\mathfrak{a}(-\lambda\_1\nu t^a), \ 0 \le t < \tau(t) + 1.$$

So, we find that

$$|u(t)|^2 \le \frac{\lambda\_1 \nu f\_0}{(\lambda\_1 \nu)^2 - 2L\_\%^2} + ME\_u(\lambda^\* t^\mu) + CE\_u(-\lambda\_1 \nu t^\mu), \text{ for all } t \ge 0.$$

By the norm of *CH*, we conclude that

$$\|\|u\_t\|\|\_{\mathcal{C}\_H}^2 \le \frac{\lambda\_1 \nu f\_0}{(\lambda\_1 \nu)^2 - 2L\_{\mathcal{S}}^2} + M E\_u(\lambda^\* t^a) + \mathcal{C} E\_u(-\lambda\_1 \nu t^a), \text{ for all } t \ge 0, \theta \in [-h, 0].$$

Since *λ*∗ and −*λ*1*<sup>ν</sup>* are strictly negative, by the property of Mittag–Leffler function [2], we obtain

$$||\boldsymbol{\mu\_{1}}||\_{\mathcal{C}\_{H}}^{2} \leq \frac{\lambda\_{1}\nu f\_{0}}{(\lambda\_{1}\nu)^{2} - 2L\_{\mathcal{S}}^{2}} + \mathsf{C}\frac{\mathsf{C}\_{\boldsymbol{\alpha}}}{t^{\boldsymbol{\alpha}}}, \text{ as } t \to +\infty, \mathsf{C}$$

where *Cα* > 0 is a constant independent of *t*. Therefore, there exists *T* > 0 large enough, such that for all *t* ≥ *T*, the solution of problem (*P*) satisfies

$$\left\|\mu\_{\rm f}\right\|\_{C\_{H}}^{2} \leq \frac{\lambda\_{1}\nu f\_{0}}{(\lambda\_{1}\nu)^{2} - 2L\_{\mathcal{S}}^{2}} + 1, \; t \geq T.$$

Denote by *BCH* = *B*(0, 1 *λ*1*<sup>ν</sup> f*0 (*<sup>λ</sup>*1*<sup>ν</sup>*)<sup>2</sup>−2*L*2*g* + 1) the absorbing set in phase space *CH*, which implies that system (*P*) is dissipative. The proof is complete.
