Global Existence for Two Singular One-Dimensional Nonlinear Viscoelastic Equations with respect to Distributed Delay Term

In this current work, we are interested in a system of two singular one-dimensional nonlinear equations with a viscoelastic, general source and distributed delay terms. The existence of a global solution is established by the theory of potential well, and by using the energy method with the function of Lyapunov, we prove the general decay result of our system.


Introduction
We are interested in the following system: with where Q = ð0, LÞ × ð0, TÞ, L < ∞, T < ∞, g 1 ð:Þ, g 2 ð:Þ: ℝ + ⟶ ℝ + , μ 1 , μ 3 > 0, the second integral represents the distributed delay and μ 2 , μ 4 : ½τ 1 , τ 2 ⟶ ℝ are bounded functions, where τ 1 , τ 2 are two real numbers satisfying 0 ≤ τ 1 < τ 2 , and f 1 ð:, :Þ, f 2 ð:,:Þ: ℝ 2 ⟶ ℝ are defined functions later. Three decades ago, these problems that arise in onedimensional elasticity have been studied and developed with regard to viscosity with long-term memory. And it has been studied in many fields of science, engineering, medical sciences, and chemistry, as well as population and other matters; see, for example, . Recently, in the absence of delay (μ i = 0, i = 1::4), problem (1) was studied in [25], and also later in [26], the authors considered problem (1) with localized frictional damping term. We also know that delay, especially distributed delay, is a phenomenon in our life and is almost found in various fields, and its inclusion in any problem makes it more important. The distributed delay in many works has been studied and many authors have taken care of it, for example, [5,9,27,28]. Based on all this and the results of the research papers [14,15,17,[28][29][30]31], the introduction of the term distributed delay as a damping mechanism in problem (1) makes it a new problem from what has been previously studied.
And we have divided this paper into the following. We present in the second section the definitions, basics, and theories of function spaces that are required throughout the rest of the paper. In Section 3, we present the energy function while proving to be decreasing. And in the final section, the general decay is obtained by applying the energy method and the function of Lyapunov.

Preliminaries
Let L p x = L p x ðð0, LÞÞ be the weighted Banach space equipped with the norm H = L 2 x ðð0, LÞÞ be the Hilbert space of square integral functions having the finite norm and K = L 2 x ðð0, LÞ × ð0, 1Þ × ðτ 1 , τ 2 ÞÞ be the Hilbert space equipped with the norm z k k K,μ 2 = x is the Hilbert space equipped with the norm Theorem 1 [27]. For 2 < p < 4 and ∀v in V 0 , we have where C * is a constant depending on L and p only.
As in [18], introducing the new variables Problem (1) arrives at where With the initial data and boundary conditions We have the following assumptions: (G1) g i ðtÞ: ℝ + ⟶ ℝ + are C 1 , nonincreasing functions satisfying (G2) ∃ξðtÞ > 0 a differentiable function, such that and ξðtÞ satisfies for some l < 1
Proof. We prove inequality for f 1 and the same result also holds for f 2 .
It is clear that By Young's inequality, with q = 2r + 3 r + 1 , we get Therefore, Hence, by Poincaré's inequality and (11), we obtain The proof of lemma is complete.

Journal of Function Spaces
By (24) and (54), we get Hence, This proves that IðtÞ > 0, ∀t ∈ ½0, T m Þ. By repeating the procedure, T m is extended to t * .

Decay of Solutions
In this section, the decay result is showed by using several lemmas.
Proof. Using the inequality of Young and the Poincaré-type inequality and 0 < ξðtÞ ≤ ξð0Þ, we find where C p > 0.
Proof. It suffices to note that using Hȯlder's inequality for This completes the proof.
Lemma 11. Suppose that v ∈ L ∞ ðð0, TÞ ; HÞ be such that v x ∈ L ∞ ðð0, TÞ ; HÞ and g be a continuous function on ½0, T and assume ρ > 1. Then, ∃C > 0 so that Proof. By using (82) for θ = 1 gives Journal of Function Spaces to obtain (93). Hence, this ends the proof.
Proof. Direct calculation gives 10 Journal of Function Spaces by using d dt As we have ðu, v, z, yÞ the solution of (11), we find By Young's inequality and (14) and (15), we arrive to Similarly, we get with So Then, Thus, where Similarly, we have