Abstract

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.

1. Introduction

We are interested in the following system: with where , , , , , , the second integral represents the distributed delay and are bounded functions, where are two real numbers satisfying , and , are defined functions later.

Three decades ago, these problems that arise in one-dimensional 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, [1–24]. Recently, in the absence of delay (), 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, 28–30, 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.

2. Preliminaries

Let be the weighted Banach space equipped with the norm

be the Hilbert space of square integral functions having the finite norm and be the Hilbert space equipped with the norm

is the Hilbert space equipped with the norm

Theorem 1 [27]. For and in , we have where is a constant depending on and only.

As in [18], introducing the new variables yields

Problem (1) arrives at where

With the initial data and boundary conditions

We have the following assumptions:

(G1) are , nonincreasing functions satisfying

(G2) a differentiable function, such that and satisfies for some

And also, where , fixed, , such that

(G3) we take where and .

We have where

(G4) satisfying

Theorem 2. Assume (14) and . Then, , and problem (1) has a unique local solution for small enough.

Lemma 3. For , such that , we have

Proof. It is clear that by using the Minkowski inequality we get Also, Hlder’s and Young’s inequalities give us By applying the embedding and (25), (27) gives (15).

Lemma 4. such that

Proof. We prove inequality for and the same result also holds for .
It is clear that By Young’s inequality, with we get Therefore, Hence, by Poincaré’s inequality and (11), we obtain The proof of lemma is complete.

The energy function (see, e.g., [8, 19] and reference therein) is defined by where

Lemma 5. Let be the solution of system (11); then, is a nonincreasing function, that is, where

Proof. Multiplying equation (11)1,2 by , and integrating over , we find Using integration by parts, we get Now, multiplying equation (11)3 by and integrating over , we get Similarly, by multiplying equation (11)4 by and integrating over , we get Using Young’s and Cauchy-Schwartz inequalities, we have Similarly, we get By combining (39), (40), (41), (42), (43), (45), (46), (47), (48), (49), and (50) in (38), we get (34) and (36).

3. Global Existence

In this section, we showed the global existence of the solutions of the system (11).

First, introducing the following notation note that

Lemma 6. Assume that (24), (14), (15), (16), (17), and (22) hold, and , and satisfying

Then, such that where

Proof. As , then by continuity of , such that , ; this implies that we have a maximum time value noting such that This, with (51), (52), and (14), we have Hence, By (24) and (54), we get Hence, This proves that . By repeating the procedure, is extended to .

Theorem 7. Let (14), (15), (16), (17), (22), and (24) hold. Then, , , and satisfying (54) the solution of system (11) is bounded and global.

Proof. To prove that is bounded independently of , using (36) yields Using (52), we find By using (62) in (63), we get and using (14), (15), and (54) in (64), we get So where Hence, the solution of system (11) is bounded and global.

4. Decay of Solutions

In this section, the decay result is showed by using several lemmas.

As, we let where , and

Lemma 8. There exist , such that for , , and small enough.

Proof. Using the inequality of Young and the Poincaré-type inequality and , we find where .
A combination of (73), (74), (75), (76), and (77) in (68) gives Then, , for , , and small enough, such that Similarly, thanks to the inequalities of Young and Poincaré-type and using gives and By combining (80), (81), (82), (83), and (84) in (68), we find Then, , for , , and small enough, such that This completes the proof.

Lemma 9. For and , we have .

Proof. It suffices to note that using Hȯlder’s inequality for This completes the proof.

Lemma 10. Let be such that and be a continuous function on and suppose that and . Then, so that

Proof. By applying Lemma 8 with gives We also have by combining (82) and (83). This completes the proof.

Lemma 11. Suppose that be such that and be a continuous function on and assume . Then, so that

Proof. By using (82) for gives where to obtain (93). Hence, this ends the proof.

Lemma 12. Suppose that satisfies (15) and (52) hold. Then, the functional , given by (69), satisfies

For any .

Proof. The derivation of (11) gives By Young’s and Poincaré inequalities and (14) and (15), we find Similarly, we get . As we have and similarly, we find By using Young’s and Poincaré’s inequalities and (22) gives Similarly, we get In a combination of (98), (99), (100), (101), (102), (103), and (104) in (97), we obtain by choosing , so that ; hence, and ,and ; therefore, and .
Then, (96) is proved.

Lemma 13. Assuming that satisfies (15), (14), and (15) and (22) and (52) hold. Then, the functional given by (70) satisfies along the solution of (11) for any .

Proof. Direct calculation gives by using As we have the solution of (11), we find By Young’s inequality and (14) and (15), we arrive to Similarly, we get with So where Then, Thus, where Similarly, we have A combination of (110), (111), (112), (113), (114), (115), (117), (118), (119), (120), (121), (123), (124), and (125) into (109) gives (106).

Lemma 14. Let be the solution of (11). Then, for , the functional satisfies where and .

Proof. By differentiating and using equations (11)3 and (11)4, we get

Using the equality , , and , for any , we find

As is an increasing function, we have , for any .

Then, setting and (22), we obtain (126).

Theorem 15. Let , , and be defined and satisfy (163). Assume that satisfies (24), (14), (15), (16), (17), and (22) hold. Then, for each , and such that the solution of (11) satisfies , we have the following inequality for the energy function

Proof. As is continuous and , hence ; we have By using (36), (96), (106), (126), and (130) and , we get By choosing , , and so small that Then, At this point, we choose small enough, such that Then, Now, , and are fixed. Then, we select , and so small that (72) and (162) remain correct and Hence, by using (15) gives, for some , We choose , , and so small that By (134), we get With , and fixed, we pick , and such that We will make Then, we select , , and so small that (72) and (137) remain correct and

Next, as (137) is showed, according to the different ranges of , we give the following two cases.

Case 1. .
By choosing , and , (137) gives, for is constant so that, Therefore, with the help of the LHS of (72) and (143), we obtain By integration of (144) over gives Therefore, (129)1 is proved by (72) as well.

Case 2. .
We use (11), which gives We have, for , For , we have and (15), we find From (72) and (55) gives Similarly, we have for some . Hence, , we find We choose and and (144); we get, for some , By combining (72), (137), and (151), we find By integrating (153) gives Hence, From and as , we find Also, we use (24), and we get Hence, we find From which is bounded, using (72), (156), and (158) to get Therefore, using (55) and Lemma 10 gives This means for some .
Then, combining (137), (161), and (162) yields for some
As in [1], we obtain and some
Combining (163), (164), and (72), we find for some
By integrating (163) over , we get Hence, (129)2 is showed by (72) as well.

Data Availability

No data were used to support this study.

Conflicts of Interest

This work does not have any conflicts of interest.

Acknowledgments

The researchers would like to thank the Deanship of Scientific Research, Qassim University, for funding the publication of this project.