Int. J. Anal. Appl. (2022), 20:10 Well-Posedness and Stability for a System of Klein-Gordon Equations Naaima Latioui1,∗, Amar Guesmia1, Amar Ouaoua2 1Laboratory of Applied Mathematics and History and Didactics of Mathematics "LAMAHIS", Department of Mathematics, University 20 août 1955 Skikda, Algeria 2Laboratory of Applied Mathematics and History and Didactics of Mathematics "LAMAHIS", Department of Technologic, University 20 août 1955 Skikda, Algeria ∗Corresponding author: loubnalatioui@gmail.com, n.latioui@univ-Skikda.dz Abstract. In this paper, we study the weak existence of solution for a non-linear hyperbolic coupled system of Klein-Gordon equations with memory and source terms using the Faedo-Galerkin method techniques and compactness results, we have demonstrated the uniqueness of the solution by using the classical technique. In addition, we show that the solution remains stable over time. The reaction of the proper Lyapunov function is the primary tool of the proof. 1. Introduction In this paper, we consider a non-linear hyperbolic system of Klein-Gordon equations, defined as the following  utt − ∆u −αut + k ∗ ∆u −div(|v| 2∇u) + u|∇v|2 = 0 in Ω × (0,T ), vtt − ∆v −βvt + l ∗ ∆v −div(|u|2∇v) + v|∇u|2 = 0 in Ω × (0,T ), (1.1) with boundary conditions u(x,t) = v(x,t) = 0 on Γ × (0,T ), (1.2) ut(x,t) = vt(x,t) = 0 on Γ × (0,T ), (1.3) Received: Dec. 3, 2021. 2010 Mathematics Subject Classification. 81Q05, 37L45, 93D20. Key words and phrases. Klein-Gordon system; Faedo-Galerkin method; Lyapunov function. https://doi.org/10.28924/2291-8639-20-2022-10 ISSN: 2291-8639 © 2022 the author(s). https://doi.org/10.28924/2291-8639-20-2022-10 2 Int. J. Anal. Appl. (2022), 20:10 and Initial conditions u(x, 0) = u0(x), v(x, 0) = v0(x) on Ω, (1.4) ut(x, 0) = u1(x), vt(x, 0) = v1(x) on Ω. (1.5) Where Ω is a bounded domain of Rn (n ≥ 1) with smooth boundary Γ and let T > 0, α and β are non-positive constants, and (n∗w)(t) = ∫ t 0 n(t − s)w(s)ds. (1.6) Several authors have studied the Klein-Gordon non-linear system among them Medeiros & M. Miranda [8] considered the non-linear system  ∂2u ∂t2 − ∆u + u −|v|ρ|u|ρu = f1, ∂2v ∂t2 − ∆v + v −|u|ρ|v|ρv = f2, (1.7) they prove the existence and uniqueness of weak global solutions in Ω×[0,T ], where ρ is a real number meeting a specific condition and Ω is any domain of Rn. D. Andrade & A. Mognon [2], considered the non-linear system with memory term  utt − ∆u + f (u,v) + k ∗ ∆u = 0, vtt − ∆v + g(u,v) + l ∗ ∆v = 0, (1.8) for x ∈ Ω and t > 0 where f (u,v) = |u|ρ−2u|v|ρ, and g(u,v) = |v|ρ−2v|u|ρ, with ρ > 0 if n = 1, 2 and 1 < ρ ≤ n− 1 n− 2 if n ≥ 3, they use the argument from Komornik and Zuazua [6] to prove the existence of weak and strong solutions in Ω × (0,T ) given initial and boundary conditions. A. T. Louredo & M. M. Miranda [7], considered the non-linear system  u ′′ − ∆u + αv2u = 0, v ′′ − ∆v + αu2v = 0, (1.9) with the nonlinear boundary conditions, ∂u ∂ν + h1(.,u ′) = 0 on Γ1 × (0,∞), ∂v ∂ν + h2(.,v ′) = 0 on Γ1 × (0,∞), and boundary conditions u = v = 0 on (Γ/Γ1)×(0,∞), where Ω is a bounded open set of Rn (n ≤ 3), α > 0 a real number, Γ1 is a subset of the border Γ of Ω and hi a real function defined on Γ1×(0,∞). They use the Galerkin approach to demonstrate the existence of global solutions. Int. J. Anal. Appl. (2022), 20:10 3 K. Zennir & A. Guesmia [10], considered the non-linear κth-order with non-linear sources and memory terms   u′′1 + (−1) κ∆κu1 + m 2 1u1 + α1(t) ∫ t 0 g1(t − s)∆κu1(x,s)ds + |u′1| r−2|u′1| = |u1|p−2u1|u2|p, u′′2 + (−1) κ∆κu2 + m 2 2u2 + α2(t) ∫ t 0 g2(t − s)∆κu2(x,s)ds + |u′2| r−2|u′2| = |u2|p−2u2|u1|p, (1.10) using the potential well method, they verify the existence of global solutions in the a bounded domain Ω of Rn, where mi = 1, 2 are non-negative constants, r,p ≥ 2,κ ≥ 1. C. L. Frota & A. Vicente [5], studied the non-linear system of Klien-Gordon with acoustic boundary conditions   u ′′ − ∆u + |v|ρ+2|uρ|u = f1 in Ω × (0,T ), v ′′ − ∆v + |u|ρ+2|vρ|v = f1 in Ω × (0,T ), (1.11) they demonstrate the existence of both global and weak solutions, as well as their uniqueness. Our objective is to prove that the problem (1.1)-(1.5) has a weak and unique solution such that the kernel terms k, l have some hypothesis as well as using some ideas from articles ( [2]) and ( [9]). 2. Preliminaries Let Ω be a domain in Rn with smooth boundary Γ let T > 0. The inner product and norm in L2(Ω) are denoted by 〈u,v〉 = ∫ Ω u(x)v(x)dx, |u|2 = (∫ Ω |u(x)|2dx )1 2 . (2.1) The norm in H10 (Ω) is denoted by ‖u‖H10 (Ω) = (∫ Ω |∇u|2dx )1 2 . (2.2) We assume that k, l: R+ −→R+ are non-increasing differentiable functions satisfying : l1 = ( 1 − ∫ t 0 l(s)ds ) > 0 and k1 = ( 1 − ∫ t 0 k(s)ds ) > 0, (2.3) and k′(t) ≤−k(t), l′(t) ≤−l(t). (2.4) If w = w(t,x) is a function in L2(0.T ; H10 (Ω)) and k is continuous we put: (k ◦w)(t) = ∫ t 0 k(t − s)|∇w(t) −∇w(s)|22ds. 4 Int. J. Anal. Appl. (2022), 20:10 Lemma 2.1. [2] w ∈ C1((0,T ); H10 (Ω)), k ∈ C 1(0,∞) ∫ t 0 k(t − s) 〈 ∇w(s)∇w ′(t) 〉 ds = − 1 2 d dt (k ◦w)(t) + 1 2 d dt (∫ t 0 k(s)ds ) |∇w(t)|22 (2.5) +(k′ ◦w)(t) −k(t)|∇w(t)|22. Lemma 2.2. [9] (Young’s Inequality) Let a, b ≥ 0 and 1 q + 1 p = 1 for 1 < p,q < +∞, then one has the inequality ab ≤ δaq + c(δ)bp, where δ > 0 is an arbitrary constant, and c(δ) is a positive constant depending on δ. Lemma 2.3. [1] (Sobolev-Poincaré inequality) Let s be a number with 2 ≤ s < +∞ if n ≤ 2 and 2 ≤ s ≤ 2n n−2 if n > 2. Then there is a constant C depending on Ω and s such that ‖u‖s ≤ C‖∇u‖2, u ∈ H10. Theorem 2.1. Let u0, v0 ∈ L2(Ω) and u1, v1 ∈ L1(Ω). Then, under assumptions on two functions k and l, the problem (1.1)-(1.5) has a local solution (u(x,t),v(x,t)) such that u,v ∈ L∞(0,T ; H10 (Ω)) ∩L ∞(0,T ; L2(Ω)), (2.6) ut,vt ∈ L∞(0,T ; L2(Ω)). (2.7) Theorem 2.2. Let u,v :→ L2(Ω) be functions in the class (2.6) and (2.7) satisfying from (1.1) to (1.5), with u,v ∈ H2(Ω). Then the solution (u,v) obtained in Theorem (2.1) is unique. 3. Global Existence Step 1: Approximate solution. Using the Faedo-Galerkin process, we will determine the existence of a local solution to the problem (1.1)-(1.5) in this section. Let {wi} be a basis for both H2(Ω)∩H10 (Ω) and L2(Ω) for each positive integer m we put V = span{w1,w2, . . . ,wm}. we look for an approximate solution in the form um(t) = m∑ i=1 uimwi and vm(t) = m∑ i=1 vimwi, Int. J. Anal. Appl. (2022), 20:10 5 satisfying the approximate problem∫ Ω {umtt − ∆u m −αumt }widx − ∫ t 0 k(t − s)〈∇um(s),∇wi〉ds (3.1) + ∫ Ω |vm|2∇um∇widx + ∫ Ω um|∇vm|widx = 0,∫ Ω {vmtt − ∆v m −βvmt }widx − ∫ t 0 l(t − s)〈∇vm(s),∇wi〉ds (3.2) + ∫ Ω vm|∇um|widx + ∫ Ω |um|2∇vm∇wi = 0, with initial conditions satisfying u m(0) = um0 , ∑m i=1 aimwi = u m 0 → u0, v m(0) = vm0 , ∑m i=1 bimwi = v m 0 → v0 in L 2(Ω), umt (0) = u m 1 , ∑m i=1 a 1 imwi = u m 1 → u1, v m t (0) = v m 1 , ∑m i=1 b 1 imwi = v m 1 → v1 in L 1(Ω). (3.3) Since the vectors {wi} are linearly independent, this means det(wi,wj) 6= 0, the latter ensuring that the problem admits a local solution (um(t),vm(t)) in the interval [0,Tm]. Step 2: A priori estimate. Our system’s energy functional E(t) is given by 2E(t) = |umt | 2 L2(Ω) + |v m t | 2 L2(Ω) + (k ◦u m)(t) + (l ◦vm)(t) (3.4) + ( 1 − ∫ t 0 k(s)ds ) |∇um|22 + ( 1 − ∫ t 0 l(s)ds ) |∇vm|22 + |v m∇um|22 + |u m∇vm|22. After that, we multiply (3.1) by ut, (3.2) by vt, and use identity (2.5) to get d dt E(t) = (k′ ◦um)(t) + (l′ ◦vm)(t) −k(t)|∇um|22 − l(t)|∇v m|22 + α|u m t | 2 2 + β|v m t | 2 2 ≤ 0. (3.5) We found that d dt E(t) is a non-positive function, this last indicates that E(t) is a non-increasing function, meaning there exists a positive constant C1, independent of t and m such that |umt | 2 2 + |v m t | 2 2 + |∇u m|22 + |∇v m|22 + |u m∇vm|22 + |v m∇um|22 ≤ C1. (3.6) From this estimation, deduce that Tm = T. In addition, we get  um,vm is bounded in L ∞ (0,T ; H10 (Ω)), um,vm is bounded in L ∞ (0,T ; L2(Ω)), umt ,v m t is bounded in L ∞ (0,T ; L2(Ω)). (3.7) By the Holder inequality, the embedding H10 (Ω) ↪→ L 6(Ω) and (3.7), we obtain |um|∇vm|2|22 ≤‖u m‖2L6(Ω)‖∇v m‖4L6(Ω) ≤ C1, (3.8) ||vm|2∇um|22 ≤‖v m‖4L6(Ω)‖∇u m‖2L6(Ω) ≤ C2 ∀(u m,vm) in H2(Ω). (3.9) 6 Int. J. Anal. Appl. (2022), 20:10 Therefore (um|∇vm|2) is bounded in L∞(0,T ; L2(Ω)), (3.10) (|vm|2∇um) is bounded in L∞(0,T ; L2(Ω)). (3.11) Analogously (vm|∇um|2) is bounded in L∞(0,T ; L2(Ω)), (3.12) (|um|2∇vm) is bounded in L∞(0,T ; L2(Ω)). (3.13) Step 3: passage to the limit. From (3.7), (3.10), (3.11), (3.12) and (3.13) there exists a sub- sequence of (um) and a subsequence of (vm), denoted by same symbols, such that  um → u and vm → v weak star in L∞(0,T ; H10 (Ω)), um → u and vm → v weak star in L∞(0,T ; L2(Ω)), umt → ut and vmt → vt weak star in L∞(0,T ; L2(Ω)), um|∇vm|2 → χ1 weak star in L∞(0,T ; L2(Ω)), vm|∇um|2 → χ2 weak star in L∞(0,T ; L2(Ω)), |vm|2∇um → χ3 weak star in L∞(0,T ; L2(Ω)), |um|2∇vm → χ4 weak star in L∞(0,T ; L2(Ω)). (3.14) From (3.14) and Aubin-Lions compactness Lemma in ( [3]), we obtain um → u, vm → v strongly in L∞(0,T ; L 2 (Ω)), (3.15) since ∇um and ∇vm are bounded, then we have  um|∇vm|2 → u|∇v|2 strongly in L2(0,T ; L2(Ω)), vm|∇um|2 → v|∇u|2 strongly in L2(0,T ; L2(Ω)), |um|2∇vm →|u|2∇v strongly in L2(0,T ; L2(Ω)), |vm|2∇um →|v|2∇u strongly in L2(0,T ; L2(Ω)). (3.16) Then, there exists a subsequences of um and vm, which we will denote by um,vm respectively, such that   um|∇vm|2 → u|∇v|2 almoust everywhere in (0,T ) × Ω, vm|∇um|2 → v|∇u|2 almoust everywhere in (0,T ) × Ω, |um|2∇vm →|u|2∇v almoust everywhere in (0,T ) × Ω, |vm|2∇um →|v|2∇u almoust everywhere in (0,T ) × Ω. (3.17) Int. J. Anal. Appl. (2022), 20:10 7 From Lemma (3.15) in ( [11]) and (3.17) we deduce  um|∇vm|2 → u|∇v|2 weakly in L∞(0,T ; L2(Ω)), vm|∇um|2 → v|∇u|2 weakly in L∞(0,T ; L2(Ω)), |um|2∇vm →|u|2∇v weakly in L∞(0,T ; L2(Ω)), |vm|2∇um →|v|2∇u weakly in L∞(0,T ; L2(Ω)). (3.18) By the last formula (3.18) and (3.14) we get χ1 = u|∇v|2, χ2 = v|∇u|2, (3.19) χ3 = |v|2∇u, χ4 = |u|2∇v. Taking wi = 1 in (3.1) become (umtt, 1) −α(u m t , 1) + (u m|∇vm|2, 1) = 0 (3.20) |(umtt, 1)| = ∣∣α(umt , 1) − (u|∇vm|2, 1)∣∣ . Using the Cauchy Schwartz inequality, we have ‖umtt‖L1(Ω) ≤ |α||u m t |2m 1 2 (Ω) + |um|∇vm|2|2m 1 2 (Ω), such that, m(Ω) is a measure of Ω. Since, the measure of Ω is finite, and (3.14), we obtain ‖umtt‖L1(Ω) ≤ C1. (3.21) Analogously ‖vmtt‖L1(Ω) ≤ C2. (3.22) Then  u m tt is bounded in L ∞(0,T ; L1(Ω)), vmtt is bounded in L ∞(0,T ; L1(Ω)). (3.23) Similarly we have  u m tt → utt weakly star in L∞(0,T ; L1(Ω)), vmtt → vtt weakly star in L∞(0,T ; L1(Ω)). (3.24) From (3.14), (3.24) and lemma (3.1.7) in ( [12]) with B = L2(Ω) and B = L1(Ω) we get u m 0 → u(0), v m 0 → v(0) weakly star in L 2(Ω), um1 → u1(0), v m 1 → v1(0) weakly star in L 1(Ω). (3.25) 8 Int. J. Anal. Appl. (2022), 20:10 From (3.25) and (3.3) we get u(0) = u0, v(0) = v0, (3.26) u1(0) = u1, v1(0) = v1. (3.27) Setting up m →∞ and passing to the limit in (3.1), (3.2), we obtained ∫ Ω {utt − ∆u −αut}widx − ∫ t 0 k(t − s)〈∇u(s),∇wi〉ds (3.28) + ∫ Ω |v|2∇u∇widx + ∫ Ω u|∇v|widx = 0,∫ Ω {vtt − ∆v −βvt}widx − ∫ t 0 l(t − s)〈∇v(s),∇wi〉ds (3.29) + ∫ Ω v|∇u|widx + ∫ Ω |u|2∇v∇wi = 0. i = 1...,m. since (wi)∞i=1 is a base of H 1 0 (Ω), we deduce that (u,v) satisfies (1.1). The proof is complete. Lemma 3.1. Let u0,v0 ∈ H10 (Ω) and u1,v1 ∈ L 2(Ω) be given. Assume that (2.3) and (2.4) are true. Then the problem’s local solution (1.1)-(1.5) is global in time. proof. Since the map t → E(t) is a non-increasing function, i.e there exists a positive constant C1, independent of t, such that Delete m from these equations C1 ≥ 2E(t) = |ut|2L2(Ω) + |vt| 2 L2(Ω) + (k ◦u)(t) + (l ◦v)(t) (3.30) + ( 1 − ∫ t 0 k(s)ds ) |∇u|22 + ( 1 − ∫ t 0 l(s)ds ) |∇v|22 + |v∇u| 2 2 + |u∇v| 2 2 > 0, which give C1 ≥ 2E(t) ≥ |ut|2L2(Ω) + |vt| 2 L2(Ω) + ( 1 − ∫ t 0 k(s)ds ) |∇u|22 + ( 1 − ∫ t 0 l(s)ds ) |∇v|22 (3.31) + |v∇u|22 + |u∇v| 2 2 > 0, consequently, ∀t ∈ [0,T ], we have |ut|2L2(Ω) + |vt| 2 L2(Ω) + |∇u|22 + |∇v| 2 2 + |v∇u| 2 2 + |u∇v| 2 2 ≤ C1. This deduces that the solution. Int. J. Anal. Appl. (2022), 20:10 9 4. Uniqueness Let (u,v) and (u1,v1) two solutions of (1.1), we assume that U = u −u1 and V = v −v1 satisfy Utt − ∆U −αUt + k ∗ ∆U −div(|v|2∇u −|v1|2∇u1) + (u|∇v|2 −u1|∇v1|2) = 0 in Ω × (0,T ), (4.1) Vtt − ∆V −βVt + k ∗ ∆V −div(|u|2∇v −|u1|2∇v1) + (v|∇u|2 −v1|∇u1|2) = 0 in Ω × (0,T ), (4.2) with U(0) = V (0) = 0 Ut(0) = Vt(0) = 0. (4.3) Let as put 2E2(t) = |Ut|22 + |Vt| 2 2 + (k ◦U)(t) + (l ◦V )(t) (4.4) + ( 1 − ∫ t 0 k(s)ds ) |∇U|22 + ( 1 − ∫ t 0 l(s)ds ) |∇V |22. Multiplying (4.1) by Ut(t) and (4.2) by (Vt(t)) and summing up the product result we have d dt E2(t) ≤ ∫ Ω div(|v|2∇u −|v1|2∇u1)Ut − (u|∇v|2 −u1|∇v1|2)Utdx (4.5) + ∫ Ω div(|u|2∇v −|u1|2∇v1)Vt − (v|∇u|2 −v1|∇u1|2)Vtdx, d dt E2(t) ≤ ∫ Ω ∣∣|v|2∇u −|v1|2∇u1∣∣|∇Ut| + ∣∣u|∇v|2 −u1|∇v1|2∣∣|Ut|dx (4.6) + ∫ Ω ∣∣|u|2∇v −|u1|2∇v1∣∣|∇Vt| + ∣∣v|∇u|2 −v1|∇u1|2∣∣|Vt|dx, ∫ Ω (|v|2∇u −|v1|2∇u1)∇Ut + (|u|∇v|2 −u1|∇v1|2)Utdx (4.7) = ∫ Ω |v|2∇U∇Ut + ∇u1 [ |v|2 −|v1|2 ] ∇Ut + Ut|∇v|2U + u1 [ |∇v|2 −|∇v1|2 ] dx. From the mean value theorem, it follows that∫ Ω ∣∣|v|2∇u −|v1|2∇u1∣∣|∇Ut| + ∣∣u|∇v|2 −u1|∇v1|2∣∣|Ut|dx ≤ ∫ Ω |v|2|∇U||∇Ut| + 2|∇u1|2 [|v| + |v1|] |∇Ut||V | + |∇v|2|Ut||U|. Working in the same way as in argument of Lemma (2.2) in ( [2]) there exists C > 0 such that∫ Ω ∣∣|v|2∇u −|v1|2∇u1∣∣|∇Ut| + ∣∣|u|∇v|2 −u1|∇v1|2∣∣|Ut|dx ≤ C {|∇U|22 + |∇V |22 + |Ut|22} . Analogously we have∫ Ω ∣∣|u|2∇v −|u1|2∇v1∣∣|∇Vt| + ∣∣|v|∇u|2 −v1|∇u1|2∣∣|Vt|dx ≤ C {|∇V |22 + |∇U|22 + |Vt|22} , 10 Int. J. Anal. Appl. (2022), 20:10 and from (4.6) we have d dt E2(t) ≤ C { |∇U|22 + |∇V | 2 2 + |Ut| 2 2 + |Vt| 2 2 } (4.8) d dt E2(t) ≤ CE2(t). (4.9) Then, by using Gronwall’s lemma (1.3) in ( [4]) we get |∇U|22 = |∇V | 2 2 = |Ut| 2 2 = |Vt| 2 2 = 0. (4.10) This proves the uniqueness of the solution. 5. Stability Theorem 5.1. Let u0,v0 ∈ H10 (Ω) and u1,v1 ∈ L 2(Ω) be given. Assume that (2.3) and (2.4) hold. Then there exists two positive constants µ1 and µ2 independent of t such that 0 < E(t) ≤ µ1e −µ2t,∀t ≥ 0. Proof. We define the function of Laypunov, for � > 0 as follows L(t) = E(t) + � ∫ Ω utu + vtvdx. (5.1) We prove that L(t) and E(t) are equivalent, meaning that there exist two positive constants N and M depending on � such that for t ≥ 0 NE(t) ≤ L(t) ≤ ME(t). (5.2) From the Lemma (2.2), we have L(t) ≤ E(t) + � [ 1 2δ |ut|22 + δ|u| 2 2 ] + � [ 1 2δ |vt|22 + δ|v| 2 2 ] . By using the Poincaré inequality, we get L(t) ≤ E(t) + � [ 1 2δ |ut|22 + δC1|∇u| 2 2 ] + � [ 1 2δ |vt|22 + δC2|∇v| 2 2 ] . From (3.31) we have L(t) ≤ E(t) + � [ 1 δ E(t) + 2δ C1 k1 E(t) ] + � [ 1 δ E(t) + 2δ C2 l1 E(t) ] L(t) ≤ E(t) + 2� 1 δ E(t) + 2�δ C1 k1 E(t) + 2�δ C2 l1 E(t) L(t) ≤ ME(t) such that M = 1 + 2� 1 δ + 2�δ C1 k1 + 2�δ C2 l1 . Int. J. Anal. Appl. (2022), 20:10 11 On the other hand, we have L(t) ≥ E(t) − � [ 1 2δ |ut|22 + δ|u| 2 2 ] − � [ 1 2δ |vt|22 + δ|v| 2 2 ] ≥ E(t) − � [ 1 2δ |ut|22 + δC1|∇u| 2 2 ] − � [ 1 2δ |vt|22 + δC2|∇v| 2 2 ] ≥ E(t) − � [ 1 δ E(t) + 2δ C1 k1 E(t) ] − � [ 1 δ E(t) + 2δ C2 l1 E(t) ] , L(t) ≥ NE(t) such that N = 1 − 2� 1 δ − 2�δ C1 k1 − 2�δ C2 l1 . Now we have d dt L(t) = d dt E(t) + � ∫ Ω [ u2t + uttu + v 2 t + vttv ] dx (5.3) � ∫ Ω uttudx = � ∫ Ω [ u.∆u + αuut −u.k ∗ ∆u + u.div(|v|2∇u) −u|∇v|2.u ] dx ≤ � [ −|∇u|22 + α 1 2δ |ut|22 + αδ|u| 2 2 −|v∇u| 2 2 −|u∇v| 2 2 + ∫ Ω ∇u ∫ t 0 k(t − s)∇u(s)dsdx ] (5.4) ≤ � [ −|∇u|22 + α 1 2δ |ut|22 + αC1δ|∇u| 2 2 −|v∇u| 2 2 −|u∇v| 2 2 + ∫ Ω ∇u ∫ t 0 k(t − s)∇u(s)dsdx ] . Analogous � ∫ Ω vttvdx = � ∫ Ω [ v.∆v + βvvt −v.k ∗ ∆v + v.div(|u|2∇v) −v|∇u|2.v ] dx (5.5) ≤ � [ −|∇v|22 + β 1 2δ |vt|22 + βC1δ|∇v| 2 2 −|u∇v| 2 2 −|v∇u| 2 2 + ∫ Ω ∇v ∫ t 0 l(t − s)∇v(s)dsdx ] . d dt L(t) ≤ d dt E(t) + �|ut|22 + �|vt| 2 2 − �|∇u|22 + �α 1 2δ |ut|22 + �αC1δ|∇u| 2 2 − �|v∇u| 2 2 − �|u∇v| 2 2 + � ∫ Ω ∇u ∫ t 0 k(t − s)∇u(s)dsdx − �|∇v|22 + �β 1 2δ |vt|22 + �βC1δ|∇v| 2 2 − �|u∇v| 2 2 − �|v∇u| 2 2 + � ∫ Ω ∇v ∫ t 0 l(t − s)∇v(s)ds.dx. (5.6) The last term of relation (5.6) can be estimated as follow.∣∣∣∣ ∫ Ω ∇u ∫ t 0 k(t − s)∇u(s)dsdx ∣∣∣∣ ≤ ∫ Ω (∫ t 0 k(t − s)|∇u(s) −∇u(t)|ds ) dx + ∫ t 0 k(s)ds|∇u|22 (5.7) ≤ (1 + η)(1 −k1)|∇u|22 + 1 4η (k ◦∇u)(t) for η > 0. 12 Int. J. Anal. Appl. (2022), 20:10 Analogously ∣∣∣∣ ∫ Ω ∇v ∫ t 0 l(t − s)∇v(s)dsdx ∣∣∣∣ ≤ (1 + η)(1 − l1)|∇v|22 + 1 4η (l ◦∇v)(t) for η > 0. So d dt L(t) ≤ (k′ ◦u)(t) + (l′ ◦v)(t) −k(t)|∇u|22 − l(t)|∇v| 2 2 + α|ut| 2 2 + β|vt| 2 2 + �|ut| 2 2 + �|vt| 2 2 − �|∇u|22 + �α 1 2δ |ut|22 + �αC1δ|∇u| 2 2 − �|v∇u| 2 2 − �|u∇v| 2 2 + �(1 + η)(1 −k1)|∇u| 2 2 + � 1 4η (k ◦∇u)(t) − �|∇v|22 + �β 1 2δ |vt|22 + �βC1δ|∇v| 2 2 − �|u∇v| 2 2 − �|u∇u| 2 2 + �(1 + η)(1 − l1)|∇v|22 + � 1 4η (l ◦∇v)(t), (5.8) so d dt L(t) ≤ ( α + � + �α 1 2δ ) |ut|22 + ( β + � + �β 1 2δ ) |vt|22 + (−k(t) − � + �αC1δ + �(1 + η)(1 −k1)) |∇u|22 + (−l(t) − � + �βC1δ + �(1 + η)(1 − l1))|∇v|22 + (−2�− 1)|v∇u| 2 2 + (−2�− 1)|v∇u| 2 2 − (k ◦u)(t) − (l ◦v)(t) + |u∇v|22 + |v∇u| 2 2 + � 1 4η (k ◦∇u)(t) + � 1 4η (l ◦∇v)(t), (5.9) so d dt L(t) ≤ γE(t) + λ, (5.10) We choosing � small enough, such that γ =Min(α + � + �α 1 2δ ; β + � + �β 1 2δ ;−k(t) − � + �αC1δ + �(1 + η)(1 −k1) (5.11) − l(t) − � + �βC1δ + �(1 + η)(1 − l1); (−2�− 1);−1) < 0, and λ = |u∇v|22 + |v∇u| 2 2 + � 1 4η (k ◦∇u)(t) + � 1 4η (l ◦∇v)(t). (5.12) From (5.2), we have d dt L(t) ≤ γ M L(t) + λ, (5.13) by integrating the previous differential inequality (5.13) between 0 and t, we obtain the following estimate for the function L L(t) ≤ ce γ M t − λM γ , ∀t ≥ 0, (5.14) by using (5.2), we conclude E(t) ≤ c1e γ M t − λM γN , ∀t ≥ 0. (5.15) Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication of this paper. Int. J. Anal. Appl. (2022), 20:10 13 References [1] R.A. Adams, J.J.F. Fournier, Sobolev spaces, Academic Press, New York (2003). [2] D. Andrade, A Mognon, Global solutions for a system of Klein-Gordon equations with memory, Bol. Soc. Paran. Mat. Ser. 3, 21 (2003), 127–136. [3] S. Brrimi, S.A. Messaoudi, Exponential decay of solutions to a viscoelastic equation with nonlinear localized damping, Electron. J. Differ. Equ. 2004 (2004), No. 88, pp. 1–10. [4] Y. Boukhatem and B. Benabdrrahmane, A. 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