Int. J. Anal. Appl. (2023), 21:52 Well-Posedness and Exponential Stability of the Von Kármán Beam With Infinite Memory Abdelkader Dibes1, Lamine Bouzettouta2,∗, Manel Abdelli2, Salah Zitouni1 1Mohamed-Cherif Messaadia University, Souk Ahras, Algeria 2Laboratory of Applied Mathematics and History and Didactics of Mathematics (LAMAHIS) University of 20 August 1955, Skikda, Algeria ∗Corresponding author: lami_750000@yahoo.fr, l.bouzettouta@univ-skikda.dz Abstract. In the present work, we consider a one-dimensional Von kármán beam with infinite memory, we establish the well-posedness of the system using semigroup theory and prove the exponential stability under some conditions on the kernel of the infinite memory term. 1. Introduction Many studies have looked at nonlinear dynamical elasticity systems modeled by Von kármán equa- tions, wich is one of the basic equations in mathematical models of physics (see [14,20–22]). Their importance stems from the fact that many physical phenomena connected to the theory of oscilla- tion are described by non-linear dynamic elastic models. This nonlinear elastic systems incorporating wave equations govern the propagation of waves, oscillations, and vibrations of membranes, plates, and shells. The entire Von kármán’s model in contrast to other fundamental models like the Euler-Bernoulli, Raleigh, or Timoshenko is appropriate for considering both transverse and longitudinal displacements for vibrating slender bodies with large deflection (for more discussion see [5–7]). In [18,19] Lagnese et al investigated a one-dimensional Von kármán system listed below{ ρAutt − [ EAux + 1 2 w2x ] x , in (0,L) ×R+, ρAwtt −EI (wxx)xx − [ EA ( ux + 1 2 w2x ) wx ] x , in (0,L) ×R+. (1.1) Received: Apr. 9, 2023. 2020 Mathematics Subject Classification. 35B40, 93D23, 74F05, 93D15. Key words and phrases. Von kármán beam; exponential stability; semi-group; Lyapunov functional; infinite memory. https://doi.org/10.28924/2291-8639-21-2023-52 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-52 2 Int. J. Anal. Appl. (2023), 21:52 Here E is the Young’s modulus, A is the cross-sectional area of the beam, L is the beam length, ρA represents the weight per unit length and EI is the beam stiffness or flexural rigidity. A substantial range of literature use this type of model, where they addressing the problems of existence, uniqueness and asymptotic behavior in time when some damping effects are considered , (See Refs. [4, 17, 25] and the references therein for more information). In (1998) the system (1.1) was stabilized by Benabdallah and Teniou [14] by fusing the system using two heat equations: both the longitudinal and transverse components, respectively, they showed the unique solution decay exponentially by using Lyapunov functions. Many articles have looked into the stabilization of systems using boundary damping , see Favini et al [10], Puel and Tucsnak [24], and the references therein (see [3,8,15,26]). In [9] Djebabla and Tatar (2013) by linking the system, take into account the following full Von kármán beam in one dimension. (namely, the longitudinal component) with only one heat equation according to Green and Naghdi’s theory [11–13]  utt −d1 [( ux + 1 2 (wx) 2 )] x + γθtx = 0, in (0,L) ×R+, wtt −d1 [( ux + 1 2 (wx) 2 ) wx ] x + d2wxxxx + δwt = 0, in (0,L) ×R+, θtt −θxx + µ1θt + γutx = 0, in (0,L) ×R+, where d1, d2, δ, µ1 and γ are positive constants, with the boundary conditions{ u = 0, w = 0, θx = 0, x = 0,L, t > 0, wx = 0, x = 0,L, t > 0, (1.2) and the initial data{ u (0, .) = u0, ut (0, .) = u1, w (0, .) = w0, wt (0, .) = w1, θ (0, .) = θ0, θt (0, .) = θ1, (1.3) they succeeded an exponential decay result by using Lyapunov functions. A natural weak damping term can be thought of as the integral in the infinite memory term. It appears as a memory component in the form of a convolution in the fundamental equation between constraint and deformation. In this context, Khochemane et all In [16] think about infinite memory in a porous-elastic system and a nonlinear damping term:{ ρutt −µuxx −bφx = 0, x ∈ (0, 1) ,t > 0, Jφtt −δφxx + bux + aφ + ∫∞ 0 g(s)φxx(t − s)ds + α(t)f (φ) = 0, x ∈ (0, 1) ,t > 0, where φ and u represent, respectively,the volume fraction and displacement of the solid elastic material, the parameter ρ is the mass density, and J equals the product of the equilibrated inertia by the mass density, and the function g is the relaxation function, the term α(t)f (φ) is the nonlinear damping term. Int. J. Anal. Appl. (2023), 21:52 3 In this paper, we study the following Von kármán system with infinite memory  wtt −d1 [( ux + 1 2 (wx) 2 ) wx ] x + d2wxxxx + µwt (x,t) = 0, utt −d1 [( ux + 1 2 (wx) 2 )] x − ∫∞ 0 g(s)uxx(t − s)ds = 0. (1.4) in Ω ×(0,∞), where Ω = [0,L] and d1,d2,d and µ are positive constants and the function g is called the relaxation function. We complement system (1.4) with boundary conditions{ w(0,t) = w(1,t) = ux(0,t) = ux(1,t), t > 0 wx = 0 at x = 0, L f or any t > 0, (1.5) and the initial data   u(0, .) = u0, ut(0, .) = u1, w(0, .) = w0, wt(0, .) = w1, wt(x,t) = f0(x,t) in ∈ (0,L), (1.6) 2. Preliminaries First, we introduce the following hypothesis that has been considered in many works such that [1,2] (H1) g : R+ → R+ is a C1 function satisfying g(0) > 0, δ − ∫ ∞ 0 g(s)ds = l > 0, ∫ ∞ 0 g(s)ds = g0. (H2) There exists a non-increasing differentiable function α,ξ : R+ → R+ such that g ′ (t) ≤−ξ(t)g(t), ∀t ≥ 0. Now, to prove that systems (1.4), (1.5), (1.6) are well posed using the semigroup theory we introduce the following new variable:  ηt(x,s) = u(x,t) + u(x; t − s) uxx(x,t − s) = ηtxx(x,s) −uxx(x,t) ηtt + η t s = ut Therefore, the problem (1.4) is equivalent to  wtt −d1 [( ux + 1 2 (wx) 2 ) wx ] x + d2wxxxx + µwt (x,t) = 0, utt −d1 [( ux + 1 2 (wx) 2 )] x + g0uxx − ∫∞ 0 g(s)ηtxxds = 0, ηtt + η t s −ut = 0. (2.1) With g0 = ∫ ∞ 0 g(s)ds. For any regular solution of (2.1), the energy E (t), defined by E (t) = 1 2 ∫ L 0 { w2t + u 2 t + d2w 2 xx + d1 ( ux + 1 2 (wx) 2 )2} dx + 1 2 (g ◦ux)(t). (2.2) 4 Int. J. Anal. Appl. (2023), 21:52 3. Well-posedness of the problem In this section, we prove the existence and uniqueness of solutions for (2.1) using the semigroup theory [23]. First, we introduce the vector function Let U = ( w,wt,u,ut,η t )T , then Ut = ( wt,wtt,ut,utt,η t t )T . Introducing the vector function v = wt, and ψ = ut, system (2.1) can be written as{ Ut = AU + F (U) , U (0) = (w0,w1,u0,u1,θ0,q0, f0) , (3.1) and the linear operator A is defined by: A   w v u ψ ηt   =   v −d2wxxxx −µv ψ luxx + ∫∞ 0 g(s)ηtxxds ψ −ηts   , (3.2) and F (U) =   0 d1 [( ux + 1 2 (wx) 2 ) wx ] x 0 d1 2 (wx) 2 x 0   , U0 =   w0 w1 u0 u1 θ0   . (3.3) It is clear that F (U) is a continuous and uniformly Lipschitz operator. And H is the energy space given by H := {[ H4 (0,L) ∩H20 (0,L) ] ×H10 (0,L) × [ H2 (0,L) ∩H20 (0,L) ] ×H10 (0,L) ×Lg } . Lg = { ϕ : R+ → H10 (0,L) , ∫ L 0 ∫ ∞ 0 g(s)ϕ2xdsdx ≺∞ } , We equip the space Lg with the inner product through 〈u,v〉Lg = ∫ L 0 ∫ ∞ 0 g(s)ux(s)vx(s)dsdx, and we define on the Hilbert space H the inner product, for U = ( w,v,u,ψ,ηt )T , Ũ =( w̃, ṽ, ũ, ψ̃, η̃t )T 〈 U,Ũ 〉 = ∫ L 0 vṽdx + ∫ L 0 ψψ̃dx + d2 ∫ L 0 wxxw̃xxdx + l ∫ L 0 uxũxdx + 〈 ηt, η̃t 〉 . Int. J. Anal. Appl. (2023), 21:52 5 The domain of A is given by D (A) = { ( w,v,u,ψ,ηt )T ∈ [H4 (0,L) ∩H20 (0,L)]×H10 (0,L) × [ H2 (0,L) ∩H20 (0,L) ] ×H10 (0,L) ×Lg } . (3.4) Clearly, D (A) is dense in H. The next result is our first main goal in this paper Theorem 3.1. Let U ∈ H, for any initial datum U0 ∈ H there exists a unique solution U ∈ C ([0,∞) ,H) for problem. Moreover, if U0 ∈ D (A) , then U ∈ C ([0,∞) ,D (A)) ∩C1 ([0,∞) ,H) . Proof. We will show that the operator A generates a C0-semigroup in H. In this step, we prove that the operator A is dissipative. Let U = ( w,v,u,ψ,ηt )T firstly we have: 〈AU,U〉 = 〈   v −d2wxxxx −µv ψ luxx + ∫∞ 0 g(s)ηtxxds ψ −ηts   ,   w v u ψ ηt   〉 = ∫ L 0 v (−d2wxxxx −µv) dx −d2 ∫ L 0 ϕwxxxxdx + d1 ∫ L 0 ψuxxdx (3.5) + ∫ L 0 ψ ( luxx + ∫ ∞ 0 g(s)ηtxxds ) dx + d2 ∫ L 0 wxxvxxdx + l ∫ L 0 ψxuxdx + 〈 ηt,ψ−ηts 〉 . Using integration by parts, we obtain 〈AU,U〉 = −d2 ∫ L 0 vwxxxxdx −µ ∫ L 0 v2dx + l ∫ L 0 ψuxxdx + ∫ L 0 ψ ∫ ∞ 0 g(s)ηtxxdsdx + d2 ∫ L 0 vxxwxxdx + l ∫ L 0 ψxuxdx + ∫ L 0 ∫ ∞ 0 g(s)ηtx ( ψ −ηtsx ) dsdx, thus, 〈AU,U〉 = −µ ∫ L 0 v2dx + 1 2 (g ′ ◦ux)(t). Consequently, the operator A is dissipative. Now, we will prove that the operator λI−A is surjective. 6 Int. J. Anal. Appl. (2023), 21:52 For this purpose, let (f1, f2, f3, f4, f5) T ∈ H, we seek U = ( w,v,u,ψ,ηt )T ∈ D (A) , solution of the following system of equations  w −v = f1, v + d2wxxxx + µv = f2, u −ψ = f3, ψ − luxx − ∫∞ 0 g(s)ηtxxds = f4, ηt + ηts −ψ = f5, (3.6) we obtain   ηt = e−s ∫L 0 eς (ψ + f5(ς)) dς, v = w − f1, ψ = u − f3,{ w − f1 + d2wxxxx + µw −µf1 = f2, u − f3 − luxx − ∫∞ 0 g(s) ( e−s ∫ s 0 eς (ψ + f5(ς)) dς ) xx ds = f4, then − ∫ ∞ 0 g(s) ( e−s ∫ L 0 eς (ψ + f5(ς)) dς ) xx ds = − ∫ ∞ 0 g(s) ( e−s ∫ s 0 eς (u − f3 + f5(ς)) dς ) xx ds = − ∫ ∞ 0 g(s)e−s ∫ s 0 eς (uxx + (−f3 + f5)xx) dςds = − ∫ ∞ 0 g(s)e−s ∫ s 0 eςuxxdςds − ∫ ∞ 0 g(s)e−s ∫ s 0 eς (−f3 + f5)xx dςds = −uxx ∫ ∞ 0 g(s) ( 1 −e−s ) ds − ∫ ∞ 0 g(s)e−s ∫ s 0 eς (−f3 + f5)xx dςds, we obtain{ (µ + 1) w + d2wxxxx = f2 + (µ + 1) f1, u − [ l + ∫∞ 0 g(s) ( 1 −e−s ) ds ] uxx = f3 + f4 + ∫∞ 0 g(s)e−s ∫ s 0 eς (−f3 + f5)xx dςds, where { h1 = f2 + (µ + 1) f1, h2 = f3 + f4 + ∫∞ 0 g(s)e−s ∫ s 0 eς (−f3 + f5)xx dςds. Now, we consider the following variational formulation B ((w,u) ; (w1,u1)) = L (w1,u1) , Int. J. Anal. Appl. (2023), 21:52 7 where B : [ H20 (0, 1) ×H 1 0 (0, 1) ]2 → R is the bilinear form defined by B ((w,u) ; (w1,u1)) = (µ + 1) ∫ L 0 ww1dx + d2 ∫ L 0 wxxw1xxdx + ∫ L 0 uu 1 dx + [ l + ∫ ∞ 0 g(s) ( 1 −e−s ) ds ]∫ L 0 uxu1xdx, and L : H20 (0, 1) ×H 1 0 (0, 1) → R is the linear functional given by L (w1,u1) = ∫ L 0 h1w1dx + ∫ L 0 h2u1dx, now for V = H20 (0, 1) ×H 1 0 (0, 1) equipped with the norm ‖(w,u)‖2V = ‖w‖ 2 2 + ‖u‖ 2 2 + ‖wxx‖ 2 2 + ‖ux‖ 2 2 , we have B ((w,u) ; (w,u)) = ∫ L 0 w2dx + ∫ L 0 w2xxdx + ∫ L 0 u2dx + [ l + ∫ ∞ 0 g(s) ( 1 −e−s ) ds ]∫ L 0 u2xdx ≥‖w‖22 + ‖wxx‖ 2+ 2 ‖u‖ 2 2 + [ l + ∫ ∞ 0 g(s) ( 1 −e−s ) ds ] ‖ux‖22 ≥ M0 ( ‖w‖22 + ‖u‖ 2 2 + ‖wxx‖ 2 2 + ‖ux‖ 2 2 ) = M0‖(w,u)‖2V , thus B is coercive. On the other hand, by using Cauchy-Schwarz and Poincaré inequalities, we obtain |B ((w,u) ; (w1,u1))| ≤ ‖w‖2‖w1‖2 + d2‖wxx‖2‖w1xx‖2 + ‖u‖2‖u1‖2 + [ l + ∫ ∞ 0 g(s) ( 1 −e−s ) ds ] ‖ux‖2‖u1x‖2 ≤ ξ (‖w‖2 + ‖u‖2 + ‖wxx‖2 + ‖ux‖2) × (‖w1‖2 + ‖u1‖2 + ‖w1xx‖2 + ‖u1x‖2) ≤ ξ‖(w,u)‖V ‖(w1,u1)‖V . Similarly, we can show that |L (w1,u1)| ≤ κ‖(w1,u1)‖V , consequently, by the Lax-Milgram lemma the system has unique solution (w,u) ∈ H20 (0, 1) ×H 1 0 (0, 1) , satisfying B ((w,u) ; (w1,u1)) = L (w1,u1) ,∀(w1,u1) ∈ V, the substitution of w and u yields (v,ψ) ∈ H20 (0, 1) ×H 1 0 (0, 1) . 8 Int. J. Anal. Appl. (2023), 21:52 Similarly, now by taking u1 = 0 ∈ H20 (0, 1)∫ L 0 u1udx + [ l + ∫ ∞ 0 g(s) ( 1 −e−s ) ds ]∫ L 0 uxu1xdx = ∫ L 0 h2u1dx, hence, we obtain [ l + ∫ ∞ 0 g(s) ( 1 −e−s ) ds ]∫ L 0 uxu1xdx = ∫ L 0 (h2 −u) u1dx, by noting that (h2 −u) ∈ L2(0, 1), we obtain u ∈ H2 (0, 1) ∩H10 (0, 1) and consequentely the form∫ L 0 [ − [ l + ∫ ∞ 0 g(s) ( 1 −e−s ) ds ] uxx −h2 + u ] u1dx = 0,∀u1 ∈ H10 (0, 1) , therefore, we obtain − [ l + ∫ ∞ 0 g(s) ( 1 −e−s ) ds ] uxx + u = h2, similarly, if we take w1 = 0 ∈ H20 (0, 1) using twice integration by parts we obtain (µ + 1) ∫ L 0 ww1dx + d2 ∫ L 0 wxxw1xxdx = ∫ L 0 h1w1dx d2 ∫ L 0 wxxw1xxdx = ∫ L 0 (h1 − (µ + 1) w) w1dx∫ L 0 (d2wxxxx + (µ + 1) w −h1) w1dx = 0,∀w1 ∈ H20 (0, 1) . Therefore, we obtain d2wxxxx + (µ + 1) w = h1 ∈ L2(0, 1). Consequentely, we get w ∈ H4 (0, 1) ∩H20 (0, 1). Hence, there exists a unique U ∈ D(A) such that λI −A is satisfied. Therefore, the operator A is dissipative. From where, we conclude that A is a maximal monotone operator. Now, we prove that the operator F is locally Lipschitz in H. In fact, if U = ( w,v,u,ψ,ηt )T , Ũ = ( w̃, ṽ, ũ, ψ̃, η̃ t )T belong to H, we have ∥∥∥F (U) −F(Ũ)∥∥∥2 H = d1 ( |h|2 + |g|2 ) , (3.7) where h = [( ux + 1 2 w2x ) wx − ( ũx + 1 2 w̃2x ) w̃x ] x and g = 1 2 ( w2x − w̃2x ) x . Adding and subtracting the term ( ux + 1 2 w2x ) w̃x inside the norm |h| , we gets |h| ≤ ‖wx − w̃x‖L∞(0,L) ∣∣∣∣ux + 12w2x ∣∣∣∣ + ‖w̃x‖L∞(0,L) |ux − ũx| + 1 2 ‖w̃x‖L∞(0,L) |wx + w̃x|‖wx − w̃x‖L∞(0,L) . (3.8) Using the embedding of H1 (0,L) into L∞ (0,L) one has from (3.8) that |h| ≤ k ( ‖U‖H , ∥∥∥Ũ∥∥∥ H )∥∥∥U − Ũ∥∥∥ H (3.9) Int. J. Anal. Appl. (2023), 21:52 9 Using once again the embedding of H 1(0,L) into L∞ (0,L), one also sees that |g| ≤ k ( ‖U‖H , ∥∥∥Ũ∥∥∥ H )∥∥∥U − Ũ∥∥∥ H . (3.10) Combining (3.7), (3.9) and (3.10), it follows that F(U) is locally Lipschitz continuous in H. Finally, by using the regularity theory to solve linear elliptic equations ensures the presence of unique solution U ∈ D (A). Consequently, A is a dissipative operator. Hence, the result of Theorem 3.1 follows from the Lumer-Phillips theorem. � 4. Stability result In this section, we use the multiplier technique, to study the stability result for the energy of solution of the system (2.1). First, we state and prove the following lemma. Lemma 4.1. Let ( w,v,u,ψ,ηt ) be the solution of (2.1). Then the energy functional E (t), defined by (2.2) satisfies E ′ (t) ≤−µ ∫ L 0 w2t dx + 1 2 (g ′ ◦ux)(t). (4.1) Proof. Multiplying the first equation by wt , the second equation by ut, integrating over (0, 1), and summing them up we obtain d 2dt ∫ L 0 w2t dx + d1 ( ux + 1 2 (wx) 2 ) wxwxtdx + d2d 2dt ∫ L 0 (wxx) 2 dx + µ ∫ L 0 w2t dx = 0 d 2dt ∫ L 0 u2tdx + d1 ( ux + 1 2 (wx) 2 ) uxtdx − ∫ L 0 ut ∫ ∞ 0 g(s)ηtxxdsdx = 0 We estimate the last term as follows: − ∫ L 0 ut ∫ ∞ 0 g(s)ηtxxdsdx = − ∫ L 0 ( ηtt + η t s )∫ ∞ 0 g(s)ηtxxdsdx = − ∫ ∞ 0 g(s) ∫ L 0 ηtxx ( ηtt + η t s ) dxds = − ∫ ∞ 0 g(s) ∫ L 0 ηtxxη t tdxds − ∫ ∞ 0 g(s) ∫ L 0 ηtxxη t sdxds = ∫ ∞ 0 g(s) ∫ L 0 ηtxη t txdxds + ∫ ∞ 0 g(s) ∫ L 0 ηtxη t sxdxds = d 2dt ∫ L 0 ∫ ∞ 0 g(s) ( ηtx )2 dsdx + 1 2 ∫ L 0 ∫ ∞ 0 g(s) d ds ( ηtx )2 dsdx = d 2dt ∫ L 0 ∫ ∞ 0 g(s) [ux(t) −ux(t − s)] 2 dsdx − 1 2 ∫ L 0 ∫ ∞ 0 g ′ (s) ( ηtx )2 dsdx = d 2dt ∫ L 0 ∫ ∞ 0 g(s) [ux(t) −ux(t − s)] 2 dsdx − 1 2 ∫ L 0 ∫ ∞ 0 g ′ (s) [ux(t) −ux(t − s)] 2 dsdx = d 2dt (g ◦ux)(t) − 1 2 (g ′ ◦ux)(t). 10 Int. J. Anal. Appl. (2023), 21:52 � Lemma 4.2. Let ( w,v,u,ψ,ηt ) be the solution of (2.1). Then the functional I1(t) = ∫ L 0 ( utu + 1 2 wtw + µ 4 w2 ) dx, t ≥ 0. (4.2) satisfies, the estimate I′1(t) = −d1 ∫ L 0 ( ux + 1 2 (wx) 2 )2 dx + ∫ L 0 u2tdx − d2 2 ∫ L 0 w2xxdx + g0 ∫ L 0 u2xdx + 1 2 ∫ L 0 w2t dx + ∫ L 0 u ∫ ∞ 0 g(s)ηtxxdsdx (4.3) Proof. Using Young’s inequalitie we get∫ L 0 u ∫ ∞ 0 g(s)ηtxxdsdx = − ∫ L 0 ux ∫ ∞ 0 g(s)(ux(t) −ux(t − s))dsdx ≤ ε1 ∫ L 0 u2xdx + 1 4ε1 ∫ L 0 (∫ ∞ 0 g(s)(ux(t) −ux(t − s)) )2 dsdx ≤ ε1 ∫ L 0 u2xdx + g0 4ε1 (g ◦ux)(t), (4.4) I′1(t) ≤−d1 ∫ L 0 ( ux + 1 2 (wx) 2 )2 dx − d2 2 ∫ L 0 (wxx) 2 dx + (g0 + ε1) ∫ L 0 u2xdx + ∫ L 0 u2tdx + 1 2 ∫ L 0 w2t dx. + g0 4ε1 (g ◦ux)(t), � Lemma 4.3. Let ( w,v,u,ψ,ηt ) be the solution of (2.1). Then the functional I2(t) := − ∫ L 0 ut ∫ ∞ 0 g(s)(ux(t) −ux(t − s))dsdx t ≥ 0, (4.5) satisfies, the estimate I′2(t) ≤− g0 2 ∫ L 0 u2tdx + d1 2 ∫ L 0 ( ux + 1 2 (wx) 2 )2 dx + (g1 + d1) (g ◦ux(t)) + g0 2 ∫ L 0 u2xdx t ≥ 0, + g(0) 2g0 (g ′ ◦ux)(t). (4.6) Int. J. Anal. Appl. (2023), 21:52 11 Proof. firste we note that d dt ∫ ∞ 0 g(s)(u(t) −u(t − s))ds = d dt ∫ t −∞ g(t − s)(u(t) −u(s))ds = ∫ t −∞ g ′ (t − s)(u(t) −u(s))ds + ∫ t −∞ g(t − s)u(t)ds = g0ut(t) + ∫ ∞ 0 g ′ (s)(u(t) −u(t − s))ds, I′2(t) = − ∫ L 0 utt ∫ ∞ 0 g(s)(u(t) −u(t − s))dsdx − ∫ L 0 ut d dt ∫ ∞ 0 g(s)(u(t) −u(t − s))dsdx = − ∫ L 0 ( d1 [( ux + 1 2 (wx) 2 )] x −g0uxx + ∫ ∞ 0 g(s)ηtxxds )(∫ ∞ 0 g(s)(u(t) −u(t − s))ds ) dx −g0 ∫ L 0 u2tdx − ∫ L 0 ut ∫ ∞ 0 g ′ (s)(u(t) −u(t − s))dsdx = −d1 ∫ L 0 [( ux + 1 2 (wx) 2 )] x ∫ ∞ 0 g(s)(u(t) −u(t − s))dsdx + g0 ∫ L 0 uxx ∫ ∞ 0 g(s)(u(t) −u(t − s))dsdx − ∫ L 0 (∫ ∞ 0 g(s)ηtxxds )(∫ ∞ 0 g(s)(u(t) −u(t − s))ds ) dx −g0 ∫ L 0 u2tdx − ∫ L 0 ut ∫ ∞ 0 g ′ (s)(u(t) −u(t − s))dsdx = d1 ∫ L 0 ( ux + 1 2 (wx) 2 )∫ ∞ 0 g(s)(ux(t) −ux(t − s))dsdx −g0 ∫ L 0 ux ∫ ∞ 0 g(s)(ux(t) −ux(t − s))dsdx + ∫ L 0 (∫ ∞ 0 g(s)ηtxds )(∫ ∞ 0 g(s)(ux(t) −ux(t − s))ds ) dx −g0 ∫ L 0 u2tdx − ∫ L 0 ut ∫ ∞ 0 g ′ (s)(u(t) −u(t − s))dsdx. By recalling Young’s inequality, we get for any ε2 > 0, d1 ∫ L 0 ( ux + 1 2 (wx) 2 )∫ ∞ 0 g(s)(ux(t) −ux(t − s))dsdx ≤ d1ε2 ∫ L 0 ( ux + 1 2 (wx) 2 )2 dx + d1 4ε2 ∫ L 0 (∫ ∞ 0 g(s)(ux(t) −ux(t − s))ds )2 dx ≤ d1ε2 ∫ L 0 ( ux + 1 2 (wx) 2 )2 dx + g0d1 4ε2 (g ◦ux(t)) , 12 Int. J. Anal. Appl. (2023), 21:52 −g0 ∫ L 0 ux ∫ ∞ 0 g(s)(ux(t) −ux(t − s))dsdx ≤ g0ε3 ∫ L 0 u2xdx + g0 4ε3 (g ◦ux)(t), − ∫ L 0 ut ∫ ∞ 0 g ′ (s)(u(t) −u(t − s))dsdx = − ∫ L 0 √ g0ut ∫ ∞ 0 (√ 1 g0 g ′ (s)(u(t) −u(t − s)) ) dsdx ≤ g0 2 ∫ L 0 u2tdx + g(0) 2g0 (g ′ ◦ux)(t), ∫ L 0 (∫ ∞ 0 g(s)ηtxds )(∫ ∞ 0 g(s)(ux(t) −ux(t − s))ds ) dx = ∫ L 0 (∫ ∞ 0 g(s)(ux(t) −ux(t − s))ds )2 dx ≤ d3(g ◦ux)(t). Wich complete the proof. � Now, we define the Lyapunov functional L(t)by L(t) = NE(t) + N1F1(t) + N2F2(t), whereN,N1and N2 are positive constants. Lemma 4.4. Let (u,w) be the solution of Then, there exist two positive constants C1 and C2 such that the Lyapunov functional L (t) satisfies C1E (t) ≤ L (t) ≤ C2E (t) , ∀t ≥ 0. (4.7) In other words, the functions E and L are equivalent. Proof. Firstly we note that ∫ L 0 u2xdx ≤ 2 ∫ L 0 ( ux + 1 2 ( w2x ))2 dx + 1 2 ∫ L 0 w2xdx ≤ 2 ∫ L 0 ( ux + 1 2 ( w2x ))2 dx + 1 4 ∫ L 0 w4xdx, ≤ 2 ∫ L 0 ( ux + 1 2 ( w2x ))2 dx + L 4 ∫ L 0 w2xxdx, we get |L(t) −NE(t)| ≤ N1 ∫ L 0 ∣∣∣∣utu + 12wtw + µ4 w2 ∣∣∣∣dx + N2 ∫ L 0 ∣∣∣∣ut ∫ ∞ 0 g(s)(u(t) −u(t − s))ds ∣∣∣∣dx. Int. J. Anal. Appl. (2023), 21:52 13 By using Young’s inequality, Cauchy–Schwarz inequality, and Poincaré’s inequality, we obtain |L(t) −NE(t)| ≤ [N1ε1 + N2ε3] ∫ L 0 u2tdx + N1ε2 4 ∫ L 0 w2t dx + N1ε1 2 ∫ L 0 ( ux + 1 2 ( w2x ))2 dx + [ N1L 16ε1 + N1 4ε2 + N1µ 4 ]∫ L 0 w2xxdx + d3 4ε3 (g ◦ux)(t). So |L(t) −NE(t)| ≤ c ∫ L 0 { w2t + u 2 t + d2w 2 xx + d1 ( ux + 1 2 (wx) 2 )2} dx + c(g ◦ux)(t) ≤ cE(t). then (N −c)E(t) ≤ L(t) ≤ (N + c)E(t). Consequently, By choosing N large enough, we obtain the estimate (4.7). � Now, we are ready to state and prove the main result of this section. Theorem 4.1. Let ( w,v,u,ψ,ηt ) be the solution of (2.1). Then the energy functional (2.2) satisfies, E (t) ≤ k0e−k1t, ∀t ≥ 0, (4.8) where k0 and k1 are two positive constants. Proof. L′(t) ≤− [ N2g0 2 −N1 ]∫ L 0 u2tdx − [ Nµ− N1 2 ]∫ L 0 w2t dx − [ d1N1 − N2d1 2 − 2N1 (g0 + ε1) −N2g0 ]∫ L 0 ( ux + 1 2 ( w2x ))2 dx − [ N1d2 2 − N1L 4 (g0 + ε1) − N2g0L 8 ]∫ L 0 w2xxdx − [ N1g0 4ε1 + (N2g0 + d3) ] (g ◦ux)(t) − [ N 2 + N2g(0) 2g0 ] (g′ ◦ux)(t). 14 Int. J. Anal. Appl. (2023), 21:52 By setting ε1 = 1 N1 , we obtain L′(t) ≤− [ N2g0 2 −N1 ]∫ L 0 u2tdx − [ Nµ− N1 2 ]∫ L 0 w2t dx − [ d1N1 − N2d1 2 − 2N1g0 − 1 −N2g0 ]∫ L 0 ( ux + 1 2 ( w2x ))2 dx − [ N1d2 2 − N1L 4 g0 − L 4 − N2g0L 8 ]∫ L 0 w2xxdx − [ N21g0 4 + (N2g0 + d3) ] (g ◦ux)(t) − [ N 2 + N2g(0) 2g0 ] (g′ ◦ux)(t), Next, we carefully choose our constants so that the terms inside the brackets are positive. We choose N1 large enough such that α1 = Nµ− N1 2 > 0, then we choose N2 large enough such that α2 = N2g0 2 −N1 > 0, α3 = d1N1 − N2d1 2 − 2N1g0 − 1 −N2g0 > 0, α4 = N1d2 2 − N1L 4 g0 − L 4 − N2g0L 8 > 0, α5 = N21g0 4 + (N2g0 + d3) > 0, Finally, once N1 and N2, is fixed, we choose N large enough so that α6 = N 2 + N2g(0) 2g0 > 0, we obtain L ′ (t) ≤− 1 2 ∫ 1 0 { α1w 2 t + α2u 2 t + α3 ( ux + 1 2 ( w2x ))2 + α4w 2 xx } − α5 2 (g ◦ux)(t), By (2.2), we obtain L ′ (t) ≤−σ0E (t) , ∀t ≥ 0, (4.9) for some σ0 > 0. A combination of (4.7) and (4.9) gives L ′ (t) ≤−k1L (t) , ∀t ≥ 0, (4.10) Int. J. Anal. Appl. (2023), 21:52 15 A simple integration of (4.10) over (0,t) yields L (t) ≤ L (0) e−k1t, ∀t ≥ 0. (4.11) Finally, by combining (4.7) and (4.11) we obtain (4.8), which completes the proof. � Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the publi- cation of this paper. References [1] T.A. 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