International Journal of Analysis and Applications Volume 19, Number 6 (2021), 812-825 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-19-2021-812 STABILITY RESULT FOR A WEAKLY NONLINEARLY DAMPED POROUS SYSTEM WITH DISTRIBUTED DELAY KHOUDIR KIBECHE1, LAMINE BOUZETTOUTA2,∗, ABDELHAK DJEBABLA3, FAHIMA HEBHOUB2 1University of Badji Mokhtar, Annaba, Algeria 2Laboratory Applied Mathematics and History and Didactics of Mathematic, University 20 August 1955, Skikda, Algeria 3Laboratory of Applied Mathematics, Badji Mokhtar University-Annaba, P.O. Box 12, 23000 Annaba, Algeria ∗Corresponding author: lami 750000@yahoo.fr, l.bouzettouta@univ-skikda.dz Abstract. In this paper, we consider a one-dimensional porous system damped with a single weakly non- linear feedback and distributed delay term. Without imposing any restrictive growth assumption near the origin on the damping term, we establish an explicit and general decay rate, using a multiplier method and some properties of convex functions in case of the same speed of propagation in the two equations of the system. The result is new and opens more research areas into porous-elastic system. Received June 5th, 2021; accepted June 30th, 2021; published October 28th, 2021. 2010 Mathematics Subject Classification. 35B35, 35B40, 93D20. Key words and phrases. porous system; general decay; nonlinear damping; distributed delay. ©2021 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 812 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-19-2021-812 Int. J. Anal. Appl. 19 (6) (2021) 813 1. Introduction In this paper, we consider the following porous system:  ρutt −µuxx − bφx + µ1ut + ∫ τ2 τ1 µ2(s)ut (x,t−s) ds = 0, x ∈ (0, 1), t > 0, jφtt − δφxx + bux + ξφ + α (t) g (φt) = 0, x ∈ (0, 1), t > 0, u (x, 0) = u0 (x) , ut (x, 0) = u1 (x) , x ∈ (0, 1), φ (x, 0) = φ0 (x) , φt (x, 0) = φ1 (x) , x ∈ (0, 1), ux (0, t) = ux (1, t) , φ (0, t) = φ (1, t) = 0 ut (x,−t) = f0 (x,t) in (0, 1) × (0,τ2) (1.1) Firstly, to deal with the delay term, we introduce the new variable [17] z (x,ρ,s,t) = ut (x,t−ρs) , x ∈ (0, 1) , ρ ∈ (0, 1) , ρ ∈ (τ1,τ2) , t > 0 Then we obtain szt (x,ρ,s,t) + zρ (x,ρ,s,t) = 0, x ∈ (0, 1) , ρ ∈ (0, 1) , ρ ∈ (τ1,τ2) , t > 0 Then problem (1.1) is equivalent to  ρutt −µuxx − bφx + µ1ut + ∫ τ2 τ1 µ2(s)z (x, 1, t,s) ds = 0, x ∈ (0, 1), t > 0, jφtt − δφxx + bux + ξφ + α (t) g (φt) = 0, x ∈ (0, 1), t > 0, szt (x,ρ,s,t) + zρ (x,ρ,s,t) = 0, x ∈ (0, 1) , ρ ∈ (0, 1) , ρ ∈ (τ1,τ2) , t > 0 u (x, 0) = u0 (x) , ut (x, 0) = u1 (x) , x ∈ (0, 1), φ (x, 0) = φ0 (x) , φt (x, 0) = φ1 (x) , x ∈ (0, 1), ux (0, t) = ux (1, t) , φ (0, t) = φ (1, t) = 0 z (x,ρ,s, 0) = f0 (x,ρs) , (x,ρ,s) ∈ (0, 1) × (0, 1) × (τ1,τ2) (1.2) In recent paper, Apalara in [2] considered the following on-dimensional porous system damped with a single weakly nonlinear feedback  ρutt −µuxx − bφx = 0, x ∈ (0, 1), t > 0, jφtt − δφxx + bux + ξφ + α (t) g (φt) = 0, x ∈ (0, 1), t > 0, u (x, 0) = u0 (x) , ut (x, 0) = u1 (x) , x ∈ (0, 1), φ (x, 0) = φ0 (x) , φt (x, 0) = φ1 (x) , x ∈ (0, 1), ux (0, t) = ux (1, t) , φ (0, t) = φ (1, t) = 0 Without in pasing an explicit and general decay rate, he used a multiplier method and some proprieties of convex functions in case of the same speed of propagation in the both equation of the system. The same author, in [3] considered a porous-elastic system with memory term acting only on the porous equation, with Int. J. Anal. Appl. 19 (6) (2021) 814 the mixed boundary Neumann-Direchlet conditions, he proved a general decay result, for which exponential and polynomial decay results are special cases. Back to system (1.1), it is to be noted that when µ1 = µ2 = 0 and replacing the term α (t) g (φt) by the term ∫ t 0 g(t−s)uxx (x,s) ds then (1.1) is equivalent to the well-known Timoshenko system of memory type which is exponentially stable depending of the relaxation function g and provided that the wave speeds of the system are equal (See [1, 15]). Messaoudi and Fareh [16] investigated the following system:  ρutt = µuxx + bφx −βθx, in (0, 1) × (0,∞), jφtt = αφxx − bux + ξφ + mθ + τφt, in (0, 1) × (0,∞), cφt = −qx −βutx −mφt, in (0, 1) × (0,∞), τ0qt −q + kθx = 0, in (0, 1) × (0,∞), and established, using the energy method, an exponential decay result. For more results on the subject, we refer the reader to [5, 10, 11, 19]. Concerning the weight of the delay, we assume that∫ τ2 τ1 |µ2(s)|ds < µ1 and establish the well-posedness as well as the exponential stability results of the energy E (t), defined by E (t) = 1 2 ∫ 1 0 [ ρu2t + µu 2 x + ξφ 2 + δφ2x + jφ 2 t + 2bφux ] dx + 1 2 ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 s |µ2(s)|z 2 (x,ρ,s,t) dsdρdx (1.3) 2. Preliminaries In this section, we present some materials needed in the proof of our result. We assume α and g satisfy the following hypotheses: (H1) α : R+ → R+∗ is a non-increasing differentiable function; (H2) g : R → R is a non-decreasing C0-function such that there exist positive constants c1, c2, η and G ∈ C1 ([0,∞)) , with G (0) = 0, and G is linear or strictly convex C2−function on (0,η] such that  s 2 + g2 (s) ≤ G−1 (sg (s)) for all |s| ≤ η c1 |s| ≤ |g (s)| ≤ c2 |s| for all |s| ≥ η Remark 2.1. Hypothesis (H2) implies that sg (s) > 0 for all s 6= 0. * According to our knowledge, hypothesis (H2) with η = 1 was first introduced by Lasiecka and Tataru [13]. They established a decay result, which depends on the solution of an explicit nonlinear ordinary differential equation. Furthermore, they proved that the monotonicity and continuity of g guarantee the existence of the function G defined in (H2). Int. J. Anal. Appl. 19 (6) (2021) 815 For completeness purpose we state, without proof, the existence and regularity result of system (1.1). First, we introduce the following spaces: H = H1∗ (0, 1) ×L 2 ∗ (0, 1) ×H 1 (0, 1) ×L2 (0, 1) ×L2 ((0, 1) × (0, 1) × (τ1,τ2)) , (2.1) and H̃ = φ0 ∈ [ H2∗ (0, 1) ∩H 1 ∗ (0, 1) ] ×H1∗ (0, 1) × [ H2 (0, 1) ∩H1 (0, 1) ] ×H1 (0, 1) ×L2 ((0, 1) × (0, 1) × (τ1,τ2)) , where L2∗ (0, 1) = { ψ ∈ L2 (0, 1) : ∫ 1 0 ψ (x) dx = 0 } , H1∗ (0, 1) = H 1 (0, 1) ×L2∗ (0, 1) , H2∗ (0, 1) = { ψ ∈ H2 (0, 1) : ψx (0) = ψx (1) = 0 } . For U = (u,ut,φ,φt,z) , we have the following existence and regularity result: Proposition 2.1. Assume that (H1) and (H2) are satisfied. Then for all U0 ∈ H, the system (1.1) has a unique global (weak) solution u ∈ C ( R+; H1∗ (0, 1) ) ∩C1 ( R+; L2∗ (0, 1) ) , φ ∈ C ( R+; H1 (0, 1) ) ∩C1 ( R+; L2 (0, 1) ) . Moreover, if U0 ∈ H̃, then the solution satisfies u ∈ L∞ ( R+; H2∗ (0, 1) ∩H 1 ∗ (0, 1) ) ∩W1,∞ ( R+; H1∗ (0, 1) ) ∩W2,∞ ( R+; L2∗ (0, 1) ) , φ ∈ L∞ ( R+; H2 (0, 1) ∩H10 (0, 1) ) ∩W1,∞ ( R+; H10 (0, 1) ) ∩W2,∞ ( R+; L2 (0, 1) ) Remark 2.2. This result can be proved using the theory of maximal nonlinear monotone operators (see [8]). 3. Technical Lemmas In this section, we state and prove our stability results for the energy of system (1.1) by using the multiplier technique. To achieve our goal, we need the following lemmas. Lemma 3.1. Let (u,φ,z) be the solution of(1.2), then we have E′ (t) ≤−me ∫ 1 0 u2tdx− ∫ 1 0 α (t) φtg (φt) dx ≤ 0 (3.1) Int. J. Anal. Appl. 19 (6) (2021) 816 Proof. Multiplying (1.2)1, and (1.2)2 by ut, φt respectively, and integrating over (0, 1), using integration by parts and the boundary conditions, we obtain 1 2 d dt ∫ 1 0 ( ρu2t + µu 2 x + ξφ 2 + δφ2x + jφ 2 t + 2bφux ) dx = (3.2) − ∫ 1 0 α (t) φtg (φt) dx−µ1 ∫ 1 0 u2tdx− ∫ 1 0 ut ∫ τ2 τ1 µ2(s)z (x, 1, t,s) dsdx Multiplying (1.2)3 by |µ2(s)|z, integrating the product over (0, 1)×(0, 1)×(τ1,τ2) , and recall that z (x, 0,s,t) = ut, we get 1 2 d dt ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 s |µ2(s)|z 2 (x,ρ,t,s) dsdρdx = − 1 2 ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1, t,s) dsdx + 1 2 ∫ 1 0 ut ∫ τ2 τ1 |µ2(s)|dsdx. (3.3) A combination of (3.2) and (3.3) gives E′ (t) = − ∫ 1 0 α (t) φtg (φt) dx−µ1 ∫ 1 0 u2tdx− ∫ 1 0 ut ∫ τ2 τ1 µ2(s)z (x, 1, t,s) dsdx with − ∫ 1 0 ut ∫ τ2 τ1 µ2(s)z (x, 1, t,s) dsdx ≤ 1 2 ∫ τ2 τ1 |µ2(s)| ∫ 1 0 u2tdx + 1 2 ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1, t,s) dsdx then E′ (t) ≤− ∫ 1 0 α (t) φtg (φt) dx− ( µ1 − ∫ τ2 τ1 |µ2(s)| )∫ 1 0 u2tdx taking ( µ1 − ∫ τ2 τ1 |µ2(s)| ) = me we obtain (3.1). � Lemma 3.2. Assume that (H1) and (H2) hold. Then, for all U0 ∈H, the functional F1 (t) = j ∫ 1 0 φtφdx + bρ µ ∫ 1 0 φ ∫ x 0 ut (y) dydx (3.4) satisfies, for any ε1 > 0 F ′1 (t) ≤ ( j + ε1bρ µ )∫ 1 0 φ2tdx− jδ ∫ 1 0 φ2xdx + bjε1 ∫ 1 0 u2xdx + bρ 4ε1µ ∫ 1 0 u2tdx + ( jα (0) ε1 + bj 4ε1 − ξj )∫ 1 0 φ2dx + jα (0) 4ε1 ∫ 1 0 g2 (φt) dx (3.5) Proof. Differentiating F1 (t) , taking into account (1.2) using integrating by parts, and Young’s inequality, we obtain F ′1 (t) ≤ j ∫ 1 0 φ2tdx− jδ ∫ 1 0 φ2xdx + bjε1 ∫ 1 0 u2xdx + cpbj 4ε1 ∫ 1 0 φ2xdx− ξjcp ∫ 1 0 φ2xdx −j ∫ 1 0 α (t) φg (φt) dx + bρ µ ∫ 1 0 φt ∫ x 0 ut (y) dydx + bρ µ ∫ 1 0 φ d dt (∫ x 0 ut (y) dy ) dx Int. J. Anal. Appl. 19 (6) (2021) 817 By Caucy-Schwartz inequality, it is clear that ∫ 1 0 (∫ x 0 ut (y) dy )2 dx ≤ ∫ 1 0 (∫ 1 0 utdx )2 dx ≤ ∫ 1 0 u2tdx then F ′1 (t) ≤ j ∫ 1 0 φ2tdx− jδ ∫ 1 0 φ2xdx + bjε1 ∫ 1 0 u2xdx + cpbj 4ε1 ∫ 1 0 φ2xdx− ξjcp ∫ 1 0 φ2xdx −j ∫ 1 0 α (t) φg (φt) dx + ε1bρ µ ∫ 1 0 φ2tdx + bρ 4ε1µ ∫ 1 0 (∫ x 0 ut (y) dy )2 dx + bρ µ ∫ 1 0 φ d dt (∫ x 0 ut (y) dy ) dx thus we obtain F ′1 (t) ≤ ( j + ε1bρ µ )∫ 1 0 φ2tdx− jδ ∫ 1 0 φ2xdx + bjε1 ∫ 1 0 u2xdx + bρ 4ε1µ ∫ 1 0 u2tdx + ( jα (t) ε1 + bj 4ε1 − ξj )∫ 1 0 φ2dx + jα (t) 4ε1 ∫ 1 0 g2 (φt) dx � Lemma 3.3. Assume that (H1), (H2) and (3.8) hold. Then, for all U0 ∈H, the functional F2 (t) = b ∫ 1 0 φxutdx + b ∫ 1 0 φtuxdx (3.6) satisfies, for any ε2 > 0 F ′2 (t) ≤ ( b2 ρ + ε2 bµ1 ρ + bn0 2ρ )∫ 1 0 φ2xdx− ( b2 j − bξ 4ε2j − b j α (t) )∫ 1 0 u2xdx + bµ1 4ε2ρ ∫ 1 0 u2tdx + ε2 bξ j ∫ 1 0 φ2dx + b j α (t) ∫ 1 0 g2 (φt) dx + 1 2 b ρ ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1,s,t) dsdx (3.7) Proof. Simple computaions give F ′2 (t) = b2 ρ ∫ 1 0 φ2xdx− b2 j ∫ 1 0 u2xdx + bµ ρ ∫ 1 0 uxxφxdx− bµ1 ρ ∫ 1 0 φxutdx + bδ j ∫ 1 0 φxxuxdx− bξ j ∫ 1 0 φuxdx − b ρ ∫ 1 0 φx ∫ τ2 τ1 µ2(s)ut (x, 1, t,s) dsdx− b j ∫ 1 0 α (t) uxg (φt) dx Int. J. Anal. Appl. 19 (6) (2021) 818 taking into account the fact that µ ρ = δ j (3.8) and using young’s inequality F ′2 (t) ≤ ( b2 ρ + ε2 bµ1 ρ )∫ 1 0 φ2xdx + ( bξ 4ε2j − b2 j + b j α (t) )∫ 1 0 u2xdx + bµ1 4ε2ρ ∫ 1 0 u2tdx + ε2 bξ j ∫ 1 0 φ2dx + b j α (t) ∫ 1 0 g2 (φt) dx − b ρ ∫ 1 0 φx ∫ τ2 τ1 µ2(s)z (x, 1, t,s) dsdx − b ρ ∫ 1 0 φx ∫ τ2 τ1 µ2(s)z (x, 1, t,s) dsdx ≤ 1 2 b ρ ∫ τ2 τ1 |µ2(s)|ds ∫ 1 0 φ2xdx + 1 2 b ρ ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1,s,t) dsdx F ′2 (t) ≤ ( b2 ρ + ε2 bµ1 ρ + bn0 2ρ )∫ 1 0 φ2xdx− ( b2 j − bξ 4ε2j − b j α (t) )∫ 1 0 u2xdx + bµ1 4ε2ρ ∫ 1 0 u2tdx + ε2 bξ j ∫ 1 0 φ2dx + b j α (t) ∫ 1 0 g2 (φt) dx + 1 2 b ρ ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1,s,t) dsdx with ∫ τ2 τ1 |µ2(s)|ds = n0 � Lemma 3.4. The functional F3 (t) = −ρ ∫ 1 0 utudx (3.9) satisfies, for any ε3 > 0 F ′3 (t) = + ( µ + n0cp 2 + cpbε3 )∫ 1 0 u2xdx + b 4ε3 ∫ 1 0 φ2xdx− ( ρ− µ1 4ε3 )∫ 1 0 u2tdx + 1 2 ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1, t,s) dsdx (3.10) Proof. A simple differentiation of F3 (t), using the first equation in (1.2), give F ′3 (t) = −ρ ∫ 1 0 u2tdx + µ ∫ 1 0 u2xdx +cpbε3 ∫ 1 0 u2xdx + b 4ε3 ∫ 1 0 φ2xdx +µ1ε3cp ∫ 1 0 u2xdx + µ1 4ε3 ∫ 1 0 u2tdx + ∫ 1 0 ∫ τ2 τ1 µ2(s)uut (x, 1, t,s) dsdx ∫ 1 0 u ∫ τ2 τ1 µ2(s)ut (x, 1, t,s) dsdx ≤ cp 2 ∫ τ2 τ1 |µ2(s)| ∫ 1 0 u2xdx + 1 2 ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1, t,s) dsdx Int. J. Anal. Appl. 19 (6) (2021) 819 then F ′3 (t) = + ( µ + µ1ε3cp + n0cp 2 + cpbε3 )∫ 1 0 u2xdx + b 4ε3 ∫ 1 0 φ2xdx− ( ρ− µ1 4ε3 )∫ 1 0 u2tdx + 1 2 ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1, t,s) dsdx � Lemma 3.5. The functional F4 (t) = ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 se−sρ |µ2(s)|z 2 (x,ρ,t,s) dsdρdx (3.11) satisfies, for some positive constant m1, the following estimate F ′4 (t) ≤ −m1 ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1, t,s) dsdx + ∫ τ2 τ1 |µ2(s)|ds ∫ 1 0 u2tdx −m1 ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 s |µ2(s)|z 2 (x,ρ,t,s) dsdρdx (3.12) Proof. With szt (x,ρ,t,s) + zρ (x,ρ,t,s) = 0 in (0, 1) × (0, 1) × (τ1,τ2) × (0,∞) (3.13) zt (x,ρ,t,s) = − 1 s zρ (x,ρ,t,s) Differentiating F4 (t), and using the equation (3.13), we obtain F ′4 (t) = 2 ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 se−sρ |µ2(s)|zzt (x,ρ,t,s) dsdρdx = − ∂ ∂ρ ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 e−sρ |µ2(s)|z 2 (x,ρ,t,s) dsdρdx − ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 se−sρ |µ2(s)|z 2 (x,ρ,t,s) dsdρdx F ′4 (t) = − ∫ 1 0 ∫ τ2 τ1 |µ2(s)| [ e−sρz2 (x, 1, t,s) −z2 (x, 0, t,s) ] dsdx − ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 se−sρ |µ2(s)|z 2 (x,ρ,t,s) dsdρdx Using the fact that z (x, 0, t,s) = ut and e −s ≤ e−sρ ≤ 1, for all ρ ∈ [0, 1] , we obtain F ′4 (t) ≤ − ∫ 1 0 ∫ τ2 τ1 |µ2(s)|e −sρz2 (x, 1, t,s) dsdx + ∫ τ2 τ1 |µ2(s)|ds ∫ 1 0 u2tdx − ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 se−sρ |µ2(s)|z 2 (x,ρ,t,s) dsdρdx Because −se−s is an increasing function, we have −se−s ≤−se−τ2 , for all s ∈ [τ1,τ2] Int. J. Anal. Appl. 19 (6) (2021) 820 Finally, setting m1 = e −τ2 , with ∫ τ2 τ1 |µ2(s)| < µ1, we obtain F ′4 (t) ≤ −m1 ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1, t,s) dsdx + ∫ τ2 τ1 |µ2(s)|ds ∫ 1 0 u2tdx −m1 ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 s |µ2(s)|z 2 (x,ρ,t,s) dsdρdx � Lemma 3.6. Suppose (H1), (H2), and Eq. (3.8) hold. Let U0 ∈H. Then, for N,N1,N2,N3 > 0 sufficiently large, the Lyapunov functional defined by L(t) := NE (t) + N1F1 (t) + N2F2 (t) + F3 (t) + N3F4 (t) satisfies, for some positive constants d1,d2 and k1 d1L(t) ≤ E (t) ≤ d2L(t) , ∀t ≥ 0 (3.14) and L′ (t) ≤−k1E (t) + c ∫ 1 0 ( φ2t + g 2 (φt) ) dx, ∀t ≥ 0 (3.15) with L′ (t) ≤ [ bρ 4ε1µ N1 −Nme + N3µ1 + bµ1 4ε2ρ N2 − ( ρ− µ1 4ε3 )]∫ 1 0 u2tdx + ( N1 ( j + ε1bρ µ ))∫ 1 0 φ2tdx + ( bjε1N1 + ( µ + n0 2 + bε3 ) −N2 ( b2 j − bξ 4ε2j − b j α (t) ))∫ 1 0 u2xdx + ( N2 1 2ρ ( 2b2 + 2ε2bµ1 + bn0 ) − jδN1 + b 4ε3 )∫ 1 0 φ2xdx + ( ε2 bξ j N2 + N1 ( jα (t) ε1 + bj 4ε1 − ξj ))∫ 1 0 φ2dx + ( N1 jα (t) 4ε1 + b j α (t) N2 )∫ 1 0 g2 (φt) dx + ( 1 2 ( bN2 ρ + 1 ) −m1N3 )∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1,s,t) dsdx −m1N3 ∫ 1 0 ∫ 1 0 ∫ τ2 τ1 s |µ2(s)|z 2 (x,ρ,t,s) dsdρdx−N ∫ 1 0 α (t) φtg (φt) dx At this point, we have to choose our constants very carefully. First, choosing ε3 << 1, and ε1,ε2 small enough such that ε1 ≤ bρN1 4µ (Nme −N3µ1) , ε2 ≤ bµ1N2 4ρ Int. J. Anal. Appl. 19 (6) (2021) 821 Moreover, we pick Ni i = 1, 2, 3 large enough so that N2 ≥ bjε1N1 + ( µ + n0 2 + bε3 ) b2 j − bξ 4ε2j − b j α (t) and N3 ≥ ( bN2 ρ + 1 ) 2m1 . After that, we can choose N large enough such that N > 1 me [ bρN1 4ε1µ + N3µ1 + N2bµ1 4ε2ρ − ( ρ− µ1 4ε3 )] . Consequently, there exists a positive constant η1 such that (3.15) becomes d dt L(t) ≤ −c1 ∫ 1 0 ( u2t + ux + ϕ 2 x + φ 2 ) dx + c2 ∫ 1 0 ( φ2t + g 2 (φt) ) dx −c3 ∫ 1 0 ∫ τ2 τ1 |µ2(s)|z 2 (x, 1,s,t) dsdx. (3.16) In this section, we state and prove our stability result. 4. Stability Result Theorem 4.1. Suppose (H1), (H2), and (3.8) hold. Let U0 ∈H. there exist positive constants a1,a2,a3 and η0 such that the solution of (1.2) satisfies E (t) ≤ a1G−11 ( a2 ∫ t 0 α (s) ds + a3 ) , t ≥ 0, (4.1) where G−11 = ∫ 1 t 1 G0 (s) ds and G0 (s) = tG ′ (η0t) . Remark 4.1. G1 strictly decreases and is convex on (0, 1] and lim t→0 G1 (t) = +∞. Proof. We multiply (3.15) by α (t) to get α (t)L′ (t) ≤−k1α (t) E (t) + cα (t) ∫ 1 0 ( φ2t + g 2 (φt) ) dx. (4.2) Now, we discuss two cases: Case I: G is linear on [0,η]. In this case, using (H2) and Eq.(3.1), we deduce that α (t)L′ (t) ≤−k1α (t) E (t) + cα (t) ∫ 1 0 ( φ2t + g 2 (φt) ) dx = −k1α (t) E (t) − cE′ (t) , which can be rewritten as (α (t)L(t) + cE (t))′ −α′ (t)L(t) ≤−k1α (t) E (t) . Int. J. Anal. Appl. 19 (6) (2021) 822 Using (H1), we obtain (α (t)L(t) + cE (t))′ ≤−k1α (t) E (t) . By exploiting (3.14), it can easily be shown that S0 (t) := α (t)L(t) + cE (t) ∼ E (t) . (4.3) So, for some positive constant λ1, we obtain S′0 (t) + λ1α (t)S0 (t) ≤ 0, ∀t ≥ 0 (4.4) The combination of Eq. (4.3) and (4.4), gives E (t) ≤ E (0) e−λ1 ∫ t 0 α(s)ds = E (0) G−11 ( λ1 ∫ t 0 α (s) ds ) . (4.5) Case II: G is nonlinear on [0,η]. In this case, we first choose 0 < η1 < η such that sg (s) ≤ min{η,G (η)} , ∀|s| ≤ η1. (4.6) Using (H2) along with fact that g is continuous and |g (s)| > 0, for s 6= 0, it follows that  s 2 + g2 (s) ≤ G−1 (sg (s)) , ∀|s| ≤ η1 c1 |s| ≤ |sg (s)| ≤ c2 |s| , ∀|s| ≥ η1 (4.7) To estimate the last integral in Eq. (4.2), we consider the following partition of (0, 1): I1 = {x ∈ (0, 1) : |φt| ≤ η1} , I2 = {x ∈ (0, 1) : |φt| > η1} . Now, with I (t) defined by I (t) = ∫ I1 φtg (φt) dx, we have, using Jensen inequality (note that G−1 is concave and recall (4.6)) G−1 (I (t)) ≥ c ∫ I1 G−1 (φtg (φt)) dx. (4.8) The combination of Eq. (4.7) and (4.8) yields α (t) ∫ 1 0 ( φ2t + g 2 (φt) ) dx = α (t) ∫ I1 ( φ2t + g 2 (φt) ) dx + α (t) ∫ I2 ( φ2t + g 2 (φt) ) dx ≤ α (t) ∫ I1 G−1 (φtg (φt)) dx + cα (t) ∫ I2 φtg (φt) dx ≤ cα (t) G−1 (I (t)) − cE′ (t) . (4.9) So, by substituting (4.9) into (4.2) and using (4.3) and (H1), we have S′0 (t) ≤−k1α (t) E (t) + cα (t) G −1 (I (t)) (4.10) Int. J. Anal. Appl. 19 (6) (2021) 823 Now, for η1 < η and δ0 > 0, using (4.10) and the fact that E ′ ≤ 0, G′ > 0, G′′ > 0 on (0,η) , we find that the functional S1, defined by S1 (t) := G′ ( η0 E (t) E (0) ) S0 (t) + δ0E (t) , satisfies, for some b1, b2 > 0, b1S1 (t) ≤ E (t) ≤ b2S1 (t) (4.11) and S′0 (t) : = η0 E′ (t) E (0) G′′ ( η0 E (t) E (0) ) S0 (t) + G′ ( η0 E (t) E (0) ) S′0 (t) + δ0E ′ (t) ≤ −k1α (t) E (t) G′ ( η0 E (t) E (0) ) + cα (t) G′ ( η0 E (t) E (0) ) G−1 (I (t)) + δ0E ′ (t) (4.12) Let G∗ be the convex conjugate of G defined by G∗ (s) = s ( G ′ )−1 (s) −G [( G ′ )−1 (s) ] , if s ∈ (0,G′ (η)] , satisfying the following general Young’s inequality AB ≤ G∗ (A) + G (B) , if A ∈ (0,G′ (η)] , B ∈ (0,η] . With A = G′ ( η0 E (t) E (0) ) and B = G−1 (I (t)) , using (4.6), we obtain cα (t) G′ ( η0 E (t) E (0) ) G−1 (I (t)) ≤ cα (t) G∗ ( G′ ( η0 E (t) E (0) )) + cα (t) I (t) . By exploiting (3.1) and the fact that G∗ (s) ≤ s (G′)−1 (s) , we get cα (t) G′ ( η0 E (t) E (0) ) G−1 (I (t)) ≤ cα (t) η0 E (t) E (0) G′ ( η0 E (t) E (0) ) − cE′ (t) (4.13) By substituting (4.12) into Eq. (4.13), we obtain S′1 (t) ≤−kα (t) E (t) E (0) G′ ( η0 E (t) E (0) ) = −k1α (t) G0 ( E (t) E (0) ) (4.14) where k > 0 and G0 (t) = tG ′ (η0t) . Note that G′0 (t) = G ′ (η0t) + η0tG ′′ (η0t) . So, using the strict convexity of G on (0,η] , we find that G0 (t) ,G ′ 0 (t) > 0 on (0, 1] .With S (t) := b1S1(t) E(0) it is obvious that S (t) ≤ E(t) E(0) ≤ 1. Now, using (4.11) and (4.14), we have S (t) ∼ E (t) (4.15) Int. J. Anal. Appl. 19 (6) (2021) 824 and, for some a2 > 0 S′ (t) ≤−a2α (t) G0 (S (t)) . (4.16) Inequality (4.16) implies that d dt G1 (S (t)) ≥ a2α (t) , where G1 (t) = ∫ t 1 1 G0 (s) ds. Thus, by integrating over [0, t] , we obtain, for some a3 > 0, S (t) ≤ G−11 ( a2 ∫ t 0 α (s) ds + a3 ) . 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