() Nonlinear Analysis 63 (2005) 198–224 www elsevier com/locate/na On a shock problem involving a linear viscoelastic bar Nguyen Thanh Longa,∗, Le Van Utb, Nguyen Thi Thao Trucc aDepartment of Mathema[.]
Nonlinear Analysis 63 (2005) 198 – 224 www.elsevier.com/locate/na On a shock problem involving a linear viscoelastic bar Nguyen Thanh Longa,∗ , Le Van Utb , Nguyen Thi Thao Trucc a Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University HoChiMinh City, 227 Nguyen Van Cu Str., Dist.5, HoChiMinh City, Vietnam b Department of Natural Sciences, Can Tho In-serviced University, 256 Nguyen Van Cu Street, Ninh Kieu District, Can Tho City, Vietnam c Department of Mathematics, School of Education, Can Tho University, 3-2 Street, Ninh Kieu District, Can Tho City, Vietnam Received June 2004; accepted May 2005 Abstract The paper deals with the initial-boundary value problem for the linear wave equation utt − (t)uxx + Ku + ut = f (x, t), < x < 1, < t < T , u(0, t) = 0, −(t)ux (1, t) = Q(t), u(x, 0) = u0 (x), ut (x, 0) = u1 (x), (1) where K, are given constants and u0 , u1 , f , are given functions, the unknown function u(x, t) and the unknown boundary value Q(t) satisfy the following linear integral equation t Q(t) = K1 (t)u(1, t) + 1 (t)ut (1, t) − g(t) − k(t − s)u(1, s) ds, (2) where g, k, K1 , 1 are given functions The paper consists of four parts In Part we prove a theorem of global existence and uniqueness of a weak solution (u, Q) of problem (1)–(2) The proof is based on a Galerkin type approximation associated to various energy estimates-type bounds, weakconvergence and compactness arguments In Part 2, it is devoted to the study of the regularity of the solution (u, Q) with respect to the functions (, 1 , f, g, k, K1 ) In Part we prove that the solution ∗ Corresponding author E-mail addresses: longnt@hcmc.netnam.vn (N.T Long), utlev@yahoo.com (L.V Ut), ntthaotruc@ctu.edu.vn (N.T.T Truc) 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd All rights reserved doi:10.1016/j.na.2005.05.007 N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 199 (u, Q) of this problem is stable with respect to the data (K, , , 1 , f, g, k, K1 ) Finally, in Part we obtain an asymptotic expansion of the solution (u, Q) of this problem up to order N + in (K, ), for (K, ) sufficiently small 䉷 2005 Elsevier Ltd All rights reserved MSC: 35L20; 35L70 Keywords: Shock; Integral equations; Energy-type estimates; Compactness; Regularity of solutions; Stability of the solutions; Asymptotic expansion Introduction In this paper, we consider the following problem: Find a pair (u, Q) of functions satisfying utt − (t)uxx + F (u, ut ) = f (x, t), < x < 1, < t < T , (1.1) u(0, t) = 0, (1.2) −(t)ux (1, t) = Q(t), (1.3) u(x, 0) = u0 (x), (1.4) ut (x, 0) = u1 (x), where F (u, ut )=Ku+ ut , with K, are given constants and u0 , u1 , f, are given functions satisfying conditions specified later, and the unknown function u(x, t) and the unknown boundary value Q(t) satisfy the following integral equation t Q(t) = K1 (t)u(1, t) + 1 (t)ut (1, t) − g(t) − k(t − s)u(1, s) ds, (1.5) where g, k, K1 , 1 are given functions In [8] Santos studied the asymptotic behavior of solution of problem (1.1), (1.2), (1.4) associated with a boundary condition of memory type at x = as follows t g(t − s)(s)ux (1, s) ds = 0, t > (1.6) u(1, t) + / (0) , 1 (t) = g(0) are positive Santos transformed (1.6) into (1.3), (1.5) with K1 (t) = gg(0) constants In the case of 1 (t) ≡ 0, K1 (t) = h 0, (t) = 1, the problem (1.1)–(1.5) is formed from the problem (1.1)–(1.4) wherein, the unknown function u(x, t) and the unknown boundary value Q(t) satisfy the following Cauchy problem for ordinary differential equation: // Q (t) + 2 Q(t) = hutt (1, t), < t < T , (1.7) Q(0) = Q0 , Q/ (0) = Q1 , where h 0, > 0, Q0 , Q1 are given constants [6] 200 N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 In [1], An and Trieu have studied a special case of problem (1.1)–(1.4), (1.7) with u0 = u1 = Q0 = and F (u, ut ) = Ku + ut , with K 0, are given constants In the later case the problem (1.1)–(1.4) and (1.7) is a mathematical model describing the shock of a rigid body and a linear viscoelastic bar resting on a rigid base [1] From (1.7) we represent Q(t) in terms of Q0 , Q1 , , h, utt (1, t) and then by integrating by parts, we have t Q(t) = hu(1, t) − g(t) − k(t − s)u(1, s) ds, (1.8) where g(t) = −(Q0 − hu0 (1)) cos t − (Q1 − hu1 (1)) sin t, k(t) = h sin t (1.9) (1.10) In [2] Bergounioux et al studied problem (1.1), (1.4) with the mixed boundary conditions (1.2), (1.3) standing for t ux (0, t) = hu(0, t) + g(t) − k(t − s)u(0, s) ds, (1.11) ux (1, t) + K1 u(1, t) + 1 ut (1, t) = 0, (1.12) where g(t) = (Q0 − hu0 (0)) cos t + (Q1 − hu1 (0)) sin t, k(t) = h sin t, (1.13) (1.14) where h 0, > 0, Q0 , Q1 , K, , K1 , 1 are given constants The present paper consists of four main parts In Part we prove a theorem of global existence and uniqueness of a weak solution (u, Q) of problem (1.1)–(1.5) The proof is based on a Galerkin type approximation associated to various energy estimates-type bounds, weak-convergence and compactness arguments The main difficulty encountered here is the boundary condition at x = In order to solve this particular difficulty, stronger assumptions on the initial conditions u0 and u1 will be made We remark that the linearization method in the papers [3,7] can not be used in [2,5,6] Part is devoted to the study of the regularity of the solution (u, Q) with respect to the functions (, 1 , f, g, k, K1 ) In Part we prove that the solution (u, Q) of this problem is stable with respect to the data (K, , , 1 , f, g, k, K1 ) Finally, in Part we obtain an asymptotic expansion of the solution (u, Q) of this problem up to order N + in (K, ), for (K, ) sufficiently small The results obtained here may be considered as generalizations of those in An and Trieu [1] and in Long and Dinh [2,3,5–8] The existence and uniqueness theorem Put = (0, 1), QT = × (0, T ), T > We omit the definitions of usual function spaces: C m (), Lp (), W m,p () N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 201 We denote W m,p = W m,p (), Lp = W 0,p (), H m = W m,2 (), p ∞, m = 0, 1, The norm in L2 is denoted by · We also denote by ·, · the scalar product in L2 or pair of dual scalar product of a continuous linear functional with an element of a function space We denote by · X the norm of a Banach space X and by X / the dual space of X We denote by Lp (0, T ; X), p ∞ for the Banach space of the real functions u : (0, T ) → X measurable, such that uLp (0,T ;X) = T p u(t)X dt 1/p < ∞ for p < ∞, and uL∞ (0,T ;X) = ess sup u(t)X 0 0, f, ft ∈ L2 (QT ) Then, we have the following theorem (2.3) 202 N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 Theorem Let (H1 ) − (H6 ) hold Then, for every T > 0, there exists a unique weak solution (u, Q) of problem (1.1)–(1.5) such that u ∈ L∞ (0, T ; V ∩ H ), ut ∈ L∞ (0, T ; H ), utt ∈ L∞ (0, T ; L2 ), (2.4) u(1, ·) ∈ H (0, T ), Q ∈ H (0, T ) Remark It follows from (2.1), (2.4) that the component u in the weak solution (u, Q) of problem (1.1)–(1.5) satisfies u ∈ C (0, T ; V ) ∩ C (0, T ; L2 ) ∩ L∞ (0, T ; V ∩ H ) Proof of Theorem The proof consists of Steps 1–4 Step 1: The Galerkin approximation Let {wj } be a denumerable base of V ∩ H We find the approximate solution of problem (1.1)–(1.5) in the form um (t) = m cmj (t)wj , (2.5) j =1 where the coefficient functions cmj satisfy the system of ordinary differential equation // / um (t), wj + (t) umx (t), wj x + Qm (t)wj (1) + F (um (t), um (t)), wj = f (t), wj , j m, t / Qm (t) = K1 (t)um (1, t) + 1 (t)um (1, t) − k(t − s)um (1, s) ds − g(t), (2.6) (2.7) m mj wj → u0 u (0) = u = m 0m / um (0) = u1m = j =1 m j =1 mj wj → u1 strongly in H , (2.8) strongly in H From the assumptions of Theorem 1, system (2.6)–(2.8) has solution (um , Qm ) on some interval [0, Tm ] The following estimates allow one to take Tm = T for all m Step 2: A priori estimates A priori estimates I Substituting (2.7) into (2.6), then multiplying the jth equation of / (2.6) by cmj (t) and summing up with respect to j, we get d / / / Sm (t) = − K um (t)2 − 2um (t)2 + / (t)umx (t)2 + K1 (t)u2m (1, t) dt t / / +2 k(t − s)um (1, s) ds + g(t) um (1, t) + f (t), um (t) , (2.9) where / Sm (t) ≡ um (t)2 + (t)umx (t)2 + K1 (t)u2m (1, t) t / +2 1 (s)|um (1, s)|2 ds (2.10) N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 203 Integrating by parts with respect to the time variable, we get t / Sm (t) = Sm (0) + Ku0m 2 − 2g(0)u0m (1) − Kum (t)2 − 2 um (s)2 ds t t / + / (s)umx (s)2 ds + [K1 (s) − 2k(0)]u2m (1, s) ds 0 t + um (1, t) k(t − s)um (1, s) ds + g(t)um (1, t) t −2 um (1, ) d g / () + k / ( − s)um (1, s) ds 0 t / + f (s), um (s) ds (2.11) Using the inequality 2ab a + b , ∀a, b ∈ R, ∀ > 0, (2.12) and the following inequalities / Sm (t) um (t)2 t / + 0 umx (t) + 20 um (1, s) ds, (2.13) |um (1, t)| um (t)C () umx (t) Sm (t) , 0 (2.14) we shall estimate, respectively the following terms on the right-hand side of (2.11) t t / 2 um (t) = um (0) + um (s) ds 2u0m + 2t Sm (s) ds, (2.15) 0 t t / (s)umx (s)2 ds |/ (s)|Sm (s) ds, 0 t t / (K / (s) − 2k(0))u2 (1, s) ds |K1 (s) − 2k(0)|Sm (s) ds, m (2.16) 0 (2.17) t t t k () d Sm (s) ds, um (1, t) k(t − s)um (1, s) ds Sm (t) + 0 0 (2.18) g (t), 0 t t t / Sm () d + g ()um (1, ) d |g / ()|2 d, 0 0 2|g(t)um (1, t)| Sm (t) + (2.19) (2.20) 204 N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 t / um (1, ) d k ( − s)um (1, s) ds 0 t t t |k / ()|2 d Sm (s) ds, Sm () d + t 0 0 t t t / f (s), um (s) ds f (s)2 ds + Sm (s) ds 0 (2.21) (2.22) Combining (2.11), (2.15)–(2.22), we obtain Sm (t) Sm (0) + 3|K|u0m 2 + 2|g(0)u0m (1)| t / g (t) + |g (s)| ds + t t0 / (|/ (s)| + |K1 (s) − 2k(0)|)Sm (s) ds + f (s) ds + 0 0 t t / (k () + t|k ()| ) d Sm (s) ds + 0 0 t + [2 + + 2|| + 2t|K|] Sm (s) ds + 2Sm (t) (2.23) Choosing = 1/4, it follows from (2.23), that Sm (t) 2Sm (0) + 6|K|u0m 2 + 4|g(0)u0m (1)| t t / + g (t) + |g (s)| ds + f (s)2 ds 0 0 t / (|/ (s)| + |K1 (s) − 2k(0)|)Sm (s) ds + 0 t t / + (k () + t|k ()| ) d Sm (s) ds 0 0 t + (3 + 4|| + 4t|K|) Sm (s) ds (2.24) Using assumptions (H1 ).(H3 ), (H5 ), it follows from (2.8), (2.10), that 2Sm (0) + 6|K|u0m 2 + 4|g(0)u0m (1)| C1 for all m, where C1 is a constant depending only on (0), K1 (0), g(0), K, u0 , u1 On the other hand, we have t t (1) / g (t) + C1 + |g (s)| ds + f (s)2 ds MT 0 0 for all t ∈ [0, T ], t (2) (k () + t|k / ()|2 ) d MT (3 + 4|| + 4t|K|) + 0 for all t ∈ [0, T ], (2.25) (2.26) (2.27) N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 205 (i) where MT , i = 1, 2, is a constant depending only on T Therefore, it follows from (2.24)–(2.27), that t (1) (1) Sm (t) MT + NT (s)Sm (s) ds, (2.28) where (1) NT (s) = / (2) (|/ (s)| + |K1 (s) − 2k(0)|) + MT , 0 By Gronwall’s lemma, we obtain t (1) (1) Sm (t) MT exp NT (s) ds MT , (1) NT ∈ L1 (0, T ) for all t ∈ [0, T ] (2.29) (2.30) A priori estimates II Now differentiating (2.6) with respect to t we have /// / um (t), wj + (t) umx (t), wj x + / (t) umx (t), wj x / / // + Qm (t)wj (1) + Kum + um (t), wj = f / (t), wj , (2.31) for all j m // Multiplying the jth equation of (2.31) by cmj (t), summing up with respect to j and then integrating with respect to the time variable from to t, we have after some rearrangements Xm (t) = Xm (0) + 2/ (0) u0mx , u1mx + Ku1m 2 t / / // − 2/ (t) umx (t), umx (t) − Kum (t)2 − 2 um (s)2 ds t / / / // − [(K1 (s) + 1 (s))um (1, s) + (K1 (s) − k(0))um (1, s)]um (1, s) ds t s // / / +2 k (s − )um (1, ) d + g (s) um (1, s) ds t 0 t / / // (s) umx (s), umx (s) ds +3 / (s)umx (s)2 ds + 0 t // / + f (s), um (s) ds, (2.32) where // Xm (t) = um (t)2 / + (t)umx (t)2 +2 t // 1 (s)|um (1, s)|2 ds (2.33) Using inequality (2.12) and the following inequalities: t // / // Xm (t) um (t)2 + 0 umx (t)2 + 20 |um (1, s)|2 ds, (2.34) |um (1, t)| um (t)C () umx (t) Sm (t) 0 Xm (t) , 0 (2.36) we estimate without difficulty the following terms in the right-hand side of (2.32) as follows / | − 2/ (t) umx (t), umx (t) | Xm (t) + / um (t)2 2u1m 2 + 2t MT |/ (t)|2 , 20 t Xm (s) ds, (2.38) t / / // −2 K1 (s) + 1 (s) um (1, s)um (1, s) ds t 1 / Xm (t), |K1 (s) + 1 (s)|2 Xm (s) ds + 0 0 t −2 (K / (s) − k(0))um (1, s)u// (1, s) ds m t MT / |K1 (s) − k(0)|2 ds + Xm (t), 0 0 t // um (1, s) ds 0 s k / (s − )um (1, ) d MT Xm (t) + t 0 0 (2.37) t / |k ()| d 2 , t 1 t / // / g (s)um (1, s) ds Xm (t) + |g (s)|2 ds, 0 0 (2.39) (2.40) (2.41) (2.42) t t / |/ (s)|Xm (s) ds, (2.43) / (s)umx (s)2 ds 0 t t t / |// (s)|2 ds + // (s) umx (s), umx (s) ds MT Xm (s) ds, (2.44) 0 0 t t t / // f (s)2 ds (2.45) f / (s), um (s) ds Xm (s) ds + 0 N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 207 It follows from (2.32), (2.37)–(2.45) that Xm (t) Xm (0) + 2|/ (0)|u0m V u1m V + 3|K|u1m 2 MT t / / + MT | (t)| + |K1 (s) − k(0)|2 ds 0 0 2 t t MT |// (s)|2 ds t |k / ()| d + MT + 0 0 t / t / 2 + |g (s)| ds f (s) ds + t / |K1 (s) + 1 (s)|2 + |/ (s)| + + 2T |K| + 2|| + + 0 Xm (t) (2.46) Xm (s) ds + + 0 Choosing > with (1 + 2 ) < 1/2, hence, we obtain from (2.46), that (2) Xm (t) 2Xm (0) + 4|/ (0)|u0m V u1m V + 6|K|u1m 2 + NT (t) t (1) (s)Xm (s) ds, + N T (2.47) where (2) NT (t) = 2MT t / / |K1 (s) − k(0)|2 ds M | (t)| + T 0 20 t 2 t 2MT / |// (s)|2 ds t |k ()| d + MT + 0 0 0 t / t / + f (s)2 ds + |g (s)|2 ds, (2.48) (1) (s) = 4T |K| + 4|| + |K1 (s) + / (s)|2 + |/ (s)| + 2 + 2, N T 0 0 (1) ∈ L1 (0, T ) N T (2.49) First, we deduce from (2.8) and (H1 ).(H2 ), (H5 ), that 2Xm (0) + 4|/ (0)|u0m V u1m V + 6|K|u1m 2 // 2um (0)2 + C2 , for all m, (2.50) where C2 is a constant depending only on , u0 , u1 , K But by (2.6)–(2.8), we have // // // // um (0)2 − (0) u0mxx , um (0) + Ku0m + u1m , um (0) = f (0), um (0) (2.51) 208 N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 Therefore, // um (0) (0)u0m + |K|u0m + ||u1m + f (0) (2.52) It follows from (2.8), (2.52), that // um (0) C3 , (2.53) where C3 is a constant depending only on , u0 , u1 , f, K, Noting that H (0, T ) ֒→ C ([0, T ]), it follows from (2.48), that (2) (1) , 2C32 + C2 + NT (t) M T a.e., t ∈ [0, T ], (2.54) (1) is a constant depending only on T Therefore, it follows from (2.47), (2.50), where M T (2.53), (2.54), that t (1) + (1) (s)Xm (s) ds Xm (t) M N (2.55) T T By Gronwall’s lemma, we deduce from (2.55), that t (1) (1) T ∀t ∈ [0, T ] Xm (t) MT exp NT (s) ds M (2.56) On the other hand, we deduce from (2.7), (2.30), (2.34)–(2.36) and (2.56), that / Qm 2L2 (0,T ) T T 5M / MT 1 2∞ + |K1 (s) + 1 (s)|2 ds 0 0 T 2 5T M T 5MT T / / + |K1 (s) − k(0)| ds + |k ()| d 0 0 T +5 |g / (s)|2 ds, (2.57) where 1 ∞ = 1 L∞ (0,T ) Hence Qm H (0,T ) MT (2.58) Step 3: Limiting process From (2.13), (2.30), (2.34), (2.56) and (2.58), we deduce the existence of a subsequence of {(um , Qm )} still also so denoted, such that um → u in L∞ (0, T ; V ) weak ∗, / / in L∞ (0, T ; V ) weak ∗, um → u // // (2.59) um → u in L∞ (0, T ; L2 ) weak ∗, um (1, ·) → u(1, ·) in H (0, T ) weakly, Qm → Q in H (0, T ) weakly N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 209 By the compactness lemma of Lions [4, p 57] we can deduce from (2.59) the existence of a subsequence still denoted by {(um , Qm )} such that um → u strongly in L2 (QT ), / strongly in L2 (QT ), um → u/ (2.60) um (1, ·) → u(1, ·) strongly in H (0, T ), u/m (1, ·) → u/ (1, ·) strongly in C [0, T ], Qm → Q strongly in C [0, T ] From (2.7) and (2.60)3−4 we have Qm (t) → K1 (t)u(1, t) + 1 (t)u/ (1, t) t − k(t − s) u(1, s) ds − g(t) ≡ Q(t) (2.61) strongly in C [0, T ] Hence, we deduce from (2.60)5 and (2.61), that Q(t) = Q(t) (2.62) Passing to the limit in (2.6)–(2.8) by (2.59)–(2.60) and (2.61)–(2.62) we have (u, Q) satisfying the problem u// (t), v + (t) ux (t), vx + Q(t)v(1) + Ku(t) + u/ (t), v = f (t), v , ∀v ∈ H , u(0) = u0 , u/ (0) = u1 , Q(t) = K1 (t)u(1, t) + 1 (t)ut (1, t) − (2.63) (2.64) t k(t − s)u(1, s) ds − g(t), (2.65) in L2 (0, T ) weakly On the other hand, we have from (2.63) and assumptions (H5 ).(H6 ) that uxx = [u// + Ku(t) + u/ (t) − f ] ∈ L∞ (0, T ; L2 ) (t) (2.66) Hence u ∈ L∞ (0, T ; V ∩ H ) and the existence proof is completed Step 4: Uniqueness of the solution Let u1 , u2 be two weak solutions of problem (1.1)–(1.5) such that ui ∈ L∞ (0, T ; V ∩ H ), / // u ∈ L∞ (0, T ; H ), ui ∈ L∞ (0, T ; L2 ), (2.67) i ui (1, ·) ∈ H (0, T ), Qi ∈ H (0, T ), i = 1, Then (u, Q) with u = u1 − u2 and Q = Q1 − Q2 satisfy the variational problem // u (t), v +(t) ux (t), vx +Q(t)v(1) + Ku + u/ , v =0 ∀v ∈ H , u(0) = u/ (0) = 0, (2.68) 210 N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 where Q(t) = K1 (t)u(1, t) + 1 (t)u/ (1, t) − t k(t − s)u(1, s) ds We take v = u/ in (2.68)1 , afterwards integrating in t, we get t t / / S(t) = (s)ux (s) ds + K1 (s)u2 (1, s) ds 0 t s / +2 u (1, s) ds k(s − )u(1, ) d − Ku(t)2 0 t − 2 u/ (s)2 ds, (2.69) (2.70) where / 2 S(t)=u (t) +(t)ux (t) +K1 (t)u (1, t)+2 t 1 (s)|u/ (1, s)|2 ds (2.71) Noting that S(t) u/ (t)2 + 0 ux (t)2 + 20 |u(1, t)| u(t)C () ux (t) u(t)2 t u/ (s) ds 2 t t t |u/ (1, s)|2 ds, S(t) (t) S(t) , 0 u/ (s)2 ds We again use inequality (2.12), then, it follows from (2.70)–(2.74), that t t / [|/ (s)| + |K1 (s)|]S(s) ds + (|K|t + 2||) S(s) ds S(t) 0 0 t t t S(t) + k () d + S() d 0 0 0 (2.72) (2.73) (2.74) (2.75) Choosing > 0, such that 21 1/2, we obtain from (2.75), that S(t) t q1 (s)S(s) ds, (2.76) where 2T T / k () d, (|/ (s)| + |K1 (s)|) + 2|K|T + 4|| + 0 0 q1 ∈ L1 (0, T ) q1 (s) = By Gronwall’s lemma, we deduce that S ≡ and Theorem is completely proved (2.77) N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 211 Remark In the case of K =K(t), = (t), k =k(t, s), we consider the following problem utt − (t)uxx + K(t)u + (t)ut = f (x, t), < x < 1, < t < T , (1.1′ ) u(0, t) = 0, (1.2) −(t)ux (1, t) = Q(t), (1.3) u(x, 0) = u0 (x), (1.4) ut (x, 0) = u1 (x), where K, , u0 , u1 , f, are given functions and the unknown functions (u(x, t), Q(t)) satisfy the following integral equation: t Q(t) = K1 (t)u(1, t) + 1 (t)ut (1, t) − k(t, s)u(1, s) ds − g(t), (1.5′ ) where g, k, K1 and 1 are given functions Now, we assume that 1 ) (H K ∈ H (0, T ), 4 ) k, D1 k = (H ∈ H (0, T ), jk ∈ L2 (Q∗T ), jt k∗ ∈ L2 (0, T ), where k∗ (t) = k(t, t) and Q∗T = {(t, s) : s t T } By an argument analogous to that used to the proof of Theorem 1, we get the following result 1 ) and (H 4 ) hold Then, there exists Theorem Let T > Let (Hi ), i 6, i = 4, (H / a unique weak solution (u, Q) of problem (1.1 ), (1.2)–(1.4), (1.5/ ) such that u ∈ C (0, T ; V ) ∩ C (0, T ; L2 ) ∩ L∞ (0, T ; V ∩ H ), (2.78) ut ∈ L∞ (0, T ; H ), utt ∈ L∞ (0, T ; L2 ), u(1, ·) ∈ H (0, T ), Q ∈ H (0, T ) Regularity of solutions In this part, we study the regularity of solution of problem (1.1)–(1.5) corresponding to the data (, 1 , K, , u0 , u1 , f, g, k, K1 ) From here, we assume that (, 1 , K, , u0 , u1 , f, g, k, K1 ) satisfy assumptions (H1 ).(H6 ) Henceforth, we shall impose the following stronger assumptions: (1) (H1 ) (1) (H2 ) (1) (H3 ) (1) (H4 ) K, ∈ R, u0 ∈ H and u1 ∈ H , g, K1 , 1 ∈ H (0, T ), 1 (t) 0 > 0, K1 (t) 0, k ∈ H (0, T ), 212 (1) N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 / (H5 ) ∈ H (0, T ), (t) 0 > 0, (t)1 (t) − / (t)1 (t) 0, (1) (H6 ) f, ft , ftt ∈ L2 (0, T ; L2 ) and f (·, 0) ∈ H ≡ Formally differentiating problem (1.1) with respect to time and letting u = ut and Q(t) / of problem (P (1) ) as follows: (t)( Q(t) ) we are led to consider the solution ( u , Q) (t) u + L u ≡ utt − (t) uxx + K(t) (t) ut = f(x, t), < x < 1, < t < T , u (0, t) = 0, (P (1) ) B u ≡ −(t) ux (1, t) = Q(t), u (x, 0) = u (x), u (x, 0) = u1 (x), t t =K 1 (t) k(t, s) u(1, s) ds − g (t), Q(t) u(1, t) + 1 (t) ut (1, t) − where / / (t) = K − (t) , (t) = − , K(t) (t) (t) f / (t) f(x, t) = (t) j u, +K jt (t) (t) u0 (x) = u1 (x), u1 (x) = (0)u0xx (x) − Ku0 (x) − u1 (x) + f (x, 0), / 1 K1 (t) = K1 (t) + (t) , K1 (t) / / (t) k(t, s) = −(t) − k1 (t − s) + k(t − s), (t) t (t) with k1 (t) = k() d, / / / g K1 − k1 (t) + k(t) u0 (1) + g (t) = − (3.1) (3.2) (3.3) (1) (1) Let (, 1 , K, , u0 , u1 , f, g, k, K1 ) satisfy assumptions (H1 ).(H6 ) Then (, 1 , K, , 1 ) satisfy assumptions (Hi ), i 6, i = 4, (H 1 ) and (H 4 ) By Theorem u0 , u1 , f, g, k, K such that 2, the problem (P (1) ) there exists a unique weak solution ( u, Q) u ∈ C (0, T ; V ) ∩ C (0, T ; L2 ) ∩ L∞ (0, T ; V ∩ H ), (3.4) ut ∈ L∞ (0, T ; H ), utt ∈ L∞ (0, T ; L2 ), u(1, ·) ∈ H (0, T ), Q ∈ H (0, T ) Moreover, from the uniqueness of weak solution we have Q(t) / u = ut , Q(t) = (t) (t) From the equalities Q(t) / Q/ (t)(t) − Q(t)/ (t) = , Q(t) = (t) (t) (t) / // / / / + Q/ + − ( ) Q, + Q, Q// = Q Q/ = Q (3.5) (3.6) N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 213 it follows that Q/ , Q// ∈ L2 (0, T ), i.e., Q ∈ H (0, T ) (3.7) Then, we deduce from (3.4)–(3.5), (3.7), that u ∈ C (0, T ; V ∩ H ) ∩ C (0, T ; V ∩ H ) ∩ C (0, T ; L2 ), ut ∈ L∞ (0, T ; V ∩ H ), utt ∈ L∞ (0, T ; H ), uttt ∈ L∞ (0, T ; L2 ), u(1, ·) ∈ H (0, T ), Q ∈ H (0, T ) (3.8) We then have the following theorem (1) (1) Theorem Let (, 1 , K, , u0 , u1 , f, g, k, K1 ) satisfy assumptions (H1 ).(H6 ) Then there exists a unique weak solution (u, Q) of problem (1.1)–(1.5) satisfying (3.8) Similarly, formally differentiating problem (1.1)–(1.5) with respect to time up to order [r−1] jr u r, (r 1) and letting u[r] = jt r and Q[r] (t) = (t) dtd ( Q (t)(t) ) we are led to consider the solution (u[r] , Q[r] ) of problem (P (r) ): [r] [r] [r] + K [r] (t)u[r] + [r] (t)ut[r] = f [r] (x, t), L u ≡ utt[r] − (t)uxx < x < 1, < t < T , [r] u (0, t) = 0, [r] [r] ≡ −(t)u (1, t) = Q[r] (t), (r) Bu x (P ) [r] [r] (x), ut[r] (x, 0) = u1[r] (x), (x, 0) = u u [r] [r] [r] Q (t) = K1 (t)u (1, t) + 1 (t)ut[r] (1, t) t − k [r] (t, s) u[r] (1, s) ds − g [r] (t), where the functions K [r] , [r] , u0[r] , u1[r] , f [r] , g [r] , k [r] , K1[r] , r 1, are defined by the recurrence formulas u[0] = u0 , , u0[r] = u[r−1] r 1, j [j ] (3.9) u[0] = u1 , u1[r] = r−1 j =0 )+ , u[r−1] Cr−1 (r−1−j ) (0)u0xx − F (u[r−1] jr−1 f (·, 0), jt r−1 r 1, (3.10) j r! where Cr = j !(r−j )! , / [r] (t) = [r−1] (t) − (t) , r 1, (t) [0] (t) = , / [r−1] (t) [r] [r−1] K (t) = K (t) + (t) , (t) [0] K (t) = K, (3.11) r 1, (3.12) 214 N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 j K [r−1] (t) j f [r−1] [r−1] , r 1, −u = (t) (3.13) jt (t) jt (t) [0] f = f, 1 (t) / [r] [r−1] (t) + (t) , r 1, K1 (t) = K1 (3.14) (t) [0] K1 (t) = K1 (t), k [r] (t, s) = k [r−1] (t, t) + k [r−1] (t, t) − k [r−1] (t, s) 2 [r−1] / / (t) K1 (t) [r−1] [r−1] − (t, s)) − (t) (t, t) − k1 , r 2, (k (t) (t) s s jk [r−1] k1[r−1] (t, s) = k [r−1] (t, ) d, k2[r−1] (t, s) = (t, ) d, r 2, jt K1 (t) / / (t) t−s k [1] (t, s) = k(t, s) = −(t) − (t) k() d + k(t − s), (t) (3.15) [r−1] / / (t) [r] (t) = k [r−1] (t, t) + k [r−1] (t, t) − (t) k [r−1] (t, t) − (t) K1 g (t) (t) [r−1] / g (t) , r 2, (1) + (t) ×u[r−1] (t) K1 (t) / / (t) t [1] g (t) = −(t) − k() d + k(t) u0 (1) g (t) = (t) (t) g(t) / +(t) (t) (3.16) f [r] Assume that the data (, 1 , K, , u0 , u1 , f, g, k, K1 ) satisfy the following assumptions: (r) (H1 ) (r) (H2 ) (r) (H3 ) (r) (H4 ) (r) (H5 ) (r) K, ∈ R, u0 ∈ H r+2 and u1 ∈ H r+1 , r 1, g, K1 , 1 ∈ H r+1 (0, T ), 1 (t) 0 > 0, K1 (t) 0, r 1, k ∈ H r+1 (0, T ), r 1, / ∈ H r+2 (0, T ), (t) 0 > 0, (t)1 (t) − / (t)1 (t) 0, r 1, j f j f (H6 ) jt ∈ L2 (0, T ; L2 ), (0 r + 1) and jt (·, 0) ∈ H , r − Then (, 1 , K [r] , [r] , u0[r] , u1[r] , f [r] , g [r] , k [r] , K1[r] ) satisfy assumptions (Hi ), i 6, 1 ) and (H 4 ) Applying again Theorem for problem (P (r) ), there exists a unique i = 4, (H [r] weak solution (u , Q[r] ) satisfying (2.4) and the inclusion from Remark 1, i.e., such that [r] u ∈ C (0, T ; V ) ∩ C (0, T ; L2 ) ∩ L∞ (0, T ; V ∩ H ), (3.17) ut[r] ∈ L∞ (0, T ; H ), utt[r] ∈ L∞ (0, T ; L2 ), u[r] (1, ·) ∈ H (0, T ), Q[r] ∈ H (0, T ) N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 Moreover, from the uniqueness of weak solution we have r j u d Q[r−1] [r] [r] , (u , Q ) = jt r dt Hence we obtain from (3.17), (3.18) that u ∈ C r−1 (0, T ; V ∩ H ) ∩ C r (0, T ; H ) ∩ C r+1 (0, T ; L2 ), u(1, ·) ∈ H r+2 (0, T ), Q ∈ H r+1 (0, T ) 215 (3.18) (3.19) We then have the following theorem (r) (r) Theorem Let (H1 ).(H6 ) hold Then there exists a unique weak solution (u, Q) of problem (1.1)–(1.5) satisfying (3.19) and jr u ∈ L∞ (0, T ; V ∩ H ), jt r jr+2 u ∈ L∞ (0, T ; L2 ) jt r+2 jr+1 u ∈ L∞ (0, T ; H ), jt r+1 (3.20) Stability of the solutions In this part, we assume that the functions u0 , u1 satisfy (H2 ) By Theorem 1, problem (1.1)–(1.5) has a unique weak solution (u, Q) depending on K, , , 1 , f, g, k, K1 u = u(K, , , 1 , f, g, k, K1 ), Q = Q(K, , , 1 , f, g, k, K1 ), (4.1) where (K, , , 1 , f, g, k, K1 ) satisfy the assumptions (H1 ), (H3 ).(H6 ) and u0 , u1 are fixed functions satisfying (H2 ) We put I(0 , 0 ) = {(K, , , 1 , f, g, k, K1 ) : (K, , , 1 , f, g, k, K1 ) satisfy the assumptions (H1 ), (H3 ).(H6 )}, where 0 > 0, 0 > are given constants Then we have the following theorem Theorem Let T > Let (H1 ).(H6 ) hold Then, for every T > 0, solutions of the problems (1.1)–(1.5) are stable with respect to the data (K, , , 1 , f, g, k, K1 ), i.e., j j If (K, , , 1 , f, g, k, K1 ), (K j , j , j , 1 , f j , g j , k j , K1 ) ∈ I(0 , 0 ), such that j j j j |K − K| + | − | → 0, − H (0,T ) → 0, 1 − 1 H (0,T ) → 0, j f j − f L2 (0,T ;L2 ) + ft − ft L2 (0,T ;L2 ) → 0, g j − gH (0,T ) → 0, (4.2) j k j − kH (0,T ) → 0, K1 − K1 H (0,T ) → 0, as j → +∞, then / (uj , uj , uj (1, ·), Qj ) → (u, u/ , u(1, ·), Q) (4.3) 216 N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 in L∞ (0, T ; V ) × L∞ (0, T ; L2 ) × H (0, T ) × L2 (0, T ) strongly, as j → +∞, where j j j j uj = u(K j , j , j , 1 , f j , g j , k j , K1 ), Qj = Q(K j , j , j , 1 , f j , g j , k j , K1 ) Proof First, we note that, if the data (K, , , 1 , f, g, k, K1 ) satisfy |K| K ∗ , || ∗ , f L2 (0,T ;L2 ) + ft L2 (0,T ;L2 ) f ∗ , gH (0,T ) g ∗ , ∗ ∗ kH (0,T ) k ∗, 1 H (0,T ) 1 ,∗ H (0,T ) , K1 H (0,T ) K1 , (4.4) where K ∗ , ∗ , g ∗ , k ∗ , ∗1 , ∗ , f ∗ , K1∗ , are fixed positive constants, then, the a priori estimates of the sequences {um } and {Qm } in the proof of Theorem satisfy t / / um (t)2 + 0 umx (t)2 + 20 |um (1, s)|2 ds MT , ∀t ∈ [0, T ], (4.5) // / um (t)2 + 0 umx (t)2 + 20 t // |um (1, s)|2 ds MT , ∀t ∈ [0, T ], Qm H (0,T ) MT , (4.6) (4.7) where MT is a constant depending only on T, u0 , u1 , 0 , 0 , K ∗ , ∗ , f ∗ , g ∗ , ∗ , ∗1 , k ∗ , K1∗ (independent of K, , , 1 , f, g, k, K1 ) Hence, the limit (u, Q) in suitable function spaces of the sequence {(um , Qm )} defined by (2.6)–(2.8) is a weak solution of problem (1.1)–(1.5) satisfying the a priori estimates (4.5)–(4.7) Now, by (4.2) we can assume that, there exist positive constants K ∗ , ∗ , f ∗ , g ∗ , ∗ , ∗1 , j j ∗ k , K1∗ such that the data (K j , j , j , 1 , f j , g j , k j , K1 ) satisfy (4.4) with (K, , , 1 , j j f, g, k, K1 ) =(K j , j , j , 1 , f j , g j , k j , K1 ) Then, by the above remark, we have that the solutions (uj , Qj ) of problem (1.1)–(1.5) corresponding to (K, , , 1 , f, g, k, K1 ) =(K j , j j j , j , 1 , f j ,g j , k j , K1 ) satisfy / uj (t)2 // + 0 uj x (t) + 20 / uj (t)2 + 0 uj x (t)2 + 20 t / |uj (1, s)|2 ds MT , t Qj H (0,T ) MT , // |uj (1, s)|2 ds MT , ∀t ∈ [0, T ], ∀t ∈ [0, T ], (4.8) (4.9) (4.10) Put Kj = K j − K, j = j − , j j = j − , 1j = 1 − 1 , f = f j − f, gj = g j − g, j 1j = K j − K1 kj = k j − k, K (4.11) N.T Long et al / Nonlinear Analysis 63 (2005) 198 – 224 Then, vj = uj − u and Pj = Qj − Q satisfy the variational problem // / vj (t), v + (t) vj x (t), vx + Pj (t)v(1) + Kv j + vj , v / j uj + = fj , v − j (t) uj x (t), vx − K j uj , v , ∀v ∈ H , / vj (0) = vj (0) = 0, 217 (4.12) where Pj (t) = Qj (t) − Q(t) = K1 (t)vj (1, t) + 1 (t)vj t (1, t) − t k(t − s)vj (1, s) ds − gj (t), 1j (t)uj (1, t)−1j (t)uj t (1, t)+ gj (t)= gj (t)−K / t kj (t − s)uj (1, s) ds (4.13) (4.14) Substituting Pj (t) into (4.12), then taking v = vj in (4.12)1 , afterwards integrating in t, we obtain t t / / Sj (t) = − Kvj (t)2 − 2 vj (s)2 ds + K1 (s)|vj (1, s)|2 ds 0 t r / +2 k(r − s)vj (1, s) ds + gj (r) vj (1, r) dr t0 t / / + (s)vj x (s) ds + fj , vj (s) ds 0 t t / / / j uj + −2 j (s) uj x (s), vj x (s) ds − K j uj , vj (s) ds, (4.15) 0 where / Sj (t) = vj (t)2 + (t)vj x (t)2 + K1 (t)|vj (1, t)|2 t / 1 (s)|vj (1, s)|2 ds +2 (4.16) Using inequalities (2.12), (4.8), (4.9) and / Sj (t) vj (t)2 + 0 vj x (t)2 + 20 t / |vj (1, s)|2 ds, (4.17) then, we can prove the following inequality in a similar manner: t T Sj (t) Sj (t) + + (|K|T + 2||) + T k () d Sj (s) ds 20 0 0 t / (|K1 (s)| + |/ (s)|)Sj (s) ds + 0 T t MT + ∞ | gj (r)|2 dr + fj (s)2 ds + 4T 0 j MT √ (|Kj | + 0 | j |), (4.18) + 4T 0 for all > and t ∈ [0, T ] ... 0, are given constants In the later case the problem (1.1)–(1.4) and (1.7) is a mathematical model describing the shock of a rigid body and a linear viscoelastic bar resting on a rigid base... condition at x = In order to solve this particular difficulty, stronger assumptions on the initial conditions u0 and u1 will be made We remark that the linearization method in the papers [3,7] can... We also denote by ·, · the scalar product in L2 or pair of dual scalar product of a continuous linear functional with an element of a function space We denote by · X the norm of a Banach space