existence and asymptotic expansion for a viscoelastic problem with a mixed nonhomogeneous condition

Nonlinear Analysis 67 (2007) 842–864www.elsevier.com/locate/na

Existence and asymptotic expansion for a viscoelastic problem witha mixed nonhomogeneous condition

Nguyen Thanh Longa,∗, Le Xuan Truongb

a Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University HoChiMinh City,227 Nguyen Van Cu Street, Dist.5, HoChiMinh City, Viet Nam

b Department of Mathematics, Faculty of General Science, University of Technical Education in HoChiMinh City, 01 Vo Van Ngan Street,Thu Duc Dist., HoChiMinh City, Viet Nam

Received 8 April 2006; accepted 9 June 2006

Abstract

We study the initial-boundary value problem for a nonlinear wave equation given byut t − uxx +

∫ t

0k(t − s)uxx (s)ds + K |u|

p−2u + λ|ut |q−2ut = f (x, t), 0 < x < 1, 0 < t < T,

ux (0, t) = u(0, t), ux (1, t) + ηu(1, t) = g(t),u(x, 0) = u0(x), ut (x, 0) = u1(x),

(1)

where η ≥ 0; p ≥ 2, q ≥ 2; K , λ are given constants and u0, u1, f , g, k are given functions. In this paper, we consider threemain parts. In Part 1 we prove a theorem of existence and uniqueness of a weak solution u of problem (1). The proof is based on aFaedo–Galerkin method associated with a priori estimates, weak convergence and compactness techniques. Part 2 is devoted to thestudy of the asymptotic behavior of the solution u as η → 0+. Finally, in Part 3 we obtain an asymptotic expansion of the solutionu of the problem (1) up to order N + 1 in three small parameters K , λ, η.c© 2006 Elsevier Ltd. All rights reserved.

MSC: 35L20; 35L70

Keywords: Faedo–Galerkin method; Existence and uniqueness of a weak solution; Energy-type estimates; Compactness; Asymptotic expansion

1. Introduction

In this paper we will consider the following initial and boundary value problem:

ut t − uxx +

∫ t

0k(t − s)uxx (s)ds + F(u, ut ) = f (x, t), 0 < x < 1, 0 < t < T, (1.1)

ux (0, t) = η0u(0, t), ux (1, t) + ηu(1, t) = g(t), (1.2)u(x, 0) = u0(x), ut (x, 0) = u1(x), (1.3)

∗ Corresponding author.E-mail addresses: [email protected], [email protected] (N.T. Long), [email protected] (L.X. Truong).

0362-546X/$ - see front matter c© 2006 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2006.06.044

http://www.elsevier.com/locate/na

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http://dx.doi.org/10.1016/j.na.2006.06.044

N.T. Long, L.X. Truong / Nonlinear Analysis 67 (2007) 842–864 843

where F(u, ut ) = K |u|p−2u + λ|ut |

q−2ut , with η ≥ 0, η0 > 0; p ≥ 2, q ≥ 2; K , λ are given constants and u0, u1,f , g, k are given functions satisfying conditions specified later.

In a recent paper, [1], Berrimia and Messaoudi considered the problem

ut t − 1u +

∫ t

0k(t − s)1u(s)ds = |u|

p−2u, x ∈ Ω , t > 0, (1.4)

u = 0, on ∂Ω , (1.5)u(x, 0) = u0(x), ut (x, 0) = u1(x), x ∈ Ω , (1.6)

where p > 2 is a constant, k is a given positive function, and Ω is a bounded domain of Rn (n ≥ 1), with a smoothboundary ∂Ω . This type of problems have been considered by many authors and several results concerning existence,nonexistence, and asymptotic behavior have been established. In this regard, Cavalcanti et al. [3] studied the followingequation

ut t − 1u +

∫ t

0k(t − s)1u(s)ds + |u|

p−2u + a(x)ut = 0, in Ω × (0, ∞), (1.7)

for a : Ω → R+, a function, which may be null on a part of the domain Ω . Under the conditions that a(x) ≥ a0 > 0on ω ⊂ Ω , with ω satisfying some geometry restrictions and

−ζ1k(t) ≤ k/(t) ≤ −ζ2k(t), t ≥ 0, (1.8)

the authors established an exponential rate of decay.In [5,6], Long and Alain Pham have studied problem (1.1) and (1.3) with k ≡ 0, f (x, t) = 0.In [5], we have considered it with the mixed nonhomogeneous condition

ux (0, t) = hu(0, t) + g(t), u(1, t) = 0, (1.9)

where h > 0 is a given constant; in [6] with the more generalized boundary condition

ux (0, t) = g(t) + hu(0, t) −

∫ t

0H(t − s)u(0, s)ds, u(1, t) = 0. (1.10)

In [7], Long and Diem have studied problem (1.1) and (1.3) with k ≡ 0, and the mixed homogeneous condition

ux (0, t) − h0u(0, t) = ux (1, t) + h1u(1, t) = 0, (1.11)

where h0, h1 are given non-negative constants with h0 + h1 > 0 and a right-hand side of the form

F = F(x, t, u, ux , ut ). (1.12)

In [2] Bergounioux et al. studied problem (1.1) and (1.3) with k ≡ 0, F(u, ut ) = K u + λut , and the mixedboundary conditions (1.2) standing for

ux (0, t) = g(t) + hu(0, t) −

∫ t

0H(t − s)u(0, s)ds, (1.13)

ux (1, t) + K1u(1, t) + λ1ut (1, t) = 0, (1.14)

where h ≥ 0, K , λ, K1, λ1 are given constants and g, H are given functions.In [9], Long et al. obtained the unique existence, regularity and asymptotic expansion of the problem (1.1), (1.3),

(1.13) and (1.14) in the case of k ≡ 0, F(u, ut ) = K |u|p−2u + λ|ut |

q−2ut , with p ≥ 2, q ≥ 2; K , λ are givenconstants.

In [10], Long et al. gave the unique existence, stability, regularity in time variable and asymptotic expansion forthe solution of problem (1.1)–(1.3) when F(u, ut ) = K u + λut , u0 ∈ H2 and u1 ∈ H1. In this case, the problem(1.1)–(1.3) is the mathematical model describing a shock problem involving a linear viscoelastic bar.

In this paper, we consider three main parts. In Part 1, under conditions (u0, u1) ∈ H2× H1; f , ft ∈ L2(QT ),

k ∈ W 2,1(0, T ), g ∈ H2(0, T ); K , λ, η ≥ 0, η0 > 0; p, q ≥ 2, we prove a theorem of existence and uniqueness

844 N.T. Long, L.X. Truong / Nonlinear Analysis 67 (2007) 842–864

of a weak solution u of problem (1.1)–(1.3). The proof is based on a Faedo–Galerkin method associated to a prioriestimates, weak convergence and compactness techniques. We remark that the linearization method in the papers [7,8]cannot be used in [2,5,6,9–11]. Part 2 is devoted to the study of the asymptotic behavior of the solution u as η → 0+.Finally, in Part 3 we obtain an asymptotic expansion of the solution u of the problem (1.1)–(1.3) up to order N + 1 inthree small parameters K , λ, η. The results obtained here relatively are in part generalizations of those in [1–3,5–11].

2. The existence and uniqueness theorem of the solution

Put Ω = (0, 1), QT = Ω × (0, T ), T > 0. We omit the definitions of usual the function spaces: Cm(Ω), L p(Ω),W m,p(Ω).

We denote W m,p= W m,p(Ω), L p

= W 0,p(Ω), Hm= W m,2(Ω), 1 ≤ p ≤ ∞, m = 0, 1, . . ..

The norm in L2 is denoted by ‖ · ‖. We also denote by 〈·, ·〉 the scalar product in L2 or the scalar product of acontinuous linear functional with an element of a function space. We denote by ‖ · ‖X the norm of a Banach spaceX and by X/ the dual space of X . We denote by L p(0, T ; X), 1 ≤ p ≤ ∞ for the Banach space of the real functionsu : (0, T ) → X measurable, such that

‖u‖L p(0,T ;X) =

(∫ T

0‖u(t)‖p

X dt)1/p

< ∞ for 1 ≤ p < ∞,

and

‖u‖L∞(0,T ;X) = ess sup0<t<T

‖u(t)‖X for p = ∞.

Let u(t), u/(t) = ut (t), u//(t) = ut t (t), ux (t), uxx (t) denote u(x, t), ∂u∂t (x, t), ∂2u

∂t2 (x, t), ∂u∂x (x, t), ∂2u

∂x2 (x, t),respectively.

Without loss of generality, we can suppose that η0 = 1. For every η ≥ 0, we putaη(u, v) =

∫ 1

0ux (x)vx (x)dx + u(0)v(0) + ηu(1)v(1), for u, v ∈ H1,

‖v‖η = (aη(v, v))1/2.

(2.1)

On H1 we shall use the following equivalent norm

‖v‖1 =

(v2(0) +

∫ 1

0|vx (x)|2dx

)1/2

. (2.2)

Then we have the following lemmas.

Lemma 2.1. The imbedding H1 → C0(Ω) is compact and

‖v‖C0(Ω)≤

√2‖v‖1, for all v ∈ H1. (2.3)

Lemma 2.2. Let η ≥ 0. Then, the symmetric bilinear form aη(·, ·) defined by (2.1)1 is continuous on H1× H1 and

coercive on H1, i.e.,

(i) |aη(u, v)| ≤ Cη‖u‖1‖v‖1, for all u, v ∈ H1,(ii) aη(v, v) ≥ ‖v‖

21, for all v ∈ H1

where Cη = 1 + 2η.

The proofs of these lemmas are straightforward, and we omit the details.We also note that on H1, ‖v‖1, ‖v‖H1 =

(‖v‖

2+ ‖v/

‖2)1/2, ‖v‖η =

√aη(v, v) are three equivalent norms.

‖v‖21 ≤ ‖v‖

2η ≤ Cη‖v‖

21, for all v ∈ H1, (2.4)

13‖v‖

2H1 ≤ ‖v‖

21 ≤ 3‖v‖

2H1 , for all v ∈ H1. (2.5)


We make the following assumptions:

(H1) K , λ ≥ 0, p, q ≥ 2,(H2) u0 ∈ H2 and u1 ∈ H1,(H3) f, ft ∈ L2(QT ), for all T > 0,(H4) k ∈ W 2,1(0, T ), g ∈ H2(0, T ).

Then, we have the following theorem.

Theorem 2.1. Let (H1)–(H4) hold. Then, for every T > 0, there exists a unique weak solution u of problem (1.1)–(1.3) such that

u ∈ L∞(0, T ; H2), ut ∈ L∞(0, T ; H1), ut t ∈ L∞(0, T ; L2). (2.6)

Remark 2.1. Noting that with the regularity obtained by (2.6), it follows that the problem (1.1)–(1.3) has a uniquestrong solution u satisfying

u ∈ C0(0, T ; H1) ∩ C1(0, T ; L2) ∩ L∞(0, T ; H2). (2.7)

Proof of Theorem 2.1. The proof consists of Steps 1–4.Step 1. The Faedo–Galerkin approximation (introduced by Lions [4]). Let w j be a denumerable base of H2. We

find the approximate solution of problem (1.1)–(1.3) in the form

um(t) =

m∑j=1

cmj (t)w j , (2.8)

where the coefficient functions cmj satisfy the system of ordinary differential equations

〈u//m(t), w j 〉 + aη(um(t), w j ) −

∫ t

0k(t − s)aη(um(s), w j )ds + 〈KΨp(um(t)) + λΨq(u/

m(t)), w j 〉

= g1(t)w j (1) + 〈 f (t), w j 〉, 1 ≤ j ≤ m, (2.9)

um(0) = u0m, u/m(0) = u1m, (2.10)

where

Ψp(z) = |z|p−2z, Ψq(z) = |z|q−2z, g1(t) = g(t) −

∫ t

0k(t − s)g(s)ds, (2.11)

u0m =

m∑j=1

αmjw j → u0 strongly in H2, (2.12)

u1m =

m∑j=1

βmjw j → u1 strongly in H1. (2.13)

From the assumptions of Theorem 2.1, system (2.9) and (2.10) has solution um(t) on some interval [0, Tm]. Thefollowing estimates allow one to take Tm = T for all m.

Step 2. A priori estimates.A priori estimates I. Multiplying the j th equation of the system (2.9) by c/

mj (t), summing up with respect to j andafterwards integrating with respect to the time variable from 0 to t , we get after some rearrangements

Xm(t) = Xm(0) − 2g1(0)u0m(1) + 2g1(t)um(1, t) − 2∫ 1

0g/

1(r)um(1, r)dr

+ 2∫ t

0〈 f (s), u/

m(s)〉ds − 2k(0)

∫ t

0‖um(r)‖2

ηdr + 2∫ t

0k(t − s)aη (um(s), um(t)) ds

− 2∫ t

0dr∫ r

0k/(r − s)aη(um(s), um(r))ds, (2.14)


where

Xm(t) = ‖u/m(t)‖2

+ ‖um(t)‖2η +

2Kp

‖um(t)‖pL p + 2λ

∫ t

0‖u/

m(s)‖qLq ds. (2.15)

From the assumptions (H1), (H2), (H4) and the embedding H1(0, 1) → L p(0, 1), p ≥ 1, there exists a positiveconstant C1 such that

Xm(0) − 2g1(0)u0m(1) = ‖u1m‖2+ ‖u0m‖

2η +

2Kp

‖u0m‖pL p + 2|g1(0)||u0m(1)| ≤ C1 for all m. (2.16)

Using the inequality

2ab ≤ εa2+

1ε

b2, ∀a, b ∈ R, ∀ε > 0, (2.17)

and the following inequalities

|aη(u, v)| ≤ ‖u‖η‖v‖η, ∀u, v ∈ H1, (2.18)

|um(1, t)| ≤ ‖um(t)‖C0(Ω)≤

√2‖um(t)‖1 ≤

√2‖um(t)‖η ≤

√2Xm(t), (2.19)

we shall estimate respectively the following terms on the right-hand side of (2.14) as follows

2g1(t)um(1, t) ≤1ε

g21(t) + εu2

m(1, t) ≤1ε‖g1‖

2L∞(0,T ) + 2εXm(t), (2.20)

−2∫ t

0g/

1(r)um(1, r)dr ≤ 2∫ t

0|g/

1(r)|√

2Xm(r)dr ≤ 2‖g/

1‖2L2(0,T )

+

∫ t

0Xm(r)dr, (2.21)

2∫ t

0〈 f (s), u/

m(s)〉ds ≤ ‖ f ‖2L2(QT )

+

∫ t

0Xm(r)dr, (2.22)

−2k(0)

∫ t

0‖um(r)‖2

ηdr ≤ 2|k(0)|

∫ t

0Xm(r)dr, (2.23)

2∫ t

0k(t − s)aη(um(s), um(t))ds ≤ 2

∫ t

0|k(t − s)|‖um(s)‖η‖um(t)‖ηds

≤ 2‖um(t)‖η

∫ t

0|k(t − s)|‖um(s)‖ηds

≤ εXm(t) +1ε‖k‖

2L2(0,T )

∫ t

0Xm(s)ds, (2.24)

−2∫ t

0dr∫ r

0k/(r − s)aη(um(s), um(r))ds ≤ 2‖k/

‖L∞(0,T )

∫ t

0‖um(r)‖ηdr

∫ r

0‖um(s)‖ηds

≤ 2T ‖k/‖L∞(0,T )

∫ t

0‖um(r)‖2

ηdr ≤ 2T ‖k/‖L∞(0,T )

∫ t

0Xm(s)ds. (2.25)

Combining (2.14), (2.16) and (2.20)–(2.25) and choosing ε =16 , we obtain

Xm(t) ≤ M (1)T + N (1)

T

∫ t

0Xm(s)ds, (2.26)

whereM (1)

T = 2C1 + 12‖g1‖2L∞(0,T ) + 4‖g/

1‖2L2(0,T )

+ 2‖ f ‖2L2(QT )

,

N (1)T = 4[1 + |k(0)| + 3‖k‖

2L2(0,T )

+ T ‖k/‖L∞(0,T )].

(2.27)

By Gronwall’s lemma, we deduce from (2.26) and (2.27), that

Xm(t) ≤ M (1)T exp(t N (1)

T ) ≤ CT , for all t ∈ [0, T ], (2.28)

where CT always indicates a constant depending on T .


A priori estimates II. Now differentiating (2.9) with respect to t , we have

〈u///m(t), w j 〉 + aη(u

/m(t), w j ) −

∫ t

0k/(t − s)aη(um(s), w j )ds

− k(0)aη(um(t), w j ) + K (p − 1)〈|um |p−2u/

m, w j 〉 + λ(q − 1)〈|u/m |

q−2u//m, w j 〉

= g/

1(t)w j (1) +⟨f /(t), w j

⟩, (2.29)

for all 1 ≤ j ≤ m.Multiplying the j th equation of (2.29) by c//

mj (t), summing up with respect to j and then integrating with respectto the time variable from 0 to t , we have after some rearrangements

Ym(t) = Ym(0) + 2∫ t

0〈 f /(s), u//

m(s) 〉ds + 2g/

1(t)u/m(1, t) − 2g/

1(0)u1m(1) − 2∫ t

0g//

1 (s)u/m(1, s)ds

+ 2k(0)

∫ t

0aη(um(s), u//

m(s))ds + 2∫ t

0dr∫ r

0k/(r − s)aη(um(s), u//

m(r))ds

− 2K (p − 1)

∫ t

0〈|um(s)|p−2u/

m(s), u//m(s)〉ds, (2.30)

where

Ym(t) = ‖u//m(t)‖2

+ ‖u/m(t)‖2

η +8

q2 (q − 1) λ

∫ t

0

∥∥∥∥ ∂

∂s(|u/

m(s)|q−2

2 u/m(s))

∥∥∥∥2

ds. (2.31)

Using the following equalities

2∫ t

0dr∫ r

0k/(r − s)a(um(s), u//

m(r))ds = 2∫ t

0k/(t − s)a(um(s), u/

m(t))ds − k/(0)‖um(t)‖2η

+ k/(0)‖u0m‖2η − 2

∫ t

0dr∫ r

0k//(r − s)a(um(s), u/

m(r))ds,(2.32)

2k(0)

∫ t

0a(um(s), u//

m(s))ds = 2k(0)a(um(t), u/m(t)) − 2k(0)a (u0m, u1m) − 2k(0)

∫ t

0‖u/

m(s)‖2ηds, (2.33)

we obtain from (2.30), (2.32) and (2.33), that

Ym(t) = Ym(0) − 2g/

1(0)u1m(1) − 2k(0)aη (u0m, u1m) + k/(0)‖u0m‖2η

+ 2g/

1(t)u/m(1, t) − 2

∫ t

0g//

1 (s)u/m(1, s)ds + 2k(0)aη(um(t), u/

m(t))

− k/(0)‖um(t)‖2η − 2k(0)

∫ t

0‖u/

m(s)‖2ηds + 2

∫ t

0k/(t − s)aη(um(s), u/

m(t))ds

− 2∫ t

0dr∫ r

0k//(r − s)aη(um(s), u/

m(r))ds + 2∫ t

0〈 f /(s), u//

m(s)〉ds

− 2K (p − 1)

∫ t

0〈|um(s)|p−2u/

m(s), u//m(s)〉ds. (2.34)

From the assumptions (H1), (H2), (H4) and the imbedding H1(0, 1) → L p(0, 1), p ≥ 1, there exists positiveconstant C2 depending on T , η, p, q , u0, u1, K , λ, f , g1, such that

Ym(0) − 2g/

1(0)u1m(1) − 2k(0)aη (u0m, u1m) + k/(0)‖u0m‖2η

≤ [‖u0mxx‖ + K‖u0m‖p−1L∞ + λ‖u1m‖

q−1L∞ + ‖ f (t)‖]2

+ ‖u1m‖2η

− 2g/

1(0)u1m(1) − 2k(0)aη (u0m, u1m) + k/(0)‖u0m‖2η ≤ C2 for all m. (2.35)


Using the inequality (2.17) and the following inequality

|u/m(1, t)| ≤ ‖u/

m(t)‖C0(Ω)≤

√2‖u/

m(t)‖1 ≤√

2‖u/m(t)‖η ≤

√2Ym(t). (2.36)

We shall estimate respectively the following terms on the right-hand side of (2.34) as follows

2|g/

1(t)u/m(1, t)| ≤ 2

√2|g/

1(t)|√

Ym(t) ≤1ε

CT + εYm(t), (2.37)

−2∫ t

0g//

1 (s)u/m(1, s)ds ≤ 2

∫ t

0|g//

1 (r)|√

2Ym(r)dr ≤ CT +

∫ t

0Ym(r)dr, (2.38)

2k(0)aη(um(t), u/m(t)) ≤ 2|k(0)|‖um(t)‖η‖u/

m(t)‖η ≤1ε

CT + εYm(t), (2.39)

−k/(0)‖um(t)‖2η − 2k(0)

∫ t

0‖u/

m(s)‖2ηds ≤ CT + 2|k(0)|

∫ t

0Ym(s)ds, (2.40)

2∫ t

0k/(t − s)aη(um(s), u/

m(t))ds ≤ 2∫ t

0|k/(t − s)|‖um(s)‖ηds‖u/

m(t)‖η

≤1ε

CT + εYm(t), (2.41)

−2∫ t

0dr∫ r

0k//(r − s)aη(um(s), u/

m(r))ds ≤ 2∫ t

0dr∫ r

0|k//(r − s)|‖um(s)‖η‖u/

m(r)‖ηds

≤ 2[∫ t

0‖u/

m(r)‖2ηdr

]1/2[∫ t

0dr(∫ r

0|k//(r − s)|‖um(s)‖ηds

)2]1/2

≤ CT ‖k//‖L1(0,T )

[∫ t

0Ym(r)dr

]1/2

≤ CT +

∫ t

0Ym(r)dr, (2.42)

2∫ t

0〈 f /(s), u//

m(s)〉ds ≤ ‖ f /‖

2L2(QT )

+

∫ t

0Ym(r)dr, (2.43)

−2K (p − 1)

∫ t

0〈|um(s)|p−2u/

m(s), u//m(s)〉ds ≤ 2K (p − 1)

∫ t

0

(√2Xm(s)

)p−2‖u/

m(s)‖‖u//m(s)‖ds

≤ 2K (p − 1)CT

∫ t

0‖u//

m(s)‖ds ≤ CT +

∫ t

0Ym(r)dr. (2.44)

Combining (2.34), (2.35) and (2.37)–(2.44) and choosing ε =16 , we obtain

Ym(t) ≤ CT + 4 (2 + |k(0)|)

∫ t

0Ym(s)ds. (2.45)

By Gronwall’s lemma, we deduce from (2.45), that

Ym(t) ≤ CT exp[4t (2 + |k(0)|)] ≤ CT , (2.46)

for all t ∈ [0, T ].Step 3. Limiting process. From (2.15), (2.28), (2.31) and (2.46), we deduce the existence of a subsequence of um,

still denoted by um, such thatum → u in L∞(0, T ; H1) weak∗,

u/m → u/ in L∞(0, T ; H1) weak∗,

u//m → u// in L∞(0, T ; L2) weak∗.

(2.47)


By the compactness lemma of Lions [4, p. 57], we can deduce from (2.47) that there exists a subsequence still denotedby um, such that

um → u strongly in L2(QT ), and a.e. (x, t) ∈ QT ,

u/m → u/ strongly in L2(QT ), and a.e. (x, t) ∈ QT .

(2.48)

On the other hand, by (2.48)1 and the continuity of Ψp, we obtain

Ψp(um) → Ψp(u) a.e. (x, t) ∈ QT . (2.49)

By means of (2.19) and (2.28), it follows that

‖Ψp(um)‖L p/(QT )

≤

∫ T

0(√

2‖um(t)‖η)pdt ≤ T (

√T CT )p, (2.50)

for all m, where p/=

pp−1 .

By Lions’s lemma [4, Lemma 1.3, p. 12], it follows from (2.49) and (2.50) that

Ψp(um) → Ψp(u) in L p/

(QT ) weakly. (2.51)

By the same way, we deduce from (2.36), (2.45) and (2.48)2 that

Ψq(u/m) → Ψq(u/) in Lq/

(QT ) weakly, (2.52)

where q/=

qq−1 .

On the other hand, from (2.47)1 we obtain∫ T

0

⟨∫ t

0k(t − s)(um(s) − u(s))ds, v(t)

⟩dt =

∫ T

0

⟨um(s) − u(s),

∫ T

sk(t − s)v(t)dt

⟩ds → 0, (2.53)

for all v ∈ L1(0, T ; (H1)/).Therefore∫ t

0k(t − s)um(s)ds →

∫ t

0k(t − s)u(s)ds in L∞(0, T ; H1) weak∗. (2.54)

Passing to the limit in (2.9) and (2.10) by (2.47)1,3, (2.48), (2.51), (2.52) and (2.54), we have u satisfying theequation〈u//(t), v〉 + aη(u(t), v) −

∫ t

0k(t − s)aη(u(s), v)ds + K 〈Ψp(u(t)), v〉

+ λ〈Ψq(u/(t)), v〉 = g1(t)v(1) + 〈 f (t), v〉, for all v ∈ H1,

(2.55)

and

u(0) = u0, u/(0) = u1. (2.56)

On the other hand, we deduce from (2.47) and (2.55), that

uxx (t) −

∫ t

0k(t − s)uxx (s)ds = φ(t), (2.57)

where

φ(t) = u//(t) + K |u(t)|p−2u(t) + λ|u/|q−2u/

− f (t) ∈ L∞(0, T ; L2). (2.58)

Hence, it follows from (2.57) and (2.58), that

‖uxx (t)‖ ≤ ‖φ‖L∞(0,T ;L2) + ‖k‖L∞(0,T )

∫ t

0‖uxx (s)‖ds. (2.59)


By Gronwall’s lemma, we obtain

‖uxx (t)‖ ≤ ‖φ‖L∞(0,T ;L2) exp(t‖k‖L∞(0,T )

)≤ CT . (2.60)

Thus

u ∈ L∞(0, T ; H2) (2.61)

and the existence of a solution is proved completely.Step 4. Uniqueness of the solution. Let u1, u2 be two weak solutions of problem (1.1)–(1.3), such that

ui ∈ L∞(0, T ; H2), u/i ∈ L∞(0, T ; H1), u//

i ∈ L∞(0, T ; L2), i = 1, 2. (2.62)

Then u = u1 − u2 satisfies the variational problem〈u//(t), v〉 + aη(u(t), v) −

∫ t

0k(t − s)aη(u(s), v)ds + K 〈Ψp(u1) − Ψp(u2), v〉

+ λ〈Ψq(u/

1) − Ψq(u/

2), v〉 = 0 ∀v ∈ H1,

u(0) = u/(0) = 0,

u ∈ L∞(0, T ; H2), u/∈ L∞(0, T ; H1), u//

∈ L∞(0, T ; L2).

(2.63)

We take v = u/ in (2.63)1, and integrating with respect to t , we obtain

σ(t) = −2k(0)

∫ t

0‖u(r)‖2

ηdr + 2∫ t

0k(t − s)aη (u(s), u(t)) ds

− 2∫ t

0dr∫ r

0k/(r − s)aη (u(s), u(r)) ds − 2K

∫ t

0〈Ψp (u1(s)) − Ψp (u2(s)) , u/(s)〉ds, (2.64)

where

σ(t) = ‖u/(t)‖2+ ‖u(t)‖2

η + 2λ

∫ t

0〈Ψq(u/

1(s)) − Ψq(u/

2(s)), u/(s)〉ds. (2.65)

Using the following inequality

∀q ≥ 2, ∃Cq > 0 : (|x |q−2x − |y|

q−2 y)(x − y) ≥ Cq |x − y|q

∀x, y ∈ R, (2.66)

it follows from (2.65) that

σ(t) ≥ ‖u/(t)‖2+ ‖u(t)‖2

η + 2λCq

∫ t

0‖u/(s)‖q

Lq ds. (2.67)

By means of the following inequality

||x |δ−2x − |y|

δ−2 y| ≤ (δ − 1)Rδ−2|x − y| ∀x, y ∈ [−R, R], (2.68)

for all R > 0, δ ≥ 2, the last integral in RHS of (2.64) is estimated

−2K∫ t

0

⟨Ψp(u1(s)) − Ψp(u2(s)), u/(s)

⟩ds ≤

√2(p − 1)R p−2

S

∫ t

0σ(s)ds, (2.69)

with δ = p, R = RS =√

2 maxi=1,2 ‖ui‖L∞(0,T ;H1).From (2.64), (2.65), (2.67) and (2.69), we can prove the following inequality in a similar manner

σ(t) ≤ 2|k(0)|

∫ t

0σ(r)dr + 2

√σ(t)

∫ t

0|k(t − s)|

√σ(s)ds

+ 2∫ t

0

√σ(r)dr

∫ r

0|k/(r − s)|

√σ(s)ds +

√2K (p − 1)R p−2

S

∫ t

0σ(s)ds. (2.70)


Noting that

2√

σ(t)∫ t

0|k(t − s)|

√σ(s)ds ≤

12σ(t) + 2‖k‖

2L2(0,T )

∫ t

0σ(s)ds, (2.71)

2∫ t

0

√σ(r)

(∫ r

0|k/(r − s)|

√σ(s)ds

)dr ≤

∫ t

0σ(r)dr + T ‖k/

‖2L2(0,T )

∫ t

0σ(s)ds. (2.72)

Hence, it follows from (2.70) to (2.72) that

σ(t) ≤ ρT

∫ t

0σ(s)ds, (2.73)

where

ρT = 2[1 + 2|k(0)| + 2‖k‖2L2(0,T )

+ T ‖k/‖

2L2(0,T )

+√

2K (p − 1)R p−2S ]. (2.74)

Using Gronwall’s lemma, it follows from (2.73) that σ ≡ 0, i.e., u1 ≡ u2. Theorem 2.1 is proved completely.

Remark 2.2. (i) From the a priori estimates (2.28) and (2.46) we can see that u, ux , ut , uxt , ut t ∈ L∞(0, T ; L2) ⊂

L2(QT ). On the other hand, from (2.61) it is clear that uxx ∈ L∞(0, T ; L2) ⊂ L2(QT ). Also if the data (u0, u1) inproblem satisfy assumptions (H2), then the solution u of problem (1.1)–(1.3) belongs to H2(QT ). So the solution isalmost classical which is rather natural since the initial data u0 and u1 do not belong necessarly to C2(Ω) and C1(Ω),respectively.

(ii) In the case of p, q > 2 and K < 0, λ < 0, the question about the existence for the solutions of problem(1.1)–(1.3) is still open. However we have received the answer when k ≡ 0, p = q = 2 and K , λ ∈ R publishedin [10].

3. Asymptotic behavior of the solutions as η → 0+

In this part, we assume that (u0, u1, f, g, k, K , λ) satisfy the assumptions (H1)–(H4). Let η > 0. By Theorem 2.1,the problem (1.1)–(1.3) has a unique weak solution u depending on η:

u = uη. (3.1)

We consider the following perturbed problem, where η is a small parameter:Au ≡ ut t − uxx +

∫ t

0k(t − s)uxx (s)ds = −KΨp(u) − λΨq(ut ) + f (x, t), 0 < x < 1, 0 < t < T,

ux (0, t) = u(0, t), ux (1, t) + ηu(1, t) = g(t),u(x, 0) = u0(x), ut (x, 0) = u1(x).

(Pη)

We shall study the asymptotic expansion of the solution of problem (Pη) with respect to η.

Theorem 3.1. Let T > 0. Let (H1)–(H4) hold. Then

(i) The problem (P0) corresponding to η = 0 has only the solution u0 satisfying

u0 ∈ L∞(0, T ; H2), u /

0 ∈ L∞(0, T ; H1), u //

0 ∈ L∞(0, T ; L2). (3.2)

(ii) The solution uη converges strongly in W (QT ) to u0, as η → 0+,where

W (QT ) = v ∈ L∞(0, T ; H1): vt ∈ L∞(0, T ; L2).

Furthermore, we have the estimates

‖u/η − u/

0‖L∞(0,T ;L2) + ‖uη − u0‖L∞(0,T ;H1) ≤ CT η, (3.3)

where CT is a posistive constant depending only on T .


Proof. (i) Case η = 0. We prove in the same manner as in Theorem 2.1, that the following estimations hold a priori

‖u/m(t)‖2

+ ‖um(t)‖21 +

2Kp

‖um(t)‖pL p + 2λ

∫ t

0‖u/

m(s)‖qLq ds ≤ CT , ∀t ∈ [0, T ], ∀T > 0, (3.4)

‖u//m(t)‖2

+ ‖u/m(t)‖2

1 ≤ CT , ∀t ∈ [0, T ], ∀T > 0, (3.5)

and that the limit u0 of the sequence um defined by (2.9) and (2.10) satisfies (3.2) and the problem (P0) correspondingto η = 0. Furthermore, this solution u0 is unique.

(ii) Consider η∗ > 0 fixed and the parameter η ∈ (0, η∗). Proving in the same manner as in the case of Theorem 2.1with 0 < η < η∗, we have the following results:

‖u/η(t)‖

2+ ‖uη(t)‖2

1 +2Kp

‖uη(t)‖pL p + 2λ

∫ t

0‖u/

η(s)‖qLq ds ≤ CT , ∀t ∈ [0, T ], ∀T > 0, (3.6)

‖u//η (t)‖2

+ ‖u/η(t)‖

21 ≤ CT , ∀t ∈ [0, T ], ∀T > 0, (3.7)

where CT is a constant independent of η.Let ηm be a sequence such that ηm > 0, ηm → 0 as m → ∞. From (3.6) and (3.7), we deduce that, there exists

a subsequence of the sequence uηm still denoted by uηm , such thatuηm → u∗ in L∞(0, T ; H1) weak∗,

u/ηm

→ u/∗ in L∞(0, T ; H1) weak∗,

u//ηm

→ u//∗ in L∞(0, T ; L2) weak∗.

(3.8)

By the compactness lemma of Lions [4, p. 57] and the imbedding H1(0, T ) → C0([0, T ]), we can deduce from(3.8) the existence of a subsequence still denoted by uηm , such that

uηm → u∗ strongly in L2(QT ), and a.e. (x, t) ∈ QT ,

u/ηm

→ u/∗ strongly in L2(QT ), and a.e. (x, t) ∈ QT .

(3.9)

By passing to the limit similarly to in the proof of Theorem 2.1, we conclude that u∗ is a solution of the problem (P0)corresponding to η = 0 satisfying (3.2). From the uniqueness of the solution we have

u∗ = u0. (3.10)

Put u = uη − u0, then u satisfies

〈u//(t), v〉 + aη(u(t), v) = ηΦ(t)v(1) +

∫ t

0k(t − s)aη(u(s), v)ds

+ K 〈Ψp(uη) − Ψp (u0), v〉 + λ〈Ψq(u/η) − Ψq (u/

0), v〉 = 0 ∀v ∈ H1,

u(0) = u/(0) = 0,

u ∈ L∞(0, T ; H2), u/∈ L∞(0, T ; H1), u//

∈ L∞(0, T ; L2),

Φ(t) = −u0(1, t) +

∫ t

0k(t − s )u0(1, s)ds.

(3.11)

We take v = u/ in (3.11)1, and integrating with respect to t , we obtain

σ(t) = 2ηΦ(t)u(1, t) − 2η

∫ t

0Φ/(s)u(1, s)ds − 2k(0)

∫ t

0‖u(s)‖2

ηds

+ 2∫ t

0k(t − s)aη (u(s), u(t)) ds − 2

∫ t

0dr∫ r

0k/(r − s)aη (u(s), u(r)) ds

− 2K∫ t

0

⟨Ψp(uη(s)) − Ψp (u0(s)), u/(s)

⟩ds, (3.12)


where

σ(t) = ‖u/(t)‖2+ ‖u(t)‖2

η + 2λ

∫ t

0〈Ψq(u/

η(s)) − Ψq (u/

0(s)), u/(s)〉ds. (3.13)

Similarly as (2.67), we have

σ(t) ≥ ‖u/(t)‖2+ ‖u(t)‖2

1 + 2λCq

∫ t

0‖u/(s)‖q

Lq ds. (3.14)

From (3.12) and (3.13), we prove, in a similar manner to that in the above part, that

σ(t) ≤ 2(

8‖Φ‖2L∞(0,T ) + ‖Φ/

‖2L2(0,T )

)η2

+ 2(

4 + 2|k(0)| + 4‖k‖2L2(0,T )

+ T ‖k/‖

2L2(0,T )

) ∫ t

0σ(s)ds

+ 2K 2∫ t

0‖Ψp(uη(s)) − Ψp (u0(s))‖2ds. (3.15)

We again use inequality (2.68) with δ = p − 2, R = max‖u0‖C0(Ω),√

2CT , then, it follows from (3.15), that

σ(t) ≤ M1T η2+ M2T

∫ t

0σ(s)ds, (3.16)

where

M1T = 16‖Φ‖2L∞(0,T ) + 2‖Φ/

‖2L2(0,T )

, (3.17)

M2T = 2(

4 + 2|k(0)| + 4‖k‖2L2(0,T )

+ T ‖k/‖

2L2(0,T )

)+ 2K 2(p − 1)2 R2p−4. (3.18)

Using Gronwall’s lemma, it follows from (3.16) that

σ(t) ≤ M1T η2 exp(T M2T ). (3.19)

This implies

‖u/η − u/

0‖L∞(0,T ;L2) + ‖uη − u0‖L∞(0,T ;H1) ≤ CT η, (3.20)

where CT is a constant depending only on T . Theorem 3.1 is proved completely.

The next result gives an asymptotic expansion of the weak solution uη of order N + 1 in η, for η sufficiently small.We use the following notations. For a multi-index α = (α1, . . . , αN ) ∈ ZN

+ , and x = (x1, . . . , xN ) ∈ RN , we put

|α| = α1 + · · · + αN , α! = α1! · · ·!αN !, xα= xα1

1 . . . xαNN . (3.21)

First, we shall need the following lemma.

Lemma 3.1. Let m, N ∈ N, x = (x1, . . . , xN ) ∈ RN , and η ∈ R. Then(N∑

i=1

xiηi

)m

=

m N∑k=m

P [m][x]kη

k, (3.22)

where the coefficents P(m)[x]k , m ≤ k ≤ m N depending on x = (x1, . . . , xN ) are defined by the formula

P [m][x]k =

∑α∈A(m)

k

m!

α!xα, m ≤ k ≤ m N ,

A(m)k =

α ∈ ZN

+ : |α| = m,

N∑i=1

iαi = k

.

(3.23)

The proof of this lemma is easy, hence we omit the details.


Let u0 be a weak solution of problem (Pη) corresponding to η = 0 as in Theorem 3.1.Au0 = −KΨp (u0) − λΨq (u/

0) + f (x, t), 0 < x < 1, 0 < t < T,

u0x (0, t) = u0(0, t), u0x (1, t) = g(t),u0(x, 0) = u0(x), u/

0(x, 0) = u1(x),

u0 ∈ L∞(0, T ; H2), u/

0 ∈ L∞(0, T ; H1), u//

0 ∈ L∞(0, T ; L2).

(P0)

Let us consider the sequence of weak solutions ui , i = 1, 2, . . . , N , defined by the following problems:Aui ≡ Fi , 0 < x < 1, 0 < t < T,

ui x (0, t) = ui (0, t), ui x (1, t) = −ui−1(1, t),ui (x, 0) = u/

i (x, 0) = 0,

ui ∈ L∞(0, T ; H2), u/i ∈ L∞(0, T ; H1), u//

i ∈ L∞(0, T ; L2),

(Pi )

whereFi = −

i∑m=1

1m!

[KΨ (m)

p (u0) P [m][u ]i + λΨ (m)

q (u/

0)P [m][u/

]i

],

u = (u1, . . . , uN ) , u/= (u/

1, . . . , u/N ).

(3.24)

We also note that Fi is the first-order function with respect to ui , u/i . In fact,

Fi = −KΨ /p (u0) ui − λΨ /

q (u/

0)u/i −

i∑m=2

1m!

[KΨ (m)

p (u0) P [m][u]i + λΨ (m)

q (u/

0)P [m][u/

]i

]= −KΨ /

p (u0) ui − λΨ /q (u/

0)u/i + terms depending on uk, u/

k, k = 0, 1, . . . , i − 1. (3.25)

Let uη be a unique weak solution of problem (Pη). Then v, with

v = uη −

N∑i=0

uiηi= uη − h, (3.26)

satisfies the problemAv ≡ −K

[Ψp (v + h) − Ψp(h)

]− λ

[Ψq

(v/

+ h/)− Ψq

(v/)]

+ FN (η), 0 < x < 1, 0 < t < T,

vx (0, t) = v(0, t), vx (1, t) + ηv(1, t) = uN (1, t)ηN+1,

v(x, 0) = v/(x, 0) = 0,

v ∈ L∞(0, T ; H2), v/∈ L∞(0, T ; H1), v//

∈ L∞(0, T ; L2),

(3.27)

where

FN (η) = −K[Ψp(h) − Ψp (u0)

]− λ[Ψq(h/) − Ψq (u/

0)] −

N∑i=1

Fiηi . (3.28)

Then, we have the following lemma.

Lemma 3.2. Let p, q ≥ N + 2, N ≥ 1; K , λ ≥ 0 and (H2)–(H4) hold. Then

‖FN (η)‖L∞(0,T ;L2) ≤ CN ηN+1, ∀η ∈ (0, η∗), (3.29)

where CN is a constant depending only on η∗, N , T , p, q, K , λ and the constants ‖ui‖L∞(0,T ;H1),

‖u/i ‖L∞(0,T ;H1), (i = 0, 1, . . . , N ).


Proof. In the case of N = 1, the proof of Lemma 3.2 is easy, hence we omit the details, which we only prove withN ≥ 2. Put

h = u0 + h1, h1 =

N∑i=1

uiηi . (3.30)

By using Taylor’s expansion of the function Ψp(h) = Ψp (u0 + h1) around the point u0 up to order N , we obtain

Ψp(h) = Ψp (u0) +

N∑m=1

1m!

Ψ (m)p (u0)hm

1 +1

(N + 1)!Ψ (N+1)

p (u0 + θ1h1)hN+11 , (3.31)

where 0 < θ1 < 1. By Lemma 3.1, we obtain from (3.31), after some rearrangements in order of η, that

Ψp(h) = Ψp (u0) +

N∑k=1

[k∑

m=1

1m!

Ψ (m)p (u0)P [m]

[u]k

]ηk

+ R(1)(p, η), (3.32)

where

R(1)(p, η) =

k∑m=2

1m!

Ψ (m)p (u0)

m N∑k=N+1

P [m][u]kη

k+

1(N + 1)!

Ψ (N+1)p (u0 + θ1h1)hN+1

1 . (3.33)

Similarly, we use Taylor’s expansion of the function Ψq(h/) = Ψq (u/

0 + h/

1), up to order N , and obtain

Ψq(h/) = Ψq (u/

0) +

N∑k=1

[k∑

m=1

1m!

Ψ (m)q (u/

0)P [m][u/

]k

]ηk

+ R(2)(q, η), (3.34)

where

R(2)(q, η) =

k∑m=2

1m!

Ψ (m)q (u/

0)

m N∑k=N+1

P [m][u/

]kηk+

1(N + 1)!

Ψ (N+1)q (u/

0 + θ2h/

1)(h/

1)N+1, (3.35)

and 0 < θ2 < 1. Combining (3.24), (3.28) and (3.32)–(3.35), we then obtain

FN (η) = −K[Ψp(h) − Ψp (u0)

]− λ[Ψq(h/) − Ψq (u/

0)] −

N∑i=1

Fiηi

= −KN∑

k=1

[k∑

m=1

1m!

Ψ (m)p (u0)P [m]

[u]k

]ηk

− λ

N∑k=1

[k∑

m=1

1m!

Ψ (m)q (u/

0)P [m][u/

]k

]ηk

−

N∑i=1

Fiηi− KR(1)(p, η) − λR(2)(q, η)

= −KR(1)(p, η) − λR(2)(q, η). (3.36)

We shall estimate respectively the following terms on the right-hand side of (3.36).

Estimate — R(1)(p, η). By the boundedness of the functions uk , k = 0, 1, . . . , N , in the function spaceL∞(0, T ; H1), we obtain from (3.33), that

‖R(1)(p, η)‖L∞(0,T ;L2) ≤ ηN+1N∑

m=2

Cmp−1‖u0‖

p−m−1L∞(0,T ;H1)

m N∑k=N+1

‖F (m)[u]k‖L∞(0,T ;L2)η

k−N−1∗

+ ηN+1C N+1p−1

(N∑

i=0

‖ui‖L∞(0,T ;H1)ηi∗

)p−N−2 ( N∑i=1

‖ui‖L∞(0,T ;H1)ηi−1∗

)N+1


≤ ηN+1N∑

m=2

m N∑k=N+1

Cmp−1‖u0‖

p−m−1L∞(0,T ;H1)

∑α∈A(m)

k

m!

α!

(N∑

i=1

‖ui‖L∞(0,T ;L2)

)m

ηk−N−1∗

+ ηN+1η−N−1∗ C N+1

p−1

(N∑

i=0

‖ui‖L∞(0,T ;H1)ηi∗

)p−1

= C (1)N ηN+1, (3.37)

where Cmp−1 =

(p−1)(p−2)···(p−m)m!

and

C (1)N =

N∑m=2

m N∑k=N+1

Cmp−1‖u0‖

p−m−1L∞(0,T ;H1)

∑α∈A(m)

k

m!

α!

(N∑

i=1

‖ui‖L∞(0,T ;L2)

)m

ηk−N−1∗

+ η−N−1∗ C N+1

p−1

(N∑

i=0

‖ui‖L∞(0,T ;H1)ηi∗

)p−1

. (3.38)

Estimate — R(2)(q, η). We also obtain from (3.35) in a similar manner corresponding to the above part, that

‖R(2)(q, η)‖L∞(0,T ;L2) ≤ C (2)N ηN+1, (3.39)

where

C (2)N =

N∑m=2

m N∑k=N+1

Cmq−1‖u/

0‖q−m−1L∞(0,T ;H1)

∑α∈A(m)

k

m!

α!

(N∑

i=1

‖u/i ‖L∞(0,T ;L2)

)m

ηk−N−1∗

+ η−N−1∗ C N+1

q−1

(N∑

i=0

‖u/i ‖L∞(0,T ;H1)η

i∗

)q−1

. (3.40)

Therefore, it follows from (3.36), (3.37) and (3.39) that

‖FN (η)‖L∞(0,T ;L2) ≤ CN ηN+1, ∀η ∈ (0, η∗), (3.41)

where CN = K C (1)N + λC (2)

N and the proof of Lemma 3.2 is complete.

Next, by multiplying the two sides of (3.27)1 with v/, and after integration in t , we find without difficulty fromLemma 3.2 that

Z(t) = 2ηN+1U (t)v(1, t) − 2ηN+1∫ t

0U /(s)v(1, s)ds − 2k(0)

∫ t

0‖v(s)‖2

ηds

+ 2∫ t

0k(t − s)aη(v(s), v(t))ds − 2

∫ t

0dr∫ r

0k/(r − s)aη(v(r), v(s))ds

− 2K∫ t

0〈Ψp(v(s) + h(s)) − Ψp(h(s)), v/(s)〉ds + 2

∫ t

0〈FN (η)(s), v/(s)〉ds, (3.42)

where

Z(t) = ‖v/(t)‖2+ ‖v(t)‖2

η + 2λ

∫ t

0

⟨Ψq(v/(s) + h/(s)) − Ψq(h/(s)), v/(s)

⟩ds

≥ ‖v/(t)‖2+ ‖v(t)‖2

1, (3.43)

U (t) = uN (1, t) +

∫ t

0k(t − s )uN (1, s)ds. (3.44)

Using again Lemma 3.2, we prove, in a manner similar to that of Theorem 2.1, that

2∫ t

0

⟨FN (η)(s), v/(s)

⟩ds ≤ T C2

N η2N+2+

∫ t

0Z(s)ds. (3.45)


By using the same arguments as in the above part we can show that the weak solution u of problem (Pη) satisfies

‖u/(t)‖2+ ‖u(t)‖2

1 ≤ MT , ∀t ∈ [0, T ], (3.46)

where MT is a constant independent of η. On the other hand,

‖h‖L∞(0,T ;H1) ≤

N∑i=0

‖ui‖L∞(0,T ;H1)ηi∗ ≡ R1. (3.47)

We again use inequality (2.68) with δ = p − 2, R = maxR1, MT , then, it follows from (3.46) and (3.47) that

−2K∫ t

0

⟨Ψp(v(s) + h(s)) − Ψp(h(s)), v/(s)

⟩ds ≤ K (p − 1)2 R2p−4

∫ t

0Z(s)ds. (3.48)

Combining (3.42)–(3.45) and (3.48), we then obtain

Z(t) ≤ D1η2N+2

+ D2

∫ t

0Z(s)ds, (3.49)

for all t ∈ [0, T ], where D1 and D2 are constants independent of η defined by

T C2N + 32‖U /

‖2L1(0,T )

+ 2‖U /‖

2L2(0,T )

≤ D1, (3.50)

4 + 2|k(0)| + 4‖k‖2L∞(0,T ) + ‖k/

‖2L1(0,T )

+ K (p − 1)2 R2p−4≤ D2. (3.51)

By Gronwall’s lemma, we obtain from (3.49) that

Z(t) ≤ D1η2N+2 exp

(T D2

), ∀t ∈ [0, T ], (3.52)

or ∥∥∥∥∥u/η −

N∑i=0

u/i η

i

∥∥∥∥∥L∞(0,T ;L2)

+

∥∥∥∥∥uη −

N∑i=0

uiηi

∥∥∥∥∥L∞(0,T ;H1)

≤ DT ηN+1, (3.53)

where DT is a constant independent of η. Thus, we have the following theorem.

Theorem 3.2. Let p, q ≥ N + 2, N ≥ 1, η∗ > 0 and (H2)–(H4) hold. Then, for every η ∈ (0, η∗), problem (Pη)

has a unique weak solution uη satisfying the asymptotic estimations up to order N + 1 as in (3.53), the functions ui ,i = 0, 1, . . . , N being the weak solutions of problems (P0), (Pi ), i = 1, . . . , N, respectively.

Remark 3.1. In the case of k ≡ 0, f ≡ 0, and the boundary condition u(0, t) = 0, ux (1, t) + ηu(1, t) = g(t),standing for (1.2), we have also obtained the results of asymptotic expansion up to order 1 in η for η sufficientlysmall [5].

4. Asymptotic expansion of the solution with respect to three parameters (K, λ, η)

In this part, we assume that p, q ≥ N + 1, N ≥ 2 and (u0, u1, f, g, k) satisfy the assumptions (H2)–(H4).Let (K , λ, η) ∈ R3

+. By Theorem 2.1, the problem (1.1)–(1.3) has a unique weak solution u depending on(K , λ, η) : u = u(K , λ, η).

We consider the following perturbed problem, where K , λ, η are small parameters such that, 0 ≤ K ≤ K∗,0 ≤ λ ≤ λ∗, 0 ≤ η ≤ η∗:

Au ≡ ut t − uxx +

∫ t

0k(t − s)uxx (s)ds = −KΨp(u) − λΨq(ut )

+ f (x, t), 0 < x < 1, 0 < t < T,

ux (0, t) = u(0, t), ux (1, t) = −ηu(1, t) + g(t),u(x, 0) = u0(x), ut (x, 0) = u1(x).

(PK,λ,η)


We shall study the asymptotic expansion of the solution of problem (PK,λ,η) with respect to K , λ, η.We use the following notations. For a multi-index γ = (γ1, γ2, γ3) ∈ Z3

+, and −→K = (K , λ, η) ∈ R3+, we put

|γ | = γ1 + γ2 + γ3, γ ! = γ1!γ2!γ3!,

‖−→K ‖ =

√K 2 + λ2 + η2,

−→K γ

= K γ1λγ2ηγ3 ,

α, β ∈ Z3+, α ≤ β ⇐⇒ αi ≤ βi ∀i = 1, 2, 3.

(4.1)

First, we shall need the following lemma.

Lemma 4.1. Let m, N ∈ N and uα ∈ R, α ∈ Z3+, 1 ≤ |α| ≤ N. Then( ∑

1≤|α|≤N

uα−→K α

)m

=

∑m≤|α|≤m N

T (m)[u]α

−→K α, (4.2)

where the coefficents T (m)[u]α , m ≤ |α| ≤ m N depending on u = (uα), α ∈ Z3

+, 1 ≤ |α| ≤ N defined by therecurrence formulas

T (1)[u]α = uα, 1 ≤ |α| ≤ N ,

T (m)[u]α =

∑β∈A(m)

α

uα−β T (m−1)[u]β , m ≤ |α| ≤ m N , m ≥ 2,

A(m)α = β ∈ Z3

+ : β ≤ α, 1 ≤ |α − β| ≤ N , m − 1 ≤ |β| ≤ (m − 1)N .

(4.3)

Proof of Lemma 4.1. (i) m = 1. It is easy to see that T (1)[u]α = uα , 1 ≤ |α| ≤ N .

(ii) m ≥ 2. By using Maclaurin’s expansion of the function F(K , λ, η) ≡ F(−→K ) =

(∑1≤|α|≤N uα

−→K α)m

up toorder m N , we obtain

F(−→K ) =

∑m≤|α|≤m N

1α!

Dα F(0)−→K α, (4.4)

here we have used the notation

Dα F =∂α1+α2+α3

∂K α1∂λα2∂ηα3F, α = (α1, α2, α3) ∈ Z3

+.

Hence, it follows from (4.2) and (4.4), that

T (m)[u]α =

1α!

Dα F(0), m ≤ |α| ≤ m N . (4.5)

We rewrite the function F(−→K ) as follows

F(−→K ) =

( ∑1≤|α|≤N

uα−→K α

)( ∑m−1≤|α|≤(m−1)N

T (m−1)[u]α

−→K α

)≡ F1(

−→K )F2(−→K ). (4.6)

Differentiating the two members of (4.6), after taking −→K = 0, we obtain

Dα F(0) = Dα (F1 F2) (0) =

∑β≤α

Cβα Dα−β F1(0)Dβ F2(0). (4.7)

We also note that uα =1α!

Dα F1(0), 1 ≤ |α| ≤ N , or

Dα−β F1(0) = (α − β)!uα−β , 1 ≤ |α − β| ≤ N . (4.8)

Similarly

Dβ F2(0) = β!T (m−1)[u]β , m − 1 ≤ |β| ≤ (m − 1)N . (4.9)


Combining (4.5) and (4.7)–(4.9), we then obtain

T (m)[u]α =

∑β∈A(m)

α

uα−β T (m−1)[u]β , m ≤ |α| ≤ m N , m ≥ 2. (4.10)

The proof of Lemma 4.1 is complete.

Let u0 ≡ u0,0,0 be a unique weak solution of problem (P0,0,0) (as in Theorem 2.1) corresponding to −→K =

(K , λ, η) = (0, 0, 0), i.e.,Au0 = P0,0,0 ≡ f (x, t), 0 < x < 1, 0 < t < T,

u0x (0, t) = u0(0, t), u0x (1, t) = g(t),u0(x, 0) = u0(x), u/

0(x, 0) = u1(x),

u0 ∈ L∞(0, T ; H2), u/

0 ∈ L∞(0, T ; H1), u//

0 ∈ L∞(0, T ; L2).

(P0,0,0)

Let us consider the sequence of weak solutions uγ , γ ∈ Z3+, 1 ≤ |γ | ≤ N , defined by the following problems:

Auγ = Pγ , 0 < x < 1, 0 < t < T,

uγ x (0, t) = uγ (0, t), uγ x (1, t) = Qγ (t),

uγ (x, 0) = u/γ (x, 0) = 0,

uγ ∈ L∞(0, T ; H2), u/γ ∈ L∞(0, T ; H1), u//

γ ∈ L∞(0, T ; L2),

(Pγ )

where Pγ , Qγ , |γ | ≤ N are defined by the recurrence formulas

Qγ (t) =

g(t), |γ | = 0,

0, 1 ≤ |γ | ≤ N , γ3 = 0,

−uγ1,γ2,γ3−1(1, t), 1 ≤ |γ | ≤ N , γ3 ≥ 1,

(4.11)

Pγ =

f, |γ | = 0,

0, 1 ≤ |γ | ≤ N , γ1 = γ2 = 0,

−Ψp(u0), |γ | = 1, γ1 = 1,

−Ψq(u/

0), |γ | = 1, γ2 = 1,

−

|γ |−1∑m=1

1m!

Ψ (m)q (u/

0)T (m)[u/

]0,γ2−1,γ3 , 2 ≤ |γ | ≤ N , γ2 ≥ 1, γ1 = 0,

−

|γ |−1∑m=1

1m!

Ψ (m)p (u0)T (m)

[u]γ1−1,0,γ3 , 2 ≤ |γ | ≤ N , γ1 ≥ 1, γ2 = 0,

−

|γ |−1∑m=1

1m!

[Ψ (m)

p (u0)T (m)[u]γ1−1,γ2,γ3 + Ψ (m)

q (u/

0)T (m)[u/

]γ1,γ2−1,γ3

],

2 ≤ |γ | ≤ N , γ1 ≥ 1, γ2 ≥ 1,

(4.12)

here we have used the notation u = (uγ ), |γ | ≤ N .

Let u = uK ,λ,η be a unique weak solution of problem (PK,λ,η). Then v = u −∑

|γ |≤N uγ−→K γ

≡ u − h satisfies theproblem

Av ≡ vt t − vxx +

∫ t

0k(t − s)vxx (s)ds = −K

[Ψp(v + h) − Ψp(h)

]− λ

[Ψq(v/

+ h/) − Ψq(h/)]+ EN , 0 < x < 1, 0 < t < T,

vx (0, t) = v(0, t), vx (1, t) = −ηv(1, t) + G N ,

v(x, 0) = v/(x, 0) = 0,

v ∈ L∞(0, T ; H2), v/∈ L∞(0, T ; H1), v//

∈ L∞(0, T ; L2),

(4.13)


where

EN (x, t) = f (x, t) − KΨp(h) − λΨq(h/) −

∑|γ |≤N

Pγ−→K γ , (4.14)

G N (t) =

∑|γ |=N+1,γ3≥1

uγ1,γ2,γ3−1(1, t)−→K γ . (4.15)

Then, we have the following lemma.

Lemma 4.2. Let p, q ≥ N + 1, N ≥ 2 and (H2)–(H4) hold. Then

‖EN ‖L∞(0,T ;L2) ≤ C1N ‖−→K ‖

N+1, (4.16)

‖G N ‖H1(0,T ) ≤ C2N ‖−→K ‖

N+1, (4.17)

for all −→K = (K , λ, η) ∈ R3+, ‖

−→K ‖ ≤ ‖−→K∗‖ with −→K∗ = (K∗, λ∗, η∗), where C1N and C2N are positive constants

depending only on the constants ‖−→K∗‖, ‖uγ ‖L∞(0,T ;H1), ‖u/

γ ‖L∞(0,T ;H1), (|γ | ≤ N ).

Proof. In the case of N = 1, the proof of Lemma 4.2 is easy, hence we omit the details, which we only prove withN ≥ 2. Put

h = u0 + h1, h1 =

∑1≤|γ |≤N

uγ−→K γ . (4.18)

By using Taylor’s expansion of the function Ψp(h) = Ψp(u0 + h1) around the point u0 up to order N − 1, weobtain

Ψp(h) = Ψp(u0) +

N−1∑m=1

1m!

Ψ (m)p (u0)hm

1 +1N !

Ψ (N )p (u0 + θ1h1)hN

1 , (4.19)

where 0 < θ1 < 1. By Lemma 4.1, we obtain from (4.19), after some rearrangements in order of−→K γ , that

KΨp(h) = KΨp(u0) +

∑2≤|γ |≤N ,γ1≥1

|γ |−1∑m=1

1m!

Ψ (m)p (u0)T (m)

[u]γ1−1,γ2,γ3

−→K γ

+ R(1)(Ψp,−→K ), (4.20)

where

R(1)(Ψp,−→K ) = K

N−1∑m=1

1m!

Ψ (m)p (u0)

∑N≤|γ |≤m N

T (m)[u]γ

−→K γ

+1N !

Ψ (N )p (u0 + θ1h1)K hN

1 . (4.21)

Similarly, we use Taylor’s expansion of the function Ψq(h/) = Ψq(u/

0 + h/

1), up to order N − 1, and obtain

λΨq(h/) = λΨq(u/

0) +

∑2≤|γ |≤N ,γ2≥1

|γ |−1∑m=1

1m!

Ψ (m)q (u/

0)T (m)[u/

]γ1,γ2−1,γ3

−→K γ

+ R(2)(Ψq ,−→K ), (4.22)

where

R(2)(Ψq ,−→K ) = λ

N−1∑m=1

1m!

Ψ (m)q (u/

0)∑

N≤|γ |≤m N

T (m)[u/

]γ−→K γ

+ λ1N !

Ψ (N )q (u/

0 + θ2h/

1)(h/

1)N , (4.23)

and 0 < θ2 < 1. Combining (4.12), (4.14), (4.20) and (4.22), we then obtain

EN (x, t) = f (x, t) − KΨp(u0) − λΨq(u/

0) −

∑2≤|γ |≤N ,γ1≥1

|γ |−1∑m=1

1m!

Ψ (m)p (u0)T (m)

[u]γ1−1,γ2,γ3

−→K γ


−

∑2≤|γ |≤N ,γ2≥1

|γ |−1∑m=1

1m!

Ψ (m)q (u/

0)T (m)[u/

]γ1,γ2−1,γ3

−→K γ

−

∑|γ |≤N

Pγ−→K γ

− R(1)(Ψp,−→K ) − R(2)(Ψq ,

−→K )

= −R(1)(Ψp,−→K ) − R(2)(Ψq ,

−→K ). (4.24)

We shall estimate respectively the following terms on the right-hand side of (4.24).Estimate R(1)(Ψp,

−→K ). By the boundedness of the functions uγ , γ ∈ Z3+, |γ | ≤ N in the function space

L∞(0, T ; H1), we obtain from (4.21), that

‖R(1)(Ψp,−→K )‖L∞(0,T ;L2) ≤ |K |

N−1∑m=1

∑N≤|γ |≤m N

1m!

‖Ψ (m)p (u0)‖L∞(0,T ;H1)‖T (m)

[u]γ ‖L∞(0,T ;L2)|−→K γ

|

+1N !

K‖Ψ (N )p (u0 + θ1h1)‖L∞(0,T ;H1)‖h1‖

NL∞(0,T ;H1)

. (4.25)

Using the inequality

|−→K γ

| ≤ ‖−→K ‖

|γ |, for all γ ∈ Z3+, (4.26)

it follows from (4.25) and (4.26), that

‖R(1)(Ψp,−→K )‖L∞(0,T ;L2) ≤ C (1)

1N ‖−→K ‖

N+1, ‖−→K ‖ ≤ ‖

−→K∗‖, (4.27)

where

C (1)1N =

N−1∑m=1

Cmp−1‖u0‖

p−m−1L∞(0,T ;H1)

∑N≤|γ |≤m N

‖T (m)[u]γ ‖L∞(0,T ;L2)‖

−→K∗‖|γ |−N

+ C Np−1‖

−→K∗‖−N

( ∑|γ |≤N

‖uγ ‖L∞(0,T ;H1)‖−→K∗‖

|γ |

)p−1

, (4.28)

and −→K∗ = (K∗, λ∗, η∗), Cmp−1 =

(p−1)(p−2)···(p−m)m!

.

Estimate R(2)(Ψq ,−→K ). We obtain from (4.23) in a similar manner corresponding to the above part, that

‖R(2)(Ψq ,−→K )‖L∞(0,T ;L2) ≤ C (2)

1N ‖−→K ‖

N+1, ‖−→K ‖ ≤ ‖

−→K∗‖, (4.29)

where

C (2)1N =

N−1∑m=1

Cmq−1‖u/

0‖q−m−1L∞(0,T ;H1)

∑N≤|γ |≤m N

‖T (m)[u/

]γ ‖L∞(0,T ;L2)‖−→K∗‖

|γ |−N

+ C Nq−1‖

−→K∗‖−N

( ∑|γ |≤N

‖u/γ ‖L∞(0,T ;H1)‖

−→K∗‖|γ |

)q−1

. (4.30)

Therefore, it follows from (4.24) and (4.27)–(4.30), that

‖EN ‖L∞(0,T ;L2) ≤

(C (1)

1N + C (2)1N

)‖−→K ‖

N+1≡ C1N ‖

−→K ‖N+1. (4.31)

Hence, the first part of Lemma 4.2 is proved.With G N , then, we also obtain from (4.15), that

‖G N ‖H1(0,T ) ≤ C2N ‖−→K ‖

N+1, (4.32)


where

C2N =√

2∑

|γ |=N+1,γ3≥1

(‖uγ1,γ2,γ3−1‖L∞(0,T ;H1) + ‖u/

γ1,γ2,γ3−1‖L∞(0,T ;H1)). (4.33)

The proof of Lemma 4.2 is complete.

Next, we obtain the following theorem.

Theorem 4.1. Let p, q ≥ N + 1, N ≥ 2 and (H2)–(H4) hold. Then, for every −→K = (K , λ, η) ∈ R3+, with

0 ≤ K ≤ K∗, 0 ≤ λ ≤ λ∗, 0 ≤ η ≤ η∗, problem (PK,λ,η) has a unique weak solution u = uK ,λ,η satisfyingthe asymptotic estimations up to order N + 1 as follows∥∥∥∥∥u/

−

∑|γ |≤N

u/γ

−→K γ

∥∥∥∥∥L∞(0,T ;L2)

+

∥∥∥∥∥u −

∑|γ |≤N

uγ−→K γ

∥∥∥∥∥L∞(0,T ;H1)

≤ D∗

N ‖−→K ‖

N+1, (4.34)

for all −→K ∈ R3+, ‖

−→K ‖ ≤ ‖−→K∗‖, D∗

N is a positive constant independent of −→K , the functions uγ , |γ | ≤ N are the weaksolutions of problems (Pγ ), γ ∈ Z3

+, |γ | ≤ N.

Proof. First, we note that, if the data −→K satisfy

0 ≤ K ≤ K∗, 0 < λ ≤ λ∗, 0 ≤ η ≤ η∗, (4.35)

where K∗, λ∗, η∗ are fixed positive constants, the a priori estimates of the sequence um in the proof of Theorem 2.1satisfy

‖u/m(t)‖2

+ ‖um(t)‖21 ≤ CT , ∀t ∈ [0, T ], ∀T > 0, (4.36)

‖u//m(t)‖2

+ ‖u/m(t)‖2

1 ≤ CT , ∀t ∈ [0, T ], ∀T > 0, (4.37)

where CT is a constant depending only on T , u0, u1, f , g, k, p, q, K∗, λ∗, η∗ (independent of −→K ). Hence, the limitu in suitable function spaces of the sequence um defined by (2.9) and (2.10) is a weak solution of the problem(1.1)–(1.3) satisfying the a priori estimates (4.36) and (4.37).

By multiplying the two sides of (4.13)1 with v/, and after integration in t , we find without difficulty fromLemma 4.2, that

σ(t) = 2∫ t

0

⟨EN (s), v/(s)

⟩ds + 2ΦN (t)v(1, t) − 2

∫ t

0Φ/

N (s)v(1, s)ds

− 2k(0)

∫ t

0‖v(s)‖2

ηds + 2∫ t

0k(t − s)aη (v(s), v(t)) ds − 2

∫ t

0dr∫ r

0k/(r − s)aη (v(s), v(r)) ds

− 2K∫ t

0

⟨Ψp(v + h) − Ψp(h), v/(s)

⟩ds, (4.38)

where

ΦN (t) = G N (t) −

∫ t

0k(t − s)G N (s)ds, (4.39)

σ(t) = ‖v/(t)‖2+ ‖v(t)‖2

η + 2λ

∫ t

0

⟨Ψq(v/

+ h/) − Ψq(h/), v/⟩ds

≥ ‖v/(t)‖2+ ‖v(t)‖2

1. (4.40)

From (4.16), (4.38) and (4.40), we prove, in a similar manner to that in the above part, that

σ(t) ≤ 2T C21N ‖

−→K ‖2N+2

+ 16Φ2N (t) + 4

∫ t

0|Φ/

N (s)|2ds

+ 2(

4 + 2|k(0)| + 4‖k‖2L2(0,T )

+ T ‖k/‖

2L2(0,T )

) ∫ t

0σ(s)ds


+ 4K 2∫ t

0‖Ψp(v + h) − Ψp(h)‖2ds. (4.41)

By the embedding H1(0, T ) → C0([0, T ]), there exists a positive constant DT such that

‖ϕ‖C0([0,T ]) ≤ DT ‖ϕ‖H1(0,T ) for all ϕ ∈ H1(0, T ). (4.42)

Hence

16Φ2N (t) + 4

∫ t

0|Φ/

N (s)|2ds ≤ 4(1 + 4D2T )‖ΦN ‖

2H1(0,T )

. (4.43)

On the other hand, it follows from (4.39), that

‖ΦN ‖H1(0,T ) ≤ [1 +√

T DT (1 + |k(0)| + ‖k‖L1(0,T ) + ‖k/‖L1(0,T ))]‖G N ‖H1(0,T ). (4.44)

Therefore, we obtain from (4.17), (4.43) and (4.44), that

16Φ2N (t) + 4

∫ t

0|Φ/

N (s)|2ds ≤ C3N ‖−→K ‖

2N+2. (4.45)

By using the same arguments as in the above part we can show that the weak solution u of problem (PK,λ,η) satisfies

‖u//(t)‖2+ ‖u/(t)‖2

1 + ‖u(t)‖21 ≤ CT , ∀t ∈ [0, T ], (4.46)

where CT is a constant independent of K , λ, η. On the other hand,

‖h‖L∞(0,T ;H1) ≤

∑|γ |≤N

‖uγ ‖L∞(0,T ;H1)‖−→K∗‖

|γ |≡ R1. (4.47)

We again use inequality (2.68) with δ = p − 2, R2 =√

2 maxR1,√

CT , then, it follows from (4.47), that

4K 2∫ t

0

∥∥Ψp(v + h) − Ψp(h)∥∥2 ds ≤ 4K 2

∗ (p − 1)2 R2p−42

∫ t

0σ(s)ds. (4.48)

Combining (4.41), (4.45) and (4.48), we then obtain

σ(t) ≤ %(1)T ‖

−→K ‖2N+2

+ %(2)T

∫ t

0σ(s)ds, (4.49)

for all t ∈ [0, T ], where%

(1)T = 2T C2

1N + C3N ,

%(2)T = 2

(4 + 2|k(0)| + 4‖k‖

2L2(0,T )

+ T ‖k/‖

2L2(0,T )

)+ 4K 2

∗ (p − 1)2 R2p−42 .

(4.50)

By Gronwall’s lemma, we obtain from (4.49) that

σ(t) ≤ %(1)T ‖

−→K ‖2N+2 exp(T %

(2)T ) ≡ D(1)

T ‖−→K ‖

2N+2, ∀t ∈ [0, T ], (4.51)

for all −→K ∈ R3+, ‖

−→K ‖ ≤ ‖−→K∗‖. It follows that

‖v/(t)‖2+ ‖v(t)‖2

1 ≤ σ(t) ≤ D(1)T ‖

−→K ‖2N+2. (4.52)

Hence

‖v/‖L∞(0,T ;L2) + ‖v‖L∞(0,T ;H1) ≤ D∗

N ‖−→K ‖

N+1, (4.53)

or ∥∥∥∥∥u/−

∑|γ |≤N

u/γ

−→K γ

∥∥∥∥∥L∞(0,T ;L2)

+

∥∥∥∥∥u −

∑|γ |≤N

uγ−→K γ

∥∥∥∥∥L∞(0,T ;H1)

≤ D∗

N ‖−→K ‖

N+1. (4.54)


Remark 4.1. In the case of k ≡ 0, p = q = 2, and the boundary condition (1.10) standing for (1.2), Long et al. [10]has obtained a result about the asymptotic expansion of the solutions with respect to two parameters (K , λ) up toorder N + 1.

Acknowledgement

We wish to acknowledge the referee for constructive remarks and corrections in the manuscript.

References

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[3] M.M. Cavalcanti, V.N. Domingos Cavalcanti, J.A. Soriano, Exponential decay for the solution of semilinear viscoelastic wave equations withlocalized damping, Electron. J. Differential Equations 44 (2002) 1–14.

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existence and asymptotic expansion for a viscoelastic problem with a mixed nonhomogeneous condition

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