a wave equation associated with mixed nonhomogeneous conditions: global existence and asymptotic...

Nonlinear Analysis 66 (2007) 1526–1546www.elsevier.com/locate/na

A wave equation associated with mixednonhomogeneous conditions: Global existence and

asymptotic expansion of solutions

Nguyen Thanh Long∗, Vo Giang Giai

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

Received 13 January 2006; accepted 6 February 2006

Abstract

The paper deals with the initial–boundary value problem for the linear wave equation⎧⎨⎩utt − ux x + K u + λut = F(x, t), 0 < x < 1, 0 < t < T,

ux (0, t) = P(t), u(1, t) = 0,

u(x, 0) = u0(x), ut (x, 0) = u1(x),

(1)

where K , λ are given constants and u0, u1, F are given functions, and the unknown function u(x, t) andthe unknown boundary value P(t) satisfy the following nonlinear integral equation:

P(t) = g(t) + K1 |u(0, t)|α−2 u(0, t) + λ1 |ut (0, t)|β−2 ut (0, t) −∫ t

0k(t − s)u(0, s)ds, (2)

where K1, λ1, α, β are given constants and g, k are given functions.In this paper, we consider three main parts. In Part 1 we prove a theorem of global existence and

uniqueness of a weak solution (u, P) of problem (1.1)–(1.5). The proof is based on a Galerkin methodassociated with a priori estimates, weak convergence and compactness techniques. For the case of α = β =2, Part 2 is devoted the study of the asymptotic behavior of the solution (u, P) as λ1 → 0+. Finally, in Part3 we obtain an asymptotic expansion of the solution (u, P) of the problem (1.1)–(1.5) up to order N + 1

2 infour small parameters K , λ, K1, λ1.c© 2006 Elsevier Ltd. All rights reserved.

MSC: 35L20; 35L70

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

0362-546X/$ - see front matter c© 2006 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2006.02.007

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

mailto:[email protected]

mailto:[email protected]

http://dx.doi.org/10.1016/j.na.2006.02.007

N.T. Long, V.G. Giai / Nonlinear Analysis 66 (2007) 1526–1546 1527

Keywords: Galerkin method; Global existence and uniqueness of a weak solution; Energy-type estimates; Compactness;Asymptotic expansion

1. Introduction

In this paper, we consider the following problem: Find a pair (u, P) of functions satis-fying

utt − ux x + f (u, ut ) = F(x, t), 0 < x < 1, 0 < t < T, (1.1)

ux (0, t) = P(t), (1.2)

u(1, t) = 0, (1.3)

u(x, 0) = u0(x), ut (x, 0) = u1(x), (1.4)

where f (u, ut ) = K u + λut , where K , λ are given constants and u0, u1, F are givenfunctions satisfying conditions specified later, and the unknown function u(x, t) and the unknownboundary value P(t) satisfy the following integral equation

P(t) = g(t) + K1 |u(0, t)|α−2 u(0, t) + λ1 |ut (0, t)|β−2 ut (0, t)

−∫ t

0k(t − s)u(0, s)ds, (1.5)

where K1, λ1, α, β are given constants and g, k are given functions.In the case of α = 2, λ1 ≡ 0, K1 = h ≥ 0, 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 P(t) satisfythe following Cauchy problem for ordinary differential equation{

P ′′(t) + ω2 P(t) = hutt (0, t), 0 < t < T,

P(0) = P0, P ′(0) = P1,(1.6)

where h ≥ 0, ω > 0, P0, P1 are given constants [6].In [2], An and Trieu have studied a special case of problem (1.1)–(1.4) and (1.6) with

u0 = u1 = P0 = 0 and f (u, ut ) = K u + λut , with K ≥ 0, λ ≥ 0 given constants. In thelatter case the problem (1.1)–(1.4) and (1.6) is a mathematical model describing the shock of arigid body and a linear viscoelastic bar resting on a rigid base [2].

From (1.6) we represent P(t) in terms of P0, P1, ω, h, utt(0, t) and then by integrating byparts, we have

P(t) = g(t) + hu(0, t) −∫ t

0k(t − s)u(0, s)ds, (1.7)

where

g(t) = (P0 − hu0(0)) cosωt + 1

ω(P1 − hu1(0)) sin ωt, (1.8)

k(t) = hω sin ωt . (1.9)

In [3] Bergounioux, Long and Dinh studied problem (1.1) and (1.4) with the mixed boundaryconditions (1.2) and (1.3) standing for

ux (0, t) = g(t) + hu(0, t) −∫ t

0k(t − s)u(0, s)ds, (1.10)

1528 N.T. Long, V.G. Giai / Nonlinear Analysis 66 (2007) 1526–1546

ux(1, t) + K1u(1, t) + λ1ut (1, t) = 0, (1.11)

where h ≥ 0, K , λ, K1, λ1 are given constants and g, k are given functions.In [8], Long, Dinh and Diem obtained the unique existence, regularity and asymptotic

expansion of the problem (1.1) and (1.4) in the case of −ux(1, t) = K1u(1, t) + λut (1, t),ux(0, t) = P(t) where P(t) satisfies (1.7).

In [10], Long, Ut and Truc gave the unique existence, stability, regularity in time variableand asymptotic expansion for the solution of problem (1.1)–(1.5) when α = β = 2, f (u, ut ) =K u + λut , u0 ∈ H 2 and u1 ∈ H 1. In this case, the problem (1.1)–(1.5) is the mathematicalmodel describing a shock problem involving a linear viscoelastic bar.

In this paper, we consider three main parts. In Part 1, under conditions (u0, u1) ∈ H 1 × L2,F ∈ L2(QT ), g, k ∈ H 1(0, T ), K ∈ R, λ, K1 ≥ 0, λ1 > 0, α, β ≥ 2, we prove atheorem of global existence and uniqueness of a weak solution (u, P) of problem (1.1)–(1.5).The proof is based on a Galerkin method associated with a priori estimates, weak convergenceand compactness techniques. We remark that the linearization method in the papers [4,8] cannotbe used in [3,6,7]. For the case of α = β = 2, Part 2 is devoted the study of the asymptoticbehavior of the solution (u, P) as λ1 → 0+. Finally, in Part 3 we obtain an asymptotic expansionof the solution (u, P) of the problem (1.1)–(1.5) up to order N + 1

2 in four small parameters K ,λ, K1, λ1. The results obtained here may be considered as the generalizations of those in An andTrieu [2] and in [3,4,6–10].

2. The existence and uniqueness theorem

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

(Ω), L p (Ω), W m,p (Ω).

We define W m,p = W m,p (Ω), L p = W 0,p (Ω), H m = 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 pair ofdual scalar products of a continuous linear functional with an element of a function space. Wedenote by ‖·‖X the norm of a Banach space X and by X ′ the dual space of X . We denote byL p(0, T ; X), 1 ≤ p ≤ ∞, the Banach space of the real functions u : (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) = utt (t), ux(t), ux x(t) denote u(x, t), ∂u∂t (x, t), ∂2u

∂t2 (x, t), ∂u∂x (x, t),

∂2u∂x2 (x, t), respectively.

We put

V = {v ∈ H 1(0, 1) : v(1) = 0}, (2.1)

a(u, v) =∫ 1

0

∂u

∂x

∂v

∂xdx . (2.2)


V is a closed subspace of H 1 and on V , ‖v‖H1 and ‖v‖V = √a(v, v) = ‖vx‖ are two equivalent

norms. We then have the following lemma.

Lemma 1. The imbedding V ↪→ C0(0, 1) is compact and

‖v‖C0([0,1]) ≤ ‖v‖V for all v ∈ V . (2.3)

The proof is straightforward and we omit the details.We make the following assumptions:(H1) u0 ∈ V and u1 ∈ L2,(H2) F ∈ L2(QT ),(H3) g, k ∈ H 1(0, T ),(H4) K ∈ R, λ, K1 ≥ 0, λ1 > 0,(H5) α, β ≥ 2.Then, we have the following theorem.

Theorem 1. Let (H1)–(H5) hold. Let T > 0. Then the problem (1.1)–(1.5) has at least one weaksolution (u, P) such that⎧⎨⎩u ∈ L∞(0, T ; V ), ut ∈ L∞(0, T ; L2),

u(0, ·) ∈ W 1,β (0, T ) , P ∈ Lβ ′(0, T ) , β ′ = β

β − 1.

(2.4)

Furthermore, if α = 2 or α ≥ 3, the solution is unique.

Remark 1. In the special case α = β = 2, u0 ∈ V ∩ H 2 and u1 ∈ H 1 we have obtained someresults in the paper [10].

Proof of Theorem 1. The proof consists of Steps 1–4.

Step 1. The Galerkin approximation. Let {w j } be a denumerable base of V . We find theapproximate solution of problem (1.1)–(1.5) in the form

um(t) =m∑

j=1

cmj (t)w j , (2.5)

where the coefficient functions cmj satisfy the system of ordinary differential equation⟨u′′

m(t),w j⟩+ ⟨

umx (t),w j x⟩+ Pm(t)w j (0) + ⟨

K um(t) + λu′m(t),w j

⟩= ⟨

F (t),w j⟩, 1 ≤ j ≤ m, (2.6)

Pm(t) = g(t) + K1 |um(0, t)|α−2 um(0, t) + λ1∣∣u′

m(0, t)∣∣β−2

u′m(0, t)

−∫ t

0k(t − s)um(0, s)ds, (2.7)⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

um(0) = u0m =m∑

j=1

αmj w j → u0 strongly in H 1,

u′m(0) = u1m =

m∑j=1

βmjw j → u1 strongly in L2.

(2.8)

From the assumptions of Theorem 1, system (2.6)–(2.8) has the solution (um, Pm ), on someinterval [0, Tm]. The following estimates allow one to take Tm = T for all m.


Step 2. A priori estimates. Substituting (2.7) into (2.6), then multiplying the j th equation of (2.6)by c′

mj (t) and summing up with respect to j , afterwards, integrating by parts with respect to thetime variable from 0 to t , we get after some rearrangements

Sm(t) = Sm(0) + 2g(0)u0m(0) − 2g(t)um(0, t) + 2∫ t

0

⟨F(s), u′

m(s)⟩ds

− 2K∫ t

0

⟨um(s), u′

m(s)⟩ds − 2λ

∫ t

0

∥∥u′m(s)

∥∥2 ds

+ 2∫ t

0g′(s)um(0, s)ds + 2

∫ t

0u′

m(0, r)

(∫ r

0k(r − s)um(0, s)ds

)dr

= Sm(0) + 2g(0)u0m(0) − 2g(t)um(0, t) + I1 + I2 + I3 + I4 + I5, (2.9)

where

Sm(t) = ∥∥u′m(t)

∥∥2 + ‖umx (t)‖2 + 2

αK1 |um(0, t)|α + 2λ1

∫ t

0

∣∣u′m(0, s)

∣∣β ds. (2.10)

Using the inequality

2ab ≤ εa2 + 1

εb2, ∀a, b ∈ R, ∀ε > 0, (2.11)

and the following inequalities

|um(0, t)| ≤ ‖um(t)‖C0(Ω)≤ ‖umx (t)‖ ≤ √

Sm(t), (2.12)

‖um(t)‖2 ≤ 2 ‖u0m‖2 + 2∫ t

0

∥∥u′m(s)

∥∥2 ds ≤ 2 ‖u0m‖2 + 2∫ t

0Sm(s)ds, (2.13)

we shall estimate respectively the following terms on the right-hand side of (2.9) as follows

−2g(t)um(0, t) ≤ 1

εg2(t) + εSm(t), for all ε > 0, (2.14)

I1 = 2∫ t

0

⟨F(s), u′

m(s)⟩ds ≤

∫ t

0‖F(s)‖2 ds +

∫ 0

tSm(s)ds, (2.15)

I2 = −2K∫ t

0

⟨um(s), u′

m(s)⟩ds ≤ |K |

∫ t

0Sm(s)ds, (2.16)

I3 = −2λ

∫ t

0

∥∥u′m(s)

∥∥2ds ≤ 2 |λ|

∫ t

0Sm(s)ds, (2.17)

I4 = 2∫ t

0g′(s)um(0, s)ds ≤ ∥∥g′∥∥2

L2(0,T )+∫ t

0Sm(s)ds, (2.18)

I5 = 2∫ t

0u′

m(0, r)

(∫ r

0k(r − s)um(0, s)ds

)dr

= 2um(0, t)

(∫ t

0k(t − s)um(0, s)ds

)− 2

∫ t

0um(0, r)

(∫ r

0k ′(r − s)um(0, s)ds

)dr − 2k(0)

∫ t

0u2

m(0, r)dr

≤ εSm(t) + 1

ε

∫ t

0k2(θ)dθ

∫ t

0Sm(s)ds


+ 2

√t∫ t

0|k ′(θ)|2dθ

∫ t

0Sm(s)ds + 2 |k(0)|

∫ t

0Sm(r)dr

≤ εSm(t) +(

1

ε‖k‖2

L2(0,T )+ 2

√T∥∥k ′∥∥

L2(0,T )+ 2 |k(0)|

)∫ t

0Sm(s)ds, (2.19)

for all ε > 0. On the other hand, using (2.8) and (2.10), (H1), (H3)–(H5) and Lemma 1, we have

Sm(0) + 2g(0)u0m(0)

= ‖u1m‖2 + ‖u0mx‖2 + 2

αK1 |u0m(0)|α + 2g(0)u0m(0) ≤ C1 for all m, (2.20)

where C1 is a constant depending only on u0, u1, g, K1 and α.Combining (2.9), (2.10), (2.14)–(2.20), we obtain

Sm(t) ≤ C1 + 1

ε‖g‖2

L∞(0,T ) + ∥∥g′∥∥2L2(0,T )

+ ‖F‖2L2(QT )

+ 2εSm(t)

+(

2 + |K | + 2 |λ| + 1

ε‖k‖2

L2(0,T )+ 2

√T∥∥k ′∥∥

L2(0,T )+ 2 |k(0)|

)∫ t

0Sm(s)ds,

(2.21)

for all ε > 0. On choosing ε = 14 , it follows from (2.21) that

Sm(t) ≤ M(1)T + N (1)

T

∫ t

0Sm(s)ds, (2.22)

where⎧⎨⎩M(1)T = 2

(C1 + 4 ‖g‖2

L∞(0,T ) + ∥∥g′∥∥2L2(0,T )

+ ‖F‖2L2(QT )

),

N (1)T = 2

(2 + |K | + 2 |λ| + 4 ‖k‖2

L2(0,T )+ 2

√T∥∥k ′∥∥

L2(0,T )+ 2 |k(0)|

).

(2.23)

By Gronwall’s lemma, we deduce from (2.22), (2.23) that

Sm(t) ≤ M(1)T exp

(t N (1)

T

)≤ M(1)

T exp(

T N (1)T

)= CT , for all t ∈ [0, T ]. (2.24)

On the other hand, from the assumptions (H3), (H4), we deduce from (2.7), (2.10), (2.24) that

|Pm(t)| ≤ ‖g‖L∞(0,T ) + K1

(√CT

)α−1 +√

CT

∫ T

0|k(s)| ds + λ1

∣∣u′m(0, t)

∣∣β−1

= C(1)T + λ1

∣∣u′m(0, t)

∣∣β−1, (2.25)∥∥∥∣∣u′

m(0, ·)∣∣β−2u′

m(0, ·)∥∥∥β ′

Lβ′(0,T )

=∫ T

0

∣∣u′m(0, s)

∣∣β ds ≤ CT

2λ1. (2.26)

Hence, we deduce from (2.25), (2.26) that

‖Pm‖Lβ′(0,T )

≤ T 1/β ′C(1)

T + λ1

[∫ T

0

∣∣u′m(0, s)

∣∣β ds

]1/β ′

≤ T 1/β ′C(1)

T + λ1−1/β ′1

(CT

2

)1/β ′

= CT for all m, (2.27)

where CT is a positive constant depending only on T .


Step 3. Limiting process. From (2.10), (2.24), (2.26) and (2.27), we deduce the existence of asubsequence of {(um, Pm)} still also so denoted, such that⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

um → u in L∞(0, T ; V ) weak*,u′

m → u′ in L∞(0, T ; L2) weak*,um(0, ·) → u(0, ·) in L∞(0, T ) weakly*,u′

m(0, ·) → u′(0, ·) in Lβ(0, T ) and L2(0, T ) weakly,∣∣u′m(0, ·)∣∣β−2

u′m(0, ·) → χ in Lβ ′

(0, T ) weakly,Pm → P in Lβ ′

(0, T ) weakly.

(2.28)

By the compactness lemma of Lions [5, p. 57] and the imbedding H 1(0, T ) ↪→ C0 ([0, T ]),we can deduce from (2.28)1,2,3,4 the existence of a subsequence still denoted by {um}, such that{

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

um(0, ·) → u(0, ·) strongly in C0 ([0, T ]) .(2.29)

From (2.29)2 we have

Pm(t) = g(t) + K1 |um(0, t)|α−2 um(0, t) −∫ t

0k(t − s)um(0, s)ds → P(t), (2.30)

strongly in C0 ([0, T ]), where

P(t) = g(t) + K1 |u(0, t)|α−2 u(0, t) −∫ t

0k(t − s)u(0, s)ds. (2.31)

Hence

Pm(t) → P(t) + λ1χ(t) = P(t), (2.32)

in Lβ ′(0, T ) weakly.

Passing to the limit in (2.6)–(2.8) by (2.28)1,2,5, (2.32), we have (u, P) satisfying the equation

d

dt

⟨u′(t), v

⟩+ 〈ux(t), vx 〉 + (P(t) + λ1χ(t)

)v(0) + 〈K u + λut , v〉

= 〈F (t), v〉 ,∀ v ∈ V . (2.33)

We can prove in a manner similar to that of [6] that

u(0) = u0, u′(0) = u1. (2.34)

Then, in order to prove the existence of the solution of the problem (1.1)–(1.5), we only haveto prove that χ(t) = ∣∣u′(0, t)

∣∣β−2u′(0, t). We shall now require the following lemma.

Lemma 2. Let u be the weak solution of the following problem⎧⎪⎪⎨⎪⎪⎩utt − ux x + Φ1 = 0, 0 < x < 1, 0 < t < T,

ux (0, t) = P1(t), u(1, t) = 0,

u(x, 0) = u0(x), ut (x, 0) = u1(x),

u ∈ L∞(0, T ; V ), ut ∈ L∞(0, T ; L2).

(2.35)

Then we have1

2

∥∥u′(t)∥∥2 + 1

2‖ux (t)‖2 +

∫ t

0P1(s)u

′(0, s)ds +∫ t

0

⟨Φ1(s), u′(s)

⟩ds

≥ 1

2‖u1‖2 + 1

2‖u0x‖2 a.e. t ∈ [0, T ]. (2.36)


Furthermore, if u0 = u1 = 0 there is equality in (2.36).

The proof of Lemma 2 can be found in [1].We now return to the proof of the existence of a solution of the problem (1.1)–(1.5).It follows from (2.6)–(2.8) that

λ1

∫ t

0

∣∣u′m(0, s)

∣∣β ds =∫ t

0

⟨F(s), u′

m(s)⟩ds − 1

2

∥∥u′m(t)

∥∥2 + 1

2‖u1m‖2

− 1

2‖umx (t)‖2 + 1

2‖u0mx‖2 −

∫ t

0Pm(s)u′

m(0, s)ds

− K∫ t

0

⟨um(s), u′

m(s)⟩ds − λ

∫ t

0

∥∥u′m(s)

∥∥2ds. (2.37)

By Lemma 2 we have

λ1 lim supm→∞

∫ t

0

∣∣u′m(0, s)

∣∣β ds ≤∫ t

0

⟨F(s), u′(s)

⟩ds − 1

2lim infm→∞

∥∥u′m(t)

∥∥2 + 1

2‖u1‖2

− 1

2lim infm→∞ ‖umx(t)‖2 + 1

2‖u0x‖2 −

∫ t

0P(s)u′(0, s)ds

−K∫ t

0

⟨u(s), u′(s)

⟩ds − λ lim inf

m→∞

∫ t

0

∥∥u′m(s)

∥∥2ds

≤ 1

2‖u1‖2 + 1

2‖u0x‖2 − 1

2

∥∥u′(t)∥∥2 − 1

2‖ux (t)‖2

−∫ t

0P(s)u′(0, s)ds

−∫ t

0

⟨K u(s) + λu′(s) − F(s), u′(s)

⟩ds

+ λ1

∫ t

0χ(s)u′(0, s)ds ≤ λ1

∫ t

0χ(s)u′(0, s)ds. (2.38)

Next, consider

Xm(t) =∫ t

0

(Ψ(u′

m(0, s))− Ψ (w(s))

) (u′

m(0, s) − w(s))

ds ≥ 0, ∀w ∈ Lβ(0, T ),

(2.39)

where Ψ (z) = |z|β−2 z.It follows from (2.28)4,5 and (2.38) that

0 ≤ lim supm→∞

Xm(t) ≤∫ t

0(χ(s) − Ψ (w(s)))

(u′(0, s) − w(s)

)ds, ∀w ∈ Lβ(0, T ).

(2.40)

In (2.40), we choose w(s) = u′(0, s) − δφ(s) with δ > 0 and φ ∈ Lβ(0, T ) and use theargument of Minty and Browder (cf. Lions [5], p. 172), we deduce that χ(t) = Ψ

(u′(0, t)

) =∣∣u′(0, t)∣∣β−2

u′(0, t).The proof of existence is completed.


Step 4. Uniqueness of the solution. Let (u, P),(u, P

)be two weak solutions of problem

(1.1)–(1.5), such that⎧⎨⎩u, u ∈ L∞(0, T ; V ), ut , ut ∈ L∞(0, T ; L2),

u(0, ·), u(0, ·) ∈ W 1,β (0, T ) ,

P, P ∈ Lβ ′(0, T ) .

(2.41)

Then (v, Q) with v = u − u and Q = P − P is the weak solution of the following problem⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

vt t − vx x + Kv + λvt = 0, 0 < x < 1, 0 < t < T,

vx (0, t) = Q(t), v(1, t) = 0,

v(x, 0) = vt (x, 0) = 0,

Q(t) = P(t) − P(t) = K1 (H (u(0, t)) − H (u(0, t)))

+ λ1 (Ψ (ut (0, t)) − Ψ (ut (0, t))) −∫ t

0k(t − s)v(0, s)ds,

(2.42)

where H (z) = |z|α−2 z.By using Lemma 2 with u0 = u1 = 0, Φ1 = Kv + λvt , P1(t) = Q(t), we have

Z(t) + 2K1

∫ t

0H1(s)v

′(0, s)ds + 2λ1

∫ t

0(Ψ (ut (0, s)) − Ψ (ut (0, s))) v′(0, s)ds

= 2∫ t

0v′(0, s)ds

∫ s

0k(s − r)v(0, r)dr − 2K

∫ t

0

⟨v(s), v′(s)

⟩ds

− 2λ

∫ t

0

∥∥v′(s)∥∥2 ds, a.e. t ∈ [0, T ]. (2.43)

where{Z(t) = ∥∥v′(t)

∥∥2 + ‖vx (t)‖2 ,

H1(t) = H (u(0, t)) − H (u(0, t)) = |u(0, t)|α−2 u(0, t) − |u(0, t)|α−2 u(0, t).(2.44)

Using the inequality

∀p ≥ 2, ∃Cp > 0 :(|x |p−2 x − |y|p−2 y

)(x − y) ≥ Cp |x − y|p , ∀x, y ∈ R, (2.45)

we obtain

2λ1

∫ t

0[Ψ (ut (0, s)) − Ψ (ut (0, s))] v′(0, s)ds ≥ 2λ1Cβ

∫ t

0

∣∣v′(0, s)∣∣β ds, (2.46)

−2K∫ t

0

⟨v(s), v′(s)

⟩ds − 2λ

∫ t

0

∥∥v′(s)∥∥2 ds

≤ 2 |K |∫ t

0‖v(s)‖ ∥∥v′(s)

∥∥ ds + 2 |λ|∫ t

0

∥∥v′(s)∥∥2

ds

≤ (|K | + 2 |λ|)∫ t

0Z(s)ds. (2.47)

On the other hand, we have


2∫ t

0v′(0, s)ds

∫ s

0k(s − r)v(0, r)dr

= 2v(0, t)∫ t

0k(t − r)v(0, r)dr − 2k(0)

∫ t

0v2(0, s)ds

− 2∫ t

0v(0, s)ds

∫ t

0k ′(s − r)v(0, r)dr

≤ 1

2Z(t) +

(2 ‖k‖2

L2(0,T )+ T

∥∥k ′∥∥2L2(0,T )

+ 2 |k(0)| + 1) ∫ t

0Z(r)dr. (2.48)

It follows from (2.43), (2.44), (2.46)–(2.48) that

Z(t) + 4K1

∫ t

0H1(s)v

′(0, s)ds + 4λ1Cβ

∫ t

0

∣∣v′(0, s)∣∣β ds

≤ 2[2 ‖k‖2

L2(0,T )+ T

∥∥k ′∥∥2L2(0,T )

+ 2 |k(0)| + 1 + |K | + 2 |λ|] ∫ t

0Z(s)ds. (2.49)

Now, we consider two cases for α.

Case α = 2. H1(t) = H (u(0, t)) − H (u(0, t)) = u(0, t) − u(0, t) = v(0, t).We then have

4K1

∫ t

0H1(s)v

′(0, s)ds = 2 K1v2(0, t) ≥ 0. (2.50)

By Gronwall’s lemma, we obtain from (2.49) and (2.50) that Z ≡ 0, i.e., u ≡ u.Case α ≥ 3. By using integration by parts, it follows that

2K1

∫ t

0H1(s)v

′(0, s)ds = 2K1

∫ t

0

∫ 1

0

[d

dθH (u(0, s) + θv(0, s)) dθ

]v′(0, s)ds

= K1

[∫ 1

0H ′ (u(0, t) + θv(0, t)) dθ

]v2(0, t)

− K1

∫ t

0v2(0, s)ds

[∫ 1

0H ′′ (u(0, s) + θv(0, s)) (u ′(0, s)

+ θv′(0, s))dθ

]

≥ −K1(α − 1)(α − 2)Rα−3∫ t

0ω1(s)Z(s)ds, (2.51)

where

R = max{‖u‖L∞(0,T ;V ) , ‖u‖L∞(0,T ;V )}, ω1(s) = ∣∣u′(0, s)∣∣+ ∣∣u ′(0, s)

∣∣ . (2.52)

Using Gronwall’s lemma, it follows from (2.49)–(2.51) that Z ≡ 0, i.e., u ≡ u. Theorem 1 isproved completely. �

3. Asymptotic behavior of the solutions as λ1 → 0+

In this part, we assume that α = β = 2 and (u0, u1, F , g, k, K , λ, K1) satisfy the assumptions(H1)–(H4). Let λ1 > 0. By Theorem 1, the problem (1.1)–(1.5) has a unique weak solution


(u, P) depending on λ1:

u = uλ1, P = Pλ1 . (3.1)

We consider the following perturbed problem, where λ1 is a small parameter:⎧⎪⎪⎪⎨⎪⎪⎪⎩Au ≡ utt − ux x + K u + λut = F(x, t), 0 < x < 1, 0 < t < T,

ux (0, t) = P(t), u(1, t) = 0,

u(x, 0) = u0(x), ut (x, 0) = u1(x),

P(t) = g(t) + K1u(0, t) + λ1ut (0, t) −∫ t

0k(t − s)u(0, s)ds.

(Pλ1)

We shall study the asymptotic expansion of the solution of problem (Pλ1) with respect to λ1.

Theorem 2. Let T > 0. Let (H1)–(H4) hold. Then(i) The problem (P0) corresponding to λ1 = 0 has only the solution

(u0, P0

)satisfying{

u0 ∈ L∞(0, T ; V ), u ′0 ∈ L∞(0, T ; L2),

u0(0, ·), P0 ∈ H 1 (0, T ) .(3.2)

(ii) The solution(uλ1, Pλ1

)converges strongly in W (QT ) × L2 (0, T ) to

(u0, P0

), as λ1 →

0+,where

W (QT ) = {v ∈ L∞(0, T ; V ) : vt ∈ L∞(0, T ; L2)}.Furthermore, we have the estimates⎧⎪⎪⎨⎪⎪⎩

∥∥u′λ1

− u ′0

∥∥L∞(0,T ;L2)

+ ∥∥uλ1 − u0∥∥

L∞(0,T ;V )

+ √λ1

∥∥∥u′λ1

(0, ·) − u ′0(0, ·)

∥∥∥L2(0,T )

≤ CT√

λ1,∥∥Pλ1 − P0∥∥

L2(0,T )≤ CT

√λ1,

(3.3)

where CT is a positive constant depending only on T .

Proof. (i) Case λ1 = 0. We prove in the same manner as in [7] (Theorem 1, p. 1263) that thefollowing estimations hold a priori∥∥u′

m(t)∥∥2 + ‖umx (t)‖2 + K1u2

m(0, t) ≤ CT , ∀t ∈ [0, T ],∀T > 0, (3.4)∫ T

0

∣∣u′m(0, t)

∣∣2 dt ≤ CT ,

∫ T

0

∣∣P ′m(t)

∣∣2 dt ≤ CT , ∀T > 0, (3.5)

and that the limit(u0, P0

)of the sequence (um, Pm) defined by (2.6)–(2.8) satisfies (3.2) and the

problem (P0) corresponding to λ1 = 0. Furthermore, this solution(u0 , P0

)is unique.

(i) Consider λ0 > 0 fixed and the parameter λ1 ∈ (0, λ0). Proving in the same manner as inthe case of Theorem 1 with 0 < λ1 < λ0, we have the following results:∥∥u′

λ1(t)∥∥2 + ∥∥uλ1x (t)

∥∥2 + K1∣∣uλ1(0, t)

∣∣2+ 2λ1

∫ t

0

∣∣u′λ1

(0, s)∣∣2 ds ≤ CT , ∀t ∈ [0, T ],∀T > 0, (3.6)∥∥Pλ1

∥∥L2(0,T )

≤ CT , (3.7)√λ1∥∥u′

λ1(0, ·)∥∥

L2(0,T )≤ CT , (3.8)

where CT is a constant independent of λ1.


Let λ1m be a sequence such that λ1m > 0, λ1m → 0 as m → ∞. From (3.6)–(3.8), we deducethat, there exists a subsequence of the sequence {(uλ1m , Pλ1m )} still denoted by {(uλ1m , Pλ1m )},such that⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

uλ1m → u∗ in L∞(0, T ; V ) weak*,u′

λ1m→ u′∗ in L∞(0, T ; L2) weak*,

uλ1m (0, ·) → u∗(0, ·) in L∞(0, T ) weakly*,√λ1mu′

λ1m(0, ·) → ζ∗ in L2(0, T ) weakly,

Pλ1m → P∗ in L2(0, T ) weakly.

(3.9)

By the compactness lemma of Lions [4, p. 57] and the imbeddings H 1(0, T ) ↪→ C0 ([0, T ]),we can deduce from (3.9)1,2,3,4 the existence of a subsequence still denoted by {uλ1m }, such that{

uλ1m → u∗ strongly in L2(QT ) and a.e. in QT ,√λ1muλ1m (0, ·) → 0 strongly in C0 ([0, T ]) .

(3.10)

It follows from (3.9)4, (3.10)2 that

ζ∗ = 0. (3.11)

From (3.9)3,4,5, (3.10)2 and (3.11), we have

Pλ1m (t) = g(t) + K1uλ1m (0, t) + λ1mu′λ1m

(0, t) −∫ t

0k(t − s)uλ1m (0, s)ds

→ g(t) + K1u∗(0, t) −∫ t

0k(t − s)u∗(0, s)ds = P∗(t), (3.12)

in L2(0, T ) weakly.By passing to the limit similarly to in the proof of Theorem 1, we conclude that (u∗, P∗) is a

solution of the problem (P0) corresponding to λ1 = 0 satisfying (3.2). From the uniqueness ofthe solution we have

(u∗, P∗) = (u0, P0

). (3.13)

Put u = uλ1 − u0, P = Pλ1 − P0, then (u, P) satisfies⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Au ≡ utt − ux x + K u + λut = 0, 0 < x < 1, 0 < t < T,

ux(0, t) = P(t), u(1, t) = 0,

u(x, 0) = ut (x, 0) = 0,

P(t) = λ1u ′0(0, t) + K1u(0, t) + λ1u′(0, t) −

∫ t

0k(t − s)u(0, s)ds,

u ∈ L∞(0, T ; V ), ut ∈ L∞(0, T ; L2), u(0, ·) ∈ H 1 (0, T ) , P ∈ L2 (0, T ) .

(3.14)

Using Lemma 2 again, we prove, in a similar manner to that in the above part, that

Z(t) + 2K1u2(0, t) + 2λ1

∫ t

0

∣∣u′(0, s)∣∣2 ds + 4λ

∫ t

0

∥∥u′(s)∥∥2 ds

≤ 2λ1

∫ t

0

∣∣u′0(0, s)

∣∣2 ds + 2CT

∫ t

0Z(r)dr, (3.15)

where

Z(t) = ∥∥u′(t)∥∥2 + ‖ux (t)‖2 , (3.16)

CT = 1 + |K | + 2 |k(0)| + 2 ‖k‖2L2(0,T )

+ T∥∥k ′∥∥2

L2(0,T ). (3.17)


Using Gronwall’s lemma, it follows from (3.15)–(3.17) that

Z(t) + 2K1u2(0, t) + 2λ1

∫ t

0

∣∣u′(0, s)∣∣2 ds ≤ 2λ1

∥∥u ′0(0, ·)∥∥2

L2(0,T )exp

(2CT t

)≤ 2λ1

∥∥u ′0(0, ·)∥∥2

L2(0,T )exp

(2CT T

) = KT λ1. (3.18)

This implies∥∥u′λ1

− u ′0

∥∥L∞(0,T ;L2)

+ ∥∥uλ1 − u0∥∥

L∞(0,T ;V )

+√λ1∥∥u′

λ1(0, ·) − u ′

0(0, ·)∥∥L2(0,T )

≤ CT

√λ1, (3.19)

where CT is a constant depending only on T .On the other hand, it follows from (3.14)4, (3.15) that

‖P‖L2 ≤ λ1(∥∥u ′

0(0, ·)∥∥L2 + ∥∥u′(0, ·)∥∥L2

)+(

K1 + √T∥∥k ′∥∥

L2

)‖u(0, ·)‖L2 . (3.20)

Hence∥∥Pλ1 − P0∥∥

L2(0,T )≤ CT

√λ1, (3.21)

where CT is a constant depending only on T . Theorem 2 is proved completely. �

The next result gives an asymptotic expansion of the weak solution(uλ1, Pλ1

)of order N + 1

2in λ1, for λ1 sufficiently small.

Let(u0, P0

)be a weak solution of problem (P0) (corresponding to λ1 = 0) as in Theorem 2.⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Au0 ≡ u ′′0 − u0x x + K u0 + λu ′

0 = F(x, t), 0 < x < 1, 0 < t < T,

u0x(0, t) = P0(t), u0(1, t) = 0,

u0(x, 0) = u0(x), u ′0(x, 0) = u1(x),

P0(t) = g(t) + K1u0(0, t) −∫ t

0k(t − s )u0(0, s)ds,

u0 ∈ L∞(0, T ; V ), u ′0 ∈ L∞(0, T ; L2), u0(0, ·), P0 ∈ H 1 (0, T ) .

(P0)

Let us consider the sequence of weak solutions(ui , Pi

), i = 1, 2, . . . , N , defined by the

following problems:⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Aui ≡ u ′′i − uix x + K ui + λu ′

i = 0, 0 < x < 1, 0 < t < T,

uix (0, t) = Pi (t), ui (1, t) = 0,

ui (x, 0) = u ′i (x, 0) = 0,

Pi (t) = K1ui (0, t) + u ′i−1(0, t) −

∫ t

0k(t − s )ui (0, s)ds,

ui ∈ L∞(0, T ; V ), u ′i ∈ L∞(0, T ; L2), ui (0, ·) ∈ H 1 (0, T ) , Pi ∈ L2 (0, T ) .

(Pi )

Let(uλ1, Pλ1

)be a unique weak solution of problem (Pλ1). Then (v, Q), with

v = uλ1 −N∑

i=0

uiλi1, Q = Pλ1 −

N∑i=0

Piλi1, (3.22)


satisfies the problem⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Av ≡ vt t − vx x + Kv + λvt = 0, 0 < x < 1, 0 < t < T,

vx (0, t) = Q(t), v(1, t) = 0,

v(x, 0) = vt (x, 0) = 0,

Q(t) = λN+11 u ′

N (0, t) + K1v(0, t) + λ1v′(0, t) −

∫ t

0k(t − s)v(0, s)ds,

v ∈ L∞(0, T ; V ), vt ∈ L∞(0, T ; L2), v(0, ·) ∈ H 1 (0, T ) , Q ∈ L2 (0, T ) .

(3.23)

Using again Lemma 2, we prove, in a manner similar to that of Theorem 2, that

Z(t) + 2K1v2(0, t) + 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds + 4λ

∫ t

0

∥∥v′(s)∥∥2 ds

≤ 2λ2N+11

∫ t

0

∣∣u′N (0, s)

∣∣2 ds + 2CT

∫ t

0Z(r)dr, (3.24)

where{Z(t) = ∥∥v′(t)

∥∥2 + ‖vx (t)‖2 ,

CT = 1 + |K | + 2 |k(0)| + 2 ‖k‖2L2(0,T )

+ T∥∥k ′∥∥2

L2(0,T ).

(3.25)

Using Gronwall’s lemma, it follows from (3.24) and (3.25) that

Z(t) + 2K1v2(0, t) + 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds ≤ 2λ2N+1

1

∥∥u ′N (0, ·)∥∥2

L2(0,T )exp

(2CT t

)≤ 2λ2N+1

1

∥∥u ′N (0, ·)∥∥2

L2(0,T )exp

(2CT T

) ≡ KT λ2N+11 . (3.26)

This implies∥∥∥∥∥u′λ1

−N∑

i=0

u ′i λ

i1

∥∥∥∥∥L∞(0,T ;L2)

+∥∥∥∥∥uλ1 −

N∑i=0

uiλi1

∥∥∥∥∥L∞(0,T ;V )

+√λ1

∥∥∥∥∥u′λ1

(0, ·) −N∑

i=0

u ′i (0, ·)λi

1

∥∥∥∥∥L2(0,T )

≤ CT λN+ 1

21 , (3.27)

∥∥∥∥∥Pλ1 −N∑

i=0

Piλi1

∥∥∥∥∥L2(0,T )

≤ CT λN+ 1

21 , (3.28)

where CT is a constant depending only on T .Thus, we have the following theorem.

Theorem 3. Let (H1)–(H4) hold. Then, for every λ1 ∈ (0, λ0), problem (Pλ1) has a uniqueweak solution

(uλ1, Pλ1

)satisfying the asymptotic estimations up to order N + 1

2 as in (3.27)and (3.28), the functions

(ui , Pi

), i = 1, 2, . . . , N, being the weak solutions of problems (P0),(

P1), . . . ,

(PN), respectively. �

4. Asymptotic expansion of the solution with respect to four parameters (K, λ, K1, λ1)

In this part, we assume that α = β = 2 and (u0, u1, F , g, k) satisfy the assumptions(H1)–(H3). Let (K , λ, K1, λ1) ∈ R × R

3+. By Theorem 1, the problem (1.1)–(1.5) has a unique


weak solution (u, P) depending on (K , λ, K1, λ1):

u = u (K , λ, K1, λ1) , P = P (K , λ, K1, λ1) . (4.1)

We consider the following perturbed problem, where K , λ, K1, λ1 are small parameters suchthat, |K | ≤ K∗, 0 ≤ λ ≤ λ∗, 0 ≤ K1 ≤ K1∗, 0 ≤ λ1 ≤ λ1∗:⎧⎪⎪⎪⎨⎪⎪⎪⎩

Lu ≡ utt − ux x = −K u − λut + F(x, t), 0 < x < 1, 0 < t < T,

ux (0, t) = P(t), u(1, t) = 0,

u(x, 0) = u0(x), ut (x, 0) = u1(x),

P(t) = g(t) + K1u(0, t) + λ1ut (0, t) −∫ t

0k(t − s)u(0, s)ds.

(PK ,λ,K1,λ1 )

We shall study the asymptotic expansion of the solution of problem (PK ,λ,K1,λ1) with respect to(K , λ, K1, λ1).

We use the following notation. For a multi-index γ = (γ1, γ2, γ3, γ4) ∈ Z4+, and

−→K =

(K , λ, K1, λ1) ∈ R × R3+, we put⎧⎪⎨⎪⎩

|γ | = γ1 + γ2 + γ3 + γ4, γ ! = γ1!γ2!γ3!γ4!,∥∥∥−→K ∥∥∥ =√

K 2 + λ2 + K 21 + λ2

1,−→K

γ = K γ1λγ2 K γ31 λ

γ41 ,

α, β ∈ Z4+, α ≤ β ⇐⇒ αi ≤ βi ∀i = 1, 2, 3, 4.

(4.2)

Let (u0,0,0,0, P0,0,0,0) ≡ (u0, P0) be a unique weak solution of problem (P0,0,0,0) (as inTheorem 1) corresponding to

−→K = 0, i.e.,

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

Lu0 = F0,0,0,0 ≡ F(x, t), 0 < x < 1, 0 < t < T,

u0x(0, t) = P0(t), u0(1, t) = 0,

u0(x, 0) = u0(x), u′0(x, 0) = u1(x),

P0(t) = g(t) −∫ t

0k(t − s)u0(0, s)ds,

u0 ∈ L∞(0, T ; V ), u′0 ∈ L∞(0, T ; L2),

u0(0, ·), P0 ∈ H 1 (0, T ) .

(P0,0,0,0)

Let us consider the sequence of weak solutions(uγ , Pγ

), γ ∈ Z

4+, 1 ≤ |γ | ≤ N , defined by thefollowing problems:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

Luγ ≡ uγ t t − uγ x x = Fγ , 0 < x < 1, 0 < t < T,

uγ x (0, t) = Pγ (t), uγ (1, t) = 0,

uγ (x, 0) = u′γ (x, 0) = 0,

Pγ (t) = Cγ (t) −∫ t

0k(t − s)uγ (0, s)ds,

uγ ∈ L∞ (0, T ; V ) , u′γ ∈ L∞(0, T ; L2),

uγ (0, ·) ∈ H 1 (0, T ) , Pγ ∈ L2 (0, T ) ,

(Pγ )

where Fγ , Cγ (t), |γ | ≤ N , defined by the formulas


Fγ =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩F, |γ | = 0,

0, 1 ≤ |γ | ≤ N, γ1 = 0, γ2 = 0,

−u′γ1,γ2−1,γ3,γ4

, 1 ≤ |γ | ≤ N, γ1 = 0, γ2 ≥ 1,

−uγ1−1,γ2,γ3,γ4, 1 ≤ |γ | ≤ N, γ1 ≥ 1, γ2 = 0,

−uγ1−1,γ2,γ3,γ4 − u′γ1,γ2−1,γ3,γ4

, 1 ≤ |γ | ≤ N, γ1 ≥ 1, γ2 ≥ 1,

(4.3)

and

Cγ (t) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩g(t), |γ | = 0,

0, 1 ≤ |γ | ≤ N, γ3 = 0, γ4 = 0,

u′γ1,γ2,γ3,γ4−1(0, t), 1 ≤ |γ | ≤ N, γ3 = 0, γ4 ≥ 1,

uγ1,γ2,γ3−1,γ4(0, t), 1 ≤ |γ | ≤ N, γ3 ≥ 1, γ4 = 0,

uγ1,γ2,γ3−1,γ4(0, t) + u′γ1,γ2,γ3,γ4−1(0, t), 1 ≤ |γ | ≤ N, γ3 ≥ 1, γ4 ≥ 1.

(4.4)

Let (u, P) = (uK ,λ,K1,λ1, PK ,λ,K1,λ1

)be a unique weak solution of problem (PK ,λ,K1,λ1). Then

(v, R), with

v = u −∑

|γ |≤N

uγ−→K

γ, R = P −

∑|γ |≤N

Pγ−→K

γ, (4.5)

satisfies the problem⎧⎪⎪⎪⎨⎪⎪⎪⎩vt t − vx x + Kv + λvt = eN (x, t), 0 < x < 1, 0 < t < T,

vx (0, t) = R(t), v(1, t) = 0,

v(x, 0) = vt (x, 0) = 0,

R(t) = eN (t) + K1v(0, t) + λ1v′(0, t) −

∫ t

0k(t − s)v(0, s)ds,

(4.6)

where

eN (x, t) = −∑

|γ |=N

(K uγ + λu′

γ

)−→K

γ, (4.7)

eN (t) =∑

|γ |=N

(K1uγ (0, t) + λ1u′

γ (0, t))−→

Kγ. (4.8)

Then, we have the following lemma.

Lemma 3. Let (H1)–(H3) hold. Then

‖eN ‖L∞(0,T ;L2) ≤ D1N

∥∥∥−→K ∥∥∥N+1, (4.9)

‖eN ‖L2(0,T ) ≤ D2N

∥∥∥−→K ∥∥∥N+1, (4.10)

4

∣∣∣∣∫ t

0eN (s)v′(0, s)ds

∣∣∣∣ ≤ K1v2(0, t) + 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds

+∫ t

0|v(0, s)|2 ds + D3N

∥∥∥−→K ∥∥∥2N+1, (4.11)

where D1N , D2N and D3N are positive constants depending only on the constants∥∥∥−→K ∗

∥∥∥,∥∥uγ

∥∥L∞(0,T ;V )

,∥∥∥u′

γ

∥∥∥L∞(0,T ;L2)

,∥∥uγ (0, ·)∥∥H1(0,T )

,∥∥∥u′

γ (0, ·)∥∥∥

L2(0,T ), (|γ | = N).


Proof. (i) By the boundedness of the functions uγ , u′γ , γ ∈ Z

4+, |γ | = N in the function space

L∞(0, T ; L2), we obtain from (4.7) that

‖eN ‖L∞(0,T ;L2) ≤∑

|γ |=N

(|K |∥∥uγ

∥∥L∞(0,T ;V )

+ λ

∥∥∥u′γ

∥∥∥L∞(0,T ;L2)

) ∣∣∣−→K γ∣∣∣ . (4.12)

On the other hand, using the Holder inequality xα11 xα2

2 xα33 xα4

4 ≤ α1x1 + α2x2 + α3x3 + α4x4,∀xi ≥ 0, ∀αi ≥ 0, α1 + α2 + α3 + α4 = 1 with x1 = K 2, x2 = λ2, x3 = K 2

1 , x4 = λ21, α1 = γ1

N ,α2 = γ2

N , α3 = γ3N , α4 = γ4

N , we obtain

∣∣∣−→K γ∣∣∣ = ∣∣K γ1λγ2 K γ3

1 λγ41

∣∣ =(

K 2γ1N λ2

γ2N K

2γ3N

1 λ2

γ4N

1

) N2

≤(γ1

NK 2 + γ2

Nλ2 + γ3

NK 2

1 + γ4

Nλ2

1

) N2

≤(

K 2 + λ2 + K 21 + λ2

1

) N2 =

∥∥∥−→K ∥∥∥N, (4.13)

for all γ ∈ Z4+, |γ | = N .

Therefore, it follows from (4.12) and (4.13) that

‖eN ‖L∞(0,T ;L2) ≤ D1N

∥∥∥−→K ∥∥∥N+1, (4.14)

where

D1N =∑

|γ |=N

(∥∥uγ

∥∥L∞(0,T ;V )

+∥∥∥u′

γ

∥∥∥L∞(0,T ;L2)

). (4.15)

(ii) With eN (t), then, we obtain from (4.8), in a manner similar to that of the above part,that

‖eN ‖H2(0,T ) ≤ D2N

∥∥∥−→K ∥∥∥N+1, (4.16)

where

D2N =∑

|γ |=N

(∥∥uγ (0, ·)∥∥H1(0,T )+∥∥∥u′

γ (0, ·)∥∥∥

L2(0,T )

). (4.17)

(iii) From (4.8), we have

4∫ t

0eN (s)v′(0, s)ds = 4

∑|γ |=N

K1

∫ t

0uγ (0, s)v′(0, s)ds

−→K

γ

+ 4∑

|γ |=N

λ1

∫ t

0u′

γ (0, s)v′(0, s)ds−→K

γ

≡ J1N (t) + J2N (t). (4.18)

We shall estimate respectively the following terms on the right-hand side of (4.18).


First term J1N (t). Integrating by parts, we have

J1N (t) = 4∑

|γ |=N

K1

∫ t

0uγ (0, s)v′(0, s)ds

−→K

γ

= 4K1v(0, t)∑

|γ |=N

uγ (0, t)−→K

γ − 4K1

∑|γ |=N

∫ t

0u′

γ (0, s)v(0, s)ds−→K

γ. (4.19)

By (4.13), it follows from (4.19) that

J1N (t) ≤ K1v2(0, t) + 4K1

( ∑|γ |=N

∣∣uγ (0, t)∣∣ ∣∣∣−→K γ

∣∣∣)2

+ 4K1

∑|γ |=N

(∫ t

0

∣∣∣u′γ (0, s)

∣∣∣2 ds

)1/2(∫ t

0|v(0, s)|2 ds

)1/2 ∣∣∣−→K γ∣∣∣

≤ K1v2(0, t) +

∫ t

0|v(0, s)|2 ds

+ 4

⎡⎣( ∑|γ |=N

∥∥uγ

∥∥L∞(0,T ;V )

)2

+( ∑

|γ |=N

∥∥∥u′γ (0, ·)

∥∥∥L2(0,T )

)2 ∥∥∥−→K ∗∥∥∥⎤⎦

×∥∥∥−→K ∥∥∥2N+1

≤ K1v2(0, t) +

∫ t

0|v(0, s)|2 ds

+ 4(

1 +∥∥∥−→K ∗

∥∥∥)[ ∑|γ |=N

(∥∥uγ

∥∥L∞(0,T ;V )

+∥∥∥u′

γ (0, ·)∥∥∥

L2(0,T )

)]2

×∥∥∥−→K ∥∥∥2N+1

, (4.20)

where∥∥∥−→K ∥∥∥ ≤

∥∥∥−→K ∗∥∥∥, with

−→K ∗ = (K∗, λ∗, K1∗, λ1∗).

Second term J2N (t). Using again (4.13), we obtain

J2N (t) = 4∑

|γ |=N

λ1

∫ t

0u′

γ (0, s)v′(0, s)ds−→K

γ ≤ 4λ1

∑|γ |=N

∫ t

0

∣∣∣u′γ (0, s)v′(0, s)

∣∣∣ ds∣∣∣−→K γ

∣∣∣≤ 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds + 2λ1

( ∑|γ |=N

∥∥∥u′γ (0, ·)

∥∥∥L2(0,T )

)2 ∥∥∥−→K ∥∥∥2N

≤ 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds + 2

( ∑|γ |=N

∥∥∥u′γ (0, ·)

∥∥∥L2(0,T )

)2 ∥∥∥−→K ∥∥∥2N+1. (4.21)

Combining (4.18), (4.20) and (4.21), we then have

1544 N.T. Long, V.G. Giai / Nonlinear Analysis 66 (2007) 1526–1546∣∣∣∣4 ∫ t

0eN (s)v′(0, s)ds

∣∣∣∣ ≤ K1v2(0, t) + 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds

+∫ t

0|v(0, s)|2 ds + D3N

∥∥∥−→K ∥∥∥2N+1, (4.22)

where

D3N = 4(

1 +∥∥∥−→K ∗

∥∥∥)[ ∑|γ |=N

(∥∥uγ

∥∥L∞(0,T ;V )

+∥∥∥u′

γ (0, ·)∥∥∥

L2(0,T )

)]2

+ 2

( ∑|γ |=N

∥∥∥u′γ (0, ·)

∥∥∥L2(0,T )

)2

. (4.23)

The proof of Lemma 3 is complete. �Next, we obtain the following theorem.

Theorem 4. Let (H1)–(H3) hold. Then, for every (K , λ, K1, λ1) ∈ R × R3+, with |K | ≤ K∗,

0 ≤ λ ≤ λ∗, 0 ≤ K1 ≤ K1∗, 0 ≤ λ1 ≤ λ1∗, problem (PK ,λ,K1,λ1) has a unique weak solution(u, P) = (

uK ,λ,K1,λ1, PK ,λ, K1,λ1

)satisfying the asymptotic estimations up to order N + 1

2 asfollows∥∥∥∥∥u′ −

∑|γ |≤N

u′γ−→K

γ

∥∥∥∥∥L∞(0,T ;L2)

+∥∥∥∥∥u −

∑|γ |≤N

uγ−→K

γ

∥∥∥∥∥L∞(0,T ;V )

+√λ1

∥∥∥∥∥u′(0, ·) −∑

|γ |≤N

u′γ (0, ·)−→K γ

∥∥∥∥∥L2(0,T )

≤ D∗N

∥∥∥−→K ∥∥∥N+ 12, (4.24)

and ∥∥∥∥∥P −∑

|γ |≤N

Pγ−→K

γ

∥∥∥∥∥L2(0,T )

≤ D∗∗N

∥∥∥−→K ∥∥∥N+ 12, (4.25)

for all (K , λ, K1, λ1) ∈ R × R3+, |K | ≤ K∗, 0 ≤ λ ≤ λ∗, 0 ≤ K1 ≤ K1∗, 0 ≤ λ1 ≤ λ1∗, the

functions (uγ , Pγ ) being the weak solutions of problems (Pγ ), γ ∈ Z4+, |γ | ≤ N.

Remark 2. In [10], as in this special case for problem (1.1)–(1.5), Long, Ut, and Truc haveobtained a result about the asymptotic expansion of the solutions with respect to two parameters(K , λ) up to order N + 1.

Proof. Put{Z(t) = ∥∥v′(t)

∥∥2 + ‖vx (t)‖2 ,

CT = 1 + |K | + 2 |k(0)| + 2 ‖k‖2L2(0,T )

+ T∥∥k ′∥∥2

L2(0,T ).

(4.26)

Using Lemma 2 again, we prove, in a manner similar to that of Theorem 2, that

Z(t) + 2K1v2(0, t) + 4λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds

≤ 2CT

∫ t

0Z(s)ds + 4

∫ t

0

⟨eN (s), v′(s)

⟩ds + 4

∫ t

0eN (s)v′(0, s)ds. (4.27)


Using Lemma 3, it follows from (4.27) that

Z(t) + 2K1v2(0, t) + 4λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds

≤ 2CT

∫ t

0Z(s)ds + 4T D2

1N

∥∥∥−→K ∥∥∥2N+2 +∫ t

0

∥∥v′(s)∥∥2

ds

+ K1v2(0, t) + 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds +

∫ t

0|v(0, s)|2 ds + D3N

∥∥∥−→K ∥∥∥2N+1

= K1v2(0, t) + 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds + 2

(1 + CT

) ∫ t

0Z(s)ds

+(

4T D21N

∥∥∥−→K ∗∥∥∥+ D3N

) ∥∥∥−→K ∥∥∥2N+1, (4.28)

where∥∥∥−→K ∥∥∥ ≤

∥∥∥−→K ∗∥∥∥, with

−→K ∗ = (K∗, λ∗, K1∗, λ1∗). Hence

Z(t) + K1v2(0, t) + 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds

≤(

4T D21N

∥∥∥−→K ∗∥∥∥+ D3N

) ∥∥∥−→K ∥∥∥2N+1 + 2(1 + CT

) ∫ t

0Z(s)ds. (4.29)

Using Gronwall’s lemma, it follows from (4.29) that

Z(t) + 2K1v2(0, t) + 2λ1

∫ t

0

∣∣v′(0, s)∣∣2 ds

≤(

4T D21N

∥∥∥−→K ∗∥∥∥+ D3N

) ∥∥∥−→K ∥∥∥2N+1exp

(2(1 + CT

)T). (4.30)

Hence∥∥v′∥∥L∞(0,T ;L2)

+ ‖v‖L∞(0,T ;V ) +√λ1∥∥v′(0, ·)∥∥L2(0,T )

≤ D∗N

∥∥∥−→K ∥∥∥N+ 12, (4.31)

where

D∗N = 3

√4T D2

1N + D3N exp((

1 + CT)

T), (4.32)

or ∥∥∥∥∥u′ −∑

|γ |≤N

u′γ−→K

γ

∥∥∥∥∥L∞(0,T ;L2)

+∥∥∥∥∥u −

∑|γ |≤N

uγ−→K

γ

∥∥∥∥∥L∞(0,T ;V )

+√λ1

∥∥∥∥∥u′(0, ·) −∑

|γ |≤N

u′γ (0, ·)−→K γ

∥∥∥∥∥L2(0,T )

≤ D∗N

∥∥∥−→K ∥∥∥N+ 12. (4.33)

On the other hand, it follows from (4.6)4, (4.31) that

‖R‖L2(0,T ) ≤ ‖eN ‖L2(0,T ) +(

K1 + √T∥∥k ′∥∥

L2(0,T )

)‖v(0, ·)‖L2(0,T )

+ λ1∥∥v′(0, ·)∥∥L2(0,T )


≤[

D2N

√∥∥∥−→K ∗∥∥∥+

(K1 + √

T∥∥k ′∥∥

L2(0,T )

)D∗

N +√λ1∗ D∗

N

] ∥∥∥−→K ∥∥∥N+ 12

≡ D∗∗N

∥∥∥−→K ∥∥∥N+ 12, (4.34)

where

D∗∗N =

√∥∥∥−→K ∗∥∥∥D2N +

(K1 + √

T∥∥k ′∥∥

L2(0,T )+√

λ1∗)

D∗N , (4.35)

and hence∥∥∥∥∥P −∑

|γ |≤N

Pγ−→K

γ

∥∥∥∥∥L2(0,T )

≤ D∗∗N

∥∥∥−→K ∥∥∥N+ 12, (4.36)

Theorem 4 is proved completely. �

Acknowledgement

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

References

[1] D.D. Ang, A.P.N. Dinh, Mixed problem for some semilinear wave equation with a nonhomogeneous condition,Nonlinear Anal. 12 (1988) 581–592.

[2] N.T. An, N.D. Trieu, Shock between absolutely solid body and elastic bar with the elastic viscous frictionalresistance at the side, J. Mech. NCSR Vietnam 13 (2) (1991) 1–7.

[3] M. Bergounioux, N.T. Long, A.P.N. Dinh, Mathematical model for a shock problem involving a linear viscoelasticbar, Nonlinear Anal. 43 (2001) 547–561.

[4] A.P.N. Dinh, N.T. Long, Linear approximation and asymptotic expansion associated to the nonlinear wave equationin one dimension, Demonstratio Math. 19 (1986) 45–63.

[5] J.L. Lions, Quelques methodes de resolution des proble mes aux limites nonlineaires, Gauthier-Villars, Paris,Dunod, 1969.

[6] N.T. Long, A.P.N. Dinh, On the quasilinear wave equation: utt − �u + f (u, ut ) = 0 associated with a mixednonhomogeneous condition, Nonlinear Anal. 19 (1992) 613–623.

[7] N.T. Long, A.P.N. Dinh, A semilinear wave equation associated with a linear differential equation with Cauchydata, Nonlinear Anal. 24 (1995) 1261–1279.

[8] N.T. Long, T.N. Diem, On the nonlinear wave equation utt − uxx = f (x, t, , u, ux , ut ) associated with the mixedhomogeneous conditions, Nonlinear Anal. 29 (1997) 1217–1230.

[9] N.T. Long, A.P.N. Dinh, T.N. Diem, On a shock problem involving a nonlinear viscoelastic bar, J. Boundary ValueProblems, Hindawi Publishing Corporation 2005 (3) (2005) 337–358.

[10] N.T. Long, L.V. Ut, N.T.T. Truc, On a shock problem involving a linear viscoelastic bar, Nonlinear Anal. 63 (2)(2005) 198–224.

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