a nonlinear wave equation associated with nonlinear boundary conditions: existence and asymptotic...

Nonlinear Analysis 66 (2007) 2852–2880www.elsevier.com/locate/na

A nonlinear wave equation associated with nonlinearboundary conditions: Existence and asymptotic

expansion of solutions

Nguyen Thanh Long∗, Vo Giang Giai

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

Received 5 March 2006; accepted 13 April 2006

Abstract

Consider the initial–boundary value problem for the nonlinear wave equation⎧⎪⎪⎨⎪⎪⎩utt − ux x + K |u|p−2 u + λ |ut |q−2 ut = F(x, t), 0 < x < 1, 0 < t < T,

ux (0, t) = P(t),−ux (1, t) = |u(1, t)|p1−2 u(1, t) + |ut (1, t)|q1−2 ut (1, t),u(x, 0) = u0(x), ut (x, 0) = u1(x),

(1)

where p, p1, q1 ≥ 2, q > 1, K , λ are given constants and u0, u1, F are given functions, and the unknownfunction u(x, t) and the unknown boundary value P(t) satisfy the following nonlinear integral equation

P(t) = g(t) + K0 |u(0, t)|p0−2 u(0, t) + |ut (0, t)|q0−2 ut (0, t) −∫ t

0k(t − s)u(0, s)ds, (2)

where p0, q0 ≥ 2, K0 are given constants and g, k are given functions.In this paper, we consider three main parts. In Part 1, under the conditions (u0, u1) ∈ H1 × L2,

F ∈ L2(QT ), k ∈ W 1,1(0, T ), g ∈ Lq ′0(0, T ), λ = 1, K , K0 ≥ 0; p, p0, q0, p1, q1 ≥ 2, q > 1,

q′0 = q0

q0−1 , we prove a theorem of existence and uniqueness of a weak solution (u, P) of problem (1)and (2). The proof is based on the Faedo–Galerkin method and the weak compact method associatedwith a monotone operator. For the case of q0 = q1 = 2, p, q, p0, p1 ≥ 2, in Part 2 we prove that

the unique solution (u, P) belongs to(

L∞(0, T ; H2) ∩ C0(0, T ; H1) ∩ C1(0, T ; L2))

× H1(0, T ), with

ut ∈ L∞(0, T ; H1), utt ∈ L∞(0, T ; L2), u(0, ·), u(1, ·) ∈ H2(0, T ), if we make the assumption that

∗ 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.2006.04.013

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http://dx.doi.org/10.1016/j.na.2006.04.013

N.T. Long, V.G. Giai / Nonlinear Analysis 66 (2007) 2852–2880 2853

(u0, u1) ∈ H2 × H1 and some others. Finally, in Part 3 we obtain an asymptotic expansion of the solution(u, P) of the problem (1) and (2) up to order N + 1 in three small parameters K , λ, K0.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 consider the following problem: Find a pair (u, P) of functions satisfying

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

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

−ux(1, t) = K1 |u(1, t)|p1−2 u(1, t) + λ1 |ut (1, t)|q1−2 ut (1, t), (1.3)

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

where f (u, ut ) = K |u|p−2 u + λ |ut |q−2 ut , where p, p1, q1 ≥ 2, q > 1, K1, λ1, K , λ aregiven constants and u0, u1, F are given functions satisfying conditions specified later, and theunknown function u(x, t) and the unknown boundary value P(t) satisfy the following nonlinearintegral equation

P(t) = g(t) + K0 |u(0, t)|p0−2 u(0, t) + λ0 |ut (0, t)|q0−2 ut (0, t)

−∫ t

0k(t − s)u(0, s)ds, (1.5)

where p0, q0 ≥ 2, K0, λ0 are given constants and g, k are given functions.In [1], An and Trieu studied a special case of problem (1.1), (1.2), (1.4) and (1.5) associated

with the following homogeneous boundary condition at x = 1:

u(1, t) = 0, (1.6)

with F = u0 = u1 ≡ 0, λ0 = 0, p0 = 2, and f (u, ut ) = K u + λut , with K ≥ 0, λ ≥ 0,are given constants. In the latter case the problem (1.1), (1.2) and (1.4)–(1.6) is a mathematicalmodel describing the shock of a rigid body and a linear viscoelastic bar resting on a rigid base [1].

In [2] Bergounioux, Long and Dinh have studied a special case of problem (1.1)–(1.5) withp = q = p1 = q1 = p0 = 2, λ0 = 0; f (u, ut ) = K u + λut , where K0 ≥ 0, K , λ, K1, λ1 aregiven constants and g, k are given functions.

In [9], Long, Dinh and Diem obtained the unique existence, regularity and asymptoticexpansion of the problem (1.1)–(1.5) in the case of p = q ≥ 2, p0 = p1 = q1 = 2, λ0 = 0,u0 ∈ H 2 and u1 ∈ H 1.

In [10], Long, Ut and Truc gave the unique existence, stability, regularity in time variable andasymptotic expansion for the solution of problem (1.1), (1.2) and (1.4)–(1.6) when p = q =p0 = q0 = 2, u0 ∈ H 2 and u1 ∈ H 1.

In [11], Long and Giai obtained the unique existence and asymptotic expansion for thesolution of problem (1.1), (1.2) and (1.4)–(1.6) when p = q = 2, p0, q0 ≥ 2, (u0, u1) ∈H 1 × L2. In this case, the problem (1.1)–(1.5) is the mathematical model describing a shockproblem involving a nonlinear viscoelastic bar.

2854 N.T. Long, V.G. Giai / Nonlinear Analysis 66 (2007) 2852–2880

In this paper, we consider three main parts. In Part 1, under conditions (u0, u1) ∈ H 1 × L2,F ∈ L2(QT ), k ∈ W 1,1(0, T ), g ∈ Lq ′

0(0, T ), K1 = λ = λ0 = λ1 = 1; K ,K0 ≥ 0; p, p0, q0, p1, q1 ≥ 2, q > 1, q ′

0 = q0q0−1 , we prove a theorem of existence

and uniqueness of a weak solution (u, P) of problem (1.1)–(1.5). The proof is based on theFaedo–Galerkin method and the weak compact method associated with a monotone operator.We remark that the linearization method in the papers [3,8] cannot be used in [2,6,7]. Forthe case of q0 = q1 = 2, in Part 2 we prove that the unique solution (u, P) belongs to(L∞(0, T ; H 2) ∩ C0(0, T ; H 1) ∩ C1(0, T ; L2)

) × H 1(0, T ), with ut ∈ L∞(0, T ; H 1), utt ∈L∞(0, T ; L2), u(0, ·), u(1, ·) ∈ H 2(0, T ), if we make the assumption that (u0, u1) ∈ H 2 × H 1

and some others. Finally, in Part 3, with p1 = q1 = q0 = 2, we obtain an asymptotic expansionof the solution (u, P) of the problem (1.1)–(1.5) up to order N + 1 in three small parametersK , λ, K0. The results obtained here may be considered as generalizations of those in An andTrieu [1] and in [2,3,6–11].

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.

On H 1 we shall use the following norms

‖v‖H1 =(∫ 1

0

(v2(x) +

∣∣∣∣ ∂v

∂x(x)

∣∣∣∣2)

dx

)1/2

,

‖v‖1 =(∫ 1

0

∣∣∣∣ ∂v

∂x(x)

∣∣∣∣2 dx + v2(1)

)1/2

. (2.1)

Then we have the following lemmas.

Lemma 1. The imbedding H 1 ↪→ C0([0, 1]) is compact and

‖v‖C0(Ω)≤ √

2 ‖v‖1, for all v ∈ H 1, (2.2)

‖v‖C0(Ω)≤ √

2‖v‖H1 , for all v ∈ H 1. (2.3)


The proofs of these lemmas are straightforward, and we omit the details.

Remark 1. It follows from (2.2), (2.3) that, on H 1, ‖v‖1, ‖v‖H1 are two equivalent norms and

1√3

‖v‖H1 ≤ ‖v‖1 ≤ √3‖v‖H1 , for all v ∈ H 1. (2.4)

We make the following assumptions:

(H1) u0 ∈ H 1 and u1 ∈ L2,(H2) F ∈ L2(QT ),(H3) k ∈ L1(0, T ), g ∈ Lq ′

0 (0, T ) , q ′0 = q0

q0−1 ,(H4) K , K0 ≥ 0; K1, λ, λ0, λ1 > 0,(H5) p, p0, p1, q0, q1 ≥ 2, q > 1.

Then, we have the following theorem.

Theorem 1. Let (H1)–(H5) hold. For every T > 0, there exists a weak solution (u, P) ofproblem (1.1)–(1.5) such that⎧⎪⎨⎪⎩

u ∈ L∞(0, T ; H 1), ut ∈ L∞(0, T ; L2) ∩ Lq(QT ),

u(0, ·) ∈ W 1,q0(0, T ), u(1, ·) ∈ W 1,q1 (0, T ) ,

P ∈ Lq ′0(0, T ), q ′

0 = q0

q0 − 1.

(2.5)

Furthermore, if k ∈ W 1,1(0, T ) in (H3) and p0, p1 ∈ {2} ∪ [3,+∞), the solution is unique.

Remark 2. (i) Theorem 1 gives no conclusion about the uniqueness of the solution when2 < p0 < 3 or 2 < p1 < 3.

(ii) The corresponding results in [2] and [9] are special cases of Theorem 1 with p = q = p0 =p1 = q1 = 2, λ0 = 0, (u0, u1) ∈ H 2 × H 1 and p, q ≥ 2; p0 = p1 = q1 = 2, λ0 = 0,(u0, u1) ∈ H 2 × H 1, respectively.

Proof of Theorem 1. Without loss of generality, we can suppose that K1 = λ = λ0 = λ1 = 1.The proof consists of Steps 1–4.Step 1. The Faedo–Galerkin approximation. Let {w j } be a denumerable base of H 1. We find

the approximate solution of problem (1.1)–(1.5) in the form

um(t) =m∑

j=1

cmj (t)w j , (2.6)

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

m(t),w j⟩+ ⟨

umx (t),w j x⟩+ Pm(t)w j (0) + Qm(t)w j (1)

+⟨K |um(t)|p−2 um(t) + ∣∣u′

m(t)∣∣q−2

u′m(t),w j

⟩= ⟨

F (t),w j⟩, 1 ≤ j ≤ m, (2.7)

Pm(t) = g(t) + K0 |um(0, t)|p0−2 um(0, t) + ∣∣u′m(0, t)

∣∣q0−2u′

m(0, t)

−∫ t

0k(t − s)um(0, s)ds, (2.8)

Qm(t) = |um(1, t)|p1−2 um(1, t) + ∣∣u′m(1, t)

∣∣q1−2u′

m(1, t), (2.9)

2856 N.T. Long, V.G. Giai / Nonlinear Analysis 66 (2007) 2852–2880⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩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.

(2.10)

From the assumptions of Theorem 1, system (2.7)–(2.10) has a solution (um, Pm , Qm ) on aninterval [0, Tm] ⊂ [0, T ]. The following estimates allow one to take Tm = T for all m.

Step 2. A priori estimates. Substituting (2.8), (2.9) into (2.7), then multiplying the j th equationof (2.7) by c′

mj (t) and summing with respect to j , afterwards, integrating with respect to the timevariable from 0 to t , we get after some rearrangements

Sm(t) = Sm(0) − 2∫ t

0g(s)u′

m(0, s)ds + 2∫ t

0u′

m(0, r)

(∫ r

0k(r − s)um(0, s)ds

)dr

+ 2∫ t

0

⟨F(s), u′

m(s)⟩ds ≡ Sm(0) + I1 + I2 + I3, (2.11)

where

Sm(t) = ∥∥u′m(t)

∥∥2 + ‖umx (t)‖2 + 2K

p‖um(t)‖p

L p + 2∫ t

0

∥∥u′m(s)

∥∥2 ds

+ 2K0

p0|um(0, t)|p0 + 2

∫ t

0

∣∣u′m(0, s)

∣∣q0 ds

+ 2

p1|um(1, t)|p1 + 2

∫ t

0

∣∣u′m(1, s)

∣∣q1 ds + p1 − 2

p1. (2.12)

First, we shall need the following lemma.

Lemma 2. Let p1 ≥ 2. We have

‖ux‖2 + 2

p1|u(1)|p1 + p1 − 2

p1≥ 1

3‖u‖2

H1 , for all v ∈ H 1. (2.13)

The proof is straightforward and we omit the details. �By assumptions (H1), (H4), (H5) and the imbedding H 1 ↪→ C0

(Ω), we have

Sm(0) = ‖u1m‖2 + ‖u0mx‖2 + 2K

p‖u0m‖p

L p

+ 2K0

p0|u0m(0)|p0 + 2

p1|u0m(1)|p1 + p1 − 2

p1≤ 1

2C1, (2.14)

for all m, where C1 is a constant depending only on u0, u1, K , p, K0, p0 and p1.We shall estimate the following three integrals in the right-hand side of (2.11).First integral. From (2.12) and the Holder inequality

ab ≤ 1

qεqaq + 1

q ′ ε−q ′

bq ′, for all ε > 0, a, b ≥ 0, q > 1, q ′ = q

q − 1, (2.15)

we obtain

I1 = −2∫ t

0g(s)u′

m(0, s)ds ≤ 2εq0

q0

∫ t

0

∣∣u′m(0, s)

∣∣q0 ds + 2

q ′0ε−q ′

0 ‖g‖q ′0

Lq′0 (0,T )

≤ 1

4Sm(t) + 2

q ′0ε−q ′

0 ‖g‖q ′0

Lq′0 (0,T )

, (2.16)


where2εq0

q0= 1

2. (2.17)

Second integral. We again use inequality (2.15) with ε as in (2.17), from assumption (H3), wehave

I2 = 2∫ t

0u′

m(0, r)

(∫ r


)dr

≤∫ t

0ε2∣∣u′

m(0, r)∣∣2 dr + 1

ε2

∫ t

0dr

∣∣∣∣∫ r


∣∣∣∣2≤∫ t

0

(2εq0

q0

∣∣u′m(0, r)

∣∣q0 + q0 − 2

q0

)dr + 1

ε2

∫ t

0dr

∣∣∣∣∫ r


∣∣∣∣2≤ 1

4Sm(t) +

(1 − 2

q0

)T + 1

ε2

∫ t

0dr

∣∣∣∣∫ r


∣∣∣∣2 . (2.18)

On the other hand, by Lemma 2, we obtain from (2.3), (2.12) that

|um(0, t)| ≤ ‖um(t)‖C0(Ω)≤ √

2 ‖um(t)‖H1 ≤ √6√

Sm(t). (2.19)

By the Cauchy–Schwartz inequality, we obtain from (2.19) that(∫ r


)2

≤∫ r

0|k(r − s)| ds

∫ r

0|k(r − s)| u2

m(0, s)ds

≤ 6 ‖k‖L1(0,T )

∫ r

0|k(r − s)| Sm(s)ds. (2.20)

Inserting (2.20) into the last integral in the right-hand side of (2.18) and inverting the variablesof integration, between r and s, we get

I2 ≤ 1

4Sm(t) +

(1 − 2

q0

)T + 6

ε2‖k‖L1(0,T )

∫ t

0dr

(∫ r

0|k(r − s)| Sm(s)ds

)= 1

4Sm(t) +

(1 − 2

q0

)T + 6

ε2‖k‖L1(0,T )

∫ t

0Sm(s)ds

∫ t

s|k(r − s)| dr

≤ 1

4Sm(t) +

(1 − 2

q0

)T + 6

ε2‖k‖2

L1(0,T )

∫ t

0Sm(s)ds. (2.21)

Third integral.

I3 = 2∫ t

0

⟨F(s), u′

m(s)⟩ds ≤ ‖F‖2

L2(QT )+∫ t

0Sm(s)ds. (2.22)

Combining (2.11), (2.12), (2.16), (2.21) and (2.22), we obtain

Sm(t) ≤ M(1)T + N (1)

T

∫ t

0Sm(s)ds, (2.23)

where⎧⎪⎪⎨⎪⎪⎩M(1)

T = C1 + 2

(1 − 2

q0

)T + 2 ‖F‖2

L2(QT )+ 4

q ′0ε−q ′

0 ‖g‖q ′0

Lq′0 (0,T )

,

N (1)T = 2

(1 + 6

ε2‖k‖2

L1(0,T )

).

(2.24)


By Gronwall’s lemma, we deduce from (2.23), (2.24) 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 ]. (2.25)

On the other hand, from the assumptions (H3)–(H5), we deduce from (2.8), (2.12), (2.19) and(2.25) that∥∥∥∣∣u′

m(0, ·)∣∣q0−2u′

m(0, ·)∥∥∥q ′

0

Lq′0 (0,T )

=∫ T

0

∣∣u′m(0, s)

∣∣q0 ds ≤ CT

2, (2.26)∥∥∥∣∣u′

m(1, ·)∣∣q1−2u′

m(1, ·)∥∥∥q ′

1

Lq′1 (0,T )

=∫ T

0

∣∣u′m(1, s)

∣∣q1 ds ≤ CT

2, (2.27)∥∥∥∣∣u′

m

∣∣q−2u′

m

∥∥∥q ′

Lq′(QT )

= ∥∥u′m

∥∥qLq (QT )

≤ CT

2, (2.28)

|Pm(t)| ≤ |g(t)| + K0

(√6CT

)p0−1 +√6CT ‖k‖L1(0,T ) + ∣∣u′

m(0, t)∣∣q0−1

= C(1)T + |g(t)| + ∣∣u′

m(0, t)∣∣q0−1

. (2.29)

Hence, we deduce from (2.25), (2.29) that

‖Pm‖Lq′

0 (0,T )≤ CT for all m, (2.30)

where CT is a positive constant depending only on T .Step 3. Limiting process. From (2.12), (2.25)–(2.28) and (2.30), we deduce the existence of a

subsequence of {(um, Pm)} still also so denoted, such that⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

um → u in L∞(0, T ; H 1) weak*,u′

m → u′ in L∞(0, T ; L2) weak*,um(0, ·) → u(0, ·) in L∞(0, T ) weak*,um(1, ·) → u(1, ·) in L∞(0, T ) weak*,u′

m(0, ·) → u′(0, ·) in Lq0(0, T ) weakly,u′

m(1, ·) → u′(1, ·) in Lq1(0, T ) weakly,∣∣u′m

∣∣q−2u′

m → χ in Lq ′(QT ) weakly,∣∣u′

m(0, ·)∣∣q0−2u′

m(0, ·) → χ0 in Lq ′0(0, T ) weakly,∣∣u′

m(1, ·)∣∣q1−2u′

m(1, ·) → χ1 in Lq ′1(0, T ) weakly,

Pm → P in Lq ′0(0, T ) weakly.

(2.31)

By the compactness lemma of Lions [5, p. 57] and the imbedding W 1,r (0, T ) ↪→ C0 ([0, T ]),r > 1, we can deduce from (2.31)1,2,3,4,5,6 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 ]),um(1, ·) → u(1, ·) strongly in C0([0, T ]).

(2.32)

By means of the following inequality∣∣∣|x |γ−2 x − |y|γ−2 y∣∣∣ ≤ (γ − 1) Rγ−2|x − y|, (2.33)


for all x , y ∈ [−R, R], R > 0 and γ ≥ 2, it follows from (2.19) and (2.25) that∣∣∣|um(0, t)|p0−2 um(0, t) − |u(0, t)|p0−2 u(0, t)∣∣∣

≤ (p0 − 1)(√

6CT

)p0−2 |um(0, t) − u(0, t)|. (2.34)

Hence, it follows from (2.32)2, (2.34) that

|um(0, t)|p0−2 um(0, t) → |u(0, t)|p0−2 u(0, t) strongly in C0([0, T ]). (2.35)

From (2.32)2, (2.35) we have

Pm(t) − g(t) = K0 |um(0, t)|p0−2 um(0, t) −∫ t

0k(t − s)um(0, s)ds → P(t) − g(t),

(2.36)

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

P(t) = g(t) + K0 |u(0, t)|p0−2 u(0, t) −∫ t

0k(t − s)u(0, s)ds. (2.37)

Hence

Pm(t) → P(t) + χ0(t) = P(t), (2.38)

in Lq ′0(0, T ) weakly.

Similarly, it follows from (2.9), (2.31)9, (2.32)3, (2.33) that

|um(1, t)|p1−2 um(1, t) → |u(1, t)|p1−2 u(1, t) strongly in C0([0, T ]), (2.39)

Qm(t) = |um(1, t)|p1−2 um(1, t) + ∣∣u′m(1, t)

∣∣q1−2u′

m(1, t)

→ |u(1, t)|p1−2 u(1, t) + χ1(t), (2.40)

in Lq ′1(0, T ) weakly,

|um |p−2 um → |u|p−2 u strongly in L2(QT ) and a.e. in QT . (2.41)

Passing to the limit in (2.7) and (2.8), (2.9) by (2.31)1,2,7,8,9, (2.38), (2.40), (2.41), we have(u, P) satisfying the equation

d

dt

⟨u′(t), v

⟩ + 〈ux (t), vx 〉 +⟨K |u(t)|p−2 u(t) + χ(t), v

⟩+ (

P(t) + χ0(t))v(0)

+(|u(1, t)|p1−2 u(1, t) + χ1(t)

)v(1) = 〈F (t), v〉, (2.42)

for all v ∈ H 1. We can prove similarly to in [6] that

u(0) = u0, u′(0) = u1. (2.43)

Then, in order to prove the existence of the solution of the problem (1.1)–(1.5), we only haveto prove that χ = ∣∣u′∣∣q−2

u′, χ0(t) = ∣∣u′(0, t)∣∣q0−2

u′(0, t) and χ1(t) = ∣∣u′(1, t)∣∣q1−2

u′(1, t).We shall now require the following lemma.


Lemma 3. Let u be the weak solution of the following problem⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

u′′ − ux x + Φ1 = 0, 0 < x < 1, 0 < t < T,

ux (0, t) = G0(t), −ux(1, t) = G1(t),u(x, 0) = u0(x), u′(x, 0) = u1(x),

u ∈ L∞(0, T ; H 1), u′ ∈ L∞(0, T ; L2),

u′(0, ·) ∈ Lq0(0, T ), u′(1, ·) ∈ Lq1(0, T ),

G0 ∈ Lq ′0(0, T ), G1 ∈ Lq ′

1(0, T ),Φ1 ∈ L2(QT ).

(2.44)

Then we have

1

2

∥∥u′(t)∥∥2 +1

2‖ux (t)‖2 +

∫ t

0G0(s)u

′(0, s)ds +∫ t

0G1(s)u

′(1, s)ds

+∫ t

0

⟨Φ1(s), u′(s)

⟩ds ≥ 1

2‖u1‖2 + 1

2‖u0x‖2 a.e. t ∈ [0, T ]. (2.45)

Furthermore, if u0 = u1 = 0 there is equality in (2.45).

Proof of Lemma 3. The idea of the proof is the same as in [4, Lemma 2.1, p. 79]. Fix t1, t2,0 < t1 < t2 < T and let v(x, t) be the function defined as follows

v(x, t) = [θm(t)u′(x, t) ∗ ρk(t) ∗ ρk(t)]θm(t), (2.46)

where

(i) θm is a continuous, piecewise linear function on [0, T ] defined as follows:

θm(t) =

⎧⎪⎪⎨⎪⎪⎩0, if, t ∈ [0, T ] � [t1 + 1/m, t2 − 1/m],1, if, t ∈ [t1 + 2/m, t2 − 2/m],m(t − t1 − 1/m), if, t ∈ [t1 + 1/m, t1 + 2/m],−m(t − t2 + 1/m), if, t ∈ [t2 − 2/m, t2 − 1/m].

(ii) {ρk} is a regularizing sequence in C∞c (R), i.e.,

ρk ∈ C∞c (R), ρk(t) = ρk(−t),

∫ +∞

−∞ρk(t)dt = 1, supp ρk ⊂ [−1/k, 1/k].

(iii) (∗) is the convolution product in the time variable, i.e.,

(u ∗ ρk)(x, t) =∫ +∞

−∞u(x, t − s)ρk(s)ds.

We take the scalar product of the function v(x, t) in (2.46) with Eq. (2.44)1, then integratewith respect to the time variable from 0 to T , we have

Xmk + Ymk + Zmk = 0, (2.47)

where⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Xmk =∫ T

0dt∫ 1

0u′′(x, t)v(x, t)dx,

Ymk = −∫ T

0dt∫ 1

0ux x(x, t)v(x, t)dx,

Zmk =∫ T

0dt∫ 1

0Φ1(x, t)v(x, t)dx .

(2.48)


By using the properties of the functions θm(t) and ρk(t) we can show after some lengthycalculation:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

limk→+∞ Xmk = −

∫ T

0θm(t)θ ′

m(t)∥∥u′(t)

∥∥2dt,

limk→+∞ Ymk =

∫ T

0θ2

m(t)[G0(t)u

′(0, t) + G1(t)u′(1, t)

]dt

−∫ T

0θm(t)θ ′

m(t) ‖ux(t)‖2 dt,

limk→+∞ Zmk =

∫ T

0θ2

m(t)dt⟨Φ1(t), u′(t)

⟩dt .

(2.49)

Letting k → ∞, from (2.47) to (2.49) we obtain

1

2

∥∥u′(t2)∥∥2 + 1

2‖ux(t2)‖2 +

∫ t2

t1

[G0(t)u

′(0, t) + G1(t)u′(1, t)

]dt

+∫ t2

t1

⟨Φ1(t), u′(t)

⟩dt = 1

2

∥∥u′(t1)∥∥2 + 1

2‖ux (t1)‖2 ,

a.e., t1, t2 ∈ (0, T ), t1 < t2. (2.50)

From (2.50) we obtain (2.45) by taking t2 = t and passing to the limit as t1 → 0+ and usingthe property of weak lower semicontinuity of two functionals ‖v‖2 and ‖vx‖2. In the case ofu0 = u1 = 0, we prolong u, Φ1, G0, G1 by 0 as t < 0 and we deduce that equality (2.50) is truefor almost t1 < t2 < T . Taking t1 < 0 in (2.49), its right-hand side is 0, we take t1 → 0− andwe have equality (2.45) when u0 = u1 = 0.

The proof of Lemma 3 is completed. �

Remark 3. Lemma 3 is a relative generalization of a lemma contained in Lions’ book [5, Lemma6.1, p. 224].

We now return to the proof of the existence of a solution of the problem (1.1)–(1.5).It follows from (2.7)–(2.10) that∫ t

0

∥∥u′m(s)

∥∥qLq ds +

∫ t

0

∣∣u′m(0, s)

∣∣q0 ds +∫ t

0

∣∣u′m(1, s)

∣∣q1 ds

=∫ t

0

⟨F(s), u′

m(s)⟩ds − 1

2

∥∥u′m(t)

∥∥2 − 1

2‖umx(t)‖2 + 1

2‖u1m‖2

+ 1

2‖u0mx‖2 −

∫ t

0

(Pm(s) − g(s)

)u′

m(0, s)ds −∫ t

0g(s)u′

m(0, s)ds

− K∫ t

0

⟨|um(s)|p−2 um(s), u′

m(s)⟩

ds

−∫ t

0|um(1, s)|p1−2 um(1, s)u′

m(1, s)ds. (2.51)

By Lemma 3 we have

lim supm→∞

[∫ t

0

∥∥u′m(s)

∥∥qLq ds +

∫ t

0

∣∣u′m(0, s)

∣∣q0 ds +∫ t

0

∣∣u′m(1, s)

∣∣q1 ds

]


≤∫ t

0

⟨F(s), u′(s)

⟩ds − 1

2lim infm→∞

∥∥u′m(t)

∥∥2 − 1

2lim infm→∞ ‖umx (t)‖2

+ 1

2‖u1‖2 + 1

2‖u0x‖2 −

∫ t

0

(P(s) − g(s)

)u′(0, s)ds −

∫ t

0g(s)u′(0, s)ds

− K∫ t

0

⟨|u(s)|p−2 u(s), u′(s)

⟩ds −

∫ t

0|u(1, s)|p1−2 u(1, s)u′(1, s)ds

≤ 1

2‖u1‖2 + 1

2‖u0x‖2 − 1

2

∥∥u′(t)∥∥2 − 1

2‖ux(t)‖2

−∫ t

0

(P(s) + χ0(s)

)u′(0, s)ds −

∫ t

0

(|u(1, s)|p1−2 u(1, s) + χ1(s)

)u′(1, s)ds

−∫ t

0

⟨K |u(s)|p−2 u(s) + χ(s) − F(s), u′(s)

⟩ds

+∫ t

0

⟨χ(s), u′(s)

⟩ds +

∫ t

0χ0(s)u

′(0, s)ds +∫ t

0χ1(s)u

′(1, s)ds

≤∫ t

0

⟨χ(s), u′(s)

⟩ds +

∫ t

0χ0(s)u

′(0, s)ds +∫ t

0χ1(s)u

′(1, s)ds. (2.52)

Next, consider

ηm(t) =∫ t

0

⟨Hq

(u′

m(s))− Hq (v(s)) , u′

m(s) − v(s)⟩ds

+∫ t

0

(Hq0

(u′

m(0, s))− Hq0 (v0(s))

) (u′

m(0, s) − v0(s))

ds

+∫ t

0

(Hq1

(u′

m(1, s))− Hq1 (v1(s))

) (u′

m(1, s) − v1(s))

ds ≥ 0, (2.53)

for all (v, v0, v1) ∈ Lq(QT ) × Lq0(0, T ) × Lq1(0, T ), where Hr(z) = |z|r−2 z, r ∈ {q, q0, q1}.It follows from (2.31)2,5,6,7,8,9, (2.52) and (2.53) that

0 ≤ lim supm→∞

ηm(t) ≤∫ t

0

⟨χ(s) − Hq(v(s)), u′(s) − v(s)

⟩ds

+∫ t

0

(χ0(s) − Hq0 (v0(s))

) (u′(0, s) − v0(s)

)ds

+∫ t

0

(χ1(s) − Hq1 (v1(s))

) (u′(1, s) − v1(s)

)ds, (2.54)

for all (v, v0, v1) ∈ Lq(QT ) × Lq0(0, T ) × Lq1(0, T ).In (2.54), we choose v(s) = u′(s) − δξ , v0(s) = u′(0, s) − δξ0, v1(s) = u′(1, s) − δξ1 with

δ > 0 and (ξ, ξ0, ξ1) ∈ Lq (QT ) × Lq0(0, T ) × Lq1(0, T ) and use the argument of Minty and

Browder (cf. Lions [5], p. 172), we deduce that χ(t) = Hq(u′(t)

) = ∣∣u′(t)∣∣q−2

u′(t), χ0(t) =Hq0

(u′(0, t)

) = ∣∣u′(0, t)∣∣q0−2

u′(0, t) and χ1(t) = Hq1

(u′(1, t)

) = ∣∣u′(1, t)∣∣q1−2

u′(1, t).The proof of existence is completed.Step 4. Uniqueness of the solution. Assume now that k ∈ W 1,1(0, T ) in (H3) and p0,

p1 ∈ {2} ∪ [3,+∞). Let (u1, P1), (u2, P2) be two weak solutions of problem (1.1)–(1.5), such


that ⎧⎪⎨⎪⎩ui ∈ L∞(0, T ; H 1), u′

i ∈ L∞(0, T ; L2),

ui (0, ·) ∈ W 1,q0(0, T ), ui (1, ·) ∈ W 1,q1(0, T ),

Pi ∈ Lq ′0(0, T ), q ′

0 = q0

q0 − 1, i = 1, 2.

(2.55)

Then (v, P) with v = u1 −u2 and P = P1 − P2 is the weak solution of the following problem⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

v′′ − vx x + K[Hp (u1) − Hp (u2)

]+ Hq(u′

1

)− Hq(u′

2

) = 0,

0 < x < 1, 0 < t < T,

vx (0, t) = P(t),−vx (1, t) = Q(t),v(x, 0) = vt (x, 0) = 0,

P(t) = P1(t) − P2(t) = K0[Hp0 (u1(0, t)) − Hp0 (u2(0, t))

]+ [

Hq0

(u′

1(0, t))− Hq0

(u′

2(0, t))]−

∫ t

0k(t − s)v(0, s)ds,

Q(t) = [Hp1 (u1(1, t)) − Hp1 (u2(1, t))

]+ [Hq1

(u′

1(1, t))− Hq1

(u′

2(1, t))]

.

(2.56)

By using Lemma 3 with u0 = u1 = 0, Φ1 = K[Hp (u1) − Hp (u2)

] + Hq(u′

1

) − Hq(u′

2

),

G0(t) = P(t), G1(t) = Q(t), we have

Z(t) = −2K∫ t

0

⟨Hp (u1(s)) − Hp(u2(s)), v

′(s)⟩ds

− 2K0

∫ t

0

(Hp0 (u1(0, s)) − Hp0 (u2(0, s))

)v′(0, s)ds

− 2∫ t

0

(Hp1 (u1(1, s)) − Hp1 (u2(1, s))

)v′(1, s)ds

+ 2∫ t

0v′(0, s)ds

∫ s

0k(s − r)v(0, r)dr, a.e. t ∈ [0, T ], (2.57)

where

Z(t) = ∥∥v′(t)∥∥2 + ‖vx (t)‖2 + 2

∫ t

0

⟨Hq

(u′

1(s))− Hq

(u′

2(s)), v′(s)

⟩ds

+ 2∫ t

0

(Hq0

(u′

1(0, s))− Hq0

(u′

2(0, s)))

v′(0, s)ds

+ 2∫ t

0

(Hq1

(u′

1(1, s))− Hq1

(u′

2(1, s)))

v′(1, s)ds. (2.58)

Using the assumption k ∈ W 1,1(0, T ) and the following inequality

∀p ≥ 2, ∃Cp > 0 : (|x |p−2x − |y|p−2y)(x − y) ≥ Cp |x − y|p, ∀x, y ∈ R, (2.59)

it follows from (2.57)–(2.59) that

�(t) ≤ 1

4v2(0, t) + 2K (p − 1)R p−2

T

∫ t

0‖v(s)‖ ∥∥v′(s)

∥∥ ds

+[1 + 2 |k(0)| + 4 ‖k‖2

L2(0,T )+ ∥∥k ′∥∥2

L1(0,T )

] ∫ t

0v2(0, r)dr


− 2K0

∫ t

0

(Hp0 (u1(0, s)) − Hp0 (u2(0, s))

)v′(0, s)ds

− 2∫ t

0

(Hp1 (u1(1, s)) − Hp1 (u2(1, s))

)v′(1, s)ds, (2.60)

where

�(t) ≡ ∥∥v′(t)∥∥2 + ‖vx (t)‖2 + 2Cq

∫ t

0

∥∥v′(s)∥∥q

Lq ds + 2Cq0

∫ t

0

∣∣v′(0, s)∣∣q0 ds

+ 2Cq1

∫ t

0

∣∣v′(1, s)∣∣q1 ds ≤ Z(t), (2.61)

RT = √2 max

i=1,2‖ui‖L∞(0,T ;H1) . (2.62)

On the other hand, by using (2.3) and (2.61), we have

‖v(t)‖2 ≤ t∫ t

0

∥∥v′(s)∥∥2 ds ≤ T

∫ t

0�(s)ds, (2.63)

v2(0, t) ≤ ‖v(t)‖2C0(Ω)

≤ 2 ‖v(t)‖2H1 ≤ 2

[�(t) + T

∫ t

0�(s)ds

]. (2.64)

It follows from (2.60), (2.61), (2.63) and (2.64) that

�(t) ≤ �(1)T

∫ t

0�(s)ds − 4K0

∫ t

0

(Hp0 (u1(0, s)) − Hp0 (u2(0, s))

)v′(0, s)ds

− 4∫ t

0

(Hp1 (u1(1, s)) − Hp1 (u2(1, s))

)v′(1, s)ds, (2.65)

where

�(1)T = T + 4

(√2K (p − 1)R p−2

T + 1 + 2 |k(0)| + 4 ‖k‖2L2(0,T )

+ ∥∥k ′∥∥2L1(0,T )

)× (1 + T 2). (2.66)

Now, we consider four cases for (p0, p1).

Case 1. p0 = p1 = 2.

Hp0 (u1(0, s)) − Hp0 (u2(0, s)) = u1(0, s) − u2(0, s) = v(0, s), (2.67)

Hp1 (u1(1, s)) − Hp1 (u2(1, s)) = u1(1, s) − u2(1, s) = v(1, s). (2.68)

We then have

−4∫ t

0

(Hpi (u1(i, s)) − Hpi (u2(i, s))

)v′(i, s)ds = −2v2(i, t) ≤ 0, i = 0, 1. (2.69)

By Gronwall’s lemma, we obtain from (2.64), (2.65) and (2.69) that � ≡ 0, i.e., u1 ≡ u2.

Case 2. p0 ≥ 3, p1 = 2.

By using integration by parts, it follows that

−4K0

∫ t

0

(Hp0 (u1(0, s)) − Hp0 (u2(0, s))

)v′(0, s)ds


= −4K0

∫ t

0

[∫ 1

0

d

dθ

(Hp0 (u2(0, s) + θv(0, s))

)dθ

]v′(0, s)ds

= −2K0

[∫ 1

0H ′

p0(u2(0, t) + θv(0, t)) dθ

]v2(0, t)

+ 2K0

∫ t

0v2(0, s)ds

[∫ 1

0H ′′

p0(u2(0, s) + θv(0, s))

(u′

2(0, s) + θv′(0, s))

dθ

]

≤∫ t

0ω0(s)v

2(0, s)ds, (2.70)

where

ω0(s) = 2K0(p0 − 1)(p0 − 2)R p0−3T

(∣∣u′1(0, s)

∣∣+ ∣∣u′2(0, s)

∣∣) . (2.71)

Combining (2.64), (2.65), (2.69) with i = 1, and (2.70), we obtain

�(t) ≤∫ t

0

(�

(1)T + 2T ‖ω0‖L1(0,T ) + ω0(s)

)�(s)ds. (2.72)

By Gronwall’s lemma, we deduce from (2.72) that � ≡ 0, i.e., u1 ≡ u2.

Case 3. p0 = 2, p1 ≥ 3. Similar to case 2.

Case 4. p0 ≥ 3, p1 ≥ 3. We obtain from (2.64), (2.65) and (2.69)–(2.72) in a mannercorresponding to the above part that

�(t) ≤∫ t

0

[�

(1)T + 2T ‖ω0‖L1(0,T ) + 2T ‖ω1‖L1(0,T ) + ω0(s) + ω1(s)

]�(s)ds, (2.73)

where

ω1(s) = 2(p1 − 1)(p1 − 2)R p1−3T

(∣∣u′1(1, s)

∣∣+ ∣∣u′2(1, s)

∣∣) . (2.74)

By Gronwall’s lemma, it follows from (2.74) that � ≡ 0, i.e., u1 ≡ u2. Theorem 1 is provedcompletely. �

Next, we strengthen the hypotheses and assume that

(H ′1) u0 ∈ H 2 and u1 ∈ H 1,

(H ′2) F, Ft ∈ L2(QT ),

(H ′3) k ∈ W 1,1(0, T ), g ∈ H 1(0, T ),

(H ′4) K , λ, K0 ≥ 0, K1, λ0, λ1 > 0,

(H ′5) p, q, p0, p1 ≥ 2; q0 = q1 = 2.

Without loss of generality we can suppose that K1 = λ0 = λ1 = 1. Then, we have thefollowing theorem.

Theorem 2. Let (H ′1)–(H ′

5) hold. Then, for every T > 0, there exists a unique weak solution(u, P) of problem (1.1)–(1.5) such that{

u ∈ L∞(0, T ; H 2), ut ∈ L∞(0, T ; H 1), utt ∈ L∞(0, T ; L2),

u(0, ·), u(1, ·) ∈ H 2(0, T ), P ∈ H 1(0, T ).(2.75)


Remark 4. It follows from (2.75) that the component u in the weak solution (u, P) of problem(1.1)–(1.5) satisfies

u ∈ C0(0, T ; H 1) ∩ C1(0, T ; L2) ∩ L∞(0, T ; H 2).

Proof of Theorem 2. The proof consists of steps 1–4.Step 1. The Faedo–Galerkin approximation. Let {w j } be a denumerable base of H 2. We find

the approximate solution of problem (1.1)–(1.5) in the form

um(t) =m∑

j=1

cmj (t)w j , (2.76)

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

m(t),w j⟩+ ⟨

umx (t),w j x⟩+ Pm(t)w j (0) + Qm(t)w j (1)

+⟨K |um(t)|p−2 um(t) + ∣∣u′

m(t)∣∣q−2

u′m(t),w j

⟩= ⟨

F (t),w j⟩, 1 ≤ j ≤ m, (2.77)

Pm(t) = g(t) + K0 |um(0, t)|p0−2 um(0, t) + u′m(0, t)

−∫ t

0k(t − s)um(0, s)ds, (2.78)

Qm(t) = |um(1, t)|p1−2 um(1, t) + u′m(1, t), (2.79)⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

um(0) = u0m =m∑

j=1

αmj w j → u0 strongly in H 2,

u′m(0) = u1m =

m∑j=1

βmjw j → u1 strongly in H 1.

(2.80)

From the assumptions of Theorem 2, system (2.77)–(2.80) has a solution (um, Pm , Qm) on aninterval [0, Tm]. The following estimates allow one to take Tm = T for all m.

Step 2. A priori estimates I. Proceeding as in the proof of Theorem 1, we get, after usingassumptions

(H ′

1

)–(H ′

5

),

Sm(t) ≤ CT , for all t ∈ [0, T ], (2.81)

for all m, where

Sm(t) = ∥∥u′m(t)

∥∥2 + ‖umx (t)‖2 + 2K

p‖um(t)‖p

L p + 2λ

∫ t

0

∥∥u′m(s)

∥∥qLq ds

+ 2K0

p0|um(0, t)|p0 + 2

∫ t

0

∣∣u′m(0, s)

∣∣2 ds + 2

p1|um(1, t)|p1

+ 2∫ t

0

∣∣u′m(1, s)

∣∣2 ds + p1 − 2

p1, (2.82)

and CT is a constant depending only on T , u0, u1, K , p, K0, p0 and p1.A priori estimates II. Now differentiating (2.77) with respect to t , we have⟨

u′′′m (t),w j

⟩+ ⟨u′

mx (t),w j x⟩+ P ′

m(t)w j (0) + Q′m(t)w j (1)

+⟨K (p − 1) |um(t)|p−2 u′

m(t) + (q − 1)∣∣u′

m(t)∣∣q−2

u′′m(t),w j

⟩= 〈F ′ (t),w j 〉, (2.83)

for all 1 ≤ j ≤ m.


Multiplying the j th equation of (2.83) by c′′mj (t), summing with respect to j and then

integrating with respect to the time variable from 0 to t , we have after some rearrangements

Xm(t) = Xm(0) − 2∫ t

0g′(s)u′′

m(0, s)ds

+ 2∫ t

0u′′

m(0, r)

(k(0)um(0, r) +

∫ r

0k ′(r − s)um(0, s)ds

)dr

− 2K0(p0 − 1)

∫ t

0|um(0, s)|p0−2 u′

m(0, s)u′′m(0, s)ds

− 2(p1 − 1)

∫ t

0|um(1, s)|p1−2 u′

m(1, s)u′′m(1, s)ds

− 2K (p − 1)

∫ t

0

⟨|um(s)|p−2 u′

m(s), u′′m(s)

⟩ds + 2

∫ t

0

⟨F ′(s), u′′

m(s)⟩ds, (2.84)

where

Xm(t) = ∥∥u′′m(t)

∥∥2 + ∥∥u′mx (t)

∥∥2 + 2∫ t

0

∣∣u′′m(0, s)

∣∣2 ds + 2∫ t

0

∣∣u′′m(1, s)

∣∣2 ds

+ 8λ(q − 1)

q2

∫ t

0

∥∥∥∥ ∂

∂s

(∣∣u′m(s)

∣∣ q−22 u′

m(s)

)∥∥∥∥2

ds. (2.85)

From the assumptions(H ′

1

),(H ′

2

),(H ′

4

),(H ′

5

)and the imbedding H 1(0, 1) ↪→ L p(0, 1),

p ≥ 1, there exists positive constant D(1)T depending on u0, u1, F , u0, u1, K , λ, p, q , such that

Xm(0) = ∥∥u′′m(0)

∥∥2 + ‖u1mx‖2

≤(‖u0mx x‖ + |K | ‖u0m‖p−1

L2p−2 + |λ| ‖u1m‖q−1L2q−2 + ‖F (0)‖

)2 + ‖u1mx‖2

≤ 1

2D(1)

T , (2.86)

for all m. Using (2.2), (2.13) and (2.81) and the following inequalities

Xm(t) ≥ ∥∥u′′m(t)

∥∥2 + ∥∥u′mx (t)

∥∥2 + 2∫ t

0

∣∣u′′m(0, s)

∣∣2 ds + 2∫ t

0

∣∣u′′m(1, s)

∣∣2 ds, (2.87)

‖um(t)‖C0(Ω)≤ √

2 ‖um(t)‖H1 ≤ √6√

Sm(t) ≤√

6CT ≤ DT , (2.88)∥∥u′m(t)

∥∥C0(Ω)

≤ √2∥∥u′

m(t)∥∥

H1 ≤ √2 (Xm(t) + Sm(t))1/2

≤ √2 (Xm(t) + CT )1/2 ≤ √

2Xm(t) + DT , (2.89)

we estimate, without difficulty, the following terms in the right-hand side of (2.84) as follows

2∫ t

0

⟨F ′(s), u′′

m(s)⟩ds ≤ ∥∥F ′∥∥2

L2(QT )+∫ t

0Xm(s)ds, (2.90)

−2∫ t

0g′(s)u′′

m(0, s)ds ≤ 1

ε

∥∥g′∥∥2L2(0,T )

+ ε

∫ t

0

∣∣u′′m(0, s)

∣∣2 ds

≤ 1

ε

∥∥g′∥∥2L2(0,T )

+ ε

2Xm(t), for all ε > 0, (2.91)


−2K0(p0 − 1)

∫ t

0|um(0, s)|p0−2 u′

m(0, s)u′′m(0, s)ds

≤ 2K0(p0 − 1)(√

6CT

)p0−2∫ t

0

∣∣u′m(0, s)u′′

m(0, s)∣∣ ds

≤ 1

εK 2

0 (p0 − 1)2 (6CT )p0−2∫ t

0

∣∣u′m(0, s)

∣∣2 ds + ε

∫ t

0

∣∣u′′m(0, s)

∣∣2 ds

≤ 1

εK 2

0 (p0 − 1)2 (6CT )p0−2 T CT + ε

2Xm(t), for all ε > 0, (2.92)

−2(p1 − 1)

∫ t

0|um(1, s)|p1−2 u′

m(1, s)u′′m(1, s)ds

≤ 2(p1 − 1)(√

6CT

)p1−2∫ t

0

∣∣u′m(1, s)u′′

m(1, s)∣∣ ds

≤ 1

ε(p1 − 1)2 (6CT )p1−2

∫ t

0

∣∣u′m(1, s)

∣∣2 ds + ε

∫ t

0

∣∣u′′m(1, s)

∣∣2 ds

≤ 1

ε(p1 − 1)2 (6CT )p1−2 T CT + ε

2Xm(t), for all ε > 0, (2.93)

−2K (p − 1)

∫ t

0

⟨|um(s)|p−2 u′

m(s), u′′m(s)

⟩ds

≤ 2K (p − 1)(√

6CT

)p−2∫ t

0

∥∥u′m(s)

∥∥ ∥∥u′′m(s)

∥∥ ds

≤ K 2(p − 1)2 (6CT )p−2∫ t

0

∥∥u′m(s)

∥∥2 ds +∫ t

0

∥∥u′′m(s)

∥∥2 ds

≤ K 2(p − 1)2 (6CT )p−2 T CT +∫ t

0Xm(s)ds, (2.94)

2k(0)

∫ t

0u′′

m(0, r)um(0, r)dr ≤ 1

εk2(0)

∫ t

0|um(0, r)|2 dr + ε

∫ t

0

∣∣u′′m(0, r)

∣∣2 dr

≤ 1

εk2(0)T (6CT ) + ε

2Xm(t), (2.95)

2∫ t

0u′′

m(0, r)dr∫ r

0k ′(r − s)um(0, s)ds

≤ ε

∫ t

0

∣∣u′′m(0, s)

∣∣2 ds + 1

ε

∫ t

0dr

(∫ r

0k ′(r − s)um(0, s)ds

)2

≤ ε

2Xm(t) + 1

ε(6CT ) T

∥∥k ′∥∥2L1(0,T )

. (2.96)

In terms of (2.84)–(2.86) and (2.90)–(2.96), and on choosing ε = 15 , we obtain

Xm(t) ≤ M(2)T + 4

∫ t

0Xm(s)ds, (2.97)

where

M(2)T = D(1)

T + 2∥∥F ′∥∥2

L2(QT )+ 10

∥∥g′∥∥2L2(0,T )

+ 60T CT

(k2(0) + ∥∥k ′∥∥2

L1(0,T )

)


+ 10[

K 20 (p0 − 1)2 (6CT )p0−2 + (p1 − 1)2 (6CT )p1−2

]T CT

+ 2K 2(p − 1)2 (6CT )p−2 T CT . (2.98)

From (2.97) and (2.98) and applying Gronwall’s inequality, we obtain that

Xm(t) ≤ M(2)T exp (4T ) ≤ CT for all t ∈ [0, T ] . (2.99)

On the other hand, from the assumptions(H ′

3

),(H ′

4

), we deduce from (2.78), (2.79) and (2.99)

that ∥∥P ′m

∥∥L2(0,T )

≤ DT , (2.100)∥∥Q′m

∥∥L2(0,T )

≤ DT , (2.101)

where DT is a positive constant depending only on T .Step 3. Limiting process. From (2.81), (2.82), (2.85) and (2.99) to (2.101), we deduce the

existence of a subsequence of {(um, Pm , Qm)} still also so denoted, such that⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

um → u in L∞(0, T ; H 1) weak*,u′

m → u′ in L∞(0, T ; H 1) weak*,u′′

m → u′′ in L∞(0, T ; L2) weak*,um(0, ·) → u(0, ·) in H 2(0, T ) weakly,um(1, ·) → u(1, ·) in H 2(0, T ) weakly,Pm → P in H 1(0, T ) weakly,Qm → Q in H 1(0, T ) weakly.

(2.102)

By the compactness lemma of Lions [5, p. 57] and the imbeddings H 2(0, T ) ↪→ H 1(0, T ),H 1(0, T ) ↪→ C0 ([0, T ]), we can deduce from (2.102) the existence of a subsequence stilldenoted by {(um, Pm , Qm )} such that⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

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

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

um(0, ·) → u(0, ·) strongly in C1([0, T ]),um(1, ·) → u(1, ·) strongly in C1([0, T ]),Pm → P strongly in C0([0, T ]),Qm → Q strongly in C0([0, T ]).

(2.103)

From (2.78) and (2.103)3 we have that

Pm(t) → g(t) + K0 |u(0, t)|p0−2 u(0, t) + u′(0, t) −∫ t

0k(t − s)u(0, s)ds ≡ P(t),

(2.104)

strongly in C0([0, T ]).Combining (2.103)5 and (2.104), we conclude that

P(t) = P(t). (2.105)

Similarly, we have also

Qm(t) → |u(1, t)|p1−2 u(1, t) + u′(1, t) ≡ Q(t) = Q(t), (2.106)

strongly in C0([0, T ]).


We again use inequality (2.33), it follows from (2.81) that∣∣∣|um |p−2 um − |u|p−2 u∣∣∣ ≤ (p − 1)

(√6CT

)p−2 |um − u|. (2.107)

Hence, it follows from (2.103)1, (2.107) that

|um |p−2 um → |u|p−2 u strongly in L2(QT ). (2.108)

In the same way, we deduce from (2.33), (2.99), (2.103)2 that∣∣u′m

∣∣q−2u′

m → ∣∣u′∣∣q−2u′ strongly in L2(QT ). (2.109)

Passing to the limit in (2.77)–(2.79) by (2.102)1,3, (2.104), (2.106), (2.108) and (2.109), wehave (u, P) satisfying the problem⟨

u′′(t), v⟩+ 〈ux(t), vx 〉 + P(t)v(0) + Q(t)v(1) +

⟨K |u|p−2 u + λ

∣∣u′∣∣q−2u′, v

⟩= 〈F (t), v〉, ∀v ∈ H 1, (2.110)

u(0) = u0, u′(0) = u1, (2.111)

P(t) = g(t) + K0 |u(0, t)|p0−2 u(0, t) + u′(0, t) −∫ t

0k(t − s)u(0, s)ds, (2.112)

where

Q(t) = |u(1, t)|p1−2 u(1, t) + u′(1, t). (2.113)

On the other hand, we have from (2.102)1,2,3, (2.110) and assumption(H ′

3

)that

ux x = u′′ + K |u|p−2 u + λ∣∣u′∣∣q−2

u′ − F ∈ L∞(0, T ; L2). (2.114)

Thus u ∈ L∞(0, T ; H 2) and the existence of a solution is proved completely.Step 4. Uniqueness of the solution. Let (u1, P1), (u2, P2) be two weak solutions of problem

(1.1)–(1.5), such that{ui ∈ L∞(0, T ; H 2), u′

i ∈ L∞(0, T ; H 1), u′′i ∈ L∞(0, T ; L2),

ui (0, ·), ui (1, ·) ∈ H 2(0, T ), Pi ∈ H 1(0, T ), i = 1, 2.(2.115)

Then (u, P) with u = u1 − u2 and P = P1 − P2 satisfy the variational problem⎧⎪⎨⎪⎩⟨u′′(t), v

⟩ + 〈ux(t), vx 〉 + P(t)v(0) + Q(t)v(1) + K⟨Hp(u1) − Hp(u2), v

⟩+ λ

⟨Hq(u′

1) − Hq(u′2), v

⟩ = 0 ∀ v ∈ H 1,

u(0) = u′(0) = 0,

(2.116)

and

P(t) = K0[Hp0 (u1(0, t)) − Hp0 (u1(0, t))

]+ u′(0, t) −∫ t

0k(t − s) u(0, s)ds, (2.117)

Q(t) = Hp1 (u1(1, t)) − Hp1 (u1(1, t)) + u′(1, t). (2.118)


We take v = u′ in (2.116)1, and integrating with respect to t , we obtain

σ(t) = −2K0

∫ t

0

[Hp0 (u1(0, s)) − Hp0 (u2(0, s))

]u′(0, s)ds

− 2∫ t

0

[Hp1 (u1(1, s)) − Hp1 (u2(1, s))

]u′(1, s)ds

− 2K∫ t

0

⟨Hp (u1(s)) − Hp(u2(s)), u′(s)

⟩ds

+ 2∫ t

0u′(0, s)ds

∫ s

0k(s − r) u(0, r)dr, (2.119)

where

σ(t) = ∥∥u′(t)∥∥2 + ‖ux (t)‖2 + 2

∫ t

0

∣∣u′(0, s)∣∣2 ds + 2

∫ t

0

∣∣u′(1, s)∣∣2 ds

+ 2λ

∫ t

0

⟨Hq

(u′

1(s)) − Hq(u

′2(s)), u′(s)

⟩ds. (2.120)

Noting that

σ(t) ≥ ∥∥u′(t)∥∥2 + ‖ux(t)‖2 + 2

∫ t

0

∣∣u′(0, s)∣∣2 ds + 2

∫ t

0

∣∣u′(1, s)∣∣2 ds

+ 2λCq

∫ t

0

∥∥u′(s)∥∥q

Lq ds, (2.121)

u2(k, t) ≤ t∫ t

0

∣∣u′(k, s)∣∣2 ds ≤ 1

2tσ(t), k = 0, 1. (2.122)

We again use inequalities (2.33), (2.121) and (2.122), then, it follows from (2.119), (2.121)and (2.122) that

σ(t) ≤ D(2)T

∫ t

0σ(r)dr, (2.123)

where⎧⎪⎪⎨⎪⎪⎩D(2)

T = 4K (p − 1) T R p−2T + 3T

×[

K 20 (p0 − 1)2 R2p0−4

T + (p1 − 1)2 R2p1−4T + ‖k‖2

L1(0,T )

],

RT = √2 max

i=1,2‖ui‖L∞(0,T ;H1).

(2.124)

By Gronwall’s lemma, we deduce that σ(t) ≡ 0 and Theorem 2 is completely proved. �

Remark 5. In the case of p, q > 2 and K < 0, λ < 0, the question of existence for the solutionsof problem (1.1)–(1.5) is still open. However, we have also obtained the answer of problem (1.1),(1.2) and (1.4)–(1.6) when p = q = p0 = q0 = 2 and K , λ ∈ R published in [10].

3. Asymptotic expansion of the solution with respect to three parameters (K, λ, K1)

In this part, we consider two given functions u0, u1 as u0, u1, respectively. Then we assumethat p1 = q0 = q1 = 2; p, q , p0 ≥ N + 1, N ≥ 2, K1 = λ0 = λ1 = 1, and(u0, u1, F, g, k) satisfy the assumptions

(H ′

1

)–(H ′

3

). Let (K , λ, K0) ∈ R

3+. By Theorem 2, the


problem (1.1)–(1.5) has a unique weak solution (u, P) depending on (K , λ, K0):

u = u(K , λ, K0), P = P(K , λ, K0).

We consider the following perturbed problem, where K , λ, K1 are small parameters such that0 ≤ K ≤ K∗, 0 ≤ λ ≤ λ∗, 0 ≤ K0 ≤ K0∗:

(PK ,λ,K0

)⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Lu ≡ u′′ − ux x = −K Hp(u) − λHq(u′) + F(x, t), 0 < x < 1, 0 < t < T,

L0u ≡ ux(0, t) = P(t),L1u ≡ ux(1, t) + u(1, t) + u′(1, t) = 0,

u(x, 0) = u0(x), u′(x, 0) = u1(x),

P(t) = g(t) + K0 Hp0 (u(0, t)) + u′(0, t) −∫ t

0k(t − s)u(0, s)ds.

We shall study the asymptotic expansion of the solution of problem(PK ,λ,K0

)with respect to

(K , λ, K0).We use the following notation. For a multi-index γ = (γ1, γ2, γ3) ∈ Z

3+, and−→K =

(K , λ, K0) ∈ R3+, we put⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

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

γ = K γ1λγ2 K γ30 ,

∥∥∥−→K ∥∥∥ =√

K 2 + λ2 + K 20 ,

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

Cβα = α!

β! (α − β)! .

First, we shall need the following lemma.

Lemma 4. Let m, N ∈ N and vα ∈ R, α ∈ Z3+, 1 ≤ |α| ≤ N. Then( ∑

1≤|α|≤N

vα−→K

α

)m

=∑

m≤|α|≤mN

T (m)[v]α−→K

α, (3.1)

where the coefficients T (m)[v]α , m ≤ |α| ≤ m N, depending on v = (vα), α ∈ Z3+, 1 ≤ |α| ≤ N,

are defined by the recurrence formulas⎧⎪⎪⎨⎪⎪⎩T (1)[v]α = vα, 1 ≤ |α| ≤ N,

T (m)[v]α =∑

β∈I (m)α

vα−βT (m−1)[v]β, m ≤ |α| ≤ m N, m ≥ 2,

I (m)α = {β ∈ Z

3+ : β ≤ α, 1 ≤ |α − β| ≤ N, m − 1 ≤ |β| ≤ (m − 1)N}.(3.2)

Proof of Lemma 4. We shall prove (3.2) by induction.It is easy to see that T (1)[v]α = vα , 1 ≤ |α| ≤ N . Hence, (3.2) holds for m = 1. Suppose that

(3.1)–(3.2) holds for m − 1, we prove that (3.1)–(3.2) holds for m. In fact, we have( ∑1≤|α|≤N

vα−→K

α

)m

=( ∑

1≤|γ |≤N

vγ−→K

γ

)( ∑m−1≤|β|≤(m−1)N

T (m−1)[v]β−→K

β

)

=∑

1≤|γ |≤N

∑m−1≤|β|≤(m−1)N

vγ T (m−1)[v]β−→K

β+γ

=∑

m≤|α|≤mN

∑β∈I (m)

α

vα−βT (m−1)[v]β−→K

α. (3.3)


It follows that

T (m)[v]α =∑

β∈I (m)α

vα−β T (m−1)[v]β, m ≤ |α| ≤ m N, m ≥ 2. (3.4)

The proof of Lemma 4 is complete. �

Let (u0, P0) ≡ (u0,0,0, P0,0,0) be a unique weak solution of problem(P0,0,0

)(as in Theorem 2

corresponding to (K , λ, K0) = (0, 0, 0), i.e.,

(P0,0,0

)⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

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

L0u0 = P0(t), L1u0 = 0,

u0(x, 0) = u0(x), u′0(x, 0) = u1(x),

P0(t) = g(t) + u′0(0, t) −

∫ t

0k(t − s)u0(0, s)ds,

u0 ∈ C0(0, T ; H 1) ∩ C1(0, T ; L2) ∩ L∞(0, T ; H 2),

u′0 ∈ L∞(0, T ; H 1), u′′

0 ∈ L∞(0, T ; L2),

u0(0, ·), u0(1, ·) ∈ H 2(0, T ), P0 ∈ H 1(0, T ).

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

), γ ∈ Z

3+, 1 ≤ |γ | ≤ N , defined bythe following problems:

(Pγ

)⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Luγ = Fγ , 0 < x < 1, 0 < t < T,

L0uγ = Pγ (t), L1uγ = 0,

uγ (x, 0) = u′γ (x, 0) = 0,

Pγ (t) = Pγ (t) + u′γ (0, t) −

∫ t

0k(t − s)uγ (0, s)ds,

uγ ∈ C0(0, T ; H 1) ∩ C1(0, T ; L2) ∩ L∞(0, T ; H 2),

u′γ ∈ L∞(0, T ; H 1), u′′

γ ∈ L∞(0, T ; L2),

uγ (0, ·), uγ (1, ·) ∈ H 2(0, T ), Pγ ∈ H 1(0, T ),

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

Fγ =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

F, |γ | = 0,

0, γ1 = γ2 = 0, 1 ≤ |γ | ≤ N,

−Hp(u0), γ1 = 1, |γ | = 1,

−|γ |−1∑m=1

1

m! H (m)p (u0) T (m)[u]γ1−1,0,γ3, γ1 ≥ 1, γ2 = 0, 2 ≤ |γ | ≤ N,

−Hq(u′0), γ2 = 1, |γ | = 1,

−|γ |−1∑m=1

1

m! H (m)q

(u′

0

)T (m)[u′]0,γ2−1,γ3, γ2 ≥ 1, γ1 = 0, 2 ≤ |γ | ≤ N,

−|γ |−1∑m=1

1

m![

H (m)p (u0) T (m)[u]γ1−1,γ2,γ3 + H (m)

q

(u′

0

)T (m)[u′]γ1,γ2−1,γ3

],

γ1 ≥ 1, γ2 ≥ 1, 2 ≤ |γ | ≤ N,

(3.5)


Pγ (t) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

g(t), |γ | = 0,

0, γ3 = 0, 1 ≤ |γ | ≤ N,

Hp0(u0(0, t)), γ3 = 1, |γ | = 1,|γ |−1∑m=1

1

m! H (m)p0

(u0(0, t)) T (m)[u(0, t)]γ1,γ2,γ3−1, γ3 ≥ 1, 2 ≤ |γ | ≤ N,

(3.6)

and here we have used the notation u = (uγ ), |γ | ≤ N .Let (u, P) = (uK ,λ,K0, PK ,λ,K0) be a unique weak solution of problem

(PK ,λ,K0

). Then

(v, R), with

v = u −∑

|γ |≤N

uγ−→K

γ ≡ u − h, R = P −∑

|γ |≤N

Pγ−→K

γ, (3.7)

satisfies the problem⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Lv = −K[Hp(v + h) − Hp(h)

]− λ[Hq(v′ + h′) − Hq(h′)

]+ EN (−→K ),

0 < x < 1, 0 < t < T,

L0v = R(t), L1v = 0,

R(t) = K0[Hp0[(v + h)(0, t)] − Hp0[h(0, t)]]+ v′(0, t)

− ∫ t0 k(t − s)v(0, s)ds + EN (

−→K ),

v(x, 0) = v′(x, 0) = 0,

v ∈ C0(0, T ; H 1) ∩ C1(0, T ; L2) ∩ L∞(0, T ; H 2),

v′ ∈ L∞(0, T ; H 1), v′′ ∈ L∞(0, T ; L2),

v(0, ·), v(1, ·) ∈ H 2(0, T ), R ∈ H 1(0, T ),

(3.8)

where

EN (−→K ) = F(x, t) − K Hp(h) − λHq(h′) −

∑|γ |≤N

Fγ−→K

γ, (3.9)

EN (−→K ) = K0 Hp0[h(0, t)] −

∑1≤|γ |≤N

Pγ (t)−→K

γ. (3.10)

Then, we have the following lemma.

Lemma 5. Let p, q, p0 ≥ N + 1, N ≥ 2, p1 = q1 = q0 = 2, K1 = λ1 = λ0 = 1 and(H ′

1

)–(

H ′3

)hold. Then∥∥∥EN (

−→K )

∥∥∥L∞(0,T ;L2)

≤ C1N

∥∥∥−→K ∥∥∥N+1, (3.11)∥∥∥EN (

−→K )

∥∥∥L2(0,T )

≤ C2N

∥∥∥−→K ∥∥∥N+1, (3.12)

for all−→K = (K , λ, K0) ∈ R

3+,∥∥∥−→K ∥∥∥ ≤

∥∥∥−→K∗∥∥∥ with

−→K∗ = (K∗, λ∗, K0∗), 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. Put

h = u0 + h1, h1 =∑

1≤|γ |≤N

uγ−→K

γ. (3.13)


By using Taylor’s expansion of the function Hp(h) = Hp (u0 + h1) around the point u0 upto order N − 1, we obtain

Hp(h) = Hp(u0) +N−1∑m=1

1

m! H (m)p (u0)h

m1 + 1

N ! H (N)p (u0 + θ1h1)h

N1 , (3.14)

where 0 < θ1 < 1. By Lemma 4, we obtain from (3.14), after some rearrangements in the orderof

−→K

γ, that

K Hp(h) = K Hp(u0) +∑

2≤|γ |≤N,γ1≥1

|γ |−1∑m=1

1

m! H (m)p (u0)T (m)[u]γ1−1,γ2,γ3

−→K

γ

+ R(1)(p,−→K ), (3.15)

where

R(1)(p,−→K ) = K

N−1∑m=1

1

m! H (m)p (u0)

×∑

N≤|γ |≤mN

T (m)[u]γ −→K

γ + 1

N ! H (N)p (u0 + θ1h1)K hN

1 . (3.16)

Similarly, we use Taylor’s expansion of the functions Hq(h′) = Hq(u′

0 + h′1

), Hp0(h) =

Hp0(u0 + h1), up to order N − 1, we obtain

λHq(h′) = λHq(u′0) +

∑2≤|γ |≤N,γ2≥1

|γ |−1∑m=1

1

m! H (m)q (u′

0)T (m)[u′]γ1,γ2−1,γ3

−→K

γ

+ R(2)(q,−→K ), (3.17)

K0 Hp0(h) = K0 Hp0(u0) +∑

2≤|γ |≤N,γ3≥1

|γ |−1∑m=1

1

m! H (m)p0

(u0)T (m)[u]γ1,γ2,γ3−1−→K

γ

+ R(3)(p0,−→K ), (3.18)

where

R(2)(q,−→K ) = λ

N−1∑m=1

1

m! H (m)q (u′

0)

×∑

N≤|γ |≤mN

T (m)[u′]γ −→K

γ + λ1

N ! H (N)q (u′

0 + θ2h′1)(h

′1)

N , (3.19)

R(3)(p0,−→K ) = K0

N−1∑m=1

1

m! H (m)p0

(u0)

×∑

N≤|γ |≤mN

T (m)[u]γ −→K

γ + K01

N ! H (N)p0

(u0 + θ3h1)(h1)N , (3.20)

and 0 < θi < 1, i = 2, 3. Combining (3.5), (3.9), (3.15)–(3.20), we then obtain

EN (−→K ) = F(x, t) − K Hp(u0) − λHq(u′

0)


−∑

2≤|γ |≤N,γ1≥1

|γ |−1∑m=1

1

m! H (m)p (u0)T (m)[u]γ1−1,γ2,γ3

−→K

γ

−∑

2≤|γ |≤N,γ2≥1

|γ |−1∑m=1

1

m! H (m)q (u′

0)T (m)[u′]γ1,γ2−1,γ3

−→K

γ

−∑

|γ |≤N

Fγ−→K

γ − R(1)(p,−→K ) − R(2)(q,

−→K )

= −R(1)(p,−→K ) − R(2)(q,

−→K ). (3.21)

We shall estimate the following terms on the right-hand side of (3.21).Estimating R(1)(p,

−→K ).

By the boundedness of the functions uγ , γ ∈ Z3+, |γ | ≤ N , in the function space

L∞(0, T ; H 1), we obtain from (3.16) that∥∥∥R(1)(p,−→K )

∥∥∥L∞(0,T ;L2)

≤ |K |N−1∑m=1

∑N≤|γ |≤mN

1

m!∥∥∥H (m)

p (u0)

∥∥∥L∞(0,T ;H1)

∥∥∥T (m)[u]γ∥∥∥

L∞(0,T ;L2)

∣∣∣−→K γ∣∣∣

+ 1

N ! K∥∥∥H (N)

p (u0 + θ1h1)

∥∥∥L∞(0,T ;H1)

‖h1‖NL∞(0,T ;H1)

. (3.22)

Using the inequality∣∣∣−→K γ∣∣∣ ≤

∥∥∥−→K ∥∥∥|γ |, for all γ ∈ Z

3+, (3.23)

it follows from (3.22), (3.23) that∥∥∥R(1)(p,−→K )

∥∥∥L∞(0,T ;L2)

≤ C(1)1N

∥∥∥−→K ∥∥∥N+1,

∥∥∥−→K ∥∥∥ ≤∥∥∥−→K∗

∥∥∥ , (3.24)

where

C(1)1N =

N−1∑m=1

Cmp−1

(√2 ‖u0‖L∞(0,T ;H1)

)p−m−1

×∑

N≤|γ |≤mN

∥∥∥T (m)[u]γ∥∥∥

L∞(0,T ;L2)

∥∥∥−→K∗∥∥∥|γ |−N

+ C Np−1

∥∥∥−→K∗∥∥∥−N

( ∑|γ |≤N

√2∥∥uγ

∥∥L∞(0,T ;H1)

∥∥∥−→K∗∥∥∥|γ |

)p−1

, (3.25)

and−→K∗ = (K∗, λ∗, K0∗), Cm

p−1 = (p−1)(p−2)···(p−m)m! .

Estimating R(2)(q,−→K ). We obtain from (3.19) in a manner corresponding to the above part

that ∥∥∥R(2)(q,−→K )

∥∥∥L∞(0,T ;L2)

≤ C(2)1N

∥∥∥−→K ∥∥∥N+1,

∥∥∥−→K ∥∥∥ ≤ ‖−→K∗‖, (3.26)


where

C(2)1N =

N−1∑m=1

Cmq−1

(√2∥∥u′

0

∥∥L∞(0,T ;H1)

)q−m−1

×∑

N≤|γ |≤mN

∥∥∥T (m)[u′]γ∥∥∥

L∞(0,T ;L2)

∥∥∥−→K∗∥∥∥|γ |−N

+ C Nq−1

∥∥∥−→K∗∥∥∥−N

( ∑|γ |≤N

√2∥∥∥u′

γ

∥∥∥L∞(0,T ;H1)

∥∥∥−→K∗∥∥∥|γ |

)q−1

. (3.27)

Therefore, it follows from (3.21), (3.24)–(3.27) that∥∥∥EN (−→K )

∥∥∥L∞(0,T ;L2)

≤(

C(1)1N + C(2)

1N

) ∥∥∥−→K ∥∥∥N+1 ≡ C1N

∥∥∥−→K ∥∥∥N+1,

∥∥∥−→K ∥∥∥ ≤∥∥∥−→K∗

∥∥∥ .

(3.28)

Hence, the first part of Lemma 5 is proved.With EN (

−→K ), then, we obtain from (3.6), (3.10), (3.18), (3.20), in a similar manner to the

above, that

EN (−→K ) = R(3)(p0,

−→K ). (3.29)

Hence∥∥∥EN (−→K )

∥∥∥L2(0,T )

≤ C2N

∥∥∥−→K ∥∥∥N+1, (3.30)

where

C2N =N−1∑m=1

Cmp0−1 ‖u0(0, ·)‖p0−m−1

L∞(0,T )

∑N≤|γ |≤mN

∥∥∥T (m)[u(0, ·)]γ∥∥∥

L2(0,T )

∥∥∥−→K∗∥∥∥|γ |−N

+ C Np0−1

∥∥∥−→K∗∥∥∥−N

( ∑|γ |≤N

∥∥uγ (0, ·)∥∥L∞(0,T )

∥∥∥−→K∗∥∥∥|γ |

)p0−1

. (3.31)

The proof of Lemma 5 is complete. �Next, we obtain the following theorem.

Theorem 3. Let p, q, p0 ≥ N + 1, N ≥ 2, p1 = q1 = q0 = 2, K1 = λ1 = λ0 = 1 and(H ′

1

)–(

H ′3

)hold. Then, for every

−→K ∈ R

3+, with 0 ≤ K ≤ K∗, 0 ≤ λ ≤ λ∗, 0 ≤ K0 ≤ K0∗, problem(PK ,λ,K0

)has a unique weak solution (u, P) = (uK ,λ,K0, PK ,λ,K0) satisfying the 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)

+∥∥∥∥∥u′(0, ·) −

∑|γ |≤N

u′γ (0, ·)−→K γ

∥∥∥∥∥L2(0,T )

+∥∥∥∥∥u′(1, ·) −

∑|γ |≤N

u′γ (1, ·)−→K γ

∥∥∥∥∥L2(0,T )

≤ C∗N

∥∥∥−→K ∥∥∥N+1, (3.32)


and ∥∥∥∥∥P −∑

|γ |≤N

Pγ−→K

γ

∥∥∥∥∥L2(0,T )

≤ C∗∗N

∥∥∥−→K ∥∥∥N+1, (3.33)

for all−→K ∈ R

3+,∥∥∥−→K ∥∥∥ ≤

∥∥∥−→K∗∥∥∥, C∗

N and C∗∗N are positive constants independent of

−→K , the

functions (uγ , Pγ ) are the weak solutions of problems(Pγ

), γ ∈ Z

3+, |γ | ≤ N.

Proof. First, we note that, if the data−→K = (K , λ, K0) satisfy

0 ≤ K ≤ K∗, 0 ≤ λ ≤ λ∗, 0 ≤ K0 ≤ K0∗, (3.34)

where K∗, λ∗, K0∗ are fixed positive constants, then, the a priori estimates of the sequences {um}and {Pm} in the proof of Theorem 2 satisfy∥∥u′

m(t)∥∥2 + ‖umx (t)‖2 + 2

∫ t

0

∣∣u′m(0, s)

∣∣2 ds + 2∫ t

0

∣∣u′m(1, s)

∣∣2 ds ≤ CT ,

∀t ∈ [0, T ], (3.35)∥∥u′′m(t)

∥∥2 + ∥∥u′mx (t)

∥∥2 + 2∫ t

0

∣∣u′′m(0, s)

∣∣2 ds + 2∫ t

0

∣∣u′′m(1, s)

∣∣2 ds ≤ CT ,

∀t ∈ [0, T ], (3.36)

‖Pm‖H1(0,T ) ≤ CT , (3.37)

where CT is a constant depending only on T , u0, u1, F , g, k, p, q , p0, K∗, λ∗, K0∗ (independentof

−→K ). Hence, the limit (u, P) in suitable function spaces of the sequence {(um, Pm)} defined

by (2.77)–(2.80) is a weak solution of the problem (1.1)–(1.5) satisfying the a priori estimates(3.35)–(3.37).

On multiplying the two sides of (3.8)1 by v′, after integration in t , we find without difficultyfrom Lemma 4 that

σ(t) ≤ 2(

T C21N + 3C2

2N

) ∥∥∥−→K ∥∥∥2N+2 + 2(

2 + 3T ‖k‖2L2(0,T )

) ∫ t

0σ(s)ds

+ 2K 2∫ t

0

∥∥Hp(v + h) − Hp(h)∥∥2

ds

+ 6K 20

∫ t

0

∣∣Hp0 (v(0, s) + h(0, s)) − Hp0 (h(0, s))∣∣2 ds, (3.38)

where

σ(t) = ∥∥v′(t)∥∥2 + ‖vx (t)‖2 + v2(1, t) + 2

∫ t

0

∣∣v′(0, s)∣∣2 ds

+ 2∫ t

0

∣∣v′(1, s)∣∣2 ds + 2λ

∫ t

0

⟨Hq(v′ + h′) − Hq(h′), v′⟩ ds. (3.39)

By using the same arguments as in the above part we can show that the component u of theweak solution (u, P) of problem

(PK ,λ,K0

)satisfies∥∥u′(t)

∥∥2 + ‖ux(t)‖2 + 2∫ t

0

∣∣u′(0, s)∣∣2 ds + 2

∫ t

0

∣∣u′(1, s)∣∣2 ds ≤ CT ,

∀t ∈ [0, T ], (3.40)


where CT is a constant independent of K , λ, K1. On the other hand,

‖h‖L∞(0,T ;H1) ≤∑

|γ |≤N

∥∥uγ

∥∥L∞(0,T ;H1)

∥∥∥−→K∗∥∥∥|γ | ≡ 1√

2R1. (3.41)

We again use inequality (2.29) with γ = p − 2, R2 = max{R1,

√2(1 + T 2

)CT }, then, it

follows from (3.39) to (3.41) that

2K 2∫ t

0

∥∥Hp(v + h) − Hp(h)∥∥2

ds ≤ 2K 2∗(p − 1)2 R2p−42

∫ t

0σ(s)ds. (3.42)

Similarly,

6K 20

∫ t

0

∣∣Hp0 (v(0, s) + h(0, s)) − Hp0 (h(0, s))∣∣2 ds

≤ 6K 2∗0(p0 − 1)2 R2p0−42

∫ t

0σ(s)ds. (3.43)

Combining (3.38), (3.42) and (3.43), we then obtain

σ(t) ≤ K (1)T

∥∥∥−→K ∥∥∥2N+2 + K (2)T

∫ t

0σ(s)ds, (3.44)

for all t ∈ [0, T ], where⎧⎪⎨⎪⎩K (1)

T = 2(

T C21N + 3C2

2N

),

K (2)T = 2

[2 + 3T ‖k‖2

L2(0,T )+ K 2∗ (p − 1)2 R2p−4

2 + 3K 2∗0(p0 − 1)2 R2p0−42

].

(3.45)

By Gronwall’s lemma, we obtain from (3.44) that

σ(t) ≤ K (1)T

∥∥∥−→K ∥∥∥2N+2exp

(T K (2)

T

)≡ D(1)

T

∥∥∥−→K ∥∥∥2N+2, ∀t ∈ [0, T ], (3.46)


3+,∥∥∥−→K ∥∥∥ ≤

∥∥∥−→K∗∥∥∥. It follows that

∥∥v′(t)∥∥2 + ‖vx (t)‖2 + 2

∫ t

0

∣∣v′(0, s)∣∣2 ds + 2

∫ t

0

∣∣v′(1, s)∣∣2 ds ≤ σ(t)

≤ D(1)T

∥∥∥−→K ∥∥∥2N+2. (3.47)

Hence∥∥v′∥∥L∞(0,T ;L2)

+ ‖v‖L∞(0,T ;H1) + ∥∥v′(0, ·)∥∥L2(0,T )+ ∥∥v′(1, ·)∥∥L2(0,T )

≤ C∗N

∥∥∥−→K ∥∥∥N+1, (3.48)

or ∥∥∥∥∥u′ −∑

|γ |≤N

u′γ−→K

γ

∥∥∥∥∥L∞(0,T ;L2)

+∥∥∥∥∥u −

∑|γ |≤N

uγ−→K

γ

∥∥∥∥∥L∞(0,T ;H1)

+∥∥∥∥∥u′(0, ·) −

∑|γ |≤N

u′γ (0, ·)−→K γ

∥∥∥∥∥L2(0,T )


+∥∥∥∥∥u′(1, ·) −

∑|γ |≤N

u′γ (1, ·)−→K γ

∥∥∥∥∥L2(0,T )

≤ C∗N

∥∥∥−→K ∥∥∥N+1, (3.49)


3+,∥∥∥−→K ∥∥∥ ≤

∥∥∥−→K∗∥∥∥, where C∗

N is a constant independent of−→K .

On the other hand, it follows from (3.12), (3.48) that

‖R‖L2(0,T ) ≤ K∗0(p0 − 1)R p0−22

(∫ T

0|v(0, t)|2 ds

)1/2

+(∫ T

0

∣∣v′(0, t)∣∣2 dt

)1/2

+ √T ‖k‖L2(0,T )

(∫ T

0σ(s)ds

)1/2

+∥∥∥EN (

−→K )

∥∥∥L2(0,T )

≤ C∗∗N

∥∥∥−→K ∥∥∥N+1, (3.50)

and hence,∥∥∥∥∥P −∑

|γ |≤N

Pγ−→K

γ

∥∥∥∥∥L2(0,T )

≤ C∗∗N

∥∥∥−→K ∥∥∥N+1, (3.51)

where C∗∗N is a constant independent of

−→K . The proof of Theorem 3 is complete. �

Remark 6. In [9], as in this special case for problem (1.1)–(1.5) with p = q = p0 = q0 =p1 = q1 = 2, Long, Alain, Diem have obtained the result about the asymptotic expansion of thesolutions with respect to two parameters (K , λ) up to order N + 1.

Acknowledgements

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

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a nonlinear wave equation associated with nonlinear boundary conditions: existence and asymptotic...

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