on the nonlinear wave equation utt−b(∥ux∥2)uxx=f(x,t,u,ux,ut,∥ux∥2) associated with the...

Nonlinear Analysis 55 (2003) 493–519www.elsevier.com/locate/na

On the nonlinear wave equationutt − B(||ux||2)uxx = f(x; t; u; ux; ut; ||ux||2)

associated with the mixedhomogeneous conditions

Nguyen Thanh Longa ;∗, Bui Tien DungbaDepartment of Mathematics and Computer Science, University of Natural Science,

Vietnam National University, HoChiMinh City, 227 Nguyen Van Cu Str.,Dist. 5, HoChiMinh City, Viet Nam

bDepartment of Mathematics, University of Architecture of HoChiMinh City,196 Pasteur Str., Dist. 3, HoChiMinh City, Viet Nam

Received 10 March 2003; accepted 7 July 2003

Abstract

In this paper, we consider the following nonlinear wave equation:

(1) utt − B(‖ux‖2)uxx = f(x; t; u; ux; ut ; ‖ux‖2); x∈ (0; 1); 0¡t¡T ,(2) ux(0; t)− h0u(0; t) = u(1; t) = 0,(3) u(x; 0) = u 0(x), ut(x; 0) = u 1(x),

where B; f; u 0; u 1 are given functions. In Eq. (1), the nonlinear terms B(‖ux‖2); f(x; t; u; ux;ut ; ‖ux‖2) depending on an integral ‖ux‖2 =

∫ 10 |ux(x; t)|2 dx. In this paper, we associate with

problem (1)–(3) a linear recursive scheme for which the existence of a local and unique solutionis proved by using standard compactness argument. In case of B∈CN+1(R+); B¿ b0¿ 0; B1 ∈CN (R+); B1¿ 0; f∈CN+1([0; 1] × R+ × R3 × R+) and f1 ∈CN ([0; 1] × R+ × R3 × R+) weobtain from the following equation: utt − [B(‖ux‖2) + �B1(‖ux‖2)]uxx = f(x; t; u; ux; ut ; ‖ux‖2) +�f1(x; t; u; ux; ut ; ‖ux‖2) associated to (2), (3) a weak solution u�(x; t) having an asymptoticexpansion of order N + 1 in �, for � su:ciently small.? 2003 Elsevier Ltd. All rights reserved.

Keywords: Galerkin method; Linear recurrent sequence; Asymptotic expansion of order N + 1

∗ Corresponding author. Department of Mathematics and Computer Science, University of Natural Science,Vietnam National University, HoChiMinh City, 227 Nguyen Van Cu Str., Dist. 5, HoChiMinh City 7000,Viet Nam.

E-mail address: [email protected] (N.T. Long).

0362-546X/$ - see front matter ? 2003 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2003.07.002

mailto:[email protected]

494 N.T. Long, B.T. Dung /Nonlinear Analysis 55 (2003) 493–519

1. Introduction

In this paper, we consider a nonlinear wave equation with the KirchhoH operator

utt − B(‖∇u‖2)Iu= f(x; t; u; ux; ut ; ‖∇u‖2); x∈� = (0; 1); 0¡t¡T; (1.1)

ux(0; t)− h0u(0; t) = u(1; t) = 0; (1.2)

u(x; 0) = u 0(x); ut(x; 0) = u 1(x); (1.3)

where B; f; u 0; u 1 are given functions satisfying conditions speciJed later and h0¿ 0 isa given constant. In Eq. (1.1), the nonlinear terms f(x; t; u; ux; ut ; ‖∇u‖2) and B(‖∇u‖2)depend on the integral

‖∇u‖2 =∫�|∇u(x; t)|2 dx: (1.4)

Eq. (1.1) has its origin in the nonlinear vibration of an elastic string (KirchhoH [6]),for which the associated equation is

�hutt =

[P0 +

Eh2L

∫ L

0

∣∣∣∣ @u@y (y; t)∣∣∣∣2

dy

]uxx; 0¡x¡L; 0¡t¡T: (1.5)

Here u is the lateral deKection, � is the mass density, h is the cross section, L is thelength, E is Young’s modulus and P0 is the initial axial tension. When f = 0, theCauchy or mixed problem for (1.1) has been studied by many authors; see Ebiharaet al. [4], Pohozaev [18] and the references therein. A survey of the results about themathematical aspects of KirchhoH model can be found in Medeiros et al. [15,16].In [14] Medeiros has studied problem (1.1)–(1.3) with f = f(u) = −bu2, where b

is a given positive constant, and � is a bounded open set of R3. In [5] Hosoya andYamada have considered (1.1)–(1.3) with f=f(u)=−�|u|�u, where �¿ 0; �¿ 0 aregiven constants.In [11,12] the authors have studied the existence and uniqueness of the equation

utt + �I2u− B(‖∇u‖2)Iu+ �|ut |�−1ut = F(x; t); x∈�; t ¿ 0; (1.6)

where �¿ 0; �¿ 0; 0¡�¡ 1, are given constants, and � is a bounded open setof Rn.In [2], Alain Pham has studied the existence and asymptotic behavior as � → 0

of a weak solution of problem (1.1), (1.3) with B ≡ 1 associated with the Dirichlethomogeneous boundary condition

u(0; t) = u(1; t) = 0; (1.7)

where the nonlinear term has the form f= �f1(t; u). By a generalization of [2], AlainPham and Long [3] have considered problem (1.1), (1.3), (1.7) with B ≡ 1 and thenonlinear term having the form

f = �f1(t; u; ut): (1.8)

N.T. Long, B.T. Dung /Nonlinear Analysis 55 (2003) 493–519 495

If B� ≡ 1 and f1 ∈CN ([0;∞)×R2) satisJes f1(t; 0; 0)=0 for all t¿ 0, an asymptoticexpansion of the solution of problem (1.1), (1.3), (1.7), (1.8) up to order N + 1 in �is obtained, for � su:ciently small. This expansion extends to the partial diHerentialequation the results obtained in diHerential equations [1].In [10] Long and Diem have studied the linear recursive schemes and asymptotic

expansion associated with the nonlinear wave equation

utt − uxx = f(x; t; u; ux; ut) + �f1(x; t; u; ux; ut); (1.9)

associated with (1.3) and the mixed homogeneous condition

ux(0; t)− h0u(0; t) = u(1; t) + h1u(1; t) = 0: (1.10)

In the case of f∈C2([0; 1] × [0;∞) × R3) and f1 ∈C1([0; 1] × [0;∞) × R3), wehave been obtained an asymptotic expansion of order 2 in �, for � su:ciently small.Afterwards, this result has been extended in [13] to the nonlinear wave equation withthe KirchhoH operator

utt −[b0 + B

(∫ 1

0|ux(y; t)|2 dy

)+ �B1

(∫ 1

0|ux(y; t)|2 dy

)]uxx

=f(x; t; u; ux; ut) + �f1(x; t; u; ux; ut); (1.11)

associated with (1.3), (1.7), where b0¿ 0 is a given constant and B∈C2(R+); B1 ∈C1(R+); B¿ 0; B1¿ 0 are given functions.In this paper, we shall Jrst associate with problem (1.1)–(1.3) a linear recurrent

sequence which is bounded in a suitable space of functions. The existence of a localsolution is proved by a standard compactness argument. Note that the linearizationmethod in this paper and in the papers [3,9,10,13,17] cannot be used in the papers[4,5,11,12,14]. If B∈CN+1(R+); B1 ∈CN (R+); B¿ b0¿ 0; B1¿ 0; f∈CN+1([0; 1] ×R+ × R3 × R+) and f1 ∈CN ([0; 1] × R+ × R3 × R+), then an asymptotic expansionof order N +1 in � is obtained with a right-hand side of the form f(x; t; u; ux; ut ; ‖ux‖2)+ �f1(x; t; u; ux; ut ; ‖ux‖2) and B stand for B+ �B1, for � su:ciently small. This resultis a relative generalization of [3,8,10,13,17]

2. Preliminary results, notations

We will omit the deJnitions of the usual function spaces and denote them by thenotation Lp = Lp(0; 1); Hm = Hm(0; 1); Hm

0 = Hm0 (0; 1).

The norm in L2 is denoted by ‖ · ‖. We also denote by 〈·; ·〉 the scalar product inL2, or a pair of dual scalar products of continuous linear functional with an elementof a function space. We denote by ‖ · ‖X the norm in the Banach space X . We callX ′ the dual space of X . We denote by Lp(0; T ;X ), 16p6∞, the Banach space ofreal functions u : (0; T ) → X measurable, such that

‖u‖Lp(0;T ;X ) =(∫ T

0‖u(t)‖pX dt

)1=p¡+∞ for 16p¡∞


and

‖u‖L∞(0;T ;X ) = ess sup0¡t¡T

‖u(t)‖X for p=∞:

Let u(t); ut(t) = u(t); utt(t) = Su(t); ux(t) = ∇u(t); uxx(t) = Iu(t) denote u(x; t);(@u=@t)(x; t), (@2u=@t2)(x; t), (@u=@x)(x; t), (@2u=@x2)(x; t), respectively.

With f=f(x; t; u; v; w; z), we put D1f= @f=@x; D2f= @f=@t, D3f= @f=@u, D4f=@f=@v, D5f = @f=@w, D6f = @f=@z. We put

V = {v∈H 1(0; 1) : v(1) = 0}; (2.1)

a(u; v) =∫ 1

0ux(x)vx(x) dx + h0u(0)v(0): (2.2)

V is a closed subspace of H 1 and on V three norms ‖v‖H 1 ; ‖vx‖ and ‖v‖V =√a(v; v)

are equivalent norms.Then we have the following lemmas, the proofs of which are straightforward and

their omitted:

Lemma 1. Let h0¿ 0. Then the imbedding V ,→ C0([0; 1]) is compact and

‖v‖C0([0;1])6 ‖vx‖6 ‖v‖V1√2‖v‖H 1 6 ‖vx‖6 ‖v‖V 6max(1;

√h0)‖v‖H 1

for all v∈V: (2.3)

Lemma 2. Let h0¿ 0. Then the symmetric bilinear form a(·; ·) de6ned by (2.2) iscontinuous on V × V and coercive on V .

Lemma 3. Let h0¿ 0. Then there exists the Hilbert orthonormal base {wj} of L2

consisting of the eigenfunctions wj corresponding to the eigenvalue �j such that

0¡�16 �26 · · ·6 �j6 · · · ; limj→+∞

�j =+∞; (2.4)

a(wj; v) = �j〈wj; v〉 for all v∈V; j = 1; 2; : : : : (2.5)

Furthermore, the sequence {wj=√�j} is also the Hilbert orthonormal base of V with

respect to the scalar product a(·; ·). On the other hand, we have also wj satisfyingthe boundary value problem

−Iwj = �jwj in �;

wjx(0)− h0wj(0) = wj(1) = 0;

wj ∈V ∩ C∞([0; 1]):

(2.6)

The proof of Lemma 3 can be found in [19, p. 137, Theorem 6.2.1], with H = L2,and V; a(·; ·) as deJned by (2.1), (2.2).


3. The existence and uniqueness theorem

We make the following assumptions:

(H1) h0¿ 0;(H2) u 0 ∈V ∩ H 2; u 1 ∈V ;(H3) B∈C1(R+); B(z)¿ b0¿ 0;(H4) f∈C0([0; 1]× R+ × R3 × R+) satisJes the conditions

(H′4) f(1; t; u; v; w; z) = 0 for all t; z¿ 0 and (u; v; w)∈R3.

(H′′4 ) Dif∈C0([0; 1]× R+ × R3 × R+); i = 1; 3; 4; 5; 6.

(It is not necessary that f∈C1([0; 1]× R+ × R3 × R+):)With B and f satisfying assumptions (H3) and (H4), respectively, we introduce the

following constants, for all M ¿ 0 and T ¿ 0:

K0 = K0(M; T; f) = sup|f(x; t; u; v; w; z)|; (3.1)

K1 = K1(M; T; f) = sup

(|D1f|+

6∑i=3

|Dif|)(x; t; u; v; w; z); (3.2)

where, in each case, sup is taken over 06 t6T; 06 x6 1; |u| + |v| + |w|6M;06 z6M 2.

K0 = K0(M;B) = sup06z6M 2

B(z); (3.3)

K1 = K1(M;B) = K0(M; |B′|) = sup06z6M 2

|B′(z)|: (3.4)

For each M ¿ 0 and T ¿ 0, we put

W (M; T ) = {v∈L∞(0; T ;V ∩ H 2) : vt ∈L∞(0; T ;V ); vtt ∈L2(QT );

‖v‖L∞(0;T ;V∩H 2); ‖vt‖L∞(0;T ;V ); ‖vtt‖L2(QT )6M}; (3.5)

W1(M; T ) = {v∈W (M; T ) : vtt ∈L∞(0; T ;L2)}; (3.6)

where QT = � × (0; T ).We shall choose the Jrst term u0 = u 0. Suppose that

um−1 ∈W1(M; T ): (3.7)

We associate with problem (1.1)–(1.3) the following variational problem.Find um ∈W1(M; T ) which satisJes the linear variational problem

〈 Sum(t); v〉+ bm(t)a(um(t); v) = 〈Fm(t); v〉 for all v∈V; (3.8)

um(0) = u 0; um(0) = u 1; (3.9)

where

bm(t) = B(‖∇um−1(t)‖2);Fm(x; t) = f(x; t; um−1(t);∇um−1(t); u m−1(t); ‖∇um−1(t)‖2): (3.10)

Then, we have the following theorem.


Theorem 1. Let (H1)–(H4) hold. Then there exist positive constants M; T and thelinear recurrent sequence {um} ⊂ W1(M; T ) de6ned by (3.8)–(3.10).

Proof. The proof consists of several steps.Step 1: The Galerkin approximation (introduced by Lions [7]). Consider the basis

for V as in Lemma 3 (wj = wj=√�j). Put

u(k)m (t) =k∑j=1

c(k)mj (t)wj; (3.11)

where c(k)mj satisfy the system of linear diHerential equations

〈 Su(k)m (t); wj〉+ bm(t)a(u(k)m (t); wj) = 〈Fm(t); wj〉; 16 j6 k; (3.12)

u(k)m (0) = u 0k ; u(k)m (0) = u 1k ; (3.13)

where

u 0k → u 0 strongly in V ∩ H 2; (3.14)

u 1k → u 1 strongly in V: (3.15)

Let us suppose that um−1 satisJes (3.7). Then it is clear that system (3.12), (3.13)has a unique solution u(k)m (t) on an interval 06 t6T (k)

m 6T . The following estimatesallow one to take constant T (k)

m = T for all m and k.Step 2: A priori estimates. Put

S(k)m (t) = X (k)m (t) + Y (k)

m (t) +∫ t

0‖ Su(k)m (s)‖2 ds; (3.16)

where

X (k)m (t) = ‖u(k)m (t)‖2 + bm(t)a(u(k)m (t); u(k)m (t)); (3.17)

Y (k)m (t) = a(u(k)m (t); u(k)m (t)) + bm(t)‖Iu(k)m (t)‖2: (3.18)

Then, it follows from (3.12), (3.13) and (3.16)–(3.18) that

S(k)m (t) = S(k)m (0) +∫ t

0b′m(s)(a(u

(k)m (s); u(k)m (s)) + ‖Iu(k)m (s)‖2) ds

+2∫ t

0〈Fm(s); u(k)m (s)〉 ds+ 2

∫ t

0a(Fm(s); u(k)m (s)) ds

+∫ t

0‖ Su(k)m (s)‖2 ds

= S(k)m (0) + I1 + I2 + I3 + I4: (3.19)

We shall estimate, respectively, the following integrals on the right-hand sideof (3.19).


First integral: We have

bm(t) = B(‖∇um−1(t)‖2);b′m(t) = 2B′(‖∇um−1(t)‖2)〈∇um−1(t);∇um−1(t)〉: (3.20)

By using assumption (H3), we obtain from (3.4) and (3.7)

|b′m(t)|6 2|B′(‖∇um−1(t)‖2)|‖∇um−1(t)‖‖∇um−1(t)‖6 2M 2K1: (3.21)

Combining (3.16)–(3.18) and (3.21), we obtain

|I1|6 2M 2K1

b0

∫ t

0S(k)m (s) ds: (3.22)

Second integral: From (3.1), (3.10), (3.16) and (3.17), we deduce that

|I2|6 2∫ t

0‖Fm(s)‖‖u(k)m (s)‖ ds6 2K0

∫ t

0

√S(k)m (s) ds: (3.23)

Third integral: It follows from (3.1), (3.2), (3.7) and (3.10) that

‖Fm(s)‖2V = ‖∇Fm(s)‖2 + h0F2m(0; s)6 4K2

1 (1 + 3M 2) + h0K20 : (3.24)

Then, from (3.16), (3.18) and (3.24), we obtain

|I3|6 2∫ t

0‖Fm(s)‖V‖u(k)m (s)‖V ds

6 2[2K1

√1 + 3M 2 +

√h0K0]

∫ t

0

√S(k)m (s) ds: (3.25)

Fourth integral: Eq. (3.12) can be rewritten as follows:

〈 Su(k)m (t); wj〉 − bm(t)〈Iu(k)m (t); wj〉= 〈Fm(t); wj〉; 16 j6 k: (3.26)

Hence, it follows after replacing wj with Su(k)m (t) and integrating that∫ t

0‖ Su(k)m (s)‖2 ds6 2

∫ t

0b2m(s)‖Iu(k)m (s)‖2 ds+ 2

∫ t

0‖Fm(s)‖2 ds: (3.27)

From (3.1), (3.3), (3.7), (3.10), (3.16) and (3.18) we deduce that

I46 2K0

∫ t

0S(k)m (s) ds+ 2TK2

0 : (3.28)

Combining (3.19), (3.22), (3.23), (3.25) and (3.28), we then have

S(k)m (t)6 S(k)m (0) + 2TK20 + 2(2K1

√1 + 3M 2 + (1 +

√h0)K0)

∫ t

0

√S(k)m (s) ds

+2(K0 +

M 2K1

b0

)∫ t

0S(k)m (s) ds


6 S(k)m (0) + 2TK20 +

12T (2K1

√1 + 3M 2 + (1 +

√h0)K0)2

+ 2(1 + K0 +

M 2K1

b0

)∫ t

0S(k)m (s) ds

= S(k)m (0) + C1(M; T ) + C2(M)∫ t

0S(k)m (s) ds; (3.29)

where

C1(M; T ) = 2TK20 +

12T (2K1

√1 + 3M 2 + (1 +

√h0)K0)2;

C2(M) = 2(1 + K0 +

M 2K1

b0

): (3.30)

Now, we need an estimate on the term S(k)m (0).We have

S(k)m (0) = ‖u 1k‖2 + a(u 1k ; u 1k) + B(‖∇u 0‖2)[a(u 0k ; u 0k) + ‖Iu 0k‖2]: (3.31)

By means of (3.14), (3.15) and (3.31), we can deduce the existence of a constantM ¿ 0, independent of k and m, such that

S(k)m (0)6M 2=2 for all k and m: (3.32)

Note that, from assumption (H4) we have

limT→0+

√TKi(M; T; f) = 0; i = 0; 1: (3.33)

Then, from (3.30) and (3.33), we can always choose the constant T ¿ 0 such that

(M 2=2 + C1(M; T )) exp(TC2(M))6M 2 (3.34)

and

kT = 2

√2T

(1 +

1b0

)(M 2K1 + (1 +M)K1)

×exp((

1 +1b0

)(1 +M 2K1)T

)¡ 1: (3.35)

Finally, it follows from (3.29), (3.32), and (3.34) that

S(k)m (t)6M 2 exp(−TC2(M)) + C2(M)∫ t

0S(k)m (s) ds;

06 t6T (k)m 6T: (3.36)


By using Gronwall’s lemma we deduce from (3.36) that

S(k)m (t)6M 2 exp(−TC2(M)) exp(C2(M)t)6M 2 (3.37)

for all t ∈ [0; T (k)m ].

So we can take constant T (k)m = T for all m and k. Therefore, we have

u(k)m ∈W1(M; T ) for all m and k: (3.38)

From (3.38) we can extract from {u(k)m } a subsequence {u(ki)m } such that

u(ki)m → um in L∞(0; T ;V ∩ H 2) weak ∗; (3.39)

u(ki)m → um in L∞(0; T ;V ) weak ∗; (3.40)

Su(ki)m → Sum in L2(QT ) weak; (3.41)

where

um ∈W (M; T ): (3.42)

Passing to limit in (3.12), (3.13), by (3.39)–(3.41), we have um satisfying (3.8)–(3.10)in L2(0; T ), weak. On the other hand, it follows from (3.7), (3.8) and (3.42) that

Sum = bm(t)Ium + Fm ∈L∞(0; T ;L2);

hence um ∈W1M; T ) and the proof of Theorem 1 is complete.

Theorem 2. Let (H1)–(H4) hold. Then there exist constants M ¿ 0; T ¿ 0 satisfying(3.32), (3.34), and (3.35) such that problem (1.1)–(1.3) has a unique weak solutionu∈W1(M; T ).

On the other hand, the linear recurrent sequence {um} de6ned by (3.8)–(3.10)converges to the solution u strongly in the space

W1(T ) = {v∈L∞(0; T ;V ) : v∈L∞(0; T ;L2)}:Furthermore, we have also the estimation

‖um − u‖L∞(0;T ;V ) + ‖um − u‖L∞(0;T ;L2)6CkmT for all m; (3.43)

where

kT = 2

√2T

(1 +

1b0

)(M 2K1 + (1 +M)K1)

×exp((

1 +1b0

)(1 +M 2K1)T

)¡ 1; (3.44)

and C is a constant depending only on T; u0; u1 and kT .


Proof. (a) Existence of the solution.First, we note that W1(T ) is a Banach space with respect to the norm (see [7]):

‖v‖W1(T ) = ‖v‖L∞(0;T ;V ) + ‖v‖L∞(0;T ;L2): (3.45)

We shall prove that {um} is a Cauchy sequence in W1(T ). Let vm = um+1 − um. Thenvm satisJes the variational problem

〈 Svm(t); v〉+ bm+1(t)a(vm(t); v)− (bm+1(t)− bm(t))〈Ium(t); v〉=〈Fm+1(t)− Fm(t); v〉 for all v∈V; (3.46)

vm(0) = vm(0) = 0: (3.47)

We take v= vm in (3.46), after integrating in t

Pm(t) =∫ t

0b′m+1(s)a(vm(s); vm(s)) ds

+2∫ t

0(bm+1(s)− bm(s))〈Ium(s); vm(s)〉 ds

+2∫ t

0〈Fm+1(s)− Fm(s); vm(s)〉 ds; (3.48)

where

Pm(t) = ‖vm(t)‖2 + bm+1(t)a(vm(t); vm(t)): (3.49)

On the other hand, from (3.2), (3.4), and (3.7) we obtain

|b′m+1(t)|6 2|B′(‖∇um(t)‖2)|‖∇um(t)‖‖∇um(t)‖6 2M 2K1; (3.50)

|bm+1(t)− bm(t)|6 2MK1‖∇vm−1(t)‖6 2MK1‖vm−1‖W1(T ); (3.51)

‖Fm+1(t)− Fm(t)‖6K1[2(1 +M)‖∇vm−1(t)‖+ ‖vm−1(t)‖]6 2K1(1 +M)‖vm−1‖W1(T ): (3.52)

It follows from (3.48)–(3.52) that

‖vm(t)‖2 + b0‖vm(t)‖2V 6 Pm(t)6 2M 2K1

∫ t

0‖vm(s)‖2V ds

+4M 2K1‖vm−1‖W1(T )

∫ t

0‖vm(s)‖ ds

+4K1(1 +M)‖vm−1‖W1(T )

∫ t

0‖vm(s)‖ ds

6 2M 2K1

∫ t

0‖vm(s)‖2V ds

+4[M 2K1 + (1 +M)K1]‖vm−1‖W1(T )

∫ t

0‖vm(s)‖ ds


6 2T [M 2K1 + (1 +M)K1]2‖vm−1‖2W1(T )

+ 2(1 +M 2K1)∫ t

0(‖vm(s)‖2 + ‖vm(s)‖2V ) ds: (3.53)

Hence,

‖vm(t)‖2 + ‖vm(t)‖2V 6(1 +

1b0

)Pm(t)

6 2(1 +

1b0

)T (M 2K1 + (1 +M)K1)2‖vm−1‖2W1(T )

+ 2(1 +

1b0

)(1 +M 2K1)

×∫ t

0(‖vm(s)‖2 + ‖vm(s)‖2V ) ds: (3.54)

From (3.54), we deduce that

‖vm‖W1(T )6 kT‖vm−1‖W1(T ) for all m; (3.55)

where

kT = 2

√2T

(1 +

1b0

)(M 2K1 + (1 +M)K1)

×exp((

1 +1b0

)(1 +M 2K1)T

)¡ 1:

Hence,

‖um+p − um‖W1(T )6 ‖u1 − u0‖W1(T )kmT

1− kTfor all m;p: (3.56)

It follows from (3.73) that {um} is a Cauchy sequence in W1(T ). Therefore, thereexists u∈W1(T ) such that

um → u strongly in W1(T ): (3.57)

We also note that um ∈W1(M; T ), then from the sequence {um} we can deduce asubsequence {umj} such that

umj → u in L∞(0; T ;V ∩ H 2) weak ∗; (3.58)

umj → u in L∞(0; T ;V )weak ∗; (3.59)

Sumj → Su in L2(QT ) weak; (3.60)

u∈W (M; T ): (3.61)


We note that

‖bm(t)∇um(t)− B(‖∇u(t)‖2)∇u(t)‖6 K0‖∇um(t)−∇u(t)‖+ 2M 2K1‖∇um−1(t)−∇u(t)‖6 K0‖um − u‖W1(T ) + 2M 2K1‖um−1 − u‖W1(T ); a:e:; t ∈ (0; T ): (3.62)

It follows from (3.57) and (3.62) that

bm(t)∇um → B(‖∇u(t)‖2)∇u strongly in L∞(0; T ;L2): (3.63)

Similarly

‖Fm − f(x; t; u; ux; u; ‖ux(t)‖2)‖L∞(0;T ;L2)6 2(1 +M)K1‖um−1 − u‖W1(T ): (3.64)

Hence, from (3.57) and (3.64), we obtain

Fm → f(x; t; u; ux; u; ‖ux(t)‖2) strongly in L∞(0; T ;L2): (3.65)

Then we can take limits in (3.8)–(3.10) with m=mj → +∞, we then can deduce from(3.58)–(3.60), (3.63), and (3.65) that there exists u∈W (M; T ) satisfying the equation

〈 Su(t); v〉+ B(‖ux(t)‖2)a(u(t); v)=〈f(x; t; u; ux; u; ‖ux(t)‖2); v〉 for all v∈V; (3.66)

and the initial conditions

u(0) = u 0(0); u(0) = u 1: (3.67)

On the other hand, we have from (3.63), (3.65), and (3.66) that

Su= B(‖ux(t)‖2)uxx + f(x; t; u; ux; u; ‖ux(t)‖2)∈L∞(0; T ;L2): (3.68)

Hence, we obtain u∈W1(M; T ).The existing proof is completed.(b) Uniqueness of the solution. Let u1; u2 both be weak solutions of problems

(1.1)–(1.3) such that

ui ∈W1(M; T ); i = 1; 2: (3.69)

Then u= u1 − u2 satisJes the following variational problem:

〈 Su(t); v〉+ B1(t)a(u(t); v)− (B1(t)− B2(t))〈Iu2(t); v〉=〈F1(t)− F2(t); v〉 for all v∈V (3.70)

and the initial conditions

u(0) = u(0) = 0; (3.71)


where

Bi(t) = B(‖∇ui(t)‖2); (3.72)

F i(t) = f(t; x; ui;∇ui; u i; ‖∇ui(t)‖2); i = 1; 2:

Take v= u in (3.70), we then obtain after integrating by parts

‖u(t)‖2 + b0‖u(t)‖2V 6∫ t

0B′1(s)a(u(s); u(s)) ds

+2∫ t

0(B1(s)− B2(s))〈Iu2(s); u(s)〉 ds

+2∫ t

0〈F1(s)− F2(s); u(s)〉 ds: (3.73)

Put

Z(t) = ‖u(t)‖2 + ‖u(t)‖2V (3.74)

and

KM = 2(1 +

1b0

)[2M 2K1 + (2 +M)K1];

then it follows from (3.73), (3.74) that

Z(t)6 KM

∫ t

0Z(s) ds for all t ∈ [0; T ]: (3.75)

Using Gronwall’s lemma we deduce Z(t) = 0, i.e., v1 = v2.The proof of Theorem 2 is complete.

Remark 1. In the case of B ≡ 1; f=f(t; u; ut); f∈C1([0;∞)×R2); f(t; 0; 0)=0;∀t¿ 0,and the Dirichlet homogeneous condition standing for (1.2), we have obtained someresults in the paper [3].In the case of the function f∈C1([0; 1] × [0;∞) × R3); B ≡ 1, we have also

obtained some results in [10]. However, the result above does not use the assump-tion f∈C1([0; 1]× R+ × R3 × R+).

4. Asymptotic expansion of solutions

In this part, let (H1)–(H4). We also make the following assumptions:

(H5) B1 ∈C1(R+); B1(z)¿ 0;(H6) f1 satisJes assumption (H4).


We consider the following perturbed problem, where � is a small parameter, |�|6 1:

(P�)

utt − B�(‖ux‖2)Iu= F�(x; t; u; ux; ut ; ‖ux‖2);0¡x¡ 1; 0¡t¡T;

ux(0; t)− h0u(0; t) = u(1; t) = 0;

u(x; 0) = u 0(x); ut(x; 0) = u 1(x);

F�(x; t; u; ux; ut ; ‖ux‖2) = f(x; t; u; ux; ut ; ‖ux‖2)+ �f1(x; t; u; ux; ut ; ‖ux‖2);

B�(‖ux‖2) = B(‖ux‖2) + �B1(‖ux‖2):First, we note that if the functions u 0; u 1; B; B1; f; f1 satisfy assumptions (H1)–(H6),

then the a priori estimates of the Galerkin approximation sequence {u(k)m } in the proofof Theorem 1 for problems (1.1)–(1.3) corresponding to B=B�; f=F�; |�|6 1, satisfy

u(k)m ∈W1(M; T ); (4.1)

where M; T are constants independent of �. Indeed, in the processing we choose thepositive constants M and T as in (3.32), (3.34), (3.35), wherein Ki(M; T; f) andK i(M;B); i=0; 1, stand for Ki(M; T; f)+Ki(M; T; f1) and K i(M;B)+ K i(M;B1); i=0; 1,respectively. Hence, the limit u� in suitable function spaces of the sequence {u(k)m }as k → +∞, afterwards m → +∞, is a unique weak solution of the problem (P�)satisfying

u� ∈W1(M; T ): (4.2)

Then we can prove, in a manner similar to the proof of theorem 2, that the limit u0in suitable function spaces of the family {u�} as � → 0 is a unique weak solution ofthe problem (P0) corresponding to �= 0 satisfying

u0 ∈W1(M; T ): (4.3)

Then, we have the following theorem.

Theorem 3. Let (H1)–(H6) hold. Then there exist constants M ¿ 0 and T ¿ 0 suchthat, for every � with |�|6 1, problem (P�) has a unique weak solution u� ∈W1(M; T )satisfying the asymptotic estimation

‖u� − u0‖L∞(0;T ;V ) + ‖u � − u 0‖L∞(0;T ;L2)6C|�|; (4.4)

where C is a constant depending only on b0; h0; T;M; K1(M; T; f), K1(M;B),K0(M; T; f1), and K0(M;B1).

Proof. Put v= u� − u0. Then v satisJes the variational problem

〈 Sv(t); w〉+ B(‖∇u�(t)‖2)a(v(t); w)= (B(‖∇u�(t)‖2)− B(‖∇u0(t)‖2))〈Iu0(t); w〉+ 〈f �; w〉+ �〈f1�; w〉


+ �B1(‖∇u�(t)‖2)〈Iu�(t); w〉 for all w∈V;

v(0) = v(0) = 0; (4.5)

where

f � = f �(x; t) = f(x; t; u�;∇u�; u �; ‖∇u�‖2)− f(x; t; u0;∇u0; u 0; ‖∇u0‖2);

f1� = f1�(x; t) = f1(x; t; u�;∇u�; u �; ‖∇u�‖2):Taking w = v in (4.5), after integration by parts in t, we obtain

‖v(t)‖2 + b�(t)a(v(t); v(t)) =∫ t

0b′�(s)a(v(s); v(s)) ds+ 2

∫ t

0(B(‖∇u�(s)‖2)

−B(‖∇u0(s)‖2))〈Iu0(s); v(s)〉 ds

+2�∫ t

0B1(‖∇u�(s)‖2)〈Iu�(s); v(s)〉 ds

+2∫ t

0〈f �(s); v(s)〉 ds+ 2�

∫ t

0〈f1�(s); v(s)〉 ds; (4.6)

where

b�(t) = B(‖∇u�(t)‖2): (4.7)

Let

9(t) = ‖v(t)‖2 + ‖v(t)‖2V ; (4.8)

then, we can prove the following inequality in a similar manner:

‖v(t)‖2 + b0‖v(t)‖2V 6 �2T:2 + (:1 + :2)∫ t

09(s) ds; 06 t6T; (4.9)

where

:1 = 2(2 +M)K1(M; T; f) + 4M 2K1(M;B);

:2 = K0(M; T; f1) +MK0(M;B1): (4.10)

Next, by (4.9) and Gronwall’s lemma, we obtain

9(t)6(1 +

1b0

)�2T:2 exp

((1 +

1b0

)(:1 + :2)T

)for all t ∈ [0; T ]: (4.11)

Hence,

‖v‖L∞(0;T ;V ) + ‖v‖L∞(0;T ;L2)6C|�|; (4.12)

where C = 2√(1 + 1

b0)T:2 exp( 12 (1 +

1b0)(:1 + :2)T ) is a constant depending only on

b0; h0; T;M; K1(M; T; f), K1(M;B), K0(M; T; f1), and K0(M;B1).The proof of Theorem 3 is complete.


The next result gives an asymptotic expansion of the weak solution u� of order N+1in �, for � su:ciently small. We use the following notations:

f[u] = f(x; t; u; ux; ut ; ‖ux‖2); B[u] = B(‖ux‖2):Now, we assume that

(H7) B∈CN+1(R+); B1 ∈CN (R+); B(z)¿ b0¿ 0; B1(z)¿ 0;(H8) f∈CN+1([0; 1]× R+ × R3 × R+); f1 ∈CN ([0; 1]× R+ × R3 × R+),

satisJes the following conditions:

(H′8) D�3

3 D�44 D

�55 D

�66 f(1; t; u; v; w; z) = 0; (�3; �4; �5; �6)∈Z4

+; �3 + �4 + �5 + �66N ,(H′′

8 ) D�33 D

�44 D

�55 D

�66 f1(1; t; u; v; w; z) = 0(�3; �4; �5; �6)∈Z4

+; �3 + �4 + �5 + �66N − 1for all t; z¿ 0 and (u; v; w)∈R3.

Let u0 ∈W1(M; T ) be a weak solution of the problem (P0) corresponding to �= 0.Let us consider the weak solutions u1; u2; : : : ; uN ∈W1(M; T ) (with suitable constants

M ¿ 0 and T ¿ 0) deJned by the following problems:

(Q1)

Su 1 − B[u0]Iu1 = F1[u1]; 0¡x¡ 1; 0¡t¡T;

∇u1(0; t)− h0u1(0; t) = u1(1; t) = 0;

u1(x; 0) = u 1(x; 0) = 0;

where

F1[u1] = c1[f] + c0[f1] + (d1[B] + d0[B1])Iu0 (4.13)

with c0[f]; c1[f]; d0[B]; d1[B] are deJned as follows:

c0[f] = f[u0] ≡ f(x; t; u0;∇u0; u 0; ‖∇u0‖2); (4.14)

c1[f] = c0[D3f]u1 + c0[D4f]∇u1 + c0[D5f]u 1 + 2c0[D6f]〈∇u0;∇u1〉; (4.15)

d0[B] = B[u0] ≡ B(‖∇u0‖2) (4.16)

and

d1[B] = 2d0[B′]〈∇u0;∇u1〉 (4.17)

with 26 i6N ,

(Qi)

Su i − B[u0]Iui = F i[ui]; 0¡x¡ 1; 0¡t¡T;

∇ui(0; t)− h0ui(0; t) = ui(1; t) = 0;

ui(x; 0) = u i(x; 0) = 0; i = 1; 2; : : : ; N;

where

F i[ui] = ci[f] + ci−1[f1] +i∑

k=1

(dk [B] + dk−1[B1])Iui−k (4.18)


with ci[f] = ci[f; u0; u1; : : : ; ui]; di[B] = di[B; u0; u1; : : : ; ui]; 26 i6N deJned by therecurrence formulas

ci[f] =i−1∑k=0

i − ki

ck [D3f]ui−k + ck [D4f]∇ui−k + ck [D5f]u i−k

+2ck [D6f]i−k−1∑j=0

i − k − ji − k

〈∇uj;∇ui−k−j〉 ; 26 i6N; (4.19)

di[B] =2i

i−1∑k=0

dk [B′]i−k−1∑j=0

(i − k − j)〈∇uj;∇ui−k−j〉; 26 i6N: (4.20)

We also note that ci[f] is the Jrst-order function with respect to ui;∇ui; u i. In fact,

ci[f] = c0[D3f]ui + c0[D4f]∇ui + c0[D5f]u i + 2c0[D6f]〈∇u0;∇ui〉+ terms depending on (i; ck [D<f]; uk ;∇uk ; u k);<= 3; 4; 5; 6; k = 0; 1; : : : ; i − 1: (4.21)

Similarly

di[B] = 2c0[B′]〈∇u0;∇ui〉+ terms depending on (i; ck [B′];∇uk);k = 0; 1; : : : ; i − 1: (4.22)

Let u� ∈W1(M; T ) be a unique weak solution of the problem (P�). Then v = u� −∑Ni=0 �

iui ≡ u� − h satisJes the problem

Sv− B�[v+ h]Iv= F�[v+ h]− F�[h](B�[v+ h]− B�[h])Ih+ E�(x; t);

0¡x¡ 1; 0¡t¡T;

∇v(0; t)− h0v(0; t) = v(1; t) = 0;

v(x; 0) = v(x; 0) = 0; (4.23)

where

E�(x; t) = F�[h]− f[u0] + (B�[h]− B[u0])Ih−N∑i=1

�iF i[ui]: (4.24)

Then, we have the following lemmas.

Lemma 4. The functions ci[f]; di[B]; 06 i6N above are de6ned by the followingformulas:

ci[f] =1i!

@i

@�i(f[h])

∣∣∣∣�=0

; 06 i6N; (4.25)

di[B] =1i!

@i

@�i(B[h])

∣∣∣∣�=0

; 06 i6N: (4.26)


Proof. (i) It is easy to see that

c0[f] = f[h]|�=0 = f[u0] ≡ f(x; t; u0;∇u0; u 0; ‖∇u0‖2)With i = 1, we have

c1[f] =@@�

(f[h])∣∣∣∣�=0

: (4.27)

But

@@�

(f[h]) =D3f[h]@@�h+ D4f[h]

@@�

∇h

+D5f[h]@@�h+ D6f[h]

@@�

(‖∇h‖2): (4.28)

On the other hand, from the formulas

h=N∑i=0

�iui;@@�h=

N∑i=1

i�i−1ui;@@�

(‖∇h‖2) = 2⟨∇h; @

@�∇h

⟩;

we have

@@�h∣∣∣∣�=0

= u1;@@�

∇h∣∣∣∣�=0

=∇u1; @@� h∣∣∣∣�=0

= u 1;

@@�

(‖∇h‖2)∣∣∣∣�=0

= 2〈∇u0;∇u1〉: (4.29)

Hence, it follows from (4.27), (4.28), (4.29) that

c1[f] = c0[D3f]u1 + c0[D4f]∇u1 + c0[D5f]u 1 + 2c0[D6f]〈∇u0;∇u1〉: (4.30)

Thus, (4.15) holds.Suppose that we have deJned the functions ck [Djf]; j = 3; 4; 5; 6; k = 0; 1; : : : ; i − 1

from formulas (4.14), (4.15) and (4.25). Therefore, it follows from (4.28) that

@i

@�i(f[h]) =

@i−1

@�i−1

@@�

(f[h])

=@i−1

@�i−1

[D3f[h]

@@�h+ D4f[h]

@@�

∇h+ D5f[h]@@�h

+D6f[h]@@�

(‖∇h‖2)]

=i−1∑k=0

Cki−1

{@k

@�k(D3f[h])

@i−k

@�i−k (h) +@k

@�k(D4f[h])

@i−k

@�i−k (∇h)

+@k

@�k(D5f[h])

@i−k

@�i−k (h) +@k

@�k(D6f[h])

@i−k

@�i−k (‖∇h‖2)}: (4.31)


We also note that

@i

@�ih∣∣∣∣�=0

= i!ui; 06 i6N: (4.32)

On the other hand,

@m

@�m(‖∇h‖2) = 2

@m−1

@�m−1

⟨∇h; @

@�∇h

⟩

= 2m−1∑j=0

Cjm−1

⟨@j

@�j(∇h); @

m−j

@�m−j (∇h)⟩: (4.33)

Hence,

@m

@�m(‖∇h‖2)

∣∣∣∣�=0

= 2m−1∑j=0

Cjm−1〈j!∇uj; (m− j)!∇um−j〉

= 2m−1∑j=0

j!(m− j)!Cjm−1〈∇uj;∇um−j〉: (4.34)

It follows from (4.31), (4.32) and (4.34) that

@i

@�i(f[h])

∣∣∣∣�=0

=i−1∑k=0

k!Cki−1

{ck [D3f]

@i−k

@�i−k (h)∣∣∣∣�=0

+ ck [D4f]@i−k

@�i−k (∇h)∣∣∣∣�=0

+ ck [D5f]@i−k

@�i−k (h)∣∣∣∣�=0

+ ck [D6f]@i−k

@�i−k (‖∇h‖2)∣∣∣∣�=0

}

=i−1∑k=0

Cki−1

k!ck [D3f](i − k)!ui−k + k!ck [D4f](i − k)!∇ui−k

+ k!ck [D5f](i − k)!u i−k + 2k!ck [D6f]

×i−k−1∑j=0

j!(i − k − j)!Cji−k−1〈∇uj;∇ui−k−j〉

=i−1∑k=0

(i − k)(i − 1)!

ck [D3f]ui−k + ck [D4f]∇ui−k

+ ck [D5f]u i−k +2

i − kck [D6f]

×i−k−1∑j=0

(i − k − j)〈∇uj;∇ui−k−j〉 : (4.35)


Hence,

ci[f] =1i!

@i

@�i(f[h])

∣∣∣∣�=0

=i−1∑k=0

i − ki

ck [D3f]ui−k + ck [D4f]∇ui−k + ck [D5f]u i−k

+2

i − kck [D6f]

i−k−1∑j=0

(i − k − j)〈∇uj;∇ui−k−j〉 : (4.36)

Hence, the part 1 of Lemma 4 is proved.

(ii) In the case of B = B[h] = B(‖∇h‖2). Applying formulas (4.14), (4.15), (4.19)with f=f(z); Dif=0; i=3; 4; 5; D6f=B′ and ci[f]=di[B], we obtain formulas (4.16),(4.17), (4.20) and later part of Lemma 4 is proved.

Lemma 5. Let (H1); (H2); (H7), and (H8) hold. Then there exists a constant K suchthat

‖E�‖L∞(0;T ;L2)6 K |�|N+1; (4.37)

where K is a constant depending only on M; T; N and the constants

K i(M;B) = sup06z6M 2

|B(i)(z)|; i = 1; 2; : : : ; N + 1;

K i(M;B1) = sup06z6M 2

|B(i)1 (z)|; i = 1; 2; : : : ; N;

Ki(M; T; f) = sup∑=

|D=11 D

=33 D

=44 D

=55 D

=66 f[u]|; i = 1; 2; : : : ; N + 1;

Ki(M; T; f1) = sup∑=

|D=11 D

=33 D

=44 D

=55 D

=66 f1[u]|; i = 1; 2; : : : ; N;

where, in each case, sup is taken over 06 x6 1; 06 t6T; |u|; |ux|; |u|6M; 06 z6M 2, and the sum

∑= is taken over = = (=1; =3; : : : ; =6)∈Z5

+ satisfying |=| = =1+=3 + · · ·+ =6 = i.

Proof. In the case of N =1, the proof of Lemma 5 is easy, hence we omit the details,which we only prove with N¿ 2.

By using Taylor’s expansion of the functions f[h] and f1[h] around the point �=0up to order N + 1 and order N , respectively, we obtain from (4.25), that

f[h]− f[u0] =N∑i=1

�i

i!@i

@�i(f[h])

∣∣∣∣�=0

+�N+1

(N + 1)!@N+1

@�N+1 (f[h])∣∣∣∣�=>1�

=N∑i=1

ci[f]�i + �N+1RN+1[f; �; >1] (4.38)


and

f1[h] =N−1∑i=0

�i

i!@i

@�i(f1[h])

∣∣∣∣�=0

+�N

N !@N

@�N(f1[h])

∣∣∣∣�=>2�

=N−1∑i=0

ci[f1]�i + �NRN [f1; �; >2]; (4.39)

where ci[f]; 06 i6N are deJned by (4.14), (4.15), (4.19); RN+1[f; �; >1] andRN [f1; �; >2] are deJned as follows:

RN+1[f; �; >1] =1

(N + 1)!@N+1

@�N+1 (f[h])∣∣∣∣�=>1�

(4.40)

and

RN [f1; �; >2] =1N !

@N

@�N(f1[h])

∣∣∣∣�=>2�

(4.41)

with 0¡>i ¡ 1; i = 1; 2.Combining (4.38)–(4.41), we then obtain

F�[h]− f[u0] =f[h]− f[u0] + �f1[h]

=N∑i=1

(ci[f] + ci−1[f1])�i + �N+1RN [f;f1; �; >1; >2] (4.42)

with

RN [f;f1; �; >1; >2] = RN+1[f; �; >1] + RN [f1; �; >2]: (4.43)

Similarly, we use Taylor’s expansion around the point � = 0 up to order N + 1 ofthe functions B[h] and the function B1[h] up to order N , we obtain from (4.26), that

B[h]− B[u0] =N∑i=1

�i

i!@i

@�i(B[h])

∣∣∣∣�=0

+�N+1

(N + 1)!@N+1

@�N+1 (B[h])∣∣∣∣�=>3�

=N∑i=1

di[B]�i + �N+1RN+1[B; �; >3] (4.44)

and

B1[h] =N−1∑i=0

�i

i!@i

@�i(B1[h])

∣∣∣∣�=0

+�N

N !@N

@�N(B1[h])

∣∣∣∣�=>4�

=N−1∑i=0

di[B1]�i + �N RN [B1; �; >4]; (4.45)

where

RN+1[B; �; >3] =1

(N + 1)!@N+1

@�N+1 (B[h])∣∣∣∣�=>3�

(4.46)


and

RN [B1; �; >4] =1N !

@N

@�N(B1[h])

∣∣∣∣�=>4�

(4.47)

with 0¡>i ¡ 1; i = 3; 4.Combining (4.44)–(4.47), we then obtain

B�[h]− B[u0] = B[h]− B[u0] + �B1[h]

=N∑i=1

(di[B] + di−1[B1])�i + �N+1RN [B; B1; �; >3; >4] (4.48)

with

RN [B; B1; �; >3; >4] = RN+1[B; �; >3] + RN [B1; �; >4]; (4.49)

hence

(B�[h]− B[u0])Ih=

[N∑i=1

(di[B] + di−1[B1])�i] N∑

j=0

�jIuj

+ �N+1IhRN [B; B1; �; >3; >4]

=N 2∑i=1

[i∑

k=1

(dk [B] + dk−1[B1])Iui−k

]�i

+ �N+1IhRN [B; B1; �; >3; >4]

=N∑i=1

[i∑

k=1


]�i

+N 2∑

i=N+1

[i∑

k=1


]�i

+ �N+1IhRN [B; B1; �; >3; >4]

=N∑i=1

[i∑

k=1


]�i

+ �N+1R(1)N [B; B1; h; �; >3; >4] (4.50)

with

R(1)N [B; B1; h; �; >3; >4] =IhRN [B; B1; �; >3; >4]

+N 2∑

i=N+1

[i∑

k=1


]�i−N−1: (4.51)


Combining (4.13)–(4.18), (4.24), (4.42), (4.43), (4.50) and (4.51), we then obtain

E�(x; t) = F�[h]− f[u0] + (B�[h]− B[u0])Ih−N∑i=1

�iF i[ui]

=N∑i=1

[ci[f] + ci−1[f1] +

i∑k=1

(dk [B] + dk−1[B1])Iui−k − F i[ui]

]�i

+ �N+1(RN [f;f1; �; >1; >2] + R(1)N [B; B1; h; �; >3; >4])

= �N+1(RN [f;f1; �; >1; >2] + R(1)N [B; B1; h; �; >3; >4]): (4.52)

By the boundedness of the functions ui;∇ui; u i; i = 0; 1; 2 in the function spaceL∞(0; T ;H 1), we obtain from (4.40), (4.41), (4.43), (4.46), (4.47), (4.49), (4.51)and (4.52) that

‖E�‖L∞(0;T ;L2)6 K |�|N+1; (4.53)

where K is a constant depending only on M; T; N and the constants K i(M;B);Ki(M; T; f); i = 1; 2; : : : ; N + 1; K i(M;B1); Ki(M; T; f1); i = 1; 2; : : : ; N .The proof of Lemma 5 is complete.

Now, we consider the sequence of functions {vm} deJned by

v0 ≡ 0;

Svm − B�[vm−1 + h]Ivm = F�[vm−1 + h]− F�[h] + (B�[vm−1 + h]

−B�[h])Ih+ E�(x; t); 0¡x¡ 1; 0¡t¡T;

∇vm(0; t)− h0vm(0; t) = vm(1; t) = 0;

vm(x; 0) = vm(x; 0) = 0; m¿ 1: (4.54)

With m= 1, we have the problem

Sv1 − B�[h]Iv1 = E�(x; t); 0¡x¡ 1; 0¡t¡T;

∇v1(0; t)− h0v1(0; t) = v1(1; t) = 0;

v1(x; 0) = v1(x; 0) = 0: (4.55)

By multiplying the two sides of (4.55) by v1, we Jnd without di:culty from(4.37) that

‖v1(t)‖2 + b1; �(t)a(v1(t); v1(t))6 2K |�|N+1T‖v1‖L∞(0;T ;L2)

+∫ t

0|b′1; �(s)|a(v1(s); v1(s)) ds; (4.56)

where

b1; �(t) = B�[h] = B(‖∇h(t)‖2) + �B1(‖∇h(t)‖2):


We have

b′1; �(t) = 2(B′(‖∇h(t)‖2) + �B′1(‖∇h(t)‖2))〈∇h(t);∇h(t)〉; (4.57)

hence

|b′1; �(t)|6 2(N + 1)2M 2(K1(M;B) + K1(M;B1)) ≡ :1: (4.58)

It follows from (4.56), (4.58) that

‖v1(t)‖2 + b0‖v1(t)‖2V 6 2K |�|N+1T‖v1‖L∞(0;T ;L2) + :1

∫ t

0‖v1(s)‖2V ds: (4.59)

Using Gronwall’s lemma we obtain

‖v1‖L∞(0;T ;L2) + ‖v1‖L∞(0;T ;V )6 2(1 +

1√b0

)TK |�|N+1 exp

(:1Tb0

): (4.60)

We shall prove that there exists a constant CT , independent of m and �, such that

‖vm‖L∞(0;T ;L2) + ‖vm‖L∞(0;T ;V )6CT |�|N+1; |�|6 1 for all m: (4.61)

By multiplying the two sides of (4.54) with vm and after integration in t, we obtain

‖vm(t)‖2 + b0‖vm(t)‖2V 6∫ t

0|b′m;�(s)|‖vm(s)‖2V ds

+2∫ t

0(‖f[vm−1 + h]− f[h]‖

+ ‖f1[vm−1 + h]− f1[h]‖)‖vm‖ ds

+2∫ t

0|B[vm−1 + h]− B[h]|‖Ih‖‖vm‖ ds

+2∫ t

0|B1[vm−1 + h]− B1[h]|‖Ih‖‖vm‖ ds

+2K |�|N+1∫ t

0‖vm‖ ds; (4.62)

where

bm;�(t) = B�[vm−1 + h] = B[vm−1 + h] + �B1[vm−1 + h];

b′m;�(t) = 2(B′[vm−1 + h] + �B′1[vm−1 + h])〈∇vm−1 +∇h;∇vm−1 +∇h〉:

Hence,

|b′m;�(t)|6 2(N + 2)2M 2(K1(M;B) + K1(M;B1)) ≡ :2: (4.63)

Put

?m = ‖vm‖L∞(0;T ;L2) + ‖vm‖L∞(0;T ;V ): (4.64)

=3 = (3 + 5M)(K1(M; T; f) + K1(M; T; f1)) + 5M 2(K1(M;B) + K1(M;B1)):


It follows from (4.62), (4.63) and (4.64) that

?m6 9?m−1 + � for all m¿ 1 (4.65)

with

9= 4:3(1 +M)T [K1(M; T; f) + K1(M; T; f1)]

+ 4:3(N + 1)M 2√T [K1(M;B) + K1(M;B1)];

�= 2:3TK |�|N+1;

:3 = 2(1 +

1√b0

)exp

(:2Tb0

): (4.66)

We assume that

9¡ 1 with the suitable constant T ¿ 0: (4.67)

We shall now require the following lemma whose proof is immediate.

Lemma 6. Let the sequence {?m} satisfy

?m6 9?m−1 + � for all m¿ 1; ?0 = 0; (4.68)

where 06 9¡ 1; �¿ 0 are the given constants. Then

?m6 �=(1− 9) for all m¿ 1: (4.69)

We deduce from (4.65), (4.66) and (4.69) that

‖vm‖L∞(0;T ;L2) + ‖vm‖L∞(0;T ;V )6 �=(1− 9) = CT |�|N+1; (4.70)

where

CT =2:3TK

1− 4:3:4;

:4 = (1 +M)T (K1(M; T; f) + K1(M; T; f1))

+ (N + 1)M 2√T (K1(M;B) + K1(M;B1)): (4.71)

On the other hand, the linear recurrent sequence {vm} deJned by (4.54) convergesstrongly in the space W1(T ) to the solution v of problem (4.23). Hence, lettingm → +∞ in (4.70) gives

‖v‖L∞(0;T ;L2) + ‖v‖L∞(0;T ;V )6CT |�|N+1

or ∥∥∥∥∥u � −N∑i=0

�iu i

∥∥∥∥∥L∞(0;T ;L2)

+

∥∥∥∥∥u� −N∑i=0

�iui

∥∥∥∥∥L∞(0;T ;V )

6CT |�|N+1: (4.72)

Thus, we have the following theorem.


Theorem 4. Let (H1); (H2); (H7), and (H8) hold. Then there exist constants M ¿ 0and T ¿ 0 such that, for every �, with |�|6 1, the problem (P�) has a unique weaksolution u� ∈W1(M; T ) satisfying an asymptotic estimation up to order N + 1 asin (4.72), the functions u0; u1; : : : ; uN being the weak solutions of problems (P0);(Q1); : : : ; (QN ), respectively.

Remark 2. • In the case of B ≡ 1; B1 ≡ 0; f1 ≡ 0; f=f(t; u; ut); f∈CN+1([0;∞)×R2),and the Dirichlet homogeneous condition u(0; t) = u(1; t) = 0 standing for (1:2), wehave obtained the results above in the paper [3].• In the case of functions f∈C2([0; 1] × [0;∞) × R3); f1 ∈C1([0; 1] × [0;∞) × R3)and N = 1, we have also obtained some results concerning in the papers [9,10,13]in the cases as follows:

(a) B ≡ 1; B1 ≡ 0, (see [10]).(b) B� = b0 + B(z) + �B1(z), where b0¿ 0 is a given constant and B∈C2(R+);

B1 ∈C1(R+); B¿ 0; B1¿ 0, and (1.2) standing for the Dirichlet homogeneouscondition u(0; t) = u(1; t) = 0. (see [13]).

(c) B� = B(t; z) + �B1(t; z); B∈C2(R2+); B1 ∈C1(R2

+); B¿ b0¿ 0; B1¿ 0, and (1.2)standing for the boundary condition ux(0; t)− h0u(0; t) = ux(1; t) + h1u(1; t) = 0(see [9]).

• In the case of B� ≡ 1; F�=f(�; x; t; u; ux; ut)+�f1(�; x; t; u; ux; ut) with f∈CN+1([0; 1]×[0; 1] × [0;∞) × R3); f1 ∈CN ([0; 1] × [0; 1] × [0;∞) × R3), we have also obtainedsome results above in the paper [8].

Acknowledgements

The authors would like to thank the referee for his corrections and suggestions.

References

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[2] A.P.N. Dinh, Sur un problZeme hyperbolique faiblement non-linWeaire en dimension 1, Demonstratio Math.16 (1983) 269–289.

[3] A.P.N. Dinh, N.T. Long, Linear approximation and asymptotic expansion associated to the nonlinearwave equation in one dimension, Demonstratio Math. 19 (1986) 45–63.

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on the nonlinear wave equation utt−b(∥ux∥2)uxx=f(x,t,u,ux,ut,∥ux∥2) associated with the...

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