on the nonlinear wave equation...

Nonlinear Analysis 58 (2004) 933–959www.elsevier.com/locate/na

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

f(x; t; u; ux; ut; ‖u‖2; ‖ux‖2; ‖ut‖2) associated withthe mixed homogeneous conditions

Nguyen Thanh LongDepartment 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

Received 11 February 2004; accepted 30 May 2004

Abstract

In this paper we consider the following nonlinear wave equation:utt − B(t; ‖u‖2; ‖ux‖2; ‖ut‖2)uxx = f(x; t; u; ux; ut ; ‖u‖2; ‖ux‖2; ‖ut‖2);

x∈ (0; 1); 0¡t¡T; (1)

ux(0; t) − h0u(0; t) = ux(1; t) + h1u(1; t) = 0; (2)

u(x; 0) = u0(x); ut(x; 0) = u1(x); (3)

where h0¿ 0; h1¿ 0 are given constants and B; f; u0; u1 are given functions. In Eq. (1), thenonlinear terms B(t; ‖u‖2; ‖ux‖2; ‖ut‖2); f(x; t; u; ux; ut ; ‖u‖2; ‖ux‖2; ‖ut‖2) depending on the inte-grals ‖u‖2 =

∫� |u(x; t)|2 dx; ‖ux‖2 =

∫ 10 |ux(x; t)|2 dx and ‖ut‖2 =

∫� |ut(x; t)|2 dx. In this pa-

per we associate with problem (1)–(3) a linear recursive scheme for which the existenceof a local and unique solution is proved by using standard compactness argument. In caseof B∈CN+1(R4

+); B¿ b0¿ 0; B1 ∈CN (R4+); B1¿ 0; f∈CN+1([0; 1] × R+ × R3 × R3

+) andf1 ∈CN ([0; 1]×R+×R3×R3

+) we obtain from the following equation utt−[B(t; ‖u‖2; ‖ux‖2; ‖ut‖2)+�B1(t; ‖u‖2; ‖ux‖2; ‖ut‖2)]uxx = f(x; t; u; ux; ut ; ‖u‖2; ‖ux‖2; ‖ut‖2) + �f1(x; t; u; ux; ut ; ‖u‖2; ‖ux‖2;‖ut‖2) associated to (2), (3) a weak solution u�(x; t) having an asymptotic expansion of or-der N + 1 in �, for � su8ciently small.? 2004 Elsevier Ltd. All rights reserved.

MSC: 35L70; 35C20; 35L20

Keywords: The Kirchho<-Carrier operator; Galerkin method; Linear recurrent sequence; Asymptoticexpansion of order N + 1

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

0362-546X/$ - see front matter ? 2004 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2004.05.021

mailto:[email protected]

934 N.T. Long /Nonlinear Analysis 58 (2004) 933–959

1. Introduction

In this paper we consider a nonlinear wave equation with the Kirchho<-Carrieroperator

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

ux(0; t) − h0u(0; t) = ux(1; t) + h1u(1; t) = 0; (1.2)

u(x; 0) = u0(x); ut(x; 0) = u1(x); (1.3)

where B; f; u0; u1 are given functions satisfying conditions speciDed later and h0¿ 0;h1¿ 0 are given constants. In Eq. (1.1), the nonlinear terms f(x; t; u; ux; ut ; ‖u‖2; ‖∇u‖2;‖ut‖2) and B(t; ‖u‖2; ‖∇u‖2; ‖ut‖2) depend on the integrals

‖u‖2 =∫�

|u(x; t)|2 dx; ‖∇u‖2 =∫�

|∇u(x; t)|2 dx and

‖ut‖2 =∫�

|ut(x; t)|2 dx: (1.4)

Eq. (1.1) has its origin in the nonlinear vibration of an elastic string (Kirchho< [7]),for which the associated equation is

�hutt =

(P0 +

Eh2L

∫ L

0

∣∣∣∣ @u@y (y; t)∣∣∣∣2 dy

)uxx; (1.5)

here u is the lateral deEection, � is the mass density, h is the cross section, L is thelength, E is Young’s modulus and P0 is the initial axial tension.

In [2], Carrier also established a model of the type

utt =(P0 + P1

∫ L

0u2(y; t) dy

)uxx; (1.6)

where P0 and P1 are constants.When f=0 and B=B(‖∇u‖2) is a function depending only on ‖∇u‖2, the Cauchy

or mixed problem for (1.1) has been studied by many authors; see Ebihara, et al. [5],Pohozaev [20] and the references therein. A survey of the results about the mathematicalaspects of Kirchho< model can be found in Medeiros et al. [17,18].In [16] Medeiros has studied the 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 [6] Hosoya andYamada have considered (1.1)–(1.3) with f=f(u)=−�|u|�u, where �¿ 0; �¿ 0 aregiven constants.In [11,15] the authors have studied the existence and uniqueness of the equation

utt + �C2u− B(‖∇u‖2)Cu+ �|ut |�−1ut = F(x; t); x∈�; t ¿ 0; (1.7)

where �¿ 0; �¿ 0; 0¡�¡ 1, are given constants, and � is a bounded open setof Rn.

N.T. Long /Nonlinear Analysis 58 (2004) 933–959 935

In [3], Alain Pham has studied the existence and asymptotic behavior as � → 0of 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; (1.8)

where the nonlinear term has the form f= �f1(t; u). By a generalization of [3], AlainPham and Long [4] have considered problem (1.1), (1.3), (1.8) with B ≡ 1 and thenonlinear term having the form

f = �f1(t; u; ut): (1.9)

If B� ≡ 1 and f1 ∈CN (R+ × R2) satisDes f1(t; 0; 0) = 0 for all t¿ 0, an asymptoticexpansion of the solution of problem (1.1), (1.3), (1.8), (1.9) up to order N + 1 in �is obtained, for � su8ciently small. This expansion extends to the partial di<erentialequation the results obtained in di<erential equations [1].In [12] 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); (1.10)

associated with (1.2) and (1.3). In the case of f∈C2([0; 1]×R+×R3) and f1 ∈C1([0; 1]×R+ × R3), we have been obtained an asymptotic expansion of order 2 in �, for � su8-ciently small. Afterwards, this result has been extended in [13] to the nonlinear waveequation with the Kirchho< operator

utt − [b0 + B(‖ux‖2) + �B1(‖ux‖2)]uxx=f(x; t; u; ux; ut) + �f1(x; t; u; ux; ut); (1.11)

associated with (1.3), (1.8), 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 Drst associate with the 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 [2–4,12,13,19] cannot be used in the papers[5,6,11,15,16]. If B∈CN+1(R4

+); B1 ∈CN (R4+); B¿ b0¿ 0; B1¿ 0; f∈CN+1([0; 1]×

R+ × R3 × R3+) and f1 ∈CN ([0; 1] × R+ × R3 × R3

+), then an asymptotic expansion oforder N + 1 in � is obtained with a right-hand side of the form f(x; t; u; ux; ut ; ‖u‖2,‖ux‖2; ‖ut‖2) + �f1(x; t; u; ux; ut ; ‖u‖2; ‖ux‖2; ‖ut‖2) and B stand for B + �B1; for �su8ciently small. This result is a relative generalization of [4,9,10,12–14,19].

2. Preliminary results, notations

We will omit the deDnitions of the usual function spaces and denote them by thenotation Lp = Lp(0; 1); Hm = Hm(0; 1).Let 〈·; ·〉 be either the scalar product in L2 or the dual pairing of a continuous linear

functional and an element of a function space. The notation ‖·‖ stands for the norm inL2 and we denote by ‖·‖X the norm in the Banach space X . We call X = the dual space


of X . We denote by Lp(0; T ;X ), 16p6∞ for the Banach space of real functionsu : (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) = Ou(t), ux(t) = ∇u(t); uxx(t) = Cu(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; ux; ut ; ‖u‖2; ‖ux‖2; ‖ut‖2) = f(x; t; u; v; w; U; V;W ), we put D1f =

@f=@x; D2f=@f=@t; D3f=@f=@u; D4f=@f=@v; D5f=@f=@w; D6f=@f=@U; D7f=@f=@V and D8f = @f=@W .We put

a(u; v) =∫ 1

0ux(x)vx(x) dx + h0u(0)v(0) + h1u(1)v(1): (2.1)

In H 1 we shall use the equivalent norm ‖v‖H 1 = (v2(0)+∫ 10 |v=(x)|2 dx)1=2. Then we

have the following lemmas.

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

‖v‖C0([0;1])6√2‖v‖H 1 for all v∈H 1: (2.2)

Lemma 2. Let h0¿ 0 and h1¿ 0. Then the symmetric bilinear form a(·; ·) de3nedby (2.1) is continuous on H 1 × H 1 and coercive on H 1, i.e.,

(i) |a(u; v)|6C1‖u‖H 1‖v‖H 1 for all u; v∈H 1,(ii) a(v; v)¿C0‖v‖2H 1 for all v∈H 1,

where C0 = min{1; h0}; C1 = max{1; h0; 2h1}.

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

Lemma 3. There exists the Hilbert orthonormal base {wj} of L2 consisting ofthe eigenfunctions wj corresponding to the eigenvalue �j such that 0¡�16 �26 · · ·6 �j6 : : : ; lim

j→+∞�j =+∞;

a(wj; v) = �j〈wj; v〉 for all v∈H 1; j = 1; 2; : : :(2.3)

Furthermore, the sequence {wj=√�j} is also the Hilbert orthonormal base of H 1

with respect to the scalar product a(·; ·). On the other hand, we have also wj satisfyingthe boundary value problem{−Cwj = �jwj in (0; 1);

wjx(0) − h0 wj(0) = wjx(1) + h1wj(1) = 0; wj ∈C∞([0; 1]):(2.4)


The proof of Lemma 3 can be found in [21, p. 137, Theorem 6.2.1], with H = L2,V = H 1 and a(·; ·) as deDned by (2.1).

3. The existence and uniqueness theorem

We make the following assumptions:(A1) h0¿ 0; h1¿ 0,(A2) u0 ∈H 2; u1 ∈H 1,(A3) B∈C1(R4

+); B(t; U; V;W )¿ b0¿ 0,(A4) f∈C1([0; 1] × R+ × R3 × R3

+).We introduce the following constants that will appear in the next section. With B

and f satisfying assumptions (A3) and (A4), respectively, for all M ¿ 0 and T ¿ 0,we put

K0 = K0(M; T; f) = sup{|f(x; t; u; v; w; U; V;W )| :(x; t; u; v; w; U; V;W )∈A∗(M; T )}; (3.1)

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

{(|D1f| +

8∑i=3

|Dif|)(x; t; u; v; w; U; V;W ) :

(x; t; u; v; w; U; V;W )∈A∗(M; T )

}; (3.2)

K0 = K0(M; T; B) = sup{B(t; U; V;W ) : (t; U; V;W )∈ A∗(M; T )}; (3.3)

K1 = K1(M; T; B) = sup{(∣∣∣∣@B@t

∣∣∣∣+ ∣∣∣∣ @B@U∣∣∣∣ + ∣∣∣∣ @B@V

∣∣∣∣+ ∣∣∣∣ @B@W∣∣∣∣) (t; U; V;W ) :

(t; U; V;W )∈ A∗(M; T )}; (3.4)

where

A∗(M; T ) = {(x; t; u; v; w; U; V;W )∈R8 : 06 t6T; 06 x6 1;

|u| + |v| + |w|6M; 06U; V;W 6M 2}and

A∗(M; T ) = {(t; U; V;W )∈R4 : 06 t6T; 06U; V;W 6M 2}:For each M ¿ 0 and T ¿ 0; we put

W (M; T ) = {v∈L∞(0; T ;H 2) : vt ∈L∞(0; T ;H 1); vtt ∈L∞(0; T ;L2);

‖v‖L∞(0;T ;H 2); ‖vt‖L∞(0;T ;H 1); ‖vtt‖L∞(0;T ;L2)6M}:We shall choose the Drst term u0 = u0. Suppose that

um−1 ∈W (M; T ): (3.5)

We associate with the problem (1.1)–(1.3) the following variational problem.


Find um ∈W (M; T ) which satisDes the linear variational problem

〈 Oum(t); v〉 + bm(t)a(um(t); v) = 〈Fm(t); v〉 for all v∈H 1; (3.6)

um(0) = u0; um(0) = u1; (3.7)

where

bm(t) = B(t; ‖um−1(t)‖2; ‖∇um−1(t)‖2; ‖um−1(t)‖2); (3.8)

Fm(x; t) =f(x; t; um−1(t);∇um−1(t); u m−1(t); ‖um−1(t)‖2;‖∇um−1(t)‖2; ‖um−1(t)‖2): (3.9)

Then, we have the following theorem.

Theorem 1. Let (A1)–(A4) hold. Then there exist positive constants M; T and thelinear recurrent sequence {um} ⊂ W (M; T ) de3ned by (3.6)–(3.9).

Proof. The proof consists of several steps.Step 1: The Galerkin approximation (introduced by Lions [8]). Consider the basis

for H 1 as in Lemma 3 (wj = wj=√�j). Put

u(k)m (t) =k∑j=1

c(k)mj (t)wj; (3.10)

where c(k)mj satisfy the system of linear di<erential equations.

〈 Ou(k)m (t); wj〉 + bm(t)a(u(k)m (t); wj) = 〈Fm(t); wj〉; 16 j6 k; (3.11)

u(k)m (0) = u0k ; u(k)m (0) = u1k ; (3.12)

where

u0k → u0 strongly in H 2; (3.13)

u1k → u1 strongly in H 1: (3.14)

Let us suppose that um−1 satisDes (3.5). Then it is clear that system (3.11), (3.12)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) + Z (k)m (t); (3.15)

where

X (k)m (t) = ‖u(k)m (t)‖2 + bm(t)a(u(k)m (t); u(k)m (t)); (3.16)

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

Z (k)m (t) = ‖ Ou(k)m (t)‖2 + bm(t)a(u(k)m (t); u(k)m (t)): (3.18)


Then, it follows from (3.11), (3.12), (3.15)–(3.18), that

S(k)m (t) = S(k)m (0) +∫ t

0b=m(s){a(u(k)m (s); u(k)m (s)) + ‖Cu(k)m (s)‖2

+ a(u(k)m (s); u(k)m (s)) + 2〈Cu(k)m (s); Ou(k)m (s)〉} ds

+2∫ t

0〈Fm(s); u(k)m (s)〉 ds+ 2

∫ t

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

+2∫ t

0〈F=

m(s); Ou(k)m (s)〉 ds

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

We shall estimate respectively the following integrals on the right-hand side of(3.19).First integral. We have

b=m(t) =D2B[um−1] + 2D6B[um−1]〈um−1(t); u m−1(t)〉+2D7B[um−1]〈∇um−1(t);∇um−1(t)〉 + 2D8B[um−1]

×〈um−1(t); Oum−1(t)〉; (3.20)

here we have used the notation

DiB[um−1] = DiB(t; ‖um−1(t)‖2; ‖∇um−1(t)‖2; ‖um−1(t)‖2); i = 2; 6; 7; 8:

By using the assumption (A3), we obtain from (3.4) and (3.5)

|b=m(t)|6 (1 + 6M 2)K1: (3.21)

Combining (3.15)–(3.19) and (3.21), we obtain

|I1|6(1 +

2b0

)(1 + 6M 2)K1

∫ t

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

Second integral. It follows from (3.1), (3.9), (3.15), (3.16) and (3.19), 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. We have

@@x

Fm =D1f[um−1] + D3f[um−1]∇um−1 + D4f[um−1]Cum−1

+D5f[um−1]∇um−1; (3.24)

here we have used the notation

Dif[um−1] =Dif(x; t; um−1(t);∇um−1(t); u m−1(t); ‖um−1(t)‖2;‖∇um−1(t)‖2; ‖um−1(t)‖2); i = 1; 3; 4; 5:


It follows from (3.1), (3.2), (3.5), (3.9) and (3.24) that

‖Fm(s)‖2H 1 =∥∥∥∥ @@x Fm(s)

∥∥∥∥2 + F2m(0; s)6K2

1 (1 + 3M)2 + K20 : (3.25)

Then, from (3.15), (3.17), (3.19) and (3.25), we obtain

|I3|6 2C1

∫ t

0‖Fm(s)‖H 1‖u(k)m (s)‖H 1 ds

62C1√C0

[K1(1 + 3M) + K0]∫ t

0

√S(k)m (s) ds: (3.26)

Fourth integral. We have

F=m(t) =D2f[um−1] + D3f[um−1]um−1 + D4f[um−1]∇um−1 + D5f[um−1] Oum−1

+2D6f[um−1]〈um−1; um−1〉 + 2D7f[um−1]〈∇um−1;∇um−1〉+2D8f[um−1]〈um−1; Oum−1〉: (3.27)

It follows from (3.2), (3.5) and (3.27) that

‖F=m(t)‖6 (1 + 3M + 6M 2)K1: (3.28)

From (3.15), (3.18), (3.19) and (3.28) we deduce that

|I4|6 2∫ t

0‖F=

m(s)‖‖ Ou(k)m (s)‖ds6 2(1 + 3M + 6M 2)K1

∫ t

0

√S(k)m (s) ds: (3.29)

Combining (3.19), (3.22), (3.23), (3.26) and (3.29), we then have

S(k)m (t)6 S(k)m (0) +(1 +

2b0

)(1 + 6M 2)K1

∫ t

0S(k)m (s) ds

+2[(

1 +C1√C0

)K0 +

[1 + 3M + 6M 2 +

C1√C0

(1 + 3M)]K1

]

×∫ t

0

√S(k)m (s) ds

6 S(k)m (0) + D1(M; T ) + D2(M; T )∫ t

0S(k)m (s) ds; (3.30)

where

D1(M; T ) = T[(

1 +C1√C0

)K0 +

[1 + 3M + 6M 2 +

C1√C0

×(1 + 3M)]K1

]2; (3.31)

D2(M; T ) = 1 +(1 +

2b0

)(1 + 6M 2)K1: (3.32)


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

S(k)m (0) = B(0; ‖u0‖2; ‖∇u0‖2; ‖u1‖2)[a(u0k ; u0k) + ‖Cu0k‖2 + a(u1k ; u1k)]

+‖u0k‖2 + a(u1k ; u1k) + ‖ Ou(k)m (0)‖2: (3.33)

But by (3.11) we have

‖ Ou(k)m (0)‖2 − B(0; ‖u0‖2; ‖∇u0‖2; ‖u1‖2)〈Cu0k ; Ou(k)m (0)〉=〈f(x; 0; u0;∇u0; u1; ‖u0‖2; ‖∇u0‖2; ‖u1‖2); Ou(k)m (0)〉: (3.34)

Therefore,

‖ Ou(k)m (0)‖6 B(0; ‖u0‖2; ‖∇u0‖2; ‖u1‖2)‖Cu0k‖+‖f(x; 0; u0;∇u0; u1; ‖u0‖2; ‖∇u0‖2; ‖u1‖2)‖: (3.35)

By means of (3.13), (3.14), (3.33) and (3.35), we can deduce the existence ofa constant M ¿ 0; independent of k and m, such that

S(k)m (0)6M 2=2 for all k and m: (3.36)

Notice that, from the assumptions (A3); (A4) we have

limT→0+

√TKi(M; T; f) = lim

T→0+

√TKi(M; T; B) = 0; i = 0; 1: (3.37)

Then, from (3.31), (3.32) and (3.37), we can always choose the constant T ¿ 0 suchthat

(M 2=2 + D1(M; T )) exp(TD2(M; T ))6M 2 (3.38)

and

kT =(1 +

1√b0C0

)√T [4M 2K1 + (1 +

√2)(1 + 2M)K1]

×exp

(12T

(1 +

(1 + 6M 2)C1K1

b0C0

))¡ 1: (3.39)

Finally, it follows from (3.30), (3.36) and (3.38) that

S(k)m (t)6M 2 exp(−TD2(M; T )) + D2(M; T )∫ t

0S(k)m (s) ds;

06 t6T (k)m 6T: (3.40)

By using Gronwall’s lemma we deduce from (3.40) that

S(k)m (t)6M 2 exp(−TD2(M; T )) exp(tD2(M; T ))6M 2

for all t ∈ [0; T (k)m ]: (3.41)

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

u(k)m ∈W (M; T ) for all m and k: (3.42)


From (3.42) we can extract from {u(k)m } a subsequence {u(ki)m } such that

u(ki)m → um in L∞(0; T ;H 2) weak ∗; (3.43)

u(ki)m → um in L∞(0; T ;H 1) weak ∗; (3.44)

Ou(ki)m → Oum in L∞(0; T ;L2) weak ∗; (3.45)

um ∈W (M; T ): (3.46)

We can easily check from (3.11), (3.12), (3.43)–(3.46) that um satisDes (3.6)–(3.9)in L∞(0; T ); weak*. The proof of Theorem 1 is complete.

Theorem 2. Let (A1)–(A4) hold. Then there exist positive constants M; T satisfying(3.36), (3.38) and (3.39) such that the problem (1.1)–(1.3) has a unique weak solutionu∈W (M; T ).On the other hand, the linear recurrent sequence {um} de3ned by (3.6)–(3.9) con-

verges to the solution u strongly in the space W1(T ) = {v∈L∞(0; T ;H 1) : v∈L∞

(0; T ;L2)}.Furthermore, we have also the estimation

‖um − u‖L∞(0;T ;H 1) + ‖um − u‖L∞(0;T ;L2)6CkmT for all m; (3.47)

where the constant kT ¡ 1 is de3ned by (3.39) and C is a constant depending onlyon T; u0; u1 and kT .

Proof. (a) Existence of the solution. First, we note that W1(T ) is a Banach space withrespect to the norm (see [8]):

‖v‖W1(T ) = ‖v‖L∞(0;T ;H 1) + ‖v‖L∞(0;T ;L2): (3.48)

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

〈 Ovm(t); v〉 + bm+1(t)a(vm(t); v) − (bm+1(t) − bm(t))〈Cum(t); v〉=〈Fm+1(t) − Fm(t); v〉 for all v∈H 1; (3.49)

vm(0) = vm(0) = 0: (3.50)

We take v= vm in (3.49), after integrating in t

‖vm(t)‖2 + bm+1(t)a(vm(t); vm(t))

=∫ t

0b=m+1(s)a(vm(s); vm(s)) ds+ 2

∫ t

0(bm+1(s) − bm(s))〈Cum(s); vm(s)〉 ds

+2∫ t

0〈Fm+1(s) − Fm(s); vm(s)〉 ds: (3.51)


On the other hand, from (3.2), (3.4), (3.5) and (3.21) we get

|b=m+1(t)|6 (1 + 6M 2)K1; (3.52)

|bm+1(t) − bm(t)|6 4MK1‖vm−1‖W1(T ); (3.53)

‖Fm+1(t) − Fm(t)‖6 (1 +√2)(1 + 2M)K1‖vm−1‖W1(T ): (3.54)

It follows from (3.51)–(3.54) that

‖vm(t)‖2 + b0C0‖vm(t)‖2H 1

6 (1 + 6M 2)C1K1

∫ t

0‖vm(s)‖2H 1 ds

+2[4M 2K1 + (1 +√2)(1 + 2M)K1]‖vm−1‖W1(T )

∫ t

0‖vm(s)‖ ds

6T [4M 2K1 + (1 +√2)(1 + 2M)K1]2‖vm−1‖2W1(T )

+

[1 +

(1 + 6M 2)C1K1

b0C0

]∫ t

0(‖vm(s)‖2 + b0C0‖vm(s)‖2H 1 ) ds: (3.55)

By using Gronwall’s lemma we deduce from (3.55), that

‖vm‖W1(T )6 kT‖vm−1‖W1(T ) for all m; (3.56)

where

kT =(1 +

1√b0C0

)√T [4M 2K1 + (1 +

√2)(1 + 2M)K1]

×exp

(12T

(1 +

(1 + 6M 2)C1K1

b0C0

))¡ 1:

Hence

‖um+p − um‖W1(T )6 ‖u1 − u0‖W1(T ) kmT =(1 − kT ) for all m;p: (3.57)

It follows from (3.57) that {um} is a Cauchy sequence in W1(T ): Therefore there existsu∈W1(T ) such that

um → u strongly in W1(T ): (3.58)

We also note that um ∈W (M; T ); then from the sequence {um} we can deduce a sub-sequence {umj} such that

umj → u in L∞(0; T ;H 2) weak ∗; (3.59)

umj → u in L∞(0; T ;H 1) weak ∗; (3.60)


Oumj → Ou in L∞(0; T ;L2) weak ∗; (3.61)

u∈W (M; T ): (3.62)

We notice that∣∣∣∣∫ T

0bm(t)a(um(t); v(t)) dt −

∫ T

0B(t; ‖u(t)‖2; ‖ux(t)‖2; ‖u(t)‖2)a(u; v) dt

∣∣∣∣6C1(K0‖um − u‖W1(T ) + 4M 2K1‖um−1 − u‖W1(T ))‖v‖L1(0;T ;H 1)

for all v∈L1(0; T ;H 1): (3.63)

It follows from (3.58) and (3.63) that∫ T

0bm(t)a(um(t); v(t)) dt →

∫ T

0B(t; ‖u(t)‖2; ‖ux(t)‖2; ‖u(t)‖2)a(u; v) dt

for all v∈L1(0; T ;H 1): (3.64)

Similarly

‖Fm − f(x; t; u; ux; u; ‖u‖2; ‖ux‖2; ‖u‖2)‖L∞(0;T ;L2)

6 (1 +√2)(1 + 2M)K1‖um−1 − u‖W1(T ): (3.65)

Hence, from (3.58) and (3.65), we obtain

Fm → f(x; t; u; ux; u; ‖u‖2; ‖ux‖2; ‖u‖2) strongly in L∞(0; T ;L2): (3.66)

Then we can take limits in (3.6)–(3.9) with m=mj → +∞, we then can deduce from(3.59)–(3.61), (3.64) and (3.66) that there exists u∈W (M; T ) satisfying the equation

〈 Ou(t); v〉 + B(t; ‖u(t)‖2; ‖ux(t)‖2; ‖u(t)‖2)a(u(t); v)=〈f(x; t; u; ux; u; ‖u(t)‖2; ‖ux(t)‖2; ‖u(t)‖2); v〉 for all v∈H 1; (3.67)

and the initial conditions

u(0) = u0; u(0) = u1: (3.68)

The existence proof is completed.

(b) Uniqueness of the solution. Let u1 and u2 both be weak solutions of the problem(1.1)–(1.3) such that ui ∈W (M; T ); i = 1; 2. Then, u = u1 − u2 satisDes the followingvariational problem

〈 Ou(t); v〉 + B1(t)a(u(t); v) − (B1(t) − B2(t))〈Cu2(t); v〉 = 〈F1(t) − F2(t); v〉for all v∈H 1; (3.69)

and the initial conditions

u(0) = u(0) = 0; (3.70)


where

Bi(t) = B(t; ‖ui(t)‖2; ‖∇ui(t)‖2; ‖u i(t)‖2);Fi(t) = f(t; x; ui;∇ui; u i; ‖ui(t)‖2; ‖∇ui(t)‖2; ‖u i(t)‖2); i = 1; 2: (3.71)

Take v= u in (3.69), we then obtain after integrating by parts

‖u(t)‖2 + b0C0‖u(t)‖2H 1 6∫ t

0B=1(s)a(u(s); u(s)) ds+ 2

∫ t

0(B1(s) − B2(s))〈Cu2(s);

u(s)〉 ds+ 2∫ t

0〈F1(s) − F2(s); u(s)〉 ds: (3.72)

Put

Z(t) = ‖u(t)‖2 + b0C0‖u(t)‖2H 1 (3.73)

and

KM = (1 + 6M 2)C1K1 + 2(1 +√2)(1 +

1√b0C0

)×[2M 2K1 + (1 + 2M)K1]; (3.74)

then it follows from (3.72)–(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., u1 = u2.The proof of Theorem 2 is complete.

Remark 1.

• In the case of B ≡ 1; f=f(t; u; ut); f∈C1(R+ ×R2); f(t; 0; 0)=0 ∀t¿ 0, and theDirichlet homogeneous condition (1.8) standing for (1.2), we have obtained someresults in the paper [4].

• In the case of the function f∈C1([0; 1] × R+ × R3); B ≡ 1, we have also obtainedsome results in [12].

• In the case of B ≡ B(‖ux‖2); B∈C1(R+); B(z)¿ b0¿ 0; f=f(x; t; u; ux; ut ; ‖ux‖2),f∈C1([0; 1]×R+×R3×R+), and (1.2) standing for the condition ux(0; t)−h0u(0; t)=u(1; t) = 0, we have also obtained some results above in the paper [14].

4. Asymptotic expansion of solutions

In this part, let (A1)–(A4). We also make the following assumptions:(A5) B1 ∈C1(R4

+); B1(t; U; V;W )¿ 0,(A6) f1 ∈C1([0; 1] × R+ × R3 × R3

+).


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

(P�)

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

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

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

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

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

+�B1(t; ‖u‖2; ‖ux‖2; ‖ut‖2):First, we note that if the functions u0; u1; B; B1; f; f1 satisfy the assumptions (A1)–(A6),

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

u(k)m ∈W (M; T ); (4.1)

where M; T are constants independent of �. Indeed, in the processing we choose thepositive constants M and T as in (3.36), (3.38), (3.39), wherein Ki(M; T; f) andKi(M; T; B); i=0; 1, stand for Ki(M; T; f)+Ki(M; T; f1) and Ki(M; T; B)+Ki(M; T; 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� ∈W (M; T ): (4.2)

Then we can prove, in a manner similar to the proof of Theorem 2, that the limitu0 in suitable function spaces of the family {u�} as � → 0 is a unique weak solutionof the problem (P0) corresponding to �= 0 satisfying

u0 ∈W (M; T ): (4.3)

Then, we have the following theorem.

Theorem 3. Let (A1)–(A6) hold. Then there exist constants M ¿ 0 and T ¿ 0 suchthat, for every � with |�|6 1, problem (P�) has a unique weak solution u� ∈W (M; T )satisfying the asymptotic estimation

‖u� − u0‖L∞(0;T ;H 1) + ‖u � − u 0‖L∞(0;T ;L2)6C|�|; (4.4)

where C is a constant depending only on b0; h0; h1; T;M; K1(M; T; f), K1(M; T; B),K0(M; T; f1) and K0(M; T; B1).


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

〈 Ov(t); w〉 + b�(t)a(v(t); w)

=(b�(t) − b0(t))〈Cu0(t); w〉 + 〈f�(t) − f0(t); w〉 + �〈f1�(t); w〉+�b1�(t)〈Cu�(t); w〉 for all w∈H 1;

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

(4.5)

where

b�(t) = B(t; ‖u�(t)‖2; ‖∇u�(t)‖2; ‖u �(t)‖2);b1�(t) = B1(t; ‖u�(t)‖2; ‖∇u�(t)‖2; ‖u �(t)‖2);f�(t) = f�(x; t) = f(x; t; u�;∇u�; u �; ‖u�‖2; ‖∇u�‖2; ‖u �‖2);f1�(t) = f1�(x; t) = f1(x; t; u�;∇u�; u �; ‖u�‖2; ‖∇u�‖2; ‖u �‖2):

(4.6)

Taking w = v in (4.5), after integration by parts in t, we get

‖v(t)‖2 + b�(t)a(v(t); v(t))

=∫ t

0b=�(s)a(v(s); v(s)) ds+ 2

∫ t

0(b�(s) − b0(s))〈Cu0(s); v(s)〉 ds

+2�∫ t

0b1�(s)〈Cu�(s); v(s)〉 ds+ 2

∫ t

0〈f�(s) − f0(s); v(s)〉 ds

+2�∫ t

0〈f1�(s); v(s)〉 ds: (4.7)

Let :(t)= ‖v(t)‖2 + ‖v(t)‖2H 1 , then, we can prove the following inequality in a similarmanner

‖v(t)‖2 + b0C0‖v(t)‖2H 1 6 �2;1T + ;2

∫ t

0:(s) ds; 06 t6T; (4.8)

where

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

;2 = 1 + (1 + 6M 2)C1K1(M; T; B) + 3(1 +√2)[2M 2K1(M; T; B)

+ (1 + 2M)K1(M; T; f)]: (4.9)

Next, by (4.8) and Gronwall’s lemma, we obtain

:(t)6(1 +

1b0C0

)�2;1T exp

((1 +

1b0C0

);2T)

for all t ∈ [0; T ]: (4.10)

Hence

‖v‖L∞(0;T ;H 1) + ‖v‖L∞(0;T ;L2)6C|�|; (4.11)


where

C = 2

√(1 +

1b0C0

);1T exp

(12

(1 +

1b0C0

);2T):

The proof of Theorem 3 is complete.

The next result gives an asymptotic expansion of the weak solution u� of order N+1in �, for � su8ciently small. We use the following notations:

f[u] = f(x; t; u; ux; ut ; ‖u‖2; ‖ux‖2; ‖ut‖2); B[u] = B(t; ‖u‖2; ‖ux‖2; ‖ut‖2):Now, we assume that(A7) B∈CN+1(R4

+); B1 ∈CN (R4+); B(t; U; V;W )¿ b0¿ 0; B1(t; U; V;W )¿ 0,

(A8) f∈CN+1([0; 1] × R+ × R3 × R3+); f1 ∈CN ([0; 1] × R+ × R3 × R3

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

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

(Q1)

Ou 1 − B[u0]Cu1 = F1[u1]; 0¡x¡ 1; 0¡t¡T;

∇u1(0; t) − h0u1(0; t) = ∇u1(1; t) + h1u1(1; t) = 0;

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

where

F1[u1] = =1[f] + =0[f1] + (�1[B] + �0[B1])Cu0; (4.12)

with =0[f]; =1[f]; �0[B]; �1[B] are deDned as follows:

=0[f] = f[u0] ≡ f(x; t; u0;∇u0; u 0; ‖u0‖2; ‖∇u0‖2; ‖u 0‖2); (4.13)

=1[f] = =0[D3f]u1 + =0[D4f]∇u1 + =0[D5f]u 1 + 2=0[D6f]〈u0; u1〉+2=0[D7f]〈∇u0;∇u1〉 + 2=0[D8f]〈u 0; u 1〉; (4.14)

�0[B] = B[u0] ≡ B(t; ‖u0‖2; ‖∇u0‖2; ‖u 0‖2) (4.15)

and

�1[B] = 2�0[D6B]〈u0; u1〉 + 2�0[D7B]〈∇u0;∇u1〉 + 2�0[D8B]〈u 0; u 1〉; (4.16)

with 26 i6N ,

(Qi)

Ou i − B[u0]Cui = Fi[ui]; 0¡x¡ 1; 0¡t¡T;

∇ui(0; t) − h0ui(0; t) = ∇ui(1; t) + h1ui(1; t) = 0;

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

where

Fi[ui] = =i[f] + =i−1[f1] +i∑

k=1

(�k [B] + �k−1[B1])Cui−k ; (4.17)


with =i[f] = =i[f; u0; u1; : : : ; ui]; �i[B] = �i[B; u0; u1; : : : ; ui]; 26 i6N deDned by therecurrence formulas

=i[f] =i∑

k=0

i − ki

{=k [D3f]ui−k + =k [D4f]∇ui−k + =k [D5f]u i−k}

+2i

i−1∑k=0

i−k−1∑j=0

(i − k − j){=k [D6f]〈uj; ui−k−j〉 + =k [D7f]〈∇uj;∇ui−k−j〉

+ =k [D8f]〈u j ; u i−k−j〉}; 26 i6N; (4.18)

�i[B] =2i

i−1∑k=0

i−k−1∑j=0

(i − k − j){�k [D6B]〈uj; ui−k−j〉 + �k [D7B]〈∇uj;∇ui−k−j〉

+�k [D8B]〈u j ; u i−k−j〉}; 26 i6N: (4.19)

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

=i[f] = =0[D3f]ui + =0[D4f]∇ui + =0[D5f]u i + 2=0[D6f]〈u0; ui〉+2=0[D7f]〈∇u0;∇ui〉 + 2=0[D8f]〈u 0; u i〉 + terms depending on

×(i; =k [D>f]; uk ;∇uk ; u k); >= 3; 4; 5; 6; 7; 8; k = 0; 1; : : : ; i − 1: (4.20)

Similarly

�i[B] = 2�0[D6B]〈u0; ui〉 + 2�0[D7B]〈∇u0;∇ui〉 + 2�0[D8B]〈u 0; u i〉+ terms depending on (i; �k [D>B]; uk ;∇uk ; u k); >= 6; 7; 8;

k = 0; 1; : : : ; i − 1: (4.21)

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

iui ≡ u� − h satisDes the problem

Ov− B�[v+ h]Cv= F�[v+ h] − F�[h] + (B�[v+ h] − B�[h])Ch

+E�(x; t); 0¡x¡ 1; 0¡t¡T;

∇v(0; t) − h0v(0; t) = ∇v(1; t) + h1v(1; t) = 0;

v(x; 0) = v(x; 0) = 0; (4.22)

where

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

�iFi[ui]: (4.23)

Then, we have the following lemmas.


Lemma 4. The functions =i[f]; �i[B]; 06 i6N above are de3ned by the followingformulas:

=i[f] =1i!

@i

@�i(f[h])

∣∣∣∣�=0

; 06 i6N; (4.24)

�i[B] =1i!

@i

@�i(B[h])

∣∣∣∣�=0

; 06 i6N: (4.25)

Proof. (i) It is easy to see that

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

=1[f] =@@�

(f[h])∣∣∣∣�=0

: (4.26)

But@@�

(f[h]) =D3f[h]@@�h+ D4f[h]

@@�

∇h+ D5f[h]@@�h+ D6f[h]

@@�

(‖h‖2)

+D7f[h]@@�

(‖∇h‖2) + D8f[h]@@�

(‖h‖2): (4.27)

On the other hand, from the formulas

h=N∑i=0

�iui;@@�h=

N∑i=1

i�i−1ui;@@�

∇h=N∑i=1

i�i−1∇ui;

@@�

(‖h‖2) = 2⟨h;

@@�h⟩;

@@�

(‖∇h‖2) = 2⟨

∇h; @@�

∇h⟩;

@@�

(‖h‖2) = 2⟨h;@@�h⟩;

we have@@�h∣∣∣∣�=0

= u1;@@�

∇h∣∣∣∣�=0

= ∇u1; @@�h∣∣∣∣�=0

= u 1;

@@�

(‖h‖2)∣∣∣∣�=0

= 2〈u0; u1〉; @@�

(‖∇h‖2)∣∣∣∣�=0

= 2〈∇u0;∇u1〉;

@@�

(‖h‖2)∣∣∣∣�=0

= 2〈u 0; u 1〉: (4.28)

Hence, it follows from (4.26)–(4.28), that@@�

(f[h])∣∣∣∣�=0

= =0[D3f]u1 + =0[D4f]∇u1 + =0[D5f] u 1 + 2=0[D6f]〈u0; u1〉

+2=0[D7f]〈∇u0;∇u1〉 + 2=0[D8f]〈u 0; u 1〉 = =1[f]: (4.29)

Thus, (4.14) holds.


Suppose that we have deDned the functions =k [f]; =k [Djf]; j = 3; 4; 5; 6; 7; 8; k =0; 1; : : : ; i−1 from formulas (4.13), (4.14) and (4.24). Therefore, it follows from (4.27)that

@i

@�i(f[h]) =

@i−1

@�i−1

@@�

(f[h])

=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) +@k

@�k

(D7f[h])@i−k

@�i−k (‖∇h‖2) + @k

@�k(D8f[h])

@i−k

@�i−k (‖h‖2)}: (4.30)

We also note that

@i

@�ih∣∣∣∣�=0

= i!ui; 06 i6N: (4.31)

On the other hand,

@m

@�m(‖h‖2) = 2

@m−1

@�m−1

⟨h;

@@�h⟩= 2

m−1∑j=0

Cjm−1

⟨@j

@�j(h);

@m−j

@�m−j (h)⟩: (4.32)

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〉: (4.33)

Similarly

@m

@�m(‖∇h‖2)

∣∣∣∣�=0

= 2m−1∑j=0

j!(m− j)!Cjm−1〈∇uj;∇um−j〉; (4.34)

@m

@�m(‖h‖2)

∣∣∣∣�=0

= 2m−1∑j=0

j!(m− j)!Cjm−1〈u j ; um−j〉: (4.35)

It follows from (4.30), (4.31), (4.33)–(4.35) that

@i

@�i(f[h])

∣∣∣∣�=0

=i−1∑k=0

k!Cki−1

{=k [D3f]

@i−k

@�i−k (h)∣∣∣∣�=0

+ =k [D4f]@i−k

@�i−k (∇h)∣∣∣∣�=0

+ =k [D5f]@i−k

@�i−k (h)∣∣∣∣�=0

+ =k [D6f]@i−k

@�i−k (‖h‖2)∣∣∣∣�=0


+ =k [D7f]@i−k

@�i−k (‖∇h‖2)∣∣∣∣�=0

+ =k [D8f]@i−k

@�i−k (‖h‖2)∣∣∣∣�=0

}

=i−1∑k=0

(i − k)(i − 1)! {=k [D3f]ui−k + =k [D4f]∇ui−k

+=k [D5f]u i−k +2

i − k

i−k−1∑j=0

(i − k − j)(=k [D6f]〈uj; ui−k−j〉

+ =k [D7f]〈∇uj;∇ui−k−j〉 + =k [D8f]〈u j ; u i−k−j〉)} : (4.36)

Hence

1i!

@i

@�i(f[h])

∣∣∣∣�=0

=i−1∑k=0

i − ki

{=k [D3f]ui−k + =k [D4f]∇ui−k + =k [D5f]u i−k}

+2i

i−1∑k=0

i−k−1∑j=0

(i − k − j){=k [D6f]〈uj; ui−k−j〉

+ =k [D7f]〈∇uj;∇ui−k−j〉+ =k [D8f]〈u j ; u i−k−j〉} = =i[f]: (4.37)

Hence, the part 1 of Lemma 4 is proved.

(ii) In the case of B=B[h] =B(t; ‖h‖2; ‖∇h‖2; ‖h‖2). Applying the formulas (4.13),(4.14), (4.18) with f=f(t; U; V;W ), Dif=0, i=1; 3, 4; 5; D6f=D6B=@B=@U; D7f=D7B=@B=@V; D8f=D8B=@B=@W and =i[f]=�i[B], we obtain formulas (4.15), (4.16),(4.19) and later part of Lemma 4 is proved.

Remark 2. Lemma 4 is a generalization of a lemma contained in [14, p. 509, Lemma 4].

Lemma 5. Let (A1); (A2); (A7) and (A8) hold. Then there exists a constant K , suchthat

‖E�‖L∞(0;T ;L2)6 K |�|N+1; (4.38)

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

Ki(M; T; B) = sup06t6T;06U;V;W6M 2

∑�2+�6+�7+�8=i

|D�22 D

�66 D

�77 D

�88 B(t; U; V;W )|;

i = 1; 2; : : : ; N + 1;


Ki(M; T; B1) = sup06t6T;06U;V;W6M 2

∑�2+�6+�7+�8=i

|D�22 D

�66 D

�77 D

�88 B1(t; U; V;W )|;

i = 1; 2; : : : ; N;

Ki(M; T; f) = sup∑?

|D?11 D

?33 D

?44 D

?55 D

?66 D

?77 D

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

Ki(M; T; f1) = sup∑?

|D?11 D

?33 D

?44 D

?55 D

?66 D

?77 D

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

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

∑? is taken over ?= (?1; ?3; : : : ; ?8)∈Z7

+ satisfying |?|= ?1 +?3 + · · · + ?8 = 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 (4.24), 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

=i[f]�i + �N+1RN+1[f; �; @1]; (4.39)

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

=i[f1]�i + �NRN [f1; �; @2]; (4.40)

where =i[f]; 06 i6N are deDned by (4.13), (4.14), (4.18); RN+1[f; �; @1]and RN [f1; �; @2] are deDned as follows:

RN+1[f; �; @1] =1

(N + 1)!@N+1

@�N+1 (f[h])∣∣∣∣�=@1�

(4.41)

and

RN [f1; �; @2] =1N !

@N

@�N(f1[h])

∣∣∣∣�=@2�

; (4.42)

with 0¡@i ¡ 1; i = 1; 2:


Combining (4.39)–(4.42), we then obtain

F�[h] − f[u0] =f[h] − f[u0] + �f1[h]

=N∑i=1

(=i[f] + =i−1[f1])�i + �N+1RN [f;f1; �; @1; @2]; (4.43)

with

RN [f;f1; �; @1; @2] = RN+1[f; �; @1] + RN [f1; �; @2]: (4.44)

Similarly, we use Maclaurin’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 (4.25), 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

�i[B]�i + �N+1RN+1[B; �; @3]; (4.45)

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

�i[B1]�i + �N RN [B1; �; @4]; (4.46)

where

RN+1[B; �; @3] =1

(N + 1)!@N+1

@�N+1 (B[h])∣∣∣∣�=@3�

(4.47)

and

RN [B1; �; @4] =1N !

@N

@�N(B1[h])

∣∣∣∣�=@4�

; (4.48)

with 0¡@i ¡ 1; i = 3; 4.Combining (4.45)–(4.48), we then obtain

B�[h] − B[u0] = B[h] − B[u0] + �B1[h]

=N∑i=1

(�i[B] + �i−1[B1])�i + �N+1RN [B; B1; �; @3; @4]; (4.49)

with

RN [B; B1; �; @3; @4] = RN+1[B; �; @3] + RN [B1; �; @4]: (4.50)

Hence

(B�[h] − B[u0])Ch


=

[N∑i=1

(�i[B] + �i−1[B1])�i] N∑

j=0

�jCuj

+ �N+1ChRN [B; B1; �; @3; @4]

=N 2∑i=1

[i∑

k=1

(�k [B] + �k−1[B1])Cui−k

]�i + �N+1ChRN [B; B1; �; @3; @4]

=N∑i=1

[i∑

k=1

(�k [B] + �k−1[B1])Cui−k

]�i + �N+1R(1)

N [B; B1; h; �; @3; @4]; (4.51)

with

R(1)N [B; B1; h; �; @3; @4] =ChRN [B; B1; �; @3; @4]

+N 2∑

i=N+1

[i∑

k=1

(�k [B] + �k−1[B1])Cui−k

]�i−N−1: (4.52)

Combining (4.12)–(4.17), (4.23), (4.43), (4.44), (4.51) and (4.52), we then obtain

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

�iFi[ui]

=N∑i=1

[=i[f] + =i−1[f1] +

i∑k=1

(�k [B] + �k−1[B1])Cui−k − Fi[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]): (4.53)

By the boundedness of the functions ui;∇ui; u i; i = 0; 1; 2 in the function space L∞

(0; T ;H 1), we obtain from (4.41), (4.42), (4.44), (4.47), (4.48), (4.50), (4.52) and(4.53) that

‖E�‖L∞(0;T ;L2)6 K |�|N+1; (4.54)

where K is a constant depending only on M; T; N and the constants Ki(M; T; B), Ki(M; T; f); i = 1; 2; : : : ; N + 1, Ki(M; T; B1), Ki(M; T; f1); i = 1; 2; : : : ; N ,

The proof of Lemma 5 is complete.

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

v0 ≡ 0;

Ovm − B�[vm−1 + h]Cvm = F�[vm−1 + h] − F�[h] + (B�[vm−1 + h] − B�[h])Ch

+E�(x; t); 0¡x¡ 1; 0¡t¡T; (4.55)


∇vm(0; t) − h0vm(0; t) = ∇vm(1; t) + h1vm(1; t) = 0;

vm(x; 0) = vm(x; 0) = 0; m¿ 1:

With m= 1, we have the problem

Ov1 − B�[h]Cv1 = E�(x; t); 0¡x¡ 1; 0¡t¡T;

∇v1(0; t) − h0v1(0; t) = ∇v1(1; t) + h1v1(1; t) = 0;

v1(x; 0) = v1(x; 0) = 0: (4.56)

By multiplying the two sides of (4.56) by v1, we Dnd without di8culty from (4.38)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 (4.57)

where

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

We have

b=1; �(t) =D2B[h] + �D2B1[h] + 2(D6B[h] + �D6B1[h])〈h(t); h(t)〉

+2(D7B[h] + �D7B1[h])〈∇h(t);∇h(t)〉+2(D8B[h] + �D8B1[h])〈h(t); Oh(t)〉; (4.58)

hence

|b=1; �(t)|6 [1 + 6(N + 1)2M 2](K1(M; T; B) + K1(M; T; B1)) ≡ A1: (4.59)

It follows from (4.57), (4.59) that

‖v1(t)‖2 + b0C0‖v1(t)‖2H 1 6 2K |�|N+1T‖v1‖L∞(0;T ;L2)

+C1A1

∫ t

0‖v1(s)‖2H 1 ds: (4.60)

Using Gronwall’s lemma we obtain

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

1√b0C0

)T K |�|N+1

× exp(C1A1Tb0C0

): (4.61)

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

‖vm‖L∞(0;T ;L2) + ‖vm‖L∞(0;T ;H 1)6CT |�|N+1; |�|6 1; for all m: (4.62)


By multiplying the two sides of (4.55) with vm and after integration in t, we obtain

‖vm(t)‖2 + b0C0‖vm(t)‖2H 1

6∫ t

0|b=m;�(s)|a(vm(s); vm(s)) 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]| ‖Ch‖ ‖vm‖ ds

+2∫ t

0|B1[vm−1 + h] − B1[h]| ‖Ch‖‖vm‖ ds+ 2K |�|N+1

∫ t

0‖vm‖ ds; (4.63)

where

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

b=m;�(t) =D2B[vm−1 + h] + �D2B1[vm−1 + h]

+ 2(D6B[vm−1 + h] + �D6B1[vm−1 + h])〈vm−1 + h; vm−1 + h〉+2(D7B[vm−1 + h] + �D7B1[vm−1 + h])〈∇vm−1 + ∇h;∇vm−1 + ∇h〉+2(D8B[vm−1 + h] + �D8B1[vm−1 + h])〈vm−1 + h; Ovm−1 + Oh〉:

Hence

|b=m;�(t)|6 [1 + 6(N + 2)2M 2](K1(M; T; B) + K1(M; T; B1)) ≡ A2: (4.64)

By (4.63) and (4.64), after some lengthy calculations we can prove the followinginequality:

‖vm‖W1(T )6 :‖vm−1‖W1(T ) + � for all m¿ 1; (4.65)

where

: = (1 +√2)AT ;T ; �= AT

√TK |�|N+1;

AT =(1 +

1√b0C0

)exp(12T(1 +

C1A2b0C0

));

;T = (1 + 2M)√T [K1(M; T; f) + K1(M; T; f1)]

+ 2M 2√T [K1(M; T; B) + K1(M; T; B1)]:

We assume that

:¡ 1; with the suitable constant T ¿ 0: (4.66)

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


Lemma 6. Let the sequence {Bm} satisfy

Bm6 :Tm−1 + � for all m¿ 1; B0 = 0; (4.67)

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

Bm6 �=(1 − :) for all m¿ 1: (4.68)

Applying Lemma 6 with Bm = ‖vm‖W1(T ), it follows from (4.65), that

‖vm‖L∞(0;T ;L2) + ‖vm‖L∞(0;T ;H 1) = ‖vm‖W1(T )6 �=(1 − :) = CT |�|N+1; (4.69)

where CT = AT√TK=1 − (1 +

√2)AT ;T .

On the other hand, the linear recurrent sequence {vm} deDned by (4.55) convergesstrongly in the space W1(T ) to the solution v of problem (4.22). Hence, letting m →+∞ in (4.69), we have

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

or ∥∥∥∥∥u � −N∑i=0

�iu i

∥∥∥∥∥L∞(0;T ;L2)

+

∥∥∥∥∥u� −N∑i=0

�iui

∥∥∥∥∥L∞(0;T ;H 1)

6CT |�|N+1: (4.70)

Thus, we have the following theorem.

Theorem 4. Let (A1); (A2); (A7) and (A8) hold. Then there exist constants M ¿ 0 andT ¿ 0 such that, for every �, with |�|6 1, the problem (P�) has a unique weak solutionu� ∈W (M; T ) satisfying an asymptotic estimation up to order N + 1 as in (4.70),the functions u0; u1; : : : ; uN being the weak solutions of problems (P0); (Q1); : : : ; (QN ),respectively.

Remark 3.

• In the case of B ≡ 1; B1 ≡ 0; f1 ≡ 0; f= f(t; u; ut); f∈CN+1(R+ × R2), and theDirichlet homogeneous condition (1.8) standing for (1.2), we have obtained theresults above in the paper [4].

• In the case of functions f∈C2([0; 1] × R+ × R3); f1 ∈C1([0; 1] × R+ × R3) andN = 1, we have also obtained some results concerning in the papers [9,12,13] inthe cases as follows:(a) B ≡ 1; B1 ≡ 0; (see [12]).(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 (1.8) (see [13]).

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

+); B¿ b0¿ 0 and B1¿ 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]×R+×R3); f1 ∈CN ([0; 1]× [0; 1]×R+×R3), we have also obtainedsome results above in the paper [10].


• In the case of B� ≡ B(‖ux‖2)+�B1(‖ux‖2); F�=f(x; t; u; ux; ut ; ‖ux‖2)+�f1(x; t; u;ux; ut ; ‖ux‖2), B∈CN+1(R+), B1 ∈CN (R+), B¿ b0¿ 0, B1¿ 0, f∈CN+1([0; 1]×R+×R3×R+); f1 ∈CN ([0; 1]×R+×R3×R+), and (1.2) standing for the conditionux(0; t) − h0u(0; t) = u(1; t) = 0, we have also obtained some results above in thepaper [14].

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