Asymptotic expansion of the solution for nonlinear wave equation with the mixed homogeneous conditions

Nonlinear Analysis 45 (2001) 261 – 272

Asymptotic expansion of the solution for nonlinear wave equation with the mixed homogeneous conditions Nguyen Thanh Long Department of Applied Mathematics, Polytechnic University of Ho Chi Minh City, 268 Ly Thuong Kiet Str. Dist. 10, HoChiMinh City, Viet Nam Received 15 September 1998; accepted 2 July 1999

Keywords: Weak solution; Perturbed problem; Linear recurrent sequence; Asymptotic expansion

1. Introduction In this paper, we study an asymptotic expansion of solution in the small parameter
of the following initial and boundary value problem: utt − uxx = F
(x; t; u; ux ; ut );

x ∈ 3 = (0; 1);

0 ¡ t ¡ T;

(1.1)

ux (0; t) − h0 u(0; t) = ux (1; t) + h1 u(1; t) = 0;

(1.2)

u(x; 0) = u˜ 0 (x);

(1.3)

ut (x; 0) = u˜ 1 (x);

where h0 , h1 are the given nonnegative constants with h0 + h1 ¿ 0; u˜ 0 , u˜ 1 are the given functions with u˜ 0 ∈ H 2 (3), u˜ 1 ∈ H 1 (3) and F
is a given function depending on the small parameter
satisfying F
∈ C 1 (3 × [0; ∞) × R3 ):

(1.4)

In [1], Alain Pham has studied the existence and asymptotic behavior as
→ 0 of a weak solution of problem (1.1), (1.3) associated with the Dirichlet homogeneous boundary condition u(0; t) = u(1; t) = 0; E-mail address: [email protected] (N.T. Long). 0362-546X/01/$ - see front matter ? 2001 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 9 ) 0 0 3 3 2 - 6

(1.5)

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N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

where the nonlinear term has the form F
=
f(t; u):

(1.6)

By a generalization of [1], Alain Pham and Long [2] have considered problem (1.1), (1.3), (1.5) with the nonlinear term having the form F
=
f(t; u; ut ): N

(1.7) 2

If f ∈ C ([0; ∞) × R ) satisCes f(t; 0; 0) = 0 for all t ≥ 0, an asymptotic expansion of the solution of problem (1.1), (1.3), (1.5), (1.7) up to order N + 1 in
is obtained, for
suDciently small, that extends to partial diEerential equation (1.1) the results obtained in diEerential equations. In [4], Long and Diem have studied problem (1.1) – (1.3) with the nonlinear term F
= f(x; t; u; ux ; ut ) +
g(x; t; u; ux ; ut ): 3

2

(1.8) 1

3

In the case of f ∈ C (3 × [0; ∞) × R ) and g ∈ C (3 × [0; ∞) × R ), an asymptotic expansion of order 2 in
is obtained, for
suDciently small. The aim of this paper is to give a generalization of [1,2,4] for the problem with F
= f(
; x; t; u; ux ; ut ) +
g(
; x; t; u; ux ; ut ):

(1.9)

If the functions f ∈ C N +1 (IF × 3 × [0; ∞) × R3 ) and g ∈ C N (IF × 3 × [0; ∞) × R3 ); I = (−1; 1), then an asymptotic expansion of the solution of problem (1.1) – (1.3), (1.9) up to order N + 1 in
is obtained, for
suDciently small. This result is a relative generalization of that in [1– 4]. 2. Preliminary results, notations, function spaces We Crst set some notations 3 = (0; 1); Lp = Lp (3);

H 1 = H 1 (3);

H 2 = H 2 (3);

where H 1 ; H 2 are the usual Sobolev spaces on 3. 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 in L2 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 ); 1 ≤ p ≤ ∞ the Banach space of the measurable functions f : (0; 1) → X such that
 T 1=p p

f LP (0; T ; X ) =

f(t) X dt ¡ ∞ for 1 ≤ p ¡ ∞ 0

and

f L∞ (0; T ; X ) = ess sup f(t) X : 0¡t¡T

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

We put a(u; v) =



1

0

u
(x)v
(x) d x + h0 u(0)v(0) + h1 u(1)v(1):

In H 1 , we shall use the following equivalent norm:  1=2 1
2 2

v H 1 = |v (x)| d x + v (0) : 0

263

(2.1)

(2.2)

Then we have the following lemmas. Lemma 1. The imbedding H 1 ,→ C 0 (3) is compact and √

v C 0 (3) ≤ 2 v H 1 for all v ∈ H 1 :

(2.3)

Lemma 2. Let h0 ¿ 0; h1 ≥ 0. Then the symmetric bilinear form a(: ; :) de4ned by (2:1) is continuous on H 1 × H 1 and coercive on H 1 ; i.e; (i) |a(u; v)| ≤ C1 u H 1 v H 1 for all u; v ∈ H 1 ; 2 (ii) a(v; v) ≥ C0 v H 1 for all v ∈ H 1 ; where C0 = min{1; h0 };

C1 = max{1; h0 ; 2h1 }:

(2.4)

The proofs of these lemmas are straightforward and we omit the details. We make the following assumptions: (H1 ) h0 ¿ 0; h1 ≥ 0; (H2 ) u˜ 0 ∈ H 2 ; u˜ 1 ∈ H 1 ; (H3 ) f = F0 ∈ C 1 (3 × [0; ∞) × R3 ) (corresponding to
= 0): Given T ¿ 0; M ¿ 0, we put K0 = K0 (M; T; f) = sup |f(x; t; u; v; w)|;

(2.5)

K1 = K1 (M; T; f) = sup (|fx
| + |ft
| + |fu
| + |fv
| + |fw
|)(x; t; u; v; w);

(2.6) √ where, in each case, sup is taken over 0 ≤ x ≤ 1; 0 ≤ t ≤ T; |u|; |v|; |w| ≤ M 2. The notations u(t); u(t); ˙ uJ (t); ux = ∇u and uxx = Ku stand for u(x; t); (@u=@t)(x; t); (@2 u=@t 2 )(x; t); (@u=@x)(x; t) and (@2 u=@x2 )(x; t); respectively. For each T ¿ 0 and M ¿ 0, we set W (M; T ) = {v ∈ L∞ (0; T ; H 2 ): v˙ ∈ L∞ (0; T ; H 1 ); vJ ∈ L∞ (0; T ; L2 );

v L∞ (0; T ; H 2 ) ≤ M; v

˙ L∞ (0; T ; H 1 ) ≤ M; v

J L∞ (0; T ; L2 ) ≤ M }:

(2.7)

We shall choose the Crst term u0 ∈ W (M; T ). Suppose that um−1 ∈ W (M; T ):

(2.8)

We associate with problem (1.1) – (1.3) corresponding to F
=f the following problem.

264

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

Find um ∈ W (M; T ) which satisCes the linear variational problem uJ m ; v + a(um ; v) = F˜ m ; v for all v ∈ H 1 ; um (0) = u˜ 0 ;

(2.9)

u˙ m (0) = u˜ 1 ;

(2.10)

where F˜ m (x; t) = f(x; t; um−1 (x; t); ∇um−1 (x; t); u˙ m−1 (x; t)):

(2.11)

Then we have the following theorems. Theorem 1. Let (H1 )– (H3 ) hold. Then there exist constants M ¿ 0 and T ¿ 0 such that; for every u0 ∈ W (M; T ); there exists a linear recurrent sequence {um } ⊂ W (M; T ) de4ned by (2:9) – (2:11). Theorem 2. Let (H1 )– (H3 ) hold. Then there exist M ¿ 0 and T ¿ 0 such that problem (1:1) –(1:3) corresponding to F
= f has a unique weak solution u ∈ W (M; T ). On the other hand, the linear recurrent sequence {um } deCned by (2.9) – (2.11) converges to the solution u strongly in the following space: W1 (T ) = {v ∈ L∞ (0; T ; H 1 ): v˙ ∈ L∞ (0; T ; L2 )}:

(2.12)

Furthermore, we have also the estimation

um − u L∞ (0; T ; H 1 ) + u˙ m − u

˙ L∞ (0; T ; L2 ) ≤ CkTm where kT = 2(1 +

√

2)(1 + 1=

for all m;

 C0 )TK1 (M; T; f) ¡ 1

(2.13)

(2.14)

and C is a constant depending only on T; u0 ; u1 and kT . The proof of Theorems 1 and 2 can be found in [4]. Remark 1. The constants M ¿ 0 (suDciently large) and T ¿ 0 (suDciently small) in Theorems 1 and 2 can be chosen as follows (see [4]): 2

2

Ku˜ 0 + a(u˜ 0 ; u˜ 0 ) + u˜ 1 + 2a(u˜ 1 ; u˜ 1 ) + ( Ku˜ 0 + f(x; 0; u˜ 0 ; ∇u˜ 0 ; u˜ 1 ) )2 ¡ M 2 =4; TK(M; T; f) ≤ M

(2.15) (2.16)

and satisfy (2.14), where   2C1 K(M; T; f) = 2K0 + √ (2K1 1 + 2M 2 + K0 ) + 4K1 1 + 3M 2 : C0

(2.17)

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

265

3. Asymptotic expansion of solutions In this section, we assume that (h0 ; h1 ) and (u˜ 0 ; u˜ 1 ) satisfy assumptions (H1 ) and (H2 ), respectively. We also assume that (H4 ) f; g ∈ C 1 (IF × 3 × [0; ∞) × R3 ); I = (−1; 1): We consider the following perturbed problem, where
is a small parameter, |
| ≤ 1: utt − uxx = F
(x; t; u; ux ; ut ); x ∈ 3; 0 ¡ t ¡ T; (P
)

ux (0; t) − h0 u(0; t) = ux (1; t) + h1 u(1; t) = 0; u(x; 0) = u˜ 0 (x); ut (x; 0) = u˜ 1 (x); F
(x; t; u; ux ; ut ) = f(
; x; t; u; ux ; ut ) +
g(
; x; t; u; ux ; ut ):

First, we note that if the functions f; g satisfy (H4 ), then we can prove in a manner similar to the proof of Theorem 1 in [4], that the a priori estimates of the Galerkin (k) approximation sequence {um } for problem (1.1) – (1.3) corresponding to F
= f(
; :) +
g(
; :); |
| ≤ 1, satisfy (k) um ∈ W (M; T );

(3.1)

where M , T are constants independent of
. Besides, in the processing we choose the positive constants M and T as in (2.14) – (2.17), wherein f(x; 0; u˜ 0 ; ∇u˜ 0 ; u˜ 1 ) and Ki (M; T; f); i = 0; 1, stand for sup ( f(
; x; 0; u˜ 0 ; ∇u˜ 0 ; u˜ 1 ) + g(
; x; 0; u˜ 0 ; ∇u˜ 0 ; u˜ 1 ) )

|
|≤1

and sup (Ki (M; T; f(
; :)) + Ki (M; T; g(
; :)));

|
|≤1

i = 0; 1;

respectively. (k) Hence, the limit u
in suitable function spaces of the sequence {um } as k → + ∞, afterwards m → + ∞, is a unique weak solution of the problem (P
) satisfying u
∈ W (M; T ):

(3.2)

Then we can prove, in a manner similar to the proof of Theorem 2 in [4], that the limit u0 in suitable function spaces of the family {u
} as
→ 0, is a unique weak solution of the problem (P0 ) corresponding to
= 0 satisfying u0 ∈ W (M; T ):

(3.3)

Hence, we have the following theorem. Theorem 3. Let (H1 ); (H2 ) and (H4 ) hold. Then there exist constants M ¿ 0 and T ¿ 0 such that; for every
; with |
| ≤ 1; the 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 ) ≤ C|
|;

(3.4)

where C is a constant depending only on h0 ; h1 ; T;K 0 (M; T; g) = sup|
|≤1 K0 (M; T; g(
; :)) and K 1 (M; T; f) = sup|
|≤1 K1 (M; T; f(
; :)).

266

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

Proof. Put v = u
− u0 . Then v satisCes the variational problem v; J w + a(v; w) = f˜
; w +
g(
; :); w for all w ∈ H 1 ; v(0) = v(0) ˙ = 0;

(3.5)

f˜
= f˜
(x; t) = f(
; x; t; u
; ∇u
; u˙
) − f(0; x; t; u0 ; ∇u0 ; u˙ 0 ); g(
; :) = g(
; x; t; u
; ∇u
; u˙
): We take w = v˙ in (3.5), after integration in t, 2

2

2

v(t)

˙ + a(v(t); v(t)) ≤
2 T (K 1 (M; T; f) +K 0 (M; T; g)) + (2 + (6 + 3=C0 )  t 2 ×K 1 (M; T; f)) ( v(s)

˙ + a(v(s); v(s))) ds: (3.6) 0

Next, by (3.6) and Gronwall’s lemma, we have 2

v(t)

˙ + a(v(t); v(t)) ≤
2 C2 exp(C3 t)

for all t ∈ [0; T ];

(3.7)

where 2

2

C2 = T (K 1 (M; T; f) +K 0 (M; T; g)); C3 = 2 + (6 + 3=C0 )K 1 (M; T; f):

(3.8)

Hence, from (3.7) and (3.8) we have (3.4). The next result gives an asymptotic expansion of the weak solution u
of order N +1 in
, for
suDciently small. Now, we assume that (H5 ) f ∈ C N +1 (IF × 3 × [0; ∞) × R3 ); g ∈ C N (IF × 3 × [0; ∞) × R3 ): We use the following notations. f(
; [u]) = f(
; x; t; u; ux ; ut ); F[u] = F(x; t; u; ux ; ut ); L=

@2 @2 − 2; 2 @t @x

B0 v = vx (0; t) − h0 v(0; t); B1 v = vx (1; t) + h1 v(1; t): Let u0 ∈ W (M; T ) be a weak solution of problem (P0 ) as in Theorem 3. Let us consider the sequence of weak solutions {up }; p = 1; 2; : : : ; N; up ∈ W (M; T ) (with suitable constants M ¿ 0; T ¿ 0) deCned by the following problems: Lu1 = Fˆ 1 [u1 ]; x ∈ 3; 0 ¡ t ¡ T; (P˜ 1 )

Bi u1 = 0; i = 0; 1; u1 (x; 0) = u˙ 1 (x; 0) = 0;

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

267

where Fˆ 1 [u1 ] = g(0; [u0 ]) + f

(0; [u0 ]) + fu
(0; [u0 ])u1 +fu
x (0; [u0 ])∇u1 + fu˙
(0; [u0 ])u˙ 1 ;

(3.9)

with 2 ≤ p ≤ N , Lup = Fˆ p [up ]; x ∈ 3; 0 ¡ t ¡ T; (P˜p )

Bi up = 0; i = 0; 1; up (x; 0) = u˙ p (x; 0) = 0;

where Fˆ p [up ] = C[p; f] + C[p − 1; g]; C[p; f] =

p  k=1

 4 i=1

ki =k

with Q(p; k; k1 ; k2 ; k3 ; k4 ) =

(3.10)

1 (k) (0; [u0 ]):Q(p; k; k1 ; k2 ; k3 ; k4 ) f k4 ! k 1 k 2 k 3 k 4



(3.11)

(u1 )&1 : : : (up−k4 )&p−k4

&i ; 'i ; ( i

×(∇u1 )'1 : : : (∇up−k4 )'p−k4 :(u˙ 1 )(1 : : : (u˙ p−k4 )(p−k4 ×

1 ; &1 ! : : : &p−k4 !'1 ! : : : 'p−k4 !(1 ! : : : (p−k4 !

the &i ; 'i ; (i integers ≥ 0 satisfying p−k4



p−k4

&i = k 1 ;

i=1



p−k4



'i = k 2 ;

i=1

i=1

p−k4

(i = k 3 ;



i(&i + 'i + (i ) = p − k4 :

i=1

Here we have used the notation fk(k) = 1 k2 k 3 k4

@kf : @uk1 @uxk2 @u˙k3 @
k4

We also note that C[p; f] is the Crst-order function with respect to up ; ∇up and u˙p , respectively. In fact, C[1; f] = fu
(0; [u0 ])up + fu
x (0; [u0 ])∇up + fu˙
(0; [u0 ])u˙p

(3.12)

and with p ≥ 2 we have C[p; f] = C[1; f] +

p  k=2

 4 i=1

ki =k

1 (k) ˜ (0; [u0 ]):Q(p; k; k1 ; k2 ; k3 ; k4 ) f k4 ! k 1 k 2 k 3 k 4

(3.13)

268

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

with



˜ Q(p; k; k1 ; k2 ; k3 ; k4 ) =

(u1 )&1 : : : (up−1 )&p−1

&i ; 'i ; ( i

× (∇u1 )'1 : : : (∇up−1 )'p−1 :(u˙ 1 )(1 : : : (u˙ p−1 )(p−1 ×

1 &1 ! : : : &p−1 !'1 ! : : : 'p−1 !(1 ! : : : (p−1 !

and the &i , 'i , (i integers ≥ 0 satisfying p−1



p−1



&i = k 1 ;

i=1

p−1



'i = k 2 ;

i=1

p−1

(i = k 3 ;

i=1



i(&i + 'i + (i ) = p − k4 :

i=1

Let u
∈ W (M; T ) be a unique weak solution of the problem (P
). Then N  v = u
− h = u
− u0 −
p up p=1

satisCes the problem Lv = f(
; [v + h]) − f(
; [h]) +
(g(
; [v + h]) −g(
; [h])) + e(
; x; t); Bi v = 0;

x ∈ 3; 0 ¡ t ¡ T;

(3.14)

i = 0; 1;

v(x; 0) = v(x; ˙ 0) = 0; where e(
; x; t) = f(
; [u0 + U ]) − f(0; [u0 ]) +
(g(
; [u0 + U ]) − g(0; [u0 ])) +
(g(0; [u0 ]) − Fˆ 1 [u0 ]) −

N 


p Fˆp [up ]

(3.15)

p=2

with U=

N 


p up :

p=1

Now, we put the following notations: Kˆ k (T; f) = max sup |fk(k) (0; [u0 ])|; 1 k 2 k 3 k4 k1 +k2 +k3 +k4 =k

k = 1; 2; : : : ; N;

(3.16)

0≤x≤1;

0≤t≤T

K˜ N +1 (M; T; f) = K˜ N (M; T; g) =

max

k1 +k2 +k3 +k4 =N +1

max

k1 +k2 +k3 +k4 =N

sup|fk(N1 k+1) (
; [u])|; 2 k3 k4

sup|gk(N1 k)2 k3 k4 (
; [u])|;

(3.17) (3.18)

where,√in each case, sup is taken over −1 ≤
≤ 1, x ∈ 3, 0 ≤ t ≤ T , |u|; |ux |; |u| ˙ ≤ 2(N + 1)M .

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

269

For a multi-index & = (&1 ; &2 ; : : : ; &N ) ∈ Z+N , we put |&| = &1 + &2 + · · · + &N ; ,(&) F =

N 

i&i ;

,(&) =

&! = &1 !&2 ! : : : &N !;

N 

(i − 1)&i = |&| + ,(&); F

i=2

i=1

u = (u1 ; u2 ; : : : ; uN ); u& = (u1 )&1 :(u2 )&2 : : : (uN )&N ;

(3.19)

ux& = (∇u1 )&1 :(∇u2 )&2 : : : (∇uN )&N ; ut& = (u˙ 1 )&1 :(u˙ 2 )&2 : : : (u˙ N )&N : Then, we have the following lemma.

Lemma 3. Let (H1 ); (H2 ) and (H5 ) hold. Then there exists a constant K˜ such that ˜ N +1 ;

e(
; x; t) L∞ (0; T ; L2 ) ≤ K|
|

(3.20)

where K˜ is a constant depending only on N; M; T and the following constants Kˆ k (T; f);

Kˆ k (T; g);

k = 1; 2; : : : ; N;

k = 1; 2; : : : ; N − 1;

K˜ N +1 (M; T; f) and K˜ N (M; T; g): Proof. In the case of N = 1, the proof of Lemma 3 is easy, hence we omit the details, which we only prove with N ≥ 2. By using Taylor’s expansion of the function f(
; [u0 +U ]) round the point (0; x; t; u0 ; ∇u0 ; u˙ 0 ) up to order N + 1, we obtain after some rearrangements in order of
, that f(
; [u0 + U ]) − f(0; [u0 ]) =

N 

2

C[p; f]
p +

p=1

N 

˜ p; f]
p C[N;

p=N +1

+ RN +1 (f;
; /1 );

0 ¡ /1 ¡ 1;

(3.21)

˜ p; f] and RN +1 (f;
; /1 ) are deCned as follows where C[p; f] is deCned by (3.11), C[N; N 

˜ p; f] = C[N;

k=1

where, the sum |&| = k1 ;

 &; '; (

 4 i=1

ki =k

 u& ux' ut( 1 (k) fk1 k2 k3 k4 (0; [u0 ]): ; k4 ! &!'!(!

(3.22)

&; '; (

is taken over &; '; ( ∈ Z+N satisfying

|'| = k2 ;

|(| = k3 ;

,(& F + ' + () = p − k4

(3.23) (3.24)

270

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

and



RN +1 (f;
; /1 ) =
N +1

k1 +k2 +k3 +k4 =k



×

|&|=k1 ;|'|=k2 ;|(|=k3

fk(N1 k+1) (/1
; [u0 + /1 U ]) 2 k 3 k4 u& ux' ut( ,(&+'+() :
&!'!(!

(3.25)

Similarly, when we use Taylor’s expansion of the function g(
; [u0 + U ]) up to order N round the point (0; x; t; u0 ; ∇u0 ; u˙ 0 ), we obtain g(
; [u0 + U ]) − g(0; [u0 ]) =

N −1 

C[p; g]
p +

N (N −1)



˜ − 1; p; g]
p C[N

p=N

p=1

+ RN (g;
; /2 );

(0 ¡ /2 ¡ 1):

(3.26)

We deduce from (3.10), (3.11), (3.15), (3.21) and (3.26) that 2

e(
; x; t) =

N 

˜ p; f]
p + C[N;

N (N −1)



˜ − 1; p; g]
p+1 C[N

p=N

p=N +1

+ RN +1 (f;
; /1 ) + RN (g;
; /2 ):

(3.27)

By the boundedness of the functions ui ; ∇ui ; u˙ i ; i=0; 1; : : : ; N , in the function space L∞ (0; T ; H 1 ), we obtain from (3.22), (3.25) and (3.27) that ˜ N +1 ;

e(
; x; t) L∞ (0; T ; L2 ) ≤ K|
|

(3.28)

where K˜ = (N 2 − N − 1)

N N −1   (4NM )k ˆ (4NM )k ˆ K k (T; f) + (N 2 − 2N ) K k (T; g) k! k! k=1

+

k=1

(4NM )N ˜ (4NM )N +1 ˜ K N +1 (M; T; f) + K N (M; T; g): (N + 1)! N!

(3.29)

The proof of Lemma 3 is complete. Now, we consider the sequence of functions {vm } deCned by v0 ≡ 0; Lvm = f(
; [vm−1 + h]) − f(
; [h]) +
(g(
; [vm−1 + h]) −g(
; [h])) + e(
; x; t); Bi vm = 0;

i = 0; 1;

vm (x; 0) = v˙m (x; 0) = 0;

m ≥ 1:

x ∈ 3; 0 ¡ t ¡ T;

(3.30)

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

271

With m = 1, we have the problem Lv1 = e(
; x; t); x ∈ 3; 0 ¡ t ¡ T; Bi v1 = 0;

i = 0; 1;

(3.31)

v1 (x; 0) = v˙1 (x; 0) = 0: By multiplying the two sides of (3.31) with v˙1 , we Cnd without diDculty from (3.20) that 2 ˜ N +1 T v˙1 ∞ (3.32)

v˙1 + a(v1 ; v1 ) ≤ 2K|
| L (0; T ; L2 ) : This implies that

v1 L∞ (0; T ; H 1 ) + v˙1 L∞ (0; T ; L2 ) ≤ (1 + 1=

 ˜ |
|N +1 : C0 )2KT

(3.33)

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

vm L∞ (0; T ; H 1 ) + v˙m L∞ (0; T ; L2 ) ≤ CT |
|N +1 ;

|
| ≤ 1;

for all m:

(3.34)

By multiplying the two sides of (3.30) with v˙m and after integration in t, we obtain  t 2 [ f(
; [vm−1 + h]) − f(
; [h])

v˙m + a(vm ; vm ) ≤ 2 0

+ g(
; [vm−1 + h]) − g(
; [h]) ] ds: v˙m L∞ (0; T ; L2 ) ˜ N +1 T v˙m ∞ + 2K|
| L (0; T ; L2 ) :

(3.35)

Put Zm = vm L∞ (0; T ; H 1 ) + v˙m L∞ (0; T ; L2 ) :

(3.36)

We deduce from (3.35) that Zm ≤ 0Zm−1 + 1; with 0 = 2(1 +

√

for all m ≥ 1

(3.37)

√ 2)(1 + 1= C 0 )T sup [K1 (M; T; f(
; :)) + K1 (M; T; g(
; :))];

 ˜ |
|N +1 : 1 = 2(1 + 1= C0 )KT

|
|≤1

(3.38)

We assume that 0¡1

(3.39)

with the suitable constant T ¿ 0. We shall now require the following lemma the proof of which is immediate. Lemma 4. Let the sequence {Zm } satisfy 0 ≤ Zm ≤ 0Zm−1 + 1; m = 1; 2; : : : ; Z0 = 0;

(3.40)

where 0 ¡ 0 ¡ 1; 1 ≥ 0 are the given constants. Then Zm ≤ 1=(1 − 0);

for all m ≥ 1:

(3.41)

272

N.T. Long / Nonlinear Analysis 45 (2001) 261 – 272

We deduce from (3.36) – (3.39) and (3.41) that

vm L∞ (0; T ; H 1 ) + v˙m L∞ (0; T ; L2 ) ≤ CT |
|N +1 ; where CT = 2(1 + 1=



(3.42)

˜ C0 )KT=(1 − 0):

(3.43)

On the other hand, using Theorem 2 for problem (3.14), the linear recurrent sequence {vm } deCned by (3.30) converges strongly in the space W1 (T ) to the solution v of problem (3.14). Hence, letting m → + ∞ in (3.42), we have

v L∞ (0; T ; H 1 ) + v

˙ L∞ (0; T ; L2 ) ≤ CT |
|N +1 or

N  p u
−
up p=0

L∞ (0; T ; H 1 )

N  p + u˙
−
u˙ p p=0

≤ CT |
|N +1 :

(3.44)

L∞ (0; T ; L2 )

Thus, we have the following theorem. Theorem 4. Let (H1 ); (H2 ) and (H5 ) hold. Then there exist constants M ¿ 0 and T ¿ 0 such that; for every
; with |
| ≤ 1; the problem (P
) has a unique weak solution u
∈ W (M; T ) satisfying the asymptotic estimation up to order N + 1 as in (3:44); the functions up ; p=0; 1; : : : ; N being the weak solutions of problems (P˜p ); p=0; 1; : : : ; N; respectively. Remark 2. In the case of f=0; g=g(t; u; ut ) and the Dirichlet homogeneous condition (1.5) standing for (1.2), we have obtained the results in paper [2]. In the case of the functions f ∈ C 2 (3 × [0; ∞) × R3 );

g ∈ C 1 (3 × [0; ∞) × R3 )

which are independent of
and N = 1, we have obtained the results in [4]. Acknowledgements The author would like to thank the referee for useful comments and suggestions. References [1] Alain Pham Ngoc Dinh, Sur un problQeme hyperbolique faiblement non-linQeaire en demension 1, Demonstratio Math. 16 (1983) 269 –289. [2] Alain Pham Ngoc Dinh, Nguyen Thanh Long, Linear approximation and asymptotic expansion associated to the nonlinear wave equation in one dimension, Demonstratio Math. 19 (1986) 45 – 63. [3] J. Boujot, Alain Pham Ngoc Dinh, J.P. Veyrier, Oscillateurs harmoniques faiblement perturbQees: L’algorithme des “par de gQeants”, RAIRO, Anal. numQer. 14 (1980) 3–23. [4] Nguyen Thanh Long, Tran Ngoc Diem, On the nonlinear wave equation utt − uxx = f(x; t; u; ux ; ut ) associated with the mixed homogeneous conditions, Nonlinear Anal. 29 (1997) 1217–1230.