On a nonlinear wave equation associated with the boundary conditions involving convolution

Nonlinear Analysis 70 (2009) 3943–3965

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On a nonlinear wave equation associated with the boundary conditions involving convolution Le Thi Phuong Ngoc a , Le Nguyen Kim Hang b , Nguyen Thanh Long c,∗ a

Nhatrang Educational College, 01 Nguyen Chanh Street, Nhatrang City, Viet Nam

b

Faculty of Science, University of Agriculture and Forestry, HoChiMinh City, Linh Trung Ward, Thu Duc District, HoChiMinh City, Viet Nam

c

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

article

a b s t r a c t

info

Article history: Received 16 November 2007 Accepted 28 August 2008

The paper deals with the initial boundary value problem for the linear wave equation

 ∂  utt − (µ(x, t )ux ) + λut + F (u) = 0, 0 < x < 1, 0 < t < T ,   ∂x  Z t   µ(0, t )u (0, t ) = g (t ) + k0 (t − s)u(0, s)ds, x 0 0Z t     −µ(1, t )ux (1, t ) = g1 (t ) + k1 (t − s)u(1, s)ds,    0 u(x, 0) = u0 (x), ut (x, 0) = u1 (x),

MSC: 35L20 35L70 Keywords: Faedo–Galerkin method Existence and uniqueness of a weak solution Energy-type estimates Compactness Regularity of solutions Asymptotic expansion Stability of the solutions

(1)

where F , µ, g0 , g1 , k0 , k1 , u0 , u1 are given functions and λ is a given constant. The paper consists of four main parts. In Part 1, under conditions (u0 , u1 , g0 , g1 , k0 , k1 ) ∈ H 1 × L2 × (H 1 (0, T ))2 × (W 1,1 (0, T ))2 , µ ∈ W 1 (QRT ), µt ∈ L1 (0, T ; L∞ ), µ(x, t ) ≥ µ0 > 0, a.e. z (x, t ) ∈ QT ; the function F continuous, 0 F (s)ds ≥ −C1 z 2 − C10 , for all z ∈ R, with C1 , 0 C1 > 0 are given constants and some other conditions, we prove that, the problem (1) has a unique weak solution u. The proof is based on the Faedo–Galerkin method associated with the weak compact method. In Part 2 we prove that the unique solution u belongs to H 2 (QT ) ∩ L∞ (0, T ; H 2 ) ∩ C 0 (0, T ; H 1 ) ∩ C 1 (0, T ; L2 ), with ut ∈ L∞ (0, T ; H 1 ), utt ∈ L∞ (0, T ; L2 ), if we assume (u0 , u1 ) ∈ H 2 × H 1 , F ∈ C 1 (R) and some other conditions. In Part 3, with F ∈ C N +1 (R), N ≥ 2, we obtain an asymptotic expansion of the solution u of the problem (1) up to order N + 1 in a small parameter λ. Finally, in Part 4, we prove that the solution u of this problem is stable with respect to the data (λ, µ, g0 , g1 , k0 , k1 ). © 2008 Elsevier Ltd. All rights reserved.

1. Introduction In this paper, we consider the initial boundary value problem for the nonlinear wave equation

∂ (µ(x, t )ux ) + f (u, ut ) = 0, 0 < x < 1, 0 < t < T , ∂x Z t µ(0, t )ux (0, t ) = g0 (t ) + k0 (t − s)u(0, s)ds, utt −

0

∗

Corresponding author. E-mail addresses: [email protected], [email protected] (L.T. Phuong Ngoc), [email protected], [email protected] (L.N. Kim Hang), [email protected], [email protected] (N.T. Long). 0362-546X/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2008.08.004

(1.1) (1.2)

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−µ(1, t )ux (1, t ) = g1 (t ) +

t

Z

k1 (t − s)u(1, s)ds,

(1.3)

0

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

(1.4)

where f (u, ut ) = F (u) + λut , with λ is a constant, F , µ, g0 , g1 , k0 , k1 , u0 , u1 are given functions satisfying conditions specified later. In [1], An and Trieu studied a special case of problem (1.1) and (1.4) associated with the following boundary conditions: ux (0, t ) = g0 (t ) + h0 u(0, t ) +

t

Z

k0 (t − s)u(0, s)ds,

(1.5)

0

u(1, t ) = 0,

(1.6)

with µ ≡ 1, u0 = u1 ≡ 0, and f (u, ut ) = Ku + λut , with K ≥ 0, λ ≥ 0, h0 > 0 given constants and g0 , k0 are given functions. In the latter case, the problem (1.1) and (1.4)–(1.6) is a mathematical model describing the shock of a rigid body and a linear viscoelastic bar resting on a rigid base [1]. In [2] Bergounioux, Long and Dinh studied problem (1.1) and (1.4) with the mixed boundary conditions (1.2) and (1.3) standing for ux (0, t ) = g (t ) + hu(0, t ) −

t

Z

k(t − s)u(0, s)ds,

(1.7)

0

ux (1, t ) + K1 u(1, t ) + λ1 ut (1, t ) = 0,

(1.8)

where f (u, ut ) = Ku + λut , with K ≥ 0, λ ≥ 0, h ≥ 0, K1 ≥ 0, λ1 > 0 are given constants and g, k are given functions. In [11], Long, Dinh and Diem obtained the unique existence, regularity and asymptotic expansion of the problem (1.1), (1.4), (1.7) and (1.8) for the case of f (u, ut ) = K |u|p−2 u + λ |ut |q−2 ut , where K , λ ≥ 0, p, q ≥ 2 and (u0 , u1 ) ∈ H 2 × H 1 . In [12], Long, Ut and Truc gave the unique existence, stability, regularity in time variable and asymptotic expansion for the solution of problem (1.1) and (1.4) with the mixed boundary conditions (1.2) and (1.3) standing for u(0, t ) = 0,

(1.9)

−µ(t )ux (1, t ) = g (t ) + K1 (t )u(1, t ) + λ1 (t )ut (1, t ) −

t

Z

k(t − s)u(1, s)ds,

(1.10)

0

where µ ≡ µ(t ), u0 ∈ H 2 and u1 ∈ H 1 , f (u, ut ) = Ku + λut , with λ1 (t ) ≥ λ0 > 0, K1 (t ), g (t ), k(t ) are given functions, and K ≥ 0, λ ≥ 0, λ0 > 0 are given constants. In this case, the problem (1.1), (1.4), (1.9) and (1.10) is the mathematical model describing a shock problem involving a linear viscoelastic bar. In [14] Santos studied the asymptotic behavior of solution of problem (1.1), (1.4) and (1.9), with f (u, ut ) ≡ 0, µ = µ(t ), associated with a boundary condition of memory type at x = 1 as follows u(1, t ) +

t

Z

g (t − s)µ(s)ux (1, s)ds = 0,

t > 0.

(1.11)

0

Santos transformed (1.11) into (1.10) with K1 (t ) =

g / (0) , g (0)

λ1 ( t ) =

1 g (0)

are positive constants.


2

In this paper, we consider four main parts. In Part 1, under conditions (u0 , u1 , g0 , g1 , k0 , k1 ) ∈ H 1 × L2 × H 1 (0, T ) W

Rz

1 ,1


2

×

(0, T ) , µ ∈ W (QT ), µt ∈ L (0, T ; L ), µ(x, t ) ≥ µ0 > 0, a.e. (x, t ) ∈ QT ; the function F continuous, 1

1

∞

F (s)ds ≥ −C1 z 2 − C10 , for all z ∈ R, with C1 , C10 > 0 are given constants and some other conditions, we prove that 0 the problem (1.1)–(1.4) has a unique weak solution u. The proof is based on the Faedo–Galerkin method associated with the weak compact method. We remark that the linearization method in the paper [10] cannot be used in [9,13]. In Part 2 we prove that the unique solution u belongs to H 2 (QT ) ∩ L∞ (0, T ; H 2 ) ∩ C 0 (0, T ; H 1 ) ∩ C 1 (0, T ; L2 ), with ut ∈ L∞ (0, T ; H 1 ), utt ∈ L∞ (0, T ; L2 ), if we assume (u0 , u1 ) ∈ H 2 × H 1 , F ∈ C 1 (R) and some other conditions. In Part 3, with F ∈ C N +1 (R), N ≥ 2, we obtain an asymptotic expansion of the solution u of the problem (1.1)–(1.4) up to order N + 1 in a small parameter λ. Finally, in Part 4, we prove that the solution u of this problem is stable with respect to the data (λ, µ, g0 , g1 , k0 , k1 ). The results obtained here may be considered as the generalizations of those in [1,2,9–14]. 2. The existence and uniqueness theorem Put Ω = (0, 1), QT = Ω × (0, T ), T > 0. We omit the definitions of usual function spaces: C m Ω , Lp (Ω ), W m,p (Ω ).




We denote W m,p = W m,p (Ω ), Lp = W 0,p (Ω ), H m = W m,2 (Ω ), 1 ≤ p ≤ ∞, m = 0, 1, . . .. The norm in L2 is denoted by k · k. We also denote by h·, ·i the scalar product in L2 or pair of dual scalar products of continuous linear functionals with an element of a function space. We denote by k · kX the norm of a Banach space X and

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

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by X 0 the dual space of X . We denote by Lp (0, T ; X ), 1 ≤ p ≤ ∞, the Banach space of the real functions u : (0, T ) → X measurable, such that T

Z

p X

ku(t )k dt

kukLp (0,T ;X ) =

1/p

< ∞ for 1 ≤ p < ∞,

0

and

kukL∞ (0,T ;X ) = ess sup ku(t )kX 0
for p = ∞.

2 2 Let u(t ), u0 (t ) = ut (t ), u00 (t ) = utt (t ), ux (t ), uxx (t ) denote u(x, t ), ∂∂ut (x, t ), ∂∂ t 2u (x, t ), ∂∂ ux (x, t ), ∂∂ x2u (x, t ), respectively.

On H 1 we shall use the following norm

kvkH 1

2 !1/2

∂v

= kvk + .

∂x 2

(2.1)

Then we have the following lemma. Lemma 2.1. The imbedding H 1 ,→ C 0 ([0, 1]) is compact and

√ kvkC 0 (Ω ) ≤

2kvkH 1 ,

for all v ∈ H 1 .

The proof of this lemma is straightforward, and we omit the details.

(2.2) 

We make the following assumptions: (H1 ) u0 ∈ H 1 and u1 ∈ L2 , (H2 ) g0 , g1 ∈ H 1 , (H3 ) k0 , k1 ∈ W 1,1 (0, T ), (H4 ) µ ∈ W 1 (QT ) , µ(x, t ) ≥ µ0 > 0, µt ∈ L1 (0, T ; L∞ ), (F1 ) The function F ∈ C 0 (R) and satisfies the following conditions: there exist constants C1 , C10 > 0, such that z

Z

F (s)ds ≥ −C1 z 2 − C10 ,

for all z ∈ R,

0

(F2 ) For each M > 0, there exists a constant KM > 0, such that |F (u) − F (v)| ≤ KM |u − v|

for all u, v ∈ [−M , M] .

Remark 2.1. (i) We present an example in which the function F satisfies assumptions (F1 )–(F2 ). We consider the following function F (s) = |s|p−2 s − α s − β, where p, Rα , β are constants, with p > 1, α > 0, β > 0. Then, F satisfies assumption (F1 ). z 1 2 z − 12 β 2 , for all z ∈ R. Since 0 F (s)ds = 1p |z |p − α2 z 2 − β z ≥ − α+ 2 Furthermore, if p ≥ 2, then F satisfies assumptions (F1 )–(F2 ). (ii) The assumption (F1 ) still holds if F satisfies the following condition (e F1 ) F R∈ C 0 (R), F (0) = 0 and sF (s) > 0 for all s 6= 0. z Since 0 F (s)ds ≥ 0 for all z ∈ R. However, the function F as in the above example does not satisfy (e F1 ). Then, we have the following theorem. Theorem 2.2. Let (H1 )–(H4 ) and (F1 ) hold. Then, for every T > 0, there exists a weak solution u of problem (1.1)–(1.4) such that u ∈ L∞ (0, T ; H 1 ),

ut ∈ L∞ (0, T ; L2 ).

(2.3)

Furthermore, if F satisfies (F2 ), then the solution is unique. Proof of Theorem2.2. The proof consists of steps 1–4. Step 1. The Faedo–Galerkin approximation. Let {wj }j∈N be a denumerable base of H 1 (see [3, p. 86]). We find the approximate solution of problem (1.1)–(1.4) in the form um ( t ) =

m X j =1

cmj (t )wj ,

(2.4)

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where the coefficient functions cmj satisfy the system of ordinary differential equations



u00m (t ), wj + µ(t )umx (t ), wjx + Pm (t )wj (0) + Qm (t )wj (1) + F (um (t )) + λu0m (t ), wj = 0, 1 ≤ j ≤ m,











 Z t   k0 (t − s)um (0, s)ds, Pm (t ) = g0 (t ) + Z0 t   k1 (t − s)um (1, s)ds, Qm (t ) = g1 (t ) +

(2.5)

(2.6)

0

 m X   αmj wj → u0 strongly in H 1 , um (0) = u0m =  j =1

m X  0   βmj wj → u1 strongly in L2 . um (0) = u1m =

(2.7)

j =1

From the assumptions of Theorem 2.2, system (2.5)–(2.7) has a solution (um , Pm , Qm ) on an interval [0, Tm ] ⊂ [0, T ]. The following estimates allow one to take Tm = T for all m (see [5]). 0 Step 2. A priori estimates. Substituting (2.6) into (2.5), then multiplying the jth equation of (2.5) by cmj (t ) and summing with respect to j, and afterwards integrating with respect to the time variable from 0 to t, we get after some rearrangements

Z t Z 1 b F (u0m (x)) dx + ds µ0 (x, s)u2mx (x, s)ds 0 0 0 Z 1 Z t Z t Z t

0 2 0

u (s) ds − 2 b −2 F (um (x, t )) dx − 2λ P ( s ) u ( 0 , s ) ds − 2 Qm (s)u0m (1, s)ds m m m 0 0 0 0 Z 1 5 X b = Sm (0) + 2 F (u0m (x)) dx + Ii ,

Sm (t ) = Sm (0) + 2

1

Z

0

2

p

2

Sm (t ) = u0m (t ) + µ(t )umx (t ) ,

(2.9)

z

Z

b F (z ) =

(2.8)

i=1

F (s)ds.

(2.10)

0

We shall estimate the following five integrals on the right-hand side of (2.8). First integral I1 . By means of the following inequality 2 Sm (t ) ≥ sm (t ) = u0m (t ) + µ0 kumx (t )k2 ,



(2.11)

it follows that

Z t Z 1 Z t

0 0 2

µ (s) kumx (s)k2 ds |I1 | = ds µ (x, s)umx (x, s)ds ≤ ∞ 0 0 0 Z t

0 1

µ (s) sm (s)ds ≤ ∞ µ0 0

(2.12)

where k · k∞ = k · kL∞ (Ω ) . Second integral I2 . It follows from Lemma 2.1 and (2.11) that

kum (t )k ≤ 2



kum (0)k + 2

Z Z0

t

0

u (s) ds m

2

t

≤ 2 ku0m k + 2t sm (s)ds 0 Z t ≤ C0 + 2t sm (s)ds, 2

(2.13)

0

kum (t )k2H 1 = kumx (t )k2 + kum (t )k2 Z t 1 ≤ C0 + sm (t ) + 2t sm (s)ds, µ0 0 where C0 always indicates a constant depending on u0 .

(2.14)

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3947

It follows from the assumption (F1 ) and (2.13) that 1

Z

b F (um (x, t )) dx ≤ 2C1 kum (t )k2 + 2C10

I2 = −2 0

t

Z

sm (s)ds.

≤ C0 + 4C1 t

(2.15)

0

Third integral I3 . I3 = −2λ

t

Z

0 2

u (s) ds ≤ 2 |λ| m

t

Z

0

sm (s)ds.

(2.16)

0

Fourth integral I4 . By using integration by parts, it follows that t

Z

Pm (s)u0m (0, s)ds

I4 = −2 0

= 2g0 (0)u0m (0) − 2g0 (t )um (0, t ) + 2

t

Z

g00 (s)um (0, s)ds 0

+ 2k0 (0)

Z

t

t

Z

u2m (0, s)ds − 2um (0, t )

0

k0 (t − s)um (0, s)ds + 2 0

Z t Z 0

s



k00 (s − r )um (0, r )dr um (0, s)ds. (2.17) 0

By Lemma 2.1 and the following inequality 2ab ≤ β a2 +

1

β

b2 ,

for all a, b ∈ R, β > 0,

(2.18)

it follows from (2.17), that I4 ≤ 2 |g0 (0)u0m (0)| +

 +

2

β

1 + 4 |k0 (0)| +

≤ CT + 2β kum (t )k

2 H1

2

g02 (t ) + 2 g00 L2 (0,T ) + 2β kum (t )k2H 1 4

β

kk0 k2L2 (0,T ) + k00 L1 (0,T )

+ CT (k0 )

t

Z 0

kum (s)k2H 1 ds

t

Z 0

kum (s)k2H 1 ds

(2.19)

for all β > 0, where CT (k0 ) = 1 + 4 |k0 (0)| + β4 kk0 k2L2 (0,T ) + k00 L1 (0,T ) , and CT always indicates a constant depending on T . Fifth integral I5 . Similarly, we obtain



I5 ≤ CT + 2β kum (t )k2H 1 + CT (k1 )

t

Z 0

kum (s)k2H 1 ds,

(2.20)

where CT (k1 ) = 1 + 4 |k1 (0)| + β4 kk1 k2L2 (0,T ) + k01 L1 (0,T ) . Combining (2.8), (2.11), (2.12), (2.14)–(2.16), (2.19) and (2.20) and choosing β rearrangements



(1)

sm (t ) ≤ MT

+ 2Sm (0) + 4

1

Z

b F (u0m (x)) dx + 0

t

Z

=

1 8

µ0 , we obtain after some

(1)

NT (s)sm (s)ds,

(2.21)

0

where

 (1) M = 2 [4β C0 + (CT (k0 ) + CT (k1 )) C0 T + 2CT + C0 ],     T 2 0 1 (1) 2

NT (s) = µ (s) ∞ + 4 (2C1 T + |λ| + 4β T ) + 2 (CT (k0 ) + CT (k1 )) + (CT (k0 ) + CT (k1 )) T ,  µ0 µ0   (1) 1 NT ∈ L (0, T ) .

(2.22)

From the assumptions (H1 ) and (H4 ), the continuity of F , and the imbedding H 1 (0, 1) ,→ C 0 ([0, 1]), there exists a e (1) depending on u0 , u1 , µ, F , such that positive constant M T (1)

MT

+ 2Sm (0) + 4

1

Z

b e (1) , F (u0m (x)) dx ≤ M T

0

for all m.

(2.23)

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L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

By Gronwall’s lemma, we deduce from (2.22) to (2.23) that (1)

e exp sm (t ) ≤ M T

Z

t

(1)

NT (s)ds



≤ CT ,

for all t ∈ [0, T ].

(2.24)

0

On the other hand, from the assumptions (H1 )–(H3 ) and the imbeddings H 1 (0, 1) ,→ C 0 ([0, 1]), W 1,1 (0, T ) ,→ C ([0, T ]), we deduce from (2.6), (2.7), (2.11), (2.14) and (2.24) that 0

kPm kH 1 (0,T ) ≤ CT for all m,

(2.25)

kQm kH 1 (0,T ) ≤ CT for all m.

(2.26)

On the other hand, we deduce from (2.24) that

|um (x, t )| ≤

√

2 kum (t )kH 1 ≤

√

2 kum kL∞ (0,T ;H 1 ) ≤ C1T .

(2.27)

Hence, by the assumption F ∈ C 0 (R), we deduce from (2.27) that

|F (um (x, t ))| ≤ sup |F (z )| ≡ C2T ,

a.e. (x, t ) ∈ QT ,

(2.28)

|z |≤C1T

i.e.,kF (um )kL∞ (QT ) ≤ C2T , for all m.

Step 3. Limiting process. From (2.11), (2.24)–(2.26) and (2.28), we deduce the existence of a subsequence of {(um , Pm , Qm )} still also so denoted, such that

 um → u in L∞ (0, T ; H 1 ) weak* ,    u0m → u0 in L∞ (0, T ; L2 ) weak* ,    um (0, ·) → u(0, ·) in L∞ (0, T ) weak*, um (1, ·) → u(1, ·) in L∞ (0, T ) weak* ,   Pm → P in H 1 (0, T ) weakly,    1   Qm → Q in H (0∞, T ) weakly, F (um ) → χ in L (QT ) weak*.

(2.29)

By the compactness lemma of Lions [8, p. 57] and the imbedding H 1 (0, T ) ,→ C 0 ([0, T ]), we can deduce from (2.29)1,2,5,6 the existence of a subsequence still denoted by {um }, such that

 um → u strongly in L2 (QT ) and a.e. in QT , P → P strongly in C 0 ([0, T ]) ,  m Qm → Q strongly in C 0 ([0, T ]) .

(2.30)

By means of the continuity of F , we have F (um ) → F (u)

a.e. (x, t ) in QT .

(2.31)

Using Lions’s Lemma [8, Lemma 1.3, p. 12], it follows from (2.28) and (2.31) that F (um ) → F (u)in L2 (QT ) weakly.

(2.32)

Hence, it follows from (2.29)7 and (2.32) that

χ = F (u) a.e. (x, t ) in QT . On the other hand, with e Pm (t ) = T

Z

e Pm (t )ψ(t )dt =

0

T

Z

0

(2.33)

Rt

ψ(t )dt

k0 (t − s)um (0, s)ds, from (2.29)3 , we obtain t

Z

0

k0 (t − s)um (0, s)ds 0

T

Z

um (0, s)ds

= 0

k0 (t − s)ψ(t )dt s

T

Z

u(0, s)ds

→

T

Z

0

T

Z

k0 (t − s)ψ(t )dt ,

(2.34)

s

for all ψ ∈ L1 (0, T ). Therefore

e Pm (t ) →

t

Z

k0 (t − s)u(0, s)ds, 0

in L∞ (0, T ) weak*.

(2.35)

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3949

Hence, it follows from (2.30)2 , and (2.35), that Pm (t ) = g0 (t ) + e Pm (t ) → g0 (t ) +

t

Z

k0 (t − s)u(0, s)ds ≡ P (t ),

(2.36)

0

strongly in C 0 ([0, T ]). Similarly, it follows from (2.6)2 , (2.29)4 , and (2.30)3 , that Qm (t ) = g1 (t ) +

t

Z

k1 (t − s)um (1, s)ds → g1 (t ) +

t

Z

k1 (t − s)u(1, s)ds ≡ Q (t ),

(2.37)

0

0

strongly in C 0 ([0, T ]). Passing to the limit in (2.5)–(2.7) by (2.29)1,2,7 , (2.33), (2.36) and (2.37), we have u satisfying the equation



 d 0 u (t ), v + hµ(t )ux (t ), vx i + P (t ) v(0) + Q (t ) v(1) + F (u(t )) + λu0 (t ), v = 0, dt

(2.38)

for all v ∈ H 1 , where

 Z t   k0 (t − s)u(0, s)ds, P (t ) = g0 (t ) + Z0 t   k1 (t − s)u(1, s)ds. Q (t ) = g1 (t ) +

(2.39)

0

We can prove in a similar manner as in [9] that u(0) = u0 ,

u0 (0) = u1 .

(2.40)

In order to prove the uniqueness of the solution, we only have to prove the following lemma. Lemma 2.2. Let u be the weak solution of the following problem

 ∂ 00   u − ∂ x (µ(x, t )ux ) = Φ , 0 < x < 1, 0 < t < T ,  µ(0, t )ux (0, t ) = G0 (t ), −µ(1, t )ux (1, t ) = G1 (t ),   u(x, 0) = u0 (x), u0 (x, 0) = u1 (x), u ∈ L∞ (0, T ; H 1 ),   G0 , G1 ∈ H 1 (0, T ) , Φ ∈ L1 (0, T ; L2 ).

(2.41)

u0 ∈ L∞ (0, T ; L2 ),

Then we have

Z Z 1 1 t µ(x, t )u2x (x, t )dx − ds µ0 (x, s)u2x (x, s)dx 2 0 0 0 Z t Z t G00 (s)u(0, s)ds − G01 (s)u(1, s)ds + G1 (t )u(1, t ) + G0 (t )u(0, t ) −

1 0 2

u (t ) + 1 2 2

Z

1

1

Z

0

≥

1 2

k u1 k 2 +

2

0

1

µ(x, 0)u20 (x)dx + G1 (0)u0 (1) + G0 (0)u0 (0) + 0

t

Z



 Φ (s), u0 (s) ds,

a.e. t ∈ [0, T ].

(2.42)

0

Furthermore, if u0 = u1 = 0 there is equality in (2.42). Proof of Lemma2.2. The idea of the proof is the same as in [6, Lemma 2.1, p. 79]. Fix t1 , t2 , 0 < t1 < t2 < T and let v(x, t ) be the function defined as follows

v(x, t ) = [θm (t )u0 (x, t ) ∗ ρk (t ) ∗ ρk (t )]θm (t ),

(2.43)

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

 0,   1, θm (t ) = m(t − t1 − 1/m),  −m(t − t2 + 1/m),

if, t ∈ [0, T ] r [t1 + 1/m, t2 − 1/m], if, t ∈ [t1 + 2/m, t2 − 2/m], if, t ∈ [t1 + 1/m, t1 + 2/m], if, t ∈ [t2 − 2/m, t2 − 1/m].

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

ρk ∈ Cc∞ (R),

ρk (t ) = ρk (−t ),

Z

+∞

ρk (t )dt = 1, −∞

supp ρk ⊂ [−1/k, 1/k].

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L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

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

(u ∗ ρk )(x, t ) =

+∞

Z

u(x, t − s)ρk (s)ds. −∞

We take the scalar product of the function v(x, t ) in (2.43) with Eq. (2.41)1 , then integrate with respect to the time variable from 0 to T , and we have Xmk + Ymk = Zmk ,

(2.44)

where

 Z T Z 1   u00 (x, t )v(x, t )dx, dt X =  mk   0Z 0Z   1 T ∂ dt Ymk = − (µ(x, t )ux (x, t )) v(x, t )dx,   Z T 0 Z 10 ∂ x     Zmk = Φ (x, t )v(x, t )dx. dt

(2.45)

0

0

By using the properties of the functions θm (t ) and ρk (t ) we can show after some lengthy calculation:

 Z T

2   lim X = − θm θm0 u0 (t ) dt ,  mk  k→+∞   Z0 T Z T Z T    lim Ymk = − θm2 G00 (t )u(0, t )dt − θm2 G01 (t )u(1, t )dt − 2 θm θm0 G0 (t )u(0, t )dt k→+∞ 0 0 0 RT R1 RT R1 RT    −2 0 θm θm0 G1 (t )u(1, t )dt − 12 0 θm2 dt 0 µ0 (x, t )u2x (x, t )dx − 0 θm θm0 dt 0 µ(x, t )u2x (x, t )dx,   Z Z T 1     lim Zmk = θm2 dt Φ (x, t )u0 (x, t )dx. k→+∞

0

(2.46)

0

Letting k → ∞, from (2.44)–(2.46) we obtain



1 0

u (t2 ) 2 − 1 u0 (t1 ) 2 − 2 2

Z

t2 t1

Z

1 2 t2

=

1

Z

µ(x, t2 )u2x (x, t2 )dx − 0

 Φ (t ), u0 (t ) dt ,

t2

G01 (t )u(1, t )dt + G0 (t2 )u(0, t2 ) − G0 (t1 )u(0, t1 )

t1

+ G1 (t2 )u(1, t2 ) − G1 (t1 )u(1, t1 ) − +

Z

G00 (t )u(0, t )dt −

1 2

1 2

Z

t2

Z

1

dt t1

µ0 (x, t )u2x (x, t )dx

0

1

Z

µ(x, t1 )u2x (x, t1 )dx 0

a.e., t1 , t2 ∈ (0, T ), t1 < t2 .

(2.47)

t1

From (2.47) we obtain (2.42) by taking t2 = t and passing to the limit as t1 → 0+ and using the property of weak lower semicontinuity of the functional v 7−→ kvk2 . In the case of u0 = u1 = 0, we prolong u, µ, Φ , G0 , G1 by 0 as t < 0 and we deduce equality (2.47) is true for almost t1 < t2 < T . Taking t1 < 0 in (2.47), its right-hand side is 0, we take t1 → 0− and we have equality (2.42) when u0 = u1 = 0. The proof of Lemma 2.2 is completed.  Remark 2.2. (i) Lemma 2.2 is a relative generalization of a lemma contained in Lions’s book [8, Lemma 6.1, p. 224]. (ii) We remark that the conditions u0 (0, ·) ∈ Lq0 (0, T ), u0 (1, ·) ∈ Lq1 (0, T ) made in [13, Lemma 2.4, p. 1799] are not needed in the Lemma 2.2. We now return to the proof of the uniqueness of a solution of the problem (1.1)–(1.4). Step 4. Uniqueness of the solution. Assume now that (F2 ) hold. Let u1 , u2 be two weak solutions of problem (1.1)–(1.4), such that ui ∈ L∞ (0, T ; H 1 ),

u0i ∈ L∞ (0, T ; L2 ),

i = 1, 2.

(2.48)

Then v = u1 − u2 is the weak solution of the following problem

 ∂   v 00 − (µ(x, t )vx ) + λv 0 + F (u1 ) − F (u2 ) = 0, 0 < x < 1, 0 < t < T ,   ∂  µ(0, t )vx (0, t ) = P (t ), −µ(1, t )vx (1, t ) = Q (t ), x 0 v(x, 0) = v ( x, 0) = 0,  Z t Z t     P ( t ) = k0 (t − s)v(0, s)ds, Q (t ) = k1 (t − s)v(1, s)ds. 0

0

(2.49)

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3951

  By using Lemma 2.2 with u0 = u1 = 0, Φ = − λv 0 + F (u1 ) − F (u2 ) , G0 (t ) = P (t ), G1 (t ) = Q (t ), we have Z t Z t Z 1 P 0 (s)v(0, s)ds µ0 (x, s)vx2 (x, s)dx − 2P (t )v(0, t ) + 2 ds %(t ) ≤ Z (t ) = 0 0 0 Z t Z t

0 2 0

v (s) ds Q (s)v(1, s)ds − 2λ − 2Q (t )v(1, t ) + 2 0

0 t

Z

F (u1 ) − F (u2 ), v 0 (s) ds, a.e. t ∈ [0, T ],



−2



(2.50)

0

where

Z 1

0 2

µ(x, t )vx2 (x, t )dx, Z (t ) = v (t ) + 0

 2 %(t ) = v 0 (t ) + µ0 kvx (t )k2 .  

(2.51)

On the other hand, by using Lemma 2.1 and (2.51), we have

kv(t )k2 ≤ t

t

Z

0 2

v (s) ds ≤ T

Z

0

t

%(s)ds,

(2.52)

0 2 H1

v 2 (i, t ) ≤ kv(t )k2C 0 (Ω ) ≤ 2 kv(t )k = 2 kv(t )k2 + kvx (t )k   Z t 2 ≤ µ0 T %(s)ds + %(t ) , i = 0, 1. µ0 0 Using the assumptions (F2 ), we deduce from (2.51) and (2.53) that Z t Z t



 kv(s)kC 0 (Ω ) v 0 (s) ds −2 F (u1 ) − F (u2 ), v 0 (s) ds ≤ 2KM 0 s

≤2

(2.53)

0

2

s

t

Z

µ0

≤2


2

2

µ0

KM

  Z s µ0 T %(τ ) dτ + %(s) ds

0

0

K M 1 + µ0 T 2




t

Z

%(s)ds,

(2.54)

0

√ where M = 2 maxi=1,2 kui kL∞ (0,T ;H 1 ) . On the other hand, by the assumptions (H2 ) and (H3 ), we deduce from (2.53) that

− 2P (t )v(0, t ) + 2

t

Z

P 0 (s)v(0, s)ds ≤ β kv(t )k2H 1 + C (k0 , β)

0



Z

t

kv(s)k2H 1 ds,

0

(2.55)



for all β > 0, where C (k0 , β) = 4 β1 kk0 k2L2 (0,T ) + |k0 (0)| + k00 L1 (0,T ) . Similarly with Q (t ), we obtain

− 2Q (t )v(1, t ) + 2



t

Z

Q (s)v(1, s)ds ≤ β kv(t )k 0

0

2 H1



+ C (k1 , β)

t

Z 0

kv(s)k2H 1 ds,

(2.56)



for all β > 0, where C (k1 , β) = 4 β1 kk1 k2L2 (0,T ) + |k1 (0)| + k01 L1 (0,T ) . µ Combining (2.50)–(2.56), with β = 40 , we then obtain

%(t ) ≤

t

Z



(1)

RT (s)%(s)ds,

(2.57)

0

where (1)

RT (s) =

2 0

µ (s) + 2 (C (k0 , β) + C (k1 , β)) ∞



µ0

" + 4 β T + |λ| +

s

2

µ0

K M 1 + µ0 T

#
2

1

µ0 ,

+ T2

(1)

RT



∈ L1 (0, T ).

By Gronwall’s lemma, it follows from (2.58) that % ≡ 0, i.e., u1 ≡ u2 . Theorem 2.2 is proved completely. 

(2.58)

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L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3. The regularity of solutions In this section, we study the regularity of solution of problem (1.1)–(1.4) corresponding to (u0 , u1 ) ∈ H 2 × H 1 . Henceforth, we strengthen the hypotheses and assume that:

(H10 ) (H20 ) (H30 ) (H40 ) (F00 )

u0 ∈ H 2 and u1 ∈ H 1 , g0 , g1 ∈ H 2 , k0 , k1 ∈ W 2,1 ,


µ ∈ C 1 QT , µtt ∈ L1 (0, T ; L∞ ), µ(x, t ) ≥ µ0 > 0, ∀(x, t ) ∈ QT , F ∈ C 1 (R) satisfies (F1 ).

Then, we have the following theorem. Theorem 3.1. Let (H1 )0 –(H40 ) and (F00 ) hold. Then, for every T > 0, there exists a unique weak solution u of problem (1.1)–(1.4) such that


utt ∈ L∞ 0, T ; L2 .


ut ∈ L ∞ 0 , T ; H 1 ,


u ∈ L∞ 0, T ; H 2 ,

(3.1)

Remark 3.1. (i) Noting that with the regularity obtained by (3.1), it follows that the problem (1.1)–(1.4) has a unique strong solution u that satisfies u ∈ C 0 0, T ; H 1 ∩ C 1 0, T ; L2 ∩ L∞ 0, T ; H 2 ,








ut ∈ L∞ 0, T ; H 1 .




(3.2)

(ii) From (3.1) we can see that u, ux , ut , uxx , uxt , utt ∈ L∞ 0, T ; L ⊂ L2 (QT ). Also if the data (u0 , u1 ) in problem satisfy assumption (H10 ), then the weak solution u of problem (1.1)–(1.4) belongs to H 2 (QT ). So the solution is almost classical

which is rather natural since the initial data u0 and u1 do not belong necessarily to C 2 Ω and C 1 Ω , respectively. (iii) In [4], Browder has studied the operator differential equation utt + Au + M (u) = 0, t > 0, with the Cauchy initial conditions u(0+ ) = u0 , ut (0+ ) = u1 , where A is a positive densely defined self-adjoint
linear operator in a Hilbert space H with A1/2 its positive square root. M (u) is a (possibly) nonlinear function from D A1/2 to H and some other conditions. In general, the results in the Theorem 3.1 and [4,7] overlap and do not include each other as particular cases.


2

Proof of Theorem3.1. The proof consists of Steps 1–4. Step1. The Faedo–Galerkin approximation. Let {wj } be a denumerable base of H 2 . We find the approximate solution um (t ) of problem (1.1)–(1.4) in the form (2.4), where the coefficient functions cmj satisfy the system of ordinary differential equations (2.5) and (2.6), and

 m X   u ( 0 ) = u = αmj wj → u0 strongly in H 2 ,  m 0m  j=1

(3.3)

m X  0   u ( 0 ) = u = βmj wj → u1 strongly in H 1 . 1m  m j =1

Step 2. A priori estimates I. Proceeding as in the proof of Theorem 2.2, we get, after using assumptions (H10 )–(H40 ) and (F1 ), 2 sm (t ) = u0m (t ) + µ0 kumx (t )k2 ≤ CT ,



(3.4)

for all t ∈ [0, T ] and for all m, and CT always indicates a bound depending on T . A priori estimates II. Now differentiating (2.5) with respect to t, we have



0 0 0 u000 m (t ), wj + µ(t )umx (t ) + µ (t )umx (t ), wjx + Pm (t )wj (0)







 + Qm0 (t )wj (1) + F 0 (um (t ))u0m (t ) + λu00m (t ), wj = 0,

(3.5)

for all 1 ≤ j ≤ m. 00 Multiplying the jth equation of (3.5) by cmj (t ), summing up with respect to j and then integrating with respect to the time variable from 0 to t, we have after some rearrangements Xm (t ) = Xm (0) + 2 µ0 (0)u0mx , u1mx + 3





t

Z 0

Z +2 0

t



µ00 (s)umx (s), u0mx (s) ds − 2λ 

1

Z ds

2

 µ0 (x, s) u0mx (x, s) dx − 2 µ0 (t )umx (t ), u0mx (t )

0 t

Z

00 2

u (s) ds m

0

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965 t

Z



−2

F 0 (um (s))u0m (s), u00m (s) ds − 2



t

Z

0 Pm (s)u00m (0, s)ds − 2

Qm0 (s)u00m (1, s)ds 0

0

0

3953

t

Z

7  X

Ji , = Xm (0) + 2 µ0 (0)u0mx , u1mx +

(3.6)

i =1

where

2

p

2

Xm (t ) = u00m (t ) + µ(t )u0mx (t ) .



(3.7)

From the assumptions (H10 ), (H40 ) and (F00 ) and the imbedding H 1 (0, 1) ,→ C 0 ([0, 1]), there exists a positive constant e D(1) depending on u0 , u1 , µ, F , such that Xm (0) + 2 µ0 (0)u0mx , u1mx ≤ kµ(0)u0mxx + µx (0)u0mx − F (u0m ) − λu1m k2





p

2



+ µ(0)u1mx + 2 µ/ (0)u0mx , u1mx ≤ e D(1) ,

(3.8)

for all m. Using (H10 )–(H40 ), (F00 ), (3.7) and (3.8) and the following inequalities 2 2 Xm (t ) ≥ u00m (t ) + µ0 u0mx (t ) = xm (t ),





√

2 kum (t )kH 1 ≤

kum (t )kC 0 (Ω ) ≤

(3.9)

√

2 kum kL∞ (0,T ;H 1 ) ≤ CT ,

(3.10)

1/2 √ √

1 ≤ 2 u0m (t ) H 1 ≤ 2 xm (t ) + sm (t ) µ0  1/2 s p √ 1 2 xm (t ) + CT ≤ xm (t ) + DT , ≤ 2 µ0 µ0 

0

u (t ) 0 m C (Ω )

(3.11)

we estimate, without difficulty the following terms on the right-hand side of (3.6) as follows t

Z

1

Z

J1 = 3

ds 0

0

≤ 3 µ0 L∞ (Q

T

J2 =

≤ J3 =

≤

2 µ0 (x, s) u0mx (x, s) dx Z t Z t

0

u (s) 2 ds ≤ CT xm (s)ds, mx ) 0

(3.12)

0



−2 µ0 (t )umx (t ), u0mx (t ) ≤ 2 µ0 L∞ (Q ) kumx (t )k u0mx (t ) T β xm (t ) + CT , Z t Z t

00

00  0

µ (s) ∞ kumx (s)k u0 (s) ds 2 µ (s)umx (s), umx (s) ds ≤ 2 mx L 0 Z 0 t CT + xm (s)ds,



(3.13)

(3.14)

0

J4 = −2λ

t

Z

00 2

u (s) ds ≤ 2 |λ|

t

Z

xm (s)ds,

m

0 t

Z



J5 = −2

(3.15)

0

F 0 (um (s))u0m (s), u00m (s) ds ≤ 2KT

0



t

Z

00

u (s) ds ≤ CT + m

≤ 2KT CT

0 00

u (s) u (s) ds m m 0

t

Z

t

Z

xm (s)ds,

(3.16)

0

0

where KT = sup|z |≤CT F 0 (z ) . With J6 , by using integration by parts, we rewrite



Z J6 = −2



t

0 Pm (s)u00m (0, s)ds

0

= 2 Pm0 (0)u01m (0) − 2Pm0 (0)u0m (0, t ) − 2

t

Z 0

= J6(1) + J6(2) + J6(3) + J6(4) .



00 Pm (s)ds u0m (0, t ) + 2

Z

t

00 Pm (s)u0m (0, s)ds

0

(3.17)

3954

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

We shall estimate the following integrals on the right-hand side of (3.17). (1) First integral J6 . By means of (3.3), it follows that (1)

J6

= 2 Pm0 (0)u01m (0) ≤ C0 .

(3.18)

(2)

Second integral J6 . By means of the following inequality

√ 0

u (0, t ) ≤ u0 (t ) 0 ≤ 2 u0m (t ) H 1 ≤ m m C (Ω )

s

2 p xm (t ) + DT ,

µ0

(3.19)

it follows that

√

(2) J6 = −2Pm0 (0)u0m (0, t ) ≤ 2 2 g00 (0) + k0 (0)u0m (0) u0m (t ) H 1 s ! 2 p ≤ C0 xm (t ) + DT ≤ β xm (t ) + DT . µ0

(3.20)

(3)

Third integral J6 . By (2.6), (3.4) and Lemma 2.1, we have

√ √ Z t 00

00 00 √

k (t − s) kum (s)kH 1 ds P (t ) ≤ g (t ) + 2 |k0 (0)| u0 (t ) 1 + 2 k0 (0) kum (t )kH 1 + 2 m 0 0 m 0 H 0 Z t √ √

0

0 √ 00 k (s) ds ≤ 2 |k0 (0)| u (t ) 1 + 2CT k (0) + 2CT m

0

H

0

0

√

2 |k0 (0)| u0m (t ) H 1 .

≤ CT +

(3.21)

Hence

Z t  √ Z t 00 0

(3) 00 P (s) ds u (t ) 1 Pm (s)ds u0m (0, t ) ≤ 2 2 J6 = −2 m m H 0 0 Z t

2 00 2 2 P (s) ds ≤ β u0m (t ) H 1 + t β 0 m Z th √

0

2

i2 2

≤ β um (t ) H 1 + t CT + 2 |k0 (0)| u0m (s) H 1 ds β 0 Z t ≤ CT + β xm (t ) + CT xm (s)ds.

(3.22)

0

(4)

Fourth integral J6 . By (3.21) and Lemma 2.1, we obtain

Z t √ (4) 00 0 ≤2 2 J = 2 P ( s ) u ( 0 , s ) ds 6 m m 0 √ Z t 00 0 P (s) u (s) 1 ds ≤2 2 m

m

≤2 2

√ CT +

0

0

H

0

√ Z th

Z t P 00 (s)u0 (0, s)ds m m

i

2 |k0 (0)| u0m (s) H 1 u0m (s) H 1 ds



t

Z

xm (s)ds.

≤ CT + CT

(3.23)

0

It follows from (3.17), (3.18), (3.20), (3.22) and (3.23), that

|J6 | ≤ CT + 2β xm (t ) + 2CT

t

Z

xm (s)ds.

(3.24)

0

Similarly, the last term on the right-hand side of (3.6) is estimated

Z t Z t | J7 | = −2 Qm0 (s)u00m (1, s)ds ≤ CT + 2β xm (t ) + 2CT xm (s)ds. 0

0

(3.25)

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3955

Combining (3.6)–(3.9), (3.2)–(3.16), (3.24) and (3.25), we obtain after some rearrangements xm ( t ) ≤ e D(1) + 5CT + 5β xm (t ) + (2 + 5CT + 2 |λ|)

Z

t

xm (s)ds

0

≤ CT + 5β xm (t ) + CT

t

Z

xm (s)ds,

(3.26)

0

where CT is a positive constant depending only on T . Choosing β > 0, with β < 1/10, from (3.26) and applying Gronwall’s inequality, we obtain that xm (t ) ≤ 2CT exp(2CT t ) ≤ 2CT exp(2CT T ) ≤ CT ,

for all t ∈ [0, T ] .

(3.27)

On the other hand, from the assumptions (H3 ) and (H4 ), we deduce from (2.6), (3.4) and (3.27) that 0

0

kPm kW 2,∞ (0,T ) ≤ CT ,

(3.28)

kQm kW 2,∞ (0,T ) ≤ CT ,

(3.29)

where CT is a positive constant depending only on T . Step 3. Limiting process. From (3.4) and (3.27)–(3.29), we deduce the existence of a subsequence of {(um , Pm , Qm )} still also so denoted, such that

 um → u in L∞ (0, T ; H 1 ) weak*,   u0 → u0 in L∞ (0, T ; H 1 ) weak*,   m   00 00 ∞ 2   u  m → u in L (0, T ; L ) weak*, um (0, ·) → u(0, ·) in W 1,∞ (0, T ) weak*,    um (1, ·) → u(1, ·) in W 1,∞ (0, T ) weak*,    2,∞   Pm → P in W (0, T ) weak*,  Qm → Q in W 2,∞ (0, T ) weak*.

(3.30)

By the compactness lemma of Lions [8, p. 57] and the imbedding H 1 (0, T ) ,→ C 0 ([0, T ]), we can deduce from (3.30) the existence of a subsequence still denoted by {(um , Pm , Qm )} such that

 um → u strongly in L2 (QT ), and a.e. in QT ,     u0m → u0 strongly in L2 (QT ), and a.e. in QT ,    u (0, ·) → u(0, ·) strongly in C 0 ([0, T ]), m

 um (1, ·) → u(1, ·) strongly in C 0 ([0, T ]),     P → P strongly in C 1 ([0, T ]) ,    m Qm → Q strongly in C 1 ([0, T ]) .

(3.31)

From (2.6)1 and (3.31)3 we have that Pm (t ) → g0 (t ) +

t

Z

k0 (t − s)u(0, s)ds,

(3.32)

0

strongly in C 0 ([0, T ]). Combining (3.31)5 and (3.32), we conclude that P (t ) = g0 (t ) +

t

Z

k0 (t − s)u(0, s)ds.

(3.33)

0

Similarly, we have also Qm (t ) → g1 (t ) +

t

Z

k1 (t − s)u(1, s)ds ≡ Q (t ),

(3.34)

0

strongly in C 0 ([0, T ]) and in W 2,∞ (0, T ) weak*. We use the following inequality

|F (um ) − F (u)| ≤ KT |um − u| , where KT = sup|z |≤C F 0 (z ) , it follows from (3.31)1 and (3.35) that

(3.35)

T

F (um ) → F (u) strongly in L2 (QT ).

(3.36)

3956

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

Passing to the limit in (2.5)–(2.7) by (3.30)1,2,3 , and (3.33), (3.34) and (3.36), we have u satisfying the problem



u00 (t ), v + hµ(t )ux (t ), vx i + P (t )v(0) + Q (t )v(1) + F (u) + λu0 , v = 0,





u(0) = u0 ,



for all v ∈ H 1 ,

u (0) = u1 , 0

(3.37) (3.38)

where

 Z t   k0 (t − s)u(0, s)ds, P ( t ) = g 0 ( t ) + Z0 t   Q ( t ) = g 1 ( t ) + k1 (t − s)u(1, s)ds.

(3.39)

0

On the other hand, we have from (3.30)1,2,3 , (3.36) and (3.37) and assumption (H50 ), that uxx =

1

u00 + F (u) + λu0 − µx ux

µ(x, t )




∈ L∞ (0, T ; L2 ).

(3.40)

Thus u ∈ L∞ (0, T ; H 2 ) and the existence of solution is proved completely. Step 4. Uniqueness of the solution. Let u1 , u2 be two weak solutions of problem (1.1)–(1.4), such that


ui ∈ L∞ 0, T ; H 2 ,


u0i ∈ L∞ 0, T ; H 1 ,


u00i ∈ L∞ 0, T ; L2 ,

i = 1, 2.

(3.41)

Then u = u1 − u2 satisfy the variational problem u00 (t ), v + hµ(t )ux (t ), vx i + P (t )v(0) + Q (t )v(1) + F (u1 ) − F (u2 ) + λu0 , v = 0, u(0) = u/ (0) = 0,







∀v ∈ H 1 ,



(3.42)

and

 Z t   k0 (t − s)u(0, s)ds, P ( t ) = Z0 t   Q ( t ) = k1 (t − s)u(1, s)ds.

(3.43)

0

We take v = u0 in (3.42)1 , and integrating with respect to t, we obtain

σ (t ) ≤ e KM

t

Z

σ (s)ds,

(3.44)

0

where

       1 1 0 2

µ ∞ e  |λ| KM = 2T 2 KT + 2 + 1 + 2 + 4 β T + 2 T + C T , k , k , (β, ) 0 1  L (QT )  µ0 µ0     



  C (β, T , k0 , k1 ) = |k0 (0)| + |k1 (0)| + 1 kk0 k22 + kk1 k2L2 (0,T ) + k00 L1 (0,T ) + k01 L1 (0,T ) , L (0,T ) β K = sup F 0 (z ) ,  MT = ku1 kL∞ (0,T ;H 1 ) + ku2 kL∞ (0,T ;H 1 ) , T    |z |≤MT    1  β = µ0 .

(3.45)

4

By Gronwall’s lemma, we deduce that σ (t ) ≡ 0 and Theorem 3.1 is completely proved.



4. Asymptotic expansion of the solution with respect to the parameter λ In this part, we consider two given functions u0 , u1 as e u0 , e u1 , respectively. Then we assume that (e u0 ,e u1 , g0 , g1 , k0 , k1 , µ, F ) satisfy the assumptions (H10 )–(H40 ) and (F00 ). Let λ ∈ R. By Theorem 3.1, the problem (1.1)–(1.4) has a unique weak solution u = uλ depending on λ:

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3957

We consider the following perturbed problem, where λ is small parameter such that, |λ| ≤ λ∗ :

 ∂ 0 00   Lu ≡ u − ∂ x (µ(x, t )ux ) = −F (u) − λu ,    L0 u ≡ µ(0, t )ux (0, t ) = P (t ),     L1 u ≡ −µ(1, t )ux (1, t 0) = Q (t ), u0 (x), u1 (x), (e Pλ ) u(x, 0) = e Z t u (x, 0) = e    k0 (t − s)u(0, s)ds, P (t ) = g0 (t ) +     Z0 t    Q ( t ) = g 1 ( t ) + k1 (t − s)u(1, s)ds.

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

0

We shall study the asymptotic expansion of the solution of problem (e Pλ ) with respect to λ. We use the following notation. For a multi-index α = (α1 , . . . , αN ) ∈ ZN+ , andx = (x1 , . . . , xN ) ∈ RN , we put



α α |α| = α1 + · · · + αN , α! = α1 ! . . . αN !, xα = x1 1 . . . xNN , N α, β ∈ Z+ , β ≤ α ⇐⇒ βi ≤ αi , ∀i = 1, . . . , N .

Now, we assume that

(FN0 )F ∈ C N +1 (R) , N ≥ 2. First, we shall need the following lemma. Lemma 4.1. Let m, N ∈ N, x = (x1 , . . . , xN ) ∈ RN , and λ ∈ R. Then N X i =1

!m xi λi

=

mN X

P [m] [x]k λk ,

(4.1)

k=m

where the coefficients P (m) [x]k , m ≤ k ≤ mN depending on x = (x1 , . . . , xN ) are defined by the formula

 X m!   xα , m ≤ k ≤ mN , P [m] [x]k =   α!  (m) ( α∈Ak ) N X   (m) N   iαi = k . Ak = α ∈ Z+ : |α| = m,

(4.2)

i =1

The proof of this lemma is easy, hence we omit the details.



Let U0 be a unique weak solution of problem (e P0 ) (as in Theorem 3.1) corresponding to λ = 0, i.e.,

LU0 = −F (U0 ), 0 < x < 1, 0 < t < T ,   L0 U0 = P0 (t ), L1 U0 = Q0 (t ),   0   U0 (x, 0) = e u0 (x), u1 (x),  Z t U 0 ( x, 0 ) = e    P ( t ) = g ( t ) + k0 (t − s)U0 (0, s)ds, 0 0 (e P0 ) Z0 t     Q0 (t ) = g1 (t ) + k1 (t − s)U0 (1, s)ds,   
0 1

 0 1  U0 ∈ C 0, T ; H ∩ C 0, T ; L2 ∩ L∞ 0, T ; H 2 , 

U00 ∈ L∞ 0, T ; H 1 , U000 ∈ L∞ 0, T ; L2 . Let us consider the sequence of weak solutions Uγ , γ = 1, 2, . . . , N, defined by the following problems:

 LUγ = Fγ , 0 < x < 1, 0 < t < T ,    L1 Uγ = Qγ (t ), L0 Uγ = Pγ (t ),    Uγ (x, 0) = Uγ0 (x, 0) = 0,  Z t    P ( t ) = g ( t ) + k0 (t − s)Uγ (0, s)ds, γ 0 (e Pγ ) 0 Z t     Qγ (t ) = g1 (t ) + k1 (t − s)Uγ (1, s)ds,   
0 1

 0 1   Uγ ∈ C 0, T ; H
∩ C 0, T ; L2 ∩ L∞ 0,
T ; H 2 ,   0 Uγ00 ∈ L∞ 0, T ; L2 , Uγ ∈ L∞ 0, T ; H 1 ,

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L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

where Fγ , γ = 1, 2, . . . , N, are defined by the recurrence formulas Fγ = −Uγ0 −1 −

γ X 1 (k) F (U0 )P [k] [U ]γ , k ! k=1

γ = 1, 2, . . . , N ,

(4.3)

where we have used the notation U = (U1 , U2 , . . . , UN ). We also note that Fγ is the first-order function with respect to Uγ . In fact, Fγ = −Uγ0 −1 − F 0 (U0 )Uγ −

γ X 1 (k) F (U0 )P [k] [U ]γ k ! k=2

= −F 0 (U0 )Uγ − Uγ0 −1 + terms depending on (Ui , i = 0, 1, . . . , γ − 1). PN Let u = uλ be a unique weak solution of problem (e Pλ ). Then v = u − γ =0 Uγ λγ ≡ u − h, satisfies the problem  ∂  Lv ≡ v 00 − (µ(x, t )vx ) = −F (v + h) + F (h) − λv 0 + EN (λ, x, t ),   ∂ x  Z t   L v ≡ µ(0, t )v (0, t ) = k (t − s)v(0, s)ds, 0

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

0

x

(4.4)

(4.5)

0Z t     L v ≡ −µ( 1 , t )v ( 1 , t ) = k1 (t − s)v(1, s)ds,  1 x   0 0 v(x, 0) = v (x, 0) = 0,

where EN (λ, x, t ) = −λh0 − F (h) + F (U0 ) −

N X

Fγ λγ .

(4.6)

γ =1

Then, we have the following lemma. Lemma 4.2. Let (H10 ) − (H40 ), (F00 ) and (FN0 ) hold. Then

kEN (λ, ·, ·)kL∞ (0,T ;L2 ) ≤ e CN |λ|N +1 ,

(4.7)

for all λ ∈ R, |λ| ≤ λ∗ , with λ∗ > 0, where e CN is a positive constant depending only on the constants λ∗ , UN0 L∞ (0,T ;L2 ) ,





Uγ ∞ L

(0,T ;H 1 )

, γ = 0, 1 . . . , N.

Proof of Lemma4.2. Put h = U 0 + h1 ,

h1 =

N X

Uγ λγ .

(4.8)

γ =1

By using Taylor’s expansion of the function F (h) = F (U0 + h1 ) around the point U0 up to order N, we obtain F (h) = F (U0 + h1 ) = F (U0 ) +

N X 1 (γ ) γ (1) F (U0 )h1 + e RN , γ ! γ =1

(4.9)

where (1) e RN =

1

(1)

(N + 1)!

F (N +1) (U0 + θ1 h1 )hN1 +1 = λN +1 RN ,

(4.10)

with 0 < θ1 < 1. By Lemma 4.1, we obtain from (4.9), after some rearrangements in order of λγ that F (h) − F (U0 ) =

N X 1 (γ ) γ (1) F (U0 )h1 + λN +1 RN γ ! γ =1

=

γN N X X 1 (γ ) (1) F (U0 ) P [γ ] [U ]k λk + λN +1 RN γ ! k=γ γ =1

=

N X k   X 1 (γ ) (1) (2) F (U0 )P [γ ] [U ]k λk + λN +1 RN + RN , γ! k=1 γ =1

(4.11)

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3959

where (2) e RN =

γN N X X 1 (γ ) F (U0 ) P [γ ] [U ]k λk−N −1 . γ! γ =1 k=N +1

(4.12)

Combining (4.3) and (4.6) and (4.10)–(4.12), we then obtain EN (λ, x, t ) = −λh0 − F (h) + F (U0 ) −

N X

Fγ λγ

γ =1

# " N N k   X X X 1 (γ ) (1) (2) [γ ] 0 k F (U0 )P [U ]k λk − λN +1 RN + RN − Fk λk =− Uk−1 λ − γ ! k=1 k=1 k=1 γ =1 ( " #) N k   X X 1 =− Fk + Uk0 −1 + F (γ ) (U0 )P [γ ] [U ]k λk − λN +1 UN0 + R(N1) + R(N2) γ! k=1 γ =1   = −λN +1 UN0 + R(N1) + R(N2) . N +1 X

(4.13)

By the boundedness of the functions Uγ , Uγ0 , γ = 0, 1, . . . , N, in the function space L∞ (0, T ; H 1 ), we obtain from (4.10) and (4.12) that



(1)

RN ∞ L



(2)

RN ∞ L

(0,T ;L2 )

≤

(0,T ;L2 )

≤

N X |λ∗ |i−1 kUi kL∞ (0,T ;H 1 ) sup F (N +1) (z )

1

(N + 1)! |z |≤r (1)

2

(4.14)

γN X

sup F (γ ) (z )



(1)

γ =1

(1) =e CN ,

i=1

∗

N √ γ X

! N +1

k=N +1

|z |≤r∗

X 1 αN (2) 1 kU1 kαL∞ |λ |k−N −1 = e × . . . kUN kL∞ CN , (0,T ;H 1 ) (0,T ;H 1 ) ∗ α! (γ )

(4.15)

α∈Ak

(1)

where r∗

√ =

2 kU0 kH 1 +

P N

i=1

|λ∗ |i kUi kL∞ (0,T ;H 1 )

N +1

.

Hence, it follows from (4.13)–(4.15) that

  (2) (1) |λ|N +1 ≡ e kEN (λ, ·, ·)kL∞ (0,T ;L2 ) ≤ UN0 L∞ (0,T ;L2 ) + e CN CN |λ|N +1 . CN + e The proof of Lemma 4.2 is complete.

(4.16)



Next, we obtain the following theorem. Pλ ) has a unique weak Theorem 4.3. Let (H10 )–(H40 ), (F00 ) and (FN0 ) hold. Then, for every λ ∈ R, |λ| ≤ λ∗ , with λ∗ > 0, problem (e solution u = uλ satisfying the asymptotic estimation up to order N + 1 as follows

N

0 X 0 γ Uγ λ

u −

γ =0

L∞ (0,T ;L2 )

N

X

+ u − Uγ λγ

γ =0

≤e CN∗ |λ|N +1 ,

(4.17)

L∞ (0,T ;H 1 )

Pγ ), γ = 0, 1, . . . , N. where e CN∗ is a positive constant independent of λ, the functions Uγ are the weak solutions of problems (e Proof. First, we note that, if |λ| ≤ λ∗ , where λ∗ is a fixed positive constant, then the a priori estimates of the sequence {um } in the proof of Theorem 3.1 satisfy

0

u (t ) 2 + µ0 kumx (t )k2 ≤ CT , ∀t ∈ [0, T ], m

00 2

u (t ) + µ0 u0 (t ) 2 ≤ CT , ∀t ∈ [0, T ], m mx

(4.18) (4.19)

where CT is a constant depending only on T , e u0 , e u1 , g0 , g1 , k0 , k1 , µ, F , λ∗ (independent of λ). Hence, the limit u in suitable function spaces of the sequence {um } defined by (2.5)–(2.7) is a weak solution of the problem (1.1)–(1.4) satisfying the a priori estimates (4.18) and (4.19).

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L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

By multiplying the two sides of (4.5)1 with v 0 , and after integration in t, we find without difficulty from Lemma 4.2 that

σ (t ) ≤

1 0

µ

µ0

t

Z L∞ (QT ) t

Z +2 |λ|

σ (s)ds − 2

t

Z

P (s)v 0 (0, s)ds − 2

0 2

v (s) ds + 2

t

Z

0

Q (s)v 0 (1, s)ds 0

0

0

t

Z

kF (v + h) − F (h)k v 0 (s) ds + 2e CN |λ|N +1

t

Z

0

v (s) ds,

(4.20)

0

0

where



2  σ (t ) = Z v 0 (t ) + µ0 kvx (t )k2 ,   t   k0 (t − s)v(0, s)ds, P (t ) = Z0 t     Q ( t ) = k1 (t − s)v(1, s)ds.

(4.21)

0

Using the following inequalities

Z t

Z t 

0 

v 0 (s)ds ≤

v (s) ds, kv( t )k =  

 0  Z0 t Z t 

0 2

v (s) ds ≤ t kv(t )k2 ≤ t σ (s)ds,   0 0 Z  t  1  kv(t )k2 1 = kv(t )k2 + kvx (t )k2 ≤ t σ (t ), σ (s)ds + H µ0 0

(4.22)

we obtain from (4.21)2 , that t

Z

P (s)v 0 (0, s)ds ≤ β kv(t )k2H 1 + 4



1

kk0 k2L2 (0,T ) + |k0 (0)| + k00 L1 (0,T ) β    Z t β 1 ≤ σ (t ) + β T + D (k0 , β) + T2 σ (r )dr µ0 µ0 0 Z t β = σ (t ) + e D (T , k0 , β) σ (r )dr , µ0 0

−2 0

t

Z 0

kv(s)k2H 1 ds

(4.23)

where

   1  2  e +T , D (T , k0 , β) = β T + D (k0 , β) µ0  

1   kk0 k2L2 (0,T ) + |k0 (0)| + k00 L1 (0,T ) . D (k0 , β) = β

(4.24)

Similarly t

Z

Q (s)v 0 (1, s)ds ≤

−2 0

β σ (t ) + e D (T , k1 , β) µ0

t

Z

σ (r )dr ,

(4.25)

0

where

   1  2  e +T , D (T , k1 , β) = β T + D (k1 , β) µ0  

1   kk1 k2L2 (0,T ) + |k1 (0)| + k01 L1 (0,T ) . D (k1 , β) = β

(4.26)

Combining (4.20), (4.23) and (4.25), we then obtain

σ (t ) ≤

2β

σ (t ) + T e CN2 |λ|2N +2 µ0  Z t 1 0

µ ∞ e e |λ|) + + D T , k , β) + D T , k , β) + 2 1 + σ (r )dr ( ( ( 0 1 L (QT ) µ0 0 Z t kF (v + h) − F (h)k2 ds. +2 0

(4.27)

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3961

On the other hand N X



U γ ∞

khkL∞ (0,T ;H 1 ) ≤

L

γ =0

(0,T ;H 1 )

|λ∗ |γ ≡ R(∗1) .

(4.28)

Then, it follows from (4.28) that t

Z

kF (v + h) − F (h)k ds ≤ 2

2

2K∗2T

t

Z

kv(s)k ds ≤ 2 2

≤ K∗2T T 2

K∗2T sds

t

Z

0 2

v (r ) dr ≤ K 2 T 2 ∗T

0

s

Z

0 2

v (r ) dr

0

0

0

0

t

Z

t

Z

σ (r )dr ,

(4.29)

0

(2) √ (1) where K∗T = sup|z |≤R(2) F / (z ) , R∗ = R∗ + CT . Hence, it follows from (4.27) and (4.29) that ∗ σ (t ) ≤

2β

µ0

(1) σ (t ) + T e CN2 |λ|2N +2 + KT

t

Z

σ (r )dr ,

(4.30)

0

where (1)

KT

=

1 0

µ

Choosing β =

+e D (T , k0 , β) + e D (T , k1 , β) + 2 (1 + |λ|) + K∗2T T 2 .

L∞ (QT )

µ0

1 4

(4.31)

µ0 , by Gronwall’s lemma, we obtain from (4.30) that

(1) (2) σ (t ) ≤ 2T e CN2 |λ|2N +2 exp(2TKT ) = KT |λ|2N +2 ,

(4.32)

for all λ ∈ R, |λ| ≤ λ∗ . This implies that

0

v ∞ L

(0,T ;L2 )

CN∗ |λ|N +1 , + kvkL∞ (0,T ;H 1 ) ≤ e

(4.33)

or

N

0 X 0 γ Uγ λ

u −

γ =0

L∞ (0,T ;L2 )

N

X

γ Uγ λ + u −

γ =0

Theorem 4.3 is completely proved.

≤e CN∗ |λ|N +1 .

(4.34)

L∞ (0,T ;H 1 )



5. Stability of the solutions In this part, we assume that the functions u0 , u1 and F satisfy (H10 ) and (F00 ). By Theorem 3.1 the problem (1.1)–(1.4) has a unique weak solution u depending on λ, µ, g0 , g1 , k0 , k1 . u = u(λ, µ, g0 , g1 , k0 , k1 ),

(5.1)

where (g0 , g1 , k0 , k1 , µ) satisfy the assumptions (H2 )–(H4 ), λ ∈ R; and u0 , u1 , F are fixed functions satisfying (H1 ) and (F00 ). We put =(µ0 ) = {(g0 , g1 , k0 , k1 , µ) : (g0 , g1 , k0 , k1 , µ) satisfy the assumptions (H20 )–(H40 )}, where µ0 > 0 is given constant. Then we have the following theorem. 0

0

0

Theorem 5.1. Let T > 0. Let (H10 )–(H40 ) and (F00 ) hold. Then, solutions of the problems (1.1)–(1.4) are stable with respect to the data (λ, µ, g0 , g1 , k0 , k1 ), i.e., j j j j If (λ, µ, g0 , g1 , k0 , k1 ), (λj , µj , g0 , g1 , k0 , k1 ) ∈ R × =(µ0 ), such that



 j λ − λ + µj − µ C 1 (Q ) → 0, 

T

 

j

+ g1j − g1 2 → 0,

g0 − g0 2 H (0,T ) H (0,T )

  j

 + kj1 − k1 2,1 → 0,

k0 − k0 2,1 (0,T )

W

W

(5.2)

(0,T )

as j → +∞, then



uj − u ∞ L

(0,T ;H 1 ) j

+ u0j − u0 L∞ (0,T ;L2 ) → 0, j

j

j

where uj = u(λj , µj , g0 , g1 , k0 , k1 ).

as j → +∞,

(5.3)

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L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

Proof. First, we note that, if the data (λ, µ, g0 , g1 , k0 , k1 ) satisfy

 kµkC 1 (QT ) ≤ µ∗ , |λ| ≤ λ∗ , kg0 k 2 + kg 1 k 2 ≤ g ∗, kk kH 2(,01,T ) + kk Hk (02,T,1) ∗ 0 W 1 W (0,T ) (0,T ) ≤ k ,

(5.4)

where λ∗ , µ∗ , g ∗ , k∗ are fixed positive constants, then, the a priori estimates of the sequence {um } in the proof of the Theorem 3.1 satisfy

0

u (t ) 2 + µ0 kumx (t )k2 ≤ CT , ∀t ∈ [0, T ], m

00 2

u (t ) + µ0 u0 (t ) 2 ≤ CT , ∀t ∈ [0, T ], m mx

(5.5) (5.6)

where CT is a constant depending only on T , u0 , u1 , µ0 , λ , µ , g , k (independent of λ, µ, g0 , g1 , k0 , k1 ). Hence, the limit u in suitable function spaces of the sequence {um } defined by (2.5), (2.6) and (3.3) is a weak solution of the problem (1.1)–(1.4) satisfying the a priori estimates (5.5) and (5.6). j j j j Now, by (5.2) we can assume that, there exist positive constants λ∗ , µ∗ , g ∗ , k∗ , such that the data (λj , µj , g0 , g1 , k0 , k1 ) j

j

∗

∗

j

j

∗

∗

satisfy (5.4) with (λ, µ, g0 , g1 , k0 , k1 ) = (λj , µj , g0 , g1 , k0 , k1 ). Then, by the above remark, we have that the solutions uj of j

j

j

j

problem (1.1)–(1.4) corresponding to (λ, µ, g0 , g1 , k0 , k1 ) = (λj , µj , g0 , g1 , k0 , k1 ) satisfy

0 2

u (t ) + µ0 ujx (t ) 2 ≤ CT , ∀t ∈ [0, T ], j

00 2

u (t ) + µ0 u0 (t ) 2 ≤ CT , ∀t ∈ [0, T ]. j jx

(5.7) (5.8)

Put

 λj = λj − λ, e µj = µj − µ, e j j j j e g = g0 − g0 , e g1 = g1 − g1 , e0j j j j e k0 = k0 − k0 , k1 = k1 − k1 .

(5.9)

Then, vj = uj − u satisfy the variational problem

 00



 µj (t )ujx (t ), vx + e Pj (t )v(0) + e Q j (t )v(1) hvj (t ), vi + µ(t )vjx (t ), vx + e +hF (uj ) − F (u) + λvj0 + e λj u0j , vi = 0, ∀ v ∈ H 1 ,  vj (0) = vj0 (0) = 0,

(5.10)

 Z t  j  Pj (t ) = b g0 (t ) + k0 (t − s)vj (0, s)ds, e Z0 t  j  Q j (t ) = b g1 (t ) + k1 (t − s)vj (1, s)ds, e

(5.11)

 Z t  j j j  e g0 (t ) + k0 (t − s)uj (0, s)ds, g0 (t ) = e b Z0 t  j j j  e k1 (t − s)uj (1, s)ds. g1 (t ) = e g1 (t ) + b

(5.12)

where

0

0

Substituting e Pj (t ), e Q j (t ) into (5.10)1 , then taking v = vj0 in (5.10)1 , afterwards integrating in t, we obtain Sj (t ) =

Z

t

1

Z

 µ0j (x, s)vjx2 (x, s)dx − 2 e µj (t )ujx (t ), vjx (t ) + 2 0 0 Z t Z t 0 e e −2 Pj (s)vj (0, s)ds − 2 Q j (s)vj0 (1, s)ds ds

0

− 2λ

Z tD 0

E
0 e µj ujx , vjx ds

0 t

Z

0 2

v (s) ds − 2e λj j

t

Z

0

0

u0j (s), vj0 (s) ds − 2



t

Z



F (uj ) − F (u), vj0 (s) ds,



(5.13)

0

where

p

2

2

Sj (t ) = vj0 (t ) + µ(t )vjx (t ) .



(5.14)

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3963

Using Eq. (5.13) and the inequalities (2.18), (5.7) and (5.8) and 2 2 Sj (t ) ≥ vj0 (t ) + µ0 vjx (t ) = σj (t ),





vj (t ) 2 ≤ t

t

Z



0 2

v (s) ds ≤ t

(5.15)

t

Z

σj (s)ds,

j

(5.16)

0

0





vj (t ) 2 1 = vjx (t ) 2 + vj (t ) 2 H Z t 1 σj (s)ds, σj (t ) + t ≤ µ0 0

(5.17)

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

σj (t ) ≤



1 0

µ ∞ 3 + 2 |λ| + L (QT ) µ

Z

t

2

2 1

e σj (s)ds + βσj (t ) + TCT e λj + µj (t )ujx (t ) βµ 0 0 0 Z T Z t Z t Z t

2


1 0

F (uj ) − F (u) 2 ds, e e + µj ujx ds − 2 Pj (s)vj0 (0, s)ds − 2 Q j (s)vj0 (1, s)ds +

e µ0 0 0 0 0

(5.18)

for all β > 0 and t ∈ [0, T ]. We shall estimate

respectively,

2 the following five last terms on the right-hand side of (5.18). 1 Estimating βµ e µj (t )ujx (t ) . Using the inequality (5.7), we have 0

2

2 1 CT 2 1 2

e

e

e (5.19) µj L∞ (Q ) . µj (t )ujx (t ) ≤ µj L∞ (Q ) ujx (t ) ≤ T T βµ0 βµ0 βµ20 RT
0

2 Estimating µ1 0 e µ j ujx ds. Using the inequalities (5.7) and (5.8) then, we have 0 Z T Z T

0 2 0 2 
0 1 2

2

e µj ujx ds ≤ µj ujx + e µj ujx ds

e µ0 0 µ0 0 

0 2 2TCT  2

e

e

∞

∞ + µ ≤ µ j j L (QT ) L (QT ) µ20 4TCT 2

e ≤ µj C 1 (Q ) . (5.20) T µ20 Rt Pj (s)vj0 (0, s)ds. Using the inequalities (2.18), (5.7), (5.8), (5.15) and (5.17), then we can prove the Estimating −2 0 e

following inequality in a similar manner

Z t  0 2 Z t

j

2 2 j 2 0 b

ds + DT (β, k0 )

vj (s) 2 1 ds, e −2 Pj (s)vj (0, s)ds ≤ 2β vj (t ) H 1 + b g0 (t ) + 2 g ( s ) 0 H β 0 0 0

where DT (β, k0 ) = 1 + 4 |k0 (0)| + β4 kk0 k2L2 (0,T ) + 4 k00 L1 (0,T ) . t

Z

(1)

By the imbedding H 1 (0, T ) ,→ C 0 ([0, T ]), there exists a positive constant CT

khk2C 0 ([0,T ]) +

T

Z

0 2 h (s) ds ≤ C (1) khk2 1 T

0

H (0,T )

for all h ∈ H 1 (0, T ) .

(5.21)

such that (5.22)

Hence, it follows from (5.17), (5.21) and (5.22) that t

Z −2 0

  Z t

2

2

1

j

vj (s) 2 1 ds e + 1 CT(1) b g0 + DT (β, k0 ) Pj (s)vj0 (0, s)ds ≤ 2β vj (t ) H 1 + 2 H H 1 (0,T ) β 0      Z t

2 2β 1 1

j ≤ σj (t ) + 2 + 1 CT(1) b g0 + 2β T + + T 2 DT (β, k0 ) σj (s)ds. H 1 (0,T ) µ0 β µ0 0 (5.23)

Estimating −2 t

Z −2 0

Rt 0

e Q j (s)v (1, s)ds. Similarly 0

j

     Z t

2 2β 1 1

j e Q j (s)vj0 (1, s)ds ≤ σj (t ) + 2 + 1 CT(1) b g1 + 2β T + + T 2 DT (β, k1 ) σj (s)ds, H 1 (0,T ) µ0 β µ0 0 (5.24)

where DT (β, k1 ) = 1 + 4 |k1 (0)| + β k k 4

k1 2L2 (0,T )

+ 4 k0 1

1 L (0,T ) .

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L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

Rt

F (uj ) − F (u) 2 ds. It follows from (5.7) and (5.16) that Estimating 0

t

Z

t

Z



F (uj ) − F (u) 2 ds ≤ T 2 k2

σj (s)ds,

T

0

(5.25)

0

where kT = sup|z |≤√CT F 0 (z ) . Combining (5.18) and (5.19) and (5.23)–(5.25), we then obtain



σj (t ) ≤





t

  4 σj (s)ds + β 1 + σj (t ) µ0 µ0 0      

2 2 2 CT 1

2 1

j (1) j e

+ TCT λj + 2 + 4T e µj C 1 ( Q ) + 2 + 1 CT g0 + b g1

b T H 1 (0,T ) H 1 (0,T ) β µ0 β  Z t   1 σj (s)ds, + T 2 (DT (β, k0 ) + DT (β, k1 )) + T 2 k2T + 4β T + µ0 0 3 + 2 |λ| +

Z

1 0

µ

L∞ (QT )

(5.26)

for all β > 0 and t ∈ [0, T ].



Choosing β > 0, with β 1 + µ4 0



< 21 , by Gronwall’s lemma, we deduce that

 

2

2 2 2

j

j b , λj + e µj C 1 (Q ) + b g1 g0 + σj (t ) ≤ CT(2) e

T H 1 (0,T ) H 1 (0,T )

(5.27)

for all j and t ∈ [0, T ]. This implies that

0

v ∞ j

L

(0,T ;L2 ) (3)

for all j, where CT

+ vj ∞ L

(0,T ;H 1 )

(3)

≤ CT









j

j

e λj + e µj C 1 (Q ) + b g0 + b g1 , T H 1 (0,T ) H 1 (0,T )

(5.28)

is a constant depending only on λ, µ, g0 , g1 , k0 , k1 .

On the other hand, using the imbeddings H 1 (0, T ) ,→ C 0 ([0, T ]), W 1,1 (0, T ) ,→ C 0 ([0, T ]), it follows from (5.7) and (5.12) that



p √

j

j

j g0 ≤ 2 e g0 + 2TCT e k0 ,

b 1 1 H (0,T ) H (0,T ) W 1,1 (0,T )



p √

j

j

j g1 + 2 e g1 2TCT e k1 ≤ .

b 1 1 1,1 H (0,T )

H (0,T )

W

(0,T )

(5.29) (5.30)

Finally, by (5.2) and (5.9) and the estimates (5.28)–(5.30), we deduce that (5.3) holds. Theorem 5.1 is completely proved.  Remark 5.1. Put


2
2 e H1 = R × H 1 (0, T ) × W 1,1 (0, T ) ,
2
2 e H2 = R × H 2 (0, T ) × W 2,1 (0, T ) , 

e1 = u ∈ L∞ 0, T ; H 1 : ut ∈ L∞ 0, T ; L2 , W 
e2 = u ∈ L∞ 0, T ; H 2 : ut ∈ W e1 . W

(5.31) (5.32) (5.33) (5.34)

Let u0 , u1 , µ, F , be fixed functions satisfying (H10 ), (H40 ) and (F00 ). By Theorems 2.2 and 3.1, the problem (1.1)–(1.4) has a unique weak solution u depending on λ, g0 , g1 , k0 , k1 , u = S (λ, g0 , g1 , k0 , k1 ),

(5.35)

satisfying

e1 , u = S (λ, g0 , g1 , k0 , k1 ) ∈ W

for all (λ, g0 , g1 , k0 , k1 ) ∈ e H1 ,

(5.36)

e2 , u = S (λ, g0 , g1 , k0 , k1 ) ∈ W

for all (λ, g0 , g1 , k0 , k1 ) ∈ e H2 .

(5.37)

L.T. Phuong Ngoc et al. / Nonlinear Analysis 70 (2009) 3943–3965

3965

Then we have the following theorem. Theorem 5.2. Let T > 0. Let (H10 ), (H40 ) and (F00 ) hold. Then,

e1 is a completely continuous operator. S :e H2 → W

(5.38)

e1 is a continuous operator. Proof. First, by Theorem 5.1, we have S : e H2 → W e1 is compact operator. Next, we shall prove that S : e H2 → W j j j j Let {(λj , g0 , g1 , k0 , k1 )} ⊂ e H2 be a bounded sequence in e H2 . By the compactness of the imbedding e H2 ,→ e H1 , we deduce j

j

j

j

the existence of a subsequence of {(λj , g0 , g1 , k0 , k1 )} still also so denoted, and (λ, g0 , g1 , k0 , k1 ) ∈ e H2 , such that

(λj , g0j , g1j , kj0 , kj1 ) → (λ, g0 , g1 , k0 , k1 ), in e H1 strongly. j j g0

Put uj = S (λ , , of Theorem 5.1, that

0

u − u0 ∞ j



j g0

b



j g1

b

L

H 1 (0,T )

j g1

,

,

j k1

(5.39)

), u = S (λ, g0 , g1 , k0 , k1 ). Using (5.28)–(5.30) and (5.39), we prove, in a manner similar to that





j (4) j

b ≤ C u − u λ − λ + g +

j T 0 L∞ (0,T ;H 1 ) (0,T ;L2 )

H 1 (0,T )



≤ CT(4) g0j − g0 (4)

H 1 (0,T )

j k0

≤ CT



j

g1 − g1



H 1 (0,T )



+ kj0 − k0



H 1 (0,T )



+ kj1 − k1

W 1,1 (0,T )

W 1,1 (0,T )



j + b g1

 H 1 (0,T )

,

(5.40)

,

(5.41)

,

(5.42)

(4)

for all j, where CT is a constant depending only on F , λ, µ, g0 , g1 , k0 , k1 . e1 strongly. Finally, by the estimates (5.40)–(5.42), we deduce that uj → u in W Theorem 5.2 is completely proved.  Acknowledgement The authors wish to express their sincere thanks to the referee for the helpful comments. References [1] N.T. An, N.D. Trieu, Shock between absolutely solid body and elastic bar with the elastic viscous frictional resistance at the side, J. Mech. NCSR. Vietnam 13 (2) (1991) 1–7. [2] M. Bergounioux, N.T. Long, A.P.N. Dinh, Mathematical model for a shock problem involving a linear viscoelastic bar, Nonlinear Anal. 43 (2001) 547–561. [3] H. Brézis, Analyse fonctionnelle. Théorie et Applications, Masson Paris, 1983. [4] F.E. Browder, On nonlinear wave equations, Math. Z. 80 (1962) 249–264. [5] E.L.A. Coddington, N. Levinson, Theory of Ordinary Differential Equations, McGraw-Hill, 1955, p. 43. [6] J.L. Lions, W.A. Strauss, Some nonlinear evolution equations, Bull. Soc. Math., France 93 (1965) 43–96. [7] J.L. Lions, Equations differentielles operationelles et problèmes aux limites, Springer-Verlag, Berlin, 1961, p. 96, Section 7. [8] J.L. Lions, Quelques méthodes de résolution des problèmes aux limites nonlinéaires, Dunod, Gauthier-Villars, Paris, 1969. [9] N.T. Long, A.P.N. Dinh, On the quasilinear wave equation: utt − ∆u + f (u, ut ) = 0 associated with a mixed nonhomogeneous condition, Nonlinear Anal. 19 (1992) 613–623. [10] N.T. Long, T.N. 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. [11] N.T. Long, A.P.N. Dinh, T.N. Diem, On a shock problem involving a nonlinear viscoelastic bar, J. Boundary Value Problems, Hindawi Publishing Corporation 2005 (3) (2005) 337–358. [12] N.T. Long, L.V. Ut, N.T.T. Truc, On a shock problem involving a linear viscoelastic bar, Nonlinear Anal. 63 (2) (2005) 198–224. [13] N.T. Long, V.G. Giai, A nonlinear wave equation associated with nonlinear boundary conditions: Existence and asymptotic expansion of solutions, Nonlinear Anal. TMA 66 (12) (2007) 2852–2880. [14] M.L. Santos, Asymptotic behavior of solutions to wave with a memory condition at the boundary, Electronic J. Differential Equations (ISSN: 1072-6691) 73 (2001) 1–11. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu.