A general mathematical model for the collision between a free-fall hammer of a pile-driver and an elastic pile

Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

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A general mathematical model for the collision between a free-fall hammer of a pile-driver and an elastic pile Út V. Lê Department of Mathematical Sciences, P.O. Box 3000, FI-90014 University of Oulu, Finland

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Article history: Received 6 July 2009 Accepted 2 November 2009

abstract In this paper we consider a general mathematical model for the collision between the freefall hammer of a pile-driver and an elastic pile whose ends are furnished with a bearing. When the free-fall hammer collides with the pile, the displacement of a cross-sectional area of the pile is the weak solution of an initial-boundary value problem involving a linear wave equation with memory boundary conditions. We generalize this problem into a nonlinear one with more general boundary conditions. Then we obtain the unique solvability and the regularity of the weak solution of this nonlinear problem. The unique solvability is shortly discussed in regard to the Galerkin method. The regularity result is obtained by a combination of a fixed-point technique and an energy method, and the convenience of this procedure is also pointed out. © 2009 Elsevier Ltd. All rights reserved.

Keywords: Asymptotic expansion Contraction procedure Memory boundary conditions Regularity Stability Wave equation Well-posedness

1. Introduction 1.1. Known results and motivation In the interest of a mechanical motivation, a simple example for the problem in this paper, we consider a model demonstrating the collision between the free-fall hammer of a pile-driver and an elastic pile with a bearing (see [1–3]). The pile is furnished with a bearing at the end which the free-fall hammer directly hits, called the first bearing (the other bearing, if any, is called the second bearing) (see Fig. 1). For this model, we suppose that the force of friction at the side of the pile is visco-elastic, and is approximated to the bulk force. If u is the displacement of a cross-sectional area of the pile, then u is a function of the length x of the pile driven into the ground, and the colliding time t. Then the force of friction in a factor of the pile is defined by



K1 u + λ1

∂u ∂t



r F

,

where K1 and λ1 are, respectively, the resisting coefficients of the elastic and viscous forces at the side of the pile; r and F are the circumference and the area of the cross section of the pile, respectively. Applying d’Alembert’s principle for a factor of the pile, after some computations, we obtain the equation of the motion of the pile as follows: 2 ∂ 2u ∂u 2∂ u − a + Ku + λ = 0, (1.1) 2 2 ∂t ∂x ∂t q rK where a = ρE is the elastic wave speed of the pile, E is the elastic modulus, ρ is the material density of the pile, K = F1 ,

and λ =

r λ1 . F

The terms Ku and λ ∂∂ut are, respectively, called the internal source term and the internal damping of the wave Eq.

(1.1). Shortly we call the term Ku + λ ∂∂ut the internal damping-source term. Associating (1.1) with initial data E-mail addresses: [email protected], [email protected]. 1468-1218/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2009.11.002

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Fig. 1.

u(x, 0) =

∂u (x, 0) = 0 ∂t

and boundary conditions E

∂u (0, t ) = −h0 (t ), ∂x

u(1, t ) = 0

(the length of the bar is supposed to be 1, and h0 is either a given or unknown function varying from different applied models), we get the mathematical model for the collision when a free hammer collides with an elastic pile concerning the visco-elastic resisting force:

∂ 2u ∂ 2u ∂u − a2 2 + Ku + λ = 0, 0 < x < 1, t ≥ 0, (1.2) 2 ∂t ∂x ∂t ∂u E (0, t ) = −h0 (t ), t ≥ 0, (1.3) ∂x u(1, t ) = 0, t ≥ 0, (1.4) ∂u u(x, 0) = (x, 0) = 0, 0 ≤ x ≤ 1. (1.5) ∂t The homogeneous boundary condition at x = 1 implies that the elastic pile hits a hard underlay. In addition, if a damper or a shock absorber of this model is considered, then h0 (t ) in the Eq. (1.3) is an elastic force arising from the initial shock, and is obviously unknown. Thanks to the mechanical principles, h0 (t ) can be considered as the solution of the following Cauchy problem of an ordinary differential equation: h000 (t ) +

K0 M0

h0 (t ) −

h0 (0) = 0,

K0 ∂ 2 u F ∂t2

(1.6)

K0 V0

, (1.7) F where K0 and M0 are, respectively, the hardness and the weight of the bearing, V0 is the velocity of the free hammer at the time of collision, t = 0. Define s g0 (t ) :=

h00 (0) =

(0, t ) = 0,

M0 K0 V0 K0

F

s sin

K0 M0

! t

,

(1.8)

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K0

k0 (t ) :=

s

F

K0 M0

s sin

K0 M0

! ,

t

(1.9)

one can show from (1.6) and (1.7) that K0

h0 ( t ) = g 0 ( t ) −

F

u(0, t ) −

t

Z

k0 (t − s)u(0, s)ds.

(1.10)

0

In this boundary condition, the term − F0 u(0, t ) is the so-called boundary source term of the wave equation (1.1). Because of the convolution (k0 ∗ u(0, ·)) (t ), the integral equation (1.10) is called a memory condition at the boundary x = 0. Thus (1.2)–(1.5) and (1.10) formulate a mathematical model for the collision between the free-fall hammer and a visco-elastic pile. When the boundary value (1.4) becomes non-homogeneous, and is given by K

∂u ∂u (1, t ) + C1 u(1, t ) + C2 (1, t ) = 0 ∂x ∂t

(1.11)

for C1 and C2 non-negative constants, Bergounioux et al. [4] studied the unique solvability of this problem by the Galerkin method. In addition Takačis [5,6] investigated the solvability of this problem in the field of Mikusiński operators, and the exact solution was given. In fact, initial-boundary problems of wave equations with memory boundary conditions have been widely studied. For a more general memory term than (1.10), given at the end x = 1, in [7] Santos considered the following problem:

∂ 2u ∂ 2u − µ(t ) 2 = 0, 0 < x < 1, t ≥ 0, (1.12) 2 ∂t ∂x u(0, t ) = 0, t ≥ 0, (1.13) Z t ∂u u(1, t ) + a(t − s)µ(s) (1, s)ds = 0, t ≥ 0, (1.14) ∂x 0 ∂u (x, 0) = u1 (x), 0 < x < 1. (1.15) u(x, 0) = u0 (x), ∂t He obtained the asymptotic behavior of the solution u(x, t ) with respect to the time variable, where a, µ, u0 and u1 are given functions. It is worth noting that the memory condition (1.14) is equivalent to

− µ(t )

∂u k(0) 1 ∂u k(t )u0 (1) (1, t ) = u(1, t ) + (1, t ) − − ∂x a(0) a(0) ∂ t a(0)

t

Z

k(t − s)u(1, s)ds,

(1.16)

0

where k(t ) satisfies the integral equation k(t ) +

t

Z

1 a(0)

a0 (t − s)k(s)ds =

0

1 a(0)

a0 (t ).

Also for the homogeneous boundary condition at x = 0 and the initial conditions (1.15), Rivera and Andrade [8] studied the following nonlinear problem:

∂ 2u ∂ − σ ∂t2 ∂x



∂u ∂x

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



= f (x, t ),

t ≥ 0, a(t − s)σ



0

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

(1.17) (1.18)

t

Z

0 < x < 1, t > 0,

 ∂u (1, s) ds = g (t ), ∂x

∂u (x, 0) = u1 (x), ∂t

t ≥ 0,

0 < x < 1,

(1.19) (1.20)

where σ , f , a, g, u0 and u1 are given functions. Applying the Volterra’s inverse operator for the integral equation (1.19), it follows that   ∂u 1 ∂u 1 0 −σ (1, t ) = k(0)u(1, t ) + (1, t ) − u0 (1)k(t ) − g (t ) ∂x a(0) ∂ t a(0)

Z −

t

k(t − s)g (s)ds + 0

0

t

Z

k0 (t − s)u(1, s)ds, 0

where k(t ) is the resolvent kernel of a0 , and is the solution of k(t ) +

1 a(0)

Z 0

t

a0 (t − s)k(s)ds = −

1 a(0)2

a0 (t ).

(1.21)

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It is clear that the integral equations (1.14) and (1.19) are similar,
but the problem {(1.17)–(1.20)} is more complicated than the problem {(1.12)–(1.15)} due to the nonlinear term σ ∂∂ ux . In [8] the authors obtained the global existence of smooth solutions and the decay rate of the solutions to the problem. This improves the previous results by Qin [9,10] for the case f = 0. With a full internal damping-source term, in [11] Nguyen et al. considered the following problem:

∂ 2u ∂ 2u ∂u − µ(t ) 2 + Ku + λ = f (x, t ), 0 < x < 1, 0 < t < T , 2 ∂t ∂x ∂t u(0, t ) = 0, 0 < t < T , ∂u −µ(t ) (1, t ) = Q (t ), 0 < t < T , ∂x ∂u (x, 0) = u1 (x), 0 < x < 1, u(x, 0) = u0 (x), ∂t Z t ∂u Q (t ) = K1 (t )u(1, t ) + λ1 (t ) (1, t ) − g (t ) − k(t − s)u(1, s)ds ∂t 0

(1.22) (1.23) (1.24) (1.25) (1.26)

for K and λ given real constants, where µ, u0 , u1 , K1 , λ1 , g and k are given functions. This problem is a mathematical model for a shock problem involving a linear visco-elastic bar generalized from {(1.1)–(1.5), (1.10)}, also from {(1.12)–(1.15)}, and from {(1.17)–(1.20)}. The results in [11] include the unique existence, the regularity, the stability, and the asymptotic expansion (with respect to K and λ) of the solution u(x, t ). In addition the asymptotic behavior of the weak solution of this problem was studied in [12]. In [13,14] Lê investigated a generalization of the problem {(1.23)–(1.26)} when the linear internal dampingsource term Ku + λ ∂∂ut is replaced by a nonlinear one given by p−2

K |u|

q−2 ∂u ∂u u + λ ∂t ∂t

with p, q ≥ 2, K and λ given constants. Moreover some related problems were also studied in [15–17]. In particular, in [16, 17] I studied the well-posedness of

    ∂ 2u ∂ ∂u ∂u − µ( x , t ) + G ( u ) + H = 0, 0 < x < 1, 0 < t < T , (1.27) ∂t2 ∂x ∂x ∂t Z t ∂u k0 (t − s)u(0, s)ds, (1.28) µ(0, t ) (0, t ) = g0 (t ) + ∂x 0 Z t ∂u k1 (t − s)u(1, s)ds, (1.29) −µ(1, t ) (1, t ) = g1 (t ) + ∂x 0 ∂u (x, 0) = u1 (x), (1.30) u(x, 0) = u0 (x), ∂t where µ, G, H, g0 , g1 , k0 , k1 , u0 and u1 are given functions. In addition the asymptotic expansion with respect to a ‘‘small’’ damping-source term and the regularity of the weak solution were also given. However it is noted that the source G and the damping H in those papers are just Lipschitz. Now let us consider a more general model than {(1.2)–(1.7)}. We suppose the pile is furnished with a bearing at each end. Then we have the model with the first and second bearings (see Fig. 2). If the underlay is not hard, then the displacement of the cross-sectional area of the pile at x = 1, when the second bearing hits the underlay, is not vanished. By the same arguments for the models in [1–3], we associate the Eq. (1.1) with the boundary value at x = 1 as follows: E

∂u (1, t ) = −h1 (t ), ∂x

(1.31)

where h1 is also an elastic force arising from the collision between the pile and the ‘‘soft’’ underlay, and satisfies the Cauchy problem h001 (t ) +

K1 M1

h1 (1) = 0,

h1 (t ) −

K1 ∂ 2 u F ∂t2

h01 (1) =

(1, t ) = 0,

K1 V1 F

,

(1.32) (1.33)

where K1 and M1 are, respectively, the hardness and the weight of the second bearing, V1 is the velocity of the free hammer when it collides with the underlay. Finally the Eqs. (1.2)–(1.7) and (1.31)–(1.33) formulate the mathematical model of the

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Fig. 2.

collision between a free-fall hammer of a pile-driver and an elastic pile whose ends are furnished with a bearing: 2 ∂ 2u ∂u 2∂ u − a + Ku + λ = 0, 0 < x < 1, t ≥ 0, ∂t2 ∂ x2 ∂t ∂u E (0, t ) = −h0 (t ), t ≥ 0, ∂x ∂u (1, t ) = −h1 (t ), t ≥ 0, E ∂x ∂u (x, 0) = 0, 0 ≤ x ≤ 1, u(x, 0) = ∂t K0 ∂ 2 u K0 h0 (t ) − (0, t ) = 0, h000 (t ) + M0 F ∂t2

h0 (0) = 0, h001 (t ) +

K1 M1

h1 (1) = 0,

h00 (0) = h1 (t ) −

K0 V 0 F

K1 ∂ 2 u F ∂t2

h01 (1) =

,

(1, t ) = 0,

K1 V 1 F

.

(1.34) (1.35) (1.36) (1.37) (1.38) (1.39) (1.40) (1.41)

In this paper we are studying a more general problem than both {(1.27)–(1.30)} and {(1.34)–(1.41)}, the so-called a general mathematical model for the collision between a free-fall hammer of a pile-driver and an elastic pile. For specificity, we would like to move to Section 1.2. 1.2. Statement of the problem and new results 1.2.1. Statement of the problem From the above motivations, we consider a problem: find a trio of functions (u(x, t ), B0 (t ), B1 (t )) such that

β   ∂u ∂u ∂ 2u ∂ ∂u α | | − a ( x , t ) + b ( x , t ) u u + c ∂ t ∂ t = d(x, t ), 2 ∂t ∂x ∂x

(1.42)

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∂u (0, t ) = B0 (t ), (1.43) ∂x ∂u (1.44) −a(1, t ) (1, t ) = B1 (t ), ∂x ∂u (x, 0) = u1 (x) (1.45) u(x, 0) = u0 (x), ∂t for (α, β, c ) ∈ R2∗ , R∗ ≡ R+ ∪ {0}, and (x, t ) ∈ ΩT ≡ (0, 1) × (0, T ); where a, b, d, u0 and u1 are given functions. The unknown function u(x, t ) and the unknown boundary values B0 (t ), B1 (t ) satisfy the following Cauchy problems of ordinary a(0, t )

differential equations: B000 (t ) + p20 B0 (t ) − q0 (0)

B0 (0) = B0 ,

∂ 2u (0, t ) = 0, ∂t2 (00)

B00 (0) = B0

(1.46)

,

(1.47)

and B001 (t ) + p21 B1 (t ) − q1 (1)

B1 (1) = B1 ,

∂ 2u (1, t ) = 0, ∂t2 (11)

B01 (1) = B1 (0)

(1.48)

,

(1.49)

(00)

(1)

(11)

where p0 > 0, q0 ≥ 0, B0 , B0 , p1 > 0, q1 ≥ 0, B1 and B1 are given constants. This problem is a general mathematical model from the investigation of the collision between the free-fall hammer of a pile-driver and an elastic pile whose ends are furnished with a bearing. Recalling the techniques to deduce the integral equation (1.10) from (1.6) and (1.7), we define G0 (t ) := B00 − q0 u0 (0) cos(p0 t ) +




1 p0

B00 0 − q0 u1 (0) sin (p0 t ) ,




K0 (t ) := p0 q0 sin (p0 t ) ,

(1.50) (1.51)

G1 (t ) := B11 − q1 u0 (1) cos(p1 t ) +




1 p1


B11 1 − q1 u1 (1) sin (p1 t ) ,

K1 (t ) := p1 q1 sin (p1 t ) .

(1.52) (1.53)

Then we get from (1.46)–(1.53) that B0 (t ) = G0 (t ) + q0 u(0, t ) −

t

Z

K0 (t − s)u(0, s)ds,

(1.54)

K1 (t − s)u(1, s)ds.

(1.55)

0

B1 (t ) = G1 (t ) + q1 u(0, t ) −

t

Z 0

To be more general, we suppose G0 , G1 , K0 and K1 are given, and satisfy some conditions specified later. Thus we are actually investigating the following problem:

β   ∂u ∂u ∂ 2u ∂ ∂u α | | − a ( x , t ) + b ( x , t ) u u + c ∂ t ∂ t = d(x, t ), 2 ∂t ∂x ∂x ∂u (0, t ) = B0 (t ), ∂x ∂u −a(1, t ) (1, t ) = B1 (t ), ∂x ∂u u(x, 0) = u0 (x), (x, 0) = u1 (x), ∂t Z t B0 (t ) = G0 (t ) + q0 u(0, t ) − K0 (t − s)u(0, s)ds, a(0, t )

(1.56) (1.57) (1.58) (1.59) (1.60)

0

B1 (t ) = G1 (t ) + q1 u(1, t ) −

t

Z

K1 (t − s)u(1, s)ds

(1.61)

0

for (x, t ) ∈ ΩT , where u(x, t ), B0 (t ) and B1 (t ) are unknown. The given data include functions a, b, d, u0 , u1 , G0 , G1 , K0 and K1 , and non-negative constants α , β , c, q0 and q1 .

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In this paper we consider a superlinear source b(x, t )|u|α u which is not Lipschitz (with respect to u). In addition, while

β

the damping in (1.27) is linear, here it is given by the superlinear term c ∂∂ut ∂∂ut (certainly this damping is not Lipschitz, either). These generalizations require a different technique to deal with the regularity result of the problem {(1.56)–(1.61)}, as compared with that in [16]. Furthermore, if q0 = q1 = 0, the integral equations {(1.60), (1.61)} becomes {(1.28), (1.29)}. These show that the problem which we are studying is more general than both the problem {(1.27)–(1.30)} in [16,17] and the problem {(1.34)–(1.41)} modelized in Section 1.1. ∂ u(0,t ) ∂ u(1,t ) It is worth noting that the disappearance of boundary dampings ∂ t and ∂ t in the integral equation (1.60) and (1.61) actually gives more difficulties than those in [4,15–17,11,7]. To solve these difficulties, when considering the solvability (more precisely the existence of the solution in the case of which (u0 , u1 ) ∈ H 2 (Ω ) × H 1 (Ω ) or the regularity) and other investigations in the problem {(1.56)–(1.61)}, better smoothness of the given data at those boundaries must be modified; obviously, more elegant estimates are required. 1.2.2. New results The main goal of this project is to generalize [16,17]. Concretely we obtain the following results for the problem {(1.56)– (1.61)}, called the given problem: # The well-posedness and the regularity of the weak solution. # The low-frequency asymptotic expansion of the weak solution with respect to four parameters b, c, q0 and q1 , when b(x, t ) is identified by a real constant b.

We are mainly interested in the existence of the smooth solution (but not the classical solution) of the given problem, since it is applicable to other investigations in this problem. In order to get this, we apply a fixed-point technique and an energy method. This procedure makes Gronwall’s inequality (see [18–20]) still available for the a priori estimates in this paper. If the Galerkin method is itself applied here, then one must use an abstract result regarding the maximal solution of a Volterra integral equation (see [21, Section 5.3, p. 322–324]) to obtain the a priori estimates; and we will specify this interesting difference in another project. Therefore thanks to the visibility of Gronwall’s inequality, our procedure brings more convenience in studying the given problem. 2. Definitions and preliminaries 2.1. Definitions We omit the definitions of usual function spaces C m , Lp , W m,p , H m , where p ∈ [1, +∞], and m ∈ N. Here are some other notations used in this paper:

    

Ω = (0, 1), ΩT ≡ Ω × (0, T ) and R∗ = R+ ∪ {0}. Let X be a function space, X 2 ≡ X × X . h·, ·i: the scalar product in L2 (Ω ). k · kX : the norm of a Banach space X . C ([0, T ]; X ): the set of all continuous functions

v : [0, T ] → X with

kvkC ([0,T ];X ) := max kv(·, t )kX < ∞. 0 ≤t ≤T

 C ([0, T ]; X ): the set of all differential functions 1

v : [0, T ] → X with

 kvkC 1 ([0,T ];X ) := max

0≤t ≤T



∂v

< ∞. kv(·, t )kX + (·, t )

∂t

X

 L (0, T ; X ), 1 ≤ p ≤ ∞, T > 0: the Banach space of the real measurable functions w : (0, T ) → X such that !1/p Z T p kwkLp (0,T ;X ) = kw(·, t )kX dt < ∞ for 1 ≤ p < ∞, p

0

and

kwkL∞ (0,T ;X ) = ess sup kw(·, t )kX for p = ∞. 0 ≤t ≤T

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 W m,p (0, T ; X ), p ∈ [1, +∞], m ∈ N: the Sobolev spaces of all functions w ∈ Lp (0, T ; X ) such that w (m) exists in the weak sense and belongs to Lp (0, T ; X ), and  !1/p m  Z TX  p  ( i )  kw (·, t )kX dt 1 ≤ p < ∞,  0 i=0 kwkW m,p (0,T ;X ) := m X    ess sup kw (i) (·, t )kX p = ∞.   0≤t ≤T

i =0

m,2

 H (0, T ; X ) ≡ W (0, T ; X ).  Suppose Y is a Banach space of real functions defined on Ω , and X (0, T ; Y ) is a Banach space of real functions w : (0, T ) → Y . Let f = f (x, t ), (x, t ) ∈ ΩT such that f ∈ X (0, T ; Y ). We understand that f (t ) ≡ f (·, t ) ∈ Y for t ∈ (0, T ).  Let (f , g ) ∈ (X , Y ) for X and Y two Banach spaces. Then (f , g ) is so-called continuous on (X , Y ) if f and g are, respectively, m

continuous on X and on Y . Definition 1. A trio of functions





2 (u, B0 , B1 ) ∈ H 1 0, T ; L2 (Ω ) ∩ L∞ 0, T ; H 1 (Ω ) × L2 (0, T ) is a weak solution of the given problem if (u(x, t ), B0 (t ), B1 (t )) satisfies the following variational problem:

   ∂u ∂u (t ), v + a(t ) (t ), v 0 + B0 (t )v(0) + B1 (t )v(1) dt ∂ t ∂x * + ∂ u β ∂ u α + b(t ) |u(t )| u(t ) + c (t ) (t ), v = hd(t ), vi , ∂t ∂t d



∂u (x, 0) = u1 (x), ∂t

u(x, 0) = u0 (x), a(0, t )

(2.1)

(2.2)

∂u (0, t ) = B0 (t ) = G0 (t ) + q0 u(0, t ) − ∂x

t

Z

K0 (t − s)u(0, s)ds,

(2.3)

0

Z t ∂u (1, t ) = B1 (t ) = G1 (t ) + q1 u(1, t ) − K1 (t − s)u(1, s)ds ∂x 0

 d ∂u for each v ∈ H 1 (Ω ), a.e. time 0 ≤ t ≤ T , and dt (t ), v is the derivative in sense of distribution on (−∞, T ) of ∂t    ∂ u (t ), v , t > 0, ∂t  0, t < 0. −a(1, t )

(2.4)

If such functions u, B0 and B1 exist, we can say that the given problem is weakly solvable in H 1 0, T ; L2 (Ω ) ∩ L∞ 0, T ; H 1 (Ω )









2 × L2 (0, T ) .

2.2. Preliminaries In H 1 (Ω ) we use the norm

kukH 1 (Ω ) =

q

kuk2L2 (Ω ) + ku0 k2L2 (Ω ) ,

u ∈ H 1 (Ω ).

Then we have the following lemma: Lemma 1. The embedding H 1 (Ω ) ,→ C 0 (Ω ) is compact, and

√ kukC 0 (Ω ) ≤

2kukH 1 (Ω )

(2.5)

for all u ∈ H 1 (Ω ). Let x = (xi )i=1,4 , y = (yi )i=1,4 ∈ Z4+ × Z4+ . We denote




y ≤ x ⇔ xi ≤ yi ,

i = 1, 4,

x1 x2 x3 x4 := x1 + x2 + x3 + x4 , x1 x2 x3 x4 := x21 + x22 + x23 + x24 . 2

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Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

Now the following lemma holds: Lemma 2. Let m ∈ Z+ and (n, i, j, k, l) ∈ N5 . Then the equality

m

 X

X

aijkl ε i δ j ηk θ l  =

 1≤ijkl≤n



aijkl

 m

ε i δ j ηk θ l

(2.6)

m≤ijkl≤mn

holds for ε, δ, η, θ , aijkl ∈ R5 , and






aijkl

 m

=

 aijkl , 1 ≤ ijkl ≤ n, m = 1,   X   (p,q,r ,s)∈[Zijkl ]m     m ≤ ijkl ≤ mn,  

where the family Zijkl



Zijkl

 m



a(i−p)(j−q)(k−r )(l−s) apqrs

m

 m−1

,

(2.7)

m ≥ 2,

is defined by

:= (p, q, r , s) ∈ Z4+ : (p, q, r , s) ≤ (i, j, k, l), 1 ≤ O ≤ n, m − 1 ≤ pqrs ≤ (m − 1)n 



for O = (i − p)(j − q)(k − r )(l − s). 3. How are the given data? 3.1. For the unique solvability Assumption 1. In order to study the unique solvability of the given problem, we state the following assumptions for the given data: B B B B B B B B

(Aa ): a ∈ W 1,1 (0, T ; L∞ (Ω )), a(x, t ) ≥ a0 > 0. (Abc ): b ∈ L∞ (ΩT ) , c ∈ R∗ . (Aαβ ): (α, β) ∈ R2∗ . (Ad ): d ∈ L2 (ΩT ). (AG01 ): (G0 , G1 ) ∈ H 1 (0, T ) × H 1 (0, T ). (Aq01 ): (q0 , q1 ) ∈ R2∗ . (AK01 ): (K0 , K1 ) ∈ H 1 (0, T ) × H 1 (0, T ). (Au ): (u0 , u1 ) ∈ H 1 (Ω ) × L2 (Ω ).

3.2. For the regularity and the stability Assumption 2. To obtain the regularity and the stability of the weak solution of the given problem, beside the conditions (Aαβ ) for α and β , and (Aq01 ) for q0 and q1 , the smoothness of the other given data must be strengthened as follows: ,→


,→

Aa : a, ∂∂ t 2a

,→

Abc : b ∈ W 1,∞ (0, T ; L∞ (Ω )), c ∈ R∗ .

,→

Ad : d, ∂∂dt ∈ L2 (ΩT ) × L2 (ΩT ).

B B B

,→

B

,→

B

,→

B



2




∈ C 1 ΩT × L1 (0, T ; L∞ (Ω )), a(x, t ) ≥ a0 > 0.

,→


,→
,→




AG01 : (G0 , G1 ) ∈ H 2 (0, T ) × H 2 (0, T ). ,→




AK01 : (K0 , K1 ) ∈ H 2 (0, T ) × H 2 (0, T ).




,→


Au : (u0 , u1 ) ∈ H 2 (Ω ) × H 1 (Ω ), and u0 , q0 , q1 , G0 and G1 satisfy the compatibility conditions a(0, 0)u00 (0) = G0 (0) + q0 u0 (0),

−a(1, 0)u00 (1) = G1 (0) + q1 u0 (1). 3.3. For the low-frequency asymptotic expansion Assumption 3. For the investigation of the low-frequency asymptotic expansion, Assumption 2 is still used here, but (Aαβ ) must be more limited. In addition the function coefficient b(x, t ) is identified by a constant b. Then we state

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

←-

B ←-

B

2939

←-


Abc : (b, c ) ∈ R × R∗ . ←-

Aαβ : (α, β) ∈ {x ∈ R+ : x ≥ n + 1}2 for n ∈ Z+ given.




4. Main results Now we would like to specify the new results introduced in Section 1.2.2. First of all the unique solvability of the given problem is given by Theorem 1. If Assumption 1 holds, then the given problem admits a unique weak solution in H 1 0, T ; L2 (Ω ) ∩ L∞ 0, T ; H 1 (Ω )









2 × L2 (0, T ) .

Secondly, on account of some strengthened assumptions the smoothness of the weak solution, existed in Theorem 1, is improved as follows: Theorem 2. Suppose Assumption 2 is fulfilled. Then the given problem has a unique weak solution (u(x, t ), B0 (t ), B1 (t )) satisfying




2
(u, B0 , B1 ) ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) × H 1 (0, T ) . Then the stability of the unique weak solution of the given problem with respect to the given data is considered in ,→


Theorem 3. Let u0 and u1 be fixed given functions satisfying Au . The unique weak solution u(x, t ), B0 (t ), B1 (t ) of the given





problem for which Assumption 2 satisfies is stable in the sense: u, ∂∂ut , B0 , B1 is continuous on L∞ 0, T ; H 1 (Ω ) × L∞ (0, T ;
2 L2 (Ω )) × L2 (0, T ) with respect to the given data








  ∂d , (q0 , q1 ), (G0 , G1 , K0 , K1 ) ∈ C 1 (ΩT ) ∂t


2
4 0, T ; L∞ (Ω ) × L∞ (ΩT ) × L2 (ΩT ) × R2∗ × H 2 (0, T ) .

a, (b, c ) , d,

× W1

Finally we approximate the unique solution of the given problem and its boundary values by nth-order polynomials of four ‘‘small’’ parameters b, c, q0 and q1 , for n ∈ Z+ , in the following theorem: Theorem 4. Let Assumption 3 be satisfied. Then the unique solution (u, B0 , B1 ) of the given problem and its boundary values obey the low-frequency asymptotic expansion with respect to four parameters b, c, q0 and q1 up to order n + 1 as follows:



X ∂ uijkl

∂u

i j k l

b c q0 q1

∂t − ∂ t

∞ 0≤ijkl≤n



X

i j k l

uijkl b c q0 q1 + u −

∞

0≤ijkl≤n

L ( )



  n+2 1 X

2

B0 − B0ijkl bi c j qk0 ql1 ≤ C2 bcq0 q1 ,

2

0≤ijkl≤n L (0,T )



 n+2 1  X

2 1 i j k l

B1 − B b c q q ≤ C bcq q 3 0 1 ijkl 0 1

2 0≤ijkl≤n 0,T ;L2 (Ω )

L

  n+2 1 2 ≤ C1 bcq0 q1 , (

L (0,T )

such that Cr , r = 1, 4, are non-negative constants independent of b, c, q0 and q1 , where uijkl (x, t ), B0ijkl (t ), B1ijkl (t )

(4.1)

)

0,T ;H 1 (Ω )




are the unique weak solution of well-defined solvable problems (♥ijkl ) for 0 ≤ ijkl ≤ n.

(4.2)

(4.3)

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Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

In Theorem 4, the family of the solvable problems (♥ijkl ), 0 ≤ i + j + l ≤ n, is defined as follows:

   ∂ 2u ∂ ∂ u0000 0000  − a ( x , t ) = d(x, t ),   ∂t2 ∂x ∂x     ∂ u0000  0)   (0, t ) = B(0000 (t ), a(0, t )   ∂ x    ∂ u0000  (1) 
−a(1, t ) ∂ x (1, t ) = B0000 (t ), ♥0000 ∂ u0000   (x, 0) = u1 (x), u0000 (x, 0) = u0 (x),    ∂t  Z t    (0)  K0 (t − s)u0000 (0, s)ds, B0000 (t ) = G0 (t ) −   0  Z  t   B(1) (t ) = G (t ) − K1 (t − s)u0000 (1, s)ds 1 0000

(4.4)

0

and

   ∂ 2u ∂ ∂ uijkl ijkl  ≡ ♦ijkl ,   ∂ t 2 − ∂ x a(x, t ) ∂ x     ∂ uijkl  0)   a(0, t ) (0, t ) = B(ijkl (t ),   ∂ x    ∂ uijkl  (1) 
−a(1, t ) ∂ x (1, t ) = Bijkl (t ), ♥ijkl  uijkl (x, 0) = ∂ uijkl (x, 0) = 0, 1≤ijkl≤n    ∂t Z   t   (0)  B ( t ) = ∇ ( t ) − K0 (t − s)uijkl (0, s)ds,  ijkl  ijkl  0  Z  t   B(1) (t ) = 4 (t ) − K1 (t − s)uijkl (1, s)ds, ijkl ijkl

(4.5)

0

where ♦ijkl , ∇ijkl and 4ijkl , 1 ≤ ijkl ≤ n, are defined by the following recursive schemes:

♦00kl = 0 for 1 ≤ kl ≤ n, ♦100 = − |u0000 |α u0000 ,

(4.6)

♦0100

∂ u0000 β ∂ u000 = − , ∂t ∂t

(4.7)

ijkl−1

X 1   (|u0000 |α u0000 )(r ) u(i−1)0kl r for 2 ≤ ikl ≤ n and i ≥ 1, r! r =1 !(r )  β  ijkl −1 X 1 ∂ u0000 ∂ u0000 ∂ u0(j−1)kl =− for 2 ≤ jkl ≤ n and j ≥ 1, r! ∂t ∂t ∂t r r =1 ( !(r )  ) ijkl −1 X ∂ u0000 β ∂ u0000   ∂ ui(j−1)kl 1 (r ) α = − (|u0000 | u0000 ) u(i−1)jkl r − r! ∂t ∂t ∂t r r =1

♦i0kl = −

(4.8)

♦0jkl

(4.9)

♦ijkl

for 2 ≤ ijkl ≤ n and (1, 1) ≤ (i, j),

(4.10)

∇ij0l (t ) = 4ijk0 (t ) = 0 for 1 ≤ ijl, ijk ≤ n,

(4.11)

∇ijkl (t ) = uij(k−1)l (0, t ) for 1 ≤ ijkl ≤ n and k ≥ 1,

(4.12)

4ijkl (t ) = uijk(l−1) (1, t ) for 1 ≤ ijkl ≤ n and l ≥ 1,  

(4.13)

where uijkl

r

is defined as that in Lemma 2.

4.1. The existence and uniqueness: The proof of Theorem 1

β

Proof. The main tool in this proof is the Galerkin method. Either without the internal damping c ∂∂ut ∂∂ut or β = 0, this proof is basically similar to the proof of [14, Theorem 1]. Now with the presence of this internal nonlinear damping, the most

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

2941

difficulty in this proof is to deal with passing to the limit when applying the Galerkin method and the monotonicity method. To treat this, it is referred to [22, Chapter 1] and [23, Part II]. Hence we would like to omit this proof’s details here.  Remark 1. It is noted that our main interest is to study the smooth solution of the given problem, because this result is then applied to investigate the stability and the low-frequency asymptotic expansion of the weak solution of this problem (as introduced). Before dealing with the smoothness or the regularity of the weak solution of the given problem, we would like to prove the following lemma regarding the regularity of solution of this problem in the case b(x, t ) = 0 for all (x, t ) ∈ ΩT : Lemma 3. If b(x, t ) = 0 for all (x, t ) ∈ ΩT , and Assumption 2 holds, then the given problem has a unique weak solution (u(x, t ), B0 (t ), B1 (t )) satisfying




2
(u, B0 , B1 ) ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) × H 1 (0, T ) . Proof. From {(1.56)–(1.61)}, when b(x, t ) = 0, we have the following problem:

β   ∂u ∂u ∂ 2u ∂ ∂u − = d(x, t ), a(x, t ) + c 2 ∂t ∂x ∂x ∂t ∂t

(4.14)

∂u (0, t ) = B0 (t ), ∂x ∂u −a(1, t ) (1, t ) = B1 (t ), ∂x ∂u (x, 0) = u1 (x), u(x, 0) = u0 (x), ∂t Z t B0 (t ) = G0 (t ) + q0 u(0, t ) − K0 (t − s)u(0, s)ds, a(0, t )

(4.15) (4.16) (4.17) (4.18)

0

B1 (t ) = G1 (t ) + q1 u(1, t ) −

t

Z

K1 (t − s)u(1, s)ds

(4.19)

0

for (x, t ) ∈ ΩT , where u(x, t ), B0 (t ) and B1 (t ) are unknown. The given data include the functions a, b, d, u0 , u1 , G0 , G1 , K0 and K1 , and the non-negative constants α , β , c, q0 and q1 . For this proof, we are applying the Galerkin method as in [22,23]. The starting point is to look for the approximate solution of the problem {(4.14)–(4.19)} in the form un (x, t ) =

n X

ϕi (x)ψni (t )

(4.20)

i=1

for (ϕi )i=1,2,... a denumerable basis of H 2 (Ω ), where the coefficient functions ψni satisfy the following system of ordinary differential equations:

+    * ∂ un β ∂ un ∂ un ∂ un 0 (t ), ϕi + a(t ) (t ), ϕi + c (t ) (t ), ϕi dt ∂ t ∂x ∂t ∂t d



+ B0n (t )ϕi (0) + B1n (t )ϕi (1) = hd(t ), ϕi i ,

1 ≤ i ≤ n,

(4.21)

which

∂ un (0, t ) = B0n (t ), ∂x ∂ un −a(1, t ) (1, t ) = B1n (t ), ∂x a(0, t )

B0n (t ) = G0 (t ) + q0 un (0, t ) −

(4.22) (4.23) t

Z

K0 (t − s)un (0, s)ds,

(4.24)

K1 (t − s)un (1, s)ds,

(4.25)

0

B1n (t ) = G1 (t ) + q1 un (1, t ) −

t

Z 0

un (0) ≡ u0n =

n X i=1

ψ0i ϕi → u0

strongly in H 2 (Ω ),

ψ0i ∈ R,

(4.26)

2942

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956 n X ∂ un e0i ϕi → u1 (0) ≡ u1n = ψ ∂t i =1

strongly in H 1 (Ω ),

e0i ∈ R. ψ

(4.27)

On account of Assumption 2 for the given data a, c, d, u0 , u1 , G0 , G1 , q0 , q1 , K0 and K1 , the system {(4.21)–(4.27)} has a solution

(ψn1 , ψn2 , . . . , ψnn ) on some interval [0, Tn ) regarding [24] (with a more general system). Such a solution can be extended to the closed interval [0, T ] by using the coming a priori estimates. Indeed, by considering the non-negative functional



2

Z t

∂ un 2

p

∂ un β+2 ∂ un 2 2



n (t ) := (t ) (t ) + a(t ) + q0 un (0, t ) + q1 un (1, t ) + 2c

∂ s (s) β+2 ds ∂t ∂ x L2 (Ω ) 0 L2 (Ω ) L (Ω )

(4.28)

for each n ∈ Z+ , we deduce from (4.21)–(4.27) that

 2 +  Z t ∂ un ∂a ∂ un d(s), (s), ( s) ds + 2 (s) ds n (t ) = n (0) + ∂s ∂x ∂s 0 0  Z t Z s ∂ un −2 G0 (s) − K0 (s − τ )un (0, τ )dτ (0, s)ds ∂s 0 0   Z t Z s ∂ un −2 G1 (s) − K1 (s − τ )un (1, τ )dτ (1, s)ds. ∂s 0 0 Z t*

(4.29)

Recalling Assumption 2 in regard with the given data a, c, d, u0 , u1 , G0 , G1 , q0 , q1 and K0 , and Lemma 1, we have the following estimates from (4.28) and (4.29) as follows:

FG It is clear that

p

2

n (0) = ku1n k2L2 (Ω ) + a(0)u00n (0)

L2 (Ω )

,→


+ q0 u20n (0) + q1 u20n (1).

,→


(1)

On account of (4.26), (4.27), Aa , (Ad ) and Au , there exists a non-negative constant C independent of t and n such that (1)

n (0) ≤ C

for all n ∈ N.

(4.30)

,→


FG In view of Aa , we deduce that

 2 + Z t* Z t

∂a ∂a ∂ un 1

n (s)ds. (s), ( s) (s) ds ≤ ∂s ∂x a0 0 ∂ s L∞ (Ω ) 0

(4.31)

,→
FG With help of Ad , we get  Z t Z t Z t ∂ un 2 d(s), (s) ds ≤ kd(s)k2L2 (Ω ) ds + n (s)ds. ∂s 0 0 0

(4.32)

,→
FG From AG01 , (4.26) and (4.27), it follows that Z t Z t ε ∂ un 1 −2 G0 (s) (0, s)ds ≤ C (2) + kG0 k2H 1 (0,T ) + n (t ) + 2ε T n (s)ds ∂s ε a0 0 0

(4.33)

and

Z −2 0

t

G1 (s)

∂ un 1 ε (1, s)ds ≤ C (3) + kG1 k2H 1 (0,T ) + n (t ) + 2ε T ∂s ε a0

(2)

(3)

t

Z

n (s)ds 0

for ε > 0, where C and C are non-negative constants independent of t and n, and are defined by (2)

C ≥ |G0 (0)u0n (0)| + 2ε ku0n k2L2 (Ω ) , (3)

C ≥ |G1 (0)u0n (1)| + 2ε ku0n k2L2 (Ω ) .

(4.34)

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

2943

FG Now we compute that  Z s Z t Z t ∂ un K0 (s)un (0, s)ds K0 (s − τ )un (0, τ )dτ ds = 2u0n (0) (0, s) 2 ∂s 0 0 0 Z t Z t K0 (t − s)un (0, s)ds u2n (0, s)ds + 2un (0, t ) − 2K0 (0) 0

0 t

Z

un (0, s)

−2

s

Z

K0 (s − τ )un (0, τ )dτ 0



ds.

(4.35)

0

0

,→




From (4.26) and AK01 , we have 2u0n (0)

t

Z

K0 (s)un (0, s)ds ≤ u20n (0)kK0 k2L2 (0,T ) + 4T ku0n k2L2 (Ω ) +

0

t

Z

− 2K0 (0) 2un (0, t )



t

K0 (t − s)un (0, s)ds ≤ 4 ε +

T kK0 k2L2 (0,T )

+

2ε a0

n (t ) + 2

t

Z

un (0, s)

−2

2T 2 +

ε

a0

K00 (s − τ )un (0, τ )dτ



0



+ 2 2T 2 +

1



a0

1 + 2T kK00 k2L2 (0,T )

n (s)ds,

(4.36)

0

t

Z

a0

a0

n (s)ds,

(4.37)

0

ku0n k2L2 (Ω )

!Z  kK k2 0 L2 (0,T )

1

s

Z

0

2ε T +



1

t

Z

!

ε

0

2

4T 2 +

u2n (0, s)ds ≤ 8T |K0 (0)|ku0n k2L2 (Ω ) + 4|K0 (0)| 2T 2 +

0

Z



t

n (s)ds,

(4.38)

0



ds ≤ 4T ku0n k2L2 (Ω ) 1 + 2T kK00 k2L2 (0,T )



t

Z

n (s)ds.

(4.39)

0

Therefore combining (4.35)–(4.39), we get t

Z 2 0

∂ un (0, s) ∂s

(4)

s

Z

K0 (s − τ )un (0, τ )dτ



(4)

ds ≤ C +

0

2ε a0

(6)

n (t ) + C

t

Z

n (s)ds

(4.40)

0

(6)

for C and C non-negative constants independent of t and n such that



(4)

C ≥ u20n (0)kK0 k2L2 (0,T ) + 4 2T + ε + 4T |K0 (0)| +

T kK0 k2L2 (0,T )

ε

 + 2T 2 kK00 k2L2 (0,T ) ku0n k2L2 (Ω )

and (6)

C =

2 a0



+ 4T (T + ε) + 2 2T + 2

1



a0

1 + 2|K0 (0)| +

kK0 k2L2 (0,T ) ε

! + 2T kK0 kL2 (0,T ) . 0 2

FG By the same techniques to obtain (4.40) from (4.35), we also have Z s  Z t Z t ∂ un 2ε (7) (5) 2 (1, s) K1 (s − τ )un (1, τ )dτ ds ≤ C + n (t ) + C n (s)ds ∂s a0 0 0 0 (5)

(4.41)

(7)

for ε > 0, where C and C are non-negative constants independent of t and n, and satisfy (5)

C ≥

u20n

(1)k k

K1 2L2 (0,T )



+ 4 2T + ε + 4T |K1 (0)| +

T kK1 k2L2 (0,T )

ε

 + 2T kK1 kL2 (0,T ) ku0n k2L2 (Ω ) 2

0 2

and (7)

C =

2 a0

By choosing ε = (8)



+ 4T (T + ε) + 2 2T + a0 8

n (t ) ≤ C +

2

1 a0



1 + 2|K1 (0)| +

kK1 k2L2 (0,T ) ε

! + 2T kK1 kL2 (0,T ) . 0 2

in (4.35), (4.38), (4.40) and (4.41), we deduce from (4.28)–(4.41) that t

Z

(9)

C (s) n (s)ds, 0

(4.42)

2944

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

where (8)

C = 2

5 X

(j)

C +

j =1

16  a0

 kG0 k2H 1 (0,T ) + kG1 k2H 1 (0,T ) + 2kdk2L2 (Ω ) , T

and is a non-negative constant independent of t and n; in addition, (9)

(6)

(7)

C (t ) := 2 2 + C + C +

32T a0

!



+ (t ) a0 ∂ t L∞ (Ω ) 1 ∂a

is an L1 -function on (0, T ). Applying Gronwall’s inequality, we derive from (4.42) that (8)

n (t ) ≤ C



(9) exp C 1



L (0,T )

≡ C

(4.43)

for C non-negative independent of t and n, and all n ∈ N. Now by differentiating (4.21) with respect to the time variable, and after some technical arrangements (see also [14]), we obtain

   ∂ a ∂ un ∂ 2 un ∂ a ∂ un ∂ 2 un ~n (t ) = ~n (0) + 2 (0) (0), (0) − 2 (t ) (t ), (t ) ∂t ∂x ∂ x∂ t ∂t ∂x ∂ x∂ t + *  2 2  Z t Z t 2 ∂a ∂ un ∂ a ∂ un ∂ 2 un +3 (s), (s) ds + 2 ( s ) ( s ), ( s ) ds ∂s ∂ x∂ s ∂ s2 ∂x ∂ x∂ s 0 0  2 Z t Z s ∂ un d 0 K0 (s − τ )un (0, τ )dτ −2 G0 (s) − (0, s)ds ds 0 ∂ s2 0  2  Z t Z s Z t d ∂ un ∂d ∂ 2 un −2 G01 (s) − K1 (s − τ )un (1, τ )dτ ( 1 , s ) ds + 2 ( s ), ( s ) ds, ds 0 ∂ s2 ∂s ∂ s2 0 0 

(4.44)

where ~n (t ) is a non-negative functional defined by

2

2

p ∂ un

∂ 2 un ∂ 2 un



~n (t ) :=

∂ t 2 (t ) 2 + a(t ) ∂ x∂ t (t ) 2 + q0 ∂ t (0, t ) L (Ω ) L (Ω )





2

+ 2 Z t * 2 ∂ un ∂ un β ∂ 2 un ∂ un + q1 (1, t ) + 2c (β + 1) ∂ s (s) ∂ s2 (s), ∂ s2 (s) ds ∂t 0

(4.45)

for each n ∈ Z+ . With help of Assumption 2 for the given data a, c, d, u0 , u1 , G0 , G1 , q0 , q1 and K0 , and Lemma 1, we have the following a priori estimates for the right-hand side of (4.44) in regard with (4.45) and (4.28)–(4.43): ,→
,→
,→
,→
FG On account of (4.21)–(4.23), (4.26)–(4.27), and the assumptions Aa , Abc , Ad and Au , there exists non-negative (1)

constant C~ independent of t and n such that

2



p

∂ un

∂a

00



( 0 ) ≤ a ( t ) u ( 0 ) ku00 kL2 (Ω ) +

0n 2

2

∂t2

∂x 0 L ( Ω ) L (Ω ) C (Ω ) β+1

+ c ku1n kL2(1+β) (Ω ) + kd(0)kL2 (Ω ) ≤ C~(1) for all n ∈ Z+ . Therefore we have

2

p

2

∂ un 2

0

~n (0) = ( 0 ) a ( 0 ) u +

2 + q0 |u1n (0)|2 + q1 |u1n (1)|2 1n

2 L (Ω ) ∂t2 L (Ω )

p

2

≤ C~(1) + a(0)u01n 2 + q0 |u1n (0)|2 + q1 |u1n (1)|2 ≤ C~(2) L (Ω )

(4.46)

(2)

for all n ∈ Z+ and C~ a non-negative constant independent of t and n. ,→


,→


FG In view of Aa , Au and (4.26)–(4.27), we deduce the existence of a constant C~(3) ≥ 0 such that

 

∂a

0

∂ a ∂ un ∂ 2 un (3)

u 2 u0 2 2 (0) (0), (0) ≤ 2 ( 0 ) 0n L (Ω ) 1n L (Ω ) ≤ C~ .

∂t ∂x ∂ x∂ t ∂t 0 C (Ω )

(4.47)

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

2945

,→
FG In regard with Aa , we compute that

2  δ ∂ a ∂ un ∂ 2 un C ∂ a

+ −2 (t ) (t ), (t ) ≤ ~n (t ) for δ > 0, ∂t ∂x ∂ x∂ t a0 δ ∂ t C 0 (Ω T ) a0

 2 2 + Z t* Z t ∂a ∂ un 3 ∂a

~n (s)ds, 3 (s), ( s) ds ≤ ∂s ∂ x∂ s a0 ∂ t C 0 (Ω T ) 0 0 

(4.48)

(4.49)

and

Z t 2 0



 Z t 2

∂ a ∂ 2 a ∂ un ∂ 2 un C ∂ 2 a 1



( s ) ( s ), ( s ) ds ≤ + ( s )

∞ ~n (s)ds. ∂ s2 ∂x ∂ x∂ s a0 ∂ t 2 L1 (0,T ;L∞ (Ω )) a0 0 ∂ s2 L (Ω )

,→
FG Thanks to AG01 , we get Z t

2 ∂ 2 un G00 (s) 2 (0, s)ds ≤ 2 G00 (0) |u1n (0)| + G00 L2 (0,T ) −2 ∂s 0  Z t

1 1 2δ 2 + G00 C 0 ([0,T ]) + 4(δ + T )ku1n k2L2 (Ω ) + ~n (t ) + 2 + 2T (δ + T ) ~n (s)ds, δ a0 a0 0

(4.50)

(4.51)

and



2 ∂ 2 un (1, s)ds ≤ 2 G01 (0) |u1n (1)| + G01 L2 (0,T ) 2 ∂s 0  Z t 1 0 2 1 2δ 2

+ G + 4(δ + T )ku1n kL2 (Ω ) + ~n (t ) + 2 + 2T (δ + T ) ~n (s)ds 0 δ 1 C ([0,T ]) a0 a0 0 Z

−2

t

G01 (s)

for δ > 0. FG Integrating by parts, it yields  2 Z t Z s d ∂ un −2 K0 (s − τ )un (0, τ )dτ (0, s)ds ds 0 ∂ s2 0   Z t ∂ un 0 =2 (0, t ) K0 (0)un (0, t ) − K0 (t )u0n (0) + K0 (t − s)un (0, s)ds ∂t 0 Z t Z t ∂ un 2 ∂ un −2 K0 (0) (0, s) ds + 4u0n (0) K00 (s) (0, s)ds ∂s ∂s 0 0 Z s  Z t Z t ∂ un ∂ un − 2K00 (0) (0, s)ds − 2 (0, s) K000 (s − τ )un (0, τ )dτ ds. un (0, s) ∂s ∂s 0 0 0 ,→
,→


(4.52)

(4.53)

With regard to the assumptions Aa and AK01 , and (4.26)–(4.27), we obtain the following estimates for the terms of the right-hand side of (4.53): 2K0 (0)un (0, t )

2K02 (0) ∂ un (0, t ) ≤ ∂t δ



 + 2T 2 C + 2ku0n k2L2 (Ω ) a0 Z t 4δ + 2δku1n k2L2 (Ω ) + ~n (s)ds, ~n (t ) + 4T δ C

a0

(4.54)

0

Z t u20n (0) kK0 k2C 0 ([0,T ]) ∂ un 2δ (0, t ) ≤ 4δku1n k2L2 (Ω ) + + ~n (t ) + 4T δ ~n (s)ds, −2u0n (0)K0 (t ) ∂t δ a0 0 Z t ∂ un 2δ 2 (0, t ) K00 (t − s)un (0, s)ds ≤ 4δku1n k2L2 (Ω ) + ~n (t ) ∂t a0 0

2    Z t 2T K00 L2 (0,T ) 1 + 2ku0n k2L2 (Ω ) + C 2T + + 4T δ ~n (s)ds, δ a0 0 2  Z t Z t ∂ un ds ≤ 8T |K0 (0)|2 ku1n k22 + 4 |K0 (0)|2 2T + 1 −2K0 (0) ( 0 , s ) ~n (s)ds, ∂s L (Ω ) a0 0 0

(4.55)

(4.56)

(4.57)

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Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

2  Z t 2u20n (0) K00 L2 (0,T ) ∂ u 1 n 0 2 K0 (s) ~n (s)ds, 4u0n (0) (0, s)ds ≤ 8T δku1n kL2 (Ω ) + + 4δ 2T + ∂s δ a0 0 0    Z t 0 2 ∂ un 1 2 0 2 2 un (0, s) −2K0 (0) (0, s)ds ≤ 4T ku1n kL2 (Ω ) + 2T K0 (0) 2ku0n kL2 (Ω ) + C 2T + ∂s a0 0 Z t  1 ~n (s)ds + 2 2T 2 + t

Z

a0

(4.58)

(4.59)

0

and

Z

∂ un (0, s) ∂s

t

−2 0

Z

K000 (s − τ )un (0, τ )dτ

 ds

0

u1n 2L2 (Ω )

≤ 4T k

s

k

    Z t

00 2 1 1 2 2 2

+ 2T K0 L2 (0,T ) 2ku0n kL2 (Ω ) + C 2T + ~n (s)ds + 2 2T + 2

a0

a0

(4.60)

0

for δ > 0. Combining (4.53)–(4.60), it follows that

Z t −2

ds

0

s

Z

d

K0 (s − τ )un (0, τ )dτ

∂ 2 un 8δ (0, s)ds ≤ C~(4) + ~n (t ) + C~(6) 2 ∂s a0



0

t

Z

~n (s)ds,

(4.61)

0

(6)

(4)

where C~ and C~ are non-negative constants independent of t and n, and satisfy C~(4) := 4ku1n k2L2 (Ω ) 3δ + 2T 1 + δ + K02 (0)





2 + 2T 2 K 00 2



L (0,T )

0

2 ku0n k2L2 (Ω ) + C



+

u20n (0) 



2 kK0 k2C 0 ([0,T ]) + 2 K00 L2 (0,T ) 

δ 1

2T 2 +

a0

     2 1  2

2 1 T K00 (0) + + 2 2 ku0n k2L2 (Ω ) + C 2T 2 + K0 (0) + T K00 L2 (0,T ) a0 δ

(4.62)

and



(6)

C~ := 4

1 a0

+ T (3δ + 2T ) + δ +

K02

(0)




 2T +



1 a0

.

FG By the similar procedure to get (4.61), we also obtain  2 Z t Z s Z t d ∂ un 8δ (5) (7) −2 K1 (s − τ )un (1, τ )dτ ( 1 , s ) ds ≤ C + ~ ( t ) + C ~n (s)ds n ~ ~ ds 0 ∂ s2 a0 0 0 (5)

(4.63)

(4.64)

(7)

for δ > 0, where C~ and C~ are non-negative constants independent of t and n such that C~(5) ≥ 4ku1n k2L2 (Ω ) 3δ + 2T 1 + δ + K12 (0)

2 + 2T 2 K 00 2

L (0,T )

1



2 ku0n k2L2 (Ω ) + C





+

u20n (1) 

δ

2T 2 +

1



2 kK1 k2C 0 ([0,T ]) + 2 K10 L2 (0,T ) 

a0

     2 1  2

2 1 + 2 2 ku0n k2L2 (Ω ) + C 2T 2 + T K10 (0) + K1 (0) + T K10 L2 (0,T ) a0 δ

(4.65)

and (7)



C~ = 4

1 a0

+ T (3δ + 2T ) + δ +

K12

 
1 (0) 2T + . a0

,→
FG On account of Ad , it follows

2  Z t Z t

∂d ∂d ∂ 2 un

(s), 2 (s) ds ≤ 2 + ~n (s)ds.

∂t 2 ∂s ∂s 0 0 L (ΩT )

From (4.44)–(4.52), (4.61)–(4.67) and with the choice of δ = ~n (t ) ≤ 2



C~(8) +

t

Z 0

C~(9) (s) ~n (s)ds

a0 , 42

(4.66)

(4.67) we conclude that

 (4.68)

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

(8)

2947

(9)

for C~ ∈ R∗ and C~ ∈ L1 (0, T ) such that

2

2

2

∂d

ku1n k2L2 (Ω ) + G00 L2 (0,T ) + G01 L2 (0,T ) +

∂t 2 42 L (ΩT ) j =2 

0 2 0 0
42  0 2



G0 C 0 ([0,T ]) + G1 C 0 ([0,T ]) + 2 G0 (0) |u1n (0)| + G1 (0) |u1n (1)| + a0!

2

2 2

∂ a C 42 ∂ a



+ + , a0 a0 ∂ t C 0 (Ω T ) ∂ t 2 L1 (0,T ;L∞ (Ω )) !



2

  

∂a

∂ a a0  1 1 (9) (6) (7)



+ 2T T + + 3 + (t ) C~ ( t ) = 1 + C~ + C~ + 4 . a0 42 a0 ∂ t C 0 (Ω T ) ∂ t 2 L∞ (Ω ) C~(8) ≥

5 X



C~(j) + 8 T +

a0 

(4.69)

(4.70)

Due to Gronwall’s inequality [18–20], the linear integral inequality (4.68) admits the following solution:



~n (t ) ≤ 4C~(8) exp C~(9) L1 (0,T )



≡ C~ ,

(4.71)

wherein C~ is obviously a non-negative constant independent of t and n, and all n ∈ N.

,→


,→




,→




From Lemma 1, with help of the a priori estimates (4.43) and (4.71), and the assumptions Aa , AG01 and AK01 , and (4.26)–(4.27), we deduce from (4.24) that

k

B0n 2L2 (0,T )

G0 2L2 (0,T )

k

≤ 4k k

+ 16T



q20

K0 2L2 (0,T )

+Tk k



k

u0n 2L2 (Ω )

k



2

+ C 2T +

1 a0



≤ CB(010)

(4.72)

and

 

2

2 ≤ 16 G00 L2 (0,T ) + 16u20n (0) kK0 k2L2 (0,T ) + 64Tq20 ku1n k2L2 (Ω ) + 32T ku0n k2L2 (Ω ) K02 (0) + T K00 L2 (0,T )    

2 1  + 32 2T 2 + 2TC~ q20 + C K02 (0) + T K00 L2 (0,T ) ≤ CB(011) (4.73)

0 2

B 2

0n L (0,T )

a0

(11)

(10)

for CB0 and CB0 estimates:

non-negative constants independent of t and n. Additionally we also derive from (4.25) the following

kB1n k2L2 (0,T ) ≤ CB(112) , (12)

where CB0

(12)

CB1

(12)

and CB0

0 2

B 2

1n L (0,T )

≤ CB(113) ,

(4.74)

are non-negative constants independent of t and n, and satisfy

    1 ≥ 4 kG1 k2L2 (0,T ) + 16T q21 + T kK1 k2L2 (0,T ) ku0n k2L2 (Ω ) + C 2T 2 + a0

(4.75)

and (13)

CB1

 

2

2 ≥ 16 G01 L2 (0,T ) + 16u20n (1) kK1 k2L2 (0,T ) + 64Tq21 ku1n k2L2 (Ω ) + 32T ku0n k2L2 (Ω ) K12 (0) + T K10 L2 (0,T )    

2 1  + 32 2T 2 + 2TC~ q21 + C K12 (0) + T K10 L2 (0,T ) . a0

(4.76)

Applying the Banach–Alaoglu theorem and Lemma 1, we conclude from (4.28), (4.43), (4.45) and (4.71)–(4.74) that there exists a subsequence of {(un , B0n , B1n )} still denoted by {(un , B0n , B1n )} such that

u → u n   ∂u  ∂ un  →    ∂t ∂t    ∂ 2u  ∂ 2 un → ∂t2 ∂t2   un (0, ·) → u(0, ·)     un (1, ·) → u(1, ·)   B → e  B0  0n B1n → e B1

in L∞ 0, T ; H 1 (Ω ) weakly? ,




in L∞ 0, T ; H 1 (Ω ) weakly? ,




in L∞ 0, T ; L2 (Ω ) weakly? ,




in W 1,∞ (0, T ) weakly? , in W 1,∞ (0, T ) weakly? , in H 1 (0, T ) weakly, in H 1 (0, T ) weakly.

(4.77)

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Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

Due to [22, 1.5], and the compact embeddings H 1 (0, T ) ,→ C 0 ([0, T ]) and W 1,∞ (0, T ) ,→ C 0 ([0, T ]) (see [25, Theorem VIII.7]), it follows from (4.77) that there exists a subsequence of {(un , B0n , B1n )} which is still denoted by {(un , B0n , B1n )} admitting

 un → u    ∂u  ∂ un   →   ∂t ∂t un (0, ·) → u(0, ·)   un (1, ·) → u(1, ·)    e   B0n → B0 B1n → e B1

strongly in L2 (ΩT ), strongly in L2 (ΩT ), strongly in C 0 ([0, T ]) , strongly in C 0 ([0, T ]) , strongly in C 0 ([0, T ]) , strongly in C 0 ([0, T ]) .

(4.78)

On account of the inequality

β |x| x − |y|β y ≤ (β + 1) Rβ |x − y| for all (x, y) ∈ [−R, R]2 and (β, R) ∈ R2∗ , we get

β ∂ u β ∂ u ∂ u ∂ u ∂ u n n β ∂ un − − ≤ (β + 1)R ∂t ∂t ∂t ∂t ∂t ∂t

(4.79)

regarding {(4.27), (4.46), (4.71)}, where

s  R≥

2 2ku1n k2L2 (Ω ) + C~

 2T 2 +

1 a0



.

Then (4.78) 1 and (4.79) lead to

β ∂u ∂u ∂ un β ∂ un 2 ∂ t ∂ t → ∂ t ∂ t strongly in L (ΩT ).

(4.80)

From (4.24), (4.25) and (4.78) 3,4 , we deduce that B0n (t ) → G0 (t ) + q0 u(0, t ) −

t

Z

K0 (t − s)u(0, s)ds ≡ B0 (t ) strongly in C 0 ([0, T ])

(4.81)

K1 (t − s)u(1, s)ds ≡ B1 (t ) strongly in C 0 ([0, T ]) .

(4.82)

0

and B1n (t ) → G1 (t ) + q1 u(1, t ) −

t

Z 0

Therefore (4.78) 5,6 and (4.81)–(4.82) yield

e B0 (t ) ≡ B0 (t ),

e B1 (t ) ≡ B1 (t ) for t ∈ [0, T ].

(4.83)

With help of {(4.77), (4.78) and (4.80)–(4.83)}, passing to the limit in (4.21)–(4.27) leads to the existence of (u(x, t ), B0 (t ), B1 (t )) satisfying




2 (u, B0 , B1 ) ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) × H 1 (0, T )

(4.84)

and



+    * ∂ u β ∂ u ∂ 2u ∂u 0 (t ), ϕ + a(t ) (t ), ϕ + c (t ) (t ), ϕ + B0 (t )ϕ(0) + B1 (t )ϕ(1) = hd(t ), ϕi ∂t2 ∂x ∂t ∂t

(4.85)

for ϕ ∈ H 1 (Ω ) and a.e. time 0 ≤ t ≤ T , and a(0, t )

∂u (0, t ) = B0 (t ), ∂x

(4.86)

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

−a(1, t )

∂u (1, t ) = B1 (t ), ∂x

B0 (t ) = G0 (t ) + q0 u(0, t ) −

2949

(4.87) t

Z

K0 (t − s)u(0, s)ds,

(4.88)

K1 (t − s)u(1, s)ds,

(4.89)

0

B1 (t ) = G1 (t ) + q1 u(1, t ) −

t

Z 0

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

∂u (x, 0) = u1 (x). ∂t

(4.90)

In addition, from (4.84)–(4.87) and Assumption 2 for {a, c , β, d, u0 , u1 } we derive that

2

∂ u

a0

∂x



( t )

2

L2 (Ω )



2

∂ u

∂a

∂u

(t )

(t ) ( t ) ≤ +

∂t2 2

∂ x ∞ ∂ x 2 + kd(t )kL2 (Ω ) ≤ C L (Ω ) L (Ω ) L (Ω )

for C a non-negative constant independent of t and n, and a.e. t ∈ [0, T ]. This estimate yields


u ∈ L∞ 0, T ; H 2 (Ω ) .

(4.91)

To obtain the uniqueness of the weak solution (u(x, t ), B0 (t ), B1 (t )), we would like to refer the reader to [14, p. 81-82]; in addition, the uniqueness result for a ‘‘more general case’’ will be specified in Section 4.2. In short the proof of this lemma is completed.  Remark 2. It is clear that Gronwall’s inequality [18–20] still helps in Lemma 3. Nevertheless, if one applies the Galerkin method for the solvability of the given problem in the case of b(x, t ) 6= 0 (specially b(x, t ) < 0), then Gronwall’s inequality cannot be available anymore. To overcome this inconvenience, a combination of a fixed-point technique and an energy method is used in the following proof. 4.2. The regularity: The proof of Theorem 2 Proof. The proof of Theorem 2 includes 3 steps as follows: # The existence of u ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) satisfying the variational problem {(2.1)–(2.4)}.







# The existence of boundary values (B0 , B1 ) ∈ H 1 (0, T ) × H 1 (0, T ) obeying the boundary conditions of the variational problem {(2.1)–(2.4)}. # The uniqueness of (u, B0 , B1 ).

Step 1. The existence of u ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) satisfying the variational problem {(2.1)–(2.4)} For convenience, with every T > 0 let us consider a Banach space H(T ) defined as follows:







H(T ) = y ∈ H 1 0, T ; L2 (Ω ) ∩ L∞ 0, T ; H 1 (Ω )









endowed with a norm

kyk2H(T ) = kyk2H 1 (0,T ;L2 (Ω )) + kyk2L∞ (0,T ;H 1 (Ω )) for y ∈ H(T ). To begin, we introduce an operator  on H(T ). For a given function u ∈ H(T ), we define d (x, t ) := b(x, t )|u(x, t )|α u(x, t ) − d(x, t ),

(4.92)

where (x, t ) ∈ Ω T . On account of (Abc ), (Aαβ ) and (Ad ), it follows from (4.92) that



∂d d, ∂t



∈ L2 (ΩT ) × L2 (ΩT ) .

(4.93)

Associating (4.93) with the given problem, let us consider the following problem:

β   ∂y ∂y ∂ 2y ∂ ∂y − a ( x , t ) + c ∂ t ∂ t + d = 0 in ΩT , 2 ∂t ∂x ∂x ∂y (0, t ) = By0 (t ), t ∈ (0, T ), ∂x ∂y −a(1, t ) (1, t ) = By1 (t ), t ∈ (0, T ), ∂x a(0, t )

(4.94) (4.95) (4.96)

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Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

∂y (x, 0) = u1 (x), x ∈ Ω , ∂t Z t y K0 (t − s)y(0, s)ds, B0 (t ) = G0 (t ) + q0 y(0, t ) − y(x, 0) = u0 (x),

(4.97) (4.98)

0 y

B1 (t ) = G1 (t ) + q1 y(1, t ) −

t

Z

K1 (t − s)y(1, s)ds

(4.99)

0 y

y

for the trio of unknown functions y(x, t ), B0 (t ), B1 (t ) and for a, u0 , u1 , G0 , G1 , q0 , q1 , K0 and K1 given. Let us recall Lemma 3 for the problem {(4.94)–(4.99)} as follows:




Lemma 4. With the use of (4.93), if a, u0 , u1 , G0 , G1 , q0 ,
q1 , K0 and K1 satisfy the corresponding conditions in Assumption 2, then y y there exists a unique weak solution y(x, t ), B0 (t ), B1 (t ) of the problem {(4.94)–(4.99)} such that




2
(y, By0 , By1 ) ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) × H 1 (0, T ) .

(4.100)

Moreover y(x, t ) admits the finite energy estimates



2

Z t

p

∂ y β+2

∂ y 2 2 2

a(t ) ∂ y (t )

(s)

(t ) + + q y ( 0 , t ) + q y ( 1 , t ) + 2c 0 1

∂ s β+2 ds ≤ d0

∂t 2 ∂ x L2 (Ω ) 0 L (Ω ) L (Ω )

(4.101)

and



2 2

2 2

p ∂y



∂ y 2

a(t ) ∂ y (t )



+ + q ( t ) ( 0 , t ) 0



∂t2 2 ∂ x∂ t ∂t L (Ω ) L2 (Ω ) + 2 Z t * ∂ y β ∂ 2 y ∂y ∂ 2y + q1 (1, t ) + 2c (β + 1) ∂ s (s) ∂ s2 (s), ∂ s2 (s) ds ≤ d1 ∂t 0

(4.102)

for t ∈ [0, T ], where d0 and d0 are non-negative constants independent of t. From (4.92) and Lemma 4, for every u ∈ H(T ) there uniquely exists y ∈ H(T ), which solves the problem {(4.94)–(4.99)}, or shortly we have u ∈ H(T ) 7→ y ∈ H(T ) uniquely. So the operator  : H(T ) → H(T ), u := y

is well defined. From Lemma 4 and Definition 1, it follows that y(x, t ) solves the following variational problem:

* +    ∂ y β ∂ y ∂y ∂y y y 0 (t ), v + a(t ) (t ), v + B0 (t )v(0) + B1 (t )v(1) + c (t ) (t ), v + hd(t ), vi = 0, dt ∂ t ∂x ∂t ∂t d



y(x, 0) = u0 (x), a(0, t )

∂y (x, 0) = u1 (x), ∂t

∂y (0, t ) = By0 (t ) = G0 (t ) + q0 y(0, t ) − ∂x

−a(1, t )

(4.104) t

Z

K0 (t − s)y(0, s)ds,

Z

t

K1 (t − s)y(1, s)ds

0

for each v ∈ H 1 (Ω ) and a.e. time 0 ≤ t ≤ T . Now we are proving that  is a contraction for T > 0 small enough. Take

and put y := u,

y := u.

(4.105)

0

∂y (1, t ) = By1 (t ) = G1 (t ) + q1 y(1, t ) − ∂x

(u, u) ∈ H(T ) × H(T ),

(4.103)

(4.106)

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

2951

Then it is clear that y satisfies the problem {(4.103)–(4.106)}, while y satisfies

* +    ∂ y β ∂ y ∂y ∂y y y 0 (t ), v + a(t ) (t ), v + B0 (t )v(0) + B1 (t )v(1) + c (t ) (t ), v + hd(t ), vi = 0, dt ∂ t ∂x ∂t ∂t d



y(x, 0) = u0 (x), a(0, t )

∂y (x, 0) = u1 (x), ∂t

(4.108)

∂y y (0, t ) = B0 (t ) = G0 (t ) + q0 y(0, t ) − ∂x

−a(1, t )

(4.107)

t

Z

K0 (t − s)y(0, s)ds,

(4.109)

0

∂y y (1, t ) = B1 (t ) = G1 (t ) + q1 y(1, t ) − ∂x

t

Z

K1 (t − s)y(1, s)ds

(4.110)

0

for each v ∈ H 1 (Ω ) and a.e. time 0 ≤ t ≤ T , where α

d(t ) := b(x, t ) |u(x, t )| u(x, t ) − d(x, t ),



∂d d, ∂t



∈ L2 (ΩT ) × L2 (ΩT )

(4.111)

and





2 y y (y, B0 , B1 ) ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) × H 1 (0, T ) ,



2

Z t

p

∂ y 2

∂ y β+2 ∂y 2 2

(t )



∂ t 2 + a(t ) ∂ x (t ) 2 + q0 y (0, t ) + q1 y (1, t ) + 2c

∂ s (s) β+2 ds ≤ d0 , 0 L (Ω ) L (Ω ) L (Ω )



2

2

2 2 2 p

∂y

∂ y ∂ y





∂ t 2 (t ) 2 + a(t ) ∂ x∂ t (t ) 2 + q0 ∂ t (0, t ) L (Ω ) L (Ω ) + 2 Z t * 2 ∂ y β ∂ 2 y ∂y ∂ y (s) + q1 (1, t ) + 2c (β + 1) ∂ s ∂ s2 (s), ∂ s2 (s) ds ≤ d1 ∂t 0 for t ∈ [0, T ], where d0 and d0 are non-negative constants independent of t. From (4.103)–(4.111), we deduce that

* + β β      ∂ y ∂y ∂y ∂y ∂y ∂ y ∂ y ∂ y (t ) − (t ), v + a(t ) (t ) − (t ) , v 0 + c (t ) (t ) − (t ) (t ), v dt ∂ t ∂t ∂x ∂x ∂t ∂t ∂t ∂t     y y + By0 (t ) − B0 (t ) v(0) + By1 (t ) − B1 (t ) v(1) + hd(t ) − d(t ), vi = 0 d



(4.112)

for each v ∈ H 1 (Ω ) and a.e. time 0 ≤ t ≤ T . By defining the non-negative functional

2

  2

∂y

p

∂y ∂y ∂y

(t ) := (t ) − (t ) + a( t ) (t ) − (t )

2 ∂t ∂t ∂ x ∂ x L2 (Ω ) L (Ω ) + q0 |y(0, t ) − y(0, t )|2 + q1 |y(1, t ) − y(1, t )|2 ,

(4.113)

we derive from (4.112) that

Z t*

 2 + ∂y ∂a ∂y (t ) ≤ (s), (s) − (s) ds ∂s ∂x ∂x 0 + Z t * ∂ y β ∂ y ∂ y β ∂ y ∂ y ∂ y (s) − 2c ∂ s ∂ s (s) − ∂ s (s) ∂ s (s), ∂ s (s) − ∂ s (s) ds 0  Z s  Z t ∂y ∂y +2 (0, s) − (0, s) K0 (s − τ ) [y(0, τ ) − y(0, τ )] dτ ds ∂s ∂s 0 0  Z s  Z t ∂y ∂y +2 (1, s) − (1, s) K1 (s − τ ) [y(1, τ ) − y(1, τ )] dτ ds ∂s ∂s 0 0  Z t ∂y ∂y −2 (s) − (s) ds. d(s) − d(s), ∂s ∂s 0

(4.114)

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Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

On account of Assumption 2 for {a, β, K0 , K1 , } and {(4.92), (4.93)}, we deduce from (4.113) and (4.114) the following estimates: ,→


^ In view of Aa , we get

Z t*

 2 + Z t

∂a ∂a ∂y ∂y 1

(s) (s), (s) − (s) ds ≤ ∂s ∂x ∂x a0 0 ∂ s L∞ (Ω )

0

(s)ds.

(4.115)

^ Applying the inequality

β |x| x − |y|β y ≤ (β + 1) Rβ |x − y| for all (x, y) ∈ [−R, R]2 and (β, R) ∈ R2∗ , we obtain

β β ∂ y ∂y ∂y ∂y ∂y ∂y (x, t ) − (x, t ) (x, t ) ≤ (β + 1)Rβ (x, t ) − (x, t ) (x, t ) ∂t ∂t ∂t ∂t ∂t ∂t for (x, t ) ∈ ΩT , where R is chosen as follows:



∂v

R≥

∂t ∞ L (ΩT ) with all v ∈ C 1 [0, T ]; H 1 (Ω ) . Then we get the following estimate:




+ Z t Z t * ∂ y β ∂ y ∂ y β ∂ y ∂y ∂y β (s)ds. ( s ) ( s ) − ( s ) ( s ), ( s ) − ( s ) ds ≤ 2c (β + 1 ) R − 2c ∂s ∂s ∂s ∂s ∂s ∂s 0 0 ,→

(4.116)

^ From AG01 , there exists ε > 0 such that




Z t 2 0

 Z s  ∂y ∂y 2ε (0, s) − (0, s) K0 (s − τ ) (y(0, τ ) − y(0, τ )) dτ ds ≤ ∂s ∂s a0 0

(t ) + C (0)

 Z s  ∂y ∂y 2ε (1, s) − (1, s) K1 (s − τ ) (y(1, τ ) − y(1, τ )) dτ ds ≤ ∂s ∂s a0 0

(t ) + C (1)

t

Z

(s)ds

(4.117)

(s)ds

(4.118)

0

and

Z t 2 0

t

Z 0

for C

C

(0)

(1)

=

=



2

+ 4T (T + ε) + 2 2T +

a0



2

2

+ 4T (T + ε) + 2 2T +

a0

2

1



a0 1



a0

1 + 2|K0 (0)| +

1 + 2|K1 (0)| +

kK0 k2L2 (0,T ) ε kK1 k2L2 (0,T ) ε

! + 2T kK0 kL2 (0,T ) , 0 2

(4.119)

! + 2T kK1 kL2 (0,T ) . 0 2

(4.120)

^ In view of (4.92), (4.93) and (4.111), it follows that

Z t −2 0

 Z t Z t ∂y ∂y 2 kd(s) − d(s)kL2 (Ω ) ds + d(s) − d(s), (s) − (s) ds ≤ (s)ds. ∂s ∂s 0 0

Consequently, by choosing ε =

(t ) ≤ 2

kd(s) − d(s)k2L2 (Ω ) ds +

0

1+C

(0)

(4.113)–(4.121) yield

T

Z

for t ∈ [0, T ] and C

a0 , 8

(2)

(4.121)

t

Z

C

(2)

(s)

(s)ds

 (4.122)

0

(t ) equal to

+ C (1) + 2c (β + 1)Rβ +

1 ∂a



(t )

∞ . a0 ∂ t L (Ω )

(4.123)

,→


According to Aa , (4.119) and (4.120), we deduce from (4.123) that C

(2)

∈ L1 (0, T ).

(4.124)

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

2953

As a result, in regard with Gronwall’s inequality [18–20] we get the solutions of the linear integral inequality (4.122) as follows:



(t ) ≤ 4 exp C (2) 1

T

Z

L (0,T )

0

kd(s) − d(s)k2L2 (Ω ) ds for t ∈ [0, T ].

(4.125)

Since (u, u) ∈ H(T ) × H(T ), there exists Cα ∈ R∗ such that

(u(x, t ), u(x, t )) ∈ [−Cα , Cα ]2 . Then we deduce from (4.113) and (4.125) that

ky − yk2H(T ) ≤ C 2 ku − uk2H(T )

(4.126)

for all (u, u) ∈ H(T ) × H(T ), and

s C :=

2Cαα kbkL∞ (ΩT )



(α + 1) exp C (2) 1



L (0,T )

2T 3

+

3T 2

+

T a0



.

(4.127)

Therefore we obtain

ku − ukH(T ) ≤ C ku − ukH(T )

(4.128)

for all (u, u) ∈ H(T ) × H(T ). As a result we deduce from (4.128) that  is a contracted operator provided T > 0 is so small that 0 < C < 1. Applying Banach’s fixed-point theorem, there exists u ∈ H(T ) such that u = u,

and obviously u solves the variational problem {(2.1)–(2.4)}. Therefore, with regard to (4.92)–(4.99) we have that u ∈ H(T ) weakly satisfies

β   ∂u ∂u ∂ 2u ∂ ∂u − a(x, t ) + c + d = 0 in ΩT , ∂t2 ∂x ∂x ∂t ∂t ∂u (0, t ) = B0 (t ), t ∈ (0, T ), ∂x ∂u −a(1, t ) (1, t ) = B1 (t ), t ∈ (0, T ), ∂x ∂u (x, 0) = u1 (x), x ∈ Ω , u(x, 0) = u0 (x), ∂t Z t B0 (t ) = G0 (t ) + q0 u(0, t ) − K0 (t − s)u(0, s)ds, a(0, t )

(4.129) (4.130) (4.131) (4.132) (4.133)

0

B1 (t ) = G1 (t ) + q1 u(1, t ) −

t

Z

K1 (t − s)u(1, s)ds,

(4.134)

0

where d (x, t ) := b(x, t )|u(x, t )|α u(x, t ) − d(x, t )

(4.135)

for (x, t ) ∈ ΩT . Since d satisfies (4.93), in regard with Lemma 4 and Assumption 2 we have the following estimates for u(x, t ):



2

Z t

∂ u 2

p

∂ u β+2 ∂u 2 2

(t )



∂ t 2 + a(t ) ∂ x (t ) 2 + q0 u (0, t ) + q1 u (1, t ) + 2c

∂ s (s) β+2 ds ≤ d0 0 L (Ω ) L (Ω ) L (Ω )

(4.136)

and

2

2 2 2

∂ u 2

p

∂u

a(t ) ∂ u (t ) (0, t ) ( t ) + q + 0

∂t2 2

∂ x∂ t ∂t L (Ω ) L2 (Ω ) * + 2 Z t 2 ∂u ∂ u β ∂ 2 u ∂ u (s) + q1 (1, t ) + 2c (β + 1) ∂ s ∂ s2 (s), ∂ s2 (s) ds ≤ d1 ∂t 0

(4.137)

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Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

for t ∈ [0, T ], where d0 and d1 are non-negative constants independent of t. It is clear that the problem (4.129)–(4.135) and the problem (1.56)–(1.61) are identical. Then from (4.136), (4.137) and the technique to obtain (4.91) in Lemma 3, we deduce that u ∈ H(T ) ∩ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) ,







or shortly u ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) .







(4.138)

It is noted that in the arguments for the contractedness of , we need an (or some) ideal candidate T > 0 such that 0 < C < 1. This seems that the existence of u ∈ H(T ) is only ‘‘local’’ with respect to the time if T > 0 is given. In this case, it is essential to extend u over ΩT . To do this, let us recall an interesting argument by Lawrence C. Evans [26, 9.2]. Indeed we start this procedure by selecting T1 > 0 such that

s C =

2Cαα kbkL∞ (ΩT )



(α + 1) exp C (2) 1



L (0,T )

2T13 + 3T12 +

T1



a0

< 1.

Applying Banach’s fixed-point theorem, we find a solution u ∈ H(T1 ) of the variational problem {(2.1)–(2.4)}. Then u also satisfies u ∈ H 2 0, T1 ; L2 (Ω ) ∩ H 1 ΩT1 ∩ L∞ 0, T1 ; H 2 (Ω ) .










We can continue, upon redefining T1 if necessary, by assuming



u(T1 ),

 ∂u (T1 ) ∈ H 2 (Ω ) × H 1 (Ω ). ∂t

We can then repeat the above argument to extend u or u(x) to the time interval [T1 , 2T1 ]. Continuing, after finitely many steps we construct a solution of the variational problem {(2.1)–(2.4)} existing on the full interval [0, T ], more precisely on ΩT . Thus the first step of the proof is fulfilled. Step 2. The regularity of the boundary values (
B0 , B1 )
From Step 1, we have u ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) solves the variational problem {(2.1)–(2.4)}. ,→


,→

,→

On account of the assumptions Aa , (Aq01 ), AG01 and AK01 , we deduce from (2.3) and (2.4) regarding {(4.43), (4.71)} that







    1 kB0 k2L2 (0,T ) ≤ 4 kG0 k2L2 (0,T ) + 16T q20 + T kK0 k2L2 (0,T ) ku0 k2L2 (Ω ) + C 2T 2 + ≤ CB0

(4.139)

a0

and

 

2

2 ≤ 16 G00 L2 (0,T ) + 16u20 (0) kK0 k2L2 (0,T ) + 64Tq20 ku1 k2L2 (Ω ) + 32T ku0 k2L2 (Ω ) K02 (0) + T K00 L2 (0,T )    

2 1  2 + 32 2T + 2TC~ q20 + C K02 (0) + T K00 L2 (0,T ) ≤ CB00 (4.140)

0 2

B 2

0 L (0,T )

a0

for CB0 and CB0 non-negative constants independent of t. Additionally we also have 0

B1 2L2 (0,T )

k k

0 2

B 2

≤ CB1 ,

1 L (0,T )

≤ CB01 ,

(4.141)

where CB1 and CB0 are non-negative constants independent of t, and are yielded as below: 1

CB1 := 4 k

G1 2L2 (0,T )

k

+ 16T



q21

K1 2L2 (0,T )

+Tk k



u0 2L2 (Ω )

k k



2

+ C 2T +

1

 (4.142)

a0

and



2 2 CB0 := 16 G01 L2 (0,T ) + 16u20 (1) kK1 k2L2 (0,T ) + 64Tq21 ku1 k2L2 (Ω ) + 32T ku0 k2L2 (Ω ) K12 (0) + T K10 L2 (0,T )







1



2

+ 32 2T +

1 a0





2 2TC~ q21 + C K12 (0) + T K10 L2 (0,T )





.

These estimates show that (B0 , B1 ) ∈ H 1 (0, T ) × H 1 (0, T ). Hence we complete this step. From Step 1 and Step 2, we obtain the existence of a trio of functions





2 (u, B0 , B1 ) ∈ H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω ) × H 1 (0, T )

(4.143)

Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

2955

satisfying the variational problem {(2.1)–(2.4)}. This is certainly the weak solution of the given problem. The rest of the proof is the uniqueness of this solution specified in the last step. Step 3. The uniqueness of (u, B0 , B1 )
Let two trios of functions (u, B0 , B1 ) and u, B0 , B1 be in H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω )









2 × H 1 (0, T )

satisfying the variational problem {(2.1)–(2.4)}. If we consider the following non-negative functional:

2

2

∂u

∂u ∂u ∂u

(t ) := ( t ) − ( t ) ( t ) − ( t ) + a 0

∂t ∂ t L2 (Ω ) ∂x ∂ x L2 (Ω ) + q0 |u(0, t ) − u(0, t )|2 + q1 |u(1, t ) − u(1, t )|2 ,

(4.144)

then we deduce from {(2.1)–(2.4)}, after some rearrangements, that

+  2 + Z t * ∂ u β ∂ u ∂ u β ∂ u ∂a ∂u ∂u ∂ u ∂ u (s) ds + 2c (t ) ≤ (s), (s) − ( s) ∂ s ∂ s (s) − ∂ s (s) ∂ s (s), ∂ s (s) − ∂ s (s) ds ∂s ∂x ∂x 0 0  Z s  Z t ∂u ∂u +2 (0, s) − (0, s) K0 (s − τ ) [u(0, τ ) − u(0, τ )] dτ ds ∂s ∂s 0 0  Z s  Z t ∂u ∂u +2 (1, s) − (1, s) K1 (s − τ ) [u(1, τ ) − u(1, τ )] dτ ds ∂s ∂s 0 0  Z t
∂u ∂u α (s) − (s) ds. (4.145) −2 b(s) |u(s)|α u(s) − |u(s)| u(s) , ∂s ∂s 0 Z t*

Recalling the procedure to obtain (4.125) from (4.113) and (4.114), we compute from (4.144) and (4.145) in regard with Assumption 2 for {a, b, α, β, K0 , K1 } that

(t ) ≤ 4 exp kC kL1 (0,T )




t

Z

(s)ds

(4.146)

0

for C (t ) = C

(0)

+C

(1)

β

2α

+ 2c (β + 1)R + 2(α + 1)Cα kbk

where the non-negative constants C (4.146) and (4.147) yield

(0)

and C

(1)

2 L∞ (ΩT )



2

2T +

1 a0



1 ∂a



+ (t )

∞ , a0 ∂ t L (Ω )

(4.147)

are specified in Step 1. Then with help of Gronwall’s inequality [18–20],

(t ) = 0 for t ∈ (0, T ), and immediately it follows that u ≡ u,

B0 ≡ B0 ,

B1 ≡ B1 .

Therefore we derive that the solution of the variational problem {(2.1)–(2.4)} or the weak solution of the given problem in H 2 0, T ; L2 (Ω ) ∩ H 1 (ΩT ) ∩ L∞ 0, T ; H 2 (Ω )




is unique. This completes the proof of Theorem 2.






2 × H 1 (0, T )



Remark 3. It is noted that (4.136) and (4.137) can be obtained thanks to direct energy estimates for the problem (4.129)– (4.135) in regard with Assumption 2 and the technique to deduce the boundedness of the non-negative functional ~n (t ), given by (4.71), in the proof of Lemma 3. Remark 4. The estimates (4.139)–(4.143) can be directly obtained from (4.72)–(4.76) in the proof of Lemma 3 by omitting the index n. Together with the previous remark, the crucial role of Lemma 3 is more vivid. To avoid a lengthy literature, we would like to continue the proofs of Theorem 3 and of Theorem 4 in [27].

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Ú.V. Lê / Nonlinear Analysis: Real World Applications 11 (2010) 2930–2956

Acknowledgements The regularity result in this paper was the subject of my talk at the seminar of PDE, Institute of Mathematics, Helsinki University of Technology; I would like to thank Professor Juha Kinnunen for the invitation. In addition Professor Peter Hästö had given many useful comments on the prepared version of the talk, which benefited me in this paper; I truly appreciate that. Moreover it is an honor for me to receive helpful questions regarding Lemma 1 and the existence of the smooth solution from Professor Mikael Lindström and Doctor Amir Sanatpour; those questions helped me to clarify some arguments in this paper. Furthermore I am grateful to Professor Valeriy Serov for Russian–English translations related to [1,3]. Finally there could not have been two interesting pictures of a pile-driver in the introduction if I had not obtained the help of my flatmate Janne Torvela; I am indebted to him for that kindness. 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