Open Access

Rothe-Galerkin's method for a nonlinear integrodifferential equation

Boundary Value Problems20122012:10

DOI: 10.1186/1687-2770-2012-10

Received: 29 September 2011

Accepted: 8 February 2012

Published: 8 February 2012

Abstract

In this article we propose approximation schemes for solving nonlinear initial boundary value problem with Volterra operator. Existence, uniqueness of solution as well as some regularity results are obtained via Rothe-Galerkin method.

Mathematics Subject Classification 2000: 35k55; 35A35; 65M20.

Keywords

Rothe's method a priori estimate integrodifferential equation Galerkin method weak solution

1 Introduction

The aim of this work is the solvability of the following equation
t β ( u ) t Δ a ( u ) d ( t , x , u , a ( u ) ) + K ( u ) = f ( t , x , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ1_HTML.gif
(1.1)
where (t, x) (0, T) × Ω = Q T , with the initial condition
β ( u ( 0, x ) ) = β ( u 0 ( x ) ) , x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ2_HTML.gif
(1.2)
and the boundary condition
u ( t , x ) = 0 , ( t , x ) ( 0 , T ) × Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ3_HTML.gif
(1.3)
The memory operator K is defined by
K ( t ) u , v = Ω 0 t k ( t , s ) g ( s , x , u ( s , x ) ) v ( t , x ) d s d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ4_HTML.gif
(1.4)

Let us denote by (P), the problem generated by Equations (1.1)-(1.3). The problem (P) has relevant interest applications to the porous media equation and to integro-differential equation modeling memory effects. Several problems of thermoelasticity and viscoelasticity can also be reduced to this type of problems. A variety of problems arising in mechanics, elasticity theory, molecular dynamics, and quantum mechanics can be described by doubly nonlinear problems.

The literature on the subject of local in time doubly nonlinear evolution equations is rather wide. Among these contributions, we refer the reader to [1] where the authors studied the convergence of a finite volume scheme for the numerical solution for an elliptic-parabolic equation. Using Rothe method, the author in [2] studied a nonlinear degenerate parabolic equation with a second-order differential Volterra operator. In [3] the solutions of nonlinear and degenerate problems were investigated. In general, existence of solutions for a class of nonlinear evolution equations of second order is proved by studying a full discretization.

The article is organized as follows. In Section 2, we specify some hypotheses, precise sense of the weak solution, then we state the main results and some Lemmas that needed in the sequel. In Section 3, by the Rothe-Galerkin method, we construct approximate solutions to problem (P). Some a priori estimates for the approximations are derived. In Section 4, we prove the main results.

2 Hypothesis and mean results

To solve problem (P), we assume the following hypotheses:

(H1) The function β : is continuous, nondecreasing, β (0) = 0, β (u0) L2 (Ω) and satisfies |β(s)|2C1B* (a (s)) + C2, s .

(H2) a : is continuous, strictly increasing function, a (0) = 0 and a ( u 0 ) H 0 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq1_HTML.gif.

(H3) d : (0, T) × Ω × × N N is continuous, elliptic i.e., d0 > 0 such that d (t, x, z, ξ) ξd0 |ξ| p for ξ N and p ≥ 2, strongly monotone i.e.,

(d (t, x, η, ξ1) - d (t, x, η, ξ2)) (ξ1 - ξ2) ≥ d1 |ξ1 - ξ2| p for ξ1, ξ2 N , d1 > 0 and satisfies | d ( t , x , z , ξ ) | C 1 + | ξ | p - 1 + ( B * ( a ( z ) ) ) p - 1 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq2_HTML.gif for any (t, x) (0, T) × Ω, z , ξ N .

(H4) f : (0, T) × Ω × is continuous such that
| f ( t , x , z ) | C 1 + ( B * ( a ( z ) ) ) p - 1 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equa_HTML.gif

for any (t, x) (0, T) × Ω, z .

The functions g and k given in (1.4) satisfy the following hypotheses (H5) and (H6), respectively:

(H5) g : (0, T) × Ω × N N is continuous and satisfies |g (t, x, ξ)| ≤ C (1 + |ξ| p -1) and |g (t, x, ξ1) - g (t, x, ξ2)| ≤ d1 |ξ1 - ξ2| p -1.

(H6) k : (0, T) × (0, T) → is weak singular, i.e. |k (t, s)| ≤ |t - s|-γω(t, s) for 0 γ 1 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq3_HTML.gif and the function ω : [0, T ] × [0, T] → is continuous.

(H7) For p = 2, we have
| d ( t , x , η 1 , ξ 1 ) - d ( t , x , η 2 , ξ 2 ) | C ( | a ( η 1 ) - a ( η 2 ) | + | ξ 1 - ξ 2 | ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equb_HTML.gif
and
| f ( t , x , η 1 ) - ( t , x , η 2 ) | C | a ( η 1 ) - a ( η 2 ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equc_HTML.gif

where (t, x) (0, T) × Ω, η1, η2 , ξ1, ξ2 N .

As in [3] we define the function B* by
B * ( s ) : = β ( a - 1 ( s ) ) s - 0 s β ( a - 1 ( z ) ) d z for s { y : y = a ( z ) , z } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equd_HTML.gif

We are concerned with a weak solution in the following sense:

Definition 1 By a weak solution of the problem (P) we mean a function u : Q T such that:
  1. (1)

    β (u) L2 (Q T ), t (β (u) - Δa (u)) L q ((0, T), W-1,q(Ω)), a (u) L p ((0, T), W 0 1 , p ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq4_HTML.gif, a (u) L ((0, T), H 0 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq5_HTML.gif.

     
  2. (2)
    v L p ((0, T), W 0 1 , p ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq6_HTML.gif, v t L2 ((0, T), H 0 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq7_HTML.gif and v (T) = 0 we have
    - Q T β ( u ) t v d x d t - Q T a ( u ) t v d x d t + Q T d ( t , x , u , a ( u ) ) v d x d t + Q T β ( u 0 ) v t d x d t + Q T a ( u 0 ) v t d x d t + 0 T K ( u ) , v d t = Q T f ( t , x , u ) v d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ5_HTML.gif
    (2.1)
     

The main result of this article is the following theorem.

Theorem 2 Under hypotheses (H1) - (H6), there exists a weak solution u for problem (P) in the sense of Definition 1. In addition, if (H7) is also satisfied, then u is unique.

The proof of this theorem will be done in the last section. In the sequel, we need the

following lemmas:

Lemma 3 [3] Let J : N N be continuous and for any R > 0, (J (x), x) ≥ 0 for all |x| = R. Then there exists an y N such that y ≠ 0, |y| ≤ R and J (y) = 0.

Lemma 4 [4] Assume that ∂ t (β (u) - Δa(u)) L q ((0, T), W-1,q(Ω)), a(u) L p (0, T), W 0 1 , p ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq8_HTML.gif, a(u) L((0, T), H 0 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq9_HTML.gif, B* L((0, T), L1(Ω)), β(u0) L2(Ω) and a ( u 0 ) H 0 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq10_HTML.gif. Then for almost all t (0, T), we have
0 t ( t ( β ( u ) - Δ a ( u ) ) , a ( u ) ) d t = Ω B * ( a ( u ( t ) ) ) d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Eque_HTML.gif
+ 1 2 Ω | a ( u ( t ) ) | 2 d x - Ω B * ( a ( u 0 ) ) d x - 1 2 Ω | a ( u 0 ) | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equf_HTML.gif

3 Discretization scheme and a priori estimates

To solve problem (P) by Rothe-Galerkin method, we proceed as follows. We divide the interval I = [0, T] into n subintervals of the length h = T n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq11_HTML.gif and denote u i = u (t i ), with t i = ih, i = 1, ..., n, then problem (P) is approximated by the following recurrent sequence of time-discretized problems
1 h ( β ( u i ) - β ( u i - 1 ) ) - 1 h Δ ( a ( u i ) - a ( u i - 1 ) ) - d ( t i , x , u i - 1 , a ( u i ) ) - f ( t i , x , u i - 1 ) + K ( û i - 1 ) = 0 u i ( x ) = 0 on  Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ6_HTML.gif
(3.1)

where û i - 1 = u j - 1 , t [ t j - 1 , t j ) , j = 1 , . . . , i - 1 u i - 1 , t [ t i - 1 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq12_HTML.gif

Hence, we obtain a system of elliptic problems that can be solved by Galerkin method.

Let φ1, . . . , φ m , . . . be a basis in W 0 1 , p ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq13_HTML.gif and let V m be a subspace of W 0 1 , p ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq14_HTML.gif generated by the m first vectors of the basis. We search for each m * the functions { u i m } i = 1 n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq15_HTML.gif such that a ( u i m ) = k = 1 m a i k m e k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq16_HTML.gif and satisfying
Ω 1 h ( β ( u i m ) - β ( u i - 1 m ) ) ξ d x + Ω 1 h ( a ( u i m ) - a ( u i - 1 m ) ) ξ d x + Ω d ( t i , x , u i - 1 m , a ( u i m ) ) ξ d x + K ( û i - 1 m ) , ξ - Ω f ( t i , x , u i - 1 m ) ξ d x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ7_HTML.gif
(3.2)

Remark 5 In what follows we denote by C a nonnegative constant not depending on n, m, j and h.

Theorem 6 There exists a solution u i m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq17_HTML.gif in V m of the family of discrete Equation (3.2).

Proof. We proceed by recurrence, suppose that u 0 m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq18_HTML.gif is given and that u i - 1 m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq19_HTML.gif is known. Define the continuous function J hm : m m by:
J h m ( r ) = 1 h Ω ( β ( v ) e j + a ( v ) e j ) d x 1 h Ω ( β ( u i 1 m ) e j + a u i 1 m ) e j d x + Ω d ( t i , x , u i 1 m , a ( v ) ) e j d x + K ( u ^ i 1 m ) , ξ Ω f ( t i , x , u i 1 m ) e j d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ8_HTML.gif
(3.3)
where a ( v ) = j = 1 m r j e j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq20_HTML.gif. We shall prove that J hm satisfies the following estimates
J h m ( r ) r 1 h Ω ( B * ( a ( v ) ) + 1 2 | a ( v ) | 2 ) d x 1 h Ω ( B * ( a ( u i 1 m ) ) + 1 2 | a ( u i 1 m ) | 2 ) d x + d 0 Ω | a ( v ) | p d x C δ Ω | a ( v ) | p d x C ( δ ) . C ( γ ) k = 1 i h Ω | u k 1 m | p d x C δ 0 Ω | a ( v ) | p d x C ( δ 0 ) Ω | f ( t i , x , u i 1 m ) | q d x C Ω | a ( v ) | 2 d x + C Ω | a ( v ) | p d x C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ9_HTML.gif
(3.4)
Indeed, from hypothesis (H1) and the definition of B* we deduce
1 h Ω ( β ( v ) - β ( u i - 1 m ) ) a ( v ) d x 1 h Ω B * ( a ( v ) ) d x - 1 h Ω B * ( a ( u i - 1 m ) ) d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ10_HTML.gif
(3.5)
the hypotheses on a and d imply
Ω d ( t i , x , u i - 1 m , a ( v ) ) a ( v ) d x d 0 Ω | a ( v ) | p d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ11_HTML.gif
(3.6)
using the identity
2 ( x , x - y ) = x 2 - y 2 + x - y 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ12_HTML.gif
(3.7)
we obtain
1 h Ω ( a ( v ) a ( u i 1 m ) ) a ( v ) d x 1 2 h Ω | a ( v ) | 2 d x 1 2 h Ω | a ( u i 1 m ) | 2 d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equg_HTML.gif
applying Holder and δ-inequalities to the integral operator, it yields
K ( û i - 1 m ) , a ( v ) C δ Ω 0 t i k ( t i , s ) g ( s , x , û i - 1 m ) d s q d x + C δ Ω | a ( v ) | p d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ13_HTML.gif
(3.8)
the first integral in (3.8) can be estimated as
0 t i k ( t i , s ) g ( s , x , û i - 1 m ) d s k = 1 i h | u k - 1 m | p 1 q k = 1 i h ( t i - t k ) - γ p 1 p + C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ14_HTML.gif
(3.9)
Since γ < 1 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq21_HTML.gif, then
k = 1 i h ( t i - t k ) - γ p 1 1 - γ p = : C ( γ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equh_HTML.gif
for the function f we have
Ω f ( t i , x , u i - 1 m ) a ( v ) d x C δ Ω | a ( v ) | p d x + C ( δ ) Ω B * ( a ( u i m ) ) d x + C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ15_HTML.gif
(3.10)

Therefore (3.4) holds. Then for |r| big enough, J hm (r) r ≥ 0. Taking into account that J hm is continuous, Lemma 3 states that J hm has a zero. Since the function a is strictly increasing then there exists v = u i m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq22_HTML.gif solution of (3.2). ■

Now we derive the following estimates.

Lemma 7 There exists a constant C > 0 such that
max 1 i n Ω B * ( a ( u i m ) ) d x C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ16_HTML.gif
(3.11)
max 1 i n Ω | a ( u i m ) | 2 d x C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ17_HTML.gif
(3.12)
i = 1 n h Ω | a ( u i m ) | p d x C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ18_HTML.gif
(3.13)
Proof. Testing Equation (3.2) with the function a ( u i m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq23_HTML.gif, then summing on i it yields
i = 1 j Ω 1 h ( β ( u i m ) - β ( u i - 1 m ) ) a ( u i m ) d x + i = 1 j Ω 1 h ( a ( u i m ) - a ( u i - 1 m ) ) a ( u i m ) d x + i = 1 j Ω d ( t i , x , u i - 1 m , a ( u i m ) ) a ( u i m ) d x + i = 1 j K û i - 1 m , a ( u i m ) - i = 1 j Ω f ( t i , x , u i - 1 m ) a ( u i m ) d x = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ19_HTML.gif
(3.14)
From the definition of B* we obtain
i = 1 j Ω 1 h ( β ( u i m ) - β ( u i - 1 m ) ) a ( u i m ) d x Ω B * ( u j m ) d x - Ω B * ( u 0 m ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ20_HTML.gif
(3.15)
Using the identity (3.7) for the second integral in (3.14), we get
i = 1 i Ω 1 h ( a ( u i m ) - a ( u i - 1 m ) ) a ( u i m ) d x 1 2 Ω | a ( u j m ) | 2 d x - 1 2 Ω | a ( u 0 m ) | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ21_HTML.gif
(3.16)
The hypotheses on d imply
i = 1 j Ω d ( t i , x , u i - 1 m , a ( u i m ) ) a ( u i m ) d x d 0 i = 1 j Ω | a ( u j m ) | p d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ22_HTML.gif
(3.17)
The memory operator can be estimated as
i = 1 j K ( û i - 1 m ) , a ( u i m ) C δ i = 1 j Ω 0 t i k ( t i , s ) g ( s , x , û i - 1 m ) d s q d x + C δ i = 1 j Ω | a ( u i m ) | p d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equi_HTML.gif
Using similar steps as in the proof of Theorem 6 we obtain
0 t i k ( t i , s ) g ( s , x , û i - 1 m ) d s k = 1 i h | u k - 1 m | p 1 q k = 1 i h ( t i - t k ) - γ p 1 p + C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equj_HTML.gif
Applying Poincaré inequality, we get
i = 1 j Ω f ( t i , x , u i - 1 m ) a ( u i m ) d x C ( δ ) i = 1 j Ω B * ( u i m ) d x + C δ i = 1 j Ω | a ( u i m ) | p d x + C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ23_HTML.gif
(3.18)
Substituting inequalities (3.15)-(3.18) in (3.14) it yields
Ω B * ( u j m ) d x + 1 2 Ω | a ( u j m ) | 2 d x + ( d 0 - C δ ) i = 1 j Ω | a ( u j m ) | p d x Ω B * ( u 0 m ) d x + 1 2 Ω | a ( u 0 m ) | 2 d x + C . C ( γ ) i = 1 j h k = 1 i h Ω | a ( u j m ) | p d x + C ( δ ) i = 1 j h Ω B * ( u i m ) d x + C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ24_HTML.gif
(3.19)

Choosing δ conveniently and applying the discrete Gronwall inequality, we achieve the proof of Lemma 7. ■

Lemma 8 There exists a constant C > 0 independent on m, n, h, i, and j such that
j = 1 n - k h Ω ( β ( u j + k m ) - β ( u j m ) ) ( a ( u j + k m ) - a ( u j m ) ) d x c h k , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ25_HTML.gif
(3.20)
j = 1 n - k h Ω | a ( u j + k m ) - a ( u j m ) | 2 d x c h k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ26_HTML.gif
(3.21)

Proof. Summing Equation (3.2) for i = j + 1, j + k, choosing a ( u j + k m ) - a ( u j m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq24_HTML.gif as test

function, then summing the resultant equations for j = 1 . . . , n - k, we get
j = 1 n - k Ω 1 h ( β ( u j + k m ) - β ( u j m ) ) ( a ( u j + k m ) - a ( u j m ) ) d x + j = 1 n - k Ω 1 h | a ( u j + k m ) - a ( u j m ) | 2 d x + j = 1 n - k i = j + 1 j + k Ω d ( t i , x , u i - 1 m , a ( u i m ) ) ( a ( u j + k m ) - a ( u j m ) ) d x + j = 1 n - k i = j + 1 j + k K ( û i - 1 m ) , a ( u j + k m ) - a ( u j m ) - j = 1 n - k i = j + 1 j + k Ω f ( t i , x , u i - 1 m ) ( a ( u j + k m ) - a ( u j m ) ) d x - j = 1 n - k i = j + 1 j + k Ω f ( t i , x , u i - 1 m ) ( a ( u j + k m ) - a ( u j m ) ) d x = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ27_HTML.gif
(3.22)
The third and fifth integrals in (3.22) can be estimated as
j = 1 n k i = j + 1 j + k Ω d ( t i , x , u i 1 m , a ( u i m ) ) ( a ( u j + k m ) a ( u j m ) ) d x C i = 1 n Ω | d ( t i , x , u i 1 m , a ( u i m ) ) | q d x + C k Ω ( | a ( u j + k m ) | p + | a ( u j m ) | p ) d x , j = 1 n k i = j + 1 j + k Ω f ( t i , x , u i 1 m ) ( a ( u : i + k m ) a ( u : i m ) ) d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ28_HTML.gif
(3.23)
C i = 1 n Ω | f ( t i , x , u i - 1 m ) | q d x + C k Ω ( | a ( u j + k m ) | p + | a ( u j m ) | p ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ29_HTML.gif
(3.24)
From hypotheses on d and f it yields
i = 1 n Ω | d ( t i , x , u i 1 m , a ( u i m ) ) | q d x C + C i = 1 n Ω | a ( u i m ) | p d x + C i = 1 n Ω B * ( a ( u i m ) ) d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ30_HTML.gif
(3.25)
i = 1 n Ω | f ( t i , x , u i - 1 m ) | q d x C + C i = 1 n Ω B * ( a ( u i m ) ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ31_HTML.gif
(3.26)
The operator K can be estimated as previously. Therefore we get
j = 1 n - k Ω 1 h ( β ( u j + k m ) - β ( u j m ) ) ( a ( u j + k m ) - a ( u j m ) ) d x + j = 1 n - k Ω 1 h | a ( u j + k m ) - a ( u j m ) | 2 d x i = 1 n Ω | a ( u i m ) | p d x + C i = 1 n Ω B * ( a ( u i m ) ) d x + C ( γ ) i = 1 n Ω | a ( u i m ) | p d x + C k j = 1 n - k Ω | a ( u j + k m ) | p + | a ( u j m ) | p ) d x + C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ32_HTML.gif
(3.27)

Using the estimates of previous Lemma we obtain the desired results. ■

Notation 9 Let us introduce the step functions
ū n m ( t , x ) = u m ( t i , x ) , i = 1 , n ¯ ū n m ( 0 , x ) = u 0 m ( x ) ū n , h m ( t , x ) = u n m ( t - h , x ) , t [ h , T ] ū n , h m ( t , x ) = u 0 m ( x ) , t [ 0 , h ] d n ( t , x , s , z ) = d ( t i , x , s , z ) , t ( t i - 1 , t i ] , i = 1 , n , ¯ d n ( 0 , x , s , z ) = d ( 0 , x , s , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equk_HTML.gif
Corollary 10 There exists a constant C independent of n, m, j and h such that
sup 0 t T Ω B * ( a ( ū n m ( t , x ) ) ) d x C , sup 0 t T Ω | a ( ū n m ( t , x ) ) | 2 d x C Q τ | a ( ū n m ( t , x ) ) | p d x d t C , 0 T - τ Ω | a ( u ¯ n m ( t + τ , x ) ) - a ( u ¯ n m ( t , x ) ) | 2 d x d t C τ 0 T - τ Ω ( β ( u ¯ n m ( t + τ , x ) ) - β ( u ¯ n m ( t , x ) ) ) × ( a ( u ¯ n m ( t + τ , x ) ) - a ( u ¯ n m ( t , x ) ) ) d x C τ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ33_HTML.gif
(3.28)

for k = 0 , n - 1 ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq25_HTML.gif and τ (kh, (k + 1) h).

Remark 11 (1) Corollary 10 and hypothesis (H3) imply
d n ( t , x , ū n , h m ( t , x ) , a ( ū n m ) ) L q ( Q T ) N C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equl_HTML.gif
(2)From Equation (3.2) we get
h ( β ( ū n m ) - Δ a ( ū n m ) ) L q ( ( 0 , T ) , H - 1 , q ( Ω ) ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equm_HTML.gif
(3) The estimate of B* in Corollary 10 and hypothesis (H1) give
β ( ū n m ) L 2 ( Q T ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equn_HTML.gif
(4) For the memory operator we have
K ( û n - 1 m ) L q ( ( 0 , T ) , H - 1 , q ( Ω ) ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equo_HTML.gif

4 Convergence results and existence

Now we attend to the question of convergence and existence. From Corollary 10, Remark 11 and Kolomogorov compactness criterion, one can cite the following:

Corollary 12 There exist subsequences with respect to n and m for ( ū n m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq26_HTML.gif that we will note again ( ū n m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq27_HTML.gif such that
a ( ū n m ) α i n L ( ( 0 , T ) , H 0 1 ( Ω ) ) a ( ū n m ) α i n L p ( ( 0 , T ) , W 0 1 , p ( Ω ) ) β ( ū n m ) b i n L 2 ( Q T ) h ( β ( ū n m ) - Δ a ( u ¯ n m ) ) z i n L q ( ( 0 , T ) , H - 1 , q ( Ω ) ) d n ( t , x , ū n , h m ( t , x ) , a ( ū n m ) ) χ i n L q ( Q T ) N K ( û n - 1 m ) μ i n L q ( ( 0 , T ) , H - 1 , q ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equp_HTML.gif

when m, n → ∞.

Proof of Theorem 2. We have to show that the limit function satisfies all the conditions of Definition 1. Using Corollary 10 (third and fourth inequalities) and Kolmogorov compactness criterion [[5], p. 72] it yields a ( ū n m ) α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq28_HTML.gif in L2(Q T ). Since a is strictly increasing then u n m u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq29_HTML.gif almost everywhere in Q T . From the continuity of a it yields a ( ū n m ) a ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq30_HTML.gif almost everywhere in Q T and α = a (u), consequently a ( ū n m ) a ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq31_HTML.gif a.e. in L2(Q T ). Applying Poincaré inequality and the fourth estimate in (3.28) we obtain
a ( ū n m ) - a ( ū n , h m ) L 2 ( ( 0 , T ) , H 0 1 ( Ω ) ) 2 C n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equq_HTML.gif
then ū n , h m u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq32_HTML.gif a.e. in Q T . Analogously b ( ū n m ) b ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq33_HTML.gif a.e. in L2(Q T ). According to the hypothesis (H4) we get f n ( t , x , ū n , h m ) L q ( Q T ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq34_HTML.gif and consequently f n ( ū n , h m ) - f ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq35_HTML.gif in L q (Q T ). For B* we can easily prove that B*(u) L((0, T), L1(Ω)). Based on the foregoing points, Equation (3.2) involves
0 T z , v d t + Q T χ v d x d t + 0 T μ , v d t = 0 T Ω f n ( t , x , u ) v d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ34_HTML.gif
(4.1)
Rewriting the discrete derivative with respect to t and taking into account a ( ū n m ( 0 ) ) = a ( u 0 m ) a ( u 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq36_HTML.gif in H 0 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq37_HTML.gif we obtain
Q T 1 h ( β ( ū n m ( t ) ) - β ( ū n m ( t - h ) ) ) v d x d t + Q T 1 h ( a ( ū n m ( t ) ) - a ( ū n m ( t - h ) ) ) v d x d t = - Q T ( β ( ū n m ( t ) ) - h v + a ( ū n m ( t ) ) - h v ) d x d t + Ω ( β ( ū n m ( 0 ) ) v ( 0 ) + a ( ū n m ( 0 ) ) v ( 0 ) ) d x - Q T β ( u ) v t d x d t - Q T a ( u ) v t d x d t + Q T β ( u 0 ) v t d x d t + Q T a ( u 0 ) v t d x d t = 0 T z , v d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ35_HTML.gif
(4.2)
v L p ((0, T), W 0 1 , p ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq38_HTML.gif, v t L2((0, T), H 0 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq39_HTML.gif and v(T) = 0. Since v belongs to a dense subspace in L p ((0, T), W 0 1 , p ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq40_HTML.gif and using the second estimate in Remark 11 we get
z = t ( β ( u ) - Δ a ( u ) ) L q ( ( 0 , T ) , W - 1 , q ( Ω ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equr_HTML.gif
Now we prove that
a ( ū n m ) a ( u ) in  L p ( ( 0 , T ) , W 0 1 , p ( Ω ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equs_HTML.gif
In fact, taking in (3.2) the function ξ = a ( ū n m ) - a ( v ̄ n m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq41_HTML.gif as test function and integrating on the interval (0, τ), where a ( v ̄ n m ) L p ( ( 0 , T ) , V m ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq42_HTML.gif is the approximate of a(u) in L p ( 0 , T ) , W 0 1 , p ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq43_HTML.gif, constant on each interval ((k - 1) h, kh), we obtain
Q τ h β ( ū n m ) ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t + Q τ h a ( ū n m ) ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t + Q τ d n ( t , x , ū n , h m , a ( ū n m ) ) ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t + 0 τ K ( û n - 1 m ) , a ( ū n m ) - a ( v ̄ n m ) d t = Q τ f n ( t , x , ū n , h m ) ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ36_HTML.gif
(4.3)
Lemma 4 implies
0 τ Ω h β ( ū n m ) ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t + 0 τ Ω h a ( ū n m ) ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t 1 h τ - h τ Ω B * ( a ( ū n m ) ) d x d t + 1 2 h τ - h τ Ω | a ( ū n m ) | 2 d x d t - Ω B * ( a ( u ( τ ) ) ) d x - 1 2 Ω | a ( u ( τ ) ) | 2 d x + c ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equt_HTML.gif
From Fatou Lemma we deduce
lim  inf m ; n 1 h τ - h τ Ω B * ( a ( ū n m ) ) + 1 2 | a ( ū n m ) | 2 d x d t Ω B * ( a ( u ( τ ) ) ) d x + 1 2 Ω | a ( u ( τ ) ) | 2 d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equu_HTML.gif
consequently
lim  inf m ; n 0 τ Ω [ h β ( ū n m ) ( a ( ū n m ) - a ( v ̄ n m ) ) + h a ( ū n m ) ( a ( ū n m ) - a ( v ̄ n m ) ) ] d x d t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equv_HTML.gif
Taking into account the convergence of a ( ū n m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq44_HTML.gif to a(u) in L2(Q T ), the convergence of a ( v ̄ n m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq45_HTML.gif to a(u) in L p ( 0 , T ) , W 0 1 , p ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq46_HTML.gif, the continuity of d, the weak convergence of d in L q (Q T ) N and the dominated convergence theorem, we obtain
d n ( t , x , ū n , h m , a ( u ) ) d ( t , x , u , a ( u ) ) in L q ( Q T ) N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equw_HTML.gif
In addition to monotonicity of d gives
0 τ Ω d n ( t , x , ū n , h m , a ( ū n m ) ) ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t = 0 τ Ω ( d n ( t , x , ū n , h m , a ( ū n m ) ) - d n ( t , x , ū n , h m , a ( v ̄ n m ) ) ) × ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t + 0 τ Ω d n ( t , x , ū n , h m , a ( v ̄ n m ) ) ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t d 1 0 τ Ω | a ( ū n m ) - a ( v ̄ n m ) | p d x d t - c ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equx_HTML.gif
as previously using hypotheses (H5) and (H6), the operator memory can be estimated as
0 τ K ( û n - 1 m ) , a ( ū n m ) - a ( v ̄ n m ) d t C δ 0 τ Ω 0 t | a ( ū n m ) - a ( v ̄ n m ) | p d x d t + C δ || a ( ū n m ) - a ( v ̄ n m ) || L p ( ( 0 , T ) , W 0 1 , p ( Ω ) ) p + C ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equy_HTML.gif
For f n we have
0 τ Ω f n ( t , x , ū n , h m ) ( a ( ū n m ) - a ( v ̄ n m ) ) d x d t C ε , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equz_HTML.gif
regrouping the estimates of all terms of Equation (4.3) we obtain
( d 1 - C δ ) 0 τ Ω | a ( ū n m ) - a ( v ̄ n m ) | p d x d t C 0 τ Ω 0 t | a ( ū n m ) - a ( v ̄ n m ) | p d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equaa_HTML.gif
Gronwall inequality implies
0 τ Ω | a ( ū n m ) - a ( v ̄ n m ) | p d x d t C ε , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equab_HTML.gif
hence we get
a ( ū n m ) a ( u ) in L p ( ( 0 , T ) , W 0 1 , p ( Ω ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equac_HTML.gif
Following the Proof of Theorem 2: From the continuity of d and g it yields
d n ( t , x , ū n , h m , a ( ū n m ) ) d ( t , x , u , a ( u ) ) a .e . Q T g n ( t , x , û n - 1 m ) g ( t , x , u ) a .e . Q T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equad_HTML.gif

The weak convergences of d n ( t , x , ū n , h m , a ( ū n m ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq47_HTML.gif and K ( û n - 1 m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq48_HTML.gif and the almost everywhere convergences imply that χ = d(t, x, u, a(u)) and µ = K(u). So u is the weak solution of the problem (P) in the sense of Definition 1.

Now we prove the uniqueness of the weak solution. We assume that the problem (P) has two solutions u1 and u 2 L 2 ( ( 0 , T ) , H 0 1 ( Ω ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq49_HTML.gif. Taking into account that β ( u 0 1 ) = β ( u 0 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq50_HTML.gif and a ( u 0 1 ) = a ( u 0 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_IEq51_HTML.gif, we get
0 T Ω ( ( β ( u 1 ) - β ( u 2 ) ) v t + ( a ( u 1 ) - a ( u 2 ) ) v t ) d x d t + 0 T Ω ( d ( t , x , u 1 , a ( u 1 ) ) - d ( t , x , u 2 , a ( u 2 ) ) ) v d x d t + 0 T K ( u 1 ) - K ( u 2 ) , v d t = 0 T Ω ( f ( t , x , u 1 ) - f ( t , x , u 2 ) ) v d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equ37_HTML.gif
(4.4)
Choosing in (4.4) the test function
v s ( t ) = t s ( a ( u 1 ( τ ) ) - a ( u 2 ( τ ) ) ) d τ , t < s 0 , t s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equae_HTML.gif
and since v s (s) = 0 then integrating by parts it yields
0 s β ( u 1 ) - β ( u 2 ) , a ( u 1 ) - a ( u 2 ) d t + 0 s Ω | a ( u 1 ) - a ( u 2 ) | 2 d x d t δ 0 s Ω | a ( u 1 ) - a ( u 2 ) | 2 d x d t + C δ 0 s Ω | v s | 2 d t d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equaf_HTML.gif
On the other hand, we have
0 s Ω | v s | 2 d t d x C 0 s 0 t Ω | a ( u 1 ( x , τ ) ) - a ( u 2 ( x , τ ) ) | 2 d x d τ d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equag_HTML.gif
Applying Gronwall lemma we get
0 s Ω | a ( u 1 ) - a ( u 2 ) | 2 d x d t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-10/MediaObjects/13661_2011_Article_112_Equah_HTML.gif

consequently u1u2. This achieves the Proof of Theorem 2.

Declarations

Acknowledgements

The authors would like to express their thanks to the referees for their helpful comments and suggestions. This work was supported by the National Research Project (PNR, Code8/u160/829).

Authors’ Affiliations

(1)
Laboratory of advanced materials, Badji Mokhtar University

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© Chaoui and Guezane-Lakoud; licensee Springer. 2012

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