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Rothe-Galerkin's method for a nonlinear integrodifferential equation

Abderrazek Chaoui* and Assia Guezane-Lakoud

Author Affiliations

Laboratory of advanced materials, Badji Mokhtar University, Annaba, Algeria

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Boundary Value Problems 2012, 2012:10  doi:10.1186/1687-2770-2012-10

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/10


Received:29 September 2011
Accepted:8 February 2012
Published:8 February 2012

© 2012 Chaoui and Guezane-Lakoud; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M1">View MathML</a>

(1.1)

where (t, x) ∈ (0, T) × Ω = QT, with the initial condition

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M2">View MathML</a>

(1.2)

and the boundary condition

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M3">View MathML</a>

(1.3)

The memory operator K is defined by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M4">View MathML</a>

(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 : ℝ → ℝ is continuous, nondecreasing, β (0) = 0, β (u0) ∈ L2 (Ω) and satisfies |β(s)|2 C1B* (a (s)) + C2, ∀s ∈ ℝ.

(H2) a : ℝ → ℝ is continuous, strictly increasing function, a (0) = 0 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M5">View MathML</a>.

(H3) d : (0, T) × Ω × ℝ × ℝN → ℝN is continuous, elliptic i.e., ∃d0 > 0 such that d (t, x, z, ξ) ξ d0 |ξ|p for ξ ∈ ℝN 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 ∈ ℝN, d1 > 0 and satisfies <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M6">View MathML</a> for any (t, x) ∈ (0, T) × Ω, ∀z ∈ ℝ, ξ ∈ ℝN.

(H4) f : (0, T) × Ω × ℝ → ℝ is continuous such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M7">View MathML</a>

for any (t, x) ∈ (0, T) × Ω, ∀z ∈ ℝ.

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

(H5) g : (0, T) × Ω × ℝN → ℝN 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) → ℝ is weak singular, i.e. |k (t, s)| ≤ |t - s|-γω(t, s) for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M8">View MathML</a> and the function ω : [0, T ] × [0, T] → ℝ is continuous.

(H7) For p = 2, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M9">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M10">View MathML</a>

where (t, x) ∈ (0, T) × Ω, η1, η2 ∈ ℝ, ξ1, ξ2 ∈ ℝN.

As in [3] we define the function B* by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M11">View MathML</a>

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 : QT → ℝ such that:

(1) β (u) ∈ L2 (QT), t (β (u) - Δa (u)) ∈ Lq ((0, T), W-1,q(Ω)), a (u) ∈ Lp ((0, T), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M12">View MathML</a>, a (u) ∈ L((0, T), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M13">View MathML</a>.

(2) ∀v Lp ((0, T), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M14">View MathML</a>, vt L2 ((0, T), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M15">View MathML</a>and v (T) = 0 we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M16">View MathML</a>

(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 : ℝN → ℝN be continuous and for any R > 0, (J (x), x) ≥ 0 for all |x| = R. Then there exists an y ∈ ℝN such that y ≠ 0, |y| ≤ R and J (y) = 0.

Lemma 4 [4]Assume that ∂t(β (u) - Δa(u)) ∈ Lq((0, T), W-1,q(Ω)), a(u) ∈ Lp (0, T), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M17">View MathML</a>, a(u) ∈ L((0, T), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M18">View MathML</a>, B* ∈ L((0, T), L1(Ω)), β(u0) ∈ L2(Ω) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M19">View MathML</a>. Then for almost all t ∈ (0, T), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M20">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M21">View MathML</a>

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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M22">View MathML</a> and denote ui = u (ti), with ti = ih, i = 1, ..., n, then problem (P) is approximated by the following recurrent sequence of time-discretized problems

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M23">View MathML</a>

(3.1)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M24">View MathML</a>

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

Let φ1, . . . , φm, . . . be a basis in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M25">View MathML</a> and let Vm be a subspace of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M26">View MathML</a> generated by the m first vectors of the basis. We search for each m ∈ ℕ* the functions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M27">View MathML</a> such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M28">View MathML</a> and satisfying

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M29">View MathML</a>

(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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M30">View MathML</a>in Vm of the family of discrete Equation (3.2).

Proof. We proceed by recurrence, suppose that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M31">View MathML</a> is given and that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M32">View MathML</a> is known. Define the continuous function Jhm : ℝm → ℝm by:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M33">View MathML</a>

(3.3)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M34">View MathML</a>. We shall prove that Jhm satisfies the following estimates

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M35">View MathML</a>

(3.4)

Indeed, from hypothesis (H1) and the definition of B* we deduce

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M36">View MathML</a>

(3.5)

the hypotheses on a and d imply

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M37">View MathML</a>

(3.6)

using the identity

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M38">View MathML</a>

(3.7)

we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M39">View MathML</a>

applying Holder and δ-inequalities to the integral operator, it yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M40">View MathML</a>

(3.8)

the first integral in (3.8) can be estimated as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M41">View MathML</a>

(3.9)

Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M42">View MathML</a>, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M43">View MathML</a>

for the function f we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M44">View MathML</a>

(3.10)

Therefore (3.4) holds. Then for |r| big enough, Jhm(r) r ≥ 0. Taking into account that Jhm is continuous, Lemma 3 states that Jhm has a zero. Since the function a is strictly increasing then there exists <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M45">View MathML</a> solution of (3.2). ■

Now we derive the following estimates.

Lemma 7 There exists a constant C > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M46">View MathML</a>

(3.11)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M47">View MathML</a>

(3.12)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M48">View MathML</a>

(3.13)

Proof. Testing Equation (3.2) with the function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M49">View MathML</a>, then summing on i it yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M50">View MathML</a>

(3.14)

From the definition of B* we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M51">View MathML</a>

(3.15)

Using the identity (3.7) for the second integral in (3.14), we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M52">View MathML</a>

(3.16)

The hypotheses on d imply

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M53">View MathML</a>

(3.17)

The memory operator can be estimated as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M54">View MathML</a>

Using similar steps as in the proof of Theorem 6 we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M55">View MathML</a>

Applying Poincaré inequality, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M56">View MathML</a>

(3.18)

Substituting inequalities (3.15)-(3.18) in (3.14) it yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M57">View MathML</a>

(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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M58">View MathML</a>

(3.20)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M59">View MathML</a>

(3.21)

Proof. Summing Equation (3.2) for i = j + 1, j + k, choosing <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M60">View MathML</a> as test

function, then summing the resultant equations for j = 1 . . . , n - k, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M61">View MathML</a>

(3.22)

The third and fifth integrals in (3.22) can be estimated as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M62">View MathML</a>

(3.23)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M63">View MathML</a>

(3.24)

From hypotheses on d and f it yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M64">View MathML</a>

(3.25)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M65">View MathML</a>

(3.26)

The operator K can be estimated as previously. Therefore we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M66">View MathML</a>

(3.27)

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

Notation 9 Let us introduce the step functions

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M67">View MathML</a>

Corollary 10 There exists a constant C independent of n, m, j and h such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M68">View MathML</a>

(3.28)

for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M69">View MathML</a> and τ ∈ (kh, (k + 1) h).

Remark 11 (1) Corollary 10 and hypothesis (H3) imply

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M70">View MathML</a>

(2)From Equation (3.2) we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M71">View MathML</a>

(3) The estimate of B* in Corollary 10 and hypothesis (H1) give

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M72">View MathML</a>

(4) For the memory operator we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M73">View MathML</a>

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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M74">View MathML</a>that we will note again <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M75">View MathML</a>such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M76">View MathML</a>

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 onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M77">View MathML</a> in L2(QT). Since a is strictly increasing then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M78">View MathML</a> almost everywhere in QT. From the continuity of a it yields <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M79">View MathML</a> almost everywhere in QT and α = a (u), consequently <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M80">View MathML</a> a.e. in L2(QT). Applying Poincaré inequality and the fourth estimate in (3.28) we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M81">View MathML</a>

then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M82">View MathML</a> a.e. in QT. Analogously <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M83">View MathML</a> a.e. in L2(QT). According to the hypothesis (H4) we get <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M84">View MathML</a> and consequently <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M85">View MathML</a> in Lq(QT). For B* we can easily prove that B*(u) ∈ L((0, T), L1(Ω)). Based on the foregoing points, Equation (3.2) involves

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M86">View MathML</a>

(4.1)

Rewriting the discrete derivative with respect to t and taking into account <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M87">View MathML</a> in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M88">View MathML</a> we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M89">View MathML</a>

(4.2)

v Lp ((0, T), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M90">View MathML</a>, vt L2((0, T), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M91">View MathML</a> and v(T) = 0. Since v belongs to a dense subspace in Lp ((0, T), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M92">View MathML</a> and using the second estimate in Remark 11 we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M93">View MathML</a>

Now we prove that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M94">View MathML</a>

In fact, taking in (3.2) the function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M95">View MathML</a> as test function and integrating on the interval (0, τ), where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M96','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M96">View MathML</a> is the approximate of a(u) in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M97">View MathML</a>, constant on each interval ((k - 1) h, kh), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M98','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M98">View MathML</a>

(4.3)

Lemma 4 implies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M99','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M99">View MathML</a>

From Fatou Lemma we deduce

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M100">View MathML</a>

consequently

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M101','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M101">View MathML</a>

Taking into account the convergence of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M102','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M102">View MathML</a> to a(u) in L2(QT), the convergence of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M103','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M103">View MathML</a> to a(u) in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M104','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M104">View MathML</a>, the continuity of d, the weak convergence of d in Lq(QT)N and the dominated convergence theorem, we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M105','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M105">View MathML</a>

In addition to monotonicity of d gives

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M106','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M106">View MathML</a>

as previously using hypotheses (H5) and (H6), the operator memory can be estimated as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M107','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M107">View MathML</a>

For fn we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M108','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M108">View MathML</a>

regrouping the estimates of all terms of Equation (4.3) we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M109','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M109">View MathML</a>

Gronwall inequality implies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M110','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M110">View MathML</a>

hence we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M111">View MathML</a>

Following the Proof of Theorem 2: From the continuity of d and g it yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M112','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M112">View MathML</a>

The weak convergences of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M113','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M113">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M114','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M114">View MathML</a> 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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M115','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M115">View MathML</a>. Taking into account that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M116','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M116">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M117','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M117">View MathML</a>, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M118','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M118">View MathML</a>

(4.4)

Choosing in (4.4) the test function

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M119','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M119">View MathML</a>

and since vs(s) = 0 then integrating by parts it yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M120','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M120">View MathML</a>

On the other hand, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M121','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M121">View MathML</a>

Applying Gronwall lemma we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M122','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/10/mathml/M122">View MathML</a>

consequently u1 u2. This achieves the Proof of Theorem 2.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

The authors declare that the work was realized in collaboration with same responsibility. All authors read and approved the final manuscript.

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).

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  4. Showalter, RE: Monotone operators in Banach space and nonlinear partial differential equations. Math Surv Monogr. 49, 113–142 (1996)

  5. Brezis, H: Analyse Fonctionnelle, Théorie et Applications. Masson, Paris (1983)