Table of Contents
International Journal of Stochastic Analysis
Volume 2010, Article ID 519684, 16 pages
Research Article

The Rothe's Method to a Parabolic Integrodifferential Equation with a Nonclassical Boundary Conditions

Department of Mathematics, University of Larbi Ben M'Hidi, Oum El Bouagui 04000, Algeria

Received 26 February 2009; Revised 17 August 2009; Accepted 10 December 2009

Academic Editor: Alexander M. Krasnosel'skii

Copyright © 2010 Abdelfatah Bouziani and Rachid Mechri. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


This paper is devoted to prove, in a nonclassical function space, the weak solvability of parabolic integrodifferential equations with a nonclassical boundary conditions. The investigation is made by means of approximation by the Rothes method which is based on a semidiscretization of the given problem with respect to the time variable.

1. Introduction

The purpose of this paper is to study the solvability of the following equation: 𝜕𝑣(𝜕𝜕𝑡𝑥,𝑡)2𝑣𝜕𝑥2(𝑥,𝑡)=𝑡0𝑎(𝑡𝑠)𝑘([]𝑠,𝑣(𝑥,𝑠))𝑑𝑠+𝑔(𝑥,𝑡),(𝑥,𝑡)(0,1)×0,𝑇,(1.1) with the initial condition 𝑣(𝑥,0)=𝑉0(𝑥),𝑥(0,1),(1.2) and the integral conditions 10[],𝑣(𝑥,𝑡)𝑑𝑥=𝐸(𝑡),𝑡0,𝑇10[],𝑥𝑣(𝑥,𝑡)𝑑𝑥=𝐺(𝑡),𝑡0,𝑇(1.3) where 𝑣 is an unknown function, 𝐸, 𝐺, and 𝑉0 are given functions supposed to be sufficiently regular, while 𝑘 and 𝑎 are suitably defined functions satisfying certain conditions to be specified later and 𝑇 is a positive constant.

Since 1930, various classical types of initial boundary value problems have been investigated by many authors using Rothe time-discretization method; see, for instance, the monographs by Rektorys [1] and Kačur [2] and references cited therein. The linear case of our problem, that is, 𝑡0𝑎(𝑡𝑠)𝑘(𝑠,𝑣(𝑥,𝑠))𝑑𝑠=0, appears, for instance, in the modelling of the quasistatic flexure of a thermoelastic rod (see [3]) and has been studied, firstly, by the second author with a more general second-order parabolic equation or a 2m-parabolic equation in [35] by means of the energy-integrals method and, secondly, by the two authors via the Rothe method [68]. For other models, we refer the reader, for instance, to [912], and references therein.

The paper is organized as follows. In Section 2, we transform problem (1.1)–(1.3) to an equivalent one with homogeneous integral conditions, namely, problem (2.3). Then, we specify notations and assumptions on data before stating the precise sense of the desired solution. In Section 3, by the Rothe discretization in time method, we construct approximate solutions to problem (2.3). Some a priori estimates for the approximations are derived in Section 4, while Section 5 is devoted to establish the existence and uniqueness of the solution.

2. Preliminaries, Notation, and Main Result

It is convenient at the beginning to reduce problem (1.1)–(1.3) with inhomogeneous integral conditions to an equivalent one with homogeneous conditions. For this, we introduce a new unknown function 𝑢 by setting

[]𝑢(𝑥,𝑡)=𝑣(𝑥,𝑡)𝑅(𝑥,𝑡),(𝑥,𝑡)(0,1)×0,𝑇,(2.1) where

𝑅(𝑥,𝑡)=6(2𝐺(𝑡)𝐸(𝑡))𝑥2(3𝐺(𝑡)2𝐸(𝑡)).(2.2) Then, the function 𝑢 is seen to be the solution of the following problem:

𝜕𝑢(𝜕𝜕𝑡𝑥,𝑡)2𝑢𝜕𝑥2(𝑥,𝑡)=𝑡0[],𝑎(𝑡𝑠)𝑘(𝑠,𝑢(𝑥,𝑠))𝑑𝑠+𝑓(𝑥,𝑡),(𝑥,𝑡)(0,1)×0,𝑇𝑢(𝑥,0)=𝑈0(𝑥),𝑥(0,1),10[],𝑢(𝑥,𝑡)𝑑𝑥=0,𝑡0,𝑇10[],𝑥𝑢(𝑥,𝑡)𝑑𝑥=0,𝑡0,𝑇(2.3) where 𝑓(𝑥,𝑡)=𝑔(𝑥,𝑡)𝜕𝑅(𝑥,𝑡),𝑈𝜕𝑡0(𝑥)=𝑉0(𝑥)𝑅(𝑥,0),𝑘(𝑠,𝑢(𝑥,𝑠))=𝑘(𝑠,𝑢(𝑥,𝑠))𝑅(𝑥,𝑡).(2.4) Hence, instead of looking for the function 𝑣, we search for the function 𝑢. The solution of problem (1.1)–(1.3) will be simply given by the formula 𝑣(𝑥,𝑡)=𝑢(𝑥,𝑡)+𝑅(𝑥,𝑡).

We introduce the function spaces, which we need in our investigation. Let 𝐿2(0,1) and 𝐿2(0,𝑇;𝐿2(0,1)) be the standard function spaces. We denote by 𝐶0(0,1) the linear space of continuous functions with compact support in (0,1). Since such functions are Lebesgue integrable, we can define on 𝐶0(0,1) the bilinear form given by ((𝑢,𝑣))=10𝑥𝑢𝑥𝑣𝑑𝑥,(2.5) where

𝑥𝑢=𝑥0𝑢(𝜁,)𝑑𝜁.(2.6) The bilinear form (2.5) is considered as a scalar product on 𝐶0(0,1) for which 𝐶0(0,1) is not complete.

Definition 2.1. We denote by 𝐵12(0,1) a completion of 𝐶0(0,1) for the scalar product (2.5), which is denoted by (,)𝐵12(0,1), called the Bouziani space or the space of square integrable primitive functions on (0,1). By the norm of function 𝑢 from 𝐵12(0,1), we understand the nonnegative number 𝑢𝐵12(0,1)=(𝑢,𝑢)𝐵12(0,1)=𝑥𝑢,(2.7)where 𝑣 denotes the norm of 𝑣 in 𝐿2(0,1).

For 𝑢𝐿2(0,1), we have the elementary inequality 𝑢𝐵12(0,1)12𝑢.(2.8)

We denote by 𝐿2(0,𝑇;𝐵12(0,1)) the space of functions which are square integrable in the Bochner sense, with the scalar product


Since the space 𝐵12(0,1) is a Hilbert space, it can be shown that 𝐿2(0,𝑇;𝐵12(0,1)) is a Hilbert space as well. The set of all continuous abstract functions in [0,𝑇] equipped with the norm

sup0𝜏𝑇𝑢(,𝜏)𝐵12(0,1)(2.10) is denoted 𝐶(0,𝑇;𝐵12(0,1)). Let 𝑉 be the set which we define as follows:


Since 𝑉 is the null space of the continuous linear mapping 𝑙: 𝐿2(0,1)2, 𝜑𝑙(𝜑)=(10𝜑(𝑥)𝑑𝑥,10𝑥𝜑(𝑥)𝑑𝑥), it is a closed linear subspace of 𝐿2(0,1), consequently 𝑉 is a Hilbert space endowed with the inner product (,). Strong or weak convergence is denoted by or , respectively. The letter 𝐶 will stand for a generic positive constant which may be different in the same discussion.

Lemma 2.2 (Gronwall's lemma). (a1) Let 𝑥(𝑡)0, (𝑡), 𝑦(𝑡) be real integrable functions on the interval [𝑎,𝑏]. If 𝑦(𝑡)(𝑡)+𝑡𝑎𝑥(𝑠)𝑦(𝑠)𝑑𝑠,𝑡(𝑎,𝑏),(2.12) then 𝑦(𝑡)(𝑡)+𝑡𝑎(𝑠)𝑥(𝑠)exp𝑡𝑎𝑥(𝜏)𝑑𝜏𝑑𝑠,𝑡(0,𝑇).(2.13)
In particular, if 𝑥(𝑡)𝐶 is a constant and (𝑡) is nondecreasing, then 𝑦(𝑡)(𝑡)𝑒𝑐(𝑡𝑎),𝑡(0,𝑇).(2.14)
(a2) Let {𝑎𝑖}𝑖be a sequence of real nonnegative numbers satisfying 𝑎𝑖𝐴+𝐵𝑖1𝑘=1𝑎𝑘,𝑖=1,2,,(2.15)where𝐴, 𝐵, and are positive constants, such that 𝐵<1. Then 𝑎𝑖[]𝐴exp𝐵(𝑖1),(2.16)takes place for all 𝑖=1,2,.

In the sequel, we make the following assumptions.

(𝐻1) Functions 𝑓[0,𝑇]𝐿2(0,1) and 𝑎[0,𝑇] are Lipschitz continuous, that is,


(𝐻2) The mapping 𝑘[0,𝑇]×𝑉𝐿2(0,1) is Lipschitz continuous in both variables, that is,

𝑙3+;𝑘𝑡(𝑡,𝑢)𝑘,𝑢𝑙3||𝑡𝑡||+𝑢𝑢,(2.18) for all 𝑡,𝑡𝐼, 𝑢,𝑢𝑉, and satisfies

𝑙4,𝑙5+;𝑘(𝑡,𝑢)𝐵12(0,1)𝑙4𝑢𝐵12(0,1)+𝑙5,(2.19) for all 𝑡𝐼 and all 𝑢𝑉, where 𝑙4 and 𝑙5 are positive constants.

(𝐻3)𝑈0𝐻2(0,1) and


We will be concerned with a weak solution in the following sense.

Definition 2.3. A function 𝑢𝐼𝐿2(0,1) is called a weak solution to problem (2.3) if the following conditions are satisfied:(i)𝑢𝐿(𝐼,𝑉)𝐶(𝐼,𝐵12(0,1)),(ii)𝑢 is strongly differentiable a.e. in 𝐼 and 𝑑𝑢/𝑑𝑡𝐿(𝐼,𝐵12(0,1)),(iii)𝑢(0)=𝑈0 in 𝑉,(iv)the identity 𝑑𝑢𝑑𝑡(𝑡),𝑣𝐵12(0,1)=+(𝑢(𝑡),𝑣)𝑡0𝑎(𝑡𝑠)𝑘(𝑠,𝑢(𝑠))𝑑𝑠,𝑣𝐵12(0,1)+(𝑓(𝑡),𝑣)𝐵12(0,1)(2.21)holds for all 𝑣𝑉 and a.e. 𝑡[0,𝑇].

To close this section, we announce the main result of the paper.

Theorem 2.4. Under assumptions (𝐻1)-(𝐻3), problem (2.3) admits a unique weak solution 𝑢, in the sense of Definition (2.3).

3. Construction of an Approximate Solution

In order to solve problem (2.3) by the Rothe method, we proceed as follows. Let 𝑛 be a positive integer, we divide the time interval 𝐼=[0,𝑇] into 𝑛 subintervals 𝐼𝑛𝑗=[𝑡𝑛𝑗1,𝑡𝑛𝑗], 𝑗=1,,𝑛, where 𝑡𝑛𝑗=𝑗𝑛 and 𝑛=𝑇/𝑛. Then, for each 𝑛1, problem (2.3) may be approximated by the following recurrent sequence of time-discretized problems. Successively, for 𝑗=1,,𝑛, we look for functions 𝑢𝑛𝑗𝑉 such that 𝑢𝑛𝑗𝑢𝑛𝑗1𝑛𝑑2𝑢𝑛𝑗𝑑𝑥2=𝑛𝑗1𝑖=0𝑎𝑡𝑛𝑗𝑡𝑛𝑖𝑘𝑡𝑛𝑖,𝑢𝑛𝑖+𝑓𝑛𝑗,(3.1)10𝑢𝑛𝑗(𝑥)𝑑𝑥=0,(3.2)10𝑥𝑢𝑛𝑗(𝑥)𝑑𝑥=0,(3.3) starting from

𝑢𝑛0=𝑈0,𝛿𝑢𝑛0=𝑑2𝑑𝑥2𝑈0+𝑓(0),(3.4) where 𝑢𝑛𝑗(𝑥)=𝑢(𝑥,𝑡𝑛𝑗), 𝛿𝑢𝑛𝑗=(𝑢𝑛𝑗𝑢𝑛𝑗1)/𝑛, 𝑓𝑛𝑗(𝑥)=𝑓(𝑥,𝑡𝑛𝑗). For this, multiplying for all 𝑗=1,,𝑛, (3.1) by 2𝑥𝑣=𝑥0(𝜉0𝑣(𝜏)𝑑𝜏)𝑑𝜉 and integrating over (0,1), we get

10𝛿𝑢𝑛𝑗(𝑥)2𝑥𝑣𝑑𝑥10𝑑2𝑢𝑛𝑗𝑑𝑥2(𝑥)2𝑥𝑣𝑑𝑥=𝑛10𝑗1𝑖=0𝑎𝑡𝑛𝑗𝑡𝑛𝑖𝑘𝑡𝑛𝑖,𝑢𝑛𝑖2𝑥𝑣𝑑𝑥+10𝑓𝑛𝑗2𝑥𝑣𝑑𝑥.(3.5) Noting that, using a standard integration by parts, we have

21𝑣=10(1𝜉)𝑣(𝜉)𝑑𝜉=10𝑣(𝜉)𝑑𝜉10𝜉𝑣(𝜉)𝑑𝜉=0,𝑣𝑉.(3.6) Carrying out some integrations by parts and invoking (3.6), we obtain for each term in (3.5) 10𝛿𝑢𝑛𝑗2𝑥𝑣𝑑𝑥=𝛿𝑢𝑛𝑗,𝑣𝐵12(0,1),10𝑑2𝑢𝑛𝑗𝑑𝑥2(𝑥)2𝑥𝑢𝑣𝑑𝑥=𝑛𝑗,,𝑣𝑛10𝑗1𝑖=0𝑎𝑡𝑛𝑗𝑡𝑛𝑖𝑘𝑡𝑛𝑖,𝑢𝑛𝑖(𝑥)2𝑥𝑣𝑑𝑥=𝑛𝑗1𝑖=0𝑎𝑡𝑛𝑗𝑡𝑛𝑖𝑘𝑡𝑛𝑖,𝑢𝑛𝑖,𝑣𝐵12(0,1),(3.7) and for the last one

10𝑓𝑛𝑗(𝑥)2𝑥𝑓𝑣(𝑥)𝑑𝑥=𝑛𝑗,𝑣𝐵12(0,1).(3.8) By virtue of (3.7) and (3.8), (3.5) becomes 𝛿𝑢𝑛𝑗,𝑣𝐵12(0,1)+𝑢𝑛𝑗,𝑣=𝑛𝑗1𝑖=0𝑎𝑡𝑛𝑗𝑡𝑛𝑖𝑘𝑡𝑛𝑖,𝑢𝑛𝑖,𝑣𝐵12(0,1)+𝑓𝑛𝑗,𝑣𝐵12(0,1),(3.9) or

𝑢𝑛𝑗,𝑣𝐵12(0,1)+𝑛𝑢𝑛𝑗,𝑣=2𝑛𝑗1𝑖=0𝑎𝑡𝑛𝑗𝑡𝑛𝑖𝑘𝑡𝑛𝑖,𝑢𝑛𝑖,𝑣𝐵12(0,1)+𝑛𝑓𝑛𝑗,𝑣𝐵12(0,1)+𝑢𝑛𝑗1,𝑣𝐵12(0,1).(3.10) Let 𝜂(,)𝑉×𝑉and 𝐿𝑗()𝑉 be two functions defined by


It is easy to see that the bilinear form 𝜂(,) is continuous on 𝑉 and V-elliptic, and the form 𝐿𝑗() is continuous for each 𝑗=1,,𝑛. Then, Lax-Milgram lemma guarantees the existence and uniqueness of 𝑢𝑛𝑗, for all 𝑗=1,,𝑛.

4. A Priori Estimates

Lemma 4.1. There exists 𝐶>0 such that, for all 𝑛1 and all 𝑗=1,,𝑛, the solution 𝑢𝑗 of the discretized problem (3.1)–(3.4) satisfies the estimates 𝑢𝑛𝑗𝐶,(4.1)𝛿𝑢𝑛𝑗𝐵12(0,1)𝐶.(4.2)

Proof. Testing the difference (3.9)𝑗1-(3.9)𝑗 with 𝑣=𝛿𝑢𝑛𝑗(𝑉), taking into account assumptions (𝐻1)-(𝐻3) and the Cauchy-Schwarz inequality, we obtain 𝛿𝑢𝑛𝑗𝐵12(0,1)+𝑢𝑛𝑗𝑢𝑛𝑗1𝐵12(0,1)𝛿𝑢𝑛𝑗1𝐵12(0,1)+𝐶132𝑛𝑗2𝑖=0𝑢𝑛𝑖𝐵12(0,1)+𝐶13𝑛+𝐶13𝑛𝑢𝑛𝑗1𝐵12(0,1),(4.3)where 𝐶1𝑙=3max2𝜁,𝑇𝑙2𝜁+𝑀1𝜁+𝑙1,𝑀1=max𝑡𝐼||𝑎||𝑙(𝑡),𝜁=max4,𝑙5.(4.4)Multiplying the left-hand side of the last inequality with (1(𝐶1/3)𝑛)(<1) and adding the terme 23𝐶1𝑛𝑢𝑛𝑗𝑢𝑛𝑗1𝐵12(0,1)𝛿𝑢𝑛𝑗𝐵12(0,1)(<0),(4.5)we get 1𝐶1𝑛𝛿𝑢𝑛𝑗𝐵12(0,1)+𝑢𝑛𝑗𝐵12(0,1)𝑢𝑛𝑗1𝐵12(0,1)+𝛿𝑢𝑛𝑗1𝐵12(0,1)+𝐶12𝑛𝑗2𝑖=0𝑢𝑛𝑖𝐵12(0,1)+𝐶1𝑛.(4.6) Applying the last inequality recursively, it follows that 1𝐶1𝑛𝑗𝛿𝑢𝑛𝑗𝐵12(0,1)+𝑢𝑛𝑗𝐵12(0,1)𝑢𝑛0𝐵12(0,1)+𝛿𝑢𝑛0𝐵12(0,1)+𝐶1𝑇+𝑇𝐶1𝑛𝑗2𝑖=0𝑢𝑛𝑖𝐵12(0,1),(4.7) or, by virtue of Lemma 2.2, there exists 𝑛0 such that 𝛿𝑢𝑛𝑗𝐵12(0,1)+𝑢𝑛𝑗𝐵12(0,1)𝐶2,𝑛𝑛0,(4.8)where 𝐶2=exp𝑇𝐶1+1𝛿𝑢𝑛0𝐵12(0,1)+𝑢𝑛0𝐵12(0,1)+𝑇𝐶1×expexp𝑇𝐶1+1𝑇𝐶1,(4.9)and so our proof is complete.

We address now the question of convergence and existence.

5. Convergence and Existence

Now let us introduce the Rothe function 𝑢𝑛(𝑡)𝐼𝑉 obtained from the functions 𝑢𝑗 by piecewise linear interpolation with respect to time

𝑢𝑛(𝑡)=𝑢𝑛𝑗1+𝛿𝑢𝑛𝑗𝑡𝑡𝑛𝑗1,in𝐼𝑛𝑗,(5.1) as well the step functions ̃𝑢𝑛(𝑡), ̂𝑢𝑛(𝑡), 𝑓𝑛(𝑡), and ̃𝑘(𝑡,̃𝑢𝑛(𝑡)) defined as follows:


Corollary 5.1. There exist 𝐶>0 such that the estimates 𝑢𝑛(𝑡)𝐶,̃𝑢𝑛(𝑡)𝐶,𝑡𝐼,(5.5)𝑑𝑢𝑛𝑑𝑡(𝑡)𝐵12(0,1)𝐶,fora.e.𝑡𝐼,(5.6)̃𝑢𝑛(𝑡)𝑢𝑛(𝑡)𝐵12(0,1)𝐶𝑛,̂𝑢𝑛(𝑡)𝑢𝑛(𝑡)𝐵12(0,1)𝐶𝑛̃𝑘,𝑡𝐼,(5.7)𝑛(𝑡)𝐶,𝑡𝐼,(5.8) hold for all 𝑛.

Proof. For the inequalities (5.5), (5.6), and (5.7) see [6, Corollary  4.2.], whereas for the last inequality, assumption (𝐻2) and estimate (4.1) guarantee the desired result.

Proposition 5.2. The sequence (𝑢𝑛)𝑛 converges in the norm of the space 𝐶(𝐼,𝐵12(0,1)) to some function 𝑢𝐶(𝐼,𝐵12(0,1)) and the error estimate 𝑢𝑛𝑢𝐶(𝐼,𝐵12(0,1))𝐶𝑛(5.9) takes place for all 𝑛𝑛0.

Proof. By virtue of (5.2), (5.3), and (5.4) the variational equation (3.9) may be rewritten in the form 𝑑𝑢𝑛𝑑𝑡(𝑡),𝑣𝐵12(0,1)+̃𝑢𝑛=̃𝑘(𝑡),𝑣𝑛(𝑡),𝑣𝐵12(0,1)+𝑓𝑛(𝑡),𝑣𝐵12(0,1),(5.10) for a.e. 𝑡[0,𝑇].In view of (5.10), using (5.6) and (5.8) with the fact that 𝑓𝑛(𝑡)𝐵12(0,1)𝑀2=max𝑡𝐼(𝑓𝑡)𝐵12(0,1)<,(5.11)we obtain ||̃𝑢𝑛||̃𝑘(𝑡),𝑣𝑛(𝑡)𝐵12(0,1)+𝑓𝑛(𝑡)𝐵12(0,1)+𝑑𝑢𝑛𝑑𝑡(𝑡)𝐵12(0,1)𝑣𝐵12(0,1)𝐶𝑣𝐵12(0,1)[].,a.e.𝑡0,𝑇(5.12) Now, for 𝑛, 𝑚 being two positive integers, testing the difference (5.10)𝑛-(5.10)𝑚 with 𝑣=𝑢𝑛(𝑡)𝑢𝑚(𝑡) which is in 𝑉, with the help of the Cauchy-Schwarz inequality and taking into account that 2𝑑𝑑𝑡𝑢(𝑡),𝑢(𝑡)𝐵12(0,1)=𝑑𝑑𝑡𝑢(𝑡)2𝐵12(0,1)[],a.e.𝑡0,𝑇,(5.13)and, by virtue of (5.12) we obtain after some rearrangements 12𝑑𝑑𝑡𝑢𝑛(𝑡)𝑢𝑚(𝑡)2𝐵12(0,1)+̃𝑢𝑛(𝑡)̃𝑢𝑚(𝑡)2𝑢𝐶𝑚(𝑡)̃𝑢𝑚(𝑡)𝐵12(0,1)+𝐶̃𝑢𝑛(𝑡)𝑢𝑛(𝑡)𝐵12(0,1)+̃𝑘𝑛̃𝑘(𝑡)𝑚(𝑡)𝐵12(0,1)𝑢𝑛(𝑡)𝑢𝑚(𝑡)𝐵12(0,1)+𝑓𝑛𝑓(𝑡)𝑚(𝑡)𝐵12(0,1)𝑢𝑛(𝑡)𝑢𝑚(𝑡)𝐵12(0,1)[].,a.e.𝑡0,𝑇(5.14) To derive the required result, we need to estimate the third and the last terms in the right-hand side, for this, let 𝑡 be arbitrary but fixed in (0,𝑇], without loss of generality we can suppose that there exist three positive integers 𝑝,𝑞 and 𝛽, such that 𝑡𝑡𝑛𝑝1,𝑡𝑛𝑝𝑡𝑚𝑞1,𝑡𝑚𝑞,𝑛=𝛽𝑚,𝑡𝑛𝑝=𝑡𝑚𝑞.(5.15)Hence, using (5.4) we can write ̃𝑘𝑛̃𝑘(𝑡)𝑚(𝑡)𝐵12(0,1)=𝑚𝑝1𝑗=0𝛽(𝑗+1)1𝑖=𝑗𝛽𝑎𝑡𝑛𝑝𝑡𝑛𝑗𝑘𝑡𝑛𝑗,𝑢𝑛𝑗𝑡𝑎𝑚𝑞𝑡𝑚𝑖𝑘𝑡𝑚𝑖,𝑢𝑚𝑖𝐵12(0,1).(5.16)By virtue of assumption (𝐻1) and the fact that |𝑎(𝑡𝑛𝑝𝑡𝑛𝑗)𝑎(𝑡𝑚𝑞𝑡𝑚𝑖)|𝐶𝑛, there exist 𝜀𝑛[0,𝐶𝑛] such that ̃𝑘𝑛̃𝑘(𝑡)𝑚(𝑡)𝐵12(0,1)𝑚𝑝1𝑗=0𝛽(𝑗+1)1𝑖=𝑗𝛽𝐶𝑛𝜀𝑛𝑘𝑡𝑛𝑗,𝑢𝑛𝑗𝐵12(0,1)+||𝑎𝑡𝑚𝑞𝑡𝑚𝑖||𝑘𝑡𝑛𝑗,𝑢𝑛𝑗𝑡𝑘𝑚𝑖,𝑢𝑚𝑖𝐵12(0,1).(5.17)Therefore, recalling assumptions (𝐻1), (𝐻2) and having in mind that 𝜀𝑛[0,𝐶𝑛], we estimate ̃𝑘𝑛̃𝑘(𝑡)𝑚(𝑡)𝐵12(0,1)𝑚𝑝1𝑗=0𝛽(𝑗+1)1𝑖=𝑗𝛽𝐶𝑛+𝐶𝑛+𝑢𝑛𝑗𝑢𝑚𝑖𝐵12(0,1),(5.18)from where, we derive for all 𝑠(𝑡𝑚𝑖,𝑡𝑚𝑖+1]̃𝑘𝑛̃𝑘(𝑡)𝑚(𝑡)𝐵12(0,1)𝑚𝑝1𝑗=0𝛽(𝑗+1)1𝑖=𝑗𝛽𝐶𝑛+𝐶𝑛+̃𝑢𝑛(𝑠)𝑢𝑛(𝑠)𝐵12(0,1)+𝑢𝑛(𝑠)𝑢𝑚(𝑠)𝐵12(0,1)+𝑢𝑚(𝑠)̃𝑢𝑚(𝑠)𝐵12(0,1).(5.19)Taking the supremum with respect to 𝑠 from 0 to 𝑡 in the right-hand side, invoking the fact that 𝑠(𝑡𝑚𝑖,𝑡𝑚𝑖+1](𝑡𝑛𝑗1,𝑡𝑛𝑗] and estimate (5.7), we obtain ̃𝑘𝑛̃𝑘(𝑡)𝑚(𝑡)𝐵12(0,1)𝑚𝑞1𝑖=0𝐶𝑛+𝐶sup0𝑠𝑡𝑢𝑛(𝑠)𝑢𝑚(𝑠)𝐵12(0,1),(5.20)so that ̃𝑘𝑛̃𝑘(𝑡)𝑚(𝑡)𝐵12(0,1)𝐶𝑛+𝐶sup0𝑠𝑡𝑢𝑛(𝑠)𝑢𝑚(𝑠)𝐵12(0,1).(5.21) Let 𝑡(𝑡𝑛𝑝1,𝑡𝑛𝑝](𝑡𝑚𝑞1,𝑡𝑚𝑞], from assumption (𝐻1) it follows that 𝑓𝑛(𝑓𝑡)𝑚(𝑡)𝐵12(0,1)=𝑓𝑡𝑛𝑝𝑡𝑓𝑚𝑞𝐵12(0,1)𝑙1||𝑡𝑛𝑝𝑡𝑚𝑞||𝑙1𝑛.(5.22) Ignoring the second term in the left-hand side of (5.14) which is clearly positive and using estimates (5.5), (5.7), (5.21), and (5.22) yield 𝑑𝑑𝑡𝑢𝑛(𝑡)𝑢𝑚(𝑡)2𝐵12(0,1)𝐶𝑛+𝑚+𝐶sup0𝑠𝑡𝑢𝑛(𝑠)𝑢𝑚(𝑠)2𝐵12(0,1)[],a.e.𝑡0,𝑇.(5.23)Integrating this inequality with respect to time from 0 to 𝑡 and invoking the fact that 𝑢𝑛(0)=𝑢𝑚(0)=𝑈0, we get 𝑢𝑛(𝑡)𝑢𝑚(𝑡)2𝐵12(0,1)𝐶𝑛+𝑚+𝐶𝑡0sup0𝜉𝑡𝑢𝑛(𝜉)𝑢𝑚(𝜉)2𝐵12(0,1)𝑑𝜉,(5.24)whence sup0𝑠𝑡𝑢𝑛(𝑠)𝑢𝑚(𝑠)2𝐵12(0,1)𝐶𝑛+𝑚+𝐶𝑡0sup0𝜉𝑡𝑢𝑛(𝜉)𝑢𝑚(𝜉)2𝐵12(0,1)𝑑𝜉.(5.25)Accordingly, by Gronwall's lemma we obtain sup0𝑠𝑡𝑢𝑛(𝑠)𝑢𝑚(𝑠)2𝐵12(0,1)𝐶𝑛+𝑚[]exp(𝐶𝑡),𝑡0,𝑇,(5.26)consequently sup0𝑠𝑡𝑢𝑛(𝑠)𝑢𝑚(𝑠)𝐵12(0,1)𝐶𝑛+𝑚(5.27) takes place for all 𝑛,𝑚. This implies that (𝑢𝑛(𝑡))𝑛 is a Cauchy sequence in the Banach space 𝐶(𝐼,𝐵12(0,1)), and hence it converges in the norm of this latter to some function 𝑢𝐶(𝐼,𝐵12(0,1)). Besides, passing to the limit 𝑚 in (5.27), we obtain the desired error estimate, which finishes the proof.

Now, we present some properties of the obtained solution.

The limit-function 𝑢 from Proposition  5.2, possesses the following properties:

(i)𝑢𝐶(𝐼,𝐵12(0,1))𝐿(𝐼,𝑉)),(ii)𝑢 is strongly differentiable a.e. in 𝐼 and 𝑑𝑢/𝑑𝑡𝐿(𝐼,𝐵12(0,1)),(iii)̃𝑢𝑛(𝑡)𝑢(𝑡) in 𝐵12(0,1) for all 𝑡𝐼,(iv)𝑢𝑛(𝑡), ̃𝑢𝑛(𝑡)𝑢(𝑡) in 𝑉 for all 𝑡𝐼,(v)(𝑑𝑢𝑛/𝑑𝑡)(𝑡)(𝑑𝑢/𝑑𝑡)(𝑡) in 𝐿2(𝐼,𝐵12(0,1)).

Proof. On the basis of estimates (5.5) and (5.6), uniform convergence statement from Proposition  5.2, and the continuous embedding 𝑉𝐵12(0,1), the assertions of the present theorem are a direct consequence of [2, Lemma  1.3.15].

Theorem 5.3. Under Assumptions (𝐻1)-(𝐻3), (2.3) admits a unique weak solution, namely, the limit function 𝑢 from Proposition  5.2, in the sense of Definition  2.3.

Proof. We have to show that the limit function 𝑢 satisfies all the conditions (i), (ii), (iii), and (iv) of Definition  2.3. Obviously, in light of the properties of the function 𝑢 listed in Theorem 5.3, the first two conditions of Definition  2.3 are already seen. On the other hand, since 𝑢𝑛𝑢 in 𝐶(𝐼,𝑉) as 𝑛 and, by construction, 𝑢𝑛(0)=𝑈0, it follows that 𝑢(0)=𝑈0, so the initial condition is also fulfilled, that is, Definition  2.3(iii) takes place. It remains to see that the integral identity (2.21) is obeyed by 𝑢. For this, integrating (5.10) over (0,𝑡) and using the fact that 𝑢𝑛(0)=𝑈0, we get 𝑢𝑛(𝑡)𝑈0,𝑣𝐵12(0,1)+𝑡0̃𝑢𝑛(𝜏),𝑣𝑑𝜏=𝑡0̃𝑘𝜏,̃𝑢𝑛(𝜏),𝑣𝐵12(0,1)𝑑𝜏+𝑡0𝑓𝑛(𝜏),𝑣𝐵12(0,1)𝑑𝜏,(5.28)consequently, after some rearrangements 𝑢𝑛(𝑡)𝑈0,𝑣𝐵12(0,1)+𝑡0̃𝑢𝑛(=𝜏),𝑣𝑑𝜏𝑡0𝜏0𝑎(𝜏𝑠)𝑘(𝑠,𝑢(𝑠))𝑑𝑠,𝑣𝐵12(0,1)𝑑𝜏+𝑡0(𝑓(𝜏),𝑣)𝐵12(0,1)+𝑑𝜏𝑡0̃𝑘𝜏,̃𝑢𝑛(𝜏)𝜏0𝑎(𝜏𝑠)𝑘(𝑠,𝑢(𝑠))𝑑𝑠,𝑣𝐵12(0,1)+𝑑𝜏𝑡0𝑓𝑛(𝜏)𝑓(𝜏),𝑣𝐵12(0,1)𝑑𝜏.(5.29) Let ̂𝑠𝑛𝐼𝐼 and ̂𝑠𝑛𝐼𝐼 denote the functions ̂𝑠𝑛𝑡(𝑡)=0,for𝑡=0,𝑛𝑗1𝐼,in𝑛𝑗,̃𝑠𝑛𝑡(𝑡)=0,for𝑡=0,𝑛𝑗𝐼,in𝑛𝑗.(5.30) To investigate the desired result, we prove some convergence statements. Using (5.2), (5.4), and (5.30) we have for all𝑡(𝑡𝑛𝑗1,𝑡𝑛𝑗]̃𝑘𝑡,̃𝑢𝑛(𝑡)𝑡0=𝑎(𝑡𝑠)𝑘(𝑠,𝑢(𝑠))𝑑𝑠𝑡𝑛𝑗0𝑎𝑡𝑛𝑗̂𝑠𝑛𝑘(𝑠)̂𝑠𝑛(𝑠),̂𝑢𝑛(𝑠)𝑎(𝑡𝑠)𝑘(𝑠,𝑢(𝑠))𝑑𝑠+𝑡𝑛𝑗𝑡𝑎(𝑡𝑠)𝑘(𝑠,𝑢(𝑠))𝑑𝑠.(5.31) Taking into account (5.5), (5.9), and assumptions (𝐻1), (𝐻2) it follows that 𝑎𝑡𝑛𝑗̂𝑠𝑛𝑘(𝑠)̂𝑠𝑛(𝑠),̂𝑢𝑛(𝑠)𝑎(𝑡𝑠)𝑘(𝑠,𝑢(𝑠))𝐵12(0,1)𝐶𝑛.(5.32) Thanks to (5.31) and (5.32) we obtain ̃𝑘𝑡,̃𝑢𝑛(𝑡)𝑡0𝑎(𝑡𝑠)𝑘(𝑠,𝑢(𝑠))𝑑𝑠𝐵12(0,1)𝐶𝑛.(5.33) On the other hand, in view of the assumed Lipschitz continuity of 𝑓, we have 𝑓𝑛(𝜏)𝑓(𝜏)𝐵12(0,1)𝑓̃𝑠𝑛(𝜏)𝑓(𝜏)𝐵12(0,1)𝑙1𝑛.(5.34) Now, the sequences {(̃𝑢𝑛(𝜏),𝑣)}, 𝑓{(𝑛(𝜏),𝑣)𝐵12(0,1)}, and ̃{(𝑘(𝜏,̃𝑢𝑛(𝜏)),𝑣)𝐵12(0,1)} are uniformly bounded with respect to both 𝜏 and 𝑛, so the Lebesgue theorem of majorized convergence is applicable to (5.29). Thus, having in mind (5.7), (5.9), (5.33), and (5.34), we derive that 𝑢(𝑡)𝑈0,𝑣𝐵12(0,1)+𝑡0(=𝑢(𝜏),𝑣)𝑑𝜏𝑡0𝜏0𝑎(𝜏𝑠)𝑘(𝑠,𝑢(𝑠))𝑑𝑠,𝑣𝐵12(0,1)𝑑𝜏+𝑡0(𝑓(𝜏),𝑣)𝐵12(0,1)𝑑𝜏(5.35) takes place for all 𝑣𝑉 and 𝑡[0,𝑇]. Finally, differentiating (5.35) with respect to 𝑡, we get 𝑑𝑑𝑡𝑢(𝑡),𝑣𝐵12(0,1)=+(𝑢(𝑡),𝑣)𝑡0𝑎(𝑡𝑠)𝑘(𝑠,𝑢(𝑠))𝑑𝑠,𝑣𝐵12(0,1)+(𝑓(𝑡),𝑣)𝐵12(0,1)[].,a.e.𝑡0,𝑇(5.36)The uniqueness may be argued in the usual manner. Indeed, exploiting an idea in [11], consider 𝑢1 and 𝑢2 two different solutions of (2.3), and define 𝑤=𝑢1𝑢2 then, we have 𝑑𝑑𝑡𝑤(𝑡),𝑣𝐵12(0,1)+(𝑤(𝑡),𝑣)=𝑡0𝑘𝑎(𝑡𝑠)𝑠,𝑢1(𝑠)𝑘𝑠,𝑢2(𝑠)𝑑𝑠,𝑣𝐵12(0,1).(5.37)Choosing 𝑣=𝑤(𝑡) as a test function, with the aid of Cauchy-Schwarz inequality and assumption (𝐻1), we obtain 12𝑑(𝑑𝑡𝑤𝑡)2𝐵12(0,1)+𝑤(𝑡)2𝐶𝑡0𝑘𝑠,𝑢1(𝑠)𝑘𝑠,𝑢2(𝑠)𝐵12(0,1)𝑑𝑠𝑤(𝑡)𝐵12(0,1).(5.38) Let 𝜉[0,𝑝] such that 𝑤(𝜉)𝐵12(0,1)=max[]𝑠0,𝑝𝑤(𝑠)𝐵12(0,1),(5.39) integrating (5.38) over (0,𝑝), 0𝑝𝑇, using (5.39), and invokingassumption (𝐻2), we get 𝑝012𝑑𝑑𝑡𝑤(𝑡)2𝐵12(0,1)+𝑤(𝑡)2𝑑𝑡𝐶𝑝2𝑤(𝜉)2𝐵12(0,1),(5.40)consequently, with the fact that 𝑤(0)=0𝑝012𝑑𝑑𝑡𝑤(𝑡)2𝐵12(0,1)+𝑤(𝑡)2𝑑𝑡𝐶𝑝2𝜉0𝑑𝑑𝑡𝑤(𝑡)2𝐵12(0,1)𝑑𝑡.(5.41) Choosing 𝑝 as a constant verifying the condition 𝛼,𝑇=𝛼𝑝,𝐶𝑝212,(5.42)we have, by virtue of (5.41) 𝑝012𝑑𝑑𝑡𝑤(𝑡)2𝐵12(0,1)𝑑𝑡+𝑝0𝑤(𝑡)2𝑑𝑡𝜉012𝑑𝑑𝑡𝑤(𝑡)2𝐵12(0,1)𝑑𝑡,(5.43)taking into account that 𝜉𝑝, we obtain []𝑤(𝑡)=0,on0,𝑝.(5.44)Following the same lines as for [0,𝑝], we deduce that []𝑤(𝑡)=0,on𝑖𝑝,(𝑖+1)𝑝,𝑖=1,2,3,,(5.45)therefore, we derive 𝑤(𝑡)0, on [0,𝑇], then 𝑢1𝑢2. This achives the proof.


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