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Journal of Applied Mathematics and Stochastic Analysis
Volume 2009 (2009), Article ID 975601, 13 pages
http://dx.doi.org/10.1155/2009/975601
Research Article

A Boundary Value Problem with Multivariables Integral Type Condition for Parabolic Equations

1Laboratory Equations Différentielles, Departement of Mathematics, University Mentouri Constantine, Constantine 25017, Algeria
2Department of Mathematics, Science University of 08 Mai 45, P.O. Box 401, Guelma 24000, Algeria

Received 27 January 2009; Revised 14 May 2009; Accepted 13 October 2009

Academic Editor: Sergiu Aizicovici

Copyright © 2009 A. L. Marhoune and F. Lakhal. 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.

Abstract

We study a boundary value problem with multivariables integral type condition for a class of parabolic equations. We prove the existence, uniqueness, and continuous dependence of the solution upon the data in the functional wieghted Sobolev spaces. Results are obtained by using a functional analysis method based on two-sided a priori estimates and on the density of the range of the linear operator generated by the considered problem.

1. Introduction

Certain problems of modern physics and technology can be effectively described in terms of nonlocal problems with integral conditions for partial differential equations.These nonlocal conditions arise mainly when the data on the boundary cannot be measured directly. Motivated by this, we consider in the rectangular domain Ω=(0,1)×(0,𝑇), the following nonclassical boundary value problem of finding a solution 𝑢(𝑥,𝑡) such that 𝑢=𝜕𝑢𝜕𝜕𝑡𝑎(𝑡)2𝑢𝜕𝑥2=𝑓(𝑥,𝑡),(1.1) where the function 𝑎(𝑡) and its derivative are bounded on the interval [0,𝑇]

0<𝑐0𝑎(𝑡)𝑐1,0<𝑐2𝑑𝑎(𝑡)𝑑𝑡𝑐3,(1.2)𝑙𝑢=𝑢(𝑥,0)=𝜑(𝑥),𝑥(0,1),(1.3)𝑢(0,𝑡)=𝑢(𝛽,𝑡)=𝑢(𝛾,𝑡)=𝑢(1,𝑡),𝑡(0,𝑇),(1.4)𝛼0𝑢(𝑥,𝑡)𝑑𝑥+2𝛾𝛽𝑢(𝑥,𝑡)𝑑𝑥+1𝛿𝑢(𝑥,𝑡)𝑑𝑥=0,0<𝛼<𝛽<𝛾<𝛿<1,𝛼=1𝛿=𝛾𝛽,𝑡(0,𝑇).(1.5) Here, we assume that the known function 𝜑 satisfies the conditions given in (1.4) and (1.5), that is,

𝜑(0)=𝜑(𝛽)=𝜑(𝛾)=𝜑(1),𝛼0𝜑(𝑥)𝑑𝑥+2𝛾𝛽𝜑(𝑥)𝑑𝑥+1𝛿𝜑(𝑥)𝑑𝑥=0.(1.6) When considering the classical solution of the problem (1.1)–(1.5), along with (1.5), there should be the fulfilled conditions:

𝑎𝜑(0)(0)+𝜑(𝛽)𝜑(𝛾)𝜑(1)=𝑓(1,0)+𝑓(𝛾,0)𝑓(𝛽,0)𝑓(0,0),𝑎(0)𝛼0𝜑(𝑥)𝑑𝑥+2𝛾𝛽𝜑(𝑥)𝑑𝑥+1𝛿𝜑(=𝑥)𝑑𝑥𝛼0𝑓(𝑥,0)𝑑𝑥+2𝛾𝛽𝑓(𝑥,0)𝑑𝑥+1𝛿𝑓(𝑥,0)𝑑𝑥.(1.7)

Mathematical modelling of different phenomena leads to problems with nonlocal or integral boundary conditions. Such a condition occurs in the case when one measures an averaged value of some parameter inside the domain. This amounts to the specification of the energy or mass contained in a portion of the conductor or porous medium as a function of time. This problems arise in plasma physics, heat conduction, biology and demography, as well as modelling of technological process, see, for example, [15]. Boundary-value problems for parabolic equations with integral boundary condition are investigated by Batten [6], Bouziani and Benouar [7], Cannon [8, 9], Cannon, Perez Esteva and van der Hoek [10], Ionkin [11], Kamynin [12], Shi and Shillor [13], Shi [4], Marhoune and Bouzit [14], Marhoune and Hameida [15], Yurchuk [16], and many references therein. The problem with one-variable (resp., two-variable) boundary integral type condition is studied in [5] and by Marhoune and Latrous [17] (resp., in Marhoune [2]).

Mention that in the cited paper [16], the author proved the existence, uniqueness, and continuous dependence of a stronge solution in weighted Sobolev spaces to the problem

𝜕𝑢=𝜕𝜕𝑡𝜕𝑥𝑎(𝑥,𝑡)𝜕𝑢𝜕𝑥+𝑓(𝑥,𝑡),(1.8) under the following conditions:

𝑢𝑢(𝑥,0)=0,0𝑥1,(0,𝑡)=0,0<𝑡𝑇,10𝑢(𝑥,𝑡)𝑑𝑥=0.(1.9) This last integral condition in the form

10𝑢(𝑥,𝑡)𝑑𝑥=𝑚(𝑡),0<𝑡𝑇,(1.10) arises, for example, in biochemistry in which 𝑚 is a constant, and in this case is known as the conservation of protein [18]. Further, in [5], the author studied a similar problem with the weak integral condition

𝛼0𝑢(𝑥,𝑡)𝑑𝑥=0,0<𝛼<1.(1.11) The same problem with the new integral condition

𝛼0𝑢(𝑥,𝑡)𝑑𝑥+1𝛽𝑢(𝑥,𝑡)𝑑𝑥=0,𝛼+𝛽=1,(1.12) was investigated in [2]. The present paper is an extension in the same direction. By constructing a suitable multiplicator, we will try to establish existence and uniqueness of solution of problem (1.1)–(1.5). Note that the multivariables integral type condition (1.5) is considerably much weaker and better than that used in [2]. In fact, some physical problems have motivated specialists to consider nonlocal integral condition (1.5), which tells us the integral total effect of the solution 𝑢 over several independent portions [0,𝛼], [𝛽,𝛾], and [𝛿,1] of interval 𝐼=(0,1) at certain time 𝑡 that give this effect over the entire or part of this interval.

We associate with (1.1)–(1.5) the operator 𝐿=(,𝑙), defined from 𝐸 into 𝐹, where 𝐸 is the Banach space of functions 𝑢𝐿2(Ω), satisfying (1.4) and (1.5), with the finite norm 𝑢2𝐸=Ω𝑝(𝑥)3|||𝜕𝑢|||𝜕𝑡2+||||𝜕2𝑢𝜕𝑥2||||2𝑑𝑥𝑑𝑡+sup0𝑡𝑇10𝑝(𝑥)2|||𝜕𝑢|||𝜕𝑥2𝑑𝑥+sup0𝑡𝑇𝛼0|𝑢|2𝑑𝑥+sup0𝑡𝑇𝛾𝛽|𝑢|2𝑑𝑥+sup0𝑡𝑇1𝛿|𝑢|2𝑑𝑥,(1.13) and 𝐹 is the Hilbert space of vector-valued functions =(𝑓,𝜑) obtained by completion of the space 𝐿2(Ω)×𝑊22(0,1) with respect to the norm

2𝐹=(𝑓,𝜑)2𝐹=Ω𝑝(𝑥)3||𝑓||2𝑑𝑥𝑑𝑡+10𝑝(𝑥)2|||𝑑𝜑|||𝑑𝑥2𝑑𝑥+𝛼0||𝜑||2𝑑𝑥+𝛾𝛽||𝜑||2𝑑𝑥+1𝛿||𝜑||2𝑑𝑥,(1.14) where

𝑥𝑝(𝑥)=2]],,𝑥0,𝛼(𝛾𝛽)2[][],,𝑥𝛼,𝛽𝛾,𝛿(𝛾𝑥)2+(𝑥𝛽)2[],,𝑥𝛽,𝛾(1𝑥)2[[.,𝑥𝛿,1(1.15) Using the energy inequalities method proposed in [16], we establish two-sided a priori estimates. Then, we prove that the operator 𝐿 is a linear homeomorphism between the spaces 𝐸 and 𝐹.

2. Two-Sided A Priori Estimates

Theorem 2.1. For any function 𝑢𝐸, one has the a priori estimate 𝐿𝑢2𝐹𝑐4𝑢2𝐸,(2.1) where the constant 𝑐4 is independent of 𝑢. In fact, 𝑐4=2max(1,𝑐21).

Proof. Using (1.1) and initial condition (1.3), we obtain Ω𝑝(𝑥)3||||𝑢2𝑑𝑥𝑑𝑡2Ω𝑝(𝑥)3|||𝜕𝑢|||𝜕𝑡2+𝑐21||||𝜕2𝑢𝜕𝑥2||||2𝑑𝑥𝑑𝑡,10𝑝(𝑥)2|||𝑑𝜑|||𝑑𝑥2𝑑𝑥sup0𝑡𝑇10𝑝(𝑥)2|||𝜕𝑢|||𝜕𝑥2𝑑𝑥,𝛼0||𝜑||2𝑑𝑥sup0𝑡𝑇𝛼0|𝑢|2𝑑𝑥,𝛾𝛽||𝜑||2𝑑𝑥sup0𝑡𝑇𝛾𝛽|𝑢|2𝑑𝑥,1𝛿||𝜑||2𝑑𝑥sup0𝑡𝑇1𝛿|𝑢|2𝑑𝑥.(2.2) Combining the inequalities in (2.2), we obtain (2.1) for 𝑢𝐸.

Theorem 2.2. For any function 𝑢𝐸, one has the a priori estimate 𝑢2𝐸𝑐5𝐿𝑢2𝐹,(2.3) with the constant 𝑐5=exp(𝑐𝑇)max49,2𝑐1min13/32,𝑐0,𝑐20,/2(2.4) and 𝑐 is such that 𝑐𝑐0𝑐3.(2.5)

Before proving this theorem, we need the following lemma.

Lemma 2.3 (see [19]). For 𝑢𝐸, one has 𝑏𝑎||||𝑏𝑥𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2𝑑𝑥4𝑏𝑎(𝑥𝑎)2|||𝜕𝑢|||𝜕𝑡2𝑑𝑥,𝑏𝑎||||𝑥𝑎𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2𝑑𝑥4𝑏𝑎(𝑏𝑥)2|||𝜕𝑢|||𝜕𝑡2𝑑𝑥.(2.6)

Proof of Theorem 2.2. Define 𝑥𝑀𝑢=2𝜕𝑢𝜕𝑡+2𝑥𝛼𝑥𝜕𝑢(𝜉,𝑡)]],𝜕𝑡𝑑𝜉,𝑥0,𝛼(𝛾𝛽)2𝜕𝑢[][],𝜕𝑡,𝑥𝛼,𝛽𝛾,𝛿(𝛾𝑥)2𝜕𝑢𝜕𝑡+(𝑥𝛽)2𝜕𝑢𝜕𝑡+2(𝛾𝑥)𝑥𝛽𝜕𝑢(𝜉,𝑡)𝜕𝑡𝑑𝜉+2(𝑥𝛽)𝛾𝑥𝜕𝑢(𝜉,𝑡)[],𝜕𝑡𝑑𝜉,𝑥𝛽,𝛾(1𝑥)2𝜕𝑢𝜕𝑡+2(1𝑥)𝑥𝛿𝜕𝑢(𝜉,𝑡)[[.𝜕𝑡𝑑𝜉,𝑥𝛿,1(2.7) We consider for 𝑢𝐸 the quadratic formula Re𝜏010exp(𝑐𝑡)𝑢𝑀𝑢𝑑𝑥𝑑𝑡,(2.8) with the constant 𝑐 satisfying (2.5), obtained by multiplying (1.1) by exp(𝑐𝑡)𝑀𝑢, by integrating over Ω𝜏, where Ω𝜏=(0,1)×(0,𝜏), with 0𝜏𝑇, and by taking the real part. Integrating by parts in (2.8) by report to 𝑥 with the use of boundary conditions (1.4) and (1.5), we obtain Re𝜏010exp(𝑐𝑡)𝑢=𝑀𝑢𝑑𝑥𝑑𝑡𝜏010|||𝑝(𝑥)exp(𝑐𝑡)𝜕𝑢|||𝜕𝑡2𝑑𝑥𝑑𝑡+𝜏0𝛼0||||exp(𝑐𝑡)𝛼𝑥𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2+𝑑𝑥𝑑𝑡𝜏0𝛾𝛽||||exp(𝑐𝑡)𝑥𝛽𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2+||||𝛾𝑥𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2+𝑑𝑥𝑑𝑡𝜏01𝛿||||exp(𝑐𝑡)𝑥𝛿𝜕𝑢(𝜉,𝑡)||||𝜕𝑡𝑑𝜉2𝑑𝑥𝑑𝑡+Re𝜏010exp(𝑐𝑡)𝑝(𝑥)𝑎𝜕𝑢𝜕𝜕𝑥2𝑢𝜕𝑥𝜕𝑡𝑑𝑥𝑑𝑡+2Re𝜏0𝛼0𝜕exp(𝑐𝑡)𝑎𝑢𝑢𝜕𝑡𝑑𝑥𝑑𝑡+4Re𝜏0𝛾𝛽𝜕exp(𝑐𝑡)𝑎𝑢𝑢𝜕𝑡𝑑𝑥𝑑𝑡+2Re𝜏01𝛿𝜕exp(𝑐𝑡)𝑎𝑢𝑢𝜕𝑡𝑑𝑥𝑑𝑡.(2.9) On the other hand, by using the elementary inequalities we get Re𝜏010exp(𝑐𝑡)𝑢𝑀𝑢𝑑𝑥𝑑𝑡𝜏010𝑝|||(𝑥)exp(𝑐𝑡)𝜕𝑢|||𝜕𝑡2𝑑𝑥𝑑𝑡+Re𝜏010exp(𝑐𝑡)𝑝(𝑥)𝑎𝜕𝑢𝜕𝜕𝑥2𝑢𝜕𝑥𝜕𝑡𝑑𝑥𝑑𝑡+2Re𝜏0𝛼0𝜕exp(𝑐𝑡)𝑎𝑢𝑢𝜕𝑡𝑑𝑥𝑑𝑡+4Re𝜏0𝛾𝛽𝜕exp(𝑐𝑡)𝑎𝑢𝑢𝜕𝑡𝑑𝑥𝑑𝑡+2Re𝜏01𝛿𝜕exp(𝑐𝑡)𝑎𝑢𝑢𝜕𝑡𝑑𝑥𝑑𝑡.(2.10) Again, integrating by parts the second, third, fourth, and fifth terms of the right-hand side of the inequality (2.10) by report to 𝑡 and taking into account the initial condition (1.3) and (2.5) gives Re𝜏010exp(𝑐𝑡)𝑝(𝑥)𝑎𝜕𝑢𝜕𝜕𝑥2𝑢𝜕𝑥𝜕𝑡𝑑𝑥𝑑𝑡10exp(𝑐𝜏)2|||𝑝(𝑥)𝑎(𝜏)𝜕𝑢(𝑥,𝜏)|||𝜕𝑥21𝑑𝑥210𝑝|||(𝑥)𝑎(0)𝑑𝑙𝑢|||𝑑𝑥2𝑑𝑥,Re𝜏0𝛼0𝜕exp(𝑐𝑡)𝑎𝑢𝑢𝜕𝑡𝑑𝑥𝑑𝑡𝛼0exp(𝑐𝜏)2||||𝑎(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥𝛼0𝑎(0)2||||𝑙𝑢2𝑑𝑥;Re𝜏0𝛾𝛽𝜕exp(𝑐𝑡)𝑎𝑢𝑢𝜕𝑡𝑑𝑥𝑑𝑡𝛾𝛽exp(𝑐𝜏)2||||𝑎(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥𝛾𝛽𝑎(0)2||||𝑙𝑢2𝑑𝑥,Re𝜏01𝛿𝜕exp(𝑐𝑡)𝑎𝑢𝑢𝜕𝑡𝑑𝑥𝑑𝑡1𝛿exp(𝑐𝜏)2||||𝑎(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥1𝛿𝑎(0)2||||𝑙𝑢2𝑑𝑥.(2.11) Using (2.11) in (2.10), we get Re𝜏010exp(𝑐𝑡)𝑢𝑀𝑢𝑑𝑥𝑑𝑡+𝛼0||||𝑎(0)𝑙𝑢2𝑑𝑥+2𝛾𝛽||||𝑎(0)𝑙𝑢2+𝑑𝑥1𝛿||||𝑎(0)𝑙𝑢21𝑑𝑥+210|||𝑝(𝑥)𝑎(0)𝑑𝑙𝑢|||𝑑𝑥2𝑑𝑥𝜏010|||𝑝(𝑥)exp(𝑐𝑡)𝜕𝑢|||𝜕𝑡2+1𝑑𝑥𝑑𝑡210|||exp(𝑐𝜏)𝑝(𝑥)𝑎(𝜏)𝜕𝑢(𝑥,𝜏)|||𝜕𝑥2𝑑𝑥+𝛼0||||exp(𝑐𝜏)𝑎(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥+2𝛾𝛽||||exp(𝑐𝜏)𝑎(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥+1𝛿||||exp(𝑐𝜏)𝑎(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥.(2.12) By using the 𝜀-inequalities on the first integral in the left-hand side of (2.12) and Lemma 2.3, we obtain 1532𝜏010|||𝑝(𝑥)exp(𝑐𝑡)𝜕𝑢|||𝜕𝑡21𝑑𝑥𝑑𝑡+210|||exp(𝑐𝜏)𝑝(𝑥)𝑎(𝜏)𝜕𝑢(𝑥,𝜏)|||𝜕𝑥2+𝑑𝑥𝛼0||||exp(𝑐𝜏)𝑎(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥+2𝛾𝛽||||exp(𝑐𝜏)𝑎(𝜏)𝑢(𝑥,𝜏)2+𝑑𝑥1𝛿||||exp(𝑐𝜏)𝑎(𝜏)𝑢(𝑥,𝜏)2𝑑𝑥16𝜏010||||𝑝(𝑥)exp(𝑐𝑡)𝑢2+𝑑𝑥𝑑𝑡𝛼0||||𝑎(0)𝑙𝑢21𝑑𝑥+210|||𝑝(𝑥)𝑎(0)𝑑𝑙𝑢|||𝑑𝑥2𝑑𝑥+2𝛾𝛽||||𝑎(0)𝑙𝑢2𝑑𝑥+1𝛿||||𝑎(0)𝑙𝑢2𝑑𝑥.(2.13) Now, from (1.1), we have 𝑐206𝜏010||||𝜕𝑝(𝑥)exp(𝑐𝑡)2𝑢𝜕𝑥2||||2𝑑𝑥𝑑𝑡𝜏010𝑝(𝑥)3||||exp(𝑐𝑡)𝑢2𝑑𝑥𝑑𝑡+𝜏010𝑝(𝑥)3|||exp(𝑐𝑡)𝜕𝑢|||𝜕𝑡2𝑑𝑥𝑑𝑡.(2.14) Combining inequalities (2.13) and (2.14), we get exp(𝑐𝑇)1332𝜏010𝑝(𝑥)3|||𝜕𝑢|||𝜕𝑡2𝑑𝑥𝑑𝑡+𝑐010𝑝(𝑥)2|||𝜕𝑢(𝑥,𝜏)|||𝜕𝑥2𝑑𝑥+𝑐0𝛼0||||𝑢(𝑥,𝜏)2𝑑𝑥+2𝑐0𝛾𝛽||||𝑢(𝑥,𝜏)2𝑑𝑥+𝑐01𝛿||||𝑢(𝑥,𝜏)2𝑐𝑑𝑥+206𝜏010𝑝(𝑥)3||||𝜕2𝑢𝜕𝑥2||||2𝑑𝑥𝑑𝑡49Ω𝑝(𝑥)3||||𝑢2𝑑𝑥𝑑𝑡+𝑐110𝑝(𝑥)2|||𝑑𝑙𝑢|||𝑑𝑥2𝑑𝑥+𝑐1𝛼0||||𝑙𝑢2𝑑𝑥+2𝑐1𝛾𝛽||||𝑙𝑢2𝑑𝑥+𝑐11𝛿||||𝑙𝑢2𝑑𝑥.(2.15) As the right-hand side of (2.15) is independent of 𝜏, by replacing the left-hand side by its upper bound with respect to 𝜏 in the interval [0,𝑇], we obtain the desired inequality.

3. Solvability of the Problem

From estimates (2.1) and (2.3), it follows that the operator 𝐿𝐸𝐹 is continuous and its range is closed in 𝐹. Therefore, the inverse operator 𝐿1 exists and is continuous from the closed subspace 𝑅(𝐿) onto 𝐸, which means that 𝐿 is an homeomorphism from 𝐸 onto 𝑅(𝐿). To obtain the uniqueness of solution, it remains to show that 𝑅(𝐿)=𝐹. The proof is based on the following lemma.

Lemma 3.1. Let 𝐷0(𝐿)={𝑢𝐸𝑙𝑢=0}.(3.1) If for 𝑢𝐷0(𝐿) and some 𝑤𝐿2(Ω), one has Ω𝑞(𝑥)𝑢𝑤𝑑𝑥𝑑𝑡=0,(3.2) where ]],[],[[,𝑞(𝑥)=𝑥,𝑥0,𝛼𝛾𝛽,𝑥𝛼,𝛿1𝑥,𝑥𝛿,1(3.3) then 𝑤=0.

Proof. From (3.2) we have Ω𝑞(𝑥)𝜕𝑢𝜕𝑡𝑤𝑑𝑥𝑑𝑡=Ω𝜕𝑞(𝑥)𝑎(𝑡)2𝑢𝜕𝑥2𝑤𝑑𝑥𝑑𝑡.(3.4) Now, for given 𝑤(𝑥,𝑡), we introduce the function 𝑣(𝑥,𝑡)=𝑤(𝑥,𝑡)𝛼𝑥𝑤(𝜉,𝑡)𝜉]],𝑤[],𝑑𝜉,𝑥0,𝛼(𝑥,𝑡),𝑥𝛼,𝛿𝑤(𝑥,𝑡)𝑥𝛿𝑤(𝜉,𝑡)𝜉[[.𝑑𝜉,𝑥𝛿,1(3.5) Integrating by parts with respect to 𝜉, we obtain 𝑞(𝑥)𝑤=𝑥𝑣+𝛼𝑥]],[][],𝑣(𝜉,𝑡)𝑑𝜉,𝑥0,𝛼(𝛾𝛽)𝑣,𝑥𝛼,𝛽𝛾,𝛿(𝛾𝛽)𝑣+𝛾𝛽[],𝑣(𝜉,𝑡)𝑑𝜉,𝑥𝛽,𝛾(1𝑥)𝑣+𝑥𝛿[[,𝑣(𝜉,𝑡)𝑑𝜉,𝑥𝛿,1(3.6) which implies that 𝛼0𝑣(𝜉,𝑡)𝑑𝜉+2𝛾𝛽𝑣(𝜉,𝑡)𝑑𝜉+1𝛿𝑣(𝜉,𝑡)𝑑𝜉=0.(3.7) Then, from (3.4), we obtain Ω𝜕𝑢𝜕𝑡𝑁𝑣𝑑𝑥𝑑𝑡=Ω𝐴(𝑡)𝑢𝑣𝑑𝑥𝑑𝑡,(3.8) where 𝜕𝑁𝑣=𝑞(𝑥)𝑣,𝐴(𝑡)𝑢=𝜕𝑥𝑞(𝑥)𝑎(𝑡)𝜕𝑢.𝜕𝑥(3.9) If we introduce the smoothing operators with respect to 𝑡 [16], 𝜀1=(𝐼+𝜀𝜕/𝜕𝑡)1 and (𝜀1), then these operators provide the solutions of the respective problems: 𝜀𝑑𝑔𝜀(𝑡)𝑑𝑡+𝑔𝜀(𝑡)=𝑔(𝑡),𝑔𝜀(𝑡)𝑡=0=0,(3.10)𝜀𝑑𝑔𝜀(𝑡)𝑑𝑡+𝑔𝜀(𝑡)=𝑔(𝑡),𝑔𝜀(𝑡)𝑡=𝑇=0,(3.11) and also have the following properties: for any 𝑔𝐿2(0,𝑇), the functions 𝑔𝜀=(𝜀1)𝑔 and 𝑔𝜀=(𝜀1)𝑔 are in 𝑊12(0,𝑇) such that 𝑔𝜀(𝑡)𝑡=0=0 and 𝑔𝜀(𝑡)𝑡=𝑇=0. Morever, 𝜀1 commutes with 𝜕/𝜕𝑡, so 𝑇0|𝑔𝜀𝑔|2𝑑𝑡0 and 𝑇0|𝑔𝜀𝑔|2𝑑𝑡0 for 𝜀0.
Putting 𝑢=𝑡0exp(𝑐𝜏)𝑣𝜀(𝑥,𝜏)𝑑𝜏 in (3.8), where the constant 𝑐 satisfies 𝑐𝑐0𝑐3𝜀𝑐23/𝑐00, and using (3.11), we obtain Ωexp(𝑐𝑡)𝑣𝜀𝑁𝑣𝑑𝑥𝑑𝑡=Ω𝜕𝐴(𝑡)𝑢exp(𝑐𝑡)𝑢𝜕𝑡𝑑𝑥𝑑𝑡𝜀Ω𝜕𝐴(𝑡)𝑢𝑣𝜀𝜕𝑡𝑑𝑥𝑑𝑡.(3.12)
Integrating by parts each term in the right-hand side of (3.12) and taking the real parts yield 2ReΩ𝜕𝐴(𝑡)𝑢exp(𝑐𝑡)𝑢𝜕𝑡𝑑𝑥𝑑𝑡=10|||𝑎(𝑇)𝑞(𝑥)exp(𝑐𝑇)𝜕𝑢(𝑥,𝑇)|||𝜕𝑡2+𝑑𝑥Ω𝑞(𝑥)exp(𝑐𝑡)𝑐𝑎(𝑡)𝑑𝑎(𝑡)|||𝑑𝑡𝜕𝑢|||𝜕𝑥2𝑑𝑥𝑑𝑡;(3.13)Re𝜀Ω𝜕𝐴(𝑡)𝑢𝑣𝜀𝜀𝜕𝑡𝑑𝑥𝑑𝑡=𝑅𝑒Ω𝑑𝑎(𝑡)𝑑𝑡𝑞(𝑥)𝜕𝑢𝜕𝜕𝑥𝑣𝜀𝜕𝑥𝑑𝑥𝑑𝑡+𝜀Ω𝜕𝑎(𝑡)exp(𝑐𝑡)𝑞(𝑥)𝑣𝜀𝜕𝑥𝑑𝑥𝑑𝑡.(3.14) Using 𝜀-inequalities, we obtain Re𝜀Ω𝜕𝐴(𝑡)𝑢𝑣𝜀𝜕𝑡𝑑𝑥𝑑𝑡𝜀𝑐232𝑐0Ω|||𝑞(𝑥)exp(𝑐𝑡)𝜕𝑢|||𝜕𝑥2𝑑𝑥𝑑𝑡.(3.15) Combining (3.13) and (3.15), we get ReΩexp(𝑐𝑡)𝑣𝜀𝑁𝑣𝑑𝑥𝑑𝑡Ω𝑞(𝑥)exp(𝑐𝑡)𝑐𝑐0𝑐3𝜀𝑐23𝑐0|||𝜕𝑢|||𝜕𝑥2𝑑𝑥𝑑𝑡0.(3.16) From (3.16), we deduce that ReΩexp(𝑐𝑡)𝑣𝜀𝑁𝑣𝑑𝑥𝑑𝑡0.(3.17) Then, for 𝜀0, we obtain ReΩexp(𝑐𝑡)𝑣𝑁𝑣𝑑𝑥𝑑𝑡=Ω𝑞(𝑥)exp(𝑐𝑡)|𝑣|2𝑑𝑥𝑑𝑡0.(3.18) We conclude that 𝑣=0, hence 𝑤=0, which ends the proof of the the lemma.

Theorem 3.2. The range 𝑅(𝐿) of 𝐿 coincides with 𝐹.

Proof. Since 𝐹 is a Hilbert space, we have 𝑅(𝐿)=𝐹 if and only if the relation Ω𝑝(𝑥)3𝑢𝑓𝑑𝑥𝑑𝑡+10𝑝(𝑥)2𝑑𝑙𝑢𝑑𝑑𝑥𝜑𝑑𝑥𝑑𝑥+𝛼0𝑙𝑢𝜑𝑑𝑥+𝛾𝛽𝑙𝑢𝜑𝑑𝑥+1𝛿𝑙𝑢𝜑𝑑𝑥=0,(3.19) for arbitrary 𝑢𝐸 and (𝑓,𝜑)𝐹, implies that 𝑓=0 and 𝜑=0.
Putting 𝑢𝐷0(𝐿) in (3.19), we conclude from Lemma 3.1 that 𝜃𝑓=0, where ]],[],[[,𝜃𝑓=𝑥𝑓,𝑥0,𝛼(𝛾𝛽)𝑓,𝑥𝛼,𝛿(1𝑥)𝑓,𝑥𝛿,1(3.20) then 𝑓=0.
Taking 𝑢𝐸 in (3.19) yields 10𝑝(𝑥)2𝑑𝑙𝑢𝑑𝑑𝑥𝜑𝑑𝑥𝑑𝑥+𝛼0𝑙𝑢𝜑𝑑𝑥+𝛾𝛽𝑙𝑢𝜑𝑑𝑥+1𝛿𝑙𝑢𝜑𝑑𝑥=0.(3.21) The range of the operator 𝑙 is everywhere dense in Hilbert space with the norm 10𝑝(𝑥)2|||𝑑𝜑|||𝑑𝑥2𝑑𝑥+𝛼0||𝜑||2𝑑𝑥+𝛾𝛽||𝜑||2𝑑𝑥+1𝛿||𝜑||2𝑑𝑥1/2,(3.22) hence, 𝜑=0.

Acknowledgment

The authors would like to thank the referee for helpful suggestions and comments.

References

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