JAMSAJournal of Applied Mathematics and Stochastic Analysis1687-21771048-9533Hindawi Publishing Corporation97560110.1155/2009/975601975601Research ArticleA Boundary Value Problem with Multivariables Integral Type Condition for Parabolic EquationsMarhouneA. L.1LakhalF.2AizicoviciSergiu1Laboratory Equations DifférentiellesDepartement of MathematicsUniversity Mentouri ConstantineConstantine 25017Algeriaumc.edu.dz2Department of MathematicsScience University of 08 Mai 45P.O. Box 401Guelma 24000Algeriauniv-guelma.dz2009212200920092701200914052009131020092009Copyright © 2009This 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.

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=ut-a(t)2ux2=f(x,t), where the function a(t) and its derivative are bounded on the interval [0,T]:

0<c0a(t)c1,0<c2da(t)dtc3,lu=u(x,0)=φ(x),x(0,1),u(0,t)=u(β,t)=u(γ,t)=u(1,t),  t(0,T),0αu(x,t)dx+2βγu(x,t)dx+δ1u(x,t)dx=0,  0<α<β<γ<δ<1,α=1-δ=γ-β,  t(0,T). 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),0αφ(x)dx+2βγφ(x)dx+δ1φ(x)dx=0. When considering the classical solution of the problem (1.1)–(1.5), along with (1.5), there should be the fulfilled conditions:

a(0){φ(0)+φ(β)-φ(γ)-φ(1)}=f(1,0)+f(γ,0)-f(β,0)-f(0,0),a(0){0αφ(x)dx+2βγφ(x)dx+δ1φ(x)dx}=0αf(x,0)dx+2βγf(x,0)dx+δ1f(x,0)dx.

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, . Boundary-value problems for parabolic equations with integral boundary condition are investigated by Batten , Bouziani and Benouar , Cannon [8, 9], Cannon, Perez Esteva and van der Hoek , Ionkin , Kamynin , Shi and Shillor , Shi , Marhoune and Bouzit , Marhoune and Hameida , Yurchuk , and many references therein. The problem with one-variable (resp., two-variable) boundary integral type condition is studied in  and by Marhoune and Latrous  (resp., in Marhoune ).

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

ut=x(a(x,t)ux)+f(x,t), under the following conditions:

u(x,0)=0,0x1,u(0,t)=0,0<tT,01u(x,t)dx=0. This last integral condition in the form

01u(x,t)dx=m(t),0<tT, arises, for example, in biochemistry in which m is a constant, and in this case is known as the conservation of protein . Further, in , the author studied a similar problem with the weak integral condition

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

0αu(x,t)dx+β1u(x,t)dx=0,α+β=1, was investigated in . 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 . 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 uE2=Ωp(x)3[|ut|2+|2ux2|2]dxdt+sup0tT01p(x)2|ux|2dx+sup0tT0α|u|2dx+sup0tTβγ|u|2dx+sup0tTδ1|u|2dx, 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

F2=(f,φ)F2=Ωp(x)3|f|2dxdt+01p(x)2|dφdx|2dx+0α|φ|2dx+βγ|φ|2dx+δ1|φ|2dx, where

p(x)={x2,x]0,α],(γ-β)2,x[α,β][γ,δ],(γ-x)2+(x-β)2,x[β,γ],(1-x)2,x[δ,1[. Using the energy inequalities method proposed in , 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 EstimatesTheorem 2.1.

For any function uE, one has the a priori estimate LuF2c4uE2, where the constant c4 is independent of u. In fact, c4=2max(1,c12).

Proof.

Using (1.1) and initial condition (1.3), we obtain Ωp(x)3|u|2dxdt2Ωp(x)3[|ut|2+c12|2ux2|2]dxdt,01p(x)2|dφdx|2dxsup0tT01p(x)2|ux|2dx,0α|φ|2dxsup0tT0α|u|2dx,βγ|φ|2dxsup0tTβγ|u|2dx,δ1|φ|2dxsup0tTδ1|u|2dx. 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 uE2c5LuF2, with the constant c5=exp(cT)max(49,2c1)min(13/32,c0,c02/2), and c is such that cc0c3.

Before proving this theorem, we need the following lemma.

Lemma 2.3 (see [<xref ref-type="bibr" rid="B18">19</xref>]).

For uE, one has ab|xbu(ξ,t)tdξ|2dx4ab(x-a)2|ut|2dx,ab|axu(ξ,t)tdξ|2dx4ab(b-x)2|ut|2dx.

Proof of Theorem <xref ref-type="statement" rid="thm2">2.2</xref>.

Define Mu={x2ut+2xxαu(ξ,t)tdξ,x]0,α],(γ-β)2ut,x[α,β][γ,δ],(γ-x)2ut+(x-β)2ut+2(γ-x)βxu(ξ,t)tdξ+2(x-β)xγu(ξ,t)tdξ,x[β,γ],(1-x)2ut+2(1-x)δxu(ξ,t)tdξ,x[δ,1[. We consider for uE the quadratic formula Re0τ01exp(-ct)uMu¯dxdt, 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 Re0τ01exp(-ct)uMu¯dxdt=0τ01p(x)exp(-ct)|ut|2dxdt+0τ0αexp(-ct)|xαu(ξ,t)tdξ|2dxdt+0τβγexp(-ct)[|βxu(ξ,t)tdξ|2+|xγu(ξ,t)tdξ|2]dxdt+0τδ1exp(-ct)|δxu(ξ,t)tdξ|2dxdt+Re0τ01exp(-ct)p(x)aux2u¯xtdxdt+2Re0τ0αexp(-ct)auu¯tdxdt+4Re0τβγexp(-ct)auu¯tdxdt+2Re0τδ1exp(-ct)auu¯tdxdt. On the other hand, by using the elementary inequalities we get Re0τ01exp(-ct)uMu¯dxdt0τ01p(x)exp(-ct)|ut|2dxdt+Re0τ01exp(-ct)p(x)aux2u¯xtdxdt+2Re0τ0αexp(-ct)auu¯tdxdt+4Re0τβγexp(-ct)auu¯tdxdt+2Re0τδ1exp(-ct)auu¯tdxdt. 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 Re0τ01exp(-ct)p(x)aux2u¯xtdxdt01exp(-cτ)2p(x)a(τ)|u(x,τ)x|2dx-1201p(x)a(0)|dludx|2dx,Re0τ0αexp(-ct)auu¯tdxdt0αexp(-cτ)2a(τ)|u(x,τ)|2dx-0αa(0)2|lu|2dx;Re0τβγexp(-ct)auu¯tdxdtβγexp(-cτ)2a(τ)|u(x,τ)|2dx-βγa(0)2|lu|2dx,Re0τδ1exp(-ct)auu¯tdxdtδ1exp(-cτ)2a(τ)|u(x,τ)|2dx-δ1a(0)2|lu|2dx. Using (2.11) in (2.10), we get Re0τ01exp(-ct)uMu¯dxdt+0αa(0)|lu|2dx+2βγa(0)|lu|2dx+δ1a(0)|lu|2dx+1201p(x)a(0)|dludx|2dx0τ01p(x)exp(-ct)|ut|2dxdt+1201exp(-cτ)p(x)a(τ)|u(x,τ)x|2dx+0αexp(-cτ)a(τ)|u(x,τ)|2dx+2βγexp(-cτ)a(τ)|u(x,τ)|2dx+δ1exp(-cτ)a(τ)|u(x,τ)|2dx. By using the ε-inequalities on the first integral in the left-hand side of (2.12) and Lemma 2.3, we obtain 15320τ01p(x)exp(-ct)|ut|2dxdt+1201exp(-cτ)p(x)a(τ)|u(x,τ)x|2dx+0αexp(-cτ)a(τ)|u(x,τ)|2dx+2βγexp(-cτ)a(τ)|u(x,τ)|2dx+δ1exp(-cτ)a(τ)|u(x,τ)|2dx160τ01p(x)exp(-ct)|u|2dxdt+0αa(0)|lu|2dx+1201p(x)a(0)|dludx|2dx+2βγa(0)|lu|2dx+δ1a(0)|lu|2dx. Now, from (1.1), we have c0260τ01p(x)exp(-ct)|2ux2|2dxdt0τ01p(x)3exp(-ct)|u|2dxdt+0τ01p(x)3exp(-ct)|ut|2dxdt. Combining inequalities (2.13) and (2.14), we get exp(-cT)(13320τ01p(x)3|ut|2dxdt+c001p(x)2|u(x,τ)x|2dx+c00α|u(x,τ)|2dx+2c0βγ|u(x,τ)|2dx+c0δ1|u(x,τ)|2dx+c0260τ01p(x)3|2ux2|2dxdt)49Ωp(x)3|u|2dxdt+c101p(x)2|dludx|2dx+c10α|lu|2dx+2c1βγ|lu|2dx+c1δ1|lu|2dx. 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 L:EF is continuous and its range is closed in F. Therefore, the inverse operator L-1 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)={uE:lu=0}. If for uD0(L) and some wL2(Ω), one has Ωq(x)uw¯dxdt=0, where                                                                                                                             q(x)={x,x]0,α],γ-β,x[α,δ],1-x,x[δ,1[, then w=0.

Proof.

From (3.2) we have Ωq(x)utw¯dxdt=Ωq(x)a(t)2ux2w¯dxdt. Now, for given w(x,t), we introduce the function v(x,t)={w(x,t)-xαw(ξ,t)ξdξ,x]0,α],w(x,t),x[α,δ],w(x,t)-δxw(ξ,t)ξdξ,x[δ,1[.     Integrating by parts with respect to ξ, we obtain q(x)w={xv+xαv(ξ,t)dξ,x]0,α],(γ-β)v,x[α,β]    [γ,δ],(γ-β)v+βγv(ξ,t)dξ,x[β,γ],(1-x)v    +δxv(ξ,t)dξ,x[δ,1[, which implies that 0αv(ξ,t)dξ+2βγv(ξ,t)dξ+δ1v(ξ,t)dξ=0. Then, from (3.4), we obtain -ΩutNv¯dxdt=ΩA(t)uv¯dxdt, where Nv=q(x)v,A(t)u=-x(q(x)a(t)ux). If we introduce the smoothing operators with respect to t , ε-1=(I+ε/t)-1 and (ε-1)*, then these operators provide the solutions of the respective problems: εdgε(t)dt+gε(t)=g(t),gε(t)t=0=0,-εdgε*(t)dt+gε*(t)=g(t),gε*(t)t=T=0, and also have the following properties: for any gL2(0,T), the functions gε=(ε-1)g and gε*=(ε-1)*g are in W21(0,T) such that gε(t)t=0=0 and gε*(t)t=T=0. Morever, ε-1 commutes with /t, so 0T|gε-g|2dt0 and 0T|gε*-g|2dt0 for ε0.

Putting u=0texp(cτ)vε*(x,τ)dτ in (3.8), where the constant c satisfies cc0-c3-εc32/c00, and using (3.11), we obtain -Ωexp(ct)vε*Nv¯dxdt=ΩA(t)u  exp(-ct)u¯tdxdt-εΩA(t)uvε*¯tdxdt.

Integrating by parts each term in the right-hand side of (3.12) and taking the real parts yield 2ReΩA(t)u  exp(-ct)u¯tdxdt=01a(T)q(x)exp(-cT)|u(x,T)t|2dx+Ωq(x)exp(-ct)(ca(t)-da(t)dt)|ux|2dxdt;Re(-εΩA(t)uvε*¯tdxdt)=Re(εΩda(t)dtq(x)uxvε*¯xdxdt)+εΩa(t)exp(ct)q(x)vε*¯xdxdt. Using ε-inequalities, we obtain Re(-εΩA(t)uvε*¯tdxdt)-εc322c0Ωq(x)exp(-ct)|ux|2dxdt. Combining (3.13) and (3.15), we get Re(Ωexp(ct)vε*Nv¯dxdt)-Ωq(x)exp(-ct)(cc0-c3-εc32c0)|ux|2dxdt0. From (3.16), we deduce that Re(Ωexp(ct)vε*Nv¯dxdt)0. Then, for ε0, we obtain ReΩexp(ct)vNv¯dxdt=Ωq(x)exp(ct)|v|2dxdt0. 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)3uf¯dxdt+01p(x)2dludxdφ¯dxdx+0αluφ¯dx+βγluφ¯dx+δ1luφ¯dx=0, 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,x]0,α],(γ-β)f,x[α,δ],(1-x)f,x[δ,1[, then f=0.

Taking uE in (3.19) yields 01p(x)2dludxdφ¯dxdx+0αluφ¯dx+βγluφ¯dx+δ1luφ¯dx=0. The range of the operator l is everywhere dense in Hilbert space with the norm [01p(x)2|dφdx|2dx+0α|φ|2dx+βγ|φ|2dx+δ1|φ|2dx]1/2, hence, φ=0.

Acknowledgment

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

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