# A Mixed Problem with an Integral Two-Space-Variables Condition for Parabolic Equation with The Bessel Operator

Authors: Bouziani Abdelfatah; Oussaeif Taki-Eddine; Ben Aoua Leila
Journal: Journal of Mathematics
 (2013)

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## Abstract

We study a mixed problem with an integral two-space-variables condition for parabolic equation with the Bessel operator. The existence and uniqueness of the solution in functional weighted Sobolev space
are proved. The proof is based on a priori estimate “energy inequality” and the density of the range of the operator generated by the problem considered.


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## Body

## 1. Introduction

The importance of boundary value problems with integral boundary condition has been pointed out by Samarskiĭ [1] and problems with integral conditions for parabolic equations were treated by Kamynin [2], Ionkin [3], Yurchuk [4], Benouar and Yurchuk [5], Bouziani [6], Bouziani and Benouar [7, 8], and Mesloub and Bouziani [9]. Other parabolic problems also arise in plasma physics by Samarskiĭ [1], heat conduction by Cannon [10], Ionkin [3], dynamics of ground waters by Nakhushev [11], Vodakhova [12], Kartynnik [13], and Lin [14]. Regular case of this problem is studied in [15]. The problem where the equation contains an operator of the form (∂/∂x)(a(x,t)(∂u/∂x)), instead of Bessel operator, is treated in [16]. Similar problems for second-order parabolic equations are investegated by the potential method in [17]. Later problems with an integral two-space-variables condition for parabolic equations were treated by Marhoune [18], Marhoune and Lakhal [19]. Motivated by this, we study a mixed problem with an integral two-space-variables condition for parabolic equation with the Bessel operator.


## 2. Setting of the Problem

In the rectangular domainΩ=(0,1)×(0,T), with T<∞, we consider the following equation:
(1)ℒu=∂u∂t-(∂2u∂x2+1x∂u∂x)=f(x,t),
with the initial data
(2)ℓu=u(x,0)=φ(x),x∈(0,1),
Neumann boundary condition
(3)∂u∂x(1,t)=0,
and the integral condition
(4)∫0α‍xu(x,t)dx+∫β1‍xu(x,t)dx=0,β>α>0,α+β=1,t∈(0,T),
where φ is a known function.We shall assume that the functionsφ and f  satisfy the compatibility condition with (4), that is,
(5)∫0α‍xφ(x)dx+∫β1‍xφ(x)dx=0,(6)∫0α‍xf(x,t)dx+∫β1‍xf(x,t)dx=0.
The presence of integral terms in boundary condition can, in general, greatly complicate the application of standard functional or numerical techniques, specially the integral two-space-variables condition. Then to avoid this difficulty, we introduce a technique for transfering this problem to another classically less complicated one that does not contain integral conditions. For that, we establish the following lemma.Lemma 1.
Problem (1)–(4) is equivalent to the following problem (PR):
(PR)ℒu=∂u∂t-(∂2u∂x2+1x∂u∂x)=f(x,t),ℓu=u(x,0)=φ(x),x∈(0,1),α∂u∂x(α,t)=β∂u∂x(β,t),∂u∂x(1,t)=0.Proof.
Letu(x,t) be a solution of (1)–(4), we prove that
(7)∂u∂x(α,t)=∂u∂x(β,t).
So, multiplying (1) by x and integrating with respect to x over (0,α) and (β,1) and taking into account (4) and (6), we obtain
(8)x∂u∂x|x=0x=αdx+x∂u∂x|x=βx=1dx=0,this impliesα∂u∂x(α,t)+∂u∂x(1,t)-β∂u∂x(β,t)=0.
Then, from (3), we obtain
(9)α∂u∂x(α,t)=β∂u∂x(β,t).
Let now u(x,t) be a solution of (PR), we are bound to prove that
(10)∫0α‍xu(x,t)dx+∫β1‍xu(x,t)dx=0.
So, multiplying (1) by x and integrating with respect to x over (0,α) and (β,1) and taking into account
(11)α∂u∂x(α,t)=β∂u∂x(β,t),∂u∂x(1,t)=0,
we obtain
(12)∂∂t∫0α‍xu(x,t)dx=∫0α‍xf(x,t)dx,∂∂t∫β1‍xu(x,t)dx=∫β1‍xf(x,t)dx,
combining the two preceding equations, and from (6) we get
(13)∫0α‍xu(x,t)dx+∫β1‍xu(x,t)dx=0.


## 3. A Priori Estimate

The method used here is one of the most efficient functional analysis methods in solving partial differential equations with integral conditions, the so-called a priori estimate method or the energy-integral method. This method is essentially based on the construction of multiplicators for each specific given problem, which provides the a priori estimate from which it is possible to establish the solvability of the posed problem. More precisely, the proof is based on an energy inequality and the density of the range of the operator generated by the abstract formulation of the stated problem. But here we use the energy inequality method for the equivalent problem(PR) given in Lemma 1; so to investigate the posed problem, we introduce the needed function spaces.We introduce function spaces needed in our investigation. We denote byLρ2(Ω) the weighted Lebesgue space that consists of all measurable functions u equipped with the finite norm
(14)∥u∥Lρ2(Ω)=(∫Ω‍ρ(x)|u(x,·)|2dx)1/2,
where ρ(x)=(x2+x). If ρ(x)=1, Lρ2(Ω) are identified with the standard spaces L2(Ω).In this paper, we prove the existence and the uniqueness for solution of the problem (1)–(4) as a solution of the operator equation
(15)Lu=ℱ,
where L=(ℒ,ℓ), with domain of definition B consisting of functions u∈Lρ2(Ω) such that ∂u/∂t,  ∂u/∂x,  ∂2u/∂x2,  ∂2u/∂t∂x∈Lρ2(Ω), and u satisfy condition (4); the operator L is considered from B to F, where B  is the Banach space consisting of all functions u(x,t) having a finite norm
(16)∥u∥B2=∫Ω‍(x2+x)(∂u∂t)2dxdt+sup0≤τ≤T∫01‍(x2+x)(∂u(x,τ)∂x)2dx,
and F is the Hilbert space consisting of all elements ℱ=(f,φ) for which the norm
(17)∥ℱ∥F2=∫Ω‍f2dxdt+∫01‍(x2+x)(∂φ∂x)2dx
is finite.Theorem 2.
For any functionu∈B, one has the estimate
(18)∥u∥B≤c∥Lu∥F,
where c is a positive constant independent of u.Proof.
Multiplying (1) by the following Mu:
(19)Mu={(x2-x)∂u∂x+(x2+x)∂u∂t,0≤x≤α,-x∂u∂x+(x2+x)∂u∂t,α≤x≤β,(x2-x)∂u∂x+(x2+x)∂u∂t,β≤x≤1,
and integrating over Ωτ, where Ωτ=(0,1)×(0,τ),
(20)∫Ωτ‍ℒu·Mudxdt={∫Ωατ(x2-x)∂u∂x∂u∂tdxdt+∫Ωατ(1-x)(∂u∂x)2dxdt+∫Ωατ(x2+x)(∂u∂t)2dxdt-∫Ωατ(x-x2)∂u∂x∂2u∂x2dxdt-∫Ωατ(x+1)∂∂x(x∂u∂x)∂u∂tdxdt=∫Ωατf·((x2-x)∂u∂x+(x2+x)∂u∂t(x2-x)∂u∂x)dxdt,Ωατ,∫Ωα,βτ(x2+x)(∂u∂t)2dxdt-∫Ωα,βτx∂u∂x∂u∂tdxdt+∫Ωα,βτ(∂u∂x)2dxdt+∫Ωα,βτx∂u∂x∂2u∂x2dxdt-∫Ωα,βτ(x+1)∂∂x(x∂u∂x)∂u∂tdxdt=∫Ωα,βτf·(-x∂u∂x+(x2+x)∂u∂t)dxdt,Ωα,βτ,∫Ωβτ(x2+x)(∂u∂t)2dxdt+∫Ωβτ(x2-x)∂u∂x∂u∂tdxdt+∫Ωβτ(1-x)(∂u∂x)2dxdt+∫Ωβτ(x-x2)∂u∂x∂2u∂x2dxdt-∫Ωβτ(x+1)∂∂x(x∂u∂x)∂u∂tdxdt=∫Ωβτf·((x2-x)∂u∂x+(x2+x)∂u∂t)dxdt,Ωβτ.
Integration by parts, we obtain(21)-∫Ωατ‍(x-x2)∂u∂x∂2u∂x2dxdt=∫0τ‍(α2-α22)(∂u∂x(α,t))2dt-∫Ωατ‍(12-x)(∂u∂x)2dxdt,-∫Ωατ‍(x+1)∂∂x(x∂u∂x)∂u∂tdxdt=-∫0τ‍(α2+α)∂u∂x(α,t)∂u∂t(α,t)dt+∫Ωατ‍x∂u∂x∂u∂tdxdt+12∫0α‍(x2+x)(∂u(·,τ)∂x)2dx-12∫0α‍(x2+x)(∂φ∂x)2dx,∫Ωα,βτ‍x∂u∂x∂2u∂x2dxdt=β2∫0τ‍(∂u∂x(β,t))2dt-α2∫0τ‍(∂u∂x(α,t))2dt-12∫Ωα,βτ‍(∂u∂x)2dxdt,-∫Ωα,βτ‍(x+1)∂∂x(x∂u∂x)∂u∂tdxdt=-∫0τ‍(β2+β)∂u∂x(β,t)∂u∂t(β,t)dt+∫0τ‍(α2+α)∂u∂x(α,t)∂u∂t(α,t)dt+∫Ωα,β‍x∂u∂x∂u∂tdxdt+12∫αβ‍(x2+x)(∂u(·,τ)∂x)2dx-12∫αβ‍(x2+x)(∂φ∂x)2dx,∫Ωβτ‍(x-x2)∂u∂x∂2u∂x2dxdt=∫0τ‍(β22-β2)(∂u∂x(β,t))2dt-∫Ωβτ‍(12-x)(∂u∂x)2dxdt,-∫Ωβτ‍(x+1)∂∂x(x∂u∂x)∂u∂tdxdt=∫0τ‍(β2+β)∂u∂x(β,t)∂u∂t(β,t)dt+∫Ωβτ‍x∂u∂x∂u∂tdxdt+12∫β1‍(x2+x)(∂u(·,τ)∂x)2dx-12∫β1‍(x2+x)(∂φ∂x)2dx,
and by using the Cauchy's ε-inequality, we get
(22)∫Ωατ‍x2∂u∂x∂u∂tdxdt≤ε2∫Ωατ‍x2(∂u∂x)2dxdt+12ε∫Ωατ‍x2(∂u∂t)2dxdt≤14∫Ωατ‍(∂u∂x)2dxdt+14(α2+α)∫Ωατ‍(x2+x)(∂u∂x)2dxdt+(α2+α)2∫Ωατ‍(x2+x)(∂u∂t)2dxdt,∫Ωατ‍f·((x2-x)∂u∂x+(x2+x)∂u∂t)dxdt≤(α2+α)2∫Ωατ‍f2dxdt+14∫Ωατ‍(∂u∂x)2dxdt+14(α2+α)∫Ωατ‍(x2+x)(∂u∂x)2dxdt+12(α2+α)∫Ωατ‍(x2+x)(∂u∂t)2dxdt,∫Ωα,βτ‍f·(-x∂u∂x+(x2+x)∂u∂t)dxdt≤((β-α)((β2-α2)+(β-α))2+12)×∫Ωα,βτ‍f2dxdt+14∫Ωα,βτ‍(∂u∂x)2dxdt+14((β2-α2)+(β-α))×∫Ωα,βτ‍(x2+x)(∂u∂x)2dxdt+12∫Ωα,βτ‍(x2+x)(∂u∂t)2,-∫Ωβτ‍x2∂u∂x∂u∂tdxdt≤ε2∫Ωβτ‍x(∂u∂x)2dxdt+12ε∫Ωβτ‍x(∂u∂t)2dxdt≤ε(1-β)2∫Ωβτ‍(∂u∂x)2dxdt+12ε∫Ωβτ‍(x2+x)(∂u∂t)2dxdt≤14∫Ωβτ‍(∂u∂x)2dxdt+(1-β)∫Ωβτ‍(x2+x)(∂u∂t)2dxdt,∫Ωβτ‍f·((x2-x)∂u∂x+(x2+x)∂u∂t)dxdt≤((1-β)2+(1-β))2∫Ωβτ‍f2+14∫Ωβτ‍(∂u∂x)2+14((1-β)2+(1-β))∫Ωβτ‍(x2+x)(∂u∂x)2+12((1-β)2+(1-β))∫Ωβτ‍(x2+x)(∂u∂t)2.
Substituting (21)–(22) into ∫Ωτ‍ℒu·Mudxdt, we obtain
(23)(1-α2+α2-12(α2+α))∫Ωατ‍(x2+x)(∂u∂t)2dxdt+12∫0α‍(x2+x)(∂u(·,τ)∂x)2dx∫0τ‍(α2-α22)(∂u∂x(α,t))2dt-∫0τ‍(α2+α)∂u∂x(α,t)∂u∂t(α,t)dt≤(α2+α)2∫Ωατ‍f2dxdt+12∫0α‍(x2+x)(∂φ∂x)2dx+12(α2+α)∫Ωατ‍(x2+x)(∂u∂x)2dxdt,Ωατ,(24)12∫Ωα,βτ‍(x2+x)(∂u∂t)2+12∫αβ‍(x2+x)(∂u(·,τ)∂x)2dx-∫0τ‍(β2+β)∂u∂x(β,t)∂u∂t(β,t)dt+∫0τ‍(α2-α22)(∂u∂x(α,t))2dtβ2∫0τ‍(∂u∂x(β,t))2dt-α2∫0τ‍(∂u∂x(α,t))2dt≤((β-α)2(β+α+1)2+12)∫Ωα,βτ‍f2dxdt+12∫αβ‍(x2+x)(∂φ∂x)2dx+14((β2-α2)+(β-α))×∫Ωα,βτ‍(x2+x)(∂u∂x)2dxdt,Ωα,βτ,(25)(β-12((1-β)2+(1-β)))∫Ωβτ‍(x2+x)(∂u∂t)2dxdt+12∫β1‍(x2+x)(∂u(·,τ)∂x)2dx∫0τ‍(β22-β2)(∂u∂x(β,t))2dt+∫0τ‍(β2+β)∂u∂x(β,t)∂u∂t(β,t)dt≤((1-β)2+(1-β))2∫Ωβτ‍f2dxdt+12∫β1‍(x2+x)(∂φ∂x)2dx+14((1-β)2+(1-β))×∫Ωβτ‍(x2+x)(∂u∂x)2dxdt,Ωβτ.
Now, by using the conditions (PR) in (23), (24), and (25), respectively, we obtain
(26)(1-α2+α2-12(α2+α))∫Ωατ‍(x2+x)(∂u∂t)2dxdt+12∫0α‍(x2+x)(∂u(·,τ)∂x)2dx≤(α2+α)2∫Ωατ‍f2dxdt+12∫0α‍(x2+x)(∂φ∂x)2dx+12(α2+α)∫Ωατ‍(x2+x)(∂u∂x)2dxdt,Ωατ,(27)12∫Ωα,βτ‍(x2+x)(∂u∂t)2+12∫αβ‍(x2+x)(∂u(·,τ)∂x)2dx≤((β-α)2(β+α+1)2+12)∫Ωα,βτ‍f2dxdt+12∫αβ‍(x2+x)(∂φ∂x)2dx+14((β2-α2)+(β-α))×∫Ωα,βτ‍(x2+x)(∂u∂x)2dxdt,Ωα,βτ,(28)(β-12((1-β)2+(1-β)))∫Ωβτ‍(x2+x)(∂u∂t)2dxdt+12∫β1‍(x2+x)(∂u(·,τ)∂x)2dx≤((1-β)2+(1-β))2∫Ωβτ‍f2dxdt+12∫β1‍(x2+x)(∂φ∂x)2dx+14((1-β)2+(1-β))×∫Ωβτ‍(x2+x)(∂u∂x)2dxdt.Ωβτ.
Or in Ωτ, by combining (26), (27), and (28), we get
(29)min(1-α2+α2-12(α2+α),12,β-12((1-β)2+(1-β)))×∫Ωτ‍(x2+x)(∂u∂t)2dxdt+12∫01‍(x2+x)(∂u(·,τ)∂x)2dx≤max((α2+α)2,((β-α)2(β+α+1)2+12),((1-β)2+(1-β))2((β-α)2(β+α+1)2+12))×∫Ωτ‍f2dxdt+12∫01‍(x2+x)(∂φ∂x)2dx+max(12(α2+α),14((β2-α2)+(β-α)),14((1-β)2+(1-β)))×∫Ωτ‍(x2+x)(∂u∂x)2dxdt.
Using Lemma 1 in [20], one has
(30)(∫Ω‍(x2+x)(∂u∂t)2dxdt+∫01‍(x2+x)(∂u(·,τ)∂x)2dx)≤k(∫Ω‍f2dxdt+∫01‍(x2+x)(∂φ∂x)2dx),
where
(31)k=max((α2+α)2,((β-α)2(β+α+1)2+12),((1-β)2+(1-β))2)((β-α)2(β+α+1)2+12))×(12,β-(1/2((1-β)2+(1-β))))min(121-α2+α2-12(α2+α),12,β-12((1-β)2+(1-β))))-1×e(max(1/2(α2+α),1/4((β2-α2)+(β-α)),1/4((1-β)2+(1-β)))T).
The right-hand side of (30) is independent of τ, hence replacing the left-hand side by its upper bound with respect to τ from 0 to T, we obtain the desired inequality, where c=(k)1/2.Corollary 3.
A solution of the problem (1)–(4) is unique if it exists and depends continuously on ℱ∈F.


## 4. Existence of Solution

To show the existance of solutions, we prove thatR(L) is dense in Ffor all u∈B and for arbitrary ℱ=(f,φ)∈F.Theorem 4.
Suppose the conditions of Theorem2 are satisfied. Then the problem (1)–(4) admits a unique solution u=L-1ℱ.Proof.
First we prove thatR(L) is dense in Ffor the special case where D(L)≡B is reduced to D0(L), where D0(L)={u,u∈D(L):ℓu=0}.Proposition 5.
Let the conditions of Theorem4 be satisfied. if for ω∈L2(Ω) and for all u∈D0(L), one has
(32)∫Ω‍ℒu·ωdxdt=0,
then ω vanishes almost everywhere in Ω.Proof.
The scalar product ofF is defined by
(33)(Lu,ω)F=∫Ω‍ℒu·ωdxdt+∫01‍(x2+x)(∂ℓu∂x)(∂ω0∂x)dx.
the equality (32) can be written as follows:
(34)∫Ω‍∂u∂t·ωdxdt=∫Ω‍1x∂∂x(x∂u∂x)·ωdxdt,
if we put
(35)u=ℑt(ektz)=∫0t‍ekτz(x,τ)dτ,
where k is a strictly positive constant and z,∂z/∂x,(∂/∂x)(∂ℑt(ektz)/∂x)∈L2(Ω), then, u satisfies the boundary conditions in (PR). As a result of (34), we obtain the equality
(36)∫Ω‍ektzωdxdt=∫Ω‍1x∂∂x(x∂ℑt(ektz)∂x)·ωdxdt.
The left-hand side of (36) shows that the mapping
(37)L2(Ω)∋z⟶∫Ω‍1x∂∂x(x∂ℑt(ektz)∂x)·ωdxdt
is a continuous linear functional of z. From the right-hand side of (36) there follows that is true if the function ω has the following properties:
(38)1x∂ω∂x,∂ℑt⋆ω∂x,1x∂ℑt⋆ω∂x,∂2ℑt⋆ω∂x2,ℑt⋆(ωx2)∈L2(Ω),∂ω∂x(α,0)=∂ω∂x(β,0)=∂ω∂x(1,0)=0,∂ω∂x(1,t)=0.
In terms of the given function ω and from the equality (36), we give the function ω in terms of z as follows:
(39)ω={-(x-x22)∂z∂x,0≤x≤α,-x∂z∂x,α≤x≤β,-(x-x22)∂z∂x,β≤x≤1,
and z satisfies the same conditions of the function u in (PR):
(40)α∂z∂x(α,t)=β∂z∂x(β,t),(41)∂z∂x(1,t)=0,z(α,t)=z(β,t)=0.
Replacing ω in (36) by its representation (39) and integrating by parts each term of (36) and by taking the conditions (40) and (41), we obtain
(42)∫Ωτ‍ektzωdxdt={-12∫0τ‍(α-α22)ℑtekt(∂z(α,t)∂x)2dxdt-12∫Ωατ‍(1-x)ℑtekt(∂z∂x)2dxdt,Ωατ=(0,α)×(0,τ),-β2∫0τ‍ℑtekt(∂z(β,t)∂x)2dxdt+α2∫0τ‍ℑtekt(∂z(α,t)∂x)2dxdt-12∫Ωα,βτ‍ℑtekt(∂z∂x)2dxdt,Ωα,βτ=(α,β)×(0,τ),12∫0τ‍(β-β22)ℑtekt(∂z(β,t)∂x)2dxdt-12∫Ωβτ‍(1-x)ℑtekt(∂z∂x)2dxdt,Ωβτ=(β,1)×(0,τ).
So, we obtain(43)∫Ωτ‍ektzωdxdt={-12∫Ωατ‍(1-x)ℑtekt(∂z∂x)2dxdt,Ωατ=(0,α)×(0,τ),-12∫Ωα,βτ‍ℑtekt(∂z∂x)2dxdt,Ωα,βτ=(α,β)×(0,τ),-12∫Ωβτ‍(1-x)ℑtekt(∂z∂x)2dxdt,Ωβτ=(β,1)×(0,τ).
Now, by combining overΩτ, we get
(44)12∫Ωατ‍(1-x)ektz2dxdt+12∫Ωα,βτ‍ektz2dxdt+12∫Ωβτ‍(1-x)ektz2dxdt=α24∫0τ‍ℑtekt(∂z(α,t)∂x)2dxdt-β24∫0τ‍ℑtekt(∂z(β,t)∂x)2dxdt-12∫Ωατ‍(1-x)ℑtekt(∂z∂x)2dxdt-12∫Ωα,βτ‍ℑtekt(∂z∂x)2dxdt-12∫Ωβτ‍(1-x)ℑtekt(∂z∂x)2dxdt,
by taking into accont the boundary condition (40) of the function z yields
(45)12(∫Ωατ‍(1-x)ektz2dxdt+∫Ωα,βτ‍ektz2dxdt+∫Ωβτ‍(1-x)ektz2dxdt)=-12(∫Ωατ‍(1-x)ℑtekt(∂z∂x)2dxdt+∫Ωα,βτ‍ℑtekt(∂z∂x)2dxdt∫Ωβτ‍(1-x)ℑtekt(∂z∂x)2dxdt)≤0.
Then, we obtain(46)12(∫Ωατ‍(1-x)ektz2dxdt+∫Ωα,βτ‍ektz2dxdt+∫Ωβτ‍(1-x)ektz2dxdt)≤0.
And thus z=0 in Ω, then ω=0 in Ω. This proves Proposition 5.We return to the proof of Theorem4. We have already noted that it is sufficient to prove that the set R(L) dense in F.Suppose that for someW=(ω,ω0)∈R(L)⊥ and for all u∈D(L)≡B, it holds
(47)(Lu,ω)F=∫Ω‍ℒu·ωdxdt+∫01‍(x+1)(∂ℓu∂x)(∂ω0∂x)dx.Then we must prove thatW=0. Putting u∈D0(L) in (47), we have
(48)∫Ω‍ℒu·ωdxdt=0,u∈D0(L).
Hence Proposition 5 implies that ω=0. Thus (47) takes the form
(49)∫01‍(x+1)(∂ℓu∂x)(∂ω0∂x)dx=0,u∈D(L).
Since the range of the trace operator ℓ is everywhere dense in the Hilbert space F with the norm
(50)(∫01‍[(x+1)(∂ℓu∂x)2])1/2,
the equality (49) implies that ω0=0. Hence W=0 implies (R(L)¯=F). Therefore, the proof of Theorem 4 is complete.



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*Source: 457631-2013-05-02.xml*