# Mixed Problem with an Integral Two-Space-Variables Condition for a Class of Hyperbolic Equations

Authors: Taki-Eddine Oussaeif; Abdelfatah Bouziani
Journal: International Journal of Analysis
 (2013)

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## Abstract

This paper is devoted to the proof of the existence and uniqueness of the classical solution of mixed problems which combine Neumann condition and integral two-space-variables condition for a class of hyperbolic equations. 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 integral boundary conditions for evolution problems have various applications in chemical engineering, thermoelasticity, underground water flow, and population dynamics.Cannon was the first who drew attention to these problems with an integral one-space-variable condition [1], and their importance has been pointed out by Samarskii [2]. The existence and uniqueness of the classical solution of mixed problem combining a Dirichlet and integral condition for the equation of heat demonstrated by cannon [1] using the potential method.Always using the potential method, Kamynin established in [3] the existence and uniqueness of the classical solution of a similar problem with a more general representation.Subsequently, more works related to these problems with an integral one-space-variable have been published, among them, we cite the work of Benouar and Yurchuk [4], Cannon and Van Der Hoek [5, 6], Cannon-Esteva-Van Der Hoek [7], Ionkin [8], Jumarhon and McKee [9], Kartynnik [10], Lin [11], Shi [12] and Yurchuk [13]. In these works, mixed problems related to one-dimensional parabolic equations of second order combining a local condition and an integral condition was discussed. Also, by referring to the articles of Bouziani [14–16] and Bouziani and Benouar [17–19], the authors have studied mixed problems with integral conditions for some partial differential equations, specially hyperbolic equation with integral condition which has been investigated in Bouziani [20].The present paper is devoted to the study of problems with a boundary integral two-space-variables condition for second-order hyperbolic equation.


## 2. Setting of the Problem

In the rectangleΩ=(0,1)×(0,T), with T<∞, we consider the hyperbolic equation:
(1)ℒu=∂2u∂t2-∂∂x(a(x,t)∂u∂x)=f(x,t),
where the coefficient a(x,t) is a real-valued function belonging to C1(Ω¯) such that
(2)0<c0≤a(x,t)≤c1,∂a(x,t)∂t≤c2.
in the rest of the paper, k, ci, i=1,…,12, denote strictly positive constants.we adjoin to (1) the initial conditions
(3)ℓ1u=u(x,0)=φ(x),x∈(0,1),ℓ2u=∂u∂t(x,0)=ψ(x),x∈(0,1),
the Neumann condition
(4)∂u∂x(0,t)=0,
and the integral condition
(5)∫0αu(x,t)dx+∫β1u(x,t)dx&=0,α>0,β>0,α+β=1,t∈(0,T),
where φ and ψ are known functions.We will assume that the functionφ and f  satisfy a compatibility conditions with (5), that is,
(6)∫0αφ(x)dx+∫β1φ(x)dx=0,(7)∫0αf(x,t)dx+∫β1f(x,t)dx=0.The presence of integral terms in boundary conditions can, in general, greatly complicate the application of standard functional or numerical techniques specially the integral two-space-variables condition. In order to avoid this difficulty, we introduce a technique to transfer this problem to another classically less complicated one which does not contain integral conditions. For that, we establish the following lemma.Lemma 1.
Problem (1)–(5) is equivalent to the following problem (PR):
(8)(PR)={ℒu=∂2u∂t2-∂∂x(a(x,t)∂u∂x)=f(x,t),0<c0≤a(x,t)≤c1,∂a(x,t)∂t≤c2,ℓ1u=u(x,0)=φ(x),x∈(0,1),ℓ2u=∂u∂t(x,0)=ψ(x),x∈(0,1),∂u∂x(0,t)=0,∂u∂x(α,t)=0,∂u∂x(β,t)=0,∂u∂x(1,t)=0.Proof.
Letu(x,t) be a solution of (1)–(5), we prove that
(9)∂u∂x(0,t)=0,∂u∂x(α,t)=0,∂u∂x(β,t)=0,∂u∂x(1,t)=0.
So, by integrating (1) with respect to x over (0,α) and (β,1) and taking into account (6) and (7), we obtain
(10)∫0α∂∂x(a(x,t)∂u∂x)dx+∫β1∂∂x(a(x,t)∂u∂x)dx=0,
and we get
(11)a(α,t)∂u∂x(α,t)-a(0,t)∂u∂x(0,t)+a(1,t)∂u∂x(1,t)-a(β,t)∂u∂x(β,t)=0.
The integral condition (5) has nothing to do with the coefficient a(x,t). Then, (11) imposes
(12)∂u∂x(α,t)=∂u∂x(0,t)=∂u∂x(1,t)=∂u∂x(β,t)=0.
Let nowu(x,t) be a solution of (PR), then we are bound to prove that
(13)∫0αu(x,t)dx+∫β1u(x,t)dx=0.
So, by integrating (1) with respect to x over (0,α) and taking into account that
(14)∂u∂x(0,t)=0,∂u∂x(α,t)=0,
we obtain
(15)∂2∂t2∫0αu(x,t)dx=∫0αf(x,t)dx,
and by integrating (1) with respect to x over (β,1) and taking in consideration
(16)∂u∂x(β,t)=0,∂u∂x(1,t)=0,
we obtain
(17)∂2∂t2∫β1u(x,t)dx=∫β1f(x,t)dx.
By combining the two preceding (15) and (17) and taking into account (7), we get
(18)∫0αu(x,t)dx+∫β1u(x,t)dx=0.


## 3. A Priori Estimate and Its Consequences

In this paper, we prove the existence and the uniqueness of the solution of the problem (1)–(5) and of the operator equation
(19)Lu=ℱ,
where L=(ℒ,ℓ1,ℓ2) with domain of difinition B consisting of functions u∈L2(Ω) such that ∂u/∂t, ∂u/∂x, ∂2u/∂x2∈L2(Ω), and u satisfies conditions (3) and (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
(20)∥u∥B2=∫01(∂u∂t)2dx+∫01(∂u∂x)2dx,
and F is the Hilbert space consisting of all elements ℱ=(f,φ,ψ) for which the norm
(21)∥ℱ∥F2=∫01f2dxdt+∫01ψ2(x)dx+∫01(∂φ∂x)2dx
is finite.Theorem 2.
For any functionu∈B, we have the inequality
(22)∥u∥B≤c∥Lu∥F,
where c is a positive constant independent of u.Proof.
Multiplying the (1) by the following Mu:
(23)Mu={Mu1=2(1-x)∂u∂t,0≤x≤α,Mu2=2∂u∂t,α≤x≤β,Mu3=2(2-x)∂u∂t,β≤x≤1
and integrating over Ωτ, where Ωτ=(0,1)×(0,τ),(1)
by integrating overΩατ=Ωα=(0,α)×(0,τ). Consequently,
(24)∫Ωαℒu·Mu1dxdt=2∫Ωα(1-x)∂u∂t∂2u∂t2dxdt-2∫Ωα(1-x)∂u∂t∂∂x(a(x,t)∂u∂x)dxdt=2∫Ωα(1-x)∂u∂tfdxdt.
Employing integration by parts in (24), and taking into account the boundary conditions in (PR), we obtain
(25)2∫Ωα(1-x)∂u∂t∂2u∂t2dxdt=∫0α(1-x)(∂u∂t)2dx-∫0α(1-x)ψ2(x)dx,-2∫Ωα‍(1-x)∂u∂t∂∂x(a(x,t)∂u∂x)dxdt=-∫0τ(2(1-x)a(x,t)∂u∂t∂u∂x)|x=0x=αdt+2∫Ωα(1-x)a(x,t)∂u∂x∂2u∂x∂tdxdt-2∫Ωαa(x,t)∂u∂x∂u∂tdxdt=-∫0α(1-x)a(x,0)(∂φ∂x)2dx+∫0α(1-x)a(x,τ)(∂u∂x)2dx-∫Ωα(1-x)∂a(x,t)∂t(∂u∂x)2dxdt-2∫Ωαa(x,t)∂u∂x∂u∂tdxdt.
Using the Cauchy’s inequality, we get(26)2∫Ωα(1-x)∂u∂tfdxdt≤∫Ωαf2dxdt+∫Ωα(∂u∂t)2dxdt.
Substituting (25) and (26) into (24), we obtain
(27)∫0α(1-x)(∂u∂t)2dx+∫0α(1-x)a(x,τ)(∂u∂x)2dx≤∫Ωαf2dxdt+∫Ωα(∂u∂t)2dxdt+∫0α(1-x)a(x,0)(∂φ∂x)2dx+∫0α(1-x)ψ2(x)dx+∫Ωα(1-x)∂a(x,t)∂t(∂u∂x)2dxdt+2∫Ωαa(x,t)∂u∂x∂u∂tdxdt.
Then, using the Cauchy’s inequality and according to conditions (2), we get
(28)β∫0α(∂u∂t)2dx+βc0∫0α(∂u∂x)2dx≤∫Ωαf2dxdt+(c1+1)∫Ωα(∂u∂t)2dxdt+∫0αψ2(x)dx+(c1+c2)∫Ωα(∂u∂x)2dxdt+c1∫0α(∂φ∂x)2dx,
so, we obtain
(29)∫0α(∂u∂t)2dx+∫0α(∂u∂x)2dx≤c3(∫Ωαf2dxdt+∫0αψ2(x)dx+∫0α(∂φ∂x)2dx+∫Ωα(∂u∂t)2dxdt)+c4∫Ωα(∂u∂x)2dxdt,
where
(30)c3=c1+1min(β,βc0),c4=c1+c2min(β,βc0).
By virtue of Lemma  7.1 in [21] and by using it twice, we find
(31)∫0α(∂u∂t)2dx+∫0α(∂u∂x)2dx≤c5(∫Ωαf2dxdt+∫0αψ2(x)dx+∫0α(∂φ∂x)2dx),
where
(32)c5=c3exp(c4T)exp(c3Texp(c4T)).(2)
by integrating overΩα,βτ=Ωα,β=(α,β)×(0,τ). Consequently,
(33)∫Ωα,βℒu·Mu2dxdt=2∫Ωα,β∂2u∂t2∂u∂tdxdt-2∫Ωα,β∂∂x(a(x,t)∂u∂x)∂u∂tdxdt=2∫Ωα,β‍f∂u∂tdxdt.
Employing integration by parts in (33) and taking into account the boundary conditions in (PR), we obtain
(34)2∫Ωα,β∂2u∂t2∂u∂tdxdt=∫αβ(∂u∂t)2dx-∫αβψ2(x)dx,-2∫Ωα,β∂∂x(a(x,t)∂u∂x)(∂u∂t)dxdt=-∫0τ(2a(x,t)∂u∂t∂u∂x)|x=αx=βdt+2∫Ωα,βa(x,t)∂u∂x∂2u∂x∂tdxdt=-∫αβa(x,0)(∂φ∂x)2dx+∫αβa(x,τ)(∂u∂x)2dx-∫Ωα,β∂a(x,t)∂t(∂u∂x)2dxdt.
By virtue of the Cauchy’s inequality, we obtain(35)∫Ωα,βf∂u∂tdxdt≤2∫Ωα,βf2dxdt+12∫Ωα,β(∂u∂t)2dxdt.
Substituting (34) and (35) into (33) and according to conditions (2), we get
(36)∫αβ(∂u∂t)2dx+∫αβ(∂u∂x)2dx≤c6(∫Ωα,βf2dxdt+∫αβψ2(x)dx+∫αβ(∂φ∂x)2dx+∫Ωα,β(∂u∂t)2dxdt)+c7∫Ωα,β(∂u∂x)2dxdt,
where
(37)c6=c2min(1,c0),c7=max(2,c1)min(1,c0).
By virtue of Lemma  7.1 in [21] and by using it twice, we find
(38)∫αβ(∂u∂t)2dx+∫αβ(∂u∂x)2dx≤c8(∫Ωα,βf2dxdt+∫αβψ2(x)dx+∫αβ(∂φ∂x)2dx),
where
(39)c8=c6exp(c7T)exp(c6Texp(c7T)).(3)
by integrating overΩβτ=Ωβ=(β,1)×(0,τ). Consequently,
(40)∫Ωβℒu·Mu3dxdt=2∫Ωβ(2-x)∂u∂t∂2u∂t2dxdt-2∫Ωβ(2-x)∂u∂t∂∂x(a(x,t)∂u∂x)dxdt=2∫Ωβ(2-x)∂u∂tfdxdt.
Employing integration by parts in (40) and taking into account the boundary conditions in (PR), we obtain
(41)2∫Ωβ(2-x)∂u∂t∂2u∂t2dxdt=∫β1(2-x)(∂u∂t)2dx-∫β1(2-x)ψ2(x)dx,-2∫Ωβ‍(2-x)∂u∂t∂∂x(a(x,t)∂u∂x)dxdt=-2∫0τ((2-x)a(x,t)∂u∂t∂u∂x)|x=0x=αdt+2∫Ωβ(2-x)a(x,t)∂u∂x∂2u∂x∂tdxdt-2∫Ωβa(x,t)∂u∂x∂u∂tdxdt=-∫β1(2-x)a(x,0)(∂φ∂x)2dx+∫β1(2-x)a(x,τ)(∂u∂x)2dx-∫Ωβ(2-x)∂a(x,t)∂t(∂u∂x)2dxdt-2∫Ωβa(x,t)∂u∂x∂u∂tdxdt.
Using the Cauchy’s inequality, we get(42)2∫Ωα(2-x)∂u∂tfdxdt≤2∫Ωαf2dxdt+2∫Ωα(∂u∂t)2dxdt.
Substituting (41) and (42) into (40), we obtain
(43)∫β1(2-x)(∂u∂t)2dx+∫β1(2-x)a(x,τ)(∂u∂x)2dx≤2∫Ωβf2dxdt+2∫Ωβ(∂u∂t)2dxdt+∫β1(2-x)a(x,0)(∂φ∂x)2dx+∫β1‍(2-x)ψ2(x)dx+∫Ωβ(2-x)∂a(x,t)∂t(∂u∂x)2dxdt+2∫Ωβa(x,t)∂u∂x∂u∂tdxdt.
Then, using the Cauchy’sε-inequality and according to conditions (2), we get
(44)∫β1(∂u∂t)2dx+c0∫β1(∂u∂x)2dx≤2∫Ωβf2dxdt+(c1+2)∫Ωβ(∂u∂t)2dxdt+2∫β1ψ2(x)dx+(c1+2c2)∫Ωβ(∂u∂x)2dxdt+2c1∫β1(∂φ∂x)2dx,
so, we obtain
(45)∫β1(∂u∂t)2dx+∫β1(∂u∂x)2dx≤c9(∫Ωβf2dxdt+∫β1ψ2(x)dx+∫β1(∂φ∂x)2dx+∫Ωβ(∂u∂t)2dxdt)+c10∫Ωβ(∂u∂x)2dxdt,
where
(46)c9=2c1+2min(1,c0),c10=c1+2c2min(1,c0).
By virtue of Lemma  7.1 in [21] and by using it twice, we find
(47)∫β1(∂u∂t)2dx+∫β1(∂u∂x)2dx≤c11(∫Ωβf2dxdt+∫β1ψ2(x)dx+∫β1(∂φ∂x)2dx),
where
(48)c11=c9exp(c10T)exp(c9Texp(c10T)).
Combining inequalities (31), (38), and (47), we obtain
(49)∫01(∂u∂t)2dx+∫01(∂u∂x)2dx≤c12(∫Ωf2dxdt+∫01ψ2(x)dx+∫01(∂φ∂x)2dx),
where
(50)c12=max(c5,c8,c11).
The right-hand side of (49) 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=(c12)1/2.Corollary 3.
A solution of the problem (1)–(5) is unique if it exists and depends continuously on ℱ∈F.


## 4. Solvability of the Problem

To show the existence of solutions, we prove thatR(L) is dense in F for all u∈B and for arbitrary ℱ=(f,ϕ,ψ)∈F.Theorem 4.
Suppose the conditions of Theorem2 are satisfied. Then, the problem (1)–(5) admits a unique solution u=L-1ℱ.Proof.
First, we prove thatR(L) is dense in F for the special case where D(L)≡B is reduced to D0(L), where D0(L)={u,u∈D(L):ℓ1u=0,ℓ2u=0}.Proposition 5.
Let the conditions of Theorem4 be satisfied. If for ω∈L2(Ω) and for all u∈D0(L), we have
(51)∫Ωℒu·ωdxdt=0
then, ω vanishes almost everywhere in Ω.Proof.
The scalar product ofF is defined by
(52)(Lu,ω)F=∫Ωℒu·ωdxdt+∫01(∂ℓ1u∂x)(∂ω0∂x)dx+∫01(ℓ2u)(ω1)dx.
the equality (51) can be written as follows:(53)∫Ω∂2u∂t2·ωdxdt=∫Ω∂∂x(a(x,t)∂u∂x)·ωdxdt.
If we put(54)u=ℑt(ektz)=∫0tekτz(x,τ)dτ,
where k is a constant such that kc0-c2≥0 and z,∂z/∂x,(∂/∂x)(a(∂ℑt(ektz)/∂x))∈L2(Ω), then, u satisfies the boundary conditions in (PR). As a result of (53), we obtain the equality
(55)∫Ω∂(ektz)∂tωdxdt=∫Ω∂∂x(a∂ℑt(eτz)∂x)·ωdxdt.
The left-hand side of (55) shows that the mapping
(56)L2(Ω)∋z→∫0T∂∂x(a∂ℑt(eτz)∂x)·ωdt
is a continuous linear functional of z. From the right-hand side of (55), It is true if the function ω has the following properties:
(57)∂ω∂x,ℑt⋆(aω),ℑt⋆(a∂ω∂x),∂∂xℑt⋆(aω),∂∂xℑt⋆(a∂ω∂x)∈L2(Ω),ω(x,0)=0,ω(0,t)=0,ω(1,t)=0;ℑt⋆(aω)|x=1=0.
In terms of the given functionω and from the equality (55), we give the function ω in terms of z as follows:
(58)ω(x,t)=z(x,t),
and z satisfies the same conditions of the function ω.
Replacingω in (55) by its representation (58), we obtain
(59)∫Ω∂(ektz)∂tzdxdt=∫Ω∂∂x(a∂(∫0tekτz(x,τ)dτ)∂x)zdxdt.
Integrating by parts the right-hand side of (59) with respect to x and t and by taking the conditions of the function z yields
(60)∫Ω∂(ektz)∂tzdxdt=∫0T(a(x,t)∂ℑt(ekτz)∂x)(z)|x=0x=1dt=-∫Ωa(x,t)∂(ℑt(ekτz))∂x∂z∂xdxdt=-12∫01e-kta(x,t)(∂ℑt(ekτz)∂x)2|t=0t=Tdx-12∫Ωe-kt(ka(x,t)-∂a(x,t)∂t)×(∂ℑt(ekτz)∂x)2dxdt.
According to condition (2), we obtain
(61)∫Ω∂(ektz)∂tzdxdt≤-12(kc0-c2)∫Ωe-kt(∂ℑt(ekτz)∂x)2dxdt≤0.
Integrating by parts the left-hand side of (61) with respect to t, and by taking the conditions of the function z, we obtain
(62)∫Ω∂(ektz)∂tzdxdt=∫Ω(ekt∂z∂tz+kektz2)dxdt=12∫01ektz(x,T)2dx+32∫Ωkektz2dxdt.
Then, by (61) and (62) we obtain
(63)12∫01ekt(z(x,T))2dx+32∫Ωkektz2dxdt≤0.
And thus z=0 in Ω, hence 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,ω1)∈R(L)⊥ and for all u∈D(L)≡B it holds
(64)(Lu,ω)F=∫Ωℒu·ωdxdt+∫01(∂ℓ1u∂x)(∂ω0∂x)dx+∫01(ℓ2u)(ω1)dx=0.Then, we must prove thatW=0. Putting u∈D0(L) in (64), we have
(65)∫Ωℒu·ωdxdt=0,u∈D0(L).Hence, Proposition5 implies that ω=0. Thus, (64) takes the form
(66)∫01(∂ℓ1u∂x)(∂ω0∂x)dx+∫01(ℓ2u)(ω1)dx=0,u∈D(L).Since the ranges of the trace operatorsℓ1 and ℓ2 are independents and the ranges of values ℓ1 and ℓ2 are everywhere dense in the Hilbert space F with the norm
(67)(∫01[(∂ℓ1u∂x)2+(ℓ2u)2]dx)1/2,
then the equality (66) implies that ω0=0, ω1=0 (we recall satisfies a compatibility conditions). Hence W=0 implies (R(L)¯=F). Therefore, the proof of Theorem 4 is complete.



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