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A priori estimate for resolving the boundary fractional problem

  • The energy inequality method (or a priori estimation) known in classical cases has been adopted for fractional evolution equations associated with initial conditions and boundary integral conditions. We prove the existence and uniqueness of the solution to the problem described in the following.

    Citation: Hacene Mecheri, Maryam G. Alshehri. A priori estimate for resolving the boundary fractional problem[J]. AIMS Mathematics, 2023, 8(1): 765-774. doi: 10.3934/math.2023037

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  • The energy inequality method (or a priori estimation) known in classical cases has been adopted for fractional evolution equations associated with initial conditions and boundary integral conditions. We prove the existence and uniqueness of the solution to the problem described in the following.



    Recently fractional calculus regained ground with its use in various fields, such as physical and chemical processes that occur in media with fractal geometry, as well as in the mathematical modeling of economic and social-biological phenomena [1, Chap. 5, 6, 11]. The reader can also find in the work of Alikhanov [1], Ladyzhenskaya [5], and Nakhushev [8] other rich domains in terms of applications of fractional calculus such as Abel's integral equation [2], fractional-order capacitor models [6,7], electroelectrical chemistry [3,4,8,9,10,11]. Alikhanov [1] used the method of energy inequalities to obtain a priori estimates for solutions of boundary value problems for the diffusion-wave equation with the Caputo fractional derivative. Mesloub's [6] paper deals with a fractional two-times evolution equation associated with initial and purely boundary integral conditions. Our work was inspired by the latter but with a fractional one-time evolution equation with boundary integral conditions. In the present paper, we use the method of energy inequalities to obtain a priori estimates to prove the uniqueness and existence of the solution using some functional inequalities, the Caputo derivative and Caputo integrals. In Section 1, we formulate and set our nonlocal initial boundary value problem. In Section 2, we introduce some useful preliminaries and notations, and reformulate the problem (1.1)–(1.2) as a problem with homogeneous boundary conditions, (1.3) and (1.4) to simplify the calculations. In Section 3, we apply the a priori estimation method to prove the uniqueness of the solution. We choose a certain functional differential operator multiplier (3.3). In Section 4, we discuss the existence of a solution. We always apply the same method.

    In the bounded domain, QT=(0,T)×(0,α) such that 0<α<,0<T<.

    We consider the time fractional order problem with the Caputo-derivative.

    cβtut(a(x,t)ux)x+b(x,t)ux+c(x,t)u=g(x,t). (1.1)

    The functions a(x,t),b(x,t) and c(x,t),\ satisfy the conditions

    P1:{0a(x,t)a1,0ax(x,t)a2,0b(x,t)b1,0bx(x,t)b2,0c(x,t)c,}(x,t)QT,

    where

    aiIR+,¯i=12,bkIR+,¯k=12,cIR+.

    With problem (1.1), we associate the initial conditions

    u(x,0)=φ(x)0,0<x<α, (1.2)

    and the boundary conditions

    α0u(x,t)dx=0,  α0xu(x,t)dx=0,    u(α,t)dx=0. (1.3)
    α0φ(x)dx=0, (1.4)
    α0xφ(x)dx=0. (1.5)

    For a function f(t), which has absolutely continuous derivatives up to order (n -1), the Riemann-Liouville fractional derivative of arbitrary order n1βn is defined as follows [6]:

    Dβf(t)=1Γ(nβ)ntnt0f(s)(ts)βn+1ds, (2.1)

    where Γ is the well-known gamma function. The following formula is true:

    Dβf(t)=n1k=0tkβΓ(kβ+1)fk(0)+1Γ(nβ)t0f(s)(ts)βn+1ds. (2.2)

    The Riemann-Liouville fractional integral of order\ β, \ 0βn,\ is defined by

    Dβtf(t)=1Γ(β)t0f(n)(s)(ts)βn+1ds=Dβf(t). (2.3)

    The Riemann-Liouville fractional derivative is singular at the origin because of (2.3). Because of this fact, fractional differential equations in the sense of the Riemann-Liouville fractional derivative require initial conditions of a special form lacking clear physical interpretation. These shortcomings do not occur with the regularized Caputo derivative. The Caputo derivative of order β, n-1βn,nIN, is defined as the integral part of (2.3); that is,

    cDβf(t)=1Γ(nβ)t0f(s)(ts)βn+1ds=Dβf(t)=Dβf(t)n1k=0tkβΓ(kβ+1)fk(0). (2.4)

    In the case of our problem, the Caputo and Riemann-Liiouville derivatives of order β,0β1, and their linked relationship formulas are defined as follows [4] $:

    cβtu(x,t)=1Γ(1β)t0u(x,s)s(ts)βds=βtu(x,t)u(x,0)tβΓ(1β), (2.5)
    βtu(x,t)=1Γ(1β)t0u(x,s)(ts)βds. (2.6)

    (1) The Cauchy εinequality:

    αβε2α2+12εβ2. (2.7)

    (2) Poincarˊe-type inequalities [3] $:

    xu2L2(QT)α2u2L2(QT), (2.8)
    mu2L2(QT)a2m1xu2L2(QT)mIN.

    (3) Lemma. For any function f(t) absolutely continuous on [0, T], owe have the inequality

    cDβtf2(t) 2f(t)cDβtf(t). (2.9)

    Because the proofs are based on a priori estimates, we first write (1.1)–(1.3) in an equivalent operator form in order to establish the existence and uniqueness of the solution. The solution to (1.1)–(1.3) can be regarded as the solution of the operator equation

    Lu=H, (2.10)

    where L=(L,l1), and operator L acts from B to F, with the domain of definition

    D(L)={uL2(QT),βtut(x,t),ux,utL2(QT)α0ut(x,t)dx=0,t(0,T)u(x,0)=φ(x)0α0φ(x)dx=0} (2.11)

    where B is the Banach space of functions u endowed by the finite norm

    u2B=T0(βtxut2L2(0.α)+u2L2(0.α))dt. (2.12)

    F is a Hilbert space equipped with the scalar product:

    (H,H)F=T0((H,H)L2(0.α))dt. (2.13)

    The associated norm is

    H2F=t0HL2(0.α)dt=T0(g2L2(0.α)+φ2L2(0.α))dt. (2.14)

    Theorem 1. In QT and for sufficiently small ε, there exists a positive constant K independent of u such that

    T0(βtxut2L2(0.α)+u2L2(0.α))dtKT0(H2L2(0.α))dt, (3.1)

    for all uD(L), where

    H2L2(QT)=(g2L2(QT)+φ2L2(0,α))

    and

    K=max(ε,a1)3min(2(ab2+a1+αc+b1+αε),(a1ε+cε+b1ε+b2ε)), (3.2)
    xut=x0ut(η,t)dη. (3.3)

    Proof. Let uD(L), and consider the equality

    (cβtut((a(x,t)ux)x+b(x,t)ux+c(x,t)u) . 2xut)L2(QT)=(Lu . 2xut)L2(QT)
    =(Lu . 2xut)L2(QT)
    =(g . 2xut)L2(QT), (3.4)

    where

    2xut=x0η0ut(ζ,t)dζ^dη.

    Upon integration by parts and conditions (1.1)–(1.3), we evaluate each term of (3.4) to obtain

    (cβtut,2xut)L2(QT)  =T0α0 cβtut 2xutdx= T0x(cβtut)2xutx=αx=0dt
    +T0α0 cx(βtut)xutdx=T0α0 cβt (xut)xutdxdt. (3.5)

    Using the same method, we have

    ((a(x,t)ux)x,2xut)L2(QT)=T0α0(a(x,t)ux)x2xutdxdt
    =T0(a(x,t)ux2xut)x=αx=0dtT0α0a(x,t)uxxutdxdt
    =T0α0a(x,t)uxxutdx=T0a(x,t)uxutx=αx=0dt
    +T0α0a(x,t)uutdxdt+T0α0ax(x,t)uxutdxdt
    =12α0a(x,t)u2(x,T)dx12α0a(x,t)φ2(x)dx+T0α0ax(x,t)uxutdxdt (3.6)
    (b(x,t)ux,2xut)L2(QT)=T0α0b(x,t)ux2xutdxdt.
    =T0b(x,t)u2xutx=αx=0dt+T0α0bx(x,t)u2xutdxdt+T0α0b(x,t)uxutdxdt
    =T0α0bx(x,t)u2xutdxdt+T0α0b(x,t)uxutdxdt. (3.7)
    (c(x,t)u,2xut)L2(QT)=T0α0c(x,t)u2xutdxdt. (3.8)

    Substituting (3.5), (3.6), (3.7) and (3.8) into (3.5), we get

    T0α0 cβt (xut)xutdxdt+12α0a(x,t)u2(x,T)dx
    =12α0a(x,t)φ2(x)dxT0α0bx(x,t)u2xutdxdtT0α0b(x,t)uxutdxdt
    T0α0c(x,t)u2xutdxdtT0α0ax(x,t)uxutdxdt+(g,2xut)L2(QT). (3.9)

    By using poincarˊe-type inequalities and a Cauchy-ε inequality estimate last five terms of the left side of (3.9), we get

    T0α0bx(x,t)u2xutdxdtb2ε2u2L2(QT)+αb22εut2L2(QT), (3.10)
    T0α0b(x,t)uxutdxdtb1ε2u2L2(QT)+b12εxut2L2(QT), (3.11)
    T0α0c(x,t)u2xutdxdtc1ε2u2L2(QT)+αc2εxut2L2(QT), (3.12)
    T0α0ax(x,t)uxutdxdta1ε2u2L2(QT)+a12εxut2L2(QT), (3.13)
    (g,2xut)L2(QT)ε2g2L2(QT)+α4εxut2L2(QT). (3.14)

    Combining of equalities (3.10), (3.14) and (3.9) yields

    T0α0 cβt (xut)xutdxdt+(1(αb2+a1+αc+b1+α2ε))xut2L2(QT)
    (a1ε+c1ε+b1ε+b2ε2)u2L2(QT)ε2g2L2(QT)+a12φ2L2(0,α). (3.15)

    Consequently,

    2T0α0 (cβt xut)xutdxdt+u2L2(QT)kH2L2(QT), (3.16)

    where

    H2L2(QT)=(g2L2(QT)+φ2L2(0,α))
    K=max(ε,a1)min(2(αb2+a1+αc+b1+αε), (a1ε+cε+b1ε+b2ε)).

    Estimating the first term on the left side of (3.15), we get

    cβtxut2L2(QT)=T0α0 cβt (xut)2dxdt
    2T0a0 cβt (xut)(xut)dxdt. (3.17)

    Thus, inequality (3.16) takes the form

    βtxut2L2(QT)+u2L2(QT)KH2L2(QT), (3.18)

    where

    K=max(ε,a1)min(2(αb2+a1+αc+b1+αε), (a1ε+cε+b1ε+b2ε)).

    To establish the existence of the solution to problems (1.1)–(1.6), we argueusing density argument. That is, we show that ImL, the image of the operator L is dense in the space L2(Q) for every element u in the Banach space B. For this, we consider the following theorem:

    Theorem 2. For all functions uB, and for some function zL2(QT),

    (Lu.Z)L2(QT)=0. (4.1)

    Then z is zero a.e in the domain QT.

    Proof. We see from (4.1) that

    (cβtut(a(x,t)ux)x+b(x,t)ux+c(x,t)u,z)L2(QT)=0. (4.2)

    Because (4.1) holds for any function u in B, it can be expressed in a particular form. Assume that a function σ(x,t) satisfies the conditions (1.3)–(1.4) such that

    σ,(xtσ(x,s))x,cβtσL2(QT).

    From the previous discussion, we introduce the function

    u=t0s0σ(x,s)dsdt=2tσ. (4.3)

    Equation (4.2) reduces to

    (cβttσ(a(x,t)2tσx)x+b(x,t)tσ+c(x,t)2tσ,z)L2(QT)=0, (4.4)

    where

    z(x,t)=x0ξ0t0ut(η,s)dsdηdζ=2xtσ. (4.5)

    In this case, equation (4.4) can be written in the form

    (βttσ,2xtut)L2(QT)((a(x,t)2tσx)x,2xtσ)L2(QT)
    +(b(x,t)tσ,2xtσ)L2(QT)+(c(x,t)2tσ,2xtσ)L2(QT)=0. (4.6)

    Integrating by parts, and keeping in mind that the function satisfies (1.2)–(1.4), we deduce the following expressions for each term in (4.5)

    (βttσ,2xtσ)L2(QT)=T0α0 (βt tσ)2xtσdxdt
    =T0x(cβt tσ)2xtσx=αx=0dt+T0α0(cxβttσ)xtσdxdt
    =(βt(xtσ),xtσ)L2(QT). (4.7)
    ((a(x,t)2tσx)x,2xtσ)L2(QT)=T0α0(a(x,t)2tσx)x(2xtσ)dxdt
    =T0a(x,t)2tσx(2xtσ)x=αx=0dtT0α0a(x,t)2tσxxtσdxdt
    =T0α0a(x,t)2tσxxtσdxdt
    =T0a(x,t)2tσ(xtσ)x=αx=0dt+T0α0a(x,t)(2tσ)tσdxdt
    +T0α0ax(x,t)(2tσ)xtσdxdt
    =12α0a(x,t)(2tσ)2t=Tt=0dx+T0α0ax(x,t)(2tσ)xtσdxdt
    =12α0a(x,t)(2tσ(x,T))2+T0α0ax(x,t)(2tσ)xtσdxdt. (4.8)
    (b(x,t)tσ,2xtσ)L2(QT)=T0α0b(x,t)tσ(2xtσ)dxdt
    T0b(x,t)xtσ(2xtσ)x=αx=0dt+T0α0b(x,t)(xtσ)2dxdt
    +T0α0b(x,t)xtσ(2xtσ)dxdt
    =T0α0b(x,t)(xtσ)2dxdt+T0α0b(x,t)xtσ(2xtσ)dxdt. (4.9)
    (c(x,t)2tσ,2xtσ)L2(QT)=T0α0c(x,t)2tσ(2xtσ)dxdt. (4.10)

    Substituting formulas (4.7) and (4.10) into (4.6), we get

    (βt(xtσ),xtσ)L2(QT)
    =12α0a(x,t)(2tσ(x,T))2T0α0ax(x,t)(2tσ)xtσdxdt
    T0α0b(x,t)(xtσ)2dxdtT0α0b(x,t)xtσ(2xtσ)dxdt
    +T0α0c(x,t)2tσ(2xtσ)dxdt. (4.11)

    By using Poincaré-type inequalities and the Cauchy-ε inequality estimate for the last four terms of left side of (4.12), we get

    T0α0ax(x,t)(2tσ)xtσdxdta2ε22tσ2L2(QT)+a22εxtσ2L2(QT). (4.12)
    T0α0b(x,t)(xtσ)2dxdt b1xtσ2L2(QT) (4.13)
    T0α0b(x,t)xtσ(2xtσ)dxdtb1ε2xtσ2L2(QT)+b1α4εxtσ2L2(QT). (4.14)
    T0α0c(x,t)2tσ(2xtσ)dxdtc1ε22tσ2L2(QT)+c1α2εxtσ2L2(QT). (4.15)

    Combining equalities (4.12)–(4.15) and (4.12) yields

    (βt(xtσ),xtσ)L2(QT)(b1+(2a2+b1α+2c1α4ε))xtσ2L2(QT)
    (a2ε+c1ε2)2tσ2L2(QT)12a(x,t)(2tσ(x,T))2L2(0.α). (4.16)

    We have, by lemma 3,

    βt(xtσ)22(βt(xtσ),xtσ)L2(QT). (4.17)

    Consequently, in light of (4.18) and (4.17), inequality (4.17) becomes

    βt(xtσ)2(2b1+(2a2+b1α+2c1α2ε))xtσ2L2(QT)
    (a2ε+c1ε)2tσ2L2(QT)a(x,t)(2tσ(x,T)2L2(0.α). (4.18)

    Take ε, such that

    (2b1+(2a2+b1α+2c1α2ε))0.

    Consequently,

    βt(xtσ)2
    (a2ε+c1ε)2tσ2L2(QT)a(x,t)(2tσ(x,T)2L2(0.α)0. (4.19)

    We conclude that z is zero a.e in the domain QT.

    The authors declare no conflicts of interest.



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