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Research article Special Issues

A priori estimate for resolving the boundary fractional problem

  • Received: 20 February 2022 Revised: 12 June 2022 Accepted: 20 June 2022 Published: 12 October 2022
  • MSC : 76D03, 76N10

  • 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) (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

    \begin{equation} Lu = H, \end{equation} (2.10)

    where L = (\mathfrak{L}, l_{1}) , and operator L acts from B to F, with the domain of definition

    \begin{equation} D(L) = \left\{\begin{array}{c} u\in L^{2}(Q^{T}), \partial _{t}^{\beta }u_{t}(x, t), u_{x}, u_{t}\in L^{2}(Q^{T}) \\ \int_{0}^{\alpha}u_{t}(x, t)dx = 0, \forall t\in (0, T)\\ u(x, 0) = \varphi (x)\geq 0 \\ \int_{0}^{\alpha}\varphi (x)dx = 0 \end{array} \right\} \end{equation} (2.11)

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

    \begin{equation} \left\Vert u\right\Vert _{B}^{2} = \int_{0}^{T}\left( \partial_{t}^{\beta }\left\Vert \Im _{x}u_{t}\right\Vert_{L^{2}(0.\alpha)}^{2}+\left\Vert u\right\Vert _{L^{2}(0.\alpha)}^{2}\right) dt. \end{equation} (2.12)

    F is a Hilbert space equipped with the scalar product:

    \begin{equation} \left( H, H^{\ast }\right) _{F} = \int_{0}^{T}\left(( H, H^{\ast })_{L^{2}(0.\alpha)}\right) dt. \end{equation} (2.13)

    The associated norm is

    \begin{equation} \left\Vert H\right\Vert _{F}^{2} = \int^{t}_{0}\|H\|_{{L^{2}(0.\alpha)}}dt = \int_{0}^{T} (\|g\|^{2}_{L^{2}(0.\alpha)} +\|\varphi\|^{2}_{L^{2}(0.\alpha)})dt. \end{equation} (2.14)

    Theorem 1. In Q^{T} and for sufficiently small \varepsilon , there exists a positive constant K independent of u such that

    \begin{equation} \int_{0}^{T}\left( \partial _{t}^{\beta }\left\Vert \Im_{x}u_{t}\right\Vert _{L^{2}(0.\alpha)}^{2}+\left\Vert u\right\Vert _{L^{2}(0.\alpha)}^{2}\right) dt\leq K\int_{0}^{T}\left( \left\Vert H\right\Vert _{L^{2}(0.\alpha)}^{2}\right) dt, \end{equation} (3.1)

    for all u\in D\left(L\right) , where

    \begin{equation*} \left\Vert H\right\Vert_{L^{2}(Q^{T})}^{2} = \left( \left\Vert g\right\Vert_{L^{2}(Q^{T})}^{2}+\left\Vert \varphi \right\Vert _{L^{2}(0, \alpha)}^{2}\right) \end{equation*}

    and

    \begin{equation} K = \frac{\max (\varepsilon , a_{1})}{3\min (2-(\frac{ab_{2}+a_{1}+\alpha c+b_{1}+\alpha}{\varepsilon }), (a_{1}\varepsilon +c\varepsilon +b_{1}\varepsilon +b_{2}\varepsilon ))}, \end{equation} (3.2)
    \begin{equation} \Im _{x}u_{t} = \int_{0}^{x}u_{t}(\eta , t)d\eta . \end{equation} (3.3)

    Proof. Let u\in D(L), and consider the equality

    \begin{equation*} (^{c}\partial _{t}^{\beta }u_{t}-((a(x, t)\frac{\partial u}{\partial x})_{x}+b(x, t)u_{x}+c(x, t)u) \ . \ -\Im_{x} ^{2}u_{t})_{L^{2}(Q^{T})} = (\mathcal{L}u \ . \ - \Im_{x} ^{2}u_{t})_{L^{2}(Q^{T})} \end{equation*}
    \begin{equation*} = (\mathcal{L}u \ . \ - \Im_{x} ^{2}u_{t})_{L^{2}(Q^{T})} \end{equation*}
    \begin{equation} = ( g \ . \ - \Im_{x} ^{2}u_{t})_{L^{2}(Q^{T})}, \end{equation} (3.4)

    where

    \begin{equation*} \Im _{x}^{2}u_{t} = \int_{0}^{x}\int_{0}^{\eta }u_{t}(\zeta , t)d\zeta \widehat{}d\eta . \end{equation*}

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

    \begin{equation*} \left( ^{c}\partial _{t}^{\beta }u_{t}, -\Im _{x}^{2}u_{t}\right)_{_{L^{2}(Q^{T})}}\ \ = -\int_{0}^{T}\int_{0}^{\alpha}\ ^{c}\partial _{t}^{\beta }u_{t}\ \Im _{x}^{2}u_{t}dx = -\ \int_{0}^{T}\Im _{x}\left( ^{c}\partial _{t}^{\beta }u_{t}\right) \Im _{x}^{2}u_{t}\mid _{x = 0}^{x = \alpha}dt \end{equation*}
    \begin{equation} +\int_{0}^{T}\int_{0}^{\alpha}\ \Im_{x}^{c}(\partial _{t}^{\beta }u_{t})\Im _{x}u_{t}dx = \int_{0}^{T}\int_{0}^{\alpha}\ ^{c}\partial _{t}^{\beta }\ \left( \Im_{x}u_{t}\right) \Im _{x}u_{t}dxdt. \end{equation} (3.5)

    Using the same method, we have

    \begin{equation*} \left( -(a(x, t)\frac{\partial u}{\partial x})_{x}, -\Im _{x}^{2}u_{t}\right)_{_{L^{2}(Q^{T})}} = \int_{0}^{T}\int_{0}^{\alpha}(a(x, t)\frac{\partial u}{ \partial x})_{x}\Im _{x}^{2}u_{t}dxdt \end{equation*}
    \begin{equation*} = \int_{0}^{T}\left( a(x, t)\frac{\partial u}{\partial x}\Im_{x}^{2}u_{t}\right) \mid _{x = 0}^{x = \alpha}dt-\int_{0}^{T}\int_{0}^{\alpha}a(x, t)\frac{\partial u}{\partial x}\Im _{x}u_{t}dxdt \end{equation*}
    \begin{equation*} = -\int_{0}^{T}\int_{0}^{\alpha}a(x, t)\frac{\partial u}{\partial x}\Im_{x}u_{t}dx = -\int_{0}^{T}a(x, t)u\Im _{x}u_{t}\mid _{x = 0}^{x = \alpha}dt \end{equation*}
    \begin{equation*} +\int_{0}^{T}\int_{0}^{\alpha}a(x, t)u{u}_{t}dxdt +\int_{0}^{T}\int_{0}^{\alpha}a_{x}(x, t)u\Im _{x}u_{t}dxdt \end{equation*}
    \begin{equation} = \frac{1}{2}\int_{0}^{\alpha}a(x, t)u^{2}(x, T)dx-\frac{1}{2}\int_{0}^{\alpha}a(x, t)\varphi^{2}(x)dx+\int_{0}^{T}\int_{0}^{\alpha}a_{x}(x, t)u\Im_{x}u_{t}dxdt \end{equation} (3.6)
    \begin{equation*} \left( b(x, t)u_{x}, -\Im _{x}^{2}u_{t}\right)_{_{L^{2}(Q^{T})}} = -\int_{0}^{T}\int_{0}^{\alpha}b(x, t)u_{x}\Im_{x}^{2}u_{t}dxdt. \end{equation*}
    \begin{equation*} = -\int_{0}^{T}b(x, t)u\Im _{x}^{2}u_{t}\mid_{x = 0}^{x = \alpha}dt+\int_{0}^{T}\int_{0}^{\alpha}b_{x}(x, t)u\Im_{x}^{2}u_{t}dxdt +\int_{0}^{T}\int_{0}^{\alpha}b(x, t)u\Im _{x}u_{t}dxdt \end{equation*}
    \begin{equation} = \int_{0}^{T}\int_{0}^{\alpha}b_{x}(x, t)u\Im_{x}^{2}u_{t}dxdt+\int_{0}^{T}\int_{0}^{\alpha}b(x, t)u\Im _{x}u_{t}dxdt. \end{equation} (3.7)
    \begin{equation} \left( c(x, t)u, -\Im _{x}^{2}u_{t}\right)_{_{L^{2}(Q^{T})}} = \int_{0}^{T}\int_{0}^{\alpha}c(x, t)u\Im_{x}^{2}u_{t}dxdt. \end{equation} (3.8)

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

    \begin{equation*} \int_{0}^{T}\int_{0}^{\alpha}\ ^{c}\partial _{t}^{\beta }\ \left( \Im_{x}u_{t}\right) \Im _{x}u_{t}dxdt +\frac{1}{2}\int_{0}^{\alpha}a(x, t)u^{2}(x, T)dx \end{equation*}
    \begin{equation*} = \frac{1}{2}\int_{0}^{\alpha}a(x, t)\varphi^{2}(x)dx-\int_{0}^{T}\int_{0}^{\alpha}b_{x}(x, t)u\Im_{x}^{2}u_{t}dxdt-\int_{0}^{T}\int_{0}^{\alpha}b(x, t)u\Im _{x}u_{t}dxdt \end{equation*}
    \begin{equation} -\int_{0}^{T}\int_{0}^{\alpha}c(x, t)u\Im_{x}^{2}u_{t}dxdt-\int_{0}^{T}\int_{0}^{\alpha}a_{x}(x, t)u\Im_{x}u_{t}dxdt+\left( g, -\Im _{x}^{2}u_{t}\right) _{_{L^{2}(Q^{T})}}. \end{equation} (3.9)

    By using poincar\acute{e} -type inequalities and a Cauchy- \varepsilon inequality estimate last five terms of the left side of (3.9), we get

    \begin{equation} -\int_{0}^{T}\int_{0}^{\alpha}b_{x}(x, t)u\Im_{x}^{2}u_{t}dxdt\leq \frac{b_{2}\varepsilon }{2}\left\Vert u\right\Vert_{L^{2}(Q^{T})}^{2}+\frac{\alpha b_{2}}{2\varepsilon }\left\Vert \Im u_{t}\right\Vert _{L^{2}(Q^{T})}^{2}, \end{equation} (3.10)
    \begin{equation} -\int_{0}^{T}\int_{0}^{\alpha}b(x, t)u\Im _{x}u_{t}dxdt\leq \frac{b_{1}\varepsilon }{2}\left\Vert u\right\Vert _{L^{2}(Q^{T})}^{2}+\frac{b_{1}}{2\varepsilon }\left\Vert \Im _{x}u_{t}\right\Vert _{L^{2}(Q^{T})}^{2}, \end{equation} (3.11)
    \begin{equation} -\int_{0}^{T}\int_{0}^{\alpha}c(x, t)u\Im _{x}^{2}u_{t}dxdt\leq \frac{c_{1}\varepsilon }{2}\left\Vert u\right\Vert _{L^{2}(Q^{T})}^{2}+\frac{\alpha c }{2\varepsilon }\left\Vert \Im _{x}u_{t}\right\Vert _{L^{2}(Q^{T})}^{2}, \end{equation} (3.12)
    \begin{equation} -\int_{0}^{T}\int_{0}^{\alpha}a_{x}(x, t)u\Im _{x}u_{t}dxdt\leq\frac{a_{1}\varepsilon }{2}\left\Vert u\right\Vert _{L^{2}(Q^{T})}^{2}+\frac{ a_{1}}{2\varepsilon }\left\Vert \Im _{x}u_{t}\right\Vert _{L^{2}(Q^{T})}^{2}, \end{equation} (3.13)
    \begin{equation} -\left( g, \Im _{x}^{2}u_{t}\right) _{_{L^{2}(Q^{T})}}\leq \frac{\varepsilon }{2}\left\Vert g\right\Vert _{L^{2}(Q^{T})}^{2}+\frac{\alpha}{4\varepsilon }\left\Vert \Im _{x}u_{t}\right\Vert _{L^{2}(Q^{T})}^{2}. \end{equation} (3.14)

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

    \begin{equation*} \int_{0}^{T}\int_{0}^{\alpha}\ ^{c}\partial _{t}^{\beta }\ \left( \Im_{x}u_{t}\right) \Im _{x}u_{t}dxdt+(1-(\frac{\alpha b_{2}+a_{1}+\alpha c+b_{1}+\alpha}{ 2\varepsilon }))\left\Vert \Im _{x}u_{t}\right\Vert _{L^{2}(Q^{T})}^{2} \end{equation*}
    \begin{equation} -(\frac{a_{1}\varepsilon +c_{1}\varepsilon +b_{1}\varepsilon + b_{2}\varepsilon }{2})\left\Vert u\right\Vert _{L^{2}(Q^{T})}^{2}\leq \frac{ \varepsilon }{2}\left\Vert g\right\Vert _{L^{2}(Q^{T})}^{2}+\frac{a_{1}}{2}\left\Vert \varphi \right\Vert _{L^{2}(0, \alpha)}^{2}. \end{equation} (3.15)

    Consequently,

    \begin{equation} 2\int_{0}^{T}\int_{0}^{\alpha}\ \left( ^{c}\partial _{t}^{\beta }\ \Im_{x}u_{t}\right) \Im _{x}u_{t}dxdt + \left\Vert u\right\Vert _{L^{2}(Q^{T})}^{2}\leq k\left\Vert H\right\Vert_{L^{2}(Q^{T})}^{2}, \end{equation} (3.16)

    where

    \begin{equation*} \left\Vert H\right\Vert_{L^{2}(Q^{T})}^{2} = \left( \left\Vert g\right\Vert_{L^{2}(Q^{T})}^{2}+\left\Vert \varphi \right\Vert _{L^{2}(0, \alpha)}^{2}\right) \end{equation*}
    \begin{equation*} K = \frac{\max (\varepsilon , a_{1})}{\min (2-(\frac{\alpha b_{2}+a_{1}+\alpha c+b_{1}+\alpha}{\varepsilon }), \ (a_{1}\varepsilon +c\varepsilon +b_{1}\varepsilon +b_{2}\varepsilon ))}. \end{equation*}

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

    \begin{equation*} ^{c}\partial _{t}^{\beta }\left\Vert \Im _{x}u_{t}\right\Vert _{L^{2}(Q^{T})}^{2} = \int_{0}^{T}\int_{0}^{\alpha}\ ^{c}\partial _{t}^{\beta}\ \left( \Im _{x}u_{t}\right) ^{2}dxdt \end{equation*}
    \begin{equation} \leq 2\int_{0}^{T}\int_{0}^{a}\ ^{c}\partial _{t}^{\beta }\ (\Im_{x}u_{t})(\Im _{x}u_{t})dxdt. \end{equation} (3.17)

    Thus, inequality (3.16) takes the form

    \begin{equation} \partial _{t}^{\beta }\left\Vert \Im _{x}u_{t}\right\Vert_{L^{2}(Q^{T})}^{2}+\left\Vert u\right\Vert _{L^{2}(Q^{T})}^{2}\leq K\left\Vert H\right\Vert _{L^{2}(Q^{T})}^{2}, \end{equation} (3.18)

    where

    \begin{equation*} K = \frac{\max (\varepsilon , a_{1})}{\min (2-(\frac{\alpha b_{2}+a_{1}+\alpha c+b_{1}+\alpha}{\varepsilon }), \ (a_{1}\varepsilon +c\varepsilon +b_{1}\varepsilon +b_{2}\varepsilon ))}. \end{equation*}

    To establish the existence of the solution to problems (1.1)–(1.6), we argueusing density argument. That is, we show that Im L , the image of the operator L is dense in the space L^{2}(Q) for every element u in the Banach space B. For this, we consider the following theorem:

    Theorem 2. For all functions u\in B, and for some function z\in L^{2}(Q^{T}) ,

    \begin{equation} (Lu.Z)_{L^{2}(Q^{T})} = 0. \end{equation} (4.1)

    Then z is zero a.e in the domain Q^{T} .

    Proof. We see from (4.1) that

    \begin{equation} (^{c}\partial _{t}^{\beta }u_{t}-(a(x, t)\frac{\partial u}{\partial x})_{x}+b(x, t)u_{x}+c(x, t)u, z)_{_{_{L^{2}(Q^{T})}}} = 0. \end{equation} (4.2)

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

    \begin{equation*} \sigma , \left( x\Im _{t}\sigma \left( x, s\right) \right) _{x}, ^{c}\partial_{t}^{\beta }\sigma \in L^{2}(Q^{T}). \end{equation*}

    From the previous discussion, we introduce the function

    \begin{equation} u = \int_{0}^{t}\int_{0}^{s}\sigma \left( x, s\right) dsdt = \Im _{t}^{2}\sigma . \end{equation} (4.3)

    Equation (4.2) reduces to

    \begin{equation} (^{c}\partial _{t}^{\beta }\Im _{t}\sigma -(a(x, t)\frac{\partial \Im _{t}^{2}\sigma }{\partial x})_{x}+b(x, t)\Im _{t}\sigma +c(x, t)\Im _{t}^{2}\sigma , z)_{_{_{L^{2}(Q^{T})}}} = 0, \end{equation} (4.4)

    where

    \begin{equation} z\left( x, t\right) = -\int_{0}^{x}\int_{0}^{\xi }\int_{0}^{t}u_{t}\left( \eta, s\right) dsd\eta d\zeta = -\Im _{x}^{2}\Im _{t}\sigma . \end{equation} (4.5)

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

    \begin{equation*} \left( \partial _{t}^{\beta }\Im _{t}\sigma , -\Im _{x}^{2}\Im_{t}u_{t}\right) _{_{L^{2}(Q^{T})}} -\left( (a(x, t)\frac{\partial \Im _{t}^{2}\sigma }{\partial x})_{x}, -\Im_{x}^{2}\Im _{t}\sigma \right) _{_{L^{2}(Q^{T})}} \end{equation*}
    \begin{equation} +\left( b(x, t)\Im_{t}\sigma , -\Im _{x}^{2}\Im _{t}\sigma \right)_{{_{L^{2}(Q^{T})}}}+\left( c(x, t)\Im _{t}^{2}\sigma , -\Im _{x}^{2}\Im _{t}\sigma \right)_{{_{L^{2}(Q^{T})}}} = 0. \end{equation} (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)

    \begin{equation*} -\left( \partial _{t}^{\beta }\Im _{t}\sigma , \Im _{x}^{2}\Im _{t}\sigma\right) _{_{L^{2}(Q^{T})}} = -\int_{0}^{T}\int_{0}^{\alpha}\ \left(\partial _{t}^{\beta }\ \Im _{t}\sigma \right) \Im _{x}^{2}\Im _{t}\sigma dxdt \end{equation*}
    \begin{equation*} = -\int_{0}^{T}\Im _{x}\left( ^{c}\partial _{t}^{\beta }\ \Im _{t}\sigma\right) \Im _{x}^{2}\Im _{t}\sigma \mid_{x = 0}^{x = \alpha}dt +\int_{0}^{T}\int_{0}^{\alpha}(\Im _{x}^{c}\partial_{t}^{\beta }\Im _{t}\sigma )\Im _{x}\Im _{t}\sigma dxdt \end{equation*}
    \begin{equation} = \left( \partial _{t}^{\beta }(\Im _{x}\Im _{t}\sigma ), \Im _{x}\Im_{t}\sigma \right) _{_{L^{2}(Q^{T})}}. \end{equation} (4.7)
    \begin{equation*} \left( (a(x, t)\frac{\partial \Im _{t}^{2}\sigma }{\partial x})_{x}, \Im_{x}^{2}\Im _{t}\sigma \right)_{_{L^{2}(Q^{T})}} = \int_{0}^{T}\int_{0}^{\alpha}(a(x, t)\frac{\partial \Im_{t}^{2}\sigma }{\partial x})_{x}(\Im _{x}^{2}\Im _{t}\sigma )dxdt \end{equation*}
    \begin{equation*} = \int_{0}^{T}a(x, t)\frac{\partial \Im _{t}^{2}\sigma }{\partial x}(\Im_{x}^{2}\Im _{t}\sigma )\mid_{x = 0}^{x = \alpha}dt -\int_{0}^{T}\int_{0}^{\alpha}a(x, t)\frac{\partial \Im_{t}^{2}\sigma }{\partial x}\Im _{x}\Im _{t}\sigma dxdt \end{equation*}
    \begin{equation*} = -\int_{0}^{T}\int_{0}^{\alpha}a(x, t)\frac{\partial \Im _{t}^{2}\sigma }{\partial x}\Im _{x}\Im _{t}\sigma dxdt \end{equation*}
    \begin{equation*} = -\int_{0}^{T}a(x, t)\Im _{t}^{2}\sigma (\Im _{x}\Im _{t}\sigma )\mid_{x = 0}^{x = \alpha}dt +\int_{0}^{T}\int_{0}^{\alpha}a(x, t)(\Im _{t}^{2}\sigma)\Im _{t}\sigma dxdt \end{equation*}
    \begin{equation*} +\int_{0}^{T}\int_{0}^{\alpha}a_{x}(x, t)(\Im _{t}^{2}\sigma )\Im_{x}\Im _{t}\sigma dxdt \end{equation*}
    \begin{equation*} = \frac{1}{2}\int_{0}^{\alpha}a(x, t)(\Im _{t}^{2}\sigma )^{2}\mid_{t = 0}^{t = T}dx +\int_{0}^{T}\int_{0}^{\alpha}a_{x}(x, t)(\Im_{t}^{2}\sigma )\Im _{x}\Im _{t}\sigma dxdt \end{equation*}
    \begin{equation} = \frac{1}{2}\int_{0}^{\alpha}a(x, t)(\Im _{t}^{2}\sigma(x, T))^{2} +\int_{0}^{T}\int_{0}^{\alpha}a_{x}(x, t)(\Im_{t}^{2}\sigma )\Im _{x}\Im _{t}\sigma dxdt. \end{equation} (4.8)
    \begin{equation*} \left( b(x, t)\Im _{t}\sigma , -\Im _{x}^{2}\Im _{t}\sigma \right)_{{_{L^{2}(Q^{T})}}} = -\int_{0}^{T}\int_{0}^{\alpha}b(x, t)\Im _{t}\sigma (\Im _{x}^{2}\Im_{t}\sigma )dxdt \end{equation*}
    \begin{equation*} -\int_{0}^{T}b(x, t)\Im _{x}\Im _{t}\sigma (\Im _{x}^{2}\Im _{t}\sigma )\mid_{x = 0}^{x = \alpha}dt +\int_{0}^{T}\int_{0}^{\alpha}b(x, t)(\Im _{x}\Im _{t}\sigma)^{2}dxdt \end{equation*}
    \begin{equation*} +\int_{0}^{T}\int_{0}^{\alpha}b(x, t)\Im _{x}\Im _{t}\sigma (\Im_{x}^{2}\Im _{t}\sigma )dxdt \end{equation*}
    \begin{equation} = \int_{0}^{T}\int_{0}^{\alpha}b(x, t)(\Im _{x}\Im _{t}\sigma)^{2}dxdt +\int_{0}^{T}\int_{0}^{\alpha}b(x, t)\Im _{x}\Im _{t}\sigma (\Im_{x}^{2}\Im _{t}\sigma )dxdt. \end{equation} (4.9)
    \begin{equation} \left( c(x, t)\Im _{t}^{2}\sigma , -\Im _{x}^{2}\Im _{t}\sigma \right)_{{_{L^{2}(Q^{T})}}} = -\int_{0}^{T}\int_{0}^{\alpha}c(x, t)\Im _{t}^{2}\sigma (\Im _{x}^{2}\Im_{t}\sigma )dxdt. \end{equation} (4.10)

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

    \begin{equation*} \left( \partial _{t}^{\beta }(\Im _{x}\Im _{t}\sigma ), \Im _{x}\Im_{t}\sigma \right) _{_{L^{2}(Q^{T})}} \end{equation*}
    \begin{equation*} = -\frac{1}{2}\int_{0}^{\alpha}a(x, t)(\Im _{t}^{2}\sigma(x, T))^{2} -\int_{0}^{T}\int_{0}^{\alpha}a_{x}(x, t)(\Im_{t}^{2}\sigma )\Im _{x}\Im _{t}\sigma dxdt \end{equation*}
    \begin{equation*} -\int_{0}^{T}\int_{0}^{\alpha}b(x, t)(\Im _{x}\Im _{t}\sigma)^{2}dxdt -\int_{0}^{T}\int_{0}^{\alpha}b(x, t)\Im _{x}\Im _{t}\sigma (\Im_{x}^{2}\Im _{t}\sigma )dxdt \end{equation*}
    \begin{equation} +\int_{0}^{T}\int_{0}^{\alpha}c(x, t)\Im _{t}^{2}\sigma (\Im _{x}^{2}\Im_{t}\sigma )dxdt. \end{equation} (4.11)

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

    \begin{equation} -\int_{0}^{T}\int_{0}^{\alpha}a_{x}(x, t)(\Im _{t}^{2}\sigma )\Im_{x}\Im _{t}\sigma dxdt \leq \frac{a_{2}\varepsilon }{2}\left\Vert \Im_{t}^{2}\sigma \right\Vert _{L^{2}(Q^{T})}^{2} +\frac{a_{2}}{2\varepsilon }\left\Vert \Im _{x}\Im _{t}\sigma \right\Vert _{L^{2}(Q^{T})}^{2}. \end{equation} (4.12)
    \begin{equation} \int_{0}^{T}\int_{0}^{\alpha}b(x, t)(\Im _{x}\Im _{t}\sigma )^{2}dxdt \leq\ b_{1}\left\Vert \Im _{x}\Im _{t}\sigma \right\Vert _{L^{2}(Q^{T})}^{2} \end{equation} (4.13)
    \begin{equation} \int_{0}^{T}\int_{0}^{\alpha}b(x, t)\Im _{x}\Im _{t}\sigma (\Im_{x}^{2}\Im _{t}\sigma )dxdt \leq \frac{b_{1}\varepsilon }{2}\left\Vert \Im_{x}\Im _{t}\sigma \right\Vert _{L^{2}(Q^{T})}^{2} +\frac{b_{1}\alpha}{4\varepsilon }\left\Vert \Im _{x}\Im _{t}\sigma \right\Vert _{L^{2}(Q^{T})}^{2}. \end{equation} (4.14)
    \begin{equation} \int_{0}^{T}\int_{0}^{\alpha}c(x, t)\Im _{t}^{2}\sigma (\Im _{x}^{2}\Im_{t}\sigma )dxdt\leq \frac{c_{1}\varepsilon }{2}\left\Vert \Im _{t}^{2}\sigma \right\Vert _{L^{2}(Q^{T})}^{2} +\frac{c_{1}\alpha}{2\varepsilon }\left\Vert \Im _{x}\Im _{t}\sigma \right\Vert _{L^{2}(Q^{T})}^{2}. \end{equation} (4.15)

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

    \begin{equation*} \left( \partial _{t}^{\beta }(\Im _{x}\Im _{t}\sigma ), \Im _{x}\Im_{t}\sigma \right) _{_{L^{2}(Q^{T})}} -(b_{1}+(\frac{2a_{2}+b_{1}\alpha+2c_{1}\alpha}{4\varepsilon }))\left\Vert \Im _{x}\Im _{t}\sigma\right\Vert _{L^{2}(Q^{T})}^{2} \end{equation*}
    \begin{equation} \leq (\frac{a_{2}\varepsilon +c_{1}\varepsilon }{2})\left\Vert \Im_{t}^{2}\sigma \right\Vert _{L^{2}(Q^{T})}^{2} -\frac{1}{2}\left\Vert \sqrt{a(x, t)}(\Im _{t}^{2}\sigma (x, T))\right\Vert _{L^{2}(0.\alpha)}^{2}. \end{equation} (4.16)

    We have, by lemma 3,

    \begin{equation} \partial _{t}^{\beta }(\Im _{x}\Im _{t}\sigma )^{2} \leq 2\left( \partial_{t}^{\beta }(\Im _{x}\Im _{t}\sigma ), \Im _{x}\Im _{t}\sigma \right)_{_{L^{2}(Q^{T})}}. \end{equation} (4.17)

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

    \begin{equation*} \partial _{t}^{\beta }(\Im _{x}\Im _{t}\sigma )^{2}-(2b_{1}+(\frac{2a_{2}+b_{1}\alpha +2c_{1}\alpha}{2\varepsilon }))\left\Vert \Im _{x}\Im _{t}\sigma\right\Vert _{L^{2}(Q^{T})}^{2} \end{equation*}
    \begin{equation} \leq (a_{2}\varepsilon +c_{1}\varepsilon )\left\Vert \Im _{t}^{2}\sigma\right\Vert _{L^{2}(Q^{T})}^{2} -\left\Vert \sqrt{a(x, t)}(\Im _{t}^{2}\sigma\left( x, T\right) \right\Vert _{L^{2}(0.\alpha)}^{2}. \end{equation} (4.18)

    Take \varepsilon, such that

    \begin{equation*} (2b_{1}+(\frac{2a_{2}+b_{1}\alpha+2c_{1}\alpha}{2\varepsilon }))\geq 0. \end{equation*}

    Consequently,

    \begin{equation*} \partial _{t}^{\beta }(\Im _{x}\Im _{t}\sigma )^{2} \end{equation*}
    \begin{equation} \leq (a_{2}\varepsilon +c_{1}\varepsilon )\left\Vert \Im _{t}^{2}\sigma\right\Vert _{L^{2}(Q^{T})}^{2} -\sqrt{a(x, t)}\left\Vert (\Im _{t}^{2}\sigma\left( x, T\right) \right\Vert _{L^{2}(0.\alpha)}^{2}\leq 0. \end{equation} (4.19)

    We conclude that z is zero a.e in the domain Q^{T}.

    The authors declare no conflicts of interest.



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