A priori estimate for resolving the boundary fractional problem

Hacene Mecheri; Maryam G. Alshehri; Hacene Mecheri; Maryam G. Alshehri

doi:10.3934/math.2023037

AIMS Mathematics

2023, Volume 8, Issue 1: 765-774. doi: 10.3934/math.2023037

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A priori estimate for resolving the boundary fractional problem

Hacene Mecheri ^{1
,
,},
Maryam G. Alshehri ^{2
,
,}

1.
Mathematics and Informatics Department, LAMIS, Laboratory Tebessa University, Algeria
2.
Mathematics Department, Faculty of Science, University of Tabuk, 71491, Saudi Arabia

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.

Keywords:

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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Abstract

1. Introduction

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.

1.1. Position of the problem

In the bounded domain, ${Q^{T} = (0, T)\times (0, \alpha)}$ such that ${0 < \alpha < \infty, 0 < T < \infty}$ .

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

$\begin{equation} ^{c}\partial _{t}^{\beta }u_{t}-(a(x, t)\frac{\partial u}{\partial x})_{x}+b(x, t)u_{x}+c(x, t)u = g(x, t). \end{equation}$

(1.1)

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

$P_{1}:\left\{ \begin{array}{c} 0\leq a(x, t)\leq a_{1}\rm{, }0\leq a_{x}(x, t)\leq a_{2}\rm{ , } \\ 0\leq b(x, t)\leq b_{1}, 0\leq b_{x}(x, t)\leq b_{2}, \\ 0\leq c(x, t)\leq c\rm{, } \end{array} \right\} \forall \left( x, t\right) \in Q^{T},$

where

$\begin{eqnarray*} a_{i} &\in &IR_{\ast }^{+}, \forall \overline{i = 1-2}, b_{k}\in IR_{\ast}^{+}, \forall \overline{k = 1-2}, \\ c &\in &IR_{\ast }^{+}. \end{eqnarray*}$

With problem (1.1), we associate the initial conditions

$\begin{equation} u(x, 0) = \varphi (x)\geq 0, \qquad 0 < x < \alpha, \end{equation}$

(1.2)

and the boundary conditions

$\begin{equation} \int^{\alpha}_{0}u(x, t)dx = 0, \ \ \int^{\alpha}_{0}xu(x, t)dx = 0, \ \ \ \ u(\alpha, t)dx = 0. \end{equation}$

(1.3)

$\begin{equation} \int^{\alpha}_{0}\varphi(x)dx = 0, \end{equation}$

(1.4)

$\begin{equation} \int^{\alpha}_{0}x\varphi(x)dx = 0. \end{equation}$

(1.5)

2. Prelemanaries

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

$\begin{equation} D^{\beta }f(t) = \frac{1}{\Gamma (n-\beta )}\frac{\partial ^{n}}{\partial t^{n} }\int_{0}^{t}\frac{f(s)}{\left( t-s\right) ^{\beta -n+1}}ds, \end{equation}$

(2.1)

where $\Gamma$ is the well-known gamma function. The following formula is true:

$\begin{equation} D^{\beta }f(t) = \sum\limits^{n-1}_{k = 0}\frac{t^{k-\beta}}{{\Gamma (k-\beta +1)}}{f^{k}(0)}+\frac{1}{\Gamma (n-\beta )}\int_{0}^{t}\frac{f(s)}{\left( t-s\right) ^{\beta -n+1}}ds. \end{equation}$

(2.2)

The Riemann-Liouville fractional integral of order\ $\beta,$ \ $0\leq \beta\leq n,$ \ is defined by

$\begin{equation} D_{t}^{-\beta }f(t) = \frac{1}{\Gamma (\beta )}\int_{0}^{t}\frac{f^{\left(n\right) }(s)}{\left( t-s\right) ^{\beta -n+1}}ds = D^{\beta }f(t). \end{equation}$

(2.3)

The Riemann-Liouville fractional derivative is singular at the origin because of $\left(2.3\right)$ . 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 $\beta,$ n-1 $\leq \beta \leq n, n\in IN,$ is defined as the integral part of $\left(2.3\right)$ ; that is,

$\begin{equation} ^{c}D^{\beta }f(t) = \frac{1}{\Gamma (n-\beta )}\int_{0}^{t}\frac{f(s)}{\left( t-s\right) ^{\beta -n+1}}ds = D^{\beta }f(t) = D^{\beta }f(t)-\sum\limits_{k = 0}^{n-1}\frac{t^{k-\beta }}{\Gamma (k-\beta +1)}f^{k}(0). \end{equation}$

(2.4)

In the case of our problem, the Caputo and Riemann-Liiouville derivatives of order $\beta, \; 0 \; \leq \beta \leq 1$ , and their linked relationship formulas are defined as follows ^[4] $:

$\begin{equation} ^{c}\partial _{t}^{\beta }u(x, t) = \frac{1}{\Gamma (1-\beta )}\int_{0}^{t} \frac{\frac{\partial u(x, s)}{\partial s}}{\left( t-s\right) ^{\beta }}ds = \partial _{t}^{\beta }u(x, t)-\frac{u(x, 0)}{t^{\beta }\Gamma (1-\beta )}, \end{equation}$

(2.5)

$\begin{equation} \partial _{t}^{\beta }u(x, t) = \frac{1}{\Gamma (1-\beta )}\int_{0}^{t}\frac{u\left( x, s\right) }{\left( t-s\right) ^{\beta }}ds. \end{equation}$

(2.6)

(1) The Cauchy $\varepsilon -$ inequality:

$\begin{equation} \alpha \beta \leq \frac{\varepsilon }{2}\alpha ^{2}+\frac{1}{2\varepsilon }\beta ^{2}. \end{equation}$

(2.7)

(2) $Poincar\acute{e}$ -type inequalities ^[3] $:

$\begin{equation} \left\Vert \Im _{x}u\right\Vert _{L^{2}(Q^{T})}^{2}\leq \frac{\alpha}{2}\left\Vert u\right\Vert _{L^{2}(Q^{T})}^{2}, \end{equation}$

(2.8)

$\begin{equation*} \left\Vert \Im ^{m}u\right\Vert _{L^{2}(Q^{T})}^{2}\leq \frac{a}{2}\left\Vert \Im _{x}^{m-1}u\right\Vert _{L^{2}(Q^{T})}^{2}\forall m\in IN^{\ast }. \end{equation*}$

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

$\begin{equation} ^{c}D_{t}^{\beta }f^{2}(t)\leqslant\ 2f(t)^{c}D_{t}^{\beta }f(t). \end{equation}$

(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)

3. Uniqueness of the solution

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

4. Solvability of the posed problem

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 $Q^{T}.$

Conflict of interest

The authors declare no conflicts of interest.

References

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通讯作者: 陈斌, bchen63@163.com

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沈阳化工大学材料科学与工程学院沈阳 110142

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AIMS Mathematics

A priori estimate for resolving the boundary fractional problem

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Abstract

1. Introduction

1.1. Position of the problem

2. Prelemanaries

3. Uniqueness of the solution

4. Solvability of the posed problem

Conflict of interest

References

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AIMS Mathematics

A priori estimate for resolving the boundary fractional problem

Related Papers:

Abstract

1. Introduction

1.1. Position of the problem

2. Prelemanaries

3. Uniqueness of the solution

4. Solvability of the posed problem

Conflict of interest

References

Reader Comments

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