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Theory and Modern Applications

On a nonlocal problem for parabolic equation with time dependent coefficients


This paper is devoted to the study of existence and uniqueness of a mild solution for a parabolic equation with conformable derivative. The nonlocal problem for parabolic equations appears in many various applications, such as physics, biology. The first part of this paper is to consider the well-posedness and regularity of the mild solution. The second one is to investigate the existence by using Banach fixed point theory.

1 Introduction

Let T be a positive number and \(\Omega \subset \mathbb{R}^{d}\) (\(d \ge 1\)) be a bounded domain with the smooth boundary Ω. In this paper, we consider the nonlocal value problem for parabolic equation as follows:

$$ \textstyle\begin{cases} {\frac{{} ^{\mathscr{C}} \partial ^{\beta} }{\partial t^{{\beta} }}} u(x,t) - \psi (t) \Delta u = F(x,t, u(x,t)), & x \in \Omega , t\in (0,T), \\ u(x,t) =0, & x \in \partial {\Omega} ,t\in (0,T), \\ a u(x,0) + bu (x,T) = \varphi (x), & x \in \Omega , \end{cases} $$

where the symbol \(\frac{{} ^{\mathscr{C}} \partial ^{\beta} v(t)}{\partial t^{\beta} }\) is called the conformable derivative which is defined clearly in Sect. 2. The function F represents external forces, and the function φ is the input datum which will be defined later. The function ψ is called time dependent coefficient.

There are applications of conformable derivative in various models, for example, the harmonic oscillator, the damped oscillator, and the forced oscillator (see [1]), electrical circuits (see [2]), chaotic systems in dynamics (see [3]), quantum mechanics [4]. Based on important notes in the article [5], we observe and think that studying the ODE problem with the compliance derivative is very different from studying the PDE problem with a suitable derivative. The positional results and methods for the ODE and PDEs models are not the same and completely different. In order for the reader to have more access to this kind of fractional diffusion equations with conformable derivative, we refer to [2, 617]. In addition, we can find the topics of initial and final problems, which are studied by many authors, in [1827].

Our paper is one of the first results on the nonlocal value problem given with parabolic equations with conformable derivative. In the linear part, we use the techniques of Hilbert scales space. In the nonlinear part, to establish the existence and uniqueness of the solution, we must use the Banach fixed mapping theorem combined with some techniques to evaluate inequality, some Sobolev embedding. One of the most difficult points is finding the right functional spaces for the solution. Another highlight in the results is to demonstrate the convergence of the mild solution as the parameter b approaches 0.

2 Preliminaries

Conformable derivative model: Let the function \(v:[0,\infty ) \to D \), where D is a Banach space. If the limitation

$$\begin{aligned} \frac{{} ^{\mathscr{C}} \partial ^{\beta} v(t)}{\partial t^{\beta} } :=\lim_{\varepsilon \to 0} \frac{v( t+\varepsilon t^{1-\beta} ) - v(t)}{\varepsilon} \quad \text{in } D \end{aligned}$$

for each \(t>0\) exists, then it is called be the conformable derivative of order \(\beta \in (0,1]\) of v. We can refer the reader to [11, 28].

Let \(\mathcal{A}\) be a linear, self-adjoint, unbounded, and positive definite operator. Assume that \(\mathcal{A}\) has the eigenvalues \(\lambda _{n}\) (\(n \in \mathbb{N}^{*}\)):

$$ 0 < \lambda _{1} \le \lambda _{2} \le \cdots \quad \text{with } \lambda _{n} \to \infty \text{ for } n \to \infty $$

and the corresponding eigenelements \(e_{n}\) which form an orthonormal basis. For \(\nu >0\), we introduce fractional powers of \(\mathcal{A}\) as follows:

$$ D\bigl(\mathcal{A}^{\nu }\bigr) = \Biggl\{ g \in L^{2}(\Omega ): \sum_{n=1}^{\infty } \bigl\vert \langle g, e_{n} \rangle \bigr\vert ^{2}\lambda _{n}^{2 \nu }< \infty \Biggr\} . $$

The space \(D(\mathcal{A}^{\nu }) \) is a Banach space in the following with the corresponding norm:

$$ \Vert g \Vert _{D(\mathcal{A}^{\nu })} := \Biggl( \sum_{n=1}^{\infty } \bigl\vert \langle g, e_{n} \rangle \bigr\vert ^{2} \lambda _{n}^{2 \nu } \Biggr) ^{\frac{1}{2}},\quad g \in D\bigl( \mathcal{A}^{\nu }\bigr). $$

The information for negative fractional power \(\mathcal{A}^{-\nu }\) can be defined by [29]. For any \(m >0\) and a Banach space \(\mathcal{B}\), we introduce the following space:

$$\begin{aligned} C^{m} \bigl( [0,T]; \mathcal{B} \bigr)= \biggl\{ v \in C \bigl( [0,T]; \mathcal{B} \bigr): \sup_{0 \le s< t \le T} \frac{ \Vert v(\cdot,t)-v(\cdot,s) \Vert _{\mathcal{B} } }{ \vert t-s \vert ^{m}}< \infty \biggr\} , \end{aligned}$$

corresponding to the following norm:

$$ \Vert v \Vert _{ C^{m} ( [0,T]; \mathcal{B} )}= \sup_{0 \le s< t \le T} \frac{ \Vert v(\cdot,t)-v(\cdot,s) \Vert _{\mathcal{B} } }{ \vert t-s \vert ^{m} }. $$

Let \(0< m <1\), and we recall the following space:

$$\begin{aligned} \mathcal{C}^{m} ( (0,T]; \mathcal{B})= \Bigl\{ v \in C (\bigl(0,T]; L^{2}( \Omega ) \bigr): \sup_{0< t \le T} t^{m} \bigl\Vert v(\cdot,t) \bigr\Vert _{\mathcal{B}} < \infty \Bigr\} , \end{aligned}$$

with the norm \(\|v\|_{\mathcal{C}^{m} ( (0,T]; \mathcal{B} )}:=\sup_{0< t \le T} t^{m} \|v(\cdot,t)\|_{\mathcal{B}}\).

3 Inhomogeneous problem

In this section, we consider the nonlocal value problem for equation as follows:

$$ \textstyle\begin{cases} \frac{{}^{\mathscr{C} }\partial ^{\beta }}{\partial t^{\beta }} u(x,t) - \psi (t) \Delta u = F(x,t), & x \in \Omega , t\in (0,T), \\ u(x,t) =0, & x \in \partial {\Omega },t\in (0,T), \\ a u(x,0) + bu (x,T) = \varphi (x), & x \in \Omega , \end{cases} $$

where F is defined later.

3.1 Existence and uniqueness of the mild solution

In this subsection, we state the existence and uniqueness of the mild solution.

Theorem 3.1

Let \(a_{0} \le \psi (t) \le b_{0}\), where \(a_{0}\), \(b_{0}\) are constants, \(\varphi \in D(\mathcal{A}^{\nu +\theta })\) and \(F \in L^{\infty } (0,T;D(\mathcal{A}^{\nu +\theta }) ) \), where \(\nu >0\), \(0<\theta <1\). Then we have the following regularity:

$$\begin{aligned} \bigl\Vert u(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{\nu })} &\lesssim t^{-\beta \theta } \bigl( \Vert \varphi \Vert _{D(\mathcal{A}^{\nu +\theta })}+ \Vert F \Vert _{L^{2r} (0,T;D(\mathcal{A}^{\nu +\theta }) )} \bigr) \\ &\quad {} + \Vert F \Vert _{L^{\infty } (0,T;D(\mathcal{A}^{\nu +\theta }) )},\quad t>0. \end{aligned}$$


By a simple calculation, we get the following equality:

$$\begin{aligned} u_{n}(t)&= \bigl\langle u(\cdot,t) , e_{n} \bigr\rangle \\ &=\exp \biggl( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds \biggr) \bigl\langle u( \cdot,0) , e_{n} \bigr\rangle \\ &\quad {}+ \int _{0}^{t} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{ \frac{s^{\beta }}{\beta }}^{\frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(u) (s) \,ds, \end{aligned}$$

where \(F_{n} = \langle F, e_{n} \rangle \). Replacing t with T in the above expression, we get

$$\begin{aligned} u_{n}(T)&= \bigl\langle u(\cdot,t) , e_{n} \bigr\rangle \\ &= \exp \biggl( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds \biggr) \bigl\langle u( \cdot,0) , e_{n} \bigr\rangle \\ &\quad {}+ \int _{0}^{T} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{ \frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(u) (s) \,ds. \end{aligned}$$

The condition \(au(x,0)+b u(x,T)= \varphi (x)\) gives the following result:

$$\begin{aligned} & \biggl( a+ b \exp \biggl( -\lambda _{n} \int _{0}^{ \frac{T^{\beta }}{\beta }} \psi (s) \,ds \biggr) \biggr) u_{n}(0) \\ &\quad {}+ b \int _{0}^{T} s^{ \beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{ \frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(u) (s) \,ds= \varphi _{n}. \end{aligned}$$

By switching sides and combining them, we obtain the formula for \(u_{n}(0)\) as follows:

$$\begin{aligned} u_{n}(0)= \frac{\varphi _{n}- b\int _{0}^{T} s^{\beta -1} \exp ( -\lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr ) F_{n}(u)(s) \,ds }{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) }. \end{aligned}$$

Inserting (8) into the above formula (5) and after the reduced transformation, we arrive at

$$\begin{aligned} u_{n}(t)&= \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \varphi _{n} \\ &\quad {}- b \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \int _{0}^{T} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{ \frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(u) (s) \,ds \\ &\quad {}+ \int _{0}^{t} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{ \frac{s^{\beta }}{\beta }}^{\frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(u) (s) \,ds. \end{aligned}$$

By the properties of Fourier series, we get that

$$\begin{aligned} &u(x,t) \\ &\quad {= \underbrace{\sum_{n=1}^{\infty } \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \varphi _{n} e_{n}(x)}_{J_{1}(x,t)}} \\ &\qquad {}- \underbrace{ b \sum_{n=1}^{\infty } \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \biggl( \int _{0}^{T} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(s) \,ds \biggr) e_{n}(x)}_{{J_{2}(x,t)}} \\ &\qquad {}+ \sum_{n=1}^{\infty } \biggl( \int _{0}^{t} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(s) \,ds \biggr) e_{n}(x). \end{aligned}$$

Using the inequality

$$ \exp \biggl( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds \biggr) \le \exp \biggl( -\lambda _{n} {\frac{a_{0} t^{\beta }}{\beta }} \biggr) \le C_{\theta }a_{0}^{\mu } \beta ^{-\theta } \lambda _{n}^{\theta }t^{- \beta \theta }, $$

we get

$$\begin{aligned} \bigl\Vert J_{1}(\cdot,t) \bigr\Vert ^{2}_{D(\mathcal{A}^{\nu })}&= \sum_{n=1}^{\infty }\lambda _{n}^{2\nu } \biggl( \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \biggr)^{2} \varphi _{n}^{2} \\ &\lesssim t^{-2\beta \theta } \sum_{n=1}^{\infty } \lambda _{n}^{2\nu +2 \theta } \varphi _{n}^{2}= t^{-2\beta \theta } \Vert \varphi \Vert _{D( \mathcal{A}^{\nu +\theta })}^{2}. \end{aligned}$$

Hence, we find that

$$\begin{aligned} \bigl\Vert J_{1}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{\nu })} \lesssim t^{-\beta \theta } \Vert \varphi \Vert _{D(\mathcal{A}^{\nu +\theta })}. \end{aligned}$$

We continue to estimate \(J_{2}\) as follows:

$$\begin{aligned} & \bigl\Vert J_{2}(\cdot,t) \bigr\Vert ^{2}_{D(\mathcal{A}^{\nu })} \\ &\quad =b^{2} \sum_{n=1}^{\infty } \lambda _{n}^{2\nu } \biggl( \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \biggr)^{2} \\ &\qquad {}\times \biggl( \int _{0}^{T} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(s) \,ds \biggr)^{2} \\ &\quad \lesssim t^{-2\beta \theta } \sum_{n=1}^{\infty } \lambda _{n}^{2\nu +2 \theta } \biggl( \int _{0}^{T} s^{\beta -1} \bigl\vert F_{n}(s) \bigr\vert ^{2} \,ds \biggr)= t^{-2\beta \theta } \biggl( \int _{0}^{T} s^{\beta -1} \bigl\Vert F( \cdot,s) \bigr\Vert ^{2}_{D(\mathcal{A}^{\nu +\theta })} \,ds \biggr). \end{aligned}$$

The Hölder inequality implies that

$$\begin{aligned} \int _{0}^{T} s^{\beta -1} \bigl\Vert F( \cdot,s) \bigr\Vert ^{2}_{D(\mathcal{A}^{ \nu +\theta })} \,ds \le \biggl( \int _{0}^{T} s^{(\beta -1)r^{*} \,ds} \biggr)^{\frac{1}{r^{*}}} \biggl( \int _{0}^{T} \bigl\Vert F(\cdot,s) \bigr\Vert ^{2r}_{D( \mathcal{A}^{\nu +\theta })} \,ds \biggr)^{1/r}, \end{aligned}$$

where \(\frac{1}{r}+ \frac{1}{r^{*}}=1\). Let us choose \(r > \frac{1}{\beta }\), we find that

$$\begin{aligned} \int _{0}^{T} s^{\beta -1} \bigl\Vert F( \cdot,s) \bigr\Vert ^{2}_{D(\mathcal{A}^{ \nu +\theta })} \,ds \lesssim \Vert F \Vert ^{2}_{L^{2r} (0,T;D( \mathcal{A}^{\nu +\theta }) )}. \end{aligned}$$

Inserting the latter estimate into (13), we arrive at

$$\begin{aligned} \bigl\Vert J_{2}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{\nu })} \lesssim t^{-\beta \theta } \Vert F \Vert _{L^{2r} (0,T;D(\mathcal{A}^{\nu +\theta }) )}. \end{aligned}$$

The quantity \(J_{3}\) is bounded by

$$\begin{aligned} & \bigl\Vert J_{3}(\cdot,t) \bigr\Vert ^{2}_{D(\mathcal{A}^{\nu })} \\ &\quad = \sum_{n=1}^{\infty }\lambda _{n}^{2\nu } \biggl( \int _{0}^{t} s^{ \beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{ \frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(s) \,ds \biggr)^{2}. \end{aligned}$$

It follows from the inequality

$$ \exp \biggl( -2\lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{ \frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) \le \exp \biggl( -\lambda _{n} {\frac{a_{0} (t^{\beta }-s^{\beta })}{\beta }} \biggr) \le C_{\theta }a_{0}^{\mu } \beta ^{-\theta } \lambda _{n}^{\theta } \bigl( t^{\beta } -s^{\beta } \bigr)^{-\theta } $$

for any \(0\le s \le t\), that

$$\begin{aligned} \bigl\Vert J_{3}(\cdot,t) \bigr\Vert ^{2}_{D(\mathcal{A}^{\nu })} &\lesssim \sum_{n=1}^{\infty } \lambda _{n}^{2\nu +2\theta } \biggl( \int _{0}^{t} s^{\beta -1} \bigl( t^{\beta } -s^{\beta } \bigr)^{-\theta } F_{n}^{2}(s) \,ds \biggr) \\ &\lesssim \int _{0}^{t} s^{\beta -1} \bigl( t^{\beta } -s^{\beta } \bigr)^{-\theta } \bigl\Vert F(\cdot,s) \bigr\Vert ^{2}_{D(\mathcal{A}^{\nu + \theta })} \,ds \\ &= \Vert F \Vert ^{2}_{L^{\infty } (0,T;D(\mathcal{A}^{\nu +\theta }) )} \int _{0}^{t} s^{\beta -1} \bigl( t^{\beta } -s^{\beta } \bigr)^{- \theta } \,ds. \end{aligned}$$

Next, we continue to compute the integral term. Set the variable \(\vartheta = s^{\beta }\). Then we get \(d \vartheta = \beta s^{\beta -1} \,ds\). Then it follows from \(0< \theta <1\) that

$$\begin{aligned} \int _{0}^{t} s^{\beta -1} \bigl( t^{\beta } -s^{\beta } \bigr)^{- \theta } \,ds= \frac{1}{\beta } \int _{0}^{t^{\beta }} \bigl( t^{\beta } - \vartheta \bigr)^{-\theta } \,d \vartheta = \frac{ t^{ \beta (1-\theta )}}{\beta (1-\theta )} \le \frac{ T^{ \beta (1-\theta )}}{\beta (1-\theta )}. \end{aligned}$$

This together with (18) yields that

$$\begin{aligned} \bigl\Vert J_{3}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{\nu })} \lesssim \Vert F \Vert _{L^{\infty } (0,T;D(\mathcal{A}^{\nu +\theta }) )} . \end{aligned}$$

Combining (12), (16), and (20), we deduce that

$$\begin{aligned} & \bigl\Vert u(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{\nu })} \\ &\quad \le \sum_{j=1}^{3} \bigl\Vert J_{j}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{\nu })} \\ &\quad \lesssim t^{-\beta \theta } \bigl( \Vert \varphi \Vert _{D( \mathcal{A}^{\nu +\theta })}+ \Vert F \Vert _{L^{2r} (0,T;D(\mathcal{A}^{ \nu +\theta }) )} \bigr)+ \Vert F \Vert _{L^{\infty } (0,T;D( \mathcal{A}^{\nu +\theta }) )}. \end{aligned}$$


4 The mild solution for nonlinear problem

By the previous section, we define the following definition of a mild solution of the problem as follows.

Definition 4.1

The function u is called a mild solution of the problem if u belongs to the space \(L^{\infty }(0,T;L^{2}(\Omega ))\) and it also satisfies equality (1).

We recall that the formula of the solution u is performed as the following form:

$$\begin{aligned} u(x,t)&=\sum_{n=1}^{\infty } \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \varphi _{n} e_{n}(x) \\ &\quad {}- b \sum_{n=1}^{\infty } \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \\ &\quad {}\times \biggl( \int _{0}^{T} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(u) (s) \,ds \biggr) e_{n}(x) \\ &\quad {}+ \sum_{n=1}^{\infty } \biggl( \int _{0}^{t} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(u) (s) \,ds \biggr) e_{n}(x) . \end{aligned}$$

Theorem 4.1

Let \(\varphi \in L^{2}(\Omega )\), and there exists a constant \(K_{f} \ge 0\) such that

$$\begin{aligned} \bigl\Vert F (v_{1}) (\cdot ,t) - F (v_{2}) (\cdot ,t) \bigr\Vert _{L^{2}( \Omega )} \le K_{f} \bigl\Vert v_{1}(\cdot ,t) - v_{2}(\cdot ,t) \bigr\Vert _{L^{2}(\Omega )}. \end{aligned}$$

If the condition \(1> (\frac{b}{a}+1 ) \frac{K_{f} T^{\beta }}{\beta } \) is true, then problem (22) has a mild solution u which belongs to the space \(L^{\infty } ( 0,T;L^{2}(\Omega ) )\).


Set the following:

$$\begin{aligned} \mathscr{H} (u) (t) = \mathscr{H}_{0} + \mathscr{H}_{1} (u) (t)+ \mathscr{H}_{2} (u) (t), \end{aligned}$$


$$ \mathscr{H}_{0}= \sum_{n=1}^{\infty } \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \varphi _{n} e_{n}(x) $$


$$\begin{aligned} \mathscr{H}_{1} (u) &= - b \sum_{n=1}^{\infty } \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \\ &\quad {} \times \biggl( \int _{0}^{T} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(u) (s) \,ds \biggr) e_{n}(x) \end{aligned}$$


$$\begin{aligned} \mathscr{H}_{2,\beta } (u) (t)= \sum_{n=1}^{\infty } \biggl( \int _{0}^{t} s^{ \beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{ \frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}(u) (s) \,ds \biggr) e_{n}(x). \end{aligned}$$

We have to show that the equation \(\mathscr{H} u=u\) has a unique solution. Applying the Hölder inequality, we find that

$$\begin{aligned} & \bigl\Vert \mathscr{H}_{2,\beta } (u) (t)- \mathscr{H}_{2,\beta } (v) (t) \bigr\Vert _{L^{2}(\Omega )}^{2} \\ &\quad = \sum_{n=1}^{\infty } \biggl[ \int _{0}^{t} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) \bigl( F_{n}(u) (s)- F_{n}(v) (s) \bigr) \,ds \biggr]^{2} \\ &\quad \le \sum_{n=1}^{\infty } \biggl[ \int _{0}^{t} s^{\beta -1} \,ds \biggr] \biggl[ \int _{0}^{t} s^{\beta -1} \exp \biggl( -2 \lambda _{n} \int _{ \frac{s^{\beta }}{\beta }}^{\frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) \bigl( F_{n}(u) (s)- F_{n}(v) (s) \bigr)^{2} \,ds \biggr] \\ &\quad \le \frac{T^{\beta }}{\beta } \biggl[ \int _{0}^{t} s^{\beta -1} \bigl\Vert F(u) ( \cdot,s) - F(v) (\cdot,s) \bigr\Vert ^{2}_{L^{2}(\Omega )} \,ds \biggr] \\ &\quad \le \frac{T^{\beta }}{\beta } K_{f}^{2} \Vert u-v \Vert _{L^{\infty }( 0,T;L^{2}( \Omega ))}^{2} \biggl[ \int _{0}^{t} s^{\beta -1} \,ds \biggr]= \frac{K_{f}^{2} T^{2\beta }}{\beta ^{2}} \Vert u-v \Vert _{L^{\infty }( 0,T;L^{2}( \Omega ))}^{2} , \end{aligned}$$

and using the inequality

$$ \biggl\vert -b \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \biggr\vert \le \frac{b}{a}, $$

we get that

$$\begin{aligned} & \bigl\Vert \mathscr{H}_{1,\beta } (u) (t)- \mathscr{H}_{1,\beta } (v) (t) \bigr\Vert _{L^{2}(\Omega )}^{2} \\ &\quad \le \frac{b^{2}}{a^{2}} \sum_{n=1}^{\infty } \biggl( \int _{0}^{T} s^{ \beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{ \frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) \bigl( F_{n}(u) (s)- F_{n}(v) (s) \bigr) \,ds \biggr)^{2} \\ &\quad \le \frac{b^{2}}{a^{2}} \sum_{n=1}^{\infty } \biggl[ \int _{0}^{t} s^{ \beta -1} \,ds \biggr] \biggl( \int _{0}^{T} s^{\beta -1} \exp \biggl( -2 \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) \bigl( F_{n}(u) (s)- F_{n}(v) (s) \bigr)^{2} \,ds \biggr) \\ &\quad \le \frac{b^{2}}{a^{2}} \frac{T^{\beta }}{\beta } \biggl( \int _{0}^{T} s^{ \beta -1} \bigl( F_{n}(u) (s)- F_{n}(v) (s) \bigr)^{2} \,ds \biggr) \le \frac{b^{2}}{a^{2}} \frac{K_{f}^{2} T^{2\beta }}{\beta ^{2}} \Vert u-v \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))}^{2} . \end{aligned}$$

Combining (27) and (28) leads to

$$\begin{aligned} & \bigl\Vert \mathscr{H}_{\beta } (u) (t)- \mathscr{H}_{\beta } (v) (t) \bigr\Vert _{L^{2}( \Omega )} \\ &\quad = \bigl\Vert \mathscr{H}_{1,\beta } (u) (t)- \mathscr{H}_{1,\beta } (v) (t) \bigr\Vert _{L^{2}(\Omega )}+ \bigl\Vert \mathscr{H}_{2,\beta } (u) (t)- \mathscr{H}_{2,\beta } (v) (t) \bigr\Vert _{L^{2}(\Omega )} \\ &\quad \le \frac{K_{f} T^{\beta }}{\beta } \biggl( 1+\frac{b}{a} \biggr) \Vert u-v \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))} . \end{aligned}$$

Since the left-hand side of the above observation is independent of t, we deduce that

$$\begin{aligned} \bigl\Vert \mathscr{H}_{\beta } (u)- \mathscr{H}_{\beta } (v) \bigr\Vert _{L^{\infty }( 0,T;L^{2}( \Omega ))} \le \frac{K_{f} T^{\beta }}{\beta } \biggl( 1+\frac{b}{a} \biggr) \Vert u-v \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))}. \end{aligned}$$

Let us choose T such that

$$ T \le \biggl( \frac{\beta a }{(a+b) K_{f}} \biggr)^{1/\beta }, $$

we know that \(\mathscr{H} \) is a contraction mapping. Next, we continue to show that if \(v \in L^{\infty }( 0,T;L^{2}(\Omega ))\), then \(\mathscr{H}_{\beta } (v) \in L^{\infty }( 0,T;L^{2}(\Omega ))\). We only check that if \(v=0\), then

$$\begin{aligned} \mathscr{H}_{\beta } (v) (\cdot,t)= \sum_{n=1}^{\infty } \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \varphi _{n} e_{n}(x), \end{aligned}$$

which allows us to obtain that

$$\begin{aligned} \bigl\Vert \mathscr{H}_{\beta } (v) (\cdot,t) \bigr\Vert ^{2}_{L^{2}(\Omega )} &= \sum_{n=1}^{\infty } \biggl( \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \biggr)^{2} \varphi _{n}^{2} \\ &\le \frac{1}{a^{2}}\sum_{n=1}^{\infty } \varphi _{n}^{2} \\ & = \frac{1}{a^{2}} \Vert \varphi \Vert ^{2}_{L^{2}(\Omega )}. \end{aligned}$$

The latter estimate leads to \(\mathscr{H}_{\beta } : L^{\infty }( 0,T;L^{2}(\Omega )) \to L^{\infty }( 0,T;L^{2}( \Omega ))\). By applying the Banach fixed point theory, we can deduce that our problem (1) has a unique solution \(u \in L^{\infty }( 0,T;L^{2}(\Omega ))\). □

Theorem 4.2

Let \(w_{a}\) be the solution of the initial value problem

$$ \textstyle\begin{cases} \frac{{ }^{\mathscr{C} }\partial ^{\beta }}{\partial t^{\beta }} u(x,t) - \psi (t) \Delta u = F(x,t), & x \in \Omega , t\in (0,T), \\ u(x,t) =0, & x \in \partial {\Omega },t\in (0,T), \\ a u(x,0) = \varphi (x), & x \in \Omega . \end{cases} $$

Then we have

$$\begin{aligned} \lim_{b \to 0} \Vert w_{a,\beta }-u_{a,b, \beta } \Vert _{L^{\infty }( 0,T;L^{2}( \Omega ))}=0 \end{aligned}$$


$$\begin{aligned} \Vert w_{a,\beta }-u_{a,b, \beta } \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))} \le \frac{ b \Vert \varphi \Vert _{L^{2}(\Omega )}}{ a^{2} ( 1- (\frac{b}{a}+1 ) \frac{K_{f} T^{\beta }}{\beta } ) }. \end{aligned}$$


It is easy to see that the mild solution of problem (33) is given by

$$\begin{aligned} &w_{a,\beta }(x,t) \\ &\quad =\sum_{n=1}^{\infty }\frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a } \varphi _{n} e_{n}(x) \\ &\qquad {}- b \sum_{n=1}^{\infty } \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a } \\ &\qquad {}\times\biggl( \int _{0}^{T} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{ \frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}( w_{a,\beta }) (s) \,ds \biggr) e_{n}(x) \\ &\qquad {}+ \sum_{n=1}^{\infty } \biggl( \int _{0}^{t} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) F_{n}( w_{a,\beta }) (s) \,ds \biggr) e_{n}(x). \end{aligned}$$

This together with (22) yields

$$\begin{aligned} &u_{a,b, \beta }(x,t)- w_{a,\beta }(x,t) \\ &\quad = \sum_{n=1}^{\infty } \biggl( \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a } - \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \biggr) \varphi _{n} e_{n}(x) \\ &\qquad {}- b \sum_{n=1}^{\infty } \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds )}{ a } \\ &\qquad {}\times \biggl[ \int _{0}^{T} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) \Big( F_{n}( w_{a,\beta }) (s) -F_{n}(u_{a,b, \beta }) (s )\Big) \,ds \biggr] e_{n}(x) \\ &\qquad {}+ \sum_{n=1}^{\infty } \biggl[ \int _{0}^{t} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) \Big( F_{n}( w_{a,\beta }) (s) - F_{n}\bigl( u_{a,b, \beta})(s) \Big) \,ds \biggr] e_{n}(x) \\ &\quad = A_{1}+ A_{2}+ A_{3}. \end{aligned}$$

The term \(A_{1}\) is bounded by

$$\begin{aligned} \Vert A_{1} \Vert ^{2}_{L^{2}(\Omega )}&= \frac{b^{2}}{a^{2}} \sum_{n=1}^{\infty } \biggl( \frac{\exp ( -\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds ) \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) }{ a+ b \exp ( -\lambda _{n} \int _{0}^{\frac{T^{\beta }}{\beta }} \psi (s) \,ds ) } \biggr)^{2} \varphi _{n}^{2} \\ &\le \frac{b^{2}}{a^{4}} \sum_{n=1}^{\infty } \varphi _{n}^{2}= \frac{b^{2}}{a^{4}} \Vert \varphi \Vert ^{2}_{L^{2}(\Omega )}. \end{aligned}$$

Using \(\exp ( -2\lambda _{n} \int _{0}^{\frac{t^{\beta }}{\beta }} \psi (s) \,ds ) \le 1\), the term \(A_{2}\) is bounded by

$$\begin{aligned} & \Vert A_{2} \Vert ^{2}_{L^{2}(\Omega )} \\ &\quad = \frac{b^{2}}{a^{2}} \sum_{n=1}^{\infty } \biggl[ \int _{0}^{T} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) \bigl( F_{n}( w_{a,\beta }) (s) -F_{n}(u_{a,b, \beta } )\bigr) (s ) \,ds \biggr]^{2} \\ &\quad \le \frac{b^{2}}{a^{2}} \sum_{n=1}^{\infty } \biggl( \int _{0}^{T} s^{ \beta -1} \,ds \biggr) \\ &\qquad {}\times \int _{0}^{T} s^{ \beta -1} \exp \biggl( -2 \lambda _{n} \int _{\frac{s^{\beta }}{\beta }}^{\frac{T^{\beta }}{\beta }} \psi (r) \,dr \biggr) \Big( F_{n}( w_{a,\beta }) (s) -F_{n}(u_{a,b, \beta } ) (s) \Big)^{2} \,ds \\ &\quad \le \frac{b^{2}}{a^{2}} \frac{T^{\beta }}{\beta } \biggl[ \int _{0}^{t} s^{ \beta -1} \bigl\Vert F( w_{a,\beta }) (\cdot,s) - F(u_{a,b, \beta }) (\cdot,s) \bigr\Vert ^{2}_{L^{2}( \Omega )} \,ds \biggr] \\ &\quad \le \frac{b^{2}}{a^{2}} \frac{K_{f}^{2} T^{2\beta }}{\beta ^{2}} \Vert w_{a, \beta }-u_{a,b, \beta } \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))}^{2}. \end{aligned}$$

The term \(A_{3}\) can be estimated as follows:

$$\begin{aligned} \Vert A_{3} \Vert ^{2}_{L^{2}(\Omega )} & =\sum _{n=1}^{\infty } \biggl[ \int _{0}^{t} s^{\beta -1} \exp \biggl( - \lambda _{n} \int _{ \frac{s^{\beta }}{\beta }}^{\frac{t^{\beta }}{\beta }} \psi (r) \,dr \biggr) \bigl( F_{n}( w_{a,\beta }) (s) - F_{n}\bigl( u_{a,b, \beta }(s) \bigr) \,ds \biggr]^{2} \\ &\le \sum_{n=1}^{\infty } \biggl[ \int _{0}^{T} s^{\beta -1} \,ds \biggr] \biggl[ \int _{0}^{t} s^{\beta -1} \bigl\Vert F( w_{a,\beta }) (\cdot,s) - F(u_{a,b, \beta }) (\cdot,s) \bigr\Vert ^{2}_{L^{2}(\Omega )} \,ds \biggr] \\ &\le \frac{K_{f}^{2} T^{2\beta }}{\beta ^{2}} \Vert w_{a,\beta }-u_{a,b, \beta } \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))}^{2}. \end{aligned}$$

Therefore, we find that

$$\begin{aligned} \Vert A_{3} \Vert _{L^{2}(\Omega )} \le \frac{K_{f} T^{\beta }}{\beta } \Vert w_{a,\beta }-u_{a,b, \beta } \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))}. \end{aligned}$$

Some above observations lead to

$$\begin{aligned} & \Vert w_{a,\beta }-u_{a,b, \beta } \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))} \\ &\quad \le \Vert A_{1} \Vert _{L^{2}(\Omega )} + \Vert A_{2} \Vert _{L^{2}( \Omega )}+ \Vert A_{3} \Vert _{L^{2}(\Omega )} \\ &\quad \le \frac{b}{a^{2}} \Vert \varphi \Vert _{L^{2}(\Omega )}+ \biggl( \frac{b}{a}+1 \biggr) \frac{K_{f} T^{\beta }}{\beta } \Vert w_{a,\beta }-u_{a,b, \beta } \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))}. \end{aligned}$$

Hence, we find that

$$\begin{aligned} \Vert w_{a,\beta }-u_{a,b, \beta } \Vert _{L^{\infty }( 0,T;L^{2}(\Omega ))} \le \frac{ b \Vert \varphi \Vert _{L^{2}(\Omega )}}{ a^{2} ( 1- (\frac{b}{a}+1 ) \frac{K_{f} T^{\beta }}{\beta } ) }. \end{aligned}$$

It is easy to see that

$$\begin{aligned} \lim_{b \to 0} \Vert w_{a,\beta }-u_{a,b, \beta } \Vert _{L^{\infty }( 0,T;L^{2}( \Omega ))}=0. \end{aligned}$$


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This research is supported by Industrial University of Ho Chi Minh City (IUH) under grant number 66/HÐ-ÐHCN. The authors would like to thank the reviewers and editor for their constructive comments and valuable suggestions that improved the quality of this paper.


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Phuong, N.D., Binh, H.D., Long, L.D. et al. On a nonlocal problem for parabolic equation with time dependent coefficients. Adv Differ Equ 2021, 209 (2021).

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