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On a time fractional diffusion with nonlocal in time conditions

Abstract

In this work, we consider a fractional diffusion equation with nonlocal integral condition. We give a form of the mild solution under the expression of Fourier series which contains some Mittag-Leffler functions. We present two new results. Firstly, we show the well-posedness and regularity for our problem. Secondly, we show the ill-posedness of our problem in the sense of Hadamard. Using the Fourier truncation method, we construct a regularized solution and present the convergence rate between the regularized and exact solutions.

Introduction

Time-fractional partial differential equations are well known to describe modeling of anomalously slow transport processes. These models are often expressed in the form of fractional diffusion or subdiffusion equations which have many applications in various kinds of research areas, e.g., thermal diffusion in fractal domains [1] and protein dynamics [2], we can refer for more details to [36]. By replacing many differential operators of fractional order with different type of PDEs of integer order, we formulate various types of boundary value problems with fractional order. Let us refer to many papers [724] and the references therein.

Let T be a positive number. Let Ω be a bounded and smooth enough domain in \(\mathbb{R}^{N} \). In our paper, we study the fractional diffusion equation with integral boundary conditions in the time variable on \([0, T]\) as follows:

$$ \textstyle\begin{cases} \partial _{t}^{\alpha }u(x,t) = \mathcal{A} u(x,t) + F(x,t),\quad (x,t) \in \Omega \times (0,T), \\ \frac{\partial u}{\partial n}(t,x) = 0,\quad x \in \partial \Omega , t \in (0,T], \\ a u(x,T)+ b \int _{0}^{T} u(t,x)\,dt = f(x),\quad x \in \Omega , \end{cases} $$
(1.1)

where Ω is a boundary of Ω and \(a^{2}+b^{2}>0\), \(a\ge 0\), \(b\ge 0\). Here, F is the source function, f is the given input data which is defined later, and \(\mathcal{A}\) is a linear, unbounded, self-adjoint, and positive definite operator. The second equation in (1.1) is called the Neumann boundary condition. Our problem is to identify the function \(u(\cdot,t)\) for \(0\le t < T\). Now, we explain the reason of the third equation which appears in our problem (1.1). As we know, most of some models on PDEs use an initial condition. However, in practice, some other models have to use nonlocal conditions, for example, including integrals over time intervals. Nonlocal conditions express and explain some full details about natural events because they consider additional information in the initial conditions. There are few papers on boundary conditions connecting the solution at different times, for instance, at initial time and at terminal time. The review of nonlocal initial conditions or nonlocal final conditions comes from real-life processes. For example, when the initial temperature or the final temperature for heat equation is not given immediately, but there is information regarding the temperature over a given period of time that can be described by a nonlocal initial condition. PDEs with nonlocal conditions were considered in many works, for example, see [25] for reaction-diffusion equations and [2635] for some other PDEs. As we said before, there are not any results for considering our model (1.1) with the nonlocal final condition and the integral condition

$$ a u(x,T)+ b \int _{0}^{T} u(x,t)\,dt = f(x). $$
(1.2)

The nonlocal integral condition (1.2) is motivated by the paper of Dokuchaev [36] where he investigated the well-posedness of Problem (1.1) in the integer order of derivative \(\alpha =1\). For more clarity, we can refer to some related works on our above models.

  • If \(a=1\), \(b=0\), Problem (1.1) is called backward in time problem and is well known to be ill-posed in the sense of Hadamard. Some papers [9, 37] gave the ill-posedness and regularization results.

  • If \(a=0\), \(b=1\), and \(\alpha =1\), Problem (1.1) and some similar models have been recently considered by Dokuchaev [36, 38], Volodymyr et al. [39], and Pulkina et al. [40] (see also the references therein).

It is obvious that our model in this paper is more complex than many previous models. In this paper, we consider the model with the nonlocal final and integral conditions for time fractional PDE. The main technique here is to use Fourier series in order to deal with the explicit solution for our problem. Now, we explain in more detail our studies that are the new and strong points of this article.

Two main results are described in this paper as follows:

  • Our paper may be the first study on the existence and regularity of the solution of model (1.1) on Sobolev spaces.

  • Our new second result is to give a regularization result when the noisy data \(f^{\delta }\in L^{q} (\Omega )\) for \(q\ge 1\). For case of \(L^{2}\), we can use Parseval’s equality directly to obtain stability estimates. However, Parseval’s equality is impossible to apply for the \(L^{p}\) case with \(p \neq 2\), which leads to the fact that some techniques here are more complex than the \(L^{2}\) estimate. Our new techniques in this section are based on applying some Sobolev embedding. Based on the ideas of [7], we develop some similar techniques to deal with the \(L^{p}\) case. Let us emphasize that a regularized problem is not considered in [36].

This paper is organized as follows. In Sect. 2, we introduce some preliminaries and a mild solution. The main result in this section is to show the well-posedness of the mild solution. In Sect. 3, we show the ill-posedness and give a regularized solution.

Well-posedness and regularity

Preliminaries

Definition 2.1

(see [3])

The Mittag-Leffler function \(E_{\alpha ,\beta }(\cdot)\) is

$$ E_{\alpha ,\beta }(z) = \sum_{{ k=0}}^{\infty } \frac{z^{k}}{\Gamma (\alpha k +\beta )}, \quad \alpha > 0, \beta \in \mathbb{R}, z \in {\mathbb{C}}. $$
(2.1)

Lemma 2.1

(see [41])

Let \(0 < \beta < 1\). There exist two constants \(M_{1}>0\), \(M_{2}>0\) such that, for any \(z>0\),

$$ \frac{M_{1}}{1+z} \le {E_{\beta , 1}(-z)} \le \frac{M_{2} }{1+z}. $$
(2.2)

Lemma 2.2

(see [41])

Let \(0 < \alpha < 1\) and \(\lambda >0\). Then

  1. (i)

    \(\partial _{t} ( {E_{\alpha } (-\lambda t^{\alpha }) }) = -\lambda t^{ \alpha -1} E_{\alpha ,\alpha }(-\lambda t^{\alpha })\) for \(t > 0\);

  2. (ii)

    \(\partial _{t} ( t^{\alpha -1} E_{\alpha ,\alpha }(-\lambda t^{\alpha }) ) = t^{\alpha -2} E_{\alpha ,\alpha -1} (-\lambda t^{\alpha })\) for \(t > 0 \).

Next, let us recall that \(\mathcal{A}\), see [3]. We have the following equality:

$$ \mathcal{A} \phi _{k} (x) = -\lambda _{k} \phi _{k}(x),\quad x \in \Omega ;\qquad \phi _{k} =0, \quad x \in \partial \Omega , k \in \mathbb{N}, $$
(2.3)

where \(\{\lambda _{k} \}_{k=1}^{\infty }\) satisfies that

$$\begin{aligned} 0 < \lambda _{1} \le \lambda _{2} \le \cdots\le \lambda _{k} \le \cdots , \end{aligned}$$
(2.4)

and \(\lim_{k \to \infty } \lambda _{k} = \infty \). For any \(q \ge 0\), we also define the space

$$ D^{q}(\Omega ) = \Biggl\{ u \in L^{2}(\Omega ) : \sum _{k=1}^{\infty } \lambda _{k}^{2q} \bigl\vert \langle u,\phi _{k} \rangle \bigr\vert ^{2} < + \infty \Biggr\} , $$
(2.5)

then \(D^{q}(\Omega )\) is a Hilbert space endowed with the norm

$$ \Vert u \Vert _{D^{q}(\Omega )} = \Biggl( \sum_{k=1}^{\infty } \lambda _{k}^{2q} \bigl\vert \langle u,\phi _{k}\rangle \bigr\vert ^{2} \Biggr)^{\frac{1}{2}} \cdot $$
(2.6)

Lemma 2.3

The following inclusions hold true:

$$ \left . \textstyle\begin{array}{l} L^{p}(\Omega ) \hookrightarrow D(\mathcal{A}^{\sigma }), \quad \textit{if } {-} \frac{d}{4} < \sigma \le 0, p \ge \frac{2d}{d-4\sigma }, \\ D(\mathcal{A}^{\sigma }) \hookrightarrow L^{p}(\Omega ),\quad \textit{if } 0\le \sigma < \frac{d}{4}, p \le \frac{2d}{d-4\sigma }. \end{array}\displaystyle \right \} $$
(2.7)

The mild solution of our problem

Theorem 2.1

Let \(0< \beta <\min (1,\frac{1}{2\alpha })\) and \(s\ge \beta +1\), \(\alpha >0 \) and

$$ G \in L^{\infty }\bigl(0,T; D\bigl(\mathcal{A}^{s-\beta }\bigr) \bigr) \cap L^{2}\bigl(0,T; D\bigl( \mathcal{A}^{s-\beta -1}\bigr) \bigr). $$

Then Problem (1.1) has a unique mild solution \(u \in L^{p}(0,T; D(\mathcal{A}^{s}))\) such that

$$\begin{aligned} \Vert u \Vert _{L^{p}(0,T; D(\mathcal{A}^{s}))} &\lesssim \Vert f \Vert _{D(\mathcal{A}^{s})} + \Vert G \Vert _{L^{\infty }(0,T; D(\mathcal{A}^{s-\beta }))} \\ &\quad {} + \Vert G \Vert _{L^{2}(0,T; D(\mathcal{A}^{s-\beta -1}))} + \Vert G \Vert _{L^{\infty }(0,T; L^{2}(\Omega ))} \end{aligned}$$
(2.8)

for \(1< p< \frac{1}{\alpha }\).

Proof

Assume that Problem (1.1) has a unique solution u which is given by

$$\begin{aligned} u(x,t) = \sum_{k=1}^{\infty } \bigl\langle u(\cdot,t), \phi _{k} \bigr\rangle \phi _{k}(x). \end{aligned}$$
(2.9)

From the results in Yamamoto [42], we know that

$$\begin{aligned} \bigl\langle u(\cdot,t), \phi _{k} (x) \bigr\rangle &= E_{\alpha ,1} \bigl(-\lambda _{k } t^{ \alpha }\bigr) \bigl\langle u(\cdot,0), \phi _{k} (x) \bigr\rangle \\ &\quad {}+ \int _{0}^{t} (t-\tau )^{\alpha -1} E_{\alpha ,\alpha } \bigl(- \lambda _{n } (t-\tau )^{\alpha } \bigr) { G_{k}} (\tau ) \,d\tau . \end{aligned}$$
(2.10)

The nonlocal condition \(a u(x,T)+ b \int _{0}^{T} u(x,t)\,dt = f(x) \) gives that

$$\begin{aligned} &a\sum_{k=1}^{\infty } E_{\alpha ,1} \bigl(-\lambda _{k } T^{ \alpha }\bigr) \bigl\langle u(\cdot,0), \phi _{k} \bigr\rangle \phi _{k}(x) \\ &\qquad {}+ a\sum_{k=1}^{\infty } \biggl( \int _{0}^{T} (T-\tau )^{ \alpha -1} E_{\alpha ,\alpha } \bigl(-\lambda _{n } (T-\tau )^{ \alpha } \bigr) { G_{k}} (\tau ) \,d\tau \biggr) \phi _{k}(x) \\ &\qquad {}+ b \int _{0}^{T} \Biggl( \sum _{k=1}^{\infty } E_{\alpha ,1} \bigl(- \lambda _{k } t^{\alpha }\bigr) \bigl\langle u(\cdot,0), \phi _{k} (x) \bigr\rangle \phi _{k}(x) \Biggr)\,dt \\ &\qquad {}+b \int _{0}^{T} \Biggl( \sum _{k=1}^{\infty } \int _{0}^{t} (t- \tau )^{\alpha -1} E_{\alpha ,\alpha } \bigl(-\lambda _{n } (t- \tau )^{\alpha } \bigr) { G_{k}} (\tau ) \,d\tau \bigr\rangle \phi _{k}(x) \Biggr) \,dt \\ &\quad = \sum_{k=1}^{\infty } \bigl\langle f(x), \phi _{k} (x) \bigr\rangle \phi _{k}(x). \end{aligned}$$
(2.11)

This implies that

$$\begin{aligned} & \biggl( a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{\alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \biggr) \bigl\langle u(\cdot,0), \phi _{k} (x) \bigr\rangle \\ &\quad{} +a \biggl( \int _{0}^{T} (T-\tau )^{\alpha -1} E_{\alpha ,\alpha } \bigl(-\lambda _{n } (T-\tau )^{\alpha } \bigr) { G_{k}} (\tau ) \,d\tau \biggr) \\ &\quad {}+ b \int _{0}^{T} \int _{0}^{t} (t-\tau )^{\alpha -1} E_{\alpha , \alpha } \bigl(-\lambda _{n } (t-\tau )^{\alpha } \bigr) { G_{k}} ( \tau ) \,d\tau \,dt= \bigl\langle f(x), \phi _{k} (x) \bigr\rangle . \end{aligned}$$
(2.12)

Therefore, we get that

$$\begin{aligned} \bigl\langle u(\cdot,0), \phi _{k} (x) \bigr\rangle &= \frac{ \langle f(x), \phi _{k} \rangle-a\int _{0}^{T} (T-\tau )^{\alpha -1} E_{\alpha ,\alpha } (-\lambda _{n } (T-\tau )^{\alpha } ) { G_{k}} (\tau ) \,d\tau }{ a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt } \\ &\quad {}- \frac{ b \int _{0}^{T} \int _{0}^{t} (t-\tau )^{\alpha -1} E_{\alpha ,\alpha } (-\lambda _{n } (t-\tau )^{\alpha } ) { G_{k}} (\tau ) \,d\tau \,dt}{ a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt } . \end{aligned}$$
(2.13)

This together with (2.9) implies that

$$\begin{aligned} u(x,t)&= \underbrace{\sum_{k=1}^{\infty } \frac{ E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \langle f(x), \phi _{k} (x) \rangle }{ a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt } \phi _{k}(x)}_{ \mathcal{H}_{1}(x,t)} \\ &\quad {}- \underbrace{\sum_{k=1}^{\infty } \frac{ aE_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \int _{0}^{T} (T-\tau )^{\alpha -1} E_{\alpha ,\alpha } (-\lambda _{n } (T-\tau )^{\alpha } ) { G_{k}} (\tau ) \,d\tau }{ a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt } \phi _{k}(x)}_{ \mathcal{H}_{2}(x,t)} \\ &\quad {}- \underbrace{\sum_{k=1}^{\infty } \frac{b E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \int _{0}^{T} \int _{0}^{t} (t-\tau )^{\alpha -1} E_{\alpha ,\alpha } (-\lambda _{k } (t-\tau )^{\alpha } ) G_{k} (\tau ) \,d\tau \,dt}{ a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt } \phi _{k}(x)}_{ \mathcal{H}_{3}(x,t)} \\ &\quad {}+ \underbrace{\sum_{k=1}^{\infty } \int _{0}^{t} (t-\tau )^{\alpha -1} E_{\alpha ,\alpha } \bigl(-\lambda _{k} (t-\tau )^{\alpha } \bigr) G_{k} (\tau ) \,d\tau \phi _{k}(x) }_{ \mathcal{H}_{4}(x,t)}. \end{aligned}$$
(2.14)

First, we need to show that

$$ \frac{ \overline{C}_{1}}{\lambda _{k}} \le a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{\alpha ,1} \bigl(-\lambda _{k } t^{ \alpha }\bigr) \,dt \le \frac{ \overline{C}_{2}}{\lambda _{k}}. $$
(2.15)

Indeed, from Lemma 2.1, we have the following estimate:

$$ \int _{0}^{T} E_{\alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \le D_{2} \int _{0}^{T} \frac{d t}{1 + \lambda _{k} t^{\alpha }} \le \frac{D_{2}}{\lambda _{k}} \int _{0}^{T} \frac{d t}{t^{\alpha }} = \frac{D_{2} T^{1-\alpha }}{(1-\alpha )\lambda _{k}}. $$
(2.16)

Due to the inequality

$$ E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) \le \frac{D_{2}}{1+ \lambda _{k} T^{\alpha }} \le \frac{D_{2}}{ \lambda _{k} T^{\alpha }}, $$

we find the following bound:

$$ a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{ \alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \le \frac{aD_{2}}{ \lambda _{k} T^{\alpha }}+ \frac{b D_{2}}{ \lambda _{k} T^{\alpha }}= \frac{ \overline{C}_{2}}{\lambda _{k}}. $$
(2.17)

Due to the fact that \(1+\lambda _{k} t^{\alpha }\le \lambda _{k} ( \frac{1}{\lambda _{1}} +t^{\alpha })\), we also get

$$\begin{aligned} \int _{0}^{T} E_{\alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt &\ge D_{1} \int _{0}^{T} \frac{d t}{1 + \lambda _{n} t^{\alpha }} \ge \frac{D_{1}}{\lambda _{k}} \int _{0}^{T} \frac{d t}{\frac{1}{\lambda _{1}} + t^{\alpha }} \\ &\ge \frac{D_{1}}{\lambda _{n}} \int _{0}^{T} \frac{d t}{\frac{1}{\lambda _{1}} + T^{\alpha }}= \frac{D_{1}}{\lambda _{k}} \frac{T}{\frac{1}{\lambda _{1}} + T^{\alpha }}. \end{aligned}$$
(2.18)

Hence, we find that

$$ a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{ \alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \ge \frac{\overline{C}_{1}}{\lambda _{k}}. $$
(2.19)

Step 1. Estimate \(\|\mathcal{H}_{1}(\cdot,t)\|_{D(\mathcal{A}^{s})}\). By using Paseval’s equality and recalling that

$$ E_{\alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \le \frac{D_{2}}{1+ \lambda _{k} t^{\alpha }} \le D_{2} \lambda _{k}^{-1} t^{- \alpha }, $$
(2.20)

we obtain

$$\begin{aligned} \bigl\Vert \mathcal{H}_{1}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})}^{2}&= \sum_{k=1}^{ \infty } \lambda _{k}^{2s} \biggl[ \frac{ E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \langle f(x), \phi _{k} (x) \rangle }{ a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt } \biggr]^{2} \\ &\le \sum_{k=1}^{\infty } \lambda _{k}^{2s} D_{2}^{2} \lambda _{k}^{-2} t^{-2\alpha } \biggl( \frac{\lambda _{k}}{\overline{C}_{1}} \biggr)^{2} \bigl\langle f(x), \phi _{k} (x) \bigr\rangle ^{2} . \end{aligned}$$
(2.21)

Hence, there exists \(D_{3}= D_{2} (\overline{C}_{1})^{-1}\) such that

$$ \bigl\Vert \mathcal{H}_{1}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})} \le D_{3} t^{-\alpha } \Vert f \Vert _{D(\mathcal{A}^{s})}. $$
(2.22)

Step 2. Estimate \(\|\mathcal{H}_{2}(\cdot,t)\|_{D(\mathcal{A}^{s})}\). First, we see that, for any \(0<\beta <1\),

$$ E_{\alpha ,\alpha } \bigl(-\lambda _{k } (T-\tau )^{\alpha } \bigr) \le \frac{D_{2}}{ 1+\lambda _{k } (T-\tau )^{\alpha } } \le D_{2} \lambda _{k}^{-\beta } (T-\tau )^{-\alpha \beta }. $$
(2.23)

This implies that

$$\begin{aligned} \int _{0}^{T} (T-\tau )^{\alpha -1} E_{\alpha ,\alpha } \bigl(- \lambda _{n } (T-\tau )^{\alpha } \bigr) G_{k} (\tau ) \,d\tau \le D_{2} \lambda _{k}^{-\beta } \int _{0}^{T} (T-\tau )^{\alpha -1-\alpha \beta } G_{k} (\tau ) \,d\tau . \end{aligned}$$
(2.24)

Thanks to inequality (2.20) we know that

$$\begin{aligned} & \bigl\Vert \mathcal{H}_{2}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})}^{2} \\ &\quad = \sum_{k=1}^{\infty } \lambda _{k}^{2s} \biggl[ \frac{ aE_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \int _{0}^{T} (T-\tau )^{\alpha -1} E_{\alpha ,\alpha } (-\lambda _{k } (T-\tau )^{\alpha } ) { G_{k}} (\tau ) \,d\tau }{ a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt } \biggr]^{2} \\ &\quad \le a^{2} \sum_{k=1}^{\infty } \lambda _{k}^{2s} D_{2}^{2} \lambda _{k}^{-2} t^{-2\alpha } \biggl( \frac{\lambda _{k}}{\overline{C}_{1}} \biggr)^{2} \biggl( \int _{0}^{T} (T-\tau )^{\alpha -1} E_{\alpha ,\alpha } \bigl(-\lambda _{k } (T- \tau )^{\alpha } \bigr) { G_{k}} (\tau ) \,d\tau \biggr)^{2} \\ &\quad \le D_{4}^{2} t^{-2\alpha } \sum _{k=1}^{\infty } \lambda _{k}^{2s-2 \beta } \biggl( \int _{0}^{T} (T-\tau )^{\alpha -1-\alpha \beta } G_{k} (\tau ) \,d\tau \biggr)^{2}. \end{aligned}$$
(2.25)

Using Hölder’s inequality, we obtain

$$\begin{aligned} & \biggl( \int _{0}^{T} (T-\tau )^{\alpha -1-\alpha \beta } G_{k} ( \tau ) \,d\tau \biggr)^{2} \\ &\quad \le \biggl( \int _{0}^{T} (T-\tau )^{\alpha -1-\alpha \beta } \,d\tau \biggr) \biggl( \int _{0}^{T} (T-\tau )^{\alpha -1-\alpha \beta } \bigl\vert G_{k} (\tau ) \bigr\vert ^{2} \,d\tau \biggr) \\ &\quad = \frac{T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } \biggl( \int _{0}^{T} (T-\tau )^{\alpha -1-\alpha \beta } \bigl\vert G_{k} (\tau ) \bigr\vert ^{2} \,d\tau \biggr). \end{aligned}$$
(2.26)

Combining (2.25) and (2.26), we find that

$$\begin{aligned} \bigl\Vert \mathcal{H}_{2}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})}^{2} &\le \frac{D_{4}^{2} T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } t^{-2 \alpha } \Biggl( \int _{0}^{T} (T-\tau )^{\alpha -1-\alpha \beta } \sum _{k=1}^{\infty } \lambda _{k}^{2s-2\beta } \bigl\vert G_{k} (\tau ) \bigr\vert ^{2} \,d\tau \Biggr) \\ &= \frac{D_{4}^{2} T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } t^{-2 \alpha } \biggl( \int _{0}^{T} (T-\tau )^{\alpha -1-\alpha \beta } \bigl\Vert G( \tau ) \bigr\Vert _{D(\mathcal{A}^{s-\beta })}^{2} \,d\tau \biggr) \\ &\le \frac{D_{4}^{2} T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } t^{-2 \alpha } \Vert G \Vert ^{2}_{L^{\infty }(0,T; D(\mathcal{A}^{s-\beta }))} \biggl( \int _{0}^{T} (T-\tau )^{\alpha -1-\alpha \beta } \,d\tau \biggr). \end{aligned}$$
(2.27)

The latter estimate leads to

$$ \bigl\Vert \mathcal{H}_{2}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})} \le \frac{D_{4} T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } t^{- \alpha } \Vert G \Vert _{L^{\infty }(0,T; D(\mathcal{A}^{s-\beta }))} . $$
(2.28)

Step 3. Estimate \(\|\mathcal{H}_{3}(\cdot,t)\|_{D(\mathcal{A}^{s})}\). By using inequality (2.20), we obtain that

$$\begin{aligned} & \bigl\Vert \mathcal{H}_{3}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})}^{2} \\ &\quad = \sum_{k=1}^{\infty } \lambda _{k}^{2s} \biggl[ \frac{ aE_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \int _{0}^{T} \int _{0}^{t} (t-\tau )^{\alpha -1} E_{\alpha ,\alpha } (-\lambda _{k } (t-\tau )^{\alpha } ) G_{k} (\tau ) \,d\tau \,dt}{ a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt } \biggr]^{2} \\ &\quad \le a^{2} \sum_{k=1}^{\infty } \lambda _{k}^{2s} D_{2}^{2} \lambda _{k}^{-2} t^{-2\alpha } \biggl( \int _{0}^{T} \int _{0}^{t} (t- \tau )^{\alpha -1} E_{\alpha ,\alpha } \bigl(-\lambda _{k } (t- \tau )^{\alpha } \bigr) G_{k} (\tau ) \,d\tau \,dt \biggr)^{2}. \end{aligned}$$
(2.29)

Next, by applying Hölder’s inequality and (2.23) and noting that \(0< \beta <\min (1,\frac{1}{2\alpha })\), we find that

$$\begin{aligned} & \biggl\vert \int _{0}^{T} \int _{0}^{t} (t-\tau )^{\alpha -1} E_{\alpha , \alpha } \bigl(-\lambda _{k } (t-\tau )^{\alpha } \bigr) G_{k} ( \tau ) \,d\tau \,dt \biggr\vert \\ &\quad \le D_{2} \lambda _{k}^{-\beta } \int _{0}^{T} \int _{0}^{t} (t-\tau )^{- \alpha \beta } G_{k}(\tau ) \,d\tau \,dt \\ &\quad \le D_{2} \lambda _{k}^{-\beta } \int _{0}^{T} \biggl( \int _{0}^{t} (t-r)^{-2 \alpha \beta } \,d\tau \biggr)^{1/2} \biggl( \int _{0}^{t} G_{k}^{2}( \tau ) \,d\tau \biggr)^{1/2} \,dt \\ &\quad \le D_{2} \lambda _{k}^{-\beta } \sqrt{ \frac{T^{1-2\alpha \beta }}{1-2\alpha \beta }} \int _{0}^{T} \biggl( \int _{0}^{t} G_{k}^{2}( \tau ) \,d\tau \biggr)^{1/2} \,dt \\ &\quad \le D_{2} T \lambda _{k}^{-\beta } \sqrt{ \frac{T^{1-2\alpha \beta }}{1-2\alpha \beta }} \biggl( \int _{0}^{T} G_{k}^{2}( \tau ) \,d\tau \biggr)^{1/2}. \end{aligned}$$
(2.30)

By two above observations, we deduce that

$$\begin{aligned} \bigl\Vert \mathcal{H}_{3}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})}^{2} &\le D_{5}^{2} t^{-2 \alpha } \sum_{k=1}^{\infty } \lambda _{k}^{2s-2-2\beta } \biggl( \int _{0}^{T} G_{k}^{2}( \tau ) \,d\tau \biggr) \\ &\le D_{5}^{2} t^{-2\alpha } \Vert G \Vert ^{2}_{L^{2}(0,T; D( \mathcal{A}^{s-\beta -1}))}. \end{aligned}$$
(2.31)

Hence, we get immediately that

$$ \bigl\Vert \mathcal{H}_{3}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})} \le D_{5} t^{-\alpha } \Vert G \Vert _{L^{2}(0,T; D(\mathcal{A}^{s-\beta -1}))}. $$
(2.32)

Step 4. Estimate \(\|\mathcal{H}_{4}(\cdot,t)\|_{D(\mathcal{A}^{s})}\). From inequality (2.23), we know that

$$ \int _{0}^{t} (t-\tau )^{\alpha -1} E_{\alpha ,\alpha } \bigl(- \lambda _{n } (t-\tau )^{\alpha } \bigr) G_{k} (\tau ) \,d\tau \le D_{2} \lambda _{k}^{-\beta } \int _{0}^{t} (t-\tau )^{\alpha -1-\alpha \beta } G_{k} (\tau ) \,d\tau . $$
(2.33)

Hence, by applying Hölder’s inequality, we get the following bound:

$$\begin{aligned} \bigl\Vert \mathcal{H}_{4}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})}^{2} &= \sum_{k=1}^{ \infty } \biggl[ \int _{0}^{t} (t-\tau )^{\alpha -1} E_{\alpha , \alpha } \bigl(-\lambda _{k} (t-\tau )^{\alpha } \bigr) G_{k} ( \tau ) \,d\tau \biggr]^{2} \\ &\le D_{2}^{2} \sum_{k=1}^{\infty } \lambda _{k}^{-2\beta } \biggl[ \int _{0}^{t} (t-\tau )^{\alpha -1-\alpha \beta } G_{k} ( \tau ) \,d\tau \biggr]^{2} \\ &\le D_{2}^{2} \sum_{k=1}^{\infty } \lambda _{k}^{-2\beta } \biggl[ \int _{0}^{t} (t-\tau )^{\alpha -1-\alpha \beta } \,d\tau \biggr] \biggl[ \int _{0}^{t} (t-\tau )^{\alpha -1-\alpha \beta } \bigl\vert G_{k} (\tau ) \bigr\vert ^{2} \,d\tau \biggr] \\ &\le \frac{ D_{2}^{2} T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } \sum_{k=1}^{\infty } \Biggl( \int _{0}^{t} (t-\tau )^{ \alpha -1-\alpha \beta } \sum _{k=1}^{\infty } \lambda _{k}^{-2 \beta } \bigl\vert G_{k} (\tau ) \bigr\vert ^{2} \,d\tau \Biggr) \\ &\le D_{2}^{2} \biggl( \frac{ T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } \biggr)^{2} \Vert G \Vert _{L^{\infty }(0,T; D(\mathcal{A}^{-\beta }))}^{2}. \end{aligned}$$
(2.34)

Therefore, we obtain

$$ \bigl\Vert \mathcal{H}_{4}(\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})} \le D_{2} \biggl( \frac{ T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } \biggr) \Vert G \Vert _{L^{\infty }(0,T; D(\mathcal{A}^{-\beta }))}. $$
(2.35)

Combining four steps as above, we deduce that

$$\begin{aligned} \bigl\Vert u (\cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})} &\le \sum _{j=1}^{4} \bigl\Vert \mathcal{H}_{j}( \cdot,t) \bigr\Vert _{D(\mathcal{A}^{s})} \\ &\le D_{3} t^{-\alpha } \Vert f \Vert _{D(\mathcal{A}^{s})}+ \frac{D_{4} T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } t^{- \alpha } \Vert G \Vert _{L^{\infty }(0,T; D(\mathcal{A}^{s-\beta }))} \\ &\quad {}+ D_{5} t^{-\alpha } \Vert G \Vert _{L^{2}(0,T; D(\mathcal{A}^{s- \beta -1}))}+ D_{2} \biggl( \frac{ T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } \biggr) \Vert G \Vert _{L^{\infty }(0,T; L^{2}(\Omega ))}. \end{aligned}$$
(2.36)

Let us choose \(1< p < \frac{1}{\alpha }\). The latter estimate leads to

$$\begin{aligned} \Vert u \Vert _{L^{p}(0,T; D(\mathcal{A}^{s}))}&= \biggl( \int _{0}^{T} \bigl\Vert u (\cdot,t) \bigr\Vert ^{p}_{D(\mathcal{A}^{s})} \,dt \biggr)^{1/p} \\ &\le D_{3} \Vert f \Vert _{D( \mathcal{A}^{s})} \biggl( \int _{0}^{T} t^{-\alpha p} \,dt \biggr)^{1/p} \\ &\quad {}+ \frac{D_{4} T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } \Vert G \Vert _{L^{\infty }(0,T; D(\mathcal{A}^{s-\beta }))} \biggl( \int _{0}^{T} t^{-\alpha p} \,dt \biggr)^{1/p} \\ &\quad {}+ D_{5} \Vert G \Vert _{L^{2}(0,T; D(\mathcal{A}^{s-\beta -1}))} \biggl( \int _{0}^{T} t^{-\alpha p} \,dt \biggr)^{1/p} \\ &\quad {} + D_{2} \biggl( \frac{ T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta } \biggr) \Vert G \Vert _{L^{\infty }(0,T; L^{2}(\Omega ))} \biggl( \int _{0}^{T} dt \biggr)^{1/p}. \end{aligned}$$
(2.37)

Noting that the proper integral \(\int _{0}^{T} t^{-\alpha p} \,dt\) is convergent, we deduce that

$$\begin{aligned} \Vert u \Vert _{L^{p}(0,T; D(\mathcal{A}^{s}))} &\lesssim \Vert f \Vert _{D(\mathcal{A}^{s})} + \Vert G \Vert _{L^{\infty }(0,T; D(\mathcal{A}^{s-\beta }))} \\ &\quad {}+ \Vert G \Vert _{L^{2}(0,T; D(\mathcal{A}^{s-\beta -1}))} + \Vert G \Vert _{L^{\infty }(0,T; L^{2}(\Omega ))} . \end{aligned}$$
(2.38)

 □

Identification of the initial value in the case \(G=0\)

In this section, we consider the problem of recovering the initial data \(u(0,x)\) in the case \(G=0\).

The ill-posedness

Theorem 3.1

The solution of Problem (1.1) is instability with respect to the \(L^{2}\) norm in the case \(t=0\).

Proof

Let \(u_{0}(x)= u(x,0)\). Let us consider the following operator \(\mathcal{M}:L^{2}(\Omega ) \to L^{2}(\Omega ) \):

$$\begin{aligned} \mathcal{M} u_{0}(x) &= \sum _{k=1}^{\infty } \biggl[a E_{ \alpha ,1} \bigl(- \lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{ \alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \biggr] \bigl\langle u_{0}(x), \phi _{k} (x) \bigr\rangle \phi _{k} (x) \\ & = \int _{\Omega }p(x,\nu ) u_{0}(\nu )\,d\nu , \end{aligned}$$
(3.1)

where we denote

$$ p(x,\nu ) = \sum_{k=1}^{\infty } \biggl[a E_{\alpha ,1} \bigl(- \lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{\alpha ,1} \bigl(- \lambda _{k } t^{\alpha }\bigr) \,dt \biggr] \phi _{k} (x) \phi _{k} (\nu ). $$
(3.2)

It is obvious that \(p(x,\nu ) = p(\nu ,x) \), we see that \(\mathcal{M}\) is a self-adjoint operator. We will show that it is a compact operator. Consider the finite rank operator as follows:

$$ \mathcal{M}_{K} u_{0}(x) = \sum _{k=1}^{K} \biggl[a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{\alpha ,1} \bigl(- \lambda _{k } t^{\alpha }\bigr) \,dt \biggr] \langle u_{0}, \phi _{k} \rangle \phi _{k} (x) . $$
(3.3)

It is obvious that \(\mathcal{M}_{K}\) is a finite rank operator. It follows from (3.1) and (3.3) that

$$\begin{aligned} & \Vert \mathcal{M}_{K} u_{0} -\mathcal{M} u_{0} \Vert _{L^{2}( \Omega )}^{2} \\ &\quad = \sum_{k=K+1}^{\infty } \biggl[a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{\alpha ,1} \bigl(-\lambda _{k } t^{ \alpha }\bigr) \,dt \biggr] \bigl\langle u_{0}(x), \phi _{k} (x) \bigr\rangle ^{2} \\ &\quad = \vert \overline{C}_{2} \vert ^{2} \sum _{j=K+1}^{\infty } \frac{1}{\lambda _{k}^{2}} \bigl\langle f(x), \varphi _{n} (x) \bigr\rangle ^{2} \le \frac{ \vert \overline{C}_{2} \vert ^{2} }{\lambda _{K}^{2}} \sum _{j=K+1}^{ \infty } \bigl\langle u_{0}(x), \phi _{k} (x) \bigr\rangle ^{2}, \end{aligned}$$
(3.4)

where we note that

$$ a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{ \alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \le \frac{ \overline{C}_{2}}{\lambda _{k}}. $$

Hence, we obtain that

$$ \Vert \mathcal{M}_{K} u_{0} -\mathcal{M} u_{0} \Vert _{L^{2}(\Omega )} \le \frac{ \overline{C}_{2}}{\lambda _{K}}. \Vert u_{0} \Vert _{L^{2}( \Omega )}. $$
(3.5)

This leads to \(\|\mathcal{M}_{K} - \mathcal{M}\|_{L^{2}(\Omega )} \to 0\) in the sense of operator norm in \(L(L^{2}(\Omega );L^{2}(\Omega ))\) as \(K \to \infty \). Moreover, \(\mathcal{M}\) is a compact operator. From (3.1), we get an operator equation as follows:

$$ \mathcal{M}u_{0} (x)=f(x), $$
(3.6)

and by Kirsch [43] we conclude that the problem is ill-posed. Next, we continue to give an example for ill-posedness. Taking the input final data \(f^{j} (x) = \frac{ \phi _{j}(x) }{\sqrt{\lambda _{j}}} \). Then the initial data with respect to \(f^{j}\) is

$$\begin{aligned} u_{0}^{j}(x) &= \sum_{k=1}^{\infty } \frac{ E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \langle f^{j}(x), \phi _{k} (x) \rangle }{ a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt } \phi _{k}(x) \\ & = \frac{\phi _{j}(x) }{ \sqrt{\lambda _{j}} ( a E_{\alpha ,1} (-\lambda _{k } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{k } t^{\alpha }) \,dt ) } \cdot \end{aligned}$$
(3.7)

It is obvious that

$$ \lim_{j \to +\infty } \bigl\Vert f^{j} \bigr\Vert _{L^{2}(\Omega )} = \lim_{j \to +\infty } \frac{1}{\sqrt{\lambda _{j}} }=0. $$
(3.8)

An error in the \(L^{2}\) norm of the initial data is as follows:

$$ \bigl\Vert u_{0}^{j} \bigr\Vert _{L^{2}(\Omega )}= \biggl\Vert \frac{\phi _{j}(x) }{ \sqrt{\lambda _{j}} ( a E_{\alpha ,1} (-\lambda _{j } T^{\alpha }) + b \int _{0}^{T} E_{\alpha ,1} (-\lambda _{j } t^{\alpha }) \,dt ) } \biggr\Vert _{L^{2}(\Omega )} \ge \frac{\sqrt{\lambda _{j}}}{\overline{C}_{2} }. $$
(3.9)

It is easy to see that

$$ \lim_{j \to +\infty } \bigl\Vert u_{0}^{j} \bigr\Vert _{L^{2}(\Omega )} > \lim_{j \to +\infty } \frac{\sqrt{\lambda _{j}}}{\overline{C}_{2}} =+ \infty . $$
(3.10)

Combining (3.8) and (3.10), we conclude that the solution is instability. □

Regularization and \(L^{p}\) error estimate

In this subsection, we construct a regularized solution and investigate the error between the regularized solution and the exact solution. Let us assume that \(f_{\delta }\) is noisy data and satisfies that

$$ \Vert f_{\delta }-f \Vert _{L^{q}(\Omega )} \le \delta $$
(3.11)

for any \(q\ge 1\).

Theorem 3.2

Let \(f_{\delta }\) be as in (3.11). Let \(u_{0}\) be the function which belongs to \(D(\mathcal{A}^{\sigma })\) for any \(\sigma >0\). Let us give a regularized solution as follows:

$$ u_{0}^{\delta }(x)= \sum_{k=1}^{ \lambda _{k} \le M_{\delta }} \biggl[a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha } \bigr) + b \int _{0}^{T} E_{ \alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \biggr]^{-1} \bigl\langle f^{\delta }, \phi _{k} \bigr\rangle \phi _{k} (x). $$
(3.12)

Let us choose

$$ M_{\delta }= \biggl( \frac{1}{\delta } \biggr)^{\frac{1-\mu }{m-h+1}},\quad 0< \mu < 1, $$

where

$$ -\frac{d}{4} < h \le \min \biggl(0, \frac{(q-2)d}{4q}\biggr),\quad 0\le m < \frac{d}{4}. $$
(3.13)

Then we get the following estimate:

$$ \bigl\Vert u_{0}^{\delta }-u_{0} \bigr\Vert _{L^{\frac{2d}{d-4m}}(\Omega )} \lesssim C_{h,q} \delta ^{\mu }+ \delta ^{\frac{\sigma (1-\mu )}{m-h+1}} \Vert u_{0} \Vert _{D( \mathcal{A}^{\sigma })}. $$
(3.14)

Proof

Since the Sobolev embedding \(L^{q}(\Omega ) \hookrightarrow D(\mathcal{A}^{h})\), we find that there exists a positive constant \(C:=C_{h,q}\)

$$ \bigl\Vert f^{\delta }-f \bigr\Vert _{D(\mathcal{A}^{h}) } \le C_{h,q} \bigl\Vert f^{\delta }-f \bigr\Vert _{L^{q}( \Omega )} \le C_{h,q} \delta . $$
(3.15)

For \(m>0\), we consider the term \(\|u_{0}^{\epsilon }-u_{0}\|_{D(\mathcal{A}^{m})}\). Using the triangle inequality, we obtain

$$ \bigl\Vert u_{0}^{\delta }-u_{0} \bigr\Vert _{D(\mathcal{A}^{m})} \le \bigl\Vert u_{0}^{\delta }- \overline{u}_{0}^{\delta } \bigr\Vert _{D(\mathcal{A}^{m})}+ \bigl\Vert \overline{u}_{0}^{\delta }-u_{0} \bigr\Vert _{D(\mathcal{A}^{m})}, $$
(3.16)

where

$$ \overline{u}_{0}^{\delta }(x) = \sum _{k=1}^{ \lambda _{k} \le M_{\delta }} \biggl[a E_{\alpha ,1} \bigl(- \lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{\alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \biggr]^{-1} \langle f, \phi _{k} \rangle \phi _{k} (x). $$

In the following, we first consider the term \(\| u_{0}^{\delta }-\overline{u}_{0}^{\delta }\|_{D(\mathcal{A}^{m})}\) for any \(0< m< \frac{d}{4}\). Indeed, we get

$$\begin{aligned} & \bigl\Vert u_{0}^{\delta }-\overline{u}_{0}^{\delta } \bigr\Vert _{D(\mathcal{A}^{m})}^{2} \\ &\quad = \sum_{k=1}^{\lambda _{k} \le \mathcal{M}_{\delta }} \lambda _{k}^{2m} \biggl[a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{ \alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \biggr]^{-2} \bigl\langle f^{\delta }-f, \phi _{k} \bigr\rangle ^{2} \\ &\quad \le \sum_{k=1}^{\lambda _{k} \le \mathcal{M}_{\delta }} \lambda _{k}^{2m-2h} \biggl[a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{\alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \biggr]^{-2} \lambda _{k}^{2h} \bigl\langle f^{\delta }-f, \phi _{k} \bigr\rangle ^{2}. \end{aligned}$$
(3.17)

From the fact that

$$ a E_{\alpha ,1} \bigl(-\lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{ \alpha ,1} \bigl(-\lambda _{k } t^{\alpha }\bigr) \,dt \ge \frac{\overline{C}_{1}}{\lambda _{k}}, $$

we get that

$$ \bigl\Vert u_{0}^{\delta }- \overline{u}_{0}^{\delta } \bigr\Vert _{D(\mathcal{A}^{m})}^{2} \le ( \mathcal{M}_{\delta } )^{2m-2h+2} \bigl\Vert f^{\delta }-f \bigr\Vert _{D( \mathcal{A}^{h}) }^{2} \le C_{h,q}^{2} \delta ^{2} ( \mathcal{M}_{\delta } )^{2m-2h+2} . $$
(3.18)

Next, we continue to get the following estimate:

$$\begin{aligned} \bigl\Vert \overline{u}_{0}^{\delta }-u_{0} \bigr\Vert _{D(\mathcal{A}^{m})}^{2}& = \sum _{k=1}^{ \lambda _{k} \rangle M_{\delta }} \biggl[a E_{\alpha ,1} \bigl(- \lambda _{k } T^{\alpha }\bigr) + b \int _{0}^{T} E_{\alpha ,1} \bigl(- \lambda _{k } t^{\alpha }\bigr) \,dt \biggr]^{-2} \langle f, \phi _{k} \rangle ^{2} \\ &= \sum_{k=1}^{ \lambda _{k} \rangle M_{\delta }} \langle u_{0}, \phi _{k} \rangle ^{2}= \sum_{k=1}^{ \lambda _{k} \rangle M_{\delta }} \lambda _{k}^{-2 \sigma } \lambda _{k}^{2\sigma } \langle u_{0}, \phi _{k} \rangle ^{2} \\ &\le ( \mathcal{M}_{\delta } )^{-2\sigma } \sum _{k=1}^{ \lambda _{k} \rangle M_{\delta }} \lambda _{k}^{2\sigma } \langle u_{0}, \phi _{k} \rangle ^{2} \le ( \mathcal{M}_{\delta } )^{-2\sigma } \Vert u_{0} \Vert _{D(\mathcal{A}^{\sigma })}^{2}. \end{aligned}$$
(3.19)

From the Sobolev embedding \(D(\mathcal{A}^{m}) \hookrightarrow L^{\frac{2d}{d-4m}}(\Omega )\) and combining (3.18) and (3.19), we get that

$$\begin{aligned} \bigl\Vert u_{0}^{\delta }-u_{0} \bigr\Vert _{L^{\frac{2d}{d-4m}}(\Omega )}&\le C \bigl\Vert u_{0}^{\delta }-u_{0} \bigr\Vert _{D(\mathcal{A}^{m})} \le \bigl\| u_{0}^{\delta }- \overline{u}_{0}^{\delta }\bigr\| _{D(\mathcal{A}^{m})}+ \bigl\| \overline{u}_{0}^{\delta }-u_{0} \bigr\| _{D( \mathcal{A}^{m})} \\ &\le C_{h,q} \delta ( \mathcal{M}_{\delta } )^{m-h+1}+ ( \mathcal{M}_{\delta } )^{-\sigma } \Vert u_{0} \Vert _{D(\mathcal{A}^{ \sigma })}. \end{aligned}$$
(3.20)

The proof of Theorem 3.2 is completed. □

Conclusions

In this paper, we focus on the fractional diffusion equation with nonlocal integral condition. By using the mild solution in a Fourier series form and the Mittag-Leffler function, we show two results as follows. First of all, we show the properties of the well-posedness and regularity of the mild solution to this problem. Next, we present that our problem is ill-posed (in the sense of Hadamard). In addition, we construct a regularized solution and present the convergence rate between the regularized and exact solutions by the Fourier truncation method.

Availability of data and materials

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Acknowledgements

This research is supported by Industrial University of Ho Chi Minh City (IUH) under grant number 66/HÐ-ÐHCN.

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The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.

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Correspondence to Nguyen Duc Phuong.

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Tuan, N.H., Triet, N.A., Luc, N.H. et al. On a time fractional diffusion with nonlocal in time conditions. Adv Differ Equ 2021, 204 (2021). https://doi.org/10.1186/s13662-021-03365-1

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MSC

  • 35R11
  • 35B65
  • 26A33

Keywords

  • Fractional diffusion equation
  • Well-posed problem
  • Ill-posed problem
  • Regularization
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