Recovering a Space-Dependent Source Term in the Fractional Diffusion Equation with the Riemann–Liouville Derivative

Abstract

This research determines an unknown source term in the fractional diffusion equation with the Riemann–Liouville derivative. This problem is ill-posed. Conditional stability for the inverse source problem can be given. Further, a fractional Tikhonov regularization method was applied to regularize the inverse source problem. In the theoretical results, we propose a priori and a posteriori regularization parameter choice rules and obtain the convergence estimates.

1. Introduction

In recent years, fractional calculus has become an important tool in mathematical modeling and has attracted the attention of researchers in various fields of science and engineering [1,2]. Research into the fractional diffusion equation has become a hot spot, attracting the interests of many researchers in fields such as elastic material mechanics, hydrology, random walking, biomedical, physics, medicine, and social sciences [3,4,5,6,7,8].

The direct problems for the time-fractional diffusion equation have been studied for many years, for example, the maximum principle, uniqueness results, existence results, numerical solutions, and analytic solutions [9,10,11,12,13,14,15,16,17]. In addition, various inverse problems of fractional diffusion equations have been researched extensively, such as inverse source problems [18,19], backward problems [20,21], the Cauchy problem [22,23], the inversion for parameter, or fractional order [24,25,26,27,28,29].

Let $\mathsf{\Omega }$ be a bounded domain in ${\mathbb{R}}^d$, $(d=1,2,3)$ with the sufficiently smooth boundary $\partial \mathsf{\Omega }$. We consider an unknown source issue for the fractional diffusion equation with the Riemann–Liouville derivative

$$\left\{\begin{array}{cc}{\partial }_{t}u(x,t)={\partial }_t^{1-\alpha }Au(x,t)+F(x,t),\hfill & (x,t)\in \mathsf{\Omega }\times (0,T),\hfill \\ u(x,t)=0,\hfill & x\in \partial \mathsf{\Omega },t\in (0,T],\hfill \\ u(x,T)=g\left(x\right),\hfill & x\in \overline{\mathsf{\Omega }},\hfill \end{array}\right.$$

(1)

where $T>0$ is a given time. The symbol ${\partial }_t^{1-\alpha }$ is the Riemann–Liouville derivative of the order of $1-\alpha \in (0,1)$ defined in [30]

$${\partial }_t^{1-\alpha }u(x,t)=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}\frac{d}{dt}{\int }_0^t{(t-s)}^{\alpha -1}u(x,s)ds,t>0,$$

(2)

where $\mathsf{\Gamma }(·)$ is the Gamma function. The operator A is a symmetric uniformly elliptic operator defined on $D(A):=H^2(\mathsf{\Omega })\cap H_0^1(\mathsf{\Omega })$ in [31]

$$Au(x,t)=\sum _{i=1}^d\frac{\partial }{\partial x_i}\left(\sum _{j=1}^d{a}_{ij}\left(x\right)\frac{\partial }{\partial x_j}u(x,t)\right)+b\left(x\right)u(x,t),x\in \mathsf{\Omega }.$$

(3)

Moreover, the coefficients in (3) satisfy

$$a_{ij}=a_{ji},1\le i,j\le d,$$

(4)

$$\mu \sum _{i=1}^d{\xi }_i^2\le \sum _{i,j=1}^d{a}_{ij}\left(x\right){\xi }_i{\xi }_j,x\in \overline{\mathsf{\Omega }},\xi \in {\mathbb{R}}^d,\mu >0,$$

(5)

$$a_{ij}\in C^1\left(\overline{\mathsf{\Omega }}\right),b\in C\left(\overline{\mathsf{\Omega }}\right),b\left(x\right)\le 0,x\in \overline{\mathsf{\Omega }}.$$

(6)

The purpose of our article is to determine the source term $F(x,t)$ from the measured data $u(x,t)=g\left(x\right)$. The measurement is always noise-contaminated; thus, we have the measurement data $g^{\delta }\in L^2\left(\mathsf{\Omega }\right)$ satisfying

$$\parallel g^{\delta }-g\parallel \le \delta ,$$

(7)

where the constant $\delta$ denotes the noise level, and $\parallel ·\parallel$ denotes the $L^2$-norm.

For $\alpha =1$, the problem (1) is an inverse source problem of the classical diffusion equation

$$\left\{\begin{array}{cc}{\partial }_{t}u(x,t)=Au(x,t)+F(x,t),\hfill & (x,t)\in \mathsf{\Omega }\times (0,T),\hfill \\ u(x,t)=0,\hfill & x\in \partial \mathsf{\Omega },t\in (0,T],\hfill \\ u(x,T)=g\left(x\right),\hfill & x\in \overline{\mathsf{\Omega }}.\hfill \end{array}\right.$$

(8)

Obviously, Problem (8) has been researched extensively, see [32,33,34,35,36,37] for details. Recently, the source term identifications of fractional diffusion equations have been studied in different ways. If $F(x,t)=f\left(x\right)$, Zhang and Xu [18] proved the unique result of the source term identification problem using Laplace transform and analytic continuation. In [38,39,40], the authors studied inverse source problems of time-fractional diffusion equation using different methods, such as the Tikhonov regularization method, the simplified Tikhonov regularization method, the modified quasi-boundary value method, and the quasi-reversibility method. In [41], the authors recovered a space-dependent source term of a time-fractional diffusion equation using an iterative regularization method. In [42], the authors considered an inverse space-dependent source problem for a time-fractional diffusion equation via a new fractional Tikhonov regularization method. If $F(x,t)=f\left(t\right)$, Zhang and Wei [43] studied an inverse time-dependent source problem for the time-fractional diffusion equation by a truncation method. Yang and his group [44,45,46] considered the inverse time-dependent source problem for a fractional diffusion equation via several methods, such as the mollification regularization method, the quasi-reversibility regularization method, and the Fourier regularization method. If $F(x,t)=f\left(x\right)h\left(t\right)$, Nguyen et al. [47] applied the Tikhonov method to solve the inverse source problem of a time-fractional diffusion equation. In [48], the authors used the integral equation method and the standard Tikhonov regularization method to identify a time-dependent source term for a time-fractional diffusion equation. In [49,50], the authors investigated a source term identification problem in a time-fractional diffusion equation by using the Landweber iterative regularization method. In [51], the authors solved the inverse space-dependent source term in a time-fractional diffusion equation by using generalized and revised generalized Tikhonov regularization methods. In [52], the authors identified the source function in the time-fractional diffusion equation with non-local in-time conditions by using the modified fractional Landweber method.

Inverse source problems have applications in geophysical prospecting and pollutant detection [53,54]. As far as we know, there are few articles on inverse source problems of the fractional diffusion equation with the Riemann–Liouville derivative; see [55,56]. In this article, we use a new fractional Tikhonov regularization method to solve the inverse source problem (1).

The fractional Tikhonov regularization method was firstly proposed in [57]. Compared with the Tikhonov regularization method, the fractional Tikhonov regularization method has a better numerical effect. It was also used to solve some ill-posed problems, such as the inverse source problem of the time-fractional diffusion equation [42], the inverse time-fractional diffusion problem in a two-dimensional space [58], the initial value problem for a time-fractional diffusion equation [59], the backward problem for the space fractional diffusion equation [60], and the Cauchy problem of the Helmholtz equation [61].

The outline of this manuscript is as follows. In Section 2, we provide some preliminaries. The ill-posedness and conditional stability results are given in Section 3. In Section 4, we propose a fractional Tikhonov regularization method and obtain the convergence results based on a priori and a posteriori choice rules. The conclusion is given in Section 5.

2. Preliminaries

Throughout this article, we use the following definitions and lemmas.

Definition 1.

Let${\lambda }_p,e_p$be the eigenvalues and corresponding eigenvectors of the operator A in$\mathsf{\Omega }$. The family of eigenvalues${\left\{{\lambda }_p\right\}}_{p=1}^{\infty }$satisfy$0<{\lambda }_1\le {\lambda }_2\le \cdots \le {\lambda }_p\le \cdots$, where${\lambda }_p\to \infty$as$p\to \infty$:

$$\left\{\begin{array}{cc}A{e}_p\left(x\right)=-{\lambda }_p{e}_p\left(x\right),\hfill & x\in \mathsf{\Omega },\hfill \\ e_p\left(x\right)=0,\hfill & x\in \partial \mathsf{\Omega }.\hfill \end{array}\right.$$

(9)

Definition 2.

Let$(·,·)$be an inner product in$L^2\left(\mathsf{\Omega }\right)$. The notation${\parallel ·\parallel }_X$stands for the norm in the Banach space. For any$k\ge 0$, we define the space

$$H^k\left(\mathsf{\Omega }\right):=\left\{u\in L^2\left(\mathsf{\Omega }\right)|\sum _{p=1}^{\infty }{\lambda }_p^{2k}{\left|(u,e_p)\right|}^2<+\infty \right\}$$

(10)

equipped with the norm

$${\parallel u\parallel }_{H^k\left(\mathsf{\Omega }\right)}={\left(\sum _{p=1}^{\infty }{\lambda }_p^{2k}{\left|(u,e_p)\right|}^2\right)}^{\frac{1}{2}}.$$

(11)

Definition 3

([30]). The Mittag–Leffler function $E_{\alpha ,\beta }(·)$ is

$$E_{\alpha ,\beta }\left(z\right)=\sum _{k=0}^{\infty }\frac{z^k}{\mathsf{\Gamma }(\alpha k+\beta )},z\in \mathbb{C},$$

(12)

where $\alpha >0$ and $\beta \in \mathbb{R}$ are arbitrary constants.

Lemma 1

([30]). For $\alpha >0$ and $\beta \in \mathbb{R}$, one has

$$E_{\alpha ,\beta }\left(z\right)=z{E}_{\alpha ,\alpha +\beta }\left(z\right)+\frac{1}{\mathsf{\Gamma }\left(\beta \right)},z\in \mathbb{C}.$$

(13)

Lemma 2

([31]). Let $\lambda >0$, $\alpha >0$, then we have

$$\frac{d^m}{d{t}^m}{E}_{\alpha ,1}(-\lambda t^{\alpha })=-\lambda t^{\alpha -m}{E}_{\alpha ,\alpha -m+1}(-\lambda t^{\alpha }),t>0,$$

(14)

$$\frac{d}{dt}\left(t{E}_{\alpha ,2}(-\lambda t^{\alpha })\right)=E_{\alpha ,1}(-\lambda t^{\alpha }),t>0.$$

(15)

Lemma 3

([30]). Let $0<\alpha <1$ and $\lambda ,a>0$, then we have

$$\frac{d}{dt}\left(E_{\alpha ,1}(-\lambda t^{\alpha })\right)=-\lambda t^{\alpha -1}{E}_{\alpha ,\alpha }(-\lambda t^{\alpha }),fort>0,$$

(16)

$$\frac{d}{dt}\left(t^{\alpha -1}{E}_{\alpha ,\alpha }(-\lambda t^{\alpha })\right)=t^{\alpha -2}{E}_{\alpha ,\alpha -1}(-\lambda t^{\alpha }),fort>0,$$

(17)

$${\int }_0^{\infty }{e}^{-st}{E}_{\alpha ,1}(-a{t}^{\alpha })dt=\frac{s^{\alpha -1}}{s^{\alpha }+a},for\Re \left(s\right)>a^{\frac{1}{\alpha }}.$$

(18)

Lemma 4

([30]). Let $0<{\alpha }_0<{\alpha }_1<1$. Then there exists positive constants $M_1$, $M_2$, $M_3$, depending only on ${\alpha }_0,{\alpha }_1$, such that for all $\alpha \in [{\alpha }_0,{\alpha }_1]$,

$$\frac{M_1}{1-z}\le E_{\alpha ,1}\left(z\right)\le \frac{M_2}{1-z},E_{\alpha ,\beta }\left(x\right)\le \frac{M_3}{1-z},forallz\le 0,\alpha \in \mathbb{R}.$$

(19)

Lemma 5

([55]). Let $0<\alpha <1$ and ${\lambda }_p>0$. Then

$$\frac{Q_{\alpha }^{-}({\lambda }_1,M_1)}{{\lambda }_p}\le {\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })d\tau \le \frac{Q_{\alpha }^{+}\left(M_2\right)}{{\lambda }_p},$$

(20)

where

$$Q_{\alpha }^{-}({\lambda }_1,M_1)=\frac{M_{1}T{\lambda }_1}{1+{\lambda }_1{T}^{\alpha }},Q_{\alpha }^{+}\left(M_2\right)=\frac{M_2{T}^{1-\alpha }}{1-\alpha },$$

(21)

$M_1 and M_2$ are positive constants.

3. Ill-Posedness and Conditional Stability

First, we introduce the mild solution of the following problem:

$$\left\{\begin{array}{cc}{\partial }_{t}u(x,t)={\partial }_t^{1-\alpha }Au(x,t)+F(x,t),\hfill & (x,t)\in \mathsf{\Omega }\times (0,T),\hfill \\ u(x,t)=0,\hfill & x\in \partial \mathsf{\Omega },t\in (0,T],\hfill \\ u(x,0)=0,\hfill & x\in \mathsf{\Omega },\hfill \\ u(x,T)=g\left(x\right),\hfill & x\in \overline{\mathsf{\Omega }}.\hfill \end{array}\right.$$

(22)

Here, $F(x,t)=\sigma \left(t\right)f\left(x\right)$.

Throughout this work, we assume that there exists a positive constant E, such that

$${\parallel f\parallel }_{H^k}\le E,$$

(23)

for a positive real number, k. Here,

$$0<{\sigma }_0\le \sigma \left(t\right)\le {\sigma }_1,\forall t\in [0,T].$$

(24)

Now, according to the reference [56], we know

$$f\left(x\right)=\sum _{p=1}^{\infty }\frac{(g\left(x\right),e_p\left(x\right))e_p\left(x\right)}{{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau }.$$

(25)

From [62], we know that $E_{\alpha ,1}(-x)$ is a completely monotonic decreasing function for $x\ge 0$. Furthermore, $\frac{1}{{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau }\to \infty$ with respect to ${\lambda }_p$. So the small error in the measured data $g^{\delta }\left(x\right)$ will be amplified by the factor $\frac{1}{{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau }$. Thus, we call $\frac{1}{{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau }$ the magnifying factor of the problem.

We define

$$Kf=\sum _{p=1}^{\infty }\left[{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau \right](f\left(x\right),e_p\left(x\right))e_p\left(x\right)={\int }_{\mathsf{\Omega }}k(x,\xi )f\left(\xi \right)d\xi =g\left(x\right),x\in \mathsf{\Omega }.$$

(26)

Here,

$$k(x,\xi )=\sum _{p=1}^{\infty }\left({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau \right)e_p\left(x\right)e_p\left(\xi \right).$$

Obviously, according to reference [55], we know that the operator K is a self-adjoint compact operator and an injective. By Kirsch [63], we conclude that the problem (22) is ill-posed.

The following theorem gives conditional stability for the inverse source problem.

Theorem 1.

([55]) Let $E>0$ and $k>0$, suppose ${\parallel f\parallel }_{H^k\left(\mathsf{\Omega }\right)}\le E$ holds, then

$$\parallel f\parallel \le D\left(k\right)E^{\frac{1}{k+1}}{\parallel Kf\parallel }^{\frac{k}{k+1}},$$

(27)

where

$$D\left(k\right)={\left(\frac{1}{{\sigma }_0{Q}_{\alpha }^{-}({\lambda }_1,M_1)}\right)}^{\frac{k}{k+1}}.$$

(28)

Remark 1.

Essentially, Theorem 1 provides the following condition stability estimate

$$\parallel f_1-f_2\parallel \le D\left(k\right)\parallel f_1-f_2{\parallel }_{H^k\left(\mathsf{\Omega }\right)}^{\frac{1}{k+1}}{\parallel K{f}_1-K{f}_2\parallel }^{\frac{k}{k+1}}.$$

4. Fractional Tikhonov Regularization Method and Convergence Estimates

In this section, we prove the convergence estimates for the fractional Tikhonov regularization method under a priori and a posteriori choice rules, respectively.

The fractional Tikhonov regularized solution (with exact data) is given by

$$f_{\omega }\left(x\right)=\sum _{p=1}^{\infty }\frac{{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma -1}}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g\left(x\right),e_p\left(x\right))e_p\left(x\right),\frac{1}{2}\le \gamma \le 1,$$

(29)

and the fractional Tikhonov regularized solution (with noisy data) is given by

$$f_{\omega }^{\delta }\left(x\right)=\sum _{p=1}^{\infty }\frac{{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma -1}}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g^{\delta }\left(x\right),e_p\left(x\right))e_p\left(x\right),\frac{1}{2}\le \gamma \le 1,$$

(30)

where $\omega >0$ is a regularization parameter, $\gamma$ is called the fractional parameter.

• Case 1. When $\gamma =\frac{1}{2}$, it is the quasi-boundary value method [55].

• Case 2. When $\gamma =1$, it is the classical Tikhonov regularization method [56].

Before proving the main results, several useful and important lemmas are given.

Lemma 6.

For$\frac{1}{2}\le \gamma \le 1$, we can obtain

$$T\left(b\right)=\frac{b^{2\gamma -1}}{\omega +b^{2\gamma }}\le c_1\left(\gamma \right){\omega }^{-\frac{1}{2\gamma }},$$

(31)

where$c_1\left(\gamma \right)=\frac{{(2\gamma -1)}^{\frac{2\gamma -1}{2\gamma }}}{2\gamma }>0$.

Proof. 

For $\frac{1}{2}\le \gamma \le 1$, then ${\lim }_{b\to 0}T\left(b\right)={\lim }_{b\to \infty }T\left(b\right)=0$. Thus, there exists a $b^{\ast }={\left[(2\gamma -1)\omega \right]}^{\frac{1}{2\gamma }}\ge 0$, which is a global maximizer, such that $T^{\prime }\left(b^{\ast }\right)=0$. So we have

$$T\left(b\right)\le T\left(b^{\ast }\right)=c_1\left(\gamma \right){\omega }^{-\frac{1}{2\gamma }},$$

(32)

where $c_1\left(\gamma \right)=\frac{{(2\gamma -1)}^{\frac{2\gamma -1}{2\gamma }}}{2\gamma }>0$. □

Lemma 7.

For constants$\omega >0,a>0,b\ge {\lambda }_1>0$, we have

$$G\left(b\right)=\frac{\omega b^{2\gamma -k}}{\omega b^{2\gamma }+a^{2\gamma }}\le \left\{\begin{array}{cc}{c}_2(a,k,\gamma ){\omega }^{\frac{k}{2\gamma }},\hfill & 0<k<2\gamma ,\hfill \\ c_3(a,k,\gamma ,{\lambda }_1)\omega ,\hfill & k\ge 2\gamma ,\hfill \end{array}\right.$$

where$c_2(a,k,\gamma )=\frac{1}{2\gamma a^k}{(2\gamma -k)}^{1-\frac{k}{2\gamma }}{k}^{\frac{k}{2\gamma }}$and$c_3(a,k,\gamma ,{\lambda }_1)=\frac{1}{a^{2\gamma }{\lambda }_1^{k-2\gamma }}$.

Proof. 

If $0<k<2\gamma$, we know that ${\lim }_{b\to 0}G\left(b\right)={\lim }_{b\to \infty }G\left(b\right)=0$. We have

$$0\le \underset{b\in (0,+\infty )}{\sup }G\left(b\right)\le G\left(b_0\right).$$

(33)

Let $G^{\prime }\left(b_0\right)=0$, we have $b_0=a{(\frac{2\gamma -k}{\omega k})}^{\frac{1}{2\gamma }}>0$, then we obtain

$$G\left(b\right)\le G\left(b_0\right)=\frac{\omega a^{-k}{(\frac{2\gamma -k}{\omega k})}^{1-\frac{k}{2\gamma }}}{\frac{2\gamma }{k}}:=c_2(a,k,\gamma ){\omega }^{\frac{k}{2\gamma }}.$$

(34)

If $k>2\gamma$, then we have

$$G\left(b\right)\le \frac{\omega b^{2\gamma -k}}{a^{2\gamma }}=\frac{1}{a^{2\gamma }{b}^{k-2\gamma }}\omega \le \frac{1}{a^{2\gamma }{\lambda }_1^{k-2\gamma }}\omega :=c_3(a,k,\gamma ,{\lambda }_1)\omega .$$

(35)

□

Lemma 8.

For constants$\omega >0,a>0,b\ge {\lambda }_1>0$, then we have

$$L\left(b\right)=\frac{\omega b^{2\gamma -k-1}}{\omega b^{2\gamma }+a^{2\gamma }}\le \left\{\begin{array}{cc}{c}_4(a,k,\gamma ){\omega }^{\frac{k+1}{2\gamma }},\hfill & 0<k<2\gamma -1,\hfill \\ c_5(a,k,\gamma ,{\lambda }_1)\omega ,\hfill & k\ge 2\gamma -1,\hfill \end{array}\right.$$

where$c_4(a,k,\gamma )=\frac{1}{2\gamma a^{k+1}}{(2\gamma -k-1)}^{1-\frac{k+1}{2\gamma }}{(k+1)}^{\frac{k+1}{2\gamma }}$and$c_5(a,k,\gamma ,{\lambda }_1)=\frac{1}{a^{2\gamma }{\lambda }_1^{k+1-2\gamma }}$.

Proof. 

The proof is similar to Lemma 7, so we omit it. □

4.1. A Priori Convergence Estimate

Theorem 2.

Assume that conditions (7) and (23) hold. Let$f\left(x\right)$be the exact solution of problem (22), and$f_{\omega }^{\delta }\left(x\right)$be the fractional Tikhonov regularized solution of problem (22).

(a) If$0<k\le 2\gamma$, and if we choose

$$\omega ={\left(\frac{\delta }{E}\right)}^{\frac{2\gamma }{k+1}},$$

(36)

then we can obtain the following convergence estimates

$$\parallel f_{\omega }^{\delta }\left(x\right)-f\left(x\right)\parallel \le (c_1+c_2){\delta }^{\frac{k}{k+1}}{E}^{\frac{1}{k+1}}.$$

(37)

(b) If$k\ge 2\gamma$, and if we choose

$$\omega ={\left(\frac{\delta }{E}\right)}^{\frac{2\gamma }{2\gamma +1}},$$

(38)

then we can obtain the following convergence estimates

$$\parallel f_{\omega }^{\delta }\left(x\right)-f\left(x\right)\parallel \le (c_1+c_3){\delta }^{\frac{2\gamma }{2\gamma +1}}{E}^{\frac{1}{2\gamma +1}},$$

(39)

where$c_1,c_2, and c_3$are defined in Lemma 6 and Lemma 7.

Proof. 

According to the triangle inequality, we have

$$\parallel f_{\omega }^{\delta }\left(x\right)-f\left(x\right)\parallel \le \parallel f_{\omega }^{\delta }\left(x\right)-f_{\omega }\left(x\right)\parallel +\parallel f_{\omega }\left(x\right)-f\left(x\right)\parallel .$$

(40)

Now, using (7) and Lemma 6, we estimate the first term

$$\begin{array}{ccc}\parallel f_{\omega }^{\delta }\left(x\right)-f_{\omega }\left(x\right)\parallel & =& \parallel \sum _{p=1}^{\infty }\frac{{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma -1}}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g^{\delta }\left(x\right)-g\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel \hfill \\ & \le & \delta \underset{p}{\sup }\frac{{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma -1}}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}\hfill \end{array}$$

$$\begin{array}{c}\le c_1\delta {\omega }^{-\frac{1}{2\gamma }}.\end{array}$$

(41)

For the second term, using the Parseval identity and (23), we obtain

$$\begin{array}{c}\parallel f_{\omega }{\left(x\right)-f\left(x\right)\parallel }^2\\ =\sum _{p=1}^{\infty }{\left(\frac{{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma -1}}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}-\frac{1}{{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau }\right)}^2{|(g\left(x\right),e_p\left(x\right))|}^2\hfill \\ =\sum _{p=1}^{\infty }{\left(\frac{-\omega }{(\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }){\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau }\right)}^2{|(g\left(x\right),e_p\left(x\right))|}^2\hfill \\ =\sum _{p=1}^{\infty }\frac{{\omega }^2{|(g\left(x\right),e_p\left(x\right))|}^2}{{[\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }]}^2{|{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau |}^2}\hfill \\ =\sum _{p=1}^{\infty }\frac{{\omega }^2{\lambda }_p^{-2k}{\lambda }_p^{2k}{|(g\left(x\right),e_p\left(x\right))|}^2}{{[\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }]}^2{|{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau |}^2}\hfill \\ \le \underset{p}{\sup }{|M\left(p\right)|}^2\sum _{p=1}^{\infty }\frac{{\lambda }_p^{2k}{|(g\left(x\right),e_p\left(x\right))|}^2}{|{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha }){\sigma \left(\tau \right)d\tau |}^2}\hfill \end{array}$$

$$\begin{array}{c}\le \underset{p}{\sup }{\left|M\left(p\right)\right|}^2{\parallel f\parallel }_{H^k\left(\mathsf{\Omega }\right)}^2.\end{array}$$

(42)

Here,

$$\begin{array}{cc}\hfill M\left(p\right)& =\frac{\omega {\lambda }_p^{-k}}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}\hfill \\ \hfill & \le \frac{\omega {\lambda }_p^{-k}}{\omega +{({\sigma }_0{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })d\tau )}^{2\gamma }}\hfill \\ \hfill & \le \frac{\omega {\lambda }_p^{2\gamma -k}}{\omega {\lambda }_p^{2\gamma }+{({\sigma }_0{Q}_{\alpha }^{-}({\lambda }_1,M_1))}^{2\gamma }}.\hfill \end{array}$$

(43)

Using Lemma 7, we have

$$M\left(p\right)\le \left\{\begin{array}{cc}{c}_2{\omega }^{\frac{k}{2\gamma }},\hfill & {\displaystyle 0<k<2\gamma },\hfill \\ c_3\omega ,\hfill & k\ge 2\gamma .\hfill \end{array}\right.$$

(44)

Combining (42) and (44), we have

$$\parallel f_{\omega }\left(x\right)-f\left(x\right)\parallel \le \left\{\begin{array}{cc}{c}_2{\omega }^{\frac{k}{2\gamma }}E,\hfill & 0<k<2\gamma ,\hfill \\ c_3\omega E,\hfill & k\ge 2\gamma .\hfill \end{array}\right.$$

(45)

Therefore, combining (40), (41) and (45), we have

$$\parallel f_{\omega }^{\delta }\left(x\right)-f\left(x\right)\parallel \le c_1\delta {\omega }^{-\frac{1}{2\gamma }}+\left\{\begin{array}{cc}{c}_2{\omega }^{\frac{k}{2\gamma }}E,\hfill & 0<k<2\gamma ,\hfill \\ c_3\omega E,\hfill & k\ge 2\gamma .\hfill \end{array}\right.$$

(46)

If we choose the regularization parameter by (36) and (38), we can have the following convergence estimates

$$\parallel f_{\omega }^{\delta }\left(x\right)-f\left(x\right)\parallel \le \left\{\begin{array}{cc}(c_1+c_2){\delta }^{\frac{k}{k+1}}{E}^{\frac{1}{k+1}},\hfill & 0<k<2\gamma ,\hfill \\ (c_1+c_3){\delta }^{\frac{2\gamma }{2\gamma +1}}{E}^{\frac{1}{2\gamma +1}},\hfill & k\ge 2\gamma .\hfill \end{array}\right.$$

(47)

□

4.2. A Posteriori Convergence Estimate

In this subsection, we give the convergence estimate based on the a posteriori choice rule. According to Morozov’s discrepancy principle [63], we choose the regularization parameter $\omega$ as the solution of the following equation:

$$\parallel K{f}_{\omega }^{\delta }\left(x\right)-g^{\delta }\left(x\right)\parallel =\parallel \sum _{p=1}^{\infty }\frac{{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g^{\delta }\left(x\right),e_p\left(x\right))e_p\left(x\right)-g^{\delta }\left(x\right)\parallel =\tau \delta ,$$

(48)

where $\tau >1$ is a constant.

Lemma 9.

Let

$$\rho \left(\omega \right)=\parallel \sum _{p=1}^{+\infty }\frac{{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g^{\delta }\left(x\right),e_p\left(x\right))e_p\left(x\right)-g^{\delta }\left(x\right)\parallel ,$$

then the following results hold:

(a)$\rho \left(\omega \right)$is a continuous function;

(b)${\lim }_{\omega \to 0}\rho \left(\omega \right)=0;$

(c)${\lim }_{\omega \to +\infty }\rho \left(\omega \right)=\parallel g^{\delta }\parallel ;$

(d)$\rho \left(\omega \right)$is a strictly increasing function over$(0,+\infty )$.

The proof is obvious, and we omit it here.

Remark 2.

According to Lemma 9, we know that there exists a unique solution for (48) if $0<\tau \delta <\parallel g^{\delta }\parallel$.

Lemma 10.

If$\omega$is the solution of (48), we can obtain the following inequality

$${\omega }^{-\frac{1}{2\gamma }}\le \left\{\begin{array}{cc}{(\frac{c_{4}q}{\tau -1})}^{\frac{1}{k+1}}{(\frac{E}{\delta })}^{\frac{1}{k+1}},\hfill & 0<k<2\gamma -1,\hfill \\ {(\frac{c_{5}q}{\tau -1})}^{\frac{1}{2\gamma }}{(\frac{E}{\delta })}^{\frac{1}{2\gamma }},\hfill & k\ge 2\gamma -1,\hfill \end{array}\right.$$

(49)

where $c_4,c_5$ are defined in Lemma 8.

Proof. 

From (48), there holds

$$\begin{array}{ccc}\tau \delta & =& \parallel \sum _{p=1}^{+\infty }\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g^{\delta }\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel \hfill \\ & \le & \parallel \sum _{p=1}^{+\infty }\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g^{\delta }\left(x\right)-g\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel \hfill \\ & & +\parallel \sum _{p=1}^{+\infty }\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel \hfill \end{array}$$

$$\begin{array}{c}\le \delta +\parallel \sum _{p=1}^{+\infty }\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel .\hfill \end{array}$$

(50)

So, we have

$$(\tau -1)\delta \le \parallel \sum _{p=1}^{+\infty }\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel .$$

(51)

Using condition (23), we have

$$\begin{array}{ccc}& & \parallel \sum _{p=1}^{+\infty }\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel \\ & \le & \parallel \sum _{p=1}^{+\infty }\frac{\omega {\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau ·{\lambda }_p^{-k}}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}\frac{{\lambda }_p^k(g\left(x\right),e_p\left(x\right))e_p\left(x\right)}{{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau }\parallel \hfill \\ & =& {\left\{\sum _{p=1}^{+\infty }{\left[\frac{\omega {\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau ·{\lambda }_p^{-k}}{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}\right]}^2{\left[\frac{{\lambda }_p^k(g\left(x\right),e_p\left(x\right))e_p\left(x\right)}{{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau }\right]}^2\right\}}^{\frac{1}{2}}\hfill \\ & \le & \underset{p}{\sup }\frac{\omega {\sigma }_1{Q}_{\alpha }^{+}(M_2){\lambda }_p^{-k-1}}{\omega +{(\frac{{\sigma }_0{Q}_{\alpha }^{-}({\lambda }_1,M_1)}{{\lambda }_p})}^{2\gamma }}{\left(\sum _{p=1}^{+\infty }\frac{{\lambda }_p^{2k}{(g\left(x\right),e_p\left(x\right))}^2}{{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^2}\right)}^{\frac{1}{2}}\hfill \\ & \le & \left\{\begin{array}{cc}{c}_{4}q{\omega }^{\frac{k+1}{2\gamma }}{\parallel f\parallel }_{H^k\left(\mathsf{\Omega }\right)},\hfill & 0<k<2\gamma -1,\hfill \\ c_{5}q\omega {\parallel f\parallel }_{H^k\left(\mathsf{\Omega }\right)},\hfill & k\ge 2\gamma -1,\hfill \end{array}\right.\hfill \end{array}$$

$$\le \left\{\begin{array}{cc}{c}_{4}q{\omega }^{\frac{k+1}{2\gamma }}E,\hfill & 0<k<2\gamma -1,\hfill \\ c_{5}q\omega E,\hfill & k\ge 2\gamma -1,\hfill \end{array}\right.$$

(52)

where $q={\sigma }_1{Q}_{\alpha }^{+}\left(M_2\right)$. Therefore, combining (50)–(52), we have

$${\omega }^{-\frac{1}{2\gamma }}\le \left\{\begin{array}{cc}{(\frac{c_{4}q}{\tau -1})}^{\frac{1}{k+1}}{(\frac{E}{\delta })}^{\frac{1}{k+1}},\hfill & 0<k<2\gamma -1,\hfill \\ {(\frac{c_{5}q}{\tau -1})}^{\frac{1}{2\gamma }}{(\frac{E}{\delta })}^{\frac{1}{2\gamma }},\hfill & k\ge 2\gamma -1.\hfill \end{array}\right.$$

□

Theorem 3.

Suppose the conditions (7) and (23) hold, and take the solution of (48) as the regularization parameter, then

(a) If$0<k<2\gamma -1$, the following error estimate holds

$$\parallel f_{\omega }^{\delta }(x)-f(x)\parallel \le (c_1{(\frac{c_{4}q}{\tau -1})}^{\frac{1}{k+1}}+c_6){\delta }^{\frac{k}{k+1}}{E}^{\frac{1}{k+1}}.$$

(53)

(b) If$k\ge 2\gamma -1$, the following error estimate holds

$$\parallel f_{\omega }^{\delta }\left(x\right)-f\left(x\right)\parallel \le c_1{(\frac{c_{5}q}{\tau -1})}^{\frac{1}{2\gamma }}{\delta }^{\frac{2\gamma -1}{2\gamma }}{E}^{\frac{1}{2\gamma }}+c_6{\delta }^{\frac{k}{k+1}}{E}^{\frac{1}{k+1}},$$

(54)

where$c_1$is defined in Lemma 6,$c_4,c_5$are defined in Lemma 8,$c_6={(\frac{\tau +1}{{\sigma }_0{Q}_{\alpha }^{-}({\lambda }_1,M_1)})}^{\frac{k}{k+1}}$.

Proof. 

Due to the triangle inequality, we have

$$\parallel f_{\omega }^{\delta }\left(x\right)-f\left(x\right)\parallel \le \parallel f_{\omega }^{\delta }\left(x\right)-f_{\omega }\left(x\right)\parallel +\parallel f_{\omega }\left(x\right)-f\left(x\right)\parallel .$$

(55)

Now, using (41) and Lemma 10, we estimate the first term

$$\parallel f_{\omega }^{\delta }\left(x\right)-f_{\omega }\left(x\right)\parallel \le \left\{\begin{array}{cc}{c}_1{(\frac{c_{4}q}{\tau -1})}^{\frac{1}{k+1}}{\delta }^{\frac{k}{k+1}}{E}^{\frac{1}{k+1}},\hfill & 0<k<2\gamma -1,\hfill \\ c_1{(\frac{c_{5}q}{\tau -1})}^{\frac{1}{2\gamma }}{\delta }^{\frac{2\gamma -1}{2\gamma }}{E}^{\frac{1}{2\gamma }},\hfill & k\ge 2\gamma -1.\hfill \end{array}\right.$$

(56)

In the following, we estimate the second term from Theorem 1. We know

$$\parallel f_{\omega }\left(x\right)-f\left(x\right)\parallel \le D\left(k\right){\tilde{E}}^{\frac{1}{k+1}}\parallel K(f_{\omega }\left(x\right)-f\left(x\right)){\parallel }^{\frac{k}{k+1}}=D\left(k\right){\tilde{E}}^{\frac{1}{k+1}}{\parallel K{f}_{\omega }\left(x\right)-Kf\left(x\right)\parallel }^{\frac{k}{k+1}}.$$

(57)

Here, $\tilde{E}$ is an upper bound of $\parallel f_{\omega }{-f\parallel }_{H^k\left(\mathsf{\Omega }\right)}$.

Now, we estimate

$$\begin{array}{ccc}\parallel K{f}_{\omega }\left(x\right)-Kf\left(x\right)\parallel & =& \parallel \sum _{p=1}^{\infty }\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel \hfill \\ & \le & \parallel \sum _{p=1}^{\infty }\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g\left(x\right)-g^{\delta }\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel \hfill \\ & & +\parallel \sum _{p=1}^{\infty }\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}(g^{\delta }\left(x\right),e_p\left(x\right))e_p\left(x\right)\parallel \hfill \end{array}$$

$$\begin{array}{c}\le (\tau +1)\delta .\hfill \end{array}$$

(58)

Moreover, we know

$$\begin{array}{ccc}\parallel f_{\omega }{\left(x\right)-f\left(x\right)\parallel }_{H^k\left(\mathsf{\Omega }\right)}^2& =& \sum _{p=1}^{\infty }{\left(\frac{\omega }{\omega +{({\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha })\sigma \left(\tau \right)d\tau )}^{2\gamma }}\right)}^2\frac{{\lambda }_p^{2k}{|(g\left(x\right),e_p\left(x\right))|}^2}{|{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha }){\sigma \left(\tau \right)d\tau |}^2}\hfill \end{array}$$

$$\begin{array}{c}\le \sum _{p=1}^{\infty }\frac{{\lambda }_p^{2k}{|(g\left(x\right),e_p\left(x\right))|}^2}{|{\int }_0^T{E}_{\alpha ,1}(-{\lambda }_p{(T-\tau )}^{\alpha }){\sigma \left(\tau \right)d\tau |}^2}={\parallel f\parallel }_{H^k\left(\mathsf{\Omega }\right)}^2\le E^2.\hfill \end{array}$$

(59)

From (57), we know $\parallel f_{\omega }{-f\parallel }_{H^k\left(\mathsf{\Omega }\right)}\le \tilde{E}$. Here, we can use E instead of $\tilde{E}$. Combining (55)–(59), the following convergence estimates hold

$$\parallel f_{\omega }^{\delta }\left(x\right)-f\left(x\right)\parallel \le \left\{\begin{array}{cc}(c_1{(\frac{c_{4}q}{\tau -1})}^{\frac{1}{k+1}}+c_6){\delta }^{\frac{k}{k+1}}{E}^{\frac{1}{k+1}},\hfill & 0<k<2\gamma -1,\hfill \\ c_1{(\frac{c_{5}q}{\tau -1})}^{\frac{1}{2\gamma }}{\delta }^{\frac{2\gamma -1}{2\gamma }}{E}^{\frac{1}{2\gamma }}+c_6{\delta }^{\frac{k}{k+1}}{E}^{\frac{1}{k+1}},\hfill & k\ge 2\gamma -1,\hfill \end{array}\right.$$

where $c_6={(\frac{\tau +1}{{\sigma }_0{Q}_{\alpha }^{-}({\lambda }_1,M_1)})}^{\frac{k}{k+1}}$. □

5. Conclusions

In this article, a fractional Tikhonov regularization method for the inverse source problem of the fractional diffusion equation with the Riemann–Liouville derivative is given, and we overcome its ill-posedness and prove the conditional stability result. Furthermore, the convergence estimates were obtained under a priori and a posteriori regularization parameter choice rules. In future work, we will focus on solving such an inverse source problem by using other regularization methods.