The Fractional Tikhonov Regularization Method to Identify the Initial Value of the Nonhomogeneous Time-Fractional Diffusion Equation on a Columnar Symmetrical Domain

Abstract

In this paper, the inverse problem for identifying the initial value of a time fractional nonhomogeneous diffusion equation in a columnar symmetric region is studied. This is an ill-posed problem, i.e., the solution does not depend continuously on the data. The fractional Tikhonov regularization method is applied to solve this problem and obtain the regularization solution. The error estimations between the regularization solution and the exact solution are also obtained under the priori and the posteriori regularization parameter choice rules, respectively. Some examples are given to show this method’s effectiveness.

1. Introduction

In this paper, we consider a time fractional diffusion equation with a nonhomogeneous term on a columnar symmetric region:

$$\left\{\begin{array}{cc}{D}_t^{\alpha }u(r,t)-\frac{1}{r}{u}_r-u_{rr}=f(r,t),\hfill & 0<t<T,0<r<r_0,\hfill \\ u(r,0)=\phi (r),\hfill & 0\le r\le r_0,\hfill \\ u(r_0,t)=0,\hfill & 0\le t\le T,\hfill \\ li{m}_{r\to 0}u(r,t)bounded,\hfill & 0<t<T,\hfill \\ u(r,T)=g(r),\hfill & 0\le r\le r_0,\hfill \end{array}\right.$$

(1)

where $D_t^{\alpha }$ is the Caputo time-fractional derivative, which is defined as:

$$D_t^{\alpha }u(r,t)=\left\{\begin{array}{c}\frac{1}{\Gamma (1-\alpha )}{\int }_0^t\frac{u_{\tau }(r,\tau )}{{(t-\tau )}^{\alpha }}d\tau ,0<\alpha <1,\hfill \\ u_t(r,t),\alpha =1.\hfill \end{array}\right.$$

If $\phi (r)$ is known, this problem is called the positive problem. In our problem, $\phi (r)$ is not known, and the additional data $u(r,T)=g(r)$ are applied to identify the initial value $\phi (r)$. This is called the inverse problem of identifying the initial value problem. Since the measured data $g(r)$ have disturbance, the measurement data with an error denoted $g^{\delta }(r)$ satisfy:

$$\parallel g^{\delta }(·)-g(·)\parallel \le \delta ,$$

(2)

where $0<\delta <1$ is the noise level. Throughout this paper, $L^2[0,r_0;r]$ denotes the Hilbert space of the Lebesgue measurable function $f(r)$ with weight r in $[0,r_0]$. $(·,·)$ and $\parallel ·\parallel$ denotes the inner product and norm on $L^2[0,r_0;r]$, i.e.,

$$\parallel f(r)\parallel =({\int }_0^{r_0}{r\left|f(r)\right|}^2{dr)}^{\frac{1}{2}},(f,g)={\int }_0^{r_0}rf(r)g(r)dr.$$

(3)

For the inverse problem of a diffusion equation on a spherically symmetric domain, many different regularization methods were applied to deal with it by Cheng et al., for example see the references [1,2,3,4]. As the high-dimensional inverse heat conduction problem was researched in [1,2,3,4], the authors selected the priori regularization parameter. The priori regularization parameter relies on the priori bound. However, in practice, it is difficult for the researcher to obtain the exact priori bound. For the posteriori regularization parameter choice rule, one can see [5,6,7,8,9,10,11,12]. In [5,6,7,8,9,10,11,12], the authors considered that the equation is in integer order. If the equation is the fractional order, the equation is called the fractional diffusion equation. Recently, many researchers have presented research results regarding the fractional diffusion equation for its many applications in various areas of engineering. In [13,14,15,16], the authors considered the anomalous diffusion equation in the mathematical theory and associated numerical method. However, in the process of solving a practical problem through the fractional diffusion equation model, the researcher may encounter the difficulty of the boundary condition, the unknown source term, or the diffusion coefficient of the fractional diffusion equation being unknown. This problem is called the inverse problem of the fractional diffusion equation. Nowadays, many researches use many regularization methods to solve different types of inverse problems of fractional diffusion equations. For $0<\alpha \le 1$, one can see [17,18,19,20,21,22,23], for $1<\alpha \le 2$, one can see [24,25,26]. However, for the inverse problem for the time-fractional diffusion equation, many fractional diffusion equations are only suitable for a one-dimensional situation. Up until now, the research on high-dimensional aspects of the inverse problem for the time-fractional diffusion equation has been very difficult, and there are little research results. In [27], the authors applied the quasi-boundary regularization method to identify the initial value of a time-fractional diffusion equation on a spherically symmetric domain. In [28], the authors used the Tikhonov regularization method to identify the space-dependent source for a time-fractional diffusion equation on a columnar symmetric domain. In [29], the authors utilized the Landweber iterative method to solve an inverse source problem of a time-fractional diffusion-wave equation on a spherically symmetric domain.

In this paper, we aim to identify the initial value of the time-fractional diffusion equation with a nonhomogeneous term on a columnar symmetrical domain. We use the fractional Tikhonov regularization method to solve this inverse problem. Comparing with the classical Tikhonov method, we learn that the fractional Tikhonov method overcomes saturation, as can be seen in the error estimate.

The structure of this paper is as follows. In the second section, we present some auxiliary mathematical conclusions. In the third section, we give the exact solution and the regularization solution. In the fourth section, we give the error estimation between the exact solution and the regularization solution under the priori regularization parameter choice rule. In the fifth section, we give the error estimation between the exact solution and the regularization solution under the posteriori regularization parameter choice rule. In the sixth section, some examples are given to show two regularization methods’ effectiveness. In the seventh and eighth sections, we present a brief discussion and summary of the whole paper.

2. Preliminary Results

We first present some auxiliary results that will be used later in this section.

Definition 1

([13]). The Mittag–Leffler function is

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

(4)

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

Lemma 1

([19]). Assume ${\lambda }_n\ge {\lambda }_1>0$, then there exist two positive constants $C_1$ and $C_2$ such that

$$\frac{C_2}{{\lambda }_n^2}\le E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })\le \frac{C_1}{{\lambda }_n^2}.$$

(5)

Lemma 2.

For $\eta \ge {\lambda }_1>0$, $0<\beta <1$, $0<\mu <1$ is a constant, we obtain

$$F(\eta )=\frac{{\eta }^2}{C_2^{\beta +1}+{\eta }^{2\beta +2}\mu }\le {(\frac{1}{C_2})}^{\beta }\frac{{\beta }^{\frac{\beta }{\beta +1}}}{\beta +1}{\mu }^{\frac{-1}{1+\beta }}.$$

(6)

Proof. 

Let $F^{\prime }({\eta }^{*})=0$; we obtain ${\eta }^{*}={(\frac{C_2^{\beta +1}}{\mu \beta })}^{\frac{1}{2\beta +2}}$. Thus,

$$F(\eta )\le F({\eta }^{*})\le {(\frac{1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}{\mu }^{\frac{-1}{1+\beta }}.$$

□

Lemma 3.

For $\eta \ge {\lambda }_1>0$, $0<\beta <1$, we obtain

$$F_1(\eta )=\frac{\mu {\eta }^{2\beta +2-p}}{C_2^{\beta +1}+\mu {\eta }^{2\beta +2}}\le \left\{\begin{array}{c}{\mu }^{\frac{p}{2\beta +2}}{C}_3,0<p<2\beta +2,\hfill \\ \mu C_4,p\ge 2\beta +2,\hfill \end{array}\right.$$

(7)

where $C_3={\left[\frac{(2\beta +2-p)C_2^{\beta +1}}{p}\right]}^{\frac{-p}{2\beta +2}}$, $C_4=\frac{1}{C_2^{\beta +1}{\lambda }_1^{p-2\beta -2}}$.

Proof. 

For $0<p<2\beta +2$, we can obtain ${\eta }^{*}={\left[\frac{(2\beta +2-p)C_2^{\beta +1}}{p\mu }\right]}^{\frac{1}{2\beta +2}}$ to make $F_1^{{}^{\prime }}({\eta }^{*})=0$.

$$\begin{array}{c}\hfill F_1(\eta )\le F_1({\eta }^{*})\le \frac{\mu {({\eta }^{*})}^{2\beta +2-p}}{\mu {({\eta }^{*})}^{2\beta +2}}={\mu }^{\frac{p}{2\beta +2}}{\left[\frac{(2\beta +2-p)C_2^{\beta +1}}{p}\right]}^{\frac{-p}{2\beta +2}}.\end{array}$$

For $p\ge 2\beta +2$,

$$F_1(\eta )\le \frac{\mu {\eta }^{2\beta +2-p}}{C_2^{\beta +1}}\le \mu \frac{1}{C_2^{\beta +1}{\lambda }_1^{p-2\beta -2}}.$$

□

Lemma 4.

For $\eta \ge {\lambda }_1>0$, $0<\beta <1$, we obtain

$$F_2(\eta )=\frac{{\eta }^{2\beta -p}\mu }{C_2^{\beta +1}+{\eta }^{2\beta +2}\mu }\le \left\{\begin{array}{c}{\mu }^{\frac{p+2}{2\beta +2}}{C}_5,0<p<2\beta ,\hfill \\ \mu C_6,p\ge 2\beta ,\hfill \end{array}\right.$$

(8)

where $C_5={\left[\frac{(2\beta -p)C_2^{\beta +1}}{p+2}\right]}^{\frac{-p-2}{2\beta +2}}$, $C_6=\frac{1}{C_2^{\beta +1}{\lambda }_1^{p-2\beta }}$.

Proof. 

For $0<p<2\beta$, we obtain ${\eta }^{*}={\left[\frac{(2\beta -p)C_2^{\beta +1}}{(p+2)\mu }\right]}^{\frac{1}{2\beta +2}}$ to make $F_2^{{}^{\prime }}({\eta }^{*})=0$. Thus,

$$\begin{array}{c}\hfill F_2(\eta )\le F_2({\eta }^{*})\le \frac{{({\eta }^{*})}^{2\beta -p}\mu }{{({\eta }^{*})}^{2\beta +2}\mu }={\mu }^{\frac{p+2}{2\beta +2}}{\left[\frac{(2\beta -p)C_2^{\beta +1}}{p+2}\right]}^{\frac{-p-2}{2\beta +2}}.\end{array}$$

For $p\ge 2\beta$,

$$F_2(\eta )\le \frac{{\eta }^{2\beta -p}\mu }{C_2^{\beta +1}}\le \mu \frac{1}{C_2^{\beta +1}{\lambda }_1^{p-2\beta }}.$$

□

3. The Exact Solution and Regularization Strategies

Using the method of separation of variables and properties of the Mittag–Leffler function, we obtain the solution of (1) as follows:

$$u(r,t)=\sum _{n=1}^{\infty }({\int }_0^t{s}^{\alpha -1}{E}_{\alpha ,\alpha }(-{(\frac{{\lambda }_n}{r_0})}^2{s}^{\alpha })f_n(t-s)ds+E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{t}^{\alpha }){\phi }_n){\varpi }_n(r),$$

(9)

where ${\varpi }_n=\frac{\sqrt{2}}{r_0{J}_1({\lambda }_n)}{J}_0(\frac{{\lambda }_{n}r}{r_0})$, $J_0,J_1$ are a zero-order and a first-order Bessel functions, respectively, and ${\left\{{\lambda }_n\right\}}_{n=1}^{\infty }$ is the monotonically increasing sequences without positive roots of $J_0(x)=0$ and satisfies

$$0<{\lambda }_1<{\lambda }_2<{\lambda }_3<\cdots <{\lambda }_n<\cdots ,li{m}_{n\to \infty }{\lambda }_n=\infty .$$

(10)

where $f_n(t)=(f(r,t),{\varpi }_n(r))$ and ${\phi }_n=(\phi (r),{\varpi }_n(r))$ are the fourier coefficients of function $f(r,t)$ and $\phi (r)$.

Using $u(r,T)=g(r)$, we obtain the initial function

$$\phi (r)=\sum _{n=1}^{\infty }\frac{v_n}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{\varpi }_n(r),$$

(11)

where

$$v_n=g_n-{\int }_0^T{s}^{\alpha -1}{E}_{\alpha ,\alpha }(-{(\frac{{\lambda }_n}{r_0})}^2{s}^{\alpha })f_n(T-s)ds$$

(12)

and $g_n=(g(r),{\varpi }_n(r))$.

Define an another operator $K:\phi \to v$ as follows:

$$K\phi (r)=v(r),$$

(13)

where K is a linear self-adjoint operator, and its singular value is

$${\sigma }_n=E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }).$$

(14)

Next, we define the a priori bound of $\phi (r)$,

$${\parallel \phi (r)\parallel }_{H^p}\le E,p>0,$$

(15)

where $E>0$ is a constant, and ${\parallel ·\parallel }_{H^p}$ is defined by

$${\parallel \phi (r)\parallel }_{H^p}=(\sum _{n=1}^{\infty }{\lambda }_n^{2p}|(\phi (r),{\varpi }_n(r)){{|}^2)}^{\frac{1}{2}}.$$

(16)

Theorem 1

(Conditional stable result). When ${\parallel \phi (r)\parallel }_{H^p}\le E$, we have the following results

$$\parallel \phi (r)\parallel \le C_2^{\frac{-p}{p+2}}{E}^{\frac{2}{p+2}}{\parallel v\parallel }^{\frac{p}{p+2}}.$$

(17)

Proof. 

Using (11), Lemma 1 and the H$\ddot{o}$lder inequality, we obtain

$$\begin{array}{cc}\hfill {\parallel \phi (r)\parallel }^2=& \parallel \sum _{n=1}^{\infty }\frac{v_n}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{\varpi }_n{(r)\parallel }^2=\sum _{n=1}^{\infty }{\left(\frac{v_n}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^2\hfill \\ \hfill & =\sum _{n=1}^{\infty }\frac{v_n^{\frac{4}{p+2}}}{E_{\alpha ,1}^2(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{v}_n^{\frac{2p}{p+2}}\le {\left(\sum _{n=1}^{\infty }\frac{v_n^2}{E_{\alpha ,1}^{p+2}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^{\frac{2}{p+2}}{\left(\sum _{n=1}^{\infty }{v}_n^2\right)}^{\frac{p}{p+2}}\hfill \\ \hfill & ={\left(\sum _{n=1}^{\infty }{\phi }_n^2{\lambda }_n^{2p}\frac{1}{E_{\alpha ,1}^p(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\lambda }_n^{2p}}\right)}^{\frac{2}{p+2}}{\parallel v\parallel }^{\frac{2p}{p+2}}\hfill \\ \hfill & \le {(\underset{n\ge 1}{sup}\frac{1}{E_{\alpha ,1}^p(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\lambda }_n^{2p}})}^{\frac{2}{p+2}}{E}^{\frac{4}{p+2}}{\parallel v\parallel }^{\frac{2p}{p+2}}\hfill \\ \hfill & \le C_2^{\frac{-2p}{p+2}}{E}^{\frac{4}{p+2}}{\parallel v\parallel }^{\frac{2p}{p+2}}.\hfill \end{array}$$

Thus, we obtain

$$\parallel \phi (r)\parallel \le C_2^{\frac{-p}{p+2}}{E}^{\frac{2}{p+2}}{\parallel v\parallel }^{\frac{p}{p+2}}.$$

To sum up, we use the fractional Tikhonov regularization method to give the regularization solution of the unknown initial function $\phi (r)$. Due to [30], we obtain the fractional Tikhonov regularization solution of problem (1) as follows:

$${\phi }^{\mu ,\delta }(r)=\sum _{n=1}^{\infty }\frac{E_{\alpha ,1}^{\beta }(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n^{\delta }{\varpi }_n(r),$$

(18)

where $0<\beta <1$ is called the fractional parameter and

$$v_n^{\delta }=g_n^{\delta }-{\int }_0^T{s}^{\alpha -1}{E}_{\alpha ,\alpha }(-{(\frac{{\lambda }_n}{r_0})}^2{s}^{\alpha })f_n(T-s)ds$$

(19)

The fractional Tikhonov regularization solution according to the exact data $g(r)$ is as follows:

$${\phi }^{\mu }(r)=\sum _{n=1}^{\infty }\frac{E_{\alpha ,1}^{\beta }(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n{\varpi }_n(r).$$

(20)

According to (12), (19) and (2), we obtain

$$\begin{array}{cc}\hfill \parallel v^{\delta }(r)-v(r)\parallel & =\parallel ((g_n^{\delta }-{\int }_0^T{s}^{\alpha -1}{E}_{\alpha ,\alpha }(-{(\frac{{\lambda }_n}{r_0})}^2{s}^{\alpha })f_n(T-s)ds)\hfill \\ \hfill & -(g_n-{\int }_0^T{s}^{\alpha -1}{E}_{\alpha ,\alpha }(-{(\frac{{\lambda }_n}{r_0})}^2{s}^{\alpha })f_n(T-s)ds)){\varpi }_n(r)\parallel \hfill \\ \hfill & =\parallel (g_n^{\delta }-g_n){\varpi }_n(r)\parallel \hfill \\ \hfill & =\parallel g^{\delta }(r)-g(r)\parallel \hfill \\ \hfill & \le \delta .\hfill \end{array}$$

Thus, we obtain

$$\parallel v^{\delta }(r)-v(r)\parallel \le \delta .$$

(21)

□

4. The Priori Error Estimate

Theorem 2.

The exact solution is given by (11). The regularization solution is given by (18). Assume that (2) and the priori bound (16) hold. If the regularization parameter μ is chosen by

$$\mu =\left\{\begin{array}{c}{(\frac{\delta }{E})}^{\frac{2\beta +2}{p+2}},0<p<2\beta +2,\hfill \\ {(\frac{\delta }{E})}^{\frac{\beta +1}{\beta +2}},p\ge 2\beta +2,\hfill \end{array}\right.$$

(22)

then we obtain

$$\parallel {\phi }^{\mu ,\delta }(r)-\phi (r)\parallel \le \left\{\begin{array}{c}[{(\frac{C_1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}+C_3]{\delta }^{\frac{p}{p+2}}{E}^{\frac{2}{p+2}},0<p<2\beta +2,\hfill \\ [{(\frac{C_1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}+C_4]{\delta }^{\frac{\beta +1}{\beta +2}}{E}^{\frac{1}{\beta +2}},p\ge 2\beta +2.\hfill \end{array}\right.$$

(23)

Proof. 

By the triangle inequality property of the norm, we obtain

$$\parallel {\phi }^{\mu ,\delta }(r)-\phi (r)\parallel \le \parallel {\phi }^{\mu ,\delta }(r)-{\phi }^{\mu }(r)\parallel +\parallel {\phi }^{\mu }(r)-\phi (r)\parallel .$$

We first estimate $\parallel {\phi }^{\mu ,\delta }(r)-{\phi }^{\mu }(r)\parallel$. According to Lemmas 1 and 2, we obtain

$$\begin{array}{cc}\hfill \parallel {\phi }^{\mu ,\delta }(r)-{\phi }^{\mu }(r)\parallel & =\parallel \sum _{n=1}^{\infty }\frac{E_{\alpha ,1}^{\beta }(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }(v_n^{\delta }-v_n){\varpi }_n(r)\parallel \hfill \\ \hfill & \le \underset{n\ge 1}{sup}(\frac{E_{\alpha ,1}^{\beta }(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu })\parallel \sum _{n=1}^{\infty }(v_n^{\delta }-v_n){\varpi }_n(r)\parallel \hfill \\ \hfill & \le \underset{n\ge 1}{sup}(\frac{{(\frac{C_1}{{\lambda }_n^2})}^{\beta }}{{(\frac{C_2}{{\lambda }_n^2})}^{\beta +1}+\mu })\delta =C_1^{\beta }\underset{n\ge 1}{sup}\frac{{\lambda }_n^2}{C_2^{\beta +1}+{\lambda }_n^{2\beta +2}\mu }\hfill \\ \hfill & \le {(\frac{C_1}{C_2})}^{\beta }\frac{{\beta }^{\frac{\beta }{\beta +1}}}{\beta +1}{\mu }^{\frac{-1}{1+\beta }}.\hfill \end{array}$$

(24)

Using Lemma 1, we appraise the second term in the triangle inequality,

$$\begin{array}{cc}\hfill \parallel {\phi }^{\mu }(r)-\phi (r)\parallel & =\parallel \sum _{n=1}^{\infty }(\frac{E_{\alpha ,1}^{\beta }(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n-\sum _{n=1}^{\infty }\frac{1}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{v}_n){\varpi }_n(r)\parallel \hfill \\ \hfill & =\parallel \sum _{n=1}^{\infty }\frac{\mu {\lambda }_n^{-p}}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\frac{v_n{\lambda }_n^p}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\parallel \hfill \\ \hfill & \le (\underset{n\ge 1}{sup}\frac{\mu {\lambda }_n^{-p}}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu })E\le (\underset{n\ge 1}{sup}\frac{\mu {\lambda }_n^{-p}}{{(\frac{C_2}{{\lambda }_n^2})}^{\beta +1}+\mu })E.\hfill \end{array}$$

(25)

Using Lemma 3, we obtain

$$\parallel {\phi }^{\mu }(r)-\phi (r)\parallel \le \left\{\begin{array}{c}E{\mu }^{\frac{p}{2\beta +2}}{C}_3,0<p<2\beta +2,\hfill \\ E\mu C_4,p\ge 2\beta +2.\hfill \end{array}\right.$$

(26)

Using (22), we obtain

$$\parallel {\phi }^{\mu }(r)-\phi (r)\parallel \le \left\{\begin{array}{c}{C}_3{\delta }^{\frac{p}{p+2}}{E}^{\frac{2}{p+2}},0<p<2\beta +2,\hfill \\ C_4{\delta }^{\frac{\beta +1}{\beta +2}}{E}^{\frac{1}{\beta +2}},p\ge 2\beta +2.\hfill \end{array}\right.$$

(27)

According to (24) and (27), we have

$$\parallel {\phi }^{\mu ,\delta }(r)-\phi (r)\parallel \le \left\{\begin{array}{c}[{(\frac{C_1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}+C_3]{\delta }^{\frac{p}{p+2}}{E}^{\frac{2}{p+2}},0<p<2\beta +2,\hfill \\ [{(\frac{C_1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}+C_4]{\delta }^{\frac{\beta +1}{\beta +2}}{E}^{\frac{1}{\beta +2}},p\ge 2\beta +2.\hfill \end{array}\right.$$

□

5. The Posteriori Error Estimation

In this section, we use Morozov’s discrepancy principle to choose the posteriori regularization parameter, which only depends only on the measurable data, i.e., the regularization parameter satisfies the following equation:

$$\parallel K{\phi }^{\mu ,\delta }-v^{\delta }\parallel =\tau \delta ,$$

(28)

where $\tau >1$ is a constant and $\parallel v^{\delta }\parallel >\tau \delta$.

Lemma 5.

Let $\rho (\mu )=\parallel K{\phi }^{\mu ,\delta }-v^{\delta }\parallel$; then, we have the following conclusions:

• (a)$\rho (\mu )$ is a continuous function;
• (b)$\underset{\mu \to 0}{lim}\rho (\mu )=0;$
• (c)$\underset{\mu \to \infty }{lim}\rho (\mu )=\parallel v^{\delta }\parallel ;$
• (d)$\rho (\mu )$ is a strictly increasing function.

Proof. 

The proof of Lemma 5 is obvious, we deleted the proof. □

Lemma 6.

If μ is chosen by (28), then the regularization parameter μ satisfies

$${\mu }^{-1}\le \left\{\begin{array}{c}{(\frac{E}{\delta })}^{\frac{2\beta +2}{p+2}}{(\frac{C_7}{\tau -1})}^{\frac{2\beta +2}{p+2}},0<p<2\beta ,\hfill \\ \frac{E}{\delta }\frac{C_8}{\tau -1},p\ge 2\beta ,\hfill \end{array}\right.$$

(29)

where $C_7=C_1{C}_5$, $C_8=C_1{C}_6$.

Proof. 

Due to (2), the priori bound (16), the fractional Tikhonov regularization solution (18), and Lemmas 1 and 4, we have:

$$\begin{array}{cc}\hfill & \tau \delta =\parallel \sum _{n=1}^{\infty }(\frac{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n^{\delta }-v_n^{\delta }){\varpi }_n(r)\parallel \hfill \\ \hfill & =\parallel \sum _{n=1}^{\infty }\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n^{\delta }{\varpi }_n(r)\parallel \hfill \\ \hfill & \le \parallel \sum _{n=1}^{\infty }\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }(v_n^{\delta }-v_n){\varpi }_n(r)\parallel +\parallel \sum _{n=1}^{\infty }\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n{\varpi }_n(r)\parallel \hfill \\ \hfill & \le \delta +\parallel \sum _{n=1}^{\infty }\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{\lambda }_n^{-p}{\lambda }_n^p{\phi }_n{\varpi }_n(r)\parallel \hfill \\ \hfill & \le \delta +(\underset{n\ge 1}{sup}\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{\lambda }_n^{-p})E\hfill \\ \hfill & \le \delta +(\underset{n\ge 1}{sup}\frac{\mu \frac{C_1}{{\lambda }_n^2}}{{(\frac{C_2}{{\lambda }_n^2})}^{\beta +1}+\mu }{\lambda }_n^{-p})E\hfill \\ \hfill & \le \delta +C_{1}E\left\{\begin{array}{c}{C}_5{\mu }^{\frac{p+2}{2\beta +2}},0<p<2\beta ,\hfill \\ C_6\mu ,p\ge 2\beta .\hfill \end{array}\right.\hfill \end{array}$$

(30)

Then,

$${\mu }^{-1}\le \left\{\begin{array}{c}{(\frac{E}{\delta })}^{\frac{2\beta +2}{p+2}}{(\frac{C_7}{\tau -1})}^{\frac{2\beta +2}{p+2}},0<p<2\beta ,\hfill \\ \frac{E}{\delta }\frac{C_8}{\tau -1},p\ge 2\beta ,\hfill \end{array}\right.$$

where $C_7=C_1{C}_5$, $C_8=C_1{C}_6$. □

Lemma 7.

The fractional Tikhonov regularization solution is given by (20) and (18); then we obtain the following error estimation:

$$\parallel {\phi }^{\mu ,\delta }(r)-{\phi }^{\mu }(r)\parallel \le \left\{\begin{array}{c}{C}_9{E}^{\frac{2}{p+2}}{\delta }^{\frac{p}{p+2}},0<p<2\beta ,\hfill \\ C_{10}{E}^{\frac{1}{\beta +1}}{\delta }^{\frac{\beta }{\beta +1}},p\ge 2\beta ,\hfill \end{array}\right.$$

(31)

where $C_9={(\frac{C_1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}{(\frac{C_7}{\tau -1})}^{\frac{2}{p+2}}$, $C_{10}={(\frac{C_1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}\frac{C_8}{\tau -1}$.

Proof. 

According to (24), we have

$$\parallel {\phi }^{\mu ,\delta }(r)-{\phi }^{\mu }(r)\parallel \le C_1^{\beta }\delta {(\frac{1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}{\mu }^{\frac{-1}{1+\beta }}.$$

Due to Lemma 6, we obtain

$$\begin{array}{cc}\hfill & \parallel {\phi }^{\mu ,\delta }(r)-{\phi }^{\mu }(r)\parallel \le C_1^{\beta }\delta {(\frac{1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}{\mu }^{\frac{-1}{1+\beta }}\hfill \\ \hfill & \le \left\{\begin{array}{c}{(\frac{C_1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}{(\frac{C_7}{\tau -1})}^{\frac{2}{p+2}}{E}^{\frac{2}{p+2}}{\delta }^{\frac{p}{p+2}},0<p<2\beta ,\hfill \\ {(\frac{C_1}{C_2})}^{\beta }{\beta }^{\frac{\beta }{\beta +1}}\frac{C_8}{\tau -1}{E}^{\frac{1}{\beta +1}}{\delta }^{\frac{\beta }{\beta +1}},p\ge 2\beta .\hfill \end{array}\right.\hfill \end{array}$$

□

Lemma 8.

${\phi }^{\mu }(r)$ is given by (18), and $\phi (r)$ is the exact solution of (1). Suppose conditions (15) and (2) hold. Then the estimation between ${\phi }^{\mu }(r)$ and $\phi (r)$ is

$$\parallel {\phi }^{\mu }(r)-\phi (r)\parallel \le \left\{\begin{array}{c}{C}_{11}{\delta }^{\frac{p}{p+2}}{E}^{\frac{2}{p+2}},0<p<2\beta ,\hfill \\ C_{12}{\delta }^{\frac{\beta }{1+\beta }}{E}^{\frac{1}{1+\beta }},p\ge 2\beta ,\hfill \end{array}\right.$$

(32)

where $C_{11}={(\sqrt{(2{\tau }^2+2)})}^{\frac{p}{p+2}}{C}_2^{\frac{-p}{p+2}}$, $C_{12}={(\sqrt{(2{\tau }^2+2)})}^{\frac{\beta }{\beta +1}}{C}_2^{\frac{-\beta }{\beta +1}}$.

Proof. 

$$\begin{array}{cc}\hfill \parallel {\phi }^{\mu }(r)-\phi (r)\parallel & =\parallel \sum _{n=1}^{\infty }\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{\phi }_n{\varpi }_n(r)\parallel \hfill \\ \hfill & =\parallel \sum _{n=1}^{\infty }\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\frac{{\phi }_n}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{\varpi }_n(r)\parallel .\hfill \end{array}$$

(33)

If $0<p<2\beta$, using the H$\ddot{o}$lder inequality and Lemma 1, we have

$$\begin{array}{cc}\hfill & \parallel {\phi }^{\mu }{(r)-\phi (r)\parallel }^2=\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\frac{{\phi }_n}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^2\hfill \\ \hfill & =\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^{\frac{2p}{p+2}}{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^{\frac{4}{p+2}}{\left(\frac{1}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^2\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill & \le {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^{\frac{2p}{p+2}\frac{p+2}{p}}\right)}^{\frac{p}{p+2}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^{\frac{4}{p+2}\frac{p+2}{2}}{\left(\frac{1}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^{2\frac{p+2}{2}}\right)}^{\frac{2}{p+2}}\hfill \\ \hfill & ={\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2\right)}^{\frac{p}{p+2}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2{\left(\frac{1}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^{p+2}\right)}^{\frac{2}{p+2}}\hfill \\ \hfill & ={\left(\sum _{n=1}^{\infty }{\left(\frac{\mu v_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2\right)}^{\frac{p}{p+2}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2{\left(\frac{{\phi }_n}{{(E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }))}^{\frac{p+2}{2}}}\right)}^2\right)}^{\frac{2}{p+2}}\hfill \\ \hfill & ={\left(\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }(v_n-v_n^{\delta }+v_n^{\delta })\right)}^2\right)}^{\frac{p}{p+2}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2{\left(\frac{{\phi }_n}{{(E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }))}^{\frac{p}{2}}}\right)}^2\right)}^{\frac{2}{p+2}}\hfill \\ \hfill & \le {\left(2\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }(v_n-v_n^{\delta })\right)}^2+2\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n^{\delta }\right)}^2\right)}^{\frac{p}{p+2}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{{\phi }_n}{{(\frac{C_2}{{\lambda }_n^2})}^{\frac{p}{2}}}\right)}^2\right)}^{\frac{2}{p+2}}\hfill \\ \hfill & \le {\left(2\sum _{n=1}^{\infty }{\left(v_n-v_n^{\delta }\right)}^2+2\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n^{\delta }\right)}^2\right)}^{\frac{p}{p+2}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }\left(C_2^{\frac{-2p}{p+2}}({\lambda }_n^{2p}{\phi }_n^2\right)\right)}^{\frac{2}{p+2}}\le {(\sqrt{(2{\tau }^2+2)}\delta )}^{\frac{2p}{p+2}}{C}_2^{\frac{-p}{p+2}}{E}^{\frac{4}{p+2}}.\hfill \end{array}$$

Then,

$$\parallel {\phi }^{\mu }(r)-\phi (r)\parallel \le {(\sqrt{(2{\tau }^2+2)})}^{\frac{p}{p+2}}{C}_2^{\frac{-p}{p+2}}{\delta }^{\frac{p}{p+2}}{E}^{\frac{2}{p+2}}.$$

(35)

If $p\ge 2\beta$, combining Lemma 1 with estimate (33), we obtain

$$\begin{array}{cc}\hfill & \parallel {\phi }^{\mu }{(r)-\phi (r)\parallel }^2=\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\frac{{\phi }_n}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^2\hfill \\ \hfill & =\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^{\frac{2\beta }{\beta +1}}{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^{\frac{2}{\beta +1}}{\left(\frac{1}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^2\hfill \\ \hfill & \le {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^{\frac{2\beta }{\beta +1}\frac{\beta +1}{\beta }}\right)}^{\frac{\beta }{\beta +1}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^{\frac{2}{\beta +1}\frac{\beta +1}{1}}{\left(\frac{1}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^{2\frac{\beta +1}{1}}\right)}^{\frac{1}{\beta +1}}\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill & ={\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2\right)}^{\frac{\beta }{\beta +1}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }){\phi }_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2{\left(\frac{1}{E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}\right)}^{2(\beta +1)}\right)}^{\frac{1}{\beta +1}}\hfill \\ \hfill & ={\left(\sum _{n=1}^{\infty }{\left(\frac{\mu v_n}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2\right)}^{\frac{\beta }{\beta +1}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })}{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2{\left(\frac{{\phi }_n}{{(E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }))}^{\beta +1}}\right)}^2\right)}^{\frac{1}{\beta +1}}\hfill \\ \hfill & ={\left(\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }(v_n-v_n^{\delta }+v_n^{\delta })\right)}^2\right)}^{\frac{\beta }{\beta +1}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }\right)}^2{\left(\frac{{\phi }_n}{{(E_{\alpha ,1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha }))}^{\beta }}\right)}^2\right)}^{\frac{1}{\beta +1}}\hfill \\ \hfill & \le {\left(2\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }(v_n-v_n^{\delta })\right)}^2+2\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n^{\delta }\right)}^2\right)}^{\frac{\beta }{\beta +1}}\hfill \\ \hfill & \times {\left(\sum _{n=1}^{\infty }{\left(\frac{{\phi }_n}{{(\frac{C_2}{{\lambda }_n^2})}^{\beta }}\right)}^2\right)}^{\frac{1}{\beta +1}}\hfill \\ \hfill & \le {\left(2\sum _{n=1}^{\infty }{\left(v_n-v_n^{\delta }\right)}^2+2\sum _{n=1}^{\infty }{\left(\frac{\mu }{E_{\alpha ,1}^{\beta +1}(-{(\frac{{\lambda }_n}{r_0})}^2{T}^{\alpha })+\mu }{v}_n^{\delta }\right)}^2\right)}^{\frac{\beta }{\beta +1}}{\left(\sum _{n=1}^{\infty }{C}_2^{-2\beta }{\lambda }_n^{4\beta }{\phi }_n^2\right)}^{\frac{1}{\beta +1}}\hfill \\ \hfill & \le {(\sqrt{(2{\tau }^2+2)}\delta )}^{\frac{2\beta }{\beta +1}}{C}_2^{\frac{-p}{p+2}}{\left(\sum _{n=1}^{\infty }{C}_2^{-2\beta }{\lambda }_n^{2p}{\phi }_n^2\right)}^{\frac{1}{\beta +1}}\hfill \\ \hfill & \le {(\sqrt{(2{\tau }^2+2)}\delta )}^{\frac{2\beta }{\beta +1}}{C}_2^{\frac{-2\beta }{\beta +1}}{E}^{\frac{2}{\beta +1}}.\hfill \end{array}$$

Then,

$$\parallel {\phi }^{\mu }(r)-\phi (r)\parallel \le {(\sqrt{(2{\tau }^2+2)})}^{\frac{\beta }{\beta +1}}{C}_2^{\frac{-\beta }{\beta +1}}{\delta }^{\frac{\beta }{\beta +1}}{E}^{\frac{1}{\beta +1}}.$$

(37)

□

Theorem 3.

The fractional Tikhonov regularization solution is written by (20) and (18), and the exact solution is given by (11); then, the posteriori error estimation is

$$\parallel {\phi }^{\mu ,\delta }(r)-\phi (r)\parallel \le \left\{\begin{array}{c}(C_9+C_{11})E^{\frac{2}{p+2}}{\delta }^{\frac{p}{p+2}},0<p<2\beta ,\hfill \\ (C_{10}+C_{12})E^{\frac{1}{\beta +1}}{\delta }^{\frac{\beta }{1+\beta }},p\ge 2\beta .\hfill \end{array}\right.$$

(38)

Proof. 

By the triangle inequality and Lemmas 7 and 8, we obtain (38). □

6. Numerical Examples

In order to aid readers to clearly see the effectiveness of fractional Tikhonov regularization in solving Problem (1), this section will use the finite difference method to discretize Problem (1), and finally give the graph of the regular approximate solution. Take $T=1,r_0=\pi$, the time and space grids are divided into $\Delta r=\frac{\pi }{M+1}$, $\Delta t=\frac{1}{N+1}$, $r(i)=(i-1)\Delta r(i=1,2,3,\cdots ,M+1)$, $t(n)=(n-1)\Delta t(n=1,2,3,\cdots ,N+1)$, and the exact solution of (1) can be approximated by $u(i,n)\approx u(r_i,t_n)$. Finally, Equation (1) is discretized to

$$\frac{{(\Delta t)}^{-\alpha }}{\Gamma (2-\alpha )}\sum _{j=0}^{n-1}{b}_j(u_i^{n-1}-u_i^{n-j-1})-\frac{u_{i+1}^n-u_i^n}{r_i\Delta r}-\frac{u_{i+1}^n-2u_i^n+u_{i-1}^n}{{(\Delta r)}^2}=f(i,n),$$

(39)

where $b_j={(j+1)}^{1-\alpha }-j^{1-\alpha }$. The error level could be expressed as

$$g^{\delta }=g+\epsilon g(2randn(size(g))-1),$$

(40)

and the absolute error between the exact and the regular solution is

$$e_a=\sqrt{\frac{1}{M+1}\sum _{i=1}^{M+1}{({\phi }_i-{\phi }_i^{\mu ,\delta })}^2}.$$

(41)

Because the priori bound E is difficult to gain, in the following examples we only present the numerical results under the posteriori regularization choice rule. The nonhomogeneous term $f(r,t)=sin(r,t)$. Let $M=100$, $N=1000$, $\beta =0$ and the error level $\epsilon =0.0001,0.00001$; then, the time fractional order is to choose among $\alpha =0.05,0.15,0.25$.

$$\parallel K{\phi }^{m,\delta }-v^{\delta }\parallel -\tau \delta \le 1e-5.$$

From

Table 1. The absolute error of fractional Tikhonov regularization for different $\alpha$ and $\epsilon =0.0001$ for Example 1.

$\alpha$	0.05	0.15	0.25	0.35	0.45	0.55	0.65	0.75
$e_a$	0.2844	0.2907	0.3234	0.3792	0.4393	0.4761	0.5335	1.3396

, we can see that when we fix $\epsilon =0.01$, as $\alpha$ increases, the absolute error also increases. On the other hand, from

Table 2. The absolute error of fractional Tikhonov regularization for different $\epsilon$ and $\alpha =0.05$ for Example 1.

$\epsilon$	0.0001	0.00008	0.00006	0.00004	0.00002	0.00001
$e_a$	0.2857	0.2839	0.2821	0.2816	0.2812	0.2811

, when we fix $\alpha =0.05$, as $\epsilon$ decreases, the absolute error also decreases.

Figure 1 shows the comparison of the exact solution $\phi (r)$ and the fractional Tikhonov regularization solution ${\phi }^{\mu ,\delta }(r)$ in Example 1 for the relative error levels $\epsilon =0.0001,0.00001$ with the various values $\alpha =0.05,0.15,0.25$. Figure 2 shows the comparison of the exact solution $\phi (r)$ and the fractional Tikhonov regularization solution ${\phi }^{\mu ,\delta }(r)$ in Example 2 for the relative error levels $\epsilon =0.0001,0.00001$ with the various values $\alpha =0.05,0.15,0.25$. Figure 3 shows the comparison of the exact solution $\phi (r)$ and the fractional Tikhonov regularization solution ${\phi }^{\mu ,\delta }(r)$ in Example 3 for the relative error levels $\epsilon =0.0001,0.00001$ with the various values $\alpha =0.05,0.15,0.25$. From Figure 1, Figure 2 and Figure 3, we can see that the smaller $\epsilon$ and $\alpha$, the better the fitting effect.

Example 1.

Let $\phi (r)=rsin(r)$.

Example 2.

Let $\phi (r)=\left\{\begin{array}{c}2r,0\le r\le \frac{\pi }{2},\hfill \\ 2\pi -2r,\frac{\pi }{2}\le r\le \pi .\hfill \end{array}\right.$

Example 3.

Let $\phi (r)=\left\{\begin{array}{c}0,0\le r\le \frac{\pi }{2},\hfill \\ 1,\frac{\pi }{2}\le r\le \pi .\hfill \end{array}\right.$

7. Discussion

In this paper, we do not give the uniqueness of this inverse problem; we assume the solution of this inverse problem is uniqueness. About the proof of the uniqueness of this inverse problem, one can see [31,32].

8. Conclusions

In this paper, we identify the initial value problem of time fractional nonhomogeneous diffusion equation in a columnar symmetrical region. We chose the fractional Tikhonov regularization method to solve this problem. At the same time, we give a priori error estimate and a posteriori error estimate for this regularization methods. We show that the fractional Tikhonov regularization method is effective for solving this problem through various examples.