Regularization Error Analysis for a Sideways Problem of the 2D Nonhomogeneous Time-Fractional Diffusion Equation

Abstract

The inverse and ill-posed problem of determining a solute concentration for the two-dimensional nonhomogeneous fractional diffusion equation is investigated. This model is much worse than its homogeneous counterpart as the source term appears. We propose a modified kernel regularization technique for the stable numerical reconstruction of the solution. The convergence estimates under both a priori and a posteriori parameter choice rules are proven.

1. Introduction

Fractional partial differential equations arose from the studies of Lérvy motion [1], continuous random walk [2] and high-frequency financial data [3], which have a wide range of applications in some scientific fields, such as biochemistry, physics, bioengineering, fluid mechanics, protein and tissue dynamics, control theory, electron transportation, viscoelasticity, image processing and so on [4,5,6,7,8,9,10,11,12]. In particular, fractional diffusion equations play an extremely important role in the research of some anomalous diffusion processes, especially the subdiffusion processes. Various anomalous diffusions can be modeled by the following 1D fractional diffusion equation:

$$\left\{\begin{array}{cc}\frac{{\partial }^{\beta }u}{\partial t^{\beta }}-u_{xx}=f(x,t),\hfill & x>0,t>0,\hfill \\ u(x,0)=0,\hfill & x>0,\hfill \\ u(1,t)=\phi \left(t\right),\hfill & t>0,\hfill \\ u(x,t){\mid }_{x\to \infty }bounded,\hfill \end{array}\right.$$

(1)

where x is the space variable and t is the time variable, $u(x,t)$ is the diffusion concentration at time t, $\phi \left(t\right)$ is the data given in the accessible boundary $x=1$, the function $f(x,t)$ is the source term, and $\frac{{\partial }^{\beta }u}{\partial t^{\beta }}$ is the Caputo fractional derivative of order $\beta$ $(0<\beta \le 1)$ defined by [13]

$$\begin{array}{cc}\hfill & \frac{{\partial }^{\beta }u}{\partial t^{\beta }}=\frac{1}{\mathrm{\Gamma }(1-\beta )}{\int }_0^t\frac{\partial u(x,s)}{\partial s}\frac{ds}{{(t-s)}^{\beta }},0<\beta <1,\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill & \frac{{\partial }^{\beta }u}{\partial t^{\beta }}=\frac{\partial u(x,t)}{\partial t},\beta =1,\hfill \end{array}$$

(3)

where

$$\begin{array}{c}\hfill \mathrm{\Gamma }(1-\beta )={\int }_0^{+\infty }{x}^{-\beta }{e}^{-x}dx,0<\beta <1.\end{array}$$

For the homogeneous fractional diffusion equation, i.e., $f(x,t)=0$ in the first equation in (1), we refer to [14,15,16,17,18,19,20,21,22] and the references therein. As to the nonhomogeneous system, there are also a few articles, e.g., Tuan [23] proposes a truncation regularization method to obtain a regularized solution, error estimates are established under some a priori assumptions for the exact solution. Meanwhile, authors use numerical examples to show efficiency of their method, but this article does not give an error estimate at the endpoint. Ngoc [24] considers the model (1) with a nonlinear source, where the given source f depends not only on the independent variable $(x,t)$, but also on the dependent variable u. Under some a prior assumptions, a new regularization method is proposed to stabilize the ill-posed problem.

Compared to the 1D setting, the literature of fractional diffusion equation in 2D or higher dimensional setting is much more scarce. Some articles [25,26,27,28] study the following 2D homogeneous fractional diffusion equation:

$$\left\{\begin{array}{cc}\frac{{\partial }^{\nu }u}{\partial t^{\nu }}-u_{xx}-u_{yy}=0,\hfill & x>0,y>0,t>0,\hfill \\ u(1,y,t)=g(y,t),\hfill & y\ge 0,t\ge 0,\hfill \\ u(x,y,0)=0,\hfill & x>0,y\ge 0\hfill \\ u(x,0,t)=0,\hfill & x>0,t\ge 0\hfill \\ u_y(x,0,t)=0,\hfill & x>0,t\ge 0,\hfill \\ u(x,y,t){\mid }_{x\to \infty }bounded,\hfill \end{array}\right.$$

(4)

where x, y are the space variables and t is the time variable, $u(x,y,t)$ is the diffusion concentration at time t, $\frac{{\partial }^{\nu }u}{\partial t^{\nu }}$ is the Caputo fractional derivative of order $\nu$ $(0<\nu \le 1)$ defined by

$$\begin{array}{cc}\hfill & \frac{{\partial }^{\nu }u}{\partial t^{\nu }}=\frac{1}{\mathrm{\Gamma }(1-\nu )}{\int }_0^t\frac{\partial u(x,y,s)}{\partial s}\frac{ds}{{(t-s)}^{\nu }},0<\nu <1,\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill & \frac{{\partial }^{\nu }u}{\partial t^{\nu }}=\frac{\partial u(x,y,t)}{\partial t},\nu =1.\hfill \end{array}$$

(6)

Nevertheless, numerous ill-posed problems with important application backgrounds such as inverse heat conduction problems are set in higher-dimensional nonhomogeneous case, which is more challenging. This motivates us to discuss the following nonhomogeneous fractional diffusion equation:

$$\left\{\begin{array}{cc}\frac{{\partial }^{\gamma }u}{\partial t^{\gamma }}-u_{xx}-u_{yy}=f(x,y,t),\hfill & x>0,y>0,t>0,\hfill \\ u(1,y,t)=g(y,t),\hfill & y\ge 0,t\ge 0,\hfill \\ u(x,y,0)=0,\hfill & x>0,y\ge 0\hfill \\ u(x,0,t)=0,\hfill & x>0,t\ge 0\hfill \\ u_y(x,0,t)=0,\hfill & x>0,t\ge 0,\hfill \\ u(x,y,t){\mid }_{x\to \infty }bounded,\hfill \end{array}\right.$$

(7)

where $\frac{{\partial }^{\gamma }u}{\partial t^{\gamma }}$ is the same as (5) and (6), and is yet to be investigated. It should be noted that in many real situations, the given data $g(y,t)$ and $f(x,y,t)$ are inevitably contaminated by inherent measurement errors in the form of the input noisy data $g^{\delta }(y,t)$ and $f^{\delta }(x,y,t)$ satisfying

$$max(\parallel g-g^{\delta }\parallel ,\parallel f-f^{\delta }{\parallel }_{L^2(0,1;L^2\left({\mathbb{R}}^2\right))})\le \delta .$$

(8)

where $\parallel ·\parallel$ denotes the $L^2\left({\mathbb{R}}^2\right)$-norm and $\delta >0$ represents the level of noise. Luan [29] uses a kernel regularization scheme to determine the temperature distribution and thermal flux structure of the investigated problem (7) when $\gamma =1$, and given H$\ddot{o}$lder-type convergence estimates under some appropriate a priori assumptions. Furthermore, this work presents a robust algorithm based on 2D fast Fourier transform and demonstrates the theoretical results numerically.

Most of the work mentioned above uses a priori method to select regularization parameters, however, the a priori bound E cannot be known accurately in practice generally, and working with a wrong constant E is likely to lead to a meaningless regularized solution. Thus, a posteriori regularization parameter choice rule, which is independent of a priori bound E, is more commonly used in practical application.

In the present paper, we proposed a new regularization method to stabilize the problem (7). Moreover, we shall give error estimates in the whole domain, that is, including the internal $0<x<1$ and the boundary $x=1$, under both a priori and a posteriori regularization parameter choice rules. Finally, we draw a conclusion to our method. The modified kernel method and model (1) are not new work, the original contributions presented here are the extension of the problem to nonhomogeneous and two dimensions.

2. Mathematical Analysis

In order to simplify the discussion, our theoretical analysis will be performed in $L^2\left({\mathbb{R}}^2\right)$. Let $\widehat{g}$ denote the Fourier transform of $g(y,t)$ defined by

$$\widehat{g}(\xi ,\tau )=:\frac{1}{2\pi }{\int }_{R^2}g(y,t)e^{-i(\xi y+\tau t)}dydt,\xi ,\tau \in \mathbb{R},$$

and ${\parallel ·\parallel }_p$ denotes the norm in Sobolev space $H_p\left({\mathbb{R}}^2\right)$ defined by

$${\parallel u(0,·,·)\parallel }_p:=({\int }_{R^2}{(1+{\xi }^2+{\tau }^2)}^p|\widehat{u}(0,\xi ,\tau ){{|}^{2}d\xi d\tau )}^{\frac{1}{2}}.$$

Besides, we introduce the following norm

$${\parallel f(x,·,·)\parallel }_{L^2(0,1;H_p\left({\mathbb{R}}^2\right))}=({\int }_0^1{\parallel f(x,·,·)\parallel }_p^2{dx)}^{\frac{1}{2}}.$$

When $p=0$, ${\parallel ·\parallel }_p=\parallel ·\parallel$ denotes the $L^2\left(\mathbb{R}\right)$ norm, i.e.,

$${\parallel f(x,·,·)\parallel }_{L^2(0,1;L^2\left({\mathbb{R}}^2\right))}=({\int }_0^1{\parallel f(x,·,·)\parallel }^2{dx)}^{\frac{1}{2}}.$$

Then the unique solution to problem (7) can be formulated, in frequency space, as

$$\widehat{u}(x,\xi ,\tau )=e^{\theta (\xi ,\tau )(1-x)}\widehat{g}(\xi ,\tau )+{\int }_x^1\widehat{f}(s,\xi ,\tau )\frac{sinh\left(\theta \right(\xi ,\tau \left)\right(s-x\left)\right)}{\theta (\xi ,\tau )}ds,0\le x<1,$$

(9)

and equivalently

$$u(x,y,t)=\frac{1}{2\pi }{\int }_{{\mathbb{R}}^2}(e^{\theta (\xi ,\tau )(1-x)}\widehat{g}(\xi ,\tau )+{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(\xi ,\tau \left)\right(s-x\left)\right)}{\theta (\xi ,\tau )}ds)e^{i(\xi y+\tau t)}d\xi d\tau ,0\le x<1.$$

(10)

where

$$\theta (\xi ,\tau ):=\sqrt{{\left(i\tau \right)}^{\gamma }+{\xi }^2},$$

(11)

$${\left(i\tau \right)}^{\gamma }={\left|\tau \right|}^{\gamma }(cos\frac{\gamma \pi }{2}+isign\left(\tau \right)sin\frac{\gamma \pi }{2}),\forall \tau \in \mathbb{R}.$$

(12)

Comparing the problem (7) considered in this article with (1) and (4), it is irrefutable that (7) is much more ill-posed [30]. Thus, we must use some regularization methods to deal with it. Therefore, we have to solve it using some regularization method. To obtain a stable approximate solution of (7), we apply a modified "kernel" method in the next section.

Remark 1.

Let $\widehat{g}(\xi ,\tau )\in C\left({\mathbb{R}}^2\right)$ is bounded, then the formal solution given by (10) is the bounded solution of problem (7).

Remark 2.

When $f=0$, (9) and (10) give the following solution:

$$\widehat{u}(x,\xi ,\tau )=e^{\theta (\xi ,\tau )(1-x)}\widehat{g}(\xi ,\tau ),x\in [0,1),$$

$$u(x,y,t)=\frac{1}{2\pi }{\int }_{{\mathbb{R}}^2}{e}^{\theta (\xi ,\tau )(1-x)}\widehat{g}(\xi ,\tau )e^{i(\xi y+\tau t)}d\xi d\tau ,x\in [0,1),$$

which is the solution of problem (4) (see [31]). Hence, the solution of the nonhomogeneous problem is consistent with its homogeneous system.

Lemma 1

([29]). For arbitrary $z\in \mathbb{C}$, $x\in [0,1)$ and $s\in (x,1]$, we have

$$\begin{array}{c}\hfill |\frac{sinh\left(\right(s-x\left)z\right)}{z}|\le (s-x)e^{(s-x)\Re \left(z\right)}.\end{array}$$

where $\Re \left(z\right)$ denotes the real parts of z.

Throughout this paper, we denote the real part and imaginary part of $\theta (\xi ,\tau )$ as follows

$$\begin{array}{cc}\hfill & c:=\Re \left(\theta (\xi ,\tau )\right)=\sqrt{\frac{\sqrt{{\left|\tau \right|}^{2\gamma }{sin}^2\frac{\gamma \pi }{2}+({\xi }^2+{\left|\tau \right|}^{\gamma }cos\frac{\gamma \pi }{2}{)}^2}+({\xi }^2+{\left|\tau \right|}^{\gamma }cos\frac{\gamma \pi }{2})}{2}}\hfill \\ \hfill d:& =\Im \left(\theta \right(\xi ,\tau \left)\right)\hfill \\ \hfill & ={sign\left(sign\left(\tau \right)\right|\tau |}^{\gamma }sin\frac{\gamma \pi }{2})\sqrt{\frac{\sqrt{{\left|\tau \right|}^{2\gamma }{sin}^2\frac{\gamma \pi }{2}+({\xi }^2+{\left|\tau \right|}^{\gamma }cos\frac{\gamma \pi }{2}{)}^2}-({\xi }^2+{\left|\tau \right|}^{\gamma }cos\frac{\gamma \pi }{2})}{2}}.\hfill \end{array}$$

(13)

3. Regularization and Error Estimate

In this section, we will give a modified “kernel” regularization method, and error estimates under both a priori and a posteriori regularization parameter choice rules are proven.

Motivated by [14], where the authors eliminate the high frequency effect through modifying the “kernel”, here we use a similar method that retains part of the high frequency components information on the basis of

$${\widehat{u}}_{\alpha }^{\delta }(x,\xi ,\tau )=\left\{\begin{array}{cc}{e}^{\theta (\xi ,\tau )(1-x)}{\widehat{g}}^{\delta }(\xi ,\tau )+{\int }_x^1{\widehat{f}}^{\delta }(s,\xi ,\tau )\frac{sinh\left(\theta \right(\xi ,\tau \left)\right(s-x\left)\right)}{\theta (\xi ,\tau )}ds,\hfill & e^{c(1-x)}\le \alpha \left(x\right),\hfill \\ \alpha \left(x\right)e^{-c(1-x)}[e^{\theta (\xi ,\tau )(1-x)}{\widehat{g}}^{\delta }(\xi ,\tau )+{\int }_x^1{\widehat{f}}^{\delta }(s,\xi ,\tau )\frac{sinh\left(\theta \right(\xi ,\tau \left)\right(s-x\left)\right)}{\theta (\xi ,\tau )}ds],\hfill & e^{c(1-x)}>\alpha \left(x\right).\hfill \end{array}\right.$$

where $\alpha \left(x\right)>1$ can be considered as a regularization parameter.

Remark 3.

Clearly, the blasting components of high frequency in the data have been suppressed by introducing the factor $\alpha \left(x\right)e^{-c(1-x)}$. Therefore, stability can be ensured even if there are high frequency components of the data.

Remark 4.

The regularized solution approaches the exact solution if $\alpha \to 0$ as $\delta \to 0$.

3.1. An a Priori Parameter Choice

To obtain the convergence estimates of the regularization method, we assume that the following two conditions hold:

$${\int }_0^1{\left|\widehat{f}(s,\xi ,\tau )\right|}^{2}ds<C_1{e}^{-3\sqrt{2}c},\forall \xi \in \mathbb{R},\tau \in \mathbb{R},$$

(14)

$$\parallel u(0,·,·)\parallel \le E.$$

(15)

Lemma 2.

If condition (14) and (15) hold, $P(\xi ,\tau )=\widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds,$ then

$$\parallel P(\xi ,\tau )\parallel \le E+M_1,$$

where $M_1$ is a constant.

Proof. 

By the triangle inequality, (15), H$\ddot{o}$lder inequality, (14) and Lemma 1, we obtain

$$\begin{array}{cc}\hfill \parallel P(\xi ,\tau )\parallel \le & \parallel \widehat{u}(0,\xi ,\tau )\parallel +\parallel {\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds\parallel \hfill \\ \hfill \le & E+{\left[{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }({\int }_0^1|\frac{sinh\left(\theta s\right)}{\theta }{|}^{2}ds{\int }_0^1|\widehat{f}{|}^{2}ds)d\xi d\tau \right]}^{\frac{1}{2}}\hfill \\ \hfill \le & E+{\left[{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }({\int }_0^1{s}^2{e}^{2cs}ds{\int }_0^1|\widehat{f}{|}^{2}ds)d\xi d\tau \right]}^{\frac{1}{2}}\hfill \\ \hfill \le & E+{\left[{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\frac{1}{3}{C}_1{e}^{-(3\sqrt{2}-2)c}d\xi d\tau \right]}^{\frac{1}{2}}.\hfill \end{array}$$

It is easy to know that the generalized integral on the right-hand side of the last inequality converges, and here we introduce the notation

$$M_1:={\left[{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\frac{1}{3}{C}_1{e}^{-(3\sqrt{2}-2)c}d\xi d\tau \right]}^{\frac{1}{2}}.$$

Therefore,

$$\parallel P(\xi ,\tau )\parallel \le E+M_1.$$

□

Theorem 1.

Let $\widehat{u}(x,\xi ,\tau )$ given by (9) be the exact solution of problem (7) in the frequency space, ${\widehat{u}}^{\delta }(x,\xi )$ be the regularized solution, condition (8), (14) and (15) hold. If the regularization parameter α is selected as

$$\alpha \left(x\right)=x{\left(\frac{E}{\delta }\right)}^{1-x}.$$

(16)

Then for a fixed $0<x<1$, we get

$$\parallel u_{\alpha }^{\delta }(x,·,·)-u(x,·,·)\parallel \le \left(1+\frac{\sqrt{3}}{3}{(1-x)}^{\frac{3}{2}}\right)x{E}^{1-x}{\delta }^x+(1-x)E^{-x}{\delta }^x(E+2M_1),$$

(17)

where $M_1$ is given in Lemma 2.

Proof. 

Using the triangle inequality, we obtain

$$\parallel u_{\alpha }^{\delta }(x,·,·)-u(x,·,·)\parallel \le \underset{{\mathcal{R}}_1}{\underbrace{\parallel u_{\alpha }^{\delta }(x,·,·)-u_{\alpha }(x,·,·)\parallel }}+\underset{{\mathcal{R}}_2}{\underbrace{\parallel u_{\alpha }(x,·,·)-u(x,·,·)\parallel }}$$

(18)

The proof now naturally falls into two steps.

Step 1. Estimate the term ${\mathcal{R}}_1$ in (18).

Using Parseval’s identity and the triangle inequality

$$\begin{array}{cc}\hfill {\mathcal{R}}_1=& \parallel {\widehat{u}}_{\alpha }^{\delta }(x,·,·)-{\widehat{u}}_{\alpha }(x,·,·)\parallel \hfill \\ \hfill =& \parallel min\{1,\alpha \left(x\right)e^{-c(1-x)}\}[e^{\theta (1-x)}({\widehat{g}}^{\delta }-\widehat{g})+{\int }_x^1({\widehat{f}}^{\delta }-\widehat{f})\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds]\parallel \hfill \\ \hfill \le & \underset{\stackrel{ ¯}{{\mathcal{R}}_1}}{\underbrace{\parallel min\{1,\alpha \left(x\right)e^{-c(1-x)}\}{e}^{\theta (1-x)}({\widehat{g}}^{\delta }-\widehat{g})\parallel }}+\underset{\stackrel{ ¯}{{\mathcal{R}}_2}}{\underbrace{\parallel min\{1,\alpha \left(x\right)e^{-c(1-x)}\}{\int }_x^1({\widehat{f}}^{\delta }-\widehat{f})\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds\parallel }}.\hfill \end{array}$$

By (13) and (8), we get

$$\begin{array}{c}\hfill \stackrel{ ¯}{{\mathcal{R}}_1}\le {\parallel \alpha \left(x\right)e^{-c(1-x)}{e}^{(c+di)(1-x)}({\widehat{g}}^{\delta }-\widehat{g})\parallel }_{e^{c(1-x)}>\alpha \left(x\right)}\le \delta \alpha \left(x\right).\end{array}$$

Using H$\ddot{o}$lder inequality, Lemma 1 and (8) in order

$$\begin{array}{cc}\hfill \stackrel{ ¯}{{\mathcal{R}}_2}\le & \parallel \alpha \left(x\right)e^{-c(1-x)}{\int }_x^1({\widehat{f}}^{\delta }-\widehat{f})\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds{\parallel }_{e^{c(1-x)}>\alpha \left(x\right)}\hfill \\ \hfill \le & \parallel \alpha \left(x\right)e^{-c(1-x)}({\int }_x^1|{\widehat{f}}^{\delta }-\widehat{f}{|}^{2}ds{{)}^{\frac{1}{2}}{\left({\int }_0^1\right|\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }{|}^{2}ds)}^{\frac{1}{2}}\parallel }_{e^{c(1-x)}>\alpha \left(x\right)}\hfill \\ \hfill \le & \parallel \alpha \left(x\right)({\int }_x^1|{\widehat{f}}^{\delta }-\widehat{f}{|}^{2}ds){\left({\int }_x^1{(s-x)}^{2}ds\right)}^{\frac{1}{2}}{\parallel }_{e^{c(1-x)}>\alpha \left(x\right)}\hfill \\ \hfill \le & \frac{\sqrt{3}}{3}{(1-x)}^{\frac{3}{2}}\delta \alpha \left(x\right).\hfill \end{array}$$

Therefore,

$$\begin{array}{c}\hfill {\mathcal{R}}_1\le \delta \alpha \left(1+\frac{\sqrt{3}}{3}{(1-x)}^{\frac{3}{2}}\right).\end{array}$$

(19)

Step 2. Estimate the term ${\mathcal{R}}_2$ in (18).

Again, by Parseval’s identity and the triangle inequality,

$$\begin{array}{cc}\hfill {\mathcal{R}}_2=& \parallel {\widehat{u}}_{\alpha }(x,·,·)-\widehat{u}(x,·,·)\parallel \hfill \\ \hfill =& \parallel min\{1,\alpha \left(x\right)e^{-c(1-x)}\}(e^{\theta (1-x)}\widehat{g}+{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds)-(e^{\theta (1-x)}\widehat{g}+{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds)\parallel \hfill \\ \hfill \le & \underset{\stackrel{ ˜}{{\mathcal{R}}_1}}{\underbrace{\parallel (1-min\{1,\alpha \left(x\right)e^{-c(1-x)}\})e^{\theta (1-x)}\widehat{g}\parallel }}+\underset{\stackrel{ ˜}{{\mathcal{R}}_2}}{\underbrace{\parallel (1-min\{1,\alpha \left(x\right)e^{-c(1-x)}\}){\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds\parallel }}.\hfill \end{array}$$

We first estimate $\stackrel{ ˜}{{\mathcal{R}}_1}$.

Note that

$$\widehat{g}=e^{-\theta }[\widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds],$$

(20)

and by (13), we have

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_1}=& \parallel (1-min\{1,\alpha e^{-c(1-x)}\})e^{-\theta x}[\widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds]\parallel \hfill \\ \hfill \le & \underset{e^{c(1-x)}>\alpha }{sup}F\left(c\right)\parallel \widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds\parallel ,\hfill \end{array}$$

where

$$F\left(c\right)=(1-\alpha \left(x\right)e^{-c(1-x)})e^{-cx}.$$

Differentiating $F\left(c\right)$ in the variable c, then the zero point $c_{*}$ of $F^{\prime }\left(c\right)=0$ satisfies

$$e^{-c_{*}(1-x)}=\frac{x}{\alpha }.$$

Therefore, $F\left(c\right)$ attains its maximum

$$F\left(c_{*}\right)=(1-x){\left(\frac{x}{\alpha }\right)}^{\frac{x}{1-x}}.$$

(21)

Combining (21) and Lemma 2, we get

$$\begin{array}{c}\hfill \stackrel{ ˜}{{\mathcal{R}}_1}\le (1-x){\left(\frac{x}{\alpha }\right)}^{\frac{x}{1-x}}(E+M_1).\end{array}$$

(22)

We are now in a position to estimate $\stackrel{ ˜}{{\mathcal{R}}_2}.$ From H$\ddot{o}$lder inequality, Lemma 1, (14) and (21), we obtain

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_2}\le & \parallel (1-\alpha e^{-c(1-x)})e^{-cx}{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds{e}^{cx}{\parallel }_{e^{c(1-x)}>\alpha }\hfill \\ \hfill \le & \underset{e^{c(1-x)}>\alpha }{sup}F\left(c\right)[{\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }({\int }_x^1|\widehat{f}{|}^{2}ds{\left)\left({\int }_x^1{(s-x)}^2{e}^{2c(s-x)}ds\right)e^{2cx}d\xi d\tau \right]}^{\frac{1}{2}}\hfill \\ \hfill \le & \underset{e^{c(1-x)}>\alpha }{sup}F\left(c\right){\left[{\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }\frac{1}{3}{C}_1{(1-x)}^3{e}^{-(3\sqrt{2}-2)c}d\xi d\tau \right]}^{\frac{1}{2}}\hfill \\ \hfill \le & M_1(1-x){\left(\frac{x}{\alpha }\right)}^{\frac{x}{1-x}}.\hfill \end{array}$$

(23)

Therefore,

$$\begin{array}{c}\hfill {\mathcal{R}}_2\le (1-x){\left(\frac{x}{\alpha }\right)}^{\frac{x}{1-x}}(E+2M_1).\end{array}$$

(24)

Inserting (19), (24) into (18) and noting that the parameter $\alpha \left(x\right)$ is selected as in (16), we have (17), which is the desired conclusion. □

Remark 5.

The regularization parameter α does not depend on the location x usually. This restriction has been canceled in the Theorem 1 and instead select α dynamically.

We find that the Theorem 1 is not valid for the location at $x=0$, this is common in the theory of ill-posed problems. To retain the continuous dependence of the solution at this boundary, instead of (14) and (15), we introduce some stronger assumptions

$${\parallel u(0,·,·)\parallel }_p\le E,p\ge 0.$$

(25)

$${(1+{\xi }^2+{\tau }^2)}^p{\int }_0^1{\left|\widehat{f}(s,\xi ,\tau )\right|}^{2}ds<C_2{e}^{-3\sqrt{2}c},\forall \xi \in \mathbb{R},\tau \in \mathbb{R}.$$

(26)

Lemma 3.

Let condition (25) and (26) hold, $\stackrel{ ˜}{P}(\xi ,\tau )={(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}[\widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds],$ then

$$\parallel \stackrel{ ˜}{P}(\xi ,\tau )\parallel \le E+M_1.$$

The proof is similar to the proof of Lemma 2.

Lemma 4

([26]). For arbitrary $\xi ,\tau \in \mathbb{R}$, and c is given by (13), then the following inequality holds:

$$\underset{e^{c(1-x)}>\alpha }{sup}{(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}\le max\{{(\frac{1}{2(1-x)}ln\alpha )}^{-p},{(\frac{1}{2(1-x)}ln\alpha )}^{-\frac{2p}{\gamma }}\}.$$

Theorem 2.

Let $\widehat{u}(x,\xi ,\tau )$ given by (9) be the exact solution of problem (7) in the frequency space, ${\widehat{u}}_{\alpha }^{\delta }(x,\xi ,\tau )$ be the regularized solution, condition (8), (25) and (26) hold, and the parameter α is chosen as

$$\alpha ={\left(\frac{E}{\delta }\right)}^{\frac{1}{2}}.$$

(27)

Then for $p>0$, we obtain

$$\parallel u_{\alpha }^{\delta }(0,·,·)-u(0,·,·)\parallel \le (1+\frac{\sqrt{3}}{3})E^{\frac{1}{2}}{\delta }^{\frac{1}{2}}+{(\frac{1}{4}ln\frac{E}{\delta })}^{-p}(E+2M_1).$$

(28)

Proof. 

Using the triangle inequality, we have

$$\parallel u_{\alpha }^{\delta }(0,·,·)-u(0,·,·)\parallel \le \underset{{\mathcal{R}}_3}{\underbrace{\parallel u_{\alpha }^{\delta }(0,·,·)-u_{\alpha }(0,·,·)\parallel }}+\underset{{\mathcal{R}}_4}{\underbrace{\parallel u_{\alpha }(0,·,·)-u(0,·,·)\parallel }}$$

(29)

The proof now naturally falls into two steps.

Step 1. Estimate the term ${\mathcal{R}}_3$ in (29).

Taking a similar procedure of the estimate of ${\mathcal{R}}_1$, we have

$$\begin{array}{cc}\hfill {\mathcal{R}}_3=& \parallel {\widehat{u}}_{\alpha }^{\delta }(0,·,·)-{\widehat{u}}_{\alpha }(0,·,·)\parallel \hfill \\ \hfill \le & \parallel min\{1,\alpha e^{-c}\}{e}^{\theta }({\widehat{g}}^{\delta }-\widehat{g})\parallel +\parallel min\{1,\alpha e^{-c}\}{\int }_0^1({\widehat{f}}^{\delta }-\widehat{f})\frac{sinh\left(\theta s\right)}{\theta }ds\parallel \hfill \\ \hfill \le & (1+\frac{\sqrt{3}}{3})\delta \alpha .\hfill \end{array}$$

By (27), we get

$$\begin{array}{c}\hfill {\mathcal{R}}_3\le (1+\frac{\sqrt{3}}{3})E^{\frac{1}{2}}{\delta }^{\frac{1}{2}}.\end{array}$$

(30)

Step 2. Estimate the term ${\mathcal{R}}_4$ in (29).

Taking a similar procedure of the estimate of ${\mathcal{R}}_2$, we have

$$\begin{array}{cc}\hfill {\mathcal{R}}_4=& \parallel {\widehat{u}}_{\alpha }(0,·,·)-\widehat{u}(0,·,·)\parallel \hfill \\ \hfill \le & \underset{\stackrel{ ˜}{{\mathcal{R}}_3}}{\underbrace{\parallel (1-min\{1,\alpha e^{-c}\})e^{\theta }\widehat{g}\parallel }}+\underset{\stackrel{ ˜}{{\mathcal{R}}_4}}{\underbrace{\parallel (1-min\{1,\alpha e^{-c}\}){\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds\parallel }}.\hfill \end{array}$$

Using (13), (20), Lemmas 3 and 4, (27) and note that $0<\gamma \le 1$, we get

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_3}=& \parallel (1-min\{1,\alpha e^{-c}\})[\widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds]\parallel \hfill \\ \hfill \le & \underset{e^c>\alpha }{sup}(1-\alpha e^{-c}){(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}\parallel [\widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds]{(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}\parallel \hfill \\ \hfill \le & max\{{(\frac{1}{2}ln\alpha )}^{-p},{(\frac{1}{2}ln\alpha )}^{-\frac{2p}{\gamma }}\}(E+M_1)\hfill \\ \hfill =& {(\frac{1}{4}ln\frac{E}{\delta })}^{-p}(E+M_1).\hfill \end{array}$$

Again, using Lemmas 3 and 4, (27) and note that $0<\gamma \le 1$, we get

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_4}\le & \underset{e^c>\alpha }{sup}(1-\alpha e^{-c}){(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}\parallel {\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds{(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}\parallel \hfill \\ \hfill \le & max\{{(\frac{1}{2}ln\alpha )}^{-p},{(\frac{1}{2}ln\alpha )}^{-\frac{2p}{\gamma }}\}{M}_1\hfill \\ \hfill =& {(\frac{1}{4}ln\frac{E}{\delta })}^{-p}{M}_1.\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill {\mathcal{R}}_4\le {(\frac{1}{4}ln\frac{E}{\delta })}^{-p}(E+2M_1).\end{array}$$

(31)

By substituting (30) and (31) into (29), we arrive at the final conclusion (28). □

3.2. An a Posteriori Parameter Choice

In this subsection, we consider an a posterior parameter choice in Morozov’s discrepancy principle. Morozov’s discrepancy principle for our case is to find $\alpha$ such that

$$\parallel k_{\alpha }{\widehat{g}}^{\delta }-{\widehat{g}}^{\delta }\parallel =\eta \delta ,$$

(32)

where $\eta >1$ is a constant, and

$$k_{\alpha }(x,\xi ,\tau )=\left\{\begin{array}{cc}1,\hfill & e^{c(1-x)}\le \alpha \left(x\right),\hfill \\ \alpha \left(x\right)e^{-(1-x)c},\hfill & e^{c(1-x)}>\alpha \left(x\right),\hfill \end{array}\right.$$

c and d is given by (13).

According to the following Lemma, then there exists a unique regularization parameter $\alpha >0$ for (32) if $(e^{-c(1-x)}-1){\widehat{g}}^{\delta }>\eta \delta >0$.

Lemma 5.

Set $\rho \left(\alpha \right)=\parallel k_{\alpha }{\widehat{g}}^{\delta }-{\widehat{g}}^{\delta }\parallel$. If $(e^{-c(1-x)}-1){\widehat{g}}^{\delta }>\eta \delta >0$, then $\rho \left(\alpha \right)$ is a continuous strictly decreasing function and satisfies

(a) ${lim}_{\alpha \to 1}\rho \left(\alpha \right)=\parallel (e^{-c(1-x)}-1)g^{\delta }\parallel$;

(b) ${lim}_{\alpha \to \infty }\rho \left(\alpha \right)=0$.

Proof. 

The above results are straightforward by virtue of

$$\rho \left(\alpha \right)=\parallel (min\{1,\alpha e^{-c(1-x)}\}-1){\widehat{g}}^{\delta }\parallel .$$

□

For the convenience of calculation, we write

$$R_{\alpha }:=k_{\alpha }-1.$$

(33)

Lemma 6.

Let $R_{\alpha }$ is given by (33), then we have

$$\parallel R_{\alpha }\widehat{g}\parallel \le (\eta +1)\delta .$$

Proof. 

By the triangle inequality, we have

$$\parallel R_{\alpha }\widehat{g}\parallel \le \parallel R_{\alpha }(\widehat{g}-{\widehat{g}}^{\delta })\parallel +\parallel R_{\alpha }{\widehat{g}}^{\delta }\parallel \le (\eta +1)\delta .$$

□

Lemma 7.

If the a priori condition (25), (26) and the noise assumption (8) hold, then we have the following inequality

$$\alpha \left(x\right)\le {\left(\frac{E+M_1}{(\eta -1)\delta }\right)}^{1-x}{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p(1-x)}(1+o\left(1\right)).$$

Proof. 

From (32), the triangle inequality, Lemmas 3 and 4, there holds

$$\begin{array}{cc}\hfill \eta \delta & \le \parallel R_{\alpha }({\widehat{g}}^{\delta }-\widehat{g})\parallel +\parallel R_{\alpha }\widehat{g}\parallel \hfill \\ \hfill & \le \delta +\parallel (k_{\alpha }-1){(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}{e}^{-\theta }[\widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds]{(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}\parallel \hfill \\ \hfill & \le \delta +\underset{e^{c(1-x)}>\alpha }{sup}|(1-\alpha e^{-c(1-x)}){(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}{e}^{-c}|\parallel [\widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds]{(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}\parallel \hfill \\ \hfill & \le \delta +{\alpha }^{-\frac{1}{1-x}}max\{{(\frac{1}{2(1-x)}ln\alpha )}^{-p},{(\frac{1}{2(1-x)}ln\alpha )}^{-\frac{2p}{\gamma }}\}(E+M_1).\hfill \end{array}$$

This implies that

$$(\eta -1)\delta \le {\alpha }^{-\frac{1}{1-x}}{(\frac{1}{2(1-x)}ln\alpha )}^{-p}(E+M_1).$$

(34)

Let

$$(\eta -1)\delta ={\alpha }_{*}^{-\frac{1}{1-x}}{(\frac{1}{2(1-x)}ln{\alpha }_{*})}^{-p}(E+M_1),$$

(35)

we obtain [26]

$${\alpha }_{*}={\left(\frac{E+M_1}{(\eta -1)\delta }\right)}^{1-x}{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p(1-x)}(1+o\left(1\right)).$$

(36)

Note that for $\alpha >1$, $0\le x<1$, the function

$$H\left(x\right)={\alpha }^{-\frac{1}{1-x}}{(\frac{1}{2(1-x)}ln\alpha )}^{-p}(E+M_1)$$

decreases monotonically with respect to $\alpha$, hence, we get $\alpha \le {\alpha }_{*}$. □

Theorem 3.

Let $\widehat{u}(x,\xi ,\tau )$ given by (9) be the exact solution of problem (7) in the frequency space, ${\widehat{u}}_{\alpha }^{\delta }(x,\xi ,\tau )$ be the regularized solution, condition (8), (25) and (26) hold, and the regularization parameter α is selected by (32). Then, for a fixed $0\le x<1$ and $p>0$, let $R=\parallel u_{\alpha }^{\delta }(x,·,·)-u(x,·,·)\parallel$, we have

$$\begin{array}{c}\hfill R\le {\delta }^x{(E+M_1)}^{1-x}{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p(1-x)}(M_3+o\left(1\right))+{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p}(M_4+o\left(1\right)),\end{array}$$

(37)

where $M_3$ and $M_4$ are constants dependent on η and x.

Proof. 

Using the triangle inequality, we have

$$\parallel u_{\alpha }^{\delta }(x,·,·)-u(x,·,·)\parallel \le \underset{{\mathcal{R}}_5}{\underbrace{\parallel u_{\alpha }^{\delta }(x,·,·)-u_{\alpha }(x,·,·)\parallel }}+\underset{{\mathcal{R}}_6}{\underbrace{\parallel u_{\alpha }(x,·,·)-u(x,·,·)\parallel }}$$

(38)

The proof now naturally falls into two steps.

Step 1. Estimate the term ${\mathcal{R}}_5$ in (38).

Taking a similar procedure of the estimate of ${\mathcal{R}}_1$, and by Lemma 7, we get

$$\begin{array}{c}\hfill {\mathcal{R}}_5\le \delta \alpha (1+\frac{\sqrt{3}}{3}{(1-x)}^{\frac{3}{2}})\le (1+\frac{\sqrt{3}}{3}{(1-x)}^{\frac{3}{2}}){\left(\frac{E+M_1}{(\eta -1)}\right)}^{1-x}{\delta }^x{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p(1-x)}(1+o\left(1\right)).\end{array}$$

(39)

Step 2. Estimate the term ${\mathcal{R}}_6$ in (38).

Taking a similar procedure of the estimate of ${\mathcal{R}}_2$, we have

$$\begin{array}{c}\hfill {\mathcal{R}}_6=\parallel {\widehat{u}}_{\alpha }(x,·,·)-\widehat{u}(x,·,·)\parallel \le \underset{\stackrel{ ˜}{{\mathcal{R}}_5}}{\underbrace{\parallel R_{\alpha }{e}^{\theta (1-x)}\widehat{g}\parallel }}+\underset{\stackrel{ ˜}{{\mathcal{R}}_6}}{\underbrace{\parallel R_{\alpha }{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds\parallel }}.\end{array}$$

Using (9) and the triangle inequality, we get

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_5}=& \parallel R_{\alpha }[\widehat{u}(x,\xi ,\tau )-{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds]\parallel \hfill \\ \hfill \le & \parallel R_{\alpha }\widehat{u}\parallel +\parallel R_{\alpha }{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds\parallel \hfill \\ \hfill \le & \underset{\stackrel{ ˜}{{\mathcal{R}}_{51}}}{\underbrace{\parallel R_{\alpha }\widehat{u}-k_{\alpha }{R}_{\alpha }\widehat{u}\parallel }}+\underset{\stackrel{ ˜}{{\mathcal{R}}_{52}}}{\underbrace{\parallel k_{\alpha }{R}_{\alpha }\widehat{u}\parallel }}+\underset{\stackrel{ ˜}{{\mathcal{R}}_6}}{\underbrace{\parallel R_{\alpha }{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds\parallel }}.\hfill \end{array}$$

We first estimate $\stackrel{ ˜}{{\mathcal{R}}_{51}}$. Again, using the triangle inequality and (9), we get

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_{51}}=& \parallel R_{\alpha }(e^{\theta (1-x)}\widehat{g}+{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds)(1-k_{\alpha })\parallel \hfill \\ \hfill \le & \underset{\stackrel{ ˜}{{\mathcal{R}}_{54}}}{\underbrace{\parallel R_{\alpha }(e^{\theta (1-x)}-k_{\alpha }{e}^{\theta (1-x)})\widehat{g}\parallel }}+\underset{\stackrel{ ˜}{{\mathcal{R}}_{55}}}{\underbrace{\parallel R_{\alpha }{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds(1-k_{\alpha })\parallel }}.\hfill \end{array}$$

From (20), Lemmas 3 and 4, there holds

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_{54}}\le & \parallel (e^{\theta (1-x)}-k_{\alpha }{e}^{\theta (1-x)}){(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}[\widehat{u}(0,\xi ,\tau )-{\int }_0^1\widehat{f}\frac{sinh\left(\theta s\right)}{\theta }ds]e^{-\theta }{(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}\parallel \hfill \\ \hfill \le & \underset{\xi ,\tau \in \mathbb{R}}{sup}\left|(e^{\theta (1-x)}-k_{\alpha }{e}^{\theta (1-x)}){(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}{e}^{-\theta }\right|(E+M_1)\hfill \\ \hfill \le & \underset{e^{c(1-x)}>\alpha }{sup}\left|\frac{e^{c(1-x)}-\alpha \left(x\right)}{e^{c(1-x)}}{(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}{e}^{-cx}\right|(E+M_1)\hfill \\ \hfill \le & {(\frac{1}{2(1-x)}ln\alpha )}^{-p}{\alpha }^{-\frac{x}{1-x}}(E+M_1)\hfill \\ \hfill \le & {[\frac{1}{2}ln\left(\frac{E+M_1}{(\eta -1)\delta }\right)+ln{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-\frac{p}{2}}]}^{-p}\left[{\left(\frac{E+M_1}{(\eta -1)\delta }\right)}^{-x}{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{px}\right](E+M_1)\hfill \\ \hfill =& {(\eta -1)}^x{\delta }^x{(E+M_1)}^{1-x}(\frac{1}{2}ln{\left(\frac{E+M_1}{(\eta -1)\delta }\right)}^{-p(1-x)}(1+o\left(1\right)).\hfill \end{array}$$

Using H$\ddot{o}$lder inequality, Lemmas 1 and 4 and (14), we get

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_{55}}\le & \parallel {\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds{(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}{R}_{\alpha }{(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}\parallel \hfill \\ \hfill \le & \underset{e^{c(1-x)}>\alpha }{sup}\left|{(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}\right|\parallel {\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds{(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}\parallel \hfill \\ \hfill \le & {(\frac{1}{2(1-x)}ln\alpha )}^{-p}{\left[{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\left({\int }_x^1\right|\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }{|}^{2}ds){(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}({\int }_0^1|\widehat{f}{|}^{2}ds)d\xi d\tau \right]}^{\frac{1}{2}}\hfill \\ \hfill \le & {(\frac{1}{2(1-x)}ln\alpha )}^{-p}[{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{C}_1{e}^{2c(1-x)}\left({\int }_x^1{(s-x)}^{2}ds\right)e^{-3\sqrt{2}c}{d\xi d\tau ]}^{\frac{1}{2}}\hfill \\ \hfill \le & {(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p}[{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\frac{1}{3}{C}_1{(1-x)}^3{e}^{-3\sqrt{2}c}{e}^{2c(1-x)}{d\xi d\tau ]}^{\frac{1}{2}}(1+o\left(1\right)).\hfill \end{array}$$

Since the generalized integral on the right-hand side of the last inequality converges, we introduce the notation

$$M_2:=[{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\frac{1}{3}{C}_1{(1-x)}^3{e}^{-3\sqrt{2}c}{e}^{2c(1-x)}{d\xi d\tau ]}^{\frac{1}{2}}.$$

Thence,

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_{55}}\le & {(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p}{M}_2(1+o\left(1\right)).\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_{51}}\le & {(\eta -1)}^x{\delta }^x{\left(E+M_1\right)}^{1-x}(\frac{1}{2}ln{\left(\frac{E+M_1}{(\eta -1)\delta }\right)}^{-p(1-x)}(1+o\left(1\right))\hfill \\ \hfill & +{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p}{M}_2(1+o\left(1\right)).\hfill \end{array}$$

(40)

Next we estimate $\stackrel{ ˜}{{\mathcal{R}}_{52}}.$ Using (9), the triangle inequality, and note that $|k_{\alpha }|\le \alpha \left(x\right)e^{-c(1-x)}$, we get

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_{52}}\le & \alpha \left(x\right)e^{-c(1-x)}\parallel R_{\alpha }[\widehat{g}{e}^{\theta (1-x)}+{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds]\parallel \hfill \\ \hfill \le & \underset{\stackrel{ ˜}{{\mathcal{R}}_{56}}}{\underbrace{\alpha \left(x\right)\parallel R_{\alpha }\widehat{g}\parallel }}+\underset{\stackrel{ ˜}{{\mathcal{R}}_{57}}}{\underbrace{\alpha \left(x\right)e^{-c(1-x)}\parallel R_{\alpha }{\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds\parallel }}.\hfill \end{array}$$

By Lemmas 6 and 7, we get

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_{56}}\le & (\eta +1){(\eta -1)}^{x-1}{\delta }^x{\left(E+M_1\right)}^{1-x}{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p(1-x)}(1+o\left(1\right)).\hfill \end{array}$$

Using Lemmas 3 and 4, we have

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_{57}}\le & \underset{e^{c(1-x)>\alpha \left(x\right)}}{sup}|\alpha \left(x\right)e^{-c(1-x)}{(1+{\xi }^2+{\tau }^2)}^{-\frac{p}{2}}|\parallel {\int }_x^1\widehat{f}\frac{sinh\left(\theta \right(s-x\left)\right)}{\theta }ds{(1+{\xi }^2+{\tau }^2)}^{\frac{p}{2}}\parallel \hfill \\ \hfill \le & {(\frac{1}{2(1-x)}ln\alpha )}^{-p}{M}_2\hfill \\ \hfill \le & {(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p}{M}_2(1+o\left(1\right)).\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_{52}}\le & (\eta +1){(\eta -1)}^{x-1}{\delta }^x{\left(E+M_1\right)}^{1-x}{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p(1-x)}(1+o\left(1\right))\hfill \\ \hfill & +{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p}{M}_2(1+o\left(1\right)).\hfill \end{array}$$

(41)

To reach the conclusion, it is necessary to estimate $\stackrel{ ˜}{{\mathcal{R}}_6}$. Taking a similar procedure of the estimate of $\stackrel{ ˜}{{\mathcal{R}}_{57}}$, we get

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_6}\le & {(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p}{M}_2(1+o\left(1\right)).\hfill \end{array}$$

(42)

By (40)–(42), we have

$$\begin{array}{cc}\hfill \stackrel{ ˜}{{\mathcal{R}}_5}\le & 2\eta {(\eta -1)}^{x-1}{\delta }^x{\left(E+M_1\right)}^{1-x}{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p(1-x)}(1+o\left(1\right))\hfill \\ \hfill & +3·{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p}{M}_2(1+o\left(1\right)).\hfill \end{array}$$

(43)

Therefore,

$$\begin{array}{cc}\hfill {\mathcal{R}}_6\le & 2\eta {(\eta -1)}^{x-1}{\delta }^x{\left(E+M_1\right)}^{1-x}{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p(1-x)}(1+o\left(1\right))\hfill \\ \hfill & +4·{(\frac{1}{2}ln\frac{E+M_1}{(\eta -1)\delta })}^{-p}{M}_2(1+o\left(1\right)).\hfill \end{array}$$

(44)

By substituting (39) and (44) into (38), we arrive at the final conclusion (37). □

4. Conclusions

The fractional diffusion equations are extremely important in some industrial applications, so plenty of scholars have been involved in this theme in recent decades. For the homogeneous case, theoretical concepts and computational implementation are plentifully discussed and developed well. However, there are very few theoretical stability and convergence estimates on the nonhomogeneous system. We study a sideways problem of the 2D nonhomogeneous fractional diffusion equation, and design a modified kernel regularization method to overcome the ill-posedness. Besides, we derive the convergence estimates for the regularized solution by adopting an a priori and a posteriori choice rules of regularized parameter. The general source term is a function of location x and time t, which may also depend on solute concentration $u(x,t)$, that is, $f=f(x,t,u)$. Obviously, in this case, some new and strong techniques are needed to estimate the errors. In addition, we can also apply a similar method to inverse problems for space-fractional diffusion equations. This requires further exploration.