A Quasi-Boundary Value Method for Solving a Backward Problem of the Fractional Rayleigh–Stokes Equation

Abstract

In this paper, we study a backward problem for a fractional Rayleigh–Stokes equation by using a quasi-boundary value method. This problem is ill-posed; i.e., the solution (if it exists) does not depend continuously on the data. To overcome its instability, a regularization method is employed, and convergence rate estimates are derived under both a priori and a posteriori criteria for selecting the regularization parameter. The theoretical results demonstrate the effectiveness of the proposed method in deriving stable and accurate solutions.

1. Introduction

In this article, we focus on the Rayleigh–Stokes equation for a heated generalized second-grade fluid incorporating a time-fractional derivative

$$\left\{\begin{array}{cc}{\partial }_{t}v(x,t)-(1+\gamma {\partial }_t^{\alpha })\Delta v(x,t)=0,\hfill & (x,t)\in \Omega \times (0,T),\hfill \\ v(x,t)=0,\hfill & x\in \partial \Omega ,t\in (0,T],\hfill \\ v(x,0)=\phi \left(x\right),\hfill & x\in \Omega ,\hfill \end{array}\right.$$

(1)

with the final condition

$$v(x,T)=g\left(x\right),x\in \Omega .$$

(2)

Here, $\Delta$ represents the Laplacian operator, $\Omega \in {\mathbb{R}}^d(d=1,2,3)$ is a smooth domain with a boundary of $\partial \Omega$, and $T>0$ is a fixed time. Here, the constant $\gamma >0$ is a positive parameter, $\phi$ denotes the initial data, the function g belongs to the space $L^2(\Omega )$ and represents the final data, ${\partial }_t$ denotes the time derivative $\partial /\partial t$, and ${\partial }_t^{\alpha }$ is the Riemann–Liouville fractional derivative of the order $\alpha \in (0,1)$, given by [1]

$${\partial }_t^{\alpha }\varphi \left(t\right)={\displaystyle \frac{d}{dt}}{\int }_0^t{\omega }_{1-\alpha }(t-s)\varphi \left(s\right)ds,{\omega }_{\alpha }\left(t\right)={\displaystyle \frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}},$$

(3)

where $\Gamma (·)$ denotes the Gamma function.

The fractional Rayleigh–Stokes Equation (1) is crucial for modeling the dynamics of certain non-Newtonian fluids, as discussed in [2]. The direct problems have been analyzed in many studies. For example, the author of [3] investigated the exact analytic solutions to the Rayleigh–Stokes problem in a generalized Oldroyd-B fluid. In [4], the authors studied the problem for a generalized second-grade fluid in a porous half-space with a heated flat plate. In [5], Bazhlekova et al. studied the Rayleigh–Stokes problem for a generalized second-grade fluid incorporating a time-fractional Riemann–Liouville derivative, providing an analysis in continuous, semi-discrete, and fully discrete settings. Various numerical methods have been employed to address the forward problem regarding the Rayleigh–Stokes equation. For instance, the authors of [6] studied the variable-order Rayleigh–Stokes problem for a heated generalized second-grade fluid involving a fractional derivative, developing both implicit and explicit numerical methods to address the problem. Zaky [7] proposed efficient algorithms utilizing the Legendre–tau approximation to address one- and two-dimensional fractional Rayleigh–Stokes problems regarding a generalized second-grade fluid. Dehghan [8] devised a numerical method for the 2D Rayleigh–Stokes model involving a fractional derivative applied to irregular domains, including circular, L-shaped, and unit square geometries with circular or square holes. For other numerical methods, see reference [9,10].

Certain parameters in the modeling equation are challenging to obtain directly from observations. To address this issue, inverse problem techniques have been employed. However, research on inverse problems for the Rayleigh–Stokes equation remains relatively limited. In [11], the authors addressed an inverse problem centered on identifying the order of the fractional derivative in the Rayleigh–Stokes equation. Nguyen et al. [12] addressed an inverse source problem for the Rayleigh–Stokes problem, with Gaussian random noise included in the modeling. In [13], a filter regularization approach was utilized to tackle the inverse source problem of the Rayleigh–Stokes equation. Using the Tikhonov regularization method, Binh et al. [14] determined a source term for the Rayleigh–Stokes problem. In [15], the authors investigated an inverse problem of reconstructing the source function for the Rayleigh–Stokes equation with nonlocal time conditions, employing a fractional Tikhonov regularization method for the solution. Ke et al. [16] focused on solving the inverse problem of determining a source term in a Rayleigh–Stokes equation, incorporating a nonlinear perturbation with potential superlinear growth behavior. In [17], the authors utilized the trigonometric method in nonparametric regression, combined with the Fourier-truncated expansion method, to investigate a problem of determining a function for the fractional Rayleigh–Stokes equation with a nonlinear source term. In [18], Liu et al. focused on applying a fractional Tikhonov regularization method to determine a time-independent source term in the Rayleigh–Stokes equation with a fractional derivative.

This study focuses on the investigation of the backward problem for the Rayleigh–Stokes equation. The mathematical problem is formulated as follows [19]:

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

(4)

The aim of the backward problem is to determine the initial data $\phi \left(x\right)$ from the final condition $v(x,T)=g\left(x\right)$. Noisy data $g^{\delta }$, satisfying

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

(5)

where $\parallel ·\parallel$ is the $L^2(\Omega )$-norm and $\delta >0$ is a noisy level, are used in practice. Due to the ill-posedness of the problem, regularization methods are required to stabilize the solution.

In this paper, we propose a quasi-boundary value method to resolve the ill-posedness and ensure solution stability. This method, also dubbed the non-local boundary value problem method in [20], replaces the final or boundary condition with an approximate condition. It was recently used to solve several inverse problems, including the space-fractional backward diffusion problem [21], the Cauchy problem of an elliptic equation [22], the backward problem of a space-time fractional diffusion equation [23], the backward time-fractional diffusion problem [24], and the Cauchy problem of a 3D elliptic PDE with variable coefficients [25].

The layout of our work is as follows. In the following section, we demonstrate the ill-posedness of our backward problem and give a conditional stability estimate. Section 3 focuses on the regularization method and presents the main results of this paper, namely, the error estimates obtained using a priori and a posteriori parameter choice strategies. The paper concludes with a summary of the work performed in the final section.

2. Preliminaries

2.1. The Formula of the Initial Value $\phi \left(x\right)$

Let $\{{\lambda }_k,e_k\}$ denote the Dirichlet eigenvalue and corresponding eigenvector pair of the negative Laplacian operator $-\Delta$ in domain $\Omega$:

$$\left\{\begin{array}{cc}-\Delta e_k\left(x\right)={\lambda }_k{e}_k\left(x\right),\hfill & x\in \Omega ,\hfill \\ e_k\left(x\right)=0,\hfill & x\in \partial \Omega .\hfill \end{array}\right.$$

(6)

The eigenvalues satisfy

$$0<{\lambda }_1\le {\lambda }_2\le \dots \le {\lambda }_k\le \dots$$

with ${\lambda }_k\to \infty$ as $k\to \infty$. The corresponding eigenfunctions are as follows: $e_k\in H^2(\Omega )\cap H_0^1(\Omega )$.

For $m>0$, we define

$$H^m(\Omega ):=\left\{f\in L^2(\Omega )|\sum _{k=1}^{\infty }{\lambda }_k^m{\left|\langle f,e_k\rangle \right|}^2<+\infty \right\},$$

(7)

which is modified by the norm

$${\parallel f\parallel }_{H^m(\Omega )}={\left(\sum _{k=1}^{\infty }{\lambda }_k^m{\left|\langle f,e_k\rangle \right|}^2\right)}^{{\displaystyle \frac{1}{2}}}.$$

According to the result of theorem 2.1 in [5], for the direct problem (1), there exists a solution $v(x,t)\in C([0,T];L^2(\Omega ))\bigcap C([0,T];H^2(\Omega )\bigcap H_0^1(\Omega ))$,

$$v(x,t)=\sum _{k=1}^{\infty }{\phi }_k{v}_k\left(t\right)e_k\left(x\right).$$

(8)

Here, ${\phi }_k=\langle \phi \left(x\right),e_k\left(x\right)\rangle$ is the Fourier coefficient, and the function $v_k\left(t\right)$ satisfies

$$v_k\left(t\right)={\int }_0^{\infty }{e}^{-st}{B}_k\left(s\right)ds,$$

(9)

where

$$B_k\left(s\right)={\displaystyle \frac{\gamma }{\pi }}{\displaystyle \frac{{\lambda }_k{s}^{\alpha }sin\alpha \pi }{{(-s+{\lambda }_k\gamma s^{\alpha }cos\alpha \pi +{\lambda }_k)}^2+{({\lambda }_k\gamma s^{\alpha }sin\alpha \pi )}^2}}.$$

(10)

By incorporating $t=T$ into condition (2), we have

$$g\left(x\right)=\sum _{k=1}^{\infty }{\phi }_k{v}_k\left(T\right)e_k\left(x\right):=K\phi \left(x\right),$$

(11)

or, equivalently,

$$g_k={\phi }_k{v}_k\left(T\right),$$

(12)

where $g_k=\langle g\left(x\right),e_k\left(x\right)\rangle$ is the Fourier coefficient. The backward problem can be converted to solve the following integral equation:

$$K\phi \left(x\right):={\int }_{\Omega }k(x,\xi )\phi \left(\xi \right)d\xi =g\left(x\right),$$

(13)

where the kernel function is

$$k(x,\xi )=\sum _{k=1}^{\infty }{v}_k\left(T\right)e_k\left(x\right)e_k\left(\xi \right).$$

From (12), we can obtain the solution to the backward problem (4) as follows:

$$\phi \left(x\right)=\sum _{k=1}^{\infty }{\displaystyle \frac{g_k}{v_k\left(T\right)}}{e}_k\left(x\right).$$

(14)

In order to analyze the ill-posedness of the backward problem and explain the conditional stability results, the following lemmas are useful for the whole paper.

Lemma 1

([5]). The functions $v_k\left(t\right),k=1,2,\dots$, have the following properties:

(a) 

$v_k\left(0\right)=1,0<v_k\left(t\right)\le 1,t\ge 0;$

(b) 

$v_k\left(t\right)$ are completely monotone for $t\ge 0$;

(c) 

$|{\lambda }_k{v}_k\left(t\right)|\le cmin\{t^{-1},t^{\alpha -1}\},t>0$;

(d) 

${\int }_0^T\left|v_k\left(t\right)\right|dt<{\displaystyle \frac{1}{{\lambda }_k}},T>0$,

where the constant c does not depend on k and t.

Lemma 2

([26]). Let us assume that $\alpha \in (0,1)$. The following estimate holds for all $t\in [0,T]$

$$v_k\left(t\right)\ge {\displaystyle \frac{C(\gamma ,\alpha ,{\lambda }_1)}{{\lambda }_k}},$$

(15)

where

$$C(\gamma ,\alpha ,{\lambda }_1)={\displaystyle \frac{\gamma sin\alpha \pi }{3\pi }}{\int }_0^{\infty }{\displaystyle \frac{e^{-sT}{s}^{\alpha }ds}{\frac{s^2}{{\lambda }_1^2}+1+{\gamma }^2{s}^{2\alpha }}}.$$

Proof. 

By using the inequality ${(a+b+c)}^2\le 3(a^2+b^2+c^2)$ for any real numbers $a,b,c$, we have

$${(-s+{\lambda }_k\gamma s^{\alpha }cos\alpha \pi +{\lambda }_k)}^2+{({\lambda }_k\gamma s^{\alpha }sin\alpha \pi )}^2\le 3(s^2+{\lambda }_k^2+{\lambda }_k^2{\gamma }^2{s}^{2\alpha }).$$

From (9) and (10), we get

$$v_k\left(t\right)\ge {\displaystyle \frac{\gamma }{\pi }}sin\alpha \pi {\int }_0^{\infty }{\displaystyle \frac{e^{-sT}{\lambda }_k{s}^{\alpha }ds}{3(s^2+{\lambda }_k^2+{\lambda }_k^2{\gamma }^2{s}^{2\alpha })}}\ge {\displaystyle \frac{\gamma sin\alpha \pi }{3\pi {\lambda }_k}}{\int }_0^{\infty }{\displaystyle \frac{e^{-sT}{s}^{\alpha }ds}{\frac{s^2}{{\lambda }_1^2}+1+{\gamma }^2{s}^{2\alpha }}}.$$

(16)

The proof is thus completed. □

Lemma 3. 

For any $m>0, \beta >0, A>0$, and $0<{\lambda }_1\le s$, the following inequality holds:

$$F\left(s\right)={\displaystyle \frac{\beta s^{1-\frac{m}{2}}}{A+\beta s}}\le \left\{\begin{array}{cc}{A}_1{\beta }^{{\displaystyle \frac{m}{2}}},\hfill & 0<m\le 2,\hfill \\ A_2\beta ,\hfill & m>2.\hfill \end{array}\right.$$

(17)

Here, $A_1={\displaystyle \frac{{\left(\frac{(2-m)A}{m}\right)}^{1-\frac{m}{2}}}{\frac{2}{m}A}}$, and $A_2={\displaystyle \frac{{\lambda }_1^{1-\frac{m}{2}}}{A}}$.

Proof. 

For $0<m\le 2$, ${lim}_{s\to 0}F\left(s\right)=0$ and ${lim}_{s\to \infty }F\left(s\right)=0$; thus, we can obtain

$$F\left(s\right)\le \underset{s\ge {\lambda }_1}{sup}F\left(s\right)\le F\left(s^{*}\right).$$

where $s^{*}$ such that $F^{\prime }\left(s^{*}\right)$ = 0. It is easy to prove that $s^{*}={\displaystyle \frac{(2-m)A}{m\beta }}$. Therefore, we have

$$F\left(s\right)\le F\left(s^{*}\right)={\displaystyle \frac{\beta {\left(\frac{(2-m)A}{m\beta }\right)}^{1-\frac{m}{2}}}{\frac{2-m}{m}A+A}}=A_1(m,A){\beta }^{{\displaystyle \frac{m}{2}}}.$$

(18)

For $m>2$,

$$F\left(s\right)={\displaystyle \frac{\beta s^{1-\frac{m}{2}}}{A+\beta s}}={\displaystyle \frac{\beta }{(A+\beta s)s^{\frac{m}{2}-1}}}\le {\displaystyle \frac{\beta }{A}}{\lambda }_1^{1-{\displaystyle \frac{m}{2}}}=A_2(m,{\lambda }_1,A)\beta .$$

(19)

Here, $A_1={\displaystyle \frac{{\left(\frac{(2-m)A}{m}\right)}^{1-\frac{m}{2}}}{\frac{2}{m}A}}$, and $A_2={\displaystyle \frac{{\lambda }_1^{1-\frac{m}{2}}}{A}}$. The proof of Lemma is thus completed. □

Lemma 4. 

For any $m>0, \beta >0, A>0$, and $0<{\lambda }_1\le s$, the following inequality holds:

$$G\left(s\right)={\displaystyle \frac{\beta s^{\frac{2-m}{4}}}{A+\beta s}}\le \left\{\begin{array}{cc}{A}_3{\beta }^{{\displaystyle \frac{2+m}{4}}},\hfill & 0<m\le 2,\hfill \\ A_4\beta ,\hfill & m>2.\hfill \end{array}\right.$$

(20)

Here, $A_3={\displaystyle \frac{{\left(\frac{(2-m)A}{(2+m)}\right)}^{\frac{2-m}{4}}}{\frac{4}{2+m}A}}$, and $A_4={\displaystyle \frac{{\lambda }_1^{\frac{2-m}{4}}}{A}}$.

Proof. 

For $0<m\le 2$, ${lim}_{s\to 0}G\left(s\right)=0$ and ${lim}_{s\to \infty }G\left(s\right)=0$; thus, we can obtain

$$G\left(s\right)\le \underset{s\ge {\lambda }_1}{sup}G\left(s\right)\le G\left(s_0\right).$$

where $s_0$ such that $G^{\prime }\left(s_0\right)$=0. It is easy to prove that $s_0={\displaystyle \frac{(2-m)A}{(2+m)\beta }}$. Therefore, we have

$$G\left(s\right)\le G\left(s_0\right)={\displaystyle \frac{\beta {\left(\frac{(2-m)A}{(2+m)\beta }\right)}^{\frac{2-m}{4}}}{\frac{2-m}{2+m}A+A}}=A_3(m,A){\beta }^{{\displaystyle \frac{2+m}{4}}}.$$

(21)

For $m>2$,

$$G\left(s\right)={\displaystyle \frac{\beta s^{\frac{2-m}{4}}}{A+\beta s}}={\displaystyle \frac{\beta }{(A+\beta s)s^{\frac{m-2}{4}}}}\le {\displaystyle \frac{\beta }{A}}{\lambda }_1^{{\displaystyle \frac{2-m}{4}}}=A_4(m,{\lambda }_1,A)\beta .$$

(22)

Here, $A_3={\displaystyle \frac{{\left(\frac{(2-m)A}{(2+m)}\right)}^{\frac{2-m}{4}}}{\frac{4}{2+m}A}}$, and $A_4={\displaystyle \frac{{\lambda }_1^{\frac{2-m}{4}}}{A}}$. The proof of Lemma is thus completed. □

2.2. The Ill-Posedness and Conditional Stability of the Backward Problem

Now, we will show that the considered backward problem is ill-posed. From Lemma 1, we can find that

$${\displaystyle \frac{1}{v_k\left(T\right)}}\ge {\displaystyle \frac{{\lambda }_k}{cmin\{T^{-1},T^{\alpha -1}\}}}.$$

Since ${\lambda }_k\to \infty (k\to \infty )$, we can see that ${\displaystyle \frac{1}{v_k\left(T\right)}}\to \infty .$ Because of Equation (14), we know that the small error of $g^{\delta }\left(x\right)$ will be amplified by the factor ${\displaystyle \frac{1}{v_k\left(T\right)}}$. Thus, the backward problem is an ill-posed problem.

In the following, we review a conditional stability result of the backward problem of Problem (4).

Theorem 1

([27]). Let $\varphi \in H^m(\Omega )$ be such that

$${\parallel \varphi \parallel }_{H^m(\Omega )}\le E$$

(23)

for some $E>0$. Consequently, we have the following estimate

$${\parallel \varphi \parallel }_{L^2(\Omega )}\le C^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)E^{{\displaystyle \frac{2}{m+2}}}{\parallel g\parallel }^{{\displaystyle \frac{m}{m+2}}}.$$

(24)

3. The Quasi-Boundary Value Method and Convergence Rates

In this section, the quasi-boundary value method is employed by introducing a modified boundary condition, $v_{\beta }^{\delta }(x,T)+\beta v_{\beta }^{\delta }(x,0)=g^{\delta }\left(x\right)$, which replaces the original condition $v(x,T)=g\left(x\right)$. Let $v_{\beta }^{\delta }(x,t)$ denote the solution to the regularized problem formulation:

$$\left\{\begin{array}{cc}{\partial }_t{v}_{\beta }^{\delta }(x,t)-(1+\gamma {\partial }_t^{\alpha })\Delta v_{\beta }^{\delta }(x,t)=0,\hfill & (x,t)\in \Omega \times (0,T),\hfill \\ v_{\beta }^{\delta }(x,t)=0,\hfill & x\in \partial \Omega ,t\in (0,T],\hfill \\ v_{\beta }^{\delta }(x,T)+\beta v_{\beta }^{\delta }(x,0)=g^{\delta }\left(x\right),\hfill & x\in \Omega ,\hfill \end{array}\right.$$

(25)

where $\beta >0$ is a regularization parameter.

By separating variables, we can find that $v_{\beta }^{\delta }(x,t)$ has the following form

$$v_{\beta }^{\delta }(x,t)=\sum _{k=1}^{\infty }{C}_k{v}_k\left(t\right)e_k\left(x\right).$$

(26)

From $v_{\beta }^{\delta }(x,T)+\beta v_{\beta }^{\delta }(x,0)=g^{\delta }\left(x\right)$, we get

$$C_k{v}_k\left(T\right)+\beta C_k=g_k^{\delta },$$

where $g_k^{\delta }=\langle g^{\delta }\left(x\right),e_k\left(x\right)\rangle$. Thus, $C_k={\displaystyle \frac{g_k^{\delta }}{\beta +v_k\left(T\right)}}$. By substituting $C_k$ into (26), we get

$$v_{\beta }^{\delta }(x,t)=\sum _{k=1}^{\infty }{\displaystyle \frac{g_k^{\delta }}{\beta +v_k\left(T\right)}}{v}_k\left(t\right)e_k\left(x\right).$$

(27)

Denote the following expression accordingly:

$$v_{\beta }(x,t)=\sum _{k=1}^{\infty }{\displaystyle \frac{g_k}{\beta +v_k\left(T\right)}}{v}_k\left(t\right)e_k\left(x\right).$$

(28)

Specifically, we assume

$${\phi }_{\beta }^{\delta }\left(x\right)=v_{\beta }^{\delta }(x,0)=\sum _{k=1}^{\infty }{\displaystyle \frac{g_k^{\delta }}{\beta +v_k\left(T\right)}}{e}_k\left(x\right).$$

(29)

$${\phi }_{\beta }\left(x\right)=v_{\beta }(x,0)=\sum _{k=1}^{\infty }{\displaystyle \frac{g_k}{\beta +v_k\left(T\right)}}{e}_k\left(x\right).$$

(30)

In the subsequent analysis, we establish two distinct convergence estimates for $\parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel$ by employing a priori and a posteriori strategies for selecting the regularization parameter.

3.1. The Convergence Estimate Under an a Priori Parameter Choice Rule

The convergence analysis between the quasi-boundary-value-regularized solution and the exact solution can be derived under a priori regularization parameter choice criteria.

Lemma 5. 

Suppose the noise assumption (5) holds; thus, we can derive the following estimate:

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-{\phi }_{\beta }\left(x\right)\parallel \le {\displaystyle \frac{\delta }{\beta }}.$$

(31)

Proof. 

From (27), (28), and (5), we have

$$\begin{array}{ccc}\hfill \parallel {\phi }_{\beta }^{\delta }\left(x\right)-{\phi }_{\beta }{\left(x\right)\parallel }^2& =& \parallel \sum _{k=1}^{\infty }{\displaystyle \frac{g_k^{\delta }}{\beta +v_k\left(T\right)}}{e}_k\left(x\right)-\sum _{k=1}^{\infty }{\displaystyle \frac{g_k}{\beta +v_k\left(T\right)}}{e}_k\left(x\right){\parallel }^2\hfill \\ & =& \sum _{k=1}^{\infty }{\left({\displaystyle \frac{g_k^{\delta }-g_k}{\beta +v_k\left(T\right)}}\right)}^2\le {\delta }^2{\left(\underset{k\ge 1}{sup}A\left(k\right)\right)}^2,\hfill \end{array}$$

(32)

where

$$A\left(k\right)={\displaystyle \frac{1}{\beta +v_k\left(T\right)}}\le {\displaystyle \frac{1}{\beta }}.$$

Thus, we get

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-{\phi }_{\beta }\left(x\right)\parallel \le {\displaystyle \frac{\delta }{\beta }}.$$

The proof is thus completed. □

Lemma 6. 

Suppose a priori condition (23) holds; thus, we have the following estimate:

$$\parallel {\phi }_{\beta }\left(x\right)-\phi \left(x\right)\parallel \le \left\{\begin{array}{cc}{C}_{1}E{\beta }^{{\displaystyle \frac{m}{2}}},\hfill & 0<m\le 2,\hfill \\ C_{2}E\beta ,\hfill & m>2.\hfill \end{array}\right.$$

(33)

Here, $C_1={\displaystyle \frac{{\left(\frac{(2-m)C(\gamma ,\alpha ,{\lambda }_1)}{m}\right)}^{1-\frac{m}{2}}}{\frac{2}{m}C(\gamma ,\alpha ,{\lambda }_1)}}$, and $C_2={\displaystyle \frac{{\lambda }_1^{1-\frac{m}{2}}}{C(\gamma ,\alpha ,{\lambda }_1)}}$.

Proof. 

Based on (14) and (28), we know that

$$\begin{array}{ccc}\hfill \parallel {\phi }_{\beta }{\left(x\right)-\phi \left(x\right)\parallel }^2& =& \parallel \sum _{k=1}^{\infty }{\displaystyle \frac{g_k}{\beta +v_k\left(T\right)}}{e}_k\left(x\right)-\sum _{k=1}^{\infty }{\displaystyle \frac{g_k}{v_k\left(T\right)}}{e}_k\left(x\right){\parallel }^2\hfill \\ & =& \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{(\beta +v_k\left(T\right))v_k\left(T\right)}}\right)}^2{g}_k^2\hfill \\ & =& \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta {\lambda }_k^{\frac{m}{2}}{\lambda }_k^{-\frac{m}{2}}}{(\beta +v_k\left(T\right))v_k\left(T\right)}}\right)}^2{g}_k^2\hfill \\ & \le & {\left(\underset{k\ge 1}{sup}B\left(k\right)\right)}^2\sum _{k=1}^{\infty }{\displaystyle \frac{{\lambda }_k^m{g}_k^2}{v_k^2\left(T\right)}},\hfill \end{array}$$

(34)

where

$$B\left(k\right)={\displaystyle \frac{\beta {\lambda }_k^{-\frac{m}{2}}}{\beta +v_k\left(T\right)}}.$$

(35)

Now, we can estimate $B\left(k\right)$; by using Lemma 2, we are left with

$$B\left(k\right)\le {\displaystyle \frac{\beta {\lambda }_k^{-\frac{m}{2}}}{\beta +\frac{C(\gamma ,\alpha ,{\lambda }_1)}{{\lambda }_k}}}={\displaystyle \frac{\beta {\lambda }_k^{1-\frac{m}{2}}}{\beta {\lambda }_k+C(\gamma ,\alpha ,{\lambda }_1)}}.$$

(36)

By applying Lemma 3, we can obtain the following:

$$B\left(k\right)\le {\displaystyle \frac{\beta {\lambda }_k^{1-\frac{m}{2}}}{\beta {\lambda }_k+C(\gamma ,\alpha ,{\lambda }_1)}}\le \left\{\begin{array}{cc}{C}_1{\beta }^{{\displaystyle \frac{m}{2}}},\hfill & 0<m\le 2,\hfill \\ C_2\beta ,\hfill & m>2.\hfill \end{array}\right.$$

(37)

Here, $C_1={\displaystyle \frac{{\left(\frac{(2-m)C(\gamma ,\alpha ,{\lambda }_1)}{m}\right)}^{1-\frac{m}{2}}}{\frac{2}{m}C(\gamma ,\alpha ,{\lambda }_1)}}$, and $C_2={\displaystyle \frac{{\lambda }_1^{1-\frac{m}{2}}}{C(\gamma ,\alpha ,{\lambda }_1)}}$. Therefore, we are left with

$$\parallel {\phi }_{\beta }\left(x\right)-\phi \left(x\right)\parallel \le \left\{\begin{array}{cc}{C}_{1}E{\beta }^{{\displaystyle \frac{m}{2}}},\hfill & 0<m\le 2,\hfill \\ C_{2}E\beta ,\hfill & m>2.\hfill \end{array}\right.$$

Thus, the proof is complete. □

Theorem 2.

Suppose a priori condition (23) and the noise assumption (5) hold; thus, the following applies:

(1)

If $0<m\le 2$ and we set $\beta ={\left({\displaystyle \frac{\delta }{E}}\right)}^{{\displaystyle \frac{2}{m+2}}}$, we are left with the following convergence estimate:

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel \le (1+C_1){\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}.$$

(38)

(2)

If $m>2$ and we set $\beta ={\left({\displaystyle \frac{\delta }{E}}\right)}^{{\displaystyle \frac{1}{2}}}$, we are left with the following convergence estimate:

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel \le (1+C_2){\delta }^{{\displaystyle \frac{1}{2}}}{E}^{{\displaystyle \frac{1}{2}}},$$

(39)

where $C_1={\displaystyle \frac{{\left(\frac{(2-m)C(\gamma ,\alpha ,{\lambda }_1)}{m}\right)}^{1-\frac{m}{2}}}{\frac{2}{m}C(\gamma ,\alpha ,{\lambda }_1)}}$, and $C_2={\displaystyle \frac{{\lambda }_1^{1-\frac{m}{2}}}{C(\gamma ,\alpha ,{\lambda }_1)}}$.

Proof. 

Using the triangle inequality and Lemmas 5 and 6 yields

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel \le {\displaystyle \frac{\delta }{\beta }}+\left\{\begin{array}{cc}{C}_{1}E{\beta }^{{\displaystyle \frac{m}{2}}},\hfill & 0<m\le 2,\hfill \\ C_{2}E\beta ,\hfill & m>2.\hfill \end{array}\right.$$

(40)

Choose the regularization parameter $\beta$ as follows:

$$\beta =\left\{\begin{array}{cc}{\left({\displaystyle \frac{\delta }{E}}\right)}^{{\displaystyle \frac{2}{m+2}}},\hfill & 0<m\le 2,\hfill \\ {\left({\displaystyle \frac{\delta }{E}}\right)}^{{\displaystyle \frac{1}{2}}},\hfill & m>2.\hfill \end{array}\right.$$

(41)

Then, we have

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel \le \left\{\begin{array}{cc}(1+C_1){\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}},\hfill & 0<m\le 2,\hfill \\ (1+C_2){\delta }^{{\displaystyle \frac{1}{2}}}{E}^{{\displaystyle \frac{1}{2}}},\hfill & m>2.\hfill \end{array}\right.$$

(42)

Here, $C_1={\displaystyle \frac{{\left(\frac{(2-m)C(\gamma ,\alpha ,{\lambda }_1)}{m}\right)}^{1-\frac{m}{2}}}{\frac{2}{m}C(\gamma ,\alpha ,{\lambda }_1)}}$, and $C_2={\displaystyle \frac{{\lambda }_1^{1-\frac{m}{2}}}{C(\gamma ,\alpha ,{\lambda }_1)}}$. The proof is thus complete. □

3.2. The Convergence Estimate Under an A Posteriori Parameter Choice Rule

In this subsection, we establish a convergence estimate between the quasi-boundary-value-regularized solution and the exact solution under a posteriori regularization parameter selection.

We apply a modified version of the discrepancy principle as presented in [28], which is formulated as follows:

$${\parallel \beta (\beta I+K)}^{-1}(K{\phi }_{\beta }^{\delta }\left(x\right)-g^{\delta }\left(x\right))\parallel =\tau \delta ,$$

(43)

where $\tau >1$ is a constant. Based on the following lemma, it is established that a unique solution for (43) exists under the condition that $\parallel g^{\delta }\parallel >\tau \delta >0.$

Lemma 7. 

Let ${\rho \left(\beta \right)=\parallel \beta (\beta I+K)}^{-1}(K{\phi }_{\beta }^{\delta }\left(x\right)-g^{\delta }\left(x\right))\parallel$. If $\parallel g^{\delta }\parallel >\tau \delta >0$; thus, the following results hold:

(1) 

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

(2) 

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

(3) 

${lim}_{\beta \to +\infty }=\parallel g^{\delta }\left(x\right)\parallel$;

(4) 

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

Proof. 

The proof is expressed as follows:

$$\rho \left(\beta \right)=\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k^{\delta }{e}_k\left(x\right)\parallel ={\left(\sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^4{\left(g_k^{\delta }\right)}^2\right)}^{{\displaystyle \frac{1}{2}}}.$$

The four results of the Lemma are obvious. □

Theorem 3. 

Under the condition that the a priori assumption (23) and noise assumption (5) hold and there exists a constant $\tau >1$ such that $\parallel g^{\delta }\parallel \ge \tau \delta >0$, the regularization parameter $\beta >0$ is determined via the discrepancy principle (43). Thus, the following convergence estimates hold:

(1)

If $0<m\le 2$, the convergence estimate is given by

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel \le \left({\left({\displaystyle \frac{{\left(C_3\right)}^2}{\tau -1}}\right)}^{{\displaystyle \frac{2}{m+2}}}+C_5{(\tau +1)}^{{\displaystyle \frac{m}{m+2}}}\right){\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}},$$

(44)

where $C_3={\displaystyle \frac{{(cmin\{T^{-1},T^{\alpha -1}\})}^{\frac{1}{2}}{\left(\frac{(2-m)C(\gamma ,\alpha ,{\lambda }_1)}{(2+m)}\right)}^{\frac{2-m}{4}}}{\frac{4}{2+m}C(\gamma ,\alpha ,{\lambda }_1)}}$, and $C_5=C^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)$.

(2)

If $m>2$, the convergence estimate is given by

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel \le {\left({\displaystyle \frac{{\left(C_4\right)}^2}{\tau -1}}\right)}^{{\displaystyle \frac{1}{2}}}{\delta }^{{\displaystyle \frac{1}{2}}}{E}^{{\displaystyle \frac{1}{2}}}+C_6{(\tau +1)}^{{\displaystyle \frac{m}{m+2}}}{\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}},$$

(45)

where $C_4={\displaystyle \frac{{(cmin\{T^{-1},T^{\alpha -1}\})}^{\frac{1}{2}}{\lambda }_1^{\frac{2-m}{4}}}{C(\gamma ,\alpha ,{\lambda }_1)}}$, and $C_6={\left({\displaystyle \frac{\beta +1}{\beta }}\right)}^{{\displaystyle \frac{m}{2}}-1}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)$.

Proof. 

By using the triangle inequality, we obtain

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel \le \parallel {\phi }_{\beta }^{\delta }\left(x\right)-{\phi }_{\beta }\left(x\right)\parallel + \parallel {\phi }_{\beta }\left(x\right)-\phi \left(x\right)\parallel .$$

(46)

We will now proceed to estimate the first term. Similar to (31),

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-{\phi }_{\beta }\left(x\right)\parallel \le {\displaystyle \frac{\delta }{\beta }}.$$

(47)

Using (5) and (43), we can find that

$$\begin{array}{ccc}\hfill \tau \delta & =& \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k^{\delta }{e}_k\left(x\right)\parallel \hfill \\ & \le & \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2(g_k^{\delta }-g_k)e_k\left(x\right)\parallel +\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k{e}_k\left(x\right)\parallel \hfill \\ & \le & \delta +\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k{e}_k\left(x\right)\parallel \hfill \\ & =& \delta +\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{v}_k\left(T\right){\lambda }_k^{-{\displaystyle \frac{m}{2}}}{\lambda }_k^{{\displaystyle \frac{m}{2}}}{\displaystyle \frac{g_k}{v_k\left(T\right)}}{e}_k\left(x\right)\parallel \hfill \\ & =& \delta +\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{v}_k\left(T\right){\lambda }_k^{-{\displaystyle \frac{m}{2}}}{\lambda }_k^{{\displaystyle \frac{m}{2}}}{\phi }_k{e}_k\left(x\right)\parallel \hfill \\ & \le & \delta +E\underset{k\ge 1}{sup}{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{v}_k\left(T\right){\lambda }_k^{-{\displaystyle \frac{m}{2}}}\hfill \\ & =& \delta +E\underset{k\ge 1}{sup}{\left(D\left(k\right)\right)}^2.\hfill \end{array}$$

(48)

Here, $D\left(k\right)={\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}{\left(v_k\left(T\right)\right)}^{{\displaystyle \frac{1}{2}}}{\lambda }_k^{-{\displaystyle \frac{m}{4}}}$.

By using Lemmas 1 and 2, we have

$$D\left(k\right)={\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}{\left(v_k\left(T\right)\right)}^{{\displaystyle \frac{1}{2}}}{\lambda }_k^{-{\displaystyle \frac{m}{4}}}\le {\displaystyle \frac{\beta {\left(\frac{cmin\{T^{-1},T^{\alpha -1}\}}{{\lambda }_k}\right)}^{\frac{1}{2}}{\lambda }_k^{-\frac{m}{4}}}{\beta +\frac{C(\gamma ,\alpha ,{\lambda }_1)}{{\lambda }_k}}}={\displaystyle \frac{\beta {(cmin\{T^{-1},T^{\alpha -1}\})}^{\frac{1}{2}}{\lambda }_k^{\frac{2-m}{4}}}{\beta {\lambda }_k+C(\gamma ,\alpha ,{\lambda }_1)}},$$

(49)

and by using Lemma 4, we get

$$D\left(k\right)\le \left\{\begin{array}{cc}{C}_3{\beta }^{{\displaystyle \frac{2+m}{4}}},\hfill & 0<m\le 2,\hfill \\ C_4\beta ,\hfill & m>2.\hfill \end{array}\right.$$

(50)

Then,

$$(\tau -1)\delta \le \left\{\begin{array}{cc}{\left(C_3\right)}^{2}E{\beta }^{{\displaystyle \frac{2+m}{2}}},\hfill & 0<m\le 2,\hfill \\ {\left(C_4\right)}^{2}E{\beta }^2,\hfill & m>2.\hfill \end{array}\right.$$

(51)

Therefore,

$${\displaystyle \frac{1}{\beta }}\le \left\{\begin{array}{cc}{\left({\displaystyle \frac{{\left(C_3\right)}^2}{\tau -1}}\right)}^{{\displaystyle \frac{2}{m+2}}}{\left({\displaystyle \frac{E}{\delta }}\right)}^{{\displaystyle \frac{2}{m+2}}},\hfill & 0<m\le 2,\hfill \\ {\left({\displaystyle \frac{{\left(C_4\right)}^2}{\tau -1}}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\displaystyle \frac{E}{\delta }}\right)}^{{\displaystyle \frac{1}{2}}},\hfill & m>2.\hfill \end{array}\right.$$

(52)

Combing (47) and (52), we obtain

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-{\phi }_{\beta }\left(x\right)\parallel \le {\displaystyle \frac{\delta }{\beta }}\le \left\{\begin{array}{cc}{\left({\displaystyle \frac{{\left(C_3\right)}^2}{\tau -1}}\right)}^{{\displaystyle \frac{2}{m+2}}}{\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}},\hfill & 0<m\le 2,\hfill \\ {\left({\displaystyle \frac{{\left(C_4\right)}^2}{\tau -1}}\right)}^{{\displaystyle \frac{1}{2}}}{\delta }^{{\displaystyle \frac{1}{2}}}{E}^{{\displaystyle \frac{1}{2}}},\hfill & m>2.\hfill \end{array}\right.$$

(53)

Now, we estimate the second term of (46). From (30), we have

$$\begin{array}{ccc}\hfill \parallel {\phi }_{\beta }\left(x\right)-\phi \left(x\right)\parallel & =& \parallel \sum _{k=1}^{\infty }\left({\displaystyle \frac{-\beta }{\beta +v_k\left(T\right)}}\right){\phi }_k{e}_k\left(x\right)\parallel \hfill \\ & =& \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta v_k\left(T\right)}{\beta +v_k\left(T\right)}}\right)}^{{\displaystyle \frac{m}{2}}}{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^{1-{\displaystyle \frac{m}{2}}}{\displaystyle \frac{{\phi }_k}{{\left(v_k\left(T\right)\right)}^{\frac{m}{2}}}}{e}_k\left(x\right)\parallel \hfill \\ & \le & \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta v_k\left(T\right)}{\beta +v_k\left(T\right)}}\right)}^{{\displaystyle \frac{m+2}{2}}}{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^{1-{\displaystyle \frac{m}{2}}}{\displaystyle \frac{{\phi }_k}{{\left(v_k\left(T\right)\right)}^{\frac{m}{2}}}}{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}\hfill \\ & & ·\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^{1-{\displaystyle \frac{m}{2}}}{\displaystyle \frac{{\phi }_k}{{\left(v_k\left(T\right)\right)}^{\frac{m}{2}}}}{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{2}{m+2}}}.\hfill \end{array}$$

(54)

Here, according to Lemma 1, we get

$$\begin{array}{cc}\hfill {\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^{1-{\displaystyle \frac{m}{2}}}& \le \left\{\begin{array}{cc}1,\hfill & 0<m\le 2,\hfill \\ {\left({\displaystyle \frac{\beta +1}{\beta }}\right)}^{{\displaystyle \frac{m}{2}}-1},\hfill & m>2.\hfill \end{array}\right.\hfill \end{array}$$

(55)

We will proceed to estimate Formula (54) by considering two separate cases.

Case 1. If $0<m\le 2$, by using (54), (55), Lemma 2, and a priori bound condition (23), we are left with

$$\begin{array}{ccc}\hfill \parallel {\phi }_{\beta }\left(x\right)-\phi \left(x\right)\parallel & \le & \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta v_k\left(T\right)}{\beta +v_k\left(T\right)}}\right)}^{{\displaystyle \frac{m+2}{2}}}{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^{1-{\displaystyle \frac{m}{2}}}{\displaystyle \frac{{\phi }_k}{{\left(v_k\left(T\right)\right)}^{\frac{m}{2}}}}{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}\hfill \\ & & ·\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^{1-{\displaystyle \frac{m}{2}}}{\displaystyle \frac{{\phi }_k}{{\left(v_k\left(T\right)\right)}^{\frac{m}{2}}}}{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{2}{m+2}}}\hfill \\ & \le & \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{v}_k\left(T\right){\phi }_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}\parallel \sum _{k=1}^{\infty }{\displaystyle \frac{{\phi }_k}{{\left(v_k\left(T\right)\right)}^{\frac{m}{2}}}}{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{2}{m+2}}}\hfill \\ & =& \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{v}_k\left(T\right){\phi }_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}{\left(\sum _{k=1}^{\infty }{\displaystyle \frac{{\phi }_k^2}{{\left(v_k\left(T\right)\right)}^m}}\right)}^{{\displaystyle \frac{1}{m+2}}}\hfill \\ & \le & \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{v}_k\left(T\right){\phi }_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}{\left(\sum _{k=1}^{\infty }{\lambda }_k^m{\phi }_k^2\right)}^{{\displaystyle \frac{1}{m+2}}}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)\hfill \\ & =& \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}{\left(\sum _{k=1}^{\infty }{\lambda }_k^m{\phi }_k^2\right)}^{{\displaystyle \frac{1}{m+2}}}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)\hfill \\ & =& \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}\parallel \sum _{k=1}^{\infty }{\lambda }_k^{{\displaystyle \frac{m}{2}}}{\phi }_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{2}{m+2}}}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)\hfill \\ & \le & \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)\hfill \\ & \le & {\left(\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2(g_k-g_k^{\delta })e_k\left(x\right)\parallel +\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k^{\delta }{e}_k\left(x\right)\parallel \right)}^{{\displaystyle \frac{m}{m+2}}}\hfill \\ & & ·E^{{\displaystyle \frac{2}{m+2}}}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)\hfill \\ & \le & {\left((\tau +1)\delta \right)}^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)\hfill \\ & =& {\left({\displaystyle \frac{\tau +1}{C(\gamma ,\alpha ,{\lambda }_1)}}\right)}^{{\displaystyle \frac{m}{m+2}}}{\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}.\hfill \end{array}$$

(56)

Therefore, for $0<m\le 2$, we have

$$\parallel {\phi }_{\beta }\left(x\right)-\phi \left(x\right)\parallel \le C_5{(\tau +1)}^{{\displaystyle \frac{m}{m+2}}}{\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}.$$

(57)

By combining (46), (53), and (57), we obtain

$$\begin{array}{ccc}\hfill \parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel & \le & {\left({\displaystyle \frac{{\left(C_3\right)}^2}{\tau -1}}\right)}^{{\displaystyle \frac{2}{m+2}}}{\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}+C_5{(\tau +1)}^{{\displaystyle \frac{m}{m+2}}}{\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}\hfill \\ & =& \left({\left({\displaystyle \frac{{\left(C_3\right)}^2}{\tau -1}}\right)}^{{\displaystyle \frac{2}{m+2}}}+C_5{(\tau +1)}^{{\displaystyle \frac{m}{m+2}}}\right){\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}.\hfill \end{array}$$

(58)

Here, $C_3={\displaystyle \frac{{(cmin\{T^{-1},T^{\alpha -1}\})}^{\frac{1}{2}}{\left(\frac{(2-m)C(\gamma ,\alpha ,{\lambda }_1)}{(2+m)}\right)}^{\frac{2-m}{4}}}{\frac{4}{2+m}C(\gamma ,\alpha ,{\lambda }_1)}}$, and $C_5=C^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)$.

Case 2. If $m>2$, by using (54), (55), Lemma 2, and a priori bound condition (23), we get

$$\begin{array}{ccc}\hfill \parallel {\phi }_{\beta }\left(x\right)-\phi \left(x\right)\parallel & \le & \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta v_k\left(T\right)}{\beta +v_k\left(T\right)}}\right)}^{{\displaystyle \frac{m+2}{2}}}{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^{1-{\displaystyle \frac{m}{2}}}{\displaystyle \frac{{\phi }_k}{{\left(v_k\left(T\right)\right)}^{\frac{m}{2}}}}{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}\hfill \\ & & ·\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^{1-{\displaystyle \frac{m}{2}}}{\displaystyle \frac{{\phi }_k}{{\left(v_k\left(T\right)\right)}^{\frac{m}{2}}}}{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{2}{m+2}}}\hfill \\ & \le & \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{v}_k\left(T\right){\phi }_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}\parallel \sum _{k=1}^{\infty }{\displaystyle \frac{{\phi }_k}{{\left(v_k\left(T\right)\right)}^{\frac{m}{2}}}}{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{2}{m+2}}}{\left({\displaystyle \frac{\beta +1}{\beta }}\right)}^{{\displaystyle \frac{m}{2}}-1}\hfill \\ & \le & \parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{m}{m+2}}}\parallel \sum _{k=1}^{\infty }{\lambda }_k^{{\displaystyle \frac{m}{2}}}{\phi }_k{e}_k\left(x\right){\parallel }^{{\displaystyle \frac{2}{m+2}}}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1){\left({\displaystyle \frac{\beta +1}{\beta }}\right)}^{{\displaystyle \frac{m}{2}}-1}\hfill \\ & \le & {\left(\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2(g_k-g_k^{\delta })e_k\left(x\right)\parallel +\parallel \sum _{k=1}^{\infty }{\left({\displaystyle \frac{\beta }{\beta +v_k\left(T\right)}}\right)}^2{g}_k^{\delta }{e}_k\left(x\right)\parallel \right)}^{{\displaystyle \frac{m}{m+2}}}\hfill \\ & & ·E^{{\displaystyle \frac{2}{m+2}}}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1){\left({\displaystyle \frac{\beta +1}{\beta }}\right)}^{{\displaystyle \frac{m}{2}}-1}\hfill \\ & \le & {\left({\displaystyle \frac{\beta +1}{\beta }}\right)}^{{\displaystyle \frac{m}{2}}-1}{\left((\tau +1)\delta \right)}^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)\hfill \\ & =& {\left({\displaystyle \frac{\beta +1}{\beta }}\right)}^{{\displaystyle \frac{m}{2}}-1}{\left({\displaystyle \frac{\tau +1}{C(\gamma ,\alpha ,{\lambda }_1)}}\right)}^{{\displaystyle \frac{m}{m+2}}}{\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}.\hfill \end{array}$$

(59)

Therefore, for $m>2$, we have

$$\parallel {\phi }_{\beta }\left(x\right)-\phi \left(x\right)\parallel \le C_6{(\tau +1)}^{{\displaystyle \frac{m}{m+2}}}{\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}.$$

(60)

By combining (46), (53), and (60), we obtain

$$\parallel {\phi }_{\beta }^{\delta }\left(x\right)-\phi \left(x\right)\parallel \le {\left({\displaystyle \frac{{\left(C_4\right)}^2}{\tau -1}}\right)}^{{\displaystyle \frac{1}{2}}}{\delta }^{{\displaystyle \frac{1}{2}}}{E}^{{\displaystyle \frac{1}{2}}}+C_6{(\tau +1)}^{{\displaystyle \frac{m}{m+2}}}{\delta }^{{\displaystyle \frac{m}{m+2}}}{E}^{{\displaystyle \frac{2}{m+2}}}.$$

(61)

Here, $C_4={\displaystyle \frac{{(cmin\{T^{-1},T^{\alpha -1}\})}^{\frac{1}{2}}{\lambda }_1^{\frac{2-m}{4}}}{C(\gamma ,\alpha ,{\lambda }_1)}}$, and $C_6={\left({\displaystyle \frac{\beta +1}{\beta }}\right)}^{{\displaystyle \frac{m}{2}}-1}{C}^{-{\displaystyle \frac{m}{m+2}}}(\gamma ,\alpha ,{\lambda }_1)$.

Thus, the proof is complete. □

4. Conclusions

This paper introduces a quasi-boundary value method for addressing the inverse problem of the fractional Rayleigh–Stokes equation. Under a priori assumptions about the exact solution, error estimates were derived for both a priori and a posteriori parameter selection rules. The theoretical analysis confirms that the proposed method successfully stabilizes the ill-posed backward problem. Compared to the convergence order of ${\delta }^{{\displaystyle \frac{k}{k+1}}}$ in reference [19], we obtain a different convergence order of ${\delta }^{{\displaystyle \frac{m}{m+2}}}$. However, a drawback is that the quasi-boundary value method exhibits a saturation effect. The next steps will be divided into three parts: (1) assessing the effectiveness and stability of the algorithm numerically; (2) developing a novel algorithm to eliminate the saturation effect; and (3) addressing the multi-parameter identification problem.