Simultaneous Inversion of the Space-Dependent Source Term and the Initial Value in a Time-Fractional Diffusion Equation

Shuang Yu; Zewen Wang; Hongqi Yang

doi:10.1515/cmam-2022-0058

Article

Simultaneous Inversion of the Space-Dependent Source Term and the Initial Value in a Time-Fractional Diffusion Equation

Shuang Yu , Zewen Wang and Hongqi Yang

Published/Copyright: February 8, 2023

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From the journal Computational Methods in Applied Mathematics Volume 23 Issue 3

Abstract

The inverse problem for simultaneously identifying the space-dependent source term and the initial value in a time-fractional diffusion equation is studied in this paper. The simultaneous inversion is formulated into a system of two operator equations based on the Fourier method to the time-fractional diffusion equation. Under some suitable assumptions, the conditional stability of simultaneous inversion solutions is established, and the exponential Tikhonov regularization method is proposed to obtain the good approximations of simultaneous inversion solutions. Then the convergence estimations of inversion solutions are presented for a priori and a posteriori selections of regularization parameters. Finally, numerical experiments are conducted to illustrate effectiveness of the proposed method.

Keywords: Time-Fractional Diffusion Equation; Simultaneous Inversion; Source Term; Initial Value; Exponential Tikhonov Regularization Method

MSC 2010: 35R30; 35R11; 47A52

Funding source: Guangdong Province Key Laboratory of Computational Science

Award Identifier / Grant number: 2020B1212060032

Funding source: National Natural Science Foundation of China

Award Identifier / Grant number: 11571386

Award Identifier / Grant number: 11961002

Award Identifier / Grant number: 12171248

Funding source: Natural Science Foundation of Jiangxi Province

Award Identifier / Grant number: 20212ACB201001

Funding statement: The work of S. Yu is supported in part by the NSF of China under grant 11961002. The work of S. Yu and H. Yang is supported in part by the Key-Area Research and Development Program of Guangdong Province (No. 2021B0101190003), by Guangdong Province Key Laboratory of Computational Science at the Sun Yat-sen University (2020B1212060032), by the NSF of China under grant 11571386. The work of Z. Wang is supported in part by the NSF of China under grants 11961002 and 12171248, and by Jiangxi Provincial Natural Science Foundation (20212ACB201001).

A Proofs of Theorems 1–3

Proof of Theorem 1

From (3.3) and the Hölder inequality, we have

\begin{matrix} {∥ f ∥}^{2} & = + \infty \sum n = 1 {(\frac{λ_{n}}{E_{α, 1} (- λ_{n} T_{2}^{α}) - E_{α, 1} (- λ_{n} T_{1}^{α})} (E_{α, 1} (- λ_{n} T α 2) g_{1, n} - E_{α, 1} (- λ_{n} T α 1) g_{2, n}))}^{2} \leq {(+ \infty \sum n = 1 {(\frac{λ_{n} E_{α, 1} (- λ_{n} T_{2}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α}) - E_{α, 1} (- λ_{n} T_{1}^{α})})}^{γ + 2} {(g_{1, n} - \frac{E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α})} g_{2, n})}^{2})}^{\frac{2}{γ + 2}} \times {(+ \infty \sum n = 1 {(g_{1, n} - \frac{E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α})} g_{2, n})}^{2})}^{\frac{γ}{γ + 2}} \leq {(+ \infty \sum n = 1 \frac{1}{{(1 - \frac{E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α})})}^{γ}} λ_{n}^{γ} f_{n}^{2})}^{\frac{2}{γ + 2}} {({(+ \infty \sum n = 1 {(g_{1, n})}_{1, n}^{2})}^{\frac{1}{2}} + {(+ \infty \sum n = 1 {(\frac{E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α})} g_{2, n})}^{2})}^{\frac{1}{2}})}^{\frac{2 γ}{γ + 2}} . \end{matrix}

$∥ f ∥ 2 = ∑ n = 1 + ∞ ( λ n E α , 1 ⁢ ( − λ n ⁢ T 2 α ) − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ⁢ ( E α , 1 ⁢ ( − λ n ⁢ T 2 α ) ⁢ g 1 , n − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ⁢ g 2 , n ) ) 2 ≤ ( ∑ n = 1 + ∞ ( λ n ⁢ E α , 1 ⁢ ( − λ n ⁢ T 2 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ) γ + 2 ⁢ ( g 1 , n − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) ⁢ g 2 , n ) 2 ) 2 γ + 2 × ( ∑ n = 1 + ∞ ( g 1 , n − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) ⁢ g 2 , n ) 2 ) γ γ + 2 ≤ ( ∑ n = 1 + ∞ 1 ( 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) ) γ ⁢ λ n γ ⁢ f n 2 ) 2 γ + 2 ⁢ ( ( ∑ n = 1 + ∞ ( g 1 , n ) 2 ) 1 2 + ( ∑ n = 1 + ∞ ( E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) ⁢ g 2 , n ) 2 ) 1 2 ) 2 ⁢ γ γ + 2 .$

From Lemmas 1–3, we get

\frac{E_{α, 1} (- λ_{1} T_{1}^{α})}{E_{α, 1} (- λ_{1} T_{2}^{α})} \leq \frac{E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α})} and \frac{E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α})} \leq lim n \to \infty \frac{E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α})} = {(\frac{T_{2}}{T_{1}})}^{α} .

$E α , 1 ⁢ ( − λ 1 ⁢ T 1 α ) E α , 1 ⁢ ( − λ 1 ⁢ T 2 α ) ≤ E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) and E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) ≤ lim n → ∞ E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) = ( T 2 T 1 ) α .$

Obviously, $λ_{n}^{γ} \leq exp (λ_{n}^{γ})$ $λ n γ ≤ exp ⁡ ( λ n γ )$ holds. By the assumption of a priori condition (3.4) and the notation

C_{3} := {(\frac{1}{\frac{E_{α, 1} (- λ_{1} T_{1}^{α})}{E_{α, 1} (- λ_{1} T_{2}^{α})} - 1})}^{\frac{2 γ}{γ + 2}},

$C 3 := ( 1 E α , 1 ⁢ ( − λ 1 ⁢ T 1 α ) E α , 1 ⁢ ( − λ 1 ⁢ T 2 α ) − 1 ) 2 ⁢ γ γ + 2 ,$

we have

{∥ f ∥}^{2} \leq C_{3} M^{\frac{4}{γ + 2}} {(∥ g_{1} ∥ + {(\frac{T_{2}}{T_{1}})}^{α} ∥ g_{2} ∥)}^{\frac{2 γ}{γ + 2}} .

$∥ f ∥ 2 ≤ C 3 ⁢ M 4 γ + 2 ⁢ ( ∥ g 1 ∥ + ( T 2 T 1 ) α ⁢ ∥ g 2 ∥ ) 2 ⁢ γ γ + 2 .$

Similarly, we get

\begin{matrix} {∥ φ ∥}^{2} & = + \infty \sum n = 1 {(\frac{(1 - E_{α, 1} (- λ_{n} T α 1)) g_{2, n} - (1 - E_{α, 1} (- λ_{n} T α 2)) g_{1, n}}{E_{α, 1} (- λ_{n} T_{2}^{α}) - E_{α, 1} (- λ_{n} T_{1}^{α})})}^{2} \leq {(+ \infty \sum n = 1 {(\frac{1 - E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α}) - E_{α, 1} (- λ_{n} T_{1}^{α})})}^{γ + 2} {(g_{2, n} - \frac{1 - E_{α, 1} (- λ_{n} T_{2}^{α})}{1 - E_{α, 1} (- λ_{n} T_{1}^{α})} g_{1, n})}^{2})}^{\frac{2}{γ + 2}} \times {(+ \infty \sum n = 1 {(g_{2, n} - \frac{1 - E_{α, 1} (- λ_{n} T_{2}^{α})}{1 - E_{α, 1} (- λ_{n} T_{1}^{α})} g_{1, n})}^{2})}^{\frac{γ}{γ + 2}} \leq {(+ \infty \sum n = 1 {(\frac{1 - E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α}) - E_{α, 1} (- λ_{n} T_{1}^{α})})}^{γ} φ_{n}^{2})}^{\frac{2}{γ + 2}} {({(+ \infty \sum n = 1 {(g_{2, n})}_{2, n}^{2})}^{\frac{1}{2}} + {(+ \infty \sum n = 1 {(\frac{1 - E_{α, 1} (- λ_{n} T_{2}^{α})}{1 - E_{α, 1} (- λ_{n} T_{1}^{α})} g_{1, n})}^{2})}^{\frac{1}{2}})}^{\frac{2 γ}{γ + 2}} . \end{matrix}

$∥ φ ∥ 2 = ∑ n = 1 + ∞ ( ( 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ) ⁢ g 2 , n − ( 1 − E α , 1 ⁢ ( − λ n ⁢ T 2 α ) ) ⁢ g 1 , n E α , 1 ⁢ ( − λ n ⁢ T 2 α ) − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ) 2 ≤ ( ∑ n = 1 + ∞ ( 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ) γ + 2 ⁢ ( g 2 , n − 1 − E α , 1 ⁢ ( − λ n ⁢ T 2 α ) 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ⁢ g 1 , n ) 2 ) 2 γ + 2 × ( ∑ n = 1 + ∞ ( g 2 , n − 1 − E α , 1 ⁢ ( − λ n ⁢ T 2 α ) 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ⁢ g 1 , n ) 2 ) γ γ + 2 ≤ ( ∑ n = 1 + ∞ ( 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ) γ ⁢ φ n 2 ) 2 γ + 2 ⁢ ( ( ∑ n = 1 + ∞ ( g 2 , n ) 2 ) 1 2 + ( ∑ n = 1 + ∞ ( 1 − E α , 1 ⁢ ( − λ n ⁢ T 2 α ) 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ⁢ g 1 , n ) 2 ) 1 2 ) 2 ⁢ γ γ + 2 .$

In addition, from Lemma 4, there exists a constant $C > 0$ $C > 0$ such that

\begin{matrix} {(+ \infty \sum n = 1 {(\frac{1 - E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α}) - E_{α, 1} (- λ_{n} T_{1}^{α})})}^{γ} φ_{n}^{2})}^{\frac{2}{γ + 2}} & \leq {(+ \infty \sum n = 1 {(\frac{Γ (1 - α) (1 + λ_{n} T_{2}^{α})}{C_{1} (1 - \frac{E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α})})})}^{γ} φ_{n}^{2})}^{\frac{2}{γ + 2}} \leq {(+ \infty \sum n = 1 {(\frac{Γ (1 - α) (C + T_{2}^{α})}{C_{1} (1 - \frac{E_{α, 1} (- λ_{n} T_{1}^{α})}{E_{α, 1} (- λ_{n} T_{2}^{α})})})}^{γ} λ_{n}^{γ} φ_{n}^{2})}^{\frac{2}{γ + 2}} . \end{matrix}

$( ∑ n = 1 + ∞ ( 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) ) γ ⁢ φ n 2 ) 2 γ + 2 ≤ ( ∑ n = 1 + ∞ ( Γ ⁢ ( 1 − α ) ⁢ ( 1 + λ n ⁢ T 2 α ) C 1 ⁢ ( 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) ) ) γ ⁢ φ n 2 ) 2 γ + 2 ≤ ( ∑ n = 1 + ∞ ( Γ ⁢ ( 1 − α ) ⁢ ( C + T 2 α ) C 1 ⁢ ( 1 − E α , 1 ⁢ ( − λ n ⁢ T 1 α ) E α , 1 ⁢ ( − λ n ⁢ T 2 α ) ) ) γ ⁢ λ n γ ⁢ φ n 2 ) 2 γ + 2 .$

Denoting

C_{4} := {(\frac{Γ (1 - α) (C + T_{2}^{α})}{C_{1} (\frac{E_{α, 1} (- λ_{1} T_{1}^{α})}{E_{α, 1} (- λ_{1} T_{2}^{α})} - 1)})}^{\frac{2 γ}{γ + 2}},

$C 4 := ( Γ ⁢ ( 1 − α ) ⁢ ( C + T 2 α ) C 1 ⁢ ( E α , 1 ⁢ ( − λ 1 ⁢ T 1 α ) E α , 1 ⁢ ( − λ 1 ⁢ T 2 α ) − 1 ) ) 2 ⁢ γ γ + 2 ,$

by the inequality $λ_{n}^{γ} \leq exp (λ_{n}^{γ})$ $λ n γ ≤ exp ⁡ ( λ n γ )$ , we have the estimate

{∥ φ ∥}^{2} \leq C_{4} M^{\frac{4}{γ + 2}} {(∥ g_{2} ∥ + \frac{1}{1 - E_{α, 1} (- λ_{1} T_{1}^{α})} ∥ g_{1} ∥)}^{\frac{2 γ}{γ + 2}} .

$∥ φ ∥ 2 ≤ C 4 ⁢ M 4 γ + 2 ⁢ ( ∥ g 2 ∥ + 1 1 − E α , 1 ⁢ ( − λ 1 ⁢ T 1 α ) ⁢ ∥ g 1 ∥ ) 2 ⁢ γ γ + 2 .$

The proof is completed. ∎

Proof of Theorem 2

By the triangular inequality, we have

(A.1)

∥ ∥ f_{β, γ}^{δ} - f ∥ ∥ \leq ∥ ∥ f_{β, γ}^{δ} - f_{β, γ} ∥ ∥ + ∥ ∥ f_{β, γ} - f ∥ ∥,

$∥ f β , γ δ − f ∥ ≤ ∥ f β , γ δ − f β , γ ∥ + ∥ f β , γ − f ∥ ,$

(A.2)

∥ ∥ φ_{β, γ}^{δ} - φ ∥ ∥ \leq ∥ ∥ φ_{β, γ}^{δ} - φ_{β, γ} ∥ ∥ + ∥ ∥ φ_{β, γ} - φ ∥ ∥ .

$∥ φ β , γ δ − φ ∥ ≤ ∥ φ β , γ δ − φ β , γ ∥ + ∥ φ β , γ − φ ∥ .$

We first estimate $∥ ∥ f_{β, γ}^{δ} - f ∥ ∥$ $∥ f β , γ δ − f ∥$ . For the first term on the right-hand side of inequality (A.1), there exists a constant $C > 0$ $C > 0$ such that

(A.3)

\begin{matrix} {∥ ∥ f_{β, γ}^{δ} - f_{β, γ} ∥ ∥}^{2} & \leq {∥ ∥ ∥ ∥ + \infty \sum n = 1 (\frac{β exp (λ_{n}^{γ}) m_{1, n}^{1}}{β exp (λ_{n}^{γ}) {(m_{1, n}^{1})}^{2} + β^{2} exp (2 λ_{n}^{γ})}) (g_{1, n}^{δ} - g_{1, n}) ∥ ∥ ∥ ∥}^{2} + {∥ ∥ ∥ ∥ + \infty \sum n = 1 (\frac{β exp (λ_{n}^{γ}) m_{1, n}^{2}}{β exp (λ_{n}^{γ}) {(m_{1, n}^{2})}^{2} + β^{2} exp (2 λ_{n}^{γ})}) (g_{2, n}^{δ} - g_{2, n}) ∥ ∥ ∥ ∥}^{2} \leq sup n {∣ ∣ ∣ \frac{λ_{n}}{C^{2} + β λ_{n}^{γ + 2}} ∣ ∣ ∣}^{2} (\infty \sum n = 1 {| g_{1, n}^{δ} - g_{1, n} |}^{2} + \infty \sum n = 1 {| g_{2, n}^{δ} - g_{2, n} |}^{2}) \leq {∣ ∣ ∣ \frac{{(1 + γ)}^{\frac{1 + γ}{2 + γ}}}{(2 + γ) C^{\frac{2 (1 + γ)}{2 + γ}}} β^{- \frac{1}{2 + γ}} ∣ ∣ ∣}^{2} (2 δ^{2}) . \end{matrix}

$∥ f β , γ δ − f β , γ ∥ 2 ≤ ∥ ∑ n = 1 + ∞ ( β ⁢ exp ⁡ ( λ n γ ) ⁢ m 1 , n 1 β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 1 ) 2 + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ) ⁢ ( g 1 , n δ − g 1 , n ) ∥ 2 + ∥ ∑ n = 1 + ∞ ( β ⁢ exp ⁡ ( λ n γ ) ⁢ m 1 , n 2 β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 2 ) 2 + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ) ⁢ ( g 2 , n δ − g 2 , n ) ∥ 2 ≤ sup n | λ n C 2 + β ⁢ λ n γ + 2 | 2 ⁢ ( ∑ n = 1 ∞ | g 1 , n δ − g 1 , n | 2 + ∑ n = 1 ∞ | g 2 , n δ − g 2 , n | 2 ) ≤ | ( 1 + γ ) 1 + γ 2 + γ ( 2 + γ ) ⁢ C 2 ⁢ ( 1 + γ ) 2 + γ ⁢ β − 1 2 + γ | 2 ⁢ ( 2 ⁢ δ 2 ) .$

Denote

$C 6 := 2 ⁢ ( 1 + γ ) 1 + γ 2 + γ ( 2 + γ ) ⁢ C 2 ⁢ ( 1 + γ ) 2 + γ .$

From inequality (A.3), we obtain

(A.4)

$∥ f β , γ δ − f β , γ ∥ ≤ C 6 ⁢ δ β 1 2 + γ .$

On the other hand, we get

$exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 ) + ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 β ⁢ exp ⁡ ( λ n γ ) + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ≥ ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 m 1 , n 1 ⁢ m 2 , n 1 + m 1 , n 2 ⁢ m 2 , n 2 .$

Lemma 3 yields

$lim n → + ∞ λ n 2 ⁢ ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 m 1 , n 1 ⁢ m 2 , n 1 + m 1 , n 2 ⁢ m 2 , n 2 = ( T 1 α − T 2 α ) 2 Γ ⁢ ( 1 − α ) ⁢ T 1 α ⁢ T 2 α ⁢ ( T 1 α + T 2 α ) .$

Therefore, there exist a constant

$C 5 ∈ ( 0 , ( ( T 1 α − T 2 α ) 2 Γ ⁢ ( 1 − α ) ⁢ T 1 α ⁢ T 2 α ⁢ ( T 1 α + T 2 α ) ) 1 2 )$

satisfying

(A.5)

$( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 m 1 , n 1 ⁢ m 2 , n 1 + m 1 , n 2 ⁢ m 2 , n 2 ≥ ( C 5 λ n ) 2 ,$

with constant $C 5$ depending on parameters $T 1 , T 2$ and 𝛼. We now estimate the term $∥ f β , γ − f ∥$ . We have

$∥ f β , γ − f ∥ 2 ≤ ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ f n ⁢ X n ⁢ ( x ) β ⁢ exp ⁡ ( λ n γ ) + C 5 2 λ n 2 ∥ 2 + ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ φ n ⁢ X n ⁢ ( x ) β ⁢ exp ⁡ ( λ n γ ) + C 5 2 λ n 2 ∥ 2 ≤ ∑ n = 1 + ∞ β 2 ⁢ exp ⁡ ( λ n γ ) ( β ⁢ exp ⁡ ( λ n γ ) + C 5 2 λ n 2 ) 2 ⁢ exp ⁡ ( λ n γ ) ⁢ | f n | 2 + ∑ n = 1 + ∞ β 2 ⁢ exp ⁡ ( λ n γ ) ( β ⁢ exp ⁡ ( λ n γ ) + C 5 2 λ n 2 ) 2 ⁢ exp ⁡ ( λ n γ ) ⁢ | φ n | 2 ≤ sup n ( β ⁢ exp ⁡ ( λ n γ 2 ) β ⁢ exp ⁡ ( λ n γ ) + C 5 2 λ n 2 ) 2 ⁢ ∑ n = 1 + ∞ exp ⁡ ( λ n γ ) ⁢ | f n | 2 + sup n ( β ⁢ exp ⁡ ( λ n γ 2 ) β ⁢ exp ⁡ ( λ n γ ) + C 5 2 λ n 2 ) 2 ⁢ ∑ n = 1 + ∞ exp ⁡ ( λ n γ ) ⁢ | φ n | 2 .$

From Lemma 5 and $0 < β < 1$ , we get

$∥ f β , γ − f ∥ ≤ C 7 ⁢ M ⁢ β 1 4 , where ⁢ C 7 = max ⁡ { 3 3 4 4 ⁢ ( 2 C 5 ) , λ N 0 2 ⁢ exp ⁡ ( λ N 0 2 2 ) ⁢ ( 2 C 5 2 ) } .$

This inequality, combined with estimate (A.4), yields the estimate

$∥ f β , γ δ − f ∥ ≤ C 6 ⁢ δ β 1 2 + γ + C 7 ⁢ M ⁢ β 1 4 .$

By choosing the regularization parameter $β = δ 8 + 4 ⁢ γ 6 + γ$ , we get

$∥ f β , γ δ − f ∥ ≤ ( C 6 + C 7 ⁢ M ) ⁢ δ 2 + γ 6 + γ .$

Similarly, when $0 < β < 1$ , $γ > 0$ , there exists $C 8$ , and selecting the regularization parameter $β = δ 8 + 4 ⁢ γ 6 + γ$ , inequality (A.2) yields

$∥ φ β , γ δ − φ ∥ ≤ ( C 6 + C 8 ⁢ M ) ⁢ δ 2 + γ 6 + γ .$

The proof is completed. ∎

Proof of Theorem 3

For $τ ⁢ δ$ defined in (4.7), there is the following inequality estimation:

$τ ⁢ δ = ∥ K 1 , 1 ⁢ ( f β δ ) + K 2 , 1 ⁢ ( φ β δ ) − g 1 δ ∥ + ∥ K 1 , 2 ⁢ ( f β δ ) + K 2 , 2 ⁢ ( φ β δ ) − g 2 δ ∥$

$= ∥ ∑ n = 1 + ∞ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 2 ) 2 + ( m 2 , n 2 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 1 , n δ + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 2 ⁢ m 1 , n 1 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 2 , n δ ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ⁢ X n ⁢ ( x ) ∥$

$+ ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 1 ⁢ m 1 , n 2 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 1 , n δ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 2 , n δ ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ⁢ X n ⁢ ( x ) ∥$

$≤ 4 ⁢ δ + ∑ i = 1 2 ( ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ g i , n ⁢ X n ⁢ ( x ) β ⁢ exp ⁡ ( λ n γ ) + ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ∥ + ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ g i , n ⁢ X n ⁢ ( x ) β ⁢ exp ⁡ ( λ n γ ) + ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 m 1 , n 1 ⁢ m 2 , n 1 + m 1 , n 2 ⁢ m 2 , n 2 ∥ ) .$

For

$C 5$ defined in (A.5), we obtain

$( τ − 4 ) 2 ⁢ δ 2 ≤ 2 ⁢ ( ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ m 1 , n 1 ⁢ f n ⁢ X n ⁢ ( x ) β ⁢ exp ⁡ ( λ n γ ) + ( C 5 λ n ) 2 ∥ 2 + ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ m 2 , n 1 ⁢ φ n ⁢ X n ⁢ ( x ) β ⁢ exp ⁡ ( λ n γ ) + ( C 5 λ n ) 2 ∥ 2 ) + 2 ⁢ ( ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ m 1 , n 2 ⁢ f n ⁢ X n ⁢ ( x ) β ⁢ exp ⁡ ( λ n γ ) + ( C 5 λ n ) 2 ∥ 2 + ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ m 2 , n 2 ⁢ φ n ⁢ X n ⁢ ( x ) β ⁢ exp ⁡ ( λ n γ ) + ( C 5 λ n ) 2 ∥ 2 ) ≤ 4 ⁢ ( ( sup n β ⁢ exp ⁡ ( 1 ) ⁢ λ n 1 − γ 2 β ⁢ exp ⁡ ( 0 ) ⁢ λ n 2 + ( C 5 ) 2 ) 2 ⁢ ∑ n = 1 + ∞ λ n γ ⁢ | f n | 2 + ( sup n β ⁢ exp ⁡ ( 1 ) ⁢ λ n 2 − γ 2 β ⁢ exp ⁡ ( 0 ) ⁢ λ n 2 + ( C 5 ) 2 ) 2 ⁢ ∑ n = 1 + ∞ λ n γ ⁢ | φ n | 2 ) .$

Considering different ranges of 𝛾, we have the following estimates:

$sup n β ⁢ exp ⁡ ( 1 ) ⁢ λ n 1 − γ 2 β ⁢ λ n 2 + ( C 5 ) 2 ≤ { exp ⁡ ( 1 ) ⁢ ( 2 − γ ) 2 − γ 4 ⁢ ( 2 + γ ) 2 + γ 4 4 ⁢ ( C 5 ) 2 + γ 4 ⁢ β 2 + γ 4 , 0 < γ < 2 , exp ⁡ ( 1 ) ⁢ λ 1 1 − γ 2 ( C 5 ) 2 ⁢ β , γ ≥ 2 ,$

$sup n β ⁢ exp ⁡ ( 1 ) ⁢ λ n 2 − γ 2 β ⁢ λ n 2 + ( C 5 ) 2 ≤ { exp ⁡ ( 1 ) ⁢ γ γ 4 ⁢ ( 4 − γ ) 4 − γ 4 4 ⁢ ( C 5 ) γ 2 ⁢ β γ 4 , 0 < γ < 4 , exp ⁡ ( 1 ) ⁢ λ 1 2 − γ 2 ( C 5 ) 2 ⁢ β , γ ≥ 4 .$

From

$λ n γ ≤ exp ⁡ ( λ n γ )$ and denoting

$E 1 = ( ( exp ⁡ ( 1 ) ⁢ ( 2 − γ ) 2 − γ 4 ⁢ ( 2 + γ ) 2 + γ 4 2 ⁢ ( C 5 ) 3 ⁢ γ + 6 4 ) 2 + ( exp ⁡ ( 1 ) ⁢ γ γ 4 ⁢ ( 4 − γ ) 4 − γ 4 2 ⁢ ( C 5 ) γ ) 2 ) 1 2 , E 2 = ( ( 2 ⁢ exp ⁡ ( 1 ) ⁢ λ 1 1 − γ 2 ( C 5 ) 4 ) 2 + ( exp ⁡ ( 1 ) ⁢ γ γ 4 ⁢ ( 4 − γ ) 4 − γ 4 2 ⁢ ( C 5 ) γ ) 2 ) 1 2 , E 3 = ( ( 2 ⁢ exp ⁡ ( 1 ) ⁢ λ 1 1 − γ 2 ( C 5 ) 4 ) 2 + ( 2 ⁢ exp ⁡ ( 1 ) ⁢ λ 1 2 − γ 2 ( C 5 ) 4 ) 2 ) 1 2 ,$

we get

$( τ − 4 ) ⁢ δ ≤ { E 1 ⁢ β 2 + γ 4 ⁢ M , 0 < γ < 2 , E 2 ⁢ β ⁢ M , 2 ≤ γ < 4 , E 3 ⁢ β ⁢ M , γ ≥ 4 .$

Therefore, denoting $C 9 = max ⁡ { E 1 , E 2 , E 3 }$ , we have the estimate

$1 β 1 2 + γ ≤ { ( C 9 ⁢ M τ − 4 ) 4 ( 2 + γ ) 2 ⁢ δ − 4 ( 2 + γ ) 2 , 0 < γ < 2 , ( C 9 ⁢ M τ − 4 ) 1 2 + γ ⁢ δ − 1 2 + γ , γ ≥ 2 .$

Then

$∥ f β , γ δ − f β , γ ∥ ≤ { C 6 ⁢ ( C 9 ⁢ M τ − 4 ) 4 ( 2 + γ ) 2 ⁢ δ γ 2 + γ , 0 < γ < 2 , C 6 ⁢ ( C 9 ⁢ M τ − 4 ) 1 2 + γ ⁢ δ 1 + γ 2 + γ , γ ≥ 2 .$

Similarly, the estimate for initial values is

$∥ φ β , γ δ − φ β , γ ∥ ≤ { C 6 ⁢ ( C 9 ⁢ M τ − 4 ) 4 ( 2 + γ ) 2 ⁢ δ γ 2 + γ , 0 < γ < 2 , C 6 ⁢ ( C 9 ⁢ M τ − 4 ) 1 2 + γ ⁢ δ 1 + γ 2 + γ , γ ≥ 2 .$

Furthermore,

$∥ f β , γ − f ∥ Exp 2 ≤ ∑ n = 1 + ∞ ( β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ) 2 ⁢ exp ⁡ ( λ n γ ) ⁢ | f n | 2 + ∑ n = 1 + ∞ ( β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 1 ⁢ m 2 , n 1 + m 1 , n 2 ⁢ m 2 , n 2 ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ) 2 ⁢ exp ⁡ ( λ n γ ) ⁢ | φ n | 2 ≤ ∑ n = 1 + ∞ exp ⁡ ( λ n γ ) ⁢ | f n | 2 + ∑ n = 1 + ∞ exp ⁡ ( λ n γ ) ⁢ | φ n | 2 ≤ 2 ⁢ ( M ) 2 .$

Similarly, we have the same result on $∥ φ β , γ − φ ∥ Exp 2$ . On the other hand, let

$F i ⁢ ( n ) = m 1 , n i ⁢ m 2 , n i m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 , i = 1 , 2 .$

From noise assumptions (1.2), there exists $C 10$ such that

$∥ K 1 ⁢ ( f β , γ − f , φ β , γ − φ ) ∥ ≤ 4 ⁢ C 10 ⁢ δ + ∥ ∑ n = 1 + ∞ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 2 ) 2 + ( m 2 , n 2 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 1 , n δ + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 2 ⁢ m 1 , n 1 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 2 , n δ ⁢ X n ⁢ ( x ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ∥ ⁢ lim n → ∞ | F 1 ⁢ ( n ) ⁢ m 2 , n 2 m 2 , n 1 | + ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 1 ⁢ m 1 , n 2 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 1 , n δ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 2 , n δ ⁢ X n ⁢ ( x ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ∥ ⁢ lim n → ∞ | F 1 ⁢ ( n ) | + ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 2 ) 2 + ( m 2 , n 2 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 1 , n δ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 2 ⁢ m 1 , n 1 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 2 , n δ ⁢ X n ⁢ ( x ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ∥ ⁢ lim n → ∞ | F 2 ⁢ ( n ) ⁢ m 2 , n 1 m 2 , n 2 | + ∥ ∑ n = 1 + ∞ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 1 ⁢ m 1 , n 2 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 1 , n δ + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 2 , n δ ⁢ X n ⁢ ( x ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ∥ ⁢ lim n → ∞ | F 1 ⁢ ( n ) | .$

From Lemma 3, we get

Therefore, by the Morozov discrepancy principle (4.7), we get

$∥ K 1 ⁢ ( f β , γ − f , φ β , γ − φ ) ∥ ≤ ( 4 ⁢ C 10 + 2 ⁢ τ ⁢ T 1 α ⁢ T 2 α ⁢ Γ ⁢ ( 1 − α ) T 1 α − T 2 α ) ⁢ δ .$

Similarly, for the operator $K 2$ , there exists $C 11$ such that

$∥ K 2 ⁢ ( f β , γ − f , φ β , γ − φ ) ∥ ≤ 4 ⁢ C 11 ⁢ δ + ∥ ∑ n = 1 + ∞ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 2 ) 2 + ( m 2 , n 2 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 1 , n δ + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 2 ⁢ m 1 , n 1 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 2 , n δ ⁢ X n ⁢ ( x ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ∥ ⁢ lim n → ∞ | F 2 ⁢ ( n ) | + ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 1 ⁢ m 1 , n 2 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 1 , n δ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 2 , n δ ⁢ X n ⁢ ( x ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ∥ ⁢ lim n → ∞ | F 2 ⁢ ( n ) ⁢ m 1 , n 1 m 1 , n 2 | + ∥ ∑ n = 1 + ∞ β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 2 ) 2 + ( m 2 , n 2 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 1 , n δ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 2 ⁢ m 1 , n 1 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 2 , n δ ⁢ X n ⁢ ( x ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ∥ ⁢ lim n → ∞ | F 2 ⁢ ( n ) | + ∥ ∑ n = 1 + ∞ − β ⁢ exp ⁡ ( λ n γ ) ⁢ ( m 1 , n 1 ⁢ m 1 , n 2 + m 2 , n 1 ⁢ m 2 , n 2 ) ⁢ g 1 , n δ + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ) ⁢ g 2 , n δ ⁢ X n ⁢ ( x ) ( m 1 , n 1 ⁢ m 2 , n 2 − m 1 , n 2 ⁢ m 2 , n 1 ) 2 + β ⁢ exp ⁡ ( λ n γ ) ⁢ ( ( m 1 , n 1 ) 2 + ( m 1 , n 2 ) 2 + ( m 2 , n 1 ) 2 + ( m 2 , n 2 ) 2 ) + β 2 ⁢ exp ⁡ ( 2 ⁢ λ n γ ) ∥ ⁢ lim n → ∞ | F 2 ⁢ ( n ) ⁢ m 2 , n 1 m 2 , n 2 | .$

Therefore, we have

$∥ K 2 ⁢ ( f β , γ − f , φ β , γ − φ ) ∥ ≤ ( 4 ⁢ C 11 + 2 ⁢ τ ⁢ T 1 α ⁢ T 2 α ⁢ Γ ⁢ ( 1 − α ) T 1 α − T 2 α ) ⁢ δ .$

From inequality (3.5) in Theorem 1, we get

$∥ f β , γ − f ∥ ≤ ( C 1 ) γ γ + 2 ⁢ ∥ f β , γ − f ∥ Exp 2 γ + 2 ⁢ ( ∥ K 1 ⁢ ( f β , γ − f , φ β , γ − φ ) ∥ + ( T 2 T 1 ) α ⁢ ∥ K 2 ⁢ ( f β , γ − f , φ β , γ − φ ) ∥ ) γ γ + 2 ≤ C 12 ⁢ M 1 γ + 2 ⁢ δ γ γ + 2 ,$

where

$C 12 = ( 2 1 γ ⁢ C 1 ⁢ ( 4 ⁢ ( C 10 + T 2 α T 1 α ⁢ C 11 ) + 2 ⁢ τ ⁢ Γ ⁢ ( 1 − α ) ⁢ ( T 1 α + T 2 α ) ⁢ T 2 α T 1 α − T 2 α ) ) γ γ + 2 .$

And (3.6) in Theorem 1 yields

$∥ φ β , γ − φ ∥ ≤ ( C 2 ) γ γ + 2 ⁢ ∥ φ β , γ − φ ∥ Exp 2 γ + 2 ⁢ ( ∥ K 2 ⁢ ( f β , γ − f , φ β , γ − φ ) ∥ + 1 1 − E α , 1 ⁢ ( − λ 1 ⁢ T 1 α ) ⁢ ∥ K 1 ⁢ ( f β , γ − f , φ β , γ − φ ) ∥ ) γ γ + 2 ≤ C 13 ⁢ M 1 γ + 2 ⁢ δ γ γ + 2 ,$

where

$C 13 = ( 2 1 γ ⁢ C 2 ⁢ ( 4 ⁢ ( C 11 + 1 1 − E α , 1 ⁢ ( − λ 1 ⁢ T 1 α ) ⁢ C 10 ) + 2 − E α , 1 ⁢ ( − λ 1 ⁢ T 1 α ) 1 − E α , 1 ⁢ ( − λ 1 ⁢ T 1 α ) ⁢ 2 ⁢ τ ⁢ T 1 α ⁢ T 2 α ⁢ Γ ⁢ ( 1 − α ) T 1 α − T 2 α ) ) γ γ + 2 .$

Therefore, denoting

$C 14 = C 6 ⁢ ( C 9 ⁢ M τ − 4 ) 4 ( 2 + γ ) 2 and C 15 = C 6 ⁢ ( C 9 ⁢ M τ − 4 ) 1 2 + γ ,$

we have the conclusions

$∥ f β , γ δ − f ∥ ≤ { ( C 14 + C 12 ⁢ M 1 γ + 2 ) ⁢ δ γ 2 + γ , 0 < γ < 2 , ( C 15 + C 12 ⁢ M 1 γ + 2 ) ⁢ δ 1 + γ 2 + γ , γ ≥ 2 , ∥ φ β , γ δ − φ ∥ ≤ { ( C 14 + C 13 ⁢ M 1 γ + 2 ) ⁢ δ γ 2 + γ , 0 < γ < 2 , ( C 15 + C 13 ⁢ M 1 γ + 2 ) ⁢ δ 1 + γ 2 + γ , γ ≥ 2 .$

The proof is completed. ∎

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Received: 2022-03-10

Revised: 2022-12-05

Accepted: 2022-12-28

Published Online: 2023-02-08

Published in Print: 2023-07-01

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DOI: doi.org/10.1515/cmam-2022-0058

Keywords for this article

Time-Fractional Diffusion Equation; Simultaneous Inversion; Source Term; Initial Value; Exponential Tikhonov Regularization Method