# Inverse Source Problem for Sobolev Equation with Fractional Laplacian

Authors: Nguyen Duc Phuong; Van Tien Nguyen; Le Dinh Long
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, we are interested in the problem of determining the source function for the Sobolev equation with fractional Laplacian. This problem is ill-posed in the sense of Hadamard. In order to edit the instability instability of the solution, we applied the fractional Landweber method. In the theoretical analysis results, we show the error estimate between the exact solution and the regularized solution by using an a priori regularization parameter choice rule and an a posteriori regularization parameter choice rule. Finally, we investigate the convergence of the source function when fractional orderβ⟶1+.


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## Body

## 1. Introduction

LetΩ be a bounded domain in ℝNN≥1 with sufficiently smooth boundary ∂Ω. In this paper, we are interested to study the following pseudo-parabolic equation
(1)ut−aΔut+−Δβu=Fx,t,inΩ×0,T,u∂Ω=0,inΩ,where a>0 is the diffusion coefficient, F is the source function, and u desribe the distribution of the temperature at position x and time t. The paramater β is the fractional order of Laplacian operator with β≥1.Pseudo-parabolic equations or calledSobolev equation describe describing various important physical phenomena, such as heat conduction involving two temperatures [1], homogeneous liquid permeability in fractured rock [2], unidirectional propagation of long waves in a nonlinearly dispersed medium [3], and its references.Until now, the results on fractional pseudo-parabolic equations equation are not rich we can mention them in a few some few papers, for example, [1, 4–6]. From the fraction operator −Δβ appearing in the main equation which is nonlocal, many scientists believe that it describes some physical phenomena more accurately than classical integrals differential equation. Properties of fractional operator −Δβ have been described in detail in [1].For equation (1) we usually divide it into three forms.
(i)
The first type is an initial value problem, i.e., determiningu when the initial value ux,0=u0x and the source function F is known. The results in this category are vibrant and plentiful ([7, 8])(ii)
The second type is terminal value problem, i.e., recovering the functionu from the terminal value data ux,T=uTx and the source function data F. To the best of our knowledge, there are limited results for the terminal value problem. We can list some recent papers, for example, [9–13]. In general, the terminal value problem is an ill-posed problem; namely, a solution does do not exist, and if a solution exists, it does not depend continuously on the data. The results of the regularized method for this form were recently investigated by [14, 15](iii)
The last type is inverse source problem, i.e., recovering the source functionF if we know the initial value data ux,0=u0x and the terminal data ux,T=uTxThe main purpose of this paper is to determine the source functionF=ψtfx with the split form when we know that
(2)ux,T=gx,ux,0=0,x∈Ω.The question of determining the functionf when we know ψ and g will be studied carefully in this paper. It is surprising that the problem of determining the source function for the pseudo-parabolic equation has not been investigated before. We detail the objective of the problem. In practice, the given dataψδ,gδis noisy by the observed data ψ,g by level δ>0 such that
(3)ψδ−ψL∞0,T+gδ−gL2Ω≤δ.Our main task here is to construct a regularized method which looking for the functionfδ and claims claim that
(4)lim∥fδ−f∥=0,whenδ⟶0+,in the appropriate norm. It can be claimed that our paper was one of the first works on the inverse source problem for the Sobolev equation.In [7], Tuan-Long-Thinh used the Tikhonov regularization method to regularize regularized an inverse source problem for time fractional diffusion equation. They also introduced two methods, a priori and a posteriori parameter choice rules, to obtain the convergence estimate of the regularized methods. In [16], the authors studied the problem of finding the source distribution for the linear biparabolic equation when we have the final observation. Ma et al. [17] identified the unknown space-dependent source term in a time-fractional diffusion equation by applying the generalized and revised generalized Tikhonov regularization methods. There are many different regularized methods, and in this paper, we choose the fractional Landweber regularization method. The Landweber regularization method was first derived from [18] where the authors applied the filter regularization technique for solving a linear inverse problem. Up to now, the Landweber regularization method has been applied to solve many inverse problems, for example, [19–21] and references therein. This method is beneficial very useful for investigating for the linear ill-posed equation. Recently, Binh et al. [22] studied an inverse source problem for the Rayleigh–Stokes problem using the Tikhonov method.For the reader’s convenience, we would like to outline the main results and novelties of the paper briefly:(i)
The first goal of this paper is to provide the fractional Landweber method to solve this inverse space-dependent source problem for pseudo-parabolic equation. We give the ill-posedness of our inverse source problem and introduce the convergence rate of the fractional Landweber regularized solution. In addition, we obtain the convergence rate by using an apriori parameter choice rule and an a posteriori parameter choice rule. Looking back at the articles [19–21], we realize that the source functions in these papers do not depend on the time function. So, the computation is not complicated. Meanwhile, the source function of the current paper depends on the function ψ which makes the calculation more cumbersome. The presence of (3) makes our problem more clearly complex complex than [19–21]. One point to note is that the method in the article [23] can be applied to our model, but we approach it differently, in a different way.(ii)
The second interesting point in the paper is the investigation of the convergence of the source function when the order of derivative approaches 1. Comparing the difference between the source function of equation (1) with β>1 and the classical pseudo-parabolic equation β=1 will help us understand more information about problem (1).The paper is organized as follows. Section2 states some preliminary theoretical knowledge. In Section 3, we give the Fourier formula of the source function and also present the ill-posedness of our problem. The conditional stability for the source function source function is also discussed in the same section. Section 4 provides the fractional Landweber regularization method and states a convergence estimate under a priori assumption on the exact solution. The posteriori parameter choice rule is also shown in section 4. Finally, in Section 5, we prove the convergence of the source function in Hilbert scales space with the appropriate assumption of ψ and g.


## 2. Preliminary Results

Let us consider the operatorA=−Δ on V≔H01Ω∩H2Ω, and assume that the operator A has the eigenvalues λj such that 0<λ1≤λ2≤⋯≤λj≤⋯ which approach ∞ as j goes to ∞, The corresponding eigenfunctions are denoted by ej∈V. Now, let us define fractional powers of A and its domain. For all s≥0, we define by As the following operator:
(5)Asv≔∑j=1∞v,ejλjsej,v∈DAs=v∈L2Ω:∑j=1∞v,ej2λj2s<∞.The domainℍsΩ=DAs is the Banach space equipped with the norm
(6)vDAs≔∑j=1∞v,ej2λj2s1/2,v∈DAs.We introduce the following two lemmas, which are useful and helpful in the next proofs.Lemma 1.
Letψ:0,T⟶ℝ such that ψ0≤ψt≤ψ1 where ψ0 and ψ1 are positive numbers. Let us assume that β≥1. Then, the following estimates are true:
(7)∫0Texp−T−sλjβ1+aλj−1ψsds≤ψ11+aλjλjβ,1+aλjλjβ1−exp−Tλ1β1+aλ1−1ψ0≤∫0Texp−T−sλjβ1+aλj−1ψsds.Proof.
Sinceψt≤ψ1, we infer that
(8)∫0Texp−T−sλjβ1+aλj−1ψsds≤ψ1∫0Texp−T−sλjβ1+aλj−1ds=ψ11+aλjλjβ.Sinceψt≥ψ0>0, we infer that
(9)∫0Texp−T−sλjβ1+aλj−1ψsds≥ψ0∫0Texp−T−sλjβ1+aλj−1ds=1+aλjλjβ1−exp−Tλjβ1+aλj−1ψ0·Let us consider the following function:(10)Φz=zβ1+az,z>0.The derivative of it is equal to(11)Φ′z=aβzβ−1+βzβ−zβ1+az2>0.This implies thatΦ is an increasing function on 0,+∞. Therefore, we get that
(12)λjβ1+aλj−1≥λ1β1+aλ1−1.It follows from (9) that
(13)∫0Texp−T−sλjβ1+aλj−1ψsds≥1+aλjλjβ1−exp−Tλ1β1+aλ1−1ψ0.The proof of the Lemma 1 is completed.Lemma 2.
Letψ0,ψ1 be positive constants such that ψ0<ψ<ψ1. By choosing δ∈0,ψ1/4, and Bψ0,ψ1=ψ1+ψ0/4, we obtain
(14)4−1ψ0≤ψδt≤Bψ0,ψ1.Proof.
The proof is completed in [26], page 4.


## 3. Inverse Source Problem: Explicit Form and Ill-Posedness

Let us first give the explicit of Fourier form of the mild solution to problems (1) and (2). First, taking the inner product of both sides of (1) with ejx, we find that
(15)ddt∫Ωux,tejxdx+aλj∫Ωux,tejxdx+λjβ∫Ωux,tejxdx=∫ΩFx,tejxdx,and from the initial condition ux,0=0, we have that
(16)∫Ωux,Tejxdx=∫0Texp−T−sλjβ1+aλj−1∫ΩFx,sejxdxds,since Fx,s=ψsfx; we know that
(17)∫Ωfxejxdx=∫Ωgxejxdx∫0Texp−T−sλjβ1+aλj−1ψsds·Hence, the source function is defined as follows:(18)fx=∑j=1∞∫Ωgxejxdx∫0Texp−T−sλjβ1+aλj−1ψsdsejx.Let us prove the ill-posedness of inverse source problems (1) and (2). Logically, we will consider the source function problem as the problem of finding f satisfying (18). From now on, we only treat the source term (18).Theorem 3.
The problem of determiningf that satisfies (18) is ill-posed in the sense of Hadamard.Proof.
We defined a linear operatorY:L2Ω⟶L2Ω as follows:
(19)Yfx=∑j=1∞∫0Texp−T−sλjβ1+aλj−1ψsds<f,ej>ejx=∫Ωkx,ξfξdξ.Due tokx,ξ=kξ,x, we know Y is a self-adjoint operator. Next, its compactness is explained as follows. Let us define the finite rank operators YN as follows:
(20)YNfx=∑j=1N∫0Texp−T−sλjβ1+aλj−1ψsds<f,ej>ejx.By some simple calculations and using Lemma1, we have
(21)∥YNf−Yf∥L2Ω2≤ψ12λ1−1+a2⏟A12∑j=N+1+∞<f,ej>2λj2β−2.From (21), we have
(22)YNf−Yf∥L2Ω2≤A12λN2β−2f∥L2Ω2⟶0inLL2Ω;L2ΩasN⟶∞.Therefore,Y is a compact operator. The SVDs for the linear self-adjoint compact operator Y are
(23)Y=∫0Texp−T−sλjβ1+aλj−1ψsds,and corresponding eigenvectors are ej which is an orthonormal basis in L2Ω. Therefore, the inverse source problem we introduced above can be formulated as an operator equation Yfx=gx where by gx is the numerator in formula (18), and by Kirsch, we can conclude that it is ill-posed. The final time data gi=λiei, by (18), the source term corresponding to gi is
(24)fix=∑j=1∞∫0Texp−T−sλjβ1+aλj−1ψsds−1∫Ωgxejxdxejx≥1λ1+a−11−exp−Tλiβ1+aλi−1−1ψ0−1⏟A2λiβ,whereby A1 is defined in formula (21). The input final data g=0, by (18), the source term corresponding to g is f=0. We have error in L2Ω norm between gm and g(25)limi⟶+∞∥gi−g∥L2Ω=limi⟶+∞λi−1=0.Then, the error inL2 norm between fi and f is estimated as follows:
(26)fi−fL2Ω≥λiA2⟶limi⟶+∞fi−fL2Ω≥+∞.From (25) and (26), we deduce that the solution to problem (1) is unstable in L2Ω.Next, we consider stability of the inverse source problem.Theorem 4.
Iff∈DAssuch that(27)fDAs≤E,s=kβ−12≥0,then we get
(28)fL2Ω≤C1−k/k+2E2/k+2gL2Ωk/k+2,where C1=a1−exp−Tλ1β1+aλ1−1ψ0.Proof.
From (18) and the Hölder inequality, it gives
(29)fL2Ω2=∑j=1∞∫Ωgxejxdx∫0Texp−T−sλjβ1+aλj−1ψsds2≤∑j=1∞∫Ωgxejx2k/k+2∫0Texp−T−sλjβ1+aλj−1ψsds2∫Ωgxejxdx4/k+2≤∑j=1∞∫Ωfxejxdx2∫0Texp−T−sλjβ1+aλj−1ψsdsk2/k+2∑j=1∞∫Ωgxejxdx2k/k+2·
Applying Lemma1 and the priori boundary condition (27), we have
(30)∑j=1∞∫Ωfxejxdx2∫0Texp−T−sλjβ1+aλj−1ψsdsk=∑j=1∞∫Ωfxejxdx2λjkβ−1a1−exp−Tλ1β1+aλ1−1ψ0−k≤E2C1−k.
Combining (29) to (30), one has
(31)∥f∥L2Ω≤C1−k/k+2E2/k+2∥g∥L2Ωk/k+2,where C1=a1−exp−Tλ1β1+aλ1−1ψ0. The proof of this theorem is completed.


## 4. A Fractional Landweber Method and Convergent Rate

In this section, we apply the fractional Landweber regularization method to solve the inverse source problem (1) and give a convergence estimate. The construction of this method and its iterative implementation are clarified in [19]. We denote the fractional Landweber regularization solution with the observed data by
(32)fcδ,δx=∑j=1∞1−1−bλ1−1+aλjβ−12cδ∫Ωgδxejxdxejx∫0Texp−T−sλjβ1+aλj−1ψδsds,(33)fcδ,δdx=∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgδxejxdxejx∫0Texp−T−sλjβ1+aλj−1ψδsds,(34)fcδdx=∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgxejxdxejx∫0Texp−T−sλjβ1+aλj−1ψsds·It is obvious to see that formulas (33) and (34) are more complicated. For simplicity, we put Cβa,s,λj=exp−T−sλjβ1+aλj−1. Expressions (33) and (34) become
(35)fc,δdx=∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgδxejxdxejx∫0TCβa,s,λjψδsds,12<d<1,(36)fcdx=∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgxejxdxejx∫0TCβa,s,λjψsds,12<d<1,where d∈1/2,1 is called the fractional parameter and cδ≥1 is a regularization parameter and b∈0,λ1β−1/λ1−1+a2. If d=1, it is the classical Landweber method. Next, we have the following lemmas:Lemma 5.
For0<λ<1,τ>0,n∈ℕ, let rnλ≔1−λn, we get
(37)rnλλτ≤θτn+1−τ,where
(38)θτ=1,0≤τ≤1,ττ,τ>1.Proof.
Please see in [19].Lemma 6.
For1/2<d<1,cδ≥1, choosing b∈0,λ1β−1/λ1−1+a2 then 0<bλ1−1+a/λ1β−12<1, by denoting z=bλ1−1+a/λ1β−12, we have the following estimates:
(39)a1−1−zcdzb−1/2≤b1/2c1/2,b1−zczbς/2≤ς2bς/2c−ς/2.Proof.
The proof can be found in [19].


### 4.1. A Priori Parameter Choice Rule

Theorem 7.
Suppose thatf is given by (18) such that ∥f∥DAkβ−1≤E for any E>0. Let the data ψ,g,ψδ,gδ satisfy (3). If we choose cδ=E/δ2/k+1, then we obtain
(40)∥fcδ,δr−f∥L2Ωisoforderδk/k+1,where fcδ,δd is a regularized solution defined in (35).Proof.
By using the triangle inequality, we have(41)∥fcδ,δd−f∥L2Ω≤∥fcδ,δd−fcδd∥L2Ω+∥fcδd−f∥L2Ω·We receive∥fcδ,δd−fcδd∥L2Ω as follows:
(42)fcδ,δdx−fcδdx=∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgδxejxdxejx∫0TCβa,s,λjψδsds−∫Ωgxejxdxejx∫0TCβa,s,λjψsds·From (42), we get
(43)∥fcδ,δd−fcδd∥L2Ω=∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgδx−gxejxdx∫0TCβa,s,λjψδsds⏟I1+∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgxejxdx∫0TCβa,s,λjψδsds−∫Ωgxejxdx∫0TCβa,s,λjψsds⏟I2·Using (43), Lemma 6, and Lemma 1 and noting that λ1−1+a/λjβ−1−1, we provide the estimation of I1 as follows:
(44)I1≤∑j=1∞1−1−bλ1−1+aλjβ−12cδdλ1−1+aλjβ−1−1×λ1−1+aλjβ−1λjβ−1λ1−1+a4ψ0∫Ωgδx−gxejxdx1−exp−Tλ11+aλ1−1≤cδ1/2b1/24εψ01−exp−Tλ11+aλ1−1−1.Next, we have the estimation ofI2(45)I2≤∑j=1∞∫0TCβa,s,λjψs−ψδds∫0TCβa,s,λjψδsds×∫Ωgxejxdxejx∫0TCβa,s,λjψsds≤4δψ0fL2Ω.Combining (42) to (45), we derive that
(46)fcδ,δd−fcδdL2Ω≤cδ1/2b1/24δψ01−exp−Tλ11+aλ1−1−1+4δψ0fL2Ω.Next, we give(47)fcδd−fL2Ω2=∑j=1∞1−1−1−bλ1−1+aλjβ−12cδd2∫Ωgxejxdx∫0TCβa,s,λjψsds2≤∑j=1∞1−1−1−bλ1−1+aλjβ−12cδd2λj−kβ−1fDAkβ−12≤∑j=1∞1−bλ1−1+aλjβ−122cδλj−kβ−1E2.From estimate (13) and Lemma 1, we arrive at
(48)∫0Texp−T−sλjβ1+aλj−1ds≥1+aλjλjβ1−exp−Tλ1β1+aλ1−1≥aλjλjβ1−exp−Tλ1β1+aλ1−1≥aλj−β−11−exp−Tλ1β1+aλ1−1.Hence,(49)λj−β−1≤∫0Texp−T−sλjβ1+aλj−1dsa1−exp−Tλ1β1+aλ1−1≤λ1−1+aλjβ−1a1−exp−Tλ1β1+aλ1−1·The above estimate (49) implies that
(50)λj−kβ−1≤λ1−1+akλjkβ−1ak1−exp−Tλ1β1+aλ1−1k.From observation above, using Lemma6, we conclude that
(51)∥fcδd−f∥L2Ω2≤∑j=1∞1−bλ1−1+aλjβ−122cδλ1−1+aλjβ−1kE2ak1−exp−Tλ1β1+aλ1−1k≤k2bkcδ−kE2ak1−exp−Tλ1β1+aλ1−1k·From (47) and (50), we have
(52)∥fcδd−f∥L2Ω≤k2bk/2cδ−k/2E2ak/21−exp−Tλ1β1+aλ1−1k/2.Combining (68) to (52), it can be seen
(53)∥fcδd−f∥L2Ω≤cδ1/2b1/24δψ01−exp−Tλ11+aλ1−1−1+4δψ0∥f∥L2Ω+k2bk/2cδ−k/2Eak/21−exp−Tλ1β1+aλ1−1k/2·By substitutingcδ=E/δ2/k+1 in the above expression, we deduce that
(54)∥fcδ,δd−f∥L2Ω≤δk/k+1Ek/k+1L1+L2,where
(55)L1=b1/24ψ01−exp−Tλ11+aλ1−1−1+4δk/k+1ψ0−1C1−k/k+2∥g∥k/k+2,L2=k2bk/2Eak/21−exp−Tλ1β1+aλ1−1k/2·The proof of Theorem7 is completed.


### 4.2. A Posteriori Parameter Choice Rule

In order to obtain a posteriori convergence error estimate, we apply Morozov’s discrepancy principle, which is introduced in [18]. Furthermore, we learn the analysis techniques from previous papers [19–21].Let us assume thatσ>1 is a fixed constant. By a similar claim in [19–21], we provide that the general a posteriori rule in the following:
(56)∥Yfcδ,δd−gδ∥L2Ω≤σδ.From here on, in this subsection, we need to assume further to further assume thatcδ is a natural number that greater than 1. If ∥gδ∥L2Ω≥σδ, then the equation (56) exists in a unique solution.Lemma 8.
SetRcδ=∥Yfcδ,δδ−gδ∥L2Ω where 0<δ<∥gδ∥L2Ω. Then, we declare that
(a)
Rcδ is a continuous function(b)
Rcδ⟶0 as cδ⟶+∞(c)
Rcδ⟶∥gδ∥L2Ω as cδ⟶0(d)
Rcδ is a strictly increasing function for cδ∈0,+∞Proof.
The proof of Lemma8 is simple and completely similar to that in [19–21]. Hence, we omit it here.Lemma 9.
Let us assume that (56) holds. Then, cδ satisfies
(57)cδ≤2Kβ2ψ1,a,T,λ1,ψ0σ2−21/k+1k+12bE2/k+1δ−2/k+1·Proof.
From the definition ofcδ,d∈1/2,1, 0<bλ1−1+a/λjβ−12<1, ∥f∥DAkβ−1≤E, we have
(58)∥Yfcδ,δd−gδ∥L2Ω2=∥∑j=1∞1−1−1−bλ1−1+aλj1−β2cδ−1d∫Ωgδxejxdx∥L2Ω2.
Using the inequalitya+b2≤2a2+b2, we derive that
(59)Yfcδ−1,δd−gδL2Ω2≤2∑j=1∞1−1−1−bλ1−1+aλj1−β2cδ−1d∫Ωgδx−gxejxdxL2Ω2⏟O1+2∑j=1∞1−1−1−bλ1−1+aλj1−β2cδ−1d∫ΩgxejxdxL2Ω2⏟O2.(Step 1)
Due to1−1−1−bλ1−1+aλj1−β2cδ−1d≤1 and ∥gδ−g∥L2Ω≤δ, from (59), the estimate of O1 is as follows:(60)O1≤2δ2.(Step 2)
O2 can be bounded as follows:(61)O2≤2∥∑j=1∞1−bλ1−1+aλj1−β2cδ−1∫0TCβa,s,λjψsds∫Ωfxejxdx∥L2Ω2≤2∥∑j=1∞1−bλ1−1+aλj1−β2cδ−1∫0TCβa,s,λjψsdsk+1∫Ωfxejxdx∫0TCβa,s,λjψsdsk∥L2Ω2≤2∥∑j=1∞1−bλ1−1+aλj1−β2cδ−1λ1−1+aλj1−βk+1Kβψ1,a,T,λ1,ψ0E∥L2Ω2,where
(62)Kβψ1,a,T,λ1,ψ0=ψ1k+1a1−exp−Tλ1β1+aλ1−1ψ0−k.
Thanks for the two articles [24, 25], we get the following inequality:
(63)1−ωsωr≤rrs+1−r,for 0<ω<1,r>0,ands∈ℕ. Combining (58) and (63), we deduce that
(64)σ2δ2≤2δ2+2k+12bk+11cδk+1Kβ2ψ1,a,T,λ1,ψ0E2.
This implies that(65)cδ≤2Kβ2ψ1,a,T,λ1,ψ0σ2−21/k+1k+12bE2/k+1δ−2/k+1·Theorem 10.
Letfcδ,δd be the regularized solution which is defined in (33). Suppose that condition (3) is satisfied, and the parameter regularization is chosen by (56). Then we get the following estimate:
(66)∥fcδ,δd−f∥L2Ωisoforderδk/k+1·Proof.
By the triangle inequality, we receive(67)∥fcδ,δd−f∥L2Ω≤∥fcδ,δd−fcδd∥L2Ω+∥fcδd−f∥L2Ω·
Firstly, we have(68)∥fcδ,δd−fcδd∥L2Ω≤cδ1/2b1/24δψ01−exp−Tλ11+aλ1−1−1+4δψ0∥f∥L2Ω.
Secondly, we find that(69)fcδd−fL2Ω=∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgxejxdxej·∫0TCβa,s,λjψsdsL2Ω.
In view of Hölder inequality, we follow from (69) that
(70)fcδd−fL2Ω≤V1V2,where
(71)V1=∑j=1∞1−bλ1−1+aλj1−β2κδ∫Ωfxejxdxej·1/k+1,V2=∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgxejxdxej·∫0TCβa,s,λjψsdsk/k+1.To continue the proof, we divide it into two steps.(Step 1)
In the estimate ofV1, we have(72)V1≤∥∑j=1∞1−bλ1−1+bλj1−β2cδλjkβ−1∫Ωfxejxdxej·∥L2Ω1/k+1.(Step 2)
In the estimate ofV2, we have(73)V2≤∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgx−gδxejxdxej·∫0TCβa,s,λjψsdsL2Ω+∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgδxejxdxej·∫0TCβa,s,λjψsdsL2Ωk/k+1.From (73), we have
(74)V2≤∑j=1∞1−bλ1−1+aλj1−β2cδ∫Ωgx−gδxejxdxej·∫0TCβa,s,λjψsdsL2Ω+∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgδxejxdxej·∫0TCβa,s,λjψsdsL2Ωk/k+1≤δk/k+11+σk/k+1supλj>1λjβ−1ψ0λj−1+a1−exp−Tλ11+aλ1−1k/k+1.Substituting (73) into (69), it gives
(75)fcδd−fL2Ω≤δk/k+1E1/k+11+σk/k+11ψ0a1−exp−Tλ11+aλ1−1k/k+1·Substituting (65) into (68) and combining estimate (75), we conclude that
(76)fcδ,δd−fL2Ω≤δk/k+1E1/k+1R1+R2,where
(77)R1=2Kβ2ψ1,a,T,λ1,ψ0k/2k+1k+1/2b1/2b1/24σ2−2−1/2k+1ψ01−exp−Tλ11+aλ1−1+4δk/k+1ψ0−1C1−k/k+2gk/k+2,R2=1+σk/k+11ψ0a1−exp−Tλ11+aλ1−1k/k+1.Theorem10 is proven.


## 4.1. A Priori Parameter Choice Rule

Theorem 7.
Suppose thatf is given by (18) such that ∥f∥DAkβ−1≤E for any E>0. Let the data ψ,g,ψδ,gδ satisfy (3). If we choose cδ=E/δ2/k+1, then we obtain
(40)∥fcδ,δr−f∥L2Ωisoforderδk/k+1,where fcδ,δd is a regularized solution defined in (35).Proof.
By using the triangle inequality, we have(41)∥fcδ,δd−f∥L2Ω≤∥fcδ,δd−fcδd∥L2Ω+∥fcδd−f∥L2Ω·We receive∥fcδ,δd−fcδd∥L2Ω as follows:
(42)fcδ,δdx−fcδdx=∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgδxejxdxejx∫0TCβa,s,λjψδsds−∫Ωgxejxdxejx∫0TCβa,s,λjψsds·From (42), we get
(43)∥fcδ,δd−fcδd∥L2Ω=∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgδx−gxejxdx∫0TCβa,s,λjψδsds⏟I1+∑j=1∞1−1−bλ1−1+aλjβ−12cδd∫Ωgxejxdx∫0TCβa,s,λjψδsds−∫Ωgxejxdx∫0TCβa,s,λjψsds⏟I2·Using (43), Lemma 6, and Lemma 1 and noting that λ1−1+a/λjβ−1−1, we provide the estimation of I1 as follows:
(44)I1≤∑j=1∞1−1−bλ1−1+aλjβ−12cδdλ1−1+aλjβ−1−1×λ1−1+aλjβ−1λjβ−1λ1−1+a4ψ0∫Ωgδx−gxejxdx1−exp−Tλ11+aλ1−1≤cδ1/2b1/24εψ01−exp−Tλ11+aλ1−1−1.Next, we have the estimation ofI2(45)I2≤∑j=1∞∫0TCβa,s,λjψs−ψδds∫0TCβa,s,λjψδsds×∫Ωgxejxdxejx∫0TCβa,s,λjψsds≤4δψ0fL2Ω.Combining (42) to (45), we derive that
(46)fcδ,δd−fcδdL2Ω≤cδ1/2b1/24δψ01−exp−Tλ11+aλ1−1−1+4δψ0fL2Ω.Next, we give(47)fcδd−fL2Ω2=∑j=1∞1−1−1−bλ1−1+aλjβ−12cδd2∫Ωgxejxdx∫0TCβa,s,λjψsds2≤∑j=1∞1−1−1−bλ1−1+aλjβ−12cδd2λj−kβ−1fDAkβ−12≤∑j=1∞1−bλ1−1+aλjβ−122cδλj−kβ−1E2.From estimate (13) and Lemma 1, we arrive at
(48)∫0Texp−T−sλjβ1+aλj−1ds≥1+aλjλjβ1−exp−Tλ1β1+aλ1−1≥aλjλjβ1−exp−Tλ1β1+aλ1−1≥aλj−β−11−exp−Tλ1β1+aλ1−1.Hence,(49)λj−β−1≤∫0Texp−T−sλjβ1+aλj−1dsa1−exp−Tλ1β1+aλ1−1≤λ1−1+aλjβ−1a1−exp−Tλ1β1+aλ1−1·The above estimate (49) implies that
(50)λj−kβ−1≤λ1−1+akλjkβ−1ak1−exp−Tλ1β1+aλ1−1k.From observation above, using Lemma6, we conclude that
(51)∥fcδd−f∥L2Ω2≤∑j=1∞1−bλ1−1+aλjβ−122cδλ1−1+aλjβ−1kE2ak1−exp−Tλ1β1+aλ1−1k≤k2bkcδ−kE2ak1−exp−Tλ1β1+aλ1−1k·From (47) and (50), we have
(52)∥fcδd−f∥L2Ω≤k2bk/2cδ−k/2E2ak/21−exp−Tλ1β1+aλ1−1k/2.Combining (68) to (52), it can be seen
(53)∥fcδd−f∥L2Ω≤cδ1/2b1/24δψ01−exp−Tλ11+aλ1−1−1+4δψ0∥f∥L2Ω+k2bk/2cδ−k/2Eak/21−exp−Tλ1β1+aλ1−1k/2·By substitutingcδ=E/δ2/k+1 in the above expression, we deduce that
(54)∥fcδ,δd−f∥L2Ω≤δk/k+1Ek/k+1L1+L2,where
(55)L1=b1/24ψ01−exp−Tλ11+aλ1−1−1+4δk/k+1ψ0−1C1−k/k+2∥g∥k/k+2,L2=k2bk/2Eak/21−exp−Tλ1β1+aλ1−1k/2·The proof of Theorem7 is completed.


## 4.2. A Posteriori Parameter Choice Rule

In order to obtain a posteriori convergence error estimate, we apply Morozov’s discrepancy principle, which is introduced in [18]. Furthermore, we learn the analysis techniques from previous papers [19–21].Let us assume thatσ>1 is a fixed constant. By a similar claim in [19–21], we provide that the general a posteriori rule in the following:
(56)∥Yfcδ,δd−gδ∥L2Ω≤σδ.From here on, in this subsection, we need to assume further to further assume thatcδ is a natural number that greater than 1. If ∥gδ∥L2Ω≥σδ, then the equation (56) exists in a unique solution.Lemma 8.
SetRcδ=∥Yfcδ,δδ−gδ∥L2Ω where 0<δ<∥gδ∥L2Ω. Then, we declare that
(a)
Rcδ is a continuous function(b)
Rcδ⟶0 as cδ⟶+∞(c)
Rcδ⟶∥gδ∥L2Ω as cδ⟶0(d)
Rcδ is a strictly increasing function for cδ∈0,+∞Proof.
The proof of Lemma8 is simple and completely similar to that in [19–21]. Hence, we omit it here.Lemma 9.
Let us assume that (56) holds. Then, cδ satisfies
(57)cδ≤2Kβ2ψ1,a,T,λ1,ψ0σ2−21/k+1k+12bE2/k+1δ−2/k+1·Proof.
From the definition ofcδ,d∈1/2,1, 0<bλ1−1+a/λjβ−12<1, ∥f∥DAkβ−1≤E, we have
(58)∥Yfcδ,δd−gδ∥L2Ω2=∥∑j=1∞1−1−1−bλ1−1+aλj1−β2cδ−1d∫Ωgδxejxdx∥L2Ω2.
Using the inequalitya+b2≤2a2+b2, we derive that
(59)Yfcδ−1,δd−gδL2Ω2≤2∑j=1∞1−1−1−bλ1−1+aλj1−β2cδ−1d∫Ωgδx−gxejxdxL2Ω2⏟O1+2∑j=1∞1−1−1−bλ1−1+aλj1−β2cδ−1d∫ΩgxejxdxL2Ω2⏟O2.(Step 1)
Due to1−1−1−bλ1−1+aλj1−β2cδ−1d≤1 and ∥gδ−g∥L2Ω≤δ, from (59), the estimate of O1 is as follows:(60)O1≤2δ2.(Step 2)
O2 can be bounded as follows:(61)O2≤2∥∑j=1∞1−bλ1−1+aλj1−β2cδ−1∫0TCβa,s,λjψsds∫Ωfxejxdx∥L2Ω2≤2∥∑j=1∞1−bλ1−1+aλj1−β2cδ−1∫0TCβa,s,λjψsdsk+1∫Ωfxejxdx∫0TCβa,s,λjψsdsk∥L2Ω2≤2∥∑j=1∞1−bλ1−1+aλj1−β2cδ−1λ1−1+aλj1−βk+1Kβψ1,a,T,λ1,ψ0E∥L2Ω2,where
(62)Kβψ1,a,T,λ1,ψ0=ψ1k+1a1−exp−Tλ1β1+aλ1−1ψ0−k.
Thanks for the two articles [24, 25], we get the following inequality:
(63)1−ωsωr≤rrs+1−r,for 0<ω<1,r>0,ands∈ℕ. Combining (58) and (63), we deduce that
(64)σ2δ2≤2δ2+2k+12bk+11cδk+1Kβ2ψ1,a,T,λ1,ψ0E2.
This implies that(65)cδ≤2Kβ2ψ1,a,T,λ1,ψ0σ2−21/k+1k+12bE2/k+1δ−2/k+1·Theorem 10.
Letfcδ,δd be the regularized solution which is defined in (33). Suppose that condition (3) is satisfied, and the parameter regularization is chosen by (56). Then we get the following estimate:
(66)∥fcδ,δd−f∥L2Ωisoforderδk/k+1·Proof.
By the triangle inequality, we receive(67)∥fcδ,δd−f∥L2Ω≤∥fcδ,δd−fcδd∥L2Ω+∥fcδd−f∥L2Ω·
Firstly, we have(68)∥fcδ,δd−fcδd∥L2Ω≤cδ1/2b1/24δψ01−exp−Tλ11+aλ1−1−1+4δψ0∥f∥L2Ω.
Secondly, we find that(69)fcδd−fL2Ω=∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgxejxdxej·∫0TCβa,s,λjψsdsL2Ω.
In view of Hölder inequality, we follow from (69) that
(70)fcδd−fL2Ω≤V1V2,where
(71)V1=∑j=1∞1−bλ1−1+aλj1−β2κδ∫Ωfxejxdxej·1/k+1,V2=∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgxejxdxej·∫0TCβa,s,λjψsdsk/k+1.To continue the proof, we divide it into two steps.(Step 1)
In the estimate ofV1, we have(72)V1≤∥∑j=1∞1−bλ1−1+bλj1−β2cδλjkβ−1∫Ωfxejxdxej·∥L2Ω1/k+1.(Step 2)
In the estimate ofV2, we have(73)V2≤∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgx−gδxejxdxej·∫0TCβa,s,λjψsdsL2Ω+∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgδxejxdxej·∫0TCβa,s,λjψsdsL2Ωk/k+1.From (73), we have
(74)V2≤∑j=1∞1−bλ1−1+aλj1−β2cδ∫Ωgx−gδxejxdxej·∫0TCβa,s,λjψsdsL2Ω+∑j=1∞1−1−1−bλ1−1+aλj1−β2cδd∫Ωgδxejxdxej·∫0TCβa,s,λjψsdsL2Ωk/k+1≤δk/k+11+σk/k+1supλj>1λjβ−1ψ0λj−1+a1−exp−Tλ11+aλ1−1k/k+1.Substituting (73) into (69), it gives
(75)fcδd−fL2Ω≤δk/k+1E1/k+11+σk/k+11ψ0a1−exp−Tλ11+aλ1−1k/k+1·Substituting (65) into (68) and combining estimate (75), we conclude that
(76)fcδ,δd−fL2Ω≤δk/k+1E1/k+1R1+R2,where
(77)R1=2Kβ2ψ1,a,T,λ1,ψ0k/2k+1k+1/2b1/2b1/24σ2−2−1/2k+1ψ01−exp−Tλ11+aλ1−1+4δk/k+1ψ0−1C1−k/k+2gk/k+2,R2=1+σk/k+11ψ0a1−exp−Tλ11+aλ1−1k/k+1.Theorem10 is proven.


## 5. Convergence of the Source Function whenβ⟶1

In this section, we will first prove the convergence of the source function whenβ⟶1.Theorem 11.
Let the Cauchy datag∈DAr+2βΩ for any r≥0. Let the function ψ∈L∞0,T. Then, we have the following estimate:
(78)fβx−f1xDArΩ≲β−12−β−1ε∥ψ∥L∞0,T∥g∥DAr+2βΩ,where ε>0 satisfies that 2−β−1ε>0.Proof.
Using the inequalitye−m−e−n≤Cεm−nε, we find that for z>0(79)exp−hλjβ1+aλj−1−exp−hλj1+aλj−1≤Cεhελjβ−λj1+aλjε≤Cεahελj−ελjβ−λjε·
Let us recall Lemma12 which is proved in [27].Lemma 12.
Assume that0≤a≤b and 0<z. For any ε>0, there always exists C¯ε>0 such that
(a)
Ifz<1 then(80)za−zb≤C¯εb−aεza−ε(b)
Ifz≥1 then(81)za−zb≤C¯εb−aεzb+εLet us divide the set of natural number into two sets in the following:(82)ℕ=ℕ1∪ℕ2,where
(83)ℕ1=j∈ℝ,λj≤1,ℕ2=j∈ℝ,λj>1.Let us assume thatj∈ℕ1. Let us recall that the assumption β≥1. By applying Lemma 12, we know that since λj≤1 then for any θ>0(84)λjβ−λj≤C¯1,θλj1−θβ−1θ.It follows from (79) that
(85)exp−hλjβ1+aλj−1−exp−hλj1+aλj−1≤CεC¯1,θεahελj−ελj1−θεβ−1θε≤CεC¯1,θεahελj−θεβ−1θε·Let us assume thatj∈ℕ2. By applying Lemma 12, we know that since λj>1, then
(86)λjβ−λj≤C¯2,θλjβ+θβ−1θ.It follows from (79) that
(87)exp−hλjβ1+aλj−1−exp−hλj1+aλj−1≤CεC¯1,θεahελj−ελjβ+θεβ−1θε≤CεC¯1,θεahελj−ε+βε+θεβ−1θε.Let us review that(88)fβx=∑j=1∞∫Ωgxejxdx∫0Texp−T−sλjβ1+aλj−1ψsdsejx.By in view of Parseval’s equality, we find that(89)f1x=∑j=1∞∫Ωgxejxdx∫0Texp−T−sλj1+aλj−1ψsdsejx.Since two above observations, we derive that(90)fβx−f1xDArΩ2=∑j=1∞∫Ωgxejxdx∫0Texp−T−sλjβ1+aλj−1ψsds−∫Ωgxejxdx∫0Texp−T−sλj1+aλj−1ψsds2λj2r.First, ifλj≤1, then using the estimate (85), we get that
(91)∫0Texp−T−sλjβ1+aλj−1ψsds−∫0Texp−T−sλj1+aλj−1ψsds≤CεC¯1,θεaλj−θεβ−1θε∫0TT−sεdsψL∞0,T≤M1ψL∞0,Tλj−θεβ−1θε,where M1=CεC¯1,θε/aT1+ε/1+ε. By a similar explanation, if λj>1, then using the estimate (87), we get that
(92)∫0Texp−T−sλjβ1+aλj−1ψsds−∫0Texp−T−sλj1+aλj−1ψsds≤CεC¯1,θεaλj−ε+βε+θεβ−1θε∫0TT−sεds∥ψ∥L∞0,T≤M1∥ψ∥L∞0,Tλj−ε+βε+θεβ−1θε.By using Lemma1, we find that
(93)∫0Texp−T−sλjβ1+aλj−1ψsds∫0Texp−T−sλj1+aλj−1ψsds≥1−exp−Tλ1β1+aλ1−12ψ021+aλjλjβ2=M221+aλjλjβ2≥M22a2λj2−2β,where we denote
(94)M2=1−exp−Tλ1β1+aλ1−1ψ0.From some of the above observations, we get that(95)∥fβx−f1x∥DArΩ2=∑λj≤1∫Ωgxejxdx∫0Texp−T−sλjβ1+aλj−1ψsds−∫Ωgxejxdx∫0Texp−T−sλj1+aλj−1ψsds2λj2r+∑λj>1∫Ωgxejxdx∫0Texp−T−sλjβ1+aλj−1ψsds−∫Ωgxejxdx∫0Texp−T−sλj1+aλj−1ψsds2λj2r≤M1M222β−12θε∥ψ∥L∞0,T2∑λj≤1λj2θε−4+4β+2r∫Ωgxejxdx2+M1M222β−12θε∥ψ∥L∞0,T2∑λj>1λj2θε−4+4β+2r−2ε+2βε∫Ωgxejxdx2.Noting thatβ≥1, we have that
(96)1≤λjλ12βε−2ε.This implies the following estimate:(97)∥fβx−f1x∥DArΩ2≲β−12θε∥ψ∥L∞0,T2∑j=1∞λj2θε−4+4β+2r−2ε+2βε∫Ωgxejxdx2.Let us chooseθ and ε such that 2θε+2βε=4+2ε. In order to choose such number θ and ε, we need to choose ε>0 if β=1 and such that
(98)0<ε<2β−1,β>1.Letθ be such that θ=2/ε+1−β. Then since (()(97)), we deduce that
(99)∥fβx−f1x∥DArΩ≲β−1θε∥ψ∥L∞0,T∥g∥DAr+2βΩ=β−12−β−1ε∥ψ∥L∞0,T∥g∥DAr+2βΩ.



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