# An Inverse Problem for a Two-Dimensional Time-Fractional Sideways Heat Equation

Authors: Songshu Liu; Lixin Feng
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this paper, we consider a two-dimensional (2D) time-fractional inverse diffusion problem which is severely ill-posed; i.e., the solution (if it exists) does not depend continuously on the data. A modified kernel method is presented for approximating the solution of this problem, and the convergence estimates are obtained based on both a priori choice and a posteriori choice of regularization parameters. The numerical examples illustrate the behavior of the proposed method.


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## Body

## 1. Introduction

In the past decades, fractional differential equations have attracted wide attention. Various models using fractional partial differential equations have been successfully applied to describe a range of problems in mechanical engineering [1], viscoelasticity [2], electron transport [3], dissipation [4], heat conduction [5, 6], and high-frequency financial data [7].The time-fractional diffusion equation is deduced by replacing the standard time derivative with a time-fractional derivative and can be used to describe the superdiffusion and subdiffusion phenomena [8–13]. The direct problems, i.e., initial value problems and initial boundary value problems, for the time-fractional diffusion equation have been studied extensively in recent years. For instance, a well-posedness analysis [14–18] and numerical methods and simulations [19–24] are typical examples of this kind of analysis.However, in some practical situations, part of boundary data, or initial data, or diffusion coefficient, or source term, or the order of fractional derivative may not be given and we want to find them by additional measured data which will yield some inverse problems of the fractional diffusion equation. The early papers on inverse problems of time-fractional diffusion equation were provided by Murio in [25, 26] for solving the time-fractional diffusion equation by mollification methods. Cheng et al. [27] gave the uniqueness in determining the parameter α and px by means of observation data at one end point. Liu and Yamamoto [28] considered a backward problem in time for a time-fractional partial diffusion equation in one-dimensional case. Zhang and Xu [29] studied an inverse source problem in the time-fractional diffusion equation and proved uniqueness for identifying a space-dependent source term by using analytic continuation and Laplace transform. Zheng and Wei [30] applied a new regularization method to solve an inverse problem for a time-fractional diffusion equation in a one-dimensional semi-infinite domain. Jin and Rundell [31] studied an inverse problem of recovering a spatially varying potential term in the 1D time-fractional diffusion equation from the flux measurements and proposed a quasi-Newton-type reconstruction algorithm. Recently, Wei and Wang [32] proposed a modified quasi-boundary value method for an inverse source problem of the time-fractional diffusion equation. Wei et al. [33] identified a time-dependent source term in a multidimensional time-fractional diffusion equation from boundary Cauchy data.As for time-fractional inverse diffusion problem for one-dimensional (1D) models, theoretical concepts and computational implementation have been discussed by many authors, and a number of solution methods have been proposed, i.e., spectral regularization method [34], iteration regularization method [35], optimal regularization method [36], and a new dynamic method [37]. However, for the two-dimensional time-fractional inverse diffusion problem, few results are available. Xiong et al. [38] gave a conditional stability estimate for the inverse heat conduction problem in the 2D time-fractional heat equation and studied a dynamic spectral regularization method with numerical testification. However, error estimates were given in [38] by only choosing the regularization parameter by a priori choice rule.In the present paper, we will consider the 2D time-fractional inverse diffusion problem by using a modified kernel method and a posteriori parameter choice rule is given. For a priori choice and a posteriori choice of the regularization parameter, we obtain the convergence estimates, respectively.The rest of this paper is organized as follows. In Section2, we describe the 2D time-fractional inverse diffusion problem and give an analysis on the ill-posedness of this problem. In Section 3, we propose a modified kernel method and prove the convergence estimates under a priori and a posteriori parameter choice rule, respectively. Section 4 is the numerical aspect of the proposed method. Finally, we give a brief conclusion in Section 5.


## 2. Description of the 2D Time-Fractional Inverse Diffusion Problem

We consider the following 2D time-fractional inverse diffusion problem:(1)∂αu∂tα−uxx−uyy=0,x>0,y>0,t>0,(2)u1,y,t=gy,t,y≥0,t≥0,with the corresponding measured data function gδy,t and initial and boundary conditions(3)ux,y,0=0,x>0,y≥0,(4)ux,0,t=0,x>0,t≥0,(5)uyx,0,t=0,x>0,t≥0,(6)ux,y,tx⟶∞bounded,y>0,t>0,where the time-fractional derivative ∂αu/∂tα is the Caputo fractional derivative of order α0<α≤1 defined by (see [39])(7)∂αu∂tαx,y,t=1Γ1−α∫0t∂ux,y,s∂sdst−sα,0<α<1,∂αu∂tαx,y,t=∂ux,y,t∂t,α=1.Here, we wish to determine the temperatureux,y,t for 0≤x<1 from the temperature measurements gδy,t.In order to apply the Fourier transform techniques, we extend the functionsux,⋅,⋅, g⋅,⋅, and gδ⋅,⋅ to the whole plane −∞<y<+∞ and −∞<t<+∞ by setting the functions to be zero for y<0 or t<0. We assume that these functions are in L2ℝ2 and use the corresponding L2-norm, defined as follows:(8)g=∫ℝ2gy,t2dydt1/2.We also assume that the measured data functiongδy,t satisfies(9)gδ−g≤δ,where the constant δ>0 represents a bound on the measurement error.Let(10)g^ξ,η=12π∫ℝ2gy,te−iξy+ηtdydt,ξ,η∈ℝ,be the Fourier transform of a function gy,t. Taking the transform to equations (1)–(6) with respect to y and t, we can get the solution of equations (1)–(6) in the frequency domain (see [38]):(11)u^x,ξ,η=e1−xiηα+ξ2g^ξ,η,where(12)iηα=ηαcosαπ2+isignηsinαπ2.Denoteγ as follows:(13)γ≔iηα+ξ2=A+iB,where(14)A≔η2αsin2απ/2+ξ2+ηαcosαπ/22+ξ2+ηαcosαπ/22,B≔signsignηηαsinαπ2×η2αsin2απ/2+ξ2+ηαcosαπ/22−ξ2+ηαcosαπ/22.For the above problem, sincee1−xiηα+ξ2 is unbounded with respect to variables ξ and η for fixed 0≤x<1, the small error in the high-frequency components will be amplified. Therefore, the 2D time-fractional inverse diffusion problem is severely ill-posed. To solve the 2D time-fractional inverse diffusion problem, a natural way to stabilize the problem is eliminate the high frequencies or to replace the “kernel” e1−xiηα+ξ2 by a bounded approximation.


## 3. A Modified Method and Convergence Estimates

In this section, we will give a modified “kernel” method and obtain the convergence estimates. The regularization solution is given by(15)u^βδx,ξ,η=kβx,ξ,ηg^δξ,η,where(16)kβx,ξ,η=e1−xA+iB,e1−xA≤βx,βxei1−xB,e1−xA>βx,where βx>1 is the regularization parameter.To obtain the convergence estimates between the regularization solution and the exact solution, we need to assume a priori bound:(17)u0,⋅,⋅p≤E,p>0,where E>0 is a constant and ⋅p denotes the norm in Sobolov space Hpℝ2 defined by(18)ux,⋅,⋅p≔∫ℝ21+ξ2+η2pu^0,⋅,⋅2dξdη1/2.Remark 1.
Obviously, whenp=0, we can know that Hpℝ2=H0ℝ2=L2ℝ2 and formula (17) is bounded in the L2ℝ2-norm.


### 3.1. A Priori Parameter Choice

In the following, we give the convergence estimate foruβδx,⋅,⋅−ux,⋅,⋅ by using a priori choice rule for the regularization parameter.Theorem 1.
Suppose thatuβδx,y,t is the regularization solution with noisy data gδy,t and that ux,y,t is the exact solution with the exact data gy,t. Let assumption (9) be satisfied and let u0,⋅,⋅≤E. If we choose(19)βx=xδEx−1,then for every x∈0,1, we obtain the error estimate(20)uβδx,⋅,⋅−ux,⋅,⋅≤δxE1−x.Proof.
Due to Parseval’s identity and triangle inequality, we can obtain(21)uβδx,⋅,⋅−ux,⋅,⋅=u^βδx,⋅,⋅−u^x,⋅,⋅=kβx,ξ,ηg^δ−e1−xA+iBg^=kβx,ξ,ηg^δ−kβx,ξ,ηg^+kβx,ξ,ηg^−e1−xA+iBg^≤kβx,ξ,ηg^δ−kβx,ξ,ηg^+kβx,ξ,ηg^−e1−xA+iBg^=I1+I2.
ForI1, we have(22)I1=kβx,ξ,ηg^δ−kβx,ξ,ηg^≤δsupξ,η∈ℝkβx,ξ,η≤δβx.
From (11), we know(23)g^ξ,η=e−A+iBu^0,ξ,η.
We now estimateI2; note that from (23),(24)I2=kβx,ξ,ηg^−e1−xA+iBg^=e1−xA+iB−kβx,ξ,ηeA+iBu^0,ξ,η=e1−xA+iB−mine1−xA,βxe1−xiBeA+iBu^0,ξ,η=e1−xA−mine1−xA,βxe1−xiBeA+iBu^0,ξ,η≤supξ,η∈ℝe1−xA−mine1−xA,βxe1−xiBeA+iBE≤Esupξ,η∈ℝ,e1−xA>βxe1−xA−βxeA.
Let(25)φA=e1−xA−βxeA.
DifferentiatingφA and setting the derivative equal to zero, we find that(26)A∗=11−xln1xβx,is the maximum value point of the function φA. Therefore, we know(27)I2≤1/x−1βx1/xβx1/1−xE.
Combining (19), (21), (22), and (27), we have the Hölder-type error estimate (20).Remark 2.
The error estimate in Theorem1 does not give any useful information on the continuous dependence of the solution at x=0. To retain the continuous dependence of the solution at x=0, one has to introduce a stronger a priori assumption.Theorem 2.
Suppose thatuβδ0,y,t is the regularization solution with noisy data gδy,t and that u0,y,t is the exact solution with the exact data gy,t at x=0. Let assumption (9) be satisfied and let u0,⋅,⋅p≤Ep>0. If we choose(28)β=δE−1/2,then for x=0, we obtain the error estimate(29)uβδ0,y,t−u0,y,t≤δ1/2E1/2+E14lnEδ−p.Proof.
Due to Parseval’s identity and the triangle inequality, we have(30)uβδ0,y,t−u0,y,t=u^βδ0,⋅,⋅−u^0,⋅,⋅=kβ0,ξ,ηg^δ−eA+iBg^=kβ0,ξ,ηg^δ−kβ0,ξ,ηg^+kβ0,ξ,ηg^−eA+iBg^≤kβ0,ξ,ηg^δ−kβ0,ξ,ηg^+kβ0,ξ,ηg^−eA+iBg^=I3+I4,where(31)I3=kβ0,ξ,ηg^δ−kβ0,ξ,ηg^=kβ0,ξ,ηgδ−g≤δsupξ,η∈ℝkβ0,ξ,η≤δβ,(32)I4=kβ0,ξ,ηg^−eA+iBg^=eA+iB−mineA,βeiBeA+iBu^0,ξ,η=eA−mineA,βeiB1+ξ2+η2p/2eA+iB1+ξ2+η2p/2u^0,ξ,η≤supξ,η∈ℝeA−mineA,βeiB1+ξ2+η2p/2eA+iBE≤Esupξ,η∈ℝ,eA>βeA−β1+ξ2+η2p/2eA.
Using the factγ=iηα+ξ2≤ηα/2+ξ, we can get(33)A≤A2+B2=γ≤ηα/2+ξ,i.e.,A≤ηα/2+ξ.
Note that alsoβ<eA, and we have A>lnβ; therefore, we can derive(34)lnβ≤ηα/2+ξ.
Ifξ>ηα/2≥0, we know that(35)lnβ≤ηα/2+ξ≤2ξ,i.e., ξ>1/2lnβ. Therefore,(36)1+ξ2+η2−p/2≤ξ2+η2−p/2<ξ−p<12lnβ−p.
Ifηα/2≥ξ≥0, we have(37)lnβ≤ηα/2+ξ≤2ηα/2,i.e., η≥1/2lnβ2/α. Therefore,(38)1+ξ2+η2−p/2≤ξ2+η2−p/2≤η−p<12lnβ−2p/α.
Combining (32), (36), and (38), we can obtain(39)I4≤Esupξ,η∈ℝ,eA>β11+ξ2+η2p/2≤Emax12lnβ−p,12lnβ−2p/α.
Combining (28), (30), (31), and (39), we can obtain(40)uβδ0,y,t−u0,y,t≤I3+I4≤δ1/2E1/2+Emax14lnEδ−p,14lnEδ−2p/α=δ1/2E1/2+E14lnEδ−p.


### 3.2. A Posteriori Parameter Choice

In this section, we first give the following lemma.Lemma 1 (see [40]).
Let the functionfλ:0,a⟶ℝ be given by(41)fλ=λbdln1λ−c,with a constant c∈ℝ and positive constants a<1, b, and d; then, for the inverse function, we have(42)f−1λ=λ1/bdbln1λc/b1+o1,forλ⟶0.
In this following, we give the convergence estimate foruβδx,⋅,⋅−ux,⋅,⋅ by using a posteriori choice rule for the regularization parameter, i.e., Morozov’s discrepancy principle.
According to Morozov’s discrepancy principle [41], we adopt the regularization parameter β as the solution of the following equation:(43)e−A+iB1−xkβx,ξ,ηg^δ−g^δ=τδ,where τ>1 is a constant. Denote(44)Tβ=e−A+iB1−xkβx,ξ,η−1.Here and in the following,β is a function of the variable x and sometimes the function βx is also used.Lemma 2.
Letρβ≔e−A+iB1−xkβx,ξ,ηg^δ−g^δ, and then, the following results hold:(a)
ρβ is a continuous function(b)
limβ⟶1ρβ=e−A1−x−1g^δξ,η(c)
limβ⟶+∞ρβ=0(d)
ρβ is a strictly decreasing function for β∈1,+∞
The proof is obvious, and we omit it here.Remark 3.
According to Lemma2, we find that, if 0<τδ<e−A1−x−1g^δξ,η, equation (43) has a unique solution.Lemma 3.
Ifβx is the solution of equation (43), then the following inequality holds(45)βx≤τ−1δEx−112lnEτ−1δ−p1−x1+o1.Proof.
Due to the triangle inequality and (43), note that Tβ≤1, and we obtain(46)τδ=Tβg^δ≤Tβg^δ−g^+Tβg^≤δ+Tβg^.
We now estimate the second term on the right hand side of (46):(47)Tβg^=e−A+iB1−xkβx,ξ,η−1e−A+iBu^0,⋅,⋅=1−e−A+iB1−xkβx,ξ,ηeA+iB1+ξ2+η2p/21+ξ2+η2p/2u^0,⋅,⋅≤Esupξ,η∈ℝ,eA1−x>βx1−e−A1−xβxeA1+ξ2+η2p/2=Esupξ,η∈ℝ,eA1−x>βxeA1−x−βxeA2−x1+ξ2+η2p/2=Esupξ,η∈ℝ,eA1−x>βxeA1−x−βxeA1−xe−A1+ξ2+η2−p/2≤Esupξ,η∈ℝ,eA1−x>βxe−A1+ξ2+η2−p/2.
Similar to the proof of Theorem2, we can know(48)Tβg^≤Esupξ,η∈ℝ,eA1−x>βxe−A1+ξ2+η2−p/2≤Eβx−1/1−xmax121−xlnβx−p,121−xlnβx−2p/α=Eβx−1/1−x121−xlnβx−p.
Combining (46) and (48), we obtain(49)τ−1δ≤Eβx−1/1−x121−xlnβx−p.
Letτ−1δ=Eβ∗x−1/1−x1/21−xlnβ∗x−p. From Lemma 1, we have(50)β∗x=τ−1δEx−112lnEτ−1δ−p1−x1+o1.
LetGβ≔Eβx−1/1−x1/21−xlnβx−p. For β>1 and 0≤x<1, the function Gβ is strictly decreasing monotonically with β, so we have(51)βx≤β∗x.Lemma 4.
Ifβx is the solution of equation (43), then the following inequality holds:(52)Tβg^≤τ+1δ.Proof.
Using the triangle inequality and (43), we have(53)Tβg^=Tβg^−g^δ+Tβg^δ≤Tβg^−g^δ+Tβg^δ≤τ+1δ.Theorem 3.
Assume that conditions (9) and (17) hold and taking the solution of equation (43) as the regularization parameter, then there holds the following error estimate:(54)uβδx,⋅,⋅−ux,⋅,⋅≤δxE1−x12lnEτ−1δ−p1−xC+o1,where C=τ−1x−12τ+1, 0≤x<1.Proof.
Due to the Parseval formula and the triangle inequality, we obtain(55)uβδx,⋅,⋅−ux,⋅,⋅=u^βδx,⋅,⋅−u^x,⋅,⋅=kβg^δ−eA+iB1−xg^≤kβg^δ−g^+kβ−eA+iB1−xg^≤βxδ+kβe−A+iB1−x−1u^x,⋅,⋅=βxδ+Tβu^x,⋅,⋅.
From Lemma3, we can know that(56)βxδ≤Eτ−11−xδx12lnEτ−1δ−p1−x1+o1.
In the following, we estimate the second term of (55). Denote φ=Tβu^x,⋅,⋅ and take βx=E/τ−1δ1−x1/2lnE/τ−1δ−p1−x; we have(57)φ=φ−kβe−A+iB1−xφ+kβe−A+iB1−xφ≤φ−kβe−A+iB1−xφ+kβe−A+iB1−xφ=M1+M2.
Note thatu^x,⋅,⋅=eA+iB1−xg^⋅,⋅, we get(58)M1=TβeA+iB1−xg^1−kβe−A+iB1−x=TβeA+iB1−x−kβg^≤eA+iB1−x−kβg^.
DenoteM≔eA+iB1−x−kβg^, we have(59)M=eA+iB1−x−kβg^≤Esupξ,η∈ℝeA+iB1−x−kβeA+iB1+ξ2+η2p/2≤Esupξ,η∈ℝeA+iB1−x−mineA1−x,βxeiB1−xeA+iB1+ξ2+η2p/2≤Esupξ,η∈ℝ,eA1−x>βxeA1−x−βxeA1+ξ2+η2p/2=Esupξ,η∈ℝ,eA1−x>βxeA1−x−βxeA1−x1+ξ2+η2−p/2e−Ax≤Esupξ,η∈ℝ,eA1−x>βx1+ξ2+η2−p/2e−Ax.
Similar to the proof of Theorem2, we can know that(60)M≤Esupξ,η∈ℝ,eA1−x>βx1+ξ2+η2−p/2e−Ax≤Eβx−x/1−xmax121−xlnβx−p,121−xlnβx−2p/α=Eβx−x/1−x121−xlnβx−p=E121−xlnEτ−1δ1−x12lnEτ−1δ−p1−x−p×Eτ−1δ1−x12lnEτ−1δ−p1−x−x/1−x=E12lnEτ−1δ+ln12lnEτ−1δ−p/2−pEτ−1δ−x12lnEτ−1δpx=E12lnEτ−1δ−p1+ln1/2lnE/τ−1δ−p/21/2lnE/τ−1δ−pτ−1δEx12lnEτ−1δpx=τ−1xδxE1−x12lnEτ−1δ−p1−x1+o1.
ForM2, from Lemma 4 and assuming that kβ≤βx, we have(61)M2≤βxe−A+iB1−xφ=βxTβg^≤τ+1τ−1x−1δxE1−x12lnEτ−1δ−p1−x.
Combining (56)–(61) with (55), the proof of Theorem 3 is completed.


## 3.1. A Priori Parameter Choice

In the following, we give the convergence estimate foruβδx,⋅,⋅−ux,⋅,⋅ by using a priori choice rule for the regularization parameter.Theorem 1.
Suppose thatuβδx,y,t is the regularization solution with noisy data gδy,t and that ux,y,t is the exact solution with the exact data gy,t. Let assumption (9) be satisfied and let u0,⋅,⋅≤E. If we choose(19)βx=xδEx−1,then for every x∈0,1, we obtain the error estimate(20)uβδx,⋅,⋅−ux,⋅,⋅≤δxE1−x.Proof.
Due to Parseval’s identity and triangle inequality, we can obtain(21)uβδx,⋅,⋅−ux,⋅,⋅=u^βδx,⋅,⋅−u^x,⋅,⋅=kβx,ξ,ηg^δ−e1−xA+iBg^=kβx,ξ,ηg^δ−kβx,ξ,ηg^+kβx,ξ,ηg^−e1−xA+iBg^≤kβx,ξ,ηg^δ−kβx,ξ,ηg^+kβx,ξ,ηg^−e1−xA+iBg^=I1+I2.
ForI1, we have(22)I1=kβx,ξ,ηg^δ−kβx,ξ,ηg^≤δsupξ,η∈ℝkβx,ξ,η≤δβx.
From (11), we know(23)g^ξ,η=e−A+iBu^0,ξ,η.
We now estimateI2; note that from (23),(24)I2=kβx,ξ,ηg^−e1−xA+iBg^=e1−xA+iB−kβx,ξ,ηeA+iBu^0,ξ,η=e1−xA+iB−mine1−xA,βxe1−xiBeA+iBu^0,ξ,η=e1−xA−mine1−xA,βxe1−xiBeA+iBu^0,ξ,η≤supξ,η∈ℝe1−xA−mine1−xA,βxe1−xiBeA+iBE≤Esupξ,η∈ℝ,e1−xA>βxe1−xA−βxeA.
Let(25)φA=e1−xA−βxeA.
DifferentiatingφA and setting the derivative equal to zero, we find that(26)A∗=11−xln1xβx,is the maximum value point of the function φA. Therefore, we know(27)I2≤1/x−1βx1/xβx1/1−xE.
Combining (19), (21), (22), and (27), we have the Hölder-type error estimate (20).Remark 2.
The error estimate in Theorem1 does not give any useful information on the continuous dependence of the solution at x=0. To retain the continuous dependence of the solution at x=0, one has to introduce a stronger a priori assumption.Theorem 2.
Suppose thatuβδ0,y,t is the regularization solution with noisy data gδy,t and that u0,y,t is the exact solution with the exact data gy,t at x=0. Let assumption (9) be satisfied and let u0,⋅,⋅p≤Ep>0. If we choose(28)β=δE−1/2,then for x=0, we obtain the error estimate(29)uβδ0,y,t−u0,y,t≤δ1/2E1/2+E14lnEδ−p.Proof.
Due to Parseval’s identity and the triangle inequality, we have(30)uβδ0,y,t−u0,y,t=u^βδ0,⋅,⋅−u^0,⋅,⋅=kβ0,ξ,ηg^δ−eA+iBg^=kβ0,ξ,ηg^δ−kβ0,ξ,ηg^+kβ0,ξ,ηg^−eA+iBg^≤kβ0,ξ,ηg^δ−kβ0,ξ,ηg^+kβ0,ξ,ηg^−eA+iBg^=I3+I4,where(31)I3=kβ0,ξ,ηg^δ−kβ0,ξ,ηg^=kβ0,ξ,ηgδ−g≤δsupξ,η∈ℝkβ0,ξ,η≤δβ,(32)I4=kβ0,ξ,ηg^−eA+iBg^=eA+iB−mineA,βeiBeA+iBu^0,ξ,η=eA−mineA,βeiB1+ξ2+η2p/2eA+iB1+ξ2+η2p/2u^0,ξ,η≤supξ,η∈ℝeA−mineA,βeiB1+ξ2+η2p/2eA+iBE≤Esupξ,η∈ℝ,eA>βeA−β1+ξ2+η2p/2eA.
Using the factγ=iηα+ξ2≤ηα/2+ξ, we can get(33)A≤A2+B2=γ≤ηα/2+ξ,i.e.,A≤ηα/2+ξ.
Note that alsoβ<eA, and we have A>lnβ; therefore, we can derive(34)lnβ≤ηα/2+ξ.
Ifξ>ηα/2≥0, we know that(35)lnβ≤ηα/2+ξ≤2ξ,i.e., ξ>1/2lnβ. Therefore,(36)1+ξ2+η2−p/2≤ξ2+η2−p/2<ξ−p<12lnβ−p.
Ifηα/2≥ξ≥0, we have(37)lnβ≤ηα/2+ξ≤2ηα/2,i.e., η≥1/2lnβ2/α. Therefore,(38)1+ξ2+η2−p/2≤ξ2+η2−p/2≤η−p<12lnβ−2p/α.
Combining (32), (36), and (38), we can obtain(39)I4≤Esupξ,η∈ℝ,eA>β11+ξ2+η2p/2≤Emax12lnβ−p,12lnβ−2p/α.
Combining (28), (30), (31), and (39), we can obtain(40)uβδ0,y,t−u0,y,t≤I3+I4≤δ1/2E1/2+Emax14lnEδ−p,14lnEδ−2p/α=δ1/2E1/2+E14lnEδ−p.


## 3.2. A Posteriori Parameter Choice

In this section, we first give the following lemma.Lemma 1 (see [40]).
Let the functionfλ:0,a⟶ℝ be given by(41)fλ=λbdln1λ−c,with a constant c∈ℝ and positive constants a<1, b, and d; then, for the inverse function, we have(42)f−1λ=λ1/bdbln1λc/b1+o1,forλ⟶0.
In this following, we give the convergence estimate foruβδx,⋅,⋅−ux,⋅,⋅ by using a posteriori choice rule for the regularization parameter, i.e., Morozov’s discrepancy principle.
According to Morozov’s discrepancy principle [41], we adopt the regularization parameter β as the solution of the following equation:(43)e−A+iB1−xkβx,ξ,ηg^δ−g^δ=τδ,where τ>1 is a constant. Denote(44)Tβ=e−A+iB1−xkβx,ξ,η−1.Here and in the following,β is a function of the variable x and sometimes the function βx is also used.Lemma 2.
Letρβ≔e−A+iB1−xkβx,ξ,ηg^δ−g^δ, and then, the following results hold:(a)
ρβ is a continuous function(b)
limβ⟶1ρβ=e−A1−x−1g^δξ,η(c)
limβ⟶+∞ρβ=0(d)
ρβ is a strictly decreasing function for β∈1,+∞
The proof is obvious, and we omit it here.Remark 3.
According to Lemma2, we find that, if 0<τδ<e−A1−x−1g^δξ,η, equation (43) has a unique solution.Lemma 3.
Ifβx is the solution of equation (43), then the following inequality holds(45)βx≤τ−1δEx−112lnEτ−1δ−p1−x1+o1.Proof.
Due to the triangle inequality and (43), note that Tβ≤1, and we obtain(46)τδ=Tβg^δ≤Tβg^δ−g^+Tβg^≤δ+Tβg^.
We now estimate the second term on the right hand side of (46):(47)Tβg^=e−A+iB1−xkβx,ξ,η−1e−A+iBu^0,⋅,⋅=1−e−A+iB1−xkβx,ξ,ηeA+iB1+ξ2+η2p/21+ξ2+η2p/2u^0,⋅,⋅≤Esupξ,η∈ℝ,eA1−x>βx1−e−A1−xβxeA1+ξ2+η2p/2=Esupξ,η∈ℝ,eA1−x>βxeA1−x−βxeA2−x1+ξ2+η2p/2=Esupξ,η∈ℝ,eA1−x>βxeA1−x−βxeA1−xe−A1+ξ2+η2−p/2≤Esupξ,η∈ℝ,eA1−x>βxe−A1+ξ2+η2−p/2.
Similar to the proof of Theorem2, we can know(48)Tβg^≤Esupξ,η∈ℝ,eA1−x>βxe−A1+ξ2+η2−p/2≤Eβx−1/1−xmax121−xlnβx−p,121−xlnβx−2p/α=Eβx−1/1−x121−xlnβx−p.
Combining (46) and (48), we obtain(49)τ−1δ≤Eβx−1/1−x121−xlnβx−p.
Letτ−1δ=Eβ∗x−1/1−x1/21−xlnβ∗x−p. From Lemma 1, we have(50)β∗x=τ−1δEx−112lnEτ−1δ−p1−x1+o1.
LetGβ≔Eβx−1/1−x1/21−xlnβx−p. For β>1 and 0≤x<1, the function Gβ is strictly decreasing monotonically with β, so we have(51)βx≤β∗x.Lemma 4.
Ifβx is the solution of equation (43), then the following inequality holds:(52)Tβg^≤τ+1δ.Proof.
Using the triangle inequality and (43), we have(53)Tβg^=Tβg^−g^δ+Tβg^δ≤Tβg^−g^δ+Tβg^δ≤τ+1δ.Theorem 3.
Assume that conditions (9) and (17) hold and taking the solution of equation (43) as the regularization parameter, then there holds the following error estimate:(54)uβδx,⋅,⋅−ux,⋅,⋅≤δxE1−x12lnEτ−1δ−p1−xC+o1,where C=τ−1x−12τ+1, 0≤x<1.Proof.
Due to the Parseval formula and the triangle inequality, we obtain(55)uβδx,⋅,⋅−ux,⋅,⋅=u^βδx,⋅,⋅−u^x,⋅,⋅=kβg^δ−eA+iB1−xg^≤kβg^δ−g^+kβ−eA+iB1−xg^≤βxδ+kβe−A+iB1−x−1u^x,⋅,⋅=βxδ+Tβu^x,⋅,⋅.
From Lemma3, we can know that(56)βxδ≤Eτ−11−xδx12lnEτ−1δ−p1−x1+o1.
In the following, we estimate the second term of (55). Denote φ=Tβu^x,⋅,⋅ and take βx=E/τ−1δ1−x1/2lnE/τ−1δ−p1−x; we have(57)φ=φ−kβe−A+iB1−xφ+kβe−A+iB1−xφ≤φ−kβe−A+iB1−xφ+kβe−A+iB1−xφ=M1+M2.
Note thatu^x,⋅,⋅=eA+iB1−xg^⋅,⋅, we get(58)M1=TβeA+iB1−xg^1−kβe−A+iB1−x=TβeA+iB1−x−kβg^≤eA+iB1−x−kβg^.
DenoteM≔eA+iB1−x−kβg^, we have(59)M=eA+iB1−x−kβg^≤Esupξ,η∈ℝeA+iB1−x−kβeA+iB1+ξ2+η2p/2≤Esupξ,η∈ℝeA+iB1−x−mineA1−x,βxeiB1−xeA+iB1+ξ2+η2p/2≤Esupξ,η∈ℝ,eA1−x>βxeA1−x−βxeA1+ξ2+η2p/2=Esupξ,η∈ℝ,eA1−x>βxeA1−x−βxeA1−x1+ξ2+η2−p/2e−Ax≤Esupξ,η∈ℝ,eA1−x>βx1+ξ2+η2−p/2e−Ax.
Similar to the proof of Theorem2, we can know that(60)M≤Esupξ,η∈ℝ,eA1−x>βx1+ξ2+η2−p/2e−Ax≤Eβx−x/1−xmax121−xlnβx−p,121−xlnβx−2p/α=Eβx−x/1−x121−xlnβx−p=E121−xlnEτ−1δ1−x12lnEτ−1δ−p1−x−p×Eτ−1δ1−x12lnEτ−1δ−p1−x−x/1−x=E12lnEτ−1δ+ln12lnEτ−1δ−p/2−pEτ−1δ−x12lnEτ−1δpx=E12lnEτ−1δ−p1+ln1/2lnE/τ−1δ−p/21/2lnE/τ−1δ−pτ−1δEx12lnEτ−1δpx=τ−1xδxE1−x12lnEτ−1δ−p1−x1+o1.
ForM2, from Lemma 4 and assuming that kβ≤βx, we have(61)M2≤βxe−A+iB1−xφ=βxTβg^≤τ+1τ−1x−1δxE1−x12lnEτ−1δ−p1−x.
Combining (56)–(61) with (55), the proof of Theorem 3 is completed.


## 4. Numerical Aspect

In this section, we present two numerical examples intended to illustrate the behavior of the proposed method.The numerical examples are constructed in the following way. First, we present the initial datau0,y,t=fy,t of 2D time-fractional diffusion problem at x=0 and computed the function u1,y,t=gy,t by solving a direct problem, which is a well-posed problem. Then, we added a random distributed perturbation to the data function obtaining vector gδy,t, i.e.,(62)gδy,t=gy,t+εrandnsizegy,t,where(63)gy,t=gy1,t1,gy2,t2,…,gyn,tnT,yi=i−1Δy,Δy=1n−1,i=1,2,…,n,tj=j−1Δt,Δt=1n−1,j=1,2,…,n,where δ indicates the error level of g, i.e.,(64)δ≔1n2∑i=1n∑j=1ngδyi,tj−gyi,tj2.In numerical implementations, we give the datafy,t and sample at an equidistant grid in the domain 0,1×0,1 with 64×64 grid points. The function “randn⋅” generates arrays of random numbers whose elements are normally distributed with mean 0, variance σ2=1, and standard deviation σ=1; “randn sizegy,t” returns an array of random entries which has the same size as gy,t. Let RMS denotes the mean square for a sampled function φ⋅,⋅ which is defined by(65)RMSφ≔1n2∑i=1n∑j=1nφyi,tj2,where n is the total number of test points. Similarly, we can define the mean square error (RMSE) between the computed data and the exact data.Finally, we solved the 2D time-fractional inverse diffusion problem by the modified method.Example 1.
We consider a smooth function(66)u0,y,t=fy,t=16sin2πysin2πt.Example 2.
LetΩ=y,t0.1≤y≤0.3,0.1≤t≤0.3∪y,t0.6≤y≤0.8,0.6≤t≤0.8. Consider a nonsmooth function(67)fy,t=1,ify,t∈Ω,0,if else.
Figures1–3 are the comparison of the exact solution and the approximation solution for α=0.1 with noise level ε=0.001 at different points x=0.9,0.5,and0.1 under a priori regularization parameter choice rule with Example 1. Figures 4–6 are the comparison of the exact solution and the approximation solution for α=0.1 with noise level ε=0.0001 at different points x=0.9,0.5,and0.1 under a posteriori regularization parameter choice rule with Example 1.
Table1 shows the comparison of RMSE of Example 1 for different x with ε=0.0001andα=0.1. We find that the smaller the x is, the worse the computed approximation will be.
Table2 shows the comparison of RMSE of Example 1 for different ε with α=0.1 at x=0.5. The numerical error is decreasing as the level of noise becomes smaller.
Table3 shows the comparison of RMSE of Example 1 for different α with ε=0.0001 at x=0.5. The numerical accuracy is stable to the fractional order α.
Figure7 is the comparison of the exact solution and the approximation solution for α=0.1 with noise level ε=0.0001 at x=0.4 under both a priori and a posteriori regularization parameter choice rule with Example 2.Figure 1
The exact solution and its approximation solution atε=0.001 and α=0.1: (a) the exact solution at x=0.9; (b) its approximation solution at x=0.9.
(a)
(b)Figure 2
The exact solution and its approximation solution atε=0.001 and α=0.1: (a) the exact solution at x=0.5; (b) its approximation solution at x=0.5.
(a)
(b)Figure 3
The exact solution and its approximation solution atε=0.001 and α=0.1: (a) the exact solution at x=0.1; (b) its approximation solution at x=0.1.
(a)
(b)Figure 4
The exact solution and its approximation solution atε=0.0001andα=0.1: (a) the exact solution at x=0.9; (b) its approximation solution at x=0.9.
(a)
(b)Figure 5
The exact solution and its approximation solution atε=0.0001andα=0.1: (a) the exact solution at x=0.5; (b) its approximation solution at x=0.5.
(a)
(b)Figure 6
The exact solution and its approximation solution atε=0.0001andα=0.1: (a) the exact solution at x=0.1; (b) its approximation solution at x=0.1.
(a)
(b)Table 1
Error behavior of Example1 for different x with α=0.1andε=0.0001.
x
0.5
0.3
0.1
A priori choice rule
:
RMSE
4.59
×
10
−
2
1.73
×
10
−
1
6.37
×
10
−
1
A posteriori choice rule
:
RMSE
4.55
×
10
−
2
1.63
×
10
−
1
5.88
×
10
−
1Table 2
Error behavior of Example1 for different ε with α=0.1 at x=0.5.
ε
0.01
0.001
0.0001
A priori choice rule
:
RMSE
9.22
×
10
−
2
5.16
×
10
−
2
4.59
×
10
−
2
A posteriori choice rule
:
RMSE
8.94
×
10
−
2
4.90
×
10
−
2
4.55
×
10
−
2Table 3
Error behavior of Example1 for different α with ε=0.0001 at x=0.5.
α
0.1
0.3
0.8
A priori choice rule
:
RMSE
4.59
×
10
−
2
4.38
×
10
−
2
3.89
×
10
−
2
A posteriori choice rule
:
RMSE
4.55
×
10
−
2
4.35
×
10
−
2
3.88
×
10
−
2Figure 7
The exact solution and its approximation solution atε=0.0001andα=0.1: (a) the exact solution at x=0.4; (b) its approximation solution under a priori parameter choice rule at x=0.4; (c) its approximation solution under a posteriori parameter choice rule at x=0.4.
(a)
(b)
(c)


## 5. Conclusion

In this paper, we propose a modified method to solve the time-fractional inverse diffusion problem in the two-dimensional setting. For the choice of regularization parameter, we give not only a priori but also a posteriori rules. Moreover, under both a priori rule and a posteriori rule, we prove the error estimates from the viewpoint of theoretical analysis.The numerical examples are presented to illustrate the validity and effectiveness of the proposed method. In general,E>0 such that a priori condition (17) holds will not be known. But such a E has to be known if one wants to construct a priori parameter choice rule. We propose a posteriori parameter choice rule which is independent of E. So it is more implementable in practical application. Moreover, the numerical experiments show that a posteriori parameter choice rule also works well.



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