# Regularization of Inverse Initial Problem for Conformable Pseudo-Parabolic Equation with Inhomogeneous Term

Authors: L. D. Long; Reza Saadati
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The main goal of the paper is to approximate two types of inverse problems for conformable heat equation (or called parabolic equation with conformable operator); as follows, we considered two cases: the right hand side of equation such thatFx,t and Fx,t=φtfx. Up to now, there are very few surveys working on the results of regularization in Lp spaces. Our paper is the first work to investigate the inverse problem for conformable parabolic equations in such spaces. For the inverse source problem and the backward problem, use the Fourier truncation method to approximate the problem. The error between the regularized solution and the exact solution is obtained in Lp under some suitable assumptions on the Cauchy data.


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## Body

## 1. Introduction

Partial differential equations (PDEs) have applications in many branches of science and engineering; see for example [1–8]. In this paper, for s>1, we consider the initial value problem for the conformable heat equation (or called parabolic equation with conformable operator)
(1)∂Cβ∂tβyx,t−kΔyx,t+−Δsyx,t=Fx,t,x∈D,t∈0,Tyx,t=0,x∈∂D,t∈0,Tyx,T=yTx,x∈D.Here,D⊂ℝN (N≥1) is a bounded domain with the smooth boundary ∂D, and T>0 is a given positive number. Here, C∂β/∂tβ is called the conformable time derivative with order β∈0,1 (Khalil et al. [9]) for a given function f:0,∞⟶ℝ; the C∂β/∂tβ of order β∈0,1 is defined by
(2)C∂β∂tβft=limϵ⟶0ft+ϵt1−β−ftϵ,for all t>0. For some 0,t0, t0>0 and the limt⟶t0+C∂β/∂tβft exist, then C∂β/∂tβft0=limt⟶t0+C∂β/∂tβft. Some properties of C∂β/∂tβ can be found in more detail in [10, 11]. C∂β/∂tβ is a natural extension of usual derivative, it preserves basic properties of the classical derivative [10, 11], and it is a local and limit-based operator. In [10, 12, 13], we saw some applications. For the convenience of the reader, we will consider two models related to Problem (1) that most mathematicians often study.
(i)
The first part of the paper deals with the final value problem for Problem (1) with a linear source function. The new feature of this part is the appearance of observed data, namely, yT,δ,Fδ∈LpD×L∞0,T;LpD. This result is well described in Theorem 3. We investigated the problem of restoring the temperature function yx,t, in the fact that the couple yT,F are noised by the measurement data yT,δ,Fδ such that:(3)yT,δ−yTLpD≤δFδ−FL∞0,T;LpD≤δ.(ii)
The second part of the paper deals with the final value problem for Problem (1) with F is a linear source function as follows: Fx,t=Φtfx, where both functions Φ,f,g are perturbed by Φδ,fδ,gδ in Lp0,T×LpD×LpD, respectively(4)Φδ−ΦLp0,T≤δyT,δ−yTLp0,D≤δfδ−fLpD≤δ.The main contributions and novelties of this paper are stated as follows. As we know, two inverse problems are ill-posed in the sense of Hadamard. The well-posed problem satisfies three conditions above: the solution is existence, the solution is uniqueness, and the solution continues on data The problem that violates one of the above three conditions is an ill-posed problem. We need to regularize this problem, to give a good approximation. The number of works on the regularized problem with input data inL2 is quite abundant. The results of this study can be found in the following documents, attached to the regularization methods: the Tikhonov method, see [14, 15], the Fractional Tikhonov method, see [16], the fractional Landweber method, see [17, 18], the Quasi Boundary method, see [19], the truncation method, see [20], and their references.However, forp≠2, results for regularized problem in Lp are quite rare. We confirm that our paper is the first result for the inverse problem for the conformable parabolic equation when the observed data is in the Lp space with p≠2. If the data is not in L2, the use of Parseval equality is not feasible. In this case, we used the embedding between Lp and Hilbert scales spaces XsD. These results are well described in Theorem 3 and Theorem 5. The main analytical technique in our paper is to use some embeddings and some analysis estimators related to Hölder inequality. To do this, we learn many interesting techniques from N.H. Tuan [21].This paper is organized as follows. In Section2, we state some function spaces and embeddings. In Section 3, we deal with the regularized solution for the inverse source problem for (1). Section 4 gives the mild solution of backward problem in case F=0. After that, we solve two problems in the case of observed data in Lp space.


## 2. Preliminary Results

Let us recall that the spectral problem(5)−Δsemx=λmsemx,x∈Demx=0,x∈∂Dadmits the eigenvalues 0<λ1≤λ2≤⋯≤λm≤⋯ with λm⟶∞ as m⟶∞. The corresponding eigenfunctions are em∈H01D.Definition 1 (Hilbert scale space).
We recall the Hilbert scale space, which is given as follows:(6)XnD=f∈L2D,∑m=1∞λm2n∫Dfxemxdx2<∞,
for anyn≥0. It is well-known that XnD is a Hilbert space corresponding to the norm
(7)fXnD=∑m=1∞λm2n∫Dfxemxdx21/2,f∈XnD.Lemma 2 (See [22]).
The following statement is true:(8)LpD↪XμD,if−N4<μ≤0,p≥2NN−4μXμD↪LpD,if0≤μ<N4,p≤2NN−4μ.


## 3. Regularization of Backward Problem

In order to find a precise formulation for solutions, we consider the mild solution in Fourier seriesyx,t=∑m=1∞ymtemx, with ymt=∫Dyx,temxdx. Taking the inner product of the equations of Problem (1) with em gives
(9)C∂β∂tβy.,t,em+kλmC∂β∂tβy.,t,em−λmsy.,t,em=F.,t,em,t∈0,Ty.,0,em=y0,em.The first equation of (9) is a differential equation with a conformable derivative as follows:
(10)C∂β∂tβymt−λms1+kλm−1ymt=1+kλm−1Fnt.Because of the result in [23], the solution of Problem (1) is
(11)y.,t,em=exp−λms1+kλmtββy0,em+11+kλm∫0tθβ−1expλms1+kλmθβ−tββF.,θ,emdθ.Lettingt=T, we follow from (11) that
(12)∫DyTxemxdx=exp−λms1+kλmTββ∫Dy0xemxdx+11+kλm∫0Tθβ−1expλms1+kλmθβ−Tββ∫DF.,θemxdxdθ.From (12), we have
(13)∫Dy0xemxdx=exp−λms1+kλmTββ−1∫DyTxemxdx−11+kλm∫0Tθβ−1expλms1+kλmθβ−Tββ∫DFx,θemxdxdθ.Substituting (13) into (12), we obtain
(14)y.,t,em=expλms1+kλmTβ−tββ∫DyTxemxdx−11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFx,θemxdxdθThis leads to(15)yx,t=∑m=1+∞expλms1+kλmTβ−tββ∫DyTxemxdxemx−∑m=1+∞11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFx,θemxdxdθemx.


## 4. The Mild Solution of Backward Problem in CaseF=0

In this section, we investigate the existence and regularity of mild solutions of Problem (1). Firstly, we consider the following initial value problem
(16)C∂β∂tβyx,t−kΔyx,t+−Δsyx,t=0,x∈D,t∈0,Tyx,t=0,x∈∂D,t∈0,Tyx,T=yTx,x∈D,t∈0,T.According to (15), in this case, we have
(17)yx,t=∑m=1+∞expλms1+kλmTβ−tββ∫DyTxemxdxemx.


### 4.1. The Ill-Posedness of Problem (1)

In order to prove that the solution to the backward problem is unstable atFx,t=0, let us take the perturbed final datayT,jx∈L2D, by choosingyT,jx=ejxλj−1/2. For s>3/2, let us choose input final data yTx=0; we know that an error in L2D norm between two input final data as follows:
(18)yT,j−yTL2D=ejλj−1/2L2D=λj−1/2thisleadstolimj⟶∞yT,j−yTL2D=limj⟶∞λj−1/2=0.Therefore, we obtain(19)yjx,t=expλjs1+kλjTβ−tββyT,jx.First of all, we haveλjs/1+kλj=λjs−1/1/λj+k≥λjs−1/1/λj+k, and Tβ−tβ/β≥0; this implies that
(20)expλs1+kλjTβ−tββ≥expλjs−1λ1−1+kTβ−tββ.Next, using the inequalityexpx≥x, for x>0, this leads to:
(21)yj·,tL2D≥expλjs−1λ1−1+kTβ−tββ1λj1/2L2D≥λjs−3/2λ1−1+kTβ−tββ·Fors>3/2, and from (21), we get
(22)limj⟶∞yj.,tL2D≥limj⟶∞λjs−3/2λ1−1+kTβ−tββ⟶+∞.Thus, Problem (1), in general, ill-posed in the Hadamard sense in L2D-norm.


### 4.2. Regularization of inverse Problem (1) in LpD space

From (15), we know that the explicit formula of the mild solution
(23)yx,t=∑m=1+∞expλms1+kλmTβ−tββ∫DyTxemxdxemx−∑m=1+∞11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFx,θemxdxdθemx.By applying the Fourier truncation method, we have its approximation(24)yδx,t=∑m≤Mδexpλms1+kλmTβ−tββ∫DyT,δxemxdxemx−∑m≤Mδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFδx,θemxdxdθemx.Here,Mδ is parameter regularization which is defined later.Theorem 3.
Fors>1, taking yT,F∈Lp0,T×L∞0,T;LpD for any 0≤t≤T for any 1/β<p<2, assume that yT,F is observed by the couple yT,δ,Fδ such that
(25)yT,δ−yTLpD+F−FδL∞0,T;LpD≤δ,δ>0.
Let us assume thatu∈L∞0,T;Xn+σ for σ>0 and 0<n<N/4. With Mδ such that
(26)limδ⟶0Mδ=+∞,limδ⟶0Mδn+N/2p−N/4expTββMδs−1k−1δ=0.
Then, the error estimate(27)yδ−yL2N/N−4nDisofordermaxMδn+N/2p−N/4expMδs−1k−1Tββδ,Mδ−σ.Remark 4.
One choice forMδ such that
(28)Mδ=T−ββ1−αk1/s−1log1δ1/s−1,for0<α<1.then yδ−yL2N/N−4nDisofordermaxδαlog1/δn+N/2p−N/4/s−1,log1/δ−σ/s−1.Proof.
Let(29)Vδx,t=∑m≤Nδexpλms1+kλmTβ−tββ∫DyTxemxdxemx−∑m≤Mδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFx,θemxdxdθemx.
It is clear that(30)yδ·,t−y·,tXnD≤yδ·,t−Vδ·,tXnD+Vδ·,t−y·,tXnD.
We continue to consider the two components of the right hand side.
Step 1:(31)yδx,t−Vδx,t=∑λm≤Mδexpλms1+kλmTβ−tββ∫DyT,δx−yTxemxdxemx−∑λm≤Mδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFδx,θ−Fx,θemxdxdθemx=H1x,t−H2x,t.
Fors>1, it is easy to see that λms1+kλm−1≤λms−1λm−1+k−1≤λms−1k−1. The first term H1x,t on XnD is bounded by
(32)H1.,tXnD2=∑λm≤Mδλm2nexp2λms1+kλmTβ−tββ∫DyT,δx−yTxemxdx2=∑λm≤Mδλm2n+N/p−N/2λmNp−2N/2pexp2λms−1k−1Tββ∫DyT,δx−yTxemxdx2≤Mδ2n+N/p−N/2exp2Mδs−1k−1Tββ∑λm≤MδλmNp−2N/2p∫DyT,δx−yTxemxdx2≤Mδ2n+N/p−N/2exp2Mδs−1k−1TββyT,δ−yTXNp−2N/4p·
Since the Sobolev space embeddingLpD⟶XNp−2N/4pD, we have
(33)yT,δ−yTXNp−2N/4pD≤C1N,pyT,δ−yTLpD.
This follows from (32) that
(34)H1.,tXnD≤Mδn+N/2p−N/4expMδs−1k−1TββC1N,pδ.
The second termH2x,t is estimated as follows:
(35)H2.,tXnD2=∑λm≤Mδλm2n1+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFx,θ−Fδx,θemxdxdθ2
By the same arguments as above, we find that(36)H2·,tXnD2≤∑λm≤Mδλm2n∫tTθβ−1expλms−1k−1θβ−tββ∫DFx,θ−Fδx,θemxdxdθ2≤exp2Mδs−1k−1Tββ∑λm≤Mδλm2n∫tTθβ−1∫DFx,θ−Fδx,θemxdxdθ2
We can see thatλm2n+N/p−N/2≤Mδ2n+N/p−N/2 and ∫tTθβ−1dθ≤Tβ−tβ/β≤Tβ/β. From (36), using Holder’s inequality, we get
(37)∑λm≤Mδλm2n∫tTθβ−1∫DFx,θ−Fδx,θemxdxdθ2≤∑λm≤Mδλm2n+N/p−N/2λmNp−2N/2p∫tTθβ−1dθ∫tTθβ−1∫DFx,θ−Fδx,θemxdx2dθ≤β−1TβMδ2n+N/p−N/2∫tTθβ−1∑λm≤Mδλm2n+N/p−N/2∫DFx,θ−Fδx,θemxdx2dθ≤β−1TβMδ2n+N/p−N/2∫tTθβ−1Fδ·,t−F·,tXNp−2N/4p2dθ.
This latter inequality together with Sobolev embeddingLpD⟶XNp−2N/4pD gives us
(38)∑λm≤Mδλm2n∫tTθβ−1∫DFx,θ−Fδx,θemxdxdθ2≤β−1C2N,p2TβMδ2n+N/p−N/2∫tTθβ−1Fx,θ−Fδx,θLpD2dθ≤β−2C2N,p2T2βMδ2n+N/p−N/2F−FδL∞0,T;LpD≤β−2C2N,p2T2βMδ2n+N/p−N/2δ2.
Combining (36) and (38), we get
(39)H2·,tXnD2≤exp2Mδs−1k−1TββC2N,p2T2βMδ2n+N/p−N/2β−2δ2.
Taking the square root on the both sides, we have(40)H2·,tXnD≤expMδs−1k−1TββC2N,pTβMδn+N/2p−N/4β−1δ.
From (34) and (40), we deduce that
(41)yδ·,t−Vδ·,tXnD≤H1·,tXnD+H2·,tXnD≤Mδn+N/2p−N/4expMδs−1k−1TββC1N,pδ+expMδs−1k−1TββC2N,pTβMδn+N/2p−N/4β−1δ≤Mδn+N/2p−N/4expMδs−1k−1TββδC1N,p+C2N,pTββ−1.
Step 2: Estimate ofu·,t−Vδ·,tXnD.
From the definition (23) and (29), we have
(42)y·,t−Vδ·,tXnD2=∑λm>Mδλm2n∫Du·,temxdx2=∑λm>Mδλm−2σλm2n+2σ∫Du·,temxdx2≤Mδ−2σuL∞0,T;Xn+σD2.
Therefore, we get(43)y·,t−Vδ·,tXnD≤Mδ−σuL∞0,T;Xn+σD.
Combining two steps and noting thatXnD↪L2N/N−4n,0<n<N/4, we deduce that
(44)yδ·,t−y·,tLD≤C3N,nyδ·,t−Vδ·,tXnD+C3N,nVδ·,t−y·,tXnD≤C3N,nMδn+N/2p−N/4expMδs−1k−1TββδC1N,p+C2N,pTββ−1+C3N,nMδ−σuL∞0,T;Xn+σD.
The proof of Theorem3 is completed. In the following theorem, we give a regularization result in the case that F has a split form Fx,t=Φtfx.Theorem 5.
Fors>1, let us assume that the input data Φδ,gδ,fδ such that
(45)Φδ−ΦLp0,T+yT,δ−yTLpD+fδ−fLpD≤δ.
Assume thatu∈L∞0,T;Xn+σD for any σ>0, then we construct a regularized solution defined by
(46)Wδx,t=∑m≤Bδexpλms1+kλmTβ−tββ∫DyT,δxemxdxemx−∑m≤Bδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DfxemxdxΦδθdθemx.
Then, the errorWδ.,t−y.,tL2N/N−4nD is of order maxδBδ−σ,Bδn+N/2p−N/4expBδs−1k−1Tβ/βδ.Remark 6.
Bδ=T−ββk1/s−11−α1/s−1log1/δ1/s−1, then the error
(47)Wδ.,t−y.,tL2N/N−4nDisofordermaxδlog1δ−σ/s−1log1δn+N/2p−N/4/s−1δα.Proof.
SinceFx,t=Φtfx, we know that
(48)Zδx,t=∑m≤Bδexpλms1+kλmTβ−tββ∫DyTxemxdxemx−∑m≤Bδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DfxemxdxΦθdθemx.
The triangle inequality allows us to obtain that(49)Wδ·,t−y·,tXnD≤Wδ·,t−Zδ·,tXnD+Zδ·,t−y·,tXnD.
Next, we will evaluate the right side of (49), by the same way as demonstrated in (42),
(50)y·,t−Zδ·,tXnD2≤Bδ−2σuL∞0,T;Xn+σD2.
It is easy to see that(51)Wδx,t−Zδx,t=J1x,t+J2x,t+J3x,t.whereby
(52)J1x,t=∑λm≤Bδexpλms1+kλmTβ−tββ∫DyT,δx−yTxemxdxemx,J2x,t=−∑λm≤Bδ∫tTθβ−1expλms1+kλmθβ−tββΦδθdθ∫Dfδx−fxemxemx,J3x,t=∑λm≤Bδ∫tTθβ−1expλms1+kλmθβ−tββΦθ−Φδθdθ∫Dfxemxdxemx.
We will divide this review into several steps as follows:
Step 1: Estimate ofJ1·,tXnD, we obtain that
(53)J1·,tXnD≤C4N,pBδn+N/2p−N/4expBδs−1k−1Tββδ.
Step 2: Due to Parseval’s equality, the termJ2·,tXnD can be bounded as follows:
(54)J2·,tXnD=∑λm≤Bδλm2n∫tTθβ−1expλms1+kλmθβ−tββΦδθdθ2∫Dfδx−fxemxdx2.
Thank to Holder’s inequality, we derive that forp>1 and p∗=1+1/p−1, one has
(55)∫tTθβ−1expλms1+kλmθβ−tββΦδθdθ≤∫0TΦδpdθ1/p∫tTθp∗β−1expp∗λms1+kλmθβ−tββdθ1/p∗≤expλms−1k−1TββΦδLp∗0,T∫0Tθp∗β−1dθ1/p∗≤p−1pβ−1p−1/pTpβ−1/pexpλms−1k−1TββΦδLp0,T.
The latter inequality leads to(56)J2.,tXnD2=∑λm≤Bδλm2n∫tTθβ−1expλms1+kλmθβ−tββΦδθdθ2∫Dfδx−fxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pΦδθLp0,T2×∑λm≤Bδλm2n+N/p−N/2λmNp−2N/2pexp2Tβλms−1k−1β∫Dfδx−fxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pΦδθLp0,T2Bδ2n+N/p−N/2exp2TβBδs−1k−1β×∑λm≤Bδλm2n+N/p−N/2∫Dfδx−fxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pΦδθLp0,T2Bδ2n+N/p−N/2×exp2TβBδs−1k−1βfδ−fXNp−2N/4pΩ.where s>1/β. In view of Sobolev embedding LpD°XNp−2N/4pD, we derive that the following estimate
(57)J2.,tXnD≤C6p,β,T,NΦδLp0,TBδn+N/2p−N/4expTβBδs−1k−1βδ,where
(58)C6p,β,T,N=p−1pβ−1p−1/pTpβ−1/pC5N,p.
Step 3: Let us now to consider the termJ3.,tXnD. By applying Hölder’s inequality, we get that for p>1 and p∗=1/p−1·(59)∫tTθβ−1expλms1+kλmνβ−tββΦθ−Φδθdθ≤∫0TΦθ−Φδθsdν1/p∫tTθp∗β−1expp∗λms1+kλmθβ−tββdν1/p∗≤expλms−1k−1TββΦδ−ΦLp0,T∫tTνs∗β−1dν1/p∗≤p−1pβ−1p−1/pTpβ−1/pexpTβλms−1k−1βΦ−ΦδLp0,T≤p−1pβ−1p−1/pTpβ−1/pexpTβλms−1k−1βδ.
This inequality together with Parseval’s equality allows us to derive that(60)J3.,tXnD2=∑λm≤Bδλm2n∫tTθβ−1expλms1+kλmθβ−tββΦθ−Φδθdθ2∫Dfxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pδ2∑λm≤Bδλm2n+N/p−N/2λmNp−2N/2pexp2λms−1k−1Tββ∫Dfxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pδ2Bδ2n+N/p−N/2exp2TβBδs−1k−1βfXNp−2N/4pΩ.
By the fact thatLpD°XNp−2N/4pD, we deduce that
(61)J3.,tXnD≤C6p,β,T,NfLpDBδn+N/2p−N/4expTβBδs−1k−1βδ.
Combining Step 1 to Step 3, we get(62)J1.,tXnD+J2.,tXnD+J3.,tXnD≤C4N,pBδn+N/2p−N/4expBδs−1k−1Tββδ+C6p,β,T,NBδn+N/2p−N/4expTβBδs−1k−1βδΦδLp0,T+fLpD.
Finally, combining the reviews from above, we conclude that(63)Wδ·,t−y·,tXnD≤δBδ−σuL∞0,T;Xn+σD+C4N,pBδn+N/2p−N/4expBδs−1k−1Tββδ+C6p,β,T,NBδn+N/2p−N/4expTβBδs−1k−1βδΦδLp0,T+fLpD.


## 4.1. The Ill-Posedness of Problem (1)

In order to prove that the solution to the backward problem is unstable atFx,t=0, let us take the perturbed final datayT,jx∈L2D, by choosingyT,jx=ejxλj−1/2. For s>3/2, let us choose input final data yTx=0; we know that an error in L2D norm between two input final data as follows:
(18)yT,j−yTL2D=ejλj−1/2L2D=λj−1/2thisleadstolimj⟶∞yT,j−yTL2D=limj⟶∞λj−1/2=0.Therefore, we obtain(19)yjx,t=expλjs1+kλjTβ−tββyT,jx.First of all, we haveλjs/1+kλj=λjs−1/1/λj+k≥λjs−1/1/λj+k, and Tβ−tβ/β≥0; this implies that
(20)expλs1+kλjTβ−tββ≥expλjs−1λ1−1+kTβ−tββ.Next, using the inequalityexpx≥x, for x>0, this leads to:
(21)yj·,tL2D≥expλjs−1λ1−1+kTβ−tββ1λj1/2L2D≥λjs−3/2λ1−1+kTβ−tββ·Fors>3/2, and from (21), we get
(22)limj⟶∞yj.,tL2D≥limj⟶∞λjs−3/2λ1−1+kTβ−tββ⟶+∞.Thus, Problem (1), in general, ill-posed in the Hadamard sense in L2D-norm.


## 4.2. Regularization of inverse Problem (1) in LpD space

From (15), we know that the explicit formula of the mild solution
(23)yx,t=∑m=1+∞expλms1+kλmTβ−tββ∫DyTxemxdxemx−∑m=1+∞11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFx,θemxdxdθemx.By applying the Fourier truncation method, we have its approximation(24)yδx,t=∑m≤Mδexpλms1+kλmTβ−tββ∫DyT,δxemxdxemx−∑m≤Mδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFδx,θemxdxdθemx.Here,Mδ is parameter regularization which is defined later.Theorem 3.
Fors>1, taking yT,F∈Lp0,T×L∞0,T;LpD for any 0≤t≤T for any 1/β<p<2, assume that yT,F is observed by the couple yT,δ,Fδ such that
(25)yT,δ−yTLpD+F−FδL∞0,T;LpD≤δ,δ>0.
Let us assume thatu∈L∞0,T;Xn+σ for σ>0 and 0<n<N/4. With Mδ such that
(26)limδ⟶0Mδ=+∞,limδ⟶0Mδn+N/2p−N/4expTββMδs−1k−1δ=0.
Then, the error estimate(27)yδ−yL2N/N−4nDisofordermaxMδn+N/2p−N/4expMδs−1k−1Tββδ,Mδ−σ.Remark 4.
One choice forMδ such that
(28)Mδ=T−ββ1−αk1/s−1log1δ1/s−1,for0<α<1.then yδ−yL2N/N−4nDisofordermaxδαlog1/δn+N/2p−N/4/s−1,log1/δ−σ/s−1.Proof.
Let(29)Vδx,t=∑m≤Nδexpλms1+kλmTβ−tββ∫DyTxemxdxemx−∑m≤Mδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFx,θemxdxdθemx.
It is clear that(30)yδ·,t−y·,tXnD≤yδ·,t−Vδ·,tXnD+Vδ·,t−y·,tXnD.
We continue to consider the two components of the right hand side.
Step 1:(31)yδx,t−Vδx,t=∑λm≤Mδexpλms1+kλmTβ−tββ∫DyT,δx−yTxemxdxemx−∑λm≤Mδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFδx,θ−Fx,θemxdxdθemx=H1x,t−H2x,t.
Fors>1, it is easy to see that λms1+kλm−1≤λms−1λm−1+k−1≤λms−1k−1. The first term H1x,t on XnD is bounded by
(32)H1.,tXnD2=∑λm≤Mδλm2nexp2λms1+kλmTβ−tββ∫DyT,δx−yTxemxdx2=∑λm≤Mδλm2n+N/p−N/2λmNp−2N/2pexp2λms−1k−1Tββ∫DyT,δx−yTxemxdx2≤Mδ2n+N/p−N/2exp2Mδs−1k−1Tββ∑λm≤MδλmNp−2N/2p∫DyT,δx−yTxemxdx2≤Mδ2n+N/p−N/2exp2Mδs−1k−1TββyT,δ−yTXNp−2N/4p·
Since the Sobolev space embeddingLpD⟶XNp−2N/4pD, we have
(33)yT,δ−yTXNp−2N/4pD≤C1N,pyT,δ−yTLpD.
This follows from (32) that
(34)H1.,tXnD≤Mδn+N/2p−N/4expMδs−1k−1TββC1N,pδ.
The second termH2x,t is estimated as follows:
(35)H2.,tXnD2=∑λm≤Mδλm2n1+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DFx,θ−Fδx,θemxdxdθ2
By the same arguments as above, we find that(36)H2·,tXnD2≤∑λm≤Mδλm2n∫tTθβ−1expλms−1k−1θβ−tββ∫DFx,θ−Fδx,θemxdxdθ2≤exp2Mδs−1k−1Tββ∑λm≤Mδλm2n∫tTθβ−1∫DFx,θ−Fδx,θemxdxdθ2
We can see thatλm2n+N/p−N/2≤Mδ2n+N/p−N/2 and ∫tTθβ−1dθ≤Tβ−tβ/β≤Tβ/β. From (36), using Holder’s inequality, we get
(37)∑λm≤Mδλm2n∫tTθβ−1∫DFx,θ−Fδx,θemxdxdθ2≤∑λm≤Mδλm2n+N/p−N/2λmNp−2N/2p∫tTθβ−1dθ∫tTθβ−1∫DFx,θ−Fδx,θemxdx2dθ≤β−1TβMδ2n+N/p−N/2∫tTθβ−1∑λm≤Mδλm2n+N/p−N/2∫DFx,θ−Fδx,θemxdx2dθ≤β−1TβMδ2n+N/p−N/2∫tTθβ−1Fδ·,t−F·,tXNp−2N/4p2dθ.
This latter inequality together with Sobolev embeddingLpD⟶XNp−2N/4pD gives us
(38)∑λm≤Mδλm2n∫tTθβ−1∫DFx,θ−Fδx,θemxdxdθ2≤β−1C2N,p2TβMδ2n+N/p−N/2∫tTθβ−1Fx,θ−Fδx,θLpD2dθ≤β−2C2N,p2T2βMδ2n+N/p−N/2F−FδL∞0,T;LpD≤β−2C2N,p2T2βMδ2n+N/p−N/2δ2.
Combining (36) and (38), we get
(39)H2·,tXnD2≤exp2Mδs−1k−1TββC2N,p2T2βMδ2n+N/p−N/2β−2δ2.
Taking the square root on the both sides, we have(40)H2·,tXnD≤expMδs−1k−1TββC2N,pTβMδn+N/2p−N/4β−1δ.
From (34) and (40), we deduce that
(41)yδ·,t−Vδ·,tXnD≤H1·,tXnD+H2·,tXnD≤Mδn+N/2p−N/4expMδs−1k−1TββC1N,pδ+expMδs−1k−1TββC2N,pTβMδn+N/2p−N/4β−1δ≤Mδn+N/2p−N/4expMδs−1k−1TββδC1N,p+C2N,pTββ−1.
Step 2: Estimate ofu·,t−Vδ·,tXnD.
From the definition (23) and (29), we have
(42)y·,t−Vδ·,tXnD2=∑λm>Mδλm2n∫Du·,temxdx2=∑λm>Mδλm−2σλm2n+2σ∫Du·,temxdx2≤Mδ−2σuL∞0,T;Xn+σD2.
Therefore, we get(43)y·,t−Vδ·,tXnD≤Mδ−σuL∞0,T;Xn+σD.
Combining two steps and noting thatXnD↪L2N/N−4n,0<n<N/4, we deduce that
(44)yδ·,t−y·,tLD≤C3N,nyδ·,t−Vδ·,tXnD+C3N,nVδ·,t−y·,tXnD≤C3N,nMδn+N/2p−N/4expMδs−1k−1TββδC1N,p+C2N,pTββ−1+C3N,nMδ−σuL∞0,T;Xn+σD.
The proof of Theorem3 is completed. In the following theorem, we give a regularization result in the case that F has a split form Fx,t=Φtfx.Theorem 5.
Fors>1, let us assume that the input data Φδ,gδ,fδ such that
(45)Φδ−ΦLp0,T+yT,δ−yTLpD+fδ−fLpD≤δ.
Assume thatu∈L∞0,T;Xn+σD for any σ>0, then we construct a regularized solution defined by
(46)Wδx,t=∑m≤Bδexpλms1+kλmTβ−tββ∫DyT,δxemxdxemx−∑m≤Bδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DfxemxdxΦδθdθemx.
Then, the errorWδ.,t−y.,tL2N/N−4nD is of order maxδBδ−σ,Bδn+N/2p−N/4expBδs−1k−1Tβ/βδ.Remark 6.
Bδ=T−ββk1/s−11−α1/s−1log1/δ1/s−1, then the error
(47)Wδ.,t−y.,tL2N/N−4nDisofordermaxδlog1δ−σ/s−1log1δn+N/2p−N/4/s−1δα.Proof.
SinceFx,t=Φtfx, we know that
(48)Zδx,t=∑m≤Bδexpλms1+kλmTβ−tββ∫DyTxemxdxemx−∑m≤Bδ11+kλm∫tTθβ−1expλms1+kλmθβ−tββ∫DfxemxdxΦθdθemx.
The triangle inequality allows us to obtain that(49)Wδ·,t−y·,tXnD≤Wδ·,t−Zδ·,tXnD+Zδ·,t−y·,tXnD.
Next, we will evaluate the right side of (49), by the same way as demonstrated in (42),
(50)y·,t−Zδ·,tXnD2≤Bδ−2σuL∞0,T;Xn+σD2.
It is easy to see that(51)Wδx,t−Zδx,t=J1x,t+J2x,t+J3x,t.whereby
(52)J1x,t=∑λm≤Bδexpλms1+kλmTβ−tββ∫DyT,δx−yTxemxdxemx,J2x,t=−∑λm≤Bδ∫tTθβ−1expλms1+kλmθβ−tββΦδθdθ∫Dfδx−fxemxemx,J3x,t=∑λm≤Bδ∫tTθβ−1expλms1+kλmθβ−tββΦθ−Φδθdθ∫Dfxemxdxemx.
We will divide this review into several steps as follows:
Step 1: Estimate ofJ1·,tXnD, we obtain that
(53)J1·,tXnD≤C4N,pBδn+N/2p−N/4expBδs−1k−1Tββδ.
Step 2: Due to Parseval’s equality, the termJ2·,tXnD can be bounded as follows:
(54)J2·,tXnD=∑λm≤Bδλm2n∫tTθβ−1expλms1+kλmθβ−tββΦδθdθ2∫Dfδx−fxemxdx2.
Thank to Holder’s inequality, we derive that forp>1 and p∗=1+1/p−1, one has
(55)∫tTθβ−1expλms1+kλmθβ−tββΦδθdθ≤∫0TΦδpdθ1/p∫tTθp∗β−1expp∗λms1+kλmθβ−tββdθ1/p∗≤expλms−1k−1TββΦδLp∗0,T∫0Tθp∗β−1dθ1/p∗≤p−1pβ−1p−1/pTpβ−1/pexpλms−1k−1TββΦδLp0,T.
The latter inequality leads to(56)J2.,tXnD2=∑λm≤Bδλm2n∫tTθβ−1expλms1+kλmθβ−tββΦδθdθ2∫Dfδx−fxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pΦδθLp0,T2×∑λm≤Bδλm2n+N/p−N/2λmNp−2N/2pexp2Tβλms−1k−1β∫Dfδx−fxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pΦδθLp0,T2Bδ2n+N/p−N/2exp2TβBδs−1k−1β×∑λm≤Bδλm2n+N/p−N/2∫Dfδx−fxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pΦδθLp0,T2Bδ2n+N/p−N/2×exp2TβBδs−1k−1βfδ−fXNp−2N/4pΩ.where s>1/β. In view of Sobolev embedding LpD°XNp−2N/4pD, we derive that the following estimate
(57)J2.,tXnD≤C6p,β,T,NΦδLp0,TBδn+N/2p−N/4expTβBδs−1k−1βδ,where
(58)C6p,β,T,N=p−1pβ−1p−1/pTpβ−1/pC5N,p.
Step 3: Let us now to consider the termJ3.,tXnD. By applying Hölder’s inequality, we get that for p>1 and p∗=1/p−1·(59)∫tTθβ−1expλms1+kλmνβ−tββΦθ−Φδθdθ≤∫0TΦθ−Φδθsdν1/p∫tTθp∗β−1expp∗λms1+kλmθβ−tββdν1/p∗≤expλms−1k−1TββΦδ−ΦLp0,T∫tTνs∗β−1dν1/p∗≤p−1pβ−1p−1/pTpβ−1/pexpTβλms−1k−1βΦ−ΦδLp0,T≤p−1pβ−1p−1/pTpβ−1/pexpTβλms−1k−1βδ.
This inequality together with Parseval’s equality allows us to derive that(60)J3.,tXnD2=∑λm≤Bδλm2n∫tTθβ−1expλms1+kλmθβ−tββΦθ−Φδθdθ2∫Dfxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pδ2∑λm≤Bδλm2n+N/p−N/2λmNp−2N/2pexp2λms−1k−1Tββ∫Dfxemxdx2≤p−1pβ−12p−2/pT2pβ−2/pδ2Bδ2n+N/p−N/2exp2TβBδs−1k−1βfXNp−2N/4pΩ.
By the fact thatLpD°XNp−2N/4pD, we deduce that
(61)J3.,tXnD≤C6p,β,T,NfLpDBδn+N/2p−N/4expTβBδs−1k−1βδ.
Combining Step 1 to Step 3, we get(62)J1.,tXnD+J2.,tXnD+J3.,tXnD≤C4N,pBδn+N/2p−N/4expBδs−1k−1Tββδ+C6p,β,T,NBδn+N/2p−N/4expTβBδs−1k−1βδΦδLp0,T+fLpD.
Finally, combining the reviews from above, we conclude that(63)Wδ·,t−y·,tXnD≤δBδ−σuL∞0,T;Xn+σD+C4N,pBδn+N/2p−N/4expBδs−1k−1Tββδ+C6p,β,T,NBδn+N/2p−N/4expTβBδs−1k−1βδΦδLp0,T+fLpD.



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