# Two-Parameter Regularization Method for a Nonlinear Backward Heat Problem with a Conformable Derivative

Authors: Vu Ho; Donal O’Regan; Hoa Ngo Van
Journal: Complexity
 (2020)

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## Abstract

In this paper, we consider the nonlinear inverse-time heat problem with a conformable derivative concerning the time variable. This problem is severely ill posed. A new method on the modified integral equation based on two regularization parameters is proposed to regularize this problem. Numerical results are presented to illustrate the efficiency of the proposed method.


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## Body

## 1. Introduction

Partial differential equations (PDEs) with different types of boundary conditions play an essential tool in modelling natural phenomena. For time-dependent phenomena, one usually adds an initial time condition or a final time condition, which can be considered as the data. For the time-inverse problem from the final data, the main goal is to reconstruct the whole structure in previous times. These problems were widely studied in the papers by Tikhonov and Arsenin [1], Glasko [2], and the references cited therein. An example is the backward heat problem (BHP) where the goal is to recover the previous status of a physical field from the present information. It is well known that the BHP is a classical ill-posed problem, and it is quite difficult to consider since the solution does not always exist. Furthermore, even if the solution does exist, the continuous dependence of the solution on the data is not guaranteed. The BHP has been considered in the literature using different methods (see [3–10] and the references cited therein). Fu et al. [3] applied a wavelet dual least square method to investigate a BHP with constant coefficients, in [4], Hao et al. gave an approximation for this problem using a nonlocal boundary value problem method, Hao and Duc [5] used the Tikhonov regularization method to give an approximation for this problem in a Banach space, and Tautenhahn in [6] established an optimal error estimate for a backward heat equation with constant coefficients. Using the stabilized quasireversibility method, the final value problem for a class of nonlinear parabolic equations is investigated by Trong and Tuan [7], and in [8], the authors used the integral equation method to regularize the backward heat conduction problem and they obtained some error estimates. Tuan and Ngo [10] introduced the truncation method for solving the BHP and presented new error estimates for investigating the stability of the given problem. Also, the modified integral equation method and the modified quasiboundary value are extended to investigate inverse-time problems for axisymmetric backward heat equations in [11, 12] and the nonlinear spherically symmetric backward heat equation in [13].The concept of the so-called conformable derivative was proposed by Khalil et al. [14] and discussed by Atangana et al. [15] and Abdeljawad [16]. Anderson and Ulness in [17] provided a potential application of the conformable derivative in quantum mechanics, Hammad and Khalil [18] used a conformable fourier series to interpret the solution of the conformable heat equation, and Chung [19] employed the conformable derivative concept to investigate the problem of Newtonian mechanics, and the Euler–Lagrange equation was also constructed. Eslami [20] employed the Kudryashov method to obtain the traveling wave solutions to the coupled nonlinear Schrodinger equation with a conformable derivative, Çenesiz et al. [21, 22] studied Burgers’ equation, the modified Burgers’s equation, and the Burgers–Korteweg–de Vries equation with a conformable derivative version, Çenesiz et al. [23] investigated the stochastic solution of conformable Cauchy problems where the space operators may correspond to Brownian motion or a Levy process, and Vu et al. [24] employed the quasiboundary value method to regularize the inverse-time problem for the nonhomogeneous heat equation with a conformable derivative, and a Hölder-type estimation error for the whole time interval was obtained. In this paper, we consider the following backward heat equations:(1)Dtαux,t−uxxx,t=fx,t,x,t∈Ω×0,T,(2)ux,t=0,x,t∈∂Ω×0,T,(3)ux,T=gx,x∈Ω,where Ω=0,a, T is a positive number, the functions fx,t and gx are given, and Dtα is the conformable derivative of order α with respect to t defined by(4)Dtαut=limh⟶0ut+ht1−α−uth,Dtαu0=limt⟶0+Dtαut.From the information given at final timet=T, the goal of the problem is to recover the information ux,t for 0≤t<T. Unfortunately, BHP (1)–(3) is ill posed in the sense of Hadamard, i.e., it violates at least one of the following conditions:(1)
Existence. There exists a solution of the problem.(2)
Uniqueness. The solution must be unique.(3)
Stability. The solution must depend continuously on the data, i.e., any small error in given data must lead to a corresponding small error in the solution.Using Theorem1 in Section 2, we see that the solution of problems (1)–(3) is given by(5)ux,t=∑n=1∞eknbα−tα/αgn−∫tbsα−1eknsα−tα/αfnusdssinnπax,where the terms in the above equation are given in Theorem 1. We observe that expknbα−tα/α⟶n⟶∞∞, so this yields an instability of the solution of problems (1)–(3). This violates condition (3), so problems (1)–(3) are ill posed. In this paper, to stabilize problems (1)–(3), we shall apply the modified integral equation method via a two-parameter regularization to regularize problems (1)–(3). To do this, we shall replace the above instability term by the term εkn+e−knbα+γ/αtα−bα/bα+γ, where t∈0,b, ε∈0,bα+γ/α, α∈0,1 is fixed, and γ is a positive constant. From the proposed term, we use the following modified integral equation to approximate or to regularize the solution of problems (1)–(3):(6)uε,γx,t=∑n=1∞εkn+e−knbα+γ/αtα−bα/bα+γgn−∫tbsα−1eknsα−bα/αfnuε,γsdssinnπxa.In Section2, we show that problems (1)–(3) can be transformed into an integral equation (5). In Section 3.1, we prove that the regularized problem (6) is well posed in the sense of Hadamard in two cases, namely, α∈1/2,1 and α∈0,1. In Section 3.2, the error estimates between the regularized solution of problem (6) and the solution of problems (1)–(3) with the prior condition on the solution in two cases of exact data and nonexact data are presented. In particular, we show that(7)uε,γx,t−ux,t≤Dεtα+γ/bα+γbα+γα1+lnbα+γ/εαbα−tα/bα+γ,where D is specified below, ux,t is a solution of problems (1)–(3), and uε,γx,t is a solution of the regularized problem (6). In Section 4, we provide numerical tests to illustrate the theoretical results in the paper.


## 2. Statement of the Problem

Throughout this paper, we denote byL2Ω, where Ω=0,a, the Hilbert space of Lebesgue measurable functions v on Ω. ⋅,⋅ and ⋅ represent the inner product and norm on L2Ω, respectively. Specifically, the norm and inner product in L2Ω are defined as follows:(8)v≔vL2Ω=∫Ωv2xdx1/2,v1,v2=∫Ωv1xv2xdx,where v1,v2∈L2Ω. Denoting by C0,b;L2Ω the space of all continuous functions v:0,b⟶L2Ω and denoting by ⋅ the sup norm in C0,b;L2Ω defined as(9)v≔vC0,b;L2Ω=supt∈0,bv⋅,tL2Ω.The following theorem establishes the formula of the solution to problems (1)–(3).Theorem 1.
Letg∈L2Ω and let f:Ω×0,b×L2Ω⟶L2Ω. Then, the solution of the original problems (1)–(3) has the following form:(10)ux,t=∑n=1∞eknbα−tα/αgn−∫tTsα−1eknsα−tα/αfnusdssinnπax,where kn=nπ/a2 and(11)unt=2aux,t,φnx,gn=2agx,φnx,fnut=2afx,t,ux,t,φnx.Proof.
By choosing the orthogonal basisφnx≔sinnπx/a,n=1,2,…, in the Hilbert space L2Ω and by taking the inner product in L2Ω on the two sides of (1), we obtain(12)Dtαux,t,φnx−uxxx,t,φnx=fx,t,ux,t,φnx.
On the other hand, by using boundary conditions (2), we also get(13)uxxx,t,φnx=∫0auxxx,tsinnπaxdx=uxx,tsinnπax0a−nπaux,tcosnπax0a−nπa2∫0aux,tsinnπaxdx=−knux,t,φnx.
Then, it follows from (11)–(13) that(14)Dtαunt+knunt=fnut.
Solving problem (14), we get(15)unt=exp−kntαα∫0tsα−1expknsααfnusds+C.
Basing on (3), we have that(16)C=gnexpknbαα−∫0bsα−1expknsααfnusds,where gn=unb. Then, (15) yields that(17)unt=gnexpknbα−tαα−∫tbsα−1expknsα−tααfnusds.
Therefore, the representation of solution of problems (1)–(3) can be written as the infinite series:(18)ux,t=∑n=1∞untsinnπax=∑n=1∞gnexpknbα−tαα−∫tbsα−1expknsα−tααfnusdssinnπax.Remark 1.
As stated in Section1, we observe from (18) that when n tends to infinity, the term expknbα−tα/α is increasing rather quickly. Hence, the exact solution given in (18) of problems (1)–(3) is unstable. Thus, problems (1)–(3) are ill posed, and the above term is the unstable factor. So, to regularize the problem or to obtain a stable approximation for problems (1)–(3), we shall replace this unstable factor by a stable one. In this paper, the term expknbα−tα/α is replaced by a stable term which depends on two regularization parameters defined by εkn+e−knbα+γ/αtα−bα/bα+γ, where the first one ε captures the measuring error and the second one (γ) captures the regularity of the solution. Therefore, in this paper, we shall use integral equation (6) to approximate or to regularize problems (1)–(3).


## 3. Regularization and Error Estimates

Before investigating the uniqueness and stability of the solution of problem (6), we present the following two inequalities, which will be useful in the proof of the next theorems.Lemma 1.
Letα∈0,1 be fixed, and γ≥0, 0<t≤s≤b, ε∈D≔0,bα+γ/α, and z>0. Then, the following inequalities hold:(19)εz+exp−bα+γαztα−bα/bα+γ≤αεtα−bα/bα+γbα+γ1+lnbα+γ/αεbα−tα/bα+γ,and(20)expsα−bααzεz+exp−bα+γαztα−sα/bα+γ≤αεtα−sα/bα+γbα+γ1+lnbα+γ/αεsα−tα/bα+γ.Proof.
Letb>0, ε∈D, γ≥0,andz>0, and let α∈0,1 be fixed, then we observe that the function(21)fz=εz+exp−bα+γαz−1,has the maximum value at z=lnbα+γ/αε/bα+γ/α. This yields that(22)fz≤flnbα+γ/αεbα+γ/α=bα+γαε1+lnbα+γ/αε.
Then, we obtain the following inequality:(23)εz+exp−bα+γαz−bα−tα/bα+γ≤αεtα−bα/bα+γbα+γ1+lnbα+γ/αεbα−tα/bα+γ.
Furthermore, we have(24)expsα−bααzfz−bα−tα/bα+γ=expsα−bααbα+γbα+γzfzsα−bα/bα+γfztα−sα/bα+γ=εzexpbα+γαz+1sα−bα/bα+γfztα−sα/bα+γ≤fztα−sα/bα+γ.
From (23), we obtain(25)expsα−bααzfz−bα−tα/bα+γ≤αεtα−sα/bα+γbα+γ1+lnbα+γ/αεsα−tα/bα+γ.
This yields estimate (20).


### 3.1. The Well Posedness of Regularized Problem (6)

In the following theorem, we show that regularized problem (6) is well posed in the sense of Hadamard, i.e., problem (6) has a unique solution, and this solution continuously depends on the given data.Theorem 2.
Letg∈L2Ω, and let f:Ω×0,b×L2Ω⟶L2Ω satisfy the globally Lipschitz property with respect to the third variable, i.e., there exists a constant k>0 independent of x,t,v1,v2 such that(26)fx,t,v1−fx,t,v2≤kv1−v2,for allv1,v2∈L2Ω,where ⋅ is the norm in L2Ω. Then, regularized problem (6) is well posed in the sense of Hadamard provided that α∈1/2,1.Proof.
To prove the theorem, we shall divide the proof into two steps. The first step shows that regularized problem (6) has a unique solution uε,γ∈C0,b;L2Ω provided that α∈1/2,1. In the second step, the continuous dependence of the solution on the data g will be verified.Step 1. Consider the operator T:C0,b;L2Ω⟶C0,b;L2Ω given by(27)Tu∗x,t=∑n=1∞Tu∗ntsinnπax,where(28)Tu∗nt=Λ1ε,γ,t,kngn−∫tbΛ2ε,γ,s,t,knsα−1fnu∗sds,(29)Λ1ε,γ,t,kn≔εkn+exp−knbα+γαtα−bα/bα+γ,(30)Λ2ε,γ,s,t,kn≔expknsα−bααεkn+exp−knbα+γαtα−bα/bα+γ.We set(31)λε,γ=bα+γα1+lnbα+γ/αε.It follows from Lemma1 that(32)Λ1ε,γ,t,kn≤λε,γεbα−tα/bα+γ,(33)Λ2ε,γ,t,kn≤λε,γεsα−tα/bα+γ.Therefore, by Hölder’s inequality, one obtains, for0<t≤s≤b and α∈1/2,1,(34)Tu∗nt−Tu∗nt2=∫tbΛ2ε,γ,s,t,knsα−1fnu∗s−fnv∗sds2≤λ2ε,γε2Ct∫tbfnu∗s−fnv∗s2ds,whereα∈0,1 is fixed, sα−tα/bα+γ∈0,1, and Ct≔b2α−1−t2α−1/2α−1. So, from Lipschitz condition (26), one has that(35)Tu∗⋅,t−Tu∗⋅,t2=a2∑n=1∞Tu∗nt−Tv∗nt2≤a2λ2ε,γε2Ct∫tb∑n=1∞fnu∗s−fnv∗s2ds=λ2ε,γε2Ct≤Ctk2λ2ε,γε2b−tu∗−v∗2.We observe that the functionCt is decreasing on 0,b. Hence, Ct<C0 for all t∈0,b. With the same calculation, one also gets(36)T2u∗⋅,t−T2v∗⋅,t2≤λ4ε,γε4C20k4b−t22u∗−v∗2,where form≥1,(37)Tmu∗nt=Λ1gn−∫tbΛ2sα−1fnTm−1u∗sds.Using the mathematical induction method, form≥3, one obtains(38)Tmu∗⋅,t−Tmv∗⋅,t2≤λ2mε,γε2mCm0k2mb−tmm!u∗−v∗2.Now, if we assume that (38) holds for m=i, then by Hölder’s inequality and from Lemma 1, for m=i+1, we have that(39)Ti+1u∗nt−Ti+1v∗nt2≤λ2ε,γε2Ct∫tbfnTiu∗s−fnTiv∗s2ds.Then, by (38) and Lipschitz condition (26), we have that(40)Ti+1u∗⋅,t−Ti+1v∗⋅,t2=a2∑n=1∞Ti+1u∗nt−Ti+1v∗nt2≤a2λ2ε,γε2Ct∫tb∑n=1∞fnTiu∗s−fnTiv∗s2ds=λ2ε,γε2Ct∫tbfTiu∗⋅,s−fTiv∗⋅,s2ds≤λ2ε,γε2C0k2Ci0a2iε,γ/ε2ik2ib−ti+1i+1!u∗−v∗2=λ2i+2ε,γε2i+2Ci+10k2i+2b−ti+1i+1!u∗−v∗2,which is inequality (38) for m=i+1. Thus, (38) is satisfied for all m≥1. In addition, inequality (38) is similar to the estimate as follows:(41)Tmu∗−Tmv∗≤kmCm0λmε,γεmbm/2m!u∗−v∗,whereλε,γ is given by (31). Moreover, we have(42)limm⟶∞bα+γεα1+lnbα+γ/εαmkmbm/2Cm0m!=0,whereb>0 is given and α∈1/2,1 is fixed. Therefore, there exists m0∈ℕ such that Tm0 is a contraction. Thus, problem (6) has a unique solution.Step 2. Let vε,γ be a solution of (6) with the data g and wε,γ be a solution of (6) with the data h. From problem (6), we set(43)vε,γnt=Λ1gn−∫tbΛ2sα−1fnvε,γsds,wε,γnt=Λ1hn−∫tbΛ2sα−1fnwε,γsds,where gn=2/agx,sinnπ/ax,hn=2/ahx,andsinnπ/ax. In view of the inequality z1+z22≤2z12+2z22, the estimates (32) and (33), and then by Hölder’s inequality, we obtain(44)vε,γnt−wε,γnt2≤2Λ1hn−gn2+2∫tbΛ2sα−1fnwε,γs−fnvε,γsds2≤2λε,γε2bα−tα/bα+γgn−hn2+2Ct∫tbλε,γε2sα−tα/bα+γfnwε,γs−fnvε,γsds.
Using Lipschitz condition (26), we have that(45)vε,γ⋅,t−wε,γ⋅,t2≤2λε,γε2bα−tα/bα+γg−h2+2λε,γε−2tα/bα+γCtk2∫tbλε,γε2sα/bα+γwε,γ⋅,s−vε,γ⋅,s2ds.
Then, by settingHt=λε,γ/ε2tα/bα+γvε,γ⋅,t−wε,γ⋅,t2 and by using Gronwall’s inequality, one gets the following estimate:(46)vε,γ⋅,t−wε,γ⋅,t≤2bα+γεα1+lnbα+γ/εαbα−tα/bα+γexpk2C0g−h.
This verifies that the solution of (6) depends continuously on the given data. Therefore, the proof is completed.
As stated in Theorem2, problem (6) is well posed with the condition α∈1/2,1. In the following theorem, we shall extend the restriction of α from the subinterval 1/2,1⊂0,1 to the interval α∈0,1. To achieve our aim, we replace the global Lipschitz condition in Theorem 2 with a new Lipschitz condition in the next theorem.Theorem 3.
Letg∈L2Ω, and let f:Ω×0,b×L2Ω⟶L2Ω satisfy the following Lipschitz condition:(47)fx,t,v1−fx,t,v2≤ktηv1−v2,∀v1,v2∈L2Ω,where α∈0,1 is fixed, η≥1−α, k>0 is a constant, and x,t∈Ω×0,b. Then, regularized problem (27) is well posed in the sense of Hadamard.Proof.
We follow the ideas in the proof of Theorem2. We reconsider the operator T:C0,b;L2Ω⟶C0,b;L2Ω given by (27). By Hölder’s inequality, Lipschitz condition (47), and from (28), we have that, for 0<t≤s≤b and α∈0,1,(48)Tu∗⋅,t−Tu∗⋅,t2≤a2λ2ε,γε2b−t∫tbs2α−2∑n=1∞fnu∗s−fnv∗s2ds=λ2ε,γε2b−t∫tbs2α−2fu∗⋅,s−fv∗⋅,s2ds≤λ2ε,γε2k2b∫tbs2α−1+γu∗⋅,s−v∗⋅,s2ds≤Dk2bλ2ε,γε2b−tu∗−v∗2,where α∈0,1 is fixed, η>1−α, and D≔b2α−1+η. With the same calculation and using the mathematical induction method, one also gets(49)Tmu∗⋅,t−Tmv∗⋅,t2≤λ2mε,γε2mDmk2mbmb−tmm!u∗−v∗2.
Indeed, if we assume that (49) holds for m=i, then by using Lemma 1, the Hölder’s inequality, and Lipschitz condition (47), for m=i+1, we have that(50)Ti+1u∗⋅,t−Ti+1v∗⋅,t2=a2∑n=1∞Ti+1u∗nt−Ti+1v∗nt2≤a2λ2ε,γε2b−t∫tbs2α−1∑n=1∞fnTiu∗s−fnTiv∗s2ds=λ2ε,γε2b−t∫tbs2α−1fTiu∗⋅,s−fTiv∗⋅,s2ds≤λ2ε,γε2k2bDi+1a2iε,γ/ε2ik2ibib−ti+1i+1!u∗−v∗2=λ2i+2ε,γε2i+2Di+1k2i+2bi+1b−ti+1i+1!u∗−v∗2,which is inequality (49) for m=i+1. Thus, inequality (49) is satisfied for all m≥1. Furthermore, inequality (49) yields that(51)Tmu∗−Tmv∗≤kmDmbα+γεα1+lnbα+γ/εαmbmm!u∗−v∗.
Hence,(52)limm⟶∞bα+γεα1+lnbα+γ/εαmkmbmDmm!=0,where b>0 is given and α∈0,1 is fixed. So, there exists m0∈ℕ such that Tm0 is a contraction. Thus, problem (6) has a unique solution uε,γ∈C0,b;L2Ω in the case of α∈0,1. Finally, in order to show that the solution of problem (6) depends continuously on g, we assume that vε,γ and wε,γ are two solutions of problem (6) corresponding to the final values g and h, respectively. Similar to Step 2 in the proof of Theorem 2, by using Lipschitz condition (5), we also obtain(53)vε,γ⋅,t−wε,γ⋅,t2≤2λε,γε2bα−tα/bα+γg−h2+2λε,γε−2tα/bα+γk2b∫tbs2α−1+ηλε,γε2sα/bα+γwε,γ⋅,s−vε,γ⋅,s2ds.
By puttingHt=λε,γ/ε2tα/bα+γvε,γ⋅,t−wε,γ⋅,t2 and lt=s2α−1+η and by Gronwall’s inequality, we get(54)Ht≤2λε,γε2bα/bα+γexp2k2bb2α+2η−1−t2α+2η−12α+2η−1g−h2.
This yields(55)vε,γ⋅,t−wε,γ⋅,t≤2bα+γεα1+lnbα+γ/εαbα−tα/bα+γ⋅expk2bb2α+2η−1−t2α+2η−12α+2η−1g−h.
Therefore, we can conclude that the solution of (6) depends continuously on the data g. The proof is completed.


### 3.2. Error Estimates

The following theorem presents the error estimatein the case of exact data between the solution u of (1)–(3) with the conditional stable and the regularized solution uε,γ of (6) without the conditional stable.Theorem 4.
Letε∈D be given, and fandg satisfy the conditions of Theorem 3. Suppose that the unique solution u of (1)–(3) satisfies the conditional stable Cγ,t,b≔∑n=1∞kn2exp2knbα+γ/αunt2<∞,∀t∈0,b, then we obtain the following estimate:(56)uε,γ⋅,t−u⋅,t≤λε,γbα−tα/bα+γεtα+γ/bα+γMexpk2bb2α+2η−1−t2α+2η−12α+2η−1,where t∈0,b, α∈0,1 is fixed, η>1−α, M=asupt∈0,bbα−tα/bα+γ2Cγ,t,b, λε,γ≔bα+γ/α1+lnbα+γ/αε, and uε,γ is a unique solution of regularized problem (6).Proof.
Assume thatu is a unique solution of (1)–(3), and then based on Theorem 1, one observes that u is represented by(57)ux,t=∑n=1∞untsinnπax,where(58)unt=expknbα−tααgn−∫tbsα−1expknsα−tααfnusds.
Multiplying both sides of (58) with 1+εknexpknbα+γ/α−bα−tα/bα+γ, one gets(59)1+εkneknbα−tα/α−bα−tα/bα+γunt=eknbα−tα/α1+εkneknbα+γ/α−bα−tα/bα+γgn−∫tb1+εkneknbα−tα/α−bα−tα/bα+γeknbα−tα/αeknsα−bα/αsα−1fnusds=εkn+e−knbα−tα/αtα−bα/bα+γgn−∫tbεkn+e−knbα−tα/αtα−bα/bα+γeknsα−bα/αsα−1fnusds=Λ1gn−∫tbΛ2sα−1fnusds,where Λ1 and Λ2 are denoted as in the proof of Theorem 2. Moreover, from (6), we have(60)unε,γt=Λ1gn−∫tbΛ2sα−1fnuε,γsds.
Using the inequalityz1+z22≤2z12+2z22 and the inequality 1−1+z−β≤βz, where β>0, one has that(61)unt−unε,γt2≤2unt−1+εkneknbα+γ/α−bα−tα/bα+γunt2+2unε,γt−1+εkneknbα+γ/α−bα−tα/bα+γunt2≤2∫tbΛ2ε,γ,s,t,knsα−1fnuε,γs−fnusds2+21−1+εkneknbα+γ/α−bα−tα/bα+γunt2≤2∫tbΛ2ε,γ,s,t,knsα−1fnuε,γs−fnusds2+2kn2ε2bα−tαbα+γ2e2knbα+γ/αunt2.
LetCnγ,t,b≔kn2exp2knbα+γ/αunt2. By employing Hölder’s inequality and then from Lemma 1 and (61), one obtains(62)unt−unε,γt2≤2b−t∫tbΛ22ε,γ,s,t,kns2α−1fnuε,γs−fnus2ds+2ε2bα−tαbα+γ2Cnγ,t,b≤2b−t∫tbbα+γεα1+lnbα+γ/εα2sα−2tα/bα+γs2α−1fnuε,γs−fnus2ds+2ε2bα−tαbα+γ2Cnγ,t,b.
Let(63)yz=z−εα−εαlnzεα,where α∈0,1 is fixed. It is clear that(64)yz≥yεα=εα−εα−εαln1=0,for all z≥εα. Therefore, for ε∈0,bα+γ/α, one gets(65)ε≤bα+γα1+lnbα+γ/εα.
Hence,(66)ε=εtα+γ/bα+γεbα−tα/bα+γ≤εtα+γ/bα+γbα+γα1+lnbα+γ/εαbα−tα/bα+γ=εtα+γ/bα+γλε,γbα−tα/bα+γ.
It follows from (62) that(67)unε,γt−unt2≤2b−tλε,γ2bα−2tα/bα+γε2tα+γ/bα+γ∫tbλε,γ2sα−2bα/bα+γε−2sα+γ/bα+γs2α−1fnuε,γs−fnus2ds+2λε,γ2bα−2tα/bα+γε2tα+γ/bα+γbα−tαbα+γ2Cnγ,t,b.
Then, by using Lipschitz condition (47), we get the following estimate:(68)uε,γ⋅,t−u⋅,t2≤λε,γ2bα−2tα/bα+γε2tα+γ/bα+γa∑n=1∞bα−tαbα+γ2Cnγ,t,b+k2b−t∫tbλε,γ2sα−2bα/bα+γε−2sα+γ/bα+γs2α−1+ηuε,γ⋅,s−u⋅,s2ds.
By puttingHt=λε,γ2tα−2bα/bα+γε−2tα+γ/bα+γuε,γ⋅,t−u⋅,t2 and M=supt∈0,bbα−tα/bα+γ2a∑n=1∞Cnγ,t,b, we get the inequality as follows:(69)Ht≤M+2k2b∫tbs2α−1+ηHsds.
Using Gronwall’s inequality, we have(70)Ht≤Mexp2k2bb2α+2η−1−t2α+2η−12α+2η−1.
This infers that(71)uε,γ⋅,t−u⋅,t≤λε,γbα−tα/bα+γεtα+γ/bα+γMexpk2bb2α+2η−1−t2α+2η−12α+2η−1.
Based on the results of Theorems3 and 4, the following theorem presents the error estimate in the case of nonexact data between the solution u of (1)–(3) with the conditional stable and regularized solution (6) without the conditional stable.Theorem 5.
Letε∈0,bα+γ/α, where γ is a positive parameter, and let f satisfy condition (47). Assume that u is a unique solution of (1)–(3) with the data g which satisfies the conditional stable as in Theorem 4. In addition, let h be the measured data which satisfies(72)g−h≤ε.Then, there is a functionvε,γ corresponding to the measured data h such that(73)vε,γ⋅,t−u⋅,t≤Dεtα+γ/bα+γbα+γα1+lnbα+γ/εαbα−tα/bα+γ,t∈0,b,where D≔2+Mexpk2bb2α+2η−1−t2α+2η−1/2α+2η−1 and M is given as in Theorem 4.Proof.
Denote byvε,r and uε,γ the solutions of (6) with the final values h and g, respectively. We observe that(74)vε,r⋅,t−u⋅,t≤vε,r⋅,t−uε,r⋅,t+uε,r⋅,t−u⋅,t.
Based on estimates (55) and (56) of Theorems 3 and 4, we have that(75)vε,r⋅,t−uε,r⋅,t≤2λε,γεbα−tα/bα+γexpk2bb2α+2η−1−t2α+2η−12α+2η−1g−h,uε,r⋅,t−u⋅,t≤λε,γbα−tα/bα+γεtα+γ/bα+γMexpk2bb2α+2η−1−t2α+2η−12α+2η−1.
It follows from (72) that, for every t∈0,b,(76)vε,γ⋅,t−u⋅,t≤Dεtα+γ/bα+γλε,γbα−tα/bα+γ,where D≔2+Mexpk2bb2α+2η−1−t2α+2η−1/2α+2η−1. The proof is complete.


## 3.1. The Well Posedness of Regularized Problem (6)

In the following theorem, we show that regularized problem (6) is well posed in the sense of Hadamard, i.e., problem (6) has a unique solution, and this solution continuously depends on the given data.Theorem 2.
Letg∈L2Ω, and let f:Ω×0,b×L2Ω⟶L2Ω satisfy the globally Lipschitz property with respect to the third variable, i.e., there exists a constant k>0 independent of x,t,v1,v2 such that(26)fx,t,v1−fx,t,v2≤kv1−v2,for allv1,v2∈L2Ω,where ⋅ is the norm in L2Ω. Then, regularized problem (6) is well posed in the sense of Hadamard provided that α∈1/2,1.Proof.
To prove the theorem, we shall divide the proof into two steps. The first step shows that regularized problem (6) has a unique solution uε,γ∈C0,b;L2Ω provided that α∈1/2,1. In the second step, the continuous dependence of the solution on the data g will be verified.Step 1. Consider the operator T:C0,b;L2Ω⟶C0,b;L2Ω given by(27)Tu∗x,t=∑n=1∞Tu∗ntsinnπax,where(28)Tu∗nt=Λ1ε,γ,t,kngn−∫tbΛ2ε,γ,s,t,knsα−1fnu∗sds,(29)Λ1ε,γ,t,kn≔εkn+exp−knbα+γαtα−bα/bα+γ,(30)Λ2ε,γ,s,t,kn≔expknsα−bααεkn+exp−knbα+γαtα−bα/bα+γ.We set(31)λε,γ=bα+γα1+lnbα+γ/αε.It follows from Lemma1 that(32)Λ1ε,γ,t,kn≤λε,γεbα−tα/bα+γ,(33)Λ2ε,γ,t,kn≤λε,γεsα−tα/bα+γ.Therefore, by Hölder’s inequality, one obtains, for0<t≤s≤b and α∈1/2,1,(34)Tu∗nt−Tu∗nt2=∫tbΛ2ε,γ,s,t,knsα−1fnu∗s−fnv∗sds2≤λ2ε,γε2Ct∫tbfnu∗s−fnv∗s2ds,whereα∈0,1 is fixed, sα−tα/bα+γ∈0,1, and Ct≔b2α−1−t2α−1/2α−1. So, from Lipschitz condition (26), one has that(35)Tu∗⋅,t−Tu∗⋅,t2=a2∑n=1∞Tu∗nt−Tv∗nt2≤a2λ2ε,γε2Ct∫tb∑n=1∞fnu∗s−fnv∗s2ds=λ2ε,γε2Ct≤Ctk2λ2ε,γε2b−tu∗−v∗2.We observe that the functionCt is decreasing on 0,b. Hence, Ct<C0 for all t∈0,b. With the same calculation, one also gets(36)T2u∗⋅,t−T2v∗⋅,t2≤λ4ε,γε4C20k4b−t22u∗−v∗2,where form≥1,(37)Tmu∗nt=Λ1gn−∫tbΛ2sα−1fnTm−1u∗sds.Using the mathematical induction method, form≥3, one obtains(38)Tmu∗⋅,t−Tmv∗⋅,t2≤λ2mε,γε2mCm0k2mb−tmm!u∗−v∗2.Now, if we assume that (38) holds for m=i, then by Hölder’s inequality and from Lemma 1, for m=i+1, we have that(39)Ti+1u∗nt−Ti+1v∗nt2≤λ2ε,γε2Ct∫tbfnTiu∗s−fnTiv∗s2ds.Then, by (38) and Lipschitz condition (26), we have that(40)Ti+1u∗⋅,t−Ti+1v∗⋅,t2=a2∑n=1∞Ti+1u∗nt−Ti+1v∗nt2≤a2λ2ε,γε2Ct∫tb∑n=1∞fnTiu∗s−fnTiv∗s2ds=λ2ε,γε2Ct∫tbfTiu∗⋅,s−fTiv∗⋅,s2ds≤λ2ε,γε2C0k2Ci0a2iε,γ/ε2ik2ib−ti+1i+1!u∗−v∗2=λ2i+2ε,γε2i+2Ci+10k2i+2b−ti+1i+1!u∗−v∗2,which is inequality (38) for m=i+1. Thus, (38) is satisfied for all m≥1. In addition, inequality (38) is similar to the estimate as follows:(41)Tmu∗−Tmv∗≤kmCm0λmε,γεmbm/2m!u∗−v∗,whereλε,γ is given by (31). Moreover, we have(42)limm⟶∞bα+γεα1+lnbα+γ/εαmkmbm/2Cm0m!=0,whereb>0 is given and α∈1/2,1 is fixed. Therefore, there exists m0∈ℕ such that Tm0 is a contraction. Thus, problem (6) has a unique solution.Step 2. Let vε,γ be a solution of (6) with the data g and wε,γ be a solution of (6) with the data h. From problem (6), we set(43)vε,γnt=Λ1gn−∫tbΛ2sα−1fnvε,γsds,wε,γnt=Λ1hn−∫tbΛ2sα−1fnwε,γsds,where gn=2/agx,sinnπ/ax,hn=2/ahx,andsinnπ/ax. In view of the inequality z1+z22≤2z12+2z22, the estimates (32) and (33), and then by Hölder’s inequality, we obtain(44)vε,γnt−wε,γnt2≤2Λ1hn−gn2+2∫tbΛ2sα−1fnwε,γs−fnvε,γsds2≤2λε,γε2bα−tα/bα+γgn−hn2+2Ct∫tbλε,γε2sα−tα/bα+γfnwε,γs−fnvε,γsds.
Using Lipschitz condition (26), we have that(45)vε,γ⋅,t−wε,γ⋅,t2≤2λε,γε2bα−tα/bα+γg−h2+2λε,γε−2tα/bα+γCtk2∫tbλε,γε2sα/bα+γwε,γ⋅,s−vε,γ⋅,s2ds.
Then, by settingHt=λε,γ/ε2tα/bα+γvε,γ⋅,t−wε,γ⋅,t2 and by using Gronwall’s inequality, one gets the following estimate:(46)vε,γ⋅,t−wε,γ⋅,t≤2bα+γεα1+lnbα+γ/εαbα−tα/bα+γexpk2C0g−h.
This verifies that the solution of (6) depends continuously on the given data. Therefore, the proof is completed.
As stated in Theorem2, problem (6) is well posed with the condition α∈1/2,1. In the following theorem, we shall extend the restriction of α from the subinterval 1/2,1⊂0,1 to the interval α∈0,1. To achieve our aim, we replace the global Lipschitz condition in Theorem 2 with a new Lipschitz condition in the next theorem.Theorem 3.
Letg∈L2Ω, and let f:Ω×0,b×L2Ω⟶L2Ω satisfy the following Lipschitz condition:(47)fx,t,v1−fx,t,v2≤ktηv1−v2,∀v1,v2∈L2Ω,where α∈0,1 is fixed, η≥1−α, k>0 is a constant, and x,t∈Ω×0,b. Then, regularized problem (27) is well posed in the sense of Hadamard.Proof.
We follow the ideas in the proof of Theorem2. We reconsider the operator T:C0,b;L2Ω⟶C0,b;L2Ω given by (27). By Hölder’s inequality, Lipschitz condition (47), and from (28), we have that, for 0<t≤s≤b and α∈0,1,(48)Tu∗⋅,t−Tu∗⋅,t2≤a2λ2ε,γε2b−t∫tbs2α−2∑n=1∞fnu∗s−fnv∗s2ds=λ2ε,γε2b−t∫tbs2α−2fu∗⋅,s−fv∗⋅,s2ds≤λ2ε,γε2k2b∫tbs2α−1+γu∗⋅,s−v∗⋅,s2ds≤Dk2bλ2ε,γε2b−tu∗−v∗2,where α∈0,1 is fixed, η>1−α, and D≔b2α−1+η. With the same calculation and using the mathematical induction method, one also gets(49)Tmu∗⋅,t−Tmv∗⋅,t2≤λ2mε,γε2mDmk2mbmb−tmm!u∗−v∗2.
Indeed, if we assume that (49) holds for m=i, then by using Lemma 1, the Hölder’s inequality, and Lipschitz condition (47), for m=i+1, we have that(50)Ti+1u∗⋅,t−Ti+1v∗⋅,t2=a2∑n=1∞Ti+1u∗nt−Ti+1v∗nt2≤a2λ2ε,γε2b−t∫tbs2α−1∑n=1∞fnTiu∗s−fnTiv∗s2ds=λ2ε,γε2b−t∫tbs2α−1fTiu∗⋅,s−fTiv∗⋅,s2ds≤λ2ε,γε2k2bDi+1a2iε,γ/ε2ik2ibib−ti+1i+1!u∗−v∗2=λ2i+2ε,γε2i+2Di+1k2i+2bi+1b−ti+1i+1!u∗−v∗2,which is inequality (49) for m=i+1. Thus, inequality (49) is satisfied for all m≥1. Furthermore, inequality (49) yields that(51)Tmu∗−Tmv∗≤kmDmbα+γεα1+lnbα+γ/εαmbmm!u∗−v∗.
Hence,(52)limm⟶∞bα+γεα1+lnbα+γ/εαmkmbmDmm!=0,where b>0 is given and α∈0,1 is fixed. So, there exists m0∈ℕ such that Tm0 is a contraction. Thus, problem (6) has a unique solution uε,γ∈C0,b;L2Ω in the case of α∈0,1. Finally, in order to show that the solution of problem (6) depends continuously on g, we assume that vε,γ and wε,γ are two solutions of problem (6) corresponding to the final values g and h, respectively. Similar to Step 2 in the proof of Theorem 2, by using Lipschitz condition (5), we also obtain(53)vε,γ⋅,t−wε,γ⋅,t2≤2λε,γε2bα−tα/bα+γg−h2+2λε,γε−2tα/bα+γk2b∫tbs2α−1+ηλε,γε2sα/bα+γwε,γ⋅,s−vε,γ⋅,s2ds.
By puttingHt=λε,γ/ε2tα/bα+γvε,γ⋅,t−wε,γ⋅,t2 and lt=s2α−1+η and by Gronwall’s inequality, we get(54)Ht≤2λε,γε2bα/bα+γexp2k2bb2α+2η−1−t2α+2η−12α+2η−1g−h2.
This yields(55)vε,γ⋅,t−wε,γ⋅,t≤2bα+γεα1+lnbα+γ/εαbα−tα/bα+γ⋅expk2bb2α+2η−1−t2α+2η−12α+2η−1g−h.
Therefore, we can conclude that the solution of (6) depends continuously on the data g. The proof is completed.


## 3.2. Error Estimates

The following theorem presents the error estimatein the case of exact data between the solution u of (1)–(3) with the conditional stable and the regularized solution uε,γ of (6) without the conditional stable.Theorem 4.
Letε∈D be given, and fandg satisfy the conditions of Theorem 3. Suppose that the unique solution u of (1)–(3) satisfies the conditional stable Cγ,t,b≔∑n=1∞kn2exp2knbα+γ/αunt2<∞,∀t∈0,b, then we obtain the following estimate:(56)uε,γ⋅,t−u⋅,t≤λε,γbα−tα/bα+γεtα+γ/bα+γMexpk2bb2α+2η−1−t2α+2η−12α+2η−1,where t∈0,b, α∈0,1 is fixed, η>1−α, M=asupt∈0,bbα−tα/bα+γ2Cγ,t,b, λε,γ≔bα+γ/α1+lnbα+γ/αε, and uε,γ is a unique solution of regularized problem (6).Proof.
Assume thatu is a unique solution of (1)–(3), and then based on Theorem 1, one observes that u is represented by(57)ux,t=∑n=1∞untsinnπax,where(58)unt=expknbα−tααgn−∫tbsα−1expknsα−tααfnusds.
Multiplying both sides of (58) with 1+εknexpknbα+γ/α−bα−tα/bα+γ, one gets(59)1+εkneknbα−tα/α−bα−tα/bα+γunt=eknbα−tα/α1+εkneknbα+γ/α−bα−tα/bα+γgn−∫tb1+εkneknbα−tα/α−bα−tα/bα+γeknbα−tα/αeknsα−bα/αsα−1fnusds=εkn+e−knbα−tα/αtα−bα/bα+γgn−∫tbεkn+e−knbα−tα/αtα−bα/bα+γeknsα−bα/αsα−1fnusds=Λ1gn−∫tbΛ2sα−1fnusds,where Λ1 and Λ2 are denoted as in the proof of Theorem 2. Moreover, from (6), we have(60)unε,γt=Λ1gn−∫tbΛ2sα−1fnuε,γsds.
Using the inequalityz1+z22≤2z12+2z22 and the inequality 1−1+z−β≤βz, where β>0, one has that(61)unt−unε,γt2≤2unt−1+εkneknbα+γ/α−bα−tα/bα+γunt2+2unε,γt−1+εkneknbα+γ/α−bα−tα/bα+γunt2≤2∫tbΛ2ε,γ,s,t,knsα−1fnuε,γs−fnusds2+21−1+εkneknbα+γ/α−bα−tα/bα+γunt2≤2∫tbΛ2ε,γ,s,t,knsα−1fnuε,γs−fnusds2+2kn2ε2bα−tαbα+γ2e2knbα+γ/αunt2.
LetCnγ,t,b≔kn2exp2knbα+γ/αunt2. By employing Hölder’s inequality and then from Lemma 1 and (61), one obtains(62)unt−unε,γt2≤2b−t∫tbΛ22ε,γ,s,t,kns2α−1fnuε,γs−fnus2ds+2ε2bα−tαbα+γ2Cnγ,t,b≤2b−t∫tbbα+γεα1+lnbα+γ/εα2sα−2tα/bα+γs2α−1fnuε,γs−fnus2ds+2ε2bα−tαbα+γ2Cnγ,t,b.
Let(63)yz=z−εα−εαlnzεα,where α∈0,1 is fixed. It is clear that(64)yz≥yεα=εα−εα−εαln1=0,for all z≥εα. Therefore, for ε∈0,bα+γ/α, one gets(65)ε≤bα+γα1+lnbα+γ/εα.
Hence,(66)ε=εtα+γ/bα+γεbα−tα/bα+γ≤εtα+γ/bα+γbα+γα1+lnbα+γ/εαbα−tα/bα+γ=εtα+γ/bα+γλε,γbα−tα/bα+γ.
It follows from (62) that(67)unε,γt−unt2≤2b−tλε,γ2bα−2tα/bα+γε2tα+γ/bα+γ∫tbλε,γ2sα−2bα/bα+γε−2sα+γ/bα+γs2α−1fnuε,γs−fnus2ds+2λε,γ2bα−2tα/bα+γε2tα+γ/bα+γbα−tαbα+γ2Cnγ,t,b.
Then, by using Lipschitz condition (47), we get the following estimate:(68)uε,γ⋅,t−u⋅,t2≤λε,γ2bα−2tα/bα+γε2tα+γ/bα+γa∑n=1∞bα−tαbα+γ2Cnγ,t,b+k2b−t∫tbλε,γ2sα−2bα/bα+γε−2sα+γ/bα+γs2α−1+ηuε,γ⋅,s−u⋅,s2ds.
By puttingHt=λε,γ2tα−2bα/bα+γε−2tα+γ/bα+γuε,γ⋅,t−u⋅,t2 and M=supt∈0,bbα−tα/bα+γ2a∑n=1∞Cnγ,t,b, we get the inequality as follows:(69)Ht≤M+2k2b∫tbs2α−1+ηHsds.
Using Gronwall’s inequality, we have(70)Ht≤Mexp2k2bb2α+2η−1−t2α+2η−12α+2η−1.
This infers that(71)uε,γ⋅,t−u⋅,t≤λε,γbα−tα/bα+γεtα+γ/bα+γMexpk2bb2α+2η−1−t2α+2η−12α+2η−1.
Based on the results of Theorems3 and 4, the following theorem presents the error estimate in the case of nonexact data between the solution u of (1)–(3) with the conditional stable and regularized solution (6) without the conditional stable.Theorem 5.
Letε∈0,bα+γ/α, where γ is a positive parameter, and let f satisfy condition (47). Assume that u is a unique solution of (1)–(3) with the data g which satisfies the conditional stable as in Theorem 4. In addition, let h be the measured data which satisfies(72)g−h≤ε.Then, there is a functionvε,γ corresponding to the measured data h such that(73)vε,γ⋅,t−u⋅,t≤Dεtα+γ/bα+γbα+γα1+lnbα+γ/εαbα−tα/bα+γ,t∈0,b,where D≔2+Mexpk2bb2α+2η−1−t2α+2η−1/2α+2η−1 and M is given as in Theorem 4.Proof.
Denote byvε,r and uε,γ the solutions of (6) with the final values h and g, respectively. We observe that(74)vε,r⋅,t−u⋅,t≤vε,r⋅,t−uε,r⋅,t+uε,r⋅,t−u⋅,t.
Based on estimates (55) and (56) of Theorems 3 and 4, we have that(75)vε,r⋅,t−uε,r⋅,t≤2λε,γεbα−tα/bα+γexpk2bb2α+2η−1−t2α+2η−12α+2η−1g−h,uε,r⋅,t−u⋅,t≤λε,γbα−tα/bα+γεtα+γ/bα+γMexpk2bb2α+2η−1−t2α+2η−12α+2η−1.
It follows from (72) that, for every t∈0,b,(76)vε,γ⋅,t−u⋅,t≤Dεtα+γ/bα+γλε,γbα−tα/bα+γ,where D≔2+Mexpk2bb2α+2η−1−t2α+2η−1/2α+2η−1. The proof is complete.


## 4. Numerical Illustration

Consider the following linear backward heat problem involving the conformable derivative:(77)Dtαux,t−uxxx,t=1+πa2u,0≤x≤a,0<t<b,(78)u0,t=ua,t=0,0<t≤b,(79)ux,b=expbααsinπax,0≤x≤a.It is not difficult to check thatux,t=exptα/αsinπ/ax is an exact solution of (77)–(79). It is well known that errors always occur during measurement, and so in this section, we assume the measured data are perturbed by a “noise” with the level ε as follows:(80)gnoisex=expbααsinπax+∑j=1J0εcjsinjπax,where J0∈ℕ and cj is a sequence of random numbers with mean 0. Basing on regularized problem (6), we obtain the regularized solution uε,γ corresponding to the data gnoise:(81)uε,γx,t=εk1+e−k1bα+γ/αtα−bα/bα+γebα/α−∫tbsα−1ek1sα−bα/αf1uε,γsdssinπax+∑j=1J0εkj+e−kjbα+γ/αtα−bα/bα+γεcj−∫tbsα−1ekjsα−bααfjuε,γsdssinjπax.Using the concept of the relative error given by(82)REε,t=uε,γ⋅,t−u⋅,tu⋅,t,we shall present the difference between the exact solution and the approximate solution for each t∈0,b. Now, the following situations are considered:Situation 1. We focus on the regularization parameter ε. In our numerical example, we take γ=0.3, α=0.8,b=1,anda=10 and let ε1=10−3, ε2=10−5, and ε3=10−7. The graphs of the solution ut and the regularized solution uε,γ are shown in Figures 1–4 with various values of ε. In addition, we also obtain the error of the proposed method in this paper with random measured data (80) given in Table 1 at t=0.5.Furthermore, lett=0, then the graphs of the regularized solution and the exact solution are shown in Figure 5 with the various values of ε.Figure 1
The solutionu.Figure 2
The regularized solution in the case ofε1=10−3.Figure 3
The regularized solution in the case ofε2=10−5.Figure 4
The regularized solution in the case ofε3=10−7.Figure 5
The graphs ofut and uε,γt.Table 1
The error of the proposed method.
ε
u
ε
⋅
,
0.5
−
u
⋅
,
0.5
RE
ε
,
0.5
ε
=
10
−
1
0.748065408
5.160107181
ε
=
10
−
2
0.116350273
0.802576715
ε
=
10
−
3
0.02441038
0.16838125Figure 6
The regularized solutionuε,γ1 with ε=10−3 and γ1=0.Figure 7
The regularized solutionuε,γ2 with ε=10−3 and γ2=1.Figure 8
The exact solution (black) and the regularized solution withε=10−3 and γ=0 (blue), γ=0.3 (green), and γ=1 (red).Remark 2.
From Figures1–4 and Table 1, we observe that the smaller the measurement error ε is, the closer the regularized solution becomes to the exact solution.Situation 2. We focus on the parameter γ. In our numerical example, we take ε=10−6 fixed and let α=0.5,b=1,anda=10. We consider the values of γ: γ1=0andγ2=1. The graphs of the regularized solution in this case are shown in Figures 6 and 7.Furthermore, lett=0, then the graphs of the regularized solution and the exact solution are shown in Figure 8 with the various values of γ.Remark 3.
According to Figures6–8, the regularized solution will be closer to the exact one with a higher value of γ. It is very useful if we want to obtain a better approximation while the measurement process cannot be improved.


## 5. Conclusion

In this paper, we have discussed the modified integral equation method involving two regularization parameters for the backward heat problem with a conformable derivative. We have also established error estimates between exact and regularized solutions in the cases of exact data and inexact data. These estimates are supported by several numerical examples.



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