# Existence and Uniqueness of the Solution for an Inverse Problem of a Fractional Diffusion Equation with Integral Condition

Authors: Taki-Eddine Oussaeif; Benaoua Antara; Adel Ouannas; Iqbal M. Batiha; Khaled M. Saad; Hadi Jahanshahi; Awad M. Aljuaid; Ayman A. Aly
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The solvability of the fractional partial differential equation with integral overdetermination condition for an inverse problem is investigated in this paper. We analyze the direct problem solution by using the “energy inequality” method. Using the fixed point technique, the existence and uniqueness of the solution of the inverse problem on the data are established.


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## Body

## 1. Introduction

This work devoted to study the solvability of a pair of functionsu,f satisfying the following fractional parabolic problem:
(1)CDtαu−Δu+βu=ftgx,t;x∈Ω,t∈0,T,with the initial condition
(2)ux,0=φx,x∈Ω,the boundary condition
(3)ux,t=0,x,t∈∂Ω×0,T,and the nonlocal condition
(4)∫Ωvxux,tdx=θt,t∈0,T.Here,Ω is a bounded domain in ℝn with smooth boundary ∂Ω.. The functions g, φ, and θ are known functions, and β is a positive constant. And Γ· denotes the gamma function. For any positive integer 0<α<1, the left Caputo derivative is defined as
(5)CDtαux,t≔1Γ1−α∫0t∂ux,τ∂τ1t−ταdτ.Inverse parabolic equation problems occur naturally in many fields, and there is extensive literature on inverse heat equation problems (see [1–4], and references therein). The form (4) is an additional information of problem.In engineering and physics, the parameter recognition in a partial differential equation from the data of the integral overdetermination condition plays an important role [5–10]. From a physical point of view, these conditions can be interpreted by a system averaging the domain of spatial variables as measurements of the temperature ux,t.Note that nonlocal problems related with integral overdetermination [11, 12]. Studies have shown that when we deal with these kinds of nonclassical problems, classical approaches sometimes do not work [13, 14]. To date, different methods for addressing problems resulting from nonlocal problem have been suggested. The choice of approach depends on the form of nonlocal boundary value that are involved.We note that several authors have studied the inverse parabolic problem with condition of type (4) and its special solubility (see, for example, [3, 4, 15–20]). There are also several articles dedicated to the study of the existence and uniqueness of inverse problem solutions for different parabolic equations with unknown source functions. Inverse problems related by determining unknown function in source term of a parabolic equation with overdetermination condition [21, 22].In recent years, fractional differential equations have created growing interest from engineers and scientists and have great importance in modeling complex phenomena. Because FDEs have memory, nonlocal space, and time relationships, using these equations, complex phenomena can be modelled [23–28].Namely, in the present paper, a new research on the inverse problem of a fractional parabolic equation is discussed, for which the solvability of the problem (1)–(4) is reduced to the concept of a fixed point technique. This work is divided into four sections; we start with an introduction then we give some definitions of function space and important lemmas. The third section is devoted to studying the solvability of the direct fractional parabolic problem. Finally, in the last section, we prove the existence and uniqueness of the solution to the main problem.


## 2. Functional Space

Definition 1.
Let us introduce certain notations used below, we set(6)g∗t=∫Ωvxgx,tdx,QT=Ω×0,T.
We denote byC0,T,L2Ω the space is composed of all continuous functions on 0,T with values in L2Ω. For any 0<α<1, the Caputo and Riemann-Liouville derivatives are defined, respectively, as follows:
(i)
The left Caputo derivatives:(7)CDtαux,t≔1Γ1−α∫0t∂ux,τ∂τ1t−ταdτ.(ii)
The left Riemann-Liouville derivatives:(8)RDtαux,t≔1Γ1−α∂∂t∫0tux,τt−ταdτ.(iii)
The right Riemann-Liouville derivatives:(9)tRDαux,t≔1Γ1−α∂∂t∫tTux,ττ−tαdτ.Many authors believe that the Caputo version is more natural because it makes it easier to manage inhomogeneous initial conditions. Then, the following relationship is related to the two concepts (7) and (8), which can be checked by a direct calculation:
(10)RDtαux,t=CDtαux,t+ux,0Γ1−αtα.Definition 2 (see [29]).
For any realσ>0, we define the space lH0σI as the closure of C0∞I with respect to the following norm ·lH0σI:
(11)ulHσI≔uL2I2+ulH0σI21/2,where
(12)ulHσI2≔D0RtσuL2I2.Definition 3.
For any realσ>0, we define the space rH0σI as the closure of C0∞I with respect to the following norm ·rH0σI:
(13)urH0σI≔uL2I2+urH0σI21/2,where
(14)urH0σI2≔∂tRTσuL2I2.Lemma 4 (see [29, 30]).
For any realσ∈ℝ+, if u∈lHαI and v∈C0∞I, then
(15)RDtσut,vtL2I=ut,tRDσvtL2I.Lemma 5 (see [29, 30]).
For0<σ<2, σ≠1, u∈H0σ/2I, on a
(16)RDtσut=RDtσ/2RDtσ/2ut.Lemma 6 (see [29, 30]).
Forσ∈ℝ+,σ≠n+1/2, the seminorms .lHσI,.rHσI and .cHσI are equivalent. Then, we pose
(17).lHσI=~.rHσI=~.cHσI.Lemma 7 (see [29]).
For any realσ>0, the space lH0σI with respect to the norm (11) is complete.Definition 8.
We denote byL20,T,L20,d≔L2Q the space of square functions, integrated with the scalar product in the Bochner sense,
(18)u,wL20,T,L20,d=∫0Tu,·,w,·L20,ddt.
Since the spaceL20,d is a Hilbert space, it can be shown that L20,T,L20,d is a Hilbert space as well. Let C∞0,T denote the space of infinitely differentiable functions on 0,T and C0∞0,T denote the space of infinitely differentiable functions with compact support in 0,T.


## 3. Solvability of the Direct Fractional Parabolic Problem

### 3.1. Position of Problem

In the rectangular domainQ=0,d×0,T=Ω×I, with d,T<∞ and 0<α<1, we shall study the existence and uniqueness of solutions u=ux,t to the following fractional parabolic problem:
(19)DCtαux,t−∂2ux,t∂x2+bux,t=f~x,tinQ,ux,0=φx∀x∈0,d,u0,t=ud,t=0∀t∈0,T.We consider the following fractional parabolic equation of the type(20)Lu=CDtαu−∂2u∂x2+bu=f~,with the initial condition
(21)ℓu=ux,0=φx,∀x∈0,d,and Dirichlet condition
(22)u0,t=ud,t=0,∀t∈0,T,where b∈ℝ∗+;f~ and φ are known functions.We shall assume that the functionφ satisfies a compatibility conditions, i.e.,
(23)φ0=φd=0.Now, introducing a new function(24)vx,t=ux,t−Ux⇒ux,t=vx,t+Ux,where
(25)φx=Ux.So, we get(26)CDtαvx,t−∂2vx,t∂x2+bvx,t=f~x,t−Lφx=fx,tinQ,vx,0=0∀x∈0,d,v0,t=vd,t=0∀t∈0,T.Such that(27)CDtαvx,t−∂2vx,t∂x2+bvx,t=fx,t,with the initial condition
(28)ℓv=vx,0=0,∀x∈0,d,the boundary condition of Dirichlet type
(29)v0,t=vd,t=0,∀t∈0,T,where
(30)fx,t=f~x,t+∂2φx∂x2−bφx.


### 3.2. A Priori Estimate

In this section, we illustrate the existence and uniqueness of the problem’s solution (27)–(29) as a solution of the operator equation
(31)Lv=F,where L=L,ℓ, with domain of definition B consisting of functions v∈L2Q, such that v, CDtαv,∂v/∂x∈L2Q, and v verify (29).The operatorL is considered from B to F, where B is the Banach space consisting of all functions vx,t having a finite norm
(32)vB2=DCtα/2vL2Q2+vL2Q2+∂v∂xL2Q2,and F is the Hilbert space consisting of all elements Fourier=f,0 for which the norm L2Q is finite.Theorem 9.
For any functionu∈B, we have the inequality
(33)vB≤kLvL2Q,where k is a positive constant independent of v.Proof.
Multiplying equation (27) by the following function:
(34)Mv=vx,t,and integrating over Q=0,d×0,T, we get
(35)∫QLv·Mvdxdt=∫QCDtαvx,t·vx,tdxdt−∫Q∂2vx,t∂x2vx,tdxdt+∫Qb·v2x,tdxdt=∫Qfx,t·vx,tdxdt.
Asvx,0=0, so by applying Lemmas 4, 5, and 6 becomes
(36)∫QCDtαvx,t.vx,tdxdt=CDtαvx,t,vx,tL2Q=RDtα/2RDtα/2vx,t,vx,tL2QRDtα/2vx,t,tRDα/2vx,tL2Q=ucHαQ22≅ulHαQ2=DCtα/2vL2Q2,AccordingtoLemma2AccordingtoLemma1AccordingtoLemma3and by integration by parts over 0,d, we get
(37)−∫Q∂2vx,t∂x2vx,tdxdt=∫Q∂vx,t∂x2dxdt.
So, we obtain(38)∫QDRtαvx,t−∂2vx,t∂x2+bvx,t·Mvdxdt≅RDtα/2vL2Q2+∫Q∂vx,t∂x2dxdt+∫Qbv2x,tdxdt≤12ε∫Qfx,t2dxdt+ε2∫Qvx,t2dxdt.
So, we get(39)DCtα/2vL2Q2+∫Q∂vx,t∂x2dxdt+∫Qb−ε2v2x,tdxdt≤12ε∫Qfx,t2dxdt,which give
(40)DCtα/2vL2Q2+∫Q∂vx,t∂x2dxdt+∫Qb−ε2v2x,tdxdt≤12ε∫Qfx,t2dxdt.
So, we have(41)DCtα/2vL2Q2≤12εfL2Q2.
On the other hand, we have(42)∂v∂xL2Q2≤12εfL2Q2.
Also, we have(43)vL2Q2≤12εb−ε/2fL2Q2.
By combining (41), (42), and (43), for ε<b/2, we get
(44)DCtα/2vL2Q2+∂v∂xL2Q2+vL2Q2≤12ε1+1b−ε/2fL2Q2.
Finally, it follows that(45)DCtα/2vL2Q2+∂v∂xL2Q2+vL2Q2≤CfL2Q2,with
(46)C=12ε1+1b−ε/2.
Therefore, we obtain that(47)vB≤kLvF,wherek=C.
Hence, the uniqueness of the solution.Remark 10.
This inequalityvB≤kLvF gives the uniqueness of the solution, indeed:
Letv1 and v2 two solutions, so
(48)Lv1=FLv2=F⇒Lv1−v2=0,then
(49)v1−v2B≤k0F⇒v1−v2B≤0⇒v1−v2=0,which gives the uniqueness of the solution.Proposition 11.
The operatorL from B to F admits a closure.Proof.
Letvnn∈ℕ⊂DL be a sequence such that:
(50)vn⟶0inB,Lvn⟶FinF,it must be shown that
(51)f≡0.
The convergence ofvn toward 0 in B entails that
(52)vn⟶0inC0∞QT′.
As the continuity of the fractional derivation(2) and the derivation of the first order (as a particular case of the fractional derivative) of C0∞QT′ in C0∞QT′, then (52) implies
(53)Lun⟶0inC0∞QT′.
On the other hand, the convergence ofLvn to f in F=L2QT implies that
(54)Lun⟶finC0∞QT′.
By virtue of the uniqueness of the limit inC0∞QT′, we conclude between (53) and (54) that
(55)f≡0.
Hence, the operatorL is closable.Definition 12.
LetL¯ the closure of L and DL¯ the definition domain of L¯. The solution of the equation
(56)L¯v=F,is called generalized strong solution of the problem (27)–(29).Theorem9 is valid for a generalized strong solution, i.e., we have the following inequality:
(57)vB≤kL¯vF,∀v∈DL¯.Consequently, this last inequality entails the following corollaries:Corollary 13.
The strong solution of the problem (27)–(29) is unique and depends continuously on f∈F.Corollary 14.
The rangeRL¯ of the operator L¯ is equal to the closure RL¯ of RL.Proof.
Letz∈RL¯, then there exists a Cauchy sequence znn in F consists of the elements of the set RL such that
(58)limn⟶+∞zn=z.
So there is a corresponding sequencevnn⊂DL such that
(59)Lvn=zn.
From the estimate (41), we obtain
(60)vp−vqB≤kLvp−LvqF⟶0,whenp,q⟶+∞.
We can deduce thatvnn is a Cauchy sequence in B, so there is v∈B(61)limn⟶+∞vn=vinB.
By virtue of the definition ofL¯ (limn⟶+∞vn=v in B; if limn⟶+∞Lvn=limn⟶+∞zn=z, so limn⟶+∞L¯vn=z and as L¯ is closed so L¯v=z), the function v verifies that
(62)v∈DL¯,L¯v=z.Thus, z∈RL¯, then
(63)RL¯⊂RL¯.
So we conclude here thatRL¯ is closed because it is complete (any complete subspace of a metric space (not necessarily complete) is closed).
It remains to show the opposite inclusion.
Letz∈RL¯, then there is a sequence of znn in F consists of the elements of the set RL¯ such that
(64)limn⟶+∞zn=z,where z∈RL¯, because RL¯ is closed subset of a complete space F; then, RL¯ is complete.
So there is a corresponding sequencevnn⊂DL¯ such that
(65)L¯vn=zn.
From the estimate (57), we obtain
(66)vp−vqB≤kL¯vp−L¯vqF⟶0,ifp,q⟶+∞.
We can deduce thatvnn is a Cauchy sequence in B, so there is v∈B(67)limn⟶+∞vn=vinB.
Once more, there is a corresponding sequenceLvnn∈RL such that
(68)Lvn=L¯vnoverRL,∀n.
Then(69)limn⟶+∞Lvn=z.
Consequently,z∈RL¯, and then, we conclude that
(70)RL¯⊂RL¯.


### 3.3. Existence of Solution

To show the existence of solutions, we prove thatRL is dense in F for all u∈B and for arbitrary F=f,0∈F.Theorem 15.
The problem (27)–(29) admits a solution.Proof.
The scalar product ofF is defined by
(71)Lv,WF=∫QTLv·wdxdt,whereW=w,0∈DL.
If we putw∈RL⊥, we have
(72)∫QTDCtαvx,t−∂2vx,t∂x2+bvx,t·wx,tdxdt=0,where CDtαv, ∂v/∂x, v∈L2QT, with v satisfies the boundary conditions of (27)–(29). From (72), we get the equality
(73)∫QTCDtαvx,t·wx,tdxdt−∫QT∂2vx,t∂x2·wx,tdxdt+b∫QTvx,t·wx,tdxdt=0.
And from the equality (73), we give the function w in terms of v as follows:
(74)w=v,then w∈L2QT.
Replacingw in (73) by its representation (74) and integrating by parts each term of (73) and by taking the condition of v, we obtain
(75)∫QTCDtα/2vx,t2dxdt+∫QTbv2x,tdxdt≤−∫QT∂vx,t∂x2dxdt≤0,then
(76)DCtα/2vL2QT2+bvL2QT2≤0.
Hence(77)vL2QT=0.
And thus,v=0 in QT which gives w=0 in QT. This proves Theorem 15. So RL¯=F.


## 3.1. Position of Problem

In the rectangular domainQ=0,d×0,T=Ω×I, with d,T<∞ and 0<α<1, we shall study the existence and uniqueness of solutions u=ux,t to the following fractional parabolic problem:
(19)DCtαux,t−∂2ux,t∂x2+bux,t=f~x,tinQ,ux,0=φx∀x∈0,d,u0,t=ud,t=0∀t∈0,T.We consider the following fractional parabolic equation of the type(20)Lu=CDtαu−∂2u∂x2+bu=f~,with the initial condition
(21)ℓu=ux,0=φx,∀x∈0,d,and Dirichlet condition
(22)u0,t=ud,t=0,∀t∈0,T,where b∈ℝ∗+;f~ and φ are known functions.We shall assume that the functionφ satisfies a compatibility conditions, i.e.,
(23)φ0=φd=0.Now, introducing a new function(24)vx,t=ux,t−Ux⇒ux,t=vx,t+Ux,where
(25)φx=Ux.So, we get(26)CDtαvx,t−∂2vx,t∂x2+bvx,t=f~x,t−Lφx=fx,tinQ,vx,0=0∀x∈0,d,v0,t=vd,t=0∀t∈0,T.Such that(27)CDtαvx,t−∂2vx,t∂x2+bvx,t=fx,t,with the initial condition
(28)ℓv=vx,0=0,∀x∈0,d,the boundary condition of Dirichlet type
(29)v0,t=vd,t=0,∀t∈0,T,where
(30)fx,t=f~x,t+∂2φx∂x2−bφx.


## 3.2. A Priori Estimate

In this section, we illustrate the existence and uniqueness of the problem’s solution (27)–(29) as a solution of the operator equation
(31)Lv=F,where L=L,ℓ, with domain of definition B consisting of functions v∈L2Q, such that v, CDtαv,∂v/∂x∈L2Q, and v verify (29).The operatorL is considered from B to F, where B is the Banach space consisting of all functions vx,t having a finite norm
(32)vB2=DCtα/2vL2Q2+vL2Q2+∂v∂xL2Q2,and F is the Hilbert space consisting of all elements Fourier=f,0 for which the norm L2Q is finite.Theorem 9.
For any functionu∈B, we have the inequality
(33)vB≤kLvL2Q,where k is a positive constant independent of v.Proof.
Multiplying equation (27) by the following function:
(34)Mv=vx,t,and integrating over Q=0,d×0,T, we get
(35)∫QLv·Mvdxdt=∫QCDtαvx,t·vx,tdxdt−∫Q∂2vx,t∂x2vx,tdxdt+∫Qb·v2x,tdxdt=∫Qfx,t·vx,tdxdt.
Asvx,0=0, so by applying Lemmas 4, 5, and 6 becomes
(36)∫QCDtαvx,t.vx,tdxdt=CDtαvx,t,vx,tL2Q=RDtα/2RDtα/2vx,t,vx,tL2QRDtα/2vx,t,tRDα/2vx,tL2Q=ucHαQ22≅ulHαQ2=DCtα/2vL2Q2,AccordingtoLemma2AccordingtoLemma1AccordingtoLemma3and by integration by parts over 0,d, we get
(37)−∫Q∂2vx,t∂x2vx,tdxdt=∫Q∂vx,t∂x2dxdt.
So, we obtain(38)∫QDRtαvx,t−∂2vx,t∂x2+bvx,t·Mvdxdt≅RDtα/2vL2Q2+∫Q∂vx,t∂x2dxdt+∫Qbv2x,tdxdt≤12ε∫Qfx,t2dxdt+ε2∫Qvx,t2dxdt.
So, we get(39)DCtα/2vL2Q2+∫Q∂vx,t∂x2dxdt+∫Qb−ε2v2x,tdxdt≤12ε∫Qfx,t2dxdt,which give
(40)DCtα/2vL2Q2+∫Q∂vx,t∂x2dxdt+∫Qb−ε2v2x,tdxdt≤12ε∫Qfx,t2dxdt.
So, we have(41)DCtα/2vL2Q2≤12εfL2Q2.
On the other hand, we have(42)∂v∂xL2Q2≤12εfL2Q2.
Also, we have(43)vL2Q2≤12εb−ε/2fL2Q2.
By combining (41), (42), and (43), for ε<b/2, we get
(44)DCtα/2vL2Q2+∂v∂xL2Q2+vL2Q2≤12ε1+1b−ε/2fL2Q2.
Finally, it follows that(45)DCtα/2vL2Q2+∂v∂xL2Q2+vL2Q2≤CfL2Q2,with
(46)C=12ε1+1b−ε/2.
Therefore, we obtain that(47)vB≤kLvF,wherek=C.
Hence, the uniqueness of the solution.Remark 10.
This inequalityvB≤kLvF gives the uniqueness of the solution, indeed:
Letv1 and v2 two solutions, so
(48)Lv1=FLv2=F⇒Lv1−v2=0,then
(49)v1−v2B≤k0F⇒v1−v2B≤0⇒v1−v2=0,which gives the uniqueness of the solution.Proposition 11.
The operatorL from B to F admits a closure.Proof.
Letvnn∈ℕ⊂DL be a sequence such that:
(50)vn⟶0inB,Lvn⟶FinF,it must be shown that
(51)f≡0.
The convergence ofvn toward 0 in B entails that
(52)vn⟶0inC0∞QT′.
As the continuity of the fractional derivation(2) and the derivation of the first order (as a particular case of the fractional derivative) of C0∞QT′ in C0∞QT′, then (52) implies
(53)Lun⟶0inC0∞QT′.
On the other hand, the convergence ofLvn to f in F=L2QT implies that
(54)Lun⟶finC0∞QT′.
By virtue of the uniqueness of the limit inC0∞QT′, we conclude between (53) and (54) that
(55)f≡0.
Hence, the operatorL is closable.Definition 12.
LetL¯ the closure of L and DL¯ the definition domain of L¯. The solution of the equation
(56)L¯v=F,is called generalized strong solution of the problem (27)–(29).Theorem9 is valid for a generalized strong solution, i.e., we have the following inequality:
(57)vB≤kL¯vF,∀v∈DL¯.Consequently, this last inequality entails the following corollaries:Corollary 13.
The strong solution of the problem (27)–(29) is unique and depends continuously on f∈F.Corollary 14.
The rangeRL¯ of the operator L¯ is equal to the closure RL¯ of RL.Proof.
Letz∈RL¯, then there exists a Cauchy sequence znn in F consists of the elements of the set RL such that
(58)limn⟶+∞zn=z.
So there is a corresponding sequencevnn⊂DL such that
(59)Lvn=zn.
From the estimate (41), we obtain
(60)vp−vqB≤kLvp−LvqF⟶0,whenp,q⟶+∞.
We can deduce thatvnn is a Cauchy sequence in B, so there is v∈B(61)limn⟶+∞vn=vinB.
By virtue of the definition ofL¯ (limn⟶+∞vn=v in B; if limn⟶+∞Lvn=limn⟶+∞zn=z, so limn⟶+∞L¯vn=z and as L¯ is closed so L¯v=z), the function v verifies that
(62)v∈DL¯,L¯v=z.Thus, z∈RL¯, then
(63)RL¯⊂RL¯.
So we conclude here thatRL¯ is closed because it is complete (any complete subspace of a metric space (not necessarily complete) is closed).
It remains to show the opposite inclusion.
Letz∈RL¯, then there is a sequence of znn in F consists of the elements of the set RL¯ such that
(64)limn⟶+∞zn=z,where z∈RL¯, because RL¯ is closed subset of a complete space F; then, RL¯ is complete.
So there is a corresponding sequencevnn⊂DL¯ such that
(65)L¯vn=zn.
From the estimate (57), we obtain
(66)vp−vqB≤kL¯vp−L¯vqF⟶0,ifp,q⟶+∞.
We can deduce thatvnn is a Cauchy sequence in B, so there is v∈B(67)limn⟶+∞vn=vinB.
Once more, there is a corresponding sequenceLvnn∈RL such that
(68)Lvn=L¯vnoverRL,∀n.
Then(69)limn⟶+∞Lvn=z.
Consequently,z∈RL¯, and then, we conclude that
(70)RL¯⊂RL¯.


## 3.3. Existence of Solution

To show the existence of solutions, we prove thatRL is dense in F for all u∈B and for arbitrary F=f,0∈F.Theorem 15.
The problem (27)–(29) admits a solution.Proof.
The scalar product ofF is defined by
(71)Lv,WF=∫QTLv·wdxdt,whereW=w,0∈DL.
If we putw∈RL⊥, we have
(72)∫QTDCtαvx,t−∂2vx,t∂x2+bvx,t·wx,tdxdt=0,where CDtαv, ∂v/∂x, v∈L2QT, with v satisfies the boundary conditions of (27)–(29). From (72), we get the equality
(73)∫QTCDtαvx,t·wx,tdxdt−∫QT∂2vx,t∂x2·wx,tdxdt+b∫QTvx,t·wx,tdxdt=0.
And from the equality (73), we give the function w in terms of v as follows:
(74)w=v,then w∈L2QT.
Replacingw in (73) by its representation (74) and integrating by parts each term of (73) and by taking the condition of v, we obtain
(75)∫QTCDtα/2vx,t2dxdt+∫QTbv2x,tdxdt≤−∫QT∂vx,t∂x2dxdt≤0,then
(76)DCtα/2vL2QT2+bvL2QT2≤0.
Hence(77)vL2QT=0.
And thus,v=0 in QT which gives w=0 in QT. This proves Theorem 15. So RL¯=F.


## 4. Existence and Uniqueness of the Solution of Main Problem

We are finding a solution in the form of the original inverse problem.u,f=z,f+y,0 where y is the solution of the direct problem
(78)CDtαy−Δy+βy=0.x,t∈QT,(79)yx,0=φx,x∈Ω,(80)yx,t=0,x,t∈∂Ω×0,T,while the pair z,f is the solution of the inverse problem
(81)CDtαz−Δz+βz=ftgx,t.x,t∈Q,(82)zx,0=0,x∈Ω,(83)zx,t=0,x,t∈∂Ω×0,T,(84)∫Ωvxzx,tdx=Et,t∈0,T,where
(85)Et=θt−∫Ωvxyx,tdx.We will assume that the functions that appear in the problem data are measurable and fulfill the following conditions:(86)g∈C0,T,L2Ω,v∈W21Ω,E∈W220,T,gx,t⩽m;g∗t⩾p>0,forp∈ℝ,x,t∈QT,φx∈W21Ωwhereg∗isdefindin5.The correspondence betweenf and z can be seen as one way of defining the linear operator.
(87)A:L20,T⟶L20,T,with the values
(88)Aft=1g∗∫Ω▽z▽vdx.In this view, the linear equation of the second form for the function is rational to refer tof over the space L20,T:(89)f=Af+W,where
(90)W=CDtαE+βEg∗.Remark 16.
Asu,f=z,f+y,0 where y is the solution of the direct problem (78)–(80). Obviously, the previous section implies that y exists and is unique, but instead of demonstrating the solvability of the initial problem (1)–(4), we demonstrate the existence and uniqueness of the inverse problem (81)–(84) solution.Theorem 17.
Suppose the input of the inverse problem data (81)–(84) satisfies H. Then, the following assertions are valid: (i) if the inverse problem (81)–(84) is solvable, then so is equation (89). And (ii) if equation (89) has a solution and the condition of compatibility has
(91)E0=0,holds, then a solution to the inverse problem exists.Proof.
(i)
Suppose that the inverse problem (81)–(84) is solvable. We denote its solution by z,f. Multiplying the function v scalarly in L2Ω both sides of (81), we get(92)CDtα∫Ωzvdx+∫Ω▽z▽vdx+β∫Ωzvdx=∫Ωftg∗x,t.
With (84) and (88), from (92), it follows that f=Af+CDtαE+βE/g∗. This gives that f solves equation (89).
(ii)
Equation (89) has a solution in space, according to the assumption, L20,T, say f. The resulting relationship (81)–(83) can be viewed as a direct problem with a unique solution z∈W21QT when inserting this function in (81). Let us show that the z function also satisfies the condition of integral overdetermination (84). By equation (92), the function z is subject to the following relation(93)CDtαE+βE+∫Ω▽z▽vdx=ftg∗t.
Subtracting equation (92) from equation (93), we get
(94)CDtα∫Ωzvdx+β∫Ωzvdx=CDtαE+βE.
Integrating the preceding differential equation and taking into account the compatibility condition (89), we find that the overdetermination condition (84) is satisfied by z and the function pair z,f is a solution to the inverse problem (81)–(84).
This completes the theorem’s proof.Now, we are touching on some properties of operatorA.Lemma 18.
Let the conditionH hold. Then, there exists a positive ε for which A is a contracting operator in L20,T.Proof.
Obviously, (88) yields the estimate
(95)AfL20,T≤kp∫0T▽z.,τL2Ω2dτ1/2,where
(96)k=▽vL2Ω.
Multiplying both sides of (81) by z scalarly in L2QT and integrating the resulting by parts with use of (82), we get
(97)DCtα/2zL2QT2+▽zL2QT2+βzL2QT2=∫0Tft∫Ωgx,tzdxdt.
Thus, by using the Cauchy’sε-inequality, we obtain
(98)DCtα/2zL2QT2+▽zL2QT2+βzL2QT2≤mΩ2ε∫0Tft2dt+ε2zL2QT2.
Choosing0<ε<2β, we get
(99)DCtα/2zL2QT2+▽zL2QT2+β−ε2zL2QT2≤mΩ2ε∫0Tfτ2dt.
Omitting some terms on the left-hand side (99) leads to
(100)▽zL2QT2=∫0T▽z.,τL2Ω2dτ≤mΩ2ε∫0Tfτ2dt.
According to (95) and (100), we can obtain the following estimate:
(101)AfL20,T≤δ∫0Tfτ2dt,0⩽t⩽T,where
(102)δ=kmΩp2ε.
So, we obtain(103)AfL20,T≤δfL20,T.
It is obvious from the above that there is positiveε such that
(104)δ<1.
Inequality (103) shows that the operator A is a contracting mapping on L20,T.The following result may be useful with respect to the particular solvability of the inverse problem concerned.Theorem 19.
Let the compatibility condition (91) and the condition H hold. Then, the inverse problem (81)–(84) has a unique solution z,f.Proof.
This means that the equation (89) has a unique solution f in L20,T.
The existence of a solution to the inverse problem (81)–(84) is verified, according to Lemma 6.
The uniqueness of this solution has yet to be proven.
Suppose the contrary that there are two distinct solutionsz1,f1 and z2,f2 of the under consideration inverse problem.
Also, the linear operatorA is contracting on L20,Tfrom Lemma 18, which gives that f1=f2; then, by the theorem of the uniqueness of the solution of main direct problem (78)–(80), we will just have z1=z2.Corollary 20.
The solutionf to equation (91) depends continuously, under the conditions of Theorem 19, on the data W.Proof.
LetV1 and V2 twosets of data that satisfy Theorem 19’s assumptions.
Letf and g be solutions of the equation (89) corresponding to the data V1 and V2, respectively. According to (103), we have
(105)f=Af+V1,g=Ag+V2.
Let us estimate the difference first,f−g. It is easy to see with the use of (103) that
(106)f−gL20,T=Af+V1−Ag+V2L20,T=Af−g+V1−V2L20,T⩽δf−gL20,T+V1−V2L20,T,so, we get
(107)f−gL20,T⩽11−δV1−V2L20,T.


## 5. Conclusion and Perspectives

This work contains a new inverse problem by investigating the fractional derivatives where we develop the method of fixed point and energy inequality method for proving the solvability of an inverse fractional problem. We note that our work extends to the existence of open problems as a study of the nonlinear case of this problem and the numerical part.



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