MATHEMATICA MONTISNIGRI
Vol LVIII (2023)

INVERSE COEFFICIENT SUPER-LINEAR PROBLEM FOR A TIMEFRACTIONAL PARABOLIC EQUATION UNDER INTEGRALOVERDETERMINATION CONDITIONSTAKI-EDDINE OUSSAEIF 1*, AMAL BENGUESMIA 1 AND IMAD REZZOUG 1
1

Dept.of Mathematics and Informatics, University Oum El Bouaghi, Algeria.
*Corresponding author. E-mail: taki_maths@live.fr; takieddine.oussaeif@gmail.com

DOI: 10.20948/mathmontis-2023-58-2
Summary. In this paper, we investigate the inverse problem of the determining the righthand side of a nonlinear fractional parabolic equation with integral overdetermination
condition. For the solvability of the direct problem we use the energy inequality method also
known as the methode of functional analysis or the method of a priori estimates, this method
has a higher character in that the existence theorem can be derived from the solution of the
problem posed, begining with the uniqueness theorem. The most difficult aspect of this method
is selecting the functional spaces E and F and for the unique solvability of the iverse problem
is estblished under suitable conditions. The fixed point theorem is used to establish the existence
and uniqueness of the solution to the inverse problem based on the data, this inverse problem
appears frequently in engineering and physics modeling of various phenomena. Sismology,
medicine, continuous casting, fusion welding, oil and gas production during drilling and welloperation.

1 INTRODUCTION
This manuscript is devoted to investigate the solvability of determining a pair of functions{u(x,t), f(t)} satisfying the following fractional parabolic problem:𝐶 𝛼
𝜕𝑡 𝑢-∆𝑢 + 𝛽𝑢 + 𝑢2 =f(t)g(x,t),x∊𝛺, t∊ (0,T),(1)

with the initial condition
x∊𝛺,

(2)

(x,t)∊𝜕𝛺×[0,T],(3)

u(x,0)=0,
the boundary condition
u(x,t)=0,

and the nonlocal integral overdeterminantion condition∫𝛺 𝑣(𝑥 )𝑢(𝑥, 𝑡)𝑑𝑥 = 𝐸 (𝑡),t∊[0,T].

(4)

Where 𝛺 = (0, 𝑑) with 𝑑 > 0, the functions g and 𝐸 are known functions and 𝛽 is a given
positive constant.
Inverse problems for the heat equations (see[17, 25, 27], and references therein) arise
naturally in many reality phenomena. Here supplementary or additional information about thesolution to the inverse problem comes in the form of the integral condition (4).
2020 Mathematics Subject Classification: 35R30, 35K61, 35B45.
Key words and Phrases: Inverse Nonlinear Problem, Nonlocal Integral Condition, Strong Solution, Fixed Point
Theorem, Energy Inequality, Regularization.

17T-E. Oussaeif , A. Benguesmia and I.Rezzoug

In theory of PDE's in engineering and physics, the integral condition plays an important role
as a modeling tool [6, 8, 16, 19, 22, 24].
It is worth noting that nonlocal issues are associated with integral overdetermination [5, 23].
Studies have revealed that when dealing with non-classical problems, classical procedures may
not always effective. [2, 9]. To date, various strategies for dealing with problems caused by
non-local problems have been proposed. The strategy chosen is determined by type of non-local
boundary value involved.
Several writers have investigates the inverse parabolic problem with condition of type (4)and its specific solvability ( For instance, [3, 25, 26, 27, 28, 29] ).
Several articles also devoted to the investigation of the existence and uniqueness of inverse
problem solutions for various parabolic equations with unknown source functions. Inverse
difficulties involving the determination term of a parabolic equation with overdetermination
condition [10, 13]Differential equations are generated to an arbitrary order to produce fractional differential
equations (FDE's). Fractional differential equations are important in engineering, physics andapplied mathematics because they are used to model complex phenomena. As a result, in recent
years engineers and scientists have become increasingly interested in them. Complex
phenomena can be modled using FDE's because they have memory and nonlocal relation ships
in space and time [1, 11, 12, 14, 15, 18].
2 FUNCTIONAL SPACES
Definition 1
let us define the notations and rules we will employ𝑔∗ (𝑡) = ∫𝛺 𝑔(𝑥, 𝑡). 𝑣 (𝑥 )𝑑𝑥 , 𝑄 = 𝛺 × (0, 𝑇).

(5)

Let Г(.) represent the gamma function. The Caputo and Rieman Liouville derivatives respectively aredefined as follows for all 0< 𝛼 < 1(i)The left Caputo derivatives:𝑡 𝜕𝑢(𝑥,𝜏) 11𝑐 𝛼
𝜕𝑡 𝑢(𝑥, 𝑡):=𝛤(1−𝛼) ∫0 𝜕𝜏 (𝑡−𝜏)𝛼 𝑑𝜏.

(ii)

The left Riemann-Liouville derivatives:1
𝜕 𝑡 𝑢(𝑥,𝜏)𝑅 𝛼𝜕𝑡 𝑢(𝑥, 𝑡):=𝛤(1−𝛼) 𝜕𝑡 ∫0 (𝑡−𝜏)𝛼 𝑑𝜏.

(iii)

(6)

(7)

The right Riemann-Liouville derivatives:1
𝜕 𝑇 𝑢(𝑥,𝜏)𝑅 𝛼𝑡 𝜕 𝑢(𝑥, 𝑡):=𝛤(1−𝛼) 𝜕𝑡 ∫𝑡 (𝑡−𝜏)𝛼 𝑑𝜏(8)Many authors argue that the Caputo version is more natural because it is more manageable
given homogeneous initial conditions. The following relationship exists between the two
notions (6) and (7), which may be verified by a direct calculation:18

T-E. Oussaeif , A. Benguesmia and I.Rezzoug

𝑅 𝛼𝜕𝑡 𝑢 (𝑥, 𝑡)=𝑢(𝑥,0)𝑐 𝛼
𝜕𝑡 𝑢(𝑥, 𝑡)+𝛤(1−𝛼)𝑡 𝛼(9)𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝟐 [21]We define the space 𝑙 𝐻0𝛼 (𝑄) as the closure of 𝐶0∞ (𝑄) with respect to the following norm
∥ 𝑢 ∥ 𝑙𝐻0𝛼 (𝑄) for any real 𝛼 > 0 :1∥ 𝑢 ∥ 𝑙𝐻𝛼 (𝑄) ∶= (∥ 𝑢 ∥2𝐿2 (𝑄) + |𝑢|2 𝑙𝐻𝛼 (𝑄) )2 ,

(10)

0

where
|𝑢|2 𝑙𝐻 𝛼 (𝑄) =∥ 𝑅0𝜕𝑡𝛼 𝑢 ∥𝐿2(𝑄)Definition 3
The space 𝑅 𝐻0𝛼 (𝑄)is defined as the closure of 𝐶0∞(𝑄)with respect to the following norm
∥ 𝑢 ∥ 𝑅𝐻0𝛼 (𝐼) for any real 𝛼 > 0 :1∥ 𝑢 ∥ 𝑅𝐻0𝛼 (𝐼) ∶= (∥ 𝑢 ∥2𝐿2 (𝐼) + |𝑢|2 𝑅𝐻𝛼 (𝐼) )2 ,

(11)

0

where
|𝑢|2 𝑅 𝐻𝛼 (𝑄) ∶=∥ 𝑅𝑡𝜕𝑇𝛼 𝑢 ∥2𝐿2 (𝑄) ,0

Lemma 1 [21, 22]
If u ∈ 𝑙 𝐻 𝛼 (𝑄 ) 𝑎𝑛𝑑 𝑣 ∈ 𝐶0∞ (𝑄 ) 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝛼 ∈ ℝ+ , then(𝑅 𝛼𝜕𝑡 𝑢(𝑡), 𝑣(𝑡))𝐿2 (𝑄) = (𝑢(𝑡), 𝑅𝑡 𝜕 𝛼 𝑣(𝑡))𝐿2 (𝑄).

Lemma 2 [21, 22]𝛼2

For 0< 𝛼 < 2, 𝛼 ≠ 1, 𝑢 ∈ 𝐻0 (𝑄 ), 𝑤𝑒 ℎ𝑎𝑣𝑒 ∶𝑅 𝛼 ( )
𝜕𝑡 𝑢 𝑡 =

𝛼

𝑅 2
𝜕𝑡

𝛼

𝑅 2
𝜕𝑡 𝑢(𝑡).

Lemma 3 [21, 22]
The semi-norms|. | 𝑙𝐻𝛼 (𝑄) , |. | 𝑅𝐻𝛼 (𝐼) 𝑎𝑛𝑑 |. | 𝑐𝐻𝛼 (𝑄) are equivalent for1
𝛼 ∈ ℝ+, 𝛼 ≠ 𝑛 + . Then we pose2
|𝑢| 𝑙𝐻𝛼 (𝑄) ≅ |. | 𝑅 𝐻𝛼(𝑄) ≅ |. | 𝑐 𝐻𝛼 (𝑄) .
Lemma 4 [21]For each real 𝛼 > 0, 𝑡ℎ𝑒 𝑠𝑝𝑎𝑐𝑒 l H0α (Q) with respect to the norm (10) is complete.
Definition 4
𝐿2 (0, T, 𝐿2 (0, d)) is denoted by 𝐿2 (𝑄) the space of square functions, in the Bochner sense,
integrated with the scalar product𝑇(𝑢, 𝑤)𝐿2(0,𝑇,𝐿2(0,𝑑)) = ∫ (𝑢, 𝑤)𝐿2 (0,𝑑) 𝑑𝑡 .
0

19

(12)

T-E. Oussaeif , A. Benguesmia and I.Rezzoug

3

SOLVABILITY OF THE DIRECT FRACTIONAL PARABOLIC PROBLEM

3.1

Setting of the problem

We will investigate the existence and uniqueness of the strong solution u=u(x,t)(we suppose that u > 0) to the following fractional parabolic problem in the rectangulardomain Q=Ω ×(0,T), with Ω = (0,d) , T< ∞, 𝑑 > 0 and 0< α <1.
2{

𝜕 𝑢(𝑥,𝑡)𝑐 𝛼 (𝜕𝑡 𝑢 𝑥, 𝑡) − 𝜕𝑥2 + 𝛽𝑢 (𝑥, 𝑡) + 𝑢2 (𝑥, 𝑡) = 𝑓̃(𝑥, 𝑡)

𝑖𝑛 𝑄,
(P)

𝑢(𝑥, 0) = 0, 𝑥 ∈ Ω,𝑢(0, 𝑡) = 𝑢(𝑑, 𝑡) = 0, 𝑡 ∊ (0, 𝑇).

Whose nonlinear fractional parabolic equation is given as followsℒu=2𝜕 𝑢(𝑥,𝑡)𝑐 𝛼𝜕𝑡 𝑢(𝑥, 𝑡) − 𝜕𝑥2 + 𝛽𝑢(𝑥, 𝑡) + 𝑢2 (𝑥, 𝑡) = 𝑓̃ (𝑥, 𝑡),(where f̃(x, t)= f(t)g(x,t)),with the initial condition :ℓ𝑢 = 𝑢(𝑥, 0) = 0,𝑥 ∈ Ω,and Dirichlet condition:𝑢(0, 𝑡) = 𝑢(𝑑, 𝑡) = 0,𝑡 ∊ (0, 𝑇).

̃where 𝛽 ∈ ℝ+
∗ , 𝑓 is known function.
In this section, we demonstrate the existence and uniqueness of the problem's solution. (1)(3) as a solution of the operator equation
𝐿𝑢 = ℱ,(13)

where L = (ℒ, ℓ), with domain of definition D(L) =B which defined by
D(L)={𝑢\u ∈ 𝐿2 (𝑄 ) ∩ 𝐿3 (𝑄 ),𝜕𝑢𝑐 𝛼𝜕𝑡 𝑢, 𝜕𝑥 ∈ 𝐿2 (𝑄)}.

The operator L is defined from B to F, where B is the Banach space which includes allfunctions u(x,t) having a finite norme:∥ 𝑢 ∥2𝐵 =∥

𝛼
𝑑𝑢 2
𝑐 2
𝜕𝑡 𝑢 ∥2𝐿2 (𝑄) +∥∥𝐿2 (𝑄) +∥ 𝑢 ∥2𝐿2 (𝑄) +∥ 𝑢 ∥3𝐿3 (𝑄) ,𝑑𝑥

and F is the Hilbert space consisting of all elements Fourier =( 𝑓̃,0) for which the norm 𝐿2 (𝑄)is finite.
Theorem 5
For each function 𝑢 ∈ 𝐵, we have the inequality∥ 𝑢 ∥𝐵 ≤ 𝐶 ∥ 𝐿𝑢 ∥𝐿2 (𝑄) ,where C is a positive constant independent of u.
Proof 120

(14)

T-E. Oussaeif , A. Benguesmia and I.Rezzoug

We use the scalar product in 𝐿2 (𝑄 ) of ℒu and the following function :Mu=u(x,t),
where
Q=(0,d)×(0,T)
𝑐 𝛼 (𝜕𝑡 𝑢 𝑥, 𝑡). 𝑢(𝑥, 𝑡)𝑑𝑥𝑑𝑡∫ ℒ𝑢. 𝑀𝑢 𝑑𝑥𝑑𝑡 = ∫
𝑄

𝑄

−∫ (
𝑄

𝜕 2 𝑢(𝑥, 𝑡)) . 𝑢(𝑥, 𝑡)𝑑𝑥𝑑𝑡
𝜕𝑥 2
(15)

+ 𝛽 ∫ 𝑢2 (𝑥, 𝑡)𝑑𝑥𝑑𝑡
𝑄+ ∫ 𝑢3 (𝑥, 𝑡)𝑑𝑥𝑑𝑡= ∫ 𝑓̃ (𝑥, 𝑡)𝑢(𝑥, 𝑡)𝑑𝑥𝑑𝑡
𝑄

As u(x,0)=0, so by applying the lemma 1, 2 and 3 becomes:
∫Q

c α (
𝜕t u x, t). u(x, t)dxdt = (
𝛼

𝑅 2
𝜕𝑡

=(

c α
𝜕t u , u)L2 (Q)𝛼𝑅 2𝜕𝑡 𝑢, 𝑢)𝐿2 (𝑄)𝛼( According to lemma 2)𝛼=(

𝑅 2 𝑅 2𝜕𝑡 𝑢, 𝑡 𝜕 𝑢)𝐿2(𝑄)=∥𝐶 2
𝜕𝑡 𝑢 ∥2𝐿2 (𝑄)( According to lemma 1)

𝛼

𝜀𝑎2

.

𝑏2

By applying the relationship ( ǀabǀ≤ 2 + 2𝜀 ) and using the integral by parts , we find
𝛼

∥

𝑑𝑢

𝐶 2
𝜕𝑡 𝑢 ∥2𝐿2 (𝑄)𝜀+∥ 𝑑𝑥 ∥2𝐿2(𝑄) +(𝛽-2) ∥ 𝑢 ∥2𝐿2 (𝑄) +∥ 𝑢 ∥3𝐿3 (𝑄) ≤12𝜀∥ 𝑓̃ ∥2𝐿2 (𝑄) .

So, for (ε ≤ 2β) we get𝛼𝜀𝑚𝑖𝑛(1, 𝛽 − 2 )(∥𝐶 2
𝜕𝑡 𝑢 ∥2𝐿2 (𝑄)𝑑𝑢+∥ 𝑑𝑥 ∥2𝐿2 (𝑄) +∥ 𝑢 ∥2𝐿2 (𝑄) +∥ 𝑢 ∥3𝐿3(𝑄) ) ≤1
2𝜀∥ 𝑓̃ ∥2𝐿2 (𝑄) .

So, we get
𝛼

∥

𝐶 2
𝜕𝑡 𝑢 ∥2𝐿2 (𝑄)𝑑𝑢+∥ 𝑑𝑥 ∥2𝐿2(𝑄) +∥ 𝑢 ∥2𝐿2(𝑄) +∥ 𝑢 ∥3𝐿3 (𝑄) ≤ 𝑐 ∥ 𝑓̃ ∥2𝐿2 (𝑄) .

With

21

T-E. Oussaeif , A. Benguesmia and I.Rezzoug𝑐=1
𝜀2𝜀 𝑚𝑖𝑛(1, 𝛽 − 2)

As a result, we obtain that:∥ 𝑢 ∥𝐵 ≤ 𝐶 ∥ 𝐿𝑢 ∥𝐿2 (𝑄) ,with
𝐶 = √𝑐 .
Proposition 6
The operator L from B to F has a closure.
Proof 2Let (𝑢𝑛 )𝑛∈ℕ ⊂ D(L) be a sequence in which:𝑢𝑛 ⟶ 0

in B,

(16)

L𝑢𝑛 ⟶ ℱ

in F.

(17)

and
We must prove that𝑓̃ ≡ 0.
In B, the convergence of 𝑢𝑛 to 0 causes :′𝑖𝑛 (𝐶0∞ (𝑄)) .

𝑢𝑛 ⟶ 0(18)

As the continuity of the fractional derivative (as defined by corollary(2)) and the derivation′of the first order (as a special case of the fractional derivative) of (𝐶0∞(𝑄 )) in′(𝐶0∞ (𝑄)) 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑢2 , the relationship (18)
involved′𝑖𝑛 (𝐶0∞ (𝑄)) .

ℒ𝑢𝑛 ⟶ 0(19)

In addition, the convergence of L𝑣𝑛 𝑡𝑜 𝑓̃ 𝑖𝑛 𝐿2 (𝑄 ) yields:′ℒ𝑣𝑛 ⟶ 𝑓̃ 𝑖𝑛 (𝐶0∞ (𝑄)) .

(20)

Since we know that the limit in (𝐶0∞(𝑄 ))′ is unique, we can conclude from (19) and (20)that :𝑓̃ ≡ 0.
As a result, the operatsor L is closable.
Let 𝐿̅ be the closure of L, and D(𝐿̅) be the domain of definition of 𝐿̅.
Definition 5
A solution to the operator equation22T-E. Oussaeif , A. Benguesmia and I.Rezzoug

𝐿̅𝑢 = ℱ.
Is known as a strong solution to problem (1)-(3).
We may extend the a priori estimate to strong solution, i.e; we obtain the estimate∥ 𝑢 ∥𝐵 ≤ 𝐶 ∥ 𝐿̅𝑢 ∥𝐹 , ∀𝑢 ∈ 𝐷(𝐿̅).

(21)

Corollary 7
The strong solution of the problem (1)-(3) is unique and depends continuously on 𝑓̃ ∈ 𝐹.

Corollary 8
The range R(𝐿̅) of the operator 𝐿̅ is closed in F and equals the closure of R(L), that is,R(𝐿̅)=𝑅(𝐿).
Proof 3First, we demonstrate the uniqueness of the solution if it exists
Let 𝑢1 ,𝑢2 be two solution we put 𝜂 = 𝑢1 -𝑢2 , then 𝜂 satisfies:𝑐 𝛼 (𝜕𝑡 𝜂 𝑥, 𝑡) −

𝜕 2 𝜂 (𝑥, 𝑡)+ 𝛽 𝜂 (𝑥, 𝑡) + 𝑢12 (𝑥, 𝑡) − 𝑢22 (𝑥, 𝑡) = 0,
𝜕𝑥 2
𝜂 (𝑥, 0) = 0, 𝑥 ∈ 𝛺,
𝜂 (𝑥, 𝑡) = 0, (𝑥, 𝑡) ∈ 𝜕𝛺 × (0, 𝑇).

{

𝑖𝑛 𝑄,
(𝑷′ )

.
𝑐 𝛼 (
𝜕𝑡 𝜂 𝑥, 𝑡) −

𝜕 2 𝜂 (𝑥, 𝑡)+ 𝛽𝜂 (𝑥, 𝑡) + 𝑢12 (𝑥, 𝑡) − 𝑢22 (𝑥, 𝑡) = 0,
2
𝜕𝑥

Q.
𝑖𝑛 𝑄,
(22)

We use the scalar product in 𝐿2 (Ω) of (22) and 𝜂 we get :∫𝛺𝜕 2 𝜂 (𝑥,𝑡)𝑐 𝛼𝜕𝑡 𝜂(𝑥, 𝑡). 𝜂 (𝑥, 𝑡)𝑑𝑥 − ∫𝛺 ( 𝜕𝑥2 ) . 𝜂 (𝑥, 𝑡)𝑑𝑥 + 𝛽 ∫𝛺 𝜂 2 (𝑥, 𝑡)𝑑𝑥 + ∫𝛺 (𝑢12 (𝑥, 𝑡) −𝑢22 (𝑥, 𝑡))(𝑢1 (𝑥, 𝑡) − 𝑢2 (𝑥, 𝑡))𝑑𝑥 = 0.

As 𝜂( x,0)=0, so by applying the lemma 1, 2 and 3 and integreating by parts we get:
𝛼

∥

𝑐 2
𝜕𝑡 𝜂

𝑑𝜂

∥2𝐿2(𝛺) +∥ 𝑑𝑥 ∥2𝐿2 (𝛺) + ∥ 𝜂 ∥2𝐿2 (𝛺) +∫𝛺 (𝑢12 − 𝑢22 )(𝑢1 -𝑢2 )𝑑𝑥 = 0 = (23)=0Since 𝜆2 is a monotone function in 𝜆 (on Ω = (0,d)), we can conclude that the last terme of
the left hand side of (23) is positive , so it follows from (22) that:∥ 𝜂 ∥2𝐿2(𝛺)≤ 0.

23

T-E. Oussaeif , A. Benguesmia and I.Rezzoug

Which implies that for all t𝜖 (0, 𝑇) we have:𝑢1 =𝑢2 .
We''ll return to demonstrate the corollary 8
Let 𝑧 ∈ 𝑅(𝑙) so there is a sequence (zn )n∈ℕ in R(L) such as
𝑙𝑖𝑚 𝑧𝑛 = 𝑧𝑛So like as (𝑧𝑛 )𝑛∈ℕ in R(L), there exists a sequence (𝑢𝑛 )𝑛∈ℕ in D(L) such that
L𝑢𝑛 = 𝑧𝑛 .
Let ℰ, n ≥ n0
Let m, m'∈ ℕ, 𝑚 ≥ 𝑚′ such as 𝑢𝑚 , 𝑢𝑚′ be two solutionsi,e;L𝑢𝑚 = 𝑓 𝑎𝑛𝑑 𝐿𝑢𝑚′ = 𝑓
We put y= 𝑢𝑚 − 𝑢𝑚′ then y satisfies
𝜕 2 𝑦 (𝑥, 𝑡)2
2) + 𝛽𝑦 (𝑥, 𝑡) + 𝑢𝑚
− 𝑢𝑚
′ = 0,
𝜕𝑥 2
𝑦 (𝑥, 0) = 0, 𝑥 ∈ Ω,
𝑦 (𝑥, 𝑡) = 0, (𝑥, 𝑡) ∈ 𝜕𝛺 × (0, 𝑇)𝑐 𝛼 (𝜕𝑡 𝑦 𝑥, 𝑡) − (𝑖𝑛 𝑄,(𝑃′′ ){.
We apply same proceeding which we use to prove the uniqueness of the solution we get y=0
which implies that for all t𝜖(0,T) we have0 ≤∥ 𝑢𝑚 − 𝑢𝑚′ ∥≤ 0i;e
∀ 𝜀 ≥ 0, ∃𝑛0 ∈ ℕ ∖ ∀𝑚, 𝑚′ ≥ 𝑛0 ; ∥ 𝑢𝑚 − 𝑢𝑚′ ∥≤ 𝜀So we can conclude that (𝑢𝑛 )𝑛∈ℕ is a cauchy sequence in E which is a Banach space, thenthere is 𝑢 ∈ 𝐸 such that
𝑙𝑖𝑚 𝑢𝑛 = 𝑢𝑛Using the definition of 𝐿̅ (lim 𝑢𝑛 = 𝑢 in E, if lim 𝑢𝑛 = lim 𝑧𝑛 = 𝑢, 𝑡ℎ𝑒𝑛 lim 𝐿̅ 𝑢𝑛 = 𝑧 as𝑛𝑛𝑛𝑛like 𝐿̅ is closed so 𝐿̅𝑢 = 𝑧), the function u checku∈ 𝐷(𝐿̅), 𝐿̅𝑢 = 𝑧then z∈ 𝑅(𝐿̅) so
𝑅(𝐿) ⊂R(𝐿̅ )Furthermore, because R(𝐿̅)is Banach, we conclude that is closed
It is still necessary to demonstrate the opposing inclusion.
Let z∈ R(𝐿̅), then there exists a sequence of ( 𝑧𝑛 )𝑛 𝑖𝑛 𝐹

24

T-E. Oussaeif , A. Benguesmia and I.Rezzoug

Consisting of the elements of the set R(𝐿̅), such that
𝑙𝑖𝑚 𝑧𝑛 = 𝑧𝑛As a consequence, there is a corresponding sequence (𝑢𝑛 )𝑛∈ℕ such that𝑙𝑖𝑚 𝐿̅𝑢𝑛 = 𝑧𝑛𝑛

On the other hand we have (𝑢𝑛 )𝑛∈ℕ is a cauchy sequence in F (to show this we apply thesame methode wich we use in the previous step)
So there exists u∈ 𝐸:
𝑙𝑖𝑚 𝑢𝑛 = 𝑢 𝑖𝑛 𝐸𝑛then𝑙𝑖𝑚 𝐿̅ 𝑢𝑛 = 𝑧.
𝑛

As a consequence z ∈ 𝑅(𝐿) and we can conclude that
𝑅(𝐿) =R(𝐿̅ )3.3

Existence of solution

To demonstrate the existence of solutions, we demonstrate that R(L) is dense in F for allu ∈ 𝐵 and for arbitrary ℱ = (𝑓̃, 0) ∈ 𝐹.
Theorem 9
The problem (1)-(3) has a solution.
Proof 4The scalar product of F is defined by(𝐿𝑣, 𝑊)𝐹 = ∫ ℒ𝑣. 𝑤𝑑𝑥𝑑𝑡,

𝑤ℎ𝑒𝑟𝑒 𝑊 = (𝑤, 0) ∈ 𝐷(𝐿).

(24)

𝑄

) ⊥

If we put w∈ (𝑅(𝐿 ) , we get:∫𝑄𝑐 𝛼𝜕𝑡 𝑢(𝑥, 𝑡)𝑤(𝑥, 𝑡)𝑑𝑥𝑑𝑡 − ∫𝑄(

𝜕 2 𝑢 (𝑥,𝑡)𝜕𝑥 2) 𝑤(𝑥, 𝑡)𝑑𝑥𝑑𝑡 + 𝛽 ∫𝑄 𝑢(𝑥, 𝑡)𝑤(𝑥, 𝑡)𝑑𝑥𝑑𝑡 +∫𝑄 𝑢𝟐 (𝑥, 𝑡)). 𝑤(𝑥, 𝑡)𝑑𝑥𝑑𝑡 = 0.
Putting w=u we get:∫𝑄𝑐 𝛼𝜕𝑡 𝑢(𝑥, 𝑡). 𝑢 (𝑥, 𝑡)𝑑𝑥𝑑𝑡 − ∫𝑄

𝜕 2 𝑢 (𝑥,𝑡)

(

𝜕𝑥 2

) . 𝑢 (𝑥, 𝑡)𝑑𝑥𝑑𝑡 +(25)

𝛽 ∫𝑄 𝑢 2 (𝑥, 𝑡)𝑑𝑥𝑑𝑡 + ∫𝑄 𝑢3 (𝑥, 𝑡)𝑑𝑥𝑑𝑡 = 0.
Integreating by parts each term of (25) and taking account the condition of u, we obtain :25

T-E. Oussaeif , A. Benguesmia and I.Rezzoug

𝛼𝑐 2
𝜕𝑡 𝑢 ∥2𝐿2 (𝑄)∥𝑑𝑢+∥ 𝑑𝑥 ∥2𝐿2 (𝑄) + 𝛽 ∥ 𝑢 ∥2𝐿2(𝑄)+∥ 𝑢 ∥3𝐿3(𝑄)=0

So , we get
𝛼

∥

𝑐 2
𝜕𝑡 𝑢

𝑑𝑢

∥2𝐿2 (𝑄) + 𝛽 ∥ 𝑢 ∥2𝐿2 (𝑄)+∥ 𝑢 ∥3𝐿3 (𝑄) ≤ −∥ 𝑑𝑥 ∥2𝐿2(𝑄)
≤ 0.

Then
∥ 𝑢 ∥2𝐿2 (𝑄) ≤ 0.
It means that u=0 in Q which gives w=0 in Q, this proves Theorem 9.
4

EXISTENCE AND UNIQUENESS OF THE SOLUTION OF MAIN PROBLEM

We suppose that the functions appearing in the data for the problem are mesurable and satisfy
the following conditions:{𝑔 ∈ 𝐶((0, 𝑇), 𝐿2 (𝛺 )), 𝑣 ∈ 𝑊21 (𝛺 ), 𝐸 ∈ 𝑊22 (0, 𝑇),|𝑔∗ (𝑡)| ≥ 𝑟 > 0,(𝑥, 𝑡) ∈ 𝑄∥ 𝑔(𝑥, 𝑡) ∥≤ 𝑚,
𝑓𝑜𝑟 𝑟 ∈ ℝ,

(𝐻)

The relation between f and u is given by the following linear operator
A : 𝐿2 (0, 𝑇) ⟶ 𝐿2 (0, 𝑇) .

(26)

With the values
1(Af(t))= 𝑔∗ {∫𝑑𝑢 𝑑𝑣

𝛺 𝑑𝑥 𝑑𝑥

𝑑𝑥 + ∫𝛺 𝑢2 (𝑥, 𝑡). 𝑣(𝑥 )𝑑𝑥} .

(27)

As a result, the previous relationship between f and u can be represented as a second-orderlinear equation for the function f over 𝐿2 (0, 𝑇):
f=Af+W .

(28)

Where
𝐶 𝐷𝛼 𝐸+𝛽𝐸
𝑡
𝑔∗

W=

.

(29)

And
E(0)=0.
Theorem 10
Assume that the inverse problem's data functions (1)-(4) validate the condition (H). Then
we have the following statement equivalent:
(i)
(ii)

If the inverse problem (1)-(4) is solvable, then equation is also solvabl (28)
If equation (28) has a solution and the compatibility condition E(0)=0 hold, theinverse problem (1)-(4) also has a solution.

Proof.

26

T-E. Oussaeif , A. Benguesmia and I.Rezzoug(i)Suppose that the problem (1)-(4) has been solved. We designate its solution by{u,f}.

Miltiplying both side of (1) by v and integreating over Ω we get:
𝑐 𝛼∫
𝜕𝑡
𝛺

𝑑𝑢 𝑑𝑣
𝑑𝑥 + 𝛽 ∫ 𝑢(𝑥, 𝑡). 𝑣 (𝑥 )𝑑𝑥
𝛺 𝑑𝑥 𝑑𝑥𝛺𝑢(𝑥, 𝑡). 𝑣(𝑥 )𝑑𝑥 + ∫(30)+ ∫ 𝑢2 (𝑥, 𝑡). 𝑣(𝑥 )𝑑𝑥 = 𝑓(𝑡)𝑔∗ (𝑡).
𝛺Using (4) and (27), we get
f=Af+𝛽𝐸+ 𝑐 𝐷𝑡𝛼 𝐸
𝑔∗This give us that f solves equation (28).
(ii)

By assuming that equation (28) has a solution, say f, and then substituting f into
equation (1), the resulting relation (1)-(3) can be viewed as a direct problem with aunique solution.

It remains to show that u satisfies also the condition of integral overdetermination (4). By theequation (30) the function u is subject to the following relation𝑐 𝛼
𝜕𝑡 𝐸 + ∫𝛺

𝑑𝑢 𝑑𝑣
𝑑𝑥 + 𝛽𝐸 + ∫ 𝑢2 (𝑥, 𝑡). 𝑣(𝑥 )𝑑𝑥 = 𝑓 (𝑡)𝑔∗ (𝑡).
𝑑𝑥 𝑑𝑥
𝛺

(31)

Substracting equation (30) from equation (31), we obtain
𝑐 𝛼
𝜕𝑡 ∫𝛺

𝑢(𝑥, 𝑡). 𝑣(𝑥 )𝑑𝑥 + 𝛽 ∫𝛺 𝑢(𝑥, 𝑡). 𝑣(𝑥 )𝑑𝑥 =𝑐 𝛼
𝜕𝑡 𝐸 + 𝛽𝐸.

(32)Integreating the previous equation and taking account the compatibility condition E(0)=0
we conclude that u satisfies the integral condition (4)As a result, we may deduce that {u,f} is the solution of the inverse problem (1)-(4).
Now, we state some properties of the operator A.
Lemma 11
Let the condition (H) hold, then there is a positive 𝛿 for which A is a contracting operator in
𝐿2 (0, 𝑇).
Proof 6
The following estimate comes from (27)|𝐴𝑓 (𝑡)|2 ≤

2
𝑑𝑢 2
𝑑𝑣 2
[‖
‖
‖
‖
+ ‖𝑢‖3𝐿3(𝛺) ‖𝑢‖𝐿3 (𝛺) ‖𝑣‖2𝐿3 (𝛺) ]𝑟 2 𝑑𝑥 𝐿2 (𝛺) 𝑑𝑥 𝐿2 (𝛺)We assume that27

T-E. Oussaeif , A. Benguesmia and I.Rezzoug‖𝑢‖𝐿∞ (0,𝑇,𝐿3 (𝛺)) = 𝛶(𝑡) ≥ 0,then we have|𝐴𝑓(𝑡)|2 ≤

2
𝑑𝑢 2
𝑑𝑣 2
[‖ ‖
‖ ‖
+ 𝛶(𝑡)‖𝑢‖3𝐿3 (𝛺) ‖𝑣 ‖2𝐿3 (𝛺) ]2
2
2𝑟𝑑𝑥 𝐿 (𝛺) 𝑑𝑥 𝐿 (𝛺)Integreating over (0,T) we obtain
𝑇

𝑑𝑣 2

2

𝑇

∫0 |𝐴𝑓(𝑡)|2 𝑑𝑡 ≤ 𝑟 2 𝑚𝑎𝑥 (‖𝑑𝑥‖ 2

𝐿 (𝛺)

𝑑𝑢 2

, 𝛶(𝑡)‖𝑣 ‖2𝐿3(𝛺)) [∫0 ‖𝑑𝑥 ‖ 2

𝐿 (𝛺)

𝑑𝑡 +

(33)

𝑇

∫0 ‖𝑢‖3𝐿3 (𝛺) ) 𝑑𝑡]
So, we get
𝑇

𝑑𝑢 2∥ 𝐴𝑓 ∥𝐿2 (0,𝑇) ≤ 𝐾 [∫0 ‖𝑑𝑥 ‖ 2

𝐿 (𝛺)

𝑇

1

𝑑𝑡 + ∫0 ‖𝑢‖3𝐿3 (𝛺) ) 𝑑𝑡] 22
𝑑𝑣 2
𝐾 = √ 2 𝑚𝑎𝑥 (‖ ‖, 𝛶(𝑡)‖𝑣 ‖2𝐿3 (𝛺) )𝑟
𝑑𝑥 𝐿2(𝛺)Using the A priori estimate and removing some terms, we gets𝑑𝑢∥ 𝑑𝑥 ∥2𝐿2 (𝑄) +∥ 𝑢 ∥3𝐿3 (𝑄) ≤ 𝐶 ∥ 𝑓 ∥2𝐿2 (𝑄).
So, we get∥ 𝐴𝑓 ∥𝐿2 (0,𝑇) ≤ 𝛿 ∥ 𝑓 ∥𝐿2 (0,𝑇) .

(34)

With
𝛿=𝐾 √𝐶.
It is clear from the preceding that there is a positive 𝛿, such that𝛿 ≤ 1.
The inequality (34) demonstrate that operator A is a contracting mapping on 𝐿2 (0, T).
Theorem 12
Assume that the assumption (H) and the compatibility condition be satisfied, then the
inverse problem (1)-(4) has a unique solution {u,f}.
Proof 7This implies that the equation (28) has a unique solution f in 𝐿2 (0, T). According to Lemma
3 the existence of a solution to the inverse problem (1)-(4) is established. This solution's
uniqueness has yet to be established. Assume, on the other hand, that there are two destinct
solutions to the inverse problem under examination {𝑢1 , 𝑓1 }and {𝑢2 , 𝑓2 }.

28

T-E. Oussaeif , A. Benguesmia and I.Rezzoug

Also, if the linear operator A contracts on 𝐿2 (0, T) from Lemma 11, resulting in 𝑓1 = 𝑓2 ,then the theorem of the uniqueness of the solution of the main direct problem (1)-(3) yields
𝑧1 = 𝑧2 .
Corollary 13
Under the assumptions of Theorem 2, the solution f to equation (28) is continuously
dependent on the data W.
Proof 8Let us consider 𝜔 𝑎𝑛𝑑 𝜐. Two sets of data that satisfy the assumptions of Theorem 2. Let f
and g be the solutions to equation (28) for the data 𝜔 𝑎𝑛𝑑 𝜐, respectively. According to (28)f=Af+ 𝜐g=Ag+ 𝜔First, we'll calculate the difference, f-g. With the help of (34), it's clear that∥ 𝑓 − 𝑔 ∥𝐿2 (0,𝑇) =∥ (𝐴𝑓 + 𝜐) − (𝐴𝑔 + 𝜔) ∥𝐿2(0,𝑇)≤ 𝛿 ∥ 𝑓 − 𝑔 ∥𝐿2 (0,𝑇) +∥ 𝜐 − 𝜔 ∥𝐿2 (0,𝑇).
As a result, we obtain∥ 𝑓 − 𝑔 ∥𝐿2(0,𝑇) ≤51∥ 𝜐 − 𝜔 ∥𝐿2 (0,𝑇) .
1−𝛿

CONCLUSIONS

The inverse problem of determining the right hand side of nonlinear fractional parabolic
equation with integral overdetermination condition has been considered. Where the initialcondition is homogeneous. This inverse problem has been investigated from theorical point ofview. The conditions for the problem’s existence, uniqueness and continuous dependence ondata have been established in this study. Thus, many interesting perspectives of numericalanalysis can allow for the contenuation of the work done in this paper, especially in the aspectof developing effective numerical methods that are compatible with the non local conditions of
the integrative type and also to think about solving the same problem but by establishingincompatible initial conditions.

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Received, June 17, 202231