Communications on Applied Nonlinear Analysis  

ISSN: 1074-133X 

Vol 32 No.3 (2025) 
 

939 
 

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Method of a Priori Estimate in the Case of Fractional Order   

Differential Equations with Integral Conditions 

 

1Djebaili Manel 

1University Abbes-Laghrour, Khenchela, Knowledge engineering and information security (ICOSI), Faculty 

of sciences and technology (Algeria). 

manel.djebaili@univ-khenchela.dz  

 

 

Article History: 

Received:12-01-2025 

Revised:15-02-2025 

Accepted:01-03-2025 

Abstract: 

This paper concentrate on exploring the existence and uniqueness of a solution 

for a non-linear boundary value problem that has integral conditions in the case 

of fractional partial differential equations. For this we split the proof into two 

sections: linear and non-linear problem; for the associated linear problem, we 

derive the a priori bound and demonstrate the density of the operator generated 

by the problem posed; we solve the non-linear problem by introducing a 

iterative process. The results show the efficiency of energy inequality method 

in the case of time fractional order differential equations with integral 

conditions our results illustrate the existence and uniqueness of the continuous 

dependence of solution on fractional order.  

Keywords: fractional partial differential equations; integral conditions; priori 

estimate; density of the operator; non-linear problem. 

 

1.Introduction 

In mathematics, fractional calculus is a field of analysis that investigates the extension of 

differentiation and integration from integers to non-integers, commonly referred to as 

fractional orders. Fractional differentiation, in particular, has been a subject of interest for 

almost as long as the classical calculus that we know today. 

In addition, many problems in physics and modern technology are formulated using non-local 

conditions for partial differential equations, which are described by integral conditions. These 

conditions have gained significant attention due to their applications in a variety of fields, 

such as population dynamics, blood flow models, chemical engineering, and cellular systems 

(A. Bouziani, 2002; A.Bouziani,2003; A.Bouziani, N.Merazga, A.Bouziani,2003; N.Merazga, 

A. Bouziani,2005). Numerous authors have studied the existence and uniqueness of solutions 

to problems involving fractional differential equations, including initial and boundary value 

problems (A. Anguraj, P.Karthikeyan, 2010; B. Ahmad, J. Nieto,2009; M. Benchohra, J. R. 

Graef, S. Hamani,2008; M. Belmekki, M. Benchohra,2008; R. P. Agarwal, M. Benchohra, S. 

Hamani,2005; R. W. Ibrahim, S. Momani,2007; X. J. Li, C. J. Xu,2010). 

mailto:manel.djebaili@univ-khenchela.dz


Communications on Applied Nonlinear Analysis  

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For this purpose, we employed the energy inequality method, which is a useful tool for 

studying fractional and non-local classical problems. Compared to other techniques, this 

method plays an essential role in proving the existence and uniqueness of the solution.  

It depends on density arguments and certain a priori bounds. 

2. Preliminary 

Definition 2.1 (I.Podlubny, 1999)(Gamma function) 

For any complex number 𝐳 such that Re(𝐳) > 0, we define the following function called 

Gamma and denoted by the Greek letter  " 𝚪". 

𝚪 ∶  𝐑∗+ → 𝐑 

                                                     𝐳 →   𝚪(𝐳) =  ∫ 𝐭𝐳−𝟏
+∞

𝟎
𝐞𝐱𝐩−𝐭𝐝𝐭.                                       (1) 

Definition 2.2 (I.Podlubny, 1999) (The Riemann-Liouville integral) 

The Riemann-Liouville integral of order α > 0,  for an integral function S, is defined by                                        

                                                    Dt
−αS(t) =

1

Γ(α)
∫

S(τ)

(t−τ)1−α

t

a
dτ.                                              (2) 

Definition 2.3 (Haim Brezis, 1983) 

Let R be a subspace vector of the Hilbert space H, then 𝐑⏊ the orthogonal complement of R 

is defined as: 

                                                  𝐑⏊ = {𝐟 ∈ 𝐇, (𝐟, 𝐠)𝐇 = 𝟎, ∀𝐠 ∈ 𝐑}.                           

Proposition 2.1 

Let R be a subspace vector of the Hilbert space H. R is dense in H if and only if: 

                                                                𝐑⏊ = {𝟎}. 
 

Definition 2.4 (Bertram Ross, 2006) (Left Caputo derivative) 

                                             ∂0
c
t
αu(x, t) =

1

Γ(1−α)
∫
∂u(x,τ)

∂τ

t

0

1

(t−τ)α
dτ.                                        (3) 

Definition 2.5 (Stefan G Samko, Anatoly A Kilbas, Oleg I Marichev, 1993) (Mittag-

Leffler function) 

For 𝐳 ∈ 𝕔, Mittag-Leffler function 𝐄𝛂(𝐳) is defined as follows: 



Communications on Applied Nonlinear Analysis  

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                                                 Eα(z)= ∑
zk

Γ(αk+1)
∞
k=0  , α > 0.                                                        (4) 

The two-term Mittag-Leffler function is vital in the fractional calculus theory and defined as: 

                                                 Eα,β(z)= ∑
zk

Γ(αk+β)
∞
k=0  ,    (α > 0, 𝛽 > 0).                               (5) 

Cauchy-Schwarz inequality: 

                                                    ∀(f, g) ∈ 𝕃2(Ω) × 𝕃2(Ω),  

we  have: 

                                     ∫ |f(t)g(t)|Ω
dt ≤ (∫ (f(t))2Ω

dt)
1

2(∫ (g(t))2Ω
dt)

1

2,                                (6) 

Cauchy inequality: 

∀(a, b) ∈ ℝ2:     

                                                       |ab| ≤
1

2
a2 +

1

2
b2.                                                             (7) 

Cauchy inequality with 𝛜: 

Let ϵ be a strictly positive number, then∀(a, b) ∈ ℝ2: 

                                                          |𝑎𝑏| ≤
𝜀a2

2
+
b2

2𝜀
.                                                              (8) 

Poincaré inequality 

Lemma 2.1(A.A. Alikhanov, 2010) For any function S(t) that is absolutely continuous on the 

interval [0,T], the following inequality holds:  

                                            S(t) ∂t
β
S(t) ≥

1

2
∂t
β
S2(t),         0 < 𝛼 < 1.                                     (9) 

Lemma 2.2(A.A. Alikhanov, 2010) (Gnonwall Lemma) Let a non-negative, absolutely 

continuous function y(t) satisfy the inequality: 

                                         ∂t
αy(t) ≤ k1u(t) + k2(t),        0 < 𝛼 < 1.                                     (10) 

For all t ∈ [0, T], where k1 is a positive constant and k2(t) a non-negative integrale function 

over [0,T]. Then, 

                                 y(t) ≤ y(0)Eα(k1t
α) + Γ(α)Eα,α(k1t

α)Dt
−αk2(t),                              (11) 



Communications on Applied Nonlinear Analysis  

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where∶   Eα et  Eα,αare Mittag-Leffler functions. 

Lemma 2.3 (Ladyzhenskaya, 1985) Let u(t) be a non-negative, absolutely continuous 

function on [0,T], and for all t ∈ [0, T], satisfies the inequality:  

                                                       
dφ

dt
≤ C(t)φ(t) + B(t).                                                     (12) 

Such that the functions C(t) and B(t) are summable and non-negative on [0,T]. Here: 

                                                       𝜑(𝑡)

≤ 𝑒∫ 𝐶(𝜏)𝑑𝜏
𝑡

0 𝜑(0) + ∫ 𝐵(𝜉)𝑒∫ 𝐶(𝜏)𝑑𝜏
𝜉

0

𝑡

0

𝑑𝜉                       (13)  

Lemma 2.4 (Mesloub S, Mezhoudi R, Medjeden M, 2002)  For any n ∈ ℕ, we have 

                                              ‖ℑx
2nu‖

2

𝕃2(0,1)
≤ (

1

2
)
2n
‖u‖2𝕃2(0,1).                                      (14) 

Where: 

ℑx
2nu = ∫ ∫ …

ξ1

0

∫ u(η, t)
ξ2n−1

0

x

0

dηdξ2n−1…dξ1 = ∫
(x − ξ)2n−1

(2n − 1)!

x

0

u(ξ, t)dξ 

3. Problem Statement  

In a rectangular domain: Ω = (0,1) × (0, T),    0 ≤ T ≤ ∞ , consider the following fractional 

partial differential equation: 

ℒv = ∂0
c
t
δv(x, t) − α

∂2v

∂x2
− β

∂3v

∂t ∂x2
+ γv − ∫ a(t − s)v(x, s)ds = g(x, t, v, r),

t

0
   

                                                                 0 < 𝑡 < 𝑇,     0 < 𝛿 < 1                           (15)                   

 

where:                r(x, t) = ∫ g(x, s, v(x, s), r(x, s))ds
t

0
. 

Where a(t) is a function of t and satisfies the condition 0 < a0 < 𝑎(𝑡) < a1 and α, β and γ 

are strictly positive constants. 

With Initial conditions, 

ℓv = v(x, 0) = ɸ(x),        qv =
∂v(x,0)

∂t
= Ψ(x),      0 < 𝑥 < 1,                                   (16) 

and integral conditions, 

      ∫ v(x, t)dx = m(t),       
1

0 ∫ xv(x, t)dx = n(t),       
1

0
 0 < 𝑡 ≤ 𝑇,                           (17) 



Communications on Applied Nonlinear Analysis  

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where Φ,Ψ,m, n and g are known functions. 

Since the boundary conditions are non-homogeneous, we construct the function 

                                    U(x, t) = 6(2n(t) − m(t))x − 2(3n(t) − 2m(t)).  

And we introduce a new function: u(x, t) = v(x, t) − U(x, t).  Then, the problem (15)-(17) can 

be reformulated as follows: 

             ℒu = ∂0
c
t
δu(x, t) − α

∂2u

∂x2
− β

∂3u

∂t∂x2
+ γu − ∫ a(t − s)u(x, s)ds = f(x, t, u, r)

t

0
,            (18) 

where:                f(x, t, u(x, t), r(x, t)) = g(x, t, v, r) − ℒv + ∫ a(t − s)U(x, s)ds
t

0
.   

 The initial conditions 

ℓu = u(x, 0) = ɸ(x) −ℓU = φ(x),     qu =
∂u(x,0)

∂t
= Ψ(x) − qU = ψ(x), 0 < 𝑥 < 1.         (19) 

The integral conditions:      

               ∫ u(x, t)dx = 0,       
t

0 ∫ xu(x, t)dx = 0,       
t

0
 0 < 𝑡 ≤ 𝑇 .                                        (20) 

 

4.Technical tools and associated linear problem 

We define some function spaces and tools required to investigate the following linear problem 

associated with problems (18)-(20) 

                   ℒu = ∂0
c
t
δu(x, t) − α

∂2u

∂x2
− β

∂3u

∂t∂x2
+ γu − ∫ a(t − s)u(x, s)ds = f(x, t)

t

0
,               (21) 

The initial conditions 

ℓu = u(x, 0) = ɸ(x) −ℓU = φ(x),     qu =
∂u(x,0)

∂t
= Ψ(x) − qU = ψ(x),        0 < 𝑥 < 1.  (22)      

The integrals conditions:      

                      ∫ u(x, t)dx = 0,       
t

0 ∫ xu(x, t)dx = 0,       
t

0
 0 < 𝑡 ≤ 𝑇 .                                 (23) 

We will show the existence and uniqueness of the solution of problem (21) – (23) , the proof 

will be based on a priori estimates and on the density of the set of values of the operator 

generated by problem (21) – (23). 

For this, we must first convert problem (21) – (23) into an equivalent operational form: 

                                              Lu = ℱ = (f, φ,ψ).                                                                           (24) 



Communications on Applied Nonlinear Analysis  

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Where the operator L = (ℒ, ℓ, q) with   L: E → F  is defined in D(L) suchaway that:                                             

                  D(L) = {
u ∈ 𝕃2(D), ∂0

c
t
δu,

∂u

∂t
,
∂u

∂x
,
∂2u

∂x2
 ,   

∂3u

∂t ∂x2
∈ 𝕃2(D)

∫ u(x, t)dx = 0,       
t

0 ∫ xu(x, t)dx = 0,       
t

0
 0 < 𝑡 ≤ 𝑇

}                        (25) 

and u satisfies the initial condition (4.2). E is the Banach space equipped with the following 

norm: 

        ‖u‖2E = sup (Dt
δ−1‖ℑx

∂u

∂τ
‖
2

𝕃2(Ω)
+ ∫ ‖

∂u

∂τ
‖
2

𝕃2(Ω)
dτ + ∫ u2

t

0

t

0
dx),                  (26) 

and F is the Hilbert space composed of functions with the norm: 

                ‖Lu‖2F = ‖φ‖
2
𝕃2(Ω) + ‖ℑxψ‖

2
𝕃2(Ω)

+ ‖ℑxf‖
2
𝕃2(Ω).                                    (27) 

 

 

5. Prior Estimation and Uniqueness of the Solution: 

The priori estimation method, also known as the energy integral method, is one of the most 

effective functional analysis methods for solving partial differential equations with integral 

conditions, and is an important technique for proving the existence, uniqueness, and 

continuous dependence of solutions to PDE. 

Theorem 5.1 For any function  u ∈ D(L), we have the a priori estimation 

                                             ‖u‖E ≤ C ‖Lu‖F ,                                                                 (28) 

where C is a constant that is independent of u. 

Proof: We multiply (21) by  Mu = −ℑx
2 ∂u

∂t
= −∫ ∫

∂u

∂t
(ξ, t)dξdɳ

ɳ

0

t

0
 , and integrate over the 

subdomain to obtain  Ω = (0,1) × (0, τ), we obtain:                                                    

(ℒu, Mu)𝕃2(Ω) = −( ∂0
c
t
δu, ℑx

2 ∂u

∂t
)
𝕃2(Ω)

+ α(
∂2u

∂x2
, ℑx

2 ∂u

∂t
)
𝕃2(Ω)

                 

+β(
∂3u

∂t ∂x2
, ℑx

2 ∂u

∂t
)
𝕃2(Ω)

− γ(u, ℑx
2 ∂u

∂t
)
𝕃2(Ω)

 

                                             = − (∫ a(t − s)u(x, s)ds
t

0
, ℑx

2 ∂u

∂t
)
𝕃2(Ω)

− (f(x, t), ℑx
2 ∂u

∂t
)
𝕃2(Ω)

 (29) 

By integrating by parts for each term on the left-hand side of (22), and using the conditions 

(20), we obtain 



Communications on Applied Nonlinear Analysis  

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−( ∂0
c
t
δu, ℑx

2 ∂u

∂t
)
𝕃2(Ω)

 =  −∫ ∫ ( ∂0
c
t
δu ℑx

2 ∂u

∂t
)

1

0

τ

0

dxdt 

                                                                     =  ∫ ∫ ( ∂0
c
t
δℑx

∂u

∂t
) (ℑx

∂u

∂t
)

1

0

τ

0
dxdt.                     (30) 

 

α (
∂2u

∂x2
, ℑx

2 ∂u

∂t
)
𝕃2(Ω)

= α∫ ∫ (
∂2u

∂x2
ℑx
2 ∂u

∂t
)

1

0

τ

0

dxdt 

                                                                     =  α∫ ∫ u
1

0

τ

0

∂u

∂t
dxdt .                                          (31)            

 

β(
∂3u

∂t ∂x2
, ℑx

2 ∂u

∂t
)
𝕃2(Ω)

= β∫ ∫ (
∂3u

∂t ∂x2
ℑx
2 ∂u

∂t
)

1

0

τ

0

dxdt     

                                                                    = β∫ ∫ (
∂u

∂t
)
21

0

τ

0
dxdt 

                                                                    = β∫ ‖
∂u

∂t
‖
2

𝕃2(0,1)
dt

τ

0
 .                                       (32) 

 

−γ(u, ℑx
2 ∂u

∂t
)
𝕃2(Ω)

= −γ∫ ∫ (u ℑx
2 ∂u

∂t
)

1

0

τ

0

dxdt      

                                                                  =− 
γ

2
∫ (ℑx

2u(x, τ))
21

0
dx +

γ

2
∫ ( φ(x))2
1

0
dx.       (33) 

Applying the Cauchy inequalities (7) and (8), and integrating by parts for the two terms on the 

right-hand side of (29), we obtain: 

                     −(f, ℑx
2 ∂u

∂t
)
𝕃2(Ω)

≤
ε

2
∫ ∫ (ℑxf)

21

0

τ

0
dxdt +

1

2ε
∫ ∫ (ℑx

∂u

∂t
)
21

0

τ

0
dxdt                     (34) 

 

(∫ a(t − s)u(x, s)ds
t

0

, ℑx
2 ∂u

∂t
)
𝕃2(Ω)

= ∫ ∫ (∫ a(t − s)u(x, s)ds
t

0

)
1

0

τ

0

ℑx
2 ∂u

∂t
dxdt 

                                                                               ≤ a1T
2‖u‖2𝕃2(0,1) + ∫ ‖ℑx

∂u

∂t
‖
2

𝕃2(0,1)
dt,

τ

0
     (35) 

By substituting (30)-(35) into (29), and applying lemma (2.1), we obtain: 



Communications on Applied Nonlinear Analysis  

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1

2
∫ ∫ ( ∂0

c
t
δℑx

∂u

∂t
)
21

0

τ

0

dxdt + α∫ ∫ u
1

0

τ

0

∂u

∂t
dxdt + β∫ ‖

∂u

∂t
‖
2

𝕃2(0,1)
dt

τ

0

 

                            − 
γ

2
∫ (ℑx

2u(x, τ))
2

1

0

dx +
γ

2
∫ ( φ(x))2
1

0

dx 

                           ≤
1

2ε
∫ ∫ (ℑxf)

2
1

0

τ

0

dxdt +
ε

2
∫ ∫ (ℑx

∂u

∂t
)
21

0

τ

0

dxdt 

                                                      +T2‖u‖2𝕃2(0,1) + ∫ ‖ℑx
∂u

∂t
‖
2

𝕃2(0,1)
dt,

τ

0
                                    (36)    

Evaluating the first and third terms on the left-hand side, we have: 

              α∫ ∫ u
1

0

τ

0

∂u

∂t
dxdt = α∫ ∫ u

τ

0

1

0

∂u

∂t
dtdx                                                                   

                                                   = 
α

2
∫ u2(x, τ)dx −
1

0

α

2
∫ φ2(x)
1

0
dx −

α

2
∫ ∫ u2

τ

O

1

0
(x, t)dtdx   (37) 

 

                   
1

2
∫ ∫ ( ∂0

c
t
δℑx

∂u

∂t
)
21

0

τ

0
dxdt = Dt

δ−1 ‖ℑx
∂u

∂t
‖
2

𝕃2(0,1)
−

t1−δ

Γ(1−δ)
‖ℑxψ‖

2
𝕃2(0,1)

 .       (38) 

Substituting (37) and (38) as well as the conditions (22) into inequality (36), we obtain: 

Dt
δ−1 ‖ℑx

∂u

∂t
‖
2

𝕃2(0,1)
+∫ ‖

∂u

∂t
‖
2

𝕃2(0,1)
dt

τ

0

+∫ u2dx           
1

0

                 

                                      

≤ ɳ 1

(

 
 
∫ φ2(x)
1

0

dx + ∫ ∫ u2
τ

O

1

0

dtdx + ∫ ∫ (ℑxf)
2

1

0

τ

0

dxdt + ‖ℑxψ‖
2
𝕃2(0,1)

+∫ ∫ (ℑx
∂u

∂t
)
21

0

τ

0

dxdt
)

 
 
,                 (39) 

where: 

                                    ɳ1 =
max(

α+γ

2
,a1T

2+ 
α 

2
,   
1

2ε
 ,   
ε

2
 ,   

T1−δ

Γ(1−δ)
)

min(
1

2
,β ,

α+γ

2
)

 .                                                  (40) 

 

For the second term on the right-hand side of (39), we apply the lemma (2.3) by letting 

                          φ(t) = ∫ ∫ u2dxdt     ;  
1

0

t

0

∂φ

∂t
= ∫ u2

1

0
dx   ;   φ(0) = 0,                                    (41) 



Communications on Applied Nonlinear Analysis  

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this leads to 

            ∫ ∫ u2
τ

O

1

0
dtdx ≤ Tɳ1e

ɳ1T (∫ φ2(x)
1

0
dx + ∫ ∫ (ℑxf)

21

0

τ

0
dxdt + ‖ℑxψ‖

2
𝕃2(0,1)

+

                                                      ∫ ∫ (ℑx
∂u

∂t
)
21

0

τ

0
dxdt) .                                                              (42) 

Thus, the inequality (39) becomes 

Dt
δ−1 ‖ℑx

∂u

∂t
‖
2

𝕃2(0,1)
+∫ ‖

∂u

∂t
‖
2

𝕃2(0,1)
dt

τ

0

+∫ u2dx
1

0

 

                                ≤ ɳ2

(

 
 
∫ φ2(x)
1

0

dx + ∫ ∫ (ℑxf)
2

1

0

τ

0

dxdt + ‖ℑxψ‖
2
𝕃2(0,1)

+∫ ∫ (ℑx
∂u

∂t
)
21

0

τ

0

dxdt
)

 
 
 ,                   (43) 

such that: 

                                         ɳ2 = max(ɳ1 , ɳ1
2T eɳ1T).                                                                    (44) 

Finally, we apply lemma (2.2) to the last term on the right-hand side of (43) by letting: 

                           y(t) = ∫ ∫ (ℑx
∂u

∂t
)
21

0

τ

0
dxdt    ;  ∂t

δy(t) = Dδ−1 ‖ℑx
∂u

∂t
‖
2

𝕃2(0,1)
.                    (45) 

Thus, we obtain:                   

∫ ∫ (ℑx
2 ∂u

∂t
)
21

0

τ

0

dxdt ≤   ɳ2Γ(δ)Eδ,δ(ɳ2t
δ) (

T

δ Γ(δ)
‖φ‖2𝕃2(0,1) + 

T

δ Γ(δ)
‖ℑxψ‖

2
𝕃2(0,1)

+Dt
−δ−1‖ℑxf‖

2
𝕃2(Ω)

) 

                      ≤ ɳ2Γ(δ)Eδ,δ(ɳ2t
δ)max (1,

T

δ Γ(δ)
) (Dt

−δ−1‖ℑxf‖
2
𝕃2(Ω) 

                                             +‖φ‖2𝕃2(0,1)  + ‖ℑxψ‖
2
𝕃2(0,1)

).                                                    (46)                                    

Substituting equation (46) into equation (43), we obtain: 

Dt
δ−1 ‖ℑx

∂u

∂t
‖
2

𝕃2(0,1)
+∫ ‖

∂u

∂t
‖
2

𝕃2(0,1)
dt

τ

0

+∫ u2dx
1

0

 

 

               ≤ ɳ3 (∫ φ2(x)
1

0
dx + ∫ ∫ (ℑxf)

21

0

τ

0
dxdt + ‖ℑxψ‖

2
𝕃2(0,1)

+ Dt
−δ−1‖ℑxf‖

2
𝕃2(Ω)), (47) 



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Where: 

                                  ɳ3 = ɳ2max(ɳ2, ɳ2Γ(δ), Eδ,δ(ɳ2t
δ)max (1,

T

δ Γ(δ)
)).                         (48) 

On the other hand: 

                                   Dt
−δ−1‖ℑxf‖

2
𝕃2(Ω) ≤

Tδ

 Γ(1+δ)
∫ ‖ℑxf‖

2
𝕃2(0,1)dt

T

0
  .                             (49) 

Hence, inequality (47) becomes 

Dt
δ−1 ‖ℑx

∂u

∂t
‖
2

𝕃2(0,1)
+∫ ‖

∂u

∂t
‖
2

𝕃2(0,1)
dt

τ

0

+∫ u2dx
1

0

 

                                      ≤ C(∫ φ2(x)
1

0
dx + ∫ ∫ (ℑxf)

21

0

τ

0
dxdt + ‖ℑxψ‖

2
𝕃2(0,1)

),               (50) 

such that: 

                                            C = ɳ3 (1 +
Tδ

 Γ(1+δ)
).                                                                           (51) 

We observe that the right-hand side of inequality (50) is independent of τ, by taking the 

supremum of the left-hand side with respect to  τ ∈ [0, T]. We obtain the desired inequality, 

which concludes the proof. 

Proposition 5.1: The operator L which is defined from E to F has a closure. 

Theorem 5.1 holds for strong solutions, and we have the inequality: 

                                                      ‖u‖E ≤ Ć‖L̅u‖F ,                                                              (52) 

thus, we obtain: 

Corollary 5.1: The strong solution of (21)-(23) is unique if it exists, and depends 

continuously on ℱ ∈ F. 

Corollary 5.2: The set of values R(L̅) of the operator L̅is closed in F. ∎ 

 

 

6. Existence of the Solution 



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To prove the existence of the solution, we must demonstrate that: R(L) is dense in F, for 

all:  u ∈ E,  and  ℱ = (f, φ, ψ) ∈ F. 

Theorem 5.1 

For  z ∈ L2(Ω) and all u ∈ E, we have:     

                                       ∫ Lu. z dxdt = 0Ω
,                                                              (53) 

then: z disappears almost everywhere in Ω, this implies that the problem (21)-(23) has a 

unique solution. 

Proof: 

The proof of this theorem consists of choosing  z ∈ R(L)⏊, we demonstrate that: 

R(L)⏊ = {0} ⟺ R(L)̅̅ ̅̅ ̅̅ = F. 

The scalar product in  F is defined by: 

                                      (Lu, z)F = ∫ Lu. z dxdtΩ
 .                                                    (54) 

Then (53) can be written as: 

                           ∫ ( ∂0
c
t
δu(x, t) − α

∂2u

∂x2
− β

∂3u

∂t ∂x2
+ γu, z) dxdt = 0.

Ω
                         (55) 

 If  letting: 

                                u(x, t) = ℑt
2μ = ∫ ∫ μ(x, ξ)dξds

s

0

t

0
,                                          (56) 

where ∂0
c
t
δμ ,

∂2μ

∂x2
 ,
∂3μ

∂t∂x2
, μ ∈ L2(Ω) and it also satisfies the initial boundary conditions (19) 

and (20), we can write equation (55) as: 

                          ∫ ( ∂0
c
t
δℑt

2μ − α
∂2ℑt

2μ

∂x2
− β

∂3ℑt
2μ

∂t ∂x2
+ γℑt

2μ , z) dxdt = 0.
Ω

              (57) 

We can express z as a function of μ as follows: 

                                    z(x, t) = ℑtμ − ℑx
2ℑtμ.                                                               (58) 

 

 

We can then substitute (58) into (57) and do integrations by parts on each term 



Communications on Applied Nonlinear Analysis  

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∫( ∂0
c
t
δℑt

2μ . ℑtμ)
Ω

dxdt = ∫ ( ∂0
c
t
δℑtμ . ℑtμ)

Ω

dxdt                       

                                                             ≥
1

2
∫ ∂0

c
t
δ‖ℑtμ‖𝕃2(0,1)

τ

0
dt,                              (59)                      

 

 

 

−∫( ∂0
c
t
δℑt

2μ . ℑx
2ℑtμ)

Ω

dxdt = ∫ ∫ ∂0
c
t
δℑx(ℑtμ ). ℑx(ℑtμ )

1

0

τ

0

dxdt 

                                                                      ≥
1

2
∫ ∂0

c
t
δ‖ℑx(ℑtμ)‖𝕃2(0,1)

τ

0
,                             (60) 

     

                      − ∫ (α
∂2(ℑt

2μ)

∂x2
. ℑtμ)dxdtΩ

= −α∫ ∫ (
∂

∂x
(ℑt

2μ))
2

dx
1

0
dt

τ

0
,                                  (61) 

 

                 ∫ (α
∂2(ℑt

2μ)

∂x2
. ℑx

2ℑtμ)dxdt = α ∫ ∫ (ℑt
2μ)

21

0

τ

0Ω
dxdt,                                                (62) 

 

                   −∫ (β
∂3ℑt

2μ

∂t ∂x2
. ℑtμ)dxdt =Ω

β∫ ∫ (
∂

∂X
(ℑtμ))

2
1

0

τ

0
dxdt,                                       (63) 

 

  −∫ (−β
∂3ℑt

2μ

∂t ∂x2
. ℑx

2ℑtμ)dxdt =Ω
β∫ ∫ (ℑtμ)

21

0

τ

0
dxdt,                                          (64) 

 

∫ (γℑt
2μ. ℑtμ)Ω

dxdt =
γ

2
∫ ∫ (ℑt

2μ(x, t))
21

0

τ

0
dxdt −

γ

2
∫ ∫ φ2

1

0

τ

0
dxdt,                         (65) 

 

 

      −∫ (γℑt
2μ. ℑx

2ℑtμ)Ω
dxdt = −

γ

2
∫ ∫ (ℑx

2(ℑt
2μ(x, 0)))

21

0

τ

0
dxdt +                                      

 

γ

2
∫ ∫ (ℑx

2(ℑt
2μ(x, 0)))

2
1

0

τ

0

dxdt 

 

                                   = −
γ

2
∫ ∫ (ℑt

2μ(x, t))
21

0

τ

0
dxdt +

γ

2
∫ ∫ φ2

1

0

τ

0
dxdt.                   (66) 

 

 

By substituting (59)-(66) into (57), we obtain: 



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Dt
δ−1‖ℑtμ‖

2
𝕃2(0,1)

+ Dt
δ−1‖ℑxℑtμ‖

2
𝕃2(0,1)

+ 

∫ ∫ (
∂

∂X
(ℑtμ))

2
1

0

τ

0

dxdt + ∫ ∫ (ℑtμ)
2

1

0

τ

0

dxdt 

                                        ≤ ρ1 (∫ ∫ (
∂

∂x
(ℑt

2μ))
2

dx
1

0
dt

τ

0
+ ∫ ∫ (ℑt

2μ)
21

0

τ

0
dxdt),              (67)  

where: 

                                                          ρ1 =
1

(1,2β,α)
.                                                                           (68) 

We apply lemma (2.3) and we obtain: 

         φ1(t) = ∫ ∫ (
∂

∂x
(ℑt

2μ))
2

dx
1

0
dt

τ

0
   ;  

∂φ1(t)

∂t
= ∫ (

∂

∂x
(ℑt

2μ))
2

dx
1

0
  ;   φ1(0) = 0,      (69) 

so 

                                         φ1(t) ≤ ρ1Te
Tρ1 ∫ ∫ ((ℑt

2μ))
2

dx
1

0
dt

τ

0
,                                            (70) 

then, equation (67) can be transformed as follows: 

Dt
δ−1‖ℑtμ‖

2
𝕃2(0,1)

+ Dt
δ−1‖ℑxℑtμ‖

2
𝕃2(0,1)

+ 

∫ ∫ (
∂

∂x
(ℑtμ))

2
1

0

τ

0

dxdt + ∫ ∫ (ℑtμ)
2

1

0

τ

0

dxdt 

                                             ≤ ρ2 (∫ ∫ ((ℑt
2μ))

2

dx
1

0
dt

τ

0
),                                                 (71) 

such that: 

                                             ρ2 = max(ρ1
2TeTρ1 , ρ1).                                                                   (72) 

Applying lemma (2.3), we arrive at equation  

                   φ2(t) = ∫ ∫ (ℑt
2μ)

2
dx

1

0
dt

τ

0
   ;  

∂φ2(t)

∂t
= ∫ (ℑt

2μ)
2
dx

1

0
  ;   φ2(0) = 0,               (73) 

so: 

                                           φ2(t) = ∫ ∫ (ℑt
2μ)

2
dx

1

0
dt

τ

0
≤ 0.                                                    (74) 

Hence, (71) is transformed into 

Dt
δ−1‖ℑtμ‖

2
𝕃2(0,1)

+ Dt
δ−1‖ℑxℑtμ‖

2
𝕃2(0,1)

+ 



Communications on Applied Nonlinear Analysis  

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                                   ∫ ∫ (
∂

∂x
(ℑtμ))

2
1

0

τ

0
dxdt + ∫ ∫ (ℑtμ)

21

0

τ

0
dxdt ≤ 0.                                   (75) 

Setting:  μ=0 in (58) we conclude that: z = 0 inL2(Ω). Since  z ∈ R(L)⏊ , so  R(L)⏊ = {0}.  

This proves that R(L)̅̅ ̅̅ ̅̅ =F. 

This completes the proof. ∎ 

7. The study of the nonlinear problem 

This section is devoted to solving the main problems (15)–(17). Consider now the auxiliary 

problem with the homogenous equation: 

                    ℒU = ∂0
c
t
δU(x, t) − α

∂2U

∂x2
− β

∂3U

∂t∂x2
+ γU − ∫ a(t − s)U(x, s)ds = 0

t

0
,               (76) 

                    ℓU = U(x, 0) = φ(x),     qU =
∂U(x,0)

∂t
= ψ(x),                0 < 𝑥 < 1.                    (77) 

                   ∫ U(x, t)dx = 0,       
t

0 ∫ xU(x, t)dx = 0,       
t

0
             0 < 𝑡 ≤ 𝑇 .                         (78) 

If V and u are solutions of problems (18)-(20),(21)-(23), respectively, then h = u − V satisfies 

            ℒw = ∂0
c
t
δw(x, t) − α

∂2w

∂x2
− β

∂3w

∂t∂x2
+ γw − ∫ a(t − s)w(x, s)ds = χ(x, t, w,

∂w

∂x

t

0
),   (79) 

            ℓw = w(x, 0) = φ(x),     qw =
∂w(x,0)

∂t
= ψ(x),                 0 < 𝑥 < 1.                       (80) 

           ∫ w(x, t)dx = 0,       
t

0 ∫ xw(x, t)dx = 0,       
t

0
             0 < 𝑡 ≤ 𝑇 .                                (81) 

Such that the function  χ (x, t, w,
∂w

∂x
) = χ (x, t, w + U,

∂w

∂x
+
∂U

∂x
) , verifies the following 

condition: 

      |χ(x, t, w1, y1) − χ(x, t, w2, y2)| ≤ M(|w1 − w2| + |y1 − y2|),     ∀ (x, t) ∈ ℚ              (82) 

Now we will show that the solution of problems (79)-(81) is unique. We will establish a 

similar proof for problems (21)-(23). 

First we introduce the following space: 

                                        ′C1(ℚ)= {w ∈ C1(ℚ)  ,
∂w2

∂t ∂x2
 ∈ C(ℚ) }                                          (83) 

We suppose that:   w, u ∈ ′C1(ℚ) verify homogenous initiale and boundary conditions, we 

have: 



Communications on Applied Nonlinear Analysis  

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(ℒw,ℑxu) =  ( ∂0
c
t
δw,ℑxu)𝕃2(Ω) − α (

∂2w

∂x2
, ℑxu)

𝕃2(Ω)
− β(

∂3w

∂t∂x2
, ℑxu)

𝕃2(Ω)
+

γ(w,ℑxu)𝕃2(Ω) − (∫ a(t −
t

0

s)w(x, s)ds , ℑxu)
𝕃2(Ω)

  .                                                                                                    (84) 

Where: 

                                      ( ∂0
c
t
δw,ℑxu)𝕃2(Ω) = −( ∂0

c
t
δℑxw, u)𝕃2(Ω)                                         (85) 

−α(
∂2w

∂x2
, ℑxu)

𝕃2(Ω)

= −α(
∂

∂x
(
∂w

∂x
) , ℑxu)

𝕃2(Ω)
 

                                                                       =  α (
∂w

∂x
, u)

𝕃2(Ω)
                                                   (86) 

 

                                     −β(
∂3w

∂t∂x2
, ℑxu)

𝕃2(Ω)
=−β(

∂

∂t
(
∂w

∂x
) , u)

𝕃2(Ω)
                                          (87) 

 

               −(∫ a(t − s)w(x, s)ds
t

0
, ℑxu)

𝕃2(Ω)
= (∫ a(t − s)ℑxw(x, s)ds

t

0
, u)

𝕃2(Ω)
.              (88) 

We obtain: 

−( ∂0
c
t
δℑxw, u)𝕃2 + α (

∂w

∂x
, u)

𝕃2
+ β(

∂

∂t
(
∂w

∂x
) , u)

𝕃2
+ (∫ a(t − s)ℑxw(x, s)ds

t

0
, u)

𝕃2(Ω)
=

                (u, ℑxχ)𝕃2(Ω)(89) 

Such that :                                 κ(w, u) = (u, ℑxχ)𝕃2(Ω)                                                                   (90) 

 

Definition: A function w ∈ 𝕃2(0, T, H1(Ω)) is considered as the weak solution of the problem 

(79)-(81) if it satisfies (89) and (90) holds. 

We will construct an iteration sequence as follows, let: w(0) = 0, and (w(n))
n
∈ ℕ, if w(n−1) 

is given, then for n ∈ ℕ solve the following problem: 

 

 

 



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     ℒw(n) =      ∂ 0
        c

t
δw(n) − α

∂2w(n)

∂x2
− β

∂3w(n)

∂t∂x2
+ γw(n) −

∫ a(t − s)w(n)(x, s)ds = χ(x, t, w(n−1),
∂w(n−1)

∂x

t

0
),  

                                                                                                                                                (91) 

     ℓw(n) = w(n)(x, 0) = 0,     qw(n) =
∂w(n)(x,0)

∂t
= 0,              0 < 𝑥 < 1.                               (92) 

     ∫ w(n)(x, t)dx = 0,       
t

0 ∫ xw(n)(x, t)dx = 0,       
t

0
             0 < 𝑡 ≤ 𝑇 .                            (93) 

 

Theorem: For each fixed n assure that the solution of problem (91)-(93), w(n)(x, t) is unique. 

We put:    W(n)(x, t) = w(n+1)(x, t) − w(n)(x, t), we obtain:  

 

     ℒW(n) = ∂ 0
c
t
δW(n) − α

∂2W(n)

∂x2
− β

∂3W(n)

∂t∂x2
+ γW(n) − ∫ a(t − s)W(n)(x, s)ds =

t

0
N(n−1)(x, t)      

                                                                                                                                                (94)                                                                                                                                                          

    ℓW(n) = W(n)(x, 0) = 0,     qW(n) =
∂W(n)(x,0)

∂t
= 0,                   0 < 𝑥 < 1.                      (95) 

    ∫ W(n)(x, t)dx = 0,       
t

0 ∫ xW(n)(x, t)dx = 0,       
t

0
             0 < 𝑡 ≤ 𝑇 .                             (96) 

Such that:        N(n−1)(x, t) = χ (x, t, w(n),
∂w(n)

∂x
) − χ (x, t, w(n−1),

∂w(n−1)

∂x
).                          (97) 

Lemma 7.1 

Supposing that the condition (82) holds, then for the linearized problem (94)-(96), we have: 

                                                ‖W(n)‖
𝕃2(0,T,H1(Ω)

≤ C‖W(n−1)‖
𝕃2(Ω)

.                                  (98) 

Where: C > 0. 

Proof: 

We put:  MW(n) = −ℑx
2 ∂W

(n)

∂t
,  we get: 



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( ∂ 0
c
t
δW(n), −ℑx

2 ∂W
(n)

∂t
)
𝕃2(Ω)

+ α (
∂2W(n)

∂x2
, ℑx

2 ∂W
(n)

∂t
)
𝕃2(Ω)

+ β (
∂3W(n)

∂t∂x2
, ℑx

2 ∂W
(n)

∂t
)
𝕃2(Ω)

−

              γ (W(n), ℑx
2 ∂W

(n)

∂t
)
𝕃2(Ω)

+ (∫ a(t − s)W(n)(x, s)ds,
t

0
ℑx
2 ∂W

(n)

∂t
)
𝕃2(Ω)

=

                     (N(n−1)(x, t), −ℑx
2 ∂W

(n)

∂t
)
𝕃2(Ω)

                                                                                    (99)                                                                                     

After integrating by parts all terms of (99) and using conditions (95) and (96), proceding as in 

the establishment of theorem 5.1  

Dt
δ−1 ‖ℑx

∂W(n)

∂t
‖
2

𝕃2(Ω)
+(a0 +

α

2
) ‖W(n)(. , τ)‖

2

𝕃2(Ω)
≤ ∫ ‖ℑxN

(n−1)(x, t)‖
2

𝕃2(Ω)
dt

τ

0
+

      (β +
ε

2
)∫ ‖ℑx

∂W(n)(.,t)

∂t
‖
2

𝕃2(Ω)
dt

τ

0
+ (−

α

2
+ a1T

2) ∫ ‖W(n)(. , τ)‖
2

𝕃2(Ω)
dt

τ

0
                                                                                                                                                                                                                               

                                                                                                                                          (100) 

We apply  ℑx  to the equation (94), and we multiplying the resulting equation by 
 ∂W(n)

∂x
 , and 

we integrate by parts over Ω : 

∫ ∂ 0
c
t
δℑxW

(n)

Ω

.
∂W(n)

∂x
dxdt − α∫ (

∂W(n)

∂x
)

2

dxdt
Ω

− β∫
∂

∂x
(
∂2W(n)

∂t ∂x
)dxdt

Ω

+ γ∫ℑx
Ω

W(n).
∂W(n)

∂x
dxdt

− ∫ ∫ a(t − s)ℑxW
(n)(x, s).

∂W(n)

∂x
dsdxdt = ∫ℑx

Ω

t

0Ω

N(n−1)(x, t).
∂W(n)

∂x
dxdt. 

                                                                                                                                              (101)                                                                                                                                                                                   

Where: 

                              ∫ ∂ 0
c
t
δℑxW

(n)
Ω

.
∂W(n)

∂x
dxdt = −∫ ∂ 0

c
t
δW(n).

Ω
W(n)dxdt                             (102) 

                                    α∫ (
∂W(n)

∂x
)
2

dxdt =  α∫ ‖
∂W(n)(.,t)

∂x
‖
2

𝕃2(Ω)

τ

0Ω
 dt                              (103) 

                                 β∫
∂

∂x
(
∂2W(n)

∂t∂x
) dxdt

Ω
= β ‖

∂W(n)

∂x
‖
2

𝕃2(Ω)
                                            (104) 

                            γ∫ ℑxΩ
W(n).

∂W(n)

∂x
dxdt =  −γ∫ ‖W(n)‖

2

𝕃2(Ω)

τ

0
dt                                  (105) 

                    ∫ ∫ a(t − s)ℑxW
(n)(x, s).

∂W(n)

∂x
dsdxdt ≤ a1

t

0Ω
T2 ∫ ‖W(n)(. , t)‖

2

𝕃2(Ω)
dt

τ

0
      (106) 



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∫ ℑxN
(n−1)(x, t).

∂W(n)

∂xΩ

dxdt ≤   
1

2
∫ ‖N(n−1)(x, t)‖

2

𝕃2(Ω)
dt

τ

0

+
1

2
∫ ‖W(n)(. , t)‖

2

𝕃2(Ω)
dt

τ

0

 

                                                                                                                                              (107)                                                                                                                                                            

After integration by parts of all the terms of (101) and taking into consideration conditions 

(95), (96) and using inequality (8), we have : 

 

∫ ∂ 0
c
t
δW(n).

Ω
W(n)dxdt  + α ∫ ‖

∂W(n)(.,t)

∂x
‖
2

𝕃2(Ω)

τ

0
dt +  β ‖

∂W(n)

∂x
‖
2

𝕃2(Ω)
 

                            ≤
1

2
∫ ‖N(n−1)(x, t)‖

2

𝕃2(Ω)
dt

τ

0
+ (γ +

1

2
+ a1T

2)∫ ‖W(n)(. , t)‖
2

𝕃2(Ω)
dt  

τ

0
(108) 

 

After combination of inequalities (100) and (108), we obtain: 

Dt
δ−1 ‖ℑx

∂W(n)

∂t
‖

2

𝕃2(Ω)

+∫ ∂ 0
c
t
δW(n).

Ωτ
W(n)dxdt + α∫ ‖

∂W(n)(. , t)

∂x
‖

2

𝕃2(Ω)

τ

0

+   β ‖
∂W(n)

∂x
‖

2

𝕃2(Ω)

+ (a0 +
α

2
)‖W(n)(. , τ)‖

2

𝕃2(Ω)

≤ ∫ ‖N(n−1)(x, t)‖
2

𝕃2(Ω)
dt

τ

0

 

 + (β +
ε

2
) ∫ ‖ℑx

∂W(n)(.,t)

∂t
‖
2

𝕃2(Ω)
dt

τ

0
+ (a1T

2 −
α

2
+ γ +

1

2
) ∫ ‖W(n)(. , τ)‖

2

𝕃2(Ω)
dt          

τ

0
(109) 

Now we will eliminate the last term in (109) by applying the Gronwall lemma:   

Dt
δ−1 ‖ℑx

∂W(n)

∂t
‖
2

𝕃2(Ω)
+ ∫ ∂ 0

c
t
δW(n).

Ωτ
W(n)dxdt + α ∫ ‖

∂W(n)(.,t)

∂x
‖
2

𝕃2(Ω)

τ

0
+

                  β ‖
∂W(n)

∂x
‖
2

𝕃2(Ω)
+ (a0 +

α

2
) ‖W(n)(. , τ)‖

2

𝕃2(Ω)
   ≤

exp c0 {∫ ‖N
(n−1)(x, t)‖

2

𝕃2(Ω)
dt

τ

0
+ c1 ∫ ‖ℑx

∂W(n)(.,t)

∂t
‖
2

𝕃2(Ω)
dt

τ

0
}.                                        (110) 

Where: 

{
c0 =  β +

ε

2
,

c1 = a1T
2 −

α

2
+ γ +

1

2
.
 

 



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We apply Gronwall lemma to the last term of (110)  

∫ ‖ℑx
∂W(n)(. , t)

∂t
‖

2

𝕃2(Ω)

dt
τ

0

≤ Γ(δ)Eδδ(c1 exp(c0T) t
δ) exp(c0t) . Dt

−δ‖N(n−1)(x, t)‖
2

𝕃2(Ω)
 

                                                                                                                                              (111) 

On the other side, we applying the condition (82), we get : 

∫ ‖N(n−1)(x, t)‖
2

𝕃2(Ω)
dt

τ

0

≤ 2M2∫ (‖W(n−1)(. , t)‖
2

𝕃2(Ω)
+ ‖

∂W(n−1)

∂x
‖

2

𝕃2(Ω)

)dt
τ

0

 

                                                                                                                                             (112) 

 

Combining (111)-(112) and using (49), we get: 

Dt
δ−1 ‖ℑx

∂W(n)

∂t
‖

2

𝕃2(Ω)

+∫ ∂ 0
c
t
δW(n).

Ωτ
W(n)dxdt + ∫ ‖

∂W(n)(. , t)

∂x
‖

2

𝕃2(Ω)

τ

0

+  ‖
∂W(n)

∂x
‖

2

𝕃2(Ω)

 

+‖W(n)(. , τ)‖
2

𝕃2(Ω)
≤  C∗M2∫ (‖W(n−1)(. , t)‖

2

𝕃2(Ω)
+  ‖

∂W(n−1)

∂x
‖

2

𝕃2(Ω)

)dt
τ

0

 

                                                                                                                                              (113)                                                                                                                                                            

Such that: 

                               C∗ = exp(c0t) (1 + Γ(δ)Eδδ(c1 exp(c0T) t
δ)) .

Tδ

 Γ(1+δ)
                       (114) 

After discarding the first two terms on (113), we obtain: 

∫ ‖
∂W(n)(. , t)

∂x
‖

2

𝕃2(Ω)

τ

0

+  ‖
∂W(n)

∂x
‖

2

𝕃2(Ω)

+ ‖W(n)(. , τ)‖
2

𝕃2(Ω)
         

≤     C∗M2∫ (‖W(n−1)(. , t)‖
2

𝕃2(Ω)
+ ‖

∂W(n−1)

∂x
‖

2

𝕃2(Ω)

)dt
τ

0

                (115) 

                                                                                                                                                           

 

 



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Here, the RHS doesn’t depend on τ so, we can replace the LHS by upper bounds with respect 

to τ, we obtain: 

∫ ‖
∂W(n)(. , t)

∂x
‖

2

𝕃2(Ω)

T

0

+  ‖
∂W(n)

∂x
‖

2

𝕃2(Ω)

+ ‖W(n)(. , τ)‖
2

𝕃2(Ω)

≤    C∗M2∫ (‖W(n−1)(. , t)‖
2

𝕃2(Ω)
+ ‖

∂W(n−1)(. , t)

∂x
‖

2

𝕃2(Ω)

)dt
T

0

  

                                                                                                                                            (116) 

We integrate over (0,T), we get: 

∫ ‖
∂W(n)(. , t)

∂x
‖

2

𝕃2(Ω)

T

0

+∫ ‖W(n)(. , τ)‖
2

𝕃2(Ω)
dt

T

0

≤     λM2∫ (‖W(n−1)(. , t)‖
2

𝕃2(Ω)
+ ‖

∂W(n−1)(. , t)

∂x
‖

2

𝕃2(Ω)

)dt
T

0

         (117) 

                                                                                                                                                            

Such that:                                                   λ =
C∗M2T

min (1,T)
. 

We obtain the inequality: 

                                     ‖W(n)‖
2

𝕃2(0,T,H1(Ω))
≤  λ‖W(n−1)‖

2

𝕃2(0,T,H1(Ω))
.                           (118) 

Using the convergence of series criteria we conclude that ∑ W(n)∞
n−1  converges if λ < 1, in 

other words if M < √
min (1,T)

C∗T
.  Since: W(n) = w(n+1)(x, t) − w(n)(x, t), then (w(n))

n∈ℕ
 

converge to a function w ∈ 𝕃2(0, T, H1(Ω)). 

So to prove that  w  is the solution of problem (94)-(96), we have only to prove that w verifies 

(81) and (90). 

We have from problem (91)-(93), that:  

                                  κ(w(n), u) = (u, ℑxχ (x, t, w
(n−1),

∂w(n−1)

∂x
))
𝕃2(Ω)

                             (119) 

 

 



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More precisely 

κ(w(n) −w, u) + κ(w, u) = (u, ℑxχ (x, t, w
(n−1),

∂w(n−1)

∂x
) − ℑxχ (x, t, w,

∂w

∂x
))

𝕃2(Ω)

 

                                                                + (u, ℑxχ (x, t, w,
∂w

∂x
))
𝕃2(Ω)

,                                        (120) 

After using (91), then (120) becomes: 

κ(w(n) −w, u)

= −( ∂ 0
c
t
δℑx(w

(n) −w), u)
𝕃2(Ω)

+ α(
∂(w(n) −w)

∂x
, u)

𝕃2(Ω)

+ β(
∂

∂t
(
∂w(n)

∂x
) , u)

𝕃2(Ω)

+ γ(ℑx(w
(n) −w), u)

𝕃2(Ω)

+ (∫ a(t − s)ℑx(w
(n) −w)(x, s)ds

t

0

, u)
𝕃2(Ω)

                                             (121) 

                                                                                                                                                         

After applying the integration by parts, and taking in consideration conditions on:u and w, 

(121) will be transformed as: 

 

κ(w(n) −w, u)

= −( ∂ 0
c
t
δ(w(n) −w), ℑxu)𝕃2(Ω) + α(

∂(w(n) −w)

∂x
, u)

𝕃2(Ω)

+ β((
∂w(n)

∂x
) ,
∂u

∂t
)

𝕃2(Ω)

+ γ(ℑx(w
(n) − w), u)

𝕃2(Ω)

+ (∫ a(t − s)ℑx(w
(n) −w)(x, s)ds

t

0

, u)
𝕃2(Ω)

                                             (122) 

                                                                                                                                                          

We will apply the inequality of Cauchy-Schwartz and lemma (2.4), we have: 

                 κ(w(n) − w, u) ≤ ζ‖w(n) − w‖
𝕃2(0,T,H1(Ω))

. [‖u‖𝕃2(Ω) + ‖
∂u

∂t
‖
𝕃2(Ω)

]               (123) 

Such that:                        ζ = max (α +
T

2
+
γ

2
, β), 



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And from (120), we have the following estimation: 

(u, ℑxχ (x, t, w
(n−1),

∂w(n−1)

∂x
) − ℑxχ (x, t, w,

∂w

∂x
))

𝕃2(Ω)

≤
M

√2
‖w(n) −w‖

𝕃2(0,T,H1(Ω))
. ‖u‖𝕃2(Ω)                                                       (124) 

                                                                                                                                                           

When the limit n → ∞ in (122) , and we take in consideration (123) and (124), we obtain 

                                      κ(w, u) = (u, ℑxχ (x, t, w,
∂w

∂x
))
𝕃2(Ω)

.                                                 (125) 

So the problem (94)-(96) admit a weak solution. 

Now, we will prove the uniqueness of problem (79)-(81). 

Theorem: 

Under condition of lemma (76), the problem (79)-(81) admits a unique solution. 

Proof: We suppose that the problem (79)-(81) admit u1 , u2 solutions in𝕃2(0, T, H1(Ω)), and 

W = u1 − u2, and verifies: 

              ℒW = ∂ 0
c
t
δW− α

∂2W

∂x2
− β

∂3W

∂t∂x2
+ γW − ∫ a(t − s)W(x, s)ds =

t

0
N(x, t)               (126) 

              ℓW = W(x, 0) = 0,     qW =
∂W(x,0)

∂t
= 0,                      0 < 𝑥 < 1.                         (127) 

             ∫ W(x, t)dx = 0,       
t

0 ∫ xW(x, t)dx = 0,       
t

0
             0 < 𝑡 ≤ 𝑇 .                           (128) 

 

Where:              N(x, t) = χ(x, t, u1, r1) − χ(x, t, u2, r2) 

This will be done by establishing the same proof of lemma (76), we obtain: 

                                   ‖W‖
𝕃2(0,T,H1(Ω))

≤ C‖W‖
𝕃2(0,T,H1(Ω))

.                                             (129) 

Since  C < 1, then: (1 − C)‖W‖
𝕃2(0,T,H1(Ω))

≤ 0, we deduce finally that: u1 − u2 = 0, so 

u1 = u2 in 𝕃2(0, T, H1(Ω)).∎ 

 

 



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8. Conclusion 

Based on the a priori estimate method and the density of the operator generated by the 

problem, and the iterative process, we have successfully proven the problem's existence and 

uniqueness of its solution. After carrying out our experimental research, we were able to 

confirm our initial hypotheses and respond to our problem. This work could be considered as 

a contribution to the development of the functional analysis method. And to conclude we 

argue that the research carried out in this article could serve as a contribution for possible 

studies in the field of applied mathematics. 

Our study being based on the a priori estimation method is not exhaustive and opens several 

research perspectives such as: the application of the latter to ordinary physical problems and 

other purely fractional ones. 

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