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Article

On a Time-Fractional Biharmonic Nonlocal Initial Boundary-Value Problem with Frictional and Viscoelastic Damping Terms

Mathematics Department, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
*
Author to whom correspondence should be addressed.
Mathematics 2026, 14(2), 387; https://doi.org/10.3390/math14020387
Submission received: 21 December 2025 / Revised: 18 January 2026 / Accepted: 20 January 2026 / Published: 22 January 2026
(This article belongs to the Special Issue Applications of Partial Differential Equations, 2nd Edition)

Abstract

This research work investigates the existence, uniqueness, and stability of solution for a time-fractional fourth-order partial differential equation, subject to two initial conditions and four nonlocal integral boundary conditions. The equation incorporates several key components: the Caputo fractional derivative operator, the Laplace operator, the biharmonic operator, as well as terms representing frictional and viscoelastic damping. The presence of these elements, particularly the nonlocal boundary constraints, introduces new mathematical challenges that require the development of advanced analytical methods. To address these challenges, we construct a functional analytic framework based on Sobolev spaces and employ energy estimates to rigorously prove the well-posedness of the problem.

1. Introduction

Initial and boundary-value problems involving fractional integro-differential equations have become a significant and expanding field of study in applied mathematics. The combination of fractional derivatives with integral operators adds considerable complexity, but also allows for more accurate modeling of systems that exhibit memory effects. These equations are particularly effective in representing complex physical processes such as anomalous diffusion, viscoelastic behavior, and memory-dependent phenomena in various materials. Fractional integro-differential equations are mathematical models that incorporate both fractional-order derivatives and integrals. They have proven to be valuable tools for describing numerous phenomena across multiple scientific disciplines. In fact, fractional models that include memory components often offer a more realistic representation than their classical integer-order counterparts. Such equations find applications in diverse areas including bioengineering, medicine, astrophysics, chemical engineering, fluid dynamics, electromagnetism, polymer science, power electronics, thermodynamics, signal and image processing, seismology, hemodynamics, aerodynamics, neural networks, control theory, geology, anomalous transport, viscoelastic materials, optics, and chaotic systems. For further reading on these applications, see references [1,2,3,4,5,6,7,8,9].
Recent studies [10,11,12,13,14,15] have explored the existence and uniqueness of solutions for certain classes of fractional integro-differential equations.
The obtained results in this paper contribute significantly to the well-posedness theory for higher-order fractional partial differential equations that include memory effects and nonlocal constraints which in fact model viscoelastic materials and anomalous diffusion beyond the scope of classical integer-order or simpler fractional models. Our paper builds upon and extends earlier works of fractional integro-differential equations such as those in references [10,11,12,13,14,15] by integrating biharmonic and Laplace operators along with purely integral boundary conditions. This advancement is achieved through the development of new a priori estimates, overcoming challenges via novel a priori bounds, unlike earlier lower-order or local-boundary analyses.
We outline the content of this paper as follows. In Section 2, we outline the problem and provide the functional framework. Section 3 is devoted to the proving the uniqueness of the solution of the posed problem. The main results concerning the solvability of the problem are considered in Section 4.

2. Outlining the Problem and the Functional Framework

Let T > 0 , and Λ T = Ω × [ 0 , T ] , where Ω = ( 0 , a ) × ( 0 , b ) R 2 . We search for a function Ψ : Λ T R satisfying the fractional two-dimensional fourth-order integro-differential equation
L Ψ = t σ + 1 Ψ Δ Ψ + Δ 2 Ψ + γ Ψ + 0 t K ( t s ) Δ Ψ d s = H ( x , y , t ) ,
with Caputo fractional time order σ + 1 , where 0 < σ < 1 which is defined for a certain function V by [7]
t σ + 1 V ( x , t ) = 1 Γ ( 1 σ ) 0 t V s s ( x , s ) ( t s ) σ d s .
The equation is associated with the initial conditions
Γ 1 Ψ = Ψ ( x , y , 0 ) = Z 1 ( x , y ) , Γ 2 Ψ = Ψ t ( x , y , 0 ) = Z 2 ( x , y ) ,
and the nonlocal boundary conditions of integral type
0 a Ψ d x = 0 , 0 b Ψ d y = 0 , 0 a x Ψ d x = 0 , 0 b y Ψ d y = 0 .
Here H L 2 ( Λ T ) ,   Z 1 H 1 ( Ω ) ,   Z 2 L 2 ( Ω ) .
We assume that the given data verify the following compatibility conditions
0 a Z 1 d x = 0 , 0 b Z 1 d y = 0 , 0 a x Z 1 d x = 0 , 0 b y Z 1 d y = 0
0 a Z 2 d x = 0 , 0 b Z 2 d y = 0 , 0 a x Z 2 d x = 0 , 0 b y Z 2 d y = 0
For the relaxation function K ( t ) , we suppose that G : R + R + is a C 2 bounded function such that K ( t ) 0 , K ( t ) 0 .
The differential Equation (1) represents a two-dimensional fourth-order integro-differential equation with a time-fractional Caputo derivative of order σ + 1 , where 0 < σ < 1 . The model combines several key mathematical features: the nonlocal Caputo operator, the Laplace operator, the biharmonic operator, and a convolution-type memory term involving a damping kernel K ( t ) . The presence of these components, together with the nonlocal integral boundary constraints, makes the problem both mathematically rich and physically relevant. Such integral boundary conditions often appear in problems where averaged or moment-based quantities over the domain must vanish for instance, in systems with prescribed mass or momentum distributions. These conditions are particularly useful when classical Dirichlet or Neumann boundary values are not physically meaningful or cannot be directly measured. Some references on nonlocal fractional problems can be found in [16,17,18,19,20,21] and references therein. In the presented model (1)–(3), the field of functions Ψ ( x , y , t ) evolves under the influence of several physical mechanisms: a diffusive term . Ψ = Δ Ψ , which tends to smooth out spatial variations, possibly representing energy diffusion or elastic stress–strain behavior. A nonlocal memory term t σ + 1 Ψ , modeling viscoelasticity or anomalous wave propagation. A fourth-order spatial term Δ 2 Ψ , introducing higher-order stiffness or bending rigidity. A damping term γ Ψ , pulling the system toward equilibrium. The integral term 0 t K ( t s ) Δ Ψ d s models a history-dependent diffusion or deformation mechanism, where the current spatial variation via Δ Ψ is influenced by its entire past, weighted by the decaying relaxation function K . This is typical in materials with memory, biological tissues, porous media with trapping, or structural systems with viscoelastic damping. An external forcing term H ( x , y , t ) , representing external influences. The differential Equation (1) is supplemented by two initial conditions, Ψ ( x , y , 0 ) = Z 1 ( x , y ) , and Ψ t ( x , y , 0 ) = Z 2 ( x , y ) , which specify the initial configuration and initial rate of change of the system, and supplemented by four nonlocal boundary conditions (purely integral constraints) 0 a Ψ d x = 0 , 0 b Ψ d y = 0 ,   0 a x Ψ d x = 0 , 0 b y Ψ d y = 0 which ensure symmetry and conservation of spatial moments.
We first introduce the functional framework, where we seek the solution of the posed problem (1)–(3). Let ϑ and V be the Hilbert spaces of square integrable functions with scalar product, and associated norms, respectively,
( ϑ , V ) L 2 ( Ω ) = Ω ϑ V d x d y , ϑ L 2 ( Ω ) 2 = Ω ϑ 2 d x d y .
Let L 2 ( 0 , T ; Y * ) be the set of all measurable functions M : ( 0 , T ) Y * , where Y * is the Banach space with finite norm
ϑ L 2 ( 0 , T ; Y * ) 2 = 0 T ϑ ( . , . , t ) Y * 2 d t < .
The function space L 2 ( 0 , T ; Y * ) is a Hilbert space, since Y * is a Hilbert space. We design using C ( 0 , T ; Y * ) the set of all continuous functions ϑ : ( 0 , T ) Y * such that
ϑ C ( 0 , T ; Y * ) = sup 0 t T ϑ ( . , . , t ) Y * is finite .
We denote by H x 1 ( Ω ) and H y 1 ( Ω ) the Sobolev space with, respectively, the norms
ϑ H x 1 ( Ω ) 2 = ϑ L 2 ( Ω ) 2 + ϑ x L 2 ( Ω ) 2 and ϑ H y 1 ( Ω ) 2 = ϑ L 2 ( Ω ) 2 + ϑ y L 2 ( Ω ) 2
We can view the problem (1)–(3) as the problem of solving the operator equation T Ψ = V = ( L Ψ , Γ 1 Ψ , Γ 2 Ψ ) , for all Ψ D ( T ) where T is the operator defined by T = ( L , Γ 1 , Γ 2 ) and D ( T ) is its domain defined by the set
D ( T ) = Ψ L 2 ( Λ ) : t σ Ψ , Ψ x , Ψ x x , Ψ x x x , Ψ x x x x , Ψ y , Ψ y y , Ψ y y y , Ψ y y y y , Ψ t
belonging to L 2 ( Λ ) , which with Ψ verifies (2) and (3). The operator T : B H * , where B is the Banach space with finite norm
Ψ B 2 = sup 0 t T I x Ψ ( . , . t ) H y 1 ( Ω ) 2 + sup 0 t T I y Ψ ( . , . t ) H x 1 ( Ω ) 2 + sup 0 t T Ψ ( . , . t ) L 2 ( Ω ) 2   .
The space H * is the Hilbert space L 2 ( 0 , T , L 2 ( Ω ) ) × H 1 ( Ω ) × L 2 ( Ω ) of vector-valued functions W = ( H , Z 1 , Z 2 ) for which the norm
W H * 2 = H L 2 ( 0 , T ; L 2 ( Ω ) ) 2 + Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) 2   ,
is finite.
Remark 1. 
For the temporal regularity of the function Ψ to ensure that the fractional derivative t σ + 1 Ψ is well defined, the function t Ψ should be in L 2 ( 0 , T ; L 2 ( Ω ) ) (see the domain of definition), and the continuity in time is guaranteed by (2.8), meaning the function Ψ should be in C ( 0 , T ; L 2 ( Ω ) ) (see the definition of space B).

3. Uniqueness and Stability of the Solution

We will establish an a priori bound for the operator T from which the uniqueness and continuous dependence of the solution upon the initial conditions (2) follow.
Here, we present the first result of the uniqueness of the solution and its dependence on the given data
Theorem 1. 
For any function Ψ D ( T ) there exists a positive constant ω independent of Ψ such that the following a priori bound holds
sup 0 t T I x Ψ ( . , . t ) H y 1 ( Ω ) 2 + sup 0 t T I y Ψ ( . , . t ) H x 1 ( Ω ) 2 + sup 0 t T Ψ ( . , . t ) L 2 ( Ω ) 2 ω H L 2 ( 0 , T ; L 2 ( Ω ) ) 2 + Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) ) 2 ,
where
ω = β + β 2 Γ ( σ ) E σ , σ ( β T ) max T σ + 1 Γ ( σ ) , T σ σ Γ ( σ ) β = max T σ a 2 b 2 ( 1 σ ) Γ ( 1 σ ) , a 2 2 , b 2 2 , a 2 b 2 2 .
Proof. 
We first introduce the integro-differential operator
I x y 2 Ψ t = 0 x 0 y 0 ξ 0 η Ψ t ( ρ , ν , t ) d ν d ρ d η d ξ ,
and consider the identity
( t σ + 1 Ψ Δ Ψ + Δ 2 Ψ + γ Ψ + 0 t K ( t s ) Δ Ψ d s , I x y 2 Ψ t ) L 2 ( Ω ) ) = H , I x y 2 Ψ t L 2 ( Ω ) )
The identity (10) may be written as
( t σ + 1 Ψ , I x y 2 Ψ t ) L 2 ( Ω ) ) Ψ x x + Ψ y y , I x y 2 Ψ t L 2 ( Ω ) ) + Ψ x x x x + Ψ y y y y , I x y 2 Ψ t L 2 ( Ω ) ) + 2 Ψ x x y y , I x y 2 Ψ t L 2 ( Ω ) ) + γ ( Ψ , I x y 2 Ψ t ) L 2 ( Ω ) ) + 0 t K ( t s ) ( Ψ x x + Ψ y y ) d s , I x y 2 Ψ t L 2 ( Ω ) ) = H , I x y 2 Ψ t L 2 ( Ω ) )     .
Applying conditions (3), the repeated integration by parts of the terms on the left-hand side of (11) is direct, though somewhat cumbersome. Their direct results are as follows:
( t σ + 1 Ψ , I x y 2 Ψ t ) L 2 ( Ω ) ) = ( t σ I x y Ψ t , I x y Ψ t ) L 2 ( Ω ) )     ,
Ψ x x + Ψ y y , I x y 2 Ψ t L 2 ( Ω ) ) = 1 2 t I y Ψ L 2 ( Ω ) 2 + 1 2 t I x Ψ L 2 ( Ω ) 2     ,
Ψ x x x x + Ψ y y y y , I x y 2 Ψ t L 2 ( Ω ) ) = 1 2 t I y Ψ x L 2 ( Ω ) 2 + 1 2 t I x Ψ y L 2 ( Ω ) 2     ,
2 ( Ψ x x y y , I x y 2 Ψ t ) L 2 ( Ω ) ) = t Ψ L 2 ( Ω ) 2     ,
γ ( Ψ , I x y 2 Ψ t ) L 2 ( Ω ) ) = γ 2 I x y Ψ L 2 ( Ω ) 2     ,
0 t K ( t s ) ( Ψ x x + Ψ y y ) d s , I x y 2 Ψ t L 2 ( Ω ) ) = 0 t K ( t s ) Ψ x x d s , I x y 2 Ψ t L 2 ( Ω ) ) + 0 t K ( t s ) Ψ y y ) d s , I x y 2 Ψ t L 2 ( Ω ) )     .
We first evaluate the first term on the right-hand side of (17).
0 t K ( t s ) Ψ x x ( . , . s ) d s , I x y 2 Ψ t L 2 ( Ω ) ) = 0 b 0 t K ( t s ) Ψ x ( . , . s ) d s I x y 2 Ψ t 0 a d y 0 t K ( t s ) Ψ x ( . , . s ) d s , I x y y Ψ t L 2 ( Ω ) ) = 0 b 0 t K ( t s ) Ψ ( . , . s ) d s I x y y Ψ t 0 a d y + ( 0 t K ( t s ) Ψ ( . , . s ) d s , I y y Ψ t ) L 2 ( Ω ) ) = 0 a 0 t K ( t s ) I y Ψ ( . , . s ) d s I y y Ψ t | 0 b d y ( 0 t K ( t s ) I y Ψ ( . , . s ) d s , I y Ψ t ) L 2 ( Ω ) ) = ( 0 t K ( t s ) I y Ψ ( . , . s ) d s , I y Ψ t ) L 2 ( Ω ) )     .
Similarly, we have
0 t K ( t s ) Ψ x x ( . , . s ) d s , I x y 2 Ψ t L 2 ( Ω ) ) = ( 0 t K ( t s ) I x Ψ ( . , . s ) d s , I x Ψ t ) L 2 ( Ω ) )     .
Observe that
( 0 t K ( t s ) I y Ψ ( . , . s ) d s , I y Ψ t ) L 2 ( Ω ) ) = 0 t K ( t s ) Ω I y Ψ t ( . , . t ) [ I y Ψ ( . , . s ) I y Ψ ( . , . t ) ] d x d y d s 0 t K ( t s ) Ω [ I y Ψ t ( . , . t ) I y Ψ ( . , . t ) ] d x d y d s = 1 2 0 t K ( t s ) d d t Ω | I y Ψ ( . , . s ) I y Ψ ( . , . t ) | 2 d x d y d s 1 2 0 t K ( t s ) d d t Ω | I y Ψ ( . , . t ) | 2 d x d y d s = 1 2 d d t 0 t K ( t s ) Ω | I y Ψ ( . , . s ) I y Ψ ( . , . t ) | 2 d x d y d s 1 2 d d t 0 t K ( s ) Ω | I y Ψ ( . , . t ) | 2 d x d y d s + K ( t ) 2 Ω | I y Ψ ( . , . t ) | 2 d x d y 1 2 0 t K ( t s ) Ω | I y Ψ ( . , . s ) I y Ψ ( . , . t ) | 2 d x d y d s
Similarly, we have
( 0 t G ( t s ) I x Ψ ( . , . s ) d s , I x Ψ t ) L 2 ( Ω ) ) = 1 2 d d t 0 t G ( t s ) Ω | I x Ψ ( . , . s ) I x Ψ ( . , . t ) | 2 d x d y d s 1 2 d d t 0 t G ( s ) Ω | I x Ψ ( . , . t ) | 2 d x d y d s + G ( t ) 2 Ω | I x Ψ ( . , . t ) | 2 d x d y 1 2 0 t G ( t s ) Ω | I x Ψ ( . , . s ) I x Ψ ( . , . t ) | 2 d x d y d s .
Substituting formulas (11)–(21) in (10), we get
t σ I x y Ψ t , I x y Ψ t L 2 ( Ω ) ) + 1 2 t I y Ψ L 2 ( Ω ) 2 + 1 2 t I x Ψ L 2 ( Ω ) 2 + 1 2 t I y Ψ x L 2 ( Ω ) 2 + 1 2 t I x Ψ y L 2 ( Ω ) 2 + t Ψ L 2 ( Ω ) 2 + γ 2 I x y Ψ L 2 ( Ω ) 2 + 1 2 d d t 0 t K ( t s ) Ω | I y Ψ ( . , . s ) I y Ψ ( . , . t ) | 2 d x d y d s + 1 2 d d t 0 t K ( t s ) Ω | I x Ψ ( . , . s ) I x Ψ ( . , . t ) | 2 d x d y d s + K ( t ) 2 Ω | I x Ψ ( . , . t ) | 2 d x d y + K ( t ) 2 Ω | I y Ψ ( . , . t ) | 2 d x d y 1 2 0 t K ( t s ) Ω | I x Ψ ( . , . s ) I x Ψ ( . , . t ) | 2 d x d y d s 1 2 0 t K ( t s ) Ω | I y Ψ ( . , . s ) I y Ψ ( . , . t ) | 2 d x d y d s = 1 2 d d t 0 t K ( s ) Ω | I y Ψ ( . , . t ) | 2 d x d y d s + 1 2 d d t 0 t K ( s ) Ω | I x Ψ ( . , . t ) | 2 d x d y d s + H , I x y 2 Ψ t L 2 ( Ω ) )     .
Since K ( t ) 0 for all t [ 0 , T ] , we can discard the last two terms on the left-hand side of (22) and use Lemma 2 [21] to obtain the inequality
t σ I x y Ψ t L 2 ( Ω ) 2 + t I y Ψ L 2 ( Ω ) 2 + t I x Ψ L 2 ( Ω ) 2 + t I y Ψ x L 2 ( Ω ) 2 + t I x Ψ y L 2 ( Ω ) 2 + 2 t Ψ L 2 ( Ω ) 2 + γ I x y Ψ L 2 ( Ω ) 2 + d d t 0 t K ( t s ) Ω | I y Ψ ( . , . s ) I y Ψ ( . , . t ) | 2 d x d y d s + d d t 0 t K ( t s ) Ω | I x Ψ ( . , . s ) I x Ψ ( . , . t ) | 2 d x d y d s + K ( t ) Ω | I x Ψ ( . , . t ) | 2 d x d y + K ( t ) Ω | I y Ψ ( . , . t ) | 2 d x d y d d t 0 t K ( s ) Ω | I y Ψ ( . , . t ) | 2 d x d y d s + d d t 0 t K ( s ) Ω | I x Ψ ( . , . t ) | 2 d x d y d s + 2 H , I x y 2 Ψ t L 2 ( Ω ) )     .
Taking into account that
0 t t σ Ψ L 2 ( Ω ) 2 d s = D σ 1 Ψ ( x , y , t ) L 2 ( Ω ) 2 t 1 σ ( 1 σ ) Γ ( 1 σ ) Ψ ( x , y , 0 ) L 2 ( Ω ) 2     ,
By ignoring the term γ I x y Ψ L 2 ( Ω ) 2 on the left-hand side of (23), and then replacing t with s and integrating with respect to s over ( 0 , t ) , and applying the Cauchy ε inequality as well as a Poincare-type inequality [22,23] for the right-hand side of (24), we obtain the estimate
D σ 1 I x y Ψ t L 2 ( Ω ) 2 + I y Ψ ( . , . t ) H x 1 ( Ω ) 2 + I x Ψ ( . , . t ) H y 1 ( Ω ) 2 + Ψ ( . , . t ) L 2 ( Ω ) 2 + 0 t K ( t s ) Ω | I y Ψ ( . , . s ) I y Ψ ( . , . t ) | 2 d x d y d s + 0 t K ( t s ) Ω | I x Ψ ( . , . s ) I x Ψ ( . , . t ) | 2 d x d y d s T 1 σ ( 1 σ ) Γ ( 1 σ ) I x y Z 2 L 2 ( Ω ) 2 + I y Z 1 L 2 ( Ω ) 2 + I x Z 1 L 2 ( Ω ) 2 + I y Z 1 x L 2 ( Ω ) 2 + I x Z 1 y L 2 ( Ω ) 2 + Z 1 L 2 ( Ω ) 2 + H L 2 ( 0 , t ; L 2 ( Ω ) ) 2 + a b 2 I x y Ψ t L 2 ( 0 , t ; L 2 ( Ω ) ) 2     .
We again apply a Poincare-type inequality to the first five terms on the right-hand side of (25), and drop the last two terms on its left-hand side (since K 0 ), obtaining
D σ 1 I x y Ψ t L 2 ( Ω ) 2 + I y Ψ ( . , . t ) H x 1 ( Ω ) 2 + I x Ψ ( . , . t ) H y 1 ( Ω ) 2 + Ψ ( . , . t ) L 2 ( Ω ) 2 β Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) 2 + H L 2 ( 0 , t ; L 2 ( Ω ) ) 2 + β I x y Ψ t L 2 ( 0 , t ; L 2 ( Ω ) ) 2     ,
where
β = max T σ a 2 b 2 ( 1 σ ) Γ ( 1 σ ) , a 2 2 , b 2 2 , a 2 b 2 2 .
We now use Lemma 1 [21] to eliminate the last term on the right-hand side of (26). We denote by
L ( t ) = I x y Ψ t L 2 ( 0 , t ; L 2 ( Ω ) ) 2     , t σ L ( t ) = D σ 1 I x y Ψ t L 2 ( Ω ) 2     , L ( 0 ) = 0 ,
The function L ( t ) is nonnegative and absolutely continuous on [ 0 , T ] since Ψ t L 2 ( 0 , T ; L 2 ( Ω ) ) . It also satisfies
D σ 1 I x y Ψ t L 2 ( Ω ) 2 β I x y Ψ t L 2 ( 0 , t ; L 2 ( Ω ) ) 2 + β Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) 2 + H L 2 ( 0 , t ; L 2 ( Ω ) ) 2 ,
where β Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) 2 + H L 2 ( 0 , t ; L 2 ( Ω ) ) 2 is obviously an integrable nonnegative function on [ 0 , T ] .
Then
L ( t ) β Γ ( σ ) E σ , σ ( β T ) D σ Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) 2 + H L 2 ( 0 , t ; L 2 ( Ω ) ) 2 .
Thus inequality (26) becomes
D σ 1 I x y Ψ t L 2 ( Ω ) 2 + I y Ψ ( . , . t ) H x 1 ( Ω ) 2 + I x Ψ ( . , . t ) H y 1 ( Ω ) 2 + Ψ ( . , . t ) L 2 ( Ω ) 2 β Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) 2 + H L 2 ( 0 , t ; L 2 ( Ω ) ) 2 + β β Γ ( σ ) E σ , σ ( β T ) D σ Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) 2 + H L 2 ( 0 , t ; L 2 ( Ω ) ) 2 .
Observe that
D σ Z 1 H 1 ( Ω ) 2 T σ Γ ( σ ) Z 1 H 1 ( Ω ) 2 D σ Z 2 L 2 ( Ω ) 2 T σ Γ ( σ ) Z 2 L 2 ( Ω ) 2 D σ H L 2 ( 0 , t ; L 2 ( Ω ) ) 2 t σ + 1 Γ ( σ ) 0 t H L 2 ( Ω ) 2 d s T σ + 1 Γ ( σ ) H L 2 ( 0 , T ; L 2 ( Ω ) ) 2     ,
then after dropping the first term on the left-hand side of (30) it then takes the form
I y Ψ ( . , . t ) H x 1 ( Ω ) 2 + I x Ψ ( . , . t ) H y 1 ( Ω ) 2 + Ψ ( . , . t ) L 2 ( Ω ) 2 ω Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) 2 + H L 2 ( 0 , T ; L 2 ( Ω ) ) 2 ,
where
ω = β + β 2 Γ ( σ ) E σ , σ ( β T ) max T σ + 1 Γ ( σ ) , T σ σ Γ ( σ ) .
We see that the right side of inequality (32) does not depend on t; thus we can replace the left side with the upper bound with respect to t, and consequently the a priori estimate (9) follows. □
The range of the operator T : B H * is a subset of H * , so we can extend T so that the a priori estimate (9) is true for the extension and T ¯ Ψ = H * .
It is straightforward to show that the operator T : B H * has a closure. We then have the extended a priori estimate
sup 0 t T I x Ψ ( . , . t ) H y 1 ( Ω ) 2 + sup 0 t T I y Ψ ( . , . t ) H x 1 ( Ω ) 2 + sup 0 t T Ψ ( . , . t ) L 2 ( Ω ) 2 ω H L 2 ( 0 , T ; L 2 ( Ω ) ) 2 + Z 1 H 1 ( Ω ) 2 + Z 2 L 2 ( Ω ) ) 2 ,
for all Ψ D ( T ¯ ) . where T ¯ is the closure of the operator T with the domain of definition D ( T ¯ ) .

4. Solvability of Problem (1)–(3)

The estimate (34) asserts that the operator T ¯ is one to one and that T ¯ 1 is continuous from R ( T ) onto B . As a consequence, if a solution of problem (1)–(3) exists, it is unique and depends continuously on the given data H L 2 ( Λ T ) , Z 1 H 1 ( Ω ) , Z 2 L 2 ( Ω ) . We also deduce from the estimate (34) that the set R ( T ¯ ) H * is closed and R ( T ¯ ) = R ( T ) ¯ .
Theorem 2. 
For any H L 2 ( Λ T ) ,   Z 1 H 1 ( Ω ) ,   Z 2 L 2 ( Ω ) , the posed problem (1)–(3) has a unique solution Ψ =   T 1 ( Z 1 , Z 2 , H ) = T 1 ¯ ( Z 1 , Z 2 , H ) .
Proof. 
To show that problem (1)–(3) has a unique solution for all functions V = ( Z 1 , Z 2 , H ) B , we must show that the range of the operator T is dense in H * . That is, R ( T ) ¯ = H * . We first need to prove the following Proposition. □
Proposition 1. 
If for some function Y ( x , y , t ) L 2 ( Λ T ) and for all W D 0 ( T ) = W D ( T ) , Γ 1 W = 0 , Γ 2 W = 0 we have
t σ + 1 W Δ W + Δ 2 W + γ W + 0 t K ( t s ) Δ W d s , Y L 2 ( Λ T ) = 0 , W D 0 ( T )
then Y vanishes almost everywhere in Λ T .
Proof. 
We set W = t 2 θ , and Y = x y 2 ( t θ ) where θ ( x , y , t ) is a function verifying the boundary and initial conditions (2) and (3) and such that θ ,   θ x ,   θ y , θ x x , θ y y , θ x y , t x y 2 θ , t θ ,   t 2 θ ,   t β + 1 θ belong to L 2 ( Λ T ) , then consider the inner product (35) in L 2 ( Ω ) ; we have
t σ + 1 t 2 θ , x y 2 ( t θ ) L 2 ( Ω ) Δ ( t 2 θ ) , x y 2 ( t θ ) L 2 ( Ω + Δ 2 ( t 2 θ ) , x y 2 ( t θ ) L 2 ( Ω + γ t 2 θ , x y 2 ( t θ ) L 2 ( Ω + 0 t K ( t s ) Δ ( t 2 θ ) d s , x y 2 ( t θ ) L 2 ( Ω = 0 .
Upon using the boundary conditions (3),the successive integration by parts of integrals on the left-hand side of (36) are straightforward but somewhat tedious. We only provide their final results
t σ + 1 t 2 θ , x y 2 ( t θ ) L 2 ( Ω ) = t σ x y ( t θ ) , x y ( t θ ) L 2 ( Ω )     ,
Δ ( t 2 θ ) , x y 2 ( t θ ) L 2 ( Ω = 2 x 2 ( t 2 θ ) , x y 2 ( t θ ) L 2 ( Ω 2 y 2 ( t 2 θ ) , x y 2 ( t θ ) L 2 ( Ω = Ω y ( t 2 θ ) y ( t θ ) d x d y + Ω x ( t 2 θ ) x ( t θ ) d x d y = 1 2 d d t Ω [ y ( t 2 θ ) ] 2 d x d y + 1 2 d d t Ω [ x ( t 2 θ ) ] 2 d x d y ,
Δ 2 ( t 2 θ ) , x y 2 ( t θ ) L 2 ( Ω = 4 x 4 ( t 2 θ ) , x y 2 ( t θ ) L 2 ( Ω + 4 y 4 ( t 2 θ ) , x y 2 ( t θ ) L 2 ( Ω + 2 4 x 2 y 2 ( t 2 θ ) , x y 2 ( t θ ) L 2 ( Ω = Ω y ( t 2 θ x ) y ( t θ x ) d x d y + Ω x ( t 2 θ y ) x ( t θ y ) d x d y + Ω t 2 θ t θ ) d x d y = 1 2 d d t Ω [ y ( t 2 θ x ) ] 2 d x d y + 1 2 d d t Ω [ x ( t 2 θ y ) ] 2 d x d y + 1 2 d d t Ω [ t 2 θ ] 2 d x d y
γ t 2 θ , x y 2 ( t θ ) L 2 ( Ω = γ Ω x y ( t 2 θ ) x y ( t θ ) = γ 2 d d t Ω [ x y ( t 2 θ ) ] 2 d x d y ,
0 t K ( t s ) Δ ( t 2 θ ) d s , x y 2 ( t θ ) L 2 ( Ω = 0 t K ( t s ) 2 x 2 ( t 2 θ ) d s , x y 2 ( t θ ) L 2 ( Ω + 0 t K ( t s ) 2 y 2 ( t 2 θ ) d s , x y 2 ( t θ ) L 2 ( Ω = 0 t K ( t s ) y ( t 2 θ ) d s , y ( t θ ) L 2 ( Ω 0 t K ( t s ) x ( t 2 θ ) d s , y ( t θ ) L 2 ( Ω .
The first term on the right-hand side of (41) can be evaluated as
0 t K ( t s ) y ( t 2 θ ) d s , y ( t θ ) L 2 ( Ω = 0 t K ( t s ) Ω y ( t 2 θ ( . , . s ) ) y ( t θ ) d x d y d s = 0 t K ( t s ) Ω y ( t 2 θ ( . , . s ) ) d d t y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) d x d y d s = 0 t K ( t s ) Ω y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) d d t y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) d x d y d s + 0 t K ( t s ) Ω y ( t 2 θ ) d d t y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) d x d y d s = 1 2 0 t K ( t s ) Ω d d t [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s 1 2 0 t K ( t s ) Ω d d t [ y ( t 2 θ ] 2 d x d y d s . = 1 2 0 t d d t K ( t s ) Ω [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s 1 2 0 t K t ( t s ) Ω [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s 1 2 d d t Ω [ y ( t 2 θ ) ] 2 d x d y 0 t K ( s ) d s + 1 2 K ( t ) Ω [ y ( t 2 θ ) ] 2 d x d y .
By noting that
0 t d d t K ( t s ) Ω [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s = d d t 0 t K ( t s ) Ω [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s ,
then (42) reduces to
0 t K ( t s ) y ( t 2 θ ) d s , y ( t θ ) L 2 ( Ω = 1 2 d d t 0 t K ( t s ) Ω [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s 1 2 0 t K t ( t s ) Ω [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s 1 2 d d t Ω [ y ( t 2 θ ) ] 2 d x d y 0 t K ( s ) d s + 1 2 K ( t ) Ω [ y ( t 2 θ ) ] 2 d x d y .
Similarly, we have
0 t K ( t s ) x ( t 2 θ ) d s , y ( t θ ) L 2 ( Ω = 1 2 d d t 0 t K ( t s ) Ω [ x ( t 2 θ ( . , . s ) ) x ( t 2 θ ) ] 2 d x d y d s 1 2 0 t K t ( t s ) Ω [ x ( t 2 θ ( . , . s ) ) x ( t 2 θ ) ] 2 d x d y d s 1 2 d d t Ω [ x ( t 2 θ ) ] 2 d x d y 0 t K ( s ) d s + 1 2 K ( t ) Ω [ x ( t 2 θ ) ] 2 d x d y .
Combination of (36), (37), (38), (39), (40), (41), (44) and (45) and use of Lemma 2 [21] yields
t σ x y ( t θ ) L 2 ( Ω ) 2 + d d t Ω [ y ( t 2 θ ) ] 2 d x d y + d d t Ω [ x ( t 2 θ ) ] 2 d x d y + d d t Ω [ y ( t 2 θ x ) ] 2 d x d y + d d t Ω [ x ( t 2 θ y ) ] 2 d x d y + d d t Ω [ t 2 θ ] 2 d x d y + γ d d t Ω [ x y ( t 2 θ ) ] 2 d x d y + d d t 0 t K ( t s ) Ω [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s 0 t K t ( t s ) Ω [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s d d t Ω [ y ( t 2 θ ) ] 2 d x d y 0 t K ( s ) d s + K ( t ) Ω [ y ( t 2 θ ) ] 2 d x d y + d d t 0 t K ( t s ) Ω [ x ( t 2 θ ( . , . s ) ) x ( t 2 θ ) ] 2 d x d y d s 0 t K t ( t s ) Ω [ x ( t 2 θ ( . , . s ) ) x ( t 2 θ ) ] 2 d x d y d s d d t Ω [ x ( t 2 θ ) ] 2 d x d y 0 t K ( s ) d s + K ( t ) Ω [ x ( t 2 θ ) ] 2 d x d y . 0 .
Since K t < 0 and K > 0 , then (46) reduces to
t σ x y ( t θ ) L 2 ( Ω ) 2 + d d t Ω [ y ( t 2 θ ) ] 2 d x d y + d d t Ω [ x ( t 2 θ ) ] 2 d x d y + d d t Ω [ y ( t 2 θ x ) ] 2 d x d y + d d t Ω [ x ( t 2 θ y ) ] 2 d x d y + d d t Ω [ t 2 θ ] 2 d x d y + γ d d t Ω [ x y ( t 2 θ ) ] 2 d x d y + d d t 0 t K ( t s ) Ω [ y ( t 2 θ ( . , . s ) ) y ( t 2 θ ) ] 2 d x d y d s d d t Ω [ y ( t 2 θ ) ] 2 d x d y 0 t K ( s ) d s + d d t 0 t K ( t s ) Ω [ x ( t 2 θ ( . , . s ) ) x ( t 2 θ ) ] 2 d x d y d s d d t Ω [ x ( t 2 θ ) ] 2 d x d y 0 t K ( s ) d s 0 .
By replacing t by s and integrating over ( 0 , t ) , and using again the fact that K t < 0 and K > 0 , we obtain
D σ 1 x y ( t θ ) L 2 ( Ω ) 2 + Ω [ y ( t 2 θ ) ] 2 d x d y + Ω [ x ( t 2 θ ) ] 2 d x d y Ω [ y ( t 2 θ x ) ] 2 d x d y + Ω [ x ( t 2 θ y ) ] 2 d x d y + Ω [ t 2 θ ] 2 d x d y + Ω [ x y ( t 2 θ ) ] 2 d x d y T σ 1 C ( 1 σ ) Γ ( 1 σ ) x y ( t θ ) L 2 ( Ω )   | t = 0 + Ω [ y ( t 2 θ ) ] 2 d x d y t = 0 + Ω [ x ( t 2 θ ) ] 2 d x d y t = 0 + Ω [ y ( t 2 θ x ) ] 2 d x d y t = 0 + Ω [ x ( t 2 θ x ) ] 2 d x d y t = 0 + Ω [ t 2 θ ] 2 d x d y t = 0 + Ω [ x y ( t 2 θ ) ] 2 d x d y t = 0 ,
where C = min ( 1 , γ ) . Since all terms on the right-hand side of (48) vanish at t = 0 , then we conclude from (48) that Y = 0 almost everywhere in Λ T .
Now, we complete the proof of Theorem 2; we suppose that for some element W = ( ( F , W 1 , W 2 ) ) R ( T ) , and we have
( L Ψ , F ) L 2 ( 0 , T ; L 2 ( Ω ) ) + ( Γ 1 Ψ , W 1 ) H 1 ( Ω + ( Γ 2 Ψ , W 2 ) L 2 ( Ω ) = 0 .
We must prove that W = 0 . If we put Ψ D ( T ) satisfying homogeneous conditions initial conditions, we have
( L Ψ , F ) L 2 ( 0 , T ; L 2 ( Ω ) ) = 0 .
Applying Proposition 1 to (50), it follows from that F = 0 .
Thus, (49) reduces to
( Γ 1 Ψ , W 1 ) H 1 ( Ω + ( Γ 2 Ψ , W 2 ) L 2 ( Ω ) = 0 .
But since the range of the operators Γ 1 , Γ 2 are dense in the spaces H 1 ( Ω ) , L 2 ( Ω ) , respectively, then relation (51) implies W 1 = W 2 = 0 . Consequently, W = 0 and Theorem 2 follows.
Remark 2. 
(Concerning the density of D 0 ( T ) ). From (9), it follows that T is bounded below with closed range, and since the range of T | D 0 ( T ) reduces to zero in H * , then the closure of the range of T | D 0 ( T ) coincides with H * . Thus, with the a priori estimate and the closed range result, we conclude that the density of D 0 ( T ) in the space B .

5. Conclusions

This paper investigates the well-posedness of an initial boundary-value problem involving a fourth-order partial differential equation with a Caputo time-fractional derivative. Several significant components: the nonlocal Caputo fractional operator, the Laplace operator, the biharmonic operator, and a memory term defined through a convolution with a damping kernel K ( t ) . The combination of these elements, especially the nonlocal nature of the boundary conditions, introduces novel mathematical difficulties, necessitating the use of sophisticated analytical tools. To overcome these challenges, we develop a functional analytic framework rooted in Sobolev spaces and apply energy estimation techniques to rigorously establish the well-posedness of the problem.
The operator-theoretic framework combined with energy estimates in weighted Sobolev spaces employed here can be generalized to other higher-order fractional evolution equations—such as sixth-order models—or to time–space fractional variants, as long as the nonlocal boundary conditions maintain coercivity through Poincaré-type inequalities. This approach advances earlier nonlocal fractional studies by incorporating biharmonic stiffness together with viscoelastic memory effects, and it applies to multi-dimensional domains that go beyond two-dimensional rectangles. A promising direction for future research involves nonlinear extensions, particularly establishing well-posedness for semi-linear formulations that include reaction–diffusion terms to better represent real-world viscoelastic materials, as well as analyzing their asymptotic behavior and exponential decay rates.

Author Contributions

Methodology, R.A. and S.M.; Validation, R.A.; Investigation, R.A. and S.M.; Writing—original draft, R.A. and S.M.; Writing—review & editing, R.A. and S.M.; Supervision, S.M. All authors have read and agreed to the published version of the manuscript.

Funding

The second author extends his appreciation to Ongoing Research Funding Program Number (ORF-2026-975) at King Saud University, Riyadh, Saudi Arabia.

Data Availability Statement

No new data were created or analyzed in this study.

Conflicts of Interest

The authors declare no conflicts of interest.

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Alrajhi, R.; Mesloub, S. On a Time-Fractional Biharmonic Nonlocal Initial Boundary-Value Problem with Frictional and Viscoelastic Damping Terms. Mathematics 2026, 14, 387. https://doi.org/10.3390/math14020387

AMA Style

Alrajhi R, Mesloub S. On a Time-Fractional Biharmonic Nonlocal Initial Boundary-Value Problem with Frictional and Viscoelastic Damping Terms. Mathematics. 2026; 14(2):387. https://doi.org/10.3390/math14020387

Chicago/Turabian Style

Alrajhi, Rowaida, and Said Mesloub. 2026. "On a Time-Fractional Biharmonic Nonlocal Initial Boundary-Value Problem with Frictional and Viscoelastic Damping Terms" Mathematics 14, no. 2: 387. https://doi.org/10.3390/math14020387

APA Style

Alrajhi, R., & Mesloub, S. (2026). On a Time-Fractional Biharmonic Nonlocal Initial Boundary-Value Problem with Frictional and Viscoelastic Damping Terms. Mathematics, 14(2), 387. https://doi.org/10.3390/math14020387

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