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Article

Global Existence, General Decay, and Blow Up of the Solution to the Coupled p-Biharmonic Equation of Hyperbolic Type with Degenerate Damping Terms

1
Department of Mathematics, Echahid Cheikh Larbi Tebessi University, Tebessa 12000, Algeria
2
Laboratory of Mathematics, Informatics and Systems (LAMIS), Echahid Cheikh Larbi Tebessi University, Tebessa 12000, Algeria
3
Department of Mathematics and Computer Science, University of Oum El-Bouaghi, Oum El Bouaghi 04000, Algeria
4
Department of Mathematics, Henan University of Technology, Zhengzhou 450001, China
5
Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
*
Authors to whom correspondence should be addressed.
Mathematics 2025, 13(19), 3152; https://doi.org/10.3390/math13193152
Submission received: 23 August 2025 / Revised: 27 September 2025 / Accepted: 29 September 2025 / Published: 2 October 2025
(This article belongs to the Section C: Mathematical Analysis)

Abstract

In this work, we study a nonlinear system of p-Biharmonic hyperbolic equations with degenerate damping and source terms in a bounded domain. Under appropriate assumptions on the initial data and the damping terms, we establish the global existence of solutions. Furthermore, we derive a general decay result, and finally, we prove the occurrence of blow-up for solutions with negative initial energy.

1. Introduction

In this work, we consider the following coupled system of p-Biharmonic equation of hyperbolic type with degenerate damping and source terms:
u t t + Δ p 2 u + | u | k + | v | r | u t | m 1 u t + u t = F 1 ( u , v ) , x Ω , t > 0 , v t t + Δ p 2 v + | v | ϱ + | u | σ | v t | q 1 v t + v t = F 2 ( u , v ) , x Ω , t > 0 , u ( x , t ) = v ( x , t ) = u ν ( x , t ) = v ν ( x , t ) = 0 , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) , v ( x , 0 ) = v 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , v t ( x , 0 ) = v 1 ( x ) , x Ω ,
where Ω R n is a bounded domain with sufficiently smooth boundary Ω , ν denotes the unit outer normal derivative, p > 2 , m , q > 1 , k , r , ϱ , σ > 0 , F i : R 2 R are given functions to be specified later, u 0 , u 1 , v 0 , v 1 are given functions belonging to suitable spaces, Δ p 2 is the fourth-order operator called the p-Biharmonic operator, which is defined by Δ p 2 v = Δ | Δ v | p 2 Δ v .
Fourth-order differential equations arise in the study of deflections of elastic beams on nonlinear elastic foundations. Therefore, they have important applications in engineering and physical sciences [1,2,3]. In fact, the p-Biharmonic operator arises naturally in the study of nonlinear elastic plates and beams, where the restoring force depends nonlinearly on the curvature. For p = 2 , it reduces to the classical Biharmonic operator Δ 2 , which models thin elastic plates (Kirchhoff–Love theory). For p > 2 , it describes nonlinear elastic responses of materials subject to large deformations. The coupling through nonlinear terms like | u | k + | v | r | u t | m 1 u t represents interactions between the two modes of vibration and reflects nonlinear energy transfer within the system. Such terms naturally appear in the study of non-Newtonian fluid–structure interactions, nonlinear viscoelasticity, and mechanical systems where damping or friction depends on the amplitude of oscillations. Finally, the source terms model external forces acting on the system. From a dynamical perspective, they may also represent sustained energy input, which competes with the dissipative mechanisms. This interplay between damping and forcing is crucial in determining the system’s long-term behavior and may lead to stabilization, or in contrast, to finite-time blow-up phenomena.
In recent years, a great attention has been focused on the study of fourth-order differential problems involving biharmonic and p-Biharmonic operators.
To motivate this study, let us recall some results concerning single p-Biharmonic equation of hyperbolic type. When p = 2 , this type of equation reduces to the Petrovsky equation, which has been extensively studied, with results established on existence, nonexistence, and stability. For example, Messaoudi [4] investigated the following problem:
v t t + Δ 2 v + | v t | m 2 v t = | v | p 2 v .
He established an existence result and showed that the solution persists globally if m p , while it blows up in finite time if m < p and the initial energy is negative. Wu and Tsai [5] proved the global existence and blow-up of solution to the problem (2). Chen and Zhou [6] extended the blow-up results of [4,5] to a solution with positive initial energy. Li et al. [7] studied (2) with a strong damping term Δ v t and proved the global existence of the solution without requiring any relation between m and p, and further established exponential decay rates.
Concerning global existence, general decay, and blow-up results for systems of Petrovsky equations, Guesmia [8] considered the following coupled system of wave and Petrovsky equations:
u t t + Δ 2 u + a ( x ) v + g 1 ( u t ) = 0 , v t t Δ v + a ( x ) u + g 2 ( v t ) = 0 ,
Under suitable assumptions on the functions g i ( · ) and a ( · ) , he proved that this system is well-posed by using nonlinear semigroup theory, and (exponentially of polynomailly) stable by exploiting the multiplier method. Bahlil and Feng [9] studied (3) with a source term and proved the global existence of solutions by using the potential well method due to Payne and Sattinger [10] and Sattinger [11], combined with the Faedo–Galerkin method. They also established a more general energy decay of solutions by exploiting the multiplier method and some properties of convex functions. Recently, Saadaoui et al. [12] extended the results in [8].
Li et al. [13] considered the following coupled system of Petrovsky equations in a bounded domain:
u t t + Δ 2 u + | u t | p 1 u t = F 1 ( u , v ) , v t t + Δ 2 v + | v t | q 1 v t = F 2 ( u , v ) ,
with Dirichlet boundary conditions. Under suitable assumptions, they proved the global existence of solutions and established the uniform decay rates by means of Nakao’s inequality. Furthermore, they showed the blow-up of the solutions and provided lifespan estimates under different damping terms and initial energy conditions. Liu [14] studied the blow-up of solutions and derived lifespan estimates for the coupled Petrovsky system with a linear damping term in the first equation. Peyravi [15] obtained lower bounds for the blow-up time for weak solutions to (3). Recently, Nhan [16] investigated global existence, uniform decay, and blow-up of solutions for the same problem (3).
When p > 2 , Ferreira et al. [17] studied the following nonlinear beam equation with a strong damping and the p ( x ) -Biharmonic operator:
v t + Δ p ( x ) 2 v Δ v t = f ( x , t , v ) ,
and proved the existence of local solutions by using the Faedo–Galerkin, as well as the decay of energy based on Nakao’s method, under suitable assumptions on the variable exponent p ( · ) . Butakın and Pişkin [18] investigated the global existence and blow-up of solutions with negative initial energy for the p ( x ) -Biharmonic equation with variable exponent sources. Gheraibia et al. [19] studied the following equation:
v t t + Δ p 2 v μ Δ m v t + v t = ω | v | r 2 v ,
where p , m , r > 2 and μ , ω 0 . Under suitable assumption on the on the initial data, they proved the global existence, stability and blow-up results of solutions. Yu et al. [20] studied (6) with m = 2 , they established the local existence of the solutions by using the Galerkin method. Furthermore, the global existence, decay, and finite time blow-up of weak solutions in cases of E ( 0 ) < d and E ( 0 ) = d are studied. Recently, Gheraibia et al. [21] considered the following p-Biharmonic equation of hyperbolic type:
| v t | σ v t t + Δ p 2 v Δ v t = 0 ,
subject to the acoustic boundary conditions and delay term. The authors proved the global existence in time of solutions and established the decay rate result. Furthermore, the blow-up of the solutions in finite time with negative initial energy is shown.
Motivated by the above mentioned papers, our purpose in this research is to investigate the global existence, energy decay estimates, and blow-up of solutions to problem (1).
The rest of this article is organized as follows: In Section 2, we give some notations and material needed for this work. In Section 3, we establish a global existence result. In Section 4, we state and prove energy decay estimates. In Section 5, we show the blow-up of solutions.

2. Preliminaries

In this section, we present the assumptions, notations, and known results which will be used throughout this work. We denote by v q and ( · , · ) the usual L q ( Ω ) norm and the inner product in L 2 ( Ω ) , respectively. For Sobolev spaces norms, we adopt the following notations:
v W 0 1 , q = v q and v W 0 2 , q = Δ v q , 1 < q < + ,
where v = v x 1 , v x 2 , . . . , v x n , Δ v = k = 1 n 2 v x k 2 , and
L q ( Ω ) = { v : Ω R ; v is measurable and Ω | v | q < } .
Next, we consider the functions
F 1 ( u , v ) = a u + v 2 ( α + 1 ) ( u + v ) + b | u | α u | v | ( α + 2 ) , F 2 ( u , v ) = a u + v 2 ( α + 1 ) ( u + v ) + b | u | ( α + 2 ) | v | α v ,
where a , b are constants, and α satisfies
1 < α , if n = 1 , 2 , 1 < α 4 n n 2 , if n 3 .
We can easily verify that
u F 1 ( u , v ) + v F 2 ( u , v ) = 2 ( α + 2 ) F ( u , v ) ,
where
F ( u , v ) = 1 2 ( α + 2 ) a u + v 2 ( α + 2 ) + 2 b | u v | ( α + 2 ) .
Lemma 1
((Sobolev-Poincaré inequality) [22]). Let q be a number with 2 q < ( n = 1 , 2 , , p ) or 2 q < n p n p ( n p + 1 ) . Then, there is a constant c * = c * ( Ω , q , p ) such that
v q c ˜ * v p , v W 0 1 , p ( Ω ) .
Let c ˜ ς be the optimal constant of Sobolev embedding which satisfies the inequality
v q c ˜ ς Δ v q , v W 0 2 , q ( Ω ) .
Lemma 2
([23]). There exist two positive constants k 1 and k 2 such that
k 1 2 ( α + 2 ) | u | 2 ( α + 2 ) + | v | 2 ( α + 2 ) F ( u , v ) k 2 2 ( α + 2 ) | u | 2 ( α + 2 ) + | v | 2 ( α + 2 ) .
Lemma 3.
Assume that (8) holds. Then there exists a positive constant ζ such that
u + v 2 ( α + 2 ) 2 ( α + 2 ) + 2 u v α + 2 α + 2 ζ Δ u p p + Δ v p p 2 ( α + 2 ) p
Proof. 
The proof of this lemma is almost identical to that of the corresponding Lemma in [24], and is therefore omitted here.  □
Now, we define the energy function associated with problem (1) as follows:
E ( t ) = 1 2 u t 2 2 + 1 2 v t 2 2 + 1 p Δ u p p + 1 p Δ v p p Ω F ( u , v ) d x .
Lemma 4.
The functional E ( t ) defined in (15) satisfies
E ( t ) = u t 2 2 + v t 2 2 Ω | u | k + | v | r | u t | m + 1 d x Ω | u | ϱ + | v | σ | v t | q + 1 d x 0 .
Proof. 
Multiplying the first equation in (1) by u t , the second equation in (1) by v t , integrating by parts over Ω , and summing the resulting we get
Ω u t t u t d x + Ω v t t v t d x + Ω Δ p 2 u u t d x + Ω Δ p 2 v v t d x + Ω | u | k + | v | r | u t | m + 1 d x + Ω | u | ϱ + | v | σ | v t | q + 1 d x + u t 2 2 + v t 2 2 = Ω u t F 1 ( u , v ) + v t F 2 ( u , v ) d x ,
which gives
d d t 1 2 u t 2 2 + 1 2 v t 2 2 + Ω | Δ u | p 2 Δ u Δ u t d x + Ω | Δ v | p 2 Δ v Δ v t d x + Ω | u | k + | v | r | u t | m + 1 d x + Ω | u | ϱ + | v | σ | v t | q + 1 d x + u t 2 2 + v t 2 2 = Ω u t F 1 ( u , v ) + v t F 2 ( u , v ) d x .
Using the definition of F 1 and F 2 , and the Formula (10), we can easily find that
Ω u t F 1 ( u , v ) + v t F 2 ( u , v ) d x = d d t Ω F ( u , v ) d x .
Inserting this last equality in (17), using the fact that Ω | Δ u | p 2 Δ u Δ u t d x = d d t 1 p Δ u p p , and (15), we get (16).  □
Next, we state a local existence theorem for problem (1), whose proof follows the arguments in [25] (Theorem 1.3).
Theorem 1
(Local existence). Assume that (8) hold. Then, for any given u 0 , v 0 W 0 2 , p ( Ω ) and u 1 , v 1 L 2 ( Ω ) . Then, there exists a unique local solution ( u , v ) of problem (1):
u , v L [ 0 , T ] ; W 0 2 , p ( Ω ) , u t C [ 0 , T ] ; H 0 1 ( Ω ) L m + 1 Ω , v t C [ 0 , T ] ; H 0 1 ( Ω ) L q + 1 Ω .
We recall the following theorem (see [26], Theorem 8.1) that will be useful in the proof of the stability of system (1).
Theorem 2.
Let E : [ 0 , ) [ 0 , ) be a non-increasing function and assume that there exist two constants c > 0 and ϖ > 0 such that
t E 1 + ϖ ( s ) d s c E ( t ) , t 0 .
Then, we have, for all t 0 and some positive constants λ and τ,
E ( t ) λ e τ t , if ϖ = 0 , E ( t ) λ ( 1 + t ) 1 ϖ , if ϖ > 0 .

3. Global Existence

The aim of this section is to prove the global existence of solutions for problem (1). For this goal, we put the following functionals
I ( t ) = Δ u p p + Δ v p p 2 ( α + 2 ) Ω F ( u , v ) d x ,
and
J ( t ) = 1 p Δ u p p + 1 p Δ v p p Ω F ( u , v ) d x .
Then, we have
E ( t ) = 1 2 u t 2 2 + 1 2 v t 2 2 + J ( t ) .
In order to show our result, we first establish the following lemma.
Lemma 5.
Suppose that 2 ( α + 2 ) > p holds, and let u 0 , v 0 W 0 2 , p ( Ω ) and u 1 , v 1 L 2 ( Ω ) such that
I ( 0 ) : = I ( u 0 , v 0 ) > 0 and μ = ζ 2 p ( α + 2 ) ( 2 ( α + 2 ) p ) E ( 0 ) 2 ( α + 2 ) p p < 1 ,
then,
I ( t ) > 0 , t > 0 .
Proof. 
Since I ( 0 ) > 0 , then by continuity of ( u , v ) , there exists a time T 1 < T such that
I ( t ) 0 , t [ 0 , T 1 ] .
From (18), (19), and (23), we have
J ( t ) = ( 2 ( α + 2 ) p ) 2 p ( α + 2 ) Δ u p p + Δ v p p + 1 2 ( α + 2 ) I ( t ) ( 2 ( α + 2 ) p ) 2 p ( α + 2 ) Δ u p p + Δ v p p .
Applying (24), (20), and (16), we obtain
Δ u p p + Δ u p p 2 p ( α + 2 ) ( 2 ( α + 2 ) p ) J ( t ) 2 p ( α + 2 ) ( 2 ( α + 2 ) p ) E ( t ) 2 p ( α + 2 ) ( 2 ( α + 2 ) p ) E ( 0 ) , t [ 0 , T 1 ] .
Exploiting Lemmas 2 and 3, (20), and (25), we obtain
2 ( α + 2 ) Ω F ( u , v ) d x ζ Δ u p p + Δ v p p 2 ( α + 2 ) p = ζ Δ u p p + Δ v p p 1 + 2 ( α + 2 ) p p ζ 2 p ( α + 2 ) ( 2 ( α + 2 ) p ) E ( 0 ) 2 ( α + 2 ) p p Δ u p p + Δ v p p = μ Δ u p p + Δ v p p < Δ u p p + Δ v p p , t [ 0 , T 1 ] .
Therefore, we conclude that
I ( t ) > 0 , t [ 0 , T 1 ] .
By repeating the procedure, T 1 is extended to T. The proof is complete.  □
Theorem 3.
Assume that the conditions of Lemma 5 hold. Then, the solution ( u , v ) to problem (1) is global.
Proof. 
It suffices to show that u t 2 2 + v t 2 2 + Δ u p p + Δ v p p is bounded independently of t. From (16) and (24), we obtain
E ( 0 ) E ( t ) = 1 2 u t 2 2 + 1 2 v t 2 2 + J ( t ) 1 2 u t 2 2 + 1 2 v t 2 2 + ( 2 ( α + 2 ) p ) 2 p ( α + 2 ) Δ u p p + Δ v p p ,
which means,
u t 2 2 + v t 2 2 + Δ u p p + Δ v p p C E ( 0 ) ,
where C = C ( α , p ) is a positive constant. The proof is complete.  □

4. Decay Result

In this section, we state and prove the decay result of solution to problem (1) by using the Komornik’s method.
Lemma 6.
Assume that the conditions of Lemma 5 hold and
m a x k + m + 1 , 2 r , 2 ( m + 1 ) , ϱ + q + 1 , 2 σ , 2 ( q + 1 ) < n p n p , if n p + 1 .
Then, for any η 1 > 0 and η 2 > 0 , there exist positive constants a i , i = 1 , . . . , 5 such that
Ω u | u | k + | v | r | u t | m 1 u t d x η 1 m + 1 m + 1 c ˜ * c ˜ ς k + m + 1 a 0 + c ˜ * c ˜ ς 2 ( m + 1 ) a 2 2 Δ u p p + η 1 m + 1 2 ( m + 1 ) c ˜ * c ˜ ς 2 r a 1 Δ v p p + 2 m m + 1 η 1 m + 1 m Ω | u | k + | v | r | u t | m + 1 d x ,
and
Ω v | v | ϱ + | u | σ | v t | q 1 v t d x η 2 q + 1 q + 1 c ˜ * c ˜ ς ϱ + q + 1 a 4 + c ˜ * c ˜ ς 2 ( q + 1 ) a 5 2 Δ v p p + η 2 q + 1 2 ( q + 1 ) c ˜ * c ˜ ς 2 σ a 3 Δ u p p + 2 q q + 1 η 2 q + 1 q Ω | v | ϱ + | u | σ | v t | q + 1 d x .
Proof. 
We have
Ω u | u | k + | v | r | u t | m 1 u t d x Ω | u | k + 1 | u t | m d x + Ω | u | | v | r | u t | m d x .
By using Hölder’s, Young’s, and Sobolev–Poincaré inequalities, we get, for any η 1 > 0 ,
Ω | u | k + 1 | u t | m d x Ω | u | k | u t | m + 1 d x m m + 1 Ω | u | k + m + 1 d x 1 m + 1 m m + 1 η 1 m m + 1 Ω | u | k | u t | m + 1 d x + η 1 m + 1 m + 1 u k + m + 1 k + m + 1 m m + 1 η 1 m + 1 m Ω | u | k + | v | r | u t | m + 1 d x + η 1 m + 1 m + 1 c ˜ * c ˜ ς k + m + 1 Δ u p k + m + 1 m m + 1 η 1 m + 1 m Ω | u | k + | v | r | u t | m + 1 d x + η 1 m + 1 m + 1 c ˜ * c ˜ ς k + m + 1 a 0 Δ u p p ,
and
Ω | u | | v | r | u t | m d x Ω | v | r | u t | m + 1 d x m m + 1 Ω | v | r | u | m + 1 d x 1 m + 1 m m + 1 η 1 m + 1 m Ω | u | k + | v | r | u t | m + 1 d x + η 1 m + 1 2 ( m + 1 ) v 2 r 2 r + u 2 ( m + 1 ) 2 ( m + 1 ) m m + 1 η 1 m + 1 m Ω | u | k + | v | r | u t | m + 1 d x + η 1 m + 1 2 ( m + 1 ) c ˜ * c ˜ ς 2 r Δ v p 2 r + η 1 m + 1 2 ( m + 1 ) c ˜ * c ˜ ς 2 ( m + 1 ) Δ u p 2 ( m + 1 ) m m + 1 η 1 m + 1 m Ω | u | k + | v | r | u t | m + 1 d x + η 1 m + 1 2 ( m + 1 ) c ˜ * c ˜ ς 2 r a 1 Δ v p p + η 1 m + 1 2 ( m + 1 ) c ˜ * c ˜ ς 2 ( m + 1 ) a 2 Δ u p p ,
where
a 0 = C E ( 0 ) k + m + 1 p p , a 1 = C E ( 0 ) 2 r p p , a 2 = C E ( 0 ) 2 ( m + 1 ) p p .
Inserting (31) and (32) in (30), we get (28).
In a similar way, we obtain (29) with
a 3 = C E ( 0 ) ϱ + q + 1 p p , a 4 = C E ( 0 ) 2 σ p p , a 5 = C E ( 0 ) 2 ( q + 1 ) p p .
Theorem 4.
Assume that the conditions of Lemma 6 hold. Then, there exist two positive constant λ and τ such that
E ( t ) λ e τ t , t > 0 , if ϖ = 0 , E ( t ) λ ( 1 + t ) 1 / ϖ , t > 0 , if ϖ > 0 .
Proof. 
Multiplying the first equation in (1) by u E ϖ ( t ) , the second equation in (1) by v E ϖ ( t ) ( ϖ > 0 will be precised later), integrating over Ω × ( S , T ) , and summing up, we get
S T E ϖ ( t ) Ω [ u t t u + v t t v + u Δ p 2 u + v Δ p 2 v + u t u + v t v + u | u | k + | v | r | u t | m 1 u t d x + v | v | ϱ + | u | σ | v t | q 1 v t ] d x d t 2 ( α + 2 ) S T E ϖ ( t ) Ω F ( u , v ) d x d t = 0 ,
which gives
S T E ϖ ( t ) Ω [ d d t ( u t u + v t v ) | u t | 2 | v t | 2 + | Δ u | p + | Δ v | p v + u t u + v t v + u | u | k + | v | r | u t | m 1 u t d x + v | v | ϱ + | u | σ | v t | q 1 v t ] d x d t 2 ( α + 2 ) S T E ϖ ( t ) Ω F ( u , v ) d x d t = 0 ,
Using the definition of E ( t ) , we see that
p S T E ϖ + 1 ( t ) d t = S T E ϖ ( t ) Ω d d t ( u t u + v t v ) d x d t + p 2 + 1 S T E ϖ ( t ) u t 2 2 + v t 2 2 d t S T E ϖ ( t ) Ω u | u | k + | v | r | u t | m 1 u t d x + v | v | ϱ + | u | σ | v t | q 1 v t d x d t S T E ϖ ( t ) Ω u t u + v t v d x d t + 2 ( α + 2 ) p S T E ϖ ( t ) Ω F ( u , v ) d x d t ,
On the other hand, we have
E ϖ ( t ) Ω d d t ( u t u + v t v ) d x = d d t E ϖ ( t ) Ω ( u t u + v t v ) d x ϖ E ( t ) E ϖ 1 ( t ) Ω ( u t u + v t v ) d x
Then, inequality (35) becomes
p S T E ϖ + 1 ( t ) d t = ϖ S T E ( t ) E ϖ 1 ( t ) Ω ( u t u + v t v ) d x d t S T d d t E ϖ ( t ) Ω ( u t u + v t v ) d x d t + p 2 + 1 S T E ϖ ( t ) u t 2 2 + v t 2 2 d t S T E ϖ ( t ) Ω u t u + v t v d x d t S T E ϖ ( t ) Ω u | u | k + | v | r | u t | m 1 u t d x + v | v | ϱ + | u | σ | v t | q 1 v t d x d t + 2 ( α + 2 ) p S T E ϖ ( t ) Ω F ( u , v ) d x d t .
In what follows, we will estimate the right-hand side terms in (37). Set β = p 2 ( p 1 ) . By Young’s inequality, we have
| ϖ S T E ( t ) E ϖ 1 ( t ) Ω ( u t u + v t v ) d x d t | ϖ S T E ( t ) E ϖ 1 ( t ) Ω 1 p ( | u | p + | v | p ) + p 1 p ( | u t | p p 1 + | v t | p p 1 ) d x d t c ϖ S T E ( t ) E ϖ 1 ( t ) E ( t ) + E β ( t ) d t c ϖ S T E ( t ) E ϖ ( t ) d t c ϖ S T E ( t ) E ϖ + β 1 ( t ) d t c ϖ E ϖ + 1 ( S ) + E ϖ + β ( S ) c ϖ E ϖ + β ( S ) .
The second term can be estimated as follows:
S T d d t E ϖ ( t ) Ω ( u t u + v t v ) d x d t | Ω E ϖ ( T ) ( u ( T ) u t ( T ) + v ( T ) v t ( T ) ) d x | + | Ω E ϖ ( S ) ( u ( S ) u t ( S ) + v ( S ) v t ( S ) ) d x | E ϖ ( T ) Ω 1 p ( | u ( T ) | p + | v ( T ) | p ) + p 1 p ( | u t ( T ) | p p 1 + | v t ( T ) | p p 1 ) d x + E ϖ ( S ) Ω 1 p ( | u ( S ) | p + | v ( S ) | p ) + p 1 p ( | u t ( S ) | p p 1 + | v t ( S ) | p p 1 ) d x c E ϖ + 1 ( T ) + E ϖ + β ( T ) + c E ϖ + 1 ( S ) + E ϖ + β ( S ) c E ϖ + β ( S ) .
Now, it is easy to see that
p 2 + 1 S T E ϖ ( t ) u t 2 2 + v t 2 2 d t p 2 + 1 S T E ϖ ( t ) ( E ) ( t ) d t p 2 + 1 E ϖ + 1 ( S ) c E ϖ + β ( S ) .
Furthermore, by Young’s inequality we have, for any ε > 0 ,
| S T E ϖ ( t ) Ω u t u + v t v d x d t | S T E ϖ ( t ) Ω ε ( | u ( t ) | p + | v ( t ) | p ) + C ε ( | u t ( t ) | p p 1 + | v t ( t ) | p p 1 ) d x d t c ε S T E ϖ + 1 ( t ) d t + c C ε S T ( E ( t ) ) β E ϖ ( t ) d t
Now, by the help of Young’s inequality and by choosing ϖ = p 2 p , the term S T ( E ( t ) ) β E ϖ ( t ) d t can be estimated as
S T ( E ( t ) ) β E ϖ ( t ) d t ε 1 S T E ϖ + 1 ( t ) d t + C ε 1 S T ( E ( t ) ) d t ,
for any ε 1 > 0 .
Then, we get
| S T E ϖ ( t ) Ω u t u + v t v d x d t | c ε S T E ϖ + 1 ( t ) d t + c C ε ε 1 S T E ϖ + 1 ( t ) d t + c C ε C ε 1 S T ( E ( t ) ) d t c ε S T E ϖ + 1 ( t ) d t + c C ε ε 1 S T E ϖ + 1 ( t ) d t + c C ε C ε 1 E ( S ) .
Using Lemma 6, we infer that
S T E ϖ ( t ) Ω u | u | k + | v | r | u t | m 1 u t d x + v | v | ϱ + | u | σ | v t | q 1 v t d x d t c η 1 m + 1 m + 1 + η 2 q + 1 q + 1 S T E ϖ + 1 ( t ) d t + c η 1 , η 2 , m , q S T E ϖ ( t ) ( E ( t ) ) d t c η 1 m + 1 m + 1 + η 2 q + 1 q + 1 S T E ϖ + 1 ( t ) d t + c η 1 , η 2 , m , q E ϖ + β ( S ) .
By using (25) and (26), we get
2 ( α + 2 ) p S T E ϖ ( t ) Ω F ( u , v ) d x d t 2 ( α + 2 ) p 2 ( α + 2 ) μ S T E ϖ ( t ) Δ u p p + Δ v p p d t 2 ( α + 2 ) p 2 ( α + 2 ) 2 p ( α + 2 ) 2 ( α + 2 ) p μ S T E ϖ + 1 ( t ) d t p μ S T E ϖ + 1 ( t ) d t .
Inserting (38)–(43) in (37) and using the fact that ϖ + β > 1 , we find
( 1 μ ) p c ε c η 1 m + 1 m + 1 + η 2 q + 1 q + 1 c C ε ε 1 S T E ϖ + 1 ( t ) c E ( S ) .
Since μ < 1 , we choose ε , η 1 and η 2 small enough such that
κ = ( 1 μ ) p c ε c η 1 m + 1 m + 1 + η 2 q + 1 q + 1 > 0 .
After that, we take ε 1 small enough such that
κ c C ε ε 1 > 0 .
By these choices, we obtain
S T E ϖ + 1 ( t ) c E ( S ) .
By letting T + and applying Theorem 2, we get the desired result. This completes the proof.  □

5. Blow Up

In this section, we will prove that the solution of (1) blows up in finite time when the initial energy E ( 0 ) is negative. Throughout this section, c 0 denotes a positive constant, which may vary from line to line. The main result of this section is stated as follows:
Theorem 5.
Assume that
2 ( α + 2 ) max p , k + m + 1 , r + m + 1 , σ + q + 1 , ϱ + q + 1 ,
and E ( 0 ) < 0 holds. Then, the solution of problem (1) blows up in finite time T * .
Proof. 
Set
H ( t ) = E ( t ) .
From (15) and (16), we have
H ( t ) = E ( t ) 0 ,
and
0 < H ( 0 ) H ( t ) Ω F ( u , v ) d x , t [ 0 , T ) .
Next, we define
Ψ ( t ) = H 1 γ ( t ) + ϵ Ω ( u u t + v v t ) d x + ϵ 2 u 2 2 + v 2 2 ,
where ϵ > 0 is a small constant to be chosen later, and
0 < γ < min ( α + 1 ) 2 ( α + 2 ) , 2 α + 3 k m 2 m ( α + 2 ) , 2 α + 3 r m 2 m ( α + 2 ) , 2 α + 3 σ q 2 q ( α + 2 ) , 2 α + 3 ϱ q 2 q ( α + 2 ) .
Differentiating (48) with respect to t and using (1), we have
Ψ ( t ) = ( 1 γ ) H γ ( t ) H ( t ) + ϵ u t 2 2 + ϵ v t 2 2 ϵ Δ u p p ϵ Δ v p p ϵ Ω u | u | k + | v | r | u t | m 1 u t d x + 2 ϵ ( α + 2 ) Ω F ( u , v ) d x ϵ Ω v | v | ϱ + | u | σ | v t | q 1 v t d x .
Applying (15), (45), and Lemma 2, we find
Ψ ( t ) ( 1 γ ) H γ ( t ) H ( t ) + N 2 + 1 ϵ u t 2 2 + v t 2 2 + ϵ N p 1 Δ u p p + Δ v p p + k 1 ϵ 1 N 2 ( α + 2 ) u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) ϵ Ω u | u | k + | v | r | u t | m 1 u t d x ϵ Ω v | v | ϱ + | u | σ | v t | q 1 v t d x + N ϵ H ( t ) ,
where p < N < 2 ( α + 2 ) .
Applying Hölder’s inequality, we get
u | u t | m 1 u t η 1 m + 1 m + 1 | u | m + 1 + m m + 1 η 1 m + 1 m | u t | m + 1 ,
which implies that
Ω u | u | k + | v | r | u t | m 1 u t d x η 1 m + 1 m + 1 Ω | u | k + | v | r | u | m + 1 d x + m m + 1 η 1 m + 1 m Ω | u | k + | v | r | u t | m + 1 d x .
A similar inequality can be obtained for the penultimate term in (51). Then, (51) becomes
Ψ ( t ) ( 1 γ ) H γ ( t ) H ( t ) + N 2 + 1 ϵ u t 2 2 + v t 2 2 + N ϵ H ( t ) + ϵ N p 1 Δ u p p + Δ v p p + ϵ k 1 1 N 2 ( α + 2 ) u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) ϵ η 1 m + 1 m + 1 Ω | u | k + | v | r | u | m + 1 d x ϵ m m + 1 η 1 m + 1 m Ω | u | k + | v | r | u t | m + 1 d x ϵ η 2 q + 1 q + 1 Ω | v | ϱ + | u | σ | v | q + 1 d x ϵ q q + 1 η 2 q + 1 q Ω | v | ϱ + | u | σ | v t | q + 1 d x .
Taking η 1 = μ 1 H γ ( t ) m m + 1 and η 2 = μ 2 H γ ( t ) q q + 1 where μ 1 and μ 2 are positive constants to be specified later, we see that
Ψ ( t ) ( 1 γ ) ϵ ω H γ ( t ) H ( t ) + N ϵ H ( t ) + N 2 + 1 ϵ u t 2 2 + v t 2 2 + ϵ N p 1 Δ u p p + Δ v p p + ϵ k 1 1 N 2 ( α + 2 ) u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) ϵ μ 1 m H γ m ( t ) m + 1 Ω | u | k + | v | r | u | m + 1 d x ϵ μ 2 q H γ q ( t ) q + 1 Ω | v | ϱ + | u | σ | v | q + 1 d x ,
where ω = μ 1 m m + 1 + μ 2 q q + 1 . Exploiting Hölder’s inequality, we obtain
Ω | u | k + | v | r | u | m + 1 d x Ω | u | k + m + 1 d x + Ω | v | r | u | m + 1 d x Ω | u | k + m + 1 d x + r r + m + 1 δ 1 r + m + 1 r Ω | v | r + m + 1 d x + m + 1 r + m + 1 δ 1 r + m + 1 m + 1 Ω | u | r + m + 1 d x = u k + m + 1 k + m + 1 + r r + m + 1 δ 1 r + m + 1 r v r + m + 1 r + m + 1 + m + 1 r + m + 1 δ 1 r + m + 1 m + 1 u r + m + 1 r + m + 1 .
Similar to (55), we have
Ω | v | ϱ + | u | σ | v | q + 1 d x v ϱ + q + 1 ϱ + q + 1 + σ σ + q + 1 δ 1 σ + q + 1 σ u σ + q + 1 σ + q + 1 + q + 1 σ + q + 1 δ 1 σ + q + 1 q + 1 v σ + q + 1 σ + q + 1 .
Combining (55) and (56) with (54), we get
Ψ ( t ) ( 1 γ ) ϵ ω H γ ( t ) H ( t ) + N 2 + 1 ϵ u t 2 2 + v t 2 2 + ϵ N p 1 Δ u p p + Δ v p p + ϵ k 1 1 N 2 ( α + 2 ) u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) ϵ μ 1 m m + 1 H γ m ( t ) r r + m + 1 δ 1 r + m + 1 r v r + m + 1 r + m + 1 + m + 1 r + m + 1 δ 1 r + m + 1 m + 1 u r + m + 1 r + m + 1 ϵ μ 2 q q + 1 H γ q ( t ) σ σ + q + 1 δ 1 σ + q + 1 σ u σ + q + 1 σ + q + 1 + q + 1 σ + q + 1 δ 1 σ + q + 1 q + 1 v σ + q + 1 σ + q + 1 ϵ μ 2 q q + 1 H γ q ( t ) v ϱ + q + 1 ϱ + q + 1 ϵ μ 1 m m + 1 H γ m ( t ) u k + m + 1 k + m + 1 + N ϵ H ( t ) .
Exploiting (44), (47), and (13), we infer that
H γ m ( t ) u k + m + 1 k + m + 1 c 0 u k + m + 1 k + m + 1 u 2 ( α + 2 ) 2 γ m ( α + 2 ) + v 2 ( α + 2 ) 2 γ m ( α + 2 ) c 0 u 2 ( α + 2 ) 2 γ m ( α + 2 ) + k + m + 1 + u k + m + 1 k + m + 1 v 2 ( α + 2 ) 2 γ m ( α + 2 ) .
Applying (49) and the following algebraic inequality
A ν A + 1 ( 1 + 1 β ) A + β , A 0 , 0 < ν 1 , β 0 ,
we have, for all t 0 ,
u 2 ( α + 2 ) 2 γ m ( α + 2 ) + k + m + 1 d u 2 ( α + 2 ) 2 ( α + 2 ) + H ( 0 ) d u 2 ( α + 2 ) 2 ( α + 2 ) + H ( t ) ,
where d = 1 + 1 H ( 0 ) . By using Young’s inequality, Sobolev embedding, and (49), we get
u k + m + 1 k + m + 1 v 2 ( α + 2 ) 2 γ m ( α + 2 ) c 0 u 2 ( α + 2 ) k + m + 1 v 2 ( α + 2 ) 2 γ m ( α + 2 ) c 0 u 2 ( α + 2 ) 2 γ m ( α + 2 ) + k + m + 1 2 ( α + 2 ) + v 2 ( α + 2 ) 2 γ m ( α + 2 ) + k + m + 1 2 ( α + 2 ) 2 ( α + 2 ) c 0 u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) .
Using the same arguments, we can derive analogous inequalities for the other terms of (57), which yield to
Ψ ( t ) ( 1 γ ) ϵ ω H γ ( t ) H ( t ) + N 2 + 1 ϵ u t 2 2 + v t 2 2 + ϵ N p 1 Δ u p p + Δ v p p + ϵ [ k 1 1 N 2 ( α + 2 ) c 0 μ 1 m 1 + r r + m + 1 δ 1 r + m + 1 r + m + 1 r + m + 1 δ 1 r + m + 1 m + 1 c 0 μ 2 q 1 + σ σ + q + 1 δ 1 σ + q + 1 σ + q + 1 σ + q + 1 δ 1 σ + q + 1 q + 1 ] u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) + ϵ [ N c 0 μ 1 m 1 + r r + m + 1 δ 1 r + m + 1 r + m + 1 r + m + 1 δ 1 r + m + 1 m + 1 c 0 μ 2 q 1 + σ σ + q + 1 δ 1 σ + q + 1 σ + q + 1 σ + q + 1 δ 1 σ + q + 1 q + 1 ] H ( t ) .
At this point, we carefully choose our constants. First, we take μ 1 and μ 2 sufficiently large such that
k 1 1 N 2 ( α + 2 ) c 0 μ 1 m 1 + r r + m + 1 δ 1 r + m + 1 r + m + 1 r + m + 1 δ 1 r + m + 1 m + 1 c 0 μ 2 q 1 + σ σ + q + 1 δ 1 σ + q + 1 σ + q + 1 σ + q + 1 δ 1 σ + q + 1 q + 1 > 0 ,
and
N c 0 μ 1 m 1 + r r + m + 1 δ 1 r + m + 1 r + m + 1 r + m + 1 δ 1 r + m + 1 m + 1 c 0 μ 2 q 1 + σ σ + q + 1 δ 1 σ + q + 1 σ + q + 1 σ + q + 1 δ 1 σ + q + 1 q + 1 > 0 .
Once μ 1 and μ 2 are fixed, we select ϵ > 0 small enough so that
( 1 γ ) ϵ ω > 0 and Ψ ( 0 ) = H 1 γ ( 0 ) + ϵ Ω u 0 u 1 + v 0 v 1 d x + ϵ 2 u 0 2 2 + v 0 2 2 > 0 .
Consequently, inequality (62) becomes
Ψ ( t ) ϖ 1 u t 2 2 + v t 2 2 + Δ u p p + Δ v p p + u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) + H ( t ) ,
where ϖ 1 is a positive constant. Thus, we have
Ψ ( t ) Ψ ( 0 ) > 0 , t 0 .
On the other hand, from (48) we have
Ψ 1 1 γ ( t ) c 0 H ( t ) + Ω u u t d x 1 1 γ + Ω v v t d x 1 1 γ + u 2 2 1 γ + v 2 2 1 γ .
Applying Hölder’s and Young’s inequalities, we have
Ω u u t d x 1 1 γ c 0 u 2 ( α + 2 ) 1 1 γ u t 2 1 1 γ c 0 u 2 ( α + 2 ) ν 1 γ + u t 2 θ 1 γ .
By taking θ = 2 ( 1 γ ) , which gives ν 1 γ = 2 1 2 γ , we deduce
Ω u u t d x 1 1 γ + Ω v v t d x 1 1 γ c 0 u t 2 2 + u 2 ( α + 2 ) 2 1 2 γ + v t 2 2 + v 2 ( α + 2 ) 2 1 2 γ = c 0 u t 2 2 + u 2 ( α + 2 ) 2 ( α + 2 ) 1 ( 1 2 γ ) ( α + 2 ) + v t 2 2 + v 2 ( α + 2 ) 2 ( α + 2 ) 1 ( 1 2 γ ) ( α + 2 ) .
By the help of (49), we have 1 ( α + 2 ) ( 1 2 γ ) 1 and 1 ( α + 2 ) ( 1 γ ) 1 . Hence, using the algebraic inequality (59), we obtain
u 2 ( α + 2 ) 2 ( α + 2 ) 1 ( 1 2 γ ) ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) 1 ( 1 2 γ ) ( α + 2 ) d u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) + 2 H ( t ) ,
and
u 2 2 1 γ + v 2 2 1 γ d u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) + 2 H ( t ) .
By substituting (66)–(68) into (65), we get
Ψ 1 1 γ ( t ) c 0 u t 2 2 + v t 2 2 + u 2 ( α + 2 ) 2 ( α + 2 ) + v 2 ( α + 2 ) 2 ( α + 2 ) + H ( t ) .
It follows from (63) and (69) that
Ψ ( t ) κ Ψ 1 1 γ ( t ) , t > 0 ,
where κ is a positive constant. A simple integration of (70) over ( 0 , t ) yields
Ψ γ 1 γ ( t ) 1 Ψ γ 1 γ ( 0 ) κ γ t 1 γ .
Consequently, the solution of problem (1) blows up in finite time T * and T * 1 γ κ γ Ψ γ 1 γ ( 0 ) .  □

6. Conclusions

In this work, we have investigated the qualitative behavior of solutions to a nonlinear system of p-Biharmonic hyperbolic equations with degenerate damping and source terms. Our analysis established the global existence of solutions under suitable assumptions on the nonlinearities and the initial data. Moreover, by using integral inequalities, we obtained an exponential stability result. On the other hand, we also identified conditions under which finite-time blow-up occurs. The results presented here enrich the theory of higher-order nonlinear coupled hyperbolic equations with nonstandard damping mechanisms and provide a rigorous framework for understanding the interplay between stability and instability in such models.
Possible extensions of this work include the study of more general boundary conditions, the inclusion of memory-type or viscoelastic damping, and the treatment of variable exponent p ( x ) -Biharmonic operators, which may better reflect heterogeneous physical media. Another interesting direction would be to explore numerical simulations that illustrate the qualitative behaviors predicted by our theoretical results.

Author Contributions

Writing—original draft, N.B. and B.G.; Writing—review and editing, H.Z. and Z.H.; Supervision, H.Z. All authors have read and agreed to the published version of the manuscript.

Funding

Zayd Hajjej is supported by Ongoing Research Funding program (ORF-2025-736), King Saud University, Riyadh, Saudi Arabia.

Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

Conflicts of Interest

The authors declare that there is no conflict of interest.

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Boumaza, N.; Gheraibia, B.; Zhang, H.; Hajjej, Z. Global Existence, General Decay, and Blow Up of the Solution to the Coupled p-Biharmonic Equation of Hyperbolic Type with Degenerate Damping Terms. Mathematics 2025, 13, 3152. https://doi.org/10.3390/math13193152

AMA Style

Boumaza N, Gheraibia B, Zhang H, Hajjej Z. Global Existence, General Decay, and Blow Up of the Solution to the Coupled p-Biharmonic Equation of Hyperbolic Type with Degenerate Damping Terms. Mathematics. 2025; 13(19):3152. https://doi.org/10.3390/math13193152

Chicago/Turabian Style

Boumaza, Nouri, Billel Gheraibia, Hongwei Zhang, and Zayd Hajjej. 2025. "Global Existence, General Decay, and Blow Up of the Solution to the Coupled p-Biharmonic Equation of Hyperbolic Type with Degenerate Damping Terms" Mathematics 13, no. 19: 3152. https://doi.org/10.3390/math13193152

APA Style

Boumaza, N., Gheraibia, B., Zhang, H., & Hajjej, Z. (2025). Global Existence, General Decay, and Blow Up of the Solution to the Coupled p-Biharmonic Equation of Hyperbolic Type with Degenerate Damping Terms. Mathematics, 13(19), 3152. https://doi.org/10.3390/math13193152

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