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Article

Existence of Weak Solutions for the Class of Singular Two-Phase Problems with a ψ-Hilfer Fractional Operator and Variable Exponents

1
Laboratory of Mathematics, Informatics and Systems (LAMIS), Echahid Cheikh Larbi Tebessi University, Tebessa 12002, Algeria
2
Department of Mathematics, College of Science, Qassim University, Buraydah 52571, Saudi Arabia
3
Institute of Energy Infrastructure (IEI), Department of Civil Engineering, College of Engineering, Universiti Tenaga Nasional (UNITEN), Putrajaya Campus, Jalan IKRAM-UNITEN, Kajang 43000, Selangor, Malaysia
4
Mathematics Research Center, Near East University TRNC, Mersin 10, Nicosia 99138, Turkey
5
Department of Management Information Systems, College of Business and Economics, Qassim University, Buraydah 51452, Saudi Arabia
6
Department of Mathematics and Statistics, Faculty of Management Technology and Information Systems, Port Said University, Port Said 42511, Egypt
*
Author to whom correspondence should be addressed.
Fractal Fract. 2024, 8(6), 329; https://doi.org/10.3390/fractalfract8060329
Submission received: 9 May 2024 / Revised: 28 May 2024 / Accepted: 29 May 2024 / Published: 30 May 2024

Abstract

:
In this paper, we prove the existence of at least two weak solutions to a class of singular two-phase problems with variable exponents involving a ψ-Hilfer fractional operator and Dirichlet-type boundary conditions when the term source is dependent on one parameter. Here, we use the fiber method and the Nehari manifold to prove our results.

1. Introduction and Background

Numerous researchers have explored the existence of solutions to boundary value problems from various perspectives by utilizing a range of mathematical tools and methodologies [1,2]. In this context, partial differential equations and variational problems using a two-phase operator have attracted many researchers [3,4,5]. This research has shed light on various fields of application, including but not limited to anisotropic materials, Lavrentiev’s phenomenon, and elasticity theory. The study of mathematical problems involving variable exponents lies in the modeling of many physical applications: for example, image processing (see Radulescu [6]), space technology, the field of robotics, and electrorheological fluids. Winslow’s [7] study of electrofluids, which was highlighted at the beginning of the last century, has very important properties related to an electric field having an effect on the viscosity of a liquid. Furthermore, it was discovered that viscosity is inversely proportional to the strength of the electric field. This is called the Winslow effect. For more information, see the work by Halsey [8]. A summary of Radulescu’s [6] work on electrorheological fluids and image restoration via Gaussian smoothing has also been provided by Chen et al. [9].
This paper deals with the p ξ -Laplacian fractional singular two-phase equation:
R p . , q . κ . = g x ξ x α + ϱ ξ x τ 1 , in Δ ξ > 0 , in Δ ξ = 0 , on Δ
where Δ = 0 , T × 0 , T , ϱ > 0 is the control parameter, and R p x , q x κ x denotes the two-phase operator given by
R p x , q x κ x : = H D T γ , δ , ψ H D 0 + γ , δ , ψ ξ x p x 2 H D 0 + γ , δ , ψ ξ x + κ x H D 0 + γ , δ , ψ ξ x q x 2 H D 0 + γ , δ , ψ ξ x ,
where H D T γ , δ , ψ . and H D 0 + γ , δ , ψ . are the ψ -Hilfer fractional operators of order 1 p x < γ x < 1 and type δ   0 δ 1 .
In addition, we make the following assumptions:
A
a The functions p . , q . , τ . C Δ ¯ verify the following assumptions:
1 < q q + < p p + < τ τ + < + ,
where
p + = sup x Δ ¯ p x , p = inf x Δ ¯ p x , q + = sup x Δ ¯ q x , q = inf x Δ ¯ p x , τ + = sup x Δ ¯ τ x , τ = inf x Δ ¯ τ x ,
p q + < 3 2 .
b   κ x : Δ ¯ 1 ; + is a Lipschitz continuous function.
c   g x L Δ and g x 0 for all x Δ .
d   0 < α x C Δ ¯ ,   0 < α α + < 1 .
In the literature, many researchers have worked on the problems of two-phase operators with different assumptions. In [4], Liu and Dai proved the existence and multiplicity of solutions to the two-phase problem of the form
d i v u p 2 u + a x u q 2 u = f x , u , in Δ , u = 0 , on Δ ,
where Δ is a bounded domain with a smooth boundary, N 2 ,   1 < p < q ,   q p < 1 + 1 N ,   a . : Δ ¯ 0 ; + is Lipschitz continuous, and f fulfills certain conditions. You can also see the work presented by the researchers in [10,11] with different ideas. In [12], the existence of positive solutions to a class of two-phase Dirichlet equations that have combined effects of the singular term and the parametric linear term is studied. The reader can be referred to many other papers that discuss two-phase problems [13,14]. It is also well-known that the Nehari manifold method is a significant analytical tool in the field of nonlinear analysis and partial differential equations, particularly for its utility in variational problems [15,16].
Recently, fractional differential equation modeling has led to significant development in several fields due to the important results obtained (see [17,18]). This is due to the fact that fractional differential equations have several applications in many models, for example, in physics, engineering, mechanics, and medicine, which has led to great interest in these equations from a mathematical viewpoint (for example, [4,19]). In [20], a fractional boundary value problem has been investigated for the existence of the solutions with the help of critical point theory. The authors of [18] introduced the ψ -Hilfer fractional operator with several examples. In [21], the authors constructed a H p α , β , ψ 0 , T , R space, and a variational approach is used to address a complex problem involving the ψ -Hilfer fractional operator.
In [22], the authors discussed the existence and nonexistence of weak solutions to a nonlinear problem with a fractional p-Laplacian operator problem:
H D T α , β , ψ D 0 + α , β , ψ ξ t p 2 D 0 + α , β , ψ ξ t = λ ξ t p 2 ξ t + b x ξ t q 2 ξ t , I 0 + β β 1 , ψ ξ 0 = I T β β 1 , ψ ξ T ,
where 1 p < α < 1 ,   0 β 1 ,   1 < q < p 1 < ,   b L 0 , T , and λ > 0 . The researchers used the Nehari manifold technique and combined it with fiber maps in this work.
The existence of solutions for a singular double-phase problem involving a ψ -Hilfer fractional operator has been established through the utilization of the Nehari manifold [23]. In the reference [24], Ezati and Nyamoradi, using the genus properties of critical point theory, studied the existence and multiplicity of solutions of the Kirchhoff equation ψ -Hilfer fractional operator p -Laplacian.
Many models of fractional differential equations have been worked on by researchers using variational problems that include fractional operators: for example, Nyamoradi and Tayyebi [25], Ghanmi and Zhang [26], Kamache et al. [27], and Sousa et al. [21,23]. For example, in [27], Kamache et al. discussed a class of perturbed nonlinear fractional p-Laplacian differential systems and proved the existence of three weak solutions by using the variational method and Ricceri’s critical point theorems. On the other hand, in [23], Sousa et al. investigated a new class of two-phase single p-Laplacian equation problems with a fractional ψ -Hilfer operator incorporated from a parametric term. Using the fiber method with a Nehari manifold, they proved that there are at least two weak solutions to such problems when the parameter is small enough. Sousa et al. in [28] presented existence and multiplicity results for a new mean curvature operator that includes a ψ -Hilfer fractional operator, variable exponents, and appropriate fractional spaces by using the Nehari manifold technique.
Motivated by the above works, we study the existence and multiplicity of solutions for a class of fractional singular two-phase problem involving a ψ -Hilfer fractional operator with variable exponents by using the fiber method and the Nehari manifold in the Sobolev spaces with variable exponents. The novelty of this work lies in its focus on a fractional singular two-phase problem involving a ψ -Hilfer fractional operator with a variable exponent. The study stands out by employing the fiber method along with the Nehari manifold within the context of Sobolev spaces with variable exponents.
Now, we give the definition of the weak solution to Problem (1):
Definition 1. 
A positive function ξ H q ξ γ , δ , ψ Δ is called a weak solution for Problem (1) if
Δ H D 0 + γ , δ , ψ ξ x p x 2 H D 0 + γ , δ , ψ ξ x + κ x H D 0 + γ , δ , ψ ξ x q x 2 H D 0 + γ , δ , ψ ξ x × H D 0 + γ , δ , ψ ζ x d x = Δ g x ξ x α ζ x d x + ϱ Δ ξ x τ 1 ζ x d x .
is satisfied for every ζ H q ξ γ , δ , ψ Δ .
Let the energy function related to Problem (1) be
J ϱ γ , δ , ψ ξ = Δ 1 p x H D 0 + γ , δ , ψ ξ x p x d x + Δ κ x q x H D 0 + γ , δ , ψ ξ x q x d x Δ 1 1 α x ξ x 1 α x d x Δ ϱ τ x ξ x τ x d x .
We mention here our main results of this work:
Theorem 1. 
Assuming A is true. A ϱ 0 > 0 exists such that, for all ϱ 0 , ϱ 0 , Equation (1) has at least two positive weak solutions ξ , η H p ξ γ , δ , ψ Δ such that J ϱ γ , δ , ψ ξ < 0 J ϱ γ , δ , ψ η .
The rest of the paper is organized as follows: In Section 2, we present some preliminary results for Lebesgue spaces with variable exponents, and the fractional space H p ξ γ , δ , ψ Δ and its properties are defined (see [28]). In Section 3, the existence and multiplicity of solutions to Equation (1) is proven.

2. Preliminaries

Let Δ = 0 , T × 0 , T be a bounded domain in R 2 . We denote by ( Δ ) the space of all measurable functions. Consider
C + Δ = q ξ q ξ C Δ ¯ , q ξ > 1 for all ξ Δ ¯ .
The weighted variable exponent Lebesgue space L κ x q ξ Δ is defined by
L κ x q ξ Δ : = ξ : ( Δ ) R ; Δ κ x ξ x q ξ d ξ < + ,
endowed with
ξ q ξ , κ x = inf λ > 0 ; Δ κ x ξ x λ q ξ d ξ 1 .
We can define a weighted modular on L κ x q ξ Δ as the mapping ρ q ξ , κ x : L κ x q ξ Δ R as follows
ρ q ξ , κ x ξ = Δ κ x ξ x q ξ d ξ .
Definition 2 
([28]). Let 0 < γ 1 ,   0 δ 1 , and q C + Δ . The left-sided ψ-fractional derivative space H q ξ γ , δ , ψ Δ is defined with the following norm:
f H q ξ γ , δ , ψ Δ = inf λ > 0 : Δ f ξ λ q ξ d ξ + Δ H D 0 + γ , δ , ψ f ξ λ q ξ d ξ 1 ,
where H D 0 + γ , δ , ψ is the ψ-Hilfer fractional operator with 0 < γ 1 and type δ 0 δ 1 , which is given by
H q ξ γ , δ , ψ Δ = f L q ξ Δ ; H D 0 + γ , δ , ψ f L q ξ Δ , f = 0 , in Δ .
For further consideration, the reader can refer to [28] and the references therein.
Proposition 1 
([23]). Let 0 < γ 1 ,   0 δ 1 , and q ξ > 1 . The ψ-Hilfer fractional derivative space H q ξ γ , δ , ψ Δ is a reflexive and separable Banach space.
The following results are very useful for use in the rest of our paper.
Proposition 2 
([4]). Assume a of A is achieved and let z L q ξ Δ , ξ n L q ξ Δ ,   n N ; then, make following assertions:
1   z H < 1   resp . > 1 , = 1 if and only if ρ q ξ < 1 resp . > 1 , = 1 .
2 For z L q ξ Δ { 0 } ,   z q ξ = λ if and only if ρ q ξ z λ = 1 .
3 If z q ξ < 1 , then z q ξ q + < ρ q ξ z < z q ξ q .
4 If z H > 1 , then z q ξ q < ρ q ξ z < z q ξ q + .
5   z n z q ξ 0 if and only if ρ q ξ z n z 0 .
To compare the functionals . q . and ρ q . . , the following inequality is satisfied:
ρ p . ξ K p + C + 1 p + ρ q . H D 0 + γ , δ , ψ ξ p + q .
where C is the Poincaré constant, and 0 < K = K N , λ , Δ , q + , q . For further consideration, we refer the reader to the work [29].
Proposition 3 
([29]). Let q x ,   p x C Δ ¯ L Δ . Assume that
q x < 2 , p x < τ x < 2 q x 2 q x , for q Δ . ¯
Then, the embedding
H q ξ γ , δ , ψ Δ L p x Δ
and
H q ξ γ , δ , ψ Δ L τ x Δ
are continuous and compact.

3. Main Results

In this section, we prove Theorem 1 based on the Nehari manifold method, which is consistent with our problem.
Since the presence of the singular term g x ξ 1 α x in (1) implies that the energy functional J ϱ γ , δ , ψ associated with Problem (1) as defined in (5) is not C 1 , we consider the Nehari manifold of the energy functional J ϱ γ , δ , ψ :
E ϱ = ξ H q ξ γ , δ , ψ Δ { 0 } : H D 0 + γ , δ , ψ ξ x L p x p x + Δ κ x H D 0 + γ , δ , ψ ξ x q x d x = Δ g x ξ x 1 α x d x + ϱ ξ τ x τ x .
Since E ϱ H q ξ γ , δ , ψ Δ and it contains the non-trivial solutions of (1), we consider following decomposition of E ϱ :
E ϱ + = { ξ E ϱ : Δ p x α x 1 H D 0 + γ , δ , ψ ξ x p x d x + Δ q x + α x 1 κ x H D 0 + γ , δ , ψ ξ x q x d x ϱ Δ τ x + α x 1 ξ τ x d x > 0 }
E ϱ 0 = { ξ E ϱ : Δ p x α x 1 H D 0 + γ , δ , ψ ξ x p x d x + Δ q x + α x 1 κ x H D 0 + γ , δ , ψ ξ x q x d x = ϱ Δ τ x + α x 1 ξ τ x }
E ϱ = { ξ E ϱ : Δ p x α x 1 H D 0 + γ , δ , ψ ξ x p x d x + Δ q x + α x 1 κ x H D 0 + γ , δ , ψ ξ x q x d x ϱ Δ τ x + α x 1 ξ τ x d x < 0 }
Note that since κ . : Δ ¯ 1 ; + , κ 0 > 0 : κ x > κ 0 for x Δ .
Lemma 1. 
Assuming A is achieved, then J ϱ γ , δ , ψ E ϱ + is coercive.
Proof. 
Let ξ E ϱ with ξ > 1 . Using the definition of E ϱ , we have
ϱ Δ ξ x τ x d x = Δ H D 0 + γ , δ , ψ ξ x p x d x + Δ κ x H D 0 + γ , δ , ψ ξ x q x d x Δ g x ξ x 1 α x d x .
Hence, from (2), Propositions 1, 2, and using Inequality (6), we obtain
J ϱ γ , δ , ψ ξ = Δ 1 p x H D 0 + γ , δ , ψ ξ x p x d x + Δ κ x q x H D 0 + γ , δ , ψ ξ x q x d x Δ 1 1 α x g x ξ 1 α x d x Δ ϱ τ x ξ τ x d x 1 p + Δ H D 0 + γ , δ , ψ ξ x p x d x + 1 q + Δ κ x H D 0 + γ , δ , ψ ξ x q x d x 1 1 α + Δ g x ξ x 1 α x d x ϱ τ Δ ξ τ x d x = 1 p + Δ H D 0 + γ , δ , ψ ξ x p x d x + 1 q + Δ κ x H D 0 + γ , δ , ψ ξ x q x d x
1 1 α + Δ g x ξ x 1 α x d x 1 τ Δ H D 0 + γ , δ , ψ ξ x p x d x + Δ κ x H D 0 + γ , δ , ψ ξ x q x d x Δ g x ξ x 1 α x d x 1 C 1 p + 1 τ + κ 0 K p + C + 1 p + 1 q + 1 τ ϱ p . ξ + 1 τ 1 1 α + Δ g x ξ x 1 α x d x C 1 ξ p + + C 2 ξ 1 α + ,
for some C 1 , C 2 > 0 with q + < p + < τ , and C is the Poincaré constant.
From the conditions 0 < α + < 1 and 1 α + < 1 < p + , we get the result. □
Lemma 2. 
Assuming A is achieved and E ϱ + , if ς ϱ + = inf E ϱ + J ϱ γ , δ , ψ , then ς ϱ + < 0 .
Proof. 
Let ξ E ϱ + . First, by the definition of E ϱ + , we get
ϱ ξ τ x τ x < p + + α + 1 τ + α 1 H D 0 + γ , δ , ψ ξ x p x + q + + α + 1 τ + + α 1 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x .
Since E ϱ + E ϱ , we get
J ϱ γ , δ , ψ ξ = Δ 1 p x H D 0 + γ , δ , ψ ξ x p x d x + Δ κ x q x H D 0 + γ , δ , ψ ξ x q x d x Δ 1 1 α x ξ 1 α x d x Δ ϱ τ x ξ τ x d x 1 p Δ H D 0 + γ , δ , ψ ξ x p x d x + 1 q Δ κ x H D 0 + γ , δ , ψ ξ x q x d x 1 1 α + Δ g x ξ x 1 α x d x ϱ τ + Δ ξ τ x d x = 1 p Δ H D 0 + γ , δ , ψ ξ x p x d x + 1 q Δ κ x H D 0 + γ , δ , ψ ξ x q x d x 1 1 α + Δ H D 0 + γ , δ , ψ ξ x p x d x + Δ κ x H D 0 + γ , δ , ψ ξ x q x d x ϱ Δ ξ τ x d x ϱ τ + Δ ξ τ x d x
= 1 p 1 1 α + Δ H D 0 + γ , δ , ψ ξ x p x d x + 1 q 1 1 α + Δ κ x H D 0 + γ , δ , ψ ξ x q x d x + ϱ 1 1 α + + 1 τ + Δ ξ τ x d x 1 p 1 1 α + + 1 1 α + + 1 τ + p + + α + 1 τ + α 1 Δ H D 0 + γ , δ , ψ ξ x p x d x + 1 q 1 1 α + + 1 1 α + 1 τ + q + + α + 1 τ + α 1 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x p + + α + 1 1 α 1 τ + 1 p Δ H D 0 + γ , δ , ψ ξ x p x d x + κ 0 q + + α + 1 1 α 1 τ + 1 q Δ H D 0 + γ , δ , ψ ξ x q x d x < 0 .
Since q < p < τ + ,   J ϱ γ , δ , ψ E ϱ + < 0 , and so ς ϱ + < 0 .
Lemma 3. 
Assuming A is achieved, then there exists ϱ > 0 such that E ϱ 0 = for all ϱ 0 , ϱ .
Proof. 
Suppose that for every ϱ > 0 , ϱ 0 , ϱ with E ϱ 0 . In this sense, for ϱ > 0 , we have ξ E ϱ 0 such that
p + + α + 1 Δ H D 0 + γ , δ , ψ ξ x p x d x + Δ q + + α + 1 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x = ϱ Δ τ + α 1 ξ τ x d x .
Since ξ E ϱ , we have
τ + + α + 1 Δ H D 0 + γ , δ , ψ ξ x p x d x + Δ τ + + α + 1 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x = τ + α 1 Δ g x ξ x 1 α x d x + ϱ Δ τ + α 1 ξ τ x d x ,
Subtracting (7) from (8), we find
p + τ + Δ H D 0 + γ , δ , ψ ξ x p x d x + q + τ + Δ κ x H D 0 + γ , δ , ψ ξ x q x d x = τ + α 1 Δ g x ξ x 1 α x d x .
According to Theorem 13.17 of Hewitt–Stromberg [30] and Propositions 1 and 2, we deduce from (8) that
min ξ p + , ξ q + C 3 ξ 1 α
for some C 3 > 0 . Since, 1 α < q + < p + ,
ξ C 4
for some C 4 > 0 .
From (7) and Proposition 2, we have
ξ τ 1 ϱ C 5 min ξ p + , ξ q +
for some C 5 > 0 . Consequently,
ξ 1 ϱ C 5 1 τ p + or ξ 1 ϱ C 5 1 τ q + .
If ϱ 0 + , then ξ + due to the inequality q + < p + < τ + . This contradicts (10). □
Lemma 4. 
Suppose that A is satisfied. Then there exists ϱ ¯ 0 , ϱ such that E ϱ + for all ϱ 0 , ϱ ¯ . In addition, for any ϱ 0 , ϱ ¯ , there exists ξ E ϱ + such that J ϱ γ , δ , ψ ξ = ς ϱ + < 0 and ξ 0 for all x Δ .
Proof. 
Let ξ H q ξ γ , δ , ψ Δ { 0 } and the function ϕ ξ : 0 , R be defined by
ϕ ξ s = s p τ Δ H D 0 + γ , δ , ψ ξ x p x d x s τ α + 1 Δ g x ξ x 1 α x d x .
Since τ p < τ + α 1 , we can find s ¯ 0 > 0 such that
ϕ ξ s ¯ 0 = max s > 0 ϕ ξ s .
Thus, ϕ ξ s ¯ 0 = 0 , and therefore,
p τ s ¯ 0 p τ 1 Δ H D 0 + γ , δ , ψ ξ x p x d x + τ + α 1 s ¯ 0 τ α Δ g x ξ x 1 α x d x = 0
Hence,
s ¯ 0 = τ + α 1 Δ g x ξ x 1 α x d x p τ Δ H D 0 + γ , δ , ψ ξ x p x d x 1 p + α 1
And so
ϕ ξ s ¯ 0 = τ p Δ H D 0 + γ , δ , ψ ξ x p x d x τ p p + α 1 τ + α 1 Δ g x ξ x 1 α x d x τ p p + α 1 Δ H D 0 + γ , δ , ψ ξ x p x d x τ p Δ H D 0 + γ , δ , ψ ξ x p x d x τ + α 1 p + α 1 τ + α 1 Δ g x ξ x 1 α x d x τ + α 1 p + α 1 Δ g x ξ x 1 α x d x
= τ p τ p p + α 1 Δ H D 0 + γ , δ , ψ ξ x p x d x τ + p 1 p + α 1 τ + α 1 τ p p + α 1 Δ g x ξ x 1 α x d x τ p p + α 1 τ p τ + α 1 p + α 1 Δ H D 0 + γ , δ , ψ ξ x p x d x τ + α 1 p + α 1 τ + α 1 τ + α 1 p + α 1 Δ g x ξ x 1 α x d x τ p p + α 1 = τ + α 1 τ p . τ p τ + α 1 p + α 1 τ + α 1 τ + α 1 p + α 1 τ p τ + α 1 p + α 1 τ + α 1 τ + α 1 p + α 1
× Δ H D 0 + γ , δ , ψ ξ x p x d x τ + p 1 p + α 1 Δ g x ξ x 1 α x d x τ p p + α 1 = p + α 1 τ p . τ p + τ + α 1 τ + α 1 p + α 1 Δ H D 0 + γ , δ , ψ ξ x p x d x τ + p 1 p + α 1 Δ g x ξ x 1 α x d x τ p p + α 1
Let S be the best Sobolev constant; then by Proposition 2, we have for ξ < 1
S ξ p + Δ H D 0 + γ , δ , ψ ξ x p x d x .
Moreover, we have
Δ g x ξ x 1 α x d x C 6 ξ 1 α , for all C 6 > 0 .
Further, (12) and (13) above imply
ϕ ξ s ¯ 0 ϱ Δ ξ x τ d x = p + α 1 τ p . τ p + τ + α 1 τ + α 1 p + α 1 Δ H D 0 + γ , δ , ψ ξ x p x d x τ + p 1 p + α 1 Δ g x ξ x 1 α x d x τ p p + α 1 ϱ Δ ξ x τ d x p + α 1 τ p . τ p τ + α 1 τ + α 1 p + α 1 S τ + p 1 p + α 1 ξ p + τ + p 1 p + α 1 C 6 ξ 1 α τ p p + α 1 ϱ C 7 ξ τ + C 8 ϱ C 7 ξ τ + , for some C 7 , C 8 > 0 .
Therefore, ϱ ¯ 0 , ϱ independent of ξ such that
ϕ ξ s ¯ 0 ϱ Δ ξ x τ d x > 0 , for all ϱ 0 , ϱ ¯ .
Let χ ξ : 0 , + R be given by
χ ξ s = s p τ Δ H D 0 + γ , δ , ψ ξ x p x d x + s q τ Δ κ x H D 0 + γ , δ , ψ ξ x q x d x s τ α + 1 Δ g x ξ x 1 α x d x ,
for all s > 0 .
Since
τ p < τ q < τ + α 1 ,
we can find s 0 > 0 such that
χ ξ s 0 = max s > 0 χ ξ s
Because χ ξ ϕ ξ and by (14), we can find ϱ ¯ 0 , ϱ independent of ξ such that
χ ξ s 0 ϱ Δ ξ x τ x d x > 0 , for all ϱ 0 , ϱ ¯ .
So there is s 1 < s 0 < s 2 such that
χ ξ s 1 = ϱ Δ ξ x τ x d x = χ ξ s 2 ,
and
χ ξ s 2 < 0 < χ ξ s 1 ,
where
χ ξ s = p τ s p τ 1 Δ H D 0 + γ , δ , ψ ξ x p x d x + q τ s q τ 1 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x τ α + 1 s τ α Δ g x ξ x 1 α x d x ,
We will now analyze the function of fibers ψ ξ : 0 , + R defined as
ψ ξ s = J ϱ γ , δ , ψ s ξ for all s > 0 .
Since ψ ξ C 2 0 , + , we get
ψ ξ s 1 = s 1 p 1 Δ H D 0 + γ , δ , ψ ξ x p x d x + s 1 q 1 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x s 1 α Δ g x ξ x 1 α x d x ϱ s 1 τ 1 Δ ξ x τ x d x ,
and
ψ ξ s 1 = p 1 s 1 p 2 Δ H D 0 + γ , δ , ψ ξ x p x d x + q 1 s 1 q 2 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x + α s 1 α 1 Δ g x ξ x 1 α x d x ϱ τ 1 s 1 τ 2 Δ ξ x τ x d x .
Using (15) and (16), we get
s 1 p τ Δ H D 0 + γ , δ , ψ ξ x p x d x + s 1 q τ Δ κ x H D 0 + γ , δ , ψ ξ x q x d x s 1 τ α + 1 Δ g x ξ x 1 α x d x = ϱ Δ ξ x τ x d x .
This implies that by multiplying by α s 1 τ 2 ,   τ 1 s 1 τ 2 , respectively, that
α s 1 p 2 Δ H D 0 + γ , δ , ψ ξ x p x d x + α s 1 q 2 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x α ϱ s 1 τ 2 Δ ξ x τ x d x = α s 1 τ 1 Δ g x ξ x 1 α x d x
and
τ 1 s 1 p 2 Δ H D 0 + γ , δ , ψ ξ x p x d x τ 1 s 1 q 2 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x τ 1 s 1 α 1 Δ g x ξ x 1 α x d x = ϱ τ 1 s 1 τ 2 Δ ξ x τ x d x .
Substituting (19) into (18), we find
ψ ξ s 1 = p + α 1 s 1 p 2 Δ H D 0 + γ , δ , ψ ξ x p x d x + q + α 1 s 1 q 2 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x ϱ τ + α 1 s 1 α 1 Δ g x ξ x 1 α x d x .
In the same way, we substitute (20) into (18) and find
ψ ξ s 1 = s 1 1 τ χ ξ s 1 > 0 .
From (7) and (22), we conclude that
ψ ξ s 1 = p + α 1 s 1 p Δ H D 0 + γ , δ , ψ ξ x p x d x + q + α 1 s 1 q Δ κ x H D 0 + γ , δ , ψ ξ x q x d x ϱ τ + α 1 s 1 τ Δ ξ x τ x d x .
Thus,
s 1 ξ E ϱ + for all ϱ 0 , ϱ ¯ .
Hence, E ϱ + .
Similarly, it is shown that E ϱ for s 2 .
Let ξ n n N   E ϱ + be a minimizing sequence; i.e.,
J ϱ γ , δ , ψ ξ n ς ϱ + < 0 as n + .
According to the validity of E ϱ + E ϱ and Lemma 1, we have that
ξ n n N H q ξ γ , δ , ψ Δ is bounded .
Therefore, we may assume that
ξ n ξ in H q ξ γ , δ , ψ Δ and ξ n ξ in L τ x Δ .
From (23), we know that
J ϱ γ , δ , ψ ξ lim n inf J ϱ γ , δ , ψ ξ n < 0 = J ϱ γ , δ , ψ 0
Consequently, ξ 0 .
Arguing by contradiction, suppose that ξ n ξ in H q ξ γ , δ , ψ Δ . Then
lim n inf Δ H D 0 + γ , δ , ψ ξ n x p x d x > Δ H D 0 + γ , δ , ψ ξ x p x d x .
From (24), we have
lim n inf ψ ξ n s 1 = lim n inf s 1 p 1 Δ H D 0 + γ , δ , ψ ξ n x p x d x + s 1 q 1 Δ κ x H D 0 + γ , δ , ψ ξ n x q x d x s 1 α Δ g x ξ n x 1 α x d x ϱ s 1 τ 1 Δ ξ n x τ x d x , > s 1 p 1 Δ H D 0 + γ , δ , ψ ξ x p x d x + s 1 q 1 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x s 1 α Δ g x ξ x 1 α x d x ϱ s 1 τ 1 Δ ξ x τ x d x , .
According to (15) and (16), we obtain that
lim n inf ψ ξ n s 1 > ψ ξ s 1 = 0 ,
so n 0 N such that
ψ ξ n s 1 > 0 for all n > n 0 .
Since ξ n E ϱ + E ϱ and
ψ ξ n = s τ 1 χ ξ n ϱ Δ ξ n x τ x d x ,
therefore,
ψ ξ n < 0 for all s 0 , 1 and ψ ξ n 1 = 0
Then, by (26), we have s 1 > 0 .
Then, the function ψ ξ is decreasing on 0 , s 1 . Hence, from (26), we have
J ϱ γ , δ , ψ s 1 ξ J ϱ γ , δ , ψ s ξ < ς ϱ + .
Remember that s 1 ξ E ϱ + . Hence, by (27), we get
ς ϱ + J ϱ γ , δ , ψ s 1 ξ < ς ϱ + ,
i.e., this is a contradiction. Thus,
ξ n ξ in H q ξ γ , δ , ψ Δ ,
and we have
J ϱ γ , δ , ψ ξ n J ϱ γ , δ , ψ ξ
This indicates
J ϱ γ , δ , ψ ξ = ς ϱ + .
Since ξ n E ϱ + for all n N , we have
p + α 1 Δ H D 0 + γ , δ , ψ ξ n x p x d x + q + α 1 Δ κ x H D 0 + γ , δ , ψ ξ n x q x d x > ϱ τ + α 1 Δ ξ n x τ x d x .
When n + , we have
p + α 1 Δ H D 0 + γ , δ , ψ ξ x p x d x + q + α 1 Δ κ x H D 0 + γ , δ , ψ ξ x q x d x > ϱ τ + α 1 Δ ξ x τ x d x .
Recall that ϱ 0 , ϱ ¯ and ϱ ¯ ϱ . And using Lemma 2, we conclude that the equality in (28) is incorrect. And so we have ξ > 0 for all x Δ with ξ 0 . Here we complete the proof. □
Lemma 5. 
Suppose that A is satisfied, suppose z H q ξ γ , δ , ψ Δ , and let ϱ 0 , ϱ ¯ . Then there exists β > 0 such that for all s 0 , β , we have
J ϱ γ , δ , ψ ξ J ϱ γ , δ , ψ ξ + s z .
Proof. 
Let ϑ z : 0 , + R be defined by
ϑ z s = p + α 1 Δ H D 0 + γ , δ , ψ ξ x + s z x p x d x + q + α 1 Δ κ x H D 0 + γ , δ , ψ ξ x + s z x q x d x > ϱ τ + α 1 Δ ξ x + s z x τ x d x .
Since ξ E ϱ + , we have ϑ z 0 > 0 . From the continuous ϑ z . , we find β > 0 as
ϑ z s > 0 for all s 0 , β .
Thus, ξ + s z E ϱ + for all s 0 , β . Using Lemma 4, it can be deduced that
ς ϱ + = J ϱ γ , δ , ψ ξ J ϱ γ , δ , ψ ξ + s z for all s 0 , β .
We finish the proof. □
Proposition 4. 
Suppose that A is satisfied, and let ϱ 0 , ϱ ¯ . Then ξ represents a weak solution of (1).
Proof. 
Let z H q ξ γ , δ , ψ Δ . Using Lemma 5, we have for all s 0 , β
J ϱ γ , δ , ψ ξ + s z J ϱ γ , δ , ψ ξ > 0 ,
i.e.,
1 1 α Δ g x ξ x + s z x 1 α x ξ x 1 α x d x 1 p Δ H D 0 + γ , δ , ψ ξ x + s H D 0 + γ , δ , ψ z x p x H D 0 + γ , δ , ψ ξ x p x d x + 1 q Δ κ x H D 0 + γ , δ , ψ ξ x + s H D 0 + γ , δ , ψ z x q x H D 0 + γ , δ , ψ ξ x q x d x ϱ q Δ ξ x + s z x τ x ξ x τ x d x .
If (30) is divided by s, then s 0 + . This yields
Δ g x ξ x α x z x d x Δ H D 0 + γ , δ , ψ ξ x p x 2 H D 0 + γ , δ , ψ ξ x . H D 0 + γ , δ , ψ z x d x + Δ κ x H D 0 + γ , δ , ψ ξ x q x 2 H D 0 + γ , δ , ψ ξ x . H D 0 + γ , δ , ψ z x d x ϱ Δ ξ x τ x 1 . z d x .
Since z H q ξ γ , δ , ψ Δ is arbitrary, then that equality must hold, and so ξ is a weak solution of (1) for all ϱ 0 , ϱ ¯ .
We will achieve the second weak solution when the parameter ϱ > 0 is small enough by using a manifold E ϱ .
Lemma 6. 
Suppose that A is satisfied. Then there exists ϱ ¯ 0 0 , ϱ ¯ such that for all ϱ 0 , ϱ ¯ 0 , we have J ϱ γ , δ , ψ E ϱ 0 .
Proof. 
Let ξ E ϱ . Using Lemma 4, Proposition 3, and the definition of E ϱ , we have
p + α 1 H D 0 + γ , δ , ψ ξ p x p + + q + α 1 H D 0 + γ , δ , ψ ξ κ x , q x q + < ϱ τ + α 1 ξ τ x τ .
Then
ϱ τ + α 1 ξ τ x τ > κ 0 q + α 1 H D 0 + γ , δ , ψ ξ q x q + > κ 0 q + α 1 . C ξ q x q + ,
with C as the Poincaré constant. Hence, by Proposition 3, we have
ξ τ x τ q + > κ 0 q + α 1 . C . C 8 ϱ τ + α 1 ,
where C 8 is the embedding constant in Proposition 3. Hence,
ξ τ x > κ 0 q + α 1 . C . C 8 ϱ τ + α 1 1 τ q + .
Assume that the Lemma is not satisfied. Using the contradiction, we can find ξ E ϱ such that J ϱ γ , δ , ψ < 0 ; thus,
1 p H D 0 + γ , δ , ψ ξ x p x p + + 1 q H D 0 + γ , δ , ψ ξ κ x , q x q + 1 1 α Δ g x ξ 1 α x d x ϱ τ ξ τ x τ < 0 .
Since ξ E ϱ , we define
H D 0 + γ , δ , ψ ξ κ x , q x q + = H D 0 + γ , δ , ψ ξ x p x p + + Δ g x ξ 1 α x d x + ϱ ξ τ x τ .
Using (32) and (33), we obtain
1 p 1 q H D 0 + γ , δ , ψ ξ x p x p + + 1 q 1 1 α Δ g x ξ 1 α x d x + ϱ 1 q 1 τ + ξ τ x τ < 0 ,
This means that
ϱ 1 q 1 τ + ξ τ x τ < 1 1 α 1 q C 9 ξ τ x 1 α for C 9 > 0 .
Thus, since q < p < τ + , we have
ξ τ x τ + α 1 C 9 q + α 1 ϱ 1 α τ + + q ,
i.e.,
ξ τ x C 10 1 ϱ 1 τ + α 1 for C 10 > 0 .
Substituting (34) into (31), we find
C 11 1 ϱ 1 τ q + C 10 1 ϱ 1 τ + α 1
when
C 11 = κ 0 . C . C 8 q + α 1 τ + α 1 1 τ q + .
Hence,
0 < C 11 C 10 < ϱ 1 τ q + 1 τ + α 1 = ϱ q + + + α 1 τ q + τ + α 1 0 as ϱ 0 + .
This is a contradiction, since 1 < q + < τ and α . 0 , 1 .
From here, we can have ϱ ¯ 0 0 , ϱ ¯ such that for all ϱ 0 , ϱ ¯ 0 , we have J ϱ γ , δ , ψ E ϱ 0 .
Lemma 7. 
Suppose that A is satisfied; let ϱ 0 , ϱ ¯ 0 . Then there exists 0 η E ϱ such that
ς ϱ = inf E ϱ J ϱ γ , δ , ψ = J ϱ γ , δ , ψ η > 0 .
Proof. 
Following the same approach as Lemma 4. If η n n N E ϱ is a minimizing sequence, by using Lemma 1, we have that η n n N H q ξ γ , δ , ψ Δ is bounded. Then, we suppose that
η n η weakly in H q ξ γ , δ , ψ Δ and η n η in L τ x Δ .
From (15) and (16), we can find 0 < s 2 such that
ψ η n s 2 < 0 and ψ η s 2 = ϱ η τ x τ .
As stated in the proof of Lemma 4, by using (35), we find
η E ϱ , η 0 , ς ϱ = J ϱ γ , δ , ψ η .
Lemma 8. 
Suppose that A is satisfied and ϱ 0 , ϱ ¯ 0 . Then η is a weak solution of (1).
Proof. 
We follow the same steps for the proofs obtained in Lemma 5 and Proposition 4. □
According to the previous results, Problem (1) has at least two positive solutions ξ , η H q ξ γ , δ , ψ Δ such that
J ϱ γ , δ , ψ ξ < 0 J ϱ γ , δ , ψ η for all ϱ 0 , ϱ ¯ 0 , ϱ ¯ 0 > 0 .

Author Contributions

Conceptualization, R.G., R.J., S.B. and T.R.; Methodology, T.B. and R.J.; Validation, R.G., S.B. and T.R.; Investigation, R.G.; Writing—original draft, R.G.; Writing—review & editing, T.B., R.J., S.B. and T.R.; Visualization, T.B.; Supervision, S.B. All authors have read and agreed to the published version of the manuscript.

Funding

The researchers would like to thank the Deanship of Graduate Studies and Scientific Research at Qassim University for financial support (QU-APC-2024-9/1).

Institutional Review Board Statement

Not applicable.

Data Availability Statement

There are no data associated with the current study.

Acknowledgments

The researchers would like to thank the Deanship of Graduate Studies and Scientific Research at Qassim University for financial support (QU-APC-2024-9/1).

Conflicts of Interest

The authors declare there are no conflicts of interest.

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Bouali, T.; Guefaifia, R.; Jan, R.; Boulaaras, S.; Radwan, T. Existence of Weak Solutions for the Class of Singular Two-Phase Problems with a ψ-Hilfer Fractional Operator and Variable Exponents. Fractal Fract. 2024, 8, 329. https://doi.org/10.3390/fractalfract8060329

AMA Style

Bouali T, Guefaifia R, Jan R, Boulaaras S, Radwan T. Existence of Weak Solutions for the Class of Singular Two-Phase Problems with a ψ-Hilfer Fractional Operator and Variable Exponents. Fractal and Fractional. 2024; 8(6):329. https://doi.org/10.3390/fractalfract8060329

Chicago/Turabian Style

Bouali, Tahar, Rafik Guefaifia, Rashid Jan, Salah Boulaaras, and Taha Radwan. 2024. "Existence of Weak Solutions for the Class of Singular Two-Phase Problems with a ψ-Hilfer Fractional Operator and Variable Exponents" Fractal and Fractional 8, no. 6: 329. https://doi.org/10.3390/fractalfract8060329

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