Existence of Weak Solutions for a New Class of Fractional P-Laplacian Boundary Value Systems

Article Existence of Weak Solutions for a New Class of Fractional p-Laplacian Boundary Value Systems

Fares Kamache 1,†, Raﬁk Guefaiﬁa 1,†, Salah Boulaaras 2,3,*,† and Asma Alharbi 2,†

1 Laboratory of Mathematics, Informatics and systemes (LAMIS), University of Larbi Tebessi, 12000 Tebessa, Algeria; [email protected] (F.K.); raﬁk.guefaiﬁ[email protected] (R.G.) 2 Department of Mathematics, College of Sciences and Arts, Al-Rass, Qassim University, 51452 Qassim, Saudi Arabia; [email protected] 3 Laboratory of Fundamental and Applied Mathematics of Oran (LMFAO), University of Oran 1 Ahmed Ben Bella, 31000 Oran, Algeria * Correspondence: [email protected] or [email protected]; Tel.: +966-559-618-327 † These authors contributed equally to this work.

Received: 2 March 2020; Accepted: 28 March 2020; Published: 31 March 2020

Abstract: In this paper, at least three weak solutions were obtained for a new class of dual non-linear dual-Laplace systems according to two parameters by using variational methods combined with a critical point theory due to Bonano and Marano. Two examples are given to illustrate our main results applications.

Keywords: nonlinear fractional; dirichlet boundary value problems; p-Laplacian type; variational method; critical point theory

1. Introduction Fractional differential equations have proved to be valuable tools in modeling many phenomena in various fields of physics, chemistry, biology, engineering and economics. There was a significant development in fractional differential equations. We can see the studies of Miller and Ross [1], Samko et al. [2], Podlupny [3], Hilfer [4], Kelpas et al. [5] and papers [6–16] and references therein. The critical point theory was very useful in determining the existence of solutions to complete differential equations with certain boundary conditions; see for example, in the extensive literature on the subject, classical books [17–19] and references appearing there. However, so far, some problems have been created for fractional marginal value problems (briefly BVP) by exploiting this approach, where it is often very difficult to create a suitable space and a suitable function for fractional problems. In [20], the authors investigated the following nonlinear fractional differential equation depending on two parameters:

 α α D i a (t) D i u (t) = λF (t, u (t) , u (t) , ..., u (t))  t T i 0 t i ui 1 2 n   +µGui (t, u1 (t) , u2 (t) , ..., un (t)) + hi (ui) a.e [0, T] , (1)     ui (0) = ui (T) = 0

αi αi for 1 ≤ i ≤ n, where αi ∈ (0; 1], 0Dt and tDT are the left and right Riemann-Liouville fractional ∞ derivatives of order αi respectively, ai ∈ L ([0, T]) with

ai0 = ess inf ai > 0 for 1 ≤ i ≤ n, λ, µ [0,T]

Mathematics 2020, 8, 475; doi:10.3390/math8040475 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 475 2 of 18 are positive parameters, F, G : [0, T] × Rn −→ R are measurable functions with respect to t ∈ [0, T] n 1 n for every (x1, ..., xn) ∈ R and are C with respect to (x1, ..., xn) ∈ R for a.e. t ∈ [0, T] , Fui and Gui denotes the partial derivative of F and G with respect to ui, respectively, and hi: R → R are Lipschitz continuous functions with the Lipschitz constants Li > 0 for 1 ≤ i ≤ n, i.e.,

|hi (x1) − hi (x2)| ≤ Li |x1 − x2| for every x1, x2 ∈ R, and hi (0) = 0 for 1 ≤ i ≤ n. Motivated by [21,22], using a three critical points theorem obtained in [23] which we recall in the next section (Theorem 2.6), we ensure the existence of at least three solutions for this system. For example, according to some assumptions, in [24], by using variational methods the authors obtained the existence of at least one weak solution for the following p−Laplacian fractional differential equation [24]  Dα φ ( Dαu (t)) = λ f (t, u (t)) a.e. t ∈ [0, T] ,  t T p 0 t (2)  u (0) = u (T) = 0,

α α where 0Dt and tDT are the left and right Riemann-Liouville fractional derivatives with 0 < α ≤ 1, p−2 respectively, the function φp (s) = |s| s, p > 1. Taking a class of fractional differential equation with p−Laplacian operator as a model, Li et al. investigated the following equation recently [25]   α 1 α  tD p−2 ϕp (0D u (t)) + λu (t)  T w(t) t   (3) = f t, u,c Dαu (t) + h (u (t)) a.e. t ∈ [0, T] ,  0 t    u (0) = u (T) = 0

1 with p < α ≤ 1, λ is a non-negative real parameter. The functions p−2 ϕp (s) = |s| s, p ≥ 2, f : [0; T] × R × R → R is continuous h : R → R is a Lipschitz continuous function. By using the mountain pass theorem combined with iterative technique, the authors obtained the existence of at least one solution for problem (3). In this paper, we are interested in ensuring the existence of three weak solutions for the following system   αi 1 αi p−2  tDT p−2 φp 0Dt ui (t) + µ |ui (t)| ui (t)  wi(t)   (4) = λF (t, u (t) , u (t) , ..., u (t)) a.e. t ∈ [0, T] ,  ui 1 2 n     ui (0) = ui (T) = 0, where p−2 ∞ φp (s) = |s| s, p > 1, wi (t) ∈ L [0, T]

0 αi αi with wi = ess inf[0,T] wi (t) > 0,0 Dt and tDT are the left and right Riemann-Liouville fractional derivatives of order 0 < αi ≤ 1 respectively, for 1 ≤ i ≤ n, λ is positive parameter, and F : n n [0, T] × R → R is measurable function with respect to t ∈ [0, T] for every (x1, x2, ..., xn) ∈ R 1 n and are C with respect to (x1, x2, ..., xn) ∈ R . for a.e. t ∈ [0, T], Fui denote the partial derivative of F with respect to ui,respectively, (H0) αi ∈ (0; 1] for 1 ≤ i ≤ n. Mathematics 2020, 8, 475 3 of 18

n (H1) F :[0, T] × R → R be a function such that F (., u1, u2, ..., un) is continuous in [0, T] for every n 1 2 (u1, u2, ..., un) ∈ R , F (t, ., ., ..., .) is a C function in R . In the present paper, motivated by [26,27], using a three critical points theorem obtained in [23], which we recall in the next section (Theorem 2.6), we ensure the existence of at least three solutions for system (4). This theorem has been successfully employed to establish the existence of at least three solutions for perturbed boundary value problems in the papers ([26–30]). This paper is organized as follows. In Section2, we present some necessary preliminary facts that will be needed in the paper. In Section3, we prove our main result. In Section4, we give two numerical examples in order to support the theory of our contribution.

2. Preliminaries In this section, we ﬁrst introduce some necessary deﬁnitions and preliminary facts are introduced for fractional calculus which are used in this paper.

For [0, T] ⊆ R, let C ([0, T] , R) be the real space of all continuous functions with norm kxk∞ = max |x(t)| ,and Lp ([0, T] , R)(1 ≤ p < ∞) be the space of functions for which the pth power of the t∈[0,T] 1 T ! p R p absolute value is Lebesgue integrable with norm kxkLp = |x(t)| dt 0

Definition 1 (Kilbas et al. [5]). Let u be a function defined on [a, b] . The left and right Riemann-Liouville fractional derivatives of order α > 0 for a function u are defined by

t n n Z α d α−n 1 d n−α−1 aDt u (t) := Dt u (t) = (t − s) u (s) ds dtn a Γ (n − α) dtn a

and b n n n Z α n d α−n (−1) d n−α−1 tDb u (t) := (−1) D u (t) = (t − s) u (s) ds dtn t b Γ (n − α) dtn t for every t ∈ [a, b] , provided the right-hand sides are pointwise deﬁned on [a, b] , where n − 1 ≤ α < n and n ∈ N. Here, Γ (α) is the standard gamma function given by

+∞ Z Γ (α) := zα−1e−zdz. 0

− Setting ACn ([a, b] , R) the space of functions u : [a, b] → R such that u ∈ Cn 1 ([a, b] , R) and − − u(n 1) ∈ ACn ([a, b] , R) . Here, as usual, Cn 1 ([a, b] , R) denotes the set of mappings being (n − 1) times continuously differentiable on [a, b] . In particular, we denote AC ([a, b] , R) := AC1 ([a, b] , R) .

Deﬁnition 2 ([25]). Let 0 < αi ≤ 1, for 1 ≤ i ≤ n, 1 < p < ∞. The fractional derivative space

p = ( ) ∈ p αi ∈ p Eαi u t L ([0, T] , R)a Dt u (t) L ([0, T] , R) .

∈ p p Then, for any u Eαi , we can deﬁne the weighted norm for Eαi as

1  T T  p Z Z p α p kxk = |u(t)| dt + w (t) D i u (t) dt . (5) αi  i a t  0 0 Mathematics 2020, 8, 475 4 of 18

Deﬁnition 3 ([31]). We mean by a weak solution of system (4), any u = (u1, u2, ..., un) ∈ X such that for all v = (v1, v2, ..., vn) ∈ X

T Z n 1 ( ) αi ( ) αi ( ) ∑ p−2 φp wi t 0 Dt ui t 0 Dt vi t dt = w (t) 0 i 1 i T Z n p−2 +µ ∑ |ui (t)| ui (t) vi (t) dt = 0 i 1 T Z n − ( ( ) ( ) ( )) ( ) = λ ∑ Fui t, u1 t , u2 t , ..., un t vi t dt 0. = 0 i 1

< ≤ ≤ ≤ < < ∈ p Lemma 1 ([31]). Let 0 αi 1, for 1 i n , 1 p ∞.For any u Eαi we have

Tαi ≤ αi kuikLp 0Dt ui Lp . (6) Γ (αi + 1)

1 1 Moreover, if αi > p and p + q = 1, then

Tαi k k ≤ αi ui ∞ 1 0Dt ui Lp . (7) q Γ (αi) Γ ((αi − 1) q + 1)

From Lemma 1, we easily observe that

1/p b ! p αi R αi T wi(t) aDt u (t) dt a kuikLp ≤ (8) Γ (αi + 1) for 0 < αi ≤ 1 , and !1/p αi−1 b p R αi p T wi(t) aDt u (t) dt a k k ≤ ui ∞ 1 1 (9) 0 p q Γ (αi) wi Γ ((αi − 1) q + 1) 1 1 for αi > p and p + q = 1.

By using (8), the norm of (5) is equivalent to

1  T  p Z α p p kxk = w (t) D i u (t) dt , ∀u ∈ E . (10) αi  i a t  αi 0

p ≤ ≤ Throughout this paper, we let X be the Cartesian product of the n spaces Eαi for 1 i n, i.e., = p × p × × p X Eα1 Eα2 ... Eαn equipped with the norm

n kuk = ku k p , u = (u , u , ..., un) , ∑ i Eα 1 2 i=1 i

n p where kuikE is deﬁned in (10) Obviously, X is compactly embedded in C ([0, T] , R) . αi

Lemma 2 ([32]). For 0 < αi ≤ 1 and 1 < p < ∞, the fractional derivative space X is a reﬂexive separable Banach space. Mathematics 2020, 8, 475 5 of 18

Lemma 3 ([33]). Let A : X → X∗ be a monotone, coercive and hemicontinuous operator on the real, separable, reﬂexive Banach space X. Assume {w1, w2...}is a basis in X. Then the following assertion holds: If A is strictly monotone, then the inverse operator A−1 : X∗ → X exists. This operator is strictly monotone, demicontinuous and bounded. If A is uniformly monotone, then A−1 is continuous. If A is strongly monotone, then it is Lipschitz continuous.

Theorem 1 ([34]). Let X be a reﬂexive real Banach space; Φ : X → R be a coercive ,continuously Gateaux differentiable sequentially weakly lower semicontinuous functional whose Gateaux derivative admits a continuous ∗ inverse on X , bounded on bounded subsets of X, Ψ : X → R a continuously Gateaux differentiable functional whose Gateaux derivative is compact such that

Φ (0) = Ψ (0) = 0.

Assume that there exists r > 0 and x ∈ X, with r < Φ (x) , such that ( ) Ψ(u) < Φ(x) a1 sup r Ψ(x) . Φ(u)≤r   ( ) ∈ = Φ(x) r − a2 For each λ Λλ  Ψ(x) , sup Ψ(u)  , the functional Φ λΨ is coercive. Φ(u)≤r

Then, for any λ ∈ Λλ, the functional Φ − λΨ has at least three critical point in X.

3. The Main Results In the present section ,the existence of multiple solutions for system (4) is examined by using Theorem 1. First and foremost, we deﬁne the functionals Φ, Ψ : X → R as

T n 1 R αi p p Φ (u) = p ∑ wi (t) 0Dt ui (t) + µ |ui (t)| dt, 0 i=1 (11)

u = (u1, u2, ..., un) ∈ X and T Z Ψ (u) = F (t, u1 (t) , u2 (t) , ..., un (t)) dt (12) 0

Lemma 4. Let 0 < αi ≤ 1, u = (u1, u2, ..., un) ∈ X. Functionals Φ and Ψ are deﬁned in (11) and (12). Then, Φ : X → R is a coercive, continuously Gâteaux differentiable and sequentially weakly lower semicontinuous ∗ functionl whose Gâteaux derivative admits a continuous inverse on X , and Ψ : X → R is a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact.

Proof. For each u = (u1, u2, ..., un) ∈ X , deﬁne Φ, Ψ : X → R as

T Z n 1 αi p p Φ (u) = ∑ wi (t) 0Dt ui (t) + µ |ui (t)| dt p = 0 i 1 and T Z Ψ (u) = F (t, u1 (t) , u2 (t) , ..., un (t)) dt. 0 Mathematics 2020, 8, 475 6 of 18

Clearly, Φ and Ψ are continuously Gateaux differentiable functionals whose Gateaux derivatives at the point u ∈ X are given by

T n 0 R 1 αi αi Φ (u)(v) = ∑ − φp w (t) D u (t) D v (t) dt w (t)p 2 i 0 t i 0 t i 0 i=1 i (13) T n R p−2 +µ ∑ |ui (t)| ui (t) vi (t) dt 0 i=1 for every v = (v1, v2, ..., vn) ∈ X. 1 p In addition, according to (11), one has Φ (u) ≥ p kukX , which means that Φ is a coercive functional. 0 ∗ Next, we claim that Φ admits a continuous inverse on X . Let u = (u1, u2, ..., un) ∈ X, v = (v1, v2, ..., vn) ∈ X. Recalling (13),we get

T n 0 0 R 1 αi αi hΦ (u) − Φ (v) , u − vi = ∑ − φp w (t) D u (t) D (u − v) dt w (t)p 2 i 0 t i 0 t 0 i=1 i T n R p−2 + µ ∑ |ui (t)| ui (t)(u − v) dt 0 i=1 (14) T n R 1 αi αi − ∑ − φp w (t) D v (t) D (u − v) dt w (t)p 2 i 0 t i 0 t 0 i=1 i T n R p−2 + µ ∑ |vi (t)| vi (t)(u − v) dt. 0 i=1

According to the well-known inequality

p−2 p−2 |s1| s1 − |s2| s2 (s1 − s2)

 p (15)  |s1 − s2| , p ≥ 2 2 ≥ |s1−s2|  2−p , 1 < p ≤ 2. (|s1|+|s2|)

We have αi αi φp wi (t)0 Dt ui (t) − φp wi (t)0 Dt vi (t)

 1 αi αi p wi (t) D ui (t) − wi (t) D vi (t) , p ≥ 2  wi(t) 0 t 0 t  α α 2 ≥ w (t) D i u (t)−w (t) D i v(t) 1 i 0 t i i 0 t  2−p , 1 < p < 2.  wi(t) αi αi  wi(t)0Dt ui(t) + wi(t)0Dt vi(t) Hence, when 1 < p < 2, one has

T n R αi αi p ∑ wi (t)0 Dt ui (t) − wi (t)0 Dt ui (t) dt 0 i=1 p α α 2 ! 2 T n ( ) i ( )− ( ) i ( ) R wi t 0Dt ui t wi t 0Dt vi t ≤ ∑ 2−p dt (16) αi αi 0 i=1 wi(t) wi(t)0Dt ui(t) + wi(t)0Dt vi(t) ! 2−p T n p 2 R 2−p αi αi p ∑ wi (t) wi (t)0 Dt ui (t) + wi (t)0 Dt vi (t) dt 0 i=1 Mathematics 2020, 8, 475 7 of 18 which means that 2 T n ( ) αi ( )− ( ) αi ( ) R wi t 0Dt ui t wi t 0Dt vi t ∑ 2−p dt αi αi 0 i=1 wi(t) wi(t) D ui(t) + wi(t) D vi(t) 0 t 0 t (17) 2(p−1) p−2 p−2 0 p 2 (w1) 2 p p p ≥ ∼ kui − vikα kuikα + kvikα . 0 i i i w1 Then, we deduce

T n R ( ) αi ( ) − ( ) αi ( ) αi ( − ) ∑ φp wi t 0 Dt ui t φp wi t 0 Dt vi t 0 Dt u v dt 0 i=1 2(p−1) p−2 (18) p−2 0 p 2 (w1) 2 p p p ≥ ∼ kui − vikα kuikα + kvikα > 0. 0 i i i w1

When p ≥ 2, we get

T n R ( ) αi ( ) − ( ) αi ( ) αi ( − ) ∑ φp wi t 0 Dt ui t φp wi t 0 Dt vi t 0 Dt u v dt 0 i=1 (19) p−2 ≥ w0 ku − v kp > 0. 1 i i αi

Then, combining with (18), yields

T Z n αi αi αi αi ∑ φp wi (t)0 Dt ui (t) − φp wi (t)0 Dt vi (t) 0Dt ui −0 Dt vi dt > 0. (20) = 0 i 1

For every 1 < p < ∞ Further, denote

T T Z n Z n p−2 p−2 A = ∑ |ui (t)| ui (t)(u − v) dt + ∑ |vi (t)| vi (t)(u − v) dt. = = 0 i 1 0 i 1

Then, reapplying inequality (15), we always have

A ≥ ku − v kp > 0, for p ≥ 2 i i αi

and p−2 p−2 2 p p p A ≥ 2 kui − vikLp kuikLp + kvikLp > 0 for1 < p < 2. That is, A > 0 for every 1 < p < ∞. Therefore, by using (14) and (20), the following inequality holds Φ0 (u) − Φ0 (v) , u − v > 0 0 which means that Φ is strictly monotone. Furthermore, in view of X being reﬂexive, for un → u in X 0 0 ∗ strongly, as n → ∞, one has Φ (un) * Φ (u) in X as n → ∞. Thus, we say that Φ0 is demicontinuous. Then, according to lemma 2 and 3, we obtain that the − inverse operator (Φ0) 1 of Φ0 exists and is continuous. Moreover , let T Z n p α p p kuk = w (t) D i u (t) + µ |u (t)| dt µ,αi ∑ i 0 t i i = 0 i 1 owing to the sequentially weakly lower semicontinuity of kukp we observe that Φ is sequentially µ,αi weakly lower semicontinuous in X. Mathematics 2020, 8, 475 8 of 18

Considering the functional Ψ, we will point out that Ψ is a Gâteaux differentiable, sequentially weakly upper semicontinuous functional on X. Indeed, for un ⊂ X, assume that un * u in X, i.e., un uniformly converges to u on [0, T] as n → ∞. By using reverse Fatou’s lemma, one has

T Z lim inf Ψ (un) ≤ lim inf F (t, un (t)) dt n→+∞ n→+∞ 0

T Z = F (t, u1 (t) , u2 (t) , ..., un (t)) dt = Ψ (u) , 0 whereas u = (u1, u2, ..., un) ∈ X, which implies that Ψ is sequentially weakly upper semicontinuous. Furthermore, since F is continuously differentiable with respect to u and v for almost every t ∈ [0, T] , 0 0 then based on the Lebesgue control convergence theorem, we obtain that Ψ (un) → Ψ (u) strongly , 0 0 that is Ψ is strongly continuous on X. Hence, we conﬁrm that Ψ is a compact operator. 0 Moreover, it is easy to prove that the functional with the Gâteaux derivative Ψ (u) ∈ X∗at the point u ∈ X T Z n 0 ( )( ) = ( ( ) ( ) ( )) ( ) Ψ u v ∑ Fui t, u1 t , u2 t , ..., un t vi t dt (21) = 0 i 1 for any v = (v1, v2, ..., vn) ∈ X.The proof is completed.

In order to facilitate the proof of our main result, some notations are given. Putting    Tpαi−1  k : = max p , 1≤i≤n  p 0 q  (Γ (αi)) wi ((αi − 1) q + 1) ( ) ∼ Tpαi k : = max p . 1≤i≤n 0 (Γ (αi + 1)) wi

Deﬁne ( n ) n 1 p π (σ) = u = (u1, u2, ..., un) ∈ R : ∑ |ui| < σ . p i=1

1 Theorem 2. Let p < αi ≤ 1, for 1 ≤ i ≤ n. Assume that there exists a positive constant r and a function u = (u1, u2, ..., un) ∈ X such that

(i) n n p p ku k + µ ku k p > pr; ∑ i αi ∑ i L i=1 i=1 (ii) T T R ( ) R sup F t, u1, u2, ..., un dt p F (t, u1, u2, ..., un) dt 0 u∈π(kr) 0 < n n ; r p p ku k + µ ku k p ∑ i αi ∑ i L i=1 i=1 Mathematics 2020, 8, 475 9 of 18

(iii) RT sup F (t, u1, u2, ..., un) dt F (t, u1, u2, ..., un) 0 u∈π(kr) lim inf n ≤ ∼ . |u|→∞ p ∑ |xi| prk i=1 Then, setting   n n p p ku k + µ ku k p  ∑ i αi ∑ i L   i=1 i=1 r  Λ =  ,   RT RT   p F (t, u1, u2, ..., un) dt sup F (t, u1, u2, ..., un) dt  0 0 u∈π(kr)

for each λ ∈ Λ system (4) admits at least three weak solutions in X.

Proof. Considering Theorem 1 and Lemma 5, in order to obtain that system (4) possesses at least three weak solutions in X, we only need to guarantee the assumptions (a1) and (a2) of Theorem 1 are satisﬁed. Choose u0 = (u01, u02, ..., u0n) and u1 = (u11, u12, ..., u1n) with (u01, u02, ..., u0n) = (0, 0, ..., 0) . Due to (12) and (i) , we get Ψ (u0) = 0 and Φ (u1) > r > 0, which satisfy the requirement of Theorem 1. Then, combining (11) and (9) , yields

{u = (u1, u2, ..., un) ∈ X : Φ (u) ≤ r} ( n n ) 1 p µ p = u = (u , u , ..., u ) ∈ X : ku k + ku k p ≤ r 1 2 n ∑ i αi ∑ i L p i=1 p i=1 ( ) 1 n ⊆ u = (u , u , ..., u ) ∈ X : ku kp ≤ r 1 2 n ∑ i αi p i=1  p  n p 0 q  (Γ (αi)) w1((αi − 1) q + 1)  ⊆ u = (u , u , ..., un) ∈ X : ku k ≤ r 1 2 ∑ pαi−1 i ∞  i=1 pT  ( n ) p ⊆ u = (u1, u2, ..., un) ∈ X : ∑ |ui| ≤ kpr i=1 which implies that

T Z sup Ψ (u) = sup F (t, u1, u2, ..., un) dt Φ(u)≤r Φ(u)≤r 0 T Z ≤ sup F (t, u1, u2, ..., un) dt. Φ(u)≤r 0 Mathematics 2020, 8, 475 10 of 18

Then, the following inequality is obtained under condition (ii)

RT sup F (t, u1, u2, ..., un) dt 0 u∈π(kr) sup Ψ (u) ≤ Φ(u)≤r r RT n p ∑ F (t, u1, u1, ..., un) dt 0 i=1 ≤ n n p p ku k + µ ku k p ∑ i αi ∑ i L i=1 i=1 Ψ (u ) = 1 . Φ (u1)

Thus the hypothesis (a1) of Theorem 1 holds. On the other hand, taking (iii) into account , there exist constants C, ε ∈ R with

RT sup F (t, u1, u2, ..., un) dt 0 u∈π(kr) C < , r such that n C p F (t, u1, u2, ..., un) ≤ ∼ ∑ |ui| + ε (22) pk i=1

for any t ∈ [0, T] and u = (u1, u2, ..., un) ∈ X, when C > 0 by using (11), (22) and (8) yields

T n n Z 1 p µ p Φ (u) − λΨ (u) = ku k + ku k p − λ F (t, u , u , ..., u ) dt ∑ i αi ∑ i L 1 2 n p = p = i 1 i 1 0 T 1 n Z ≥ ku kp − λ F (t, u , u , ..., u ) dt ∑ i αi 1 2 n p = i 1 0 T n Z n 1 p λC p ≥ ku k − ku k p dt − λTε ∑ i αi ∼ ∑ i L p = = i 1 pk 0 i 1 n n α ! 1 p λC T i p ≥ kuik − kuik − λTε p ∑ αi ∼ ∑ ( ( + ))p 0 αi i=1 pk i=1 Γ αi 1 wi 1 n λC n ≥ ku kp − ku kp − λTε ∑ i αi ∑ i αi p i=1 p i=1  

n !   1 p  r  ≥ ku k 1 − C  − λTε. ∑ i αi T p i=1  R   sup F (t, u1, u2, ..., un) dt  0 u∈π(kr)

That is lim Φ (u) − λΨ (u) = +∞. kukX →+∞

Furthermore, analogous to the case of C > 0 , we can deduce that Φ (u) − λΨ (u) → +∞ as kukX → +∞ with C ≤ 0. Hence, all the hypotheses of Theorem 1 hold, then , system (4) admits at least three weak solutions in X. The proof is completed. For simplicity, before giving a corollary of Theorem 2, some notations are presented. Mathematics 2020, 8, 475 11 of 18

1 Let 0 < h < 2 we put

 hT (1−h)T  n (1−α )p n h 1−α ip R i R i 1−αi  ∑ wi (t) t dt + ∑ wi (t) t − (t − hT) dt  1  0 i=1 hT i=1  Ai (αi, h) =  T  (23) ( ) n h 1−α ip hT  R i 1−αi 1−αi   + ∑ wi (t) t − (t − hT) − (t − ((1 − h) T)) dt.  (1−h)T i=1

1 n Corollary 1. Let p < αi ≤ 1. Assume that there exist τ > 0 and θ = (θ1, θ2, ..., θn) ∈ R with θ1 > 0, θ2 > n p ∑ Ai(αi,h)θi i=1 0, ..., θn > 0 and τ ≤ k p , such that (i)0

F (t, u1, u2, ..., un) ≥ 0 for

(t, u1, u2, ..., un) ∈ ([0, hT] ∪ [(1 − h) T, T] × [0, θ])

(ii)0

RT sup F (t, u1, u2, ..., un) dt 0 u∈π(kr) r (1−h)T R p F (t, Γ (2 − α1) θ1, Γ (2 − α2) θ2, ..., Γ (2 − αn) θn) dt hT < ∼ ; n p k 1 + µk ∑ Ai (αi, h) θi i=1

(iii)0 F (t, u1, u2, ..., un) lim inf n ≤ 0 |u|→∞ p ∑ |ui| i=1 for each ∼  n p  1+µk ∑ Ai(αi,h)θi i=1  (1−h)T ,   R   p F(t,Γ(2−α1)θ1,Γ(2−α2)θ2,...,Γ(2−αn)θn)dt  0   λ ∈ Λ =  hT  . (24)      τ   RT  k sup F(t,u1,u2,...,un)dt 0 u∈π(kr) Thus, system (4) admits at least three weak solutions in X.

Proof. Choose  Γ (2 − αi) θi  t , t ∈ [0, hT[  hT Ui(t) = Γ (2 − αi) θi, t ∈ [hT, (1 − h) T]  Γ (2 − α ) θ  i i (T − t ) , t ∈ ](1 − h) T, T] . hT Mathematics 2020, 8, 475 12 of 18

p Obviously Ui (0) = Ui (T) = 0, U (t) ∈ L [0, T]. Owing to Deﬁnition 1 , we derive,  a (t) , t ∈ [0, hT[ ,  1 αi 0Dt U(t) = a2 (t) , t ∈ [hT, (1 − h) T] ,   a3 (t) , t ∈ ](1 − h) T, T] , where θ θ h i a (t) = i t1−αi , a (t) = i t1−αi − (t − (hT)1−αi 1 hT 2 hT and θ h i a (t) = i t1−αi − (t − (hT)1−αi − (t − (T − hT)1−αi . 3 hT That is

T Z n p α p kUk = w (t) D i U (t) dt αi ∑ i 0 t i = 0 i 1 hT (1−h)T T Z Z Z n αi p = + + ∑ wi (t) 0Dt U (t) dt i=1 0 hT (1−h)T n p = ∑ Ai (αi, h) θi , i=1 where (23) is used. Hence U = (U1, U2, ..., Un) ∈ X. τ Take r = , then k

n p ∑ Ai (αi, h) θi = rk = τ ≤ k i 1 p n p ∑ kUkα = i = k i 1 ≤ kΦ (U) p for every U = (U1, U2, ..., Un) ∈ X which means that

n n 1 p µ p r ≤ ku k + ku k p . ∑ i αi ∑ i L p i=1 p i=1

Thus, the assumption (ii) of Theorem 2 holds. On the other hand, based on (7) and (23), yields

n n 1 p µ Tαi p Φ (U) ≤ ∑ ku k + ∑ p ku k p i αi p (Γ(α +1)) w0 i αi i=1 i=1 i i n n ≤ 1 ku kp + µ k˜ ku kp p ∑ i αi p ∑ i αi (25) i=1 i=1 n ˜ p (1+µk) ∑ Ai(αi,h)θi i=1 ≤ p .

0 Then, from (25) and (ii) , we can obtain the following inequality Mathematics 2020, 8, 475 13 of 18

RT RT sup F (t, u1, u2, ..., un) dt k sup F (t, u1, u2, ..., un) dt 0 u∈π(kr) 0 u∈π(kr) = r τ (1−h)T R kp F (t, Γ (2 − α1) θ1, Γ (2 − α2) θ2, ..., Γ (2 − αn) θn) dt hT < ∼ n p k 1 + µk ∑ Ai (αi, h) θi i=1 (1−h)T R F (t, Γ (2 − α1) θ1, Γ (2 − α2) θ2, ..., Γ (2 − αn) θn) dt ≤ hT Φ (U) RT k F (t, u1, u2, ..., un) dt 0 ≤ n n p p ku k + µ ku k p ∑ i αi ∑ i L i=1 i=1 which means that the hypothesis (ii) of Theorem 2 is satisﬁed. 0 0 Furthermore, the condition (iii) of Theorem 2 holds under (iii) since Λ ⊆ Λ Theorem 2 is successfully employed to ensure the existence of at least three weak solutions for system (4), the proof is completed.

4. Numerical Examples Now, we give the following two examples to illustrate the applications of our result:

2 Example 1. Let p = 2, α1 = 0.8, α2 = 0.65, µ = 1, w1(t) = 1 + t , w2(t) = 0.5 + t, T = 1. Then, system (4) gets the following form  D0.8 1 + t2 D0.8u (t) + u (t) = λF (t, u (t) , u (t)) , t ∈ [0, 1] ,  t 1 0 t 1 1 u1 1 2   0.65 + 0.65 + = ∈ tD1 (0.5 t)0 Dt u2 (t) u2 (t) λFu2 (t, u1 (t) , u2 (t)) , t [0, 1] ,    u1 (o) = u1(1) = 0, u2 (o) = u2(1) = 0.

Taking U1 (t) = Γ (1.2) t (1 − t) , U2 (t) = Γ (1.35) t (1 − t) and 2 F (t, u1 (t) , u2 (t)) = 1 + t G (u1, u2) , where  2 u2 + u2 , u2 + u2 ≤ 1 ,  1 2 1 2 G (u1, u2) = 1 1  2 2 2 2 2 3 2 2 10 u1 + u2 − 9 u1 + u2 , u1 + u2 > 1 .

0 0 Clearly, F (t, 0, 0) = 0, w1 = 1 and w2 = 0.5 for any t ∈ [0, 1]. Mathematics 2020, 8, 475 14 of 18

By the direct calculation, we have ( ) 1 1 max , (Γ (0.8))2 (2 × 0.8 − 1) (Γ (0.65))2 (2 × 0.65 − 1)

= k ≈ 3.4764, ( ) 1 1 max , = k˜ ≈ 2.4684 (Γ (0.8) + 1)2 (Γ (0.65) + 1)2 × 0.5 and

2Γ (1.2) D0.8U (t) = t0.2 − t1.2, 0 t 1 Γ (2.2)

2Γ (1.35) D0.65U (t) = t0.35 − t1.35. 0 t 2 Γ (2.35)

So that

2 2 kU1(t)k0.8 ≈ 0.19333, kU2(t)k0.65 ≈ 0.078559

2 2 kU1(t)kL2 ≈ 0.028101, kU2(t)kL2 ≈ 0.026716.

Take r = 1 × 10−4. We easily obtain that

1 2 2 1 2 2 kU (t)k + kU (t)k + kU (t)k 2 + kU (t)k 2 ≈ 0.1632 > r 2 1 0.8 2 0.65 2 1 L 2 L which implies that the condition (i) holds, and

R1 sup F (t, u1, u2) dt 0 (u ,u )∈π(kr) 16k2r 1 2 = ≈ 0.006445 r 3

R1 2 F (t, U1, U2) dt < 0 ≈ 0.0320085949 2 2 2 2 kU k + kU k 2 ∑ i αi ∑ i L i=1 i=1 and

R1 sup F (t, u1, u2) dt F (t, u1, u2) 0 (u1,u2)∈π(kr) 0 = lim inf ≤ ∼ → → 2 2 |u1| ∞, |u2| ∞ |u1| + |u2| prk

≈ 0.001305.

Thus, conditions (ii) and (iii) are satisﬁed. Then, in view of Theorem 2 for each λ ∈ (31.241, 155.159) , the system (4) has at least three weak solutions in X. Mathematics 2020, 8, 475 15 of 18

2 Example 2. Let p = 3, α1 = 0.8, α2 = 0.6, µ = 1, w1(t) = 1 + t , w2(t) = 0.5 + t and T = 1. Then , system (4) gets the following form  D0.8 1 + t2 D0.8u (t) + u (t) = λF (t, u (t) , u (t)) , t ∈ [0, 1] ,  t 1 0 t 1 1 u1 1 2   0.6 + 0.6 + = ∈ tD1 (0.5 t)0 Dt u2 (t) u2 (t) λFu2 (t, u1 (t) , u2 (t)) , t [0, 1] ,    u1 (o) = u1(1) = 0, u2 (o) = u2(1) = 0.

Taking U1 (t) = Γ (1.2) t (1 − t) , U2 (t) = Γ (1.4) t (1 − t) and F (t, u1 (t) , u2 (t)) = (1 + t) H (u1, u2) , where  2 u3 + u3 , u3 + u3 ≤ 1,  1 2 1 2 H (u1, u2) = 1 1  3 3 2 3 3 3 3 3 10 u1 + u2 − 9 u1 + u2 , u1 + u2 > 1.

0 0 Clearly, F (t, 0, 0) = 0, w1 = 1 and w2 = 0.5 for any t ∈ [0, 1] By the direct calculation, we have ( ) 1 1 max , (Γ (0.8))3 ((0.8 − 1) × 1.5 + 1)2 ((Γ (0.6))3 (0.6 − 1) × 1.5 + 1)2 × 0.5

= k ≈ 3.7849, ( ) 1 1 max , = k˜ ≈ 2.803 (Γ(0.8 + 1))3 (Γ(0.6 + 1))3 × 0.5 and

 ( ) t0.2 − 2Γ 1.2 t1.2, t ∈ [0, 0.6] 0.8  Γ(2.2) 0D U1(t) = t − 0.2 − 2Γ(1.2) 1.2 ∈ ( ]  t Γ(2.2) t , t 0.6, 1 ,  ( ) t0.4 − 2Γ 1.35 t1.4, t ∈ [0, 0.7] 0.6  Γ(2.4) 0D U2(t) = t − 0.4 − 2Γ(1.35) 1.4 ∈ ( ]  t Γ(2.35) t , t 0.7, 1 .

So that

3 3 kU1(t)k0.8 ≈ 0.09228, kU2(t)k0.6 ≈ 0.0212 3 3 kU1(t)kL3 ≈ 0.0053, kU2(t)kL3 ≈ 0.005

Take r = 1 × 10−5. We easily obtain that

1 3 3 1 3 3 kU (t)k + kU (t)k + kU (t)k 3 + kU (t)k 3 3 1 0.8 2 0.65 3 1 L 2 L

≈ 0.041 > r Mathematics 2020, 8, 475 16 of 18 which implies that the condition (i) holds, and

R1 sup F (t, u1, u2) dt 0 (u ,u )∈π(kr) 27k2r 1 2 = ≈ 0.00193 r 2

R1 3 F (t, U1, U2) dt < 0 ≈ 0.00656 2 2 3 3 kU k + kU k 3 ∑ i αi ∑ i L i=1 i=1 and R1 sup F (t, u1, u2) dt F (t, u , u ) 0 (u ,u )∈π(kr) = 1 2 ≤ 1 2 ≈ 0 lim inf 3 3 ∼ 0.00023. u1→+∞,u2→+∞ u + u 1 2 3rk Thus, conditions (ii) and (iii) are satisﬁed. Then, in view of Theorem 2 for each λ ∈ (152.43, 518.13), the system (4) admits at least three weak solutions in X.

5. Conclusions Fractional differential equations have been carefully investigated. Such problems were studied in many scientiﬁc and engineering applications such as models for various processes in plasma physics, biology, medical science, chemistry, chemical engineering, as well as population dynamics, and control theory. In the current contribution, motivated by work in ([21,22]) and using a three critical points theorem obtained in [23], we could ensure the existence of at least three solutions for system (4). Note that some appropriate function spaces and variational frameworks were successfully created for the system (4). Furthermore, we have given two examples to illustrate the application of Theorem 2 we have discussed for the special case of p = 2, and the discussions presented with respect to the case of p 6= 2, which highlighted the superiority of our results. In next work we will use two control parameters to study a class of perturbed nonlinear fractional p-Laplacian differential systems where we will try to prove the existence of three weak solutions by using the variational method and Ricceri’s critical points theorems respecting some necessary conditions.

Author Contributions: Conceptualization, methodology and writing-original manuscript, F.K. and R.G.; supervision, S.B. and R.G.; writing review and editing, S.B. and A.A., formal analysis, S.B. and A.A. All authors have read and agreed to the published version of the manuscript. Funding: This research received no external funding. Acknowledgments: The authors would like to thank the anonymous referees and the handling editor for their careful reading and for relevant remarks/suggestions which helped them to improve the paper. Conﬂicts of Interest: The authors declare no conﬂict of interest.

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