Existence of the positive solutions for a tripled system of fractional differential equations via integral boundary conditions

Available online at www.resultsinnonlinearanalysis.com Research Article

Existence of the positive solutions for a tripled system of fractional dierential equations via integral

boundary conditions

Hojjat Afshari^a, Hadi Shojaat^b, Mansoureh Siahkali Moradi^c

aDepartment of Mathematics, Faculty of Sciences, University of Bonab, Bonab, Iran.

bDepartment of Mathematics, Farhangian University, Qazvin, Iran.

cDepartment of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran.

Abstract

The purpose of this paper, is studying the existence and nonexistence of positive solutions to a class of a following tripled system of fractional dierential equations.















D^αu(ζ) + a(ζ)f (ζ, v(ζ), ω(ζ)) = 0, u(0) = 0, u(1) =R1

0 φ(ζ)u(ζ)dζ, D^βv(ζ) + b(ζ)g(ζ, u(ζ), ω(ζ)) = 0, v(0) = 0, v(1) =R1

0 ψ(ζ)v(ζ)dζ, D^γω(ζ) + c(ζ)h(ζ, u(ζ), v(ζ)) = 0, ω(0) = 0, ω(1) =R₁

0 η(ζ)ω(ζ)dζ,

where 0 ≤ ζ ≤ 1, 1 < α, β, γ ≤ 2, a, b, c ∈ C((0, 1), [0, ∞)), φ, ψ, η ∈ L¹[0, 1] are nonnegative and f, g, h ∈ C([0, 1] × [0, ∞) × [0, ∞), [0, ∞))and D is the standard Riemann-Liouville fractional derivative.

Also, we provide some examples to demonstrate the validity of our results.

Keywords: Tripled system, fractional dierential equation, integral boundary conditions, existence and nonexistence of positive solutions.

2010 MSC: 34A08, 34B37, 35R11

Email addresses: hojat.afshari@yahoo.com (Hojjat Afshari), hadishojaat@yahoo.com (Hadi Shojaat), mansorehmoradi@gmail.com (Mansoureh Siahkali Moradi)

Received : May 18, 2021; Accepted: August 25, 2021; Online: August 28, 2021.

1. Introduction

E. Karapnar and coauthors obtained some xed point results and applied them to proving the existence and uniqueness of positive solutions for functional boundary value problem (see [1]-[17], [15], [26]). In recent years, some systems of nonlinear fractional dierential equations were examined by many authors, [18]-[25]

and other references. In [27], Su investigated some conditions for the existence of solutions for a coupled system of two-point fractional boundary value problem.

In [31] the authors studied the existence and nonexistence of positive solutions to boundary values problem for a coupled system of nonlinear fractional dierential equations as follows:











D^αu(ζ) + a(ζ)f (ζ, v(ζ)) = 0, u(0) = 0, u(1) =R1

0 φ(ζ)u(ζ)dζ, D^βv(ζ) + b(ζ)g(ζ, u(ζ)) = 0, v(0) = 0, v(1) =R₁

0 ψ(ζ)v(ζ)dζ, (1)

where 0 ≤ ζ ≤ 1, 1 < α, β ≤ 2, a, b ∈ C((0, 1), [0, ∞)), φ, ψ ∈ L¹[0, 1] are nonnegative and f, g ∈ C([0, 1] × [0, ∞), [0, ∞))and D is the standard Riemann-Liouville fractional derivative.

In this paper we study the equations















D^αu(ζ) + a(ζ)f (ζ, v(ζ), ω(ζ)) = 0, u(0) = 0, u(1) =R1

0 φ(ζ)u(ζ)dζ, D^βv(ζ) + b(ζ)g(ζ, u(ζ), ω(ζ)) = 0, v(0) = 0, v(1) =R₁

0 ψ(ζ)v(ζ)dζ, D^γω(ζ) + c(ζ)h(ζ, u(ζ), v(ζ)) = 0, ω(0) = 0, ω(1) =R1

0 η(ζ)ω(ζ)dζ,

(2)

where 0 ≤ ζ ≤ 1, 1 < α, β, γ ≤ 2, a, b, c ∈ C((0, 1), [0, ∞)), φ, ψ, η ∈ L¹[0, 1] are nonnegative and f, g, h ∈ C([0, 1] × [0, ∞) × [0, ∞), [0, ∞))and D is the standard Riemann-Liouville fractional derivative.

Denition 1.1. [28, 29] The Riemann-Liouville fractional derivative for a continuous function f is dened by

D^νf (τ ) = 1

Γ(n − ν)( d dτ)ⁿ

Z τ 0

f (ζ)

(τ − ζ)^ν−n+1dζ, (n = [ν] + 1) where the right-hand side is point-wise dened on (0, ∞).

Denition 1.2. [28, 29] Let [a, b] be an interval in R and ν > 0. The Riemann-Liouville fractional order integral of a function f ∈ L¹([a, b], R) is dened by

I_a^νf (τ ) = 1 Γ(ν)

Z τ a

f (ζ) (τ − ζ)^1−νdζ, whenever the integral exists.

Lemma 1.3. (Nonlinear Dierentiation of Leray–Schauder Type, [32]). Let E be a Banach space with C ⊂ E closed and convex. Let U be a relatively open subset of C with 0 ∈ U and let T : U → C be a continuous and compact mapping. Then either

(a) the mapping T has a xed point in U,

or (b) there exist u ∈ ∂U and λ ∈ (0, 1) with u = λT u.

Lemma 1.4. (Fixed-Point Theorem of Cone Expansion and Compression of Norm Type, See [33]). Let P be a cone of real Banach space E, and let Ω1 and Ω2 be two bounded open sets in E such that 0 ∈ Ω1⊂ Ω₁ ⊂ Ω₂. Let operator A : P ∩(Ω2−Ω₁) → P be completely continuous operator. Suppose that one of the two conditions holds:

(i₁) kAuk ≤ kuk, for all u ∈ P ∩ ∂Ω1; kAuk ≥ kuk, for all u ∈ P ∩ ∂Ω2; (i2) kAuk ≥ kuk, for all u ∈ P ∩ ∂Ω1; kAuk ≤ kuk, for all u ∈ P ∩ ∂Ω2. Then A has at least one xed point in P ∩ (Ω2− Ω₁).

Lemma 1.5. Assume that R₀¹ζ^ν−1φ(ζ)dζ 6= 1. Then for any σ ∈ C[0, 1], the unique solution of boundary value problem

 D^νu(ζ) + σ(ζ) = 0, 0 < τ < 1, u(0) = 0, u(1) =R1

0 φ(ζ)u(ζ)dζ, is given by

u(ζ) = Z ₁

0

G_1ν(ζ, τ )σ(τ )dτ where

G_1ν(ζ, τ ) = G_2ν(ζ, τ ) + G_3ν(ζ, τ ), (ζ, τ ) ∈ [0, 1] × [0, 1], (3) with

G2ν(ζ, τ ) = 1 Γ(ν)

 ζ^ν−1(1 − τ )^ν−1− (ζ − τ )^ν−1, 0 ≤ τ ≤ ζ ≤ 1, ζ^ν−1(1 − τ )^ν−1, 0 ≤ ζ ≤ τ ≤ 1

and

G_3ν(ζ, τ ) = ^ζν−1

1−R1

0 φ(ζ)ζ^ν−1dζ

R1

0 G_2ν(ζ, τ )φ(ζ)dζ.

We call G = (G1ν, G_1ν⁰, G_1ν⁰⁰) the Green’s functions of the boundary value problem (2).

Lemma 1.6. If R₁⁰ϕ(τ )τ^ν−1dτ ∈ [0, 1), the function G1ν(τ, ζ) dened by (3) satises

(i₁) G_1ν(τ, ζ) ≥ 0 is continuous for all τ, ζ ∈ [0, 1], G1ν(τ, ζ) > 0 for all τ, ζ ∈ (0, 1);

(i₂) G_1ν(τ, ζ) ≤ G_1ν(ζ)for each τ, ζ ∈ (0, 1), and min_{τ ∈[θ,1−θ]}G1ν(τ, ζ) ≥ G1ν(ζ), where θ ∈ (0,¹₂) and

G_1ν(ζ) = G_2ν(ζ, ζ) + G_3ν(1, ζ), Υ_ν = θ^ν−1.

We will discuss the existence of positive solutions for boundary value problem (2). First of all, we dene the Banach space

X = {u(ζ)|u(ζ) ∈ C[0, 1]} endowed with the normkuk_X = max

ζ∈[0,1]|u|, Y = {v(ζ)|v(ζ) ∈ C[0, 1]} endowed with the normkvkY = max

ζ∈[0,1]|v|, Z = {ω(ζ)|ω(ζ) ∈ C[0, 1]} endowed with the normkωk_Z = max

ζ∈[0,1]

|ω|.

For (u, v, ω) ∈ X × Y × Z, let k(u, v, ω)kX×Y ×Z = max{kukX, kvkY, kωkZ}. Clearly, (X × Y × Z, k(u, v, ω)k_{X×Y ×Z})is a Banach space. Dene,

P = {(u, v, ω) ∈ X × Y × Z|u(ζ) ≥ 0, v(ζ) ≥ 0, ω(ζ) ≥ 0}, then the cone P ⊂ X × Y × Z. Let Jθ = [θ, 1 − θ]

for θ ∈ (0,¹₂)and

K =

 (u, v, ω) ∈ P, min_{τ ∈Jθ}u(τ ) ≥ Υ_αkuk,

min_{τ ∈Jθ}v(τ ) ≥ Υ_βkvk, min_{τ ∈Jθ}ω(τ ) ≥ Υ_γkωk

 , Kr =

(u, v, ω) ∈ K, k(u, v, ω)k ≤ r ,

∂K_r=

(u, v, ω) ∈ K, k(u, v, ω)k = r . From Lemma 1.5, we can obtain the following lemma.

Lemma 1.7. Suppose that f(t, v, ω), g(t, u, ω) and h(t, u, v) are continuous, then (u, v, ω) ∈ X × Y × Z is a solution of B.V.P (2) if and only if

(u, v, ω) ∈ X × Y × Z is a solution of the integral equations















u(ζ) =R1

0 G_1α(ζ, τ )a(τ )f (τ, v(τ ), ω(τ ))dτ, v(ζ) =R₁

0 G_1β(ζ, τ )b(τ )g(τ, u(τ ), ω(τ ))dτ, ω(ζ) =R1

0 G1γ(ζ, τ )c(τ )h(τ, u(τ ), v(τ ))dτ.

Let T : X × Y × Z → X × Y × Z be the operator dened as T (u, v, ω) =

Z 1 0

G_1α(ζ, τ )a(τ )f (τ, v(τ ), ω(τ ))dτ, (4)

Z 1 0

G_1β(ζ, τ )b(τ )g(τ, u(τ ), ω(τ ))dτ, Z 1

0

G_1γ(ζ, τ )c(τ )h(τ, u(τ ), v(τ ))dτ

=: (T1u(ζ), T2v(ζ), T3ω(ζ)), (5)

then by Lemma (1.7), the xed point of operator T coincides with the solution of system (2).

Lemma 1.8. Let f(τ, v, v), g(τ, u, u) and h(τ, ω, ω) be continuous on

[0, 1] × [0, +∞) × [0, +∞) → [0, +∞), then T : P → P , T : K → K dened by (4) are completely continuous.

Proof. Since Lemma (1.8) is similar to Lemma (1.8) in [2] and [30] we omit the proof Lemma (1.8).

Theorem 1.9. Assume that a(τ), b(τ) and c(τ), are continuous on

(0, 1) → [0, +∞)and f(τ, v(τ), ω(τ)), g(τ, u(τ), ω(τ)) and h(τ, u(τ), v(τ)) are continuous on [0, 1] × [0, ∞) × [0, ∞) → [0, ∞), and there exist three positive functions m(τ), n(τ) and k(τ) that satisfy

(L₁) f (τ, v₂, ω₂) − f (τ, v₁, ω₁) ≤ m(τ ) max{|v₂− v₁|, |ω₂− ω₁|}, (L2) g(τ, u2, ω2) − g(τ, u1, ω1) ≤ k(τ ) max{|u2− u₁|, |ω₂− ω₁|}, (L3) h(τ, u2, v2) − h(τ, u1, v1) ≤ n(τ ) max{|v2− v₁|, |u₂− u₁|},

for τ ∈ (0, 1), v1, v₂, ω₁, ω₂, u₁, u₂∈ (0, +∞). Then system (2) has a unique positive solution if

ρ = Z 1

0

G1α(τ )a(τ )m(τ )dτ < 1, θ =

Z 1 0

G_1β(τ )b(τ )k(τ )dτ < 1, κ =

Z 1 0

G1γ(τ )c(τ )n(τ )dτ < 1. (6)

Proof. For all (u, v, ω) ∈ P by the nonnegativeness of G(ζ, τ) and a(τ), b(τ), c(τ), f(τ, v(τ), ω(τ)), g(τ, u(τ), ω(τ)), h(τ, u(τ ), v(τ )), we have

T (u, v, ω) ≥ 0. Hence, T (P ) ⊂ P . From Lemma 1.6, we obtain kT₁v₂− T₁v₁k = max

ζ∈[0,1]

|T₁v₂− T₁v₁|

= max

ζ∈[0,1]

| Z 1

0

G1α(ζ, τ )a(τ )[f (τ, v2(τ ), ω(τ )) − f (τ, v1(τ ), ω(τ ))])dτ |

≤ Z 1

0

G_1α(τ )a(τ )m(τ ))dτ |kv₂− v₁k = ρkv₂− v₁k (7)

Similarly,

kT₂u₂− T₂u₁k ≤ θku₂− u₁k (8)

and

kT₃ω2− T₃ω1k ≤ κkω₂− ω₁k (9)

From (7), (8) to (9), we get

kT (u₂, v₂, ω₂) − T (u₁, v₁, ω₁)k ≤ max(ρ, θ, κ)k(u₂, v₂, ω₂) − (u₁, v₁, ω₁)k

From Lemma (1.8), T is completely continuous, by Banach xed point theorem, the operator T has a unique

xed point in P , which is the unique positive solution of system (2). This completes the proof.

Theorem 1.10. Assume that a(τ), b(τ) and c(τ), are continuous on

(0, 1) → [0, +∞)and f(τ, v(τ), ω(τ)), g(τ, u(τ), ω(τ)) and h(τ, u(τ), v(τ)) are continuous on [0, 1] × [0, ∞) × [0, ∞) → [0, ∞), and satisfy

(L₄) |f (τ, v(τ ), ω(τ ))| ≤ a₁(τ ) + a₂(τ ) max{|v(τ )|, |ω(τ )|}, (L5) |g(τ, u(τ ), ω(τ ))| ≤ b1(τ ) + b2(τ ) max{|u(τ )|, |ω(τ )|}, (L₆) |h(τ, u(τ ), v(τ ))| ≤ c₁(τ ) + c₂(τ ) max{|v(τ )|, |u(τ )|}, (L₇) A₁ =R1

0 G_1α(τ )a(τ )a₂(τ )dτ < 1, B₁=R1

0 G_1α(τ )a(τ )a₁(τ )dτ < ∞, (L8) A2 =R1

0 G_1β(τ )b(τ )b2(τ )dτ < 1, B2 =R1

0 G_1β(τ )b(τ )b1(τ )dτ < ∞, (L₉) A₃ =R1

0 G_1γ(τ )c(τ )c₂(τ )dτ < 1, B₃ =R1

0 G_1γ(τ )c(τ )c₁(τ )dτ < ∞. Then the system (2) has at least one positive solution (u, v, ω) in

Q =n

(u, v, ω) ∈ P : k(u, v, ω)k < min(_1−B^A1

1,_1−B^A2

2,_1−B^A3

3) o . Proof. Let Q =  (u, v, ω) ∈ P : k(u, v, ω)k < r with

r = min( A₁ 1 − B1

, A₂ 1 − B2

, A₃ 1 − B3

).

Dene the operator T : Q → P as (4). Let (u, v, ω) ∈ Q, that is, k(u, v, ω)k < r. Then

kT₁uk = max

ζ∈[0,1]

| Z ₁

0

G_1α(ζ, τ )a(τ )f (τ, v(τ ), ω(τ ))dτ |

≤ Z 1

0

G1α(τ )a(τ )(a1(τ ) + a2(τ )|v(τ )|)dτ

≤ Z 1

0

G1α(τ )a(τ )a1(τ )dτ + Z 1

0

G1α(τ )a(τ )a2(τ )dτ kv(τ )k

= B1+ A1kv(τ )k ≤ r.

Similarly, kT2vk ≤ r, kT3ωk ≤ r. So, T (u, v, ω) ≤ (r, r, r) and hence T (u, v, ω) ∈ Q. From Lemma (1.8), we have T : Q → Q is completely continuous. Consider the eigenvalue problem

(u, v, ω) = λT (u, v, ω), λ ∈ (0, 1). (10)

Under the assumption that (u, v, ω) is a solution of (10) for λ ∈ (0, 1), we have kuk = kλT₁uk = max

ζ∈[0,1]

| Z ₁

0

G_1α(ζ, τ )a(τ )f (τ, v(τ ), ω(τ ))dτ |

≤ Z 1

0

G1α(τ )a(τ )(a1(τ ) + a2(τ )|v(τ )|)dτ

≤ Z 1

0

G1α(τ )a(τ )a1(τ )dτ + Z 1

0

G1α(τ )a(τ )a2(τ )dτ kv(τ )k

= B1+ A1kv(τ )k ≤ r.

Similarly, kvk = kT2λvk ≤ r, kωk = kT3λωk ≤ r, so, k(u, v, ω)k ≤ r, which shows that (u, v, ω) ∈ ∂Q. By Lemma 1.3, T has a xed point in Q. We complete the proof of theorem 1.10.

Remark 1.11. In the following we need the following assumptions and some notations:

(B₁) a, b, c ∈ C((0, 1), [0, ∞)), a(τ) 6= 0, b(τ) 6= 0, c(τ) 6= 0 on any subinterval of (0, 1) and 0 <

Z 1 0

G1α(τ )a(τ )dτ < ∞, 0 < R1

0 G_1β(τ )b(τ )dτ < ∞ and 0 < R₀¹G_1γ(τ )c(τ )dτ < ∞ where G1α, G1β and G1γ are dened in Lemma 1.6;

(B₂) f, g, h ∈ C([0, 1] × [0, ∞) × [0, ∞), [0, ∞)) and f(ζ, 0, 0) = 0,

g(ζ, 0, 0) = 0 and h(ζ, 0, 0) = 0 uniformly with respect to ζ on [0, 1];

(B3) λ, µ, ν ∈ [0, 1) where λ, µ, ν is dened as follows:

λ = Z 1

0

φ(ζ)ζ^α−1dζ, µ = Z 1

0

ψ(ζ)ζ^β−1dζ and ν = Z 1

0

ϕ(ζ)ζ^γ−1dζ.

let

f^δ= lim sup

u→δ

ζ∈[0,1]max

f (ζ, u, u)

u , fδ= lim inf

u→δ min

ζ∈[0,1]

f (ζ, u, u)

u ,

where δ denotes 0 or ∞, and σ1 =

Z 1 0

G1α(τ )a(τ )dτ, σ2 = Z 1

0

G1β(τ )b(τ )dτ and σ3 = Z 1

0

G1γ(τ )c(τ )dτ.

Theorem 1.12. Assume that (B1)−(B₃)hold. And supposes that one of the following conditions is satised:

(H₁) f₀> ¹

Υ²_αR1−θ

θ G1α(τ )a(τ )dτ and f^∞< _σ¹

1 (particularly, f⁰= ∞ and f^∞= 0);

g₀ > ¹

Υ²_βR1−θ

θ G_1β(τ )b(τ )dτ and g^∞< _σ¹

2 (particularly, g⁰ = ∞and g^∞= 0);

h₀ > ¹

Υ²_γR1−θ

θ G1γ(τ )c(τ )dτ and h^∞< _σ¹

3 (particularly, h⁰ = ∞ and h^∞= 0).

(H₂) There exist two constants r2, R₂ with 0 < r2 ≤ R₂ such that f(ζ, ., .), g(ζ, ., .) and h(ζ, ., .) are nonde- creasing on [0, R2]for all ζ ∈ [0, 1],

f (ζ, Υαr2, Υαr2) ≥ r2

Υ²_αR_1−θ

θ G1α(τ )a(τ )dτ, g(ζ, Υ_βr₂, Υ_βr₂) ≥ r₂

Υ²_βR1−θ

θ G_1β(τ )b(τ )dτ, h(ζ, Υ_γr₂, Υ_γr₂) ≥ r₂

Υ²_γR1−θ

θ G_1γ(τ )c(τ )dτ and f(ζ, R2, R2) ≤ ^R_σ²

1, g(ζ, R2, R2) ≤ ^R_σ²

2, h(ζ, R2, R2) ≤ ^R_σ²

3 for all ζ ∈ [0, 1]. Then boundary value problem (2) has at least one positive solution.

Proof. Let T be cone preserving completely continuous that is dened by (4).

Case1. The condition (H1) holds. Considering f0 > ¹

Υ²_αR1−θ

θ G1α(τ )a(τ )dτ, there exists r1 > 0 such that f (t, v, v) = (f0− ε₁)v, for all t ∈ [0, 1], v ∈ [0, r1], where ε1> 0, satises

(f0− ε₁)Υ²_α Z 1−θ

θ

G1α(τ )a(τ )dτ ≥ 1.

Then, for t ∈ [0, 1], (u, v, ω) ∈ ∂Kr1, we get

T₁v(t) = Z 1

0

G_1α(ζ, τ )a(τ )f (τ, v(τ ), v(τ ))dτ

≥ Υ_α Z 1

0

G_1α(τ )a(τ )f (τ, v(τ ), v(τ ))dτ

≥ Υ_α Z 1

0

G_1α(τ )a(τ )(f₀− ε₁)v(τ )dτ

≥ (f₀− ε₁)Υ²_α Z 1

0

G_1α(τ )a(τ )dτ kvk

≥ kvk.

Similarly, we have T2ω(t) ≥ kωk, T3u(t) ≥ kukthat is (u, v, ω) ∈ ∂K_r1 implies that

kT (u, v, ω)k ≥ k(u, v, ω)k. (11)

On the other hand, for f^∞< 1/σ₁, there exists R1 > 0 such that f(t, v, v) = (f∞+ ε₂)v, for t ∈ [0, 1], v ∈ (R1, +∞), where ε2 > 0 satises σ1(f^∞+ ε2) = 1. Set M = maxt∈[0,1],v∈[0,R1]f (t, v, v), then f (t, v, v) = M + (f^∞+ ε₂)v. Choose

R₁ > max{r₁, R₁, M σ₁(1 − σ₁(f^∞+ ε₂))⁻¹}. Then, for t ∈ [0, 1], (u, v, ω) ∈ ∂KR1 , we get T1v(t) =

Z 1 0

G1α(ζ, τ )a(τ )f (τ, v(τ ), v(τ ))dτ

≤ Z 1

0

G1α(τ )a(τ )(M + (f^∞+ ε2))v(τ )dτ

≤ M Z 1

0

G1α(τ )a(τ )dτ + Z 1

0

G1α(τ )a(τ )(f^∞+ ε2)dτ kvk

≤ R₁− σ₁(f^∞+ ε2)R1+ (f^∞+ ε2)σ1kvk

≤ R₁.

Similarly, we have T3u(t) ≤ kuk, T2ω(t) ≤ kωk that is (u, v, ω) ∈ ∂KR1 implies that

kT (u, v, ω)k ≤ k(u, v, ω)k. (12)

Case2. The condition (H2)holds. For (u, v, ω) ∈ K, from the denition of K, we obtain that mint∈Jθu(t) ≥ Υαkuk, min_t∈Jθv(t) ≥ Υβkvk, min_t∈Jθω(t) ≥ Υγkωk.

Therefore, for (u, v, ω) ∈ ∂Kr2 , we have k(u, v, ω)k = r2 for t ∈ Jθ . From (H2), we have

T1v(t) = Z 1

0

G1α(ζ, τ )a(τ )f (τ, v(τ ), v(τ ))dτ

≥ Υ_α Z 1−θ

θ

G_1α(τ )a(τ )f (τ, v(τ ), v(τ ))dτ

≥ Υ_α r2

Υα

R1−θ

θ G1α(τ )a(τ )dτ Z 1−θ

θ

G1α(τ )a(τ )dτ

= r2.

Similarly, we have T3u(t) ≥ r2, T2ω(t) ≥ r2 that is (u, v, ω) ∈ ∂Kr2 implies that

kT (u, v, ω)k ≥ k(u, v, ω)k (13)

On the other hand, for (u, v, ω) ∈ ∂KR2 , we have that (u, v, ω) = R2 for t ∈ [0, 1], from (H2), we obtain

T₁v(t) = Z ₁

0

G_1α(ζ, τ )a(τ )f (τ, v(τ ), v(τ ))dτ

≤ Υ_α Z 1

0

G_1α(τ )a(τ )f (τ, v(τ ), v(τ ))dτ

≤ R₂ σ₁Υ_α

Z 1 0

G_1α(τ )a(τ )dτ

= R₂.

Similarly, we have T3u(t) ≤ R2, T2ω(t) ≤ R2 that is (u, v, ω) ∈ ∂KR2 implies that

kT (u, v, ω)k ≤ k(u, v, ω)k. (14)

Applying Lemma 1.4 to (11) and (12), or (13) and (14), yields that T has a xed point (u, v, ω) ∈ Kr,R

or (u, v, ω) ∈ Kri,Ri(i = 1, 2)with u(t) = Υαkuk > 0, v(t) = Υβkvk > 0and ω(t) = Υγkωk > 0. Thus it follows that boundary value problems (1.1) has a positive solution (u, v, ω). We complete the proof of Theorem 1.12.

Similarly, we have the following result.

Theorem 1.13. Assume that (B1)−(B3)hold. And supposes that the following three conditions are satised:

(H3) f⁰ < _σ¹

1 and f∞> ¹

Υ²_αR1−θ

θ G1α(τ )a(τ )dτ (particularly, f⁰= 0 and f∞= ∞);

g⁰ < _σ¹

2 and g∞> ¹

Υ²_βR1−θ

θ G1β(τ )b(τ )dτ (particularly, g⁰ = 0 and g∞= ∞);

h⁰ < _σ¹

3 and h∞> ¹

Υ²_γR1−θ

θ G1γ(τ )c(τ )dτ (particularly, h⁰ = 0 and h∞= ∞).

Then boundary value problem (2) has at least one positive solution.

Theorem 1.14. Assume that (B1) − (B₃) hold. And supposes that the following two conditions are satised:

(H₄) f₀> ¹

Υ²_αR1−θ

θ G1α(τ )a(τ )dτ and f∞> ¹

Υ²_αR1−θ

θ G1α(τ )a(τ )dτ

(particularly, f⁰ = f∞= ∞);

g₀ > ¹

Υ²_βR1−θ

θ G1β(τ )b(τ )dτ and g∞> ¹

Υ²_βR1−θ

θ G1β(τ )b(τ )dτ

(particularly, g⁰ = g∞= ∞);

h0 > ¹

Υ²_γR1−θ

θ G1γ(τ )c(τ )dτ and h∞> ¹

Υ²_γR1−θ

θ G1γ(τ )c(τ )dτ

(particularly, h⁰ = h∞= ∞).

(H₅) there exists b > 0 such that max

ζ∈[0,1],(u,v,ω)∈∂Kb

f (ζ, u, u) < b/σ₁, max

ζ∈[0,1],(u,v,ω)∈∂Kb

g(ζ, v, v) < b/σ₂

and

max

ζ∈[0,1],(u,v,ω)∈∂Kb

h(ζ, ω, ω) < b/σ₃. Then boundary value problem (2) has at least two positive solutions (u1, v₁, ω₁), (u2, v2, ω2), which satisfy

0 < k(u₁, v₁, ω₁)k < b < k(u₂, v₂, ω₂)k. (15) Proof. We consider condition (H4). Choose r, R with 0 < r < b < R.

If f0> ¹

Υ²_αR1−θ

θ G1α(τ )a(τ )dτ, g0> ¹

Υ²_βR1−θ

θ G1β(τ )b(τ )dτ and h0 > ¹

Υ²_γR1−θ

θ G1γ(τ )c(τ )dτ, then similar to the proof of (11), we have

kT (u, v, ω)k ≥ k(u, v, ω)k, for

(u, v, ω) ∈ ∂K_r. (16)

If f∞> ¹

Υ²_αR1−θ

θ G1α(τ )a(τ )dτ, g∞> ¹

Υ²_βR1−θ

θ G1β(τ )b(τ )dτ and h∞> ¹

Υ²_γR1−θ

θ G1γ(τ )c(τ )dτ, then similar to the proof of (3.6), we have

kT (u, v, ω)k ≥ k(u, v, ω)k, f or (u, v, ω) ∈ ∂K_R. (17)

On the other hand, together with (H5), (u, v, ω) ∈ ∂Kb, we have

T₁v(ζ) = Z 1

0

G_1α(ζ, τ )a(τ )f (τ, v(τ ), ω(τ ))dτ

≤ Z 1

0

G1α(τ )a(τ )f (τ, v(τ ), ω(τ ))dτ

< b σ1

Z 1 0

G1α(τ )a(τ )dτ

= b.

Similarly, we have T3u(ζ) < b, T2ω(ζ) < b, that is (u, v, ω) ∈ ∂Kb implies that

kT (u, v, ω)k < k(u, v, ω)k. (18)

Applying Lemma 1.4 to (16) − (18) yields that T has a xed point

(u1, v1, ω1) ∈ ∂Kr,b, and a xed point (u2, v2, ω2) ∈ ∂Kb,R. Thus it follows that boundary value problem (2) has at least two positive solutions (u1, v1, ω1) and (u2, v2, ω2). Noticing (18), we have (u1, v1, ω1) 6= b and (u₂, v₂, ω₂) 6= b. Therefore (15) holds, and the proof is complete.

Similarly, we have the following results.

Theorem 1.15. Assume that (B1) − (B3) hold. And supposes that the following conditions is satised:

(H6) f⁰ < 1/σ1 and f^∞< 1/σ1; g⁰ < 1/σ2 and g^∞< 1/σ2; h⁰< 1/σ3 and h^∞< 1/σ3. (H7) there exists B > 0 such that

max

ζ∈[0,1],(u,v,ω)∈∂KB

f (ζ, u, u) > B Υ_αR1−θ

θ G_1α(τ )a(τ )dτ, max

ζ∈[0,1],(u,v,ω)∈∂KB

g(ζ, v, v) > B Υ_βR1−θ

θ G_1β(τ )b(τ )dτ, max

ζ∈[0,1],(u,v,ω)∈∂KB

h(ζ, ω, ω) > B Υ_γR1−θ

θ G_1γ(τ )c(τ )dτ, Then boundary value problem (2) has at least two positive solutions (u1, v₁, ω₁), (u2, v2, ω2), which satisfy

0 < k(u₁, v₁, ω₁)k < B < k(u₂, v₂, ω₂)k.

Theorem 1.16. Assume that (B1) − (B3) hold. If there exist 3l positive numbers dk, Dk, k = 1, 2, · · · , l with

d₁< Υ_αD₁ < D₁ < d₂< Υ_αD₂ < D₂ < · · · < d_l< Υ_αD_l < D_l,

d₁< Υ_βD₁ < D₁< d₂ < Υ_βD₂ < D₂ < · · · < d_l< Υ_βD_l< D_l and

d1 < ΥγD1 < D1< d2 < ΥγD2 < D2 < · · · < d_l< ΥγD_l< D_l, such that

(H8)

f (ζ, u, u) > d_k Υ_αR1

0 G_1α(τ )a(τ )dτ, for

(ζ, u, u) ∈ [0, 1] × [Υαd_k, d_k] × [Υαd_k, d_k] and

f (ζ, u, u) = σ⁻¹₁ D_k for

(ζ, u, u) ∈ [0, 1] × [Υ_αD_k, D_k] × [Υ_αD_k, D_k], k = 1, 2, · · · , l.

Also

g(ζ, v, v) > d_k Υβ

R1

0 G1β(τ )b(τ )dτ,

for

(ζ, v, v) ∈ [0, 1] × [Υβdk, dk] × [Υβdk, dk] and

g(ζ, v, v) = σ₁⁻¹Dk

for

(ζ, v, v) ∈ [0, 1] × [Υ_βD_k, D_k] × [Υ_βD_k, D_k], k = 1, 2, · · · , l.

And also

h(ζ, ω, ω) > d_k Υ_γR1

0 G_1γ(τ )c(τ )dτ; for

(ζ, ω, ω) ∈ [0, 1] × [Υγd_k, d_k] × [Υγd_k, d_k] and

h(ζ, ω, ω) = σ₁⁻¹D_k for

(ζ, ω, ω) ∈ [0, 1] × [Υ_γD_k, D_k] × [Υ_γD_k, D_k], k = 1, 2, · · · , l.

Then boundary value problem (2) has at least l positive solutions (uk, vk, ωk) which satisfy d_k< k(u_k, v_k, ω_k)k < D_k, k = 1, 2, · · · , l.

Theorem 1.17. Assume that (B1) − (B₃) hold. If there exist 3l positive numbers dk, Dk, k = 1, 2, · · · , l with d1< D₁ < d₂ < D₂ < · · · < d_l< D_l such that

(H9) f (ζ, ., .), g(ζ, ., .) and h(ζ, ., .) are nondecreasing on [0, Dl]for all t ∈ [0, 1].

(H₁₀)

f (ζ, Υ_αd_k, Υ_αd_k) ≥ d_k Υ_αR1−θ

θ G_1α(τ )a(τ )dτ, and

f (ζ, D_k, D_k) ≤ σ₁⁻¹D_k, k = 1, 2, · · · , l.

Also

g(ζ, Υ_βd_k, Υ_βd_k) ≥ d_k Υ_βR1−θ

θ G_1β(τ )b(τ )dτ, and

g(ζ, Dk, Dk) ≤ σ⁻¹₁ Dk, k = 1, 2, · · · , l.

And also

h(ζ, Υγdk, Υγdk) ≥ dk

Υγ

R1−θ

θ G1γ(τ )c(τ )dτ; and

h(ζ, D_k, D_k) ≤ σ₁⁻¹D_k, k = 1, 2, · · · , l.

Then boundary value problem (2) has at least l positive solutions (uk, vk, ωk) which satisfy d_k< k(u_k, v_k, ω_k)k < D_k, k = 1, 2, · · · , l.

Now the nonexistence of positive solutions for boundary value problem (2).

Theorem 1.18. Suppose (B1) − (B₃) hold, f(ζ, u, u) < σ₁¹u, g(ζ, v, v) < σ¹₂v and h(ζ, ω, ω) < σ₃¹ω for all ζ ∈ [0, 1], u > 0, v > 0 and ω > 0 then boundary value problem (2) has no positive solution.

Proof. Assume to the contrary that (u, v, ω) is a positive solution of the boundary value problem (2). Then (u, v, ω) ∈ K, u > 0, v > 0 and ω > 0 for ζ ∈ [0, 1], and

kuk = max

ζ∈[0,1]|u(ζ)| = max

ζ∈[0,1]

Z 1 0

G_1α(ζ, τ )a(τ )f (τ, v(τ ), v(τ ))dτ

≤ Z ₁

0

G_1α(τ )a(τ )f (τ, v(τ ), v(τ ))dτ

<

Z ₁

0

G_1α(τ )a(τ )kvk σ₁ dτ

= 1 σ₁

Z 1 0

G_1α(τ )a(τ )dτ kvk

= kvk.

Similarly, kvk < kuk, kvk < kωk and kωk < kvk, which is a contradiction, and Theorem is received.

Theorem 1.19. Assume that (B1) − (B₃) hold, and

f (ζ, u, u) > u Υ²_αR1−θ

θ G_1α(τ )a(τ )dτ, g(ζ, v, v) > v

Υ²_βR1−θ

θ G_1β(τ )b(τ )dτ, h(ζ, ω, ω) > ω

Υ²_γR1−θ

θ G1γ(τ )c(τ )dτ,

for all t ∈ [0, 1], u > 0, v > 0, ω > 0, then boundary value problem (2) has no positive solution.

Example 1.20. Consider the system of nonlinear fractional dierential equations:















D⁵³u(τ ) +_1+τ^τ |sinv(τ )| = 0, D³²v(τ ) +_1+τ^τ |sinω(τ )| = 0, D⁴³ω(τ ) +_1+τ^τ |sinu(τ )| = 0, 0 < τ < 1,

u(0) = 0, u(1) =R1

0 τ u(τ )dτ, v(0) = 0, v(1) =R1

0 τ v(τ )dτ, ω(0) = 0, ω(1) =R1

0 τ ω(τ )dτ.

(19)

Set e(τ), f(τ), g(τ) ∈ [0, +∞) and τ ∈ [0, 1], then we have 

 τ

1 + τ|sine(τ )| − τ

1 + τ|sinf (τ )|

≤ τ 1 + τ

e(τ ) − f (τ ) , 

 τ

1 + τ|sinf (τ )| − τ

1 + τ|sing(τ )|

≤ τ 1 + τ

f (τ ) − g(τ ) , 

 τ

1 + τ|sing(τ )| − τ

1 + τ|sine(τ )|

≤ τ 1 + τ

g(τ ) − e(τ ) .

Therefore,

ρ = Z 1

0

G1α(τ )a(τ )m(τ )dτ ≤ Z 1

0

G1α(τ )dτ, θ =

Z 1 0

G1β(τ )b(τ )k(τ )dτ ≤ Z 1

0

G1β(τ )dτ, κ =

Z 1 0

G1γ(τ )c(τ )n(τ )dτ ≤ Z 1

0

G1γ(τ )dτ.

With the use of Theorem 1.4, B.V.P (19) has a unique positive solution.

Example 1.21. Consider the system of nonlinear fractional dierential equations:















D⁵³u(τ ) + [v(τ )]^a= 0, D⁵³v(τ ) + [ω(τ )]^b = 0, D⁵³ω(τ ) + [u(τ )]^c= 0, 0 < τ < 1,

u(0) = 0, u(1) =R1

0 τ u(τ )dτ, v(0) = 0, v(1) =R1

0 τ v(τ )dτ, ω(0) = 0, ω(1) =R₁

0 τ ω(τ )dτ.

(20)

Let f(τ, v, v) = va, g(τ, u, u) = ub and h(τ, ω, ω) = ωc, 0 < a, b, c < 1. It is easy to see that (B1) − (B₃) hold. By simple computation, we have f0 = g0 = h0 = ∞ and f^∞ = g^∞ = h^∞ = 0. Thus it follows that problem (20) has a positive solution by (H1).

Example 1.22. Consider the system of nonlinear fractional dierential equations:















D³²u(τ ) + [v(τ )]^a0 = 0, D³²v(τ ) + [ω(τ )]^b0 = 0, D³²ω(τ ) + [u(τ )]^c0 = 0, 0 < τ < 1,

u(0) = 0, u(1) =R1

0 τ u(τ )dτ, v(0) = 0, v(1) =R1

0 τ v(τ )dτ, ω(0) = 0, ω(1) =R1

0 τ ω(τ )dτ.

(21)

Let f(τ, v, v) = va⁰, g(τ, u, u) = ub⁰and h(τ, ω, ω) = ωc⁰, 0 < a⁰, b⁰, c⁰ < 1. It is easy to see that (B1) − (B3) hold. By simple computation, we have f⁰ = g⁰ = h⁰ = 0 and f∞ = g∞ = h∞ = ∞. Thus it follows that problem (21) has a positive solution by (H3).

References

[1] M. S. ABDO, Further results on the existence of solutions for generalized fractional quadratic functional integral equations, Journal of Mathematical Analysis and Modeling, (2020)1(1) : 33-46, doi:10.48185/jmam.v1i1.2.

[2] B. Ahmad, J. Nieto, Existence results for a coupled system of nonlinear fractional dierential equations with three-point boundary conditions, Comput. Math. Appl. 58 (2009) 18381843.

[3] H. Afshari, M. Atapour, E. Karapnar, A discussion on a generalized Geraghty multi-valued mappings and applications.

Adv. Dier. Equ. 2020, 356 (2020).

[4] H. Afshari, D. Baleanu, Applications of some xed point theorems for fractional dierential equations with Mittag-Leer kernel, Advances in Dierence Equations, 140 (2020), Doi:10.1186/s13662-020-02592-2.

[5] H. Afshari, S. Kalantari, D. Baleanu, Solution of fractional dierential equations via α − φ-Geraghty type mappings. Adv.

Dier. Equ. 2018, 347(2018), https://doi.org/10.1186/s13662-018-1807-4.

[6] H. Afshari, Solution of fractional dierential equations in quasi-b-metric and b-metric-like spaces, Adv. Dier. Equ. 2018, 285(2018), https://doi.org/10.1186/s13662-019-2227-9.

[7] H. Afshari, M. Sajjadmanesh, D. Baleanu, Existence and uniqueness of positive solutions for a new class of coupled system via fractional derivatives. Advances in Dierence Equations. 2020 Dec;2020(1):1-8, https://doi.org/10.1186/s13662-020-02568-2.

[8] H. Afshari, F. Jarad, and T., Abdeljawad, On a new xed point theorem with an application on a coupled system of fractional dierential equations. Advances in Dierence Equations 2020.1 (2020): 1-13, https://doi.org/10.1186/s13662-020-02926-0.

[9] H. Aydi, E. Karapnar, W. Shatanawi, Tripled xed point results in generalized metric spaces. J. Appl. Math. 10 (2012).

Article ID 314279.

[10] E. Karapnar, Couple xed point theorems for nonlinear contractions in cone metric spaces Computers and Mathematics With Applications Volume: 59 Issue: 12 Pages: 3656-3668 Published: JUN 2010.

[11] E. Karapnar, Fixed point theorems in cone Banach spaces, Fixed Point Theory Appl, (2009):9.

[12] E. Karapnar, H.D. Binh, N.H. Luc, and N.H., Can, On continuity of the fractional derivative of the time-fractional semilinear pseudo-parabolic systems, Advances in Dierence Equations 2021, no. 1 (2021): 1-24.

[13] E. Karapnar, S.I. Moustafa, A. Shehata, R.P. Agarwal, Fractional Hybrid Dierential Equations and Coupled Fixed-Point Results for α-Admissible F (ψ1, ψ1)-Contractions in M-Metric Spaces, Discrete Dynamics in Nature and Society, Volume 2020, Article ID 7126045, 13 pages https://doi.org/10.1155/2020/7126045,2020.

[14] C. Li, X. Luo, Y. Zhou, Existence of positive solutions of the boundary value problem for nonlinear fractional dierential equations, Comput. Math. Appl. 59 (2010) 13631375.

[15] H.R. Marasi, H. Afshari, M. Daneshbastam, C.B. Zhai, Fixed points of mixed monotone operators for existence and uniqueness of nonlinear fractional dierential equations, Journal of Contemporary Mathematical Analysis, vol. 52, p. 8C13, (2017).

[16] S. Zhang, Positive solutions to singular boundary value problem for nonlinear fractional dierential equation, Comput.

Math. Appl. 59 (2010) 13001309.

[17] Y. Zhao, et al., Positive solutions for boundary value problems of nonlinear fractional dierential equations, Appl. Math.

Comput. 217 (2011) 69506958.

[18] V. Daftardar-Gejji, Positive solutions of a system of non-autonomous fractional dierential equations, J. Math. Anal. Appl.

302 (2005) 5664.

[19] J. Henderson, et al., Positive solutions for systems of generalized three-point nonlinear boundary value problems, Comment.

Math. Univ. Carolin. 49 (2008) 7991.

[20] C. Goodrich, Existence of a positive solution to a class of fractional dierential equations, Appl. Math. Lett. 23 (2010) 10501055.

[21] H. Salem, On the existence of continuous solutions for a singular system of nonlinear fractional dierential equations, Appl.

Math. Comput. 198 (2008) 445452.

[22] X. Su, Existence of solution of boundary value problem for coupled system of fractional dierential equations, Engrg. Math.

26 (2009) 134137.

[23] C. Bai, J. Fang, The existence of a positive solution for a singular coupled system of nonlinear fractional dierential equations, Appl. Math. Comput. 150 (2004) 611621.

[24] M. Rehman, R. Khan, A note on boundary value problems for a coupled system of fractional dierential equations, Comput.

Math. Appl. 61 (2011) 26302637.

[25] W. Feng, et al., Existence of solutions for a singular system of nonlinear fractional dierential equations, Comput. Math.

Appl. 62 (2011) 13701378.

[26] H. Shojaat, H. Afshari, M.S. Asgari, A new class of mixed monotone operators with concavity and applications to fractional dierential equations, TWMS J. App. and Eng. Math. V.11, N.1, 2021, pp. 122-133.

[27] X. Su, Boundaryvalue problem for a coupled system of nonlinear fractional dierential equations, Appl. Math. Lett. 22 (2009) 6469.

[28] A.A., Kilbas, H.M., Srivastava, j.j., Trujillo, (2006), Theory and applications of fractiona dierential equations, North- Holland Mathematics Studies. 204(204) 7-10.

[29] Podlubny, I. (1999), Fractional Dierential Equations, Academic Press, New york.

[30] J. Wang, H. Xiang, Z. Liu, Positive solution to nonzero boundary values problem for a coupled system of nonlinear fractional dierential equations, Internat. J. Dier. Equ. 2010 (2010) 12. Article ID 186928.

[31] W. Yang, Positive solutions for a coupled system of nonlinear fractional dierential equations with integral boundary conditions, Computers and Mathematics with Applications 63 (2012) 288297.

[32] E. Zeidler, Nonlinear Functional Analysis and Its Applications-I: Fixed-Point Theorems, Springer, New York, NY, USA, 1986.

[33] D. Guo, V. Lakshmikantham, X. Liu, Nonlinear Integral Equations in Abstract Spaces, in: Mathematics and Its Applica- tions, vol. 373, Kluwer Academic Publishers, Dordrecht, The Netherlands, 1996.