Existence Results for Langevin Fractional Differential Inclusions Involving Two Fractional Orders with Four-Point Multiterm Fractional Integral Boundary Conditions

Definition 1. For at least n-times differentiable function g : [0, ∞) → ℝ, the Caputo derivative of fractional order q is defined as

where [q] denotes the integer part of the real number q.

Definition 2. The Riemann-Liouville fractional integral of order q is defined as

provided the integral exists.

Lemma 3. For q > 0, the general solution of the fractional differential equation D^qx(t) = 0 is given by

where c_i ∈ ℝ,   i = 1,2, …, n − 1  (n = [q] + 1).

Definition 4. A function x ∈ AC¹([0, T], ℝ) is called a solution of problem (1) if there exists a function v ∈ L¹([0, T], ℝ) with v(t) ∈ F(t, x(t)), a.e. [0, T] such that   ^cD^p(^cD^q + λ)x(t) = v(t), a.e. [0, T], and $$.

Lemma 5. Let h ∈ C[0,1]. Then the boundary value problem

has a unique solution where

Proof. As argued in [23], the solution of  ^cD^p(^cD^q + λ)x(t) = h(t) can be written as

Using the given conditions in (9) together with (8), we find that where Solving (10) for c₀ and c₁, we find that

Substituting these values in (9), we find the desired solution.

Lemma 6. One has

Proof. By using the following property of beta function

we have By a similar way, we have which completes the proof.

Definition 7. A multivalued map F : [0,1] × ℝ → 𝒫(ℝ) is said to be Carathéodory if

• (i)

  t ↦ F(t, x) is measurable for each x ∈ ℝ;

• (ii)

  x ↦ F(t, x) is upper semicontinuous for almost all t ∈ [0,1];

Further a Carathéodory function F is called L¹-Carathéodory if

• (iii)

  for each α > 0, there exists φ_α ∈ L¹([0,1], ℝ⁺) such that

  $$ ()

•  

  for all ∥x∥_∞ ≤ α and for a. e. t ∈ [0,1].

Lemma 8 (see [27], Proposition 1.2.)If G : X → 𝒫_cl(Y) is u.s.c., then Gr(G) is a closed subset of X  ×  Y; that is, for every sequence {x_n} _n∈ℕ ⊂ X and {y_n} _n∈ℕ ⊂ Y, if, when n → ∞, x_n → x_*, y_n → y_*, and y_n ∈ G(x_n), then y_* ∈ G(x_*). Conversely, if G is completely continuous and has a closed graph, then it is upper semicontinuous.

Lemma 9 (see [30].)Let X be a Banach space. Let F : [0, T] × ℝ → 𝒫_cp,c(X) be an L¹-Carathéodory multivalued map, and let Θ be a linear continuous mapping from L¹([0,1], X) to C([0,1], X). Then the operator

is a closed graph operator in C([0,1], X) × C([0,1], X).

Lemma 10 (nonlinear alternative for Kakutani maps [31]). Let E be a Banach space, C a closed convex subset of E, U an open subset of C, and 0 ∈ U. Suppose that $$ is an upper semicontinuous compact map; here 𝒫_c,cv(C) denotes the family of nonempty, compact convex subsets of C. Then either

• (i)

  F has a fixed point in $$, or

• (ii)

  there is a u ∈ ∂U and λ ∈ (0,1) with u ∈ λF(u).

Definition 11. Let A be a subset of [0,1] × ℝ. A is ℒ ⊗ ℬ measurable if A belongs to the σ-algebra generated by all sets of the form 𝒥 × 𝒟, where 𝒥 is Lebesgue measurable in [0,1] and 𝒟 is Borel measurable in ℝ.

Definition 12. A subset 𝒜 of L¹([0,1], ℝ) is decomposable if, for all u, v ∈ 𝒜, and measurable 𝒥 ⊂ [0,1] = J, the function uχ_𝒥 + vχ_J−𝒥 ∈ 𝒜, where χ_𝒥 stands for the characteristic function of 𝒥.

Lemma 13 (see [32].)Let Y be a separable metric space, and let N : Y → 𝒫(L¹([0,1], ℝ)) be a lower semicontinuous (l.s.c.) multivalued operator with nonempty closed and decomposable values. Then N has a continuous selection; that is, there exists a continuous function (single-valued) g : Y → L¹([0,1], ℝ) such that g(x) ∈ N(x) for every x ∈ Y.

Definition 14. A multivalued operator N : X → 𝒫_cl(X) is called

• (a)

  γ-Lipschitz if and only if there exists γ > 0 such that

  $$ ()

• (b)

  a contraction if and only if it is γ-Lipschitz with γ < 1.

Lemma 15 (see [34].)Let (X, d) be a complete metric space. If N : X → 𝒫_cl(X) is a contraction, then Fix N ≠ ∅.

Theorem 16. Suppose that

•  

  (H₁) the map F : [0,1] × ℝ → 𝒫(ℝ) is Carathéodory and has nonempty compact and convex values;

•  

  (H₂) there exist a continuous nondecreasing function ψ : [0, ∞)→(0, ∞) and function p ∈ L¹([0,1], ℝ⁺) such that

  $$ ()

•   

  for each (t, u)∈[0,1] × ℝ;

•  

  (H₃) there exists a number M > 0 such that

  $$ ()

with Ψ < 1, where Ω,   Ψ are defined in (17).

Then BVP (1) has at least one solution.

Proof. Let us introduce the operator N : C([0,1], ℝ) → 𝒫(C([0,1], ℝ)) as

for v ∈ S_F,x. We will show that the operator N satisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof consists of several steps. As a first step, we show that N(x)  is convex for each x ∈ C([0,1], ℝ). For that, let h₁, h₂ ∈ N(x). Then there exist v₁, v₂ ∈ S_F,x such that, for each t ∈ [0,1], we have for i = 1,  2. Let 0 ≤ ω ≤ 1. Then, for each t ∈ [0,1] and putting z(r) = ωv₁(r)+(1 − ω)v₂(r), we have

Since S_F,x is convex (F has convex values), therefore it follows that ωh₁ + (1 − ω)h₂ ∈ N(x).

Next, we show that N(x)  maps bounded sets into bounded sets in C([0,1], ℝ). For a positive number r, let B_ρ = {x ∈ C([0,1], ℝ) :  ∥x∥≤ρ} be a bounded set in C([0,1], ℝ). Then, for each h ∈ N(x), x ∈ B_ρ, there exists v ∈ S_F,x such that

Then

Now we show that Nmaps bounded sets into equicontinuous sets of C([0,1], ℝ). Let t^′, t^′′ ∈ [0,1] with t^′ < t^′′ and x ∈ B_ρ, where B_ρ, as above, is a bounded set of C([0,1], ℝ). For each h ∈ N(x), we obtain

Obviously the right-hand side of the above inequality tends to zero independently of x ∈ B_ρ as t^′′ − t^′ → 0. As N satisfies the above three assumptions, therefore it follows by Ascoli-Arzelá theorem that N : C([0,1], ℝ) → 𝒫(C([0,1], ℝ)) is completely continuous.

In our next step, we show that N has a closed graph. Let x_n → x_*,  h_n ∈ N(x_n), and h_n → h_*. Then we need to show that h_* ∈ N(x_*). Associated with h_n ∈ N(x_n), there exists $$ such that, for each t ∈ [0,1],

Thus we have to show that there exists $$ such that, for each t ∈ [0,1],

Let us consider the continuous linear operator Θ : L¹([0,1], ℝ) → C([0,1], ℝ) so that

Observe that which tends to zero as n → ∞.

Thus, it follows from Lemma 9 that Θ∘S_F is a closed graph operator. Further, we have $$. Since x_n → x_*, it follows that

for some $$.

Finally, we discuss a priori bounds on solutions. Let x be a solution of (1). Then there exists v ∈ L¹([0,1], ℝ) with v ∈ S_F,x such that, for t ∈ [0,1], we have

Using the computations proving that N(x) maps bounded sets into bounded sets and the notations (17), we have

Consequently

In view of (H₃), there exists M such that ∥x∥≠M. Let us set Note that the operator $$ is upper semicontinuous and completely continuous. From the choice of U, there is no x ∈ ∂U such that x ∈ μN(x) for some μ ∈ (0,1). Consequently, by the nonlinear alternative of Leray-Schauder type [31], we deduce that N has a fixed point $$ which is a solution of the problem (1). This completes the proof.

Theorem 17. Assume that (H₂)-(H₃) and the following conditions hold:

•  

  (H₄) F : [0,1] × ℝ → 𝒫(ℝ) is a nonempty compact-valued multivalued map such that

  • (a)

    (t, x) ↦ F(t, x) is ℒ ⊗ ℬ measurable,

  • (b)

    x ↦ F(t, x) is lower semicontinuous for each t ∈ [0,1];

•  

  (H₅) for each σ > 0, there exists φ_σ ∈ L¹([0,1], ℝ₊) such that

  $$ ()

Then the boundary value problem (1) has at least one solution on [0,1].

Proof. It follows from (H₄) and (H₅) that F is of l.s.c. type [35]. Then from Lemma 13, there exists a continuous function f : C([0,1], ℝ) → L¹([0,1], ℝ) such that f(x) ∈ ℱ(x) for all x ∈ C([0,1], ℝ).

Consider the problem

Observe that, if x ∈ AC¹([0,1]) is a solution of (43), then x is a solution to the problem (1). In order to transform the problem (43) into a fixed point problem, we define the operator $$ as

It can easily be shown that $$ is continuous and completely continuous. The remaining part of the proof is similar to that of Theorem 16. So we omit it. This completes the proof.

Theorem 18. Assume that the following conditions hold:

•  

  (H₆) F : [0,1] × ℝ → 𝒫_cp(ℝ) is such that F(·, x):[0,1] → 𝒫_cp(ℝ) is measurable for each x ∈ ℝ;

•  

  (H₇) $$ for almost all t ∈ [0,1] and $$ with m ∈ C([0,1], ℝ⁺) and d(0, F(t, 0)) ≤ m(t) for almost all t ∈ [0,1].

Then the boundary value problem (1) has at least one solution on [0,1] if $$, that is,

Proof. Observe that the set S_F,x is nonempty for each x ∈ C([0,1], ℝ) by the assumption (H₆), so F has a measurable selection (see Theorem III.6 [36]). Now we show that the operator N satisfies the assumptions of Lemma 15. To show that N(x) ∈ 𝒫_cl((C[0,1], ℝ)) for each x ∈ C([0,1], ℝ), let {u_n} _n≥0 ∈ N(x) be such that u_n → u  (n → ∞) in C([0,1], ℝ). Then u ∈ C([0,1], ℝ), and there exists v_n ∈ S_F,x such that, for each t ∈ [0,1],

As F has compact values, we pass onto a subsequence to obtain that v_n converges to v in L¹([0,1], ℝ). Thus, v ∈ S_F,x and, for each t ∈ [0,1],

Hence, u ∈ N(x).

Next we show that there exists γ < 1 such that

Let $$ and h₁ ∈ N(x). Then there exists v₁(t) ∈ F(t, x(t)) such that, for each t ∈ [0,1],

By (H₇), we have

So, there exists $$ such that

Define U : [0,1] → 𝒫(ℝ) by

Since the multivalued operator $$ is measurable (Proposition III.4 [36]), there exists a function v₂(t) which is a measurable selection for V. So $$, and for each t ∈ [0,1], we have $$.

For each t ∈ [0,1], let us define

Thus, Hence, Analogously, interchanging the roles of x and $$, we obtain

Since N is a contraction, it follows from Lemma 15 that N has a fixed point x which is a solution of (1). This completes the proof.

Remark 19. It is important to note that several new interesting special results of the present work can be obtained by fixing the parameters involved in the given problem. Some of these results are listed in the following.

• (i)

  Our results correspond to the multivalued extension of the Dirichlet problem considered in [23] for β_i = 0 = α_i,   i = 1,2, …, n.

• (ii)

  In case we take β_i = 0,   i = 1,2, …, n, we obtain the results for Langevin fractional differential inclusions with the three-point integral boundary conditions of the following form:

  $$ ()

• (iii)

  By taking β_i = 0 = α_i,   i = 2,3, …, n, and ν₁ = 1 = μ₁, we get the results for Langevin fractional differential inclusions with four-point nonlocal integral boundary conditions of the following type:

  $$ ()

•   

  Note that we obtain the typical integral boundary condition in the limit ζ, η → 1⁻.

Example 20. Consider the problem

where F : [0,1] × ℝ → 𝒫(ℝ) is a multivalued map given by For f ∈ F, we have Here p = 2/3,  q = 4/5,  λ = 1/10,  η = 1/3,  n = 4,  α₁ = 1/2,  α₂ = −2/3,  α₃ = 3/4,  α₄ = −4/5,  ν₁ = 1/4,  ν₂ = 3/4,  ν₃ = 5/4,  ν₄ = 7/4,  ζ = 1/4,  β₁ = 1/2,  β₂ = −1/3,  β₃ = 1/4,  β₄ = −1/5,  μ₁ = 1/2,  μ₂ = 3/4,  μ₃ = 3/2,  and μ₄ = 4/3. Clearly, with p(t) = 1,   ψ(∥x∥) = 3/4. Using the given data, it is found that Thus, Clearly, all the conditions of Theorem 16 are satisfied. So there exists at least one solution of the problem (59) on [0,1].

Example 21. Consider the fractional inclusion boundary value problem (59) with F : [0,1] × ℝ → 𝒫(ℝ) given by

Then, we have where m(t) = 1/(2 + t) ². With $$ and Ω = 1.764724 (from Example 20), it is found that Since all the conditions of Theorem 18 are satisfied, therefore the problem (59) with F given by (65) has at least one solution on [0,1].