Weak Solutions for Nonlinear Fractional Differential Equations in Banach Spaces

Definition 2.1. A function h : E → E is said to be weakly sequentially continuous if h takes each weakly convergent sequence in E to a weakly convergent sequence in E (i.e., for any (x_n) _n in E with x_n(t) → x(t) in (E, ω) then h(x_n(t)) → h(x(t)) in (E, ω) for each t → J).

Definition 2.2 (see [33].)The function x : J → E is said to be Pettis integrable on J if and only if there is an element x_J ∈ E corresponding to each I ⊂ J such that φ(x_I) = ∫_I φ(x(s))ds for all φ ∈ E^*, where the integral on the right is supposed to exist in the sense of Lebesgue. By definition, x_I = ∫_I x(s)ds.

Let P(J, E) be the space of all E-valued Pettis integrable functions in the interval J.

Lemma 2.3 (see [33].)If x(·) is Pettis integrable and h(·) is a measurable and essentially bounded real-valued function, then x(·)h(·) is Pettis integrable.

Definition 2.4 (see [34].)Let E be a Banach space, Ω_E the set of all bounded subsets of E, and B₁ the unit ball in E. The De Blasi measure of weak noncompactness is the map β : Ω_E → [0, ∞) defined by β(X) = inf {ϵ > 0: there exists a weakly compact subset Ω of E such that X ⊂ ϵB₁ + Ω}.

Lemma 2.5 (see [34].)The De sBlasi measure of noncompactness satisfies the following properties:

• (a)

  S ⊂ T⇒β(S) ≤ β(T);

• (b)

  β(S) = 0⇔S is relatively weakly compact;

• (c)

  β(S ∪ T) = max {β(S), β(T)};

• (d)

  $$, where $$ denotes the weak closure of S;

• (e)

  β(S + T) ≤ β(S) + β(T);

• (f)

  β(aS) = |a | α(S);

• (g)

  β( conv (S)) = β(S);

• (h)

  β(∪_|λ|≤hλS) = hβ(S).

The following result follows directly from the Hahn-Banach theorem.

Lemma 2.6. Let E be a normed space with x₀ ≠ 0. Then there exists φ ∈ E^* with ∥φ∥ = 1 and φ(x₀) = ∥x₀∥.

For completeness, we recall the definitions of the Pettis-integral and the Caputo derivative of fractional order.

Definition 2.7 (see [26].)Let h : J → E be a function. The fractional Pettis integral of the function h of order α ∈ ℝ⁺ is defined by

where the sign “∫ ” denotes the Pettis integral and Γ is the Gamma function.

Definition 2.8 (see [3].)For a function f : J → E, the Caputo fractional-order derivative of f is defined by

where n = [α] + 1 and [α] denotes the integer part of α.

Lemma 2.9 (see [28].)Let D be a closed convex and equicontinuous subset of a metrizable locally convex vector space C(J, E) such that 0 ∈ D. Assume that A : D → D is weakly sequentially continuous. If the implication

holds for every subset V of D, then A has a fixed point.

Definition 3.1. A function x ∈ C(J, E_ω) is said to be a solution of the problem (1.1) if x satisfies the equation $$ on J and satisfies the conditions u(0) = λ₁u(T) + μ₁,   u′(0) = λ₂u′(T) + μ₂.

For the existence results on the problem (1.1), we need the following auxiliary lemmas.

Lemma 3.2 (see [3], [7].)For α > 0, the general solution of the fractional differential equation $$ is given by

Lemma 3.3 (see [3], [7].)Assume that h ∈ C(0,1)∩L(0,1) with a fractional derivative of order α > 0 that belongs to C(0,1)∩L(0,1). Then

for some c_i ∈ ℝ, i = 0,1, 2, …, n − 1, where n = [α] + 1.

We derive the corresponding Green′s function for boundary value problem (1.1) which will play major role in our next analysis.

Lemma 3.4. Let ρ ∈ C(J, E) be a given function, then the boundary-value problem

has a unique solution where G(t, s) is defined by the formula Here G(t, s) is called the Green′s function of boundary value problem (3.3).

Proof. By the Lemma 3.3, we can reduce the equation of problem (3.3) to an equivalent integral equation

for some constants c₁, c₂ ∈ ℝ. On the other hand, by relations $$ and $$, for m, n > 0,   u ∈ L(0,1), we have

Applying the boundary conditions (3.3), we have

Therefore, the unique solution of problem (3.3) is

which completes the proof.

Remark 3.5. From the expression of G(t, s), it is obvious that G(t, s) is continuous on J × J. Denote

Remark 3.6. Letting ξ₁ = 1/(λ₁ − 1), ξ₂ = 1/(λ₁ − 1)(λ₂ − 1), g(t) = μ₂[λ₁T + (1 − λ₁)t]/(λ₁ − 1)(λ₂ − 1)−μ₁/(λ₁ − 1) = μ₂[λ₁T + (1 − λ₁)t]ξ₂ − μ₁ξ₁, it is obvious that g(t) is continuous in J, denoting g^* = sup {|g(t)|,   t ∈ J}.

To prove the main results, we need the following assumptions:

• (H1)

  for each t ∈ J, the function f(t, ·) is weakly sequentially continuous;

• (H2)

  for each x ∈ C(J, E), the function f(·, x(·)) is Pettis integrable on J;

• (H3)

  there exists p_f ∈ L^∞(J, ℝ⁺) such that ∥f(t, u)∥≤p_f(t)∥u∥, for a.e. t ∈ J and each u ∈ E;

• (H3)’

  there exists p_f ∈ L^∞(J, E) and a continuous nondecreasing function ψ:[0, ∞)→(0, ∞) such that ∥f(t, u)∥≤p_f(t)ψ(∥u∥), for a.e. t ∈ J and each u ∈ E;

• (H4)

  for each bounded set D ⊂ E, and each t ∈ J, the following inequality holds

  $$ (3.11)

• (H5)

  there exists a constant R > 0 such that

  $$ (3.12)

where $$.

Theorem 3.7. Let E be a reflexive Banach space and assume that (H1)–(H3) are satisfied. If

then the problem (1.1) has at least one solution on J.

Proof. Let the operator 𝒜 : C(J, E) → C(J, E) defined by the formula

where G(·, ·) is the Green′s function defined by (3.5). It is well known the fixed points of the operator 𝒜 are solutions of the problem (1.1).

First notice that, for x ∈ C(J, E), we have f(·, x(·)) ∈ P(J, E) (assumption (H2)). Since, s ↦ G(t, s) ∈ L^∞(J), then G(t, ·)f(·, x(·)) is Pettis integrable for all t ∈ J by Lemma 2.3, and so the operator 𝒜 is well defined.

Let

and consider the set Clearly, the subset D is closed, convex, and equicontinuous. We shall show that 𝒜 satisfies the assumptions of Lemma 2.9. The proof will be given in three steps.

Step  1. We will show that the operator 𝒜 maps D into itself.

Take x ∈ D, t ∈ J and assume that 𝒜x(t) ≠ 0. Then there exists ψ ∈ E^* such that ∥𝒜x(t)∥ = ψ(𝒜x(t)). Thus

Let τ₁, τ₂ ∈ J,     τ₁ < τ₂ and ∀x ∈ D, so 𝒜x(τ₂) − 𝒜x(τ₁) ≠ 0. Then there exists ψ ∈ E^*, such that ∥𝒜x(τ₂) − 𝒜x(τ₁)∥ = ψ(𝒜x(τ₂) − 𝒜x(τ₁)). Hence,

this means that 𝒜(D) ⊂ D.

Step  2. We will show that the operator 𝒜 is weakly sequentially continuous.

Let (x_n) be a sequence in D and let (x_n(t)) → x(t) in (E, w) for each t ∈ J. Fix t ∈ J. Since f satisfies assumptions (H1), we have f(t, x_n(t)) converge weakly uniformly to f(t, x(t)). Hence, the Lebesgue Dominated Convergence Theorem for Pettis integrals implies that 𝒜x_n(t) converges weakly uniformly to 𝒜x(t) in E_ω. Repeating this for each t ∈ J shows 𝒜x_n → 𝒜x. Then 𝒜 : D → D is weakly sequentially continuous.

Step  3. The implication (2.3) holds. Now let V be a subset of D such that $$. Clearly, $$ for all t ∈ J. Hence, 𝒜V(t) ⊂ 𝒜D(t), t ∈ J, is bounded in E. Thus, 𝒜V(t) is weakly relatively compact since a subset of a reflexive Banach space is weakly relatively compact if and only if it is bounded in the norm topology. Therefore,

thus, V is relatively weakly compact in E. In view of Lemma 2.9, we deduce that 𝒜 has a fixed point which is obviously a solution of the problem (1.1). This completes the proof.

Remark 3.8. In the Theorem 3.7, we presented an existence result for weak solutions of the problem (1.1) in the case where the Banach space E is reflexive. However, in the nonreflexive case, conditions (H1)–(H3) are not sufficient for the application of Lemma 2.9; the difficulty is with condition (2.3). Our results can be seen as a supplement of the results in [32] (see Remark 3.8).

Theorem 3.9. Let E be a Banach space, and assume assumptions (H1), (H2), (H3), (H4) are satisfied. If (3.13) holds, then the problem (1.1) has at least one solution on J.

Theorem 3.10. Let E be a Banach space, and assume assumptions (H1), (H2), (H3)’, (H4), (H5) are satisfied. If (3.13) holds, then the problem (1.1) has at least one solution on J.

Proof. Assume that the operator 𝒜 : C(J, E) → C(J, E) is defined by the formula (3.14). It is well known the fixed points of the operator 𝒜 are solutions of the problem (1.1).

First notice that, for x ∈ C(J, E), we have f(·, x(·)) ∈ P(J, E) (assumption (H2)). Since, s ↦ G(t, s) ∈ L^∞(J), then G(t, ·)f(·, x(·)) for all t ∈ J is Pettis integrable (Lemma 2.3) and thus the operator 𝒜 makes sense.

Let R > 0, and consider the set

Clearly the subset 𝒟 is closed, convex and equicontinuous. We shall show that 𝒜 satisfies the assumptions of Lemma 2.9. The proof will be given in three steps.

Step  1. We will show that the operator 𝒜 maps 𝒟 into itself.

Take x ∈ 𝒟, t ∈ J and assume that 𝒜x(t) ≠ 0. Then there exists ψ ∈ E^* such that ∥𝒜x(t)∥ = ψ(𝒜x(t)). Thus

Let τ₁, τ₂ ∈ J,   τ₁ < τ₂ and ∀x ∈ 𝒟, so 𝒜x(τ₂) − 𝒜x(τ₁) ≠ 0. Then there exist ψ ∈ E^* such that

Thus

this means that 𝒜(𝒟) ⊂ 𝒟.

Step  2. We will show that the operator 𝒜 is weakly sequentially continuous.

Let (x_n) be a sequence in 𝒟 and let (x_n(t)) → x(t) in (E, w) for each t ∈ J. Fix t ∈ J. Since f satisfies assumptions (H1), we have f(t, x_n(t)), converging weakly uniformly to f(t, x(t)). Hence the Lebesgue Dominated Convergence theorem for Pettis integral implies 𝒜x_n(t) converging weakly uniformly to 𝒜x(t) in E_ω. We do it for each t ∈ J so 𝒜x_n → 𝒜x. Then 𝒜 : 𝒟 → 𝒟 is weakly sequentially continuous.

Step  3. The implication (2.3) holds. Now let V be a subset of 𝒟 such that $$. Clearly, $$ for all t ∈ J. Hence, 𝒜V(t) ⊂ 𝒜𝒟(t),   t ∈ J, is bounded in E. Since function g is continuous on J, the set $$ is compact, so β(g(t)) = 0. Using this fact, assumption (H4), Lemma 2.5 and the properties of the measure β, we have for each t ∈ J

which gives

This means that

By (3.13) it follows that ∥v∥_∞ = 0, that is v(t) = 0 for each t ∈ J, and then V(t) is relatively weakly compact in E. In view of Lemma 2.9, we deduce that 𝒜 has a fixed point which is obviously a solution of the problem (1.1). This completes the proof.

Example 4.1. Let us consider the following fractional boundary value problem:

Set T = 1, f(t, u) = (2/(19 + e^t))(∥u∥/(1 + ∥u∥)), (t, u) ∈ J × E, λ₁ = λ₂ = −1, μ₁ = μ₂ = 0.

Clearly conditions (H1), (H2), and (H3) hold with p_f(t) = 2/(19 + e^t). From (3.5), we have

We have

A simple computation gives

We shall check that condition (3.13) is satisfied. Indeed

which is satisfied for some α ∈ (1,2]. Then by Theorem 3.7, the problem (4.1) has at least one solution on J for values of α satisfying (4.5).