Let be a separable Hilbert space. We denote by the unit closed ball with center at the origin in Given a set-valued map, we denote by, respectively, the domain and the graph of A defined by and. We say that an operator is monotone if for all. Moreover, an operator is maximal monotone if it is monotone and its graph is maximal in the sense of the inclusion, i.e., Gr (A) is not properly contained in the graph of any other monotone operator. We refer to [2] for more details on maximal monotone operators

The Clarke subdifferential of a locally Lipschitz function is defined by for all, where stands for the generalized directional derivative of in the direction v ∈ H defined by For a convex function the convex subdifferential of f at x ∈ H is given by for all. It is well known that for a proper, convex, and lower semicontinuous function, the convex subdifferential defines a maximal monotone operator. Moreover, for a convex and locally Lipschitz function, the convex subdifferential coincides to the Clarke subdifferential (see, e.g., [3])

The following result is an important characterization of convexity. We refer to [3, Prop. 2.2.9] for its proof

Proposition 1. Let be a Hilbert space. Let be a locally Lipschitz function in an open convex set. Then f is convex on if and only if the multifunction ∂f is monotone on, that is, if and only if for all for all

Proposition 2. Let Y be a Banach space and consider be a function such that for, the map is convex and lower semicontinuous on H. Assume that for alland, it holds

Then, for all and, the following inequality holds:

Proof. Let and. Then, according to the definition of the convex subdifferential, for all, we have

Hence, taking in the inequalities above, respectively, and summing the resulting inequalities, we get

Hence, we get the desired inequality.

Definition 1. Let X, Y be normed spaces. An operator is called a history-dependent operator if there exists L > 0 such that for all,

The following result is an essential fixed point property for history-dependent operators (see, e.g., [21, p. 118]).

Theorem 1. Let X be a Banach space and be a historydependent operator. Then F has a unique fixed point.

We end this subsection with a technical lemma related to differential inequalities.

Lemma 1. Let be two absolutely continuous functions such that

where is a nonnegative function. Then it holds

Proof. Let us consider the sets On the one hand, for, we have

which implies the desired inequality. On the other hand, for any, we can seethat the map attains a minimum. Thus, for, we have, which implies the desired inequality. The proof is thencomplete.

Let be a bounded domain in with Lipschitz boundary, and let s ∈ (0, 1) be such that N > 2s. We adopt the symbols, and to denote the fractional critical exponent. Also, we denote by the function u restricted to the domain. In what follows, we assume that function satisfies the conditions:

(HK) is such that

(i) the function belongs to

(ii) there exists a constant such that for all

(iii) for each, we have.

Consider the function spaceand. It is clear (see, e.g., [20]) that X is a Banach space endowed with the norm. We also introduce a subspace of X given by. Also, we recall the following lemma (see [20]), which will be used in Section 5.

Lemma 2. Let s ∈ (0, 1) and Ω be a bounded, open subset of R N with Lipschitz boundary and N > 2s. Then we have

(i) X₀ is a Hilbert space with the inner producto

(ii) If p ∈ [1, 2^∗_s ], then there exists a positive constant c(p) such that for all u ∈ X₀, kukLp(RN ) 6 c(p)kukX0

(iii) The embedding fromis compact if p ∈ [1, 2^∗_s ).

this subsection, we prove the existence of solutions for the following Cauchy problem:

(4)

The following result provides the well-posedness for (4)

Theorem 2. Assume that (H_A), (H_J ), and (H_ψ) hold. If , then for each f ∈ L² (I; H) and x₀ ∈ H, problem (4) has a unique solution x(·, f, x₀) ∈ W^1,2 (I; H). Moreover, the solution operator (f, x₀) → x(·, f, x₀) is Lipschitz continuous from L² (I; H) × H into C(I;H)

Proof. We will employ [22, Thm. 1] to obtain the desired conclusion. So, the proof is divided into three steps.

Step 1. For a.e. t ∈ I, the operator x → ∂J(t, x) + A(t, x) + ∂_cψ(t, x) is maximal monotne

Proof of Step 1. It follows directly from Lemma 3

Step 2. For all x ∈ H, the operator t → ∂J(t, x) + A(t, x) + ∂ψ(t, x) is measurable.

Proof of Step 2. The measurability can be obtained directly from the separability of H and hypotheses (H_A)(i), (H_J )(i), and (H_ψ)(i).

Step 3. There exist such that for all x ∈ H

Proof of Step 3. Indeed, conditions (H_J )(iii), (H_A)(iii), and (H_ψ)(iii) indicate that

where , which proves Step 3.

Therefore, by virtue of Steps 1–3 and [22, Thm. 1], the Cauchy problem (4) has a unique solution x(·, f, x₀) ∈ W^1,2 (I; H).

Furthermore, let and and set and . Then, due to the monotonicity of the set-valued map x → ∂J(t, x) + A(t, x) + ∂_cψ(t, x), for a.e. t ∈ I, it follows that

Therefore, by virtue of Lemma 1, for a.e. t ∈ I, it holds, which implies that

We conclude that the solution operator (f, x₀) → x(·, f, x₀) is Lipschitz from L² (I; H) × H into C(I; H) by using Hölder inequality

Remark 1. From the proof of Theorem 2 we can see that hypotheses (H_J )(iii), (H_A)(iii), and (H_ψ)(iii) are only used to ensure that

for some functions . So, Theorem 2 holds too if we interchange the infimum by a supremum in (H_J )(iv) and the supremum by a infimum in (H_ψ)(iii).

Remark 2. According to [22, Thm. 1] and Lemma 3, Theorem 2 still holds if we consider a set-valued operator A : I × H ⇒ H such that the map x→ A(t, x) is maximal monotone with a full domain.

In this subsection, we focus our attention on the study of the existence of solutions for the following Cauchy problem involving history-dependent operators:

where R and S are two history-dependent operators, i.e., (HRS ) is satisfied

Theorem 3. Assume that (H_A), (H_J ), (H), and (H_RS) hold. If , then for each f ∈ L² (I; H) and x₀ ∈ H, problem (5) has a unique solution x(·, f, x₀) ∈ W^1,2 (I; H). Moreover, the solution operator is Lipschitz continuous from L² (I; H) × H into C(I;H)

Proof. Fix v ∈ L² (I; H) and let us consider the intermediate problema

(6)

Our aim is to prove that (6) has a unique fixed point in W^1,2 (I; H), which is clearly a solution of (5). The proof is divided into several steps

Step 1. For v ∈ L² (I; H), problem (6) has a unique solution x(v) ∈ W^1,2 (I; H).

Proof of Step 1. It follows directly from Theorem 2

We now denote by F : L² (I; H) → W^1,2 (I; H) the operator, which associates to any v ∈ L² (I; H) for the unique solution x(v) ∈ W^1,2 (I; H) of (6).

Step 2. The operator F is history-dependent. More precisely, for all v₁, v₂ ∈ L² (I; H),we have

Proof of Step 2. Denote Let us considersuch thatfor all t ∈ I and i= 1,2, and ̇x_i(t) + ξ_i (t) ∈for allt∈Iandi= 1,2. Defineh(t) :=. Thenhis

where, and we have used the monotonicity of the set-valued map, Proposition 2, and hypotheses (HRS). Therefore, by virtueof Lemma 1, for a.e.t∈I, we concludeHds, which implies that

which proves Step 2

Step 3. Problem (5) has a unique solution x^∗ ∈ W^1,2 (I; H). Proof of Step 3. Since F is a history-dependent operator, employing Theorem 1 implies that the operator F : L² (I; H) → W^1,2 (I; H) has a unique fixed point x∗ , which clearly solves (5).

To end the proof, let and and consider and. Then, by virtue of the monotonicity of the operator A(t, x) for a.e. t ∈ I, it follows that

Hence

So, we have

Applying Grönwall’s inequality, it finds therefore

where the constant M > 0 only depends on c_R, c_S , β, and T. Consequently, we have

which proves the Lipschitzianity of the solution operator.

In this subsection, we consider the existence of periodic solutions for the following differential inclusion problem:

(7)

Theorem 4. Assume that (H_A), (H_J ), and (H_ψ) hold. If , then problem (7) has a unique solution x_π ∈ W^1,2 (I; H)

Proof. Let us consider the operator F : H → H defined by F(x₀) = x(T; x₀), where x(·, x₀) is the unique solution of (4) with the initial condition . Our goal is to prove that F has a unique fixed point in H. Let . Keeping in mind that and the operator is m strongly monotone with m := m_A − m_J , it gives

Employing Grönwall’s inequality yields t ∈ I. Then we have where κ := e^−mT < 1. Therefore, the operator F has a unique fixed point x0,T ∈ H by the contractive fixed point theory. It is clear that x_π := x(·, x₀,T ) is the unique solution of (7)

Likewise, we can consider the existence of antiperiodic solutions to the following differential inclusion problema:

(8)

Using the same argument given in the proof of Theorem 4, it is easy to conclude the following result

Theorem 5. Assume that (HA), (HJ ), and (Hψ) hold. If mA > mJ , then problem (8) has a unique solution x−π ∈ W1,2 (I; H).

We end this section by showing that the unique solution to (7) can be obtained asymptotically from any solution of (4)

Theorem 6. Assume, in addition to (H_A), (H_J ), (H_ψ), that and for allandbe the unique solution of (4) with the initial condition and define for all t ∈ I and . Then, for any , where xπ is the unique periodic solution of (7)

Proof. Set , and let us consider and n ∈ N. Then, for a.e. t ∈ I, it gives −mh(t), where we used the monotonicity of the map . Therefore, for all t ∈ I, it is true for all t ∈ I, where we have used the Grönwall inequality. Thus, by using the same inequalities, for all t ∈ I, we have

which shows that in C(I; H).

Remark 3. As showed in the previous proof, converges to exponentially.

In the subsection, we are interested in the study of an evolutionary inclusion problem with a generalized nonlocal space-fractional Laplace operator and a Clarke generalized subgradient operator. Let be a bounded domain in R N with Lipschitz boundary, s ∈ (0, 1) be such that, and . More precisely, the classical form of the evolutionary inclusion problem under consideration is formulated as follows

Problem 1. Find function such that

where the operator L_K stands for the generalized nonlocal space-fractional Laplace operator defined as follows: dy for . , for all

We first impose the following assumptions for the data of Problem 1.

(Hj ) is such that and

(i) for each the function is measurable on Ω × I;

(ii) for a.e. (x, t) ∈ Ω × I, the functional ) is locally Lipschitz continuous;

(iii) there exist satisfying for all ξ ∈ ∂j(x, t, r) and all (x, t, r) ∈ Ω × I × R;

(iv) there exists such that for all

(H0) .

The weak solutions to Problem 1 are understood as follows.

Definition 3. We say that is a weak solution to Problem 1 if and the following inequality holds for all v ∈ X0:

Let us define function for all (. For function J, we have the following lemm

Lemma 4. Assume that (H_j ) is fulfilled. Then the following statements hold:

(i) is measurable on I for all u ∈ X₀.

(ii) For is locally Lipschitz.

(iii) For all , we have and .

(iv) There exists a constant such that for all (.

(v) For any and , the inequality is satisfied

Proof. Statements (i)–(iv) are the direct consequences of [13, Thm. 3.47]. It remains us to prove conclusion (v). Let be arbitrary. Statement (iii) indicates

where the last inequality is obtained by using Lemma 2. Therefore, the desired inequality is valid.

The existence and uniqueness of weak solutions to Problem 1 is provided by the following result

Theorem 7. Assume that H(K), H(j), H(0), and (2) hold, then Problem 1 has a unique weak solution), and the solution operator is Lipschitz continuous from

Proof. Let H := X₀. Consider the operator A : H → H defined for all u, v ∈ H by. We now claim that A is a continuous linear operator. For any u, v ∈ H, it has

It obvious to see that A is a linear continuous operator and . In addition, the fact for all u, v ∈ H implies that A is strongly monotone with constant m_A = 1.

Let us consider the following intermediate problem: find such that for all v ∈ H

(9)

Employing Lemma 4(iii), we can see that a solution to problem (9) is also a weak solution to Problem 1. On the other hand, it is not difficult to verify that problem (9) is equivalent to the following evolutionary inclusion proble

(10)

Observe that reads hypothesis (H_A), hence, we are now in a position to invoke Lemma 4 and Theorem 2 that problem (10) has a unique solution u(·, f, u₀) ∈ W^1,2 (I; H). So, u is also a weak solution to Problem 1. For the uniqueness part of Problem 1, it can be obtained directly by using the standard procedure (see the proof of Theorem 2). Finally, the Lipschitz continuity of the solution operator could be verified by employing the same argument with the proof of Theorem 2.

Furthermore, we are going to apply the results established in Section 4.3 to investigate the fractional evolution inclusion problem, Problem 1, with periodic and antiperiodic boundary value conditions, respectively

Problem 2. Find function such that

Likewise, the weak solutions to Problem 2 are given as follows.

Definition 4. We say that u : I → X₀ is a weak solution to Problem 2 if u(x, 0) = u(x, T) in Ω and the following inequality holds for all v ∈ X₀:

Invoking Theorem 4 and the proof of Theorem 7, we have the following existence and uniqueness result for Problem 2

Theorem 8. Assume that (H_K), (H_j ), (H₀), and (2) are satisfied, then Problem 2 has a unique solution

We end the subsection by considering the antiperiodic boundary value problem.

Problem 3. Find function such tha

Definition 5. We say that is a weak solution to Problem 3 if u(x, 0) = −u(x, T) in Ω and the following inequality holds for all v ∈ X₀:

Analogously, from Theorem 5 and the proof of Theorem 7, we have the following theorem.

Theorem 9. Assume that (H_K), (H_j ), (H₀), and (2) , then Problem 3 has a unique solution .

The semipermeability boundary conditions can describe exactly behavior of various types of membranes, natural and artificial ones, and arise in models of heat conduction, electrostatics, hydraulics and in the description of the flow of a Bingham fluid in which the solution represents temperature, electric potential, pressure, and so forth. The current subsection is devoted to exploring a comprehensive semipermeability problem of parabolic type involving Volterra-type integral terms and nonsmooth potential functions.

Let and be a bounded domain in with Lipschitz continuous boundary. Denote by ν the unit outward normal on the boundary The boundary is decomposed into three mutually disjoint and relatively open subsets Γ₁, and such thatand meas. In the sequel, we denote by and . The classical formulation of semipermeability problem is described as follows.

Problem 4. Find such that

where denotes the conormal derivative with respect to the second-order differential operator , and sgn stands for the sign functio

In order to deliver the weak formulation of Problem 4, we are now in a position to introduce the following function spaces and V = L² (Ω). However, from Korn’s inequality and the condition meas, it finds that H is a Hilbert space endowed with the inner product for all u, v ∈ H.

Also, we impose the following assumptions.

(H_a) is such that aij ∈ and there exists a constant such that

(H_j )is such that:

(i) j(·, ·, r) is measurable on Q for all r ∈ R and there exists e ∈ L² (Ω) such that j(·, ·, e(·)) ∈ L¹ (Q);

(ii) j(x, t,·) is locally Lipschitz for a.e. (x, t) ∈ Q;

(iii) and c_1j > 0;

(iv) there exists satisfying for all and for a.e. (x, t) ∈ Q.

(HF ) is such that:

(i) F(·, r) is measurable on for all r ∈ R.

(ii) there is L_F > 0 such that for all r₁, r₂ ∈ R, a.e. x ∈ .

(iii) .

(HE)

(H0)

It follows from Riesz’s representation theorem that there is a function f : I →H⁻¹ (Ω) such that Γfor all v ∈ H and all t ∈ I.

Using a standard procedure, it is not difficult to get the weak formulation of Problem 4 as follows.

Problem 5. Find a function u : I → H such that u(0) = u0 and for all v ∈ H,

where the operator A : H → H is defined

We also consider a function defined by for all (t, u) ∈ I × H. It is obvious that under hypothesis (H_j ) the following lemma is available

Lemma 5. Assume that (H_j ) is fulfilled. Then the following statements hold:

(i) t → J(t, u) is measurable on I for all u ∈ H.

(ii) For a.e. is locally Lipschitz.

(iii) For all (t, u) ∈ I × H, we havedx and ∂J(t, u) ⊂

(iv) There exists a constant such thatfor all .

(v) We have h, and η ∈ ∂J(t, v). Here is such that .

Let us define the operatorsR : L² (I; H) → L² (I; H) and S : L² (I; H) → L² (I; Y ) by Ru(t) := R t 0 E(t − s)u(s) ds and Su(t) := R t 0 |u(s)| ds for all u ∈ L 2 (I; H), where Y = L 2 (Γ₃).

Remark 4. It follows from [21, Exs. 4 and 6] that R and S are two history-dependent operators.

Moreover, let us consider the function defined by for all y ∈ Y and u ∈ H. The following result establishes the wellposedness for Problem 5.

Theorem 10. Assume that (Ha), (Hj ), (HE), (HF ), (H0), and ma > mj c 2 H are fulfilled, then Problem 5 has a unique solution u(·, f₀, f₂, u₀) ∈ W^1,2 (I; H). Moreover, the solution operator (f₀, f₂, u₀) 7→ u(·, f₀, f₂, u₀) is Lipschitz continuous from L² (I; V ) × L 2 (I;L² (Γ2)) × H into C(I; H).

Proof. We first study the intermediate problem: find u : I → H such that u(0) = u₀ and for all v ∈ H,

In fact, the inequality above could be rewritten to the following inclusion problem: find u : I → H such that u(0) = u0 and

(11)

However, Lemma 5 reveals the fact that a solution of problem (11) is also a solution of Problem 5. Based on this fact, we are going to utilize Theorem 3 for concluding the existence of solutions of Problem 5.

From hypothesis (Ha) it is not difficult to prove that A is a continuous and strongly monotone operator with constant . Notice that 1, . . . , d, we can obtain the inequality for all u ∈ H with some c_A > 0. Besides, by virtue of condition (HF ) and the definition of , it gives that ϕ satisfies condition (H) (see [21, p. 251, Thm. 113,]).

Therefore, all conditions in Theorem 3 are verified. This theorem implies that problem (11) has a unique solution u(·, f₀, f₂, u₀) ∈ W^1,2 (I; H), which is a solution to Problem 5 as well. On the other hand, from the smallness condition Hmj it follows a standard procedure to obtain that u(·, f₀, f₂, u₀) ∈ W^1,2 (I; H) is also the unique solution to Problem 5. Moreover, the Lipschitz continuity of the solution operator T : (f₀, f₂, x₀) 7→ u(·, f₀, f₂, x₀) could be verified by employing the same argument with the proof of Theorem 2.