Lemma 1 (see [22].)Let u, v : ℝ_ℱ⟶[0,1] be the fuzzy sets. Then, u = v if and only if [u]^α = [v]^α, for all α ∈ [0,1].

The following arithmetic operations on fuzzy numbers are well known and frequently used below. If u, v ∈ ℝ_ℱ, then

For u, v ∈ ℝ_ℱ, if there exists w ∈ ℝ_ℱ such that u = v + w, then w is the Hukuhara difference of u and v denoted by u⊖v.

Let us define d : ℝ_ℱ × ℝ_ℱ⟶ℝ⁺ ∪ {0} by the equation

where d_H is the Hausdorff metric defined in $$.

Theorem 1 (see [23].)(ℝ_ℱ, d) is a complete metric space.

We list the following properties of d(u, v):

for all u, v, w ∈ ℝ_ℱ and λ ∈ ℝ.

Theorem 2 (see [24].)There exists a real Banach space X such that ℝ_ℱ can be the embedded as a convex cone C with vertex 0 in X. Furthermore, the following conditions hold true:

• (i)

  The embedding j is isometric,

• (ii)

  The addition in X induces the addition in ℝ_ℱ,

• (iii)

  The multiplication by a nonnegative real number in X induces the corresponding operation in ℝ_ℱ,

• (iv)

  C − C = {a − b/a, b ∈ ℝ_ℱ} is dense in X,

• (v)

  C is closed.

Remark 1. Let $$ as $$. It verifies the following properties: $$, $$, for all u, v ∈ ℝ_ℱ, $$, since (−1)ℝ_ℱ = ℝ_ℱ.

Definition 1 (see [20].)Let F : (0, a)⟶ℝ_ℱ be a fuzzy function. q^th order “fuzzy conformable fractional derivative” of F is defined by (where the limit is taken in the metric space (ℝ_ℱ, d)).

for all t > 0, q ∈ (0,1). Let F^(q)(t) stand for T_q(F)(t). Hence,

If F is q-differentiable in some (0, a) and $$ exists, then

Definition 2. Let F ∈ C((0, a), ℝ_ℱ)∩L¹((0, a), ℝ_ℱ). Define the fuzzy fractional integral for q ∈ (0,1].

where the integral is the usual Riemann improper integral.

Definition 3. Let q ∈ (0, a],  for any a > 0. A family {T(t)}_t≥0 of operators from ℝ_ℱ is called a fuzzy fractional q-semigroup (or fuzzy q-semigroup) of operators if

• (i)

  T(0) = I, where I is the identity mapping on ℝ_ℱ,

• (ii)

  T(s + t)^(1/q) = T(s^(1/q))T(t^(1/q)), for all s, t ≥ 0.

Definition 4. A q-semigroup T(t) is called a c₀-q-semigroup if

• (a)

  The function g:[0, ∞)⟶ℝ_ℱ, defined by g(t) = T(t)(x), is continuous at t = 0, for all x ∈ ℝ_ℱ, i.e.,

• (b)

  There exist constants w ≥ 0 and M ≥ 1 such that $$, for all t ≥ 0,  x, y ∈ ℝ_ℱ.

Example 1. Define on ℝ_ℱ the linear operator $$. Then, {T(t)}_t≥0 is a fuzzy (1/2)-semigroup. Indeed

• (i)

  T(0) = I, T(0)x = x, for all x ∈ ℝ_ℱ,

• (ii)

  For t, s ≥ 0,  x ∈ ℝ_ℱ,

• (a)

  For $$, then $$, and then using Remark 1, we deduce that $$. Therefore, the Hukuhara difference $$ exists and we have

•  

  Then,

•  

  Since $$, then $$.

• (b)

  For $$. Consequently, {T(t)}_t≥0 is a fuzzy c₀-q-semigroup on ℝ_ℱ.

Definition 5. The conformable q-derivative of T(t) at t = 0 is called the q-infinitesimal generator of the fuzzy q-semigroup {T(t)}_t≥0, with domain equals

We will write A for such generator.

Lemma 2. Let A : ℝ_ℱ⟶ℝ_ℱ and A₁ = jAj⁻¹: C⟶C tow the operator.

A is the operator of the fuzzy q-semigroup {T(t)}_t≥0 on ℝ_ℱ if and only if A₁ is the operator of the q-semigroup $$ defining on the convex closed set C and T₁ = jT(t)j⁻¹.

By using Definition 5, the proof is similar to the proof of Lemma 5 in [18] and is omitted.

Theorem 3. Let {T(t)}_t≥0 be a c₀-q-semigroup with infinitesimal generator A, 0 < q ≤ 1. Then, for all x ∈ ℝ_ℱ such that T(t)x ∈ D(A), for all t ≥ 0; the mapping t⟶T(t)x is q-differentiable and

Proof. Let q ∈ (0,1] and x ∈ ℝ_ℱ, for t ≥ 0, and we have

Since T(t)x ∈ D(A), then

Now, using Theorem 2.4 in [20], we get

for some $$. If ε⟶0, then c⟶0 and $$.

Consequently,

By using L’Hopital’s Rule, we get $$.

Example 2. Let f : ℝ_ℱ⟶ℝ_ℱ be continuous on [0,1]. Define

Then, T(t) is obviously a c₀-q-semigroup of contraction on ℝ_ℱ.

Remark 2. If M = 1 and w = 0 in Definition 4, we say that {T(t)}_t≥0 is a contraction fuzzy semigroup.

•  

  For q ∈ (0,1],

•  

  T(0) = I and T(t)f ∈ ℝ_ℱ whenever f ∈ ℝ_ℱ and that

Theorem 4. Let q ∈ (0,1).

If x(t) is a solution to the fuzzy fractional initial value problem,

Then, T(t)(x₀) = x(t) is a fuzzy semigroup. Furthermore, T(t)(x₀) is q-differentiable w.r.t t and T^(q)(t)(x₀) = F(x(t)) = F(T(t)(x₀)).

Proof. Let q ∈ (0,1) and k > 0. As obtained above, y(t) = x(k + (1/q)t^q) is a solution of the fractional initial value problem y^(q)(t) = F(y(t)), y(0) = x(k). Hence,

We set k = (1/q)s, then

and T(0)(x₀) = x(0) = x₀. Being a solution to a differential equation, T(t)(x₀) is q-differentiable with respect to t and T^(q)(t)(x₀) = x^(q)(t) = F(x(t)).

Theorem 5. Let q ∈ (0,1]. Suppose that a fuzzy semigroup T(t)(x) is q-differentiable w.r.t t, for all x ∈ ℝ_ℱ. Then, T(t)(x₀) is a solution to the fractional initial value problem

where F(x(t)) = T^(q)(0)(x₀).

Proof. By the q-semigroup property and using proof of Theorem 3, we obtain

and T(0)(x₀) = x₀.

Finally, we show that the fuzzy exponential function is a generalization of the fuzzy semigroup introduced in [5].

Theorem 6. If A : ℝ_ℱ⟶ℝ_ℱ is a bounded linear operator, then the fuzzy exponential function has a power series representation

Proof. Let A : ℝ_ℱ⟶ℝ_ℱ be a bounded linear operator as defined by Gal and Gal in [5]. Then,

and hence by [6] satisfies the condition. Consequently, is a solution to the Cauchy problem x^(q)(t) = Ax(t), x(0) = x₀. Define S(t) by a power series as

Now, by Theorem 3.9 in [5], $$ in [5], so S(t) is a fuzzy semigroup, and hence by Theorem 5, S(t)(x₀) is a solution to the problem

Since a bounded linear operator is Lipschitzian, it follows by Theorem 6.1 in [25] that the problem x^(q)(t) = Ax(t), x(0) = x₀, has a unique solution. Hence, $$, for all x₀ ∈ ℝ_ℱ.