Two-point boundary value problems for first-order nonlinear fuzzy differential equation 1

Abstract

This paper provides some existence results about first-order fuzzy differential equations with two-point boundary value conditions. We firstly study a class of linear fuzzy differential equation, the results are applied to discuss some nonlinear fuzzy boundary value problems. By using contraction mapping principle in complete metric space, we obtain some existence results about first-order nonlinear fuzzy differential equation with boundary value condition x (0) = αx (T), where α ∈ R.

Keywords

Nonlinear fuzzy differential equations two-point boundary value problem contracting mapping principle

1 Introduction

In real world, many phenomena were formulated in terms of differential equations. To consider the relationship between initial value and terminal value, people may need to study boundary value problems. Shooting method and fixed point theorems were frequently used to discuss existence of solutions to boundary value problems of crisp differential equations. Basic monographs on theory of boundary value problems were introduced in [1 –5], and some new results were shown in [6 –8].

In the past twenty years, uncertainty or vagueness in real world problems have been considered in many mathematical models. For example, L.C. Barros et al. discussed fuzzy Malthusian growth model and fuzzy logistic discrete model in [9]. In their opinion, fuzzy models were more appropriate than ordinary differential equations and stochastic models considering different predatory degrees, competition, survival, and so on. On the other hand, it is almost impossible to obtain exact value of population density or size in real world, an alterative way is to present range of the population density or size as some intervals. To consider demographic fuzziness, we need to study corresponding fuzzy differential equations.

By using strongly generalized differentiability concept introduced in [10, 11], B. Bede et al. presented a variation of constants formulas for first order linear differential equations in [12]. Based on this paper, some existence theories of fuzzy differential equations were obtained. We refer to [13 –16] and the references therein.

There are some different ideas to discuss fuzzy differential equation. For example, in [17], B. Liu considered a new kind of fuzzy differential equation based on credibility measure. Utilizing concepts and theories of fuzzy process and uncertain process introduced by B. Liu, V. Lupulescu et al. discussed local existence and uniqueness of fuzzy delay differential equations in [18]. Based on generalized Hukuhara differentiability concept introduced by L. Stefanini and B. Bede in [19], some new concepts of solutions to fuzzy differential equations were discussed in[20 –22].

Boundary value problems of fuzzy differential equations have been studied frequently. By the help of some new structures and properties of fuzzy sets, M. Chen et al. studied existence of analytic solution to two-point boundary value problems in [23, 24]. In [25], A. Khastan and J.J. Nieto introduced a new concept of solutions to a two-point boundary value problem for a second order fuzzy differential equation, and provided some methods to calculate the solutions. H.V. Long et al. discussed existence and uniqueness of fuzzy solution to hyperbolic partial differential equation in [26]. The authors introduced some new weighted metrics to study the solvability for boundary valued problems of fuzzy hyperbolic equations.

A. Khastan, J.J. Nieto and R. Rodriguez-Lopez studied periodic boundary value problems for first-order linear differential equations by using contraction mapping principle in [13]. By considering two types of derivatives with a switching point, the authors provided some sufficient conditions for the existence of piecewise-defined periodic solutions. In [14], J.J. Nieto and R. Rodriguez-Lopez studied boundary value problems for linear fuzzy differential equation. They calculated the Green function and the exact solutions, then provided some existence and uniqueness theories. More results were provided by R. Rodriguez-Lopez in [27]. The same method was applied to study existence of solutions to periodic boundary value problem for impulsive linear fuzzy differential equation by J.J. Nieto, R. Rodriguez-Lopez and M. Villanueva-Pesqueira in [28].

In [29], N. Gasilov et al. provided a method to calculate solution for differential equation with fuzzy boundary values. Under their assumptions, existence of solutions to crisp boundary value problem could guarantee existence of solutions to corresponding fuzzy problems.

Sum up to all, we can find few results discussing boundary value problems of nonlinear fuzzy differential equations, which has variable parameter in boundary value condition. In [31], we studied fuzzy boundary value problem $\begin{matrix} {\begin{matrix} x^{'} (t) = f (t), t \in (0, T), \\ x (0) = α x (T), \end{matrix} \end{matrix}$ where α ∈ R \ {±1}. Some necessary conditions for the existence of solutions were provided. Based on these results and the upper and lower solutions method, we also discussed existence of solutions to

${\begin{matrix} x^{'} (t) = f (t, x (t)), t \in (0, T), \\ x (0) = α x (T), \end{matrix}$ (1) where α ∈ (0, 1) ∪ (1, + ∞), f ∈ C ([0, T] × R_F,R_F) and f (t, x) was nondecreasing or nonincreasing with respect to x, ∀t ∈ [0, T].

In the present work, we will reconsider boundary value problem (1) without monotonicity constraints on f and nonnegative constraint on α. By using contraction mapping principle, some existence results of several kinds of solution to (1) will be proved.

The present paper is organized as follows. In Section 2, some basic definitions and lemmas are introduced. In Section 3, we provide some important results about x′ (t) = u (t) , x (0) = αx (T). Then we apply Banach contraction mapping method to study existence of solution to (1) in Section 4. Finally, we present some examples in Section 5.

2 Preliminaries

Let R_F be the class of fuzzy subsets of the real axis, u : R → [0, 1], which satisfies:

∃t₀ ∈ R, u (t₀) =1 (normality).

For all s ∈ [0, 1] and t₁, t₂ ∈ R, u (st₁ + (1 - s) t₂) ≥ min {u (t₁) , u (t₂)} (convex fuzzy).

u is upper semi-continuous on R.

The closure of {t ∈ R|u (t) >0} is compact.

For all r ∈ (0, 1], let [u] ^r = {t ∈ R|u (t) ≥ r} and

[u]^{0} = \bar{{t \in R | u (t) > 0}}

, where

\bar{A}

means the closure of A. [u] ^r is also written as

[u_{-}^{r}, u_{+}^{r}]

, and we denote

diam [u]^{r} = u_{+}^{r} - u_{-}^{r}

as in [27].

Set u, v ∈ R_F and λ ∈ R. u + v, λu and u - v are defined as [u + v] ^r = [u] ^r + [v] ^r, [λu] ^r = {λt ∈ R|t ∈ [u] ^r} and u - v = u + (-1) · v as [12].

Now we introduce some lemmas.

Lemma 1. [12] For all λ, μ ∈ R and u, v ∈ R_F,

λ (u + v) = λu + λv.

(λ + μ) u = λu + μu is true only if λμ ≥ 0.

Lemma 2. [31] Suppose that λ ∈ R, v ∈ R_F \ R, then x = λ (x + v) and x = λx + v have solutions in R_F only if λ ∈ (-1, 1). Moreover, the solution of x = λ (x + v) can be written as $\begin{matrix} x = {\begin{matrix} \frac{λ}{1 - λ} v, λ \in [0, 1), \\ \frac{λ^{2}}{1 - λ^{2}} v + \frac{λ}{1 - λ^{2}} v, λ \in (- 1, 0) \end{matrix} \end{matrix}$ and the solution of x = λx + v can be written as $\begin{matrix} x = {\begin{matrix} \frac{1}{1 - λ} v, λ \in [0, 1), \\ \frac{1}{1 - λ^{2}} v + \frac{λ}{1 - λ^{2}} v, λ \in (- 1, 0) . \end{matrix} \end{matrix}$

Definition 1. [32] Let u, v ∈ R_F. z = u ⊖ v is called the H-difference of u and v, if u = v + z.

Obviously, $[u ⊖ v]^{r} = [u_{-}^{r} - v_{-}^{r}, u_{+}^{r} - v_{+}^{r}]$ , D (u $⊖ v, \tilde{0}) \to 0$ is equivalent to D (u, v) →0. In addition, u ⊖ v does not always exist for u, v ∈ R_F.

Lemma 3. Let λ ∈ R and u, v ∈ R_F \ R. Suppose that x is a solution to equation $x + u = λ x + v,$ (2) then $\begin{matrix} x = {\begin{matrix} \frac{1}{λ - 1} (u ⊖ v), λ \in (1, + \infty), \\ \frac{1}{1 - λ} (v ⊖ u), λ \in [0, 1), \\ \frac{1}{1 - λ^{2}} [λ (v ⊖ u) + v ⊖ u], λ \in (- 1, 0), \\ \frac{1}{λ^{2} - 1} [λ (u ⊖ v) + u ⊖ v], λ \in (- \infty, - 1) . \end{matrix} \end{matrix}$

Moreover, if λ = 1, then (2) has no solution unless u = v; if λ = -1, then (2) has no solution unless diam [u] ^r = diam [v] ^r, r ∈ [0, 1].

Proof. Suppose that x is a solution to (2), then ∀r ∈ [0, 1],

$[x + u]^{r} = {\begin{matrix} [λ x_{-}^{r} + v_{-}^{r}, λ x_{+}^{r} + v_{+}^{r}], λ \geq 0, \\ [λ x_{+}^{r} + v_{-}^{r}, λ x_{-}^{r} + v_{+}^{r}], λ < 0 . \end{matrix}$ (3)

Consequently, supposing λ ∈ (1, + ∞), (3) implies that $x_{-}^{r} = \frac{1}{λ - 1} (u_{-}^{r} - v_{-}^{r}), x_{+}^{r} = \frac{1}{λ - 1} (u_{+}^{r} - v_{+}^{r}),$ that is, $x = \frac{1}{λ - 1} (u ⊖ v) .$

If λ ∈ [0, 1), then we get $x_{-}^{r} = \frac{1}{1 - λ} (v_{-}^{r} - u_{-}^{r}), x_{+}^{r} = \frac{1}{1 - λ} (v_{+}^{r} - u_{+}^{r}),$ thus $x = \frac{1}{1 - λ} (v ⊖ u) .$ Similarly, we can check that $\begin{matrix} x = {\begin{matrix} \frac{1}{1 - λ^{2}} [λ (v ⊖ u) + v ⊖ u], λ \in (- 1, 0), \\ \frac{1}{λ^{2} - 1} [λ (u ⊖ v) + u ⊖ v], λ \in (- \infty, - 1) . \end{matrix} \end{matrix}$

If λ = -1, then (3) provides that $[x + u]^{r} = [- x_{+}^{r} + v_{-}^{r}, - x_{-}^{r} + v_{+}^{r}], r \in [0, 1],$ which implies that $x_{+}^{r} + x_{-}^{r} = v_{-}^{r} - u_{-}^{r} = v_{+}^{r} - u_{+}^{r}$ . Hence, x + u = (-1) x + v has no solution unless $v_{-}^{r} - u_{-}^{r} = v_{+}^{r} - u_{+}^{r}$ , that is, $v_{+}^{r} - v_{-}^{r} = u_{+}^{r} - u_{-}^{r}$ or diam [u] ^r = diam [v] ^r for all r ∈ [0, 1]. Moreover, any x ∈ R_F satisfying $x_{+}^{r} + x_{-}^{r} = v_{-}^{r} - u_{-}^{r} = v_{+}^{r} - u_{+}^{r}$ is a solution to (2).

Let D : R_F × R_F → [0, + ∞), $D (u, v) = sup_{r \in [0, 1]} \max {| u_{-}^{r} - v_{-}^{r} |, | u_{+}^{r} - v_{+}^{r} |} .$ D is the Hausdorff distance on R_F and (R_F, D) is a complete metric space.

Lemma 4.For all k ∈ R and u, v, w, e ∈ R_F,

D (u + w, v + w) = D (u, v).

D (ku, kv) = |k| · D (u, v).

D (u + v, w + e) ≤ D (u, w) + D (v, e).

Suppose that u ⊖ v and w ⊖ e exist, then D (u ⊖ v, w ⊖ e) ≤ D (u, w) + D (v, e).

Proof. (1)–(3) are from [12]. Here we just prove (4). In fact, we have $\begin{matrix} D (u ⊖ v, w ⊖ e) \\ = sup_{r \in [0, 1]} max {| u_{-}^{r} - v_{-}^{r} - (w_{-}^{r} - e_{-}^{r}) |, | u_{+}^{r} - v_{+}^{r} \\ - (w_{+}^{r} - e_{+}^{r}) |} \\ \leq sup_{r \in [0, 1]} max {| u_{-}^{r} - w_{-}^{r} | + | v_{-}^{r} - e_{-}^{r} |, | u_{+}^{r} - w_{+}^{r} | \\ + | v_{+}^{r} - e_{+}^{r} |} \\ \leq sup_{r \in [0, 1]} [max {| u_{-}^{r} - w_{-}^{r} |, | u_{+}^{r} - w_{+}^{r} |} \\ + max {| v_{-}^{r} - e_{-}^{r} |, | v_{+}^{r} - e_{+}^{r} |}] \\ \leq sup_{r \in [0, 1]} max {| u_{-}^{r} - w_{-}^{r} |, | u_{+}^{r} - w_{+}^{r} |} \\ + sup_{r \in [0, 1]} max {| v_{-}^{r} - e_{-}^{r} |, | v_{+}^{r} - e_{+}^{r} |} \\ = D (u, w) + D (v, e) . \end{matrix}$

Definition 2. A function f : [a, b] → R_F is said to be H-continuous (Hausdorff continuous) on [a, b], if ∀t ∈ [a, b] and ɛ > 0, ∃δ_t > 0, D (f (t) , f (t′)) < ɛ for t′ ∈ [a, b] and |t - t′| < δ_t.

Let C ([a, b] , R_F) be the class of Hausdorff continuous functions on [a, b] and denote $d (f, g) = max_{t \in [a, b]} D (f (t), g (t)) .$ It is well known that (C ([a, b] , R_F) , d) is a complete metric space.

Definition 3. [12] A function f : [a, b] → R_F is said to be Riemann integrable on [a, b], if there exists I_R ∈ R_F, ∀ɛ > 0, ∃ δ > 0, for any division of [a, b], ▵ : a = t₀ < t₁ < ⋯ < t_n = b with norm λ (▵) < δ, and for any points ξ_i ∈ [t_i, t_i+1] , i = 0, 1, ⋯ , n - 1, $D (\sum_{i = 0}^{n - 1} f (ξ_{i}) \cdot (t_{i + 1} - t_{i}), I_{R}) < ɛ .$ We denote by $I_{R} = \int_{a}^{b} f (t) dt$ the fuzzy Riemann integral.

Definition 4. [12] A function f : [a, b] → R_F is said to be strongly generalized differentiable at t ∈ (a, b), if there exists f′ (t) ∈ R_F such that ∀h > 0 sufficiently small, H-differences and limits in the following formulas exist with metric D:

$lim_{h \to 0 +} \frac{f (t + h) ⊖ f (t)}{h} = lim_{h \to 0 +} \frac{f (t) ⊖ f (t - h)}{h} = f^{'} (t)$ or

$lim_{h \to 0 +} \frac{f (t) ⊖ f (t + h)}{(- h)} = lim_{h \to 0 +} \frac{f (t - h) ⊖ f (t)}{(- h)} = f^{'} (t)$ or

$lim_{h \to 0 +} \frac{f (t + h) ⊖ f (t)}{h} = lim_{h \to 0 +} \frac{f (t - h) ⊖ f (t)}{(- h)} = f^{'} (t)$ or

$lim_{h \to 0 +} \frac{f (t) ⊖ f (t + h)}{(- h)} = lim_{h \to 0 +} \frac{f (t) ⊖ f (t - h)}{h} = f^{'} (t)$ .

Lemma 5. [33, 34] Suppose that g, h ∈ C ([a, b],R_F). Then we have

g is Riemann integrable on [a, b] and $G (t) = \int_{a}^{t} g (s) ds$ is strongly generalized differentiable as in Definition 4(i), G′ (t) = g (t).

$[\int_{a}^{b} g (s) ds]^{r} = [\int_{a}^{b} [g (s)]_{-}^{r} ds, \int_{a}^{b} [g (s)]_{+}^{r} ds]$ ,∀r ∈ [0, 1].

$D (\int_{a}^{b} g (s) ds, \int_{a}^{b} h (s) ds) \leq \int_{a}^{b} D (g (s), h (s)) ds$ .

Definition 5. A function f : [a, b] × R_F → R_F is said to be continuous at (t, x) ∈ [a, b] × R_F, if ∀ɛ > 0, there exists δ > 0 such that D (f (t, x) , f (t′, y)) < ɛ for t′ ∈ [a, b], |t - t′| < δ and D (x, y) < δ.

If f is continuous at any point in [a, b] × R_F, then f is said to be continuous on [a, b] × R_F, denoted by f ∈ C ([a, b] × R_F, R_F).

Lemma 6. [12] Let t₀ ∈ R and f ∈ C (R × R_F, R_F). Initial value problem y′ = f (t, y) , y (t₀) = y₀ ∈ R_F is equivalent to one of the integral equations: $y (t) = y_{0} + \int_{t_{0}}^{t} f (s, y (s)) ds$ or $y (t) = y_{0} ⊖ (- 1) \int_{t_{0}}^{t} f (s, y (s)) ds$ on some interval (t₀, t₁) ⊂ R, depending on the strongly generalized differentiability considered Definition 4(i) or (ii), respectively.

3 Existence of x′ (t) = u (t) , x (0) = αx (T)

In this section, we consider fuzzy boundary value problem

${\begin{matrix} x^{'} (t) = u (t), t \in (0, T), \\ x (0) = α x (T), \end{matrix}$ (4) where u ∈ C ([0, T] , R_F) is fixed.

Definition 6.x ∈ C ([0, T] , R_F) is called (i) or (ii)-differentiable solution to (4), if x satisfies (4) and is strongly generalized differentiable as in Definition 4(i) or (ii).

Definition 7. [30] x ∈ C ([0, T] , R_F) is called mixed solution of (4), if x satisfies (4) and is (i)-differentiable on some subintervals of [0, T] and (ii)-differentiable on the remaining parts.

Generally, (4) may have many kinds of mixed solution. Here we just consider two cases: (i)-mixed solution, which is (i)-differentiable on [0, δ), (ii)-differentiable on (δ, T] and strongly generalized differentiable at t = δ as in Definition 4(iii); (ii)-mixed solution, which is (ii)-differentiable on [0, δ), (i)-differentiable on (δ, T] and strongly generalized differentiable at t = δ as in Definition 4(iv).

Let x (0) = x₀ ∈ R_F. By Lemma 6, fuzzy differential equation x′ (t) = u (t) has a (i)-differentiable solution $x (t) = x_{0} + \int_{0}^{t} u (s) ds, t \in [0, T] .$ (5)

Suppose that $x_{0} ⊖ \int_{0}^{T} (- u (s)) ds \in R_{F}$ , then x′ (t) = u (t) also has a (ii)-differentiable solution $x (t) = x_{0} ⊖ \int_{0}^{t} (- u (s)) ds, t \in [0, T] .$ (6)

If there exists δ ∈ (0, T) such that u (δ) ∈ R, then x′ (t) = u (t) may have (i) or (ii)-mixed solutions [30]. The (i)-mixed solution can be written as

$x (t) = {\begin{matrix} x_{0} + \int_{0}^{t} u (s) ds, t \in [0, δ], \\ x_{0} + \int_{0}^{δ} u (s) ds \\ ⊖ \int_{δ}^{t} (- u (s)) ds, t \in (δ, T] \end{matrix}$ (7) and the (ii)-mixed solution can be written as

$x (t) = {\begin{matrix} x_{0} ⊖ \int_{0}^{t} (- u (s)) ds, t \in [0, δ], \\ x_{0} ⊖ \int_{0}^{δ} (- u (s)) ds \\ + \int_{δ}^{t} u (s) ds, t \in (δ, T] . \end{matrix}$ (8)

To meet boundary value condition x (0) = αx (T), (5)–(8) must satisfy the following equations respectively: $α (x_{0} + \int_{0}^{T} u (s) ds) = x_{0},$ (9) $α (x_{0} ⊖ \int_{0}^{T} (- u (s)) ds) = x_{0},$ (10) $α (x_{0} + \int_{0}^{δ} u (s) ds ⊖ \int_{δ}^{T} (- u (s)) ds) = x_{0},$ (11) $α (x_{0} ⊖ \int_{0}^{δ} (- u (s)) ds + \int_{δ}^{T} u (s) ds) = x_{0} .$ (12)

By Lemma 2 and Lemma 3, we have following results.

Theorem 1. [31] Boundary value problem (4) has (i)-differentiable solution only if α ∈ (-1, 1). Moreover, if α ∈ [0, 1), then the (i)-differentiable solution of (4) can be written as $x (t) = \frac{α}{1 - α} \int_{0}^{T} u (s) ds + \int_{0}^{t} u (s) ds$ (13)for t ∈ [0, T]. If α ∈ (-1, 0), then the (i)-differentiable solution is

$\begin{matrix} x (t) = & \frac{α^{2}}{1 - α^{2}} \int_{0}^{T} u (s) ds + \frac{α}{1 - α^{2}} \int_{0}^{T} u (s) ds \\ + \int_{0}^{t} u (s) ds, t \in [0, T] . \end{matrix}$ (14)Theorem 2. [31] Boundary value problem (4) has (ii)-differentiable solution only if α ∈ (- ∞ , -1) ∪ (1, + ∞). Moreover, if α ∈ (1, + ∞), then the (ii)-differentiable solution can be written as $x (t) = \frac{α}{1 - α} \int_{0}^{T} u (s) ds ⊖ \int_{0}^{t} (- u (s)) ds$ (15)for t ∈ [0, T]. If α ∈ (- ∞ , -1), then the (ii)-differentiable solution is

$\begin{matrix} x (t) = & \frac{α^{2}}{1 - α^{2}} \int_{0}^{T} u (s) ds + \frac{α}{1 - α^{2}} \int_{0}^{T} u (s) ds \\ ⊖ \int_{0}^{t} (- u (s)) ds, t \in [0, T] . \end{matrix}$ (16)

Now we turn to consider existence of mixed solutions to (4). Denote $\begin{matrix} I_{1} (u) = \int_{δ}^{T} (- u (s)) ds ⊖ \int_{0}^{δ} u (s) ds, \\ I_{2} (u) = \int_{0}^{δ} u (s) ds ⊖ \int_{δ}^{T} (- u (s)) ds . \end{matrix}$

Theorem 3. [31] Suppose that there exists δ ∈ (0, T) such that u (δ) ∈ R.

If α ∈ (1, + ∞) and I₁ (u) ∈ R_F, then (4) has (i)-mixed solution

$x (t) = {\begin{matrix} \frac{α}{α - 1} I_{1} (u) + \int_{0}^{t} u (s) ds, t \in [0, δ], \\ \frac{α}{α - 1} I_{1} (u) + \int_{0}^{δ} u (s) ds \\ ⊖ \int_{δ}^{t} (- u (s)) ds, t \in (δ, T] . \end{matrix}$ (17)

If α ∈ (- ∞ , -1) and I₁ (u) ∈ R_F, then (4) has (i)-mixed solution

$x (t) = {\begin{matrix} \frac{α}{α^{2} - 1} I_{1} (u) + \frac{α^{2}}{α^{2} - 1} I_{1} (u) \\ + \int_{0}^{t} u (s) ds, t \in [0, δ], \\ \frac{α}{α^{2} - 1} I_{1} (u) + \frac{α^{2}}{α^{2} - 1} I_{1} (u) \\ + \int_{0}^{δ} u (s) ds ⊖ \int_{δ}^{t} (- u (s)) ds, t \in (δ, T] . \end{matrix}$ (18)

If α ∈ (0, 1) and I₂ (u) ∈ R_F, then (4) has (i)-mixed solution

$x (t) = {\begin{matrix} \frac{α}{1 - α} I_{2} (u) + \int_{0}^{t} u (s) ds, t \in [0, δ], \\ \frac{α}{1 - α} I_{2} (u) + \int_{0}^{δ} u (s) ds \\ ⊖ \int_{δ}^{t} (- u (s)) ds, t \in (δ, T] . \end{matrix}$ (19)

If α ∈ (-1, 0) and I₂ (u) ∈ R_F, then (4) has (i)-mixed solution

$x (t) = {\begin{matrix} \frac{α}{1 - α^{2}} I_{2} (u) + \frac{α^{2}}{1 - α^{2}} I_{2} (u) \\ + \int_{0}^{t} u (s) ds, t \in [0, δ], \\ \frac{α}{1 - α^{2}} I_{2} (u) + \frac{α^{2}}{1 - α^{2}} I_{2} (u) \\ + \int_{0}^{δ} u (s) ds ⊖ \int_{δ}^{t} (- u (s)) ds, t \in (δ, T] . \end{matrix}$ (20)

Theorem 4. [31] Suppose that there exists δ ∈ (0, T) such that u (δ) ∈ R.

If α ∈ (1, + ∞) and H-difference $\int_{0}^{δ} (- u (s)) ds ⊖ α \int_{δ}^{T} u (s) ds$ exists, then (4) has (ii)-mixed solution

$x (t) = {\begin{matrix} \frac{α}{1 - α} I_{2} (u) ⊖ \int_{0}^{t} (- u (s)) ds, t \in [0, δ], \\ \frac{α}{1 - α} I_{2} (u) ⊖ \int_{0}^{δ} (- u (s)) ds \\ + \int_{δ}^{t} u (s) ds, t \in (δ, T] . \end{matrix}$ (21)

If α ∈ (- ∞ , -1) and H-difference $\int_{0}^{δ} (- u (s)) ds ⊖ α^{2} \int_{δ}^{T} u (s) ds$ exists, then (4) has (ii)-mixed solution

$x (t) = {\begin{matrix} \frac{α}{1 - α^{2}} I_{2} (u) + \frac{α^{2}}{1 - α^{2}} I_{2} (u) \\ ⊖ \int_{0}^{t} (- u (s)) ds, t \in [0, δ], \\ \frac{α}{1 - α^{2}} I_{2} (u) + \frac{α^{2}}{1 - α^{2}} I_{2} (u) \\ ⊖ \int_{0}^{δ} (- u (s)) ds + \int_{δ}^{t} u (s) ds, t \in (δ, T] . \end{matrix}$ (22)

If α ∈ [0, 1) and H-difference $α \int_{δ}^{T} u (s) ds ⊖ \int_{0}^{δ} (- u (s)) ds$ exists, then (4) has (ii)-mixed solution

$x (t) = {\begin{matrix} \frac{α}{α - 1} I_{1} (u) ⊖ \int_{0}^{t} (- u (s)) ds, t \in [0, δ], \\ \frac{α}{α - 1} I_{1} (u) ⊖ \int_{0}^{δ} (- u (s)) ds \\ + \int_{δ}^{t} u (s) ds, t \in (δ, T] . \end{matrix}$ (23)

If α ∈ (-1, 0) and H-difference $α^{2} \int_{δ}^{T} u (s) ds ⊖ \int_{0}^{δ} (- u (s)) ds$ exists, then (4) has (ii)-mixed solution

$x (t) = {\begin{matrix} \frac{α}{α^{2} - 1} I_{1} (u) + \frac{α^{2}}{α^{2} - 1} I_{1} (u) \\ ⊖ \int_{0}^{t} (- u (s)) ds, t \in [0, δ], \\ \frac{α}{α^{2} - 1} I_{1} (u) + \frac{α^{2}}{α^{2} - 1} I_{1} (u) \\ ⊖ \int_{0}^{δ} (- u (s)) ds + \int_{δ}^{t} u (s) ds, t \in (δ, T] . \end{matrix}$ (24)

When α = 1 or -1, (4) is called periodic or anti-periodic boundary value problem respectively. By Lemma 2, if α = ±1, then (4) has no (i) or (ii)-differentiable solution. Here we consider existence of mixed solutions.

Theorem 5.Suppose that there exists δ ∈ (0, T) such that u (δ) ∈ R.

If α = 1 and $\int_{0}^{δ} u (s) ds = \int_{δ}^{T} (- u (s)) ds,$ then (4) has (i) and (ii)-mixed solutions.

If α = -1 and $diam [\int_{δ}^{T} (- u (s)) ds]^{r} = diam [\int_{0}^{δ} u (s) ds]^{r}$ (25) for all r ∈ [0, 1], then (4) has (i) and (ii)-mixed solutions.

Proof. (i) If α = 1 and $\int_{0}^{δ} u (s) ds = \int_{δ}^{T} (- u (s)) ds$ holds, then (7) is well defined and satisfies x (0) = x (T) for any x₀ ∈ R_F. Hence, (7) is a (i)-mixed solution to (4).

On the other hand, referring to (8), we denote

$x (t) = {\begin{matrix} v + \int_{0}^{δ} (- u (s)) ds \\ ⊖ \int_{0}^{t} (- u (s)) ds, t \in [0, δ], \\ v + \int_{δ}^{t} u (s) ds, t \in (δ, T], \end{matrix}$ (26) where v ∈ R_F. We can check that (26) is well defined and satisfies x (0) = x (T), thus (26) is a (ii)-mixed solution to (4) with α = 1.

(ii) Referring to (7), we denote

$x (t) = {\begin{matrix} v + \int_{δ}^{T} (- u (s)) ds \\ + \int_{0}^{t} u (s) ds, t \in [0, δ], \\ v + \int_{δ}^{T} (- u (s)) ds + \int_{0}^{δ} u (s) ds \\ ⊖ \int_{δ}^{t} (- u (s)) ds, t \in (δ, T], \end{matrix}$ (27) where v ∈ R_F satisfies

$\begin{matrix} v_{-}^{r} + v_{+}^{r} \\ = & - {[\int_{0}^{δ} u (s) ds]}_{+}^{r} - {[\int_{δ}^{T} (- u (s)) ds]}_{-}^{r} \end{matrix}$ (28) for all r ∈ [0, 1]. Obviously, (27) is well defined and satisfies $x (0) = v + \int_{δ}^{T} (- u (s)) ds$ , $x (T) = v + \int_{0}^{δ} u (s) ds$ . Now we prove x (0) = - x (T).

For all r ∈ [0, 1], (28) implies that $\begin{matrix} \begin{matrix} [- x (T)]_{-}^{r} & = [- (v + \int_{0}^{δ} u (s) ds)]_{-}^{r} \\ = - v_{+}^{r} - [\int_{0}^{δ} u (s) ds]_{+}^{r} \\ = v_{-}^{r} + [\int_{δ}^{T} (- u (s)) ds]_{-}^{r} \\ = [x (0)]_{-}^{r} . \end{matrix} \end{matrix}$

On the other hand, (25) implies that

$\begin{matrix} [\int_{δ}^{T} (- u (s)) ds]_{+}^{r} - [\int_{δ}^{T} (- u (s)) ds]_{-}^{r} \\ = & [\int_{0}^{δ} u (s) ds]_{+}^{r} - [\int_{0}^{δ} u (s) ds]_{-}^{r} \end{matrix}$ (29) for all r ∈ [0, 1]. Thus we have

$\begin{matrix} [- x (T)]_{+}^{r} \\ = & - v_{-}^{r} - [\int_{0}^{δ} u (s) ds]_{-}^{r} \\ = & v_{+}^{r} + [\int_{0}^{δ} u (s) ds]_{+}^{r} + [\int_{δ}^{T} (- u (s)) ds]_{-}^{r} \\ - [\int_{0}^{δ} u (s) ds]_{-}^{r} \\ = & v_{+}^{r} + [\int_{δ}^{T} (- u (s)) ds]_{+}^{r} \\ = & [x (0)]_{+}^{r} . \end{matrix}$ (30) As a conclusion, (27) is a (i)-mixed solution to (4) with α = -1.

Now we turn to consider existence of (ii)-mixed solution. Referring to (8), we denote

$x (t) = {\begin{matrix} w + \int_{0}^{δ} (- u (s)) ds \\ ⊖ \int_{0}^{t} (- u (s)) ds, t \in [0, δ], \\ w + \int_{δ}^{t} u (s) ds, t \in (δ, T], \end{matrix}$ (31) where w ∈ R_F and satisfies

$\begin{matrix} w_{-}^{r} + w_{+}^{r} \\ = & [\int_{δ}^{T} (- u (s)) ds]_{-}^{r} + [\int_{0}^{δ} u (s) ds]_{+}^{r} \end{matrix}$ (32) for all r ∈ [0, 1].

We can check that (31) is well defined and satisfies $x (0) = w + \int_{0}^{δ} (- u (s)) ds$ , $x (T) = w + \int_{δ}^{T} u (s) ds$ .

Now we claim that x (0) = - x (T). In fact, ∀r ∈ [0, 1], (32) implies that $\begin{matrix} \begin{matrix} [- x (T)]_{-}^{r} & = [(- 1) w + (- 1) \int_{δ}^{T} u (s) ds]_{-}^{r} \\ = - w_{+}^{r} + [\int_{δ}^{T} (- u (s)) ds]_{-}^{r} \\ = w_{-}^{r} - [\int_{0}^{δ} u (s) ds]_{+}^{r} \\ = w_{-}^{r} + [\int_{0}^{δ} (- u (s)) ds]_{-}^{r} \\ = [x (0)]_{-}^{r} . \end{matrix} \end{matrix}$

In addition, (29) and (32) provide that $\begin{matrix} [- x (T)]_{+}^{r} & = - w_{-}^{r} + [\int_{δ}^{T} (- u (s)) ds]_{+}^{r} \\ = w_{+}^{r} - [\int_{δ}^{T} (- u (s)) ds]_{-}^{r} \\ - [\int_{0}^{δ} u (s) ds]_{+}^{r} + [\int_{δ}^{T} (- u (s)) ds]_{+}^{r} \\ = w_{+}^{r} - [\int_{0}^{δ} u (s) ds]_{-}^{r} \\ = w_{+}^{r} + [\int_{0}^{δ} (- u (s)) ds]_{+}^{r} \\ = [x (0)]_{+}^{r} . \end{matrix}$

Consequently, (31) satisfies x (0) = - x (T) and is a (ii)-mixed solution of (4) with α = -1.

Example 1. Consider $x^{'} (t) = (2 t - 1) \cdot c, t \in [0, 1],$ where c ∈ R_F is fixed.

Denote u (t) = (2t - 1) · c, then we have $u (\frac{1}{2}) = \tilde{0}$ and

$\int_{0}^{\frac{1}{2}} u (s) ds = \int_{\frac{1}{2}}^{1} (- u (s)) ds = - \frac{1}{4} \cdot c .$ (33) By Theorem 5(i), boundary value problem $\begin{matrix} {\begin{matrix} x^{'} (t) = (2 t - 1) \cdot c, \\ x (0) = x (1) \end{matrix} \end{matrix}$ has (i)-mixed solution (see Fig. 1) $\begin{matrix} x (t) = {\begin{matrix} v + \int_{0}^{t} (2 s - 1) cds, t \in [0, \frac{1}{2}], \\ v + \int_{0}^{\frac{1}{2}} (2 s - 1) cds \\ ⊖ \int_{\frac{1}{2}}^{t} (1 - 2 s) cds, t \in (\frac{1}{2}, 1] \end{matrix} \end{matrix}$ and (ii)-mixed solution (see Fig. 2) $\begin{matrix} x (t) = {\begin{matrix} v + \int_{0}^{\frac{1}{2}} (1 - 2 s) cds \\ ⊖ \int_{0}^{t} (1 - 2 s) cds, t \in [0, \frac{1}{2}], \\ v + \int_{\frac{1}{2}}^{t} (2 s - 1) cds, t \in (\frac{1}{2}, 1], \end{matrix} \end{matrix}$ where v ∈ R_F.

On the other hand, (33) implies that (25) is satisfied. By Theorem 5(ii), boundary value problem $\begin{matrix} {\begin{matrix} x^{'} (t) = (2 t - 1) \cdot c, \\ x (0) = - x (1) \end{matrix} \end{matrix}$ has (i)-mixed solution, which can be written as (see Fig. 3) $\begin{matrix} x (t) = {\begin{matrix} w + \int_{0}^{t} (2 s - 1) cds, t \in [0, \frac{1}{2}], \\ w + \int_{0}^{\frac{1}{2}} (2 s - 1) cds \\ ⊖ \int_{\frac{1}{2}}^{t} (1 - 2 s) cds, t \in (\frac{1}{2}, 1] . \end{matrix} \end{matrix}$

Here w is an arbitrary asymmetric fuzzy subset in R_F, that is, $\forall r \in [0, 1], [w]_{-}^{r} + [w]_{+}^{r} = 0$ .

We also can check that $\begin{matrix} x (t) = {\begin{matrix} \frac{1}{4} c - \frac{1}{4} c ⊖ \int_{0}^{t} (1 - 2 s) cds, t \in [0, \frac{1}{2}], \\ - \frac{1}{4} c + \int_{\frac{1}{2}}^{t} (2 s - 1) cds, t \in (\frac{1}{2}, 1] \end{matrix} \end{matrix}$ is a (ii)-mixed solution of x′ (t) = (2t - 1) · c and satisfies x (0) = - x (1) (see Fig. 4).

4 Existence of x′ (t) = f (t, x) , x (0) = αx (T)

By using Banach contraction mapping principle, we study existence of solutions to boundary value problem (1) in this section.

Firstly, let u ∈ C ([0, T] , R_F) be fixed, we consider

${\begin{matrix} x^{'} (t) = f (t, u (t)), t \in (0, T), \\ x (0) = α x (T), \end{matrix}$ (34) where f ∈ C ([0, T] × R_F, R_F).

Referring to (13)–(16), we define map A_α as $\begin{matrix} A_{α} u (t) = {\begin{matrix} \frac{α}{1 - α} \int_{0}^{T} f (s, u (s)) ds \\ + \int_{0}^{t} f (s, u (s)) ds, α \in (0, 1), \\ \frac{α^{2}}{1 - α^{2}} \int_{0}^{T} f (s, u (s)) ds \\ + \frac{α}{1 - α^{2}} \int_{0}^{T} f (s, u (s)) ds \\ + \int_{0}^{t} f (s, u (s)) ds, α \in (- 1, 0), \\ \frac{α}{1 - α} \int_{0}^{T} f (s, u (s)) ds \\ ⊖ \int_{0}^{t} (- f (s, u (s))) ds, α \in (1, + \infty), \\ \frac{α^{2}}{1 - α^{2}} \int_{0}^{T} f (s, u (s)) ds \\ + \frac{α}{1 - α^{2}} \int_{0}^{T} f (s, u (s)) ds \\ ⊖ \int_{0}^{t} (- f (s, u (s))) ds, α \in (- \infty, - 1) . \end{matrix} \end{matrix}$

By Theorem 1, if α ∈ (-1, 0) ∪ (0, 1), then A_αu is a (i)-differentiable solution to (34). If α ∈ (- ∞ , -1) ∪ (1, + ∞), then Theorem 2 provides that A_αu is a (ii)-differentiable solution to (34).

Apparently, if there exists u ∈ C ([0, T] , R_F) such that A_αu = u, then u is also a (i)-differentiable or (ii)-differentiable solution to (1).

Here we introduce a hypothesis which will be used frequently in this section:

(H) For all t ∈ [0, T] and v, w ∈ R_F, $D (f (t, v), f (t, w)) \leq k (t) D (v, w),$ where k ∈ C ([0, T] , [0, + ∞)) satisfies $\int_{0}^{T} k (s) ds < 1 - min {| α |, | \frac{1}{α} |}$ , α ≠ 0, ± 1.

Theorem 6.Suppose that (H) holds.

If α ∈ (-1, 0) ∪ (0, 1), then (1) has a unique (i)-differentiable solution and has no (ii)-differentiable solution.

If α ∈ (- ∞ , -1) ∪ (1, + ∞), then (1) has a unique (ii)-differentiable solution and has no (i)-differentiable solution.

Proof. Setting α ∈ (0, 1), A_α can be rewritten as $\begin{matrix} \begin{matrix} A_{α} u (t) = & \frac{1}{1 - α} \int_{0}^{t} f (s, u (s)) ds \\ + \frac{α}{1 - α} \int_{t}^{T} f (s, u (s)) ds . \end{matrix} \end{matrix}$

We claim that A_α is a contraction on C ([0, T] , R_F).

In fact, ∀t ∈ [0, T] and x, y ∈ C ([0, T] , R_F), Lemma 4, Lemma 5 and (H) provide: $\begin{matrix} \begin{matrix} D (A_{α} x (t), A_{α} y (t)) \\ \leq \frac{1}{1 - α} D (\int_{0}^{t} f (s, x (s)) ds, \int_{0}^{t} f (s, y (s)) ds) \\ + \frac{α}{1 - α} D (\int_{t}^{T} f (s, x (s)) ds, \int_{t}^{T} f (s, y (s)) ds) \\ \leq \frac{1}{1 - α} \int_{0}^{t} D (f (s, x (s)), f (s, y (s))) ds \\ + \frac{α}{1 - α} \int_{t}^{T} D (f (s, x (s)), f (s, y (s))) ds \\ \leq \frac{1}{1 - α} \int_{0}^{T} k (s) \cdot D (x (s), y (s)) ds . \end{matrix} \end{matrix}$

Hence, we get $d (A_{α} x, A_{α} y) \leq \frac{1}{1 - α} \int_{0}^{T} k (s) ds \cdot d (x, y) .$

In addition, (H) implies that $\frac{1}{1 - α} \int_{0}^{T} k (s) ds < 1$ for α ∈ (0, 1), thus A_α is a contraction on C ([0, T] , R_F).

By Banach contraction mapping principle, there exists a unique x^* ∈ C ([0, T] , R_F) such that A_αx^* = x^*, that is, x^* is a (i)-differentiable solution to (1).

Similarly, we can prove that A_α is a contraction with α ∈ (-1, 0) ∪ (- ∞ , -1) ∪ (1, + ∞).

Finally, Theorem 1 guarantees that (1) has no (ii)-differentiable solution with α ∈ (-1, 1). If α ∈ (- ∞ , -1) ∪ (1, + ∞), then Theorem 2 provides that (1) has no (i)-differentiable solution.

Now we consider existence of mixed solutions to (1). Set x ∈ C ([0, T] , R_F). Referring to (17)–(20), if α ∈ (1, + ∞), we denote $\begin{matrix} B_{α} x (t) = {\begin{matrix} \frac{α}{α - 1} I_{1} (f (\cdot, x)) \\ + \int_{0}^{t} f (s, x (s)) ds, t \in [0, δ], \\ \frac{α}{α - 1} I_{1} (f (\cdot, x)) + \int_{0}^{δ} f (s, x (s)) ds \\ ⊖ \int_{δ}^{t} (- f (s, x (s))) ds, t \in (δ, T] . \end{matrix} \end{matrix}$

If α ∈ (- ∞ , -1), we denote $\begin{matrix} B_{α} x (t) = {\begin{matrix} \frac{α}{α^{2} - 1} I_{1} (f (\cdot, x)) + \frac{α^{2}}{α^{2} - 1} I_{1} (f (\cdot, x)) \\ + \int_{0}^{t} f (s, x (s)) ds, t \in [0, δ], \\ \frac{α}{α^{2} - 1} I_{1} (f (\cdot, x)) + \frac{α^{2}}{α^{2} - 1} I_{1} (f (\cdot, x)) \\ + \int_{0}^{δ} f (s, x (s)) ds \\ ⊖ \int_{δ}^{t} (- f (s, x (s))) ds, t \in (δ, T] . \end{matrix} \end{matrix}$ When α ∈ (0, 1), we denote $\begin{matrix} B_{α} x (t) = {\begin{matrix} \frac{α}{1 - α} I_{2} (f (\cdot, x)) \\ + \int_{0}^{t} f (s, x (s)) ds, t \in [0, δ], \\ \frac{α}{1 - α} I_{2} (f (\cdot, x)) + \int_{0}^{δ} f (s, x (s)) ds \\ ⊖ \int_{δ}^{t} (- f (s, x (s))) ds, t \in (δ, T] . \end{matrix} \end{matrix}$ When α ∈ (-1, 0), we denote $\begin{matrix} B_{α} x (t) = {\begin{matrix} \frac{α}{1 - α^{2}} I_{2} (f (\cdot, x)) + \frac{α^{2}}{1 - α^{2}} I_{2} (f (\cdot, x)) \\ + \int_{0}^{t} f (s, x (s)) ds, t \in [0, δ], \\ \frac{α}{1 - α^{2}} I_{2} (f (\cdot, x)) + \frac{α^{2}}{1 - α^{2}} I_{2} (f (\cdot, x)) \\ + \int_{0}^{δ} f (s, x (s)) ds \\ ⊖ \int_{δ}^{t} (- f (s, x (s))) ds, t \in (δ, T] . \end{matrix} \end{matrix}$

Since I₁ (f (· , x)) or I₂ (f (· , x)) may not exist, B_αx is not always well defined for every x ∈ C ([0, T] , R_F). We need more conditions to guarantee existence of fixed point of B_α.

Theorem 7. Suppose that α ∈ R \ {0, ± 1} and (H) is satisfied. Moreover, there exist δ ∈ (0, T) and x ∈ C ([0, T] , R_F) such that ${B_{α}^{n} x} \subset C ([0, T], R_{F})$ and $f (δ, B_{α}^{n} x (δ)) \in R$ , n = 0, 1, 2, ⋯. Then (1) has at least one (i)-mixed solution.

Proof. Set α ∈ (0, 1). By Lemma 4, Lemma 5 and (H), ∀t ∈ [0, T], we have $\begin{matrix} \begin{matrix} D (B_{α}^{n} x (t), B_{α}^{n + 1} x (t)) \\ \leq & \frac{1}{1 - α} \int_{0}^{T} k (s) \cdot D (B_{α}^{n - 1} x (s), B_{α}^{n} x (s)) ds, \end{matrix} \end{matrix}$ that is,

$\begin{matrix} d (B_{α}^{n} x, B_{α}^{n + 1} x) \\ \leq & \frac{1}{1 - α} \int_{0}^{T} k (s) ds \cdot d (B_{α}^{n - 1} x, B_{α}^{n} x) . \end{matrix}$ (35)

Similarly, setting α ∈ (-1, 0), we can prove that

$\begin{matrix} d (B_{α}^{n} x, B_{α}^{n + 1} x) \\ \leq & \frac{1}{1 - | α |} \int_{0}^{T} k (s) ds \cdot d (B_{α}^{n - 1} x, B_{α}^{n} x) . \end{matrix}$ (36)

If α ∈ (- ∞ , -1) ∪ (1, + ∞), then we have

$\begin{matrix} d (B_{α}^{n} x, B_{α}^{n + 1} x) \\ \leq & \frac{1}{1 - | \frac{1}{α} |} \int_{0}^{T} k (s) ds \cdot d (B_{α}^{n - 1} x, B_{α}^{n} x) . \end{matrix}$ (37)

Consequently, ${B_{α}^{n} x}$ is a convergent sequence for α ∈ R \ {0, ± 1}. Suppose that $B_{α}^{n} x \to x^{*}$ as n→ + ∞, then x^* is a (i)-mixed solution to (1).

Corollary 1. Suppose that (H) is satisfied and there exists δ ∈ (0, T) such that f (δ, w) ∈ R for all w ∈ R_F. If α ∈ (- ∞ , -1) ∪ (1, + ∞) and I₁ (f (· , x)) exists for all x ∈ C ([0, T] , R_F), or α ∈ (-1, 0) ∪ (0, 1) and I₂ (f (· , x)) exists for all x ∈ C ([0, T] , R_F), then (1) has at least one (i)-mixed solution.

Proof. If α ∈ (- ∞ , -1) ∪ (1, + ∞) and I₁ (f (· , x)) exists for all x ∈ C ([0, T] , R_F), then we can check that B_α is well defined on C ([0, T] , R_F). Hence, ∀x ∈ C ([0, T] , R_F), ${B_{α}^{n} x}$ satisfies all conditions in Theorem 7. The same result holds for α ∈ (-1, 0) ∪ (0, 1) and I₂ (f (· , x)) exists for all x ∈ C ([0, T] , R_F).

For the existence of (ii)-mixed solutions, referring to (21)–(24), we denote $\begin{matrix} T_{α} x (t) = {\begin{matrix} \frac{α}{1 - α} I_{2} (f (\cdot, x)) \\ ⊖ \int_{0}^{t} (- f (s, x (s))) ds, t \in [0, δ], \\ \frac{α}{1 - α} I_{2} (f (\cdot, x)) ⊖ \int_{0}^{δ} (- f (s, x (s))) ds \\ + \int_{δ}^{t} f (s, x (s)) ds, t \in (δ, T], \end{matrix} \end{matrix}$ where α ∈ (1, + ∞).

If α ∈ (- ∞ , -1), then we denote $\begin{matrix} T_{α} x (t) = {\begin{matrix} \frac{α}{1 - α^{2}} I_{2} (f (\cdot, x)) + \frac{α^{2}}{1 - α^{2}} I_{2} (f (\cdot, x)) \\ ⊖ \int_{0}^{t} (- f (s, x (s))) ds, t \in [0, δ], \\ \frac{α}{1 - α^{2}} I_{2} (f (\cdot, x)) + \frac{α^{2}}{1 - α^{2}} I_{2} (f (\cdot, x)) \\ ⊖ \int_{0}^{δ} (- f (s, x (s))) ds \\ + \int_{δ}^{t} f (s, x_{n - 1} (s)) ds, t \in (δ, T] . \end{matrix} \end{matrix}$

When α ∈ (0, 1), we denote $\begin{matrix} T_{α} x (t) = {\begin{matrix} \frac{α}{α - 1} I_{1} (f (\cdot, x)) \\ ⊖ \int_{0}^{t} (- f (s, x (s))) ds, t \in [0, δ], \\ \frac{α}{α - 1} I_{1} (f (\cdot, x)) ⊖ \int_{0}^{δ} (- f (s, x (s))) ds \\ + \int_{δ}^{t} f (s, x (s)) ds, t \in (δ, T] . \end{matrix} \end{matrix}$

When α ∈ (-1, 0), we denote $\begin{matrix} T_{α} x (t) = {\begin{matrix} \frac{α}{α^{2} - 1} I_{1} (f (\cdot, x)) + \frac{α^{2}}{α^{2} - 1} I_{1} (f (\cdot, x)) \\ ⊖ \int_{0}^{t} (- f (s, x (s))) ds, t \in [0, δ], \\ \frac{α}{α^{2} - 1} I_{1} (f (\cdot, x)) + \frac{α^{2}}{α^{2} - 1} I_{1} (f (\cdot, x)) \\ ⊖ \int_{0}^{δ} (- f (s, x (s))) ds \\ + \int_{δ}^{t} f (s, x (s)) ds, t \in (δ, T] . \end{matrix} \end{matrix}$

Theorem 8. Suppose that α ∈ R \ {0, ± 1} and (H) is satisfied. If there exist δ ∈ (0, T) and x ∈ C ([0, T] , R_F) such that ${T_{α}^{n} x} \subset C ([0, T], R_{F})$ , $f (δ, T_{α}^{n} x (δ)) \in R$ , n = 0, 1, 2, ⋯, then (1) has at least one (ii)-mixed solution.

Proof. Setting α ∈ (0, 1), we have $\begin{matrix} T_{α}^{n} x (t) = {\begin{matrix} \frac{1}{1 - α} [α \int_{δ}^{T} f (s, T_{α}^{n - 1} x (s)) ds \\ ⊖ \int_{t}^{δ} (- α f (s, T_{α}^{n - 1} x (s))) ds \\ ⊖ \int_{0}^{t} (- f (s, T_{α}^{n - 1} x (s))) ds], t \in [0, δ], \\ \frac{1}{1 - α} [\int_{t}^{T} α f (s, T_{α}^{n - 1} x (s)) ds \\ + \int_{δ}^{t} f (s, T_{α}^{n - 1} x (s)) ds \\ ⊖ \int_{0}^{δ} (- f (s, T_{α}^{n - 1} x (s)) ds], t \in (δ, T] . \end{matrix} \end{matrix}$

Underlying Lemma 4, Lemma 5 and (H), we can check that (35) holds for T_α, thus ${T_{α}^{n} x}$ is a convergent sequence. Supposing $T_{α}^{n} x \to x^{*}$ as n→ + ∞, then x^* is a (ii)-mixed solution to (1).

Similarly, when α ∈ (- ∞ , -1) ∪ (-1, 0) ∪ (1, + ∞), we also can prove that (1) has (ii)-mixed solution.

Corollary 2.Suppose that (H) is satisfied and there exists δ ∈ (0, T) such that f (δ, w) ∈ R, ∀w ∈ R_F. Moreover, one of the following conditions is satisfied:

α ∈ (0, 1) and $\begin{matrix} α \int_{δ}^{T} f (s, x (s)) ds ⊖ \int_{0}^{δ} (- f (s, x (s))) ds \end{matrix}$ exists for all x ∈ C ([0, T] , R_F).

α ∈ (-1, 0) and $\begin{matrix} α^{2} \int_{δ}^{T} f (s, x (s)) ds ⊖ \int_{0}^{δ} (- f (s, x (s))) ds \end{matrix}$ exists for all x ∈ C ([0, T] , R_F).

α ∈ (1, + ∞) and $\begin{matrix} \int_{0}^{δ} (- f (s, x (s))) ds ⊖ α \int_{δ}^{T} f (s, x (s)) ds \end{matrix}$ exists for all x ∈ C ([0, T] , R_F).

α ∈ (- ∞ , -1) and $\begin{matrix} \int_{0}^{δ} (- f (s, x (s))) ds ⊖ α^{2} \int_{δ}^{T} f (s, x (s)) ds \end{matrix}$ exists for all x ∈ C ([0, T] , R_F).

Then (1) has at least one (ii)-mixed solution.

Proof. Suppose that one of (i)–(iv) holds, then we can check that T_α is well defined on C ([0, T] , R_F). Thus ∀x ∈ C ([0, T] , R_F), ${T_{α}^{n} x}$ satisfies all conditions in Theorem 8.

5 Examples

In this section, we show some examples. Let γ ∈ R_F.

Example 2. Consider fuzzy boundary value problem

${\begin{matrix} x^{'} (t) = \frac{t}{3} \cdot x (t) + t \cdot γ, t \in (0, 1), \\ x (0) = \frac{1}{2} \cdot x (1) . \end{matrix}$ (38)

For all t ∈ [0, 1] and v, w ∈ R_F, we have $D (\frac{t}{3} \cdot v + t \cdot γ, \frac{t}{3} \cdot w + t \cdot γ) \leq \frac{t}{3} \cdot D (v, w)$ . Moreover, $\begin{matrix} \int_{0}^{1} \frac{t}{3} dt = \frac{1}{6} < 1 - \min {\frac{1}{2}, 2} = \frac{1}{2} . \end{matrix}$

Theorem 6 implies that (38) has a unique (i)-differentiable solution. By [12], $\begin{matrix} x (t) = e^{\frac{t^{2}}{6}} \cdot \frac{3 e^{\frac{1}{6}} - 3}{2 - e^{\frac{1}{6}}} \cdot γ + (3 e^{\frac{t^{2}}{6}} - 3) \cdot γ \end{matrix}$ is a (i)-differentiable solution to (38) (see Fig. 5).

Example 3. Consider

${\begin{matrix} x^{'} (t) = - \frac{t}{3} \cdot x (t) + t \cdot γ, t \in (0, 1), \\ x (0) = - 3 \cdot x (1) . \end{matrix}$ (39)

For all t ∈ [0, 1] and v, w ∈ R_F, $D (- \frac{t}{3} \cdot v + t \cdot γ, - \frac{t}{3} \cdot w + t \cdot γ) \leq \frac{t}{3} \cdot D (v, w)$ . We also have $\begin{matrix} \int_{0}^{1} \frac{t}{3} dt = \frac{1}{6} < 1 - \min {3, \frac{1}{3}} = \frac{2}{3} . \end{matrix}$

By Theorem 6 and [12], (39) has a unique (ii)-differentiable solution (see Fig. 6) $\begin{matrix} \begin{matrix} x (t) = & \frac{9 (e^{\frac{1}{3}} - e^{\frac{1}{6}})}{9 - e^{\frac{1}{3}}} \cdot e^{- \frac{t^{2}}{6}} \cdot γ \\ - \frac{27 (e^{\frac{1 - t^{2}}{6}} - 1) + 3 e^{\frac{1}{3}} (1 - e^{- \frac{t^{2}}{6}})}{9 - e^{\frac{1}{3}}} \cdot γ . \end{matrix} \end{matrix}$

Example 4. Consider

${\begin{matrix} x^{'} (t) = f (t, x (t)), t \in (0, π), \\ x (0) = \frac{1}{5} \cdot x (π), \end{matrix}$ (40) where $\begin{matrix} f (t, x (t)) = {\begin{matrix} \frac{{cos}^{2} t}{2} \cdot [x (t) + γ], t \in [0, \frac{π}{2}], \\ - \frac{{cos}^{2} t}{2} \cdot x (π - t), t \in (\frac{π}{2}, π] . \end{matrix} \end{matrix}$

Obviously, $f (\frac{π}{2}, x (\frac{π}{2})) = \tilde{0}$ for all x ∈ C ([0, π] , R_F). Moreover, ∀t ∈ [0, π] and v, w ∈ R_F, we have $D (f (t, v), f (t, w)) \leq \frac{{cos}^{2} t}{2} \cdot D (v, w)$ and $\begin{matrix} \int_{0}^{π} \frac{{cos}^{2} t}{2} dt = \frac{π}{4} < 1 - \min {\frac{1}{5}, 5} = \frac{4}{5} . \end{matrix}$

In addition, ∀x ∈ C ([0, π] , R_F), $\begin{matrix} \begin{matrix} I_{2} (f (\cdot, x)) \\ = & \int_{0}^{\frac{π}{2}} \frac{{cos}^{2} t}{2} [x (t) + γ] dt ⊖ \int_{\frac{π}{2}}^{π} \frac{{cos}^{2} t}{2} x (π - t) dt \\ = & γ \int_{0}^{\frac{π}{2}} \frac{{cos}^{2} t}{2} dt . \end{matrix} \end{matrix}$ By Corollary 1, (40) has at least one (i)-mixedsolution.

Example 5. Consider

${\begin{matrix} x^{'} (t) = f (t, x (t)), t \in (0, π), \\ x (0) = 5 \cdot x (π), \end{matrix}$ (41) where $\begin{matrix} f (t, x (t)) = {\begin{matrix} - \frac{{cos}^{2} t}{2} \cdot [x (t) + γ], t \in [0, \frac{π}{2}], \\ \frac{{cos}^{2} t}{10} \cdot x (π - t), t \in (\frac{π}{2}, π] . \end{matrix} \end{matrix}$

Similar to Example 4, f satisfies (H1) and $f (\frac{π}{2}, w) = \tilde{0}$ for all w ∈ R_F. In addition, ∀x ∈ C ([0, π] , R_F), we also have $\begin{matrix} \begin{matrix} \int_{0}^{\frac{π}{2}} (- f (t, x (t))) dt ⊖ α \int_{\frac{π}{2}}^{π} f (t, x (t)) dt \\ = \int_{0}^{\frac{π}{2}} \frac{{cos}^{2} t}{2} \cdot [x (t) + γ] ds ⊖ 5 \int_{\frac{π}{2}}^{π} \frac{{cos}^{2} t}{10} \cdot x (π - t) dt \\ = γ \int_{0}^{\frac{π}{2}} \frac{{cos}^{2} t}{2} dt . \end{matrix} \end{matrix}$

Hence, condition (iii) in Corollary 2 is satisfied and (41) has at least one (ii)-mixed solution.

Remark. There are two kinds of special case, namely, periodic and anti-periodic boundary value problems are not discussed in Section 4. Existence results about periodic boundary value problems of fuzzy differential equation can be found in [13 , 28], while the anti-periodic issue is a new problem to fuzzy differential equation. We will discuss this problem in the future.

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