On the boundary value problems of piecewise differentialequations with left-right fractional derivatives and delay

In recent decades, fractional calculus has been widely used in various fields of science and technology, and the theoretical research of fractional differential equations has also received extensive attention, see [5–8, 10, 12–15, 17, 18, 22, 23, 25, 27, 30–34, 36] and the references therein. And the differential equation with left and right fractional derivatives have been studied extensively due to the wide application [2–4, 16, 24, 29]. In [2], the following nonlocal boundary value problems of integro-differential equations involving mixed left and right fractional derivatives and left and right fractional integrals are studied

where and are given continuous functions, and are constants. At the same time, the differential equations with delay have many successful applications in the fields of communication engineering, population control and so on; see [1, 9, 20, 21, 26, 28, 35].

Motivated by above works, we discuss the multi-point boundary value problems for piecewise differential equations with left and right fractional derivatives and dela y

where is the right Caputo fractional derivative, cis the left Caputo fractional derivative,

In boundary value problem (1), the state variable u = u(t) satisfies the different equa- tions in different time intervals, and they interact with each other through positive delay τ1 and negative delay τ₂. The parameters a and b in the boundary conditions represent the error in certain measurement. Some new results on the existence, no-existence and multiplicity for the positive solutions of the boundary value problems are obtained by us- ing Guo–Krasnoselskii’s fixed point theorem and Leggett–Williams fixed point theorem. The results for existence highlight the influence of perturbation parameters. Finally, an example is given out to illustrate our main results.

For convenience of reading, in this section, we give out some definitions about the frac- tional calculus and some lemmas.

Definition 1. (See [17].) Let α > 0, a < b ∈ R, and the left and right Riemann–Liouville fractional integral of . : [a, b] → R are defined as

respectively, for

Definition 2. (See [17]). Let , and the left Caputo fractional derivative and right Caputo fractional derivative of function are defined as

respectively, provided the right-sided integral converges, where

Lemma 1. (See [17]). then

where

Lemma 2 2.(See [11].) Let E be a Banach space and P ⊂ E is a cone. Assume that Ω₁, Ω₂ are bounded open subsets of E with θand let P be a completely continuous operator such that either.

(i)

(ii)

Then the operator T has at least one fixed point on

Lemma 3. (See [19].) Assume P is a cone in Banach space, ω is a nonnegative continu- ous concave functional on P, the constants 0 < d < q < c ≤ r. Denote be a completely continuous operator such that such that

(i)

(ii)

(iii)

Then T has at least three fixed points such that and

Lemma 4. Let then the boundary value problem

has a unique solution given by

where

Proof. From Lemma 1 the general solution of the linear differential equation .h (t)= 0 is given by

and

The general solution of the linear differential equation is given

and

By the boundary value conditions we can easily get that

By (6)–(8) and the boundary conditions we ca also get that

Thus, by substituting c0 and c1 into (6) we can get that for

and by substituting c₂ and c₃ into (7) we can get that for

Hence, u(t) satisfies equation (3) if it is the solution of the boundary value problem (2) and vice versa.

Lemma 5. Suppose are defined by (4), (5), then G_i(t, s) has the following properties, respectively

(i) G1(t, s) is continuous and 0 on (t,s)

(ii) G2(t, s) is continuous and 0 on (t,s)

Proof. (i) Obviously, G1(t, s) is continuous on For

and for 0 ≤ t ≤ s ≤ ξ

Hence, G₁(t, s) is monotone increasing for any , and

Because G₁ (0,s)then

(ii) Similar to the proof of (i), we can prove that (ii) holds.

Letis continuous in J₀Obviously, E is a Banach space with the norm

Denotemax .Set

Obviously, u = u(t) is a positive solution of (1) if and only if u is a fixed point of the operator T in P

Lemma 6. The operator T : P → P is completely continuous.

Proof. By Lemma 5 we can easily obtain that T : P → P.

Let There exists a constant such that

By the continuity of f(t, u, v), g(t, u, v) we have

and there is a constant , which makes sup and sup where A

Thus,, which implies that the operator . is a continuous operator.

Let be bounded. By the continuity of f , g, we can get that there is a constant such that |f (t, u, v) for any, and |g(t, u, v) for all

By Lemma 5 we can show that (Tu)′(t) ≥ 0 for t ∈ [0, ξ] and (Tu)′(t) ≤ 0 for . Hence ,

Consequently, T (Ω) is uniformly bounded

Since G1(t, s) is continuous, it is uniformly continuous on (t, s)

Hence, for any , there exists a constant

such that| for all

Thus, for we have

Similarly, due to that G₂(t, s) is continuous on[ξ, 1] x [ξ, 1], for above mentioned ε > 0, there exists a constant δ₂ > 0 such that for t₃, t₄, s (ξ, 1], [t₃ t ₄]< δ_2, we have Tu(t₃) Tu(t₄) < ε.

Hence, T (Ω) is equicontinuous on [0, ξ], (ξ, 1], respectively.

By Arzela–Ascoli theorem we know that operator T is a relative compactness op- erator, and because operator T is a continuous operator, it is a completely continuous operator.

Denote

where

Theorem 1. Assume that M₂ > 0 and the following conditions hold:

Then there exist constants a₀, b₀ ≥ 0 such that boundary value problem (1) has at least one positive solution for the parameters a and b with 0 ≤ a ≤ a0, 0 ≤ b ≤ b0

Proof. Because f⁰ < M₁, there exists a constant r₁ > 0 such that f (t, u, v) < M₁(u+v) for anyu + v (0, r₁). Similarly, by g⁰ < M₂there is a constant r₂ > 0 such that g(t, u, v) < M₂(u + v) for any(0, r₂).

Let

and. For any, we have ǁuǁ = r. When, for any, we have

Similarly,

In view of

we have ǁTuǁ_(ξ,1] < r = ǁuǁ.

Then for any, we get ǁTuǁ ≤ ǁuǁ.

If f_∞ > N₁, there exists a constant R₁ > 0 such that f (t, u, v) > N₁(u + v) for any

Let

For, we have ǁuǁ = R, and

Because ǁuǁ = max{ǁuǁ_[0,ξ], ǁuǁ_(ξ,1]} for t ∈ (ξ − τ0, ξ] ⊂ [0, ξ], then t + τ₁ ∈ and

So that

By Lemma 5 and (9) we can easily get that

Then for, we have

According to Lemma 2, T has at least one fixed point in

Similarly, if g_∞ >N₂, there is a constant R₂ > 0, which makes g(t, u, v) >N₂(u + v) for any

Let

We have ǁuǁ = R₀ for any and for, then

Thus

and because

then for any, we have

According to Lemma 2, T has at least one fixed point in which implies tha boundary value problem (1) has at least one positive solution.

In particular, the following result holds by Theorem 1.

Corollary 1. Assume that the following conditions hold: (H1′) f 0 = g0 =

Then there exist constants a₀, b₀ ≥ 0 such that boundary value problem (1) has at least one positive solution for

Theorem 2. Assume that the following conditions (H3) and (H4) hold: (H3) f

Then there exist constants a0, b0 > 0 such that boundary value problem (1) has at least one positive solution for

Proof. Due to f^∞ < M₃, then there exists a constant λ₁ > 0 such that f (t, u, v) < M₃(u + v) for any

Let

where

Since f is bounded on D₁, then there is L₁ > 0, which makes |f (t, u, v)| ≤ L₁ for any (t, u, v) ∈ D₁. Hence, for all (t, u, v) ∈ D, we have

Similarly, because g^∞ < M₄, then there is a constant λ₂ > 0 such that g(t, u, v) < M₄(u + v) for any t ∈ (ξ, 1], u + v ∈ [λ₂, +∞).

Let, then there is L₂ > 0 such tat |g(t, u, v)| ≤ L₂ for any (t, u, v) ∈ D. Then for any t ∈ [ξ, 1] and u, v ∈ [0, +∞], we have |g(t, u, v)| < L₂ + M4(u + v). Denote

For anyu∈, which implies Since

Then for

and

Then for any

If f₀> N₁, then there exists a constantsuch that f(t,u,v)> N₁(u+v)for any

Let

Hence, for any, we have ‖u‖=μ. It is similar to (10), for t ∈(ξ−τ₀, ξ]

Then for any

Thus, for any, there is According to Lemma 2,T has at least one fixed ponit in

Similarly, if g₀ > N₂, there is a constant λ₂ > µ₂ > 0 that makes g(t, u, v) > N2 (u+v) for any

Let

Then for any, we have ǁuǁ = µ. Similar to (11), for t ∈ (ξ, ξ + τ₀], we have

and for any,

Thus, for any, we have

According to Lemma 2, T has at least one fixed point in, which implies that boundary value problem (1) has at leas une positive solution.

In particular, the following result holds by Theorem 2.

Corollary 2. Assume that the following conditions hold:

Then there exist constants a₀, b₀ ≥ 0 such that boundary value problem (1) has at least one positive solution for

Theorem 3. Assume f_∞ > N₁ holds. Then there exists a large enough positive constant a₁>0 such that boundary value problem (1) has no positive solution for a>a₁

Proof. If f_∞ > N₁, there exists a constant R > 0 such that for any t [ξ τ₀, ξ], u + v [γR, + ), we have f (t, u, v) > N₁(u + v). Assume that for any large enough a > 0, boundary value problem (1) has a positive solution u = u(t)

Let

In fact , since

Hace

From (10) of Theorem 1 we have for any

Hence

So, which is a contradiction. Thus, there exists a constant a₁ > 0 such that the boundary value problem (1) has no positive ssolution for a>a₂

Theorem 4.Assume g_∞ > N₂ holds. Then there exist large enough positive constants a₂, b₁ > 0 such that boundary value problem (1) has no positive solution for a > a₂, b > b₁.

Proof. Similarly, if g_∞ > N₂, there exists a constant R₀ > 0 such that for any , we have g(t, u, v) > N₂(u + v). Assume that for any large enough a > 0, b > 0, the boundary value problem (1) has a positive solution u = u(t).

Let

Since Tu = u, we have

Hence, ǁuǁ > R₀. Since for any t ∈ (ξ, ξ + τ₀],

Then

So₀, which is a contradiction. Thus, there exist constants a₂, b₁ > 0 such that the boundary value problem (1) has no positive solution for a > a₂, b > b₁

In this section, we consider the multiplicity of solutions for boundary value problem (1) by using Lemma 3.

Let. Define a nonnegative continuous concave functional Obviously,for any

Theorem 5.Suppose there are three constants d, q, c with 0 < d < q < c, where min{M₁c, M₂c} ≥ Nq > 0, and the following hypotheses hold:

Then there exist constants a₀, b₀ ≥ 0 such that boundary value problem (1) has at least three positive solutions u₁, u₂, u₃ on P for , where ǁu₁ǁ < d, ω(u₂) > q, ǁu₃ǁ > d, ω(u₃) < q.

Proof. First of all, for any we have Let

Define a operator

Thenu=u(t)is a solution of (1) if and only ifuis a fixed point of the operatorT1onPc. By (H5)

Since min {M₁c,M₂c}>Nq >0, then M₂>0

and

He, we have , which implies

Similarly, by (H5), we can get that‖T1u‖< dforu∈Pd. Therefore, the condition(ii) in Lemma 3 is satisfied.

Select u(t) . Obviously, u(t) = (q + c)/2 ∈ P(ω, q, c) and.For any u ∈ P(ω, q, c), we can get that u(t) > q for t ∈ [ξ − τ₀, ξ + τ₀] and 0 6 u(t) 6 c for t ∈ [0, 1]. By Lemma 5, we can easily get that T1u is monotone increasing on [0, ξ] and T₁u is monotone decreasing on (ξ, 1]. From (H6), we get f(t, u, v) > Nq, t ∈ [ξ − τ0, ξ] and g(t, u, v) > Nq, t ∈ (ξ, ξ τ₀]

Hence

and

Therefore, for u ∈ P(ω, q, c), we have ω(T₁u(t)) > q. Hence, condition (i) in Lemma 3 holds.

Due to Lemma 3 involves paramaters d, q, c, r with 0<r. Let c = r, then by condition (i) in Lemma 3 it is clearly that for u ∈ P(ω, q, c) and, we have ω(T u) > q

Therefore, condition (iii) in Lemma 3 also satisfied. Then Lemma 3 implies that the boundary value problem (1) has at least three solutions u₁, u₂, u₃ on Pc and

In order to illustrate the applicability of our main results, the following boundary value problem is considered in this section.

Example 1. For the following boundary value problem

we can establish the following results:

(i) If [0, 0.019191], [0, 0.028873], then boundary value problem (12) has at least one positive solution.

(ii) If (2.46995 10⁶, + ), (2.22295 107, + ), then boundary value problem (12) has no positive solutions.

Proof. Boundary value problem (12) can be regarded as boundary value problem (1), where α = 3/2, β = 7/4, ξ = π/8, ρ₁ = 2, ρ2 = 0, γ₁ = γ₂ = 1/2, τ₁ = 3/5, τ₂ = 1/3, f (t, u, v) = (u + v)(1/100) sin t + (u + v) cos t and g(t, u, v) = (u + v)((1/4) cos t + (u + v)² sin t).

Let τ₀ = 1/4 < min{τ₁, τ₂}, we can easily obtain that M1 ≈ 0.811256, M2 ≈ 0.218239 > 0, N₁ ≈ 13.8342 and N2 ≈ 12.7312, f 0 ≈ 0.00382783 < M1, g 0 ≈ 0.23097 < M₂ and g_∞ = +∞ > N2

Then there exist constants r₁ = 0.151, r₂ = 0.234, r = min{r₁/2, r₂/2} = 0.0755, R0 = 7.5 × 106 , It is easy to get that a0 = (ξ α−1M1r)/2 = 0.019191, b0 = (ξ ^α−1M_1r)/Γ(α + 1) = 0.028873, a₂ = 2.46995 × 106 and b₁ = 2.22295 × 107 .

(i) According to Theorem 1, if a ∈ [0, a0] and b ∈ [0, b0], then boundary value problem (12) has at least one positive solution.

(ii) According to Theorem 4, if a ∈ (a₂, +∞) and b ∈ (b₁, +∞), then boundary value problem (12) has no positive solutions.

The authors would like to acknowledge the helpful comments of the referees on an earlier version of this paper.