TWO POINT FRACTIONAL BOUNDARY VALUE PROBLEM AT RESONANCE

Abstract

In this paper, a two-point fractional boundary value problem at resonance is considered. By using the coincidence degree theory some existence results of solutions are established.

1. Introduction

In this paper, we investigate the existence of solutions for the following twopoint fractional boundary value problem (BVP) at resonance

where f : [0, 1] × IR3 → IR is continuous, 2 < α < 3 and denotes the Caputo’s fractional derivative. The two-point boundary value problem (1.1) happens to be at resonance in the sense that its associated linear homogeneous boundary value problem

has a nontrivial solution u(t) = ct2, c ∈ IR. To solve this problem, we use the coincidence degree of Mawhin [12,13]. This method is based on an equivalent formulation in an abstract space and a theory of topological degree. This formulation generally leads to an abstract operator of the form N + L, where L is a Fredholm operator of index zero and N is generally a nonlinear operator having some compactness properties with respect to L. Fractional differential equations arise in different areas of sciences such as in rheology, fluid flows, viscoelasticity, chemical physics, and so on [1,6,8,11,14-19]. Recently, boundary value problems for fractional differential equations at nonresonance have been studied by many authors [1,3,4,10] by using fixed point theorems, lower and upper solution. Moreover, boundary value problems for differential equations at resonance have also been studied in some papers, see [2,7]. In [20], the authors studied, by using the coincidence degree theory, the following BVP of fractional equation at resonance

where denotes the Caputo’s fractional differential operator of order α, 1 < α ≤ 2 and f : [0, 1] × IR2 → IR is a continuous function.

A similar boundary value problem at resonance involving Riemann-Liouville fractional derivative is considered in [7]. The author solved the problem

by applying degree theory theorem for coincidences.

By the same method, in [5] the authors established the existence of solutions for the following third-order differential equation

where f : [0, 1] × IR2 → IR is a Caratheodory function, and η ∈ (0, 1).

The organization of this paper is as follows. We present in Section 2 some notation and some basic results involved in the reformulation of the problem. In section 3, we give the main theorem and some lemmas, then we will see that the proof of the main theorem is an immediate consequence of these lemmas and the coincidence degree of Mawhin. At the end of this section, we give an example to illustrate the main result.

 

2. Preliminaries

We begin by introducing the fundamental tools of fractional calculus and the coincidence degree theory which will be used throughout this paper.

Definition 2.1. The Riemann-Liouville fractional integral is defined by

for g ∈ C([a, b]) and α > 0.

Definition 2.2. Let f ∈ Cn([a, b]), the Caputo fractional derivative of order α ≥ 0 of f is defined by , where n = [α] + 1 ([α] is the integer part of α).

Lemma 2.3. For α > 0, g ∈ C([0, 1], IR), the homogenous fractional differential equation has a solution g(t) = c0+c1t+c2t2+…+cn−1tn−1, where ci ∈ IR, i = 0,…, n−1, here n is the smallest integer greater than or equal to α.

Definition 2.4. Let X and Y be real normed spaces. A linear mapping L : domL ⊂ X → Y is called a Fredholm mapping if the following two conditions hold:

(i) kerL has a finite dimension, and

(ii) ImL is closed and has a finite codimension.

If L is a Fredholm mapping, its index is the integer IndL = dimkerL − codimImL.

From here if L Fredholm mapping of index zero then there exist continuous projectors P : X → X and Q : Y → Y such that ImP = KerL,KerQ = ImL,X = KerL ⊕ KerP,Y = ImL ⊕ ImQ and that the mapping L|domL∩KerP : domL ∩ KerP → ImL is one-to-one and onto. We denote its inverse by KP. Moreover, since dimImQ = codimImL, there exists an isomorphism J : ImQ → KerL.

If Ω is an open bounded subset of X, and , the map N : X → Y will be called L−compact on if QN() is bounded and KP (I − Q)N : → X is compact.

Lemma 2.5 ([13]). Let L : domL ⊂ X → Y be a Fredholm operator of index zero and N : X → Y, L − compact on . Assume that the following conditions are satisfied

(1) Lu ≠ Nu for every (u, λ) ∈ [(domL KerL)] ∩ ∂Ω × (0, 1);

(2) Nx ∉ ImL for every x ∈ KerL ∩ ∂Ω;

(3) deg(JQN|KerL,KerL ∩ Ω, 0) ≠ 0, where Q : Y → Y is a projection such that ImL = KerQ.

Then the equation Lx = Nx has at least one solution in domL ∩ .

In this paper, we denote by X and Y the Banach spaces: X = C2([0, 1], IR) equipped with the norm ║u║X = max{║u║∞; |u′║∞, ║u′′║∞} and Y = C([0, 1], IR) equipped with the norm ║y║Y = ║y║∞, where ║u║∞ = maxt∈[0,1] |u(t)|.

Define the operator L : domL ⊂ X → Y by

where

Let N : X → Y be the operator

Then BVP (1.1) is equivalent to the operator equation

It is known [12] that the coincidence equation Lu = Nu is equivalent to

 

3. Main results

We can now state our result on the existence of a solution for the BVP (1.1).

Theorem 3.1. Let f : [0, 1] × IR3 → IR be continuous. Assume that

(H1) there exists nonnegative functions p, q, r, z ∈ C[0, 1] with Γ(α−1)−2q1−2r1 − 2z1 > 0 such that

|f(t, u, v, w)| ≤ p(t) + q(t)|u| + r(t)|v| + z(t)|w|, ∀t ∈ [0, 1], (u, v, w) ∈ IR3

where

(H2) there exists a constant B > 0 such that for all c ∈ IR with |2c| > B either

or

Then BVP(1.1) has at least one solution in X.

In order to prove Theorem 3.1, we need to prove some lemmas below.

Lemma 3.2. Let L be defined by (2.1), then

Proof. By Lemma 2.3, has solution

Combining with boundary condition of BVP (1.1), (3.1) holds. For y ∈ ImL, there exists u ∈ domL such that y = Lu ∈ Y . We have

Differentiating two times and using the boundary conditions for BVP (1.1), we get

On the other hand, let y ∈ Y satisfying , so y ∈ ImL. □

Lemma 3.3. If L be defined by (2.1), then L is a Fredholm operator of index zero and the linear continuous projector operators P : X → X and Q: Y → Y can be defined as

Furthermore, the operator Kp : ImL → domL ∩ KerP can be written as

Proof. Obviously, ImP = KerL and P2u = Pu. It follows from u = (u − Pu) + Pu that X = KerP +KerL. By simple calculation, we get that KerL∩KerP = 0. Hence

For y ∈ Y, it yields

Let y = (y − Qy)+Qy, where y − Qy ∈ KerQ = ImL, Qy ∈ ImQ. It follows from KerQ = ImL and Q2y = Qy that ImQ ∩ ImL = 0. Then, we have

Thus

This means that L is a Fredholm operator of index zero. From the definitions of P and KP , it is easy to see that the generalized inverse of L is KP . In fact, for y ∈ ImL, we have

Moreover, for u ∈ domL ∩ KerP, we get u(0) = u′(0) = u′′(0) = 0. By Lemma 2.3, we obtain that

This together with u(0) = u′(0) = u′′(0) = 0 yields

Combining (3.3) with (3.4), we get KP = (L|domL∩KerP)−1. □

Lemma 3.4. If Ω ⊂ X is an open bounded subset such that , then N is L − compact on .

Proof. By the continuity of f, we conclude that QN() and KP (I − Q)N() are bounded. In view of Arzelà-Ascoli theorem, we need only to prove that KP (I − Q)N() ⊂ X is equicontinuous. The continuity of f implies that exists a constant A > 0 such that |(I − Q)Nu| ≤ A, ∀u ∈ , ∀t ∈ [0, 1]. Denote KP,Q = KP (I − Q)N and let 0 ≤ t1 ≤ t2 ≤ 1, u ∈ (), then

and

Since tα, tα−1 and tα−2 are uniformly continuous on [0, 1], we get that

KP,Q(), (KP,Q)′ ()(KP,Q)′′ () ∈ C[0, 1] are equicontinuous. Hence KP,Q : → X is compact. □

Lemma 3.5. Suppose (H1) holds, then the set

is bounded.

Proof. Take u ∈ Ω1, then Nu ∈ ImL. By (3.2), it yields

By the integral mean value theorem, we conclude that there exists a constant h ∈ (0, 1) such that f(h, u(h), u′(h), u′′(h)) = 0. Then from (H2), we have |u′′(h)| ≤ B.

Since u ∈ domL, then u(0) = u′(0) = 0. Therefore

and

That is

By Lu = λNu and u ∈ domL, we have

and

Take t = h, we get

Together with |u′′(h)| ≤ B, (H1) and (3.5), we have

Then we have

Since Γ(α − 1) − 2q1 − 2r1 − 2z1 > 0, we obtain

and

Therefore, ║u║X ≤ M, consequently Ω1 is bounded. □

Lemma 3.6. Suppose (H2) holds, then the set

is bounded.

Proof. For u ∈ Ω2, we have u(t) = ct2,c ∈ IR, and Nu ∈ ImL. Then we get

this together with (H2) implies |2c| ≤ B. Thus, we have ║u║X ≤ B. Hence, Ω2 is bounded. □

Lemma 3.7. Suppose the first part of (H2) holds, then the set

is bounded. Where J : KerL → ImQis an isomorphism defined by J(ct2) = ct2; ∀c ∈ IR; t ∈ [0, 1]:

Proof. For u ∈ Ω3, we have u(t) = ct2,c ∈ IR, and

If λ = 0, then f(s, cs2, 2cs, 2c)ds = 0, thus |2c| ≤ B in view of the first part of (H2). If λ ∈ (0, 1], we can also obtain |2c| ≤ B. Otherwise, if |2c| > B, in view of the first part of (H2), one has

which contradicts (3.6). Therefore, Ω3 is bounded. □

Lemma 3.8. Suppose the second part of (H2) hold, the set

is bounded.

Proof. Using similar argument as in the proof of Lemma 3.7, we prove that Ω′3 is bounded. □

Now we are able to give the proof of Theorem 3.1, which is an immediate consequence of Lemmas 3.2-3.8 and Lemma 2.5.

Proof. Set Ω = {u ∈ X| ║u║X < max{M, B} + 1}. It follows from Lemma 3.2 and Lemma 3.3 that L is a Fredholm operator of index zero and N is L−compact on . By Lemma 3.4 and Lemma 3.5, we get that the following two conditions are satisfied

(1) Lu ≠ Nu for every (u, λ) ∈ [(domL KerL)] ∩ ∂Ω × (0, 1);

(2) Nu ∉ ImL for every u ∈ KerL ∩ ∂Ω;

Define H(u, λ) = λIdu + (1 − λ)JQNu: According to Lemma 3.6 (or Lemma 3.7), we see that H(u, λ) ≠ 0 for u ∈ KerL ∩ ∂Ω. By the degree property of invariance under a homotopy, it yields

So that, the condition (3) of Lemma 2.1 is satisfied. By Lemma 2.1, we can get that Lu = Nu has at least one solution in domL ∩ . Therefore by Lemma 2.5, the BVP (1.1) has at least one solution. The proof is complete. □

Example 3.9. Let us consider the following fractional boundary value problem

We have f(t, u, v, w) = (, then

where p(t) = 3t, q(t) = 0, . We get that q1 = 0, and

If we choose B = 15 then for |2c| > B it yields

Then, the conditions of Theorem 3.1 are satisfied, so BVP(3.7) has at least one solution.