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Existence results for a coupled system of nonlinear fractional 2m-point boundary value problems at resonance
Advances in Difference Equations volume 2011, Article number: 44 (2011)
Abstract
A 2m-point boundary value problem for a coupled system of nonlinear fractional differential equations is considered in this article. An existence result is obtained with the use of the coincidence degree theory.
MSC: 34B17; 34L09.
1. Introduction
In this article, we will consider a 2m-point boundary value problem (BVP) at resonance for a coupled system of nonlinear fractional differential equations given by
where 2 < α, β ≤ 3, 0 < ξ 1 < ⋯ < ξ m < 1, 0 < η 1 < ⋯ < η m < 1, 0 < γ 1 < ⋯ < γ m < 1, 0 < δ 1 < ⋯ < δ m < 1, a i , b i , c j , d j ∈ R, f, g : [0, 1] × R 3 → R, f, g satisfies Carathéodory conditions, and are the standard Riemann-Liouville fractional derivative and fractional integral, respectively.
Setting:
In this article, we will always suppose that the following conditions hold:
(C1):
(C2):
The subject of fractional calculus has gained considerable popularity and importance because of its frequent appearance in various fields such as physics, chemistry, and engineering. In consequence, the subject of fractional differential equations has attracted much attention. For details, refer to [1–4] and the references therein. Some basic theory for the initial value problems of fractional differential equations(FDE) involving Riemann-Liouville differential operator has been discussed by Lakshmikantham [5–7], El-Sayed et al. [8, 9], Diethelm and Ford [10], Bai [11], and so on. Also, there are some articles which deal with the existence and multiplicity of solutions for nonlinear FDE BVPs using techniques of topological degree theory. For example, Su [12] considered the BVP of the coupled system
By using the Schauder fixed point theorem, one existence result was given.
However, there are few articles which consider the BVP at resonance for nonlinear ordinary differential equations of fractional order. In [13], Zhang and Bai investigated the nonlinear nonlocal problem
where 1 < α ≤ 2, we consider the case βη α-1= 1, i.e., the resonance case.
In [14], Bai investigated the BVP at resonance
is considered, where 1 < α ≤ 2 is a real number, and are the standard Riemann-Liouville fractional derivative and fractional integral, respectively, and f : [0, 1] × R 2 → R is continuous, and e(t) ∈ L 1[0, 1], m ≥ 2, 0 < ξ i < 1, β i ∈ R, i = 1, 2, ..., m - 2, are given constants such that .
The coupled system (1.1)-(1.3) happens to be at resonance in the sense that the associated linear homogeneous coupled system
has (u(t), v(t)) = (at α-1+ bt α-2, ct β-1+ dt β-2), a, b, c, d ∈ R as a nontrivial solution.
The purpose of this article is to study the existence of solution for BVP (1.1)-(1.3) at resonance case, and establish an existence theorem under nonlinear growth restriction of f. Our method is based upon the coincidence degree theory of Mawhin.
Now, we will briefiy recall some notation and an abstract existence result.
Let Y, Z be real Banach spaces, L : domL ⊂ Y → Z be a Fredholm map of index zero and P : Y → Y, Q : Z → Z be continuous projectors such that
It follows that L| domL∩KerP : domL ∩ Ker P → ImL is invertible. We denote the inverse of the map by K p . If Ω is an open bounded subset of Y such that domL ∩ Ω ≠ ∅, the map N : Y → Z will be called L-compact on Ω if is bounded, and is compact.
The theorem that we used is Theorem 2.4 of [15].
Theorem 1.1. Let L be a Fredholm operator of index zero and N be L-compact on . Assume that the following conditions are satisfied:
-
(i)
Lx ≠ λNx ∀(x, λ) ∈ [domL\KerL ∩ ∂Ω] × [0, 1];
(ii)) Nx ∉ ImL, ∀x ∈ KerL ∩ ∂Ω;
-
(iii)
deg(JQN | KerL , KerL ∩ Ω, 0) ≠ 0;
where Q : Z → Z is a projection as above with KerQ = ImL, and J : ImQ → KerL is any isomorphism. Then, the equation Lx = Nx has at least one solution in .
The rest of this article is organized as follows. In Section 2, we give some notation and lemmas. In Section 3, we establish a theorem of existence of a solution for the problem (1.1)-(1.3).
2. Background materials and preliminaries
For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory. These definitions can be found in the recent literature [1–14, 16].
Definition 2.1. The fractional integral of order α > 0 of a function y : (0, ∞) → R is given by
provided the right side is pointwise defined on (0, ∞), where Γ(·) is the Gamma function.
Definition 2.2. The fractional derivative of order α > 0 of a function y : (0, ∞) → R is given by
Where n = [α] + 1, provided the right side is pointwise defined on (0, ∞).
Definition 2.3. We say that the map f : [0, 1] × R n → R satisfies Carathéodory conditions with respect to L 1[0, 1] if the following conditions are satisfied:
-
(i)
for each z ∈ R n, the mapping t → f (t, z) is Lebesgue measurable;
-
(ii)
for almost every t ∈ [0, 1], the mapping t → f (t, z) is continuous on R n;
-
(iii)
for each r > 0, there exists ρ r ∈ L 1 ([0, 1], R) such that, for a.e. t ∈ [0, 1] and every |z| ≤ r, we have f (t, z) ≤ ρ r (t).
Lemma 2.1. [13] Assume that u ∈ C(0, 1) ∩ L 1(0, 1) with a fractional derivative of order α > 0 that belongs to C(0, 1) ∩ L 1(0, 1). Then,
for some c i ∈ R, i = 1, 2, ..., N, where N is the smallest integer grater than or equal to α.
We use the classical Banach space C[0, 1] with the norm
L[0, 1] with the norm . For n ∈ N, we denote by AC n[0, 1] the space of functions u(t) which have continuous derivatives up to order n - 1 on [0, 1] such that u (n-1)(t) is absolutely continuous:
AC n [0, 1] = {u|[0, 1] → R and D n-1 u(t) is absolutely continuous in [0, 1]}.
Definition 2.4. Given μ > 0 and N = [μ] + 1 we can define a linear space
where x ∈ C[0, 1], c i ∈ R, i = 1, 2, ..., N - 1.
Remark 2.1. By means of the linear functional analysis theory, we can prove that with the
C μ [0, 1] is a Banach space.
Remark 2.2. If μ is a natural number, then C μ [0, 1] is in accordance with the classical Banach space C n [0, 1].
Lemma 2.2. [13] f ⊂ C μ [0, 1] is a sequentially compact set if and only if f is uniformly bounded and equicontinuous. Here, uniformly bounded means there exists M > 0, such that for every u ∈ f
and equicontinuous means that ∀ε > 0, ∃δ > 0, such that
and
Lemma 2.3. [14] Let α > 0, n = [α] + 1. Assume that u ∈ L 1 (0, 1) with a fractional integration of order n - α that belongs to AC n[0, 1]. Then, the equality
holds almost everywhere on [0, 1].
Definition 2.5. [14] Let , α > 0 denote the space of functions u(t), represented by fractional integral of order α of a summable function: .
In the following lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming that for α > 0.
Let Z 1 = L 1[0, 1], with the norm , Y 1 = C α-1[0, 1], Y 2 = C β-1[0, 1], defined by Remark 2.1, with the norm
where Y = Y 1 × Y 2 is a Banach space, with the norm
and Z = Z 1 × Z 1 is a Banach space, with the norm
Define L 1 to be the linear operator from domL 1 ∩ Y 1 to Z 1 with
and
Define L 2 to be the linear operator from domL 2 ∩ Y 2 to Z 1 with
and
Define L to be the linear operator from domL ∩ Y to Z with
and
we define N : Y → Z by setting
where N 1 : Y 2 → Z 1 is defined by
and N 2 : Y 1 → Z 2 is defined by
Then, the coupled system of BVPs (1.1) can be written as
3. Main results
Lemma 3.1. The mapping L : domL ⊂ Y → Z is a Fredholm operator of index zero.
Proof. Let , by Lemma 2.3, has solution
Combine with (1.2), so
Similarly, let , by Lemmas 2.3, 2.4, , combine with (1.3),
so
It is clear that
Let (x, y) ∈ ImL, then there exists (u, v) ∈ domL, such that (x, y) = L(u, v), that is u ∈ Y 1, and v ∈ Y 2, . By Lemma 2.3, we have
where
and by the boundary condition (1.2), we obtain c 3 = 0, c 1, c2 can be any constant, and x satisfies
Similarly, by the boundary condition (1.3), we obtain d 3 = 0, d 1, d 2 can be any constant, and y satisfies
On the other hand, suppose x, y ∈ Z 1 satisfy (3.1), (3.2), respectively, let , , then u ∈ domL 1, and v ∈ domL 2, . That is to say, (x, y) ∈ ImL. From the above argument, we obtain
Consider the continuous linear mapping A i , B i , T i , R i , Q i : Z 1 → Z 1, i = 1, 2 and Q : Z → Z defined by
and
and
Since the conditions (C1) and (C2) hold, the mapping defined by
is well-defined. It is clear that dimImQ 1 = dimImQ 2 = 2.
Recall (C1) and (C2) and note that
and similarly we can derive that
Hence, for x ∈ Z 1, it follows from the four relations above that
that is, the map Q 1 is idempotent. In fact, Q 1 is a continuous linear projector.
Similarly, the map Q 2 is a continuous linear projector.
Therefore,
It is clear that Q is a continuous linear projector.
Note (x, y) ∈ ImL implies Q(x, y) = (Q 1 x, Q 2 y) = (0, 0). Conversely, if Q(x, y) = (0, 0),
so
but
then we must have A i x = B i y = 0, i = 1, 2, that is, (x, y) ∈ ImL. In fact, KerQ = ImL.
Take (x, y) ∈ Z in the form (x, y) = ((x, y) - Q(x, y)) + Q(x, y) so that ((x, y) - Q(x, y)) ∈ KerQ = ImL, Q(x, y) ∈ ImQ. Thus, Z = ImL + ImQ. Let (x, y) ∈ ImL ∩ ImQ and assume that (x, y) = (at α-1+ bt α-2, ct β-1+ dt β-2) is not identically zero on [0, 1]. Then, since (x, y) ∈ ImL, from (3.1) and (3.2) and the condition (C2), we have
So,
but
we derive a = b = c = d = 0, which is a contradiction. Hence, ImL ∩ ImQ = {0, 0}; thus, Z = ImL ⊕ ImQ.
Now, IndL = dimKerL - codimImL = 0, and so L is a Fredholm operator of index zero.
Let P 1 : Y 1 → Y 1, P 2 : Y 2 → Y 2, P : Y → Y be defined by
and
Note that P 1, P 2, P are continuous linear projectors and
It is clear that Y = KerL ⊕ KerP.
Note that the projectors P and Q are exact. Define by K p : ImL → domL ∩ KerP by
Hence, we have
Then,
and thus
and
Hence, we have
Then,
and thus
so
For (x, y) ∈ ImL, we have
Also, if (u, v) ∈ domL ∩ KerP, we have u ∈ domL 1, , v ∈ domL 2, , so the coefficients c i , d i , i = 1, 2, 3 in the expressions then
where
where
and from the boundary value conditions (1.2), (1.3) and the fact that (u, v) ∈ domL ∩ KerP, P(u, v) = 0, we have c i = d i = 0, thus
This shows that K p = [L| domL∩KerP ]-1.
Using (3.3)-(3.5), we write
By Lemma 2.2 and a standard method, we obtain the following lemma.
Lemma 3.2. [16] K p (I - Q)N : Y → Y is completely continuous.
In this section, we shall prove existence results for (1.1)-(1.3).
First, let us set the following notations for convenience:
Assume that the following conditions on the function f(t, x, y, z), g(t, x, y, z) are satisfied:
(H1) There exist functions a i (t), b i (t), c i (t), d i (t), r i (t) ∈ L 1[0, 1], i = 1, 2 and a constant θ i ∈ [0, 1), i = 1, 2 such that for all (x, y, z) ∈ R 3, t ∈ [0, 1], one of the following inequalities is satisfied:
(H2) There exists a constant A > 0, such that for (u, v) ∈ domL\KerL satisfying
or for all t ∈ [0, 1], we have
(H3) There exists a constant B > 0 such that for every a, b, c, d ∈ R satisfying min{a 2 + b 2, c 2 + d 2} > B then either
or
Theorem 3.1 If (C1)-(C2) and (H1)-(H3) hold, then the BVP (1.1)-1.3) has at least one solution provided that
Proof. Set
Then, for (u, v) ∈ Ω1, L(u, v) = λN (u, v), thus λ ≠ 0, N(u, v) ∈ ImL = KerQ, and hence QN(u, v) = (0, 0) for all t ∈ [0, 1]. By the definition of Q, we have Q 1 N 1 v(t) = Q 2 N 2 u(t) = 0. It follows from (H2) that there exists t 0, t 1 ∈ [0, 1], such that
Now
and so
and
Similarly,
Therefore, we have noted that (I - P)(u, v) ∈ domL ∩ KerP for ∀(u, v) ∈ Ω1.
Then,
so, we have
If the first condition of (H1) is satisfied, then from (3.12), the proof can be divided into four cases:
Case 1. ||(u, v)|| Y ≤ q||N 1 v||1 + mA.
From (3.6), we have
where , and consequently, for
so
But θ 1 ∈ [0, 1) and , so there exists A 1, A 2, A 3 > 0 such
that
Therefore, for all (u, v) ∈ Ω1,
we can prove that Ω1 is also bounded.
Case 2. ||(u, v)|| Y ≤ w||N 2 u||1 + nA.
The proof is similar to that of case 1. Here, we omit it, where
Case 3. ||(u, v)|| Y ≤ n(A + ||N 2 u||1) + j||N 1 v||1.
From (3.6) and (3.7), we have