Abstract
In this paper, by using the coincidence degree theory, we consider the following boundary value problem for fractional differential equation
where denotes the Caputo fractional differential operator of order α, 2 < α ≤ 3. A new result on the existence of solutions for above fractional boundary value problem is obtained.
Mathematics Subject Classification (2000): 34A08, 34B15.
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1 Introduction
Fractional calculus is a generalization of ordinary differentiation and integration on an arbitrary order that can be noninteger. This subject, as old as the problem of ordinary differential calculus, can go back to the times when Leibniz and Newton invented differential calculus. As is known to all, the problem for fractional derivative was originally raised by Leibniz in a letter, dated September 30, 1695.
In recent years, the fractional differential equations have received more and more attention. The fractional derivative has been occurring in many physical applications such as a non-Markovian diffusion process with memory [1], charge transport in amorphous semiconductors [2], propagations of mechanical waves in viscoelastic media [3], etc. Phenomena in electromagnetics, acoustics, viscoelasticity, electrochemistry, and material science are also described by differential equations of fractional order (see [4–9]).
Recently, boundary value problems (BVPs for short) for fractional differential equations at nonresonance have been studied in many papers (see [10–16]). Moreover, Kosmatov studied the BVPs for fractional differential equations at resonance (see [17]). Motivated by the work above, in this paper, we consider the following BVP of fractional equation at resonance
where denotes the Caputo fractional differential operator of order α, 2 < α ≤ 3. f : [0, 1] × ℝ3 → ×ℝ is continuous.
The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions, and lemmas. In Section 3, we establish a theorem on existence of solutions for BVP (1.1) under nonlinear growth restriction of f, basing on the coincidence degree theory due to Mawhin (see [18]). Finally, in Section 4, an example is given to illustrate the main result.
2 Preliminaries
In this section, we will introduce notations, definitions, and preliminary facts that are used throughout this paper.
Let X and Y be real Banach spaces and let L : domL ⊂ X → Y be a Fredholm operator with index zero, and P : X → X, Q : Y → Y be projectors such that
It follows that
is invertible. We denote the inverse by K P .
If Ω is an open bounded subset of X, and , the map N : X → Y will be called L-compact on if is bounded and is compact. Where I is identity operator.
Lemma 2.1. ([18]) If Ω is an open bounded set, 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)
Lx ≠ λNx for every (x, λ) ∈ [(domL\KerL)] ∩ ∂Ω × (0, 1);
-
(2)
Nx ∉ ImL for every x ∈ KerL ∩ ∂Ω;
-
(3)
deg(QN|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 .
Definition 2.1. The Riemann-Liouville fractional integral operator of order α > 0 of a function x is given by
provided that the right side integral is pointwise defined on (0, +∞).
Definition 2.2. The Caputo fractional derivative of order α > 0 of a continuous function x is given by
where n is the smallest integer greater than or equal to α, provided that the right side integral is pointwise defined on (0, +∞).
Lemma 2.2. ([19]) For α > 0, the general solution of the Caputo fractional differential equation
is given by
where c i ∈ ℝ, i = 0, 1, 2, . . ., n - 1; here, n is the smallest integer greater than or equal to α.
Lemma 2.3. ([19]) Assume that x ∈ C(0, 1) ∩ L(0, 1) with a Caputo fractional derivative of order α > 0 that belongs to C(0, 1) ∩ L(0, 1). Then,
where c i ∈ ℝ, i = 0, 1, 2, . . ., n - 1; here, n is the smallest integer greater than or equal to α.
In this paper, we denote X = C2[0, 1] with the norm ||x|| X = max{||x||∞, ||x'||∞, ||x"||∞} and Y = C[0, 1] with the norm ||y|| Y = ||y||∞, where ||x||∞ = max t ∈[0, 1] |x(t)|. Obviously, both X and Y are Banach spaces.
Define the operator L : domL ⊂ X → Y by
where
Let N : X → Y be the Nemytski operator
Then, BVP (1.1) is equivalent to the operator equation
3 Main result
In this section, a theorem on existence of solutions for BVP (1.1) will be given.
Theorem 3.1. Let f : [0, 1] × ℝ3 → ℝ be continuous. Assume that
(H1) there exist nonnegative functions p, q, r, s ∈ C[0, 1] with Γ(α - 1) - q1 - r1 - s1 > 0 such that
where p1 = ||p||∞, q1 = ||q||∞, r1 = ||r||∞, s1 = ||s||∞.
(H2) there exists a constant B > 0 such that for all u ∈ ℝ with |u| > B either
or
Then, BVP (1.1) has at leat one solution in X.
Now, we begin with some lemmas below.
Lemma 3.1. Let L be defined by (2.1), then
Proof. By Lemma 2.2, has solution
Combining with the boundary value condition of BVP (1.1), one has (3.1) hold.
For y ∈ ImL, there exists x ∈ domL such that y = Lx ∈ Y. By Lemma 2.3, we have
Then, we have
and
By conditions of BVP (1.1), we can get that y satisfies
Thus, we get (3.2). On the other hand, suppose y ∈ Y and satisfies . Let , then x ∈ domL and . So that, y ∈ ImL. The proof is complete.
Lemma 3.2. Let 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 K P : ImL → domL ∩ KerP can be written by
Proof. Obviously, ImP = KerL and P2x = Px. It follows from x = (x - Px) + Px that X = KerP + KerL. By simple calculation, we can get that KerL ∩ KerP = {0}. Then, we get
For y ∈ Y, we have
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, K P , it is easy to see that the generalized inverse of L is K P . In fact, for y ∈ ImL, we have
Moreover, for x ∈ domL ∩ KerP, we get x(0) = x'(0) = x"(0) = 0. By Lemma 2.3, we obtain that
which together with x(0) = x'(0) = x"(0) = 0 yields that
Combining (3.3) with (3.4), we know that K P = (L|domL∩KerP)-1. The proof is complete.
Lemma 3.3. Assume Ω ⊂ X is an open bounded subset such that , then N is L-compact on .
Proof. By the continuity of f, we can get that and are bounded. So, in view of the Arzelà -Ascoli theorem, we need only prove that is equicontinuous.
From the continuity of f, there exists constant A > 0 such that |(I - Q)Nx| ≤ A, , t ∈ [0, 1]. Furthermore, denote K P,Q = K P (I - Q)N and for 0 ≤ t1 < t2 ≤ 1, , we have
and
Since tα , tα-1 and tα-2 are uniformly continuous on [0, 1], we can get that , and are equicontinuous. Thus, we get that is compact. The proof is completed.
Lemma 3.4. Suppose (H1), (H2) hold, then the set
is bounded.
Proof. Take x ∈ Ω1, then Nx ∈ ImL. By (3.2), we have
Then, by the integral mean value theorem, there exists a constant ξ ∈ (0, 1) such that f(ξ, x(ξ), x'(ξ), x"(ξ)) = 0. Then from (H2), we have |x(ξ)| ≤ B.
Then, we have
That is
From x ∈ domL, we get x'(0) = 0. Therefore,
That is
By Lx = λNx and x ∈ domL, we have
Then we get
and
From (3.5),(3.6), and (H1), we have
Thus, from Γ(α - 1) - q1 - r1 - s1 > 0, we obtain that
Thus, we get
and
Therefore,
So Ω1 is bounded. The proof is complete.
Lemma 3.5. Suppose (H2) holds, then the set
is bounded.
Proof. For x ∈ Ω2, we have x(t) = c, c ∈ ℝ, and Nx ∈ ImL. Then, we get
which together with (H2) implies |c| ≤ B. Thus, we have
Hence, Ω2 is bounded. The proof is complete.
Lemma 3.6. Suppose the first part of (H2) holds, then the set
is bounded.
Proof. For x ∈ Ω3, we have x(t) = c, c ∈ ℝ, and
If λ = 0, then |c| ≤ B because of the first part of (H2). If λ ∈ (0, 1], we can also obtain |c| ≤ B. Otherwise, if |c| > B, in view of the first part of (H2), one has
which contradicts to (3.7).
Therefore, Ω3 is bounded. The proof is complete.
Remark 3.1. Suppose the second part of (H2) hold, then the set
is bounded.
The proof of Theorem 3.1. Set Ω = {x ∈ X | ||x|| X < max{M1, B, B + M1} + 1}. It follows from Lemma 3.2 and 3.3 that L is a Fredholm operator of index zero and N is L-compact on . By Lemma 3.4 and 3.5, we get that the following two conditions are satisfied
-
(1)
Lx ≠ λNx for every (x, λ) ∈ [(domL\KerL) ∩ ∂Ω] × (0, 1);
-
(2)
Nx ∉ ImL for every x ∈ KerL ∩ ∂Ω.
Take
According to Lemma 3.6 (or Remark 3.1), we know that H(x, λ) ≠ 0 for x ∈ KerL ∩ ∂Ω. Therefore,
So that, the condition (3) of Lemma 2.1 is satisfied. By Lemma 2.1, we can get that Lx = Nx has at least one solution in . Therefore, BVP (1.1) has at least one solution. The proof is complete.
4 An example
Example 4.1. Consider the following BVP
where
Choose , , r(t) = 0, s(t) = 0, B = 10. We can get that , r1 = 0, s1 = 0 and
Then, all conditions of Theorem 3.1 hold, so BVP (4.1) has at least one solution.
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Acknowledgements
The authors would like to thank the referees very much for their helpful comments and suggestions. This research was supported by the Fundamental Research Funds for the Central Universities (2010LKSX09) and the Science Foundation of China University of Mining and Technology (2008A037).
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Hu, Z., Liu, W. Solvability for fractional order boundary value problems at resonance. Bound Value Probl 2011, 20 (2011). https://doi.org/10.1186/1687-2770-2011-20
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DOI: https://doi.org/10.1186/1687-2770-2011-20