Journal of Applied Mathematics and Computing

, Volume 41, Issue 1, pp 119–131

Existence of solutions of two-point boundary value problems for fractional p-Laplace differential equations at resonance

Authors

    • College of Mathematics and PhysicsJinggangshan University
  • Changyuan Yan
    • College of Mathematics and PhysicsJinggangshan University
  • Qing Liu
    • College of Electronics and Information EngineeringJinggangshan University
Original Research

DOI: 10.1007/s12190-012-0598-0

Cite this article as:
Tang, X., Yan, C. & Liu, Q. J. Appl. Math. Comput. (2013) 41: 119. doi:10.1007/s12190-012-0598-0

Abstract

In this paper, we consider the following two-point boundary value problem for fractional p-Laplace differential equation
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equa_HTML.gif
where \(D^{\alpha}_{0^{+}}\), \(D^{\beta}_{0^{+}}\) denote the Caputo fractional derivatives, 0<α,β≤1, 1<α+β≤2. By using the coincidence degree theory, a new result on the existence of solutions for above fractional boundary value problem is obtained. These results extend the corresponding ones of ordinary differential equations of integer order. Finally, an example is inserted to illustrate the validity and practicability of our main results.

Keywords

Caputo fractional derivativep-Laplace differential equationTwo-point boundary value problemResonanceCoincidence degree theory

Mathematics Subject Classification

34A0834B15

1 Introduction

In this paper, we are concerned with the following two-point boundary value problem for fractional p-Laplace differential equation
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ1_HTML.gif
(1.1)
where \(D^{\alpha}_{0^{+}}\), \(D^{\beta}_{0^{+}}\) denote the Caputo fractional derivatives, 0<α,β≤1, 1<α+β≤2, f is continuous functions. The boundary value problem (BVP for short) (1.1) is a resonance problem, because its associated homogeneous boundary value problem
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equb_HTML.gif
has a nontrivial solution u(t)=ctβ, cR.
The equations with p-Laplacian operator
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ2_HTML.gif
(1.2)
arise in the modeling of different physical and natural phenomena, e.g., non-Newtonian mechanics, nonlinear elasticity and glaciology, population biology, combustion theory, nonlinear flow laws, system of Monge-Kantorovich partial differential equations (see [15]).
In recent years, the fractional differential equations have received more and more attention. Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, biology, economics, control theory, signal and image processing, biophysics, blood flow phenomena, aerodynamics, fitting of experimental data, etc. involves derivatives of fractional order. Fractional differential equations also serve as an excellent tool for the description of hereditary properties of various materials and processes (see [6, 7]). Recently, many people have established the existence of solutions or positive solutions of fractional differential equations at nonresonance and at resonance, readers can see [817] and references cited therein. For example, Agarwal et al. (see [8]) considered a two-point boundary value problem given by
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ3_HTML.gif
(1.3)
where \(D^{\alpha}_{0^{+}}\) and \(D^{\beta}_{0^{+}}\) are the Riemann-Liouville fractional derivatives, 1<α<2, β>0 are real numbers, αβ≥1. They established the existence results by the fixed point theorem in a cone.
For fractional boundary value problems at resonance, Chen et al. (see [17]) considered a two-point boundary value problem with p-Laplacian operator given by
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ4_HTML.gif
(1.4)
where \(D^{\alpha}_{0^{+}}\) and \(D^{\beta}_{0^{+}}\) are Caputo fractional derivatives, 0<α,β≤1, 1<α+β≤2. They obtained the existence results by using the coincidence degree theory.

However, to the best of our knowledge, no paper has concerned the existence of solutions to the boundary value problem (1.1) at resonance. Motivated by the work above, in this paper, we consider the existence of solutions for two-point boundary value problem for fractional p-Laplace differential equation (1.1) at resonance. By using the coincidence degree theory, a new result on the existence of solutions for above fractional boundary value problem (1.1) is obtained. These results extend the corresponding ones of ordinary differential equations of integer order.

This paper is organized as follows. In Sect. 2, we introduce some basic definitions and preliminaries later used. In Sect. 3, the existence results of solutions for fractional p-Laplace differential equation (1.1) is discussed by using the coincidence degree theory. In Sect. 4, we give an example to illustrate our main results.

2 Preliminary results

The material in this section is basic in some sense. For the reader’s convenience, we present some definitions form fractional calculus theory and preliminary results.

Definition 2.1

([18])

The Riemann-Liouville fractional integral of order α (α>0) of a function f:(0,+∞)→ℝ is given by
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equc_HTML.gif
provided that the right side is pointwise defined on (0,+∞).

Definition 2.2

([18])

The Caputo fractional derivative of order α (α>0) of a continuous function f:(0,+∞)→ℝ is given by
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equd_HTML.gif
where n=[α]+1, [α] denotes the integer part of number α, provided that the right side is pointwise defined on (0,+∞).

Lemma 2.1

([19])

Letα>0. If we assumeuC(0,1)∩L(0,1), then the fractional differential equation
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Eque_HTML.gif
hasu(t)=C0+C1t+⋯+Cn−1tn−1, Ci∈ℝ, i=0,2,…,n−1, as unique solutions, where nis the smallest integer greater than or equal toα.

Lemma 2.2

([19])

Given that uC(0,1)∩L(0,1) with a fractional derivative of orderα (α>0) that belongs toC(0,1)∩L(0,1). Then\(I^{\alpha}_{0^{+}}D^{\alpha}_{0^{+}}u(t)=u(t)+C_{0} +C_{1}t+\cdots+C_{n-1}t^{n-1}\), for someCi∈ℝ, i=0,2,…,n−1, wherenis the smallest integer greater than or equal toα.

Now, we briefly recall some notation and an abstract existence result due to Mawhin [11].

Let XY be two real Banach spaces and \(L: \operatorname{dom}L\subset X\rightarrow Y \) a linear operator which is a Fredholm map of index zero, N:XY is nonlinear continuous map. If \(\dim\ker L=\dim(Y/\operatorname{Im}L)<+\infty\), and \(\operatorname{Im}L\) is a closed set of Y, then L is a Fredholm map of index zero. If L is a Fredholm map of index zero, and P:XX, Q:YY be projectors such that \(\operatorname{Im}P= \operatorname{Ker} L\), \(\operatorname{Ker}Q= \operatorname{Im}L\), \(X= \operatorname{Ker} L\oplus \operatorname{Ker} P\), \(Y= \operatorname{Im}L \oplus \operatorname{Im}Q\). It follows that \(L_{p}=L|_{\operatorname{dom}L\cap\operatorname {Ker}P}:\operatorname{dom}L\cap\operatorname{Ker} P \rightarrow\operatorname{Im}L\) is invertible. We denote the inverse by Kp. Let Ω is an open bounded subset of X, and \(\operatorname{dom}L\cap\overline{\varOmega}\neq\emptyset\), the map N:XY will be called L-compact on \(\overline{\varOmega}\) if \(QN(\overline{\varOmega})\) is bounded and \(K(I-Q)N:\overline{\varOmega}\rightarrow X\) is compact.

Theorem 2.1

([20])

LetXYbe real Banach spaces, \(L:\operatorname{dom}L\subset X\rightarrow Y\)be a Fredholm operator of index zero andN:XYbeL-compact on\(\overline{\varOmega}\). Assume that the following conditions are satisfied
  1. (1)

    LxλNx, \(\forall(x, \lambda)\in[(\operatorname{dom}L\backslash\operatorname{Ker} L)\cap \partial\varOmega]\times(0, 1)\);

     
  2. (2)

    \(Nx\notin\operatorname{Im}L\), \(\forall x\in\operatorname {Ker} L\cap\partial\varOmega\);

     
  3. (3)

    \(\deg(QN|_{\operatorname{Ker}L}, \varOmega\cap \operatorname{Ker} L, 0)\neq0\), where, Q:YYis a projection such that\(\operatorname{Im}L= \operatorname{Ker} Q\).

     
Then the equationLx=Nxhas at least one solution in\(\operatorname{dom}L\cap\overline{\varOmega}\).

In this paper, we take Y=C[0,1], with the norm ∥yY=maxt∈[0,1]|y(t)|, and \(X=\{x|x,D^{\beta}_{0^{+}}x\in Y\}\), with the norm \(\|x\|_{X}=\max\{\|x\|_{\infty}, \|D^{\beta}_{0^{+}}x\|_{\infty}\}\), where ∥x=maxt∈[0,1]|x(t)|. By means of the linear functional analysis theory, we can prove that X is a Banach space (see [21, Lemma 3.2]).

Define the operator \(L: \operatorname{dom}L\subset X\rightarrow Y\) by
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ5_HTML.gif
(2.1)
where \(\operatorname{dom}L=\{x\in X|D^{\alpha}_{0^{+}}\phi_{p}(D^{\beta}_{0^{+}}x) \in Y,x(0)=0,D^{\beta}_{0^{+}}x(0)=D^{\beta}_{0^{+}}x(1)\}\).
Let N:XY be the Nemytski operator
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ6_HTML.gif
(2.2)
Then BVP (1.1) is equivalent to the operator equation
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equf_HTML.gif

3 Some lemmas

In this section, we need the following auxiliary lemmas to prove the existence of solutions to (1.1).

Lemma 3.1

LetLbe defined by (2.1), then
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ7_HTML.gif
(3.1)
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ8_HTML.gif
(3.2)

Proof

By Lemma 2.1, \(D^{\alpha}_{0^{+}}\phi_{p}(D^{\beta}_{0^{+}}x(t))=0\) has solution
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equg_HTML.gif
Combining with the boundary value condition x(0)=0, one has that (3.1) holds.
If \(y\in\operatorname{Im}L\), then there exists a function \(x\in \operatorname{dom}L\) such that \(y(t)= D^{\alpha}_{0^{+}}\phi_{p}(D^{\beta}_{0^{+}}x(t))\). Basing on Lemma 2.1, we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equh_HTML.gif
From condition \(D^{\beta}_{0^{+}}x(0)=D^{\beta}_{0^{+}}x(1)\), we obtain that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ9_HTML.gif
(3.3)
Thus, we get (3.2).

On the other hand, suppose yY, and satisfies (3.3). Let \(x(t)=I^{\beta}_{0^{+}}\phi_{q}(I^{\alpha}_{0^{+}}y(t))\), then \(x\in\operatorname{dom}L\), and \(Lx(t)=D^{\alpha}_{0^{+}}\phi_{p}(D^{\beta}_{0^{+}}x(t))=y(t)\). So that, \(y\in\operatorname{Im}L\). The proof is complete. □

Lemma 3.2

LetLbe defined by (2.1); thenLis a Fredholm operator of index zero, and the linear continuous projector operatorsP:XXandQ:YYcan be defined as
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equi_HTML.gif
Furthermore, the operator\(K_{p}: \operatorname{Im}L\rightarrow\operatorname{dom}L\cap\operatorname {Ker}_{P}\)can be written by
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equj_HTML.gif

Proof

Obviously, \(\operatorname{Im}P = \operatorname{Ker}L\) and P2x=Px. It follows from x=(xPx)+Px that \(X= \operatorname{Ker} P+ \operatorname{Ker} L\). By simple calculation, we can get that \(\operatorname{Ker}P\cap\operatorname{Ker} L=\{0\}\). Then we get
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equk_HTML.gif
For any yY, we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ10_HTML.gif
(3.4)
Let y1=yQy, then we get from (3.4) that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equl_HTML.gif
which implies \(y_{1}\in\operatorname{Im}L\). Hence \(Y= \operatorname {Im}L+ \operatorname{Im}Q\). Since \(\operatorname{Im}L\cap\operatorname{Im}Q=\{0\}\), we have \(Y= \operatorname{Im}L \oplus \operatorname{Im}Q\). Thus
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equm_HTML.gif
This means that L is a Fredholm operator of index zero.
From the definitions of P, KP, it is easy to see that the generalized inverse of LP is KP. In fact, for \(y\in\operatorname{Im}L\), we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ11_HTML.gif
(3.5)
Moreover, for \(x\in\operatorname{dom}L\cap\operatorname{Ker} P\), we get \(x(0)=D^{\beta}_{0^{+}}x(0)=D^{\beta}_{0^{+}}x(1)=0\). By Lemma 2.1, we obtain that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equn_HTML.gif
which together with \(D^{\beta}_{0^{+}}x(0)=0\), yields that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equo_HTML.gif
Thus, we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equp_HTML.gif
which together with x(0)=0, yields that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ12_HTML.gif
(3.6)
Combining (3.5) with (3.6), we know that KP is the inverse of LP. The proof is complete. □

Lemma 3.3

AssumeΩXis an open bounded subset such that\(\operatorname{dom}L\cap\overline{\varOmega}\neq\emptyset\), thenNisL-compact on\(\overline{\varOmega}\).

Proof

For convenience, denote KP,Q=KP(IQ)N. By the continuity of f, we can get that \(QN(\overline{\varOmega})\) and \(K_{P,Q}(\overline{\varOmega})\) are bounded. Moreover, there exists a constant M>0 such that \(|I^{\alpha}_{0^{+}}(I-Q)Nx|\leq M\), \(\forall x\in\overline{\varOmega}\), t∈[0,1]. Thus, in view of the Arzela-Ascoli theorem, we need only prove that \(K_{P,Q}(\overline{\varOmega})\subset X\) is equicontinuous.

For 0≤t1<t2≤1, \(x\in\overline{\varOmega}\), we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equq_HTML.gif
Since tβ is uniformly continuous on [0, 1], we can obtain that \(K_{P,Q}(\overline{\varOmega})\subset C[0,1]\) is equicontinuous. Similar proof can show that \(I^{\alpha}_{0^{+}}(I-Q)N(\overline{\varOmega})\subset C[0,1]\) is equicontinuous. This, together with the uniformly continuity of ϕ(s) on [−T,T], yields that \(D^{\beta}_{0^{+}}(K_{P,Q})(\overline{\varOmega}) =\phi_{q}(I^{\alpha}_{0^{+}}(I-Q)N)(\overline{\varOmega})\subset C[0,1]\) is also equicontinuous. Thus, we get that \(K_{P}(I-Q)N: \overline{\varOmega}\rightarrow X\) is compact. The proof is complete. □

Lemma 3.4

Suppose (H1) and (H2) hold, then the set
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equr_HTML.gif
is bounded. (H1) and (H2) will be given in Sect4.

Proof

Take xΩ1, then Lx=λNx and \(Nx\in\operatorname {Im}L\). By (3.2), we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equs_HTML.gif
Then, by the integral mean value theorem, there exists a constant ξ∈(0,1) such that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equt_HTML.gif
So, from (H2), we get \(|D^{\beta}_{0^{+}}x(\xi)|\leq D\).
From \(x\in\operatorname{dom}L\), we get x(0)=0. Therefore
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equu_HTML.gif
That is
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ13_HTML.gif
(3.7)
By Lx=λNx, we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ14_HTML.gif
(3.8)
Take t=ξ, we get
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ15_HTML.gif
(3.9)
Together with \(|D^{\beta}_{0^{+}}x(\xi)|\leq D\), (H1) and (3.6), we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equv_HTML.gif
So, we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equw_HTML.gif
Thus, from Γ(α+1)−2(∥b+∥c)>0, we obtain that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ16_HTML.gif
(3.10)
and
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ17_HTML.gif
(3.11)
Combining (3.10) with (3.11), we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equx_HTML.gif
Therefore, Ω1 is bounded. The proof is complete. □

Lemma 3.5

Suppose (H3) hold, then the set
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equy_HTML.gif
is bounded. (H3) will be given in Sect. 4.

Proof

For xΩ2, we have x(t)=ctβ, cR and \(Nx\in\operatorname{Im}L\). Then we get
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equz_HTML.gif
which together with (H3) implies |c|≤D. Thus, we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equaa_HTML.gif
Hence, Ω2 is bounded. The proof is complete. □

Lemma 3.6

Suppose (H3) holds, then the set
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equab_HTML.gif
is bounded. (H3) will be given in Sect4.

Proof

The proof divided into two cases by the relations of f(t,u,v) and v.

Case 1. There exists a constant D>0 such that for all cR with |c|>D,
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ18_HTML.gif
(3.12)
In this case, \(\varOmega_{3}=\{x|x\in\operatorname{Ker} L, \lambda x+(1-\lambda)QNx=0,\lambda\in[0,1]\}\).
For xΩ3, we have x(t)=ctβ, cR, and
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ19_HTML.gif
(3.13)
If λ=0, then |c|≤D. If λ∈(0,1], we can also obtain |c|≤D. Otherwise, if |c|>D, by (3.12), one has
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equac_HTML.gif
which contradicts to (3.13). Therefore, Ω3 is bounded.
Case 2. There exists a constant D>0 such that for all cR with |c|>D,
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ20_HTML.gif
(3.14)
In this case, \(\varOmega_{3}=\{x|x\in\operatorname{Ker} L, -\lambda x+(1-\lambda)QNx=0,\lambda\in[0,1]\}\).
For xΩ3, we have x(t)=ctβ, cR, and
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ21_HTML.gif
(3.15)
If λ=0, then |c|≤D. If λ∈(0,1], we can also obtain |c|≤D. Otherwise, if |c|>D, by (3.14), one has
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equad_HTML.gif
which contradicts to (3.13). Therefore, Ω3 is bounded.

From the above Case 1 and Case 2, we can know that Ω3 is bounded. The proof is complete. □

4 Main result and example

In this section, we firstly investigate the existence of solutions for two-point boundary value problem for fractional p-Laplace differential equation (1.1) at resonance, which is based on the coincidence degree theory. Then, we will give an example to illustrate the validity and practicability of our main results.

Now, we begin with some theorems below.

Theorem 4.1

Letf:[0,1]×R2Rbe continuous. Assume that
(H1)
there exist nonnegative functionsa,b,cYsuch that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equae_HTML.gif
(H2)
there exists a constantD>0 such that either
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equaf_HTML.gif
or
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equag_HTML.gif
(H3)
there exists a constantD>0 such that either
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equah_HTML.gif
or
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equai_HTML.gif
Then BVP (1.1) has at least one solution, provided that\(\frac{2(\|b\|_{\infty}+\|c\|_{\infty})}{\varGamma(\alpha+1)}<1\).

Proof

Set
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equaj_HTML.gif
Obviously, Ω1Ω2Ω3Ω. It follows from Lemmas 3.2 and 3.3 that L (defined by (2.1)) is a Fredholm operator of index zero and N (defined by (2.2)) is L-compact on \(\overline{\varOmega}\). By Lemmas 3.4 and 3.5, we get that the following two conditions are satisfied
  1. (1)

    LxλNx, \(\forall(x, \lambda)\in[(\operatorname{dom} L\backslash\operatorname{Ker} L)\cap\partial \varOmega]\times(0, 1)\);

     
  2. (2)

    \(Nx\notin\operatorname{Im}L\), \(\forall x\in\operatorname {Ker} L\cap\partial\varOmega\).

     
It remains to verify the condition (3) of Theorem 2.1. In order to do that, let
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equak_HTML.gif
Basing on Lemma 3.6, we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equal_HTML.gif
Thus, by the homotopy property of degree, we have
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equam_HTML.gif
So that, the condition (3) of Theorem 2.1 is satisfied.

Consequently, by using Theorem 2.1, the operator equation Lx=Nx has at least one solution in dom\(L\cap\overline{\varOmega}\). Namely, BVP (1.1) has at least one solution in X. The proof is complete. □

Theorem 4.2

Letf:[0,1]×R2Rbe continuous. Supposed that the conditions (H2) and (H3) hold. Further, assume that
(H4)
there exist nonnegative functionsrYsuch that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equan_HTML.gif
Then BVP (1.1) has at least one solution, provided that\(\frac{4\|r\|_{\infty}}{\varGamma(\alpha+1)}<1\).

Proof

Note that \(\frac{4\|r\|_{\infty}}{\varGamma(\alpha+1)}<1\), then there is a constant ε>0 such that \(\frac{4(\|r\|_{\infty}+\varepsilon)}{\varGamma(\alpha+1)}<1\).

From (H4), there exists H>0 such that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equao_HTML.gif
Let M=maxt∈[0,1],|u|+|v|≤H|f(t,u,v)|,
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equap_HTML.gif
From Theorem 4.1, BVP (1.1) has at least one solution in X. The proof is complete. □

Corollary 4.1

Letf:[0,1]×R2Rbe continuous. Supposed that the conditions (H2) and (H3) hold. Further, assume that
(H5)
there exist nonnegative functionsrYsuch that
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equaq_HTML.gif
or
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equar_HTML.gif
Then BVP (1.1) has at least one solution, provided that\(\frac{2\|r\|_{\infty}}{\varGamma(\alpha+1)}<1\).

Now, we will give an example to illustrate our main result.

Example 4.1

Consider the following boundary value problem
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equ22_HTML.gif
(4.1)
where \(\alpha=\frac{1}{2}\), \(\beta=\frac{5}{8}\), p=3 and
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equas_HTML.gif
Choose a(t)=7, b(t)=0, \(c(t)=\frac{1}{4}\), D=5. By simple calculation, we can get that ∥b=0, \(\|c\|_{\infty}=\frac{1}{4}\), and
https://static-content.springer.com/image/art%3A10.1007%2Fs12190-012-0598-0/MediaObjects/12190_2012_598_Equat_HTML.gif
Obviously, BVP (4.1) satisfies all conditions of Theorem 4.1. Hence, it has at least one solution.

Acknowledgements

The author would like to thank the referee for his or her careful reading and some comments on improving the presentation of this paper.

Copyright information

© Korean Society for Computational and Applied Mathematics 2012