Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 37, 1-15; http://www.math.u-szeged.hu/ejqtde/

On solutions of some fractional m-point boundary value problems at resonance ^{∗}

Zhanbing Bai

College of Information Science and Engineering, Shandong University of Science and Technology, Qingdao 266510, P.R. China. E-mail: zhanbingbai@163.com

Abstract

In this paper, the following fractional order ordinary differential equation boundary value problem:

D_{0+}^{α}u(t) = f(t, u(t), D_{0+}^{α−1}u(t)) + e(t), 0 < t < 1,

I_{0+}^{2−α}u(t) |_{t=0}= 0, D_{0+}^{α−1}u(1) = ∑_{i=1}^{m−2} β_{i}D_{0+}^{α−1}u(η_{i}),

is considered, where 1 < α ≤ 2, is a real number, D_{0+}^{α} and I_{0+}^{α} are the standard Riemann-Liouville differentiation and integration, and f : [0, 1] × R^{2} → R is continuous and e ∈ L^{1}[0, 1], and η_{i} ∈ (0, 1), β_{i} ∈ R, i = 1, 2, ⋯, m − 2, are given constants such that ∑_{i=1}^{m−2} β_{i} = 1. By using the coincidence degree theory, some existence results of solutions are established.

Key Words: Fractional differential equation; m-point boundary value problem; At resonance; Coincidence degree

MR (2000) Subject Classification: 34B15

1 Introduction

The subject of fractional calculus has gained considerable popularity and impor-

tance during the past decades or so, due mainly to its demonstrated applications in

numerous seemingly and widespread fields of science and engineering. It does indeed

provide several potentially useful tools for solving differential and integral equations,

and various other problems involving special functions of mathematical physics as well

as their extensions and generalizations in one and more variables. For details, see [1-9,

13-18, 21-25] and the references therein.

Recently, m-point integer-order differential equation boundary value problems have

been studied by many authors, see [4, 12, 13, 14]. However, there are few papers consider the nonlocal boundary value problem at resonance for nonlinear ordinary differential equations of fractional order. In [6] we investigated the nonlinear nonlocal non-resonance problem

∗This work is sponsored by the Tianyuan Youth Grant of China (10626033).

D_{0+}^{α}u(t) = f(t, u(t)), 0 < t < 1,
u(0) = 0, βu(η) = u(1),

where 1 < α ≤ 2, 0 < βη^{α−1} < 1. In [7], we investigated the boundary value problem at resonance

D_{0+}^{α}u(t) = f(t, u(t), D_{0+}^{α−1}u(t)) + e(t), 0 < t < 1,

I_{0+}^{2−α}u(t) |_{t=0}= 0, u(1) = ∑_{i=1}^{m−2} β_{i}u(η_{i}),

where β_{i} ∈ R, i = 1, 2, ⋯, m − 2, 0 < η_{1} < η_{2} < ⋯ < η_{m−2} < 1 are given constants such that ∑_{i=1}^{m−2} β_{i}η_{i}^{α−1} = 1.

In this paper, the following fractional order ordinary differential equation boundary value problem:

D_{0+}^{α}u(t) = f(t, u(t), D_{0+}^{α−1}u(t)) + e(t), 0 < t < 1,  (1.1)
I_{0+}^{2−α}u(t) |_{t=0}= 0, D_{0+}^{α−1}u(1) = ∑_{i=1}^{m−2} β_{i}D_{0+}^{α−1}u(η_{i}),  (1.2)

is considered, where 1 < α ≤ 2 is a real number, D_{0+}^{α} and I_{0+}^{α} are the standard Riemann-Liouville differentiation and integration, and f : [0, 1]×R^{2} → R is continuous and e ∈ L^{1}[0, 1], η_{i} ∈ (0, 1), β_{i} ∈ R, i = 1, 2, ⋯, m − 2, are given constants such that ∑_{i=1}^{m−2} β_{i} = 1.

The m-point boundary value problem (1.1), (1.2) happens to be at resonance in the sense that its associated linear homogeneous boundary value problem

D_{0+}^{α}u(t) = 0, 0 < t < 1,

I_{0+}^{2−α}u(t) |_{t=0}= 0, D_{0+}^{α−1}u(1) = ∑_{i=1}^{m−2} β_{i}D_{0+}^{α−1}u(η_{i}),

has u(t) = ct^{α−1}, c ∈ R as a nontrivial solution.

The purpose of this paper is to study the existence of solution for boundary value problem (1.1), (1.2) at resonance case, and establish an existence theorem under nonlinear growth restrictions of f. Our method is based upon the coincidence degree theory of Mawhin [22]. Finally, we also give an example to demonstrate our result.

EJQTDE, 2010 No. 37, p. 2 Now, we briefly recall some notation and an abstract existence result. Let Y, Z be real Banach spaces, L : dom(L) ⊂ Y → Z be a Fredholm map of

index zero and P : Y → Y, Q : Z → Z be continuous projectors such that Im(P) = Ker(L), Ker(Q) = Im(L) and Y = Ker(L) ⊕ Ker(P), Z = Im(L) ⊕ Im(Q). It follows that L|_{dom(L)∩Ker(P)} : dom(L)∩Ker(P) → Im(L) is invertible. We denote the inverse of the map by K_{P}. If Ω is an open bounded subset of Y such that dom(L)∩Ω ≠ ∅, the map N : Y → Z will be called L-compact on \overline{Ω} if QN(\overline{Ω}) is bounded and K_{P} (I − Q)N : \overline{Ω} → Y is compact.

The main tool we use is the Theorem 2.4 of [22].

Theorem 1.1 Let L be a Fredholm operator of index zero and let N be L-compact on \overline{Ω}. Assume that the following conditions are satisfied:

(i) Lx ≠ λNx for every (x, λ) ∈ [(dom(L)\Ker(L)) ∩ ∂Ω] × (0, 1);

(ii) Nx ∉ Im(L) for every x ∈ Ker(L) ∩ ∂Ω;

(iii) deg (QN|_{Ker(L)}, Ω ∩ Ker(L), 0) ≠ 0, where Q : Z → Z is a projection as above with Im(L) = Ker(Q).

Then the equation Lx = Nx has at least one solution in dom(L) ∩ \overline{Ω}.

The rest of this paper is organized as follows. In section 2, we give some notations and lemmas. In section 3, we establish a theorem of existence of a solution for the problem (1.1), (1.2). In section 4, we give an example to demonstrate our result.

2 Background materials and preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory. Which can be found in [6, 16, 24].

We use the classical Banach spaces C[0, 1] with the norm ∥u∥_{∞} = max_{t∈[0,1]} |u(t)|, L^{1}[0, 1] with the norm ∥u∥_{1} = ∫_{0}^{1} |u(t)|dt. 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 2.1 The fractional integral of order α > 0 of a function y : (0, ∞) → R is given by

I_{0+}^{α}y(t) = \frac{1}{Γ(α)} ∫_{0}^{t}(t − s)^{α−1}y(s)ds

provided the right side is pointwise defined on (0, ∞).

Definition 2.2 The fractional derivative of order α > 0 of a function y : (0, ∞) → R is given by

D_{0+}^{α}y(t) = \frac{1}{Γ(n − α)} (\frac{d}{dt})^{n} ∫_{0}^{t} \frac{y(s)}{(t − s)^{α−n+1}} ds,

where n = [α] + 1, provided the right side is pointwise defined on (0, ∞).

It can be directly verified that the Riemann-Liouvell fractional integration and fractional differentiation operators of the power functions t^{μ} yield power functions of the same form. For α ≥ 0, μ > −1, there are

I_{0+}^{α}t^{μ} = \frac{Γ(μ + 1)}{Γ(μ + α + 1)}t^{μ+α}, D_{0+}^{α}t^{μ} = \frac{Γ(μ + 1)}{Γ(μ − α + 1)}t^{μ−α}.

Lemma 2.1 [17](Page 74, Lemma 2.5) 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

(I_{0+}^{α}D_{0+}^{α}u)(t) = u(t) − ∑_{i=1}^{n} \frac{((I_{0+}^{n−α}u)(t))^{(n−i)} |_{t=0}}{Γ(α − i + 1)} t^{α−i},

holds almost everywhere on [0, 1].

Now, we define another spaces which are fundamental in our work.

Definition 2.3 Given μ > 0 and N = [μ] + 1 we can define a linear space

C^{μ}[0, 1] = {u(t)|u(t) = I_{0+}^{μ}x(t) + c_{1}t^{μ−1} + ⋯ + c_{N−1}t^{μ−(N−1)}, x ∈ C[0, 1], t ∈ [0, 1]},

where c_{i} ∈ R, i = 1, …, N − 1.

Remark 2.1 By means of the linear functional analysis theory, we can prove that with the norm

∥u∥_{C^{μ}} = ∥D_{0+}^{μ}u∥_{∞} + ⋯ + ∥D_{0+}^{μ−(N−1)}u∥_{∞} + ∥u∥_{∞}

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].

EJQTDE, 2010 No. 37, p. 4 Definition 2.4 Let I_{0+}^{α}(L^{1}(0, 1)), α > 0, denote the space of functions u(t), represented by fractional integral of order α of a summable function: u = I_{0+}^{α}v, v ∈ L^{1}(0, 1).

In the following Lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming that I_{0+}^{α} = D_{0+}^{α} for α < 0.

Lemma 2.2 [16]The relation

I_{0+}^{α}I_{0+}^{β}φ = I_{0+}^{α+β}φ

is valid in any of the following cases:

1) β ≥ 0, α + β ≥ 0, φ(t) ∈ L^{1}(0, 1);

2) β ≤ 0, α ≥ 0, φ(t) ∈ I_{0+}^{−β}(L^{1}(0, 1));

3) α ≤ 0, α + β ≤ 0, φ(t) ∈ I_{0+}^{−α−β}(L^{1}(0, 1)).

Lemma 2.3 [11] (Page 74, Property 2.3) Denote by D = \frac{d}{dt}. If (D_{0+}u^{α})(t) and (D_{0+}u^{α+m})(t) all exist, then there holds (D^{m}D_{0+}^{α}u)(t) = (D_{0+}^{α+m})u(t).

Lemma 2.4 [7] 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

∥u∥_{C^{μ}} = ∥D_{0+}^{μ}u∥_{∞} + ⋯ + ∥D_{0+}^{μ−(N−1)}u∥_{∞} + ∥u∥_{∞} < M,

and equicontinuous means that ∀ε > 0, ∃δ > 0, for all t_{1}, t_{2} ∈ [0, 1], |t_{1} − t_{2}| < δ, u ∈ F, i ∈ {0, ⋯, N − 1}, there hold

|u(t_{1}) − u(t_{2})| < ε, |D_{0+}^{μ−i}u(t_{1}) − D_{0+}^{μ−i}u(t_{2})| < ε.

3 Existence result

In this section, we always suppose that 1 < α ≤ 2 is a real number and ∑_{i=1}^{m−2} β_{i} = 1. Let Z = L^{1}[0, 1]. Y = C^{α−1}[0, 1] = {u(t)|u(t) = I_{0+}^{α−1}x(t), x ∈ C[0, 1], t ∈ [0, 1]} with the norm ∥u∥_{C^{α−1}} = ∥D_{0+}^{α−1}u∥_{∞} + ∥u∥_{∞}. Then Y is a Banach space.

Given a function u such that D_{0+}^{α}u = f(t) ∈ L^{1}(0, 1) and I_{0+}^{2−α}u(t) |_{t=0}= 0, there holds u ∈ C^{α−1}[0, 1]. In fact, with Lemma 2.1, one has

u(t) = I_{0+}^{α}f(t) + c_{1}t^{α−1} + c_{2}t^{α−2}.

EJQTDE, 2010 No. 37, p. 5 Combine with I_{0+}^{2−α}u(t) |_{t=0}= 0, there is c_{2} = 0. So,

u(t) = I_{0+}^{α}f(t) + c_{1}t^{α−1} = I_{0+}^{α−1} [I_{0+}^{1}f(t) + c_{1}Γ(α)] ,

Thus u ∈ C^{α−1}[0, 1]. Define L to be the linear operator from dom(L) ∩ Y to Z with

dom(L) = {u ∈ C^{α−1}[0, 1]

|D_{0+}^{α}u ∈ L^{1}(0, 1), I_{0+}^{2−α}u(0) = 0, D_{0+}^{α−1}u(1) = ∑_{i=1}^{m−2} β_{i}D_{0+}^{α−1}u(η_{i})} ,

and

Lu = D_{0+}^{α}u, u ∈ dom(L).  (3.1)

Define N : Y → Z by

Nu(t) = f (t, u(t), D_{0+}^{α−1}u(t)) + e(t), t ∈ [0, 1].

Then boundary value problem (1.1), (1.2) can be written as

Lu = Nu.

Lemma 3.1 Let L be defined as (3.1), then

Ker(L) = {ct^{α−1}|c ∈ R}  (3.2)

and

Im(L) = {y ∈ Z |∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} y(s)ds = 0} .  (3.3)

Proof. By Lemma 2.1, Lemma 2.2, D_{0+}^{α}u(t) = 0 has solution

u(t) = \frac{(I_{0+}^{2−α}u(t))′ |_{t=0}}{Γ(α)} t^{α−1} + \frac{I_{0+}^{2−α}u(t) |_{t=0}}{Γ(α − 1)} t^{α−2}

= \frac{D_{0+}^{α−1}u(t) |_{t=0}}{Γ(α)} t^{α−1} + \frac{I_{0+}^{2−α}u(t) |_{t=0}}{Γ(α − 1)} t^{α−2}

Combine with (1.2), one has (3.2) holds.

If y ∈ Im(L), then there exists a function u ∈ dom(L) such that y(t) = D_{0+}^{α}u(t). By Lemma 2.1,

I_{0+}^{α}y(t) = u(t) − c_{1}t^{α−1} − c_{2}t^{α−2}.

where

c_{1} = \frac{D_{0+}^{α−1}u(t) |_{t=0}}{Γ(α)} , c_{2} = \frac{I_{0+}^{2−α}u(t) |_{t=0}}{Γ(α − 1)} .

By the boundary condition I_{0+}^{2−α}u(t) |_{t=0}= 0, one has c_{2} = 0. So,

u(t) = I_{0+}^{α}y(t) + c_{1}t^{α−1}

and by Lemma 2.2,

D_{0+}^{α−1}u(t) = D_{0+}^{α−1}I_{0+}y(t) + D_{0+}^{α−1}(c_{1}t^{α−1}) = I_{0+}^{1}y(t) + c_{1}Γ(α)

In view of the condition D_{0+}^{α−1}u(1) = ∑_{i=1}^{m−2} β_{i}D_{0+}^{α−1}u(η_{i}), we have

∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} y(s)ds = 0,

thus, we obtain (3.3).

On the other hand, suppose y ∈ Z and satisfies:

∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} y(s)ds = 0.

Let u(t) = I_{0+}^{α}y(t), then u ∈ dom(L) and D_{0+}^{α}u(t) = y(t). So, y ∈ Im(L). ¶

Lemma 3.2 There exists k ∈ {0, 1, ⋯, m − 2} satisfies ∑_{i=1}^{m−2} β_{i}η_{i}^{k+1} ≠ 1.

Proof. Suppose it is not true, we have

( \begin{matrix}η_{1} & η_{2} & ⋯ & η_{m−2} \\ η_{1}^{1} & η_{2}^{1} & ⋯ & η_{m−2}^{1} \\ ⋮ & ⋮ & ... & ⋮ \\ η_{1}^{m−2} & η_{2}^{m−2} & ⋯ & η_{m−2}^{m−2} \end{matrix} ) ( β_{1} ; β_{2} ; ⋮ ; β_{m−2} ) = ( 1 1 ⋮ 1 ) .

It is equal to

( \begin{matrix}η_{1} & η_{2} & ⋯ & η_{m−2} & 1 \\ η_{1}^{1} & η_{2}^{1} & ⋯ & η_{m−2}^{1} & 1 \\ ⋮ & ⋮ & ... & ⋮ & ⋮ \\ η_{1}^{m−3} & η_{2}^{m−3} & ⋯ & η_{m−2}^{m−3} & 1 \\ η_{1}^{m−2} & η_{2}^{m−2} & ⋯ & η_{m−2}^{m−2} & 1 \end{matrix} ) ( β_{1} ; β_{2} ; ⋮ ; β_{m−2} ; −1 ) = ( 0 0 ⋮ 0 0 ) .

However, it is well known that the Vandermonde Determinant is not equal to zero, so there is a contradiction. ¶

Lemma 3.3 L : dom(L) ∩ Y → Z is a Fredholm operator of index zero. Furthermore, the linear continuous projector operators Q : Z → Z and P : Y → Y can be defined by

Qu = C_{u}t^{k}, for every u ∈ Z,

EJQTDE, 2010 No. 37, p. 7

P u(t) = D_{0+}^{α−1}u(t) |_{t=0} t^{α−1}, for every u ∈ Y,

where

C_{u} = \frac{∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} u(s)ds}{(k + 1)(1 − ∑_{i=1}^{m−2} β_{i}η_{i}^{k+1})}

Here k ∈ {0, 1, ⋯, m − 2} satisfies ∑_{i=1}^{m−2} β_{i}η_{i}^{k+1} ≠ 1. And the linear operator K_{P} : Im(L) → dom(L) ∩ Ker(P) can be written by

K_{P} (y) = I_{0+}^{α}y(t).

Furthermore

∥K_{P} y∥_{C^{α−1}} ≤ (1 + \frac{1}{Γ(α)}) ∥y∥_{1}, for all y ∈ Im(L).

Proof. For y ∈ Z, let y_{1} = y − Qy, then y_{1} ∈ Im(L) (since ∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} y_{1}(s)ds = 0). Hence Z = Im(L) + {ct^{k} | c ∈ R}. Since Im(L) ∩ {ct^{k} | c ∈ R} = {0}, we have Z = Im(L) ⊕ {ct^{k} | c ∈ R}. Thus

dim Ker(L) = dim {ct^{k} | c ∈ R} = co dim Im(L) = 1.

So, L is a Fredholm operator of index zero.

With definitions of P, K_{P}, it is easy to show that the generalized inverse of L : Im(L) → dom(L) ∩ Ker(P) is K_{P}. In fact, for y ∈ Im(L), we have

(LK_{P})y = D_{0+}^{α}I_{0+}^{α}y = y,

and for u ∈ dom(L) ∩ Ker(P), we know

(K_{P} L)u(t) = I_{0+}^{α}D_{0+}^{α}u(t) = u(t) + c_{1}t^{α−1} + c_{2}t^{α−2},

where

c_{1} = \frac{D_{0+}^{α−1}u(t) |_{t=0}}{Γ(α)} , c_{2} = \frac{I_{0+}^{2−α}u(t) |_{t=0}}{Γ(α − 1)} .

In view of u ∈ dom(L) ∩ Ker(P), D_{0+}^{α−1}u(t) |_{t=0}= 0, I_{0+}^{2−α}u(t) |_{t=0}= 0, we have c_{1} = c_{2} = 0, thus

(K_{P} L)u(t) = u(t).

This shows that K_{P} = (L|_{dom(L)∩Ker(P)})^{−1}.

EJQTDE, 2010 No. 37, p. 8 Again since

∥K_{P} y∥_{C^{α−1}} = ∥I_{0+}^{α}y∥_{C^{α−1}}
= ∥D_{0+}^{α−1}I_{0+}^{α}y∥_{∞} + ∥I_{0+}^{α}y∥_{∞}
= ∥I_{0+}^{1}y∥_{∞} + ∥I_{0+}^{α}y∥_{∞}
= ∥∫_{0}^{t} y(s)ds∥_{∞} + \frac{1}{Γ(α)} ∥∫_{0}^{t}(t − s)^{α−1}y(s)ds∥_{∞}
≤ ∥y∥_{1} + \frac{1}{Γ(α)}∥y∥_{1}
= (1 + \frac{1}{Γ(α)}) ∥y∥_{1}.

The proof is complete. ¶ Lemma 3.4 [7] For given e ∈ L^{1}[0, 1], K_{P} (I −Q)N : Y → Y is completely continuous. Theorem 3.1 Let f : [0, 1] × R^{2} → R be continuous. Assume that (A_{1}) There exist functions a, b, c, r ∈ L^{1}[0, 1], and constant θ ∈ [0, 1) such that for all

(x, y) ∈ R^{2}, t ∈ [0, 1] either

|f(t, x, y)| ≤ a(t)|x| + b(t)|y| + c(t)|y|^{θ} + r(t)  (3.4)

or else

|f(t, x, y)| ≤ a(t)|x| + b(t)|y| + c(t)|x|^{θ} + r(t).  (3.5)

(A_{2}) There exists constant M > 0 such that for u ∈ dom(L), if |D_{0+}^{α−1}u(t)| > M for

all t ∈ [0, 1], then

∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} [f(s, u(s), D_{0+}^{α−1}u(s)) + e(s)] ds ≠ 0.

(A_{3}) There exists M^{∗} > 0 such that for any c ∈ R, if |c| > M^{∗}, then either

c (∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} [f(s, cs^{α−1}, cΓ(α)) + e(s)] ds) < 0.

or else

c (∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} [f(s, cs^{α−1}, cΓ(α)) + e(s)] ds) > 0.

Then, for every e ∈ L^{1}[0, 1], the boundary value problem (1.1), (1.2) has at least one solution in C^{α−1}[0, 1] provided that

∥a∥_{1} + ∥b∥_{1} < \frac{1}{\overline{C}} ,

where \overline{C} = Γ(α) + 2 + \frac{1}{Γ(α)}.

EJQTDE, 2010 No. 37, p. 9 Proof. Set

Ω_{1} = {u ∈ dom(L)\Ker(L)|Lu = λNu for some λ ∈ (0, 1)}.

Then for u ∈ Ω_{1}, Lu = λNu, and Nu ∈ Im(L), hence

∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} [f(s, u(s), D_{0+}^{α−1}u(s)) + e(s)] ds = 0.

Thus, from (A_{2}), there exists t_{0} ∈ [0, 1] such that |D_{0+}^{α−1}u(t) |_{t=t_{0}} | ≤ M. For u ∈ Ω_{1}, there holds D_{0+}^{α−1}u ∈ C^{α−1}[0, 1], D_{0+}^{α}u ∈ (L^{1}(0, 1)). By Lemma 2.3,

DD_{0+}^{α−1}u(t) = D_{0+}^{α}u(t).

So,

D_{0+}^{α−1}u(t) |_{t=0}= D_{0+}^{α−1}u(t) |_{t=t_{0}} −I_{0+}^{1}D_{0+}^{α}u(t) |_{t=t_{0}} .

Thus,

|D_{0+}^{α−1}u(t) |_{t=0}| ≤ M + ∥D_{0+}^{α}u(t)∥_{1} = M + ∥Lu∥_{1} ≤ M + ∥Nu∥_{1}.  (3.6)

Again for u ∈ Ω_{1}, u ∈ dom(L)\Ker(L), then (I − P)u ∈ dom(L) ∩ Ker(P) and LP u = 0. Thus from Lemma 3.3, we have

∥(I − P)u∥_{C^{α−1}} = ∥K_{P} L(I − P)u∥_{C^{α−1}}
≤ (1 + \frac{1}{Γ(α)}) ∥L(I − P)u∥_{1}
= (1 + \frac{1}{Γ(α)}) ∥Lu∥_{1}
≤ (1 + \frac{1}{Γ(α)}) ∥Nu∥_{1}.  (3.7)

From (3.6), (3.7), we have

∥u∥_{C^{α−1}} ≤ ∥P u∥_{C^{α−1}} + ∥(I − P)u∥_{C^{α−1}}
= (Γ(α) + 1) |D_{0+}^{α−1}u(t) |_{t=0}| + ∥(I − P)u∥_{C^{α−1}}
≤ (Γ(α) + 1)(M + ∥Nu∥_{1}) + (1 + \frac{1}{Γ(α)}) ∥Nu∥_{1}
= (Γ(α) + 1)M + (Γ(α) + 2 + \frac{1}{Γ(α)}) ∥Nu∥_{1}
= (Γ(α) + 1)M + \overline{C}∥Nu∥_{1}.  (3.8)

If (3.4) holds, then from (3.8), we get

∥u∥_{C^{α−1}} ≤ \overline{C} [∥a∥_{1}∥u∥_{∞} + ∥b∥_{1}∥D_{0+}^{α−1}u∥_{∞}
+ ∥c∥_{1}∥D_{0+}^{α−1}u∥_{∞}^{θ} + ∥r∥_{1} + ∥e∥_{1}] + (Γ(α) + 1)M.  (3.9)

Thus, from ∥u∥_{∞} ≤ ∥u∥_{C^{α−1}} and (3.9), we obtain

∥u∥_{∞} ≤ \frac{\overline{C}}{1 − \overline{C}∥a∥_{1}} [∥b∥_{1}∥D_{0+}^{α−1}u∥_{∞}
+ ∥c∥_{1}∥D_{0+}^{α−1}u∥_{∞}^{θ} + ∥r∥_{1} + ∥e∥_{1} + \frac{(Γ(α) + 1)M}{\overline{C}} ] .  (3.10)

Again, from (3.9), (3.10), one has

∥D_{0+}^{α−1}u∥_{∞} ≤ \frac{\overline{C}∥c∥_{1}}{1 − \overline{C}(∥a∥_{1} + ∥b∥_{1})}∥D_{0+}^{α−1}u∥_{∞}^{θ}
+\frac{\overline{C}}{1 − \overline{C}(∥a∥_{1} + ∥b∥_{1})} [∥r∥_{1} + ∥e∥_{1} + \frac{(Γ(α) + 1)M}{\overline{C}} ] . (3.11)

Since θ ∈ [0, 1), from the above last inequality, there exists M_{1} > 0 such that

∥D_{0+}^{α−1}u∥_{∞} ≤ M_{1},

thus from (3.10) and (3.11), there exists M_{2} > 0 such that

∥u∥_{∞} ≤ M_{2},

hence ∥u∥_{C^{α−1}} = ∥u∥_{∞} + ∥D_{0+}^{α−1}u∥_{∞} ≤ M_{1} + M_{2}. Therefore Ω_{1} ⊂ Y is bounded. If (3.5) holds, similar to the above argument, we can prove that Ω_{1} is bounded too. Let

Ω_{2} = {u ∈ Ker(L)|Nu ∈ Im(L)}.

For u ∈ Ω_{2}, there is u ∈ Ker(L) = {u ∈ dom(L)|u = ct^{α−1}, c ∈ R, t ∈ [0, 1]}, and Nu ∈ Im(L), thus

∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} [f(s, cs^{α−1}, cΓ(α)) + e(s)] ds = 0.

From (A_{2}), we get |c| ≤ \frac{M}{Γ(α)}, thus Ω_{2} is bounded in Y.

Next, according to the condition (A_{3}), for any c ∈ R, if |c| > M^{∗}, then either

c (∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} [f(s, cs^{α−1}, cΓ(α)) + e(s)] ds) < 0.  (3.12)

or else

c (∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} [f(s, cs^{α−1}, cΓ(α)) + e(s)] ds) > 0.  (3.13)

If (3.12) holds, set

Ω_{3} = {u ∈ Ker(L)| − λV u + (1 − λ)QNu = 0, λ ∈ [0, 1]},

EJQTDE, 2010 No. 37, p. 11 here V : Ker(L) → Im(Q) is the linear isomorphism given by V (ct^{α−1}) = ct^{k}, ∀c ∈ R, t ∈ [0, 1]. For u = c_{0}t^{α−1} ∈ Ω_{3},

λc_{0}t^{k} = (1 − λ) (∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} [f(s, c_{0}s^{α−1}, c_{0}Γ(α)) + e(s)] ds) .

If λ = 1, then c_{0} = 0. Otherwise, if |c_{0}| > M^{∗}, in view of (3.12), one has

c_{0}(1 − λ) (∑_{i=1}^{m−2} β_{i} ∫_{η_{i}}^{1} [f(s, c_{0}s^{α−1}, c_{0}Γ(α)) + e(s)] ds) < 0,

which contradicts to λc_{0}^{2} ≥ 0. Thus Ω_{3} ⊂ {u ∈ Ker(L) | u = ct^{α−1}, |c| ≤ M^{∗}} is bounded in Y.

If (3.13) holds, then define the set

Ω_{3} = {u ∈ Ker(L)|λV u + (1 − λ)QNu = 0, λ ∈ [0, 1]},

here V as in above. Similar to above argument, we can show that Ω_{3} is bounded too.

In the following, we shall prove that all conditions of Theorem 1.1 are satisfied. Set Ω be a bounded open set of Y such that ⋃_{i=1}^{3} \overline{Ω}_{i} ⊂ Ω. By Lemma 3.4, K_{P} (I − Q)N : \overline{Ω} → Y is compact, thus N is L−compact on \overline{Ω}. Then by above arguments, we have

(i) Lx ≠ λNx for every (x, λ) ∈ [(dom(L)\Ker(L)) ∩ ∂Ω] × (0, 1);

(ii) Nx ∉ Im(L) for every x ∈ Ker(L) ∩ ∂Ω.

Finally, we will prove that (iii) of Theorem 1.1 is satisfied. Let H(u, λ) = ±λV u + (1 − λ)QNu. According to the above argument, we know

H(u, λ) ≠ 0, for all u ∈ Ker(L) ∩ ∂Ω.

Thus, by the homotopy property of degree

deg(QN|_{Ker(L)}, Ω ∩ Ker(L), 0) = deg(H(·, 0), Ω ∩ Ker(L), 0)
= deg(H(·, 1), Ω ∩ Ker(L), 0)
= deg(±V, Ω ∩ Ker(L), 0) ≠ 0.

Then by Theorem 1.1, Lu = Nu has at least one solution in dom(L) ∩ \overline{Ω}, so that the problem (1.1), (1.2) has one solution in C^{α−1}[0, 1]. The proof is complete. ¶

EJQTDE, 2010 No. 37, p. 12

4 An example

Example 4.1 Consider the boundary value problem

D0+^{\frac{3}{2}}u(t) = \frac{1}{10} sin (u(t)) + \frac{1}{10}D0+^{\frac{1}{2}}u(t) + 3 sin (D0+^{\frac{1}{2}}u(t))^{\frac{1}{3}} + 1 + cos^{2} t, (4.1)
I0+^{\frac{1}{2}}u(0) = 0, D_{0+}^{α−1}u(1) = 6D0+^{\frac{1}{2}}u (\frac{1}{3}) − 5D0+^{\frac{1}{2}}u (\frac{1}{2}) .  (4.2)

Let β_{1} = 6, β_{2} = −5, η_{1} = \frac{1}{3}, η_{2} = \frac{1}{2} and

f(t, x, y) = \frac{sin x}{10} + \frac{y}{10} + 3 sin (y^{\frac{1}{3}} ) , e(t) = 1 + cos^{2} t,

then

β_{1} + β_{2} = 1. |f(t, x, y)| ≤ \frac{|x|}{10} + \frac{|y|}{10} + 3|y|^{\frac{1}{3}} .

Again, taking a(t) = b(t) ≡ \frac{1}{10}, then

∥a∥_{1} + ∥b∥_{1} = \frac{1}{5} < \frac{1}{Γ (\frac{3}{2}) + 2 + _{\frac{1}{Γ(\frac{3}{2})}}} ≈ \frac{1}{4}.

Finally, taking M = 52, for any u ∈ C^{\frac{1}{2}} ⋂ I0+^{\frac{3}{2}}(L^{1}[0, 1]), assume |D0+^{\frac{1}{2}}u(t)| > M holds for any t ∈ [0, 1]. Since the continuity of D0+^{\frac{1}{2}}u, then either D0+^{\frac{1}{2}}u(t) > M or D0+^{\frac{1}{2}}u(t) < −M holds for any t ∈ [0, 1]. If D0+^{\frac{1}{2}}u(t) > M holds for any t ∈ [0, 1], then

f (t, u(t), D0+^{\frac{1}{2}}u(t)) + e(t) ≥ \frac{M − 21}{10} > 0,

so

6 ∫ 1_{\frac{1}{36}} [f (s, u(s), D0+^{\frac{1}{2}}u(s)) + e(s)] ds − 5 ∫ 1_{\frac{1}{25}} [f (s, u(s), D0+^{\frac{1}{2}}u(s)) + e(s)] ds

> ∫ 1_{\frac{1}{36}} [f (s, u(s), D0+^{\frac{1}{2}}u(s)) + e(s)] ds

≥ \frac{35(M − 21)}{360} > 0.

If D0+^{\frac{1}{2}}u(t) < −M hold for any t ∈ [0, 1], then

f (t, u(t), D0+^{\frac{1}{2}}u(t)) + e(t) ≤ \frac{51 − M}{10} < 0,

so

6 ∫ 1_{\frac{1}{36}} [f (s, u(s), D0+^{\frac{1}{2}}u(s)) + e(s)] ds − 5 ∫ 1_{\frac{1}{25}} [f (s, u(s), D0+^{\frac{1}{2}}u(s)) + e(s)] ds

< ∫ 1_{\frac{1}{36}} [f (s, u(s), D0+^{\frac{1}{2}}u(s)) + e(s)] ds

≤ \frac{35(51 − M)}{360} < 0.

Thus, the condition (A_{2}) holds. Again, taking M^{∗} = \frac{52}{Γ(3/2)}, for any c ∈ R, if |c| > M^{∗},

we have

c (6 ∫ 1_{\frac{1}{36}} [f (s, cs^{\frac{1}{2}} , cΓ (\frac{3}{2})) + e(s)] ds − 5 ∫ 1_{\frac{1}{25}} [f (s, cs^{\frac{1}{2}} , cΓ (\frac{3}{2})) + e(s)] ds) > 0.

So, the condition (A_{3}) holds. Thus, with Theorem 3.1, the boundary value problem

(4.1), (4.2) has at least one solution in C^{\frac{1}{2}} [0, 1].

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(Received December 20, 2009)