The Existence of Solutions to Integral Boundary Value Problems of ...

Hindawi Journal of Function Spaces Volume 2017, Article ID 2785937, 7 pages https://doi.org/10.1155/2017/2785937

Research Article The Existence of Solutions to Integral Boundary Value Problems of Fractional Differential Equations at Resonance Yumei Zou1 and Guoping He2 1

Department of Statistics and Finance, Shandong University of Science and Technology, Qingdao 266590, China Shandong Academy of Sciences, Jinan 250014, China

2

Correspondence should be addressed to Yumei Zou; [email protected] Received 10 April 2017; Accepted 10 September 2017; Published 2 November 2017 Academic Editor: Xinguang Zhang Copyright © 2017 Yumei Zou and Guoping He. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. This paper deals with the integral boundary value problems of fractional differential equations at resonance. By Mawhin’s coincidence degree theory, we present some new results on the existence of solutions for a class of differential equations of fractional order with integral boundary conditions at resonance. An example is also included to illustrate the main results.

1. Introduction In this paper, we are concerned with the following integral boundary value problem for nonlinear fractional differential equation: 𝑝

𝑝−1

𝑝−2

−𝐷0+ 𝑥 (𝑡) = 𝑓 (𝑡, 𝑥 (𝑡) , 𝐷0+ 𝑥 (𝑡) , 𝐷0+ 𝑥 (𝑡)) , 𝑡 ∈ (0, 1) , 𝑥 (0) = 𝑥󸀠 (0) = 0,

(1)

1

𝑥 (1) = ∫ 𝑥 (𝑡) 𝑑𝐴 (𝑡) , 0

𝑝

where 2 < 𝑝 < 3, 𝐷0+ is the standard Riemann-Liouville differentiation, 𝑓 : [0, 1]×R3 → R, and 𝑓 satisfies Carath´eodory conditions; 𝐴(𝑡) is right continuous on [0, 1) and left con1 tinuous at 𝑡 = 1; ∫0 𝑥(𝑡)𝑑𝐴(𝑡) denotes the Riemann-Stieltjes integrals of 𝑥 with respect to 𝐴. Our problem are at resonance, in the sense that, under the integral boundary conditions, we 𝑝 study the linear equation −𝐷0+ 𝑥(𝑡) = 0, 𝑡 ∈ (0, 1), which has nontrivial solutions. Recently, fractional differential equations have received considerable attentions not only because of a generalization of ordinary differential equations but also because they have

played a significant role in science, engineering, economy, and other fields; see, for example, [1–3]. When 𝐴(𝑡) ≡ 0, problem (1) is nonresonant. In [4], the authors studied the existence of positive solutions for the nonresonant case by Krasnosel’skii’s fixed point theorem. In [5], the author investigated the uniqueness of solutions for the nonresonant case by use of the 𝑢0 -positive operator under a Lipschitz condition on 𝑓. In present, many papers are devoted to the integral boundary value problem for fractional differential equation under nonresonance conditions; see [4–20]. On the other hand, there are some papers studying integral boundary value problem for differential equation under resonant conditions; we refer the reader to [21–29]. Motivated by the above results, in this paper, we consider the existence of solutions for the resonance integral boundary value problem (1) under nonlinear growth restriction of 𝑓. Our method is based upon the coincidence degree theorem of Mawhin. Now, we recall the essentials of the coincidence degree theory. Let 𝑋 and 𝑌 be real Banach spaces, and let 𝐿 : dom 𝐿 ⊂ 𝑋 → 𝑌 be a Fredholm operator of index zero. If 𝑃 : 𝑋 → 𝑋 and 𝑄 : 𝑌 → 𝑌 are continuous projectors such that Im 𝑃 = Ker 𝐿, Ker 𝑄 = Im 𝐿, 𝑋 = Ker 𝐿 ⊕ Ker 𝑃, and 𝑌 = Im 𝐿 ⊕ Im 𝑄, then the inverse operator of 𝐿|dom 𝐿∩Ker 𝑃 :

2

Journal of Function Spaces

dom 𝐿 ∩ Ker 𝑃 → Im 𝐿 exists and is denoted by 𝐾𝑃 (generalized inverse operator of 𝐿). If Ω is an open bounded subset of 𝑋 such that dom 𝐿 ∩ Ω ≠ 0, the mapping 𝑁 : 𝑋 → 𝑌 will be called 𝐿-compact on Ω, if 𝑄𝑁(Ω) is bounded and 𝐾𝑃 (𝐼 − 𝑄)𝑁 : Ω → 𝑋 is compact. The abstract equation 𝐿𝑥 = 𝑁𝑥 is shown to be solvable in view of Theorem IV.13 [30]. Theorem 1 (see [30]). Let 𝐿 be a Fredholm operator of index zero and let 𝑁 be 𝐿-compact on Ω. Assume the following conditions are satisfied:

Lemma 4 (see [1]). Assume that 𝑢 ∈ 𝐿1 [0, 1]. If 𝛼, 𝛽, 𝑝 > 0, then 𝛽

𝛼+𝛽

𝛼 𝐼0+ ) 𝑢 (𝑡) = (𝐼0+ 𝑢) (𝑡) , (𝐼0+ 𝑝

𝑁−𝑝

Lemma 5 (see [1]). Assume that 𝐼0+ 𝑢 ∈ 𝐴𝐶𝑁−1 [0, 1], 𝑝 > 0. Then one has 𝑝

𝑝

(𝐼0+ 𝐷0+ ) 𝑢 (𝑡) 𝑁−1

𝑡𝑝−𝑘−1 𝑑𝑁−𝑘−1 𝑛−𝑝 ( 𝑁−𝑘−1 𝐼0+ 𝑢) (0) , Γ (𝑝 − 𝑘) 𝑑𝑡 𝑘=0

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

= 𝑢 (𝑡) − ∑

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

Then the equation 𝐿𝑥 = 𝑁𝑥 has at least one solution in dom 𝐿 ∩ Ω.

where 𝑁 = [𝑝] + 1. Lemma 6 (see [4]). Let 𝜎 ∈ 𝐶(0, 1) ∩ 𝐿(0, 1) and 2 < 𝑝 ≤ 3, then the unique solution of 𝑝

𝐷0+ 𝑥 (𝑡) + 𝜎 (𝑡) = 0,

Throughout this paper, we always suppose that

(𝐻2 ) 𝑓 : [0, 1] × R × R × R → R satisfies the Carath´eodory conditions; that is, 𝑓(⋅, 𝑢, V, 𝑤) is measurable for each fixed (𝑢, V, 𝑤) ∈ R × R × R, 𝑓(𝑡, ⋅, ⋅, ⋅) is continuous for a.e. 𝑡 ∈ [0, 1], and for each 𝑟 > 0, there exists Φ𝑟 ∈ 𝐿∞ [0, 1] such that |𝑓(𝑡, 𝑢, V, 𝑤)| ≤ Φ𝑟 (𝑡) for all |𝑢|, |V|, and |𝑤| ≤ 𝑟 and for a.e. 𝑡 ∈ [0, 1].

is given by 1

𝑥 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝜎 (𝑠) 𝑑𝑠, where 𝐺(𝑡, 𝑠) is Green’s function given by 𝐺 (𝑡, 𝑠)

Definition 2 (see [1]). The Riemann-Liouville fractional integral of order 𝑝 > 0 of a function 𝑓 : (0, ∞) → R is given by (2)

(1 − 𝑠)𝑝−1 𝑡𝑝−1 − (𝑡 − 𝑠)𝑝−1 { { , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1, { { Γ (𝑝) ={ { (1 − 𝑠)𝑝−1 𝑡𝑝−1 { { , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1. Γ (𝑝) {

Definition 3 (see [1]). The Riemann-Liouville fractional derivative of order 𝑝 > 0 of a continuous function 𝑓 : (0, ∞) → R is given by 𝑝

𝑓 (𝑠) 1 𝑑 𝑛 𝑡 𝑑𝑠, ( ) ∫ 𝑝−𝑛+1 Γ (𝑛 − 𝑝) 𝑑𝑡 0 (𝑡 − 𝑠)

(8)

Lemma 7 (see [4]). The function 𝐺(𝑡, 𝑠) defined by (8) satisfies 𝐺(𝑡, 𝑠) > 0 and 𝑡, 𝑠 ∈ (0, 1). We use the classical Banach space 𝐶[0, 1] with the norm ‖𝑥‖∞ = max𝑡∈[0,1] |𝑥(𝑡)| and 𝐿1 [0, 1] with the norm ‖𝑥‖1 = 1

∫0 |𝑥(𝑡)|𝑑𝑡. We also use the Banach space 𝑋

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

𝐷0+ 𝑓 (𝑡) =

(7)

0

In this section, first we provide recall some necessary basic definitions and lemmas of the fractional calculus theory, which will be used in this paper. For more details, we refer to books [1–3] for details.

𝑡 1 ∫ (𝑡 − 𝑠)𝑝−1 𝑓 (𝑠) 𝑑𝑠, Γ (𝑝) 0

(6)

𝑥 (1) = 0

2. Preliminaries and Lemmas

𝑝

𝑡 ∈ (0, 1) ,

𝑥 (0) = 𝑥󸀠 (0) = 0,

1

(𝐻1 ) ∫0 𝑡𝑝−1 𝑑𝐴(𝑡) = 1, 𝜅 = ∫0 𝑡𝑝 𝑑𝐴(𝑡) − 1 ≠ 0;

𝐼0+ 𝑓 (𝑡) =

(5)

𝑡 ∈ [0, 1] ,

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

1

(4)

𝑝

(𝐷0+ 𝐼0+ 𝑢) (𝑡) = 𝑢 (𝑡) .

𝑝−1

𝑝−2

𝑋 = {𝑥 : [0, 1] 󳨀→ R | 𝑥, 𝐷0+ 𝑥, 𝐷0+ 𝑥 ∈ 𝐶 [0, 1]} 𝑝−1

(9)

𝑝−2

with the norm ‖𝑥‖𝑋 = max{‖𝑥‖∞ , ‖𝐷0+ 𝑥‖∞ , ‖𝐷0+ 𝑥‖∞ }. Define 𝐿 : dom 𝐿 ⊂ 𝑋 → 𝑌 = 𝐿1 [0, 1] and 𝑁 : 𝑋 → 𝑌 as follows: 𝑝

(3)

where 𝑛 − 1 ≤ 𝑝 < 𝑛, provided that the right-hand side is pointwise defined on (0, ∞).

(𝐿𝑥) (𝑡) = −𝐷0+ 𝑥 (𝑡) , 𝑝−1

𝑝−2

(𝑁𝑥) (𝑡) = 𝑓 (𝑡, 𝑥 (𝑡) , 𝐷0+ 𝑥 (𝑡) , 𝐷0+ 𝑥 (𝑡)) , 𝑡 ∈ [0, 1] ,

(10)


3 𝑝

where dom 𝐿 = {𝑥 ∈ 𝑋 | 𝐷0+ 𝑥 ∈ 𝑌, 𝑥(0) = 𝑥󸀠 (0) = 0, 𝑥(1) = 1

∫0 𝑥(𝑡)𝑑𝐴(𝑡)}. Then integral boundary value problems (1) can be rewritten as follows: (𝐿𝑥) (𝑡) = (𝑁𝑥) (𝑡) ,

𝑥 ∈ dom 𝐿.

(11)

By a simple computation, we can obtain that 𝑢 ∈ dom 𝐿 and 𝐿𝑢 = 𝑦; that is, 𝑦 ∈ Im 𝐿. Clearly, dim Ker 𝐿 = 1 and Im 𝐿 is closed. It follows from 𝑌1 = 𝑌 \ Im 𝐿 that 𝑌1 = {𝑦1 ∈ 𝑌 : 𝑦1 =

Lemma 8. The operator 𝐿 is a Fredholm operator of index zero.

(12)

By Lemma 5, 𝑢 ∈ Ker 𝐿 means that 𝑢(𝑡) = 𝑐1 𝑡𝑝−1 + 𝑐2 𝑡𝑝−2 + 𝑐3 𝑡𝑝−3 . It follows from 𝑢(0) = 𝑢󸀠 (0) = 0 that 𝑐2 = 𝑐3 = 0. That is, Ker 𝐿 = {𝑐𝑡𝑝−1 : 𝑐 ∈ R}. Now we prove 1

Im 𝐿 = {𝑦 ∈ 𝑌 : ∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠

0

0

(13)

− ∫ ∫ (𝑡 − 𝑠)

𝑝

(14)

By the boundary condition, we obtain 𝑐2 = 𝑐3 = 0, 1

∫ 𝑢 (𝑡) 𝑑𝐴 (𝑡) 0

0

=−

𝑑𝐴 (𝑡)

(15)

∫ (1 − 𝑠) 0

1

𝑡

0

0

𝑝−1

∫0 ∫0 (𝑡 − 𝑠)

𝑦 (𝑠) 𝑑𝑠 𝑑𝐴 (𝑡) . 𝑝−1

1

𝑡

1

0

0

0

1

𝑡

0

0

which shows that 𝑦−𝑦1 ∈ Im 𝐿. This together with 𝑌1 ∩Im 𝐿 = {𝜃} implies that 𝑌 = 𝑌1 ⊕ Im 𝐿. Note that dim 𝑌1 = 1 and thus codim Im 𝐿 = 1. Therefore, 𝐿 is a Fredholm operator of index zero. The proof is completed.

(𝑄𝑦) (𝑡) = 1

𝑡

0

0

𝑡 1 ∫ (𝑡 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 Γ (𝑝) 0

𝑡𝑝−1 1 + ∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠. Γ (𝑝) 0

=

1 𝑝 (∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 𝜅 0 𝑝−1

(20)

(21)

𝑦 (𝑠) 𝑑𝑠 𝑑𝐴 (𝑡)) .

𝑡𝑝−1 1 ∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 Γ (𝑝) 0 −

𝑦(𝑠)𝑑𝑠 =

𝑦(𝑠)𝑑𝑠 𝑑𝐴(𝑡), let

𝑢 (𝑡) = −

(19)

0

On the other hand, if 𝑦 ∈ 𝑌 satisfies ∫0 (1 − 𝑠) 𝑡

1 𝑝 (∫ (1 − 𝑠)𝑝−1 𝑑𝑠 𝜅 0

1

1

1

⋅ (𝑦 (𝑠) − 𝑦1 (𝑠)) 𝑑𝑠 𝑑𝐴 (𝑡) = (1

(𝐾𝑝 𝑦) (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝑦 (𝑠) 𝑑𝑠 (16)

= ∫ ∫ (𝑡 − 𝑠)

0

Clearly, Im 𝑃 = Ker 𝐿 and Ker 𝑄 = Im 𝐿. The generalized inverse operator of 𝐿, 𝐾𝑃 : Im 𝐿 → dom 𝐿 ∩ Ker 𝑃 can be defined by

𝑦 (𝑠) 𝑑𝑠 𝑝−1

0

− ∫ ∫ (𝑡 − 𝑠)

The above equalities imply that 𝑝−1

0

and 𝑄 : 𝑌 → 𝑌 by

1 1 ∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 + 𝑐1 . Γ (𝑝) 0

1

𝑡

(𝑃𝑦) (𝑡) = 𝑦 (1) 𝑡𝑝−1

1 𝑡 1 ∫ ∫ (𝑡 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 𝑑𝐴 (𝑡) + 𝑐1 , Γ (𝑝) 0 0

𝑢 (1) = −

1

Next, define the projections 𝑃 : 𝑋 → 𝑋 by

1 𝑡 1 =− ∫ ∫ (𝑡 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 𝑑𝐴 (𝑡) Γ (𝑝) 0 0

+ 𝑐1 ∫ 𝑡

1

∫ (1 − 𝑠)𝑝−1 (𝑦 (𝑠) − 𝑦1 (𝑠)) 𝑑𝑠 − ∫ ∫ (𝑡 − 𝑠)𝑝−1

⋅ 𝑦 (𝑠) 𝑑𝑠 − ∫ ∫ (𝑡 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 𝑑𝐴 (𝑡)) = 0

𝑢 (𝑡) = −𝐼0+ 𝑦 (𝑡) + 𝑐1 𝑡𝑝−1 + 𝑐2 𝑡𝑝−2 + 𝑐3 𝑡𝑝−3 .

𝑝−1

𝑦 (𝑠) 𝑑𝑠 𝑑𝐴 (𝑡)) , 𝑦 ∈ 𝑌} ,

− ∫ ∫ (𝑡 − 𝑠)𝑝−1 𝑑𝑠 𝑑𝐴 (𝑡))) ⋅ (∫ (1 − 𝑠)𝑝−1

𝑦 (𝑠) 𝑑𝑠 𝑑𝐴 (𝑡) = 0} .

In fact, if 𝑢 ∈ dom 𝐿 and 𝐿𝑢 = 𝑦, then by Lemma 5,

1

0

(18)

where 𝜅 = ∫0 𝑡𝑝 𝑑𝐴(𝑡) − 1 ≠ 0. In fact, for each 𝑦 ∈ 𝑌, we have

−

0

𝑝−1

0

𝑝−1

1

Ker 𝐿 = {𝑐𝑡𝑝−1 : 𝑐 ∈ R} .

𝑡

𝑡

− ∫ ∫ (𝑡 − 𝑠)

Proof. Firstly, we show that

1

1

1 𝑝 (∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 𝜅 0

(22)

𝑡 1 ∫ (𝑡 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠. Γ (𝑝) 0

In fact, if 𝑦 ∈ Im 𝐿, then 𝑝

𝐿𝐾𝑝 𝑦 = −𝐷0+ (− (17)

𝑡 1 ∫ (𝑡 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 Γ (𝑝) 0

𝑡𝑝−1 1 + ∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠) = 𝑦. Γ (𝑝) 0

(23)

4

Journal of Function Spaces For 𝑥 ∈ dom 𝐿 ∩ Ker 𝑃, 𝐿𝑥 = 𝑦, we have 𝑝

−𝐷0+ 𝑥 (𝑡) = 𝑦 (𝑡) ,

Then by Lemma 7, we obtain 1 󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩𝐾𝑝 𝑦󵄩󵄩󵄩 ≤ max 𝐺 (𝑡, 𝑠) 󵄩󵄩󵄩𝑦󵄩󵄩󵄩1 󵄩∞ Γ (𝑝) 𝑡,𝑠∈[0,1] 󵄩

𝑡 ∈ (0, 1) ,

𝑥 (0) = 𝑥󸀠 (0) = 0,

(24)

=

𝑥 (1) = 0. Then by Lemma 6, we obtain 𝐾𝑝 𝐿𝑥 = 𝑥 whenever 𝑥 ∈ dom 𝐿 ∩ Ker 𝑃. Using (21) and (22), we write 1

0

1 𝑡𝑝−1 (1 − 𝑡) ⋅ (∫ (1 − 𝑠)𝑝−1 𝑁𝑦 (𝑠) 𝑑𝑠 𝜅Γ (𝑝) 0 1

𝑡

0

0

𝑝−1

− ∫ ∫ (𝑡 − 𝑠)

(29)

󵄩 󵄩󵄩 𝑝−1 󵄩󵄩𝐷0+ 𝐾𝑝 𝑦󵄩󵄩󵄩 ≤ 2 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩1 , 󵄩∞ 󵄩 󵄩󵄩 𝑝−2 󵄩 󵄩󵄩𝐷0+ 𝐾𝑝 𝑦󵄩󵄩󵄩 ≤ 2 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩1 . 󵄩 󵄩∞ It follows that 1 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝐾𝑝 𝑦󵄩󵄩󵄩 ≤ max { , 2} 󵄩󵄩󵄩𝑦󵄩󵄩󵄩1 = 2 󵄩󵄩󵄩𝑦󵄩󵄩󵄩1 . 󵄩 󵄩𝑋 Γ (𝑝)

𝐾𝑝 (𝐼 − 𝑄) 𝑁𝑦 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝑁𝑦 (𝑠) 𝑑𝑠 −

1 󵄩󵄩 󵄩󵄩 󵄩𝑦󵄩 Γ (𝑝) 󵄩 󵄩1

(25)

(30)

The proof is completed.

3. Main Results 𝑁𝑦 (𝑠) 𝑑𝑠 𝑑𝐴 (𝑡)) .

By a standard method, we obtain the following lemma.

In this section, we will use Theorem 1 to prove the existence of solutions to IBVP (1). To obtain our main theorem, we need the following conditions:

Lemma 9. 𝐾𝑝 (𝐼 − 𝑄)𝑁 : 𝑌 → 𝑌 is completely continuous.

(𝐻3 ) There exist functions 𝜌, 𝛼, 𝛽, 𝛾 ∈ 𝐿1 [0, 1] such that, for all (𝑢, V, 𝑤) ∈ R3 , 𝑡 ∈ [0, 1],

Lemma 10. For 𝑦 ∈ 𝑌,

󵄨󵄨 󵄨 󵄨󵄨𝑓 (𝑡, 𝑢, V, 𝑤)󵄨󵄨󵄨 ≤ 𝜌 (𝑡) + 𝛼 (𝑡) |𝑢| + 𝛽 (𝑡) |V| + 𝛾 (𝑡) |𝑤| . (31)

1 󵄩󵄩 󵄩󵄩 󵄩 󵄩󵄩 󵄩󵄩𝐾𝑝 𝑦󵄩󵄩󵄩 ≤ 󵄩∞ Γ (𝑝) 󵄩󵄩𝑦󵄩󵄩1 , 󵄩

(𝐻4 ) There exists a constant 𝐵 > 0 such that either for each 𝑐 ∈ R : |𝑐| > 𝐵

󵄩󵄩 𝑝−1 󵄩 󵄩󵄩𝐷0+ 𝐾𝑝 𝑦󵄩󵄩󵄩 ≤ 2 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩1 , 󵄩 󵄩∞ 󵄩󵄩 𝑝−2 󵄩 󵄩󵄩𝐷0+ 𝐾𝑝 𝑦󵄩󵄩󵄩 ≤ 2 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩1 . 󵄩 󵄩∞

𝑐𝑄𝑁 (𝑐𝑡𝑝−1 ) > 0

(26)

or for each 𝑐 ∈ R : |𝑐| > 𝐵, 𝑐𝑄𝑁 (𝑐𝑡𝑝−1 ) < 0.

Moreover, (27)

𝑝−1

𝑡𝑝−1 1 (𝐾𝑝 𝑦) (𝑡) = ∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 Γ (𝑝) 0 𝑡 1 − ∫ (𝑡 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠, Γ (𝑝) 0 1

𝑡

0

0

𝑝−1 𝐷0+ (𝐾𝑝 𝑦) (𝑡) = ∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠 − ∫ 𝑦 (𝑠) 𝑑𝑠, (28) 1

0

𝑡

− ∫ (𝑡 − 𝑠) 𝑦 (𝑠) 𝑑𝑠. 0

(𝐻5 ) There exists a constant 𝑀 > 0 such that if |𝐷0+ 𝑥(𝑡)| > 𝑀 for all 𝑡 ∈ [0, 1], then 𝑄𝑁𝑥 ≠ 0. (𝐻6 ) There exists a constant 𝑀 > 0 such that if |𝐷0+ 𝑥(𝑡)|+ 𝑝−2 |𝐷0+ 𝑥(𝑡)| > 𝑀 for all 𝑡 ∈ [0, 1], then 𝑄𝑁𝑥 ≠ 0.

Proof. It is easy to see that

𝑝−2

(33) 𝑝−1

󵄩 󵄩󵄩 󵄩󵄩𝐾𝑝 𝑦󵄩󵄩󵄩 ≤ 2 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩1 . 󵄩𝑋 󵄩

𝐷0+ (𝐾𝑝 𝑦) (𝑡) = 𝑡 ∫ (1 − 𝑠)𝑝−1 𝑦 (𝑠) 𝑑𝑠

(32)

Theorem 11. Suppose (𝐻1 )–(𝐻5 ) hold. Then IBVP (1) has at least one solution in 𝑌, provided 1 󵄩 󵄩 󵄩 󵄩 ‖𝛼‖1 + 󵄩󵄩󵄩𝛽󵄩󵄩󵄩1 + 󵄩󵄩󵄩𝛾󵄩󵄩󵄩1 < . 5

(34)

Proof. Set Ω1 = {𝑥 ∈ dom 𝐿 \ Ker 𝐿 : 𝐿𝑥 = 𝜆𝑁𝑥 for some 𝜆 ∈ [0, 1]} .

(35)

Take 𝑥 ∈ Ω1 . Since 𝐿𝑥 = 𝜆𝑁𝑥, so 𝜆 ≠ 0, 𝑁𝑥 ∈ Im 𝐿 = Ker 𝑄, and hence 𝑄𝑁𝑥 = 0.

(36)


5

Thus, from (𝐻5 ), there exists 𝑡0 ∈ [0, 1] such that 󵄨 󵄨󵄨 𝑝−1 󵄨󵄨𝐷0+ 𝑥 (𝑡0 )󵄨󵄨󵄨 < 𝑀. 󵄨 󵄨

Let

Noticing that 𝑝−1

𝑝−1

𝑡

Ω2 = {𝑥 ∈ Ker 𝐿 : 𝑁𝑥 ∈ Im 𝐿} .

(37)

𝑝

𝐷0+ 𝑥 (𝑡) = 𝐷0+ 𝑥 (𝑡0 ) + ∫ 𝐷0+ 𝑥 (𝑠) 𝑑𝑠, 𝑡0

(38)

we obtain

(46)

Then for 𝑥 ∈ Ω2 , 𝑥 = 𝑐𝑡𝑝−1 for some 𝑐 ∈ R. So, 𝑄𝑁𝑥(𝑡) = 0. By (𝐻4 ), we have |𝑐| ≤ 𝐵. Therefore, Ω2 is bounded. We define the isomorphism 𝐽 : Ker 𝐿 → Im 𝑄 by 𝐽 (𝑐𝑡𝑝−1 ) = 𝑐.

(47)

If (32) holds, then let

𝑡 󵄨󵄨 𝑝−1 󵄨 󵄨󵄨𝐷0+ 𝑥 (𝑡)󵄨󵄨󵄨 ≤ 𝑀 + ∫ |𝑁𝑥 (𝑠)| 𝑑𝑠 ≤ 𝑀 + ‖𝑁𝑥‖1 . 󵄨 󵄨 𝑡

(39)

0

Observe that (𝐼 − 𝑃)𝑥 ∈ dom 𝐿 ∩ Ker 𝑃 for all 𝑥 ∈ Ω1 . Then by Lemma 10, 󵄩 󵄩 󵄩 󵄩 ‖(𝐼 − 𝑃) 𝑥‖𝑋 = 󵄩󵄩󵄩󵄩𝐾𝑝 𝐿 (𝐼 − 𝑃) 𝑥󵄩󵄩󵄩󵄩𝑋 = 󵄩󵄩󵄩󵄩𝐾𝑝 𝐿𝑥󵄩󵄩󵄩󵄩𝑋 ≤ 2 ‖𝐿𝑥‖1 ≤ 2 ‖𝑁𝑥‖1 , 󵄩 󵄩 𝑝−1 󵄩 󵄩󵄩 𝑝−1 󵄩󵄩𝐷0+ (𝐼 − 𝑃) 𝑥󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝐷0+ 𝐾𝑝 𝐿𝑥󵄩󵄩󵄩󵄩∞ ≤ 2 ‖𝐿𝑥‖1 󵄩∞ 󵄩 󵄩 ≤ 2 ‖𝑁𝑥‖1 .

(40)

(43)

𝐻 (𝑥, 𝜆) ≠ 0 for 𝑥 ∈ Ker 𝐿 ∩ 𝜕Ω.

(52)

Thus, by the homotopy property of degree deg ( 𝑄𝑁|Ker 𝐿 , Ker 𝐿 ∩ Ω, 0) = deg (𝐻 (⋅, 0) , Ker 𝐿 ∩ Ω, 0) (44)

Applying (𝐻3 ), we have 󵄩 󵄩 󵄩 󵄩 󵄩 𝑝−1 󵄩 ‖𝑥‖𝑋 ≤ 𝑀 + 5 (󵄩󵄩󵄩𝜌󵄩󵄩󵄩1 + ‖𝛼‖1 ‖𝑥‖∞ + 󵄩󵄩󵄩𝛽󵄩󵄩󵄩1 󵄩󵄩󵄩󵄩𝐷0+ 𝑥󵄩󵄩󵄩󵄩∞ 󵄩 󵄩 󵄩 𝑝−2 󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝛾󵄩󵄩󵄩1 󵄩󵄩󵄩󵄩𝐷0+ 𝑥󵄩󵄩󵄩󵄩∞ ) ≤ 𝑀 + 5 󵄩󵄩󵄩𝜌󵄩󵄩󵄩1 + 5 (‖𝛼‖1 + 󵄩󵄩󵄩𝛽󵄩󵄩󵄩1 (45)

Therefore, Ω1 is bounded.

(51)

where 𝐽 is as above. Similar to the above argument, we can show that Ω3 is bounded too. Next, we will prove that all the assumptions of Theorem 1 are satisfied. Let Ω be given any bounded open subset of 𝑌 such that ⋃3𝑖=1 Ω𝑖 ⊂ Ω. By Lemma 9, 𝐾𝑃 (𝐼 − 𝑄)𝑁 : Ω → 𝑌 is compact; thus 𝑁 is 𝐿-compact on Ω. Clearly, assumptions (i) and (ii) of Theorem 1 are fulfilled. At last, we will prove that (iii) of Theorem 1 is satisfied. Let 𝐻(𝑥, 𝜆) = ±𝜆𝐽𝑥 + (1 − 𝜆)𝑄𝑁𝑥. According to above argument, we know

that is, for all 𝑥 ∈ Ω1 ,

󵄩 󵄩 + 󵄩󵄩󵄩𝛾󵄩󵄩󵄩1 ) ‖𝑥‖𝑋 .

(50)

(42)

= 𝑀 + 5 ‖𝑁𝑥‖1 ;

‖𝑥‖𝑋 ≤ 𝑀 + 5 ‖𝑁𝑥‖1 .

𝑐 (1 − 𝜆) 𝑄𝑁 (𝑐𝑡𝑝−1 ) > 0

= {𝑥 ∈ Ker 𝐿 : −𝜆𝐽𝑥 + (1 − 𝜆) 𝑄𝑁𝑥 = 0, 𝜆 ∈ [0, 1]} ,

‖𝑥‖𝑋 ≤ ‖(𝐼 − 𝑃) 𝑥‖𝑋 + ‖𝑃𝑥‖𝑋

1 ≤ 2 ‖𝑁𝑥‖1 + (𝑀 + 3 ‖𝑁𝑥‖1 ) Γ (𝑝) Γ (𝑝)

If 𝜆 = 1, then 𝑐 = 0. Otherwise, if |𝑐| > 𝐵, in view of (𝐻4 ), one has

Ω3

so that

1 󵄩 󵄩 ‖𝑁𝑥‖1 + |𝑥 (1)| ⋅ 󵄩󵄩󵄩󵄩𝑡𝑝−1 󵄩󵄩󵄩󵄩𝑋 Γ (𝑝)

(49)

which contradicts 𝜆𝑐2 ≥ 0. Thus Ω3 is bounded. If (33) holds, then define the set

≤ 𝑀 + 3 ‖𝑁𝑥‖1 ,

≤

(48)

For 𝑥 = 𝑐𝑡𝑝−1 ∈ Ω3 ,

(41)

󵄨 󵄨 𝑝−1 󵄨 󵄨 𝑝−1 Γ (𝑝) |𝑥 (1)| = 󵄨󵄨󵄨󵄨𝐷0+ (𝑥 (1) 𝑡𝑝−1 )󵄨󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨𝐷0+ 𝑃𝑥 (𝑡)󵄨󵄨󵄨󵄨 󵄨 𝑝−1 󵄨 󵄨 󵄨 𝑝−1 ≤ 󵄨󵄨󵄨󵄨𝐷0+ 𝑥 (𝑡)󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨𝐷0+ ((𝐼 − 𝑃) 𝑥) (𝑡)󵄨󵄨󵄨󵄨

= {𝑥 ∈ Ker 𝐿 : 𝜆𝐽𝑥 + (1 − 𝜆) 𝑄𝑁𝑥 = 0, 𝜆 ∈ [0, 1]} .

𝜆𝑐 = − (1 − 𝜆) 𝑄𝑁 (𝑐𝑡𝑝−1 ) .

Using (39) and (41), we have

󵄨 󵄨 𝑝−1 = 󵄨󵄨󵄨󵄨𝐷0+ (𝑥 (𝑡) − ((𝐼 − 𝑃) 𝑥) (𝑡))󵄨󵄨󵄨󵄨

Ω3

= deg (𝐻 (⋅, 1) , Ker 𝐿 ∩ Ω, 0)

(53)

= deg (±𝐽, Ker 𝐿 ∩ Ω, 0) ≠ 0. Then by Theorem 1, 𝐿𝑥 = 𝑁𝑥 has at least one solution in dom 𝐿 ∩ Ω, so that IBVP (1) has a solution. The proof is completed. Theorem 12. Suppose (𝐻1 )–(𝐻4 ) and (𝐻6 ) hold. Then IBVP (1) has at least one solution in 𝑌, provided 󵄩 󵄩 󵄩 󵄩 ‖𝛼‖1 + 󵄩󵄩󵄩𝛽󵄩󵄩󵄩1 + 󵄩󵄩󵄩𝛾󵄩󵄩󵄩1 < Γ (𝑝 − 1) . (54)

6


Proof. As in the proof of Theorem 11, 𝑥 ∈ Ω1 , implies from (𝐻6 ), there exist 𝑡0 ∈ [0, 1] such that 󵄨󵄨 𝑝−1 󵄨 󵄨󵄨𝐷0+ 𝑥 (𝑡0 )󵄨󵄨󵄨 < 𝑀, 󵄨 󵄨 󵄨󵄨󵄨𝐷𝑝−2 𝑥 (𝑡 )󵄨󵄨󵄨 < 𝑀. 󵄨󵄨 0+ 0 󵄨󵄨

(55)

𝑡

𝑝−1

Therefore, Ω1 is bounded. The rest of the proof repeats that of Theorem 11. 𝑝

𝐷0+ 𝑥 (𝑡) = 𝐷0+ 𝑥 (𝑡0 ) + ∫ 𝐷0+ 𝑥 (𝑠) 𝑑𝑠, 𝑡0

𝑝−2 𝐷0+ 𝑥 (𝑡)

=

𝑝−2 𝐷0+ 𝑥 (𝑡0 )

+∫

𝑡

𝑡0

󵄩󵄩 󵄩󵄩 󵄩󵄩𝜌󵄩󵄩1 + 2𝑀 . 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 ‖𝛼‖1 + 󵄩󵄩𝛽󵄩󵄩1 + 󵄩󵄩𝛾󵄩󵄩1 − Γ (𝑝 − 1)

‖𝑥‖𝑋 ≤

(61)

Noticing that 𝑝−1

󵄩󵄩 󵄩󵄩 󵄩𝜌󵄩 + 2𝑀 + 󵄩 󵄩1 , Γ (𝑝 − 1)

Example 13. Consider the IBVP (56)

+

we obtain (57)

0

𝑥 (1) =

Hence, we have 𝑡 󵄨󵄨 𝑝−2 󵄨 󵄨 𝑝−2 󵄨󵄨 󵄨󵄨𝐷0+ 𝑥 (𝑡)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨𝐷0+ 󵄨󵄨 + ∫ 𝑥 (𝑡 ) 0 󵄨 󵄨 󵄨 󵄨 𝑡

0

󵄨󵄨 𝑝−1 󵄨 󵄨󵄨𝐷0+ 𝑥 (𝑠)󵄨󵄨󵄨 𝑑𝑠 󵄨 󵄨

𝑝−2

𝑝−2

1

(58)

𝐼3−𝑝 𝑥 (0) 𝑝−3 𝑡 Γ (𝑝 − 2)

≤

𝑝−2

𝑝−1

(59)

𝑡 1 𝑝−2 ∫ (𝑡 − 𝑠)𝑝−3 𝐷0+ 𝑥 (𝑠) 𝑑𝑠. Γ (𝑝 − 2) 0

1 (2𝑀 + ‖𝑁𝑥‖1 ) . Γ (𝑝 − 1)

󵄩 𝑝−1 󵄩 󵄩 𝑝−2 󵄩 ‖𝑥‖𝑋 = max {‖𝑥‖∞ , 󵄩󵄩󵄩󵄩𝐷0+ 𝑥󵄩󵄩󵄩󵄩∞ , 󵄩󵄩󵄩󵄩𝐷0+ 𝑥󵄩󵄩󵄩󵄩∞ } ≤ max {𝑀 1 + ‖𝑁𝑥‖1 , 2𝑀 + ‖𝑁𝑥‖1 , (2𝑀 + ‖𝑁𝑥‖1 )} Γ (𝑝 − 1) 2𝑀 1 ‖𝑁𝑥‖1 + Γ (𝑝 − 1) Γ (𝑝 − 1)

󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩 𝑝−1 󵄩 󵄩 󵄩 󵄩 𝑝−2 󵄩 󵄩󵄩𝜌󵄩󵄩1 + ‖𝛼‖1 ‖𝑥‖∞ + 󵄩󵄩󵄩𝛽󵄩󵄩󵄩1 󵄩󵄩󵄩󵄩𝐷0+ 𝑥󵄩󵄩󵄩󵄩∞ + 󵄩󵄩󵄩𝛾󵄩󵄩󵄩1 󵄩󵄩󵄩󵄩𝐷0+ 𝑥󵄩󵄩󵄩󵄩∞ ≤ Γ (𝑝 − 1) +

1 2𝑀 󵄩 󵄩 󵄩 󵄩 ≤ (‖𝛼‖1 + 󵄩󵄩󵄩𝛽󵄩󵄩󵄩1 + 󵄩󵄩󵄩𝛾󵄩󵄩󵄩1 ) ‖𝑥‖𝑋 Γ (𝑝 − 1) Γ (𝑝 − 1)

−

+ (60)

1 𝑝−1 𝐷 𝑥 (𝑡) 12 0+

2+𝜋 𝑝−1 > 0, if 𝐷0+ 𝑥 (𝑡) > 15; 48 𝑝−1

𝑡

1 󵄨 𝑝−2 󵄨 ∫ (𝑡 − 𝑠)𝑝−3 󵄨󵄨󵄨󵄨𝐷0+ 𝑥 (𝑠)󵄨󵄨󵄨󵄨 𝑑𝑠 Γ (𝑝 − 2) 0

𝑝−2

𝑝−2

𝑓 (𝑡, 𝑥 (𝑡) , 𝐷0+ 𝑥 (𝑡) , 𝐷0+ 𝑥 (𝑡)) ≤

Considering (57), (58), and (60), and applying (𝐻3 ), we get

≤

1 1 1 (63) sin 𝑢 + V + arctan 𝑤, 24 12 24 so that ‖𝛼‖1 = ‖𝛾‖1 = 1/24, ‖𝛽‖1 = 1/12, and ‖𝛼‖1 + ‖𝛽‖1 + ‖𝛾‖1 = 1/6 < 1/5, which verifies (𝐻2 ) and (34). 𝑝−1 Taking 𝑀 = 15, we have |𝐷0+ 𝑥(𝑡)| > 15, 𝑓 (𝑡, 𝑥 (𝑡) , 𝐷0+ 𝑥 (𝑡) , 𝐷0+ 𝑥 (𝑡)) ≥

Thus, by (58), |𝑥 (𝑡)| ≤

(5/2) ∫0 𝑥5/2 𝑑𝑡 = 5/7; thus (𝐻1 ) is satisfied. Let 𝑓 (𝑡, 𝑢, V, 𝑤) = 𝑡 +

= (𝐼0+ 𝐷0+ 𝑥) (𝑡) =

5 1 ∫ 𝑥 (𝑡) 𝑑𝑡. 2 0 1

In view of 𝑥(0) = 0, 𝐼3−𝑝 𝑥(0) = 0. From this together with Lemma 5, for 𝑥 ∈ dom 𝐿, we have 𝑝−2

(62)

Let 𝑝 = 5/2 and 𝐴(𝑡) = (5/2)𝑡, then (5/2) ∫0 𝑥3/2 𝑑𝑡 = 1 and

≤ 2𝑀 + ‖𝑁𝑥‖1 .

𝑥 (𝑡) = (𝐼0+ 𝐷0+ 𝑥) (𝑡) +

1 𝑝−2 arctan 𝐷0+ 𝑥 (𝑡) , 24

𝑥 (0) = 𝑥󸀠 (0) = 0,

𝑡

󵄨󵄨 𝑝−1 󵄨 󵄨󵄨𝐷0+ 𝑥 (𝑡)󵄨󵄨󵄨 ≤ 𝑀 + ∫ |𝑁𝑥 (𝑠)| 𝑑𝑠 ≤ 𝑀 + ‖𝑁𝑥‖1 . 󵄨 󵄨 𝑡

1 1 𝑝−1 sin 𝑥 (𝑡) + 𝐷0+ 𝑥 (𝑡) 24 12

5/2 𝑥 (𝑡) = 𝑡 + −𝐷0+

𝑝−1 𝐷0+ 𝑥 (𝑠) 𝑑𝑠,

50 + 𝜋 < 0, 48

𝑄𝑁𝑥 (𝑡) = 1

t

0

0

1 𝑝−1 𝐷 𝑥 (𝑡) 12 0+

(64)

𝑝−1

if 𝐷0+ 𝑥 (𝑡) < −15,

1 𝑝 (∫ (1 − 𝑠)𝑝−1 𝑁𝑥 (𝑠) 𝑑𝑠 𝜅 0

− ∫ ∫ (𝑡 − 𝑠)𝑝−1 𝑁𝑥 (𝑠) 𝑑𝑠 𝑑𝐴 (𝑡)) = 1

⋅ ∫ (1 − 𝑠)𝑝−1 𝑠𝑁𝑥 (𝑠) 𝑑𝑠 = 0

𝑝−1

𝑝 𝜅

(65)

1

𝑝 ∫ (1 − 𝑠)𝑝−1 𝜅 0

𝑝−2

⋅ 𝑠𝑓 (𝑠, 𝑥 (𝑠) , 𝐷0+ 𝑥 (𝑠) , 𝐷0+ 𝑥 (𝑠)) 𝑑𝑠 ≠ 0. Hence (𝐻5 ) holds. Finally, taking 𝐵 = 10, when |𝑐| > 10, 𝑝−1

𝑝−2

𝑐𝑓 (𝑡, 𝑐𝑡𝑝−1 , 𝑐𝐷0+ 𝑡𝑝−1 , 𝑐𝐷0+ 𝑡𝑝−1 ) ≥

𝑐2 50 + 𝜋 Γ (𝑝) − |𝑐| > 0, 12 48

(66)

then we obtain 𝑐𝑄𝑁(𝑐𝑡𝑝−1 ) > 0; that is, condition (𝐻4 ) is satisfied. It follows from Theorem 11 that IBVP (62) has at least one solution.


7

Conflicts of Interest The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

[15]

[16]

The project was supported by the National Natural Science Foundation of China (11371221, 11571207, and 51774197).

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