Existence Results for a Coupled System of Nonlinear Fractional Boundary Value Problems at Resonance

Abstract

Some new Banach spaces are established. Based on those new Banach spaces and by using the coincidence degree theory, we present the existence results for a coupled system of nonlinear fractional differential equations with multipoint boundary value conditions at resonance case.

1. Introduction

Fractional differential equations (FDEs) have been of great interest for the last three decades. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, and so forth. Therefore, the theory of FDEs has been developed very quickly. Many qualitative theories of FDEs have been obtained. Many important results can be found in [1–13] and the references cited therein.

However, there are few articles which consider the boundary value problems at resonance for a coupled system of nonlinear fractional differential equations. In [14], the authors investigated the existence and uniqueness of solutions for the multipoint boundary value problems for fractional differential equations of the form 𝐷^𝛼_𝑡𝑦(𝑡)=𝑓𝑡,𝑦(𝑡),𝐷^𝛽_𝑡𝑦(𝑡)(),𝑦(0),𝑡∈0,1=0,𝐷^𝛽_𝑡𝑦(1)−^𝑚−2_𝑖=1𝜍_𝑖𝐷^𝛽_𝑡𝑦𝜉_𝑖=𝑦₀, (1) where 1<𝛼≤2,  0<𝛽<1,  0<𝜉_𝑖<1(𝑖=1,2,…,𝑚−2),  𝜍_𝑖≥0, with 𝛾=∑^𝑚−2_𝑖=1𝜍_𝑖𝜉_𝑖^{𝛼−𝛽−1}<1 and 𝐷^𝛼_𝑡 represents the standard Riemann-Liouville fractional derivative. And the analysis relies on the Schauder fixed point theorem and the Banach contraction principle.

In [15], the following coupled system of nonlinear fractional differential equation boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)=𝑓𝑡,𝑣(𝑡),𝐷^𝛽−1₀₊𝑣(𝑡)𝐷,0<𝑡<1,_𝛽0+𝑣(𝑡)=𝑔𝑡,𝑢(𝑡),𝐷^𝛼−1₀₊𝑢(𝑡)𝑢(0)(0)(1),0<𝑡<1,=𝑣=0,𝑢=𝜎₁𝑢𝜂₁,𝑣(1)=𝜎₂𝑣𝜂₂ (2) was considered. Where 1<𝛼, 𝛽≤2,  0<𝜂₁,  𝜂₂<1, 𝜎₁,𝜎₂>0, 𝜎₁𝜂₁^𝛼−1=𝜎₂𝜂₂^𝛽−1, 𝐷 is the standard Riemann-Liouville fractional derivative and 𝑓,𝑔∶[0,1]×𝑅²→𝑅 are continuous. For more works about coupled system involving fractional differential equations, for instance, see [16, 17].

In this paper, we investigate the existence of solutions for the boundary value problem with the following form: 𝐷^𝛼₀₊𝑢(𝑡)=𝑓𝑡,𝑣(𝑡),𝐷^𝑝₀₊𝑢(𝑡(),𝐷),𝑡∈0,1_𝛽0+𝑣(𝑡)=𝑔𝑡,𝑢(𝑡),𝐷^𝑞₀₊𝑣(𝑡(),𝑢(0)),𝑡∈0,1=0,𝐷^𝑝₀₊𝑢(1)=^𝑚−2_𝑖=1𝑎_𝑖𝐷^𝑝₀₊𝑢𝜉_𝑖,𝑣(0)=0,𝐷^𝑞₀₊𝑣(1)=^𝑚−2_𝑖=1𝑏_𝑖𝐷^𝑞₀₊𝑣𝜂_𝑖, (3) where 1<𝛼, 𝛽≤2, 0<𝑝, 𝑞<1, 𝛼−𝑝−1, 𝛽−𝑞−1≥0, 𝑎_𝑖,𝑏_𝑖≥0, 0<𝜉_𝑖, and  𝜂_𝑖<1(𝑖=1,2,…,𝑚−2).Moreover, at resonance case, ^𝑚−2_𝑖=1𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−1}=1,^𝑚−2_𝑖=1𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−1}=1, (4) a sufficient condition for the existence of solutions is established by the coincidence degree continuation theorem.

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 and lemmas can be found in [4–6].

Definition 1. The fractional integral of order 𝛼>0 of a function 𝑓∶(0,∞)→𝑅 is given by 𝐼^𝛼₀₊𝑓(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑓(𝑠)𝑑𝑠, (5) provided the right side is pointwise defined on (0,∞).

Definition 2. The fractional derivative of order 𝛼>0 of a function 𝑓∶(0,∞)→𝑅 is given by 𝐷^𝛼₀₊𝑓(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑓(𝑠)𝑑𝑠, (6) where 𝑛=[𝛼]+1, provided the right side is pointwise defined on (0,∞).
It can be directly verified that the Riemann-Liouville fractional integration and fractional differentiation operators of the power functions 𝑡^𝛽 yield power functions of the same form. For 𝛼≥0,  𝛽>−1, there are 𝐼^𝛼₀₊𝑡^𝛽=Γ𝛽+1Γ𝑡𝛽+𝛼+1^𝛽+𝛼,𝐷^𝛼₀₊𝑡^𝛽=Γ𝛽+1Γ𝑡𝛽−𝛼+1^𝛽−𝛼. (7)

Lemma 3. Assume that 𝛼>0, 𝑦∈𝐶(0,1)∩𝐿¹[0,1], then the differential equation 𝐷^𝛼₀₊𝑦(𝑡)=0 has solutions 𝑦(𝑡)=𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛, (8) where 𝑐_𝑖∈𝑅,  𝑖=1,2,…,𝑛,  𝑛=[𝛼]+1.

Lemma 4. Assume that 𝑦∈𝐶(0,1)∩𝐿¹[0,1] and 𝐷^𝛼₀₊𝑦∈𝐶(0,1)∩𝐿¹[0,1], then 𝐼^𝛼₀₊𝐷^𝛼₀₊𝑦(𝑡)=𝑦(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛, where 𝑐_𝑖∈𝑅,  𝑖=1,2,…,𝑛, 𝑛=[𝛼]+1.

Lemma 5. Let 𝛼>𝛽>0,𝑡∈[0,1] and 𝑦(𝑡)∈𝐿¹[0,1], and then 𝐼^𝛼₀₊𝐼^𝛽₀₊𝑦(𝑡)=𝐼^𝛼+𝛽₀₊𝑦(𝑡),𝐷^𝛼₀₊𝐼^𝛼₀₊𝑦(𝑡)(𝑡),𝐷=𝑦_𝛽0+𝐼^𝛼₀₊𝑦(𝑡)=𝐼^𝛼−𝛽₀₊𝑦(𝑡). (9) The following definitions and coincidence degree theory are fundamental in the proofs of our main results.

Definition 6. Let 𝑌 and 𝑍 be Banach spaces. A linear mapping 𝐿∶dom(𝐿)⊂𝑌→𝑍 is called a Fredholm mapping if the following two conditions hold:(1)Ker𝐿 has a finite dimension;(2)Im𝐿 is closed and has a finite codimension.If 𝐿 is a Fredholm mapping, its (Fredholm) index is the integer ind𝐿=dimKer𝐿−codimIm𝐿.
And if 𝐿 is a Fredholm mapping of index zero, there exist linear continuous projectors 𝑃∶𝑌→𝑌 and 𝑄∶𝑍→𝑍 such that Im𝑃=Ker𝐿, Ker𝑄=Im𝐿, and 𝑌=Ker𝐿⊕Ker𝑃,  𝑍=Im𝐿⊕Im𝑄. Then it follows that 𝐿|_{dom𝐿∩Ker𝑃}∶dom𝐿∩Ker𝑃→Im𝐿 is invertible. We denote the inverse of this map by 𝐾_𝑃∶Im𝐿→dom𝐿∩Ker𝑃. The generalized inverse of this map is denoted by 𝐾_𝑃,𝑄∶𝑌→dom𝐿∩Ker𝑃, 𝐾_𝑃,𝑄=𝐾_𝑃(𝐼−𝑄).

Definition 7. Assume that 𝐿∶dom𝐿⊂𝑌→𝑍 is a Fredholm mapping, and if Ω is an open bounded subset of 𝑌, the map 𝑁 will be called 𝐿-compact on Ω if 𝑄𝑁∶Ω→𝑍 is bounded and 𝐾_𝑃,𝑄𝑁∶Ω→𝑌 is compact.
If Im𝑄 is isomorphic to Ker𝐿, there exists an isomorphism 𝐽_𝑁𝐿∶Im𝑄→Ker𝐿. We have the coincidence degree continuation theorem which is proved in [18].

Theorem 8. Assume that 𝐿 is a Fredholm mapping of index zero and let 𝑁 be 𝐿-compact on Ω, where Ω is an open bounded subset of 𝑌. Suppose that the following conditions are satisfied:(1)𝐿𝑦≠𝜆𝑁𝑦 for each (𝑦,𝜆)∈[(dom𝐿⧵Ker𝐿)∩𝜕𝑄]×(0,1);(2)𝑁𝑦∉Im𝐿 for each 𝑦∈Ker𝐿∩𝜕𝑄;(3)deg(𝐽_𝑁𝐿𝑄𝑁∣_Ker𝐿,Ω∩Ker𝐿,0)≠0, where 𝑄∶𝑍→𝑍 is a continuous projection as mentioned above with Im𝐿=Ker𝑄 and 𝐽_𝑁𝐿∶Im𝑄→Ker𝐿 is any isomorphism.Then the equation 𝐿𝑦=𝑁𝑦 has at least one solution in dom𝐿∩Ω.

Let 𝑍₁=𝐿¹[0,1] with the norm ‖𝑥‖₁=∫¹₀|𝑥(𝑠)|𝑑𝑠. For 0<𝑝, 𝑞<1, we define a linear space 𝐶^𝑝[0,1],𝐶^𝑞[]∶𝑌0,1₁=𝐶^𝑝[]=𝑢(𝑡)(𝑡)0,1∣𝑢=𝐼^𝑝₀₊𝑥(𝑡)[][,𝑌,𝑥∈𝐶0,1,𝑡∈0,1]₂=𝐶^𝑞[]=𝑣(𝑡)(𝑡)0,1∣𝑣=𝐼^𝑞₀₊𝑦(𝑡),𝑦∈𝐶[0,1],𝑡∈[0,1], (10) to be accompanied by ‖𝑢‖_𝑌1=max_[]𝑡∈0,1𝐷^𝑝₀₊𝑢(𝑡)+max_[]𝑡∈0,1|𝑢(𝑡,‖𝑣‖)|_𝑌2=max_[]𝑡∈0,1𝐷^𝑞₀₊𝑣(𝑡)+max_[]𝑡∈0,1|𝑣(𝑡)|. (11)

We can prove that 𝑌=𝑌₁×𝑌₂ is a Banach space with the norm‖(𝑢,𝑣)‖_𝑌=max{‖𝑢‖_𝑌1,‖𝑢‖_𝑌2}, and 𝑍=𝑍₁×𝑍₁ is a Banach space, with the norm‖(𝑥,𝑦)‖_𝑍=max{‖𝑥‖₁,‖𝑦‖₁}.

Let 𝐿₁ to be the linear operator from dom𝐿₁∩𝑌₁ to 𝑍₁ with dom𝐿₁=𝑢∈𝐶^𝑝[0,1]∣𝐷^𝛼₀₊𝑢∈𝐿¹[],𝑢(0)0,1=0,𝐷^𝑝₀₊𝑦(1)=^𝑚−2_𝑖=1𝑎_𝑖𝐷^𝑝₀₊𝑢𝜉_𝑖. (12) And 𝐿₁𝑢=𝐷^𝛼₀₊𝑢, 𝑢∈dom𝐿₁. Let 𝐿₂ be the linear operator from dom𝐿₂∩𝑌₂ to 𝑍₁ with dom𝐿₂=𝑣∈𝐶^𝑞[0,1]∣𝐷^𝛽₀₊𝑣∈𝐿¹[],𝑣(0)0,1=0,𝐷^𝑞₀₊𝑣(1)=^𝑚−2_𝑗=1𝑏_𝑗𝐷^𝑞₀₊𝑣𝜂_𝑗. (13) And 𝐿₁𝑣=𝐷^𝛽₀₊𝑣,  𝑣∈dom𝐿₂. Define 𝐿 to be the linear operator from dom𝐿∩𝑌 to 𝑍 with dom𝐿=(𝑢,𝑣)∈𝑌∣𝑢∈dom𝐿₁,𝑣∈dom𝐿₂, (14) and 𝐿(𝑢,𝑣)=(𝐿₁𝑢,𝐿₂𝑣). We define 𝑁∶𝑌→𝑍 by setting 𝑁(𝑢,𝑣)=𝑁₁(𝑢(𝑡),𝑣(𝑡)),𝑁₂(𝑢(𝑡),𝑣(𝑡)), (15) where 𝑁₁(𝑢(𝑡),𝑣(𝑡))=𝑓(𝑡,𝑣(𝑡),𝐷^𝑝₀₊𝑢(𝑡)),   𝑁₂(𝑢(𝑡),𝑣(𝑡))=𝑔(𝑡,𝑢(𝑡),𝐷^𝑞₀₊𝑣(𝑡)).

Then the coupled system of BVPs (3) can be written as 𝐿(𝑢,𝑣)=𝑁(𝑢,𝑣).

Lemma 9 (see [19]). 𝐹⊂𝐶^𝜇[0,1],  0<𝜇<1, is a sequentially compact set if and only if 𝐹 is uniformly bounded and equicontinuous. Here uniformly bounded means that there exists 𝑀>0 such that for every 𝑢∈𝐹, 𝑦_𝐶𝜇=max_𝑡∈𝐽𝐷^𝜇₀₊𝑢(𝑡)+max_𝑡∈𝐽|𝑢(𝑡)|<𝑀, (16) and equicontinuous means that for all 𝜀>0,  ∃𝛿>0, for all 𝑡₁,𝑡₂∈[0,1],  |𝑡₁−𝑡₂|<𝛿,  𝑢∈𝐹, there hold |𝑢(𝑡₁)−𝑢(𝑡₂)|<𝜀,  |𝐷^𝛽₀₊𝑢(𝑡₁)−𝐷^𝛽₀₊𝑢(𝑡₂)|<𝜀.

Definition 10 (see [19]). One says that 𝑓∶[0,1]×𝑅²→𝑅 satisfies the Carathéodory condition if (i) for each (𝑥,𝑦)∈𝑅², the function 𝑡∈[0,1]→𝑓(𝑡,𝑥,𝑦)∈𝑅 is measurable on [0,1]; (ii) for a.e. 𝑡∈[0,1], the function (𝑥,𝑦)∈𝑅²→𝑓(𝑡,𝑥,𝑦)∈𝑅 is continuous on 𝑅²; (iii) for each 𝑟>0, there exists 𝜑_𝑟(𝑡)∈𝐿¹[0,1] such that |𝑓(𝑡,𝑥,𝑦)|≤𝜑_𝑟(𝑡) for a.e. 𝑡∈[0,1] and all (𝑥,𝑦)∈𝑅² with 𝑥²+𝑦²≤𝑟.

3. Main Results

In this section, we will give and prove our main results.

Lemma 11. Let 𝑓,𝑔∈𝑍₁,  𝐽=[0,1],  ∑^𝑚−2_𝑖=1𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−1}=1,  ∑^𝑚−2_𝑖=1𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−1}=1. Then 𝑢,𝑣∈𝑌 are, respectively, the solutions of the following boundary value problems: 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑢(0)=𝑓,𝑡∈𝐽,=0,𝐷^𝑝₀₊𝑢(1)=^𝑚−2_𝑖=1𝑎_𝑖𝐷^𝑝₀₊𝑢𝜉_𝑖,𝐷^𝛽₀₊𝑣(𝑡)(𝑡)𝑣(0)=𝑔,𝑡∈𝐽,=0,𝐷^𝑞₀₊𝑣(1)=^𝑚−2_𝑖=1𝑏_𝑖𝐷^𝑞₀₊𝑣𝜂_𝑖 (17)(18) if and only if 𝑢,𝑣 satisfied 𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝐼^𝛼₀₊𝑓(𝑡),_{𝑚−2𝑖=1}𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−11}₀(1−𝑠)^{𝛼−𝑝−1}𝑓(𝑠)=𝑑𝑠_𝜉𝑖0𝜉_𝑖−𝑠^{𝛼−𝑝−1}𝑓(𝑠)𝑣(𝑡)𝑑𝑠,=𝑐₂𝑡^𝛽−1+𝐼^𝛽₀₊𝑔(𝑡),_{𝑚−2𝑖=1}𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−11}₀(1−𝑠)^{𝛽−𝑞−1}𝑔(𝑠)𝑑𝑠=^𝜂𝑖₀𝜂_𝑖−𝑠^{𝛽−𝑞−1}𝑔(𝑠)𝑑𝑠. (19)

Proof. (𝑢,𝑣) in 𝑌₁×𝑌₂ are the solutions of the following boundary value problem: 𝐼^𝛼₀₊𝐷^𝛼₀₊𝑢(𝑡)=𝐼^𝛼₀₊𝑓(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑓(𝑠)𝐼𝑑𝑠,_𝛽0+𝐷^𝛽₀₊𝑢(𝑡)=𝐼^𝛽₀₊𝑔(𝑡)=1Γ𝛽^𝑡₀(𝑡−𝑠)^𝛽−1𝑔(𝑠)𝑑𝑠. (20) According to Lemma 4, we have 𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+𝐼^𝛼₀₊𝑓(𝑡),𝑣(𝑡)=𝑐₃𝑡^𝛽−1+𝑐₄𝑡^𝛽−2+𝐼^𝛽₀₊𝑔(𝑡). (21) By 𝑢(0)=𝑣(0)=0, we get 𝑐₂=𝑐₄=0, and then 𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝐼^𝛼₀₊𝑓(𝑡), 𝑣(𝑡)=𝑐₃𝑡^𝛽−1+𝐼^𝛽₀₊𝑔(𝑡).
Using (7) and Lemma 5, we get that 𝐷^𝑝₀₊𝑢(𝑡)=𝐷^𝑝₀₊𝐼^𝛼₀₊𝑓(𝑡)+𝑐₁Γ(𝛼)Γ𝑡𝛼−𝑝^{𝛼−𝑝−1}=𝐼^𝛼−𝑝₀₊𝑓(𝑡)+𝑐₁Γ(𝛼)Γ𝑡𝛼−𝑝^{𝛼−𝑝−1},𝐷^𝑞₀₊𝑣(𝑡)=𝐷^𝑞₀₊𝐼^𝛽₀₊𝑔(𝑡)+𝑐₃Γ𝛽Γ𝑡𝛽−𝑞^{𝛽−𝑞−1}=𝐼^𝛽−𝑞₀₊𝑔(𝑡)+𝑐₃Γ(𝛼)Γ𝑡𝛽−𝑞^{𝛽−𝑞−1}. (22) By the boundary conditions, ∑^𝑚−2_𝑖=1𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−1}=1,  ∑^𝑚−2_𝑖=1𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−1}=1, we have ^𝑚−2_𝑖=1𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−11}₀(1−𝑠)^{𝛼−𝑝−1}𝑓(𝑠)=𝑑𝑠_𝜉𝑖0𝜉_𝑖−𝑠^{𝛼−𝑝−1}𝑓(𝑠)𝑑𝑠,_{𝑚−2𝑖=1}𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−11}₀(1−𝑠)^{𝛽−𝑞−1}𝑔(𝑠)=𝑑𝑠_𝜂𝑖0𝜂_𝑖−𝑠^{𝛽−𝑞−1}𝑔(𝑠)𝑑𝑠. (23)(24) Conversely, suppose that (23) and (24) hold. In view of Lemma 5, we can easily verify that 𝑢,𝑣 are the solutions of (17) and (18), respectively. Then, we complete the proof.

Lemma 12. The mapping 𝐿∶dom(𝐿)⊂𝑌→𝑍 is a Fredholm operator of index zero.

Proof. Since Ker𝐿₁={𝑢∣𝐷^𝛼₀₊𝑢(𝑡)=0}, Ker𝐿₂={𝑣∣𝐷^𝛽₀₊𝑣(𝑡)=0}, it can be directly verified that Ker𝐿₁={𝑢∣𝑢=𝑐₁𝑡^𝛼−1,𝑐₁∈𝑅}, Ker𝐿₂={𝑣∣𝑣=𝑐₂𝑡^𝛽−1,𝑐₂∈𝑅}. It is clear that Ker𝐿={𝑐₁𝑡^𝛼−1,𝑐₂𝑡^𝛽−1∣𝑐₁,𝑐₂∈𝑅}.
Let (𝑥,𝑦)∈Im𝐿; then there exists (𝑢,𝑣)∈dom𝐿 such that (𝑥,𝑦)=𝐿(𝑢,𝑣), that is, 𝑢∈𝑌₁, 𝑥=𝐷^𝛼₀₊𝑢 and 𝑣∈𝑌₂, 𝑦=𝐷^𝛽₀₊𝑣. From these results, we can easily deduce the results Im𝐿=𝑥,𝑦∈𝑍∣𝑥satisfies(23),𝑦satisfies(24). (25)
Let 𝑡∈(0,1), ∑^𝑚−2_𝑖=0𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−1}=1, ∑^𝑚−2_𝑖=0𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−1}=1. Assume that Λ₁=^𝑚−2_𝑖=1𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−11}₀(1−𝑠)^{𝛼−𝑝−1}𝑠^𝑘−𝑑𝑠_𝜉𝑖0𝜉_𝑖−𝑠^{𝛼−𝑝−1}𝑠^𝑘Λ𝑑𝑠≠0,₂=^𝑚−2_𝑖=1𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−11}₀(1−𝑠)^{𝛽−𝑞−1}𝑠^𝑘−𝑑𝑠_𝜂𝑖0𝜂_𝑖−𝑠^{𝛽−𝑞−1}𝑠^𝑘𝑑𝑠≠0. (26) Consider the auxiliary mapping, 𝑄₁∶𝑍⟶𝑅∶𝑄₁𝑥=^𝑚−2_𝑖=1𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−11}₀(1−𝑠)^{𝛼−𝑝−1}𝑥(𝑠)−𝑑𝑠_𝜉𝑖0𝜉_𝑖−𝑠^{𝛼−𝑝−1}𝑥(𝑠),𝑄𝑑𝑠₂∶𝑍⟶𝑅∶𝑄₂𝑦=^𝑚−2_𝑖=1𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−11}₀(1−𝑠)^{𝛽−𝑞−1}𝑦(𝑠)−𝑑𝑠_𝜂𝑖0𝜂_𝑖−𝑠^{𝛽−𝑞−1}𝑦(𝑠)𝑑𝑠. (27) It is obvious that 𝑄₁,𝑄₂ are continuous linear mappings. Also we have that 𝑄₁(𝑡^𝑘)=Λ₁, 𝑄₂(𝑡^𝑘)=Λ₂. Take the mapping 𝑄∶𝑍→𝑍 defined by 𝑄(𝑥,𝑦)=(𝑄^′₁𝑥,𝑄^′₂𝑦). 𝑄^′₁𝑥=(𝑄₁𝑥/Λ₁)𝑡^𝑘, 𝑄^′₂𝑥=(𝑄₂𝑥/Λ₂)𝑡^𝑘, where 𝑥,𝑦∈𝑍. Note that (𝑥,𝑦)∈Im𝐿 implies 𝑄(𝑥,𝑦)=(𝑄₁𝑥,𝑄₂𝑦)=(0,0). Evidently, Im𝑄=𝑥,𝑦∣𝑥=𝑐₁𝑡^𝑘,𝑦=𝑐₂𝑡^𝑘,𝑐₁,𝑐₂∈𝑅, (28) and 𝑄∶𝑍→𝑍 is a continuous linear projector. In fact, for any (𝑥,𝑦)∈𝑍, we have 𝑄₁𝑄^′₁𝑥=𝑄₁𝑄₁𝑥Λ₁𝑡^𝑘=𝑄₁𝑥Λ₁𝑄₁𝑡^𝑘=𝑄₁𝑥Λ₁Λ₁=𝑄₁𝑄𝑥,₂𝑄^′₂𝑦=𝑄₂𝑄₂𝑦Λ₂𝑡^𝑘=𝑄₂𝑦Λ₂𝑄₂𝑡^𝑘=𝑄₂𝑦Λ₂Λ₂=𝑄₂𝑄𝑦,_1′2𝑥=𝑄^′₁𝑄^′₁𝑥=𝑄₁𝑄^′₁𝑥Λ₁𝑡^𝑘=𝑄₁𝑥Λ₁𝑡^𝑘=𝑄^′₁𝑄𝑥,_2′2𝑦=𝑄^′₂𝑄^′₂𝑦=𝑄₂𝑄^′₂𝑦Λ₂𝑡^𝑘=𝑄₂𝑦Λ₂𝑡^𝑘=𝑄^′₂𝑦, (29) that is to say, 𝑄∶𝑍→𝑍 is idempotent.
Observe that (𝑥,𝑦)∈Im𝐿 leads to 𝑄(𝑥,𝑦)=(0,0) and (𝑥,𝑦)∈Ker𝑄, where (0,0) is the element in 𝑍. Conversely, if (𝑥,𝑦)∈Ker𝑄, we can have that 𝑄(𝑥,𝑦)=(0,0), that is to say, (𝑥,𝑦)∈Im𝐿. So, Ker𝑄=Im𝐿.
Let (𝑥,𝑦)=(𝑥,𝑦)−𝑄(𝑥,𝑦)+𝑄(𝑥,𝑦)=(𝐼−𝑄)(𝑥,𝑦)+𝑄(𝑥,𝑦), where (𝑥,𝑦)∈𝑍 is an arbitrary element. Since 𝑄(𝑥,𝑦)∈Im𝑄 and (𝐼−𝑄)(𝑥,𝑦)∈Ker𝑄, we obtain that 𝑍=Im𝑄+Ker𝑄. Take (𝑥₀,𝑦₀)∈Im𝑄∩Ker𝑄, (𝑥₀,𝑦₀) can be written as (𝑥₀,𝑦₀)=(𝑐₁𝑡^𝑘,𝑐₂𝑡^𝑘), 𝑐₁,𝑐₂∈𝑅. For (𝑥₀,𝑦₀)∈Im𝑄 then since (𝑥₀,𝑦₀)∈Ker𝑄=Im𝐿, by (27), we get that 𝑄𝑥₀,𝑦₀=𝑄𝑐₁𝑡^𝑘,𝑐₂𝑡^𝑘=𝑐₁𝑄₁𝑡^𝑘,𝑐₂𝑄₂𝑡^𝑘=𝑐₁Λ₁,𝑐₂Λ₂=(0,0), (30) which implies that (𝑥₀,𝑦₀)=(0,0). Therefore, Im𝑄∩Ker𝑄={0,0}, and thus 𝑍=Im𝑄⊕Ker𝑄=Im𝑄⊕Im𝐿.
Now, dimKer𝐿=1=dimIm𝑄=codimKer𝑄=codimIm𝐿<+∞, and observing that Im𝐿 is closed in 𝑍, so 𝐿 is a Fredholm mapping of index zero.
Let 𝑃∶𝑌→𝑌 be defined by 𝑃₁𝑢(𝑡)=𝐷^𝑝₀₊𝑢(𝑡)|_𝑡=0𝑡^𝛼−1,𝑡∈𝐽,𝑢∈𝑌₁,𝑃₂𝑣(𝑡)=𝐷^𝑞₀₊𝑣(𝑡)|_𝑡=0𝑡^𝛽−1,𝑡∈𝐽,𝑣∈𝑌₂, (31) and 𝑃(𝑢,𝑣)=(𝑃₁𝑢,𝑃₂𝑣) it is clear that 𝑃₁, 𝑃₂, 𝑃 are linear continuous projectors and Im𝑃=Ker𝐿={𝑐₁𝑡^𝛼−1,𝑐₂𝑡^𝛽−1∣𝑐₁,𝑐₂∈𝑅}. Also, proceeding as the proof of Lemma 12, we can show that 𝑌=Im𝑃⊕Ker𝑃=Ker𝐿⊕Ker𝑃.

Lemma 13. Assume that 𝑓,𝑔∶[0,1]×𝑅²→𝑅 satisfies the Carathéodory conditions ^𝑚−2_𝑖=1𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−1}=1,^𝑚−2_𝑖=1𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−1}=1,𝛼−𝑝−1≥0,𝛽−𝑞−1≥0,Λ₁≠0,Λ₂≠0. (32) Then 𝐾_𝑃(𝐼−𝑄)𝑁∶𝑌→𝑌 is completely continuous.

Here we omit reasoning process, and it is easy to prove.

Consider the mapping 𝐾_𝑃∶Im𝐿→dom𝐿∩Ker𝑃,𝐾_𝑃(𝑥,𝑦)=(𝐼^𝛼₀₊𝑥,𝐼^𝛽₀₊𝑦), 𝐾_𝑃𝑥(𝑡)=𝐼^𝛼₀₊𝑥(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑥(𝑠)𝑑𝑠,𝑥∈Im𝐿₁,𝐾_𝑃𝑦(𝑡)=𝐼^𝛽₀₊𝑦(𝑡)=1Γ𝛽^𝑡₀(𝑡−𝑠)^𝛽−1𝑦(𝑠)𝑑𝑠,𝑦∈Im𝐿₂. (33)

For (𝑥,𝑦)∈Im𝐿, we have (𝐾_𝑃𝐿)(𝑥,𝑦)=𝐾_𝑃(𝐿(𝑥,𝑦))=(𝑥,𝑦).

(𝐿𝐾_𝑃)(𝑥,𝑦)=𝐿(𝐾_𝑃(𝑥,𝑦))=(𝑥,𝑦), for all (𝑥,𝑦)∈Im𝐿. Thus, 𝐾_𝑃=(𝐿_𝑃)⁻¹, where 𝐿_𝑃=𝐿|_{dom𝐿∩Ker𝑃}∶dom𝐿∩Ker𝑃→Im𝐿; hence, for each (𝑥,𝑦)∈Im𝐿 and 𝑡∈𝐽, we have 𝐾_𝑃(𝑥,𝑦)_𝑌=𝐼^𝛼₀₊𝑥,𝐼^𝛽₀₊𝑦_𝑌=max_𝑡∈𝐽𝐼^𝛼₀₊𝑥_𝑌1,𝐼^𝛽₀₊𝑦_𝑌2=max_𝑡∈𝐽𝐷^𝑝₀₊𝐼^𝛼₀₊𝑥+𝐼^𝛼₀₊𝑥,𝐷^𝑞₀₊𝐼^𝛽₀₊𝑦+𝐼^𝛽₀₊𝑥=max_𝑡∈𝐽1Γ𝛼−𝑝^𝑡₀(𝑡−𝑠)^{𝛼−𝑝−1}𝑥(𝑠)+1𝑑𝑠Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑥(𝑠),1𝑑𝑠Γ𝛽−𝑞^𝑡₀(𝑡−𝑠)^{𝛽−𝑞−1}𝑦(𝑠)+1𝑑𝑠Γ𝛽^𝑡₀(𝑡−𝑠)^𝛽−1𝑦(𝑠)𝑑𝑠≤max_𝑡∈𝐽1Γ+1𝛼−𝑝Γ(𝛼)‖𝑥‖₁,1Γ+1𝛽−𝑞Γ𝛽𝑦₁=max_𝑡∈𝐽𝜂₁‖𝑥‖₁,𝜂₂𝑦₁, (34) where 𝜂₁=1Γ+1𝛼−𝑝Γ(𝛼),𝜂₂=1Γ+1𝛽−𝑞Γ𝛽. (35) The operator 𝐾_𝑃(𝐼−𝑄)𝑁∶𝑌→𝑌 is continuous in view of the continuity of the function 𝑓 and 𝑔. Let Ω⊂𝑌 be bounded; that is to say, there exists a positive constant 𝑀>0 such that ||(𝑥,𝑦)||_𝑌≤𝑀 for all (𝑥,𝑦)∈Ω.

We can derive that max_{𝑥,𝑦∈Ω}𝐾_𝑃(𝐼−𝑄)𝑁𝑥,𝑦=max_{𝑥,𝑦∈Ω}𝐼^𝛼₀₊{𝑁𝑥−𝑄𝑁𝑥},𝐼^𝛽₀₊𝑁𝑦−𝑄𝑁𝑦=max_{𝑥,𝑦∈Ω}𝐼^𝛼₀₊𝑓𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥−𝑄𝑓𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥,𝐼^𝛽₀₊𝑔𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑥−𝑄𝑔𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑦≤max_{𝑥,𝑦∈Ω}1Γ(𝛼)𝑔𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑦−𝑄𝑔𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑦₁,1Γ𝛽𝑔𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑦−𝑄𝑔𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑦₁≤max_{𝑥,𝑦∈Ω}1Γ(𝛼)𝑀₁,1Γ𝛽𝑀₂. (36) Here, 𝑀₁=max_{‖𝑥‖𝐶𝑝≤𝑀}𝑓(𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥)−𝑄(𝑓(𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥))₁,𝑀₂=max_{𝑦𝐶𝑞≤𝑀}𝑓(𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑦)−𝑄(𝑓(𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑦))₁. (37) By Lemma 5, max_𝑥∈Ω𝐷^𝑝₀₊𝐾_𝑃(𝐼−𝑄)𝑁𝑥=max_𝑥∈Ω𝐷^𝑝₀₊𝐼^𝛼₀₊{𝑁𝑥−𝑄𝑁𝑥}=max_𝑥∈Ω𝐼^𝛼₀₊𝑓𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥−𝑄𝑓𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥≤1Γ𝛼−𝑝𝑓(𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥)−𝑄(𝑓(𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥))₁≤1Γ𝑀𝛼−𝑝¹. (38) Similarly, we can prove that max_𝑦∈Ω|𝐷^𝑞₀₊𝐾_𝑃(𝐼−𝑄)𝑁𝑦|≤(1/Γ(𝛽−𝑞))𝑀₂.

Hence, 𝐾_𝑃(𝐼−𝑄)𝑁(𝑥,𝑦)_𝑌≤max_{𝑥,𝑦∈Ω}1Γ+1𝛼−𝑝Γ(𝛼)𝑀₁,1Γ+1𝛽−𝑞Γ𝛽𝑀₂, (39) that is to say, 𝐾_𝑃(𝐼−𝑄)𝑁(Ω)⊂𝑌 is bounded. On the other hand, let 𝑀₃=max_{‖𝑥‖𝑌1≤𝑀}max_𝑡∈𝐽𝑓𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥−𝑄𝑓𝑡,𝑥(𝑡),𝐷^𝑝₀₊𝑥𝑀<+∞,₄=max_{𝑦𝑌2≤𝑀}max_𝑡∈𝐽𝑓𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑦−𝑄𝑓𝑡,𝑦(𝑡),𝐷^𝑞₀₊𝑦<+∞. (40) For every (𝑥,𝑦)∈Ω, 0≤𝑡₁≤𝑡₂≤1,   for all 𝜀>0, there is 𝛿=min{(Γ(𝛼−𝑝)/(𝑀₁+1))𝜀,(Γ(𝛽−𝑞)/(𝑀₂+1))𝜀}, and 𝑡₂−𝑡₁<𝛿 such that 𝐷^𝑝₀₊𝐾_𝑃(𝐼−𝑄)𝑁𝑥𝑡₂−𝐷^𝑝₀₊𝐾_𝑃(𝐼−𝑄)𝑁𝑥𝑡₁=𝐼^𝛼−𝑝₀₊𝑓𝑡₂,𝑥𝑡₂,𝐷^𝑝₀₊𝑥𝑡₂−𝑄𝑓𝑡₂,𝑥𝑡₂,𝐷^𝑝₀₊𝑥𝑡₂−𝐼^𝛼−𝑝₀₊𝑓𝑡₁,𝑥𝑡₁,𝐷^𝑝₀₊𝑥𝑡₁−𝑄𝑓𝑡₁,𝑥𝑡₁,𝐷^𝑝₀₊𝑥𝑡₁=1Γ𝛼−𝑝^𝑡2_𝑡1(𝑡−𝑠)^{𝛼−𝑝−1}×𝑓𝑠,𝑥(𝑠),𝐷^𝑝₀₊𝑥(𝑠)−𝑄𝑓𝑠,𝑥(𝑠),𝐷^𝑝₀₊𝑥(𝑠)≤1𝑑𝑠Γ𝑀𝛼−𝑝¹𝐷𝛿<𝜀,_𝑞0+𝐾_𝑃(𝐼−𝑄)𝑁𝑦𝑡₂−𝐷^𝑞₀₊𝐾_𝑃(𝐼−𝑄)𝑁𝑦𝑡₁<𝜀. (41)(42) This ensures that 𝐾_𝑃(𝐼−𝑄)𝑁(Ω) is equicontinuous too. Then an application of Ascoli-Arezela theorem ensures that 𝐾_𝑃(𝐼−𝑄)𝑁∶𝑌→𝑌 is completely continuous.

Theorem 14. Assume that 𝑓,𝑔∶[0,1]×𝑅²→𝑅 satisfy the Carathéodory conditions, 𝑡∈[0,1],  Λ₁,Λ₂≠0, 1<𝛼,  𝛽≤2,  0<𝑝,  𝑞<1,  𝛼−𝑝−1≥0,  𝛽−𝑞−1≥0, and ∑^𝑚−2_𝑖=1𝑎_𝑖𝜉_𝑖^{𝛼−𝑝−1}=1, ∑^𝑚−2_𝑖=1𝑏_𝑖𝜂_𝑖^{𝛽−𝑞−1}=1. Assume that all of the following hypotheses hold.(𝐻₁)  There exist nonnegative functions 𝑎_𝑖,𝑏_𝑖,𝑐_𝑖∈𝑍, 𝑖=1,2,𝑡∈𝐽, such that for all (𝑥,𝑦)∈𝑅², 𝑓𝑡,𝑥,𝑦≤𝑎₁(𝑡)|𝑥|+𝑏₁(𝑡)𝑦+𝑐₁(𝑡),𝑔𝑡,𝑥,𝑦≤𝑎₂(𝑡)|𝑥|+𝑏₂(𝑡)𝑦+𝑐₂(𝑡). (43) (𝐻₂)0<𝜂^′₁𝜂₁^′′<1, 0<𝜂^′₂𝜂₂^′′<1, where 𝜂₁,𝜂₂ is defined by (35) and 𝜂₁^′′=‖𝑎₁‖₁+‖𝑏₁‖₁, 𝜂₂^′′=‖𝑎₂‖₁+‖𝑏₂‖₁, 𝜂^′₁=(Γ(𝑝+1)+1)/Γ(𝛼−𝑝)+𝜂₁,𝜂^′₂=(Γ(𝑞+1)+1)/Γ(𝛽−𝑞)+𝜂₂. (𝐻₃) There exist constants 𝑇₁,𝑇₂>0 such that 𝑄₁(𝑁𝑢)≠0, 𝑄₂(𝑁𝑣)≠0, for each 𝑢∈dom𝐿₁, 𝑣∈dom𝐿₂ satisfying |𝐷^𝑝₀₊𝑢(𝑡)|>𝑇₁, |𝐷^𝑞₀₊𝑣(𝑡)|>𝑇₂,  for all 𝑡∈[0,1]. (𝐻₄) There exist constants 𝑆₁,𝑆₂>0 such that for any 𝑐₁,𝑐₂∈𝑅, if |𝑐₁|>𝑆₁, |𝑐₂|>𝑆₂, either 𝑐₁Λ₁𝑄₁𝑁𝑐₁𝑡^𝛼−1<0,𝑐₂Λ₂𝑄₂𝑁𝑐₂𝑡^𝛽−1<0, (44)or else 𝑐₁Λ₁𝑄₁𝑁𝑐₁𝑡^𝛼−1>0,𝑐₂Λ₂𝑄₂𝑁𝑐₂𝑡^𝛽−1>0. (45) Then the problem (3) has at least one solution in dom𝐿.

Proof. The proof consists of four main steps.
Step 1. Set Ω₁={(𝑢,𝑣)∈dom𝐿⧵Ker𝐿∣𝐿(𝑢,𝑣)=𝜆𝑁(𝑢,𝑣),𝜆∈[0,1]}, and prove Ω₁ is bounded. In fact, for (𝑢,𝑣)∈Ω₁, then (𝑢,𝑣)∈dom𝐿⧵Ker𝐿 and 𝐿(𝑢,𝑣)=𝜆𝑁(𝑢,𝑣), so 𝜆≠0 and 𝑁(𝑢,𝑣)∈Im𝐿=Ker𝑄⊂𝑍. Hence, 𝑄(𝑁(𝑢,𝑣))=0, that is, 𝑄₁(𝑁₁𝑢(𝑡))=0,𝑄₂(𝑁₂𝑣(𝑡))=0. From (𝐻₃), we have that 𝑡₀∈[0,1] such that |𝐷^𝑝₀₊𝑢(𝑡₀)|≤𝑇₁, |𝐷^𝑞₀₊𝑣(𝑡₀)|≤𝑇₂.
Again, for (𝑢,𝑣)∈Ω₁, we get ‖𝑃𝑢‖_𝑌1=max_𝑢∈Ω1𝐷^𝑝₀₊𝑃𝑢+max_𝑢∈Ω1|𝑃𝑢|=max_𝑢∈Ω1𝐷^𝑝₀₊𝐷^𝑝₀₊𝑢𝑡₀𝑡^𝑝+max_𝑢∈Ω1𝐷^𝑝₀₊𝑢𝑡₀𝑡^𝑝≤𝐷^𝑝₀₊𝑢𝑡₀Γ𝑝+1+𝐷^𝑝₀₊𝑢𝑡₀,‖𝑃𝑣‖_𝑌2≤𝐷^𝑞₀₊𝑣𝑡₀Γ𝑞+1+𝐷^𝑞₀₊𝑣𝑡₀. (46) In view of Lemma 5, we have 𝐷^𝑝₀₊𝑢𝑡₀=𝐷^𝑝₀₊𝑢𝑡₀−𝐼^𝛼−𝑝₀₊𝐷^𝑝₀₊𝑢𝑡₀≤𝐷^𝑝₀₊𝑢𝑡₀+1Γ𝐷𝛼−𝑝^𝑝0+𝑢(𝑡)₁≤𝑇₁+1Γ𝐿𝛼−𝑝¹𝑢₁≤𝑇₁+1Γ‖‖𝛼−𝑝𝑁𝑢₁. (47) Similarly, we can also obtain |𝐷^𝑞₀₊𝑣(𝑡₀)|≤𝑇₂+(1/Γ(𝛽−𝑞))‖𝐿₂𝑣‖₁.
Thus, ‖𝑃(𝑢,𝑣)‖_𝑌≤maxΓ𝑝+1+1𝑇₁+1Γ‖‖𝛼−𝑝𝑁𝑢₁,Γ𝑞+1+1𝑇₂+1Γ‖‖𝛽−𝑞𝑁𝑣₁. (48) According to Lemma 13, we have ‖(𝐼−𝑃)(𝑢,𝑣)‖_𝑌=𝐾_𝑃𝐿(𝐼−𝑃)(𝑢,𝑣)_𝑌≤max𝜂₁‖𝐿(𝐼−𝑃)𝑢‖₁,𝜂₂‖𝐿(𝐼−𝑃)𝑣‖₁=max𝜂₁‖𝐿𝑢‖₁,𝜂₂‖𝐿𝑣‖₁≤max𝜂₁‖𝑁𝑢‖₁,𝜂₂‖𝑁𝑣‖₁. (49) Combining (48) and (49), we can obtain ‖(𝑢,𝑣)‖_𝑌=‖(𝐼−𝑃)𝑢+𝑃𝑢‖_𝑌≤‖𝑃(𝑢,𝑣)‖_𝑌+‖(𝐼−𝑃)(𝑢,𝑣)‖_𝑌=maxΓ𝑝+1+1𝑇₁+𝜂₁‖𝑁𝑢‖₁,Γ𝑞+1+1𝑇₂+𝜂₂‖𝑁𝑣‖₁. (50) From (𝐻₁), for each 𝑢∈Ω₁, we have ‖𝑁𝑢‖₁=¹₀|(𝑠=𝑁𝑢)|𝑑𝑠₁₀𝑓𝑠,𝑢(𝑠),𝐷^𝑝₀₊𝑢(𝑠≤)𝑑𝑠₁₀𝑎₁(𝑠𝑢(𝑠+)|)|𝑑𝑠₁₀𝑏₁(𝑠)𝐷^𝑝₀₊𝑢(𝑠)𝑑𝑠+¹₀𝑐₁(𝑠≤)𝑑𝑠₁₀𝑎₁(𝑠)𝑑𝑠+¹₀𝑏₁(𝑠)𝑑𝑠‖𝑢‖_𝑌1+𝑐₁₁=𝜂₁^′′‖𝑢‖_𝑌1+𝜂₃. (51) Also, we can obtain that ‖𝑁𝑣‖₁=¹₀|𝑁𝑣(𝑠)|𝑑𝑠≤𝜂₂^′′‖𝑣‖_𝑌2+𝜂₄, (52) where 𝜂₃=‖𝑐₁‖₁,𝜂₄=‖𝑐₂‖₁. Thus, by (𝐻₂) and (50)–(52) we can derive that ‖(𝑢,𝑣)‖_𝑌Γ𝑇≤max𝑝+1+1₁+𝜂₁𝜂₃1−𝜂₁𝜂₁^′′,Γ𝑞+1+1𝑇₂+𝜂₂𝜂₄1−𝜂₂𝜂₂^′′, (53) which clearly states that Ω₁ is bounded.
Step 2. Set Ω₂={(𝑢,𝑣)∈Ker𝐿∣𝑁(𝑢,𝑣)∈Im𝐿} and prove that Ω₂ is bounded. Let (𝑢,𝑣)∈Ω₂; then (𝑢,𝑣)∈Ker𝐿=(𝑢,𝑣)∈dom𝐿∣𝑢=𝑐₁𝑡^𝛼−1,𝑣=𝑐₂𝑡^𝛽−1,∀𝑐₁,𝑐₂∈𝑅,𝑡∈(0,1). (54) Since 𝑁(𝑢,𝑣)∈Im𝐿=Ker𝑄, we have 𝑄(𝑁(𝑢,𝑣))=(0,0), that is, 𝑄₁(𝑁(𝑐₁𝑡^𝛼−1))=0, 𝑄₂(𝑁(𝑐₂𝑡^𝛽−1))=0. Taking account of (𝐻₄), |𝑐₁|≤𝑆₁, |𝑐₂|≤𝑆₂, which implies that Ω₂ is bounded.
Step 3. Set Ω₃={(𝑢,𝑣)∈Ker𝐿∣−𝜆(𝑢,𝑣)+(1−𝜆)𝐽_𝑁𝐿𝑄𝑁(𝑢,𝑣)=(0,0), 𝜆∈[0,1]}; if (35) holds, we will set Ω₃=(𝑢,𝑣)∈Ker𝐿∣𝜆(𝑢,𝑣)+(1−𝜆)𝐽_𝑁𝐿()=()[,𝑄𝑁𝑢,𝑣0,0,𝜆∈0,1]} (55) if (44) holds, where 𝐽_𝑁𝐿∶Im𝑄→Ker𝐿 is a linear isomorphism defined as 𝐽_𝑁𝐿𝑐₁𝑡^𝑘=𝑐₁𝑡^𝛼−1,𝐽_𝑁𝐿𝑐₂𝑡^𝑘=𝑐₂𝑡^𝛽−1,∀𝑐₁,𝑐₂∈𝑅,𝑡∈(0,1). (56) And then we will prove that Ω₃ is bounded. Without loss of generality, in the following part of the proof, we assume that (35) holds in (𝐻₄), for (𝑢,𝑣)∈Ω₃; we have 𝑢=𝑐₁𝑡^𝛼−1, 𝑣=𝑐₂𝑡^𝛽−1, 𝑐₁,𝑐₂∈𝑅,  𝑡∈[0,1], and 𝜆(𝑢,𝑣)=(1−𝜆)𝐽_𝑁𝐿𝑄𝑁(𝑢,𝑣) Thus, 𝜆𝑡^𝛼−1=()𝑄1−𝜆¹𝑁𝑐₁𝑡^𝛼−1𝑐₁Λ₁𝑡^𝛼−1,𝜆𝑡^𝛽−1=()𝑄1−𝜆²𝑁𝑐₂𝑡^𝛽−1𝑐₂Λ₂𝑡^𝛽−1. (57) Therefore, via (𝐻₄) and (35), we have |𝑐₁|≤𝑆₁, |𝑐₂|≤𝑆₂, which show that Ω₃ is bounded.
Step 4. Let Ω be an bounded open set such that Ω⊃⋃³_𝑖=1Ω_𝑖 and prove that deg(𝐽_𝑁𝐿𝑄𝑁|_Ker𝐿,Ω∩Ker𝐿,0)≠0. The operator 𝑁 is 𝐿-compact on Ω due to the fact that 𝑄𝑁(Ω) is bounded and 𝐾_𝑃,𝑄𝑁=𝐾_𝑃(𝐼−𝑃)𝑁∶Ω→𝑌 is compact by Lemma 13; then by Step 1 and Step 2, we have(i)𝐿(𝑢,𝑣)≠𝜆𝑁(𝑢,𝑣) for each ((𝑢,𝑣),𝜆)∈[(dom𝐿⧵Ker𝐿)∩𝜕Ω]×(0,1);(ii)𝑁(𝑢,𝑣)∉Im𝐿 for each (𝑢,𝑣)∈Ker𝐿∩𝜕Ω.
Define 𝐻((𝑢,𝑣),𝜆)=−𝜆𝐼(𝑢,𝑣)+(1−𝜆)𝐽_𝑁𝐿𝑄𝑁(𝑢,𝑣), where 𝐼 is the identity operator in 𝑌. According to the arguments in Step 3, we have 𝐻((𝑢,𝑣),𝜆)≠0, for all (𝑢,𝑣)∈Ker𝐿∩𝜕Ω, and therefore, via the homotopy property of degree, we obtain that deg𝐽_𝑁𝐿𝑄𝑁|_Ker𝐿(𝐻())(𝐻())(),Ω∩Ker𝐿,0=deg⋅,0,Ω∩Ker𝐿,0=deg⋅,1,Ω∩Ker𝐿,0=deg−𝐼,Ω∩Ker𝐿,0=−1≠0, (58) which certifies the condition (3) of Theorem 8. Then, applying Theorem 8, we conclude that the problem (3) has at least one solution in dom𝐿∩Ω. The proof is complete.

4. Conclusion

In this paper, we obtain the existence results of muti-point boundary value problem at resonance for a coupled system of nonlinear fractional differential equations by means of the coincidence degree theory.

Conflict of Interests

The authors declare that they have no competing interests.

Authors’ Contribution

Ruijuan Liu completed the proof and wrote the initial draft. Chunhai Kou provided the problem and gave some suggestions of amendment. Ruijuan Liu then finalized the paper. All authors read and approved the final paper.

Acknowledgment

This work was supported by the Natural Science Foundation of China under Grant no. 11271248.