Existence of Solutions of Fractional Differential Equation with 𝑝-Laplacian Operator at Resonance

Abstract

By using the extension of Mawhin’s continuation theorem due to Ge, we consider boundary value problems for fractional 𝑝-Laplacian equation. A new result on the existence of solutions for the fractional boundary value problem is obtained, which generalizes and enriches some known results to some extent from the literature.

1. Introduction and Preliminaries

Recently, fractional differential equations have played an important role in many fields such as physics, electrical circuits, and control theory (see [1–9]). Many scholars have paid more attention to boundary value problems for fractional differential equations (see [10–25]).

By using a fixed point theorem on a cone, Agarwal et al. (see [10]) considered a two-point boundary value problem at nonresonance given by 𝐷^𝛼₀⁺𝑥(𝑡)+𝑓𝑡,𝑥(𝑡),𝐷^𝜇₀₊𝑥(𝑡𝑥(0)(1))=0,=𝑥=0, (1) where 1<𝛼<2, 𝜇>0 are real numbers, 𝛼−𝜇≥1, and 𝐷^𝛼₀⁺ is the Riemann-Liouville fractional derivative.

By using the coincidence degree theory, Bai (see [20]) considered the following 𝑚-point fractional boundary value problems: 𝐷^𝛼₀⁺𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷₀^𝛼−1+𝑢(𝑡(𝑡)𝐼)+𝑒,0<𝑡<1,_02−𝛼+𝑢(𝑡)|_𝑡=0=0,𝑢(1)=^𝑚−2_𝑖=1𝛽_𝑖𝑢𝜂_𝑖, (2) where 1<𝛼≤2 is a real number, 𝛽_𝑖∈ℝ, 𝜂_𝑖∈(0,1) are given constants such that ∑^𝑚−2_𝑖=1𝛽_𝑖𝜂_𝑖^𝑚−1=1, and 𝐷^𝛼₀⁺,𝐼^𝛼₀⁺ are the Riemann-Liouville differentiation and integration.

The turbulent flow in a porous medium is a fundamental mechanics problem. For studying this type of problems, Leibenson (see [26]) introduced the 𝑝-Laplacian equation as follows: (𝜙_𝑝(𝑥^′(𝑡)))^′=𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡), (3) where 𝜙_𝑝(𝑠)=|𝑠|^𝑝−2𝑠, 𝑝>1. Obviously, 𝜙_𝑝 is invertible and its inverse operator is 𝜙_𝑞, where 𝑞>1 is a constant such that 1/𝑝+1/𝑞=1.

In the past few decades, many important results relative to (3) with certain boundary value conditions have been obtained. We refer the reader to [27–31] and the references cited therein. However, to the best of our knowledge, there are relatively few results on boundary value problems for fractional 𝑝-Laplacian equations.

Motivated by the work above, in this paper, we investigate the existence of solutions for boundary value problem (BVP for short) of fractional 𝑝-Laplacian equation with the following form: 𝐷^𝛽₀₊𝜙_𝑝𝐷^𝛼₀⁺𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),𝐷^𝛼₀⁺𝑥(𝑡)[],𝐷,𝑡∈0,1_𝛼0+𝑥(0)=𝐷^𝛼₀⁺𝑥(1)=𝑥^′(0)=0, (4) where 0<𝛽≤1, 1<𝛼≤2, 𝐷^𝛼₀⁺ is Caputo fractional derivative, and 𝑓∶[0,1]×ℝ²→ℝ is continuous.

BVP (4) happens to be at resonance in the sense that its associated linear homogeneous boundary value problem 𝐷^𝛽₀₊𝜙_𝑝𝐷^𝛼₀⁺𝑥(𝑡)[],𝐷=0,𝑡∈0,1_𝛼0+𝑥(0)=𝐷^𝛼₀⁺𝑥(1)=𝑥^′(0)=0 (5) has a nontrivial solution 𝑥(𝑡)=𝑐, where 𝑐∈ℝ.

For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory, which can be found, for instance, in [32–35].

Definition 1. The Riemann-Liouville fractional integral operator of order 𝛼>0 of a function 𝑥∶(0,+∞)→ℝ is given by 𝐼^𝛼₀⁺𝑥(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑥(𝑠)𝑑𝑠, (6) provided that the right side integral is pointwise defined on (0,+∞).

Definition 2. The Caputo fractional derivative of order 𝛼>0 of a continuous function 𝑥∶(0,+∞)→ℝ is given by 𝐷^𝛼₀⁺𝑥(𝑡)=𝐼₀^𝑛−𝛼+𝑑^𝑛𝑥(𝑡)𝑑𝑡^𝑛=1Γ(𝑛−𝛼)^𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑥^(𝑛)(𝑠)𝑑𝑠, (7) where 𝑛 is the smallest integer greater than or equal to 𝛼, provided that the right side integral is pointwise defined on (0,+∞).

Lemma 3. Assume that 𝐷^𝛼₀⁺𝑥∈𝐶[0,1], 𝛼>0. Then 𝐼^𝛼₀⁺𝐷^𝛼₀⁺𝑥(𝑡)=𝑥(𝑡)+𝑐₀+𝑐₁𝑡+𝑐₂𝑡²+⋯+𝑐_𝑛−1𝑡^𝑛−1, (8) where 𝑐_𝑖=−𝑥^(𝑖)(0)/𝑖!, 𝑖=0,1,2,…,𝑛−1, and here 𝑛 is the smallest integer greater than or equal to 𝛼.
Now, one briefly recalls some notations and an abstract existence result, which can be found in [36].

Definition 4. Let 𝑋 and 𝑌 be two Banach spaces with norms ‖⋅‖_𝑋 and ‖⋅‖_𝑌, respectively. A continuous operator 𝑀∶𝑋∩dom𝑀⟶𝑌 (9) is said to be quasilinear if(i)Im 𝑀∶=𝑀(𝑋∩dom𝑀) is a closed subset of 𝑌,(ii)Ker 𝑀∶={𝑋∩dom𝑀∶𝑀𝑢=0} is linearly homeomorphic to ℝ^𝑛, 𝑛<∞.

Definition 5. Let 𝑋 be a real Banach space and let 𝑋⊂𝑋. The operator 𝑃∶𝑋→𝑋 is said to be a projector provided that 𝑃²=𝑃, 𝑃(𝜆₁𝑥₁+𝜆₂𝑥₂)=𝜆₁𝑃(𝑥₁)+𝜆₂𝑃(𝑥₂) for 𝑥₁,𝑥₂∈𝑋 and 𝜆₁,𝜆₂∈ℝ. The operator 𝑄∶𝑋→𝑋 is said to be a semiprojector provided 𝑄²=𝑄.

Definition 6 (see [36]). Let 𝑋=Ker𝑀 and let 𝑋 be the complement space of 𝑋 in 𝑋, and then 𝑋=𝑋⊕𝑋. On the other hand, suppose that 𝑌 is a subspace of 𝑌 and 𝑌 is the complement space of 𝑌 in 𝑌 so that 𝑌=𝑌⊕𝑌. Let 𝑃∶𝑋→𝑋 be a projector, let 𝑄∶𝑌→𝑌 be a semiprojector, and let Ω⊂𝑋 be an open and bounded set with origin 𝜃∈Ω, where 𝜃 is the origin of a linear space.
Suppose that 𝑁_𝜆∶Ω→𝑌, 𝜆∈[0,1], is a continuous operator. Denote 𝑁₁ by 𝑁. Let Σ_𝜆={𝑥∈Ω∶𝑀𝑥=𝑁_𝜆𝑥}. 𝑁_𝜆 is said to be 𝑀-compact in Ω if there is 𝑌⊂𝑌 with dim𝑌 = dim𝑋 and an operator 𝑅∶Ω×[0,1]→𝑋 continuous and compact such that, for 𝜆∈[0,1], (𝐼−𝑄)𝑁_𝜆Ω⊂Im𝑀⊂(𝐼−𝑄)𝑌,𝑄𝑁_𝜆𝑥=𝜃,𝜆∈(0,1)⟺𝑄𝑁𝑥=𝜃,𝑅(⋅,0)isthezerooperatorand𝑅(⋅,𝜆)|_Σ𝜆=(𝐼−𝑃)|_Σ𝜆,𝑀[𝑃+𝑅(⋅,𝜆)]=(𝐼−𝑄)𝑁_𝜆. (10)(11)(12)(13)

Lemma 7 (see [36] Ge-Mawhin’s continuation theorem). Let 𝑋 and 𝑌 be two Banach spaces with norms ‖⋅‖_𝑋 and ‖⋅‖_𝑌, respectively. Ω⊂𝑋 is an open and bounded nonempty set. Suppose that 𝑀∶𝑋∩dom𝑀⟶𝑌 (14) is a quasilinear operator and 𝑁_𝜆∶Ω⟶𝑌,𝜆∈[0,1] (15) is 𝑀-compact in Ω. In addition, if(C₁)𝑀𝑥≠𝑁_𝜆𝑥, ∀(𝑥,𝜆)∈(dom𝑀∩𝜕Ω)×(0,1),(C₂)𝑄𝑁𝑥≠0, for 𝑥∈dom𝑀∩𝜕Ω,(C₃)deg(𝐽𝑄𝑁,Ker𝑀∩Ω,0)≠0, where 𝑁=𝑁₁, then the equation 𝑀𝑥=𝑁𝑥 has at least one solution in Ω.

In this paper, we take 𝑌=𝐶[0,1] with the norm ‖𝑥‖_∞=max_𝑡∈[0,1]|𝑥(𝑡)| and 𝑋={𝑥∣𝑥,𝐷^𝛼₀⁺𝑥∈𝑌} with the norm ‖𝑥‖_𝑋=max{‖𝑥‖_∞,‖𝐷^𝛼₀⁺𝑥‖_∞}. By means of the linear functional analysis theory, we can prove that 𝑋 is a Banach space.

Define the operator 𝑀∶dom𝑀⊂𝑋→𝑌 by 𝑀𝑥=𝐷^𝛽₀₊𝜙_𝑝𝐷^𝛼₀⁺𝑥, (16) where dom𝑀=𝑥∈𝑋∣𝐷^𝛽₀₊𝜙_𝑝𝐷^𝛼₀⁺𝑥𝐷∈𝑌,_𝛼0+𝑥(0)=𝐷^𝛼₀⁺𝑥(1)=𝑥^′(0)=0. (17) Define the operator 𝑁∶𝑋→𝑌 by 𝑁𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),𝐷^𝛼₀⁺𝑥(𝑡),∀𝑡∈[0,1]. (18) Then BVP (4) is equivalent to the operator equation. Consider 𝑀𝑥=𝑁𝑥,𝑥∈dom𝑀. (19)

2. Main Result

We will always assume that the nonlinearity 𝑓(𝑡,𝑢,𝑣) will be retained:(H₁)there exist nonnegative functions 𝑎,𝑏,𝑐∈𝑌 such that 𝑓(𝑡,𝑢,𝑣)≤𝑎(𝑡)+𝑏(𝑡)|𝑢|^𝑝−1+𝑐(𝑡)|𝑣|^𝑝−1[],(),∀𝑡∈0,1𝑢,𝑣∈ℝ²; (20)(H₂)there exists a constant 𝐵>0 such thateither 𝑢𝑓(𝑡,𝑢,𝑣)>0,∀𝑡∈[0,1],𝑣∈ℝ,|𝑢|>𝐵, (21) or 𝑢𝑓(𝑡,𝑢,𝑣)<0,∀𝑡∈[0,1],𝑣∈ℝ,|𝑢|>𝐵. (22) Moreover, we will always assume that 𝑓∶[0,1]×ℝ²→ℝ is continuous and 1Γ2‖𝑏‖𝛽+1(Γ(𝛼+1^∞))^𝑝−1+‖𝑐‖_∞<1. (23)

Now, we begin with some lemmas below.

Lemma 8. Let 𝑀 be defined by (16), and then Ker𝑀={𝑥∈𝑋∣𝑥(𝑡)=𝑐∈ℝ,∀𝑡∈[0,1]},Im𝑀=𝑦∈𝑌∣¹₀(1−𝑠)^𝛽−1𝑦(𝑠)𝑑𝑠=0, (24)(25) and 𝑀 is a quasilinear operator

Proof. By Lemma 3, 𝐷^𝛽₀₊𝜙_𝑝(𝐷^𝛼₀⁺𝑥(𝑡))=0 has solution: 𝑥(𝑡)=𝑐₀+𝑐₁𝑡+𝐼^𝛼₀⁺𝜙_𝑞𝑐₂=𝑐₀+𝑐₁𝜙𝑡+^𝑞𝑐₂Γ()𝑡𝛼+1^𝛼,𝑐₀,𝑐₁,𝑐₂∈ℝ, (26) which satisfies 𝐷^𝛼₀⁺𝑥(𝑡)=𝜙_𝑞𝑐₂. (27) Combining with the boundary value condition 𝐷^𝛼₀⁺𝑥(0)=0 and 𝑥^′(0)=0, we can get that (24) holds.
If 𝑦∈Im𝑀, then there exists a function 𝑥∈dom𝑀 such that 𝑦(𝑡)=𝐷^𝛽₀₊𝜙_𝑝(𝐷^𝛼₀⁺𝑥(𝑡)). Based on Lemma 3, we have 𝐷^𝛼₀⁺𝑥(𝑡)=𝜙_𝑞𝐼^𝛽₀₊𝑦(𝑡)+𝑐=𝜙_𝑞1Γ𝛽^𝑡₀(𝑡−𝑠)^𝛽−1𝑦(𝑠)𝑑𝑠+𝑐,𝑐∈ℝ. (28) From condition 𝐷^𝛼₀⁺𝑥(0)=0, one has 𝑐=0. By the condition 𝐷^𝛼₀⁺𝑥(1)=0, we obtain that ¹₀(1−𝑠)^𝛽−1𝑦(𝑠)𝑑𝑠=0. (29) Thus, we get (25).
Then we have dim Ker 𝑀=1 and 𝑀(dom𝑀∩𝑋)⊂𝑌 closed. Therefore, 𝑀 is a quasilinear operator.

Lemma 9. Let Ω⊂𝑋 be an open and bounded set; then 𝑁_𝜆 is 𝑀-compact in Ω.

Proof. Define the continuous projector 𝑃∶𝑋→𝑋 and the semiprojector 𝑄∶𝑌→𝑌: (𝑡)(0)[],(𝑡)𝑃𝑥=𝑥,∀𝑡∈0,1𝑄𝑦=𝛽¹₀(1−𝑠)^𝛽−1𝑦(𝑠)𝑑𝑠,∀𝑡∈[0,1]. (30) where 𝑋=Ker𝑀 and 𝑌=Im𝑄.
Obviously, Im𝑃=Ker𝑀 and 𝑃²𝑥(𝑡)=𝑃𝑥(𝑡). It follows from 𝑥=(𝑥−𝑃𝑥)+𝑃𝑥 that 𝑋=Ker𝑃+Ker𝑀. By a simple calculation, we can get Ker𝑀∩Ker𝑃={0}. Then we get 𝑋=Ker𝑀⊕Ker𝑃=𝑋⊕𝑋. (31)
By the definition of 𝑄, we can get 𝑄²𝑦=𝑄𝑦⋅𝛽¹₀(1−𝑠)^𝛽−1𝑑𝑠=𝑄𝑦. (32) Let 𝑦=(𝑦−𝑄𝑦)+𝑄𝑦, where 𝑦−𝑄𝑦∈Ker𝑄=Im𝑀, 𝑄𝑦∈Im𝑄. It follows from Ker𝑄=Im𝑀 and 𝑄²𝑦=𝑄𝑦 that Im𝑄∩Im𝑀={0}. Then, we have 𝑌=Im𝑄⊕Im𝑀=𝑌⊕𝑌. (33) Thus dim𝑋=dimKer𝑀=dimIm𝑄=dim𝑌. (34)
Let Ω⊂𝑋 be an open and bounded set with 𝜃∈Ω. For each 𝑥∈Ω, we can get 𝑄[(𝐼−𝑄)𝑁_𝜆𝑥]=0. Thus, (𝐼−𝑄)𝑁_𝜆𝑥∈Im𝑀=Ker𝑄. Take any 𝑦∈Im𝑀 in the type 𝑦=(𝑦−𝑄𝑦)+𝑄𝑦. Since 𝑄𝑦=0, we can get (𝐼−𝑄)𝑦∈𝑌. So (10) holds. It is easy to verify (11).
Furthermore, define 𝑅∶Ω×[0,1]→𝑋 by 𝑅(𝑡)=1𝑥,𝜆)(Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞×1Γ𝛽^𝑠₀(𝑠−𝜏)^𝛽−1×(𝐼−𝑄)𝑁_𝜆𝑥(𝜏)𝑑𝜏𝑑𝑠. (35)
By the continuity of 𝑓, it is easy to get that 𝑅(𝑥,𝜆) is continuous on Ω×[0,1]. Moreover, for all 𝑥∈Ω, there exists a constant 𝑇>0 such that |𝐼^𝛽₀₊(𝐼−𝑄)𝑁_𝜆𝑥(𝜏))|≤𝑇, so we can easily obtain that 𝑅(Ω,𝜆) is uniformly bounded. By the Arzelà-Ascoli theorem, we just need to prove that 𝑅∶Ω×[0,1]→𝑋 is equicontinuous. Furthermore, for 0≤𝑡₁<𝑡₂≤1, (𝑥,𝜆)∈Ω×[0,1], we have 𝑅(𝑥,𝜆)𝑡₂−𝑅(𝑥,𝜆)𝑡₁=𝐼^𝛼₀⁺𝜙_𝑞𝐼^𝛽₀₊(𝐼−𝑄)𝑁_𝜆𝑥𝑡₂−𝐼^𝛼₀⁺𝜙_𝑞𝐼^𝛽₀₊(𝐼−𝑄)𝑁_𝜆𝑥𝑡₁. (36)
By |𝐼^𝛽₀₊(𝐼−𝑄)𝑁_𝜆𝑥|≤𝑇, we have 𝐼^𝛼₀⁺𝜙_𝑞𝐼^𝛽₀₊(𝐼−𝑄)𝑁_𝜆𝑥𝑡₂−𝐼^𝛼₀⁺𝜙_𝑞𝐼^𝛽₀₊(𝐼−𝑄)𝑁_𝜆𝑥𝑡₁≤1Γ(𝛼)^𝑡2₀(𝑡₂−𝑠)^𝛼−1𝜙_𝑞𝐼^𝛽₀₊(𝐼−𝑄)𝑁_𝜆𝑥(𝑠)−𝑑𝑠_𝑡10(𝑡₁−𝑠)^𝛼−1𝜙_𝑞𝐼^𝛽₀₊(𝐼−𝑄)𝑁_𝜆𝑥(𝑠)≤𝜙𝑑𝑠_𝑞(𝑇)Γ(𝛼)^𝑡1₀(𝑡₂−𝑠)^𝛼−1−(𝑡₁−𝑠)^𝛼−1+𝑑𝑠_𝑡2𝑡1(𝑡₂−𝑠)^𝛼−1=𝜙𝑑𝑠_𝑞(𝑇)Γ()𝑡𝛼+1^𝛼2−𝑡^𝛼₁. (37) Since 𝑡^𝛼 is uniformly continuous on [0,1], so 𝑅(Ω,𝜆) is equicontinuous. Similarly, we can get that 𝐼^𝛽₀₊((𝐼−𝑄)𝑁_𝜆𝑥(𝜏))⊂𝐶[0,1] is equicontinuous, and considering that 𝜙_𝑞(𝑠) is uniformly continuous on [−𝑇,𝑇], we get that 𝐷^𝛼₀⁺𝑅(Ω,𝜆)=𝐼^𝛽₀₊((𝐼−𝑄)𝑁_𝜆(Ω)) is also equicontinuous. So we can obtain that 𝑅(Ω,𝜆)→𝑋 is compact.
For each 𝑥∈Σ_𝜆={𝑥∈Ω∶𝑀𝑥=𝑁_𝜆𝑥}, we have 𝐷^𝛽₀₊𝜙_𝑝(𝐷^𝛼₀⁺𝑥(𝑡))=𝑁_𝜆𝑥(𝑡)∈Im𝑀. Thus, 𝑅(𝑡)=1𝑥,𝜆)(Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞×1Γ𝛽^𝑠₀(𝑠−𝜏)^𝛽−1×(𝐼−𝑄)𝑁_𝜆𝑥(𝜏=1)𝑑𝜏𝑑𝑠.Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞×1Γ𝛽^𝑠₀(𝑠−𝜏)^𝛽−1×𝐷^𝛽₀₊𝜙_𝑝𝐷^𝛼₀⁺𝑥(𝜏)𝑑𝜏𝑑𝑠, (38) which together with 𝐷^𝛼₀⁺𝑥(0)=𝑥^′(0)=0 yields that 𝑅(𝑥,𝜆)(𝑡)=𝑥(𝑡)−𝑥(0)=[(𝐼−𝑃)𝑥](𝑡). (39) It is easy to verify that 𝑅(𝑥,0)(𝑡) is the zero operator. So (12) holds.
On the other hand, consider 𝑀[(𝑡)1𝑃𝑥+𝑅𝑥,𝜆)](=𝑀Γ(𝛼)×^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞1Γ𝛽^𝑠₀(𝑠−𝜏)^𝛽−1×(𝐼−𝑄)𝑁_𝜆𝑥(𝜏)(0=()𝑁𝑑𝜏𝑑𝑠+𝑥)𝐼−𝑄_𝜆𝑥(𝑡). (40) So (13) holds. Then we get that 𝑁_𝜆 is 𝑀-compact in Ω. The proof is complete.

Lemma 10. Suppose that (𝐻₁), (𝐻₂) hold; then the set Ω₁={𝑥∈dom𝑀⧵Ker𝑀∣𝑀𝑥=𝜆𝑁𝑥,𝜆∈(0,1)} (41) is bounded.

Proof. Take 𝑥∈Ω₁; then 𝑀𝑥=𝜆𝑁𝑥, 𝐷^𝛼₀⁺𝑥(0)=𝑥^′(0)=0, and 𝑁𝑥∈Im𝑀. By (25), we have ¹₀(1−𝑠)^𝛽−1𝑓𝑠,𝑥(𝑠),𝐷^𝛼₀⁺𝑥(𝑠)𝑑𝑠=0. (42) Then, by the integral mean value theorem, there exists a constant 𝜉∈(0,1) such that 𝑓(𝜉,𝑥(𝜉),𝐷^𝛼₀⁺𝑥(𝜉))=0. So, from (𝐻₂), we get |𝑥(𝜉)|≤𝐵. By 𝑥^′(0)=0, we get 𝑥(𝑡)=𝑥(0)+𝑥^′(0)𝑡+𝐼^𝛼₀⁺𝐷^𝛼₀⁺𝑥(𝑡)(0)+1=𝑥Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝐷^𝛼₀⁺𝑥(𝑠)𝑑𝑠. (43) Take 𝑡=𝜉; we have 𝑥(𝜉)(0)+1=𝑥Γ(𝛼)^𝜉₀(𝜉−𝑠)^𝛼−1𝐷^𝛼₀⁺𝑥(𝑠)𝑑𝑠. (44) Then we have |𝑥(0≤𝑥(𝜉+1)|)Γ(𝛼)^𝜉₀(𝜉−𝑠)^𝛼−1𝐷^𝛼₀⁺𝑥(𝑠≤𝑥(𝜉+1)𝑑𝑠)Γ(𝛼)𝐷^𝛼₀⁺𝑥_∞⋅1𝛼𝜉^𝛼1≤𝐵+Γ()𝐷𝛼+1^𝛼0+𝑥_∞. (45) So we get |𝑥(𝑡≤|𝑥(0+1)|)|Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝐷^𝛼₀⁺𝑥(𝑠≤|𝑥(0+1)𝑑𝑠)|Γ(𝛼)𝐷^𝛼₀⁺𝑥_∞⋅1𝛼𝑡^𝛼2≤𝐵+Γ()𝐷𝛼+1^𝛼0+𝑥_∞,∀𝑡∈[0,1]. (46) That is, ‖𝑥‖_∞2≤𝐵+Γ()𝐷𝛼+1^𝛼0+𝑥_∞. (47) By 𝑀𝑥=𝜆𝑁𝑥 and 𝐷^𝛼₀⁺𝑥(0)=0, we get 𝜙_𝑝𝐷^𝛼₀⁺𝑥(𝑡)=𝜆𝐼^𝛽₀₊(𝑡)=𝜆𝑁𝑥Γ𝛽^𝑡₀(𝑡−𝑠)^𝛽−1𝑓𝑠,𝑥(𝑠),𝐷^𝛼₀⁺𝑥(𝑠)𝑑𝑠. (48) So, from (𝐻₁), we have 𝜙_𝑝𝐷^𝛼₀⁺𝑥(𝑡)≤1Γ𝛽^𝑡₀(𝑡−𝑠)^𝛽−1𝑓𝑠,𝑥(𝑠),𝐷^𝛼₀⁺𝑥(𝑠)≤1𝑑𝑠Γ𝛽^𝑡₀(𝑡−𝑠)^𝛽−1𝑎(𝑠)+𝑏(𝑠)|𝑥(𝑠)|^𝑝−1+𝑐(𝑠)𝐷^𝛼₀⁺𝑥(𝑠)^𝑝−1≤1𝑑𝑠Γ𝛽‖𝑎‖_∞+‖𝑏‖_∞‖𝑥‖_∞^𝑝−1+‖𝑐‖_∞𝐷^𝛼₀⁺𝑥_∞^𝑝−1⋅1𝛽𝑡^𝛽≤1Γ‖𝑎‖𝛽+1^∞+‖𝑏‖_∞‖𝑥‖_∞^𝑝−1+‖𝑐‖_∞𝐷^𝛼₀⁺𝑥_∞^𝑝−1,∀𝑡∈[0,1], (49) which together with |𝜙_𝑝(𝐷^𝛼₀⁺𝑥(𝑡))|=|𝐷^𝛼₀⁺𝑥(𝑡)|^𝑝−1 and (47) yields that 𝐷^𝛼₀⁺𝑥_∞^𝑝−1≤1Γ𝑎‖𝛽+1‖_∞+‖𝑏‖_∞×2𝐵+𝐷Γ(𝛼+1)^𝛼0+𝑥_∞^𝑝−1+‖𝑐‖_∞𝐷^𝛼₀⁺𝑥_∞^𝑝−1. (50) In view of (23), we can see that there exists a constant 𝑀₁>0 such that 𝐷^𝛼₀⁺𝑥_∞≤𝑀₁. (51) Thus, from (47), we get ‖𝑥‖_∞≤𝐵+2𝑀₁Γ(𝛼+1)∶=𝑀₂. (52)
Combining (51) with (52), we have ‖𝑥‖_𝑋=max‖𝑥‖_∞,𝐷^𝛼₀⁺𝑥_∞≤max𝑀₁,𝑀₂∶=𝑀. (53) Therefore, Ω₁ is bounded. The proof is complete.

Lemma 11. Suppose that (𝐻₂) holds; then the set Ω₂={𝑥∈Ker𝑀∣𝑁𝑥∈Im𝑀} (54) is bounded.

Proof. For 𝑥∈Ω₂, we have 𝑥(𝑡)=𝑐, 𝑐∈ℝ and 𝑁𝑥∈Im𝑀. Then we get ¹₀(1−𝑠)^𝛽−1𝑓(𝑠,𝑐,0)𝑑𝑠=0, (55) which together with (H₂) implies |𝑐|≤𝐵. Thus, we have ‖𝑥‖_𝑋≤max{𝐵,0}=𝐵. (56) Hence, Ω₂ is bounded. The proof is complete.

Lemma 12. Suppose that the first part of (𝐻₂) holds; then the set Ω₃={𝑥∈Ker𝑀∣𝜆𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈[0,1]} (57) is bounded.

Proof. For 𝑥∈Ω₃, we have 𝑥(𝑡)=𝑐, 𝑐∈ℝ, and 𝜆𝑐+(1−𝜆)𝛽¹₀(1−𝑠)^𝛽−1𝑓(𝑠,𝑐,0)𝑑𝑠=0. (58) 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 𝜆𝑐²+(1−𝜆)𝛽¹₀(1−𝑠)^𝛽−1𝑐𝑓(𝑠,𝑐,0)𝑑𝑠>0, (59) which contradicts (58). Therefore, Ω₃ is bounded. The proof is complete.

Remark 13. If the second part of (H₂) holds, then the set Ω^′₃={𝑥∈Ker𝑀−𝜆𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈[0,1]} (60) is bounded.

Theorem 14. Let 𝑓∶[0,1]×ℝ²→ℝ be continuous. Suppose that (𝐻₁), (𝐻₂) hold. Then BVP (4) has at least one solution.

Proof. Set Ω={𝑥∈𝑋|‖𝑥‖_𝑋<max{𝑀,𝐵}+1}. It follows from Lemmas 8 and 9 that 𝑀 is a quasilinear operator and 𝑁_𝜆 is 𝑀-compact on Ω. By Lemmas 10 and 11, we get that the following two conditions are satisfied:(C₁)𝑀𝑥≠𝑁_𝜆𝑥,∀(𝑥,𝜆)∈(dom𝑀∩𝜕Ω)×(0,1),(C₂)𝑄𝑁𝑥≠0, for 𝑥∈dom𝑀∩𝜕Ω.Take 𝐻(𝑥,𝜆)=±𝜆𝑥+(1−𝜆)𝑄𝑁𝑥. (61) According to Lemma 12 (or Remark 13), we know that 𝐻(𝑥,𝜆)≠0 for 𝑥∈Ker𝑀∩𝜕Ω. Therefore deg𝑄𝑁|_Ker𝑀(𝐻())(𝐻())(),Ω∩Ker𝑀,0=deg⋅,0,Ω∩Ker𝑀,0=deg⋅,1,Ω∩Ker𝑀,0=deg±𝐼,Ω∩Ker𝑀,0≠0. (62) So the condition (𝐶₃) of Lemma 7 is satisfied. By Lemma 7, we can get that 𝑀𝑥=𝑁𝑥 has at least one solution in dom𝑀∩Ω. Therefore BVP (4) has at least one solution. The proof is complete.

3. Example

In this section, we will give an example to illustrate our main result.

Example 1. Consider the following BVP: 𝐷₀^3/4₊𝜙₃𝐷₀^3/2₊𝑥(𝑡)=−253+13𝑥²(𝑡)+𝑡𝑒^{−|𝐷0+3/2𝑥(𝑡)|}[],𝐷,𝑡∈0,1_03/2+𝑥(0)=𝐷₀^3/2₊𝑥(1)=𝑥^′(0)=0. (63) Corresponding to BVP (4), we get that 𝑝=3, 𝛼=3/2, 𝛽=3/4, and 𝑓(𝑡,𝑢,𝑣)=−253+13𝑢²+𝑡𝑒^−|𝑣|. (64) Choose 𝑎(𝑡)=10, 𝑏(𝑡)=1/3, 𝑐(𝑡)=0, 𝐵=5. By a simple calculation, we can get that ‖𝑏‖_∞=1/3,‖𝑐‖_∞=0 and 1Γ(3/4+1)2/3(Γ(3/2+1))²+0<1. (65) Obviously, BVP (63) satisfies all conditions of Theorem 14. Hence, it has at least one solution.

4. Conclusions

In this paper, the boundary value problem for 𝑝-Laplacian equation at resonance is investigated. In view of the boundary value problem (4) is equivalent to the operator equation (19); we only need to find a fixed point of the operator equation (19). Firstly, we established the sufficient conditions of existence of boundary value problem for 𝑝-Laplacian equation. Then, by using the extension of Mawhin’s continuation theorem due to Ge, we got the fixed point of operator equation (19).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This research was supported by the Fundamental Research Funds for the Central Universities (2013XK03) and the National Natural Science Foundation of China (11271364).