Solvability of a Third-Order Multipoint Boundary Value Problem at Resonance

Abstract

We discuss a third-order multipoint boundary value problem under some appropriate resonance conditions. By using the coincidence degree theory, we establish the existence result of solutions. The emphasis here is that the dimension of the linear operator is equal to two. Our results supplement other results.

1. Introduction

In this paper, we are concerned with the following third-order ordinary differential equation: 𝑥^′′′(𝑡)=𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡),𝑥^′′(𝑡)+𝑒(𝑡),𝑡∈(0,1), (1) with the boundary conditions 𝑥(0)=^𝑚−2_𝑖=1𝛼_𝑖𝑥𝜉_𝑖,𝑥^′(0)=0,𝑥(1)=𝛽𝑥𝜂, (2) where 𝑓∶[0,1]×𝑅³→𝑅 is a Carathéodory function, 𝑒∈𝐿¹[0,1], 0<𝜉₁<⋯<𝜉_𝑚−2<1, 𝛼_𝑖∈𝑅, 𝑖=1,…,𝑚−2, 𝛽≥0, and 𝜂∈(0,1).

In recent years, many authors have paid much attention to the existence of solutions for multipoint boundary value problems at resonance: we refer the readers to see [1–11]. If the linear equation 𝐿𝑥=𝑥^′′′(𝑡)=0 with boundary conditions (2) has nontrivial solutions, that is, dimKer𝐿≥1, the BVP (1)-(2) is called a resonance problem. In [5–11], the authors all discussed the case that dimKer𝐿=1. In [2, 3], the authors established the existence results for resonance boundary value problems with the case of dimKer𝐿=2. However, we will show that some conditions such as Λ≠0 assumed in [2, 3] are not necessary. We establish existence of some solutions for BVP (1)-(2) by using the coincidence degree theory of Mawhin [12] at resonance.

According to the constant 𝛽, the BVP (1)-(2) is divided into the following five resonance cases:(𝐶₁)0<𝛽<1,(1−𝛽)∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=(1−∑^𝑚−2_𝑖=1𝛼_𝑖)(𝛽𝜂²−1); (𝐶₂)𝛽=1,∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=0,∑^𝑚−2_𝑖=1𝛼_𝑖=1; (𝐶₃)1<𝛽<1/𝜂²,(1−𝛽)∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=(1−∑^𝑚−2_𝑖=1𝛼_𝑖)(𝛽𝜂²−1); (𝐶₄)𝛽=1/𝜂²,∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=0,∑^𝑚−2_𝑖=1𝛼_𝑖=1; (𝐶₅)1/𝜂²<𝛽,(1−𝛽)∑^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=(1−∑^𝑚−2_𝑖=1𝛼_𝑖)(𝛽𝜂²−1).

Du et al. [1] studied the existence results of BVP (1)-(2) under the resonance conditions (𝐶₂) and (𝐶₄), that is, dimKer𝐿=1, but they did not discuss the other three cases. In this paper, under the resonance conditions, (𝐶₁), (𝐶₃), or (𝐶₅), we could imply dimKer𝐿=2; thus we supplement the results in [1].

The layout of this paper is as follows. In Section 2, we briefly present some notations and an abstract existence result due to Mawhin. In Section 3, we study BVP (1)-(2) under the condition (𝐶₁) and obtain some existence results. In Section 4, we give an example of the existence results in Section 3.

2. Preliminary

Now, we briefly recall some notations and an abstract existence result by Mawhin [12].

Let 𝑌, 𝑍 be real Banach spaces and let 𝐿∶dom𝐿⊂𝑌→𝑍 be a linear operator which is a Fredholm map of index zero and let 𝑃∶𝑌→𝑌, 𝑄∶𝑍→𝑍 be continuous projectors such that Im𝑃=Ker𝐿,  Ker𝑄=Im𝐿  and 𝑌=Ker𝐿⊕Ker𝑃,  𝑍=Im𝐿⊕Im𝑄. It follows that 𝐿|_{dom𝐿∩Ker𝑃}: dom𝐿∩Ker𝑃→Im𝐿 is invertible; we denote the inverse of that map by 𝐾_𝑃. Let Ω be an open bounded subset of 𝑌 such that dom𝐿∩Ω≠𝜙; the map 𝑁∶𝑌→𝑍 is said to be 𝐿-compact on Ω if the map 𝑄𝑁(Ω) is bounded and 𝐾_𝑃(𝐼−𝑄)𝑁∶Ω→𝑌 is compact. For more details we refer the reader to the lecture of Mawhin [12].

The theorem we use in this paper is Theorem IV.13 of [12].

Theorem 1. Let 𝐿 be a Fredholm map of index zero and let 𝑁 be 𝐿-compact on Ω. Assume that the following conditions are satisfied. (i)𝐿𝑥≠𝑁𝑥 for every (𝑥,𝜆)∈[(dom𝐿⧵Ker𝐿)∩𝜕Ω]×(0,1).(ii)𝑁𝑥∉Im𝐿 for every 𝑥∈Ker𝐿∩𝜕Ω.(iii)deg(𝑄𝑁∣_Ker𝐿, Ω∩Ker𝐿,0)≠0, where 𝑄∶𝑍→𝑍 is a continuous projector as above with Im𝐿 = Ker𝑄.Then the abstract equation 𝐿𝑥=𝑁𝑥 has at least one solution in dom𝐿∩Ω.

In the following, we will use the classical spaces 𝐶[0,1],𝐶¹[0,1], 𝐶²[0,1], and 𝐿¹[0,1]. For 𝑥∈𝐶²[0,1], we use the norms ‖𝑥‖_∞=max_𝑡∈[0,1]|𝑥(𝑡)| and ‖𝑥‖=max{‖𝑥‖_∞,‖𝑥^′‖_∞,‖𝑥^′′‖_∞}, denote the norm in 𝐿¹[0,1] by ‖⋅‖₁, and define the Sobolev space 𝑊^3,1(0,1) as 𝑊^3,1(0,1)=𝑥∶[0,1]⟶𝑅∣𝑥,𝑥^′,𝑥^′′[]areabsolutelycontinuouson0,1with𝑥^′′′∈𝐿¹[0,1]. (3)

Let 𝑌=𝐶²[0,1],  𝑍=𝐿¹[0,1], and define the linear operator 𝐿∶dom𝐿⊂𝑌→𝑍 as 𝐿𝑥=𝑥^′′′,𝑥∈dom𝐿, where dom𝐿=𝑥∈𝑊^3,1()∶(2.0,1𝑥satisfiesboundaryconditions) (4) We define 𝑁∶𝑌→𝑍 as 𝑁𝑥=𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡),𝑥^′′(𝑡),𝑡∈(0,1). (5) Then BVP (1)-(2) can be written as 𝐿𝑥=𝑁𝑥.

3. Existence Results

Lemma 2. If the condition (𝐶₁) holds, then there exist 𝑝∈{1,2,…,𝑚−2}, 𝑞∈𝑁⁺,𝑞≥𝑝+1, such that Λ𝑝,𝑞=^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+21−𝛽𝜂^𝑝+2−^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+21−𝛽𝜂^𝑞+2≠0, (6) where 𝑁⁺={1,2,…}.

Proof. We prove that, for any 𝑙∈𝑁, there exists 𝑘_𝑙∈{𝑙(𝑚−2)+1,…,(𝑙+1)(𝑚−2)}, such that ∑^𝑚−2_𝑖=1𝛼_𝑖𝜉^𝑘𝑙_𝑖⁺²≠0.
If else, one has ∑^𝑚−2_𝑖=1𝛼_𝑖𝜉^𝑘𝑙_𝑖⁺²=0, any 𝑘_𝑙∈{𝑙(𝑚−2)+1,𝑙(𝑚−2)+2,…,(𝑙+1)(𝑚−2)}; that is, 𝜉₁^{𝑙(𝑚−2)+3}…𝜉^{𝑙(𝑚−2)+3}_𝑚−2𝜉⋮⋱⋮_{1(𝑙+1)(𝑚−2)+2}⋯𝜉^{(𝑙+2)(𝑚−2)+2}_𝑚−2𝛼₁⋮𝛼_𝑚−2=0⋮0. (7) Since 𝜉₁^{𝑙(𝑚−2)+3}⋯𝜉^{𝑙(𝑚−2)+3}_𝑚−2𝜉⋮⋱⋮_{1(𝑙+1)(𝑚−2)+2}⋯𝜉^{(𝑙+1)(𝑚−2)+2}_𝑚−2=^𝑚−2_𝑗=1𝜉_𝑗^{𝑙(𝑚−2)+3}𝜉1⋯1₁⋯𝜉_𝑚−2𝜉⋮⋱⋮_1𝑚−3⋯𝜉^𝑚−3_𝑚−2=^𝑚−2_𝑗=1𝜉_𝑗^{𝑙(𝑚−2)+3}_{1≤𝑖<𝑗≤𝑚−2}𝜉_𝑗−𝜉_𝑖≠0, (8) thus 𝛼_𝑖=0,  𝑖=1,2,…,𝑚−2.
It is clear that 1−𝛽^𝑚−2_𝑖=1𝛼_𝑖𝜉²_𝑖=0,1−^𝑚−2_𝑖=1𝛼_𝑖𝛽𝜂²−1=𝛽𝜂²−1<0, (9) which is a contradiction to the condition (𝐶₁).
Set 𝑆=𝑘_𝑙∈𝑁⁺∶1−𝛽𝜂^𝑝+2=∑^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+21−𝛽𝜂^𝑘𝑙+2∑^𝑚−2_𝑖=1𝛼_𝑖𝜉^𝑘𝑙_𝑖⁺². (10) Then 𝑆 is a finite set.
If else, there exists a monotone sequence {𝑘_𝑙𝑟}, 𝑟=1,2,…,  𝑘_𝑙𝑟<𝑘_𝑙𝑟+1, such that 1−𝛽𝜂^𝑝+2=∑^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+21−𝛽𝜂^{𝑘𝑙𝑟+2}∑^𝑚−2_𝑖=1𝛼_𝑖𝜉^𝑘𝑙𝑟_𝑖⁺². (11) From 0<𝛽𝜂^𝑝+2<1, we get ^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+2≠0. (12) Thus 1−𝛽𝜂^𝑝+2=lim_{𝑘𝑙𝑟→∞}∑^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+21−𝛽𝜂^{𝑘𝑙𝑟+2}∑^𝑚−2_𝑖=1𝛼_𝑖𝜉^𝑘𝑙𝑟_𝑖⁺²=∞. (13)
So it is a contradiction. Thus the Lemma is proved.

Lemma 3. Let (𝐶₁) hold and Λ(𝑝,𝑞)≠0; then 𝐿: dom𝐿⊂𝑌→𝑍 is a Fredholm map of index zero. Furthermore, the linear continuous projector operator 𝑄∶𝑍→𝑍 can be defined by 𝑄𝑦(𝑡)=𝑇₁𝑦𝑡^𝑝−1+𝑇₂𝑦𝑡^𝑞−1, (14) where 𝑇₁𝑝𝑦=𝑝+1𝑝+2Λ×𝑝,𝑞𝛽𝜂^𝑞+2−1𝑄₁𝑦+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2𝑄₂𝑦,𝑇₂𝑞𝑦=−𝑞+1𝑞+2Λ×𝑝,𝑞𝛽𝜂^𝑝+2−1𝑄₁𝑦+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+2𝑄₂𝑦,𝑄₁𝑦=^𝑚−2_𝑖=1𝛼_𝑖^𝜉𝑖₀^𝑠₀^𝜏₀𝑦(𝑣)𝑄𝑑𝑣𝑑𝜏𝑑𝑠,₂𝑦=¹₀^𝑠₀^𝜏₀𝑦(𝑣)𝑑𝑣𝑑𝜏𝑑𝑠−𝛽^𝜂₀^𝑠₀^𝜏₀𝑦(𝑣)𝑑𝑣𝑑𝜏𝑑𝑠. (15) And the linear operator 𝐾_𝑃: Im𝐿→dom𝐿∩Ker𝑃 can be written by 𝐾_𝑃𝑦(𝑡)=^𝑡₀^𝑠₀^𝜏₀𝑦(𝑣)𝑑𝑣𝑑𝜏𝑑𝑠,𝑦∈Im𝐿. (16) Furthermore 𝐾_𝑃𝑦≤𝑦₁,𝑦∈Im𝐿. (17)

Proof. It is clear that Ker𝐿={𝑥∈dom𝐿∶𝑥=𝑎+𝑐𝑡², 𝑎=((𝛽𝜂²−1)/(1−𝛽))𝑐, 𝑐∈𝑅}.
Now we show that Im𝐿=𝑦∈𝑍∶𝑄₁𝑦=𝑄₂𝑦=0. (18) The equation 𝑥^′′′=𝑦, (19) has a solution 𝑥(𝑡) satisfying (2) if and only if 𝑄₁𝑦=𝑄₂𝑦=0. (20)
In fact, if (19) has a solution 𝑥(𝑡) such that (2), then from (19) we have 𝑥(𝑡)=𝛽𝜂²−11−𝛽𝑐+𝑐𝑡²+^𝑡₀^𝑠₀^𝜏₀𝑦(𝑣)𝑑𝑣𝑑𝜏𝑑𝑠. (21) According to the condition (𝐶₁), we obtain 𝑄₁𝑦=𝑄₂𝑦=0. (22) On the other hand, if (20) holds, let 𝑥(𝑡)=𝛽𝜂²−11−𝛽𝑐+𝑐𝑡²+^𝑡₀^𝑠₀^𝜏₀𝑦(𝑣)𝑑𝑣𝑑𝜏𝑑𝑠, (23) where 𝑐 is an arbitrary constant; then 𝑥(𝑡) is a solution of (19) and (2). Hence (18) holds.
Set 𝑇₁𝑝𝑦=𝑝+1𝑝+2Λ×𝑝,𝑞𝛽𝜂^𝑞+2−1𝑄₁𝑦+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2𝑄₂𝑦,𝑇₂𝑞𝑦=−𝑞+1𝑞+2Λ×𝑝,𝑞𝛽𝜂^𝑝+2−1𝑄₁𝑦+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+2𝑄₂𝑦. (24) Then we define 𝑄𝑦(𝑡)=𝑇₁𝑦𝑡^𝑝−1+𝑇₂𝑦𝑡^𝑞−1. (25) It is clear that dimIm𝑄=2.
Again from 𝑇₁𝑇₁𝑦𝑡^𝑝−1=𝑝𝑝+1𝑝+2Λ×𝑝,𝑞𝛽𝜂^𝑞+2−1𝑄₁𝑇₁𝑦𝑡^𝑝−1+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2𝑄₂𝑇₁𝑦𝑡^𝑝−1=1Λ𝑝,𝑞𝛽𝜂^𝑞+2𝑝×𝑄−1𝑝+1𝑝+2₁𝑡^𝑝−1×+𝑝𝑝+1𝑝+2_{𝑚−2𝑖=1}𝛼_𝑖𝜉_𝑖^𝑞+2𝑄₂𝑡^𝑝−1×𝑇₁𝑦=𝑇₁𝑇𝑦,₁𝑇₂𝑦𝑡^𝑞−1=𝑝𝑝+1𝑝+2Λ×𝑝,𝑞𝛽𝜂^𝑞+2−1𝑄₁𝑇₂𝑦𝑡^𝑞−1+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2𝑄₂𝑇₂𝑦𝑡^𝑞−1=1Λ𝑝,𝑞𝛽𝜂^𝑞+2𝑝×𝑄−1𝑝+1𝑝+2₁𝑡^𝑞−1×+𝑝𝑝+1𝑝+2_{𝑚−2𝑖=1}𝛼_𝑖𝜉_𝑖^𝑞+2𝑄₂𝑡^𝑞−1×𝑇₂𝑦𝑇=0,₂𝑇₁𝑦𝑡^𝑝−1𝑞=−𝑞+1𝑞+2Λ×𝑝,𝑞𝛽𝜂^𝑝+2−1𝑄₁𝑇₁𝑦𝑡^𝑝−1+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+2𝑄₂𝑇₁𝑦𝑡^𝑝−1𝑞=−𝑞+1𝑞+2Λ×𝑝,𝑞𝛽𝜂^𝑝+2−1𝑄₁𝑡^𝑝−1+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+2𝑄₂𝑡^𝑝−1×𝑇₁𝑦𝑇=0,₂𝑇₂𝑦𝑡^𝑞−1𝑞=−𝑞+1𝑞+2Λ×𝑝,𝑞𝛽𝜂^𝑝+2−1𝑄₁𝑇₂𝑦𝑡^𝑞−1+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+2𝑄₂𝑇₂𝑦𝑡^𝑞−1=𝑇₂𝑦. (26) One has 𝑄²𝑦=𝑄𝑇₁𝑦𝑡^𝑝−1+𝑇₂𝑦𝑡^𝑞−1=𝑇₁𝑇₁𝑦𝑡^𝑝−1+𝑇₂𝑦𝑡^𝑞−1𝑡^𝑝−1+𝑇₂𝑇₁𝑦𝑡^𝑝−1+𝑇₂𝑦𝑡^𝑞−1𝑡^𝑞−1=𝑇₁𝑇₁𝑦𝑡^𝑝−1𝑡^𝑝−1+𝑇₁𝑇₂𝑦𝑡^𝑞−1𝑡^𝑝−1+𝑇₂𝑇₁𝑦𝑡^𝑝−1𝑡^𝑞−1+𝑇₂𝑇₂𝑦𝑡^𝑞−1𝑡^𝑞−1=𝑇₁𝑦𝑡^𝑝−1+𝑇₂𝑦𝑡^𝑞−1=𝑄𝑦. (27) Thus the operator 𝑄 is a projector.
Now we show that Ker𝑄=Im𝐿. If 𝑦∈Ker𝑄, from 𝑄𝑦=0, we have ^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2𝑄₂𝑦+𝛽𝜂^𝑞+2−1𝑄₁𝑦=0,_{𝑚−2𝑖=1}𝛼_𝑖𝜉_𝑖^𝑝+2𝑄₂𝑦+𝛽𝜂^𝑝+2−1𝑄₁𝑦=0. (28) Because of ^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2𝛽𝜂^𝑞+2−1_{𝑚−2𝑖=1}𝛼_𝑖𝜉_𝑖^𝑝+2𝛽𝜂^𝑝+2−1=−Λ𝑝,𝑞≠0, (29)𝑄₁𝑦=𝑄₂𝑦=0, which yields 𝑦∈Im𝐿. On the other hand, if 𝑦∈Im𝐿, from 𝑄₁𝑦=𝑄₂𝑦=0 and the definition of 𝑄, so 𝑄𝑦=0; thus 𝑦∈Ker𝑄. Hence, Ker𝑄=Im𝐿.
For 𝑦∈𝑍, from 𝑦=(𝑦−𝑄𝑦)+𝑄𝑦, (𝑦−𝑄𝑦)∈Ker𝑄=Im𝐿, 𝑄𝑦∈Im𝑄, we have 𝑍=Im𝐿+Im𝑄. And if 𝑦∈Im𝐿∩Im𝑄, from 𝑦∈Im𝑄, there exist constants 𝑎,𝑏∈𝑅, such that 𝑦(𝑡)=𝑎𝑡^𝑝−1+𝑏𝑡^𝑞−1.
From 𝑦∈Im𝐿, we obtain ×𝑎𝑞𝑞+1𝑞+2_{𝑚−2𝑖=1}𝛼_𝑖𝜉_𝑖^𝑝+2+𝑏𝑝𝑝+1𝑝+2^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2=0,𝑎𝑞𝑞+1𝑞+21−𝛽𝜂^𝑝+2+𝑏𝑝𝑝+1𝑝+21−𝛽𝜂^𝑞+2=0. (30) In view of 𝑞𝑞+1𝑞+2^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+2𝑝𝑝+1𝑝+2^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2𝑞𝑞+1𝑞+21−𝛽𝜂^𝑝+2𝑝𝑝+1𝑝+21−𝛽𝜂^𝑞+2=−𝑞𝑞+1𝑞+2𝑝𝑝+1𝑝+2Λ𝑝,𝑞≠0, (31) therefore (30) has a unique solution 𝑎=𝑏=0, which implies Im𝐿∩Im𝑄={0}. So we have 𝑍=Im𝐿⊕Im𝑄. Since imKer𝐿=dimIm𝑄=codimIm𝐿=2, thus 𝐿 is a Fredholm map of index zero.
Let 𝑃∶𝑌→𝑌 be defined by (𝑡)(0)+1𝑃𝑥=𝑥2𝑥^′′(0)𝑡²,𝑡∈[0,1]. (32) Then, the generalized inverse 𝐾_𝑃∶Im𝐿→dom𝐿∩Ker𝑃 can be written by 𝐾_𝑃𝑦(𝑡)=^𝑡₀^𝑠₀^𝜏₀𝑦(𝑣)𝑑𝑣𝑑𝜏𝑑𝑠,𝑦∈Im𝐿. (33)
In fact, for 𝑦∈Im𝐿, we have 𝐿𝐾_𝑃𝑦(𝑡)=𝐾_𝑃𝑦^′′′=𝑦(𝑡), (34) and for 𝑥∈dom𝐿∩Ker𝑃, we know 𝐾_𝑃𝐿𝑥(𝑡)=𝐾_𝑃𝑥^′′′(𝑡)=^𝑡₀^𝑠₀^𝜏₀𝑥^′′′(𝑣)(𝑡)−𝑥(0)+1𝑑𝑣𝑑𝜏𝑑𝑠=𝑥2𝑥^′′(0)𝑡²=𝑥(𝑡)−𝑃𝑥(𝑡). (35)
Taking note that 𝑥∈dom𝐿∩Ker𝑃, 𝑃𝑥(𝑡)=0, thus (𝐾_𝑃𝐿)𝑥(𝑡)=𝑥(𝑡).
It is clear that ‖𝐾_𝑃𝑦‖≤‖𝑦‖₁.

Theorem 4. Let the condition (𝐶₁) hold and Λ(𝑝,𝑞)≠0. Assume the following.  (𝐻₁) There exist functions 𝛼,𝛽,𝛾,𝜃∈𝐿¹[0,1], such that 𝑓𝑡,𝑥₁,𝑥₂,𝑥₃(𝑡(𝑡𝑥+𝑒)≤𝛼)₁+𝛽(𝑡)𝑥₂+𝛾(𝑡)𝑥₃(𝑡),‖𝛼‖+𝜃₁+𝛽₁+𝛾₁<1, (36)   where ∀(𝑥₁,𝑥₂,𝑥₃)∈𝑅³, 𝑡∈[0,1]. (𝐻₂) There exists a constant 𝐴>0 such that for 𝑥∈dom𝐿, if |𝑥(𝑡)|>𝐴 or |𝑥^′′(𝑡)|>𝐴 for all 𝑡∈[0,1], then 𝑄₁𝑁(𝑥(𝑡))≠0𝑜𝑟𝑄₂𝑁(𝑥(𝑡))≠0. (37) (𝐻₃) There exists a constant 𝐵>0 such that for 𝑎,𝑐∈𝑅, if |𝑎|>𝐵 or |𝑐|>𝐵, then either 𝑄₁𝑁𝑎+𝑐𝑡²+𝑄₂𝑁𝑎+𝑐𝑡²>0 (38)   or 𝑄₁𝑁𝑎+𝑐𝑡²+𝑄₂𝑁𝑎+𝑐𝑡²<0. (39)

Then BVP (1)-(2) has at least one solution in 𝐶²[0,1].

Proof. We divide the proof into the following steps.
Step 1. The set Ω₁={𝑥∈dom𝐿⧵Ker𝐿∶𝐿𝑥=𝜆𝑁𝑥forsome𝜆∈[0,1]} is bounded.
For 𝑥∈Ω₁, since 𝐿𝑥=𝜆𝑁𝑥, so 𝜆≠0, 𝑁𝑥∈ Im 𝐿; hence 𝑄₁𝑁(𝑥(𝑡))=0,𝑄₂𝑁(𝑥(𝑡))=0. (40)
From (𝐻₂), there exist 𝑡₀,𝑡₁∈[0,1] such that |𝑥(𝑡₀)|≤𝐴, |𝑥^′′(𝑡₁)|≤𝐴. 𝑥,𝑥^′, and 𝑥^′′ are absolutely continuous for all 𝑡∈[0,1], and 𝑥(𝑡)=𝑥𝑡₀+^𝑡_𝑡0𝑥^′(𝑠)𝑑𝑠,𝑥^′(𝑡)=𝑥^′(0)+^𝑡₀𝑥^′′(𝑠)𝑥𝑑𝑠_′′(𝑡)=𝑥^′′𝑡₁+^𝑡_𝑡1𝑥^′′′(𝑠)𝑑𝑠, (41) which imply ‖𝑥‖_∞≤𝐴+𝑥^′_∞,𝑥^′_∞≤𝐴+𝑥^′′′₁,𝑥^′′_∞≤𝐴+𝑥^′′′₁. (42)
From (𝐻₁), we obtain 𝑥^′′′₁=‖𝐿𝑥‖₁≤‖𝑁𝑥‖₁≤‖𝛼‖₁‖𝑥‖_∞+𝛽₁𝑥^′_∞+𝛾₁𝑥^′′_∞+‖𝜃‖₁≤‖𝛼‖₁+𝛽₁+𝛾₁𝑥^′′′₁+𝐴2‖𝛼‖₁+𝛽₁+𝛾₁+‖𝜃‖₁,𝑥^′′′₁≤11−‖𝛼‖₁+𝛽₁+𝛾₁×𝐴2‖𝛼‖₁+𝛽₁+𝛾₁+‖𝜃‖₁. (43) So there exists a constant 𝑀₁>0 such that ‖𝑥‖≤𝑀₁; that is, the set Ω₁ is bounded.
Step 2. The set Ω₂={𝑥∈Ker𝐿∶𝑁𝑥∈Im𝐿} is bounded.
For 𝑥∈Ω₂, 𝑥∈Ker𝐿 implies that 𝑥=𝑎+𝑐𝑡², 𝑎=((𝛽𝜂²−1)/(1−𝛽))𝑐, 𝑡∈[0,1], 𝑐∈𝑅. From 𝑄𝑁𝑥=0, we get 𝑄₁𝑁(𝑎+𝑐𝑡²)=𝑄₂𝑁(𝑎+𝑐𝑡²)=0. From (𝐻₃), then ‖𝑥‖≤|𝑎|+|𝑐|≤2𝐵; that is, the set Ω₂ is bounded.
Step 3. The set Ω₃={𝑥∈Ker𝐿∶𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0, 𝜆∈[0,1]} is bounded.
For any 𝑎,𝑐∈𝑅, we define the linear isomorphism 𝐽∶Ker𝐿→Im𝑄 by 𝐽𝑎+𝑐𝑡²=𝑎^′𝑡^𝑝−1+𝑏^′𝑡^𝑞−1Λ,𝑝,𝑞 (44) where 𝑎^′=𝑝𝑝+1𝑝+2𝛽𝜂^𝑞+2−1|𝑎|+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2|𝑐,𝑏_′|=−𝑞𝑞+1𝑞+2𝛽𝜂^𝑝+2−1|𝑎|+^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+2|𝑐|. (45)
Set 𝑝𝜇=𝑝+1𝑝+2Λ𝑝,𝑞𝛽𝜂^𝑞+2,𝑝−1𝜈=𝑝+1𝑝+2Λ𝑝,𝑞^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑞+2,𝑞𝜌=−𝑞+1𝑞+2Λ𝑝,𝑞𝛽𝜂^𝑝+2,𝑞−1𝜔=−𝑞+1𝑞+2Λ𝑝,𝑞^𝑚−2_𝑖=1𝛼_𝑖𝜉_𝑖^𝑝+2. (46)
For any 𝑥(𝑡)=𝑎+𝑐𝑡²∈Ω₃, we obtain 𝜇𝜆|𝑎|+(1−𝜆)𝑄₁𝑁𝑎+𝑐𝑡²+𝜈𝜆|𝑐|+(1−𝜆)𝑄₂𝑁𝑎+𝑐𝑡²𝜌𝜆|𝑎|+()𝑄=0,1−𝜆₁𝑁𝑎+𝑐𝑡²+𝜔𝜆|𝑐|+(1−𝜆)𝑄₂𝑁𝑎+𝑐𝑡²=0. (47) On account of 𝜇𝜈𝑝𝑞𝜌𝜔=−𝑝+1𝑝+2𝑞+1𝑞+2Λ𝑝,𝑞≠0, (48) therefore, we have 𝜆|𝑎|+(1−𝜆)𝑄₁𝑁𝑎+𝑐𝑡²𝜆|𝑐|+()𝑄=0,1−𝜆₂𝑁𝑎+𝑐𝑡²=0. (49)
If 𝜆=1, then 𝑎=𝑐=0. If 𝜆≠1 and |𝑎|>𝐵 or |𝑐|>𝐵, from the above equality and (38), one has 𝜆𝑎|+|𝑐()×𝑄(||)=−1−𝜆₁𝑁𝑎+𝑐𝑡²+𝑄₂𝑁𝑎+𝑐𝑡²<0, (50) which contradicts 𝜆(|𝑎|+|𝑐|)≥0; thus ‖𝑥‖≤|𝑎|+|𝑐|≤2𝐵. So the set Ω₃ is bounded.
Step 4. If (39) holds, similar to the above argument, we can prove that the set Ω₃={𝑥∈Ker𝐿∶−𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈[0,1]} (51) is bounded too, where J is defined in (44).
Now, we will prove that all conditions of Theorem 1 are satisfied.
Let Ω be an open bounded subset of 𝑌 such that ∪³_𝑖=1Ω_𝑖⊂Ω. By the Arzelá-Ascoli theorem, we can prove that 𝐾_𝑃(𝐼−𝑄)𝑁∶Ω→𝑌 is compact, so 𝑁 is 𝐿-compact on Ω.
Then by the above argument, we have(i)𝐿𝑥≠𝑁𝑥 for every (𝑥,𝜆)∈[(dom𝐿⧵Ker𝐿)∩𝜕Ω]×(0,1);(ii)𝑁𝑥∉ Im𝐿 for every 𝑥∈Ker𝐿∩𝜕Ω(iii)let 𝐻(𝑥,𝜆)=±𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥.
According to the above argument in Steps  3 and  4, we know 𝐻(𝑥,𝜆)≠0 for every 𝑥∈𝜕Ω∩Ker𝐿. Thus, by using the homotopy property of degree, we have deg𝑄𝑁|_Ker𝐿(𝐻())(𝐻())(),Ω∩Ker𝐿,0=deg⋅,0,Ω∩Ker𝐿,0=deg⋅,1,Ω∩Ker𝐿,0=deg±𝐽,Ω∩Ker𝐿,0=±1≠0. (52) Then by Theorem 1, 𝐿𝑥=𝑁𝑥 has at least one solution in dom𝐿∩Ω; that is, BVP (1)-(2) has at least one solution in 𝐶²[0,1].

4. Example

Example 1. We consider the following boundary value problem: 𝑥^′′′(𝑡)=𝑡²3+4+8𝑥11+2sin𝑥^′+14𝑡cos𝑥^′′2(),𝑥(0)=,𝑡∈0,168𝑥1592,𝑥^′(0)(1)=1=0,𝑥2𝑥13. (53)

Let 𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡),𝑥^′′(𝑡)3=4+8𝑥11+2sin𝑥^′+14𝑡cos𝑥^′′2,𝑒(𝑡)=𝑡²,𝑚=3,𝛼₁=68159,𝛽=2,𝜉₁=121,𝜂=3. (54) Then the condition (𝐶₁) holds.

From Lemma 2, one has Λ(1,2)=−697/9558≠0. By Lemma 3, we define 𝑄₁𝑦=6859₀^1/2𝑠₀^𝜏₀𝑦(𝑣)𝑄𝑑𝑣𝑑𝜏𝑑𝑠,₂𝑦=¹₀^𝑠₀^𝜏₀𝑦(𝑣)−1𝑑𝑣𝑑𝜏𝑑𝑠2₀^1/3𝑠₀^𝜏₀𝑦(𝑣)𝑇𝑑𝑣𝑑𝜏𝑑𝑠,₁𝑦=−57348−1697𝑄162¹𝑦+17𝑄236²𝑦,𝑇₂𝑦=229392−69753𝑄54¹𝑦+17𝑄118²𝑦,𝐾_𝑃𝑦(𝑡)=^𝑡₀^𝑠₀^𝜏₀𝑦(𝑣)(𝑡)=𝑇𝑑𝑣𝑑𝜏𝑑𝑠,𝑦∈Im𝐿.𝑄𝑦₁𝑦𝑡¹⁻¹+𝑇₂𝑦𝑡²⁻¹=𝑇₁𝑦+𝑇₂𝑦𝑡. (55)

Since |𝑓(𝑡,𝑥(𝑡),𝑥^′(𝑡),𝑥^′′(𝑡))+𝑒(𝑡)|≤(9/16)|𝑥(𝑡)|+𝑡²+(1/4)𝑡+4, then 𝛼(𝑡)=9/16, 𝛽(𝑡)=𝛾(𝑡)=0, 𝜃(𝑡)=𝑡²+(1/4)𝑡+4.

If 𝑥(𝑡)>1=𝐴, |𝑥^′′(𝑡)|>1=𝐴, and 𝐵=1, one has 𝑄₁𝑁𝑥=6859₀^1/2𝑠₀^𝜏₀𝑣²3+4+8𝑥11+2sin𝑥^′+14𝑣cos𝑥^′′2≥𝑑𝑣𝑑𝜏𝑑𝑠6859₀^1/2𝑠₀^𝜏₀𝑣²3+4+−1164𝑣=𝑑𝑣𝑑𝜏𝑑𝑠6859₀^1/2𝑠₀^𝜏₀1𝑣−8²+267𝑄64𝑑𝑣𝑑𝜏𝑑𝑠>0,₂𝑁𝑥=¹₀^𝑠₀^𝜏₀𝑣²3+4+8𝑥11+2sin𝑥^′+14𝑣cos𝑥^′′2−1𝑑𝑣𝑑𝜏𝑑𝑠2₀^1/3𝑠₀^𝜏₀𝑣²3+4+8𝑥11+2sin𝑥^′+14𝑣cos𝑥^′′2≥1𝑑𝑣𝑑𝜏𝑑𝑠2₀^1/3𝑠₀^𝜏₀𝑣²3+4+8𝑥11+2sin𝑥^′+14𝑣cos𝑥^′′2≥1𝑑𝑣𝑑𝜏𝑑𝑠2₀^1/3𝑠₀^𝜏₀𝑣²3+4+−1164𝑣=1𝑑𝑣𝑑𝜏𝑑𝑠2₀^1/3𝑠₀^𝜏₀1𝑣−8²+267𝑄64𝑑𝑣𝑑𝜏𝑑𝑠>0,₁𝑁2+𝑡²+𝑄₂𝑁2+𝑡²>0. (56) Then BVP (53) satisfies Theorem 4. So it has at least one solution in 𝐶²[0,1].

Remark 2. By using a similar method as employed in the above proof, we could obtain some similar results under the condition (𝐶₃) or (𝐶₅), then we omit them.

Conflict of Interests

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

Acknowledgments

This paper is sponsored by the Natural Science Foundation of China (11071205, 11101349, 61201431), the Natural Science Foundation of Jiangsu Province, and PAPD of Jiangsu Higher Education Institutions.