Positive Solutions for a Fourth-Order Boundary Value Problem

Abstract

This paper deals with the existence and multiplicity of positive solutions for the fourth-order boundary value problem 𝑢⁽⁴⁾=𝑓(𝑡,𝑢,𝑢^′,−𝑢^′′,𝑢^′′′),𝑢(0)=𝑢^′(1)=𝑢^′′′(0)=𝑢^′′(1)=0. Here 𝑓∈𝐶([0,1]×ℝ⁴₊,ℝ₊) (ℝ₊∶=[0,+∞)). We use fixed point index theory to establish our main results based on a priori estimates achieved by utilizing some integral identities and integral inequalities.

1. Introduction

The purpose of this paper is to study the existence and multiplicity of positive solutions for the fourth-order boundary value problem 𝑢⁽⁴⁾=𝑓𝑡,𝑢,𝑢^′,−𝑢^′′,𝑢^′′′,𝑢(0)=𝑢^′(1)=𝑢^′′′(0)=𝑢^′′(1)=0,(1) where 𝑓∈𝐶([0,1]×ℝ⁴₊,ℝ₊). By a positive solution of (1), we mean a function 𝑢∈𝐶⁴([0,1],ℝ₊) that solves (1) and satisfies 𝑢(𝑡)>0 for all 𝑡∈(0,1].

In recent years, fourth-order boundary value problems for nonlinear ordinary differential equations have been extensively studied by using diverse methods, including fixed point theorems on cones [1–6], the method of lower and upper solutions [7–10], the iterative method [11–14], critical point theory [15], and the shooting method [16]. It should be remarked that none of the results cited above involves derivatives of all orders in their nonlinearities. In [9], Ma and Yang studied the fourth-order four-point boundary value problem 𝑢⁽⁴⁾=𝑓𝑡,𝑢,−𝑢^′′,𝑢(0)(1)𝜂,𝑢=0,𝑢=𝑎𝑢_′′(0)=0,𝑢^′′(1)=𝑏𝑢(𝜉),(2) where 𝜂,𝜉∈(0,1), 𝑎,𝑏⩾0. The authors used the method of upper and lower solutions and a new maximum principle to establish their existence results. In [14], Pei and Chang studied a class of fourth-order boundary value problem 𝑦⁽⁴⁾,𝑦(0)(1)=𝑓𝑡,𝑦=𝑦=𝑦^′(0)=𝑦^′(1)=0.(3) By using the monotone iterative technique, the authors proved that problem (3) has at least one symmetric positive solution under certain conditions. In [17], Pang et al. studied the existence and multiplicity of nontrivial solutions for the fourth-order boundary value problem𝑥⁽⁴⁾=𝑓𝑥,−𝑥^′′,𝑥(0)=𝑥(1)=𝑥^′′(0)=𝑥^′′(1)=0,(4) where 𝑓∈𝐶(ℝ²,ℝ). Making use of the theory of Leray-Schauder degree, under appropriate conditions on the nonlinearity 𝑓, the authors proved that problem (4) has at least six different nontrivial solutions, including two positive ones and two negative ones.

In [18], Yang and Sun studied the fourth-order boundary value problem 𝑢⁽⁴⁾=𝑓𝑡,𝑢,𝑢^′,−𝑢^′′,−𝑢^′′′,𝑢(0)=𝑢^′(1)=𝑢^′′(0)=𝑢^′′′(1)=0,(5) where 𝑓∈𝐶([0,1]×ℝ⁴₊,ℝ₊). By using fixed point index theory based on a priori estimates, the authors established the main results on the existence, multiplicity and uniqueness of positive solutions for (5).

Motivated by [18], in this paper, we will discuss the existence and multiplicity of positive solutions for problem (1). As in [18], we first use the method of order reduction to transform (1) into a boundary value problem for a second-order integro-differential equation and then seek the existence and multiplicity of positive solutions for the resultant problem. To overcome the difficulties stemming from the presence of derivatives of all orders and the difference between (1) and (5) in their boundary value conditions, we have to prove that the maximum of every nonnegative concave function 𝑣 can be dominated by the integrals¹₀𝑣(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡,₁₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡(6) (see Lemmas 5 and 6 below for more details). Based on a priori estimates achieved by utilizing some integral identities and integral inequalities, we use fixed point index theory to prove the existence and multiplicity of positive solutions for (1).

This paper is organized as follows. In Section 2, we transform (1) into a boundary value problem for a second-order integro-differential equation and then establish some basic integral identities and integral inequalities that are useful in deriving the priori estimates in the next section. Our main results, namely Theorems 10–12, are stated and proved in Section 3.

2. Preliminaries

Let 𝐸∶=𝐶¹([0,1],ℝ) and ‖𝑢‖∶=max{‖𝑢‖₀,‖𝑢^′‖₀}, where ‖𝑢‖₀∶=max{|𝑢(𝑡)|∶𝑡∈[0,1]}. Furthermore, let 𝑃∶=𝑢∈𝐸∶𝑢(𝑡)⩾0,𝑢^′(𝑡)0,∀𝑡∈[0,1].(7) Clearly (𝐸,‖⋅‖) is a real Banach space and 𝑃 is a cone in 𝐸.

Let 𝑘₁(){},𝑘𝑡,𝑠∶=min𝑡,𝑠₂(𝑡,𝑠)∶=min{1−𝑠,1−𝑡}.(8)

Define the linear integral operators 𝐵₁ and 𝐵₂ by 𝐵_𝑖𝑣(𝑡)∶=¹₀𝑘_𝑖(𝑡,𝑠)𝑣(𝑠)d𝑠,𝑣∈𝑃(𝑖=1,2).(9) Substituting 𝑣(𝑡)∶=−𝑢^′′(𝑡) into (1), we transform (1) into the following second-order boundary value problem for the integro-differential equation −𝑣^′′(𝑡)=𝑓𝑡,𝐵₁𝑣(𝑡),𝐵₁𝑣^′(𝑡),𝑣(𝑡),−𝑣^′(𝑡),𝑣^′(0)=𝑣(1)=0,(10) which is equivalent to the nonlinear integral equation 𝑣(𝑡)=¹₀𝑘₂(𝑡,𝑠)𝑓𝑠,𝐵₁𝑣(𝑠),𝐵₁𝑣^′(𝑠),𝑣(𝑠),−𝑣^′(𝑠)d𝑠.(11)

Define the operator 𝐴∶𝑃→𝑃 by (𝐴𝑣)(𝑡)∶=¹₀𝑘₂(𝑡,𝑠)𝑓𝑠,𝐵₁𝑣(𝑠),𝐵₁𝑣^′(𝑠),𝑣(𝑠),−𝑣^′(𝑠)d𝑠,𝑣∈𝑃.(12)

Now 𝑓∈𝐶([0,1]×ℝ⁴₊,ℝ₊) implies that 𝐴 is a completely continuous operator. In our setting, the existence of positive solutions for (1) is equivalent to that of positive fixed points of 𝐴∶𝑃→𝑃.

To establish the priori estimates of positive solutions for some problems associated with (11), we need several integral identities and integral inequalities below.

Lemma 1. If 𝑣∈𝑃∩𝐶²[0,1],𝑣^′(0)=𝑣(1)=0, then ¹₀−𝑣^′′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡=¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡.(13)

Proof. Integrating by parts and using 𝑣^′(0)=𝑣(1)=0, we have ¹₀−𝑣^′′(𝑡)(1−𝑡)𝑒^1−𝑡=d𝑡₁₀𝑣^′(𝑡)(𝑡−2)𝑒^1−𝑡=d𝑡₁₀−𝑣^′(𝑡)𝑡𝑒^1−𝑡d𝑡−2¹₀𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡,(14)₁₀−𝑣^′(𝑡)𝑡𝑒^1−𝑡d𝑡=¹₀𝑣(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡,(15) from which (13) follows. This completes the proof.

Lemma 2. If 𝑣∈𝐶([0,1],ℝ₊), then ¹₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)𝑡𝑒^𝑡d𝑡=¹₀𝑣(𝑡)𝑡𝑒^𝑡d𝑡.(16)

Proof. Notice (𝐵₁𝑣)(0)=0. Integrating by parts, we obtain ¹₀𝐵₁𝑣(𝑡)𝑡𝑒^𝑡d𝑡=¹₀(1−𝑡)𝑒^𝑡𝐵₁𝑣^′(𝑡)d𝑡,(17) so that ¹₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)𝑡𝑒^𝑡d𝑡=¹₀(1+𝑡)𝑒^𝑡𝐵₁𝑣^′(𝑡)d𝑡.(18) Notice (𝐵₁𝑣)^′(1)=0 and (𝐵₁𝑣)^′′(𝑡)=−𝑣(𝑡). Integrating the right hand of the above by parts again gives (16). This completes the proof.

If both ℎ₁ and ℎ₂ are continuous and decreasing on [0,1], then Chebyshev's inequality for ℎ₁ and ℎ₂ is given by ¹₀ℎ₁(𝑡)ℎ₂(𝑡)d𝑡⩾¹₀ℎ₁(𝑡)d𝑡¹₀ℎ₂(𝑡)d𝑡.(19) In addition, if both ℎ₁ and ℎ₂ are continuous, with one of them decreasing on [0,1] and the other increasing on [0,1], then Chebyshev's inequality for ℎ₁ and ℎ₂ becomes ¹₀ℎ₁(𝑡)ℎ₂(𝑡)d𝑡⩽¹₀ℎ₁(𝑡)d𝑡¹₀ℎ₂(𝑡)d𝑡.(20)

Lemma 3. If 𝑣∈𝑃,𝑣(1)=0, and 𝑣^′ is decreasing on [0,1], then ¹₀𝑣(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩾(𝑒−2)¹₀−𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡.(21)

Proof. Notice 𝑣(1)=0 and 𝑣^′⩽0. Since −𝑣^′ is increasing on [0,1] and (1−𝑡)𝑒^1−𝑡 is decreasing on [0,1], it follows from Chebyshev's inequality that ¹₀−𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩽¹₀−𝑣^′(𝑡)d𝑡¹₀(1−𝑡)𝑒^1−𝑡(0).d𝑡=𝑣(22)
Notice that 𝑡𝑒^1−𝑡 is increasing on [0,1]. Chebyshev's inequality again implies ¹₀−𝑣^′(𝑡)𝑡𝑒^1−𝑡d𝑡⩾¹₀−𝑣^′(𝑡)d𝑡¹₀𝑡𝑒^1−𝑡d𝑡=(𝑒−2)𝑣(0).(23)
Note that (15) is guaranteed by 𝑣∈𝑃,𝑣(1)=0. By virtue of (15), we have ¹₀𝑣(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩾(𝑒−2)𝑣(0).(24)
Combining this with (22) yields (21). This complete the proof.

Lemma 4. If 𝑣∈𝑃,𝑣(1)=0, then 𝑣(0)⩽¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡.(25)

Proof. By (15), we have ¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡=d𝑠₁₀−𝑣^′(𝑡)(2−𝑡)𝑒^1−𝑡⩾d𝑡₁₀−𝑣^′(𝑡)d𝑡=𝑣(0).(26) This completes the proof.

Lemma 5. If 𝑣∈𝑃∩𝐶²[0,1] and 𝑣^′′⩽0, then (𝑒−2)‖𝑣‖₀⩽¹₀𝑣(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩽𝑒‖𝑣‖₀.(27)

Proof. It is easy to see that ∫¹₀𝑣(𝑡)𝑡𝑒^𝑡d𝑡⩽𝑒‖𝑣‖₀. The condition 𝑣^′⩽0 implies ‖𝑣‖₀=𝑣⩽(0). By the concavity of 𝑣, we have ¹₀𝑣(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩾¹₀(1−𝑡)𝑒^1−𝑡𝑣(())(0)𝑡⋅1+1−𝑡⋅0d𝑡⩾𝑣¹₀(1−𝑡)²𝑒^1−𝑡d𝑡=(𝑒−2)‖𝑣‖₀.(28) This completes the proof.

Lemma 6. If 𝑣∈𝑃∩𝐶²[0,1] and 𝑣^′′⩽0, then we have 3−𝑒6‖𝑣‖₀⩽¹₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡‖𝑣‖d𝑡⩽3𝑒₀.(29)

Proof. By the concavity of 𝐵₁𝑣 and 𝑣, we have ¹₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡⩾d𝑡₁₀𝐵₁𝑣(𝑡)(1−𝑡)𝑒^1−𝑡=d𝑡₁₀𝐵₁𝑣(𝑡⋅1+(1−𝑡)⋅0)(1−𝑡)𝑒^1−𝑡⩾𝐵d𝑡₁𝑣(1)¹₀𝑡(1−𝑡)𝑒^1−𝑡=()d𝑡3−𝑒¹₀(𝑠)⩾()𝑠𝑣d𝑠3−𝑒¹₀𝑠()𝑣(0)=1−𝑠d𝑠3−𝑒6‖𝑣‖₀,₁₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩽𝑒¹₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)d𝑡=𝑒¹₀¹₀𝑘₁(𝑡,𝑠)𝑣(𝑠)d𝑠+2¹₀𝑣(𝑠)‖𝑣‖d𝑠d𝑡⩽3𝑒₀.(30) This completes the proof.

Lemma 7 (see [19]). Let 𝐸 be a real Banach space and 𝑃 a cone in 𝐸. Suppose that Ω⊂𝐸 is a bounded open set and that 𝑇∶Ω∩𝑃→𝑃 is a completely continuous operator. If there exists 𝜔₀∈𝑃⧵{0} such that 𝜔−𝑇𝜔≠𝜆𝜔₀,∀𝜆⩾0,𝜔∈𝜕Ω∩𝑃,(31) then 𝑖(𝑇,Ω∩𝑃,𝑃)=0, where 𝑖 indicates the fixed point index on 𝑃.

Lemma 8 (see [19]). Let 𝐸 be a real Banach space and 𝑃 a cone in 𝐸. Suppose that Ω⊂𝐸 is a bounded open set with 0∈Ω and that 𝑇∶Ω∩𝑃→𝑃 is a completely continuous operator. If 𝜔−𝜆𝑇𝜔≠0,∀𝜆∈[0,1],𝜔∈𝜕Ω∩𝑃,(32) then 𝑖(𝑇,Ω∩𝑃,𝑃)=1.

3. Existence and Multiplicity of Positive Solutions for (1)

For simplicity, we denote by 𝑥∶=(𝑥₁,𝑥₂,𝑥₃,𝑥₄)∈ℝ⁴₊ and 𝐼_𝜌∶=[0,𝜌] for 𝜌>0 in this section. Now we list our hypotheses.(H1)𝑓∈𝐶([0,1]×ℝ⁴₊,ℝ₊). (H2) There exist 𝑎₁⩾0, 𝑏₁⩾0, 𝑐>0, such that 𝑑₁∶=𝑎₁(3−𝑒(𝑒−2))/6𝑒²+𝑏₁>1, and 𝑓(𝑡,𝑥)⩾𝑎₁𝑥₁+2𝑥₂+𝑏₁𝑥₃+2𝑥₄−𝑐(33) holds for all (𝑡,𝑥)∈[0,1]×ℝ⁴₊.(H3) There exists a function Φ∈𝐶(ℝ₊,ℝ₊) such that 𝑓(𝑡,𝑥)⩽Φ𝑥₄(34) for all 𝑥∈[0,1]×𝐼³_𝑚×ℝ₊ and ^∞₀𝜉d𝜉Φ(𝜉)+2𝑚𝑒>𝑚.(35) where 𝑚∶=𝑐/(𝑑₁−1)>0 and 𝑐,𝑑₁ are given as in (H2).(H4) There exist 𝑎₂⩾0, 𝑏₂⩾0, 𝑟>0 such that 𝑑₂∶=3𝑎₂𝑒/(𝑒−2)+𝑏₂<1, and 𝑓(𝑡,𝑥)⩽𝑎₂𝑥₁+2𝑥₂+𝑏₂𝑥₃+2𝑥₄(36) holds for all (𝑡,𝑥)∈[0,1]×𝐼³_𝑟.(H5) There exist 𝑎₃⩾0, 𝑏₃⩾0, 𝑐>0 such that 𝑑₃∶=3𝑎₃𝑒0𝑥02215(𝑒−2)+𝑏₂<1, and 𝑓(𝑡,𝑥)⩽𝑎₃𝑥₁+2𝑥₂+𝑏₃𝑥₃+2𝑥₄+𝑐(37) holds for all (𝑡,𝑥)∈[0,1]×ℝ⁴₊.(H6) There are nonnegative constants 𝑎₄⩾0, 𝑏₄⩾0, 𝑟>0, such that 𝑑₄∶=𝑎₄(3−𝑒)(𝑒−2)/6𝑒²+𝑏₄>1, and 𝑓(𝑡,𝑥)⩾𝑎₄𝑥₁+2𝑥₂+𝑏₄𝑥₃+2𝑥₄(38) holds for all (𝑡,𝑥)∈[0,1]×𝐼⁴_𝑟.(H7) There is a constant 𝜔>0 such that 𝑓(𝑡,𝑥) is increasing on 𝐼⁴_𝜔 in 𝑥, and ¹₀𝑓(𝑠,𝜔,𝜔,𝜔,𝜔)d𝑠<𝜔.(39)

Remark 9. 𝑓(𝑡,𝑥) is said to be increasing in 𝑥 if 𝑓(𝑡,𝑥)⩽𝑓(𝑡,𝑦) holds for every pair 𝑥,𝑦∈ℝ⁴₊ with 𝑥⩽𝑦 and for all 𝑡∈[0,1], where the partial ordering in ⩽ℝ⁴ is understood componentwise.

We denote Ω_𝜌∶={𝑣∈𝐸∶‖𝑣‖<𝜌} for 𝜌>0 in the sequel.

Theorem 10. If (H1)–(H4) hold, then (1) has at least one positive solution.

Proof. Let ℳ₁∶=𝑣∈𝑃∶𝑣=𝐴𝑣+𝜇𝜑,𝜇⩾0,(40) where 𝜑(𝑡)∶=(1−𝑡)𝑒^𝑡−1∈𝑃. We are now going to prove that ℳ₁ is bounded. Indeed, if 𝑣∈ℳ₁, then there exist 𝜇⩾0 such that 𝑣=𝐴𝑣+𝜇𝜑, which can be written in the form 𝑣(𝑡)=¹₀𝑘₂(𝑡,𝑠)𝑓𝑠,𝐵₁𝑣(𝑠),𝐵₁𝑣^′(𝑠),𝑣(𝑠),−𝑣^′(𝑠(𝑡),)d𝑠+𝜇𝜑(41) which is equivalent to −𝑣^′′(𝑡)=𝑓(𝑡),𝐵₁𝑣(𝑡),𝐵₁𝑣^′(𝑡),𝑣(𝑡),−𝑣^′(𝑡)+𝜇(1+𝑡)𝑒^𝑡−1.(42) By (H2), we have −𝑣^′′(𝑡)⩾𝑎₁𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)+𝑏₁𝑣(𝑡)−2𝑣^′(𝑡)−𝑐.(43) Note (27) and (29). Multiply the above by (1−𝑡)𝑒^1−𝑡, integrate over [0,1], and use Lemmas 3, 5, and 6 to obtain ¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩾𝑎₁¹₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑏₁¹₀𝑣(𝑡)2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡⩾𝑎d𝑡−𝑐₁(3−𝑒)6‖𝑣‖₀+𝑏₁¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡⩾𝑎d𝑡−𝑐₁(3−𝑒)6𝑒¹₀𝑣(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑏₁¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡−𝑐⩾𝑑₁¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡−𝑐,(44) so that ¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩽𝑚,∀𝑣∈ℳ₁.(45) Now Lemma 4 implies ‖𝑣‖₀=𝑣(0)⩽𝑚,∀𝑣∈ℳ₁,(46) where 𝑚 is given in (H3). Furthermore, this estimate leads to 𝐵₁𝑣(𝑡)₀=𝐵₁𝑣(1)=¹₀(𝑠)𝐵𝑠𝑣d𝑠⩽𝑚,₁𝑣^′(𝑡)₀=𝐵₁𝑣^′(0)=¹₀𝑣(𝑠)d𝑠⩽𝑚(47) for all 𝑣∈ℳ₁. Let .Λ∶=𝜇⩾0∶thereexistsome𝑣∈𝑃suchthat𝑣=𝐴𝑣+𝜇𝜑(48) Now (46) implies that 𝜇⩽𝑚𝑒 for all 𝜇∈Λ. By (H3), there is a function Φ∈𝐶(ℝ₊,ℝ₊) such that 𝑓𝑡,𝐵₁𝑣(𝑡),𝐵₁𝑣^′(𝑡),𝑣(𝑡),−𝑣^′(𝑡)⩽Φ−𝑣^′(𝑡),∀𝑣∈ℳ₁.(49) Thus −𝑣^′′(𝑡)=𝑓𝑡,𝐵₁𝑣(𝑡),𝐵₁𝑣^′(𝑡),𝑣(𝑡),−𝑣^′(𝑡)+𝜇(1+𝑡)𝑒^𝑡−1⩽Φ−𝑣^′(𝑡)+𝜇(1+𝑡)𝑒^𝑡−1⩽Φ−𝑣^′(𝑡)+2𝑚𝑒,∀𝑣∈ℳ₁,𝜇∈Λ,_{−𝑣′0(1)}𝜉d𝜉Φ(𝜉)⩽+2𝑚𝑒₁₀−𝑣^′(𝑡)d𝑡=𝑣(0)⩽𝑚,∀𝑣∈ℳ₁.(50) Combining the last inequality with (35), we see that there exists 𝑀>0 such that −𝑣(1)⩽𝑀 for all 𝑣∈ℳ₁. This means that ℳ₁ is bounded. Taking 𝑅>sup{‖𝑣‖∶𝑣∈ℳ₁}, we have 𝑣≠𝐴𝑣+𝜇𝜑,∀𝑣∈Ω_𝑅∩𝑃,𝜇≥0.(51)
Now Lemma 7 yields 𝑖𝐴,Ω_𝑅∩𝑃,𝑃=0.(52)
Let ℳ₂∶=𝑣∈Ω_𝑟∩𝑃∶𝑣=𝜆𝐴𝑣,𝜆∈[0,1].(53)
Now we want to prove that ℳ₂={0}. In fact, if 𝑣∈ℳ₂, then 𝑣∈Ω_𝑟∩𝑃 and there is 𝜆∈[0,1] such that 𝑣=𝜆𝐴𝑣, that is, 𝑣(𝑡)=𝜆¹₀𝑘₂()𝐵𝑡,𝑠×𝑓𝑠,₁𝑣(𝑠),𝐵₁𝑣^′(𝑠),𝑣(𝑠),−𝑣^′(𝑠)d𝑠,(54)
which can be written in the form −𝑣^′′(𝑡)=𝜆𝑓𝑡,𝐵₁𝑣(𝑡),𝐵₁𝑣^′(𝑡),𝑣(𝑡),−𝑣^′(𝑡).(55)
By (H4), we have −𝑣^′′(𝑡)⩽𝑎₂𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)+𝑏₂𝑣(𝑡)−2𝑣^′(𝑡).(56)
Note (27) and (29). Multiply (56) by (1−𝑡)𝑒^1−𝑡, integrate over [0,1], and use Lemmas 5 and 6 to obtain ¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩽𝑎₂¹₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑏₂¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩽3𝑒𝑎₂‖𝑣‖₀+𝑏₂¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡⩽d𝑡3𝑒𝑎₂𝑒−2¹₀𝑣(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑏₂¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩽𝑑₂¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡,(57) so that ∫¹₀(𝑣(𝑡)−2𝑣^′(𝑡))(1−𝑡)𝑒^1−𝑡=0, whence 𝑣(𝑡)≡0 and ℳ₂={0}, as required. A consequence of this is 𝑣≠𝜆𝐴𝑣,∀𝑣∈𝜕Ω_𝑟∩𝑃,𝜆∈[0,1].(58) Now Lemma 8 yields 𝑖𝐴,Ω_𝑟∩𝑃,𝑃=1.(59) This together with (52) implies 𝑖𝐴,Ω_𝑅⧵Ω_𝑟∩𝑃,𝑃=0−1=−1.(60) Therefore 𝐴 has at least one fixed point on (Ω_𝑅⧵Ω_𝑟)∩𝑃 and thus (1) has at least one positive solution. This completes the proof.

Theorem 11. If (H1), (H5), and (H6) hold, then (1) has at least one positive solution.

Proof. Let ℳ₃∶={𝑣∈𝑃∶𝑣=𝜆𝐴𝑣,𝜆∈[0,1]}.(61)
We now assert that ℳ₃ is bounded. Indeed, if 𝑣∈ℳ₃, then there is 𝜆∈[0,1] such that 𝑣=𝜆𝐴𝑣, which is equivalent to −𝑣^′′(𝑡)=𝜆𝑓𝑡,𝐵₁𝑣(𝑡),𝐵₁^′(𝑡),𝑣(𝑡),−𝑣^′(𝑡).(62)
Now (H5) implies −𝑣^′′(𝑡)⩽𝑎₃𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)+𝑏₃𝑣(𝑡)−2𝑣^′(𝑡)+𝑐.(63)
Note (27) and (29). Multiply the above by (1−𝑡)𝑒^1−𝑡, integrate over [0,1], and use Lemmas 5 and 6 to obtain ¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩽𝑎₃¹₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑏₃¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑐⩽3𝑒𝑎₃‖𝑣‖₀+𝑏₃¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑐⩽𝑑₃¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑐,(64) so that ∫¹₀(𝑣(𝑡)−2𝑣^′(𝑡))(1−𝑡)𝑒^1−𝑡d𝑡⩽𝑐₁ for all 𝑣∈ℳ₃, where 𝑐₁∶=𝑐/(1−𝑑₃)>0. By Lemma 4, we obtain ‖𝑣‖₀=𝑣(0)⩽¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩽𝑐₁,∀𝑣∈ℳ₃.(65)
Furthermore, this estimate leads to 𝐵₁𝑣₀=𝐵₁𝑣(1)=¹₀𝑠𝑣(𝑠)d𝑠⩽𝑐₁,𝐵₁𝑣^′₀=𝐵₁𝑣^′(0)=¹₀𝑣(𝑠)d𝑠⩽𝑐₁(66) for all 𝑣∈ℳ₃. By (63), we have −𝑣^′′(𝑡)⩽3𝑎₃+𝑏₃𝑐₁−2𝑏₃𝑣^′(𝑡)+𝑐,∀𝑣∈ℳ₃.(67)
Note 𝑣(1)=𝑣^′(0)=0 and 𝑣^′′⩽0. Integrating the above over [0,1], we obtain 𝑣^′₀=−𝑣^′(1)⩽3𝑎₃+𝑏₃𝑐₁+2𝑏₃𝑣(0)𝑎+𝑐⩽3₃+𝑏₃𝑐₁+𝑐.(68)
This proves the boundedness of ℳ₃. Taking 𝑅>sup{‖𝑣‖∶𝑣∈ℳ₃}, we have 𝑣≠𝜆𝐴𝑣,∀𝑣∈𝜕Ω_𝑅∩𝑃,𝜆∈[0,1].(69)
Now Lemma 8 yields 𝑖𝐴,Ω_𝑅∩𝑃,𝑃=1.(70) On the other hand, let ℳ₄∶=𝑣∈Ω_𝑟∩𝑃∶𝑣=𝐴𝑣+𝜆𝜑,𝜆⩾0,(71) where 𝜑∈𝑃⧵{0} is given by 𝜑(𝑡)∶=(1−𝑡)𝑒^𝑡−1. Now we want to prove that ℳ₄⊂{0}. Indeed, if 𝑣∈ℳ₄, then there exists 𝜆⩾0 such that 𝑣=𝐴𝑣+𝜆𝜑, which can be written as −𝑣^′′(𝑡)=𝑓𝑡,𝐵₁𝑣(𝑡),𝐵₁𝑣^′(𝑡),𝑣(𝑡),−𝑣^′(𝑡)+𝜆(𝑡+1)𝑒^𝑡−1.(72) By (H6), we have −𝑣^′′(𝑡)𝑎₄⩾𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)+𝑏₄𝑣(𝑡)−2𝑣^′(𝑡).(73) Note (27) and (29). Multiply the above by (1−𝑡)𝑒^1−𝑡, integrate over [0,1], and use Lemmas 3, 5, and 6 to obtain ¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡⩾𝑎₄¹₀𝐵₁𝑣(𝑡)+2𝐵₁𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑏₄¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡⩾𝑎d𝑡,₄(3−𝑒)‖𝑣‖6𝑒⁰+𝑏₄¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡⩾𝑎d𝑡,₄(3−𝑒)6𝑒¹₀𝑣(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡+𝑏₄¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡,⩾𝑑₄¹₀𝑣(𝑡)−2𝑣^′(𝑡)(1−𝑡)𝑒^1−𝑡d𝑡,(74) so that ∫¹₀(𝑣(𝑡)−2𝑣^′(𝑡))(1−𝑡)𝑒^1−𝑡d𝑡=0, whence 𝑣(𝑡)≡0. This proves ℳ₄⊂{0}, as required. As a result of this, we obtain 𝑣≠𝐴𝑣+𝜆𝜑,∀𝑣∈𝜕Ω_𝑟∩𝑃,𝜆⩾0.(75)
Lemma 7 implies 𝑖𝐴,Ω_𝑟∩𝑃,𝑃=0.(76)
Now (70) and (76) combined lead to 𝑖𝐴,Ω_𝑅⧵Ω_𝑟∩𝑃,𝑃=1−0=1.(77)
Consequently, 𝐴 has at least one fixed point on (Ω_𝑅⧵Ω_𝑟)∩𝑃. Thus (1) has at least one positive solution. This completes the proof.

Theorem 12. If (H1), (H2), (H3), (H6), and (H7) hold, then (1) has at least two positive solutions.

Proof. If 𝑣∈𝑃 and ‖𝑣‖⩽𝜔, then we have 𝐵₁𝑣₀=𝐵₁𝑣(1)=¹₀𝑠𝑣(𝑠)d𝑠⩽𝜔,𝐵₁𝑣^′₀=𝐵₁𝑣(0)=¹₀𝑣(𝑠)d𝑠⩽𝜔.(78)
By (H7), we have 𝑓(𝑡,𝑥)⩽𝑓(𝑡,𝜔,𝜔,𝜔,𝜔),∀𝑡∈[0,1],𝑥∈𝐼⁴_𝜔.(79)
Consequently, for all 𝑣∈𝜕Ω_𝜔∩𝑃, we have ‖𝐴𝑣‖₀=(0)=𝐴𝑣)(₁₀(1−𝑠)𝑓𝑠,𝐵₁𝑣(𝑠),𝐵₁𝑣^′(𝑠)𝑣(𝑠),−𝑣^′(𝑠⩽)d𝑠₁₀𝑓()‖𝑣‖,)𝑠,𝜔,𝜔,𝜔,𝜔d𝑠<𝜔=(80)(𝐴𝑣^′₀=−(𝐴𝑣)^′(1)=¹₀𝑓𝑠,𝐵₁𝑣(𝑠),𝐵₁𝑣^′(𝑠),𝑣(𝑠),−𝑣^′(𝑠)⩽d𝑠₁₀𝑓(𝑠,𝜔,𝜔,𝜔,𝜔)d𝑠<𝜔=‖𝑣‖.(81) The preceding inequalities imply ‖𝐴𝑣‖<𝜔=‖𝑣‖,∀𝑣∈𝜕Ω_𝜔∩𝑃(82) and thus 𝑣≠𝜆𝐴𝑣,∀𝑣∈𝜕Ω_𝜔∩𝑃,0⩽𝜆⩽1.(83) Now Lemma 5 yields 𝑖𝐴,Ω_𝜔∩𝑃,𝑃=1.(84)By (H2), (H3), and (H6), we know that (52) and (76) hold (see the proofs of Theorems 10 and 11). Note that we can choose 𝑅>𝜔>𝑟 in (52) and (76). Combining (52), (76), and (84), we obtain 𝑖𝐴,Ω_𝑅⧵Ω_𝜔𝑖Ω∩𝑃,𝑃=0−1=−1,𝐴,_𝜔⧵Ω_𝑟∩𝑃,𝑃=1−0=1.(85) Therefore, 𝐴 has at least two fixed points, with one on (Ω_𝑅⧵Ω_𝜔)∩𝑃 and the other on (Ω_𝜔⧵Ω_𝑟)∩𝑃. Hence (1) has at least two positive solutions. This completes the proof.