Positive Solutions for a Class of Fourth-Order Boundary Value Problems in Banach Spaces

Abstract

Using a specially constructed cone and the fixed point index theory, this work shows existence and nonexistence results of positive solutions for fourth-order boundary value problem with two different parameters in Banach spaces.

1. Introduction

In this paper, we study the existence of positive solutions for the following fourth-order boundary value problem (BVP) with singularity in Banach space (𝐸,‖⋅‖):𝑥⁽⁴⁾(𝑡)+𝛽𝑥^′′(𝑡)()()𝑥(0)(1)=𝜆𝑓𝑡,𝑥+𝜇𝑔𝑡,𝑥,𝑡∈𝐽,=𝑥=𝜃,𝑥^′′(0)=𝑥^′′(1)=𝜃, (1.1) where 𝐽=[0,1],  𝛽<𝜋².𝑔, 𝑓∈𝐶(𝐽,[0,+∞)), 𝑅⁺=[0,+∞),  and 𝜃 is the zero element of 𝐸.

Fourth-order boundary value problems are studied not only in mathematics but also in physics and many other fields. For example, some models for bridge, underground water flow, and plasma physics can be reduced to fourth-order boundary value problems. In the recent years, some authors (cf. [1–7]) studied fourth-order boundary value problems, but these papers mainly dealt with the problems in real space or the problems without singularity. As far as we know, the fourth-order ordinary differential equation with singularity has been seldom studied in Banach spaces. Also, not so much is known about the case that the nonlinear term has two different parameters or has two parts with different properties.

In [7], Yao studied the following two-point boundary value problem:𝑢⁽⁴⁾(𝑡)(𝑡)𝑓(𝑢(𝑡𝑢(0)(1)−𝜆ℎ))=0,=𝑢=𝑢^′′(0)=𝑢^′′(1)=0, (1.2) where 𝜆>0 and  𝑓∈𝐶[0,+∞). The author obtained positive solutions of BVP (1.2) in real space not abstract space.

In [4], the authors studied the following boundary value problem in real space: 𝑥⁽⁴⁾(𝑡)+𝛽𝑥^′′(𝑡)(),𝑥(0)(1)=𝜆𝑓𝑡,𝑥=𝑥=𝑥^′′(0)=𝑥^′′(1)=0, (1.3) where 𝑓∈𝐶([0,1]×[0,+∞)), 𝛽<𝜋², and 𝜆>0. With the method of monotone iterative, the existence and uniqueness of positive solution of BVP (1.3) are obtained.

Comparing with the results in [5–7], this paper has the following features. Firstly, we discuss positive solutions of BVP (1.1) in abstract space, not 𝐸=𝑅 as in [5, 7]. Secondly, BVP (1.1) has two parameters which may have different domains. Fourthly, 𝑓 and 𝑔 may have different properties, and the main tool we used is the fixed point theorem on cone. Thirdly, we talk about both the existence and nonexistence of positive solutions, but [5,9] only study the existence of positive solutions.

Basic facts about ordered Banach space 𝐸 can be found in [8]. Here we recall some of them. The cone 𝑃 in 𝐸 is said to be normal if there exists a positive constant 𝑁 such that 𝜃≤𝑥≤𝑦 implies ‖𝑥‖≤𝑁‖𝑦‖. In this paper, we always suppose that 𝑃 is normal in 𝐸, and without loss of generality, suppose that the normal constant 𝑁=1. 𝐸^∗ is dual space of 𝐸. Denote 𝑃^∗={𝜓∈𝐸^∗∣𝜓(𝑥)≥0,forall𝑥∈𝑃}, then 𝑃^∗ is a dual cone of cone 𝑃. The noncompactness of Kuratowski is denoted by 𝛼(⋅).

We study BVP (1.1) in 𝐶(𝐽,𝐸). Evidently, (𝐶(𝐽,𝐸),‖⋅‖_𝑐) is a Banach space under the norm ‖𝑥‖_𝑐=max_𝑡∈𝐽‖𝑥(𝑡)‖. 𝑥∈𝐶(𝐽,𝐸) is a positive solution of BVP(1.1) if it satisfies BVP (1.1) and 𝑥(𝑡)>𝜃, for all 𝑡∈𝐽.

2. Preliminaries and Lemmas

In this paper, we make the following conditions. (H₁)𝑓,𝑔∈𝐶(𝐽′×𝑃,𝑃), 𝑓(𝑡,𝑃_𝑟)={𝑓(𝑡,𝑢),𝑢∈𝑃_𝑟} is relatively compact and 𝑔(𝑡,𝑃_𝑟) is also relatively compact, where 𝑃_𝑟={𝑢∈𝑃∶‖𝑢‖≤𝑟}. For any 𝑢∈𝑃, there exist 𝑎_𝑖(𝑡), 𝑏_𝑖(𝑡)∈𝐿(𝐽,𝑅⁺), 𝑤₁(𝑥), 𝑤₂(𝑥)∈𝐶(𝐽,𝑅⁺) such that 𝑓(𝑡,𝑢)≤𝑎₁(𝑡)+𝑏₁(𝑡)𝑤₁(‖𝑢‖),𝑔(𝑡,𝑢)≤𝑎₂(𝑡)+𝑏₂(𝑡)𝑤₂(‖𝑢‖),a.e.𝑡∈𝐽. (2.1)(H₂) There exist a function 𝑚∈𝐿(𝐽,(0,+∞)), such that 𝜓(𝑓(𝑡,𝑢))≥𝑚(𝑡)‖𝑢‖ or 𝜓(g(𝑡,𝑢))≥𝑚(𝑡)‖𝑢‖, for 𝑡∈𝐽,𝑢∈𝑃, where 𝜓∈𝑃^∗, ‖𝜓‖=1.

Lemma 2.1 (see [8]). Let 𝐻={𝑥∣𝑥∶𝐽→𝐸,𝑥 is strong measurable} be countable, and there exists 𝑀∈𝐿[𝐽,𝑅⁺] such that ‖𝑥(𝑡)‖≤𝑀(𝑡),𝑎.𝑒.𝑡∈𝐽,𝑥∈𝐻, then 𝛼(𝐻(𝑡))∈𝐿[𝐽,𝑅⁺] and 𝛼_𝐽𝑥(𝑡)𝑑𝑡∶𝑥∈𝐻≤2_𝐽𝛼(𝐻(𝑡))𝑑𝑡. (2.2)

Lemma 2.2 (see [8]). Let 𝑃 be a cone in Banach space 𝐸. For 𝑟>0, define 𝑃_𝑟={𝑥∈𝑃∶‖𝑥‖<𝑟}. Assume that 𝑇∶𝑃_𝑟→𝑃 is a completely continuous map such that 𝑥≠𝑇𝑥 for 𝑥∈𝜕𝑃_𝑟.(i)If ‖𝑥‖≤‖𝑇𝑥‖ for 𝑥∈𝜕𝑃_𝑟, then 𝑖(𝑇,𝑃_𝑟,𝑃)=0.(ii)If ‖𝑥‖≥‖𝑇𝑥‖ for 𝑥∈𝜕𝑃_𝑟, then 𝑖(𝑇,𝑃_𝑟,𝑃)=1.

Lemma 2.3 (see [4]). For any 𝜑∈𝐶(𝐽,𝑃), the following fourth-order boundary value problem: 𝑥⁽⁴⁾(𝑡)+𝛽𝑥^′′(𝑡)(𝑡),𝑥(0)(1)=𝜑=𝑥=𝑥^′′(0)=𝑥^′′(1)=𝜃 (2.3) has the solution 𝑥(𝑡)=¹₀¹₀𝐺₁(𝑡,𝜏)𝐺₂(𝜏,𝑠)𝜑(𝑠)𝑑𝑠𝑑𝜏, (2.4) where 𝐺₂(𝑡,𝑠) is defined in Proposition 2.4 and 𝐺₁()=𝑡()𝑠()𝑡,𝑠1−𝑠,0≤𝑡≤𝑠≤1,1−𝑡,0≤𝑠≤𝑡≤1. (2.5)

Proposition 2.4. For all 𝑡,𝑠∈[0,1]×[0,1] one has the following properties: 𝑡(1−𝑡)𝐺₁(𝑠,𝑠)≤𝐺₁(𝑡,𝑠)≤𝐺₁(𝑠,𝑠)=𝑠(1−𝑠),𝐺₁(𝑡,𝑠)≤𝑡(1−𝑡);𝐺₁()≤1𝑡,𝑠4. (2.6) Let 𝜔=𝛽. 𝐺₂(𝑡,𝑠) is explicitly given by the following
If 𝛽<0, then 𝐺₂()=()𝑡,𝑠sinh𝜔𝑡sinh𝜔1−𝑠()𝜔sinh𝜔,0≤𝑡≤𝑠≤1,sinh𝜔𝑠sinh𝜔1−𝑡𝜔sinh𝜔,0≤𝑠≤𝑡≤1. (2.7)
If 𝛽=0, then 𝐺₂(𝑡,𝑠)=𝐺₁(𝑡,𝑠).
If 0<𝛽<𝜋², then 𝐺₂()=()𝑡,𝑠sin𝜔𝑡sin𝜔1−𝑠()𝜔sin𝜔,0≤𝑡≤𝑠≤1,sin𝜔𝑠sin𝜔1−𝑡𝜔sin𝜔0≤𝑠≤𝑡≤1. (2.8)

Proposition 2.5. For any 𝑡,𝑠∈(0,1), one has 𝐺₁(𝑡,𝑠)>0, 𝐺₂(𝑡,𝑠)>0, and 𝐺₁(𝑡,𝑠)≥𝛿𝐺₁(𝑢,𝑠), for 𝑡∈𝐽_𝛿,𝑠,𝑢∈[0,1], where 𝛿∈(0,1/2).

Proposition 2.6. It is easy to see that 𝐺₂()≤1𝑡,𝑠4𝑒,𝛽=0,_2𝜔4,𝛽<0,sin𝜔𝜔,𝛽>0. (2.9)

Let 𝑑=max{1/4,𝑒^2𝜔/4,sin𝜔/𝜔}, 𝑄={𝑥∈𝐶(𝐽,𝐸)∶𝑥(𝑡)≥𝜃,𝑡∈𝐽}, and 𝐾={𝑥∈𝑄∶𝑥(𝑡)≥𝛿𝑥(𝑠), 𝑡∈𝐽_𝛿,𝑠∈[0,1]}. It is easy to know that 𝐾 is a cone in 𝐶(𝐽,𝐸). Let 𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖≤𝑟}⊂𝐾, 𝐾⊂𝑄. Define 𝑇 as (𝑇𝑥)(𝑡)=¹₀¹₀𝐺₁(𝑡,𝜏)𝐺₂(𝜏,𝑠)𝜆𝑓(𝑠,𝑥(𝑠))+𝜇𝑔(𝑠,𝑥(s))𝑑𝑠𝑑𝜏, (2.10) then 𝑥 is the solution of BVP(1.1) if and only if 𝑥 is the fixed point of 𝑇.

Lemma 2.7. Suppose that condition (𝐻₁) holds, then 𝑇∶𝐾→𝐾 is completely continuous.

Proof. By the continuity of 𝐺₁(𝑡,𝑠) and (H₁), we have 𝑇𝑥∈𝑄. And for any 𝑡∈𝐽_𝛿, 𝑥∈𝐾, by Proposition 2.5, we get (𝑇𝑥)(𝑡)=𝛿(𝑇𝑥)(𝑢). (2.11)
Next we prove that 𝑇 is compact. Let 𝑉={𝑥_𝑛}^∞_𝑛=1⊂𝐾 be any bounded set, and we suppose that ‖𝑥_𝑛‖≤𝑟, for some 𝑟>0. Let 𝑀_𝑟=max_{0≤𝑣≤𝑟}{𝑤_𝑖(𝑣),𝑖=1,2}, then by condition (H₁), 𝑓𝑡,𝑥_𝑛(𝑡)≤𝑎₁(𝑡)+𝑏₁(𝑡)𝑀_𝑟,𝑔𝑡,𝑥_𝑛(𝑡)≤𝑎₂(𝑡)+𝑏₂(𝑡)𝑀_𝑟,a.e.𝑡∈𝐽,𝑥_𝑛∈𝑉. (2.12) We obtain from Lemma 2.1𝛼𝑇𝑥_𝑛(𝑡)∶𝑥_𝑛∈𝑉=𝛼¹₀𝐺₁(𝑡,𝜏)𝐺₂(𝜏,𝑠)𝜆𝑓𝑠,𝑥_𝑛(𝑠)+𝜇𝑔𝑠,𝑥_𝑛(𝑠)𝑑𝑠𝑑𝜏∶𝑥_𝑛∈𝑉≤2𝑑¹₀𝛼𝜆𝑓𝑠,𝑥_𝑛(𝑠)+𝜇𝑔𝑠,𝑥_𝑛(𝑠)∶𝑥_𝑛∈𝑉𝑑𝑠=0. (2.13) Hence, 𝑇𝑉 is relatively compact, so there exists a subsequence {𝑥_𝑛𝑖} of {𝑥_𝑛} such that {𝑇𝑥_𝑛𝑖} converges to some 𝑣∈𝐶(𝐽,𝐸). So 𝑇∶𝐾→𝐾 is completely continuous.

In this paper, for 𝑢∈𝑃 we denote 𝜓𝑓_𝜈=liminf_{‖𝑢‖→𝜈}min_𝑡∈𝐽𝜓𝑓(𝑡,𝑢)‖𝑢‖,𝜓𝑔_𝜈=liminf_{‖𝑢‖→𝜈}min_𝑡∈𝐽𝜓𝑔(𝑡,𝑢)‖𝑢‖, (2.14) where 𝜈 denotes 0 or ∞, 𝜓∈𝑃^∗, ‖𝜓‖=1.

3. Main Results

Theorem 3.1. Assume that (H₁) holds and the following conditions are satisfied: 𝜓𝑓_∞=∞,𝜓𝑔₀=∞. (3.1) Then BVP(1.1) has at least two positive solutions when 𝜆 and 𝜇 are sufficiently small.

Proof. 𝑇 is completely continuous by Lemma 2.7. For any 𝑙>0, set 𝐹(𝑙)=𝑑4max_{‖𝑥‖𝑥∈𝐾,𝑐=𝑙}¹₀𝑓((𝑠(𝑙)=𝑑𝑠,𝑥))𝑑𝑠,𝐺4max_{‖𝑥‖𝑥∈𝐾,𝑐=𝑙}¹₀𝑔(𝑠,𝑥(𝑠))𝑑𝑠. (3.2) Since (𝜓𝑓)_∞=∞, there exists 𝑟₁>0 such that 𝐹(𝑟₁)>0. If 𝐺(𝑟₁)=0, let 𝜆₁=𝑟₁/𝐹(𝑟₁); then for any 𝜆∈(0,𝜆₁) and 𝑥∈𝐾,‖𝑥‖_𝑐=𝑟₁, we have 𝑡≤𝑑‖(𝑇𝑥)()‖4¹₀((𝑠((𝑠≤𝑑𝜆𝑓𝑠,𝑥))+𝜇𝑔𝑠,𝑥))𝑑𝑠4¹₀((𝑠𝑑𝜆𝑓𝑠,𝑥))𝑑s+4¹₀𝜇𝑔(𝑠,𝑥(𝑠))𝑑𝑠≤𝜆₁𝐹𝑟₁=𝑟₁=‖𝑥‖_𝑐. (3.3)
If 𝐺(𝑟₁)≠0, let 𝜇₁=𝑟₁/2𝐺(𝑟₁) and 𝜆₁=𝑟₁/2𝐹(𝑟₁). For 𝜆∈(0,𝜆₁),𝜇∈(0,𝜇₁), and 𝑥∈𝐾,‖𝑥‖_𝑐=𝑟₁, we get 𝑡≤𝑑‖(𝑇𝑥)()‖4¹₀((𝑠𝑑𝜆𝑓𝑠,𝑥))𝑑𝑠+4¹₀((𝑠≤𝑑𝜇𝑔𝑠,𝑥))𝑑𝑠4𝜆₁¹₀((𝑠𝑑𝜆𝑓𝑠,𝑥))𝑑𝑠+4𝜇₁¹₀𝑔(𝑠,𝑥(𝑠))𝑑𝑠≤𝑟₁𝐹𝑟₁+𝜇₁𝐺𝑟₁≤𝑟₁2+𝑟₁2=𝑟₁=‖𝑥‖_𝑐, (3.4) which means that ‖𝑇𝑥‖_𝑐<𝑟₁=‖𝑥‖_𝑐, for 𝑥∈𝐾, ‖𝑥‖_𝑐=𝑟₁, and Lemma 2.2 implies that 𝑖(𝑇,𝐾_𝑟1,𝐾)=1.
Since (𝜓𝑓)_∞=∞, there exists 𝑟₂>0 such that 𝜓(𝑓(𝑡,𝑢))≥𝜀₁‖𝑢‖, 𝑢∈𝑃,𝑡∈𝐽,‖𝑢‖≥𝑟₂, where 𝜀₁>0 satisfies 𝜀₁𝜆𝛿∬_𝛿^1−𝛿𝐺₁(1/2,𝜏)𝐺₂(𝜏,𝑠)𝑑𝑠𝑑𝜏>1. Let 𝑟₂=max{2𝑟₁,𝑟₂/𝛿}, then for any 𝑡∈𝐽_𝛿,𝑥∈𝐾, ‖𝑥‖_𝑐=𝑟₂, by the definition of 𝐾 we get ‖𝑥(𝑡)‖≥𝛿‖𝑥‖_𝑐≥𝑟₂, and we have 1(𝑇𝑥)2(1≥𝜓𝑇𝑥)2≥𝜆¹₀¹₀𝐺₁12,𝜏𝐺₂(𝜏,𝑠)𝜓𝑓(𝑠,𝑥(𝑠))𝑑𝑠𝑑𝜏≥𝜆¹_0𝛿^1−𝛿𝐺₁12,𝜏𝐺₂(𝜏,𝑠)𝜓𝑓(𝑠,𝑥(𝑠))𝑑𝑠𝑑𝜏≥𝜆𝜀₁¹_0𝛿^1−𝛿𝐺₁12,𝜏𝐺₂(𝜏,𝑠)‖𝑥(𝑠)‖𝑑𝑠𝑑𝜏≥𝜆𝜀₁𝛿‖𝑥‖_𝑐¹_0𝛿^1−𝛿𝐺₁12,𝜏𝐺₂()>‖𝑥‖𝜏,𝑠𝑑𝑠𝑑𝜏_𝑐, (3.5) which implies that ‖𝑇𝑥‖_𝑐>𝑟₂, 𝑥∈𝐾_𝑟2. By Lemma 2.2 we have 𝑖(𝑇,𝐾_𝑟2,𝐾)=0.
On the other hand, since (𝜓𝑔)₀=∞, there exists 𝑟₃∶0<𝑟₃<𝑟₁ such that 𝜓(𝑔(𝑡,𝑢))≥𝜀₂‖𝑢‖, for 𝑡∈𝐽,𝑢∈𝑃,‖𝑢‖≤𝑟₃, where 𝜀₂>0 and satisfies 𝜀₂𝜇𝛿¹_0𝛿^1−𝛿𝐺₁12,𝜏𝐺₂(𝜏,𝑠)𝑑𝑠𝑑𝜏>1. (3.6)Accordingly,0<𝑟₃<𝑟₃. Then for 𝑥∈𝐾,‖𝑥‖_𝑐=𝑟₃, similar to (3.5) we get ‖𝑇𝑥‖_𝑐>𝑟₃. By Lemma 2.2 we have 𝑖(𝑇,𝐾_𝑟3,𝐾)=0. Hence, from the additivity of fixed point index, we obtain 𝑖𝐾𝑇,^𝑟2𝐾_𝑟1𝐾,𝐾=−1,𝑖𝑇,^𝑟1𝐾_𝑟3,𝐾=1. (3.7) Hence, 𝑇 has two fixed points 𝑥₁∈𝐾_𝑟2/𝐾_𝑟1 and 𝑥₂∈𝐾_𝑟1/𝐾_𝑟3. Therefore, BVP(1.1) has at least two fixed points 𝑥₁ and 𝑥₂.

Corollary 3.2. If (H₁), (𝜓𝑔)_∞=∞, and (𝜓𝑓)₀=∞ hold, then BVP(1.1) has at least two positive solutions when 𝜆 and 𝜇 are sufficiently small.

Theorem 3.3. Suppose that (H₁) and (H₂) hold, and if 𝜆 is sufficiently large, then BVP(1.1) has no positive solutions.

Proof. First by (H₂) we can suppose that 𝜓(𝑓(𝑡,𝑢))≥𝑚(𝑡)‖𝑢‖, for 𝑡∈𝐽. suppose that BVP(1.1) has a positive solution 𝑥, and choose 𝜆 sufficiently large satisfying 𝜆𝛿∬_𝛿^1−𝛿𝐺₁(1/2,𝜏)𝐺₂(𝜏,𝑠)𝑚(𝑠)𝑑𝑠𝑑𝜏>1. Then we have ‖𝑥‖_𝑐≥𝑥12𝑥1≥𝜓2≥𝜆¹₀¹₀𝐺₁12,𝜏𝐺₂(𝜏,𝑠)𝜓𝑓(𝑠,𝑥(𝑠))𝑑𝑠𝑑𝜏≥𝜆¹_0𝛿^1−𝛿𝐺₁12,𝜏𝐺₂(𝜏,𝑠)𝑚(𝑠)‖𝑥(𝑠)‖𝑑𝑠𝑑𝜏≥𝜆𝛿¹_0𝛿^1−𝛿𝐺₁12,𝜏𝐺₂(𝜏,𝑠)𝑚(𝑠)𝑑𝑠𝑑𝜏‖𝑥‖_𝑐>‖𝑥‖_𝑐, (3.8) which is a contradiction; so BVP(1.1) has no positive solutions. Similarly, if 𝜓(𝑔(𝑡,𝑢))≥𝑚(𝑡)‖𝑢‖, for 𝑡∈𝐽, and 𝜇 is sufficiently large, then BVP(1.1) has no positive solutions.

Acknowledgments

The authors thank Professor Lishan Liu and Professor Bendong Lou for many useful discussions and helpful suggestions. This work was partially supported by NSFC (10971155) and Innovation Program of Shanghai Municipal Education Commission (09ZZ33).