On a Third-Order Three-Point Boundary Value Problem

Abstract

We consider a third-order three-point boundary value problem. We introduce a generalized polynomial growth condition to obtain the existence of a nontrivial solution by using Leray-Schauder nonlinear alternative, then we give an example to illustrate our results.

1. Introduction

In this work, we study the existence of nontrivial solution for the following third-order three point boundary value problem (BVP): 𝑢^′′′(𝑡)((𝑡𝑢(0)+𝑓𝑡,𝑢))=0,0<𝑡<1,=𝛼𝑢^′(0),𝑢(1)=𝛽𝑢^′𝜂,𝑢^′(1)=0, (1.1)(1.2) where 𝜂∈(0,1),𝛼,𝛽∈ℝ,𝑓∈𝐶([0,1]×ℝ,ℝ).

The parameters 𝛼 and 𝛽 are arbitrary in ℝ such that 1+2𝛼+2𝛽𝜂−2𝛽≠0. Our aim is to give new conditions on the nonlinearity of 𝑓, then using Leray-Schauder nonlinear alternative, we establish the existence of nontrivial solution. We only assume that 𝑓(𝑡,0)≠0 and a generalized polynomial growth condition, that is, there exist two nonnegative functions 𝑘,ℎ∈𝐿¹[0,1] such that 𝑓(𝑡,𝑥)≤𝑘(𝑡)|𝑥|^𝑝+ℎ(𝑡),∀(𝑡,𝑥)∈[0,1]×ℝ, (1.3) where 𝑝∈ℝ₊, so our conditions are new and more general than [1].

Such problems arise in the study of the equilibrium states of a heated bar. Very recently, there have been several papers on third-order boundary value problems. Graef and Yang [2, 3], Guo et al. [4], Hopkins and Kosmatov [5], and Sun [6] have all considered third-order problems. Anderson [7] considered the three-point boundary value problem for (1.1) in the case 𝑡₁<𝑡<𝑡₂ and the three-point conditions 𝑢(𝑡₁)=𝑢^′(𝑡₂)=0,𝛾𝑢(𝑡₃)+𝛿𝑢^′′(𝑡₃)=0; using the Krasnoselskii and Leggett-Williams fixed-point theorems, the existence of solutions to the nonlinear problem is proved. Excellent surveys of theoretical results can be found in Agarwal [8], Agarwal et al. [9], and Ma [10]. More results can be found in [11–13].

This paper is organized as follows. First, we list some preliminary materials to be used later. Then in Section 3, we present and prove our main results which consist in existence theorems. We end our work with some illustrating examples.

2. Preliminary Lemmas

Let 𝐸=(𝐶[0,1],ℝ), with supremum norm ||𝑦||=sup_𝑡∈[0,1]|𝑦(𝑡)|, for all 𝑦∈𝐸. Now, we state two preliminary results.

Lemma 2.1. Let 𝑦∈𝐸. If 𝜁=1+2𝛼+2𝛽𝜂−2𝛽≠0, then the three-point BVP 𝑢^′′′(𝑡)(𝑡)𝑢(0)+𝑦=0,0<𝑡<1,=𝛼𝑢^′(0),𝑢(1)=𝛽𝑢^′𝜂,𝑢^′(1)=0 (2.1) has a unique solution 𝑢(𝑡)1=−2^𝑡₀(𝑡−𝑠)²𝑦(𝑠)𝛽𝑑𝑠+𝜁𝑡²−2𝑡−2𝛼^𝜂₀𝑦(𝑠)+1𝜂−𝑠𝑑𝑠𝜁¹₀(𝑡1−𝑠)²21+𝑠+2𝛼−2𝛽+(𝑡+𝛼)2𝛽𝜂−𝑠𝑦(𝑠)𝑑𝑠. (2.2)

Proof. Integrating 𝑢^′′′(𝑡)=−𝑦(𝑡) over the interval [0,𝑡], we see that 𝑢(𝑡)1=−2^𝑡₀(𝑡−𝑠)²𝑦(𝑠)1𝑑𝑠+2𝐴𝑡²+𝐵𝑡+𝐶. (2.3)
The constants 𝐴, 𝐵, and 𝐶 are given by the three-point boundary conditions (1.2).

We define the integral operator 𝑇∶𝐸→𝐸 by (𝑡)1𝑇𝑢=−2^𝑡₀(𝑡−𝑠)²𝑓((𝑠𝛽𝑠,𝑢))𝑑𝑠+𝜁𝑡²−2𝑡−2𝛼^𝜂₀𝑓((𝑠⋅1𝜂−𝑠𝑠,𝑢))𝑑𝑠𝜁¹₀(𝑡1−𝑠)²21+𝑠+2𝛼−2𝛽+(𝑡+𝛼)2𝛽𝜂−𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠. (2.4)

By Lemma 2.1, the BVP (1.1)-(1.2) has a solution if and only if the operator 𝑇 has a fixed point in 𝐸. By Ascoli-Arzela theorem, we prove that 𝑇 is a completely continuous operator. Now we cite the Leray-Schauder as nonlinear alternative.

Lemma 2.2 (see [14]). Let 𝐹 be a Banach space and Ω a bounded open subset of 𝐹, 0∈Ω. Let 𝑇∶Ω→𝐹 be a completely continuous operator. Then, either there exists 𝑥∈𝜕Ω,  𝜆>1 such that 𝑇(𝑥)=𝜆𝑥, or there exists a fixed point 𝑥^∗∈Ω.

3. Main Results

In this section, we present and prove our main results.

Theorem 3.1. It is assumed that 𝑓(𝑡,0)≠0,𝜁≠0,𝑝∈ℝ₊, and there exist two nonnegative functions 𝑘,ℎ∈𝐿¹[0,1] such that 𝑓(𝑡,𝑥)≤𝑘(𝑡)|𝑥|^𝑝(𝑡)()∈[]𝛽()+ℎ,∀𝑡,𝑥0,1×ℝ,1+1+2𝛼𝜁+1𝜁2|1+𝛼|𝛽𝜂+1+𝛽¹₀()𝑘(𝑠)11−𝑠𝑑𝑠<2,𝛽()1+1+2𝛼𝜁+1𝜁2|1+𝛼|𝛽𝜂+1+𝛽¹₀()ℎ(𝑠)11−𝑠𝑑𝑠<2. (3.1)(3.2)(3.3) Then the BVP (1.1)-(1.2) has at least one nontrivial solution 𝑢^∗∈𝐶([0,1],ℝ).

Proof. Setting 𝛽()𝑀=1+1+2𝛼𝜁+1𝜁2|1+𝛼|𝛽𝜂+1+𝛽¹₀()𝑘(𝑠)𝛽()1−𝑠𝑑𝑠,𝑁=1+1+2𝛼𝜁+1𝜁2|1+𝛼|𝛽𝜂+1+𝛽¹₀(1−𝑠)ℎ(𝑠)𝑑𝑠, (3.4) by hypothesis (3.2), we have 𝑀<1/2. Since 𝑓(𝑡,0)≠0, then there exists an interval [𝜎,𝜏]⊂[0,1] such that min_{𝜎≤𝑡≤𝑟}|𝑓(𝑡,0)|>0, then 𝑁>0. Let 𝑚=𝑀/𝑁,  Ω={𝑢∈𝐶[0,1]∶||𝑢||<1}. Assume that 𝑢∈𝜕Ω,𝜆>1 such 𝑇𝑢=𝜆𝑢, then 𝜆‖𝑢‖=‖𝑇𝑢‖=max_0≤𝑡≤1|𝑇𝑢(𝑡)|≤𝑀‖𝑢‖^𝑝+𝑁, (3.5) as ||𝑢||=1, then 𝜆≤𝑀+𝑁. (3.6)
First, if 𝑚≤1, then 𝜆≤2𝑁<1; hence, 𝜆<1, and this contradicts the fact that 𝜆>1. By Lemma 2.2, we deduce that 𝑇 has a fixed point 𝑢^∗∈Ω, and then the BVP (1.1)-(1.2) has a nontrivial solution 𝑢^∗∈𝐶([0,1],ℝ).
Second, if 𝑚≥1, then 𝜆≤2𝑀<1. By arguing as above, we complete the proof.

Let us define the following notation: 𝛽()𝑎=1+1+2𝛼𝜁+1𝜁2|1+𝛼|𝛽𝜂+1+𝛽. (3.7)

Theorem 3.2. Under the conditions of Theorem 3.1  (𝑝∈ℝ₊)  and if one of the following conditions is satisfied
(1) there exist 𝑛>1 and 𝑟>1  such that ¹₀𝑘^𝑛(𝑠)𝑑𝑠^1/𝑛<1+𝑞^1/𝑞12𝑎𝑛+1𝑞=1₁₀,ℎ^𝑟(𝑠)𝑑𝑠^1/𝑟<(1+𝑙)^1/𝑙12𝑎𝑟+1𝑙=1, (3.8)(3.9)
(2) there exist two constants 𝜇,𝜏>−1 such that 𝑘(𝑠)<𝜇+1𝜇+2𝑠2𝑎^𝜇,ℎ(𝑠)<(𝜏+1)(𝜏+2)𝑠2𝑎^𝜏,meas𝑠∈[0,1],𝑘(𝑠)<𝜇+1𝜇+2𝑠2𝑎^𝜇[](𝑠)<()>0,meas𝑠∈0,1,ℎ𝜏+1)(𝜏+2𝑠2𝑎^𝜏>0, (3.10)(3.11)(3.12)
(3) the functions 𝑘(𝑠) and ℎ(𝑠) satisfy 𝑘(𝑠)<1𝑎,ℎ(𝑠)<1𝑎,[](𝑠)<1meas𝑠∈0,1,𝑘𝑎[](𝑠)<1>0,meas𝑠∈0,1,ℎ𝑎>0, (3.13)(3.14)(3.15)
(4) the function 𝑓(𝑡,𝑥) satisfies 𝜔₁=lim_|𝑥|→∞supmax_[]𝑡∈0,1𝑓(𝑡,𝑥)|𝑥|^𝑝<1,𝜔2𝑎₂=lim_|𝑥|→∞supmax_[]𝑡∈0,1𝑓(<1𝑡,𝑥).2𝑎 (3.16)(3.17) Then the BVP (1.1)-(1.2) has at least one nontrivial solution 𝑢^∗∈𝐶([0,1],ℝ).

Proof. Let 𝑀 and 𝑁 be defined as in the proof of Theorem 3.1. To prove Theorem 3.2, we only need to prove that 𝑀<1/2 and 𝑁<1/2.
(1) By using Hölder inequality, we get 𝑀≤𝑎¹₀𝑘^𝑛(𝑠)𝑑𝑠^1/𝑛1₀(1−𝑠)^𝑞𝑑𝑠^1/𝑞. (3.18) Integrating, using (3.8), and remarking that 𝑎>1, we arrive at 𝑀<1/2.
Using Hölder inequality a second time, we get 𝑁≤𝑎¹₀ℎ^𝑟(𝑠)𝑑𝑠^1/𝑟1₀(1−𝑠)^𝑙𝑑𝑠^1/𝑙. (3.19)
Integrating and then using (3.9), we arrive at 𝑁<1/2.
(2) Taking into account (3.10), it yields 𝑀<𝜇+1𝜇+2𝑎2𝑎¹₀(1−𝑠)𝑠^𝜇=1𝑑𝑠2. (3.20) On the other hand, using (3.11), we obtain ()𝑁<𝜏+1)(𝜏+2𝑎2𝑎¹₀(1−𝑠)𝑠^𝜇=1𝑑𝑠2. (3.21)
(3) Using the same reasoning as in the proof of the second statement, we prove the third statement.
(4) From the condition 𝜔₁=lim_|𝑥|→∞supmax_𝑡∈[0,1]|𝑓(𝑡,𝑥)|/|𝑥|^𝑝, we deduce that there exists 𝑐₁>0 such that for |𝑥|>𝑐₁, we get |𝑓(𝑡,𝑥)|≤(𝜔₁+𝜀)|𝑥|^𝑝, for  all  𝜀>0. Choosing 𝜀=𝜔₁, then |𝑓(𝑡,𝑥)|≤2𝜔₁|𝑥|^𝑝. Now from the condition 𝜔₂=lim_|𝑥|→∞supmax_𝑡∈[0,1]|𝑓(𝑡,𝑥)|, we deduce that there exists 𝑐₂>0 such that for |𝑥|>𝑐₂, we have 𝑓(𝑡,𝑥)≤𝜔₂+𝜀,∀𝑥∈ℝ⧵−𝑐₂,𝑐₂, (3.22) choosing 𝜀=𝜔₂, then |𝑓(𝑡,𝑥)|≤2𝜔₂, for  all  𝑥∈ℝ⧵]−𝑐₂,𝑐₂[; consequently, 𝑓(𝑡,𝑥)≤2𝜔₁|𝑥|^𝑝+2𝜔₂,∀𝑥∈ℝ⧵]−𝑐,𝑐[, (3.23) where 𝑐=max(𝑐₁,𝑐₂), setting 𝑅=max{|𝑓(𝑡,𝑥)|∶(𝑡,𝑥)∈[0,1]×(−𝑐,𝑐)}, then for all (𝑡,𝑥)∈[0,1]×ℝ, we get |𝑓(𝑡,𝑥)|≤𝑘(𝑡)|𝑥|^𝑝+ℎ(𝑡), where 𝑘(𝑡)=2𝜔₁ and ℎ(𝑡)=2𝜔₂. Using (3.16), we obtain 𝑘(𝑡)=2𝜔₁<1𝑎, (3.24) then from (3.13), we get 𝑀<1/2. Using (3.14), we arrive at ℎ(𝑡)=2𝜔₂<1𝑎, (3.25) then 𝑁<1/2. Now applying the third statement, we achieve the proof of Theorem 3.2.

Example 3.3. Consider the three-point BVP, 𝑢^′′′+17𝑢⁵²√𝑡³2+cos𝑡+arcsin𝑡5𝑢(0)1=0,0<𝑡<1,=−2𝑢^′(0)(1)1,𝑢=−2𝑢^′13,𝑢^′(1)=0. (3.26) We have 𝑓()=𝑥𝑡,𝑥⁵7²√𝑡³2+cos𝑡+arcsin𝑡51,𝛼=𝛽=−21,𝜂=32,𝜁=3. (3.27) So, 𝑓(≤1𝑡,𝑥)7²√𝑡³2+cos𝑡|𝑥|⁵+arcsin𝑡5=𝑘(𝑡)|𝑥|⁵+ℎ(𝑡). (3.28) Applying the first statement of Theorem 3.2 for 𝑝=5,𝑛=𝑟=2, to get ¹₀𝑘²(𝑠)𝑑𝑠^1/2=149¹₀²√𝑠³2+cos𝑠²𝑑𝑠^1/2<=0.14881+𝑞^1/𝑞2𝑎=0.16496,₁₀ℎ²(𝑠)𝑑𝑠^1/2=125¹₀(arcsin𝑠)²𝑑𝑠^1/2<()=0.136731+𝑙^1/𝑙2𝑎=0.16496. (3.29) Then the BVP (3.26) has at least one nontrivial solution 𝑢^∗ in 𝐶[0,1].