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Monotone positive solution of a fourth-order BVP with integral boundary conditions
Boundary Value Problems volume 2015, Article number: 172 (2015)
Abstract
In this paper, we investigate the existence of concave and monotone positive solutions for a nonlinear fourth-order differential equation with integral boundary conditions of the form \(x^{(4)}(t)=f(t,x(t),x'(t),x''(t))\), \(t\in[0,1]\), \(x(0)=x'(1)=x'''(1)=0\), \(x''(0)=\int_{0}^{1}g(s)x''(s) \, \mathrm{d}s\), where \(f\in C([0,1]\times[0,+\infty)^{2}\times(-\infty,0],[0,+\infty))\), \(g\in C([0,1],[0,+\infty))\). By using a fixed point theorem of cone expansion and compression of norm type, the existence and nonexistence of concave and monotone positive solutions for the above boundary value problems is obtained. Meanwhile, as applications of our results, some examples are given.
1 Introduction
This paper is the follow-up of [1]. In [1], by using a fixed point theorem for the sum of two operators due to O’Regan [2], we obtained existence of solutions for a fully nonlinear fourth-order equation with integral boundary conditions of type
In this paper, we study the existence of concave and monotone positive solutions for its simplified form
subject to the integral boundary conditions
where \(f\in C([0,1]\times[0,+\infty)^{2}\times(-\infty,0],[0,+\infty))\), \(g\in C([0,1],[0,+\infty))\).
It is well known that fourth-order boundary value problems models bending equilibria of elastic beams, and have been studied extensively. Among a substantial number of works dealing with fourth-order boundary value problems, we mention [1, 3–31]. We notice that if \(g(\cdot)\equiv0\) in (1.2), the models are known as the one endpoint simply supported and the other one sliding clamped beam. The study of this class of problems was considered by some authors via various methods, we refer the reader to [4, 7, 10, 14, 15, 23, 26].
The aim of this paper is to establish the existence and nonexistence results of concave and monotone positive solutions for the problems (1.1), (1.2). Here, a solution \(x(t)\) of the BVP (1.1), (1.2) is said to be monotone and positive if \(x'(t)\geq0\) on \([0,1]\) and \(x(t)>0\) on \(t\in(0,1]\). Our main tool is the fixed point theorem of cone expansion and compression of norm type [32]. The paper [33] motivated our study.
2 Preliminary
In this section, we present some lemmas which are needed for our main results.
Throughout this paper, we assume that \(f:[0,1]\times [0,+\infty)^{2}\times(-\infty,0]\rightarrow[0,+\infty)\) and \(g: [0,1]\rightarrow[0,+\infty)\) are continuous, moreover, \(\mu:=\int_{0}^{1}g(s)\, \mathrm{d}s<1\).
Simple computations lead to the following lemma.
Lemma 2.1
For any \(h\in C[0,1]\), the BVP
has a unique solution
where
Lemma 2.2
Let \(G_{1}(t,s)\) be as in Lemma 2.1. Then
Proof
For \(0\leq t\leq s \leq1\), one has
On the other hand, for \(0\leq s\leq t \leq1\), we have \(\frac{1}{6}s^{2}+\frac{1}{6}t^{2}\leq\frac{1}{3}t\), and then
This completes the proof of the lemma. □
Lemma 2.3
If \(h\in C[0,1]\) with \(h(t)\geq0\) on \([0,1]\), then the unique solution \(x=x(t)\) of the BVP (2.1) satisfies:
-
(1)
\(x(t)\geq0\) for \(t\in[0,1]\);
-
(2)
\(x'(t)\geq0\), \(x''(t)\leq0\) for \(t\in[0,1]\), and
$$x(t)\geq\frac{1}{2}\biggl(t-\frac{1}{2}t^{2}\biggr)\bigl\Vert x''\bigr\Vert _{\infty}, \quad t\in[0,1]. $$
Proof
(1) From Lemma 2.2 and the fact
it follows that
(2) Note that whenever \((t,s)\in[0,1]\times[0,1]\),
it follows that
On the one hand, by (2.3), we have
On the other hand, in view of (2.2) and Lemma 2.2, we have
It follows from (2.4) and (2.5) that
This completes the proof of the lemma. □
Let
be endowed with the norm \(\|x\|=\max_{t\in[0,1]}|x''(t)|=:\|x''\| _{\infty}\). Then E is a Banach space. If we denote
then it is easy to see that K is a cone in E.
Now, we define an operator T on K as follows: for \(x\in K\),
By Lemma 2.3, we know that \(T(K)\subset K\) and if x is a fixed point of T, then x is a concave and monotone positive solution of the BVP (1.1), (1.2).
Lemma 2.4
\(T:K\rightarrow K\) is completely continuous.
Proof
First, we show that T is continuous. To do this, suppose \(x_{n}, x_{0}\in K\) and \(\|x_{n}-x_{0}\|\rightarrow0\) (\(n\rightarrow\infty\)). Then there exists \(M_{1}>0\) such that \(\|x_{0}\|, \|x_{n}\|\leq M_{1}\) for all \(n\in\mathbb{N}=\{ 1,2,\ldots\}\). Hence from the continuity of f on \([0,1]\times[0,M_{1}]^{2}\times[-M_{1},0]\), we have
uniformly on \([0,1]\). Also, since
we have
i.e.,
Therefore \(T:K\rightarrow K\) is continuous.
Next, we prove that T is relatively compact. With this aim, let \(D\subset K\) be a bounded set, then there exists a constant \(M_{2}>0\) such that \(\|x\|\leq M_{2}\) for all \(x\in D\). Suppose that \(\{y_{n}\}\subset T(D)\), there exist \(\{x_{n}\} \subset D\) such that \(Tx_{n}=y_{n}\). Let
For all \(n\in\mathbb{N}\), we have
and
Consequently there exists a constant \(M_{4}>0\) such that, for all \(n\in \mathbb{N}\),
By the Arzela-Ascoli theorem, we know that \(\{y_{n}''\}\) has a convergent subsequence in supremum norm, i.e., \(\{y_{n}\}\) has a convergent subsequence in E, which indicates that \(T(D)\subset K\) is relatively compact in E. This completes the proof of the lemma. □
The following fixed point theorem of cone expansion and compression of norm type plays a crucial role in our paper.
Lemma 2.5
([32])
Let E be a Banach space and let K be a cone in E. Assume that \(\Omega_{1}\) and \(\Omega_{2}\) are bounded open subsets of E such that \(\theta\in\Omega_{1}\subset\overline{\Omega}_{1}\subset\Omega_{2}\), and let \(T:K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\rightarrow K\) be a completely continuous operator such that either
-
(i)
\(\|Tx\|\leq\|x\|\) for \(x\in K\cap\partial\Omega_{1}\) and \(\|Tx\|\geq\|x\|\) for \(x\in K\cap\partial\Omega_{2}\), or
-
(ii)
\(\|Tx\|\geq\|x\|\) for \(x\in K\cap\partial\Omega_{1}\) and \(\|Tx\|\leq\|x\|\) for \(x\in K\cap\partial\Omega_{2}\).
Then T has a fixed point in \(K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\).
3 Main results
For convenience, firstly we introduce some notations:
Theorem 3.1
If \(H_{1}f^{0}<1<H_{2}f_{\infty}\), then the BVP (1.1), (1.2) has at least one concave and monotone positive solution.
Proof
Since \(H_{1}f^{0}<1\), there exists \(\varepsilon_{1}>0\) such that
By the definition of \(f^{0}\) and the continuity of f, there exists \(\rho _{1}>0\) such that, for \(t\in[0,1]\), \(x_{0}+x_{1}-x_{2}\in[0,\rho_{1}]\),
Let \(\Omega_{1}=\{x\in E:\|x\|<\rho/3\}\). For all \(x\in K\cap\partial \Omega_{1}\), from (3.1) and (3.2), we have
which implies that
On the other hand, in view of \(H_{2}f_{\infty}>1\), there exists \(\varepsilon_{2}>0\) such that
By the definition of \(f_{\infty}\), there exists \(\rho_{2}>\rho_{1}\) such that, for \(t\in[0,1]\), \(x_{0}+x_{1}-x_{2}\in[\rho_{2},+\infty)\),
Let \(\Omega_{2}=\{x\in E:\|x\|<\rho_{2}\}\). Then for all \(x\in K\cap \partial\Omega_{2}\), from Lemma 2.2, (3.4), and (3.5) it follows that
which implies that
Therefore, it follows from (3.3), (3.6), and Lemma 2.5 that the operator T has one fixed point \(x\in K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\), which is a concave and monotone positive solution of the BVP (1.1), (1.2). This completes the proof of the theorem. □
Corollary 3.1
Suppose that f is superlinear, i.e.,
Then the BVP (1.1), (1.2) has at least one concave and monotone positive solution.
Theorem 3.2
If \(H_{1}f^{\infty}<1<H_{2}f_{0}\), then the BVP (1.1), (1.2) has at least one concave and monotone positive solution.
Proof
Since \(H_{2}f_{0}>1\), there exists \(\varepsilon_{1}>0\) such that
By the definition of \(f_{0}\), there exists \(\rho_{1}>0\) such that, for \(t\in[0,1]\), \(x_{0}+x_{1}-x_{2}\in[0,\rho_{1}]\),
Let \(\Omega_{1}=\{x\in E:\|x\|<\rho_{1}\}\). Then, for all \(x\in K\cap \partial\Omega_{1}\), from Lemma 2.2, (3.7), and (3.8) it follows that
which implies that
On the other hand, in view of \(H_{1}f^{\infty}<1\), there exists \(\varepsilon_{2}>0\) such that
By the definition of \(f^{\infty}\), there exists \(\rho^{*}>3\rho_{1}\) such that, for \(t\in[0,1]\), \(x_{0}+x_{1}-x_{2}\in[\rho^{*},+\infty)\),
Let
Then for \((t,x_{0},x_{1},x_{2})\in[0,1]\times[0,+\infty)^{2}\times(-\infty,0]\) one has
Now, we choose \(\rho_{2}>\frac{1}{3}\max\{\rho^{*},\frac{\beta H_{1}}{1-H_{1}(f^{\infty}+\varepsilon_{2})}\}\) and let
For all \(x\in K\cap\partial\Omega_{2}\), from (3.10) and (3.11) it follows that
which implies that
Therefore, it follows from (3.9), (3.12), and Lemma 2.5 that the operator T has one fixed point \(x\in K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\), which is a concave and monotone positive solution of the BVP (1.1), (1.2). This completes the proof of the theorem. □
Corollary 3.2
Suppose that f is sublinear, i.e.,
Then the BVP (1.1), (1.2) has at least one concave and monotone positive solution.
Theorem 3.3
Suppose that
for \((t,x_{0},x_{1},x_{2})\in[0,1]\times[0,+\infty)^{2}\times(-\infty,0]\) with \(x_{0}+x_{1}-x_{2}>0\). Then the BVP (1.1), (1.2) has no concave and monotone positive solution.
Proof
By contradiction, assume that x is a concave and monotone positive solution of the BVP (1.1), (1.2). Then
and
Hence
which implies that
This is a contradiction. Therefore the BVP (1.1), (1.2) has no concave and monotone positive solution. This completes the proof of the theorem. □
Theorem 3.4
Suppose that
for \((t,x_{0},x_{1},x_{2})\in[0,1]\times[0,+\infty)^{2}\times(-\infty,0]\) with \(x_{0}+x_{1}-x_{2}>0\). Then the BVP (1.1), (1.2) has no concave and monotone positive solution.
Proof
Suppose on the contrary that x is a concave and monotone positive solution of the BVP (1.1), (1.2). Then from Lemma 2.2 we have
which is a contradiction. This completes the proof of the theorem. □
Finally, we give some examples to demonstrate applications of our results.
Example 3.1
Consider the fourth-order boundary value problem
Let
Then \(f\in C([0,1]\times[0,+\infty)^{2}\times(-\infty,0],[0,+\infty))\), \(g\in C([0,1],[0,+\infty))\), and \(\mu=\int_{0}^{1}g(s)\,\mathrm{d}s=\frac{1}{2}<1\). It is easy to compute that
and hence
So, it follows from Theorem 3.1 that the BVP (3.13), (3.14) has at least one concave and monotone positive solution.
Example 3.2
Consider the fourth-order boundary value problem
Let
Then \(f\in C([0,1]\times[0,+\infty)^{2}\times(-\infty,0],[0,+\infty))\), \(g\in C([0,1],[0,+\infty))\), and \(\mu=\int_{0}^{1}g(s)\,\mathrm{d}s=\frac{3}{4}<1\). It is easy to compute that
and hence
So, it follows from Theorem 3.2 that the BVP (3.15), (3.16) has at least one concave and monotone positive solution.
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This work was supported by the National Natural Science Foundation of China (11201008).
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Lv, X., Wang, L. & Pei, M. Monotone positive solution of a fourth-order BVP with integral boundary conditions. Bound Value Probl 2015, 172 (2015). https://doi.org/10.1186/s13661-015-0441-2
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DOI: https://doi.org/10.1186/s13661-015-0441-2