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Monotone iterative technique and positive solutions to a thirdorder differential equation with advanced arguments and Stieltjes integral boundary conditions
 Bo Sun^{1}Email authorView ORCID ID profile
https://doi.org/10.1186/s1366201816703
© The Author(s) 2018
 Received: 3 November 2017
 Accepted: 11 June 2018
 Published: 22 June 2018
Abstract
We treat the existence of monotonic iteration positive solutions to a thirdorder boundary value problem with advanced arguments and Stieltjes integral boundary conditions. In our work, the main tool is a monotone iterative technique. Meanwhile, at the end of this paper, an example is presented to show that this method can be well used to get the main results.
Keywords
 Stieltjes integral boundary conditions
 Monotone Iterative
 Completely continuous
MSC
 34K10
 34B15
1 Introduction
Thirdorder equations play a significant role in various aspects of applied mathematics and physics. Take draining or coating fluidflow problems for example, surface tension forces are important in these problems and thirdorder ordinary differential equations serve well to describe them. Besides, thirdorder equations also work out well in the deflection of a curved beam having a constant or varying crosssection, a threelayer beam, electromagnetic waves, or gravity driven flows, etc. To learn more about the applications of the thirdorder boundary value problems, readers can refer to [1] and related themes.
Recently, many authors have widely studied the existence of multiple solutions to some boundary value problems; to see the details, we refer the readers to [1–5] and the references therein.
Since science and technology are developing at an unprecedented speed, a lot of boundary value problems with integral boundary conditions are applied in different industries and fields, for instance, thermal conduction, chemical engineering, semiconductor, underground water flow, hydrodynamic, thermoelasticity, etc.; these can be found in [6–8] and related topics. The point is that boundary value problems with integral boundary conditions are made up of a very interesting and significant class of problems, since they include two, three, multipoint, and nonlocal boundary value problems as special cases.
Although the existence of multiple solutions to some boundary value problems with integral boundary conditions has been studied widely by many authors nowadays, we find that most authors study the secondorder and fourthorder differential equations involving integral boundary conditions; to give example, the readers can see [9–13] and the references therein. As far as we are concerned, there are few papers dealing with thirdorder differential equations with Stieltjes integral boundary conditions in the existing literature.
We cannot help pointing out that only \(x(1)\) is related to a Stieltjes integral in the above boundary condition in (1.2), and this is the specific source where our thoughts came from. That is, whether there will be some interesting findings when \(x(0)\) is also related to a Stieltjes integral.
When it comes to our work here, we will show that not only can we prove the existence of positive solutions to problem (1.3), but also a few of successive iterative schemes with either a known constant function starting point or a simple linear function one will be set up to approach the solutions. Last but not least, a perfect example is shown at the end of our paper to represent the applicability of the above mentioned methods and results. We must point out that acquiring the knowledge of how to find the solutions is perhaps the most significant skill when we turn to numerical analysis and application.
2 Preliminaries
At the beginning, we will show some important and necessary definitions here by using the theories of cones in Banach spaces.
Definition 2.1
 (i)
\(au+bv\in P\) for all \(u,v\in P\) and all \(a\geq0\), \(b\geq0\), and
 (ii)
\(u,u\in P \) implies \(u=0\).
Definition 2.2
Definition 2.3
An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.
Lemma 2.1
Proof
So, the proof is completed. □
 (H_{1})::

\(f(t,x)\in C([0,1]\times[0,+\infty)\rightarrow[0,+\infty ))\).
 (H_{2})::

\(q(t)\) is a nonnegative continuous function on \([0,1]\), \(q(t)\not\equiv0\) on any subinterval of \((0,1)\).
 (H_{3})::

\(\int^{1}_{0}d\Lambda(t)\geq0\), \(\int^{1}_{0}t\, d\Lambda (t)\geq0\), \(\int^{1}_{0}F(t,s)\, d\Lambda(t)\geq0\), \(0< s<1\).
 (H_{4})::

\(1\beta(\gamma\beta)\eta>0\).
Lemma 2.2
Since (H_{1})–(H_{4}) hold, then \(T:P\rightarrow P\) defined by (2.3) is completely continuous.
Proof
Then it follows that Tu is nonnegative on \([0,1]\).
On the other hand, we must show that \(\min_{\eta\leq t\leq1} (Tu)(t)\geq\delta\Tu\\).
Because of the concavity of Tu, we can obtain that there exists \(\sigma\in[0,1]\) such that \(\Tu\=(Tu)(\sigma)\).
It is easy to see that T is continuous. Then, let \(\Omega\subset P\) be a bounded set, the proof that TΩ is bounded and equicontinuous is easy and obvious. Then the Arzela–Ascoli theorem makes sure that TΩ is relatively compact, which means T is compact. Then we obtain that T is completely continuous.
So, based on what has been discussed above, we can arrive at the conclusion that \(T:P\rightarrow P\) is completely continuous. □
3 Main results
Theorem 3.1
 (H_{5})::

\(f(t,x_{1})\leq f(t,x_{2})\) for any \(0\leq t\leq1\), \(0\leq x_{1}\leq x_{2}\leq a\);
 (H_{6})::

\(\sup_{0\leq t\leq1} f(t,a)\leq\frac{a}{A}\), \(\inf_{\eta\leq t\leq1} f(t,\delta b)\geq\frac {b}{B}\);
 (H_{7})::

\(f(t,0)\not\equiv0\) for \(0\leq t\leq1\).
The successive iterative schemes in the theorem are \(w_{0}(t)=a\), \(w_{n+1}=Tw_{n}=T^{n}w_{0}\), \(n=0,1,2,\ldots\) , which starts off with the constant function, and \(v_{0}(t)=\frac{b}{\gamma} (\beta+(\gamma\beta)t )\), \(v_{n+1}=Tv_{n}=T^{n}v_{0}\), \(n=0,1,2,\ldots\) , which starts off with a known simple linear function.
Proof
We denote \(P[b,a]=\{u\in P \mid b\leq\u\\leq a\}\).
In the following content of proof, we will firstly prove that \(T:P[b,a]\rightarrow P[b,a]\).
Thus, we get \(b\leq\Tu\\leq a\). So, we get that \(T:P[b,a]\rightarrow P[b,a]\).
Let \(w_{0}(t)=a\), \(0\leq t\leq1\), then \(w_{0}(t)\in P[b,a]\). Let \(w_{1}=Tw_{0}\), then \(w_{1}\in P[b,a]\). We denote \(w_{n+1}=Tw_{n}\), \(n=0,1,2,\ldots\) . Then we have \(w_{n}\subseteq P[b,a]\), \(n=1,2,\ldots\) . Since T is completely continuous, we assert that \(\{w_{n}\}_{n=1}^{\infty}\) is a sequentially compact set.
On the other hand, another way to approach this is to start off with a linear function. Let \(v_{0}(t)=\frac{b}{\gamma} (\beta+(\gamma\beta)t )\), \(0\leq t\leq1\), then \(v_{0}(t)\in P[b,a]\). Let \(v_{1}=Tv_{0}\), then \(v_{1}\in P[b,a]\). We denote \(v_{n+1}=Tv_{n}\), \(n=0,1,2,\ldots\) . Then we have \(v_{n}\subseteq P[b,a]\), \(n=1,2,\ldots\) . Since T is completely continuous, we assert that \(\{v_{n}\}_{n=1}^{\infty}\) is a sequentially compact set.
Assumption (H_{7}) indicates that \(f(t,0)\not\equiv0\), \(0\leq t\leq 1\), then the zero function is not the solution of (1.3). Thus we have \(v^{\ast}>0\) for \(0< t<1\).
It is well known that each fixed point of T in P is a solution of (1.3). Hence, we assert that the boundary value problem (1.3) has at least two positive concave solutions \(w^{\ast}\) and \(v^{\ast}\).
The proof is completed. □
Remark 3.1
If \(\lim_{n\rightarrow\infty}w_{n}\neq\lim_{n\rightarrow\infty}v_{n}\), then \(w^{\ast}\) and \(v^{\ast}\) are two positive concave solutions of problem (1.3). And if \(\lim_{n\rightarrow\infty}w_{n}=\lim_{n\rightarrow\infty}v_{n}\), then \(w^{\ast}=v^{\ast}\) is a positive concave solution of problem (1.3).
The following corollary can be obtained easily.
Corollary 3.1
 (H_{8})::

\(\varlimsup_{l\rightarrow0} \inf_{\eta \leq t\leq1} \frac{f(t,l)}{l}>\frac{1}{\delta B}\), \(\varliminf_{l\rightarrow+\infty} \sup_{0\leq t\leq1} \frac{f(t,l)}{l}<\frac{1}{A}\)
(particularly, \(\varlimsup_{l\rightarrow0} \inf_{\eta\leq t\leq 1} \frac{f(t,l)}{l}=+\infty\), \(\varliminf_{l\rightarrow +\infty} \sup_{0\leq t\leq1} \frac{f(t,l)}{l}=0\).)
4 Example
In the following part, we will discuss an example and simulations. Then we will get a perfect result by using the method above.
Example 4.1
Declarations
Acknowledgements
The author would like to thank the referees for careful reading and several constructive comments and making some useful corrections that have improved the presentation of the paper.
Funding
This work was supported by the National Natural Science Foundation of China (No. 11126245), China Scholarship Council (CSC[2014]3072), and Discipline Construction Fund of Central University of Finance and Economics.
Authors’ contributions
All authors read and approved the final manuscript.
Competing interests
The author declares that they have no competing interests.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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