Iteration for a Third-Order Three-Point BVP with Sign-Changing Green's Function

Abstract

We are concerned with the following third-order three-point boundary value problem: 𝑢^‴(𝑡)=𝑓(𝑡,𝑢(𝑡)),𝑡∈[0,1],𝑢^′(0)=𝑢(1)=0,𝑢^″(𝜂)+𝛼𝑢(0)=0, where 𝛼∈[0,2) and 𝜂∈[2/3,1). Although corresponding Green's function is sign-changing, we still obtain the existence of monotone positive solution under some suitable conditions on 𝑓 by applying iterative method. An example is also included to illustrate the main results obtained.

1. Introduction

Third-order differential equations arise from a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross-section, a three-layer beam, electromagnetic waves or gravity driven flows, and so on [1].

Recently, the existence of single or multiple positive solutions to some third-order three-point boundary value problems (BVPs) has received much attention from many authors; see [2–7] and the references therein. However, all the above-mentioned papers are achieved when corresponding Green’s functions are nonnegative, which is a very important condition. A natural question is that whether we can obtain the existence of positive solutions to some third-order three-point BVPs when corresponding Green’s functions are sign-changing.

In 2008, Palamides and Smyrlis [8] studied the existence of at least one positive solution to the singular third-order three-point BVP with an indefinitely signed Green’s function 𝑢^′′′(𝑡)(𝑡)𝑓((𝑡(),𝑢(0)(1)=𝑎𝑡,𝑢)),𝑡∈0,1=𝑢=𝑢^′′𝜂=0, (1) where 𝜂∈(17/24,1). Their technique was a combination of the Guo-Krasnoselskii fixed point theorem [9, 10] and properties of the corresponding vector field.

Very recently, for the third-order three-point BVP with sign-changing Green’s function 𝑢^′′′(𝑡)((𝑡[],𝑢=𝑓𝑡,𝑢)),𝑡∈0,1_′(0)=𝑢(1)=𝑢^′′𝜂=0, (2) Sun and Zhao proved the existence of single or multiple positive solutions when 𝜂∈(1/2,1) by using the Guo-Krasnoselskii and Leggett-Williams fixed point theorems in [11, 12] and obtained the existence of a positive solution when 𝜂∈[2−√2,1) by using iterative technique in [13].

In 2013, Li et al. [14] established the existence of at least one positive solution to the following third-order three-point BVP with sign-changing Green’s function: 𝑢^′′′(𝑡)((𝑡[],𝑢=𝑓𝑡,𝑢)),𝑡∈0,1_′(0)=𝑢(1)=0,𝑢^′′𝜂+𝛼𝑢(0)=0, (3) where 𝛼∈[0,2) and 𝜂∈[(√121+24𝛼−5)/3(4+𝛼),1). The main tool used was the Guo-Krasnoselskii fixed point theorem [9, 10].

It is worth mentioning that there are other types of works on sign-changing Green’s functions which prove the existence of sign-changing solutions, positive in some cases; see Infante and Webb’s papers [15–17].

Motivated greatly by the above-mentioned works, in this paper, we will study the BVP (3) by applying iterative method. Throughout this paper, we always assume that 𝛼∈[0,2) and 𝜂∈[2/3,1). Although corresponding Green’s function is sign-changing, we still obtain the existence of monotone positive solution for the BVP (3) under some suitable conditions on 𝑓. Moreover, our iterative scheme starts off with zero function, which implies that the iterative scheme is feasible.

2. Main Results

For any 𝑦∈𝐶[0,1], we consider the BVP 𝑢^′′′(𝑡)(𝑡)[],𝑢=𝑦,𝑡∈0,1_′(0)=𝑢(1)=0,𝑢^′′𝜂+𝛼𝑢(0)=0. (4)

It follows from [14] that the expression of Green’s function 𝐺(𝑡,𝑠) of the BVP (4) is as follows: 𝐺(𝑡,𝑠)=𝑔₁(𝑡,𝑠)+𝑔₂(𝑡,𝑠)+𝑔₃𝜂,𝑡,𝑠, (5) where 𝑔₁(𝑡,𝑠)=−2−𝛼𝑡²(1−𝑠)²2(),()∈[]×[],𝑔2−𝛼𝑡,𝑠0,10,1₂(𝑡,𝑠)=()0,0≤𝑡≤𝑠≤1,𝑡−𝑠²2𝑔,0≤𝑠≤𝑡≤1,₃𝜂,𝑡,𝑠=0,𝑠≥𝜂,1−𝑡²2−𝛼,𝑠<𝜂, (6) and the 𝐺(𝑡,𝑠) has the following properties: 𝐺(𝑡,𝑠)≥0for0≤𝑠<𝜂,𝐺(𝑡,𝑠)≤0for𝜂≤𝑠≤1. (7) Moreover, for 𝑠≥𝜂, {𝐺()[(){𝐺()[()()max𝑡,𝑠∶𝑡∈0,1]}=𝐺1,𝑠=0,min𝑡,𝑠∶𝑡∈0,1]}=𝐺0,𝑠=−1−𝑠²2−𝛼≥−1−𝜂²,2−𝛼 (8) and, for 𝑠<𝜂, max{𝐺(𝑡,𝑠)∶𝑡∈[0,1]}=𝐺(0,𝑠)=2𝑠−𝑠²≤2−𝛼2𝜂−𝜂²,{𝐺()[()2−𝛼min𝑡,𝑠∶𝑡∈0,1]}=𝐺1,𝑠=0. (9)

So, if we let 𝑀=max{|𝐺(𝑡,𝑠)|∶𝑡,𝑠∈[0,1]}, then 𝑀=max1−𝜂²,2−𝛼2𝜂−𝜂²<12−𝛼.2−𝛼 (10)

Let Banach space 𝐸=𝐶[0,1] be equipped with the norm ‖𝑢‖=max_𝑡∈[0,1]|𝑢(𝑡)| and (𝑡)[.𝐾=𝑦∈𝐸∶𝑦isnonnegativeanddecreasingon0,1] (11) Then 𝐾 is a cone in 𝐸. Note that this induces an order relation “≲” in 𝐸 by defining 𝑢≲𝑣 if and only if 𝑣−𝑢∈𝐾.

In the remainder of this paper, we always assume that 𝑓∈𝐶([0,1]×[0,+∞),[0,+∞)) and satisfies the following two conditions:(𝐻₁)for each 𝑢∈[0,+∞), the mapping 𝑡↦𝑓(𝑡,𝑢) is decreasing;(𝐻₂)for each 𝑡∈[0,1], the mapping 𝑢↦𝑓(𝑡,𝑢) is increasing.

Now, we define an operator 𝑇 as follows: (𝑇𝑢)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,𝑢∈𝐾,𝑡∈[0,1]. (12) Obviously, if 𝑢 is a fixed point of 𝑇 in 𝐾, then 𝑢 is a nonnegative and decreasing solution of the BVP (3).

Lemma 1. 𝑇∶𝐾→𝐾 is completely continuous.

Proof. Let 𝑢∈𝐾. Then, for 𝑡∈[0,𝜂], we have (𝑇𝑢)(𝑡)=^𝑡₀𝑔₁()+()𝑡,𝑠𝑡−𝑠²2+1−𝑡²𝑓((𝑠+2−𝛼𝑠,𝑢))𝑑𝑠_𝜂𝑡𝑔₁(𝑡,𝑠)+1−𝑡²𝑓((𝑠+2−𝛼𝑠,𝑢))𝑑𝑠_1𝜂𝑔₁(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠, (13) which together with (𝐻₁) and (𝐻₂) implies that (𝑇𝑢)^′(𝑡)=𝛼𝑡2−𝛼^𝜂₀𝑠²𝑓((𝑠−−2𝑠𝑠,𝑢))𝑑𝑠_𝑡0𝑠𝑓(𝑠,𝑢(𝑠))𝑑𝑠−𝑡^𝜂_𝑡𝑓((𝑠+𝑠,𝑢))𝑑𝑠𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝑓((𝑠𝜂𝑠,𝑢))𝑑𝑠≤𝑓𝜂,𝑢𝛼𝑡2−𝛼^𝜂₀𝑠²−2𝑠𝑑𝑠−^𝑡₀𝑠𝑑𝑠−𝑡^𝜂_𝑡𝑑𝑠+𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝜂𝛼𝑑𝑠=𝑡𝑓𝜂,𝑢1−3𝜂3()𝑡2−𝛼−𝜂+2𝜂𝛼≤𝑡𝑓𝜂,𝑢1−3𝜂3()−𝜂2−𝛼2≤0. (14) For 𝑡∈[𝜂,1], we have (𝑇𝑢)(𝑡)=^𝜂₀𝑔₁()+()𝑡,𝑠𝑡−𝑠²2+1−𝑡²𝑓((𝑠+2−𝛼𝑠,𝑢))𝑑𝑠_𝑡𝜂𝑔₁()+()𝑡,𝑠𝑡−𝑠²2𝑓((𝑠+𝑠,𝑢))𝑑𝑠_1𝑡𝑔₁(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠, (15) which together with (𝐻₁) and (𝐻₂) shows that (𝑇𝑢)^′(𝑡)=𝛼𝑡2−𝛼^𝜂₀𝑠²𝑓((𝑠+−2𝑠𝑠,𝑢))𝑑𝑠_𝑡𝜂(𝑡−𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠−^𝜂₀((𝑠+𝑠𝑓𝑠,𝑢))𝑑𝑠𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝑓((𝑠𝜂𝑠,𝑢))𝑑𝑠≤𝑓𝜂,𝑢𝛼𝑡2−𝛼^𝜂₀𝑠²−2𝑠𝑑𝑠+^𝑡_𝜂()−𝑡−𝑠𝑑𝑠_𝜂0𝑠𝑑𝑠+𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝜂𝛼𝑑𝑠=𝑡𝑓𝜂,𝑢1−3𝜂3()+2−𝛼𝑡−2𝜂2𝜂𝛼≤𝑡𝑓𝜂,𝑢1−3𝜂3()+2−𝛼1−2𝜂2≤0. (16)
So, (𝑇𝑢)(𝑡) is decreasing on [0,1]. At the same time, since (𝑇𝑢)(1)=0, we know that (𝑇𝑢)(𝑡) is nonnegative on [0,1]. This indicates that 𝑇𝑢∈𝐾. Furthermore, although 𝐺(𝑡,𝑠) is not continuous, it follows from known textbook results, for example, see [18], that 𝑇∶𝐾→𝐾 is completely continuous.

Theorem 2. Assume that 𝑓(𝑡,0)≢0 for 𝑡∈[0,1] and there exist two positive constants 𝑎 and 𝑏 such that the following conditions are satisfied:(𝐻₃)𝑓(0,𝑎)≤(2−𝛼)𝑎;(𝐻₄)𝑏(𝑢₂−𝑢₁)≤𝑓(𝑡,𝑢₂)−𝑓(𝑡,𝑢₁)≤2𝑏(𝑢₂−𝑢₁), 0≤𝑡≤1, 0≤𝑢₁≤𝑢₂≤𝑎.
If we construct an iterative sequence 𝑣_𝑛+1=𝑇𝑣_𝑛, 𝑛=0,1,2,…, where 𝑣₀(𝑡)≡0 for 𝑡∈[0,1], then {𝑣_𝑛}^∞_𝑛=1 converges to 𝑣^∗ in 𝐸 and 𝑣^∗ is a decreasing and positive solution of the BVP (3).

Proof. Let 𝐾_𝑎={𝑢∈𝐾∶‖𝑢‖≤𝑎}. Then we may assert that 𝑇∶𝐾_𝑎→𝐾_𝑎.
In fact, if 𝑢∈𝐾_𝑎, then it follows from (𝐻₃) that 0≤(𝑇𝑢)(𝑡)=¹₀𝐺()𝑓((𝑠≤𝑡,𝑠𝑠,𝑢))𝑑𝑠₁₀|𝐺(𝑓()≤()[],𝑡,𝑠)|0,𝑎𝑑𝑠2−𝛼𝑎𝑀<𝑎,𝑡∈0,1 (17) which shows that ‖𝑇𝑢‖≤𝑎. So, 𝑇∶𝐾_𝑎→𝐾_𝑎.
Now, we prove that {𝑣_𝑛}^∞_𝑛=1 converges to 𝑣^∗ in 𝐸 and 𝑣^∗ is a decreasing and positive solution of the BVP (3).
Indeed, in view of 𝑣₀∈𝐾_𝑎 and 𝑇∶𝐾_𝑎→𝐾_𝑎, we have 𝑣_𝑛∈𝐾_𝑎, 𝑛=1,2,…. Since the set {𝑣_𝑛}^∞_𝑛=0 is bounded and 𝑇 is completely continuous, we know that the set {𝑣_𝑛}^∞_𝑛=1 is relatively compact. In what follows, we prove that {𝑣_𝑛}^∞_𝑛=0 is monotone by induction. First, it is obvious that 𝑣₁−𝑣₀=𝑣₁∈𝐾, which shows that 𝑣₀≲𝑣₁. Next, we assume that 𝑣_𝑘−1≲𝑣_𝑘. Then 𝑣_𝑘−𝑣_𝑘−1 is decreasing and 0≤𝑣_𝑘−1(𝑡)≤𝑣_𝑘(𝑡)≤𝑎, 𝑡∈[0,1]. So, it follows from (𝐻₄) that 𝑣^′_𝑘+1(𝑡)−𝑣^′_𝑘(𝑡)=𝛼𝑡2−𝛼^𝜂₀𝑠²−2𝑠𝑓𝑠,𝑣_𝑘(𝑠)−𝑓𝑠,𝑣_𝑘−1(𝑠)−𝑑𝑠_𝑡0𝑠𝑓𝑠,𝑣_𝑘(𝑠)−𝑓𝑠,𝑣_𝑘−1(𝑠)𝑑𝑠−𝑡^𝜂_𝑡𝑓𝑠,𝑣_𝑘(𝑠)−𝑓𝑠,𝑣_𝑘−1(𝑠)+𝑑𝑠𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝑓𝑠,𝑣_𝑘(𝑠)−𝑓𝑠,𝑣_𝑘−1(𝑠≤)𝑑𝑠𝑏𝛼𝑡2−𝛼^𝜂₀𝑠²−2𝑠𝑣_𝑘(𝑠)−𝑣_𝑘−1(𝑠)𝑑𝑠−𝑏^𝑡₀𝑠𝑣_𝑘(𝑠)−𝑣_𝑘−1(𝑠)𝑑𝑠−𝑏𝑡^𝜂_𝑡𝑣_𝑘(𝑠)−𝑣_𝑘−1(𝑠+)𝑑𝑠2𝑏𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝑣_𝑘(𝑠)−𝑣_𝑘−1(𝑠𝑣)𝑑𝑠≤𝑏_𝑘𝜂−𝑣_𝑘−1𝜂×𝛼𝑡2−𝛼^𝜂₀𝑠²−−2𝑠𝑑𝑠_𝑡0𝑠𝑑𝑠−𝑡^𝜂_𝑡𝑑𝑠+2𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝑣𝑑𝑠=𝑏_𝑘𝜂−𝑣_𝑘−1𝜂𝑡×𝛼−𝜂³+3𝜂²−6𝜂+23()𝑡2−𝛼−𝜂+2≤𝑏𝑣_𝑘𝜂−𝑣_𝑘−1𝜂𝑡𝛼−𝜂³+3𝜂²−6𝜂+23()−𝜂2−𝛼2≤𝑏𝑣_𝑘𝜂−𝑣_𝑘−1𝜂𝑡𝛼−3𝜂+23()−𝜂2−𝛼2,𝑣≤0,𝑡∈0,𝜂_′𝑘+1(𝑡)−𝑣^′_𝑘(𝑡)=𝛼𝑡2−𝛼^𝜂₀𝑠²−2𝑠𝑓𝑠,𝑣_𝑘(𝑠)−𝑓𝑠,𝑣_𝑘−1(𝑠)+𝑑𝑠_𝑡𝜂(𝑡−𝑠)𝑓𝑠,𝑣_𝑘(𝑠)−𝑓𝑠,𝑣_𝑘−1(𝑠)−𝑑𝑠_𝜂0𝑠𝑓𝑠,𝑣_𝑘(𝑠)−𝑓𝑠,𝑣_𝑘−1(𝑠)+𝑑𝑠𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝑓𝑠,𝑣_𝑘(𝑠)−𝑓𝑠,𝑣_𝑘−1(𝑠)≤𝑑𝑠𝑏𝛼𝑡2−𝛼^𝜂₀𝑠²−2𝑠𝑣_𝑘(𝑠)−𝑣_𝑘−1(𝑠)𝑑𝑠+2𝑏^𝑡_𝜂(𝑡−𝑠)𝑣_𝑘(𝑠)−𝑣_𝑘−1(𝑠)𝑑𝑠−𝑏^𝜂₀𝑠𝑣_𝑘(𝑠)−𝑣_𝑘−1(𝑠+)𝑑𝑠2𝑏𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝑣_𝑘(𝑠)−𝑣_𝑘−1(𝑠𝑣)𝑑𝑠≤𝑏_𝑘𝜂−𝑣_𝑘−1𝜂×𝛼𝑡2−𝛼^𝜂₀𝑠²−2𝑠𝑑𝑠+2^𝑡_𝜂(𝑡−𝑠)𝑑𝑠−^𝜂₀𝑠𝑑𝑠+2𝛼𝑡2−𝛼¹_𝜂(1−𝑠)²𝑣𝑑𝑠=𝑏_𝑘𝜂−𝑣_𝑘−1𝜂×𝛼𝑡−𝜂³+3𝜂²−6𝜂+23(2−𝛼)+𝑡²𝜂−2𝜂𝑡+²2≤𝑏𝑣_𝑘𝜂−𝑣_𝑘−1𝜂𝑡×𝛼−𝜂³+3𝜂²−6𝜂+23(2−𝛼)+𝑡−3𝜂2≤𝑏𝑣_𝑘𝜂−𝑣_𝑘−1𝜂𝑡𝛼−3𝜂+23()+2−𝛼2−3𝜂2≤0,𝑡∈𝜂,1. (18) So, 𝑣^′_𝑘+1(𝑡)−𝑣^′_𝑘(𝑡)≤0,𝑡∈[0,1]. (19) This together with 𝑣_𝑘+1(1)−𝑣_𝑘(1)=∫¹₀𝐺(1,𝑠)[𝑓(𝑠,𝑣_𝑘(𝑠))−𝑓(𝑠,𝑣_𝑘−1(𝑠))]𝑑𝑠=0 implies that 𝑣_𝑘+1(𝑡)−𝑣_𝑘(𝑡)≥0,𝑡∈[0,1]. (20) In view of (19) and (20), we know that 𝑣_𝑘+1−𝑣_𝑘∈𝐾, which indicates that 𝑣_𝑘≲𝑣_𝑘+1. Thus, we have shown that 𝑣_𝑛≲𝑣_𝑛+1, 𝑛=0,1,2…. Since {𝑣_𝑛}^∞_𝑛=1 is relatively compact and monotone, there exists 𝑣^∗∈𝐾_𝑎 such that lim_𝑛→∞𝑣_𝑛=𝑣^∗, which together with the continuity of 𝑇 and the fact that 𝑣_𝑛+1=𝑇𝑣_𝑛 implies that 𝑣^∗=𝑇𝑣^∗. This indicates that 𝑣^∗ is a decreasing and nonnegative solution of the BVP (3). Moreover, since 𝑓(𝑡,0)≢0 for 𝑡∈[0,1], we know that zero function is not a solution of the BVP (3), which shows that 𝑣^∗ is a positive solution of the BVP (3).

3. An Example

Consider the BVP 𝑢^′′′(𝑡)=1𝑢16²(𝑡)+14𝑢(𝑡)+()[],𝑢1−𝑡,𝑡∈0,1_′(0)=𝑢(1)=0,𝑢^′′𝜂+𝑢(0)=0. (21)

If we let 𝛼=1, 𝜂=7/9, and 𝑓(𝑡,𝑢)=(1/16)𝑢²+(1/4)𝑢+(1−𝑡), (𝑡,𝑢)∈[0,1]×[0,+∞), then all the hypotheses of Theorem 2 are fulfilled with 𝑎=2 and 𝑏=1/4. Therefore, it follows from Theorem 2 that the BVP (21) has a decreasing and positive solution. Moreover, the iterative scheme is 𝑣₀(𝑡)≡0 for 𝑡∈[0,1] and 𝑣_𝑛+1(𝑡)=∫^𝑡₀−2−𝑡²(1−𝑠)²2+(𝑡−𝑠)²2+1−𝑡²×1𝑣16^𝑛(𝑠)²+14𝑣_𝑛(𝑠)+(1−𝑠)𝑑𝑠+∫_𝑡^7/9−2−𝑡²(1−𝑠)²2+1−𝑡²×1𝑣16^𝑛(𝑠)²+14𝑣_𝑛(𝑠)+(1−𝑠)𝑑𝑠+∫¹_7/9−2−𝑡²(1−𝑠)²2×1𝑣16^𝑛(𝑠)²+14𝑣_𝑛(𝑠)+(71−𝑠)𝑑𝑠,𝑡∈0,9∫,𝑛=0,1,2,…,_07/9−2−𝑡²(1−𝑠)²2+(𝑡−𝑠)²2+1−𝑡²×1𝑣16^𝑛(𝑠)²+14𝑣_𝑛(𝑠)+(+∫1−𝑠)𝑑𝑠_𝑡7/9−2−𝑡²(1−𝑠)²2+(𝑡−𝑠)²2×1𝑣16^𝑛(𝑠)²+14𝑣_𝑛(𝑠)+(1−𝑠)𝑑𝑠+∫¹_𝑡−2−𝑡²(1−𝑠)²2×1𝑣16^𝑛(𝑠)²+14𝑣_𝑛(𝑠)+(71−𝑠)𝑑𝑠,𝑡∈9,1,𝑛=0,1,2,…. (22) The first, second, third, and fourth terms of this scheme are as follows: 𝑣₀(𝑡)𝑣≡0,₁(𝑡)1=−𝑡24⁴+16𝑡³−227𝑡648²+73,𝑣324₂(𝑡)=1𝑡55296¹¹−35𝑡165888¹⁰+839𝑡746496⁹−7489𝑡2239488⁸+137893𝑡40310784⁷+1415521𝑡120932352⁶−481871𝑡10077696⁵+162649𝑡10077696⁴+1987201𝑡10077696³−481465𝑡1119744²+951409,𝑣3779136₃(𝑡)=1𝑡293534171136²⁵−1𝑡10871635968²⁴+9341𝑡7925422620672²³−24949𝑡2641807540224²²+1842703𝑡35664401793024²¹−20695715𝑡106993205379072²⁰[phone]𝑡2888816545234944¹⁹−131313793𝑡962938848411648¹⁸−463466345483𝑡155996093442686976¹⁷[phone]𝑡1925877696823296¹⁶−4124079293903𝑡467988280328060928¹⁵−213831233807699𝑡4211894522952548352¹⁴[phone]𝑡9749755840167936¹³−86168085379277𝑡526486815369068544¹²−14488536542111𝑡19499511680335872¹¹[phone]𝑡175495605123022848¹⁰−42359864609123𝑡29249267520503808⁹−2156592823827731𝑡263243407684534272⁸[phone]𝑡9749755840167936⁷[phone]𝑡263243407684534272⁶−13158305753867𝑡203119913336832⁵[phone]𝑡10968475320188928⁴[phone]𝑡457019805007872³−1830285163670609𝑡4113178245070848²+1148605916775291.4503599627370496 (3)

4. Conclusion

In [14], only the existence of at least one positive solution to the BVP (3) was obtained when 𝛼∈[0,2) and 𝜂∈[(√121+24𝛼−5)/3(4+𝛼),1). In this paper, when 𝛼∈[0,2) and 𝜂∈[2/3,1), we have successfully constructed an iterative sequence, whose limit is just a desired monotone positive solution of the BVP (3). Moreover, our iterative scheme starts off with zero function, which implies that the iterative scheme is feasible.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.