# Iterative Solutions for the Differential Equation withp-Laplacian on Infinite Interval

Authors: Guoying Zhao; Ying Wang
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This paper systematically investigates a class of fourth-order differential equation withp-Laplacian on infinite interval in Banach space. By means of the monotone iterative technique, we establish not only the existence of positive solutions but also iterative schemes under the suitable conditions. At last, we give an example to demonstrate the application of the main result.


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## Body

## 1. Introduction

The partial differential equation with thep-Laplacian operator
(1)−div∇uxp−2∇ux+bxφux=0,which is often used to describe, for example, diffusion process [1], with a spatial symmetric potential b, can be reduced to rtφpy′t′+ctφpyt=0, where φpx=xp−2x,p>1. This fact leads us to study the following p-Laplacian boundary value problem (BVP):
(2)φpu″t″−k2φpu″t=atft,ut,∈0,+∞,u0=∫0∞gtutdt,limt⟶+∞u′t=ξ,u″0=limt⟶+∞u″t=0,where φpx=xp−2x,p>1, φq=φp−1, 1/p+1/q=1, k>0, ξ≥0 are real constants, g:0,+∞⟶0,+∞ is a Lebesgue integrable function with ∫0∞gtdt<1, ∫0∞tgtdt<+∞, a:0,+∞⟶0,+∞ is continuous and may be singular at t=0, and f:0,+∞×0,+∞⟶0,+∞ is a continuous function.Thep-Laplacian equation arises quite naturally in the modeling of different physical and natural phenomena. For instance, in fluid mechanics, the shear stress τ⟶ and the velocity gradient ∇pu of certain fluids obey a relation of the form τ⟶=ax∇pu, where ∇pu=∇up−2∇u. Here, the real number p>1 and p=2 (respectively, p<2,p>2) designate a Newtonian (respectively, pseudoplastic, dilatant) fluid. Given a is a constant, the resulting equations of motion then involve diva∇pu, which reduces to aΔpu=adiv∇pu. Over the last couple of decades, many important results including integral and fractional equations with p-Laplacian on certain boundary value conditions had been obtained. We refer the reader to [2–17] and the references cited therein. Liang and Zhang [18] considered the m-point BVP with a p-Laplacian operator
(3)φpu′t′+htfut=0,t∈0,+∞,u0=∑i=1m−2αiuξi,limt⟶+∞u′t=0,where ξi∈0,+∞, with 0<ξ1<ξ2<⋯<ξm−2<+∞, and αi satisfies αi∈0,+∞, 0≤∑i=1m−2αi<1, h:0,+∞⟶0,+∞ and has countably many singularities in 1,+∞, f:0,+∞⟶0,+∞ is a continuous function. The existence of positive solutions is obtained by applying the fixed-point theorem of Leggett-Williams.Using the fixed point index theory, Xu and Yang in [19] researched the existence of positive solutions for the fourth order p-Laplacian BVP
(4)u″p−1u″″=ft,u,t∈0,1,u2i0=u2i1=0,i=0,1,where p>0, f:0,1×0,+∞⟶0,+∞ is a continuous function.The motivation for the present work stems from both practical and theoretical aspects. In fact, many mathematical problems in science and engineering are set in unbounded domains, such as unsteady flow of gas through a semi-infinite porous media, the theory of drain flows, plasma physics, in determining the electrical potential in an isolated neutral atom. In all these applications, it is frequent that only positive solutions are useful. In this paper, we study the differential equation withp-Laplacian operator as BVP (2); when p=2, BVP (2) becomes the ordinary fourth-order differential equation. The results for the existence of the maximal and minimal solutions to the BVP (2) are established. In addition, we establish iterative schemes for approximating the solutions, which start from the known simple linear functions. However, to the best knowledge of the authors, there are few works in the literature dealing with the existence of positive solutions to boundary value problems of differential equation on infinite intervals with p-Laplacian operator by using iterative technique up to now. The goal of the present paper is to fill the gap in this area, so it is interesting and important to study the existence of positive solutions for BVP (2).


## 2. Preliminaries and Lemmas

The basic space used in this paper isE, where E is denoted by
(5)E=u∈C0,+∞:supt∈0,+∞∣ut∣1+t<+∞.Then,E is a Banach space equipped with the norm ∥u∥=supt∈0,+∞∣ut∣/1+t. Define a cone K in the Banach space E by K=u∈E:ut≥0,t∈0,+∞.Lemma 1 (see [20]).
Lety∈C0,+∞∩L10,+∞, then x is a solution of
(6)x″t−k2xt+yt=0,t∈0,+∞,x0=limt⟶+∞xt=0,if and only if x is a solution of
(7)xt=∫0∞Gt,sysds,where
(8)Gt,s=12ke−ksekt−e−kt,0≤t≤s<+∞,e−kteks−e−ks,0≤s≤t<+∞.Lemma 2.
Green’s functionGt,s satisfies
(1)
Gt,s≥0, for any t,s∈0,+∞(2)
Gt,s≤Gs,s≤1/2k, for any t,s∈0,+∞
In what follows, we list some conditions for convenience.
H0g:0,+∞⟶0,+∞ is a Lebesgue integrable function with ∫0∞gtdt<1, ∫0∞tgtdt<+∞.
H1f:0,+∞×0,+∞⟶0,+∞ is a continuous function, ft,0≡0 on 0,+∞ and ft,1+tx is bounded, for t∈0,+∞, x∈D, D⊂0,+∞ is a closed subinterval.
H2a:0,+∞⟶0,+∞ is continuous, at≡0 on 0,+∞ and
(9)0<∫0∞Gs,sasds<+∞,∫0∞sφq∫0∞Gs,ςaςdςds<+∞.By routine discussion, Lemma3 is valid.Lemma 3.
Assume thatH0 holds and φqx∈C0,+∞∩L10,+∞; then, u is a solution of
(10)u″t+φqxt=0,t∈0,+∞,u0=∫0∞gtutdt,limt⟶+∞u′t=ξ,if and only if u is a solution of
(11)ut=11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φqxτdτdsdt+∫0t∫s∞φqxτdτds+ξt.Letxt=−φpu″t, then BVP (2) is divided into the following two parts:
(12)x″t−k2xt+atft,ut=0,t∈0,+∞,x0=limt⟶+∞xt=0,u″t+φqxt=0,t∈0,+∞,u0=∫0∞gtutdt,limt⟶+∞u′t=ξ.From Lemmas1 and 3, under the above assumptions H0–H2, denote the operator A:K⟶E as follows:
(13)Aut=11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds+ξt,t∈0,+∞.Now, we claim thatAu is well defined for u∈K. In fact, for any u∈K, there exists r>0, such that ut/1+t≤r,t∈0,+∞. From H1 and the definition of ∥·∥, we have
(14)Sfr≔supft,1+tx:t,x∈0,+∞×0,r<+∞.Thus, byH1H2, we know
(15)∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds≤φqSfr∫0∞sφq∫0∞Gs,ςaςdςds<+∞.Since(16)∫0∞gtdt<1,∫0∞tgtdt<+∞.Together with (15), for any t∈0,+∞, we can see that
(17)∣Aut∣1+t≤11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds+ξ≤11−∫0∞gtdtξ∫0∞tgtdt+φqSfr∫0∞sφq∫0∞Gs,ςaςdςds∫0∞gtdt+φqSfr∫0∞sφq∫0∞Gs,ςaςdςds+ξ.So, by (15) and (17), we obtain
(18)supt∈0,+∞∣Aut∣1+t<+∞.On the other hand, for anyt1,t2∈0,+∞, by (13), we have
(19)Aut1−Aut2≤∫t1t2∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds+ξt1−t2⟶0,ast1⟶t2.Therefore,Au∈C0,+∞, for any u∈K. Hence, A:K⟶E is well defined. Obviously, u is a positive solution of BVP (2) if and only if u is a fixed point of A in K.The Arzela-Ascoli theorem fails to work in the Banach spaceE due to the fact that the infinite interval 0,+∞ is noncompact. The following compactness criterion will help us to resolve this problem.Lemma 4 (see [21, 22]).
LetE be defined as (5) and M be any bounded subset of E. Then, M is relatively compact in Eifxt/1+t:x∈M is equicontinuous on any finite subinterval of 0,+∞, and for any given ε>0, there exists N>0, such that xt1/1+t1−xt2/1+t2<ε uniformly with respect to x∈M, as t1,t2>N.


## 3. Main Results

Lemma 5.
Assume thatH0−H2 hold. Then, A:K⟶K is completely continuous.Proof.
It is clear thatAut≥0 for any u∈K, t∈0,+∞. Thus, AK⊆K. Now, we prove that A is continuous and compact, respectively. Let M⊂K be a bounded subset. Then, there exists R>0, such that x<R, for any x∈M. So, for any x∈M, we have
(20)∥Au∥=supt∈0,+∞11+t11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds+ξt≤11−∫0∞gtdtξ∫0∞tgtdt+φqSfR∫0∞sφq∫0∞Gs,ςaςdςds∫0∞gtdt+φqSfR∫0∞sφq∫0∞Gs,ςaςdςds+ξ<+∞,where
(21)SfR≔supft,1+tx:t,x∈0,+∞×0,R<+∞.
So,AM is bounded in E. Moreover, given T∈0,+∞, for any x∈M and t1,t2∈0,T, without loss of generality, we may assume that t1<t2. In fact,
(22)Aut11+t1−Aut21+t2≤11+t1−11+t211−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτdsdt+11+t1∫0t1∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds−11+t2∫0t2∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds+ξt11+t1−t21+t2≤11+t1−11+t211−∫0∞gtdtξ∫0∞tgtdt+φqSfR∫0∞gtdt∫0∞sφq∫0∞Gs,ςaςdςds+11+t1−11+t2φqSfR∫0t1∫s∞φq∫0∞Gτ,ςaςdςdτds+φqSfR∫t1t2∫s∞φq∫0∞Gτ,ςaςdςdτds+ξt11+t1−t21+t2.
So, for anyε>0, there exists δ>0, such that for any t1,t2∈0,T with t1−t2<δ, and for any x∈M, we have
(23)Aut11+t1−Aut21+t2<δ.
Hence,Aut/1+t:u∈M is equicontinuous on 0,T. Since T>0 is arbitrary, Aut/1+t:u∈M is locally equicontinuous on 0,+∞.
Next, we prove that we show thatA:K⟶K is equiconvergent at +∞. For any u∈M, we have
(24)limt⟶+∞Aut1+t=limt⟶+∞11+t11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds+ξt=limt⟶+∞∫t∞φq∫0∞Gτ,ςaςfς,uςdςdτ+ξ=ξ.
So, for anyu∈M, we have
(25)Aut1+t−ξ=11+t11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds+ξt−ξ≤11+t11−∫0∞gtdtξ∫0∞tgtdt+φqSfR∫0∞sφq∫0∞Gs,ςaςdςds∫0∞gtdt+φqSfR∫0∞sφq∫0∞Gs,ςaςdςds−ξ1+t⟶0,t⟶+∞.
Thus, for anyε>0, there exists N>0, for any t>N and for any x∈M, such that
(26)Aut1+t−ξ<ε2.
Consequently, for anyt1,t2>N and for any x∈M, we have
(27)Aut11+t1−ξ<ε2,Aut21+t2−ξ<ε2,t1,t2>N.
Therefore, for anyt1,t2>N and for any x∈M, we have
(28)Aut11+t1−Aut21+t2=Aut11+t1−ξ−Aut21+t2−ξ≤Aut11+t1−ξ+Aut21+t2−ξ<ε,t1,t2>N.
This implies thatAM is equiconvergent at +∞.
Letun⟶u as n⟶+∞; then, there exists r0 such that maxn∈ℕ\0∥un∥,∥u∥<r0, ℕ is a natural number set. By H2, we have
(29)∫0∞Gτ,ςaςfς,unςdς−∫0∞Gτ,ςaςfς,uςdς≤∫0∞Gτ,ςaςfς,unς−fς,uςdς≤2Sfr0∫0∞Gτ,ςaςdς<+∞,where
(30)Sfr0≔supft,1+tx:t,x∈0,+∞×0,r0<+∞.
So, for anyε>0, we can find a sufficiently large H0>0, such that
(31)Sfr0∫H0∞Gτ,ςaςdς<ε4.
It follows from the Lebesgue dominated convergence theorem and continuity off, we can get
(32)∫0H0Gτ,ςaςfς,unςdς−∫0H0Gτ,ςaςfς,uςdς≤∫0H0Gτ,ςaςfς,unς−fς,uςdς⟶0,n⟶+∞,τ∈0,+∞.
So, for the aboveε>0, there exists N>0, when n>N0, we have
(33)∫0H0Gτ,ςaςfς,unς−fς,uςdς<ε2.
Therefore, whenn>N0,
(34)∫0∞Gτ,ςaςfς,unςdς−∫0∞Gτ,ςaςfς,uςdς≤∫0∞Gτ,ςaςfς,unς−fς,unςdς≤∫0H0Gτ,ςaςfς,unς−fς,unςdς+∫H0∞Gτ,ςaςfς,unς−fς,unςdς≤ε2+2Sfr0∫H0∞Gτ,ςaςdς<ε,which implied that
(35)τφq∫0∞Gτ,ςaςfς,unςdς−τφq∫0∞Gτ,ςaςfς,uςdς⟶0,as n⟶+∞,τ∈0,+∞. Since
(36)∫0t∫s∞φq∫0∞Gτ,ςaςfς,unςdςdτds−∫0t∫s∞φq∫0∞Gτ,ςaςfς,uςdςdτds≤2φqSfr0∫0∞sφq∫0∞Gs,ςaςdςds<+∞.
Then, by the Lebesgue dominated convergence theorem, we get(37)Aun−Au=supt⟶+∞11+t11−∫0∞gtdt∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,unςdς−φq∫0∞Gτ,ςaςfς,uςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,unςdς−φq∫0∞Gτ,ςaςfς,uςdςdτds≤11−∫0∞gtdt∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,unςdς−φq∫0∞Gτ,ςaςfς,uςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,unςdς−φq∫0∞Gτ,ςaςfς,uςdςdτds≤∫0∞gtdt1−∫0∞gtdt+1∫0∞sφq∫0∞Gs,ςaςfς,unςdς−φq∫0∞Gs,ςaςfς,uςdςds⟶0,n⟶+∞.
Therefore,A:K⟶K is continuous. In conclusion, by Lemma 4, we know that A:K⟶K is completely continuous. The proof is completed.☐Theorem 6.
Assume thatH0–H2 hold and there exists d>2m which satisfies the following condition:
H3ft,x1≤ft,x2, t∈0,+∞, 0≤x1≤x2.
H4ft,1+tx≤φpd/2l, t,x∈0,+∞×0,d,
where(38)m=ξ1+∫0∞tgtdt1−∫0∞gtdt,l=11−∫0∞gtdt∫0∞sφq∫0∞Gs,ςaςdςds.
Then, BVP (2) has the maximal and minimal positive solutions μ∗ and μ¯∗ on 0,+∞, such that
(39)0<supt∈0,+∞∣μ∗t∣1+t≤d,0<supt∈0,+∞∣μ¯∗t∣1+t≤d.
Moreover, for initial values(40)μ0t=d2+ξt+∫0∞tgtdt1−∫0∞gtdt,μ¯0t=0,t∈0,+∞,define the iterative sequences μn and μ¯n by
(41)μn=11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,μn−1ςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,μn−1ςdςdτds+ξt,μ¯n=11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,μ¯n−1ςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,μ¯n−1ςdςdτds+ξt.Then,
(42)limn⟶+∞supt∈0,+∞∣μnt−μ∗t∣1+t=0,limn⟶+∞supt∈0,+∞∣μ¯nt−μ¯∗t∣1+t=0.Proof.
From Lemma5, we know that A:K⟶K is completely continuous. For any u1,u2∈K with u1≤u2, from the definition of A and H3, we know that Au1≤Au2. Let Kd=u∈K:∥u∥≤d. In what follows, we firstly prove A:Kd⟶Kd. In fact, for any u∈Kd, we have 0≤ut/1+t≤d,t∈0,+∞. By H4, we know that ft,x≤φpd/2l,t,x∈0,+∞×0,d. Also, by H4, we have
(43)μn+1t≤μnt,t∈0,+∞,n=0,1,2,⋯.
Thus, we have proved thatA:Kd⟶Kd. Let μ0t=d/2+ξt+∫0∞tgtdt/1−∫0∞gtdt,t∈0,+∞, and then, μ0t∈Kd. Let μ1=Aμ0, μ2=Aμ1=A2μ0; by Theorem 6, we have μ1,μ2∈Kd. Denote μn+1=Aμn=Anμ0,n=1,2,⋯. Since A:Kd⟶Kd, we have μn∈AKd⊂Kd. It follows from the complete continuity of A that μnn=1∞ is a sequentially compact set. By H4, we have
(44)μ1t=Aμ0t=11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,μ0ςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,μ0ςddςdτds+ξt≤11−∫0∞gtdt∫0∞gtdt∫0∞sφq∫0∞Gs,ςaςfς,μ0ςdςds+∫0∞sφq∫0∞Gs,ςaςfς,μ0ςddςds+ξ∫0∞tgtdt1−∫0∞gtdt+ξt≤d2l1−∫0∞gtdt∫0∞sφq∫0∞Gs,ςaςdςds+ξt+∫0∞tgtdt1−∫0∞gtdt≤d2+ξt+∫0∞tgtdt1−∫0∞gtdt=μ0t.
Together withH4, we also have μ2t=Aμ1t≤Aμ0t=μ1t,t∈0,+∞. By induction, we obtain
(45)μn+1t≤μnt,t∈0,+∞,n=0,1,2,⋯.
Therefore, there existsμ∗∈Kd such that μn⟶μ∗ as n⟶+∞. Applying the continuity of A and μn+1=Aμn, we get that Aμ∗=μ∗.
On the other hand, letμ¯0t=0, t∈0,+∞, and then μ¯0t∈Kd. Let μ¯1=Aμ¯0, μ¯2=Aμ¯1=A2μ¯0; then, by Theorem 6, we have μ¯1,μ¯2∈Kd. Denote μ¯n+1=Aμ¯n=Anμ¯0,=1,2,⋯. Since A:Kd⟶Kd, we have μ¯n∈AKd⊂Kd. It follows from the complete continuity of A that μ¯nn=1∞ is a sequentially compact set. Since μ¯1=Aμ¯0∈Kd, we have
(46)μ¯2t=Aμ¯1t=A0t≥0,t∈0,+∞.
By induction, we get(47)μ¯n+1t≥μ¯nt,∈0,+∞,n=1,2,⋯.
Thus, there existsμ¯∗∈K such that μ¯n⟶μ¯∗ as n⟶+∞. Applying the continuity of A and μ¯n+1=Aμ¯n, we get that Aμ¯∗=μ¯∗.
Now, we are in a position to show thatμ∗ and μ¯∗ are the maximal and minimal positive solutions of BVP (2) in 0,d/2+ξt+∫0∞tgtdt/1−∫0∞gtdt. Let u∈0,d/2+ξt+∫0∞tgtdt/1−∫0∞gtdt be any solution of BVP (2), that is, Au=u. Noting that A is nondecreasing and μ¯0t=0≤ut≤d/2+ξt+∫0∞tgtdt/1−∫0∞gtdt=μ0t, then we have μ¯1t=Aμ¯0t≤ut≤Aμ0t=μ1t, for all t∈0,+∞. By induction, we have
(48)μ¯nt≤ut≤μnt,n=1,2,3,⋯.
Sinceμ∗=limn⟶+∞μn,μ¯∗=limn⟶+∞μ¯n, it follows from (42)–(45) that
(49)μ¯0≤μ¯1≤⋯μ¯n≤⋯≤μ¯∗≤u≤μ∗≤⋯≤μn≤⋯≤μ1≤μ0.
In virtue offt,0,t∈0,+∞, then the zero function is not the solution of BVP (2). Therefore, by (46), we know that μ∗ and μ¯∗ are the maximal and minimal positive solutions of BVP (2) in 0,d/2+ξt+∫0∞tgtdt/1−∫0∞gtdt, which can be obtained by the corresponding iterative sequences μn=Aμn−1,μ¯n=Aμ¯n−1. The proof is completed.☐Remark 1.
The iterative schemes in Theorem6 start with a known simple linear function and the zero function, respectively. This is very convenient in application. So Theorem 6 is very interesting and importance. Similarly, we can obtain Theorem 7.Theorem 7.
Assume thatH0–H3 hold and there exist dn>dn−1>⋯>d2>d1>2m satisfying the following condition:
H′4ft,1+tx≤φpdj/2l, t,x∈0,+∞×0,dj, j=1,2,⋯,n.m,l are defined as Theorem 6. Then, BVP (2) has the maximal and minimal positive solutions μj∗ and μ¯j∗, on 0,+∞, such that
(50)0<supt∈0,+∞∣μj∗t∣1+t≤d,0<supt∈0,+∞∣μ¯j∗t∣1+t≤d.
Moreover, for initial values(51)μj0t=dj2+ξt+∫0∞tgtdt1−∫0∞gtdt,μ¯j0t=0,t∈0,+∞,define the iterative sequences μjn and μ¯jn by
(52)μjn=11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,μjn−1ςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,μjn−1ςdςdτds+ξt,μ¯jn=11−∫0∞gtdtξ∫0∞tgtdt+∫0∞gt∫0t∫s∞φq∫0∞Gτ,ςaςfς,μjn−1ςdςdτdsdt+∫0t∫s∞φq∫0∞Gτ,ςaςfς,μjn−1ςdςdτds+ξt.Then,
(53)limn⟶+∞supt∈0,+∞∣μjnt−μj∗t∣1+t=0,limn⟶+∞supt∈0,+∞∣μ¯jnt−μ¯j∗t∣1+t=0.Remark 2.
We note thatμ∗ and μ¯∗ in Theorem 6 may coincide; then, BVP (2) has only one solution in Kd. Similarly, positive solutions μj∗ and μ¯j∗ in Theorem 7 may also coincide.


## 4. Example

Consider the following BVP(54)u″t−1/2u″t″−u″t−1/2u″t=atft,ut,t∈0,+∞,u0=∫0∞11+t3utdt,limt⟶+∞u′t=34,″0=limt⟶+∞u″t=0.Obviously, we know thatp=3/2, k=1, ξ=3/4, gt=1/1+t3, so we can get
(55)∫0∞gtdt=∫0∞11+t3dt=12<1,∫0∞tgtdt=∫0∞t1+t3dt=12<+∞,Gt,s=12e−set−e−t,0≤t≤s<+∞,e−tes−e−s,0≤s≤t<+∞.Let(56)at=e−2t,ft,x=∣sin55t+21∣+1144x1+t3,x≤2,∣sin55t+21∣+114421+t3,x≥2.By computation, we get(57)∫0∞Gs,sasds≤12∫0∞e−2sds=14<+∞,∫0∞sφq∫0∞Gs,ςaςdςds=∫0∞sφq∫0sGs,ςaςdς+∫s∞Gs,ςaςdςds=13432<+∞.So, the conditionsH0–H3 hold. For m=3/2,l=13/216. Take d=10, it follows that
(58)f1t,1+tx≤109<φpd2l=φp102·13/216=108013,t,x∈0,+∞×0,10.Hence, conditionH4 holds, that is, all conditions of Theorem 6 are satisfied. Therefore, BVP (51) has the minimal and maximal positive solutions in 0,5+3/4t+1, which can be obtained by two explicit monotone iterative sequences.



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*Source: 8765932-2021-06-25.xml*