# Analysis on Existence of Positive Solutions for a Class Second Order Semipositone Differential Equations

Authors: Yunhai Wang; Xu Yang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we study the existence of positive solutions of the following second-order semipositone system (see equation 1). By applying a well-known fixed-point theorem, we prove that the problem admits at least one positive solution, iff is bounded below.


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## Body

## 1. Introduction

This paper is focused on the existence of positive solutions of a second-order semipositone system(1)−u″+ρu=ϕu+ft,u,ϕ,t∈0,1,−ϕ″=μu,t∈0,1,u0=u1=ϕ0=ϕ1=0,where μ is a positive constant and f satisfies the following assumption: F0f:0,1×ℝ+2⟶ℝ is continuous, and
(2)ft,u,ϕ≥−et,fort,u,ϕ∈0,1×ℝ+2,where e:0,1⟶ℝ+ is continuous and et≢0 on [0,1].The second-order elliptic systems(3)−Δu+ρu=ϕu+fu,x∈Ω,−Δϕ=μu,x∈Ω,u=ϕ=0,x∈∂Ω,have a strong physical meaning in quantum mechanics models [1, 2], in semiconductor theory [3], or in a time- and space-dependent mathematical model of nuclear reactors in a closed container [4]. To the best of our knowledge, existence and multiplicity of nontrivial solutions of BVP(1) have been widely studied by using the variational method [5], bifurcation techniques [6, 7], or fixed-point theorems [8–11]. In general, in order to ensure the positivity of the solutions of Equation (1), one of the crucial assumptions is that the nonlinearity f is nonnegative. Of course, the natural question is whether Equation (1) has a positive solution or not if f satisfies the assumption F0.On the other hand, many authors have been interested in finding the relations between the positivity of solutions and the changing sign of the nonlinearity in order to prove the existence of the positive solutions. We refer the readers to [12–16] and the references.Inspired by these references, the purpose of this paper is to find some new conditions, which are used to study the existence and multiplicity of positive solutions of the semipositone Equation (1). The main tool is the following well-known fixed-point theorem.Lemma 1 [17].
LetE be a Banach space and K be a cone in E. Assume Ωr and ΩR are open bounded subsets of E with Ωr∩K≠ϕ,Ωr∩K¯⊂ΩR∩K. Let T:ΩR∩K¯⟶K be a completely continuous operator such that
(a)
Tu≤u, for u∈∂Ωr∩K, and(b)
there exists aηt∈K\0 such that(4)u≠Tu+ληt,foru∈∂ΩR∩K,λ>0.
Then,T has a fixed point in ΩR∩K¯\Ωr∩K. The same conclusion remains valid if (a) holds on ∂ΩR∩K and (b) holds on ∂Ωr∩K.The paper is organized as follows: in Section2, we give some preliminaries, which are about the properties of the Green functions, the notations of some sets, etc.; in Section 3, we give the main results and the corresponding proof. In Section 4, some examples are given to illustrate the main results.


## 2. Preliminary

LetGt,s be the Green function of linear boundary value problem
(5)−u″+ρu=0,u0=u1=0,where ρ>−π2.Lemma 2 [18].
Letω=ρ, then Gt,s can be expressed by
(i)
Gt,s=sinhωtsinhω1−s/ωsinhω,0≤t≤s≤1,sinhωssinhω1−t/ωsinhω,0≤s≤t≤1,ifρ>0(ii)
Gt,s=t1−s,0≤t≤s≤1,s1−t,0≤s≤t≤1,ifρ=0(iii)
Gt,s=sinωtsinω1−s/ωsinω,0≤t≤s≤1,sinωssinω1−t/ωsinω,0≤s≤t≤1,if−π2<ρ>0.Lemma 3 [18].
The functionGt,s has the following properties:
(i)
Gt,s>0,∀t,s∈0,1(ii)
Gt,s≤CGs,s,∀t,s∈0,1(iii)
Gt,s≥δGt,tGs,s,∀t,s∈0,1where C=1,δ=ω/sinhω, if ρ>0; C=1,δ=1, if ρ=0; and C=1/sinω,δ=ωsinω, if −π2<ρ<0.Lemma 4.
For the functionGt,s, there exists a ξ>0 such that
(6)Gt,t≥ξ∫01Gt,sds.Proof.
(i)
Forρ>0, we have(7)∫01Gt,sds=1ω2sinhwsinhωtcoshω1−t−1+sinhω1−tcoshωt−1.
Let(8)J1t=coshω1−t−1sinhω1−t,0≤t<1,0,t=1,J2t=coshωt−1sinhωt,0<t≤1,0,t=0.
SinceJ1t is positive and continuous on [0,1] and J11=0, we have
(9)J1∗=maxt∈0,1J1t>0.
In the similar way, we also have(10)J2∗=maxt∈0,1J2t>0.
Choosingξ<ω/J1∗+J2∗. Then, for any t∈0,1, we have
(11)ξ∫01Gt,sds=ξ1ω2sinhwsinhωtcoshω1−t−1+sinhω1−tcoshωt−1=ξ1ω2sinhwsinhωtsinhω1−tcoshω1−t−1sinhω1−t+sinhωtsinhω1−tcoshωt−1sinhωt=ξω1ωsinhwsinhωtsinhω1−tJ1∗+sinhωtsinhω1−tJ2∗=ξJ1∗+J2∗ωGt,t≤Gt,t.
SinceG0,0=∫01G0,sds=G1,1=∫01G1,sds=0, then for any ξ>0, we have
(12)Gt,t=ξ∫01Gt,sds,fort=0,1.
Therefore, there exists aξ>0 such that
(13)Gt,t≥ξ∫01Gt,sds.(ii)
Forρ=0, it is obvious that
(14)∫01Gt,sds=12t21−t+12t1−t2≤t1−t2=Gt,t.(iii)
For−π2<ρ<0, we have
(15)∫01Gt,sds=1ω2sinwsinωt1−cosω1−t+sinhω1−t1−cosωt.
Using the similar discussion of Case (i), it follows that there exists aξ>0 such that
(16)Gt,t≥ξ∫01Gt,sds.
For convenience, letKt,s denote the Green function for ρ=0. Then, Equation (1) can be rewritten as
(17)−u″+ρu=μu∫01Kt,susds+ft,u∫01Kt,susds,u0=u1=0.
Furthermore, letx=u+ω, where ωt=∫01Gt,sesds is the unique solution of the linear boundary value problem
(18)−u″+ρu=et,u0=u1=0.
Then, we rewrite (17) as
(19)−x″+ρx=Ft,x−ω,x0=x1=0,where
(20)Ft,x−λω=μx−ω∫01Kt,sxs−ωsds+ft,x−ω,∫01Kt,sxs−ωsds+et.
From the above discussion, then we have the following lemma.Lemma 5.
Assume thatF0 holds. Then, ut is a positive solution of (1) if only if xt is a positive solution of the following problem:
(21)−x″+ρx=Ft,Hx−ωx−ω,with xt≥ωt. Here, Ht denotes the Heaviside function of a single real variable:
(22)Ht=1,t≥0,0,t<0.
LetE denote the Banach space C0,1 with the norm x=maxt∈0,1xt.
Define a coneK⊂E by
(23)K=xt∈C0,1:mint∈θ,1−θxt≥σx,where θ∈0,1/2,σ=mint∈θ,1−θσ/CGt,t∈0,1. Define an operator T by
(24)Txt=∫01Gt,sFs,Hx−ωx−ωds.Lemma 6.
Assume that (F0) holds. Then, TK⊆K, and T:K⟶K is completely continuous.Proof.
For anyxt∈K, from Lemma 3, it follows that
(25)Txt=∫01Gt,sFs,Hx−ωx−ωds≥δ∫01Gt,tGs,sFs,Hx−ωx−ωds=δCGt,t∫01CGs,sFs,Hx−ωx−ωds≥δCGt,t∫01Gt,sFs,Hx−ωx−ωds,which implies that TK⊆K.
Now, we show thatT:K⟶K is completely continuous.
First, we show thatT maps the bounded set into itself. Since e and f are continuous, for any given c>0, let
(26)L=maxFt,Hx−ωx−ω:0≤t≤1,0≤x≤c.
Then, forx∈K¯c, we have
(27)Txt∞=∫01Gt,sFs,Hx−ωx−ωds∞≤Lmax0≤t,s≤1Gt,s,which implies that TK¯c is uniformly bounded.
Second, fort1,t2∈0,1, we have
(28)Txt2−Txt1=∫01Gt2,s−Gt1,sFs,Hx−ωx−ωds=∫01Git2,s−Git1,sf~is,us−ωsds≤Lmax0≤t,s≤1∂Gt,s∂tt2−t1,which implies that the operator T is equicontinuous.
Thus, by applying the Arzela-Ascoli theorem [17], we obtain that TKc¯ is relatively compact, namely, the operator T is compact.
Finally, we claim thatT:Kc¯⟶K is continuous. Assume that xnn=1∞⊂Kc¯ which converges to xt uniformly on [0,1]. By Lebesgue’s dominated convergence theorem and letting n⟶∞, we have
(29)Txnt−Txt=∫01Gt,sFs,Hxn−ωxn−ω−Fs,Hxn−ωxn−ωds≤C∫01Gs,sFs,Hxn−ωxn−ω−Fs,Hx−ωx−ωds⟶0,asn⟶+∞.
So,T is continuous on Kc. The proof is completed.
At the end of this section, let(30)e∗=maxt∈0,1et>0,ω∗=maxt∈0,1ωt.
Define the height functions(31)Φ∗r=minft,u,ϕ:t,u,ϕ∈0,1×0,r×0,r6,Φ∗t,r=maxft,u,ϕ:u,ϕ∈0,r×0,r6.
In addition, we need to select some suitable open bounded sets. For anyγ>0, let
(32)Ωγ=x∈E:mint∈θ,1−θxt<σγ,Bγ=x∈E:x<γ,ΩKγ=Ωγ∩K,∂ΩKγ=∂Ωγ∩K,BKγ=Bγ∩K,∂BKγ=∂Bγ∩K.
From [19, 20], we can conclude the lemma below.Lemma 7.
(i)
ΩKγ, BKγ are open relative to K(ii)
BKσγ⊂ΩKγ⊂BKγ(iii)
x∈∂ΩKγ if and only if minθ,1−θxt=σγ(iv)
Ifx∈∂ΩKγ, then σγ≤xt≤γ, for t∈θ,1−θ


## 3. Main Results

Theorem 8.
Assume that (F0) holds. In addition, the function f satisfies the following assumption:
(F1) There exists a α>0 such that Φ∗α≥0 and
(33)μα26C∫01Gs,sds+C∫01Gs,sesds+C∫01Gs,sΦ∗s,αds<α.
Then, we have(i)
Ifσα>ω∗, then (1) has at least one partly positive solution u,ϕ, namely,
(34)ut>0,fort∈θ,1−θ(ii)
Ifαδξ>Ce∗, then (1) has at least one positive solution u,ϕ, which satisfies
(35)ut>0,fort∈0,1Proof.
For anyx∈∂BKα, it is obvious that
(36)Hx−ωx−ω≤x=α,∫01Kt,sHx−ωxs−ωsds≤∫01Ks,sHx−ωxs−ωsds≤16α.
Then, from (F1) it follows that
(37)Txt=∫01Gt,sFs,Hx−ωx−ωds=μ∫01Gt,sHx−ωx−ω∫01Ks,τHx−ωxτ−ωτdτds+∫01Gt,sfs,Hx−ωx−ω∫01Ks,τHx−ωxτ−ωτdτ+esds≤μα26C∫01Gs,sds+C∫01Gs,sesds+C∫01Gs,sfs,Hx−ωx−ω,∫01Ks,τHx−ωxτ−ωτdτds≤μα26C∫01Gs,sds+C∫01Gs,sesds+C∫01Gs,sΦ∗s,αds<α=x,which implies that (a) of Lemma 1 holds.
Let(38)Ψρ=max0≤x≤ρHx−ωx−ω⋅∫θ1−θKτ,τHx−ωxτ−ωτdτ.
From [21], we have that
(39)limρ→+∞Ψρρ=+∞.
Then, there exists aβ> with σβ>α such that
(40)Hσβ−ωσβ−ω⋅∫θ1−θKτ,τHσβ−ωσβ−ωτdτ>Λβ,where Λ satisfies
(41)ΛμσC∫θ1−θGs,sKs,sds>1.
Letηt=1; now we prove that x≠Tx+λ,forx∈∂ΩKβandλ>0. On the contrary, if there exists a pair of x0∈∂ΩKβ and λ0>0 such that x0t=Tx0t+λ0, then from (iv) of Lemma 7, it follows that
(42)σβ=σx0≤x0t≤β,fort∈θ,1−θ.
Furthermore, fort∈θ,1−θ, we have
(43)x0≥minθ≤t≤1−θx0t=minθ≤t≤1−θTx0t+λ0=minθ≤t≤1−θμ∫01Gt,sHx0−ωx0−ω∫01Ks,τHx−ωxτ−ωτdτds+minθ≤t≤1−θ∫01Gt,sfs,Hx0−ωx−ω,∫01Ks,τHx0−ωxτ−ωτdτ+esds+λ0≥minθ≤t≤1−θμ∫01Gt,sHx0−ωx0−ω∫01Ks,τHx−ωxτ−ωτdτds≥minθ≤t≤1−θμδGt,t∫01Gs,sHx0−ωx0−ω∫01Ks,τHx−ωxτ−ωτdτds≥μσC∫01Gs,sHx0−ωx0−ω∫01Ks,sKτ,τHx−ωxτ−ωτdτds=μσC∫01Gs,sKs,sHx0−ωx0−ω∫01Kτ,τHx−ωxτ−ωτdτds=μσC∫01Gs,sKs,sHx0−ωx0−ω∫01−θKτ,τHx−ωxτ−ωτdτds=μσC∫01Gs,sKs,sHσx0−ωσx0−ω∫θ1−θKτ,τHσx0−ωσx0−ωτdτds≥μσC∫01Gs,sKs,sds⋅Λβ>β=x0,which contradicts with the statement (iii) of Lemma 7. So (b) holds.
Sinceα<σβ, from Lemma 7, we have BKα¯⊂BKσβ⊂ΩKβ. Therefore, by Lemma 1, we can get that T has at least one positive fixed-point xt∈ΩKβ¯\BKα. Hence, the inequalities hold,
(44)x≥α,σα≤mint∈θ,1−θxt≤σβ.
On the other hand, sinceσx≤mint∈θ,1−θxt≤σβ, we have x≤β.
(i)
Since(45)mint∈θ,1−θxt≥σα>ω∗>ωt=∫01Gt,sesds,we have
(46)ut=xt−ωt>0,t∈θ,1−θ(ii)
From Lemmas3 and 4, we have
(47)xt≥δCGt,tt=δCGt,tα≥αδCe∗ξe∗∫01Gt,sds≥αδCe∗ξ∫01Gt,sesds=αδCe∗ξ⋅ωt>ωt,which implies that ut=xt−ωt>0
Therefore, (1) has one positive solution u,ϕ=u,∫01Kt,susds.Theorem 9.
Assume that (F0) holds. In addition, the function f satisfies the following assumptions:
(F2) There exists a α>maxω∗/σ,Ce∗/δξ such that Φ∗α≤0 and
(48)μα26C∫01Gs,sds+C∫01Gs,sesds+C∫01Gs,sΦ∗s,αds<α.
(F3) There exists a r∗∈0,σα−ω∗ such that
(49)μr∗26C∫01Gs,sds+C∫01Gs,sesds+C∫01Gs,sΦ∗s,r∗ds<r∗.
(F4) limu+ϕ→0+ft,u,ϕ/u+ϕ=+∞, uniformly for t∈0,1.
Then, (1) has at least two positive solutionui,ϕii=1,2, which satisfies
(50)0≤u1t<r∗,mint∈θ,1−θu2t>σα−ω∗.Proof.
From (F2) and Theorem 8, it follows that there exists a solution u2t≥0 and
(51)mint∈θ,1−θu2t>σα−ω∗.
Now, we apply Lemma1 to prove the existence of another solution u1t.
Sincer∗<σα−ω∗<α and Φ∗α, then we can define the operator
(52)T¯ut=∫01Gt,sFs,usds.
For anyu∈∂BKr∗, it is obvious that
(53)∫01Kt,susds≤∫01Ks,susds≤16r∗.
Then, we have(54)T¯ut=∫01Gt,sFs,usds≤μr∗26C∫01Gs,sds+C∫01Gs,sesds+C∫01Gs,sfs,Hx−ωx−ω,∫01Ks,τHx−ωxτ−ωτdτds≤μr∗26C∫01Gs,sds+C∫01Gs,sesds+C∫01Gs,sΦ∗s,r∗ds<r∗=u,which implies that (a) of Lemma 1 holds.
Sincelimu+ϕ→0+ft,u,ϕ/u+ϕ=+∞, uniformly for t∈0,1, there exists a r∗<r∗ such that
(55)ft,u,ϕ>Mu+ϕ,for0<u+ϕ≤r∗,where M satisfies
(56)Mδσ⋅max0≤t≤1Gt,t⋅∫θ1−θGs,s1+∫θ1−θKs,τdτds>1.
Letηt=1, now we prove that u≠T¯u+λ,foru∈∂ΩKr∗/2andλ>0. On the contrary, if there exists a pair of u0∈∂ΩKr∗/2andλ0>0suchthatu0t=T¯u0t+λ0, then from (iv) of Lemma 7, it follows that
(57)u0t≤r∗2,ϕ0t=∫01Kt,su0sds<r∗12.
Furthermore, we have(58)u0=T~u0t+λ0≥∫01Gt,sfs,us,∫01Ks,τuτdτds≥δmax0≤t≤1Gt,t∫01Gs,sfs,us,∫01Ks,τuτdτds≥δmax0≤t≤1Gt,t∫01Gs,sMu,s+∫01Ks,τuτdτds≥δmax0≤t≤1Gt,t∫01−θGs,sMu,s+∫01−θKs,τuτdτds≥Mδσ⋅max0≤t≤1Gt,t⋅∫01−θGs,s1+∫01−θKs,τdτds⋅u0>u0,which contradicts with the statement (iii) of Lemma 7. So (b) holds.Therefore, from Lemma1, we can get that T~ has at least one positive fixed-point u1t∈ΩKr∗¯\BKr∗/2. Hence, the inequalities hold
(59)u1≥r∗2,σr∗2≤mint∈θ,1−θu1t≤σr∗.On the other hand, sinceσu1≤mint∈θ,1−θu1t≤σr∗, we have u1≤r∗.Finally, since(60)mint∈θ,1−θu2t≤σα−ω∗>r∗,u1≤r∗,(1) has at least two positive solutions.


## 4. Examples

Example 10.
Let us consider the following system:(61)−u″+u=ϕu+14eu+14cosπϕ−14t20<t<1,−ϕ″=14u,ϕ0=ϕ1=0,ϕ″0=ϕ″1=0,where μ=1,ρ=1,f:0,1×ℝ+2⟶ℝ is continuous, and
(62)f=14eu+cosπϕ−t2≥−14t2=−etfort,u,ϕ∈0,1×ℝ+2.
It is obvious thatF0 holds. Via some computations, we have
(63)e∗=14,C=1,δ=1sinh1,J1∗=J2∗=e1/2−e−1/2e1/2+e−1/2<1.
Choosingθ=1/4∈0,1/2,ξ=1/2. Then, we have
(64)σ=mint∈1/4,3/4δCGt,t=sinh3/4sinh1/4sinh12,ω∗=sinh1−2sinh1/2sinh1.
Furthermore, we have(65)ω∗σ=sinh1−2sinh1/2sinh1sinh3/4sinh1/4<3,Ce∗δξ<sin1J1∗+J2∗<3.
Choosingα=3>maxω∗/σ,Ce∗/δξ,r∗=3/10∈0,σα−ω∗. Then,
(66)Φ∗3≥0,Φ∗t,3=14e3+1,Φ∗310≥0,Φ∗t,310=14e3/10+1.
It is easy to get(67)μα26C∫01Gs,sds+C∫01Gs,sesds+C∫01Gs,sΦ∗s,αds≤32e−12sinh1+e−12sinh1+e−12sinh114e3+1≤3,(68)μr∗26C∫01Gs,sds+C∫01Gs,sesds+C∫01Gs,sΦ∗s,r∗ds≤3200e−12sinh1+e−12sinh1+e−12sinh114e3/10+1≤310,which implies that F2 and F3 hold.Finally, it is obvious that(69)limu+ϕ→0+ft,u,ϕu+ϕ=+∞,uniformlyfort∈0,1.SoF4 holds.Therefore, by Theorem9, (61) has two positive solutions.



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*Source: 7210608-2020-08-14.xml*