# Positive Solutions for Second-Order Nonlinear Ordinary Differential Systems with Two Parameters

Authors: Lan Sun; Yukun An; Min Jiang
Journal: ISRN Applied Mathematics
 (2011)

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## Abstract

By using fixed-point theorem and under suitable conditions, we investigate the existence and multiplicity positive solutions to the following systems:u′′(t)+au(t)+bv(t)+λh1(t)f(u(t),v(t))=0,t∈[0,1],v′′(t)+cu(t)+dv(t)+μh2(t)g(u(t),v(t))=0,t∈[0,1],u(0)=u(1)=0,v(0)=v(1)=0, where a,b,c,d are four positive constants and λ>0, μ>0, f(u,v),g(u,v)∈C(R+×R+,R+) and h1,h2∈C([0,1],R+). We derive two explicit intervals of λ and μ, such that the existence and multiplicity of positive solutions for the systems is guaranteed.


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## Body

## 1. Introduction

In this paper, we are concerned with the existence and multiplicity of positive solutions to the following boundary value problem of second-order nonlinear differential systems with two parameters:(1.1)u′′(t)+au(t)+bv(t)+λh1(t)f(u(t),v(t))=0,t∈[0,1],v′′(t)+cu(t)+dv(t)+μh2(t)g(u(t),v(t))=0,t∈[0,1],u(0)=u(1)=0,v(0)=v(1)=0,
where a,b,c,d are four positive constants.In addition, we assume the following conditions throughout this paper:(H1)
h1(t),h2(t)∈C([0,1],R+) does not vanish identically on any subinterval of [0,1];(H2)
0<a<π2,0<d<π2,c>0,b>0;(H3)
f,g∈C(R+×R+,R+).The boundary value problem of ordinary differential systems has attracted much attention, see [1–6] and the references therein. Recently, Fink and Gatica [4] and Ma [7] have studied the existence of positive solutions of the following systems:
(1.2)x′′(t)+λa(t)f(x(t),y(t))=0,t∈[0,1],y′′(t)+μb(t)g(x(t),y(t))=0,t∈[0,1],x(0)=x(1)=y(0)=y(1)=0.
In [4], a multiplicity result has been established when f(0,0)>0,g(0,0)>0. In [7], a multiplicity result has been given for the more general case f(0,0)≥0,g(0,0)≥0.Also, by using Krasnosel′skill fixed-point theorem, Ru and An [8] considered the existence of positive solutions of the following systems:
(1.3)(-)pu(2p)=λa(t)f(u(t),v(t)),t∈[0,1],(-)qv(2q)=μb(t)g(u(t),v(t)),t∈[0,1],u(2i)(0)=u(2i)(1)=0,0≤i≤p-1,v(2j)(0)=v(2j)(1)=0,0≤j≤q-1,
where λ>,μ>0, p,q∈N.f,g:[0,∞)×[0,∞)→[0,∞).More recently, Dalbono and Mckenna [5] proved the existence and multiplicity of solutions to a class of asymmetric weakly coupled systems as follows:
(1.4)[u1′′(t)u2′′(t)]+[b1ɛɛb2][u1+(t)u2+(t)]=[sin⁡(t)sin⁡(t)],u1(0)=u2(0)=0=u1(π)=u2(π),
where ɛ is suitably small and the positive numbers b1, b2 satisfy
(1.5)h2<b1<(h+1)2,k2<b2<(k+1)2forsomeh,k∈N.
Applying a classical change of variables, the authors transformed the initial problem into an equivalent problem whose solutions can be characterized by their nodal properties. Meanwhile, in [5], there are two open questions. (1)
Can one replace the near-diagonal matrix with something more general and use information on the eigenvalues of matrix?(2)
Can one replace the near-diagonal matrix with something more general and use information on the eigenvalues of matrix?Inspired by the above works and the two open questions, we consider the existence and multiplicity of positive solutions to (1.1). The paper is organized as follows. In Section 2, we state some preliminaries. In Sections 3 and 4, we prove the existence and multiplicity results of (1.1).


## 2. Preliminaries

In order to prove our results, we state the well-known fixed-point theorem [9]:Lemma 2.1.
LetE be a Banach space and let K⊂E be a cone in E. Assume Ω1, Ω2 are two open subsets of E with 0∈Ω1, Ω1̅⊂Ω2, and let T:K∩(Ω2̅∖Ω1)→K be a completely continuous operator such that either(i)
∥Tu∥≤∥u∥,u∈K∩∂Ω1 and ∥Tu∥≥∥u∥,u∈K∩∂Ω2; or(ii)
∥Tu∥≥∥u∥,u∈K∩∂Ω1 and ∥Tu∥≤∥u∥,u∈K∩∂Ω2.Then, T has a fixed point in K∩(Ω2̅∖Ω1).
To be convenient, we introduce the following notations:(2.1)f0=lim⁡u+v→0f(u,v)u+v,f∞=lim⁡u+v→∞f(u,v)u+v,g0=lim⁡u+v→0g(u,v)u+v,g∞=lim⁡u+v→∞g(u,v)u+v,
and suppose that f0,g0,f∞,g∞∈[0,+∞].
Letω1=a,ω2=d,H(t)∈C[0,1], Gi(t,s),i=1,2 be Green′s function of the corresponding to linear boundary value problem:
(2.2)-u′′(t)-ωi2u(t)=H(t),u(0)=u(1)=0.
Then, the solution of (2.2) is given by
(2.3)u(t)=∫01Gi(t,s)H(s)ds.
It is well known that Gi(t,s) can be expressed by
(2.4)Gi(t,s)={sin⁡ωitsin⁡ωi(1-s)ωisin⁡ωi0≤t≤s≤1,sin⁡ωissin⁡ωi(1-t)ωisin⁡ωi0≤s≤t≤1.
In addition, it can be easily to be checked that
(2.5)Mi=max⁡t∈[0,1]Gi(t,t)>0,mi=min⁡t∈[1/4,3/4]Gi(t,t)>0,i=1,2.See [6], we have the following lemma.Lemma 2.2.
Gi(t,s) has the following properties:(i)
Gi(t,s)>0,forallt,s∈(0,1),(ii)
Gi(t,s)≤CiGi(s,s),forallt,s∈[0,1],Ci=1/sinωi,(iii)
Gi(t,s)≥δiGi(t,t)Gi(s,s),forallt,s∈[0,1],δi=ωisinωi,i=1,2.It is obvious that problem (1.1) is equivalent to the equation:
(2.6)(u(t),v(t))=(∫01G1(t,s)(λh1(s)f(u(s),v(s))+bv(s))ds,∫01G2(t,s)(μh2(s)g(u(s),v(s))+cu(s))ds),
and consequently it is equivalent to the fixed-point problem:
(2.7)(u,v)=A(u,v)
with A:C[0,1]×C[0,1]→C[0,1]×C[0,1] given by
(2.8)A(u,v)=(∫01G1(t,s)(λh1(s)f(u(s),v(s))+bv(s))ds,∫01G2(t,s)(μh2(s)g(u(s),v(s))+cu(s))ds).
For convenience, denote
(2.9)Aλ(u,v)=∫01G1(t,s)(λh1(s)f(u(s),v(s))+bv(s))ds,Aμ(u,v)=∫01G2(t,s)(μh2(s)g(u(s),v(s))+cu(s))ds.
It is obvious that A is completely continuous. Let σi=δimi/Ci=(sinωi/4)(sin3ωi/4)(sinωi), i=1,2,σ=min{σ1,σ2} and define a cone in C[0,1]×C[0,1] by
(2.10)K={(u,v)∈C[0,1]×C[0,1]:u(t)≥0,v(t)≥0,min⁡1/4≤t≤3/4(u(t)+v(t))≥σ‖(u,v)‖},
where ∥(u,v)∥=∥u∥+∥v∥=supt∈[0,1]u(t)+supt∈[0,1]v(t).Lemma 2.3.
A(K)⊂K. 						Proof.
For any(t,s)∈[1/4,3/4]×[0,1], by Lemma 2.2, we have
(2.11)Aλ(u,v)(t)=∫01G1(t,s)(λh1(s)f(u(s),v(s))+bv(s))ds≤∫01C1G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds,‖Aλ(u,v)‖≤∫01C1G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds,Aλ(u,v)(t)=∫01G1(t,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥δ1G1(t,t)∫01G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥δ1G1(t,t)C1‖Aλ(u,v)‖.
Similarly, for any (t,s)∈[1/4,3/4]×[0,1], we have
(2.12)Aμ(u,v)(t)≥δ2G2(t,t)C2‖Aμ(u,v)‖.
Then, for any (t,s)∈[1/4,3/4]×[0,1], we have
(2.13)Aλ(u,v)(t)+Aμ(u,v)(t)=∫01G1(t,s)(λh1f(u,v)+bv)ds+∫01G2(t,s)(μh2g(u,v)+cu)ds≥δ1G1(t,t)C1‖Aλ(u,v)‖+δ2G2(t,t)C2‖Aμ(u,v)‖≥σ‖A(u,v)‖.
Thus, min1/4≤t≤3/4Aλ(u,v)+Aμ(u,v)≥σ∥A(u,v)∥. Therefore, A(K)⊂K.


## 3. Existence Results

We assume the following:(H4)
0<b<2ω1cos2(ω1/2),0<c<2ω2cos2(ω2/2);(H5)
Ai=∥∫01hi(s)ds∥,Bi=∫1/43/4hi(s)ds>0,i=1,2;(H6)
Pi=CiMi=(sec2ωi/2)/4ωi,Qi=δimi2=sin(ωi/4)sin(3ωi/4),i=1,2.Theorem 3.1.
Assume(H1)–(H6) hold. Then one has the following:
(1)
If0<f0,f∞,g0,g∞<∞,2P1A1f0<Q1σB1f∞(1-2P1b), then for each λ∈(1/Q1σB1f∞,1-2P1b/2P1A1f0) and μ∈(0,(1-2P2c)/2P2A2g0), (1.1) has at least one positive solution.(2)
If0<f0,f∞,g0,g∞<∞,2P2A2g0<Q2σB2g∞(1-2P2c), then for each λ∈(0,(1-2P1b)/2P1A1f0) and μ∈(1/Q2σB2g∞,(1-2P2c)/2P2A2g0), (1.1) has at least one positive solution.(3)
Iff0=0,f∞=∞, then for each λ∈(0,∞) and μ∈(0,(1-2P2c)/2P2A2g0), (1.1) has at least one positive solution.(4)
Ifg0=0,g∞=∞, then for each λ∈(0,(1-2P1b)/2P1A1f0) and μ∈(0,∞), (1.1) has at least one positive solution.(5)
Iff0=0,f∞=∞ and g0=0,g∞=∞, then for each λ∈(0,∞) and μ∈(0,∞), (1.1) has at least one positive solution.(6)
Iff∞=∞,0<f0<∞ or g∞=∞,0<g0<∞, then for each λ∈(0,(1-2P1b)/2P1A1f0) and μ∈(0,(1-2P2c)/2P2A2g0), (1.1) has at least one positive solution.(7)
Iff0=0,0<f∞<∞ and g0=0,0<g∞<∞, then for each λ∈(1/Q1σB1f∞,∞),μ∈(0,∞) or λ∈(0,∞),μ∈((1/Q2σB2g∞)∞),(1.1) has at least one positive solution.Proof.
We only prove case (1.1). The other cases can be proved similarly. In order to apply the Lemma 2.1, we construct the sets Ω1,Ω2.
Letλ∈(1/Q1σB1f∞,(1-2P1b)/2P1A1f0),μ∈(0,(1-2P2c)/2P2A2g0), and we choose ɛ>0 such that
(3.1)1σQ1B1(f∞-ɛ)≤λ≤1-2P1b2P1A1(f0+ɛ),0<μ≤1-2P2c2P2A2(g0+ɛ).
By the definition off0 and g0, there exists R1>0, such that
(3.2)f(u,v)≤(f0+ϵ)(u+v),g(u,v)≤(g0+ϵ)(u+v),foru+v∈[0,R1].
Choosing (u,v)∈K with ∥(u,v)∥=R1, we have
(3.3)Aλ(u,v)(t)=∫01G1(t,s)(λh1(s)f(u(s),v(s))+bv(s))ds≤C1M1∫01(λh1(s)(f0+ɛ)(u(s)+v(s))+bv(s))ds≤C1M1‖(u,v)‖∫01(λh1(s)(f0+ɛ)+b)ds=P1‖(u,v)‖(λA1(f0+ɛ)+b)≤12‖(u,v)‖,
namely, ∥Aλ(u,v)(t)∥≤(1/2)∥(u,v)∥. In the same way, we also have
(3.4)‖Aμ(u,v)(t)‖≤12‖(u,v)‖.
then ∥Aλ(u,v)(t)∥≤∥(u,v)(t)∥. Thus, if we set Ω1={(u,v)∈K:∥(u,v)∥<R1}, then
(3.5)‖A(u,v)(t)‖≤‖(u,v)‖,∀(u,v)∈K∩∂Ω1.
On the other hand, by the definition of f∞, there exists R2′>0, such that f(u,v)≥(f∞-ϵ)(u+v), for u+v∈[R2′,+∞). Let R2=max{2R1,σ-1R2′} and Ω2={(u,v)∈K:∥(u,v)∥<R2}. If we choose (u,v)∈K with ∥(u,v)∥=R2, such that min1/4≤t≤3/4(u(t)+v(t))≥σ∥(u,v)∥≥R2, then we have
(3.6)Aλ(u,v)(14)=∫01G1(14,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥δ1G1(14,14)∫1/43/4G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥δ1m12∫1/43/4(λh1(s)(f∞-ɛ))(u(s)+v(s))ds≥Q1σB1λ(f∞-ɛ)‖(u,v)‖≥‖(u,v)‖.
Hence
(3.7)‖A(u,v)(t)‖≥‖Aλ(u,v)(t)‖≥‖(u,v)‖,∀(u,v)∈K∩∂Ω2.
Therefore, it follows from (3.5) to (3.7) and Lemma 2.2, A has a fixed point (u,v)∈K∩(Ω¯2∖Ω1), which is a positive solution of (1.1).Similarly, we also have the following results.Theorem 3.2.
Assume(H1)–(H6) hold. Then one has the following:(1)
If0<f0,f∞,g0,g∞<∞,2P1A1f∞<Q1σB1f0(1-2P1b), then for each λ∈(1/Q1σB1f0,(1-2P1b)/2P1A1f∞) and μ∈(0,(1-2P2c)/2P2A2g∞), (1.1) has at least one positive solution.(2)
If0<f0,f∞,g0,g∞<∞,2P2A2g∞<Q2σB2g0(1-2P2c), then for each λ∈(0,(1-2P1b)/2P1A1f∞) and μ∈(1/Q2σB2g0,(1-2P2c)/2P2A2g∞), (1.1) has at least one positive solution.(3)
Iff0=∞,f∞=0, then for each λ∈(0,∞) and μ∈(0,(1-2P2c)/2P2A2g0), (1.1) has at least one positive solution.(4)
Ifg0=∞,g∞=0, then for each λ∈(0,(1-2P1b)/2P1A1f0) and μ∈(0,∞), (1.1) has at least one positive solution.(5)
Iff0=∞,f∞=0 and g0=∞,g∞=0, then for each λ∈(0,∞) and μ∈(0,∞), (1.1) has at least one positive solution.(6)
Iff0=∞,0<f∞<∞ or g0=∞,0<g∞<∞ then for each λ∈(0,(1-2P1b)/2P1A1f∞) and μ∈(0,(1-2P2c)/2P2A2g∞), (1.1) has at least one positive solution.(7)
Iff∞=0,0<f0<∞ and g∞=0,0<g0<∞ then for each λ∈(1/Q1σB1f0,∞),μ∈(0,∞) or λ∈(0,∞),μ∈(1/Q2σB2g0), (1.1) has at least one positive solution.Proof.
We only prove case(1). The other cases can be proved similarly.
Letλ∈(1/Q1σB1f0,(1-2P1b)/2P1A1f∞),μ∈(0,(1-2P2c)/2P2A2g∞), and we choose ɛ>0 such that
(3.8)1σQ1B1(f0-ɛ)≤λ≤1-2P1b2P1A1(f∞+ɛ),0<μ≤1-2P2c2P2A2(g∞+ɛ),
by the definition off0, there exists R1>0, such that
(3.9)f(u,v)≥(f0-ϵ)(u+v),foru+v∈[0,R1].
Choosing (u,v)∈K with ∥(u,v)∥=R1, such that min1/4≤t≤3/4(u(t)+v(t))≥σ∥(u,v)∥, then we have
(3.10)Aλ(u,v)(14)=∫01G1(14,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥δ1G1(14,14)∫1/43/4G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥δ1m12∫1/43/4(λh1(s)(f0-ɛ))(u(s)+v(s))ds≥Q1σB1λ(f0-ɛ)‖(u,v)‖≥‖(u,v)‖.
So, if we set Ω1={(u,v)∈K:∥(u,v)∥<R1}, then
(3.11)‖A(u,v)(t)‖≥‖(u,v)‖,∀(u,v)∈K∩∂Ω1.
On the other hand, by the definition of f∞ and g∞, there exists R2>2R1, such that
(3.12)f(u,v)≤(f∞+ϵ)(u+v),g(u,v)≤(g∞+ϵ)(u+v)foru+v∈[R2,′+∞).
Choosing (u,v)∈K with ∥(u,v)∥=R2, we have
(3.13)Aλ(u,v)(t)=∫01G1(t,s)(λh1(s)f(u(s),v(s))+bv(s))ds≤∫01C1G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds≤C1M1∫01(λh1(s)(f∞+ɛ)(u(s)+v(s))+bv(s))ds≤P1(λA1(f∞+ɛ)+b)‖(u,v)‖≤12‖(u,v)‖.
In the same way, we also have
(3.14)‖Aμ(u,v)(t)‖≤12‖(u,v)‖.
Hence, if we set Ω2={(u,v)∈K:∥(u,v)∥<R2}, then
(3.15)‖A(u,v)(t)‖≤‖(u,v)(t)‖,∀(u,v)∈K∩∂Ω2.
Therefore, it follows from (3.11) to (3.15) and Lemma 2.2 that A has a fixed point (u,v)∈K∩(Ω¯2∖Ω1), which is a positive solution of (1.1).


## 4. Multiplicity Results

Theorem 4.1.
Assume(H1)–(H6) hold. In addition, assume that there exist three constants r,M,N, where N is sufficient small with 2P1NA1≤Q1B1M(1-2P1b),2P2NA2≤Q2B2M(1-2P2c) such that(i)
f0=f∞=0,g0=g∞=0,(ii)
f(u,v)≥Mrorg(u,v)≥Mr,  for σr≤∥(u,v)∥≤r.Then, for any λ∈(1/Q1B1M,(1-2P1b)/2P1A1N), μ∈(0,(1-2P2c)/2P2A2N), or λ∈(0,(1-2P1b)/2P1A1N), μ∈(1/Q2B2M,(1-2P2c)/2P2A2N), the problem (1.1) has at least two positive solutions.Proof.
We only prove the case ofλ∈(1/Q1σB1M,(1-2P1b)/2P1A1N), μ∈(0,(1-2P2c/2P2A2N)), The other case is similar.
Step 1.
Sincef0=g0=0, there exists r1∈(0,r) such that
(4.1)f(u,v)≤N(u+v),g(u,v)≤N(u+v),for0<u+v<r1.
Set Ω1={(u,v)∈K:∥(u,v)∥<r1},forall(u,v)∈K∩∂Ω1, then we have
(4.2)Aλ(u,v)(t)=∫01G1(t,s)(λh1(s)f(u(s),v(s))+bv(s))ds≤∫01C1M1(λh1(s)f(u(s),v(s))+bv(s))ds≤C1M1∫01(λh1(s)N(u(s)+v(s))+bv(s))ds≤P1(λA1N+b)‖(u,v)‖≤12‖(u,v)‖.
In the same way, we also have
(4.3)‖Aμ(u,v)(t)‖≤12‖(u,v)‖.
Hence,
(4.4)‖A(u,v)‖≤‖(u,v)‖,∀(u,v)∈K∩∂Ω1.
Step 2.
Sincef∞=g∞=0, there exists R¯>r such that
(4.5)f(u,v)≤N(u+v),g(u,v)≤N(u+v),foru+v≥R¯.
Similarly, set Ω2={(u,v)∈K:∥(u,v)∥<r2}, then
(4.6)‖A(u,v)‖≤‖(u,v)‖,∀(u,v)∈K∩∂Ω2.
Step 3.
LetΩ3={(u,v)∈K:∥(u,v)∥<r}, we can see that
(4.7)Aλ(u,v)(14)=∫01G1(14,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥∫01δ1G1(14,14)G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥δ1G1(14,14)∫1/43/4G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥Q1∫1/43/4(λh1(s)(f(u(s),v(s))+bv(s)))ds≥Q1λB1Mr≥r.
Then,
(4.8)‖A(u,v)‖≥‖(u,v)‖,∀(u,v)∈K∩∂Ω3.
Consequently, by Lemma2.1 and from (4.4)–(4.8), A has two fixed points: (u1,v1)∈K∩(Ω¯3∖Ω1) and (u2,v2)∈K∩(Ω¯2∖Ω3), so (1.1) has at least two positive solutions satisfying 0<(u1,v1)<r<(u2,v2).Theorem 4.2.
Assume(H1)–(H6) hold. In addition, assume that there exist constants r,M,N, where N is sufficient large with 2P1MA1≤Q1σB1N(1-2P1b),2P2MA2≤Q2σB2N(1-2P2c)  such that(i)
f0=f∞=∞ or g0=g∞=∞,(ii)
f(u,v)≤Mr or g(u,v)≤Mr,  for σr≤∥(u,v)∥≤r,then, for any λ∈(1/Q1σB1N,(1-2P1b)/2P1A1M), μ∈(0,(1-2P2c)/2P2A2M), or λ∈(0,(1-2P1b)/2P1A1M), μ∈(1/Q2σB2N,(1-2P2c)/2P2A2M), the problem (1.1) has at least two positive solutions.Proof.
We only prove the case ofλ∈(1/Q1σB1N,(1-2P1b)/2P1A1M), μ∈(0,(1-2P2c)/2P2A2M), The other case is similar.
Step 1.
Sincef0=∞, there exists r1∈(0,r) such that
(4.9)f(u,v)≥N(u+v),for0<u+v≤r1.
Set Ω1={(u,v)∈K:∥(u,v)∥<r1}, then we have
(4.10)Aλ(u,v)(14)=∫01G1(14,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥∫01δ1G1(14,14)G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥δ1G1(14,14)∫1/43/4G1(s,s)(λh1(s)f(u(s),v(s))+bv(s))ds≥Q1∫1/43/4(λh1(s)(f(u(s),v(s))+bv(s)))ds≥Q1λB1Nσ‖(u,v)‖≥r.
Hence,
(4.11)‖A(u,v)‖≥‖(u,v)‖,∀(u,v)∈K∩∂Ω1.
Step 2.
Sincef∞=∞, there exists R¯>r such that
(4.12)f(u,v)≥N(u+v),foru+v≥R¯.
Similarly, set Ω2={(u,v)∈K:∥(u,v)∥<r2}, then
(4.13)‖A(u,v)‖≥‖(u,v)‖,∀(u,v)∈K∩∂Ω2.
Step 3.
LetΩ3={(u,v)∈K:∥(u,v)∥<r},forall(u,v)∈K∩∂Ω3, we can see that
(4.14)Aλ(u,v)(t)=∫01G1(t,s)(λh1(s)f(u(s),v(s))+bv(s))ds≤∫01C1M1(λh1(s)f(u(s),v(s))+bv(s))ds≤P1(λA1M+b)‖(u,v)‖≤12‖(u,v)‖.
In the same way, we also have
(4.15)‖Aμ(u,v)(t)‖≤12‖(u,v)‖.
Hence,
(4.16)‖A(u,v)‖≤‖(u,v)‖,∀(u,v)∈K∩∂Ω3.Consequently, by Lemma2.1 and from (4.11)–(4.16), A has two fixed points: (u1,v1)∈K∩(Ω¯3∖Ω1) and (u2,v2)∈K∩(Ω¯2∖Ω3), so (1.1) has at least two positive solutions satisfying 0<(u1,v1)<r<(u2,v2).



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*Source: 612591-2011-12-15.xml*